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Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Named_Reactions/Named_Reagents | Named reagent Structure Formula Other Name CAS number Adam's catalyst NaN PtO2 Platinum(IV) oxide 1314-15-4 Appel's salt NaN C2Cl3NS2 4,5-dichloro-1,2,3-dithiazolium chloride 75318-43-3 Bestmann ylide NaN C20H15OP 2-(triphenylphosphoranylidene)ethenone 15596-07-3 Brady's reagent NaN C6H6N4O4 (2,4-dinitrophenyl)hydrazine 119-26-6 Brassard's diene NaN C11H22O3Si (E)-((1,3-diethoxybuta-1,3-dien-1-yl)oxy)trimethylsilane NaN Bredereck's reagent NaN C9H22N2O tert-Butoxy bis(dimethylamino)methane 5815-08-7 Burgess reagent NaN C8H18N2O4S 1-methoxy-N-triethylammoniosulfonyl-methanimidate 29684-56-8 Caro's acid NaN H2O5S Peroxymonosulfuric acid 7722-86-3 Collins reagent NaN C10H10N2CrO3 Dipyridine chromium(VI) oxide 20492-50-6 Comins' reagent NaN C7H3ClF6N2O4S2 2-[N,N-Bis(trifluoromethylsulfonyl)amino]-5-chloropyridine 145100-51-2 Corey-Bakshi-Shibata catalyst NaN C18H20BNO (R)-(+)-1-Methyl-3,3-diphenylhexahydropyrrolo[1,2-c][1,3,2]oxazaborole (S)-(-)-1-Methyl-3,3-diphenylhexahydropyrrolo[1,2-c][1,3,2]oxazaborole 112022-83-0 112022-81-8 Corey-Chaykovsky reagent NaN C3H8OS Dimethylsulfoxonium methylide 5367-24-8 Corey lactone NaN C8H12O4 (3aR,4S,5R,6aS)-(−)-Hexahydro-5-hydroxy-4-(hydroxymethyl)-2H-cyclopenta[b]furan-2-one (3aS,4R,5S,6aR)-(+)-Hexahydro-5-hydroxy-4-(hydroxymethyl)-2H-cyclopenta[b]furan-2-one 32233-40-2 76704-05-7 Cornforth reagent NaN C10H12N2Cr2O7 Pyridinium dichromate 20039-37-6 Danishefsky's diene NaN C8H16O2Si (E)-1-Methoxy-3-trimethylsilyloxy-1,3-butadiene 54125-02-9 Dess-Martin periodinane NaN C13H13IO8 1,1,1-Triacetoxy-1,1-dihydro-1,2-benziodoxol-3(1H)-one 87413-09-0 Echavarren's gold complex NaN C22H30AuF6NPSb (Acetonitrile)[(2-biphenyl)di-tert-butylphosphine]gold(I) hexafluoroantimonate 866641-66-9 Eschenmoser's salt NaN C3H8NI Dimethylmethylideneammonium iodide 33797-51-2 Fétizon's reagent NaN NaN Silver carbonate on Celite NaN Fort's base NaN NaN NaN NaN Frémy's salt NaN K2NO7S2 Potassium nitrosodisulfonate 14293-70-0 Garner's aldehyde NaN C11H19NO4 (S)-tert-butyl 4-formyl-2,2-dimethyloxazolidine-3-carboxylate (R)-tert-butyl 4-formyl-2,2-dimethyloxazolidine-3-carboxylate 102308-32-7 95715-87-0 Glauber's salt NaN H20O14Na2S Sodium sulfate decahydrate 7727-73-3 Gold's reagent NaN C6H14ClN3 (Dimethylaminomethyleneaminomethylene)dimethylammonium chloride 20353-93-9 Grubbs catalyst (1st generation) NaN C43H72Cl2P2Ru Benzylidene-bis(tricyclohexylphosphine)dichlororuthenium 172222-30-9 Grubbs catalyst (2nd generation) NaN C46H65Cl2N2PRu (1,3-Bis(2,4,6-trimethylphenyl)-2-imidazolidinylidene)dichloro(phenylmethylene)(tricyclohexylphosphine)ruthenium 246047-72-3 Guindon's reagent NaN C2H6BBr Dimethylboron bromide 5158-50-9 Hajos-Parrish ketone NaN C10H12O2 (S)-7a-methyl-2,3,7,7a-tetrahydro-1H-indene-1,5(6H)-dione (R)-7a-methyl-2,3,7,7a-tetrahydro-1H-indene-1,5(6H)-dione 17553-86-5 17553-89-8 Hantzsch's ester NaN C13H19NO4 Diethyl 1,4-dihydro-2,6-dimethyl-3,5-pyridinedicarboxylate 214-561-6 Hatakeyama's catalyst NaN C19H22N2O2 (9S)-3α,9-Epoxy-10,11-dihydrocinchonan-6'-ol 253430-48-7 Hermann-Beller catalyst NaN C46H46O4P2Pd2 trans-di(μ-acetato)bis[o-(di-o-tolylphosphino)benzyl]dipalladium(II) 172418-32-5 Hieber anion NaN C3FeNO4 NaN NaN Hoveyda-Grubbs catalyst (1st generation) NaN C28H45Cl2OPRu Dichloro(o-isopropoxyphenylmethylene)(tricyclohexylphosphine)ruthenium(II) 203714-71-0 Hoveyda-Grubbs catalyst (2nd generation) NaN C31H38Cl2N2ORu (1,3-Bis-(2,4,6-trimethylphenyl)-2-imidazolidinylidene)dichloro(o-isopropoxyphenylmethylene)ruthenium 301224-40-8 Hünig's base NaN C8H19N N,N-Diisopropylethylamine 7087-68-5 Kagan's reagent NaN SmI2 Samarium diiodide 32248-43-4 Kemp's triacid NaN C12H18O6 cis,cis-1,3,5-Trimethylcyclohexane-1,3,5-tricarboxylic acid 79410-20-1 Koser's reagent NaN C13H13IO4S [Hydroxy(tosyloxy)iodo]benzene 27126-76-7 Lawesson's reagent NaN C14H14O2P2S4 2,4-bis(4-methoxyphenyl)-1,3,2,4-dithiadiphosphetane 2,4-disulfide 19172-47-5 Leuchs' anhydrides NaN NaN NaN NaN Lindlar catalyst NaN NaN Pd/CaCO3 poisoned with Lead (Pb(OAc)2 or PbO) NaN Mander's reagent NaN C3H3NO2 Methyl cyanoformate 17640-15-2 Martin's sulfurane NaN C30H20F12O2S Bis[α,α-bis(trifluoromethyl)benzyloxy]diphenylsulfur 32133-82-7 Meerwein's reagent NaN C3H9OBF4 Trimethyloxonium tetrafluoroborate 420-37-1 Meldrum's acid NaN C6H8O4 2,2-Dimethyl-1,3-dioxane-4,6-dione 2033-24-1 Mosher's acid NaN C10H9F3O3 (R)-3,3,3-Trifluoro-2-methoxy-2-phenylpropanoic acid (S)-3,3,3-Trifluoro-2-methoxy-2-phenylpropanoic acid 20445-31-2 17257-71-5 Pearlman's Catalyst NaN NaN Palladium hydroxide on carbon 12135-22-7 Ohira-Bestmann reagent NaN C5H9N2O4P Dimethyl (1-diazo-2-oxopropyl)phosphonate 90965-06-3 Roche ester NaN C5H10O3 Methyl (S)-(+)-3-hydroxy-2-methylpropionate Methyl (R)-(+)-3-hydroxy-2-methylpropionate 80657-57-4 72657-23-9 Schlosser's base NaN NaN NaN NaN Schwartz reagent NaN C10H11ClZr Bis(cyclopentadienyl)zirconium(IV) chloride hydride 37342-97-5 Seyferth-Gilbert reagent NaN C3H7N2O3P Dimethyl (diazomethyl)phosphonate 27491-70-9 Shiina's reagent NaN C16H12N2O7 2-Methyl-6-nitrobenzoic anhydride 434935-69-0 Stryker's reagent NaN C108H96Cu6P6 (Triphenylphosphine)copper hydride hexamer 33636-93-0 Tebbe reagent NaN C13H18AlClTi NaN 67719-69-1 Togni's reagent NaN C10H10F3IO 3,3-Dimethyl-1-(trifluoromethyl)-1,2-benziodoxole 887144-97-0 Tröger's base NaN C17H18N2 (+)-(5R)-2,8-dimethyl-6,12-dihydro-5,11-methanodibenzo[b,f][1,5]diazocine (-)-(5S)-2,8-dimethyl-6,12-dihydro-5,11-methanodibenzo[b,f][1,5]diazocine 529-81-7 14645-24-0 Wieland–Miescher ketone NaN C11H14O2 (S)-(+)-8a-Methyl-3,4,8,8a-tetrahydro-2H-naphthalene-1,6-dione (R)-(–)-8a-Methyl-3,4,8,8a-tetrahydro-2H-naphthalene-1,6-dione 33878-99-8 100348-93-4 Wood's metal 50% Bi / 26.7% Pb / 13.3% Sn / 10% Cd by weight NaN NaN 8049-22-7 Woollins' reagent NaN C12H10P2Se4 2,4-Diphenyl-1,3,2,4-diselenadiphosphetan-2,4-diselenide 122039-27-4 Contributors OChemOnline |
Courses/Lumen_Learning/Book%3A_Western_Civilization_I_(Lumen)/05%3A_Week_3%3A_Ancient_Egypt/05.8%3A_Primary_Source_Reading_1%3A_Wenamen%E2%80%99s_Journey | Primary Source Reading: Wenamen’s Journey This translation is by Egyptologist Miriam Lichtheim from Ancient Egyptian Literature: The New Kingdom, Volume II , pp. 224-229. Year 5, fourth month of summer, day 16, the day of departure of Wenamun, the Elder of the Portal of the Temple of Amun, Lord of Thrones-of-the-Two-Lands, to fetch timber for the great noble bark of Amen-Re, King of Gods, which is upon the river and [is called] Amen-user-he. On the day of my arrival at Tanis, the place where Smendes and Tentamun are, I gave them the dispatches of Amen-Re, King of Gods. They had read them out before them and they said: “I will do, I will do as Amen-Re, King of Gods, our lord has said.” I stayed until the fourth month of summer in Tanis. Then Smendes and Tentamun sent me off with the ship’s captain Mengebet, and I went down upon the great sea of Phoenicia in the first month summer, day 1. I arrived at Dor, a Tjeker town; and Beder, its prince, had fifty loaves, one jug of wine, and one ox-haunch brought to me. Then a man of my ship fled after stealing one vessel of gold worth 5 deben, four jars of silver worth 20 deben, and a bag with 11 deben of silver; [total of what he stole]: gold 5 deben, silver 31 deben. That morning when I had risen, I went to where the prince was and said to him: “I have been robbed in your harbor. Now you are the prince of this land, you are the one who controls it. Search for my money! Indeed, the money belongs to Amen-Re, King of Gods, the lord of the lands. It belongs to Smendes; it belongs to Herihor, my lord, and [to] the other magnates of Egypt. It belongs to you; it belongs to Weret; it belongs to Mekmer; it belongs to Tjekerbaal, the prince of Byblos!” He said to me: “Are you serious? Are you joking? Indeed, I do not understand the demand you make to me. If it had been a thief belonging to my land who had gone down to your ship and had stolen your money, I would replace it for you from my storehouse, until your thief, whatever his name, had been found. But the thief who robbed you, he is yours, he belongs to your ship. Spend a few days here with me; I will search for him.” I stayed nine days moored in his harbor. Then I went to him and said to him: “Look, you have not found my money. [Let me depart] with the ship captains, with those who go to sea.” [The next eight lines are broken. Apparently, the prince advises Wenamun to wait some more, but Wenamun departs. He passes Tyre and approaches Byblos. Then he seizes thirty deben of silver from a ship he has encountered which belongs to the Tjeker, an obvious act of >piracy. He tells the owners that he will keep the money until his money has been found. Through this action he incurs the enmity of the Tjeker.] They departed and I celebrated [in] a tent on the shore of the sea in the harbor of Byblos. And [I made a hiding place for] Amun-of-the-Road and placed possessions in it. Then the prince of Byblos sent to me saying: “[Leave my] harbor!” I sent to him, saying: “Where shall [I go]? ———-. If [you have a ship to carry me], let me be taken back to Egypt.” I spent twenty-nine days in his harbor, and he spent time sending to me daily to say: “Leave my harbor!” Now while he was offering to his gods, the god took hold of a young man [of] his young men and put him in a trance. He said to him: “Bring [the] god up! Bring the envoy who is carrying him! It is Amun who sent him. It is he who made him come!” Now it was while the entranced one was entranced that night that I had found a ship headed for Egypt. I had loaded all my belongings into it and was watching for the darkness, saying: “When it descends I will load the god so that no other eye shall see him.” Then the harbor master came to me, saying: “Wait until morning, says the prince!” I said to him: “Was it not you who daily took time to come to me, saying: ‘Leave my harbor’? Do you now say: ‘Wait this night,’ in order to let the ship that I found depart, and then you will come to to say: ‘Go away’?” He went and told it to the prince. Then the prince sent to the captain of the ship, saying: “Wait until morning, says the prince.” When morning came, he sent and brought me up, while the god rested in the tent where he was on the shore of the sea. I found him seated in his upper chamber with his back against a window, and the waves of the great sea of Phoenicia broke behind his head. I said to him: “Blessings of Amun!” He said to me: “How long is it to this day since you came from the place where Amun is?” I said to him: “Five whole months till now.” He said to me: “If you are right, where is the dispatch of Amun that was in your hand? Where is the letter of the High Priest of Amun that was in your hand?” I said to him: “I gave them to Smendes and Tentamun.” Then he became very angry and said to me: “Now then, dispatches, letters you have none. Where is the ship of pinewood that Smendes gave you? Where is its Phoenician crew? Did he not entrust you to this foreign ship’s captain in order to have him kill you and have them throw you into the sea? From whom would one then seek the god? And you, from whom would one seek you?” So he said to me. I said to him: “Is it not an Egyptian ship? Those who sail under Smendes are Egyptian crews. He has no Phoenician crews.” He said to me: “Are there not twenty ships here in my harbor. that do business with Smendes? As for Sidon, that other [place] you passed, are there not another fifty ships there that do business with Werekter and haul to this house?” I was silent in this great moment. Then he spoke to me, saying: “On what business have you come?” I said to him: “I have come in quest of timber for the great noble bark of Amen-Re, King of Gods. What your father did, what the father of your father did, you too will do it.” So I said to him. He said to me: “True, they did it. If you pay me for doing it, I will do it. My relations carried out this business after Pharaoh had sent six ships laden with the goods of Egypt, and they had been unloaded into their storehouses. You, what have you brought for me?” He had the daybook of his forefathers brought and had it read before me. They found entered in his book a thousand deben of silver and all sorts of things. He said to me: “If the ruler of Egypt were the lord of what is mine and I were his servant, he would not have sent silver and gold to say: ‘Carry out the business of Amun.’ It was not a royal gift that they gave to my father! I too, I am not your servant, nor am I the servant of him who sent you! If I shout aloud to the Lebanon, the sky opens and the logs lie here on the shore of the sea! Give me the sails you brought to move your ships, loaded with logs for [Egypt]! Give me the ropes you brought [to lash the pines] that I am to fell in order to make them for you —-, ————– that I am to make for you for the sails of your ships; or the yards may be too heavy and break, and you may die [in] the midst of the sea. For Amun makes thunder in the sky ever since he placed Seth beside him! Indeed, Amun has founded all the lands. He founded them after having first founded the land of Egypt from which you have come. Thus craftsmanship came from it in order to reach the place where I am! Thus learning came from it in order to reach the place where I am! What are these foolish travels they made you do?” I said to him: “Wrong! These are not foolish travels that I am doing. There is no ship on the river that does not belong to Amun. His is the sea and his the Lebanon of which you say, ‘It is mine.’ It is a growing ground for Amen-user-he, the lord of every ship. Truly, it was Amen-Re, King of Gods, who said to Herihor, my master: ‘Send me!’ And he made me come with this great god. But look, you have let this great god spend these twenty-nine days moored in your harbor. Did you not know that he was here? Is he not he who he was? You are prepared to haggle over the Lebanon with Amun, its lord? As to your saying, the former kings sent silver and gold: If they had owned life and health, they would not have sent these things. It was in place of life and health that they sent these things to your fathers! But Amen-Re, King of Gods, he is the lord of life and health, and he was the lord of your fathers! They passed their lifetimes offering to Amun. You too, you are the servant of Amun! If you will say ‘I will do’ to Amun, and will carry out his business, you will live, you will prosper, you will be healthy; you will be beneficent to your whole land and your people. Do not desire what belongs to Amun-Re, King of Gods! Indeed, a lion loves his possessions! Have your scribe brought to me that I may send him to Smendes and Tentamun, the pillars Amun has set up for the north of his land; and they will send all that is needed. I will send him to them, saying ‘Have it brought until I return to the south; then I shall refund you all your expenses'”. So I said to him. He placed my letter in the hand of his messenger; and he loaded the keel, the prow-piece, and the stern-piece, together with four other hewn logs, seven in all, and sent them to Egypt. His messenger who had gone to Egypt returned to me in Phoenicia in the first month of winter, Smendes and Tentamun having sent: four jars and one kakmen-vessel of gold; five jars of silver; ten garments of royal linen; ten hrd-garments of fine linen; five hundred smooth linen maats; five hundred ox-hides; five hundred ropes; twenty sacks of lentils; and thirty baskets of fish. And she sent to me: five garments of fine linen; five hrd-garments of fine linen; one sack of lentils; and five baskets of fish. The prince rejoiced. He assigned three hundred men and three hundred oxen, and he set supervisors over them to have them fell the timbers. They were felled and they lay there during the winter. In the third month of summer they dragged them to the shore of the sea. The prince came out and stood by them, and he sent to me saying: “Come!” Now when I had been brought into his presence, the shadow of his sunshade fell on me. Then Penamun, a butler of his, intervened, saying “The shadow of Pharaoh, your lord, has fallen upon you.” And he was angry with him and said: “Leave him alone.” As I stood before him, he addressed me, saying: “Look, the business my fathers did in the past, I have done it, although you did not do for me what your fathers did for mine. Look, the last of your timber has arrived and is ready. Do as I wish, and come to load it. For has it not been given to you? Do not come to look at the terror of the sea. For if you look at the terror of the sea, you will see my own! Indeed, I have not done to you what was done to the envoys of Khaemwese, after they had spent seventeen years in this land. They died on the spot.” And he said to his butler: “Take him to see the tomb where they lie.” I said to him: “Do not make me see it. As for Khaemwese, the envoys he sent you were men and he himself was a man. You have not here one of his envoys, though you say: ‘Go and see your companions.’ Should you not rejoice and have a Stella [made] for yourself, and say on it: ‘Amen-Re, King of Gods sent me Amun-of-the-Road, his envoy, together with Wenamun, his human envoy, in quest of timber for the great noble bark of Amen-Re, King of Gods. I felled it; I loaded it; I supplied my ships and my crews. I let them reach Egypt so as to beg for me from Amun fifty years of life over and above my allotted fate.’ And if it comes to pass that in another day an envoy comes from the land of Egypt who knows writing and he reads out your name on the Stella, you will receive water of the west like the gods who are there.” He said to me: “A great speech of admonition is what you have said to me.” I said to him: “As to the many [things] you have said to me: if I reach the place where the High Priest of Amun is and he sees your accomplishment, it is your accomplishment that will draw profit to you.” I went off to the shore of the sea, to where the logs were lying. And I saw eleven ships that had come in from the sea and belonged to the Tjeker [who were] saying: “Arrest him! Let no ship of his leave for the land of Egypt!” Then I sat down and wept. And the secretary of the prince came out to me and said to me: “What is it?” I said to him: “Do you not see the migrant birds going down to Egypt a second time? Look at them traveling to the cool water! Until when shall I be left here? For do you not see those who have come to arrest me?” He went and told it to the prince. And the prince began to weep on account of the words said to him, for they were painful. He sent his secretary out to me, bringing two jugs of wine and a sheep. And he sent me Tentne, an Egyptian songstress who was with him, saying: “Sing for him! Do not let his heart be anxious.” And he sent to me, saying: “Eat, drink; do not let your heart be anxious. You shall hear what I will say tomorrow.” When morning came, he had his assembly summoned. He stood in their midst and said to the Tjeker: “What have you come for?” They said to him: “We have come after the blasted ships that you are sending to Egypt with our enemy.” He said to them: “I cannot arrest the envoy of Amun in my country. Let me send him off, and you go after him to arrest him.” He had me board and sent off from the harbor of the sea. And the wind drove me to the land of Alasiya. Then the town’s people came out against me to kill me. But I forced my way through them to where Hatiba, the princess of the town was. I met her coming from one of her houses to enter another. I saluted her and said to the people who stood around her: “Is there not one among you who understands Egyptian?” And one among them said: “I understand it.” I said to him: “Tell my lady that I have heard it said as far away as Thebes, the place where Amun is: ‘If wrong is done in every town, in the land of Alasiya right is done.’ Now is wrong done here too every day?” She said: “What is it you have said?” I said to her: “If the sea rages and the wind drives me to the land where you are, will you let me be received so as to kill me, though I am the envoy of Amun? Look, as for me, they would search for me till the end of time. As for this crew of the prince of Byblos, whom they seek to kill, will not their lord find ten crews of yours and kill them also?” She had the people summoned and they were reprimanded. She said to me: “Spend the night…….” CC licensed content, Shared previously Excerpt from The Report of Wenamun: Text & Commentary. Provided by : Ancient History Encyclopedia. Located at : https://www.ancient.eu/article/1087/the-report-of-wenamun-text--commentary/ . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike Public domain content Wenamuns Reise. Provided by : Wikimedia Commons. Located at : https://commons.wikimedia.org/wiki/File:Wenamuns-Reise.jpg . License : Public Domain: No Known Copyright |
Courses/Tennessee_State_University/CHEM_4210%3A_Inorganic_Chem_II_(Siddiquee)/01%3A_Basic_Inorganic_Concepts/1.02%3A_Atomic_Structure/1.2.01%3A_Historical_Development_of_Atomic_Theory | Notes about this page The page below is a brief overview on the history of atomic theory. It contains lots of video media . Alternatives pages on the history of atomic theory are: A more detailed overview that includes practice problems is here (click) . Another brief overview is here (click) . Skills to Develop By the end of this section, you will be able to: State the postulates of Dalton’s atomic theory Use postulates of Dalton’s atomic theory to explain the laws of definite and multiple proportions Outline milestones in the development of modern atomic theory Summarize and interpret the results of the experiments of Thomson, Millikan, and Rutherford A Video Introduction to Atomic Theory through the Nineteenth Century From Crash Course Chemistry Video \(\PageIndex{1}\): Lavoisier's discovery of The Law of Conservation of Matter led to the Laws of Definite and Multiple Proportions and eventually Dalton's Atomic Theory. Atomic Theory through the Nineteenth Century The earliest recorded discussion of the basic structure of matter comes from ancient Greek philosophers, the scientists of their day. In the fifth century BC, Leucippus and Democritus argued that all matter was composed of small, finite particles that they called atomos , a term derived from the Greek word for “indivisible.” They thought of atoms as moving particles that differed in shape and size, and which could join together. Later, Aristotle and others came to the conclusion that matter consisted of various combinations of the four “elements”—fire, earth, air, and water—and could be infinitely divided. Interestingly, these philosophers thought about atoms and “elements” as philosophical concepts, but apparently never considered performing experiments to test their ideas. The Aristotelian view of the composition of matter held sway for over two thousand years, until English schoolteacher John Dalton helped to revolutionize chemistry with his hypothesis that the behavior of matter could be explained using an atomic theory. First published in 1807, many of Dalton’s hypotheses about the microscopic features of matter are still valid in modern atomic theory. Here are the postulates of Dalton’s atomic theory . Matter is composed of exceedingly small particles called atoms. An atom is the smallest unit of an element that can participate in a chemical change. An element consists of only one type of atom, which has a mass that is characteristic of the element and is the same for all atoms of that element (Figure \(\PageIndex{1}\)). A macroscopic sample of an element contains an incredibly large number of atoms, all of which have identical chemical properties. Atoms of one element differ in properties from atoms of all other elements. A compound consists of atoms of two or more elements combined in a small, whole-number ratio. In a given compound, the numbers of atoms of each of its elements are always present in the same ratio (Figure \(\PageIndex{2}\)). Atoms are neither created nor destroyed during a chemical change, but are instead rearranged to yield substances that are different from those present before the change (Figure \(\PageIndex{3}\)). Figure \(\PageIndex{1}\): A pre-1982 copper penny (left) contains approximately 3 \(\times\) 10 22 copper atoms (several dozen are represented as brown spheres at the right), each of which has the same chemical properties. (credit: modification of work by “slgckgc”/Flickr) Figure \(\PageIndex{2}\): Copper(II) oxide, a powdery, black compound, results from the combination of two types of atoms—copper (brown spheres) and oxygen (red spheres)—in a 1:1 ratio. (credit: modification of work by “Chemicalinterest”/Wikimedia Commons) Figure \(\PageIndex{3}\) : When the elements copper (a shiny, red-brown solid, shown here as brown spheres) and oxygen (a clear and colorless gas, shown here as red spheres) react, their atoms rearrange to form a compound containing copper and oxygen (a powdery, black solid). (credit copper: modification of work by http://images-of-elements.com/copper.php ). Dalton’s atomic theory provides a microscopic explanation of the many macroscopic properties of matter that you’ve learned about. For example, if an element such as copper consists of only one kind of atom, then it cannot be broken down into simpler substances, that is, into substances composed of fewer types of atoms. And if atoms are neither created nor destroyed during a chemical change, then the total mass of matter present when matter changes from one type to another will remain constant (the law of conservation of matter (or mass)). Want to learn more about the Law of Conservation of Mass? Video \(\PageIndex{2}\): "We are made of star stuff" - Carl Sagan . Example \(\PageIndex{1}\): Test ing Dalton’s Atomic Theory In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one? Solution The starting materials consist of two green spheres and two purple spheres. The products consist of only one green sphere and one purple sphere. This violates Dalton’s postulate that atoms are neither created nor destroyed during a chemical change, but are merely redistributed. (In this case, atoms appear to have been destroyed.) Exercise \(\PageIndex{1}\) In the following drawing, the green spheres represent atoms of a certain element. The purple spheres represent atoms of another element. If the spheres touch, they are part of a single unit of a compound. Does the following chemical change represented by these symbols violate any of the ideas of Dalton’s atomic theory? If so, which one Answer The starting materials consist of four green spheres and two purple spheres. The products consist of four green spheres and two purple spheres. This does not violate any of Dalton’s postulates: Atoms are neither created nor destroyed, but are redistributed in small, whole-number ratios. Dalton knew of the experiments of French chemist Joseph Proust, who demonstrated that all samples of a pure compound contain the same elements in the same proportion by mass . This statement is known as the law of definite proportions or the law of constant composition . The suggestion that the numbers of atoms of the elements in a given compound always exist in the same ratio is consistent with these observations. For example, when different samples of isooctane (a component of gasoline and one of the standards used in the octane rating system) are analyzed, they are found to have a carbon-to-hydrogen mass ratio of 5.33:1, as shown in Table \(\PageIndex{1}\). Sample Carbon Hydrogen Mass Ratio A 14.82 g 2.78 g \(\mathrm{\dfrac{14.82\: g\: carbon}{2.78\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}\) B 22.33 g 4.19 g \(\mathrm{\dfrac{22.33\: g\: carbon}{4.19\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}\) C 19.40 g 3.64 g \(\mathrm{\dfrac{19.40\: g\: carbon}{3.63\: g\: hydrogen}=\dfrac{5.33\: g\: carbon}{1.00\: g\: hydrogen}}\) It is worth noting that although all samples of a particular compound have the same mass ratio, the converse is not true in general. That is, samples that have the same mass ratio are not necessarily the same substance. For example, there are many compounds other than isooctane that also have a carbon-to-hydrogen mass ratio of 5.33:1.00. Dalton also used data from Proust, as well as results from his own experiments, to formulate another interesting law. The law of multiple proportions states that when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small, whole numbers . For example, copper and chlorine can form a green, crystalline solid with a mass ratio of 0.558 g chlorine to 1 g copper, as well as a brown crystalline solid with a mass ratio of 1.116 g chlorine to 1 g copper. These ratios by themselves may not seem particularly interesting or informative; however, if we take a ratio of these ratios, we obtain a useful and possibly surprising result: a small, whole-number ratio. \[\mathrm{\dfrac{\dfrac{1.116\: g\: Cl}{1\: g\: Cu}}{\dfrac{0.558\: g\: Cl}{1\: g\: Cu}}=\dfrac{2}{1}}\] This 2-to-1 ratio means that the brown compound has twice the amount of chlorine per amount of copper as the green compound. This can be explained by atomic theory if the copper-to-chlorine ratio in the brown compound is 1 copper atom to 2 chlorine atoms, and the ratio in the green compound is 1 copper atom to 1 chlorine atom. The ratio of chlorine atoms (and thus the ratio of their masses) is therefore 2 to 1 (Figure \(\PageIndex{4}\)). Figure \(\PageIndex{4}\) : Compared to the copper chlorine compound in (a), where copper is represented by brown spheres and chlorine by green spheres, the copper chlorine compound in (b) has twice as many chlorine atoms per copper atom. (credit a: modification of work by “Benjah-bmm27”/Wikimedia Commons; credit b: modification of work by “Walkerma”/Wikimedia Commons) Example \(\PageIndex{2}\): L aws of Definite and Multiple Proportions A sample of compound A (a clear, colorless gas) is analyzed and found to contain 4.27 g carbon and 5.69 g oxygen. A sample of compound B (also a clear, colorless gas) is analyzed and found to contain 5.19 g carbon and 13.84 g oxygen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances A and B? Solution In compound A, the mass ratio of carbon to oxygen is: \[\mathrm{\dfrac{1.33\: g\: O}{1\: g\: C}} \nonumber\] In compound B, the mass ratio of carbon to oxygen is: \[\mathrm{\dfrac{2.67\: g\: O}{1\: g\: C}} \nonumber\] The ratio of these ratios is: \[\mathrm{\dfrac{\dfrac{1.33\: g\: O}{1\: g\: C}}{\dfrac{2.67\: g\: O}{1\: g\: C}}=\dfrac{1}{2}} \nonumber\] This supports the law of multiple proportions. This means that A and B are different compounds, with A having one-half as much carbon per amount of oxygen (or twice as much oxygen per amount of carbon) as B. A possible pair of compounds that would fit this relationship would be A = CO 2 and B = CO. Exercise \(\PageIndex{2}\) A sample of compound X (a clear, colorless, combustible liquid with a noticeable odor) is analyzed and found to contain 14.13 g carbon and 2.96 g hydrogen. A sample of compound Y (a clear, colorless, combustible liquid with a noticeable odor that is slightly different from X’s odor) is analyzed and found to contain 19.91 g carbon and 3.34 g hydrogen. Are these data an example of the law of definite proportions, the law of multiple proportions, or neither? What do these data tell you about substances X and Y? Answer In compound X, the mass ratio of carbon to hydrogen is \(\mathrm{\dfrac{14.13\: g\: C}{2.96\: g\: H}}\). In compound Y, the mass ratio of carbon to oxygen is \(\mathrm{\dfrac{19.91\: g\: C}{3.34\: g\: H}}\). The ratio of these ratios is \[\mathrm{\dfrac{\dfrac{14.13\: g\: C}{2.96\: g\: H}}{\dfrac{19.91\: g\: C}{3.34\: g\: H}}=\dfrac{4.77\: g\: C/g\: H}{5.96\: g\: C/g\: H}=0.800=\dfrac{4}{5}}. \nonumber\] This small, whole-number ratio supports the law of multiple proportions. This means that X and Y are different compounds. In the two centuries since Dalton developed his ideas, scientists have made significant progress in furthering our understanding of atomic theory. Much of this came from the results of several seminal experiments that revealed the details of the internal structure of atoms. Here, we will discuss some of those key developments, with an emphasis on application of the scientific method, as well as understanding how the experimental evidence was analyzed. While the historical persons and dates behind these experiments can be quite interesting, it is most important to understand the concepts resulting from their work. Atomic Theory after the Nineteenth Century If matter were composed of atoms, what were atoms composed of? Were they the smallest particles, or was there something smaller? In the late 1800s, a number of scientists interested in questions like these investigated the electrical discharges that could be produced in low-pressure gases, with the most significant discovery made by English physicist J. J. Thomson using a cathode ray tube. This apparatus consisted of a sealed glass tube from which almost all the air had been removed; the tube contained two metal electrodes. When high voltage was applied across the electrodes, a visible beam called a cathode ray appeared between them. This beam was deflected toward the positive charge and away from the negative charge, and was produced in the same way with identical properties when different metals were used for the electrodes. In similar experiments, the ray was simultaneously deflected by an applied magnetic field, and measurements of the extent of deflection and the magnetic field strength allowed Thomson to calculate the charge-to-mass ratio of the cathode ray particles. The results of these measurements indicated that these particles were much lighter than atoms (Figure \(\PageIndex{1}\)). Figure \(\PageIndex{5}\) : (a) J. J. Thomson produced a visible beam in a cathode ray tube. (b) This is an early cathode ray tube, invented in 1897 by Ferdinand Braun. (c) In the cathode ray, the beam (shown in yellow) comes from the cathode and is accelerated past the anode toward a fluorescent scale at the end of the tube. Simultaneous deflections by applied electric and magnetic fields permitted Thomson to calculate the mass-to-charge ratio of the particles composing the cathode ray. (credit a: modification of work by Nobel Foundation; credit b: modification of work by Eugen Nesper; credit c: modification of work by “Kurzon”/Wikimedia Commons). Based on his observations, here is what Thomson proposed and why: The particles are attracted by positive (+) charges and repelled by negative (−) charges, so they must be negatively charged (like charges repel and unlike charges attract); they are less massive than atoms and indistinguishable, regardless of the source material, so they must be fundamental, subatomic constituents of all atoms. Although controversial at the time, Thomson’s idea was gradually accepted, and his cathode ray particle is what we now call an electron , a negatively charged, subatomic particle with a mass more than one thousand-times less that of an atom. The term “electron” was coined in 1891 by Irish physicist George Stoney, from “ electr ic i on .” In 1909, more information about the electron was uncovered by American physicist Robert A. Millikan via his “oil drop” experiments. Millikan created microscopic oil droplets, which could be electrically charged by friction as they formed or by using X-rays. These droplets initially fell due to gravity, but their downward progress could be slowed or even reversed by an electric field lower in the apparatus. By adjusting the electric field strength and making careful measurements and appropriate calculations, Millikan was able to determine the charge on individual drops (Figure \(\PageIndex{2}\)). Figure \(\PageIndex{6}\): Millikan’s experiment measured the charge of individual oil drops. The tabulated data are examples of a few possible values. Looking at the charge data that Millikan gathered, you may have recognized that the charge of an oil droplet is always a multiple of a specific charge, 1.6 \(\times\) 10 −19 C. Millikan concluded that this value must therefore be a fundamental charge—the charge of a single electron—with his measured charges due to an excess of one electron (1 times 1.6 \(\times\) 10 −19 C), two electrons (2 times 1.6 \(\times\) 10 −19 C), three electrons (3 times 1.6 \(\times\) 10 −19 C), and so on, on a given oil droplet. Since the charge of an electron was now known due to Millikan’s research, and the charge-to-mass ratio was already known due to Thomson’s research (1.759 \(\times\) 10 11 C/kg), it only required a simple calculation to determine the mass of the electron as well. \[\mathrm{Mass\: of\: electron=1.602\times 10^{-19}\:\cancel{C}\times \dfrac{1\: kg}{1.759\times 10^{11}\:\cancel{C}}=9.107\times 10^{-31}\:kg} \tag{2.3.1}\] Scientists had now established that the atom was not indivisible as Dalton had believed, and due to the work of Thomson, Millikan, and others, the charge and mass of the negative, subatomic particles—the electrons—were known. However, the positively charged part of an atom was not yet well understood. In 1904, Thomson proposed the “plum pudding” model of atoms, which described a positively charged mass with an equal amount of negative charge in the form of electrons embedded in it, since all atoms are electrically neutral. A competing model had been proposed in 1903 by Hantaro Nagaoka , who postulated a Saturn-like atom, consisting of a positively charged sphere surrounded by a halo of electrons (Figure \(\PageIndex{3}\)). Figure \(\PageIndex{7}\) : (a) Thomson suggested that atoms resembled plum pudding, an English dessert consisting of moist cake with embedded raisins (“plums”). (b) Nagaoka proposed that atoms resembled the planet Saturn, with a ring of electrons surrounding a positive “planet.” (credit a: modification of work by “Man vyi”/Wikimedia Commons; credit b: modification of work by “NASA”/Wikimedia Commons). The next major development in understanding the atom came from Ernest Rutherford , a physicist from New Zealand who largely spent his scientific career in Canada and England. He performed a series of experiments using a beam of high-speed, positively charged alpha particles (α particles) that were produced by the radioactive decay of radium; α particles consist of two protons and two neutrons (you will learn more about radioactive decay in the chapter on nuclear chemistry). Rutherford and his colleagues Hans Geiger (later famous for the Geiger counter) and Ernest Marsden aimed a beam of α particles, the source of which was embedded in a lead block to absorb most of the radiation, at a very thin piece of gold foil and examined the resultant scattering of the α particles using a luminescent screen that glowed briefly where hit by an α particle. What did they discover? Most particles passed right through the foil without being deflected at all. However, some were diverted slightly, and a very small number were deflected almost straight back toward the source (Figure \(\PageIndex{4}\)). Rutherford described finding these results: “It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you” 1 (p. 68). Figure \(\PageIndex{8}\): Geiger and Rutherford fired α particles at a piece of gold foil and detected where those particles went, as shown in this schematic diagram of their experiment. Most of the particles passed straight through the foil, but a few were deflected slightly and a very small number were significantly deflected. Here is what Rutherford deduced: Because most of the fast-moving α particles passed through the gold atoms undeflected, they must have traveled through essentially empty space inside the atom. Alpha particles are positively charged, so deflections arose when they encountered another positive charge (like charges repel each other). Since like charges repel one another, the few positively charged α particles that changed paths abruptly must have hit, or closely approached, another body that also had a highly concentrated, positive charge. Since the deflections occurred a small fraction of the time, this charge only occupied a small amount of the space in the gold foil. Analyzing a series of such experiments in detail, Rutherford drew two conclusions: The volume occupied by an atom must consist of a large amount of empty space. A small, relatively heavy, positively charged body, the nucleus , must be at the center of each atom. This analysis led Rutherford to propose a model in which an atom consists of a very small, positively charged nucleus, in which most of the mass of the atom is concentrated, surrounded by the negatively charged electrons, so that the atom is electrically neutral (Figure \(\PageIndex{5}\)). Figure \(\PageIndex{9}\) : The α particles are deflected only when they collide with or pass close to the much heavier, positively charged gold nucleus. Because the nucleus is very small compared to the size of an atom, very few α particles are deflected. Most pass through the relatively large region occupied by electrons, which are too light to deflect the rapidly moving particles. After many more experiments, Rutherford also discovered that the nuclei of other elements contain the hydrogen nucleus as a “building block,” and he named this more fundamental particle the proton , the positively charged, subatomic particle found in the nucleus. With one addition, which you will learn next, this nuclear model of the atom, proposed over a century ago, is still used today. Another important finding was the discovery of isotopes. During the early 1900s, scientists identified several substances that appeared to be new elements, isolating them from radioactive ores. For example, a “new element” produced by the radioactive decay of thorium was initially given the name mesothorium. However, a more detailed analysis showed that mesothorium was chemically identical to radium (another decay product), despite having a different atomic mass. This result, along with similar findings for other elements, led the English chemist Frederick Soddy to realize that an element could have types of atoms with different masses that were chemically indistinguishable. These different types are called isotopes —atoms of the same element that differ in mass. Soddy was awarded the Nobel Prize in Chemistry in 1921 for this discovery. One puzzle remained: The nucleus was known to contain almost all of the mass of an atom, with the number of protons only providing half, or less, of that mass. Different proposals were made to explain what constituted the remaining mass, including the existence of neutral particles in the nucleus. As you might expect, detecting uncharged particles is very challenging, and it was not until 1932 that James Chadwick found evidence of neutrons , uncharged, subatomic particles with a mass approximately the same as that of protons. The existence of the neutron also explained isotopes: They differ in mass because they have different numbers of neutrons, but they are chemically identical because they have the same number of protons. This will be explained in more detail later in this unit. Video \(\PageIndex{2}\): An Introduction to Subatomic Particles Summary Video \(\PageIndex{3}\): A summary of discoveries in atomic theory. Video \(\PageIndex{4}\): A different summary of discoveries in atomic theory. The ancient Greeks proposed that matter consists of extremely small particles called atoms. Dalton postulated that each element has a characteristic type of atom that differs in properties from atoms of all other elements, and that atoms of different elements can combine in fixed, small, whole-number ratios to form compounds. Samples of a particular compound all have the same elemental proportions by mass. When two elements form different compounds, a given mass of one element will combine with masses of the other element in a small, whole-number ratio. During any chemical change, atoms are neither created nor destroyed. Although no one has actually seen the inside of an atom, experiments have demonstrated much about atomic structure. Thomson’s cathode ray tube showed that atoms contain small, negatively charged particles called electrons. Millikan discovered that there is a fundamental electric charge—the charge of an electron. Rutherford’s gold foil experiment showed that atoms have a small, dense, positively charged nucleus; the positively charged particles within the nucleus are called protons. Chadwick discovered that the nucleus also contains neutral particles called neutrons. Soddy demonstrated that atoms of the same element can differ in mass; these are called isotopes. Footnotes Ernest Rutherford, “The Development of the Theory of Atomic Structure,” ed. J. A. Ratcliffe, in Background to Modern Science , eds. Joseph Needham and Walter Pagel, (Cambridge, UK: Cambridge University Press, 1938), 61–74. Accessed September 22, 2014, https://ia600508.us.archive.org/3/it...e032734mbp.pdf . Glossary Dalton’s atomic theory set of postulates that established the fundamental properties of atoms law of constant composition (also, law of definite proportions) all samples of a pure compound contain the same elements in the same proportions by mass law of multiple proportions when two elements react to form more than one compound, a fixed mass of one element will react with masses of the other element in a ratio of small whole numbers law of definite proportions (also, law of constant composition) all samples of a pure compound contain the same elements in the same proportions by mass alpha particle (α particle) positively charged particle consisting of two protons and two neutrons electron negatively charged, subatomic particle of relatively low mass located outside the nucleus isotopes atoms that contain the same number of protons but different numbers of neutrons neutron uncharged, subatomic particle located in the nucleus proton positively charged, subatomic particle located in the nucleus nucleus massive, positively charged center of an atom made up of protons and neutrons Contributors Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ). Adelaide Clark, Oregon Institute of Technology Crash Course Chemistry: Crash Course is a division of Complexly and videos are free to stream for educational purposes. TED-Ed’s commitment to creating lessons worth sharing is an extension of TED’s mission of spreading great ideas. Within TED-Ed’s growing library of TED-Ed animations, you will find carefully curated educational videos, many of which represent collaborations between talented educators and animators nominated through the TED-Ed website . Feedback Have feedback to give about this text? Click here . Found a typo and want extra credit? Click here . |
Courses/Pasadena_City_College/Chem_2A_(Ku)_Textbook/13%3A_Acids_and_Bases/13.03%3A_Maintaining_pH_Using_Buffer_Solutions/13.3.01%3A_Biological_Buffers | Identify and understand the buffer systems in the body Buffer Systems in the Body The buffer systems in the human body are extremely efficient, and different systems work at different rates. It takes only seconds for the chemical buffers in the blood to make adjustments to pH. The respiratory tract can adjust the blood pH upward in minutes by exhaling CO 2 from the body. The renal system can also adjust blood pH through the excretion of hydrogen ions (H + ) and the conservation of bicarbonate, but this process takes hours to days to have an effect. The buffer systems functioning in blood plasma include plasma proteins, phosphate, and bicarbonate and carbonic acid buffers. The kidneys help control acid-base balance by excreting hydrogen ions and generating bicarbonate that helps maintain blood plasma pH within a normal range. Protein buffer systems work predominantly inside cells. Phosphate Buffer Phosphates are found in the blood in two forms: sodium dihydrogen phosphate (Na 2 H 2 PO 4 − ), which is a weak acid, and sodium monohydrogen phosphate (Na 2 HPO 4 2 - ), which is a weak base. When Na 2 HPO 4 2 - comes into contact with a strong acid, such as HCl, the base picks up a second hydrogen ion to form the weak acid Na 2 H 2 PO 4 − and sodium chloride, NaCl. When Na 2 HPO 4 2 − (the weak acid) comes into contact with a strong base, such as sodium hydroxide (NaOH), the weak acid reverts back to the weak base and produces water. Acids and bases are still present, but they hold onto the ions. HCl + Na 2 HPO 4 → NaH 2 PO 4 + NaCl (strong acid) + (weak base) → (weak acid) + (salt) NaOH + NaH 2 PO 4 → Na 2 HPO 4 + H 2 O (strong base) + (weak acid) → (weak base) + (water) Bicarbonate-Carbonic Acid Buffer Maintaining a constant blood pH is critical for the proper functioning of our body. The buffer that maintains the pH of human blood involves a carbonic acid (H 2 CO 3 ) - bicarbonate ion (HCO 3 - ) When any acidic substance enters the bloodstream, the bicarbonate ions neutralize the hydronium ions forming carbonic acid and water. Carbonic acid is already a component of the buffering system of blood. Thus hydronium ions are removed, preventing the pH of blood from becoming acidic. On the other hand, when a basic substance enters the bloodstream, carbonic acid reacts with the hydroxide ions producing bicarbonate ions and water. Bicarbonate ions are already a component of the buffer. In this manner, the hydroxide ions are removed from blood, preventing the pH of blood from becoming basic. As depicted below, in the process of neutralizing hydronium ions or hydroxide ions, the relative concentrations of carbonic acid (H 2 CO 3 ) and bicarbonate ions (HCO 3 - ) fluctuate in the bloodstream. But this slight change in the concentrations of the two components of the buffering system doesn’t have any adverse effect; the critical thing is that this buffering mechanism prevents the blood from becoming acidic or basic, which can be detrimental. The pH of blood is maintained at ~ 7.4 by the carbonic acid–bicarbonate ion buffering system. Disorders of Acid-Base Balance Respiratory Acidosis: Primary Carbonic Acid/CO 2 Excess Respiratory acidosis occurs when the blood is overly acidic due to an excess of carbonic acid, resulting from too much CO 2 in the blood. Respiratory acidosis can result from anything that interferes with respiration, such as pneumonia, emphysema, or congestive heart failure. Respiratory Alkalosis: Primary Carbonic Acid/CO 2 Deficiency Respiratory alkalosis occurs when the blood is overly alkaline due to a deficiency in carbonic acid and CO 2 levels in the blood. This condition usually occurs when too much CO 2 is exhaled from the lungs, as occurs in hyperventilation, which is breathing that is deeper or more frequent than normal. An elevated respiratory rate leading to hyperventilation can be due to extreme emotional upset or fear, fever, infections, hypoxia, or abnormally high levels of catecholamines, such as epinephrine and norepinephrine. Surprisingly, aspirin overdose—salicylate toxicity—can result in respiratory alkalosis as the body tries to compensate for initial acidosis. Respiratory Compensation Respiratory compensation for metabolic acidosis increases the respiratory rate to drive off CO 2 and readjust the bicarbonate to carbonic acid ratio to the 20:1 level. This adjustment can occur within minutes. Respiratory compensation for metabolic alkalosis is not as adept as its compensation for acidosis. The normal response of the respiratory system to elevated pH is to increase the amount of CO 2 in the blood by decreasing the respiratory rate to conserve CO 2 . There is a limit to the decrease in respiration, however, that the body can tolerate. Hence, the respiratory route is less efficient at compensating for metabolic alkalosis than for acidosis. Key Takeaway |
Courses/Williams_School/Advanced_Chemistry/18%3A_Ionic_and_Covalent_Solids_-_Structures/18.03%3A_Structures_Related_to_NaCl_and_NiAs | There are a number of compounds that have structures similar to that of NaCl, but have a lower symmetry (usually imposed by the geometry of the anion) than NaCl itself. These compounds include: FeS 2 (pyrite, "fools gold"): S 2 2 - (disulfide) and Fe 2 + CaC 2 (a salt-like carbide): Ca 2 + and linear C 2 2 - anions CaCO 3 (calcite, limestone, marble): Ca 2 + and triangular CO 3 2 - . 0 The rhombohedral unit cell of the calcite crystal structure. The hexagonal c-axis is shown. The calcite (CaCO 3 ) crystal structure is shown above. Triangular CO 3 2 - ions fill octahedral holes between the Ca 2 + ions (black spheres) in a distorted NaCl lattice. As in NaCl, each ion is coordinated by six of the other kind. From this image we can see why the CaCO 3 structure has a lower symmetry than that of NaCl. The fourfold rotation symmetry of the NaCl unit cell is lost when the spherical Cl - ions are replaced by triangular CO 3 2 - ions. Because of this symmetry lowering, transparent crystals calcite are birefringent, as illustrated below. 0 Calcite crystals are birefringent, meaning that their refractive indices are different along the two principal crystal directions. This gives rise to the phenomenon of double refraction. NiAs structure The NaCl structure can be described a face-centered cubic lattice with all of the octahedral holes filled. What if we start with a hexagonal-close packed lattice rather than a face-centered cubic lattice? 0 Nickel arsenide crystal structure. The Ni6As trigonal prisms are shaded gray. One octahedron of six As atoms surrounding a Ni atom is shown in the center of the figure. This is the structure adopted by NiAs and many other transition metal sulfides, phosphides, and arsenides. The cations are shown in gray while the anions are light blue in the figure at the right. The cations are in octahedral coordination, so each cation is coordinated to six anions. The anions are also coordinated to six cations, but they occupy trigonal prismatic sites. In terms of layer stacking, the NiAs structure is AcBcAcBc..., where the A and B sites (the hcp lattice) are occupied by the As atoms, and the c sites, which are eclipsed along the layer stacking axis, are occupied by Ni. Unlike the NaCl structure, where the anion and cation sites are interchangeable, NiAs has unique anion and cation sites. The layer stacking sequence for NiAs is shown below: ------------ A - - -c- - - - ------------ B - - -c- - - - ------------ A - - -c- - - - ------------ B - - -c- - - - The NiAs structure cannot be adopted by ionic compounds because of the eclipsing cations, because the cation-cation repulsions would be internally destabilizing for an ionic compound. This structure is mainly adopted by covalent and polar covalent MX compounds, typically with "soft" X anions (S, Se, P, As,....) and low-valent transition metal cations. For example, some compounds with the NiAs structure are: MS, MSe, MTe (M=Ti, V, Fe, Co, Ni). Often these are nonstoichiometric or complex stoichiometries with ordered vacancies (Cr 7 S 8 , Fe 7 S 8 ). |
Courses/Alma_College/Organic_Chemistry_I_(Alma_College)/05%3A_Stereochemistry_at_Tetrahedral_Centers/5.05%3A_The_Reason_for_Handedness_in_Molecules_-_Chirality | Objectives After completing this section, you should be able to determine whether or not a compound is chiral, given its Kelulé, condensed or shorthand structure, with or without the aid of molecular models. label the chiral centres (carbon atoms) in a given Kelulé, condensed or shorthand structure. Key Terms Make certain that you can define, and use in context, the key terms below. achiral chiral chiral (stereogenic) centre plane of symmetry Symmetry and Chirality Molecules that are nonsuperimposable mirror images of each other are said to be chiral (pronounced “ky-ral,” from the Greek cheir, meaning “hand”). Examples of some familiar chiral objects are your hands. Your left and right hands are nonsuperimposable mirror images. (Try putting your right shoe on your left foot—it just doesn’t work.) An achiral object is one that can be superimposed on its mirror image, as shown by the superimposed flasks 25.7. 1 b in the figure below. 25.7. 1 b . An an important questions is why is one chiral and the other not? The answer is that the flask has a plane of symmetry and your hand does not. A plane of symmetry is a plane or a line through an object which divides the object into two halves that are mirror images of each other. When looking at the flask, a line can be drawn down the middle which separates it into two mirror image halves. However, a similar line down the middle of a hand separates it into two non-mirror image halves. This idea can be used to predict chirality. If an object or molecule has a plane of symmetry it is achiral. If if lacks a plane of symmetry it is chiral. Symmetry can be used to explain why a carbon bonded to four different substituents is chiral. When a carbon is bonded to fewer than four different substituents it will have a plane of symmetry making it achiral. A carbon atom that is bonded to four different substituents loses all symmetry, and is often referred to as an asymmetric carbon. The lack of a plane of symmetry makes the carbon chiral. The presence of a single chiral carbon atom sufficient to render the molecule chiral, and modern terminology refers to such groupings as chiral centers or stereo centers. An example is shown in the bromochlorofluoromethane molecule shown in part (a) of the figure below. This carbon, is attached to four different substituents making it chiral. which is often designated by an asterisk in structural drawings. If the bromine atom is replaced by another chlorine to make dichlorofluoromethane, as shown in part (b) below, the molecule and its mirror image can now be superimposed by simple rotation. Thus the carbon is no longer a chiral center. Upon comparison, bromochlorofluoromethane lacks a plane of symmetry while dichlorofluoromethane has a plane of symmetry. Identifying Chiral carbons Identifying chiral carbons in a molecule is an important skill for organic chemists. The presence of a chiral carbon presents the possibility of a molecule having multiple stereoisomers. Most of the chiral centers we shall discuss in this chapter are asymmetric carbon atoms, but it should be recognized that other tetrahedral or pyramidal atoms may become chiral centers if appropriately substituted. Also, when more than one chiral center is present in a molecular structure, care must be taken to analyze their relationship before concluding that a specific molecular configuration is chiral or achiral. This aspect of stereoisomerism will be treated later. Because an carbon requires four different substituents to become asymmertric, it can be said, with few exceptions, that sp 2 and sp hybridized carbons involved in multiple bonds are achiral. Also, any carbon with more than one hydrogen, such as a -CH 3 or -CH 2 - group, are also achiral. Looking for planes of symmetry in a molecule is useful, but often difficult in practice. It is difficult to illustrate on the two dimensional page, but you will see if you build models of these achiral molecules that, in each case, there is at least one plane of symmetry , where one side of the plane is the mirror image of the other. In most cases, the easiest way to decide whether a molecule is chiral or achiral is to look for one or more stereocenters - with a few rare exceptions, the general rule is that molecules with at least one stereocenter are chiral, and molecules with no stereocenters are achiral. Determining if a carbon is bonded to four distinctly different substituents can often be difficult to ascertain. Remember even the slightest difference makes a substituent unique. Often these difference can be distant from the chiral carbon itself. Careful consideration and often the building of molecular models may be required. A good example is shown below. It may appear that the molecule is achiral, however, when looking at the groups directly attached to the possible chiral carbon, it is clear that they all different. The two alkyl groups are differ by a single -CH 2 - group which is enough to consider them different. Example \(\PageIndex{1}\) Predict if the following molecule would be chiral or achiral: Answer Achiral. When determining the chirality of a molecule, it best to start by locating any chiral carbons. An obvious candidate is the ring carbon attached to the methyl substituent. The question then becomes: does the ring as two different substituents making the substituted ring carbon chiral? With an uncertainty such as this, it is then helpful try to identify any planes of symmetry in the molecule. This molecule does have a plane of symmetry making the molecule achiral. The plane of symmetry would be easier see if the molecule were view from above. Typically, monosubstituted cycloalkanes have a similar plane of symmetry making them all achiral. Exercise 5.2.1 Determine if each of the following molecules are chiral or achiral. For chiral molecules indicate any chiral carbons. Answer Explanation Structures F and G are achiral. The former has a plane of symmetry passing through the chlorine atom and bisecting the opposite carbon-carbon bond. The similar structure of compound E does not have such a symmetry plane, and the carbon bonded to the chlorine is a chiral center (the two ring segments connecting this carbon are not identical). Structure G is essentially flat. All the carbons except that of the methyl group are sp 2 hybridized, and therefore trigonal-planar in configuration. Compounds C, D & H have more than one chiral center, and are also chiral. Note In the 1960’s, a drug called thalidomide was widely prescribed in the Western Europe to alleviate morning sickness in pregnant women. Thalidomide had previously been used in other countries as an antidepressant, and was believed to be safe and effective for both purposes. The drug was not approved for use in the U.S.A. It was not long, however, before doctors realized that something had gone horribly wrong: many babies born to women who had taken thalidomide during pregnancy suffered from severe birth defects. Researchers later realized the problem lay in the fact that thalidomide was being provided as a mixture of two different isomeric forms. One of the isomers is an effective medication, the other caused the side effects. Both isomeric forms have the same molecular formula and the same atom-to-atom connectivity, so they are not constitutional isomers. Where they differ is in the arrangement in three-dimensional space about one tetrahedral, sp 3 -hybridized carbon. These two forms of thalidomide are stereoisomers . If you make models of the two stereoisomers of thalidomide, you will see that they too are mirror images, and cannot be superimposed. As a historical note, thalidomide was never approved for use in the United States. This was thanks in large part to the efforts of Dr. Frances Kelsey, a Food and Drug officer who, at peril to her career, blocked its approval due to her concerns about the lack of adequate safety studies, particularly with regard to the drug's ability to enter the bloodstream of a developing fetus. Unfortunately, though, at that time clinical trials for new drugs involved widespread and unregulated distribution to doctors and their patients across the country, so families in the U.S. were not spared from the damage caused. Very recently a close derivative of thalidomide has become legal to prescribe again in the United States, with strict safety measures enforced, for the treatment of a form of blood cancer called multiple myeloma. In Brazil, thalidomide is used in the treatment of leprosy - but despite safety measures, children are still being born with thalidomide-related defects. Example 5.2.2 Label the molecules below as chiral or achiral, and locate all stereocenters. Answer Exercise 5.2.2 1) For the following compounds, star (*) each chiral center, if any. 2) Explain why the following compound is chiral. 3) Determine which of the following objects is chiral. a) A Glove. b) A nail. c) A pair of sunglasses. d) The written word "Chiral". 4) Place an "*" by all of the chrial carbons in the following molecules. a) Erythrose, a four carbon sugar. b) Isoflurane, an anestetic. Bright green = Chlorine, Pale green = Fluorine. Answer 1) 2) Though the molecule does not contain a chiral carbon, it is chiral as it is non-superimposable on its mirror image due to its twisted nature (the twist comes from the structure of the double bonds needing to be at 90° angles to each other, preventing the molecule from being planar). 3) a) Just as hands are chiral a glove must also be chiral. b) A nail has a plane of symmetry which goes down the middle making it a achiral. c) A pair of sunglasses has a plane of symmetry which goes through the nose making it achiral. d) Most written words are chiral. Look one in a mirror to confirm this. 4 a) b) Exercise 5.2.3 Circle all of the carbon stereocenters in the molecules below. Answer Exercise 5.2.4 Circle all of the carbon stereocenters in the molecules below. Answer Here are some more examples of chiral molecules that exist as pairs of enantiomers. In each of these examples, there is a single stereocenter, indicated with an arrow. (Many molecules have more than one stereocenter, but we will get to that that a little later!) Here are some examples of molecules that are achiral ( not chiral). Notice that none of these molecules has a stereocenter. It is difficult to illustrate on the two dimensional page, but you will see if you build models of these achiral molecules that, in each case, there is at least one plane of symmetry , where one side of the plane is the mirror image of the other. Chirality is tied conceptually to the idea of asymmetry, and any molecule that has a plane of symmetry cannot be chiral . When looking for a plane of symmetry, however, we must consider all possible conformations that a molecule could adopt. Even a very simple molecule like ethane, for example, is asymmetric in many of its countless potential conformations – but it has obvious symmetry in both the eclipsed and staggered conformations, and for this reason it is achiral. Looking for planes of symmetry in a molecule is useful, but often difficult in practice. In most cases, the easiest way to decide whether a molecule is chiral or achiral is to look for one or more stereocenters - with a few rare exceptions (see section 3.7B), the general rule is that molecules with at least one stereocenter are chiral, and molecules with no stereocenters are achiral. Carbon stereocenters are also referred to quite frequently as chiral carbons . When evaluating a molecule for chirality, it is important to recognize that the question of whether or not the dashed/solid wedge drawing convention is used is irrelevant. Chiral molecules are sometimes drawn without using wedges (although obviously this means that stereochemical information is being omitted). Conversely, wedges may be used on carbons that are not stereocenters – look, for example, at the drawings of glycine and citrate in the figure above. Just because you see dashed and solid wedges in a structure, do not automatically assume that you are looking at a stereocenter. Other elements in addition to carbon can be stereocenters. The phosphorus center of phosphate ion and organic phosphate esters, for example, is tetrahedral, and thus is potentially a stereocenter. We will see in chapter 10 how researchers, in order to investigate the stereochemistry of reactions at the phosphate center, incorporated sulfur and/or 17 O and 18 O isotopes of oxygen (the ‘normal’ isotope is 16 O) to create chiral phosphate groups. Phosphate triesters are chiral if the three substituent groups are different. Asymmetric quaternary ammonium groups are also chiral. Amines, however, are not chiral, because they rapidly invert, or turn ‘inside out’, at room temperature. Exercise 5.2.5 Label the molecules below as chiral or achiral, and circle all stereocenters. a) fumarate (a citric acid cycle intermediate) b) malate (a citric acid cycle intermediate) b) malate (a citric acid cycle intermediate) Answer a) achiral (no stereocenters) b) chiral c) chiral Exercise 5.2.6 Label the molecules below as chiral or achiral, and circle all stereocenters. a) acetylsalicylic acid (aspirin) b) acetaminophen (active ingredient in Tylenol) c) thalidomide (drug that caused birth defects in pregnant mothers in the 1960’s) Answer a) achiral (no stereocenters) b) achiral (no stereocenters) c) chiral Exercise 5.2.7 Draw both enantiomers of the following chiral amino acids. a) Cysteine b) Proline Answer Exercise 5.2.8 Draw both enantiomers of the following compounds from the given names. a) 2-bromobutane b) 2,3-dimethyl-3-pentanol Answer Exercise 5.2.9 Which of the following body parts are chiral? a) Hands b) Eyes c) Feet d) Ears Answer a) Hands- chiral since the mirror images cannot be superimposed (think of the example in the beginning of the section) b) Eyes- achiral since mirror images that are superimposable c) Feet- chiral since the mirror images cannot be superimposed (Does your right foot fit in your left shoe?) d) Ears- chiral since the mirror images cannot be superimposed Exercise 5.2.10 Circle the chiral centers in the following compounds. Answer Exercise 5.2.11 Identify the chiral centers in the following compounds. Answer |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Magnetic_Resonance_Spectroscopies/Nuclear_Magnetic_Resonance/NMR%3A_Structural_Assignment/NMR8._Chemical_Shift_in_1H_NMR | The trends here are exactly the same as in carbon spectra. Wherever the carbon goes, it takes the proton with it. By analogy with carbon spectra, Source: Simulated spectrum. Figure NMR12. 1 H NMR spectrum of 1-hexene. Source: Simulated spectrum. Figure NMR13. 1 H NMR spectrum of butanal. Source: Simulated spectrum. As before, there are also hydrogens on linear carbons, although they are much less common than tetrahedral or trigonal carbons. Remember, these are general rules that you should know. There will occasionally be exceptions; the proton in a carboxylic acid may be seen at 12 ppm, and the proton in chloroform shows up at 7 ppm although it is attached to a tetrahedral carbon. (World-record shifts occur for hydrogens attached to transition metals: "late" metals like ruthenium or rhodium can move hydrogen peaks all the way up to -20 ppm, but "early" metals like tantalum can move them down as far as 25 ppm.) |
Courses/Grinnell_College/CHM_363%3A_Physical_Chemistry_1_(Grinnell_College)/03%3A_Chemical_Bond/3.07%3A_Photoelectron_Spectroscopy | Demonstrate how photoelectron spectroscopy can be used to resolve the absolute energies of molecular orbitals. Photoelectron spectroscopy (PES) utilizes photo-ionization and analysis of the kinetic energy distribution of the emitted photoelectrons to study the composition and electronic state of the surface region of a sample. X-ray Photoelectron Spectroscopy (XPS) uses soft x-rays (with a photon energy of 200-2000 eV) to examine electrons in core -levels. Ultraviolet Photoelectron Spectroscopy (UPS) using vacuum UV radiation (with a photon energy of 10-45 eV) to examine electrons in valence levels. Both photoelectron spectroscopies are based upon a single photon in/electron out process. The energy of a photon of all types of electromagnetic radiation is given by the Planck–Einstein relation: \[E = h \nu \label{5.3.1} \] where \(h\) is Planck constant and \(\nu\) is th e frequency (Hz) of the radiation. UPS is a powerful technique to exam molecular electron structure since we are interested in the molecular orbitals from polyatomic molecules (especially the valence orbitals) and is the topic of this page. Photoelectron spectroscopy uses monochromatic sources of radiation (i.e. photons of fixed energy). In UPS the photon interacts with valence levels of the molecule or solid, leading to ionization by removal of one of these valence electrons. The kinetic energy distribution of the emitted photoelectrons (i.e. the number of emitted photoelectrons as a function of their kinetic energy) can be measured using any appropriate electron energy analyzer and a photoelectron spectrum can thus be recorded. The process of photoionization can be considered in several ways. One way is to look at the overall process for a species \(A\): \[A + \text{photon} \rightarrow A^+ + e^- \label{5.3.2} \] Conservation of energy then requires that (after using Equation \(\ref{5.3.1}\)): \[E(A) + h\nu = E(A^+ ) + E(e^-) \label{5.3.3} \] Since the free electron's energy is present solely as kinetic energy (\(KE\)) (i.e., there is no internal energy in a free electron) \[ E(e^-) = KE \nonumber \] Equation \(\ref{5.3.3}\) can then be rearranged to give the following expression for the KE of the photoelectron: \[KE = h\nu - \left[ E(A^+ ) - E(A) \right] \label{5.3.4} \] The final term in brackets represents the difference in energy between the ionized and neutral species and is generally called the vertical ionization energy (\(IE\)) of the ejected electron; this then leads to the following commonly quoted equations: \[KE = h\nu - IE \label{5.3.5} \] or \[IE= h\nu - KE \label{Big} \] The vertical ionization energy is a direct measure of the energy required to just remove the electron concerned from its initial level to the vacuum level (i.e., a free electron). Photoelectron spectroscopy measures the relative energies of the ground and excited positive ion states that are obtained by removal of single electrons from the neutral molecule. Equation \ref{5.3.5} may look familiar to you as it the same equation Einstein used to describe the photoelectric effect except the vertical ionization energy (\(IE\)) is substituted for workfunction \(\Phi\). Both vertical ionization energy and workfunctions are metrics for the binding energy of an electron in the sample. At a fundamental level, ionization energies are well-defined thermodynamic quantities related to the heats of protonation, oxidation/reduction chemistry, and ionic and covalent bond energies. Ionization energies are closely related to the concepts of electronegativity, electron-richness, and the general reactivity of molecules. The energies and other characteristic features of the ionization bands observed in photoelectron spectroscopy provide some of the molecular orbitals detailed and specific quantitative information regarding the electronic structure and bonding in molecules. Ionization is explicitly defined in terms of transitions between the ground state of a molecule and ion states as shown in Equation \(\ref{Big}\) and as illustrated in the Figure 10.4.2 . Nonetheless, the information obtained from photoelectron spectroscopy is typically discussed in terms of the electronic structure and bonding in the ground states of neutral molecules, with ionization of electrons occurring from bonding molecular orbitals, lone pairs, antibonding molecular orbitals, or atomic cores. These descriptions reflect the relationship of ionization energies to the molecular orbital model of electronic structure. Ionization energies are directly related to the energies of molecular orbitals by Koopmans' theorem, which states that the negative of the eigenvalue of an occupied orbital from a Hartree-Fock calculation is equal to the vertical ionization energy to the ion state formed by removal of an electron from that orbital (Figure 10.4.3 ), provided the distributions of the remaining electrons do not change (i.e., frozen). \[ I_j = - \epsilon_j \label{Koopman} \] There are many limitations to Koopmans' theorem, but in a first order approximation each ionization of a molecule can be considered as removal of an electron from an individual orbital. The ionization energies can then be considered as measures of orbital stabilities, and shifts can be interpreted in terms of orbital stabilizations or destabilizations due to electron distributions and bonding. Koopmans' theorem is implicated whenever an orbital picture is involved, but is not necessary when the focus is on the total electronic states of the positive ions. Koopmans' theorem argues that the negative of the eigenvalue of an occupied orbital from a Hartree-Fock calculation is equal to the vertical ionization energy to the ion state formed by removal of an electron from that orbital. Several different ionization energies can be defined, depending on the degree of vibrational excitation of the cations. In general, the following two types of ionization energies are considered (Figure 10.4.4 ): Adiabatic ionization energy corresponds to the ionization energy associated with this transition \[M(X, v” = 0) + h\nu \rightarrow M^+(x, v’ = 0) + e^- \nonumber \] Adiabatic ionization energy that is, the minimum energy required to eject an electron from a molecule in its ground vibrational state and transform it into a cation in the lowest vibrational level of an electronic state x of the cation. Vertical ionization energy corresponds to the ionization energy associated with this transition \[ M(X, v” = 0) + h\nu \rightarrow M^+(x, v’ = n) + e^- \nonumber \] where, the value n of the vibrational quantum number v’ corresponds to the vibrational level whose wavefunction gives the largest overlap with the v” = 0 wavefunction. This is the most probable transition and usually corresponds to the vertical transition where the internuclear separations of the ionic state are similar to those of the ground state. The geometry of an ion may be different from the neutral molecule. The measured ionzation energy in a PES experiment can refer to the vertical ionization energy, in which case the ion is in the same geometry as the neutral, or to the adiabatic ionzaiton energy, in which case the ion is in its lowest energy, relaxed geometry (mostly the former though). This is illustrated in the Figure 10.4.4 . For a diatomic the only geometry change possible is the bond length. The figure shows an ion with a slightly longer bond length than the neutral. The harmonic potential energy surfaces are shown in green (neutral) and red (ion) with vibrational energy levels. The vertical ionzation energy is always greater than the adiabatic ionzation energy. You have been exposed to three metrics of ionization energies already, which are similar, but with distinct differences: The ionization energy (also called adiabatic ionization energy ) is the lowest energy required to effect the removal of an electron from a molecule or atom, and corresponds to the transition from the lowest electronic, vibrational and rotational level of the isolated molecule to the lowest electronic, vibrational and rotational level of the isolated ion. The binding energy (also called vertical ionization energy ) is the energy change corresponding to an ionization reaction leading to formation of the ion in a configuration which is the same as that of the equilibrium geometry of the ground state neutral molecule. The workfunction is the minimum energy needed to remove an electron from a (bulk) solid to a point in the vacuum. As you remember, the molecular orbital description of hydrogen involves two \(|1s \rangle\) atomic orbitals generating a bonding \(1\sigma_g\) and antibonding \(2\sigma_u^*\) molecular orbitals. The two electrons that are responsible for the \(\ce{H_2}\) bond are occupied in the \(1\sigma_g\). The PES spectrum has a single band that corresponds to the ionization of a 1σ g electron. The multiple peaks are due to electrons ejecting from a range of stimulated vibrational energy levels. When extensive vibrational structure is resolved in a PES molecular orbital, then the removal of an electron from that molecular orbital induces a significant change in the bonding (in this case an increase in the bond length since the bond order has been reduced). Diatomic nitrogen is more complex than hydrogen since multiple molecular orbitals are occupied. Four molecular orbitals are occupied (the two \(1\pi_u\) orbitals are both occupied). The UV photoelectron spectrum of \(N_2\), has three bands corresponding to \(3σ_g\), \(1π_u\) and \(2σ_u\) occupied molecular orbitals. Both \(3σ_g\) and \(2σ_u\) are weakly bonding and antibonding. The \(1\sigma_g\) orbital is not resolved in this spectrum since the incident light \(h\nu\) used did not have sufficient energy to ionize electrons in that deeply stabilized molecular orbital. Note that extensive vibrational structure for the \(1π_u\) band indicates that the removal of an electron from this molecular orbital causes a significant change in the bonding. Hydrogen Chloride The molecular energy level diagram for \(\ce{HCl}\) is reproduced in Figure 10.4.5 Important aspects of molecular orbital diagram in Figure 10.4.5 : The H 1 s energy lies well above the Cl 2 s and 2 p atomic orbitals; The valence electron configuration can be written 3σ 2 1π 4 ; The H 1 s orbital contributes only to the σ molecular orbitals, as does one of the Cl 2 p orbitals (hence the lines in Figure 10.4.5 connecting these atomic orbitals and the 3σ and 4σ molecular orbitals); The remaining Cl 2 p orbitals (ie those perpendicular to the bond axis) are unaffected by bonding, and these form the 1π molecular orbitals; The 1π orbitals are nonbonding - they are not affected energetically by the interaction between the atoms, and are hence neither bonding nor antibonding; The 3σ orbital is weakly bonding, and largely Cl 2 p ; The 3σ* orbital is antibonding, and primarily of H 1 s character; Figure 10.4.6 shows the analogous MO diagram and photoelectron spectrum for \(\ce{HCl}\). The spectrum has two bands corresponding to non-bonding 1 p (or \(1\pi\)) molecular orbitals (with negligible vibrational structure) and the 3 s bonding molecular orbital (vibrational structure). The higher energy (more stabilized) core molecular orbitals are not observed since the incident photon energy \(h\nu\) is below their ionization energies. Water In the simplified valence bond theory perspective of the water molecule, the oxygen atom form four \(sp^3\) hybrid orbitals. Two of these are occupied by the two lone pairs on the oxygen atom, while the other two are used for bonding. Within the molecular orbital picture, the electronic configuration of the \(\ce{H_2O^{+}}\) molecule is \((1a_1)^2 (2a_1)^2 (1b_2)^2 (3a_1)^2 (1b_1)^2\) where the symbols \(a_1\), \(b_2\) and \(b_1\) are orbital labels based on molecular symmetry that will be discussed later (Figure 10.4.7 ). Within Koopmans' theorem: The energy of the 1b 1 HOMO corresponds to the ionization energy to form the \(\ce{H_2O^{+}}\) ion in its ground state \((1a_1)^2 (2a_1)^2 (1b_2)^2 (3a_1)^2 (1b_1)^1\). The energy of the second-highest molecular orbitals \(3a_1\) refers to the ion in the excited state \((1a_1)^2 (2a_1)^2 (1b_2)^2 (3a_1)^1 (1b_1)^2\). The Hartree–Fock orbital energies (with sign changed) of these orbitals are tabulated below and compared to the experimental ionization energies. Molecular orbital Hartree–Fock orbital Energies (eV) Experimental Ionization Energies (eV) 2a1 36.7 32.2 1b2 19.5 18.5 3a1 15.9 14.7 1b1 13.8 12.6 As explained above, the deviations between orbital energy and ionization energy is small and due to the effects of orbital relaxation as well as differences in electron correlation energy between the molecular and the various ionized states. The molecular orbital perspective has the lone pair in different orbitals (one in a non-bonding orbital (\(1b_1\) and one in the bonding orbitals). We tern to the photoelectron spectroscopy to help identify which theory is more accurate (i.e., describes reality better). The photoelectron spectrum of water in Figure 10.4.6 can be interpreted as having three major peaks with some fine structure arises from vibrational energy changes. The light source used in this experiment is not sufficiently energetic to ionize electrons from the lowest lying molecular orbitals. If water was formed two identical O-H bonds and two lone pairs on the oxygen atom line valence bond theory predicts, then the PES in Figure 10.4.8 would have two (degenerate) peaks, one for the two bonds and one for the two lone pairs. The photoelectron spectrum clearly shows three peaks in the positions expected for the molecular orbitals in Figure 10.4.8 . If the molecular orbitals in Figure 10.4.7 represent the real electronic structure, how do we view the bonding? These molecular orbitals are delocalized and bare little relationship to the familiar 2-center bonds used in valence bond theory. For example, the \(2a_1\) \(1b_1\) and \(3a_1\) molecular orbitals all have contributions from all three atoms, they are really 3-centered molecular orbitals. The bonds however can be thought of as representing a build up of the total electron density which loosely put is a total of all the orbital contributions. Despite this, we keep the ideas of hybridization and 2-center bonds because they are useful NOT because they represent reality Summary A photoelecton spectrum can show the relative energies of occupied molecular orbitals by ionization. (i.e. ejection of an electron). A photoelectron spectrum can also be used to determine energy spacing between vibrational levels of a given electronic state. Each orbital energy band has a structure showing ionization to different vibrational levels. |
Courses/Oregon_Institute_of_Technology/OIT_(Lund)%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/09%3A_Phosphate_Transfer_Reactions | This chapter is about the chemistry of phosphates, a ubiquitous functional group in biomolecules that is based on phosphoric acid. 9.0: Prelude to Phosphate Transfer Reactions This chapter is about the chemistry of phosphates, a ubiquitous functional group in biomolecules that is based on phosphoric acid. 9.1: Overview of Phosphate Groups Phosphate is everywhere in biochemistry. As we were reminded in the introduction to this chapter, our DNA is linked by phosphate. The function of many proteins is regulated - switched on and off - by enzymes which attach or remove a phosphate group from the side chains of serine, threonine, or tyrosine residues. 9.2: Phosphate Transfer Reactions - An Overview In a phosphate transfer reaction, a phosphate group is transferred from a phosphate group donor molecule to a phosphate group acceptor molecule. A very important aspect of biological phosphate transfer reactions is that the electrophilicity of the phosphorus atom is usually enhanced by the Lewis acid (electron-accepting) effect of one or more magnesium ions. 9.3: ATP, The Principal Phosphate Group Donor Thus far we have been very general in our discussion of phosphate transfer reactions, referring only to generic 'donor' and 'acceptor' species. It's time to get more specific. The most important donor of phosphate groups in the cell is a molecule called adenosine triphosphate, commonly known by its abbreviation ATP. 9.4: Phosphorylation of Alcohols A broad family of enzymes called kinases catalyze transfer of a phosphate group from TP to an alcohol acceptor. Mechanistically, the alcohol oxygen acts as a nucleophile, attacking the electrophilic g-phosphorus of TP and expelling ADP. 9.5: Phosphorylation of Carboxylates Thus far we have seen hydroxyl oxygens and phosphate oxygens acting as nucleophilic accepting groups in ATP-dependent phosphate transfer reactions. Carboxylate oxygens can also accept phosphate groups from ATP. This typically happens in two different ways. 9.6: Hydrolysis of Organic Phosphates While kinase enzymes catalyze the phosphorylation of organic compounds, enzymes called phosphatases catalyze dephosphorylation reactions. 9.7: Phosphate Diesters in DNA and RNA Phosphate diesters play an absolutely critical role in nature - they are the molecular 'tape' that connect the individual nucleotides in DNA and RNA via a sugar-phosphate backbone. 9.8: The Organic Chemistry of Genetic Engineering Many enzymes that catalyze reactions involving the phosphate diester bonds of DNA have been harnessed for use in genetic engineering - techniques in which we copy, snip, and splice DNA in order to create custom versions of genes. The tools of genetic engineering have become indispensable and commonplace in the past decade, and most researchers working on the biological side of chemistry use them extensively. 9.9: NMR of phosphorylated compounds Because so many biological molecules contain phosphoryl groups, it is worthwhile to look at how scientists use NMR to determine the structure of these molecules. Recall from section 5.1 that 31P , the most abundant isotope of phosphorus, is NMR active: it can be directly observed by 31P−NMR , and indirectly observed in 1H−NMR and 13C−NMR through its spin-coupling interactions with neighboring protons and carbons, respectively. 9.E: Phosphate Transfer Reactions (Exercise) 9.S: Phosphate Transfer Reactions (Summary) |
Courses/Smith_College/Organic_Chemistry_(LibreTexts)/11%3A_Reactions_of_Alkyl_Halides-_Nucleophilic_Substitutions_and_Eliminations/11.07%3A_Biological_Substitution_Reactions | Objective After completing this section, you should have an appreciation that S N 1 and S N 2 mechanisms exist and are well-known in biological chemistry. Leaving Groups in Biochemical Reactions In biological reactions, we do not often see halides serving as leaving groups (in fact, outside of some marine organisms, halogens are fairly unusual in biological molecules). More common leaving groups in biochemical reactions are phosphates, water, alcohols, and thiols. In many cases, the leaving group is protonated by an acidic group on the enzyme as bond-breaking occurs. For example, hydroxide ion itself seldom acts as a leaving group – it is simply too high in energy (too basic). Rather, the hydroxide oxygen is generally protonated by an enzymatic acid before or during the bond-breaking event, resulting in a (very stable) water leaving group. More often, however, the hydroxyl group of an alcohol is first converted enzymatically to a phosphate ester in order to create a better leaving group. This phosphate ester can take the form of a simple monophosphate (arrow 1 in the figure below), a diphosphate (arrow 2), or a nucleotide monophosphate (arrow 3). Due to resonance delocalization of the developing negative charge, phosphates are excellent leaving groups. Here’s a specific example (from DNA nucleotide biosynthesis) that we will encounter in more detail in this section : Here, the OH group on ribofuranose is converted to a diphosphate, a much better leaving group. Ammonia is the nucleophile in the second step of this S N 1-like reaction. We will learn much more about phosphates in this section . What is important for now is that in each case, an alcohol has been converted into a much better leaving group, and is now primed for a nucleophilic substitution reaction. SAM Methyltransferases Some of the most important examples of S N 2 reactions in biochemistry are those catalyzed by S-adenosyl methionine (SAM) – dependent methyltransferase enzymes. We have already seen, in chapter 6 and again in chapter 8, how a methyl group is transferred in an S N 2 reaction from SAM to the amine group on the nucleotide base adenosine: ( Nucleic Acids Res . 2000 , 28 , 3950). Another SAM-dependent methylation reaction is catalyzed by an enzyme called catechol-O-methyltransferase. The substrate here is epinephrine, also known as adrenaline. Notice that in this example, the attacking nucleophile is an alcohol rather than an amine (that’s why the enzyme is called an O-methyltransferase). In both cases, though, a basic amino acid side chain is positioned in the active site in just the right place to deprotonate the nucleophilic group as it attacks, increasing its nucleophilicity. The electrophile in both reactions is a methyl carbon, so there is little steric hindrance to slow down the nucleophilic attack. The methyl carbon is electrophilic because it is bonded to a positively-charged sulfur, which is a powerful electron withdrawing group. The positive charge on the sulfur also makes it an excellent leaving group, as the resulting product will be a neutral and very stable sulfide. All in all, in both reactions we have a reasonably good nucleophile, an electron-poor, unhindered electrophile, and an excellent leaving group. Because the electrophilic carbon in these reactions is a methyl carbon, a stepwise S N 1-like mechanism is extremely unlikely: a methyl carbocation is very high in energy and thus is not a reasonable intermediate to propose. We can confidently predict that this reaction is S N 2. Does this S N 2 reaction occur, as expected, with inversion of stereochemistry? Of course, the electrophilic methyl carbon in these reactions is achiral, so inversion is not apparent. To demonstrate inversion, the following experiment has been carried out with catechol-O-methyltransferase: Here, the methyl group of SAM was made to be chiral by incorporating hydrogen isotopes tritium ( 3 H, T) and deuterium ( 2 H, D). The researchers determined that the reaction occurred with inversion of configuration, as expected for an S N 2 displacement ( J. Biol. Chem . 1980, 255 , 9124 ). Substitution by Electrophilic Addition/Elimination The electrophilic double bond isomerization catalyzed by IPP isomerase is a highly reversible reaction, with an equilibrium IPP:DMAPP ratio of about 6:1. In the next step of isoprenoid biosynthesis, the two five-carbon isomers condense to form a 10-carbon isoprenoid product called geranyl diphosphate (GPP). This is a nice example of an electrophilic addition/elimination mechanism, which we saw in general form in this section : The first step is ionization of the electrophile - in other words, the leaving group departs and a carbocation intermediate is formed. In this case, the pyrophosphate group on DMAPP is the leaving group, and the electrophilic species is the resulting allylic carbocation. In the condensation (addition) step, the C 3 -C 4 double bond in IPP attacks the positively-charged C 1 of DMAPP, resulting in a new carbon-carbon bond and a second carbocation intermediate, this time at a tertiary carbon. In the elimination phase, proton abstraction leads to re-establishment of a double bond in the GPP product. Notice that the enzyme specifically takes the pro-R proton in this step. To continue the chain elongation process, another IPP molecule can then condense, in a very similar reaction, with C 1 of geranyl diphosphate to form a 15-carbon product called farnesyl diphosphate (FPP). How do we know that these are indeed S N 1-like mechanisms with carbocation intermediates, rather than concerted S N 2-like mechanisms? First of all, recall that the question of whether a substitution is dissociative (S N 1-like) or associative (S N 2-like) is not always clear-cut - it could be somewhere in between, like the protein prenyltransferase reaction ( section 9.3 ). The protein prenyltransferase reaction and the isoprenoid chain elongation reactions are very similar: the electrophile is the same, but in the former the nucleophile is a thiolate, while in the latter the nucleophile is a pi bond. This difference in the identity of the nucleophilic species would lead one to predict that the chain elongation reaction has more S N 1-like character than the protein prenylation reaction. A thiolate is a very powerful nucleophile, and thus is able to push the pyrophosphate leaving group off, implying some degree of S N 2 character. The electrons in a pi bond, in contrast, are only weakly nucleophilic, and thus need to be pulled in by a powerful electrophile - ie . a carbocation. So it makes perfect sense that the chain elongation reaction should more S N 1-like than S N 2-like. Is this in fact the case? We know how to answer this question experimentally - just run the reaction with fluorinated DMAPP or GPP substrates and observe how much the fluorines slow things down (see section 9.3B ). If the reaction is S N 1-like, the electron-withdrawing fluorines should destabilize the allylic carbocation intermediate and thus slow the reaction down considerably. If the mechanism is S N 2-like, the fluorine substitutions should not have a noticeable effect, because a carbocation intermediate would not be formed. When this experiment was performed with FPP synthase, the results were dramatic: the presence of a single fluorine slowed down the rate of the reaction by a factor of about 60, while two and three fluorines resulted in a reaction that was 500,000 and 3 million times slower, respectively ( J. Am. Chem. Soc . 1981 , 103 , 3926.) These results strongly suggest indicate the formation of a carbocation intermediate in an S N 1-like displacement. |
Courses/Intercollegiate_Courses/Cheminformatics/04%3A_Searching_Databases_for_Chemical_Information/4.01%3A_PubChem_Web_Interfaces_for_Text | PubChem Homepage The PubChem homepage ( https://pubchem.ncbi.nlm.nih.gov ) provides a search interface that allow users to perform any term/keyword/identifier search against all three major databases of PubChem 1 , 2 , 3 : Compound, Substance, BioAssay. If a search returns multiple hits, they are presented on an Entrez DocSum page and will be explained in more detail later in this chapter. If the search returns a single record, the user will be directed to the web page that presents information on that record. This page is called the Compound Summary , Substance Record , or BioAssay Record page, depending on the record type (i.e., compound, substance, or assay). In addition, the PubChem homepage provides launch points to various PubChem services, tools, help documents, and more. In general, the PubChem homepage is a central location for all PubChem services. Entrez Search and Retrieval System NCBI’s Entrez 4 , 5 , 6 , 7 is a database retrieval system that integrates PubChem’s three major databases as well as other NCBI’s major databases, including PubMed , Nucleotide and Protein Sequences, Protein Structures , Genome , Taxonomy , BioSystems , Gene Expression Omnibus (GEO) and many others. Entrez provides users with an integrated view of biomedical data and their relationships. This section focuses on search and retrieval of PubChem data using the Entrez system. A more detailed description on the Entrez system is given in the following documents: The Entrez Search and Retrieval System ( http://www.ncbi.nlm.nih.gov/books/NBK184582/ ) Entrez Help ( https://www.ncbi.nlm.nih.gov/books/NBK3836/ ) Entry points to Entrez One can search the PubChem databases through Entrez, by initiating a search from the NCBI home page ( http://www.ncbi.nlm.nih.gov ). By default, if a specific database is not selected in the search menu, Entrez searches all Entrez databases available, and lists the number of records in each database that are returned for this “global query”. The following link directs you to the global query result page for the term “AIDS” against all databases integrated in the Entrez system. https://www.ncbi.nlm.nih.gov/gquery/?term=AIDS Simply by selecting one of the three PubChem databases from the global query results page (under the Chemical section), one can see the query results specific to that database. Alternatively, one can start from the PubChem home page ( http://pubchem.ncbi.nlm.nih.gov ), where a search of one of the three PubChem databases may be initiated through the search box at the top. It is also possible to initiate an Entrez search against a PubChem database from the following pages: https://www.ncbi.nlm.nih.gov/pccompound/ (to search the Compound database) https://www.ncbi.nlm.nih.gov/pcsubstance/ (to search the Substance database) https://www.ncbi.nlm.nih.gov/pcassay/ (to search the BioAssay database) Entrez DocSums If an Entrez search for a query against any of the three PubChem databases returns a single record, the user will be directed to the Compound Summary, Substance Record, or BioAssay Record page for that record (depending on whether the record is a compound, substance, or assay). If it returns multiple records, Entrez will display a document summary report (also called “DocSum” page). The following link directs you to the DocSum page for a search for the term “lipitor” against the PubChem Compound database: https://www.ncbi.nlm.nih.gov/pccompound?term=lipitor In this example, the DocSum page displays a list of the compound records returned from the search. For each record, some data-specific information is provided with a link to the summary page for that record. The DocSum page contains controls to change the display type, to sort the results by various means, or to export the page to a file or printer. Additional controls that operate on a query result list are available on the right column of the DocSum page. The DocSum page for the other two PubChem databases look similar to this example for the Compound database. Entrez Indices Entrez indices, tied to individual records in an Entrez database, include information on particular aspects (often referred to as fields) of the records. These indices may have text, numeric or date values, and some indices may have multiple values for each record. The available fields and their indexed terms in any Entrez database can be found from the drop-down menus on the Advanced Search Builder page (which can be accessed by clicking the “Advanced” link next to the “Go” button on the PubChem Home page ). When the user enters a query in the Entrez search interface, the Entrez indices are matched directly to that query. By default, in an Entrez search with a simple query, all indexed fields are matched against the query, usually resulting in the largest number of returned records including many unwanted results. One can narrow the search to a particular indexed field, by adding the index name in brackets after the term itself (e.g., “ lipitor[synonym] ”). For numeric indices, a search for a range of values can be done by using minimum and maximum values separated by a colon and followed by the bracketed index name (e.g., “ 100:105[MolecularWeight] ”). Multiple indices may be searched simultaneously using Entrez’s Boolean operators (e.g., “ AND ”, “ OR ” and “ NOT ”). A complete list of the Entrez indices available for the three PubChem databases can be retrieved in the XML format, using the eInfo functionality in E-Utilities (which will be covered in Module 7): http://eutils.ncbi.nlm.nih.gov/entrez/eutils/einfo.fcgi?db=pccompound (for Compound) http://eutils.ncbi.nlm.nih.gov/entrez/eutils/einfo.fcgi?db=pcsubstance (for Substance) http://eutils.ncbi.nlm.nih.gov/entrez/eutils/einfo.fcgi?db=pcassay (for BioAssay). Additional information on the PubChem Entrez indices is available in the “Indices and Filters in Entrez” section of the help documentation: https://pubchem.ncbi.nlm.nih.gov/help.html#PubChem_Index Entrez Links Entrez links are cross links or associations between records in different Entrez databases, or within the same database. These links may be applied to an entire search result list (via the “find related data” section at the right column of a DocSum page) or to an individual record (via links at the bottom of each record presented on the DocSum page). The Entrez links provide a way to discover relevant information in other Entrez databases based on a user’s specific interests. Equivalently, one may think of this as a way to transform an identifier list from one database to another based on a particular criterion. Note that there are limits to how many records may be used as input in a link operation. To process a large amount of input records and/or to expect a large amount of output records associated with the input records, one should use the FLink tool ( https://www.ncbi.nlm.nih.gov/Structure/flink/flink.cgi ). A complete list of the Entrez links available for the three PubChem databases can be retrieved in the XML format through these links http://eutils.ncbi.nlm.nih.gov/entrez/eutils/einfo.fcgi?db=pccompound (for Compound) http://eutils.ncbi.nlm.nih.gov/entrez/eutils/einfo.fcgi?db=pcsubstance (for Substance) http://eutils.ncbi.nlm.nih.gov/entrez/eutils/einfo.fcgi?db=pcassay (for BioAssay). Entrez Filters Entrez filters are essentially Boolean bits (true or false) for all records in a database that indicate whether or not a given record has a particular property. The Entrez filters may be used to subset other Entrez searches according to this property, by adding the filter to the query string. Entrez filters are closely related to links in that the majority of Entrez filters in the PubChem databases are generated automatically based on whether PubChem records have Entrez links to a given database. However, some special filters, such as the "lipinski rule of 5" filter, or the “all” filter, are not link-based. The Entrez filters available for each Entrez database may be found on the Advanced Search Builder page by selecting “Filter” from the “All Fields” dropdown and clicking “Show index list”. More detailed description of the Entrez filters available for the three PubChem databases are given in the “Indices and Filters in Entrez” section of the help documentation: https://pubchem.ncbi.nlm.nih.gov/help.html#PubChem_Index Entrez History Entrez has a history mechanism (Entrez history) that automatically keeps track of a user’s searches, temporarily caches them (for eight hours), and allows one to combine search result sets with Boolean logic (i.e., “AND”, “OR”, and “NOT”). The Entrez history allows one to limit a search to a subset of records returned from a previous search. Use of Entrez history can help users avoid sending and receiving (potentially) very large lists of identifiers. In addition, through the Entrez history, one can use the search results as an input to various PubChem tools for further manipulation and analysis. References (1) Kim, S.; Thiessen, P. A.; Bolton, E. E.; Chen, J.; Fu, G.; Gindulyte, A.; Han, L. Y.; He, J. E.; He, S. Q.; Shoemaker, B. A.; Wang, J. Y.; Yu, B.; Zhang, J.; Bryant, S. H. Nucleic Acids Res. 2016 , 44 , D1202. (2) Wang, Y.; Bryant, S. H.; Cheng, T.; Wang, J.; Gindulyte, A.; Shoemaker, B. A.; Thiessen, P. A.; He, S.; Zhang, J. Nucleic Acids Res. 2017 , 45 , D955. (3) Kim, S. Expert Opinion on Drug Discovery 2016 , 11 , 843. (4) Schuler, G. D.; Epstein, J. A.; Ohkawa, H.; Kans, J. A. Methods Enzymol. 1996 , 266 , 141. (5) McEntyre, J. Trends in genetics : TIG 1998 , 14 , 39. (6) The Entrez Search and Retrieval System ( https://www.ncbi.nlm.nih.gov/books/ NBK184582 / ) (Accessed on. (7) Entrez Help ( https://www.ncbi.nlm.nih.gov/books/ NBK3836 / ) (Accessed on. |
Courses/Anoka-Ramsey_Community_College/Introduction_to_Chemistry/11%3A_Chemical_Bonding/11.01%3A_Valence_Electrons_and_the_Periodic_Table | ⚙️ Learning Objectives Explain the relationship between the chemical behavior of families in the periodic table and their valence electrons. Identify elements that will have the most similar properties to a given element. The chemical properties of elements are determined primarily by the number and distribution of valence electrons. Since the groups on the periodic table were originally organized based on their chemical behavior, it should be no surprise that the individual members of each group have similar electron configurations of their valence electrons. Section 4.9 described how Dmitri Mendeleev arranged the original periodic table so that elements with the most similar properties were placed in the same group. For example, the alkali metals (Group IA) are quite soft and may easily be cut apart using a dull knife. They also reacted in a similar manner with water to produce hydrogen gas. When they combined with oxygen, they were found to do so in a 2:1 ratio of atoms. Once quantum mechanics was developed in the 1920s, it was found that the alkali metals all had one valence electron, thereby explaining their similar reactions and the 2:1 atom ratio with oxygen. The halogens (Group VIIA) were all observed to be colorful, reactive elements that combined with oxygen in a 7:2 atom ratio and with the alkali metals in a 1:1 atom ratio. As a gas or vapor, the halogens all had a pungent odor. After the development of quantum mechanics, it was shown that the halogens all had seven valence electrons, supporting their original placement into the same group on Mendeleev's periodic table. The Modern Periodic Table and Chemical Bonds As described in Section 10.6 , the modern periodic table is arranged based on an atom's valence electrons. But what does this tell us about how they form chemical bonds with each other? Why do sodium atoms and chlorine atoms combine in a 1:1 ratio, while sodium and oxygen atoms combine in a 2:1 ratio? Much of this was already described in Chapter 5 in our study of chemical nomenclature. At that time, it wasn't really explained why sodium atoms always form ions with a 1+ charge, why chlorine atoms form ions with a 1– charge, or why oxygen atoms form ions with a 2– charge. By the time you complete this chapter, you will have a much better idea. Summary All of the elements in the same group on the periodic table have the same number of valence electrons and similar chemical properties. This page is shared under a CK-12 license and was authored, remixed, and/or curated by Lance S. Lund (Anoka-Ramsey Community College), Melissa Alviar-Agnew, and Henry Agnew. Original source: https://www.ck12.org/c/chemistry/ . LICENSED UNDER |
Bookshelves/Introductory_Chemistry/Fundamentals_of_General_Organic_and_Biological_Chemistry_(LibreTexts)/02%3A_Atoms_and_the_Periodic_Table/2.09%3A_Electron-Dot_Symbols | Learning Objective Draw a Lewis electron dot symbol for a given atom. In almost all cases, chemical bonds are formed by interactions of valence electrons in atoms. To facilitate our understanding of how valence electrons interact, a simple way of representing those valence electrons would be useful. A Lewis electron-dot symbol (or electron-dot symbol or a Lewis symbol) is a representation of the valence electrons of an atom that uses dots around the symbol of the element. The number of dots equals the number of valence electrons in the atom. These dots are arranged to the right and left and above and below the symbol, with no more than two dots on a side. (It does not matter what order the positions are used.) For example, the electron-dot symbol for hydrogen is simply \[\mathbf{H}\mathbf{\cdot}\] Because the side is not important, the electron-dot symbol could also be drawn as follows: \[\mathbf{\dot{H}}\; \; or\; \mathbf{\cdot}\mathbf{H}\; \; \; or\; \; \; \mathbf{\underset{.}H}\] The electron-dot symbol for helium, with two valence electrons, is as follows: \[\mathbf{He}\mathbf{:}\] By putting the two electrons together on the same side, we emphasize the fact that these two electrons are both in the 1 s subshell; this is the common convention we will adopt, although there will be exceptions later. The next atom, lithium, has an electron configuration of 1 s 2 2 s 1 , so it has only one electron in its valence shell. Its electron dot diagram resembles that of hydrogen, except the symbol for lithium is used: \[\mathbf{Li}\mathbf{\cdot}\] Beryllium has two valence electrons in its 2 s shell, so its eelectron-dot symbol is like that of helium: \[\mathbf{Be}\mathbf{:}\] The next atom is boron. Its valence electron shell is 2 s 2 2 p 1 , so it has three valence electrons. The third electron will go on another side of the symbol: \[\mathbf{\dot{Be}}\mathbf{:}\] Again, it does not matter on which sides of the symbol the electron dots are positioned. For carbon, there are four valence electrons, two in the 2 s subshell and two in the 2 p subshell. As usual, we will draw two dots together on one side, to represent the 2 s electrons. However, conventionally, we draw the dots for the two p electrons on different sides. As such, the electron-dot symbol for carbon is as follows: \[\mathbf{\cdot \dot{C}}\mathbf{:}\] With N, which has three p electrons, we put a single dot on each of the three remaining sides: \[\mathbf{\cdot}\mathbf{\dot{\underset{.}N}}\mathbf{:}\] For oxygen, which has four p electrons, we now have to start doubling up on the dots on one other side of the symbol. When doubling up electrons, make sure that a side has no more than two electrons. \[\mathbf{\cdot}\mathbf{\ddot{\underset{.}O}}\mathbf{:}\] Fluorine and neon have seven and eight dots, respectively: \[\mathbf{:}\mathbf{\ddot{\underset{.}F}}\mathbf{:}\] \[\mathbf{:}\mathbf{\ddot{\underset{.\: .}Ne}}\mathbf{:}\] With the next element, sodium, the process starts over with a single electron because sodium has a single electron in its highest-numbered shell, the n = 3 shell. By going through the periodic table, we see that the electron-dot symbol of atoms will never have more than eight dots around the atomic symbol. Example \(\PageIndex{1}\): What is the electron-dot symbol for each element? aluminum selenium Solution The valence electron configuration for aluminum is 3 s 2 3 p 1 . So it would have three dots around the symbol for aluminum, two of them paired to represent the 3 s electrons: \[\dot{Al:} \nonumber\] The valence electron configuration for selenium is 4 s 2 4 p 4 . In the highest-numbered shell, the n = 4 shell, there are six electrons. Its electron dot diagram is as follows: \[\mathbf{\cdot }\mathbf{\dot{\underset{.\: .}Se}}\mathbf{:}\nonumber\] Exercise \(\PageIndex{1}\) What is the electron-dot symbol for each element? phosphorus argon Answer \[\mathbf{\cdot }\mathbf{\dot{\underset{.}P}}\mathbf{:}\nonumber\] \[\mathbf{:}\mathbf{\ddot{\underset{.\, .}Ar}}\mathbf{:}\nonumber\] Summary Lewis electron-dot symbols use dots to represent valence electrons around an atomic symbol. |
Courses/Modesto_Junior_College/Chemistry_143%3A_Introductory_College_Chemistry_(Brzezinski)/CHEM_143%3A_Text_(Brzezinski)/07%3A_Chemical_Reactions_and_Quantities/7.04%3A_Types_of_Chemical_Reactions_-_Single_and_Double_Replacement_Reactions/7.4.03%3A_Ionic_Equations_-_A_Closer_Look | Learning Objectives Write ionic equations for chemical reactions between ionic compounds. Write net ionic equations for chemical reactions between ionic compounds. For single-replacement and double-replacement reactions, many of the reactions included ionic compounds—compounds between metals and nonmetals, or compounds that contained recognizable polyatomic ions. Now, we take a closer look at reactions that include ionic compounds. One important aspect about ionic compounds that differs from molecular compounds has to do with dissolution in a liquid, such as water. When molecular compounds, such as sugar, dissolve in water, the individual molecules drift apart from each other. When ionic compounds dissolve, the ions physically separate from each other . We can use a chemical equation to represent this process—for example, with NaCl: \[\ce{ NaCl(s) ->[\ce{H2O}] Na^{+}(aq) + Cl^{-}(aq)}\nonumber \] When NaCl dissolves in water, the ions separate and go their own way in solution; the ions are now written with their respective charges, and the (aq) phase label emphasizes that they are dissolved (Figure \(\PageIndex{1}\)). This process is called dissociation; we say that the ions dissociate . All ionic compounds that dissolve behave this way. This behavior was first suggested by the Swedish chemist Svante August Arrhenius [1859–1927] as part of his PhD dissertation in 1884. Interestingly, his PhD examination team had a hard time believing that ionic compounds would behave like this, so they gave Arrhenius a barely passing grade. Later, this work was cited when Arrhenius was awarded the Nobel Prize in Chemistry. Keep in mind that when the ions separate, all the ions separate. Thus, when CaCl 2 dissolves, the one Ca 2 + ion and the two Cl − ions separate from one another: \[CaCl_{2}(s)\overset{H_{2}O}{\rightarrow}Ca^{2+}(aq)+Cl^{-}(aq)+Cl^{-}(aq)\nonumber \] \[CaCl_{2}(s)\overset{H_{2}O}{\rightarrow}Ca^{2+}(aq)+2Cl^{-}(aq)\nonumber \] That is, the two chloride ions go off on their own. They do not remain as Cl 2 (that would be elemental chlorine; these are chloride ions), and they do not stick together to make Cl 2 − or Cl 2 2 − . They become dissociated ions in their own right. Polyatomic ions also retain their overall identity when they are dissolved. Example \(\PageIndex{1}\) Write the chemical equation that represents the dissociation of each ionic compound. KBr Na 2 SO 4 Solution KBr(s) → K + (aq) + Br − (aq) Not only do the two sodium ions go their own way, but the sulfate ion stays together as the sulfate ion. The dissolving equation is Na 2 SO 4 (s) → 2Na + (aq) + SO 4 2 − (aq) Exercise \(\PageIndex{1}\) Write the chemical equation that represents the dissociation of (NH 4 ) 2 S. Answer (NH 4 ) 2 S(s) → 2NH 4 + (aq) + S 2− (aq) When chemicals in solution react, the proper way of writing the chemical formulas of the dissolved ionic compounds is in terms of the dissociated ions, not the complete ionic formula. A complete ionic equation is a chemical equation in which the dissolved ionic compounds are written as separated ions. Solubility rules are very useful in determining which ionic compounds are dissolved and which are not. For example, when NaCl(aq) reacts with AgNO 3 (aq) in a double-replacement reaction to precipitate AgCl(s) and form NaNO 3 (aq), the complete ionic equation includes NaCl, AgNO 3 , and NaNO 3 written as separate ions: \[\ce{Na^{+}(aq) + Cl^{−}(aq) + Ag^{+}(aq) + NO3^{−}(aq) → AgCl(s) + Na^{+}(aq) + NO3^{−}(aq)}\nonumber \] This is more representative of what is occurring in the solution. Example \(\PageIndex{1}\) Write the complete ionic equation for each chemical reaction. KBr(aq) + AgC 2 H 3 O 2 (aq) → KC 2 H 3 O 2 (aq) + AgBr(s) MgSO 4 (aq) + Ba(NO 3 ) 2 (aq) → Mg(NO 3 ) 2 (aq) + BaSO 4 (s) Solution For any ionic compound that is aqueous, we will write the compound as separated ions. The complete ionic equation is K + (aq) + Br − (aq) + Ag + (aq) + C 2 H 3 O 2 − (aq) → K + (aq) + C 2 H 3 O 2 − (aq) + AgBr(s) The complete ionic equation is Mg 2 + (aq) + SO 4 2 − (aq) + Ba 2 + (aq) + 2NO 3 − (aq) → Mg 2 + (aq) + 2NO 3 − (aq) + BaSO 4 (s) Exercise \(\PageIndex{1}\) Write the complete ionic equation for \[\ce{CaCl2(aq) + Pb(NO3)2(aq) → Ca(NO3)2(aq) + PbCl2(s)}\nonumber \] Answer Ca 2 + (aq) + 2Cl − (aq) + Pb 2 + (aq) + 2NO 3 − (aq) → Ca 2 + (aq) + 2NO 3 − (aq) + PbCl 2 (s) You may notice that in a complete ionic equation, some ions do not change their chemical form; they stay exactly the same on the reactant and product sides of the equation. For example, in Na + (aq) + Cl − (aq) + Ag + (aq) + NO 3 − (aq) → AgCl(s) + Na + (aq) + NO 3 − (aq) the Ag + (aq) and Cl − (aq) ions become AgCl(s), but the Na + (aq) ions and the NO 3 − (aq) ions stay as Na + (aq) ions and NO 3 − (aq) ions. These two ions are examples of spectator ions—ions that do nothing in the overall course of a chemical reaction. They are present, but they do not participate in the overall chemistry. It is common to cancel spectator ions (something also done with algebraic quantities) on the opposite sides of a chemical equation: \[\cancel{Na^{+}(aq)}+Cl^{-}(aq)+Ag^{+}(aq)+\cancel{NO_{3}^{-}}(aq)\rightarrow AgCl(s)+\cancel{Na}^{+}(aq)+\cancel{NO}_{3}^{-}(aq)\nonumber \] What remains when the spectator ions are removed is called the net ionic equation , which represents the actual chemical change occurring between the ionic compounds: Cl − (aq) + Ag + (aq) → AgCl(s) It is important to reiterate that the spectator ions are still present in solution, but they do not experience any net chemical change, so they are not written in a net ionic equation. Example \(\PageIndex{1}\) Write the net ionic equation for each chemical reaction. K + (aq) + Br − (aq) + Ag + (aq) + C 2 H 3 O 2 − (aq) → K + (aq) + C 2 H 3 O 2 − (aq) + AgBr(s) Mg 2 + (aq) + SO 4 2 − (aq) + Ba 2 + (aq) + 2NO 3 − (aq) → Mg 2 + (aq) + 2NO 3 − (aq) + BaSO 4 (s) Solution In the first equation, the K + (aq) and C 2 H 3 O 2 − (aq) ions are spectator ions, so they are canceled: \[\cancel{K^{+}(aq)}+Br^{-}(aq)+Ag^{+}(aq)+\cancel{C_{2}H_{3}O_{2}^{-}(aq)}\rightarrow K^{+}(aq)+\cancel{C_{2}H_{3}O_{2}^{-}(aq)}+AgBr(s)\nonumber \] The net ionic equation is Br − (aq) + Ag + (aq) → AgBr(s) In the second equation, the Mg 2 + (aq) and NO 3 − (aq) ions are spectator ions, so they are canceled: \[\cancel{Mg^{2+}(aq)}+SO_{4}^{2-}(aq)+Ba^{2+}(aq)+\cancel{2NO_{3}^{-}(aq)}\rightarrow Mg^{2+}(aq)+\cancel{2NO_{3}^{-}(aq)}+BaSo_{4}(s)\nonumber \] The net ionic equation is SO 4 2 − (aq) + Ba 2 + (aq) → BaSO 4 (s) Exercise \(\PageIndex{1}\) Write the net ionic equation for CaCl 2 (aq) + Pb(NO 3 ) 2 (aq) → Ca(NO 3 ) 2 (aq) + PbCl 2 (s) Answer Pb 2 + (aq) + 2Cl − (aq) → PbCl 2 (s) Chemistry is Everywhere: Soluble and Insoluble Ionic Compounds The concept of solubility versus insolubility in ionic compounds is a matter of degree. Some ionic compounds are very soluble, some are only moderately soluble, and some are soluble so little that they are considered insoluble. For most ionic compounds, there is also a limit to the amount of compound that can be dissolved in a sample of water. For example, you can dissolve a maximum of 36.0 g of NaCl in 100 g of water at room temperature, but you can dissolve only 0.00019 g of AgCl in 100 g of water. We consider \(\ce{NaCl}\) soluble but \(\ce{AgCl}\) insoluble. One place where solubility is important is in the tank-type water heater found in many homes in the United States. Domestic water frequently contains small amounts of dissolved ionic compounds, including calcium carbonate (CaCO 3 ). However, CaCO 3 has the relatively unusual property of being less soluble in hot water than in cold water. So as the water heater operates by heating water, CaCO 3 can precipitate if there is enough of it in the water. This precipitate, called limescale , can also contain magnesium compounds, hydrogen carbonate compounds, and phosphate compounds. The problem is that too much limescale can impede the function of a water heater, requiring more energy to heat water to a specific temperature or even blocking water pipes into or out of the water heater, causing dysfunction. Another place where solubility versus insolubility is an issue is the Grand Canyon. We usually think of rock as insoluble. But it is actually ever so slightly soluble. This means that over a period of about two billion years, the Colorado River carved rock from the surface by slowly dissolving it, eventually generating a spectacular series of gorges and canyons. And all because of solubility! Key Takeaways Ionic compounds that dissolve separate into individual ions. Complete ionic equations show dissolved ionic solids as separated ions. Net ionic equations show only the ions and other substances that change in a chemical reaction. |
Courses/University_of_West_Georgia/CCHEM_1152K%3A_Survey_of_Chemistry_II/08%3A_Amines_and_Amides/8.05%3A_Physical_Properties_of_Amides | Learning Objectives Compare the boiling points of amides with alcohols of similar molar mass. Compare the solubilities in water of amides of five or fewer carbon atoms with the solubilities of comparable alkanes and alcohols in water. With the exception of formamide (HCONH 2 ), which is a liquid, all simple amides are solids (Table \(\PageIndex{1}\)). The lower members of the series are soluble in water, with borderline solubility occurring in those that have five or six carbon atoms. Like the esters, solutions of amides in water usually are neutral—neither acidic nor basic. Condensed Structural Formula Name Melting Point (°C) Boiling Point (°C) Solubility in Water HCONH2 formamide 2 193 soluble CH3CONH2 acetamide 82 222 soluble CH3CH2CONH2 propionamide 81 213 soluble CH3CH2CH2CONH2 butyramide 115 216 soluble C6H5CONH2 benzamide 132 290 slightly soluble The amides generally have high boiling points and melting points. These characteristics and their solubility in water result from the polar nature of the amide group and hydrogen bonding (Figure \(\PageIndex{1}\)). (Similar hydrogen bonding plays a critical role in determining the structure and properties of proteins, deoxyribonucleic acid [DNA], ribonucleic acid [RNA], and other giant molecules so important to life processes. Key Takeaways Most amides are solids at room temperature; the boiling points of amides are much higher than those of alcohols of similar molar mass. Amides of five or fewer carbon atoms are soluble in water?. |
Courses/University_of_Toronto/UTSC%3A_First-Year_Chemistry_Textbook_(Winter_2025)/16%3A_Solubility_and_Complex-Ion_Equilibria/16.09%3A_Qualitative_Cation_Analysis | Learning Objectives To know how to separate metal ions by selective precipitation. To understand how several common metal cations can be identified in a solution using selective precipitation. The composition of relatively complex mixtures of metal ions can be determined using qualitative analysis , a procedure for discovering the identity of metal ions present in the mixture (rather than quantitative information about their amounts). The procedure used to separate and identify more than 20 common metal cations from a single solution consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all of the metal ions are precipitated, as illustrated in Figure \(\PageIndex{1}\). Group 1: Insoluble Chlorides Most metal chloride salts are soluble in water; only \(\ce{Ag^{+}}\), \(\ce{Pb^{2+}}\), and \(\ce{Hg2^{2+}}\) form chlorides that precipitate from water. Thus the first step in a qualitative analysis is to add about 6 M \(\ce{HCl}\), thereby causing \(\ce{AgCl}\), \(\ce{PbCl2}\), and/or \(\ce{Hg2Cl2}\) to precipitate. If no precipitate forms, then these cations are not present in significant amounts. The precipitate can be collected by filtration or centrifugation. Group 2: Acid-Insoluble Sulfides Next, the acidic solution is saturated with \(\ce{H2S}\) gas. Only those metal ions that form very insoluble sulfides, such as \(\ce{As^{3+}}\), \(\ce{Bi^{3+}}\), \(\ce{Cd^{2+}}\), \(\ce{Cu^{2+}}\), \(\ce{Hg^{2+}}\), \(\ce{Sb^{3+}}\), and \(\ce{Sn^{2+}}\), precipitate as their sulfide salts under these acidic conditions. All others, such as \(\ce{Fe^{2+}}\) and \(\ce{Zn^{2+}}\), remain in solution. Once again, the precipitates are collected by filtration or centrifugation. Group 3: Base-Insoluble Sulfides (and Hydroxides) Ammonia or \(\ce{NaOH}\) is now added to the solution until it is basic, and then \(\ce{(NH4)2S}\) is added. This treatment removes any remaining cations that form insoluble hydroxides or sulfides. The divalent metal ions \(\ce{Co^{2+}}\), \(\ce{Fe^{2+}}\), \(\ce{Mn^{2+}}\), \(\ce{Ni^{2+}}\), and \(\ce{Zn^{2+}}\) precipitate as their sulfides, and the trivalent metal ions \(\ce{Al^{3+}}\) and \(\ce{Cr^{3+}}\) precipitate as their hydroxides: \(\ce{Al(OH)3}\) and \(\ce{Cr(OH)3}\). If the mixture contains \(\ce{Fe^{3+}}\), sulfide reduces the cation to \(\ce{Fe^{2+}}\), which precipitates as \(\ce{FeS}\). Group 4: Insoluble Carbonates or Phosphates The next metal ions to be removed from solution are those that form insoluble carbonates and phosphates. When \(\ce{Na2CO3}\) is added to the basic solution that remains after the precipitated metal ions are removed, insoluble carbonates precipitate and are collected. Alternatively, adding \(\ce{(NH4)2HPO4}\) causes the same metal ions to precipitate as insoluble phosphates. Group 5: Alkali Metals At this point, we have removed all the metal ions that form water-insoluble chlorides, sulfides, carbonates, or phosphates. The only common ions that might remain are any alkali metals (\(\ce{Li^{+}}\), \(\ce{Na^{+}}\), \(\ce{K^{+}}\), \(\ce{Rb^{+}}\), and \(\ce{Cs^{+}}\)) and ammonium (\(\ce{NH4^{+}}\)). We now take a second sample from the original solution and add a small amount of \(\ce{NaOH}\) to neutralize the ammonium ion and produce \(\ce{NH3}\). (We cannot use the same sample we used for the first four groups because we added ammonium to that sample in earlier steps.) Any ammonia produced can be detected by either its odor or a litmus paper test. A flame test on another original sample is used to detect sodium, which produces a characteristic bright yellow color. The other alkali metal ions also give characteristic colors in flame tests, which allows them to be identified if only one is present. Metal ions that precipitate together are separated by various additional techniques, such as forming complex ions, changing the pH of the solution, or increasing the temperature to redissolve some of the solids. For example, the precipitated metal chlorides of group 1 cations, containing \(\ce{Ag^{+}}\), \(\ce{Pb^{2+}}\), and \(\ce{Hg2^{2+}}\), are all quite insoluble in water. Because \(\ce{PbCl2}\) is much more soluble in hot water than are the other two chloride salts, however, adding water to the precipitate and heating the resulting slurry will dissolve any \(\ce{PbCl2}\) present. Isolating the solution and adding a small amount of \(\ce{Na2CrO4}\) solution to it will produce a bright yellow precipitate of \(\ce{PbCrO4}\) if \(\ce{Pb^{2+}}\) were in the original sample ( Figure \(\PageIndex{2}\)). As another example, treating the precipitates from group 1 cations with aqueous ammonia will dissolve any \(\ce{AgCl}\) because \(\ce{Ag^{+}}\) forms a stable complex with ammonia: \(\ce{[Ag(NH3)2]^{+}}\). In addition, \(\ce{Hg2Cl2}\) disproportionates in ammonia. \[\ce{2Hg2^{2+} \rightarrow Hg + Hg^{2+}} \nonumber \] to form a black solid that is a mixture of finely divided metallic mercury and an insoluble mercury(II) compound, which is separated from solution: \[\ce{Hg2Cl2(s) + 2NH3(aq) \rightarrow Hg(l) + Hg(NH_2)Cl(s) + NH^{+}4(aq) + Cl^{−}(aq)} \nonumber \] Any silver ion in the solution is then detected by adding \(\ce{HCl}\), which reverses the reaction and gives a precipitate of white \(\ce{AgCl}\) that slowly darkens when exposed to light: \[\ce{[Ag(NH3)2]^{+} (aq) + 2H^{+}(aq) + Cl^{−}(aq) \rightarrow AgCl(s) + 2NH^{+}4(aq)} \nonumber \] Similar but slightly more complex reactions are also used to separate and identify the individual components of the other groups. Summary In qualitative analysis, the identity, not the amount, of metal ions present in a mixture is determined. The technique consists of selectively precipitating only a few kinds of metal ions at a time under given sets of conditions. Consecutive precipitation steps become progressively less selective until almost all the metal ions are precipitated. Other additional steps are needed to separate metal ions that precipitate together. |
Courses/Duke_University/CHEM_310L%3A_Physical_Chemistry_I_Laboratory/CHEM310L_-_Physical_Chemistry_I_Lab_Manual/06%3A_Rotation-Vibration_Spectrum_of_HCl/6.02%3A_Background_on_Rotovibrational_Spectroscopy/6.2.01%3A_The_Electromagnetic_Spectrum_and_Molecular_Transitions | Electromagnetic radiation, or light, exhibits dual behavior as both waves and particles. Wave properties, such as refraction, best explain phenomena like light passing through different mediums. Conversely, particle behavior provides a more accurate description in certain scenarios, like absorption and emission. The nature of electromagnetic radiation remains unclear, with wave and particle models coexisting since the development of quantum mechanics in the early 20th century. Despite this ambiguity, these dual models effectively characterize electromagnetic radiation. It consists of oscillating electric and magnetic fields that travel through space at a constant velocity. In a vacuum, the speed of light (c) is \(2.99792 \times 10^8 \frac{m}{s}\). When passing through a medium other than a vacuum, its velocity (v) is slightly less than c, with the difference being <0.1%. For practical purposes, the speed of light is often approximated as \(3.00 \times 10^8 \frac{m}{s}\). The oscillations in the electric and magnetic fields are perpendicular to each other and the wave's propagation direction. An model of plane-polarized electromagnetic radiation, featuring a single oscillating electric field and a single oscillating magnetic field, is illustrated in Figure \(\PageIndex{1}\). An electromagnetic wave is characterized by several fundamental properties, including its velocity, amplitude, frequency, phase angle, polarization, and direction of propagation. For example, the amplitude of the oscillating electric field at any point along the propagating wave is \[A_\ce{t} = A_\ce{e}\sin(2π \nu t + \phi)\] where \(A_\ce{t}\) is the magnitude of the electric field at time \(t\), \(A_\ce{t}\) is the electric field’s maximum amplitude , \(\nu\) is the wave’s frequency —the number of oscillations in the electric field per unit time—and \(\phi\) is a phase angle , which accounts for the fact that A t need not have a value of zero at t = 0. The identical equation for the magnetic field is \[A_\ce{t} =A_\ce{m}\sin(2π \nu t + \phi)\] where A m is the magnetic field’s maximum amplitude. Units of energy The wavelength , λ , is defined as the distance between successive maxima (Figure \(\PageIndex{1}\)). The relationship between wavelength (in m) and frequency (in \(s^{-1}\) or \(Hz\)) is given below. Wavelength is usually expressed in nanometers (1 nm = 10 –9 m) for ultraviolet and visible light, and in microns (1 μm = 10 –6 m) for infrared light. \[λ = \dfrac{c}{\nu}\ \label{eq1}\] Wavelength is a unit used often by chemists, especially in UV-vis spectroscopy. Because it is inversely related to energy (ie smaller wavelengths are higher energy), many scientists prefer units that are directly proportional to energy. An alternate unit, the wavenumber , \(\tilde{ν}\), is used frequently, especially in infrared spectroscopy. In general it seems to be the preferred unit of physicist due to its direct correlation to energy. It is useful to know how to convert between these two commonly-used units. The unit \(\tilde{ν}\) is simply the reciprocal of wavelength. \[\tilde{ν} = \dfrac{1}{λ} \label{eq2}\] Wavenumbers are frequently used to characterize infrared radiation, with the units given in cm –1 . Keeping symbols straight Heads up! The word "frequency" refers to cycles per unit time, but can also be measured in a length scale. Frequency measured in cycles per seconds (\(s^{-1}\) can be expressed in the metric base unit Hertz, and can have the symbol \(f\) or \(\nu\). When it is expressed in a length scale, like wavenumbers (\(cm^{-1}\)), it has the symbol \(\tilde{\nu}\). When converting units from \(\nu\) (in \(s^{-1}\)) to \(\lambda\), use the speed of light to convert between time and length scales as in \ref{eq1}. When converting from \(\tilde{\nu}\) to \(\lambda\), use \ref{eq2} and convert metric units as necessary. Regions of the Spectrum and Types of Transitions Electromagnetic radiation spans a wide range of frequencies and wavelengths, prompting the division of this spectrum into distinct regions for convenience, known as the electromagnetic spectrum (Figure \(\PageIndex2\)). These divisions are based on the atomic or molecular transitions responsible for the absorption or emission of photons. It's important to note that the boundaries between these regions are not fixed, allowing for potential overlap between different spectral regions. Given the energy levels of molecules and their corresponding wavelengths, we can determine the frequency within the electromagnetic spectrum regions using the equation: \(\Delta E=E_u-E_l=h\) Here, \(h\) represents the energy change between the upper (\(u\)) and lower (\(l\)) states. The frequencies of molecular transitions fall within specific ranges. These ranges are listed in the table below in order of increasing energies. Wavelengths Frequency (\(cm^{-1}\)) Frequency (s-1) Region Transitions 1 mm - 1 m ~ 0.01 108 - 1011 microwave rotational transitions of large polyatomic molecules ~100 \(\mu\)m ~10 1012 far infrared rotational transitions of small molecules ~ 10 \(\mu\)m ~100 - 5000 ~ 1013 infrared vibrational transitions ~700 - 400 nm ~ 14,000 - 25,000 ~1014 visible and ultraviolet electronic transitions Absorption and Emission In absorption spectroscopy a photon is absorbed by an atom or molecule, which undergoes a transition from a lower-energy state to a higher-energy, or excited state (Figure \(\PageIndex{3}\)). The type of transition depends on the photon’s energy (see table above). Absorbing electromagnetic radiation can cause transitisions (excitations) in the rotational, vibrational, or electronic states depending on the energy of the photon. Likewise, relaxation of the molecule back to a lower-energy rotational, vibrational, or electronic state can result in emission of a photon (Figure \(\PageIndex{3}\)). When it absorbs electromagnetic radiation the number of photons passing through a sample decreases. The measurement of this decrease in photons, which we call absorbance , is a useful analytical signal. Note that each of the energy levels in Figure \(\PageIndex{3}\) has a well-defined value because they are quantized . Absorption occurs only when the photon’s energy, h ν , matches the difference in energy, ∆ E , between two energy levels. A plot of absorbance as a function of the photon’s energy is called an absorbance spectrum . Figure \(\PageIndex{4}\), for example, shows the visible absorbance spectrum of cranberry juice. When an atom or molecule in an excited state returns to a lower energy state, the excess energy often is released as a photon, a process we call emission (Figure \(\PageIndex{4}\)). There are several ways in which an atom or molecule may end up in an excited state, including thermal energy, absorption of a photon, or by a chemical reaction. Emission following the absorption of a photon is also called photoluminescence , and that following a chemical reaction is called chemiluminescence . A typical emission spectrum is shown in Figure \(\PageIndex{5}\). |
Courses/Westminster_College/CHE_180_-_Inorganic_Chemistry/11%3A_Chapter_11_-_Group_12/11.1%3A_Chemistry_of_Zinc | The name for zinc is of German origins, zink. It has been known since pre-historic times and compounds as well as the metal had been used for many years before anyone ever thought about elements at all! Zinc (Zn) is a blue-white metal of moderate strength, hardness and ductility. Zinc is one of the least common elements and is mostly produced through electrolysis of aqueous zinc sulfate. About one third of all metallic zinc is used to manufacture galvanized nails. Because of its low melting point and its ability to form bonds with iron or steel, it serves to coat the metal and protect it from corrosion. Metallic zinc is also used to make dry cell batteries. Physical Properties Pure zinc is a bluish-silver and ductile metal with a low melting and boiling point. Most zinc today is obtained from ZnS, extracted from zinc blende ore and roasted to remove the sulfur. Zinc can also be obtained by electrolysis of aqueous zinc sulfate, a common laboratory exercise. Figure 1: Brass die, along with zinc and copper samples. from Wikipedia Zinc plays a huge part in the production of alloys. One of the best known zinc alloy is brass, which contains between 55-95% copper. Zinc also takes part in manufacturing solder, which has a relatively low melting point. Solder is used to join electrical components, as well as pipes and other metals. Chemistry Properties Zinc is reasonably resistant to corrosion and is used as a covering for baser metals like iron ("galvanizing"). Zinc can be readily cast or molded. Zinc has many unique characteristics. For example, its vapor burns in air with a green flame, forming zinc oxide. Zinc oxide is a common zinc compound that is used in paints, cosmetics, plastics and more. Metallic zinc reacts with weak acids very slowly. Sulfur has a strong affinity for zinc. When heated, the two powders react explosively to form zinc sulfide. Zinc sulfide is used to make television screens and fluorescent light bulbs. Zinc also reacts with halogens. However, as the electronegativity decreases among the halogen group, the reactivity with zinc decreases. Thus, the most electronegative of the halogens (Fluorine) reacts with zinc violently, while the less electronegative halogen (Iodine) only generates a small amount of heat. Interestingly, properties of zinc are strongly affected by impurities such as lead, cadmium and iron. Also, zinc is most often used as a reducing agent in chemical reactions and it forms complex ions with ammonia and cyanide ions. References Gorodetsky, Malka; Singerman, Ammon. "Electroforming: An improvised experiment on electroplating." J. Chem. Educ. 1980 , 57 , 514. |
Courses/Modesto_Junior_College/Chemistry_143_-_Bunag/Old_Textbook_for_Chemistry_143_-_2019_to_2021/10%3A_Solutions_and_Colloids/10.06%3A_Colloids | Learning Objectives Describe the composition and properties of colloidal dispersions List and explain several technological applications of colloids As a child, you may have made suspensions such as mixtures of mud and water, flour and water, or a suspension of solid pigments in water, known as tempera paint. These suspensions are heterogeneous mixtures composed of relatively large particles that are visible (or that can be seen with a magnifying glass). They are cloudy, and the suspended particles settle out after mixing. On the other hand, when we make a solution, we prepare a homogeneous mixture in which no settling occurs and in which the dissolved species are molecules or ions. Solutions exhibit completely different behavior from suspensions. A solution may be colored, but it is transparent, the molecules or ions are invisible, and they do not settle out on standing. A group of mixtures called colloids (or colloidal dispersions ) exhibit properties intermediate between those of suspensions and solutions (Figure \(\PageIndex{1}\)). The particles in a colloid are larger than most simple molecules; however, colloidal particles are small enough that they do not settle out upon standing. The particles in a colloid are large enough to scatter light, a phenomenon called the Tyndall effect . This can make colloidal mixtures appear cloudy or opaque, such as the searchlight beams shown in Figure \(\PageIndex{2}\). Clouds are colloidal mixtures. They are composed of water droplets that are much larger than molecules, but that are small enough that they do not settle out. The term “colloid”—from the Greek words kolla , meaning “glue,” and eidos , meaning “like”—was first used in 1861 by Thomas Graham to classify mixtures such as starch in water and gelatin. Many colloidal particles are aggregates of hundreds or thousands of molecules, but others (such as proteins and polymer molecules) consist of a single extremely large molecule. The protein and synthetic polymer molecules that form colloids may have molecular masses ranging from a few thousand to many million atomic mass units. Analogous to the identification of solution components as “solute” and “solvent,” the components of a colloid are likewise classified according to their relative amounts. The particulate component typically present in a relatively minor amount is called the dispersed phase and the substance or solution throughout which the particulate is dispersed is called the dispersion medium . Colloids may involve virtually any combination of physical states (gas in liquid, liquid in solid, solid in gas, etc.), as illustrated by the examples of colloidal systems given in Table \(\PageIndex{1}\). Dispersed Phase Dispersion Medium Common Examples Name solid gas smoke, dust — solid liquid starch in water, some inks, paints, milk of magnesia sol solid solid some colored gems, some alloys — liquid gas clouds, fogs, mists, sprays aerosol liquid liquid milk, mayonnaise, butter emulsion liquid solid jellies, gels, pearl, opal (H2O in SiO2) gel gas liquid foams, whipped cream, beaten egg whites foam gas solid pumice, floating soaps — Preparation of Colloidal Systems We can prepare a colloidal system by producing particles of colloidal dimensions and distributing these particles throughout a dispersion medium. Particles of colloidal size are formed by two methods: Dispersion methods: that is, by breaking down larger particles. For example, paint pigments are produced by dispersing large particles by grinding in special mills. Condensation methods: that is, growth from smaller units, such as molecules or ions. For example, clouds form when water molecules condense and form very small droplets. A few solid substances, when brought into contact with water, disperse spontaneously and form colloidal systems. Gelatin, glue, starch, and dehydrated milk powder behave in this manner. The particles are already of colloidal size; the water simply disperses them. Powdered milk particles of colloidal size are produced by dehydrating milk spray. Some atomizers produce colloidal dispersions of a liquid in air. We can prepare an emulsion by shaking together or blending two immiscible liquids. This breaks one liquid into droplets of colloidal size, which then disperse throughout the other liquid. Oil spills in the ocean may be difficult to clean up, partly because wave action can cause the oil and water to form an emulsion. In many emulsions, however, the dispersed phase tends to coalesce, form large drops, and separate. Therefore, emulsions are usually stabilized by an emulsifying agent , a substance that inhibits the coalescence of the dispersed liquid. For example, a little soap will stabilize an emulsion of kerosene in water. Milk is an emulsion of butterfat in water, with the protein casein as the emulsifying agent. Mayonnaise is an emulsion of oil in vinegar, with egg yolk components as the emulsifying agents. Condensation methods form colloidal particles by aggregation of molecules or ions. If the particles grow beyond the colloidal size range, drops or precipitates form, and no colloidal system results. Clouds form when water molecules aggregate and form colloid-sized particles. If these water particles coalesce to form adequately large water drops of liquid water or crystals of solid water, they settle from the sky as rain, sleet, or snow. Many condensation methods involve chemical reactions. We can prepare a red colloidal suspension of iron(III) hydroxide by mixing a concentrated solution of iron(III) chloride with hot water: \[\mathrm{Fe^{3+}}_{(aq)}+\mathrm{3Cl^-}_{(aq)}+\mathrm{6H_2O}_{(l)}⟶\mathrm{Fe(OH)}_{3(s)}+\mathrm{H_3O^+}_{(aq)}+\mathrm{3Cl^-}_{(aq)} \label{11.6.1} \] A colloidal gold sol results from the reduction of a very dilute solution of gold(III) chloride by a reducing agent such as formaldehyde, tin(II) chloride, or iron(II) sulfate: \[\ce{Au}^{3+}+ \ce{3e}^− \rightarrow \ce{Au} \label{11.6.2} \] Some gold sols prepared in 1857 are still intact (the particles have not coalesced and settled), illustrating the long-term stability of many colloids. Soaps and Detergents Pioneers made soap by boiling fats with a strongly basic solution made by leaching potassium carbonate, \(\ce{K_2CO_3}\), from wood ashes with hot water. Animal fats contain polyesters of fatty acids (long-chain carboxylic acids). When animal fats are treated with a base like potassium carbonate or sodium hydroxide, glycerol and salts of fatty acids such as palmitic, oleic, and stearic acid are formed. The salts of fatty acids are called soaps. The sodium salt of stearic acid, sodium stearate, has the formula \(\ce{C_{17}H_{35}CO_2Na}\) and contains an uncharged nonpolar hydrocarbon chain, the \(\mathrm{C_{17}H_{35}-}\) unit, and an ionic carboxylate group, the \(-\mathrm{\sideset{ }{_{2}^{-}}{CO}}\) unit (Figure \(\PageIndex{3}\)). Deterg ents (soap substitutes) also contain nonpolar hydrocarbon chains, such as \(\mathrm{C_{12}H_{25}—}\), and an ionic group, such as a sulfate \(—\mathrm{\sideset{ }{_{3}^{-}}{OSO}}\), or a sulfonate \(—\mathrm{\sideset{ }{_{3}^{-}}{SO}}\) (Figure \(\PageIndex{4}\)). Soaps form insoluble calcium and magnesium compounds in hard water; detergents form water-soluble products—a definite advantage for detergents. The cleaning action of soaps and detergents can be explained in terms of the structures of the molecules involved. The hydrocarbon (nonpolar) end of a soap or detergent molecule dissolves in, or is attracted to, nonpolar substances such as oil, grease, or dirt particles. The ionic end is attracted by water (polar), illustrated in Figure \(\PageIndex{5}\). As a result, the soap or detergent molecules become oriented at the interface between the dirt particles and the water so they act as a kind of bridge between two different kinds of matter, nonpolar and polar. Molecules such as this are termed amphiphilic since they have both a hydrophobic (“water-fearing”) part and a hydrophilic (“water-loving”) part. As a consequence, dirt particles become suspended as colloidal particles and are readily washed away. Deepwater Horizon Oil Spill The blowout of the Deepwater Horizon oil drilling rig on April 20, 2010, in the Gulf of Mexico near Mississippi began the largest marine oil spill in the history of the petroleum. In the 87 days following the blowout, an estimated 4.9 million barrels (210 million gallons) of oil flowed from the ruptured well 5000 feet below the water’s surface. The well was finally declared sealed on September 19, 2010. Crude oil is immiscible with and less dense than water, so the spilled oil rose to the surface of the water. Floating booms, skimmer ships, and controlled burns were used to remove oil from the water’s surface in an attempt to protect beaches and wetlands along the Gulf coast. In addition to removal of the oil, attempts were also made to lessen its environmental impact by rendering it “soluble” (in the loose sense of the term) and thus allowing it to be diluted to hopefully less harmful levels by the vast volume of ocean water. This approach used 1.84 million gallons of the oil dispersant Corexit 9527, most of which was injected underwater at the site of the leak, with small amounts being sprayed on top of the spill. Corexit 9527 contains 2-butoxyethanol (C 6 H 14 O 2 ), an amphiphilic molecule whose polar and nonpolar ends are useful for emulsifying oil into small droplets, increasing the surface area of the oil and making it more available to marine bacteria for digestion (Figure \(\PageIndex{6}\)). While this approach avoids many of the immediate hazards that bulk oil poses to marine and coastal ecosystems, it introduces the possibility of long-term effects resulting from the introduction of the complex and potential toxic components of petroleum into the ocean’s food chain. A number of organizations are involved in monitoring the extended impact of this oil spill, including the National Oceanic and Atmospheric Administration. Electrical Properties of Colloidal Particles Dispersed colloidal particles are often electrically charged. A colloidal particle of iron(III) hydroxide, for example, does not contain enough hydroxide ions to compensate exactly for the positive charges on the iron(III) ions. Thus, each individual colloidal particle bears a positive charge, and the colloidal dispersion consists of charged colloidal particles and some free hydroxide ions, which keep the dispersion electrically neutral. Most metal hydroxide colloids have positive charges, whereas most metals and metal sulfides form negatively charged dispersions. All colloidal particles in any one system have charges of the same sign. This helps keep them dispersed because particles containing like charges repel each other. We can take advantage of the charge on colloidal particles to remove them from a variety of mixtures. If we place a colloidal dispersion in a container with charged electrodes, positively charged particles, such as iron(III) hydroxide particles, would move to the negative electrode. There, the colloidal particles lose their charge and coagulate as a precipitate. The carbon and dust particles in smoke are often colloidally dispersed and electrically charged. The charged particles are attracted to highly charged electrodes, where they are neutralized and deposited as dust (Figure \(\PageIndex{7}\). This is one of the important methods used to clean up the smoke from a variety of industrial processes. The process is also important in the recovery of valuable products from the smoke and flue dust of smelters, furnaces, and kilns. There are also ionic air filters designed for home use to improve indoor air quality. Gels When we make gelatin, such as Jell-O, we are making a type of colloid (Figure \(\PageIndex{8}\)). Gelatin sets on cooling because the hot aqueous mixture of gelatin coagulates as it cools and the whole mass, including the liquid, sets to an extremely viscous body known as a gel , a colloid in which the dispersing medium is a solid and the dispersed phase is a liquid. It appears that the fibers of the dispersing medium form a complex three-dimensional network, the interstices being filled with the liquid medium or a dilute solution of the dispersing medium. Because the formation of a gel is accompanied by the taking up of water or some other solvent, the gel is said to be hydrated or solvated. Pectin, a carbohydrate from fruit juices, is a gel-forming substance important in jelly making. Silica gel, a colloidal dispersion of hydrated silicon dioxide, is formed when dilute hydrochloric acid is added to a dilute solution of sodium silicate. Canned Heat is a gel made by mixing alcohol and a saturated aqueous solution of calcium acetate. Summary Colloids are mixtures in which one or more substances are dispersed as relatively large solid particles or liquid droplets throughout a solid, liquid, or gaseous medium. The particles of a colloid remain dispersed and do not settle due to gravity, and they are often electrically charged. Colloids are widespread in nature and are involved in many technological applications. Glossary amphiphilic molecules possessing both hydrophobic (nonpolar) and a hydrophilic (polar) parts colloid (also, colloidal dispersion) mixture in which relatively large solid or liquid particles are dispersed uniformly throughout a gas, liquid, or solid dispersion medium solid, liquid, or gas in which colloidal particles are dispersed dispersed phase substance present as relatively large solid or liquid particles in a colloid emulsifying agent amphiphilic substance used to stabilize the particles of some emulsions emulsion colloid formed from immiscible liquids gel colloidal dispersion of a liquid in a solid Tyndall effect scattering of visible light by a colloidal dispersion |
Courses/Prince_Georges_Community_College/CHEM_2000%3A_General_Chemistry_for_Engineers_-_F21/16%3A_Electrochemistry/16.09%3A_Corrosion-_Undesirable_Redox_Reactions | Learning Objectives To understand the process of corrosion. Corrosion is a galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. For example, when exposed to air, iron rusts, silver tarnishes, and copper and brass acquire a bluish-green surface called a patina. Of the various metals subject to corrosion, iron is by far the most important commercially. An estimated $100 billion per year is spent in the United States alone to replace iron-containing objects destroyed by corrosion. Consequently, the development of methods for protecting metal surfaces from corrosion constitutes a very active area of industrial research. In this section, we describe some of the chemical and electrochemical processes responsible for corrosion. We also examine the chemical basis for some common methods for preventing corrosion and treating corroded metals. Corrosion is a REDOX process. Under ambient conditions, the oxidation of most metals is thermodynamically spontaneous, with the notable exception of gold and platinum. Hence it is actually somewhat surprising that any metals are useful at all in Earth’s moist, oxygen-rich atmosphere. Some metals, however, are resistant to corrosion for kinetic reasons. For example, aluminum in soft-drink cans and airplanes is protected by a thin coating of metal oxide that forms on the surface of the metal and acts as an impenetrable barrier that prevents further destruction. Aluminum cans also have a thin plastic layer to prevent reaction of the oxide with acid in the soft drink. Chromium, magnesium, and nickel also form protective oxide films. Stainless steels are remarkably resistant to corrosion because they usually contain a significant proportion of chromium, nickel, or both. In contrast to these metals, when iron corrodes, it forms a red-brown hydrated metal oxide (\(\ce{Fe2O3 \cdot xH2O}\)), commonly known as rust, that does not provide a tight protective film (Figure \(\PageIndex{1}\)). Instead, the rust continually flakes off to expose a fresh metal surface vulnerable to reaction with oxygen and water. Because both oxygen and water are required for rust to form, an iron nail immersed in deoxygenated water will not rust—even over a period of several weeks. Similarly, a nail immersed in an organic solvent such as kerosene or mineral oil will not rust because of the absence of water even if the solvent is saturated with oxygen. In the corrosion process, iron metal acts as the anode in a galvanic cell and is oxidized to Fe 2 + ; oxygen is reduced to water at the cathode. The relevant reactions are as follows: at cathode: \[\ce{O2(g) + 4H^{+}(aq) + 4e^{−} -> 2H2O(l)} \nonumber \] with \(E^o_{SRP}=1.23\; V\). at anode: \[\ce{Fe(s) -> Fe^{2+}(aq) + 2e^{−}}\nonumber \] with \(E^o_{SRP} = −0.45\; V\). overall: \[\ce{2Fe(s) + O2(g) + 4H^{+}(aq) -> 2Fe^{2+}(aq) + 2H2O(l)} \label{Eq3} \] with \(E^o_{cell} = 1.68\; V\). The \(\ce{Fe^{2+}}\) ions produced in the initial reaction are then oxidized by atmospheric oxygen to produce the insoluble hydrated oxide containing \(\ce{Fe^{3+}}\), as represented in the following equation: \[\ce{4Fe^{2+}(aq) + O2(g) + (2 + 4x)H2O \rightarrow 2Fe2O3 \cdot xH2O + 4H^{+}(aq)} \label{Eq4} \] The sign and magnitude of \(E^o_{cell}\) for the corrosion process (Equation \(\ref{Eq3}\)) indicate that there is a strong driving force for the oxidation of iron by O 2 under standard conditions (1 M H + ). Under neutral conditions, the driving force is somewhat less but still appreciable (E = 1.25 V at pH 7.0). Normally, the reaction of atmospheric CO 2 with water to form H + and HCO 3 − provides a low enough pH to enhance the reaction rate, as does acid rain. Automobile manufacturers spend a great deal of time and money developing paints that adhere tightly to the car’s metal surface to prevent oxygenated water, acid, and salt from coming into contact with the underlying metal. Unfortunately, even the best paint is subject to scratching or denting, and the electrochemical nature of the corrosion process means that two scratches relatively remote from each other can operate together as anode and cathode, leading to sudden mechanical failure (Figure \(\PageIndex{2}\)). Prophylactic Protection One of the most common techniques used to prevent the corrosion of iron is applying a protective coating of another metal that is more difficult to oxidize. Faucets and some external parts of automobiles, for example, are often coated with a thin layer of chromium using an electrolytic process. With the increased use of polymeric materials in cars, however, the use of chrome-plated steel has diminished in recent years. Similarly, the “tin cans” that hold soups and other foods are actually consist of steel container that is coated with a thin layer of tin. While neither chromium nor tin metals are intrinsically resistant to corrosion, they both form protective oxide coatings that hinder access of oxygen and water to the underlying steel (iron alloy). As with a protective paint, scratching a protective metal coating will allow corrosion to occur. In this case, however, the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for \(\ce{Sn^{2+}}\) (E° = −0.14 V) and Fe 2 + (E° = −0.45 V) in Table P2 show that \(\ce{Fe}\) is more easily oxidized than \(\ce{Sn}\). As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen (Figure \(\PageIndex{3}\)). This process is seen in some older homes where copper and iron pipes have been directly connected to each other. The less easily oxidized copper acts as the cathode, causing iron to dissolve rapidly near the connection and occasionally resulting in a catastrophic plumbing failure. Cathodic Protection One way to avoid these problems is to use a more easily oxidized metal to protect iron from corrosion. In this approach, called cathodic protection, a more reactive metal such as \(\ce{Zn}\) (E° = −0.76 V for \(\ce{Zn^{2+} + 2e^{−} -> Zn}\)) becomes the anode, and iron becomes the cathode. This prevents oxidation of the iron and protects the iron object from corrosion. The reactions that occur under these conditions are as follows: \[ \underbrace{O_{2(g)} + 4e^− + 4H^+_{(aq)} \rightarrow 2H_2O_{(l)} }_{\text{reduction at cathode}}\label{Eq5} \] \[ \underbrace{Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^−}_{\text{oxidation at anode}} \label{Eq6} \] \[ \underbrace{ 2Zn_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2Zn^{2+}_{(aq)} + 2H_2O_{(l)} }_{\text{overall}}\label{Eq7} \] The more reactive metal reacts with oxygen and will eventually dissolve, “sacrificing” itself to protect the iron object. Cathodic protection is the principle underlying galvanized steel, which is steel protected by a thin layer of zinc. Galvanized steel is used in objects ranging from nails to garbage cans. In a similar strategy, sacrificial electrodes using magnesium, for example, are used to protect underground tanks or pipes (Figure \(\PageIndex{4}\)). Replacing the sacrificial electrodes is more cost-effective than replacing the iron objects they are protecting. Example \(\PageIndex{1}\) Suppose an old wooden sailboat, held together with iron screws, has a bronze propeller (recall that bronze is an alloy of copper containing about 7%–10% tin). If the boat is immersed in seawater, what corrosion reaction will occur? What is \(E^o°_{cell}\)? How could you prevent this corrosion from occurring? Given: identity of metals Asked for: corrosion reaction, \(E^o°_{cell}\), and preventive measures Strategy: Write the reactions that occur at the anode and the cathode. From these, write the overall cell reaction and calculate \(E^o°_{cell}\). Based on the relative redox activity of various substances, suggest possible preventive measures. Solution A According to Table P2 , both copper and tin are less active metals than iron (i.e., they have higher positive values of \(E^o°_{cell}\) than iron). Thus if tin or copper is brought into electrical contact by seawater with iron in the presence of oxygen, corrosion will occur. We therefore anticipate that the bronze propeller will act as the cathode at which \(\ce{O2}\) is reduced, and the iron screws will act as anodes at which iron dissolves: \[\begin{align*} & \textrm{cathode:} & & \mathrm{O_2(s)} + \mathrm{4H^+(aq)}+\mathrm{4e^-}\rightarrow \mathrm{2H_2O(l)} & & E^\circ_{\textrm{cathode}} =\textrm{1.23 V} \\ & \textrm{anode:} & & \mathrm{Fe(s)} \rightarrow \mathrm{Fe^{2+}} +\mathrm{2e^-} & & E^\circ_{\textrm{anode}} =-\textrm{0.45 V} \\ & \textrm{overall:} & & \mathrm{2Fe(s)}+\mathrm{O_2(g)}+\mathrm{4H^+(aq)} \rightarrow \mathrm{2Fe^{2+}(aq)} +\mathrm{2H_2O(l)} & & E^\circ_{\textrm{overall}} =\textrm{1.68 V} \end{align*} \nonumber \] Over time, the iron screws will dissolve, and the boat will fall apart. B Possible ways to prevent corrosion, in order of decreasing cost and inconvenience, are as follows: disassembling the boat and rebuilding it with bronze screws; removing the boat from the water and storing it in a dry place; or attaching an inexpensive piece of zinc metal to the propeller shaft to act as a sacrificial electrode and replacing it once or twice a year. Because zinc is a more active metal than iron, it will act as the sacrificial anode in the electrochemical cell and dissolve (Equation \(\ref{Eq7}\)). Exercise \(\PageIndex{1}\) Suppose the water pipes leading into your house are made of lead, while the rest of the plumbing in your house is iron. To eliminate the possibility of lead poisoning, you call a plumber to replace the lead pipes. He quotes you a very low price if he can use up his existing supply of copper pipe to do the job. Do you accept his proposal? What else should you have the plumber do while at your home? Answer a Not unless you plan to sell the house very soon because the \(\ce{Cu/Fe}\) pipe joints will lead to rapid corrosion. Answer b Any existing \(\ce{Pb/Fe}\) joints should be examined carefully for corrosion of the iron pipes due to the \(\ce{Pb–Fe}\) junction; the less active \(\ce{Pb}\) will have served as the cathode for the reduction of \(\ce{O2}\), promoting oxidation of the more active \(\ce{Fe}\) nearby. Summary Corrosion is a galvanic process that can be prevented using cathodic protection. The deterioration of metals through oxidation is a galvanic process called corrosion. Protective coatings consist of a second metal that is more difficult to oxidize than the metal being protected. Alternatively, a more easily oxidized metal can be applied to a metal surface, thus providing cathodic protection of the surface. A thin layer of zinc protects galvanized steel. Sacrificial electrodes can also be attached to an object to protect it. |
Bookshelves/General_Chemistry/Book%3A_Structure_and_Reactivity_in_Organic_Biological_and_Inorganic_Chemistry_(Schaller)/I%3A__Chemical_Structure_and_Properties/06%3A_Conformational_Analysis/6.08%3A_Diamond_Lattice_Drawings | It will be helpful for you to be able to draw cyclohexane in the chair form because it is such a common structural element in natural compounds. Let's look at the way we would most easily draw cyclohexane: it's a six membered ring, a hexagon. Notice that the opposite sides of this ring are parallel. This feature is the key to drawing a chair. Start by drawing two parallel lines of equal length, one slightly higher than the other and one slightly offset to the side. Most people find it easiest to make this first pair of lines horizontal. A second pair of parallel lines will connect to these two at an angle, but at opposite ends of the original lines, and going in opposite directions: one upward, one downward. At one end of the structure, we are beginning to draw the uplifted back of the chair, and at the other end is the downward footrest. Next, bridge the gaps at the two ends of the drawing with another pair of parallel lines. That's the carbon skeleton of cyclohexane, but for careful conformational analysis we will also need to consider the positions of the hydrogens. Remember that there are three axial hydrogens above the ring and three below. Look at the ring you have drawn: three of the vertices are pointing downward and three are pointing upward. These vertices are pointing the way to the axial hydrogens. Extend these vertices with six parallel lines, three pointing up and three pointing down, from alternating corners. You have just formed the axial C-H bonds. Now, the useful thing to know about the drawing is that lines can only be drawn in four different directions. The remaining lines to the equatorial hydrogens will have to be chosen from among these directions. We already have lines in three directions on each carbon; we just need to add a fourth to each. To decide which direction is missing, skip one bond in either direction along the ring and you will find a parallel line. Finally, remember that the equatorial hydrogens point outward, away from the ring, so keep these lines pointing away from the center of the chair. This type of drawing is called a diamond lattice drawing, because the all-carbon structure of diamond can be constructed this way by extending the axial and equatorial lines into additional chairs. Diamond lattice drawings can also be used to depict staggered conformations of aliphatic hydrocarbons, but only if you are very careful. Taking a few minutes to practice these drawings now may be helpful before you proceed to the next section. Exercise \(\PageIndex{1}\) Using the cyclohexane chair as a template, draw diamond lattice representations of the following compounds in staggered conformations. If you have to, draw a whole chair and erase parts until you have only what you need. ethane propane butane in a gauche conformation butane in an anti conformation Answer |
Courses/Howard_University/Howard%3A_Physical_Chemistry_Laboratory/01%3A_New_Page/1.16%3A_Biomaterials-_Studies_of_Protein_Structure_by_Computational_Quantum_Chemistry | CHEM 174 Physical Chemistry Laboratory II Biomaterials: Studies of Protein Structure by Computational Quantum Chemistry Introduction You already become familiar Spartan. In this lab you will use it to: 1) Calculate an IR spectra of Glycine, Alanine, Leucine and Valine 2) Calculate an IR spectra of polyglycine 3) Examine the difference between an α-helix and a β-sheet. Procedure Part 1 Glycine Example Go into Spartan and select a new simulation. Click on the Peptide tab on the model entry toolbox. Click “Gly” to bring up glycine at the top of the toolbox. Click on the main screen to bring up glycine on the main model window. Now you need to terminate the glycine so click terminate. For simplicity on the calculate be sure that the terminate option is set to yield a neutral molecule. You are ready to set up the calculation. Enter calculations and choose the Hartree-Fock 3-21G model, choose single energy and click the IR box, also keep the molecule neutral. Submit the calculation under a name that you choose. When the calculation is complete view the IR spectra and note the IR bands that were calculated. Repeat for the other amino acids Compare the glycine calculation with your spectra for solid glycine from the previous laboratory and discuss in your laboratory report. Part 2. Triglycine Repeat for triglycine. Note that it would be better if the calculation was performed for a larger peptide but triglycine is already a little time consuming. You will compare this calculation with your spectra for solid collagen. Part 3. α-Helix and β-Sheets Use Spartan to construct a peptide with at least 20 amino acids (your choice). Transfer it to the screen as an α-helix. Save an image of the structure for your report and discuss its appearance. Use Spartan to construct the same peptide and transfer it to the screen as a β-sheet. Save an image of the structure for your report and discuss its appearance. Report Use the research journal article format. Discuss your results by answering the following questions in the discussion section of your report. For glycine, compare the individually computed vibrational frequencies to those observed (from the previous laboratory) in your experimental spectrum. Which vibrational motions produced the most intense peaks in your observed glycine spectrum? Is this consistent with the theoretical prediction? Compare the theoretical frequencies of glycine to those of the other amino acids. What are the similarities and what are the differences? For all the calculated spectra what can you say about the relationship between the functional groups and the absorbance peaks? How many peaks can you identify as being common between the different amino acids? You probably noticed that the same group (e.g. C=O stretching) in different molecules might have shifted significantly. By significantly, we mean by more than 2x the spectral resolution. Which group frequency shifted the most between your molecules? In what direction, relative to the Glycine frequencies, is the shift? (Lower energy or higher energy) Offer an explanation for this behavior. Perform some statistical analysis on the vibrational frequencies of the different amino acids based on literature data (average absolute error, standard deviation, and variance). What are the similarities and differences between the calculated spectra of glycine and triglycine? What are the similarities and differences between the calculated spectra of glycine and your measured glycine IR? What are the similarities and differences between the calculated spectra of triglycine and your measured collagen IR? Clearly the spectra will not the same but can you say anything about the trend? Discuss the major differences between peptides with α-helix and β-sheet structures. Overall, discuss your impressions on the utility of IR spectroscopy as a probe of protein structure: “thumbs up”, “thumbs down”, or something in between? Reference Spartan ’04 Windows: Tutorial and User’s Guide, Wavefunction, Irvine, CA, 2004, pages 102 – 105. |
Courses/Athabasca_University/Chemistry_350%3A_Organic_Chemistry_I/13%3A_Structure_Determination-_Nuclear_Magnetic_Resonance_Spectroscopy/13.04%3A_C_NMR_Spectroscopy-_Signal_Averaging_and_FT-NMR | The 12 C isotope of carbon - which accounts for up about 99% of the carbons in organic molecules - does not have a nuclear magnetic moment, and thus is NMR-inactive. Fortunately for organic chemists, however, the 13 C isotope, which accounts for most of the remaining 1% of carbon atoms in nature, has a magnetic moment just like protons. Most of what we have learned about 1 H-NMR spectroscopy also applies to 13 C-NMR, although there are several important differences. The basics of 13 C-NMR spectroscopy The magnetic moment of a 13 C nucleus is much weaker than that of a proton, meaning that NMR signals from 13 C nuclei are inherently much weaker than proton signals. This, combined with the low natural abundance of 13 C, means that it is much more difficult to observe carbon signals: more sample is required, and often the data from hundreds of scans must be averaged in order to bring the signal-to-noise ratio down to acceptable levels. Unlike 1 H-NMR signals, the area under a 13 C-NMR signal cannot be used to determine the number of carbons to which it corresponds. This is because the signals for some types of carbons are inherently weaker than for other types – peaks corresponding to carbonyl carbons, for example, are much smaller than those for methyl or methylene (CH 2 ) peaks. Peak integration is generally not useful in 13 C-NMR spectroscopy, except when investigating molecules that have been enriched with 13 C isotope (see section 5.6B). The resonance frequencies of 13 C nuclei are lower than those of protons in the same applied field - in a 7.05 Tesla instrument, protons resonate at about 300 MHz, while carbons resonate at about 75 MHz. This is fortunate, as it allows us to look at 13 C signals using a completely separate 'window' of radio frequencies. Just like in 1 H-NMR, the standard used in 13 C-NMR experiments to define the 0 ppm point is tetramethylsilane (TMS), although of course in 13 C-NMR it is the signal from the four equivalent carbons in TMS that serves as the standard. Chemical shifts for 13 C nuclei in organic molecules are spread out over a much wider range than for protons – up to 200 ppm for 13 C compared to 12 ppm for protons (see Table 3 for a list of typical 13 C-NMR chemical shifts). This is also fortunate, because it means that the signal from each carbon in a compound can almost always be seen as a distinct peak, without the overlapping that often plagues 1 H-NMR spectra. The chemical shift of a 13 C nucleus is influenced by essentially the same factors that influence a proton's chemical shift: bonds to electronegative atoms and diamagnetic anisotropy effects tend to shift signals downfield (higher resonance frequency). In addition, sp 2 hybridization results in a large downfield shift. The 13 C-NMR signals for carbonyl carbons are generally the furthest downfield (170-220 ppm), due to both sp 2 hybridization and to the double bond to oxygen. Example 13.4.1 How many sets of non-equivalent carbons are there in each of the molecules shown in exercise 5.1? Solution Example 13.4.2 How many sets of non-equivalent carbons are there in: toluene 2-pentanone para-xylene triclosan (all structures are shown earlier in this chapter) Solution Because of the low natural abundance of 13 C nuclei, it is very unlikely to find two 13 C atoms near each other in the same molecule, and thus we do not see spin-spin coupling between neighboring carbons in a 13 C-NMR spectrum . There is, however, heteronuclear coupling between 13 C carbons and the hydrogens to which they are bound. Carbon-proton coupling constants are very large, on the order of 100 – 250 Hz. For clarity, chemists generally use a technique called broadband decoupling , which essentially 'turns off' C-H coupling, resulting in a spectrum in which all carbon signals are singlets . Below is the proton-decoupled 13 C-NMR spectrum of ethyl acetate, showing the expected four signals, one for each of the carbons. While broadband decoupling results in a much simpler spectrum, useful information about the presence of neighboring protons is lost. However, another modern NMR technique called DEPT (Distortionless Enhancement by Polarization Transfer) allows us to determine how many hydrogens are bound to each carbon. For example, a DEPT experiment tells us that the signal at 171 ppm in the ethyl acetate spectrum is a quaternary carbon (no hydrogens bound, in this case a carbonyl carbon), that the 61 ppm signal is from a methylene (CH 2 ) carbon, and that the 21 ppm and 14 ppm signals are both methyl (CH 3 ) carbons. The details of the DEPT experiment are beyond the scope of this text, but DEPT information will often be provided along with 13 C spectral data in examples and problems. |
Bookshelves/Inorganic_Chemistry/Inorganic_Chemistry_(LibreTexts)/01%3A_Introduction_to_Inorganic_Chemistry/1.02%3A_Inorganic_vs_Organic_Chemistry | The division between the fields of Inorganic and Organic chemistry has become blurred. For example, let's look at one of the major classes of catalysts used for organic synthesis reactions: organometalic catalysts (Figure \(\PageIndex{1}\)). Organometallic catalysts like these, and all organometallic compounds, contain metals that are bonded to carbon or carbon-containing molecules. So, are they "inorganic" because they contain metals, or "organic" because they contain carbon? These illustrate that clear divisions between organic and inorganic chemistry do not exist. Further, metal ions are common in biology and so the idea that metals are "inorganic" and thus classed as "non-living or non-biological" is incorrect. A canonical example is the organometallic catalyst adenosylcobalbumin, which is an important biological cofactor containing a cobalt (Co) ion (Figure \(\PageIndex{1}\), right) and a cobalt-carbon bond. Some of the subfields of Inorganic Chemistry focus on electrical conductivity of inorganic materials (i.e., conduction, superconduction, and semiconduction) and on the study of optical and electronic properties of inorganic nanomaterials. Electrical conductivity is a canonical property of metals, but carbon-based materials also demonstrate electrical conductivity. For example, carbon nanotubes conduct electricity through their extended conjugated \(\pi\) systems. Fullerenes, of which the most famous is Buckminsterfullerene, or Buckeyball (C 60 ), demonstrate interesting properties that are similar to nanoparticles, and when combined with metals and crystallized can demonstrate superconductivity. Although carbon nanotubes and fullerenes are allotropes of carbon, their material properties are somewhat foreign to many organic chemists, who traditionally have focused on smaller organic molecules having very different properties. However, these properties are familiar to inorganic chemists. Thus, inorganic chemists have embraced these molecules as "inorganic" due to the fact that they behave more like inorganic materials than smaller organic molecules. This class of carbon-based molecules serves as another example of molecules that are not perfectly matched to the traditional definitions of "organic" and "inorganic" chemistry. Certainly, the future will hold more and more examples of molecules that do not fit into the traditional disciplines of chemistry. |
Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.13%3A_Lattice_Energy_-_Estimates_from_an_Electrostatic_Model/6.13B%3A_Coulombic_Interactions_in_an_Ionic_Lattice | Welcome to the Chemistry Library. This Living Library is a principal hub of the LibreTexts project , which is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning. The LibreTexts approach is highly collaborative where an Open Access textbook environment is under constant revision by students, faculty, and outside experts to supplant conventional paper-based books. Campus Bookshelves Bookshelves Learning Objects |
Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/17%3A_The_Organic_Chemistry_of_Vitamins/17.0S%3A_17.S%3A_The_Oganic_Chemistry_of_Vitamins_(Summary) | After completing this chapter, you should be able to: Understand how pyridoxal phosphate (PLP) acts as an 'electron sink' in a variety of reactions in amino acid metabolism. Recognize and draw mechanisms for PLP-dependent transformations of the following types: racemization decarboxylation transamination retroaldol cleavage retro-Claisen cleavage \(\beta \)-elimination \(\beta \)-substitution \(\gamma \)-elimination \(\gamma \)-substitution Recognize transformations - amino acid decarboxylation and transamination, for example - in which chemical steps occur that simply don't 'make sense' unless the electron sink role of PLP is taken into account. Understand how the orientation of the substrate in relation to the plane formed by the conjugated \(\pi \) system of PLP is a major factor in catalysis of PLP-dependent reactions. Understand how thiamine diphosphate (\(ThDP\)) acts as an 'electron sink' in a variety of reactions in which a bond to a carbonyl carbon is broken, and how these steps do not 'make sense' unless the electron sink role of \(ThDP\) is taken into account. Recognize transformations for which \(ThDP\) is likely required, and be able to draw reasonable mechanisms for them. Understand how \(ThDP\) acts in tandem with lipoamide, flavin, and nicotinamide in the reaction catalyzed by pyruvate dehydrogenase. Recognize folate in its various forms - \(DHF\), \(THF\), \(f-THF\), \(CH_2-THF\), and \(CH_3-THF\) - functions in a variety of one-carbon transfer steps. Be able to recognize the oxidation state of the carbon being transferred in a folate-depenent step. |
Bookshelves/Organic_Chemistry/Intermediate_Physical_Organic_(Morsch)/03%3A_Chemical_Thermodynamics/3.02%3A_Bond_Energies/3.2.03%3A_Entropy_Changes_in_Reactions | Entropy is another important aspect of thermodynamics. Enthalpy has something to do with the energetic content of a system or a molecule. Entropy has something to do with how that energy is stored. We sometimes speak of the energy in a system as being "partitioned" or divided into various "states". How this energy is divided up is the concern of entropy. By way of analogy, picture a set of mailboxes. You may have a wall of them in your dormitory or your apartment building. The mailboxes are of several different sizes: maybe there are a few rows of small ones, a couple of rows of medium sized ones, and a row of big mailboxes on the bottom. Instead of putting mail in these boxes, we're going to use them to hold little packages of energy. Later on, you might take the energy packages out of your own mailbox and use them to take a trip to the mall or the gym. But how does the mail get to your mailbox in the first place? The energy packages don't arrive in your molecular dormitory with addresses on them. The packages come in different sizes, because they contain different amounts of energy, but other than that there is no identifying information on them. Some of the packages don't fit into some of the mailboxes, because some of the packages are too big and some of the mailboxes are smaller than the others. The energy packages need to go into mailboxes that they will fit into. Still, there are an awful lot of mailboxes that most of the energy packages could still fit into. There needs to be some system of deciding where to put all of these packages. It turns out that, in the molecular world, there is such a system, and it follows a pretty simple rule. When a whole pile of energy packages arrive, the postmaster does her best to put one package into every mailbox. Then, when every mailbox has one, she starts putting a second one into each box, and so on. It didn't have to be that way. It could have been the case that all the energy was simply put into the first couple of mailboxes and the rest were left empty. In other words, the rule could have been that all the energy must be sorted into the same place, instead of being spread around. But that's not how it is. Energy is always partitioned into the maximum number of states possible. Entropy is the sorting of energy into different modes or states. When energy is partitioned or sorted into additional states, entropy is said to increase. When energy is bundled into a smaller number of states, entropy is said to decrease. Nature's bias is towards an increase in entropy. This is a fundamental law of the universe; there is no reason that can be used to explain why nature prefers high entropy to low entropy. Instead, increasing entropy is itself the basic reason for a wide range of things that happen in the universe. Entropy is popularly described in terms of "disorder". That can be a useful idea, although it doesn't really describe what is happening energetically. A better picture of entropy can be built by looking at how a goup of molecules might sort some energy that is added to them. In other words, what are some examples of "states" in which energy can be sorted? If you get more energy -- maybe by eating breakfast -- one of the immediate benefits is being able to increase your physical activity. You have more energy to move around, to run, to jump. A similar situation is true with molecules. Molecules have a variety of ways in which they can move, if they are given some energy. They can zip around; this kind of motion is usually called translation. They can tumble and roll; this kind of motion is referred to as rotation. Also, they can wiggle, letting their bonds get longer and shorter by moving individual atoms around a little bit. This type of motion is called vibration. When molecules absorb extra energy, they may be able to sort the energy into rotational, vibrational and translational states. This only works with energy packages of a certain size; other packages would be sorted into other kinds of states. However, these are just a few examples of what we mean by states. Okay, so energy is stored in states, and it is sorted into the maximum possible number of states. But how does entropy change in a reaction? We know that enthalpy may change by breaking or forming certain bonds, but how does the energy get sorted again? The changes in internal entropy during a reaction are often very small. In other words, the energy remaining at the end of the reaction gets sorted more or less the way it was before the reaction. However, there are some very common exceptions. The most common case in which internal entropy changes a lot is when the number of molecules involved changes between the start of the reaction and the end of the reaction. Maybe two molecules react together to form one, new molecule. Maybe one molecule splits apart to make two, new molecules. If one molecule splits apart in the reaction, entropy generally increases. Two molecules can rotate, vibrate and translate (or tumble, wiggle and zip around) independently of each other. That means the number of states available for partitioning energy increases when one molecule splits into two. Entropy generally increases when a reaction produces more molecules than it started with. Entropy generally decreases when a reaction produces fewer molecules than it started with. Apart from a factor like a change in the number of molecules involved, internal entropy changes are often fairly subtle. They are not as easy to predict as enthalpy changes. Nevertheless, there may sometimes be a trade-off between enthalpy and entropy. If a reaction splits a molecule into two, it seems likely that an increase in enthalpy will be involved, so that the bond that held the two pieces together can be broken. That's not favourable. However, when that happens, we've just seen that there will be an increase in entropy, because energy can then be sorted into additional modes in the two, independent molecules. So we have two different factors to balance. There is a tool we often use to decide which factor wins out. It's called free energy, and we will look at it next. |
Courses/Lumen_Learning/Book%3A_Microeconomics-1_(Lumen)/11%3A_9-_Perfect_Competition/11.17%3A_Reading-_Price_and_Revenue_in_a_Perfectly_Competitive_Industry_and_Firm | Price and Revenue Each firm in a perfectly competitive market is a price taker; the equilibrium price and industry output are determined by demand and supply. Figure 9.1 “The Market for Radishes” shows how demand and supply in the market for radishes, which we shall assume are produced under conditions of perfect competition, determine total output and price. The equilibrium price is $0.40 per pound; the equilibrium quantity is 10 million pounds per month. Figure 9.1 The Market for Radishes. Price and output in a competitive market are determined by demand and supply. In the market for radishes, the equilibrium price is $0.40 per pound; 10 million pounds per month are produced and purchased at this price. Because it is a price taker, each firm in the radish industry assumes it can sell all the radishes it wants at a price of $0.40 per pound. No matter how many or how few radishes it produces, the firm expects to sell them all at the market price. The assumption that the firm expects to sell all the radishes it wants at the market price is crucial. If a firm did not expect to sell all of its radishes at the market price—if it had to lower the price to sell some quantities—the firm would not be a price taker. And price-taking behavior is central to the model of perfect competition. Radish growers—and perfectly competitive firms in general—have no reason to charge a price lower than the market price. Because buyers have complete information and because we assume each firm’s product is identical to that of its rivals, firms are unable to charge a price higher than the market price. For perfectly competitive firms, the price is very much like the weather: they may complain about it, but in perfect competition there is nothing any of them can do about it. This video explains how the market supply and demand curves determine the price of a good, and why firms in a perfectly competitive market are price takers. A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/micro/?p=382 Total Revenue While a firm in a perfectly competitive market has no influence over its price, it does determine the output it will produce. In selecting the quantity of that output, one important consideration is the revenue the firm will gain by producing it. A firm’s total revenue is found by multiplying its output by the price at which it sells that output. For a perfectly competitive firm, total revenue ( TR ) is the market price ( P ) times the quantity the firm produces ( Q ), or TR = P x Q The relationship between market price and the firm’s total revenue curve is a crucial one. Panel (a) of Figure 9.2 “Total Revenue, Marginal Revenue, and Average Revenue” shows total revenue curves for a radish grower at three possible market prices: $0.20, $0.40, and $0.60 per pound. Each total revenue curve is a linear, upward-sloping curve. At any price, the greater the quantity a perfectly competitive firm sells, the greater its total revenue. Notice that the greater the price, the steeper the total revenue curve is. Figure 9.2 Total Revenue, Marginal Revenue, and Average Revenue. Panel (a) shows different total revenue curves for three possible market prices in perfect competition. A total revenue curve is a straight line coming out of the origin. The slope of a total revenue curve is MR; it equals the market price (P) and AR in perfect competition. Marginal revenue and average revenue are thus a single horizontal line at the market price, as shown in Panel (b). There is a different marginal revenue curve for each price. Marginal Revenue, Price, and Demand for the Perfectly Competitive Firm We have seen that a perfectly competitive firm’s marginal revenue curve is simply a horizontal line at the market price and that this same line is also the firm’s average revenue curve. For the perfectly competitive firm, M R = P = A R . The marginal revenue curve has another meaning as well. It is the demand curve facing a perfectly competitive firm. Consider the case of a single radish producer, Tony Gortari. We assume that the radish market is perfectly competitive; Mr. Gortari runs a perfectly competitive firm. Suppose the market price of radishes is $0.40 per pound. How many pounds of radishes can Mr. Gortari sell at this price? The answer comes from our assumption that he is a price taker: He can sell any quantity he wishes at this price. How many pounds of radishes will he sell if he charges a price that exceeds the market price? None. His radishes are identical to those of every other firm in the market, and everyone in the market has complete information. That means the demand curve facing Mr. Gortari is a horizontal line at the market price as illustrated in Figure 9.3 “Price, Marginal Revenue, and Demand”. Notice that the curve is labeled d to distinguish it from the market demand curve, D , in Figure 9.1 “The Market for Radishes”. The horizontal line in Figure 9.3 “Price, Marginal Revenue, and Demand” is also Mr. Gortari’s marginal revenue curve, MR , and his average revenue curve, AR . It is also the market price, P . Of course, Mr. Gortari could charge a price below the market price, but why would he? We assume he can sell all the radishes he wants at the market price; there would be no reason to charge a lower price. Mr. Gortari faces a demand curve that is a horizontal line at the market price. In our subsequent analysis, we shall refer to the horizontal line at the market price simply as marginal revenue. We should remember, however, that this same line gives us the market price, average revenue, and the demand curve facing the firm. Figure 9.3 Price, Marginal Revenue, and Demand. A perfectly competitive firm faces a horizontal demand curve at the market price. Here, radish grower Tony Gortari faces demand curve d at the market price of $0.40 per pound. He could sell q1 or q2—or any other quantity—at a price of $0.40 per pound. More generally, we can say that any perfectly competitive firm faces a horizontal demand curve at the market price. We saw an example of a horizontal demand curve in the module on elasticity. Such a curve is perfectly elastic, meaning that any quantity is demanded at a given price. Note that Figure 9.1 shows the market (and demand curve) for a perfectly competitive industry and Figure 9.3 shows the demand curve for a perfectly competitive firm . This video demonstrates how average revenue equals marginal revenue, which equals price in a perfectly competitive market. A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/micro/?p=382 Self Check: Perfectly Competitive Firms and Industries Answer the question(s) below to see how well you understand the topics covered in the previous section. This short quiz does not count toward your grade in the class, and you can retake it an unlimited number of times. You’ll have more success on the Self Check if you’ve completed the two Readings in this section. Use this quiz to check your understanding and decide whether to (1) study the previous section further or (2) move on to the next section. https://assessments.lumenlearning.com/assessments/652 CC licensed content, Shared previously Principles of Microeconomics Section 9.2 . Authored by : Anonymous. Located at : http://2012books.lardbucket.org/books/microeconomics-principles-v2.0/s12-competitive-markets-for-goods-.html . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike Perfect competiton: Demand curve for individual producer. Authored by : lostmy1. Provided by : University of South Africa. Located at : https://youtu.be/ZhGDce_a5eE?list=PL616B7E47EF9203CC . License : CC BY: Attribution Perfect competition: Average revenue = marginal revenue = price. Authored by : lostmy1. Provided by : University of South Africa. Located at : https://youtu.be/Nm6LY0V3YuQ?list=PL616B7E47EF9203CC . License : CC BY: Attribution |
Courses/Earlham_College/CHEM_361%3A_Inorganic_Chemistry_(Watson)/02%3A_Molecular_Symmetry_and_Point_Groups/2.04%3A_Examples_and_Applications_of_Symmetry/2.4.02%3A_Chirality | Introduction Around the year 1847, the French scientist Louis Pasteur provided an explanation for the optical activity of tartaric acid salts. When he carried out a particular reaction, Pasteur observed that two types of crystals precipitated. Patiently and carefully using tweezers, Pasteur was able to separate the two types of crystals. Pasteur noticed that the types rotated plane-polarized light by the same amount but in different directions. These two compounds are called enantiomers. What is chirality? A molecule is chiral (or dissymetric ) if it is non-superimposable on its mirror image. The two mirror images of a chiral molecule are called enantiomers. Enantiomers have the same physical properties (e.g., melting point, etc.). They differ in their ability to rotate plane polarized light and in their reactivity with other chiral molecules. Due to their ability to rotate plane polarized light, they are referred to as being optically active . Using Symmetry to Determine Chirality There are some general rules of thumb that help determine whether a molecule is chiral or achiral. The point group of the molecule, and the symmetry operations within that point group, can give clues as to whether the molecule is chiral. Symmetry operations of chiral molecules A chiral molecule cannot possess a plane of symmetry (\(\sigma\)), a center of inversion (\(i\)), or an improper rotation (\(S_n\)). Due to the fact that all groups that lack both \(\sigma\) and \(i\) also lack \(S_n\), a molecule that belongs to any group that lacks \(S_n\) is chiral . An example of an inorganic coordination complex is tris(ethylenediamine)cobalt(III) (Figure \(\PageIndex{1}\)). Figure \(\PageIndex{1}\) shows the two enantiomers of tris(ethylenediamine)cobalt(III): the \(\Delta\) and \(\Lambda\) isomers. You can visualize the \(\Lambda\) and \(\Delta\) isomers by imagining that the ligands around the metal centers are blades of a fan. To push the air toward you, you would need to rotate the \(\Delta\) isomer clockwise, and the \(\Lambda\) isomer counter-clockwise. Exercise \(\PageIndex{1}\) Which point groups are possible for chiral molecules? Answer As a result of the previous discussion, there are a few classes of point groups that lack an improper axis. Those classes are \(C_{1}\), \(C_{n}\), and \(D_{n}\). |
Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/21%3A_Nuclear_Chemistry | The chemical reactions that we have considered in previous chapters involve changes in the electronic structure of the species involved, that is, the arrangement of the electrons around atoms, ions, or molecules. Nuclear structure, the numbers of protons and neutrons within the nuclei of the atoms involved, remains unchanged during chemical reactions. This chapter will introduce the topic of nuclear chemistry, which began with the discovery of radioactivity in 1896 by French physicist Antoine Becquerel and has become increasingly important during the twentieth and twenty-first centuries, providing the basis for various technologies related to energy, medicine, geology, and many other areas. 21.1: Nuclear Structure and Stability An atomic nucleus consists of protons and neutrons, collectively called nucleons. Although protons repel each other, the nucleus is held tightly together by a short-range, but very strong, force called the strong nuclear force. A nucleus has less mass than the total mass of its constituent nucleons. This “missing” mass is the mass defect, which has been converted into the binding energy that holds the nucleus together according to Einstein’s mass-energy equivalence equation, E = mc2. 21.2: Nuclear Equations Nuclei can undergo reactions that change their number of protons, number of neutrons, or energy state. Many different particles can be involved in nuclear reactions. The most common are protons, neutrons, positrons (which are positively charged electrons), alpha (α) particles (which are high-energy helium nuclei), beta (β) particles (which are high-energy electrons), and gamma (γ) rays (which compose high-energy electromagnetic radiation). 21.3: Radioactive Decay Unstable nuclei undergo spontaneous radioactive decay. The most common types of radioactivity are α decay, β decay, γ emission, positron emission, and electron capture. Nuclear reactions also often involve γ rays, and some nuclei decay by electron capture. Each of these modes of decay leads to the formation of a new stable nuclei sometimes via multiple decays before ending in a stable isotope. All nuclear decay processes follow first-order kinetics and each radioisotope has its own half-life. 21.4: Transmutation and Nuclear Energy It is possible to produce new atoms by bombarding other atoms with nuclei or high-speed particles. The products of these transmutation reactions can be stable or radioactive. A number of artificial elements, including technetium, astatine, and the transuranium elements, have been produced in this way. Nuclear power as well as nuclear weapon detonations can be generated through fission (reactions in which a heavy nucleus is split into two or more lighter nuclei and several neutrons). 21.5: Uses of Radioisotopes Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US. 21.6: Biological Effects of Radiation We are constantly exposed to radiation from naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is particularly harmful because it can ionize molecules or break chemical bonds, which damages the molecules & causes malfunctions in cell processes. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating, but potentially most damaging, and gamma rays the most penetrating. 21.E: Nuclear Chemistry (Exercises) These are homework exercises to accompany the Textmap created for "Chemistry" by OpenStax. |
Courses/El_Paso_Community_College/CHEM1306%3A_Health_Chemistry_I_(Rodriguez)/09%3A_Solutions/9.S%3A_Solutions_(Summary) | To ensure that you understand the material in this chapter, you should review the meanings of the bold terms in the following summary and ask yourself how they relate to the topics in the chapter. A solution is a homogeneous mixture. The major component is the solvent , while the minor component is the solute . Solutions can have any phase; for example, an alloy is a solid solution. Solutes are soluble or insoluble , meaning they dissolve or do not dissolve in a particular solvent. The terms miscible and immiscible , instead of soluble and insoluble, are used for liquid solutes and solvents. The statement like dissolves like is a useful guide to predicting whether a solute will dissolve in a given solvent. The amount of solute in a solution is represented by the concentration of the solution. The maximum amount of solute that will dissolve in a given amount of solvent is called the solubility of the solute. Such solutions are saturated . Solutions that have less than the maximum amount are unsaturated . Most solutions are unsaturated, and there are various ways of stating their concentrations. Mass/mass percent , volume/volume percent , and mass/volume percent indicate the percentage of the overall solution that is solute. Parts per million (ppm) and parts per billion (ppb) are used to describe very small concentrations of a solute. Molarity , defined as the number of moles of solute per liter of solution, is a common concentration unit in the chemistry laboratory. Equivalents express concentrations in terms of moles of charge on ions. When a solution is diluted, we use the fact that the amount of solute remains constant to be able to determine the volume or concentration of the final diluted solution. Dissolving occurs by solvation , the process in which particles of a solvent surround the individual particles of a solute, separating them to make a solution. For water solutions, the word hydration is used. If the solute is molecular, it dissolves into individual molecules. If the solute is ionic, the individual ions separate from each other, forming a solution that conducts electricity. Such solutions are called electrolytes . If the dissociation of ions is complete, the solution is a strong electrolyte . If the dissociation is only partial, the solution is a weak electrolyte . Solutions of molecules do not conduct electricity and are called nonelectrolytes . Solutions have properties that differ from those of the pure solvent. Some of these are colligative properties, which are due to the number of solute particles dissolved, not the chemical identity of the solute. Colligative properties include vapor pressure depression , boiling point elevation , freezing point depression , and osmotic pressure . Osmotic pressure is particularly important in biological systems. It is caused by osmosis , the passage of solvents through certain membranes like cell walls. The osmolarity of a solution is the product of a solution’s molarity and the number of particles a solute separates into when it dissolves. Osmosis can be reversed by the application of pressure; this reverse osmosis is used to make fresh water from saltwater in some parts of the world. Because of osmosis, red blood cells placed in hypotonic or hypertonic solutions lose function through either hemolysis or crenation. If they are placed in isotonic solutions, however, the cells are unaffected because osmotic pressure is equal on either side of the cell membrane. |
Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/07%3A_Chemical_Nomenclature/7.02%3A_Empirical_Formula | What might the scientists in the picture be discussing? When the French scientist Antoine Lavoisier conducted his experiments, he did not know what the products of each reaction would be. He had to isolate the material (whether he was heating mercury or measuring gases from breathing) and then study its elemental composition before he could understand the processes that were occurring. Discovering that a new compound exists is the start of a long research project. In order to make this new compound in the lab, we need to know a lot about its structure. Often, the place to start is to determine the elements in the material. Then, we can find out the relative amounts of each element to continue our evaluation of this new material. Empirical Formula An empirical formula is a formula that shows the elements in a compound in their lowest whole-number ratio. Glucose is an important simple sugar that cells use as their primary source of energy. Its molecular formula is \(\ce{C_6H_{12}O_6}\). Since each of the subscripts is divisible by 6, the empirical formula for glucose is \(\ce{CH_2O}\). When chemists analyze an unknown compound, often the first step is to determine its empirical formula. There are a great many compounds whose molecular and empirical formulas are the same. If the molecular formula cannot be simplified into a smaller whole-number ratio, as in the case of \(\ce{H_2O}\) or \(\ce{P_2O_5}\), then the empirical formula is also the molecular formula. How do we determine an empirical formula for a compound? Consider a compound composed of carbon, hydrogen, and oxygen. We can analyze the relative amounts of each element in the compound. When we get a percent figure for each element, we now know how many grams of each are in 100 grams of the original material. This allows us to determine the number of moles for each element. The ratios can then be reduced to small whole numbers to give the empirical formula. If we wanted a molecular formula, we would need to determine the molecular weight of the compound. Summary The empirical formula gives the lowest whole-number ratio of elements in a compound. The empirical formula does not show the actual number of atoms. Review Define “empirical formula.” Why is C 6 H 12 O 6 not considered to be an empirical formula for glucose? Can the empirical formula for a compound be the same as the molecular formula? What do we need to know in order to determine a molecular formula from an empirical formula? Give three examples of compounds whose empirical formulas are the same as their molecular formulas. |
Courses/Santa_Barbara_City_College/SBCC_Chem_101%3A_Introductory_Chemistry/02%3A_Measurement_and_Problem_Solving/2.E%3A_Measurement_and_Problem_Solving_(Exercises) | 2.1: Measuring Global Temperatures 2.2: Scientific Notation: Writing Large and Small Numbers 2.3: Significant Figures: Writing Numbers to Reflect Precision Define significant figures . Why are they important? Define the different types of zeros found in a number and explain whether or not they are significant. How many significant figures are in each number? 140 0.009830 15,050 221,560,000 5.67 × 10 3 2.9600 × 10 −5 How many significant figures are in each number? 1.05 9,500 0.0004505 0.00045050 7.210 × 10 6 5.00 × 10 −6 Round each number to three significant figures. 34,705 34,750 34,570 2.4: Significant Figures in Calculations 2.5: The Basic Units of Measurement 2.6: Problem Solving and Unit Conversions 2.7: Solving Multi-step Conversion Problems 2.8: Units Raised to a Power 2.9: Density 2.10: Numerical Problem-Solving Strategies and the Solution Map |
Courses/Sewanee%3A_The_University_of_the_South/Organic_Chemistry_Lab_Textbook/06%3A_Crystallization/6.03%3A_Crystallization_Theory/6.3A%3A_Purification | Crystallization is an excellent purification technique for solids because a crystal slowly forming from a saturated solution tends to selectively incorporate particles of the same type into its crystal structure (a model of a crystal lattice is in Figure 3.16a). A pure crystal is often slightly lower in energy than an impure crystal (or has a higher lattice energy), as packing identical particles into a lattice allows for maximized intermolecular forces. If an impurity is smaller than the majority of particles in a crystal lattice, there may be a "dead space" around the impurity, resulting in a region with unrealized intermolecular forces. If the impurity is larger than the other particles, it may instead disrupt intermolecular forces by forcing other particles in the lattice out of alignment. A developing solid will tend to incorporate particles of the same type in order to create the lowest energy solid, and exclude impurities that disrupt the idealized packing of the solid. This process works best if the impurity is present as a minor component of the crude solid. When higher quantities of impurities are present, the resulting crystal tends to be heterogeneous, with pure regions of compound and pure regions of "impurity" intermixed with impure regions. This sort of heterogeneity is often seen in the crystallization of minerals (Figure 3.16b). |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/05%3A_Chemical_Kinetics_Reaction_Mechanisms_and_Chemical_Equilibrium | Chemical kinetics is the study of how fast chemical reactions occur and of the factors that affect these rates. The study of reaction rates is closely related to the study of reaction mechanisms , where a reaction mechanism is a theory that explains how a reaction occurs. 5.1: Chemical Kinetics We can distinguish two levels of detail in a chemical reaction mechanism: The first is the series of elementary processes that occurs for a given net reaction. This is called the stoichiometric mechanism. Frequently it is also possible to infer the relative positions of all of the atoms during the course of a reaction. This sort of model is called an intimate mechanism or detailed mechanism. 5.2: Reaction Rates and Rate Laws hen we talk about the rate of a particular reaction, we intend to specify the amount of chemical change that occurs in unit time because of that reaction. It is usually advantageous to specify the amount of chemical change in units of moles. We can specify the amount of chemical change by specifying the number of moles of a reactant that are consumed, or the number of moles of a product that are produced, per second, by that reaction. 5.3: Simultaneous Processes The number of moles of a substance in a system can change with time because several processes occur simultaneously. Not only can a given substance participate in more than one reaction, but also the amount of it that is present can be affected by processes that are not chemical reactions. A variety of transport process can operate to increase or decrease the amount of the substance that is present in the reaction mixture. 5.4: The Effect of Temperature on Reaction Rates In practice, rate constants vary in response to changes in several factors. Indeed, they are usually the same in two experiments only if we keep everything but the reagent concentrations the same. Another way of saying this is that the rate law captures the dependence of reaction rate on concentrations, while the dependence of reaction rate on any other variable appears as a dependence of rate constants on that variable. 5.5: Other Factors that Affect Reaction Rates A reaction that occurs in one solvent usually occurs also in a number of similar solvents. For example, a reaction that occurs in water will often occur with a low molecular weight alcohol—or an alcohol-water mixture —as the solvent. Typically, the same rate law is observed in a series of solvents, but the rate constants are solvent-dependent. 5.6: Mechanisms and Elementary Processes To see what we mean by an elementary process, let us consider some possible mechanisms for the base hydrolysis of methyl iodide. In this reaction, a carbon–iodide bond is broken and a carbon–oxygen bond is formed. While any number of reaction sequences sum to this overall equation, we can write down three that are reasonably simple and plausible. 5.7: Rate Laws for Elementary Processes If we think about an elementary bimolecular reaction rate law between molecules A and B , we recognize that the reaction can occur only when the molecules come into contact. They must collide before they can react. So the probability that they react must be proportional to the probability that they collide, and the number of molecules of product formed per unit time must be proportional to the number of A−B collisions that occur in unit time. 5.8: Experimental Determination of Rate Laws The determination of a rate law is a matter of finding an empirical equation that adequately describes reaction-rate data. We can distinguish two general approaches: One approach is to measure reaction rate directly. That is, measure the reaction rate in experiments where the concentrations of reactants and products are known. The other is to measure a concentration at frequent time intervals as reaction goes nearly to completion and seek a differential equation consistent with these data. 5.9: First-order Rate Processes First-order rate processes are ubiquitous in nature - and commerce. In chemistry we are usually interested in first-order decay processes; in other subjects, first-order growth is common. We can develop our appreciation for the dynamics - and mathematics - of first-order processes by considering the closely related subject of compound interest. 5.10: Rate Laws by the Study of Initial Rates In concept, the most straightforward way to measure reaction rate directly is to measure the change in the concentration of one reagent in a short time interval immediately following initiation of the reaction. The initial concentrations are known from the way the reaction mixture is prepared. If necessary, the initial mixture can be prepared so that known concentrations of products or reaction intermediates are present. 5.11: Rate Laws from Experiments in a Continuous Stirred Tank Reactor A continuous stirred tank reactor (CSTR)—or capacity-flow reactor—is a superior method of collecting kinetic data when the rate law is complex. Unfortunately, a CSTR tends to be expensive to construct and complex to operate. 5.12: Predicting Rate Laws from Proposed Mechanisms Because a proposed mechanism can only be valid if it is consistent with the rate law found experimentally, the rate law plays a central role in the investigation of chemical reaction mechanisms. The discussion above introduces the problems and methods associated with collecting rate data and with finding an empirical rate law that fits experimental concentration-versus-time data. We turn now to finding the rate laws that are consistent with a particular proposed mechanism. 5.13: The Michaelis-Menten Mechanism for Enzyme-catalyzed Reactions The rates of enzyme-catalyzed reactions can exhibit complex dependence on the relative concentrations of enzyme and substrate. Most of these features are explained by the Michaelis-Menten mechanism, which postulates a rapid equilibration of enzyme and substrate with their enzyme-substrate complex. Transformation of the substrate occurs within this complex. The reaction products do not complex strongly with the enzyme. After the substrate has been transformed, the products diffuse away. 5.14: The Lindemann-Hinshelwood Mechanism for First-order Decay First-order kinetics for a unimolecular reaction corresponds to a constant probability that a given molecule reacts in unit time. We outline a simple mechanism that rationalizes this fact that assumes the probability of reaction is zero unless the molecule has some minimum energy. For molecules whose energy exceeds the threshold value, we assume that the probability of reaction is constant. However, when collisions are frequent, a molecule can have excess energy only for brief intervals. 5.15: Why Unimolecular Reactions are First Order The total energy of a molecule is distributed among numerous degrees of freedom. The molecule has translational kinetic energy, rotational kinetic energy, and vibrational energy. When it acquires excess energy through a collision with another molecule, the additional energy could go directly into any of these modes. However, before the molecule can react, enough energy must find its way into some rather particular mode. 5.16: The Mechanism of the Base Hydrolysis of Co(NH₃)₅Xⁿ⁺ The rate law is rarely sufficient to establish the mechanism of a particular reaction. The base hydrolysis of cobalt pentammine complexes is a reaction for which numerous lines of evidence converge to establish the mechanism. To illustrate the range of data that can be useful in the determination of reaction mechanisms, we summarize this evidence here. 5.17: Chemical Equilibrium as the Equality of Rates for Opposing Reactions The equilibrium constnat that describe the relative concentrations of the species in equilibrium can be extracted from kinetic rate laws. 5.18: The Principle of Microscopic Reversibility The equilibrium constant expression is an important and fundamental relationship that relates the concentrations of reactants and products at equilibrium. We deduce it above from a simple model for the concentration dependence of elementary-reaction rates. In doing so, we use the criterion that the time rate of change of any concentration must be zero at equilibrium. Clearly, this is a necessary condition; if any concentration is changing with time, the reaction is not at equilibrium. 5.19: Microscopic Reversibility and the Second Law The principle of microscopic reversibility requires that parallel mechanisms give rise to the same expression for the concentration-dependence of the equilibrium constant. That is, the function that characterizes the equilibrium composition must be the same for each mechanism. 5.20: Problems |
Courses/Matanuska-Susitna_College/MatSu_College-CHEM_A104_Introduction_to_Organic_and_Biochemistry/00%3A_Front_Matter/02%3A_InfoPage | This text is disseminated via the Open Education Resource (OER) LibreTexts Project ( https://LibreTexts.org ) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact [email protected] . More information on our activities can be found via Facebook ( https://facebook.com/Libretexts ), Twitter ( https://twitter.com/libretexts ), or our blog ( http://Blog.Libretexts.org ). This text was compiled on 04/21/2025 |
Courses/University_of_Illinois_UrbanaChampaign/Chem_2363A_Fundamental_Organic_Chemistry_I_(Chan)/07%3A_Cyclic_Compounds_-_Stereochemistry_of_Reactions/7.02%3A_Cycloalkanes_and_Ring_Strain | Learning Objective explain the partial rotation of carbon-carbon single bonds in rings explain ring strain and its relationship to cycloalkane stability Cycloalkanes (aka Rings) Cycloalkanes have one or more rings of carbon atoms. The simplest examples of this class consist of a single, unsubstituted carbon ring, and these form a homologous series similar to the unbranched alkanes. The IUPAC names of the first five members of this series are given in the following table. The last column gives the general formula for a cycloalkane of any size. If a simple unbranched alkane is converted to a cycloalkane two hydrogen atoms, one from each end of the chain, must be lost. Hence the general formula for a cycloalkane composed of n carbons is C n H 2n . Although a cycloalkane has two fewer hydrogens than the equivalent alkane, each carbon is bonded to four other atoms so such compounds are still considered to be saturated with hydrogen. Name Cyclopropane Cyclobutane Cyclopentane Cyclohexane Cycloheptane Cycloalkane Molecular Formula C3H6 C4H8 C5H10 C6H12 C7H14 CnH2n Structural Formula NaN NaN NaN NaN NaN (CH2)n Line Formula NaN NaN NaN NaN NaN NaN Although the customary line drawings of simple cycloalkanes are geometrical polygons, the actual shape of these compounds in most cases is very different. Cyclic systems are a little different from open-chain systems. In an open chain, any bond can be rotated 360 degrees, going through many different conformations. Complete rotation isn't possible in a cyclic system, because the parts that you would be trying to twist away from each other would still be connected together. Cyclic systems have fewer "degrees of freedom" than aliphatic systems; they have "restricted rotation". Because of the restricted rotation of cyclic systems, most of them have much more well-defined shapes than their aliphatic counterparts. Let's take a look at the basic shapes of some common rings. Many biologically important compounds are built around structures containing rings, so it's important that we become familiar with them. In nature, three- to six-membered rings are frequently encountered, so we'll focus on those. The Baeyer Theory and the Experimental Evidence of Ring Strain Many of the properties of cyclopropane and its derivatives are similar to the properties of alkenes. In 1890, the famous German organic chemist, A. Baeyer, suggested that cyclopropane and cyclobutane derivatives are different from cyclopentane and cyclohexane, because their C—C—C angles cannot have the tetrahedral value of 109.5°. At the same time, Baeyer hypothesized that the difficulties encountered in synthesizing cycloalkane rings from C7 upward was the result of the angle strain that would be expected if the large rings were regular planar polygons (see Table 12-3). Baeyer also believed that cyclohexane had a planar structure like that shown in Figure 12-2, which would mean that the bond angles would have to deviate 10.5° from the tetrahedral value. However, in 1895, the then unknown chemist H. Sachse suggested that cyclohexane exists in the strain-free chair and boat forms discussed in Section 12-3. This suggestion was not accepted at the time because it led to the prediction of several possible isomers for compounds such as chlorocyclohexane (cf. Exercise 12-4). The idea that such isomers might act as a single substance, as the result of rapid equilibration, seemed like a needless complication, and it was not until 1918 that E. Mohr proposed a definitive way to distinguish between the Baeyer and Sachse cyclohexanes. As will be discussed in Section 12-9, the result, now known as the Sachse-Mohr theory, was complete confirmation of the idea of nonplanar large rings. Compound n Angle Strain at each CH2 Heat of Combustion ΔHo (kcal/mol) Heat of Combustion ΔHo per CH2/N (kcal/mol) Total Strain (kcal/mol) ethene 2.0 109.5 337.2 168.6 22.4 cyclopropane 3.0 49.5 499.9 166.6 27.7 cyclobutane 4.0 19.5 655.9 164.0 26.3 cyclopentane 5.0 1.5 793.4 158.7 6.5 cyclohexane 6.0 10.5 944.8 157.5 0.4 cycloheptane 7.0 19.1 1108.1 158.4 6.3 cyclooctane 8.0 25.5 1268.9 158.6 9.7 cyclononane 9.0 30.5 1429.5 158.8 12.9 cyclodecane 10.0 34.5 1586.1 158.6 12.1 cyclopentadecane 15.0 46.5 2362.5 157.5 1.5 open chain alkane NaN NaN NaN 157.4 - Ring Strain in Cycloalkanes Ring Strain occurs because the carbons in cycloalkanes are sp 3 hybridized, which means that they do not have the expected ideal bond angle of 109.5 o ; this causes an increase in the potential energy because of the desire for the carbons to be at an ideal 109.5 o . An example of ring strain can be seen in the diagram of cyclopropane below in which the bond angle is 60 o between the carbons. The reason for ring strain can be seen through the tetrahedral carbon model. The C-C-C bond angles in cyclopropane (diagram above) (60 o ) and cyclobutane (90 o ) are much different than the ideal bond angle of 109.5 o . This bond angle causes cyclopropane and cyclobutane to have a high ring strain. However, molecules, such as cyclohexane and cyclopentane, would have a much lower ring strain because the bond angle between the carbons is much closer to 109.5 o . Below are some examples of cycloalkanes. Ring strain can be seen more prevalently in the cyclopropane and cyclobutane models Below is a chart of cycloalkanes and their respective heats of combustion ( Δ H comb ). The Δ H comb value increases as the number of carbons in the cycloalkane increases (higher membered ring), and the Δ H comb /CH 2 ratio decreases. The increase in Δ H comb can be attributed to the greater amount of London Dispersion forces. However, the decrease in Δ H comb /CH 2 can be attributed to a decrease in the ring strain. Certain cycloalkanes, such as cyclohexane, deal with ring strain by forming conformers . A conformer is a stereoisomer in which molecules of the same connectivity and formula exist as different isomers, in this case, to reduce ring strain. The ring strain is reduced in conformers due to the rotations around the sigma bonds. Other Types of Strain There are many different types of strain that occur with cycloalkanes. In addition to ring strain, there is also transannular strain, eclipsing, or torsional strain and bond angle strain.Transannular strain exists when there is steric repulsion between atoms. Eclipsing (torsional) strain exists when a cycloalkane is unable to adopt a staggered conformation around a C-C bond, and bond angle strain is the energy needed to distort the tetrahedral carbons enough to close the ring. The presence of angle strain in a molecule indicates that there are bond angles in that particular molecule that deviate from the ideal bond angles required (i.e., that molecule has conformers). Cyclopropane A three membered ring has no rotational freedom whatsoever, so the three carbon atoms in cyclopropane are all constrained to lie in the same plane at the corners of an equilateral triangle. The 60º bond angles are much smaller than the optimum 109.5º angles of a normal tetrahedral carbon atom, and the resulting angle strain dramatically influences the chemical behavior of this cycloalkane. Cyclopropane also suffers substantial eclipsing strain, since all the carbon-carbon bonds are fully eclipsed. Furthermore, if you look at a model you will find that the neighboring C-H bonds (C-C bonds, too) are all held in eclipsed conformations. Cyclopropane is always at maximum torsional strain. This strain can be illustrated in a line drawing of cyclopropane as shown from the side. In this oblique view, the dark lines mean that those sides of the ring are closer to you. However, the ring isn't big enough to introduce any steric strain, which does not become a factor until we reach six membered rings. Until that point, rings are not flexible enough for two atoms to reach around and bump into each other. The really big problem with cyclopropane is that the C-C-C bond angles are all too small. All the carbon atoms in cyclopropane appear to be tetrahedral. These bond angles ought to be 109 degrees. The angles in an equilateral triangle are actually 60 degrees, about half as large as the optimum angle. This factor introduces a huge amount of strain in the molecule, called ring strain. Cyclobutane Cyclobutane is a four membered ring. In two dimensions, it is a square, with 90 degree angles at each corner. Cyclobutane reduces some bond-eclipsing strain by folding (the out-of-plane dihedral angle is about 25º), but the total eclipsing and angle strain remains high. Cyclopentane has very little angle strain (the angles of a pentagon are 108º), but its eclipsing strain would be large (about 10 kcal/mol) if it remained planar. Consequently, the five-membered ring adopts non-planar puckered conformations whenever possible. However, in three dimensions, cyclobutane is flexible enough to buckle into a "butterfly" shape, relieving torsional strain a little bit. When it does that, the bond angles get a little worse, going from 90 degrees to 88 degrees. In a line drawing, this butterfly shape is usually shown from the side, with the near edges drawn using darker lines. With bond angles of 88 rather than 109 degrees, cyclobutane has a lot of ring strain, but less than in cyclopropane. Torsional strain is still present, but the neighbouring bonds are not exactly eclipsed in the butterfly. Cyclobutane is still not large enough that the molecule can reach around to cause crowding. Steric strain is very low. Cyclobutanes are a little more stable than cyclopropanes and are also a little more common in nature. Cyclopentane Cyclopentanes are even more stable than cyclobutanes, and they are the second-most common paraffinic ring in nature, after cyclohexanes. In two dimensions, a cyclopentane appears to be a regular pentagon. In three dimensions, there is enough freedom of rotation to allow a slight twist out of this planar shape. In a line drawing, this three-dimensional shape is drawn from an oblique view, just like cyclobutane. The ideal angle in a regular pentagon is about 107 degrees, very close to a tetrahedral bond angle. Cyclopentane distorts only very slightly into an "envelope" shape in which one corner of the pentagon is lifted up above the plane of the other four, and as a result, ring strain is entirely removed. The envelope removes torsional strain along the sides and flap of the envelope. However, the neighbouring carbons are eclipsed along the "bottom" of the envelope, away from the flap. There is still some torsional strain in cyclopentane. Again, there is no steric strain in this system. Rings larger than cyclopentane would have angle strain if they were planar. However, this strain, together with the eclipsing strain inherent in a planar structure, can be relieved by puckering the ring. Cyclohexane is a good example of a carbocyclic system that virtually eliminates eclipsing and angle strain by adopting non-planar conformations. Cycloheptane and cyclooctane have greater strain than cyclohexane, in large part due to transannular crowding (steric hindrance by groups on opposite sides of the ring). Exercise \(\PageIndex{1}\) 1. If cyclobutane were to be planar how many H-H eclipsing interactions would there be, and assuming 4 kJ/mol per H-H eclipsing interaction what is the strain on this “planar” molecule? 2. In the two conformations of cis -cyclopentane one is more stable than the other. Explain why this is. Answer 1. There are 8 eclipsing interactions (two per C-C bond). The extra strain on this molecule would be 32 kJ/mol (4 kJ/mol x 8). 2. The first conformation is more stable. Even though the methyl groups are cis in the model on the left, they are eclipsing due the conformation, therefore increasing the strain within the molecule. |
Bookshelves/General_Chemistry/CLUE%3A_Chemistry_Life_the_Universe_and_Everything/06%3A_Solutions/6.2%3A_Solubility%3A_why_do_some_things_form_solutions_and_others_not | Let us say you have a \(100-\mathrm{mL}\) graduated cylinder and you take \(50 \mathrm{~mL}\) of ethanol and add it to \(50 \mathrm{~mL}\) of water. You might be surprised to find that the volume of the resulting solution is less than \(100 \mathrm{~mL}\). In fact, it is about \(98 \mathrm{~mL}\), assuming good technique (no spilling). How can we explain this? Well, we can first reassure ourselves that matter has not been destroyed. If we weigh the solution, it weighs the same as \(50 \mathrm{~mL}\) of water plus \(50 \mathrm{~mL}\) of ethanol. This means that the density of the water–ethanol solution must be greater than the density of either the water or ethanol alone. At the molecular level, we can immediately deduce that the molecules are closer together in the ethanol and water mixture than they were when pure (before mixing) –try drawing a molecular level picture of this to convince yourself that this is possible. Now, if you took \(50 \mathrm{~mL}\) of oil and \(50 \mathrm{~mL}\) of water, you would find that they do not mix—no matter how hard you tried. They will always separate away from one another into two layers. What factors determine whether or not substances form solutions? First, we need to be aware that solubility is not an all-or-nothing property. Even in the case of oil and water, a very small number of oil molecules are present in the water (the aqueous phase), and a small number of water molecules are present in the oil. There are a number of ways to describe solubility. The most common way is to define the number of moles of solute per liter of solution. This is called the solution’s molarity (\(\mathrm{M}, \mathrm{~mol/L}\)). Another common way is to define the number grams of solute per mass of solution. For example: \(1 \mathrm{~mg}\) (\(10^{-3} \mathrm{~g}\)) of solute dissolved in \(1 \mathrm{~kg}\) (\(10^{3} \mathrm{~g}\)) of solution is 1 part per million (\(10^{6}\)) solute, or \(1 \mathrm{~ppm}\). As you might expect, given the temperature term in the free energy equation, solubility data are always reported at a particular temperature. If no more solute can dissolve at a given temperature, the solution is said to be saturated; if more solute can dissolve, it is unsaturated. If we look at the structure of compounds that dissolve in water, we can begin to see some trends: hydrocarbons are not very soluble in water (remember from Chapter \(4\) that these are compounds composed only of carbon and hydrogen), whereas alcohols (hydrocarbons with an \(—\mathrm{O–H}\) group attached) with up to 3 carbons are completely soluble. As the number of carbon atoms increases, the solubility of the compound in water decreases. For example, hexanol (\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\)), is only very slightly soluble in water (\(0.4 \mathrm{~g/L}\)). So perhaps the hydroxyl (\(—\mathrm{O–H}\)) group is responsible for the molecule’s solubility in water. Evidence supporting this hypothesis can be found in the fact that diols (compounds with \(2—\mathrm{O–H}\) groups) are more soluble than similar alcohols. For example, compared to hexanol, 1,6-hexanediol (\(\mathrm{HO}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\)) is quite soluble in water. More familiar water-soluble compounds such as the sugars glucose, fructose, and sucrose (a dimer of glucose and fructose – shown in the figure) are, in fact, polyalcohols. Each of their six carbons is attached to a hydroxyl group. Compound Molar Mass (\mathrm{g/mol}\)) Structure Solubility (\(\mathrm{g/L}\)) \(20 { }^{\circ}\mathrm{C}\) Propane \(44\) \(\(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{3}\) \(0.07 \mathrm{~g/L}\) Ethanol \(46\) \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\) Completely miscible Dimethyl ether \(46\) \(\(\mathrm{CH}_{3}\mathrm{OCH}_{3}\) \(328 \mathrm{~g/L}\) Pentane \(72\) \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{3}\) \(0.4 \mathrm{~g/L}\) Butanol \(74\) \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\) \(80 \mathrm{~g/L}\) Diethyl ether \(74\) \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OCH}_{2}\mathrm{CH}_{3}\) \(69 \mathrm{~g/L}\) Hexanol \(102\) \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\) \(0.4 \mathrm{~g/L}\) 1,6 Hexanediol \(226\) \(\mathrm{HO}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{CH}_{2}\mathrm{OH}\) \(500 \mathrm{~g/L}\) Glucose \(180\) \(\mathrm{C}_{6}\mathrm{H}_{12}\mathrm{O}_{6}\) \(910 \mathrm{~g/L}\) Questions Questions to Answer Make a list of substances that you know dissolve in water. Which of these dissolve: metals, ionic compounds, molecular compounds (polar, non-polar), network solids (diamond graphite)? Can you make any generalizations about which things dissolve and which don’t? What must happen in order for something to dissolve in water? How would you design an experiment to determine the solubility of a solute? How would you determine whether or not a solution was saturated? Draw a molecular level picture of a solution of ethanol and water showing why the solution is more dense than the separate liquids. Draw a molecular level picture of an oil and water mixture. Draw a molecular level picture of the process of solution When you try mixing oil and water, which layer ends up on top? Why? Question to Ponder You have a saturated solution, with some solid solute present. Do you think the solute particles that are in solution are the same ones over time? How would you determine whether they were the same? Questions for Later What would you predict for the sign of \(\Delta \mathrm{S}\) upon the formation of any solution? Why? What would you predict for the sign of \(\Delta \mathrm{H}\) upon the formation of any solution? Why? What would you predict for the sign of \(\Delta \mathrm{G}\) upon the formation of any solution? Why? |
Bookshelves/Introductory_Chemistry/Introduction_to_General_Chemistry_(Malik)/04%3A_Stoichiometry_the_quantification_of_chemical_reactions/4.04%3A_Patterns_of_chemical_reactions | General types of chemical reactions There are several ways to classify chemical reactions. The general types of chemical reactions fall in the categories of combination, decomposition, replacement, and combustion reactions, as illustrated in Fig. 4.4.1. Combination reactions A compound is synthesized or formed from two or more substances, e.g.: \[\ce{C + O2 -> CO2}\nonumber\] \[\ce{2H2 + O2 -> 2H2O}\nonumber\] \[\ce{2Mg + O2 -> 2MgO}\nonumber\] \[\ce{2Na + Cl2 -> 2NaCl}\nonumber\] \[\ce{CaO + CO2 -> CaCO3}\nonumber\] Fig. 4.4.2 shows an example of a hydrogen with oxygen combination reaction that is being developed for use as a fuel in the future. Decomposition reactions The decomposition reactions are the reverse of the combination reaction, i.e., one compound splits apart into two or more substances, usually by heating, e.g.: \[\ce{H2CO3 -> H2O + CO2}\nonumber\] \[\ce{CaCO3 ->[\Delta] CaO + CO2}\nonumber\] \[\ce{2KClO3 ->[\Delta] 2KCl + 3O2}\nonumber\] \[\ce{2H2O (l) ->[Electrolyisis] 2H2(g) + O2(g)}\nonumber\] Fig. 4.4.3 illustrates the last reaction, i.e., decomposition of water through electrolysis. Replacement or substitution reactions There are two sub-classes of this category of reactions, i.e., single replacement and double replacement reaction. Single replacement reactions involve one substance replacing a part of another, e.g.: \[\ce{Zn(s) + CuCl2(aq) -> ZnCl2(aq) + Cu(s)}\nonumber\] \[\ce{2Al(s) + 6HCl(aq) -> AlCl3(aq) + 3H2(g)}\nonumber\] Fig. 4.4.4 shows an example of a single replacement reaction of magnesium resulting in hydrogen gas formation. Double replacement reactions or metathesis involve the mutual exchange of partners between two substances, e.g. the following precipitation reactions: \[\ce{NaCl(aq) + AgNO3(aq) -> AgCl(s)(v) + NaNO3(aq)}\nonumber\] \[\ce{Na2CO3(aq) + CaCl2(aq) -> CaCO3(s)(v) + 2NaCl(aq)}\nonumber\] Combustion reactions Combustion is a reaction of a substance with oxygen, often with the formation of flame and release of much heat, e.g.: \[\ce{C8H16 + 12O2 -> 8CO2 + 8H2O + Heat}\nonumber\] \[\ce{C + O2 -> CO2 + Heat}\nonumber\] \[\ce{2H2 + O2 -> 2H2O + Heat}\nonumber\] \[\ce{2Mg + O2 -> 2MgO + Heat}\nonumber\] Fig. 4.4.5 shows the above reaction, i.e., combustion of Mg in air. Usually, combustion is considered as the reaction of a substance containing carbon and hydrogen with oxygen resulting in carbon dioxide, water, flame, and heat, e.g., burning methane on a kitchen stove: \[\ce{CH4 + 3O2 -> CO2 + 2H2O + Heat}\nonumber\] Classification of chemical reaction The chemical reactions are generally classified based on what exchanges during the reaction. These include; the transfer of electrons in oxidation-reduction reactions, transfer of protons in acid-base reactions, and a part of reactants mutually exchanges in precipitation reactions, as described below. Oxidation-reduction reactions The oxidation-reduction or redox reaction involves the exchange of electrons. For example, reactions between a metal and nonmetal involve the transfer of electrons from the metal to the nonmetal forming an ionic bond, as shown in Fig. 4.4.6. Acid-base reactions The acid-base reactions involve the transfer of protons from an acid to a base, as shown in Fig. 4.4.7. Precipitation reactions These are double displacement reactions in water that results in the precipitation of one of the products, as shown in Fig. 4.4.8. The precipitation reactions and the acid-base reactions are described in the later chapters. The oxidation-reduction reactions are discussed in the following section. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/26%3A_Chemical_Equilibrium/26.01%3A_Equilibrium_Results_when_Gibbs_Energy_is_Minimized | Many important chemical reactions -if not most - are performed in solution rather than between solids or gases. Solid state reactions are often very slow and not all chemical species can be put in the vapor phase because they decompose before evaporating. Often we are not concerned with the temporal aspects of a reaction. That can be technologically very important but it is the domain of kinetics - a different branch of Physical Chemistry - rather than classical thermodynamics.. The latter is more concerned with the endpoint. This is the thermodynamically speaking the (stable) equilibrium, but chemically it can either represent a completed reaction or a chemical equilibrium . Unfortunately, of the three main aggregation states: gas – liquid – solid, the structure of liquids is least understood and one of the most complex liquids is also one of the most extensively used ones: water. It is vital to many branches of chemistry varying from geochemistry to environmental chemistry to biochemistry. We shall make just a small inroad into its complexity. Extent of reaction To describe the progress of a reaction we define the extent of reaction . It is usually denoted by the Greek letter \(ξ\). Consider a generic reaction: \[v_AA + v_BB \rightleftharpoons v_YY + v_ZZ \nonumber \] Using stoichiometry we can define the extent by considering how the number of moles (or molar amounts) of each species changes during the reaction: reactants \(n_A= n_{A,0} - v_A ξ\) \(n_B= n_{B,0} - v_B ξ\) products \(n_Y= n_{Y,0} + v_Y ξ\) \(n_Z= n_{Z,0} + v_Z ξ\) The dimension of ξ is [mol] because the stoichiometric coefficients v i are dimensionless integers. If the reaction goes to completion for one of the reactants -the limiting reactant - n A or B =n limiting will go to zero. If we start with \(n_{limiting}= v_{limiting}\) moles, the value of ξ starts at 0 (no products) and goes to 1 at completion (limiting reactant depleted). When approaching an equilibrium \(ξ\) will not go beyond \(ξ_{eq}\). Measuring ξ The extent of reaction is what is the central subject of reaction kinetics. Its value is typically measured as a function of time indirectly by measuring a quantity q that is linearly dependent on ξ(t): \[q( ξ ) = aξ +b \nonumber \] Consider the situation at the extremes \(ξ=0\) and \(ξ=1\): \[q_0= a.0+b= b \nonumber \] \[q_1= a.1+b= a+b \nonumber \] \[q_1-q_0= a \nonumber \] Thus, \(ξ\) can be found from \[\dfrac{q(t)-q0}{q_1-q_0}=\dfrac{q(t)-b}{a} \nonumber \] The nature of \(q\) can vary widely from UV/Vis absorption, conductivity, gravimetric to caloric data. In practice, \(q_0\) at \(ξ=0\) is often hard to observe because it takes time to mix the reactants, particularly in solutions, and q 1 at ξ=1 may never be reached if the reaction goes to equilibrium. Nevertheless the values of a and b can often be found from the available data by fitting techniques. In (equilibrium, static) thermodynamics we are only concerned with the endpoint: \(ξ=1\): the reaction runs to completion \(ξ=ξ_{eq}\): the react ion goes to a state of chemical equilibrium Thermodynamic Potentials As we have seen we can write any change in the Gibbs free energy due to changes in the molar amounts of the species involved in the reaction (at \(T\), \(P\) constant) as: \[dG =\sum \dfrac{∂G}{∂n_i} dn_i = \sum μ_idn_i \nonumber \] where \(μ\) is the thermodynamic potential , often called chemical potential when dealing with reactions. From the definition of ξ we can see by differentiation that \(d n_A=- v_Adξ\) \(d n_B=- v_Adξ\) \(d n_Y= v_Ydξ\) \(d n_Z= v_Zdξ\) This allows us to unify the changes in the molar amount of all the species into one single variable \(dξ\). We get: \[dG = \left[ \sum -v_{i,reactants} μ_{i,reactants} + \sum+v_{i,products} μ_{j,products} \right]dξ \nonumber \] or \[ \left (\dfrac{∂G}{∂ξ} \right)_{T,P} = -\sum v_{i,r}μ_{i,r}+ \sum v_{i,p}μ_{j,p} \nonumber \] This quantity is also written as: \[ \left( \dfrac{∂G}{∂ξ} \right)_{T,P} =Δ_rG \nonumber \] This quantity gives the change in Gibbs free energy for the reaction (as written!!) for Δξ=1 mole. (Units are [J/mol] therefore). This is the change in Gibbs energy (the slope of \(G\) vs \(\xi\)) when the extent of reaction changes by one mole with a fixed composition. Equilibrium results when Gibbs energy is at a minimum with respect to the extent of reaction. Gas Reactions Let us assume that our reaction is entirely between gas species and that the gas is sufficiently dilute that we can use the ideal gas law. Then we can write for each species: \[μ_i= μ_i^o+RT \ln \dfrac{P_i}{P_i^o} \nonumber \] We can then split up the Δ r G expression in two parts: \[Δ_rG = Δ_rG^o + RT\ln Q \nonumber \] The standard potentials: \[Δ_rG^o = -\sum v_i,_r μ^o_{i,r} + \sum v_{i,p} μ^o_{j,p} \nonumber \] and the logarithmic terms: \[RT \ln Q= - v_A RT \ln \left( \dfrac{P_A}{P_A^o} \right)- v_B RT\ln \left( \dfrac{P_B}{P_B^o} \right) + v_YRT \ln \left( \dfrac{P_Y}{P_Y^o} \right) + v_ZRT \ln \left( \dfrac{P_Z}{P_Z^o} \right) \nonumber \] We can combine all the logarithmic terms into Q, called the reaction quotient . The stoichiometric coefficients become exponents and the reactants' factors will be 'upside down' compared to the products, because of the properties of logarithms: \[a \ln x = \ln x^a \nonumber \] \[- a \ln x = \ln \left( \dfrac{1}{x^a} \right) \nonumber \] We have kept the standard pressures \(P_i^o\) in the expression, but often they are omitted. They are usually all 1 bar, but in principle we could choose 1 bar for A 1 Torr for B an 1 psi for the products. It creates a valid (though ridiculous) definition of what o stands for. (Of course the value of \(Δ_rG^o\) does depend on that choice!). We could write \[RT \ln Q = RT \ln \dfrac{Q_P}{Q^o} \nonumber \] \(Q^o\) is typically unity in magnitude but it cancels the dimensions of \(Q_P\). That means that \(Q\) and \(Q_P\) are equal in magnitude and we can get \(Q\) from \(Q_P\) by simply dropping the dimensions. \(Q\) is dimensionless but \(Q_P\) usually is not. Often this fine distinction is simply not made and \(Q^o\) is omitted , we get: \[Δ_rG = Δ_rG^o + RT\ln \dfrac{P_Y^{v_Y}P_Z^{v_Z} }{P_A^{v_A}P_B^{v_B}} \nonumber \] Notice the difference between \(Δ_rG\) which denotes the conditions (e.g. pressures) of your reaction and \(Δ_rG^o\) denotes standard conditions. |
Courses/University_of_North_Texas/UNT%3A_CHEM_1410_-_General_Chemistry_for_Science_Majors_I/Text/03%3A_Using_Chemical_Equations_in_Calculations/3.09%3A_Hess'_Law/Carbide_Cannon | Enthalpy of reactions: Assuming you want about a 1 kJ explosion, how much CaC 2 would you add to a carbide cannon? (1) 2 C 2 H 2 + 5 O 2 → 4 CO 2 + 5 H 2 O ΔH ? given (2) C + O 2 → CO 2 ΔH f = -393.5 kJ/mol (3) H 2 + 1/2 O 2 → H 2 O (g) Δ H f = -241.82 kJ/mol (4) 2 C + H 2 → C 2 H 2 (g) ΔH f = +226.7 Calculated enthalpy change for combustion of 2 mol of acetylene, ΔH = -2511 kJ This is 48.2 kJ/g for 1 kJ, we need 1 kJ/48.2 kJ/g = 0.02 g of acetylene CaC 2 + 2 H 2 O → HCCH + Ca(OH) 2 this requires 0.0008 mol of CaC 2 , which is 0.05 g. Do it with a Bangsite Cannon! |
Courses/Howard_University/Howard%3A_Physical_Chemistry/14%3A_Nuclear_Magnetic_Resonance | Nuclear magnetic resonance (NMR ) is a versatile and highly-sophisticated spectroscopic technique which has been applied to a growing number of diverse applications in science, technology and medicine. This chapter will consider, for the most part, magnetic resonance involving protons. Magnetic Properties of Nuclei In all our previous work, it has been sufficient to treat nuclei as structureless point particles characterized fully by their mass and electric charge. On a more fundamental level, as was discussed in Chap. 1, nuclei are actually composite particles made of nucleons (protons and neutrons) and the nucleons themselves are made of quarks. The additional properties of nuclei which will now become relevant are their spin angular momenta and magnetic moments. Recall that electrons possess an intrinsic or spin angular momentum s which can have just two possible projections along an arbitrary spacial direction, namely \( \pm \frac{1}{2} \hbar \). Since \( \hbar \) is the fundamental quantum unit of angular momentum, the electron is classified as a particle of spin one-half . The electron’s spin state is described by the quantum numbers \( s=1\) and \( m_s = \pm 1 \) . A circulating electric charge produces a magnetic moment \( \vec{\mu} \) proportional to the angular momentum J . Thus \[ \vec{\mu} = \gamma \vec{J} \label{1}\] where the constant of proportionality γ is known as the magnetogyric ratio . The z -component of \( \vec{\mu} \) has the possible values \[\vec{\mu}_z = \gamma\hbar m_J \mbox{where} m_J = -J, -J+1,..., +J \label{2}\] determined by space quantization of the angular momentum J . The energy of a magnetic dipole in a magnetic field B is given by \[E= -\vec{\mu} \cdot \vec{B} = -\vec{\mu}_z B \label{3}\] where magnetic field defines the z -axis. The SI unit of magnetic field (more correctly, magnetic induction ) is the tesla , designated T. Electromagnets used in NMR produce fields in excess of 10 T. Small iron magnets have fields around .01 T, while some magnets containing rare-earth elements such as NIB (niobium-iron-boron) reach 0.2 T. The Earth’s magnetic field is approximately 5 × 10 −5 T (0.5 gauss in alternative units), dependent on geographic location. At the other extreme, a neutron star, which is really a giant nucleus, has a field predicted to be of the order of 10 8 T. The energy relation (3) determines the most conveniently units for magnetic moment, namely joules per tesla, J T − 1 . For orbital motion of an electron, where the angular momentum is l , the magnetic moment is given by \[ \vec{\mu}_z = -\frac{e{\hbar}}{2m} m_l = -\vec{\mu}_B m_l \label{4}\] where the minus sign reflects the negative electric charge. The Bohr magneton is defined by \[\vec{\mu}_B = -\frac{e{\hbar}}{2m} = 9.274\times10^{-24} JT^{-1} \label{5}\] The magnetic moment produced by electron spin is written \[\vec{\mu}_z = -g \vec{\mu}_B m_s \label{6}\] with introduction of the g-factor . Eq (4) implies g = 1 for orbital motion. For electron spin, however, g = 2 (more exactly, 2.0023). The factor 2 compensates for m s = 1 such that spin and l = 1 orbital magnetic moments 2 are both equal to one Bohr magneton . Many nuclei possess spin angular momentum, analogous to that of the electron. The nuclear spin, designated I , has an integral or half-integral value: 0 , 1 , 1 , 3 , and so on. Table 1 lists some nuclei of importance in chemical applications of NMR. The proton and the neutron both are spin 2 particles, like the electron. Complex nuclei have angular momenta which are resultants of the spins of their component nucleons. The deuteron 2 H , with I = 1, evidently has parallel proton and neutron spins. The 4 He nucleus has I = 0, as do 12 C , 16 O , 20 Ne , 28 Si and 32 S . These nuclei contain filled shells of protons and neutrons with the vector sum of the component angular momenta equal to zero, analogous to closed shells of electrons in atoms and molecules. In fact, all even-even nuclei have spins of zero. Nuclear magnetic moments are of the order of a nuclear magneton \[ \vec{\mu}_N = \frac{e\hbar}{2M} = 5.051\times 10^{-27} JT^{-1} \label{7}\] where M is the mass of the proton. The nuclear magneton is smaller than the Bohr magneton by a factor m/M ≈ 1836. Table 1 : Some common nuclei in NMR spectroscopy In analogy with Equations \(\ref{2}\) and \(\ref{6}\), nuclear moments are represented by \[ \vec{\mu}_z = g_I \vec{\mu}_N m_I = \hbar \gamma_I m_I \label{8}\] where g I is the nuclear g -factor and γ I , the magnetogyric ratio. Most nuclei have positive g -factors, as would be expected for a rotating positive electric charge. It was long puzzling that the neutron, although lacking electric charge, has a magnetic moment. It is now understood that the neutron is a composite of three charged quarks, udd . The negatively-charged d - quarks are predominantly in the outermost regions of the neutron, thereby producing a negative magnetic moment, like that of the electron. The g - factor for 17 O , and other nuclei dominated by unpaired neutron spins, is consequently also negative. Nuclear Magnetic Resonance The energy of a nuclear moment in a magnetic field, according to Equation \(\ref{3}\), is given by \[ E_{m_I} = -\hbar \gamma_I m_I B \label{9}\] For a nucleus of spin \(I\), the energy of a nucleus in a magnetic field is split into \(2I+1\) Zeeman le vels. A proton and other nuclei with spin \( \frac{1}{2}\) particles will have just two possible levels: \[E_{\pm \frac{1}{2}} = \pm \dfrac{1}{2} \hbar \gamma B \label{10}\] with the α -spin state ( \( m_I = -\frac{1}{2} \)) lower in energy than the β -spin state (\( m_I = +\frac{1}{2} \)) by \[\Delta E = \hbar \gamma B \label{11}\] Fig. 1 shows the energy of a proton as a function of magnetic field. In zero field ( B = 0), the two spin states are degenerate. In a field B , the energy splitting corresponds to a photon of energy \(\Delta E = \hbar \omega = h \nu \) where \[ \omega_L = \gamma_B\space \mbox{or}\space \nu_L = \gamma_B \label{12}\] known as the Larmor frequency of the nucleus. For the proton in a field of 1 T, \(\nu_L \) = 42 . 576 MHz, as the proton spin orientation flips from \( +\frac{1}{2} \) to \( -\frac{1}{2} \) . This transition is in the radiofrequency region of the electromagnetic spectrum. NMR spectroscopy consequently exploits the technology of radiowave engineering. Figure 1 . Energies of spin \( \frac{1}{2} \) in magnetic field showing NMR transition at Larmor frequency \( \nu_L \) . A transition cannot occur unless the values of the radiofrequency and the magnetic field accurately fulfill Eq (12). This is why the technique is categorized as a resonance phenomenon. If some resonance condition is not satisfied, no radiation can be absorbed or emitted by the nuclear spins. In the earlier techniques of NMR spectroscopy, it was found more convenient keep the radiofrequency fixed and sweep over values of the magnetic field B to detect resonances. These have been largely supplanted by modern pulse techniques, to be described later. The transition probability for the upward transition (absorption) is equal to that for the downward transition (stimulated emission). (The contribution of spontaneous emission is neglible at radiofrequencies.) Thus if there were equal populations of nuclei in the α and β spin states, there would be zero net absorption by a macroscopic sample. The possibility of observable NMR absorption depends on the lower state having at least a slight excess in population. At thermal equilibrium, the ratio of populations follows a Boltzmann distribution \[ \frac{N_{\beta}}{N_\alpha} = \frac{e^{\frac{-E_\beta}{kT}}}{e^{\frac{-E_\alpha}{kt}}} = e^{- \frac{\hbar \gamma B}{kT}} \label{13}\] Thus the relative population difference is given by \[ \dfrac{\Delta N}{N_\alpha} = \frac{N_\alpha - N_{\beta}}{N_\alpha + N_{\beta}} \approx \frac{ \hbar \gamma B}{2kT} \label{14}\] Since nuclear Zeeman energies are so small, the populations of the α and β spin states differ very slightly. For protons in a 1 T field, ∆ N /N ≈ 3 × 10 − 6 . Although the population excess in the lower level is only of the order of parts per million, NMR spectroscopy is capable of detecting these weak signals. Higher magnetic fields and lower temperatures are favorable conditions for enhanced NMR sensitivity. The Chemical Shift NMR has become such an invaluable technique for studying the structure of atoms and molecules because nuclei represent ideal noninvasive probes of their electronic environment. If all nuclei of a given species responded at their characteristic Larmor frequencies, NMR might then be useful for chemical analysis, but little else. The real value of NMR to chemistry comes from minute differences in resonance frequencies dependent on details of the electronic structure around a nucleus. The magnetic field induces orbital angular momentum in the electron cloud around a nucleus, thus, in effect, partially shielding the nucleus from the external field B . The actual or local value of the magnetic field at the position of a nucleus is expressed \[ B_{loc} = (1- \sigma)B \label{15}\] where the fractional reduction of the field is denoted by σ , the shielding constant , typically of the order of parts per million. The actual resonance frequency of the nucleus in its local environment is then equal to \[ \nu = (1- \sigma) \frac{\gamma B}{2 \pi} \label{16}\] A classic example of this effect is the proton NMR spectrum of ethanol CH 3 CH 2 OH , shown in Fig. 2. The three peaks, with intensity ratios 3:2:1 can be identified with the three chemically-distinct environments in which the protons find themselves: three methyl protons ( CH 3 ), two methylene protons ( CH 2 ) and one hydroxyl proton (OH). Figure 2 . Oscilloscope trace showing the first NMR spectrum of ethanol, taken at Stanford University in 1951. Courtesy Varian Associates, Inc. The variation in resonance frequency due to the electronic environment of a nucleus is called the chemical shift . Chemical shifts on the delta scale are defined by \[ \delta = {\frac{ \nu - \nu^{O}}{\nu^{O}}} \times 10^{6} \label{17}\] where ν o represents the resonance frequency of a reference compound, usually tetramethylsilane Si( CH 3 ) 4 , which is rich in highly-shielded chemically-equivalent protons, as well as being unreactive and soluble in many liquids. By definition δ = 0 for TMS and almost everything else is “downfield” with positive values of δ . Most compounds have delta values in the range of 0 to 12 (hydrogen halides have negative values, e.g. δ ≈ − 13 for HI). The hydrogen atom has δ ≈ 13 while the bare proton would have δ ≈ 31. Conventionally, the δ -scale is plotted as increasing from right to left, in the opposite sense to the magnitude of the magnetic field. Nuclei with larger values of δ are said to be more deshielded , with the bare proton being the ultimate limit. Fig. 3 shows some typical values of δ for protons in some common organic compounds. Figure 3 . Ranges of proton chemical shifts for common functional groups. From P. Atkins, Physical Chemistry , (Freeman, New York, 2002). Fig. 4 shows a high-resolution NMR spectrum of ethanol, including a δ-scale. The “fine structure” splittings of the three chemically-shifted components will be explained in the next Section. The chemical shift of a nucleus is very difficult to treat theoretically. However, certain empirical regularities, for example those represented in Fig. 3, provide clues about the chemical environment of the nucleus. We will not consider these in any detail except to remark that often increased deshielding of a nucleus (larger δ ) can often be attributed to a more electronegative neighboring atom. For example the proton in the ethanol spectrum (Fig. 4) with δ ≈ 5 can be identified as the hydroxyl proton, since the oxygen atom can draw significant electron density from around the proton. Figure 4 . High-resolution NMR spectrum of ethanol showing δ scale of chemical shifts. The line at δ = 0 corresponds to the TMS trace added as a reference. Neighboring groups can also contribute to the chemical shift of a given atom, particularly those with mobile π -electrons. For example, the ring current in a benzene ring acts as a secondary source of magnetic field. Depending on the location of a nucleus, this can contribute either shielding or deshielding of the external magnetic field, as shown in Fig. 5. The interaction of neighboring groups can be exploited to obtain structural information by using lanthanide shift reagents . Lanthanides (elements 58 through 71) contain 4f-electrons , which are not usually involved in chemical bonding and can give large paramagnetic contributions. Lanthanide complexes which bind to organic molecules can thereby spread out pro- ton resonances to simplify their analysis. A popular chelating complex is Eu(dpm) 3 , tris(dipivaloylmethanato)europium, where dpm is the group ( CH 3 ) 3 C–CO =CH–CO–C( CH 3 ) 3 . Figure 5 . Magnetic field produced by ring current in benzene, shown as red loops. Where the arrows are parallel to the external field B , including protons directly attached to the ring, the effect is deshielding . However, any nuclei located within the return loops will experience a shielding effect. Spin-Spin Coupling Two of the resonances in the ethanol spectrum shown in Fig. 4 are split into closely-spaced multiplets —one triplet and one quartet. These are the result of spin-spin coupling between magnetic nuclei which are relatively close to one another, say within two or three bond separations. Identical nuclei in identical chemical environments are said to be equivalent . They have equal chemical shifts and do not exhibit spin-spin splitting. Nonequiv alent magnetic nuclei, on the other hand, can interact and thereby affect one another’s NMR frequencies. A simple example is the HD molecule, in which the spin- 1proton can interact with the spin-1 deuteron, even though the atoms are chemically equivalent. The proton’s energy is split into two levels by the external magnetic field, as shown in Fig. 1. The neighboring deuteron, itself a magnet, will also contribute to the local field at the pro- ton. The deuteron’s three possible orientations in the external field, with M I = −1 , 0 , +1, with different contributions to the magnetic field at the proton, as shown in Fig. 6. The proton’s resonance is split into three evenly spaced, equally intense lines (a triplet), with a separation of 42.9 Hz. Corespondingly the deuteron’s resonance is split into a 42.9 Hz doublet by its interaction with the proton. These splittings are independent of the external field B , whereas chemical shifts are proportional to B . Fig. 6 represents the energy levels and NMR transitions for the proton in HD. Figure 6 . Nuclear energy levels for proton in HD molecule. The two Zeeman levels of the proton when B > 0 are further split by interaction with the three possible spin orientations of the deuteron M d = − 1 , 0 , +1. The proton NMR transition, represented by blue arrows, is split into a triplet with separation 42.9 Hz. Nuclear-spin phenomena in the HD molecule can be compactly represented by a spin Hamiltonian \[ \hat{H} = -\hbar \gamma_H M_H(1- \sigma_H) - \hbar \gamma_D M_D(1-\sigma_D)B + h J_{HD} I_H \cdot I_D \label{18}\] The shielding constants σ H and σ D are, in this case, equal since the two nuclei are chemically identical. For sufficiently large magnetic fields B , the last term is effectively equal to hJ HD M H M D . The spin-coupling constant J can be directly equated to the splitting expressed in Hz. We consider next the case of two equivalent protons, for example, the CH 2 group of ethanol. Each proton can have two possible spin states with \( M_I = \pm\frac{1}{2} \) , giving a total of four composite spin states. Just as in the case of 2 electron spins, these combine to give singlet and triplet nuclear-spin states with M = 0 and 1, respectively. Also, just as for electron spins, transitions between singlet and triplet states are forbidden. The triplet state allows NMR transitions with \( \Delta M = \pm 1 \) to give a single resonance frequency, while the singlet state is inactive. As a consequence, spin-spin splittings do not occur among identical nuclei. For example, the H 2 molecule shows just a single NMR frequency. And the CH 2 protons in ethanol do not show spin- spin interactions with one another. They can however cause a splitting of the neighboring CH 3 protons . Fig. 7 (left side) shows the four possible spin states of two equivalent protons, such as those in the methylene group CH 2 , and the triplet with intensity ratios 1:2:1 which these produce in nearby protons. Also shown (right side) are the eight possible spin states for three equivalent protons, say those in a methyl group CH 3 , and the quartet with intensity ratios 1:3:3:1 which these produce. In general, n equivalent protons will give a splitting pattern of n + 1 lines in the ratio of binomial coefficients 1: n : n ( n − 1) / 2 . . . The tertiary hydrogen in isobutane ( CH 3 ) 3 CH ∗ , marked with an asterisk, should be split into 10 lines by the 9 equivalent methyl protons. Figure 7 . Splitting patterns from methylene and methyl protons. The NMR spectrum of ethanol CH 3 CH 2 OH (Fig. 4) can now be interpreted. The CH 3 protons are split into a 1:2:1 triplet by spin-spin interaction with the neighboring CH 2 . Conversely, the CH 2 protons are split into a 1:3:3:1 quartet by interaction with the CH 3 . The OH (hydroxyl) proton evidently does not either cause or undergo spin-spin splitting. The explanation for this is hydrogen bonding, which involves rapid exchange of hydroxyl protons among neighboring molecules. If this rate of exchange is greater than or comparable to the NMR radiofrequency, then the splittings will be “washed out.” Only one line with a motion-averaged value of the chemical shift will be observed. NMR has consequently become a useful tool to study intramolecular motions. |
Courses/University_of_California_Davis/Chem_205%3A_Symmetry_Spectroscopy_and_Structure/03%3A_Vibrational_Spectroscopy/3.06%3A_IR_and_Raman_Activity | Selection Rule for IR Recall, that for IR activity, we require that \[\left(\frac{d \mu}{d q}\right)_{q_{\text {eq }}} \neq 0 \nonumber \] If motion along the normal coordinate does not change the dipole moment, the transition is forbidden. In order for \(μ\) to be changed by the normal coordinate, at lease on of the following must be non-zero! \[\langle v=0|\hat{x}| v \neq 0\rangle \nonumber \] \[\langle v=0|\hat{y}| v \neq 0\rangle \nonumber \] \[\langle v=0|\hat{z}| v \neq 0\rangle \nonumber \] integrated over \(q\), the normal coordinate. This is since \(μ\) is a vector that transforms as x, y, z in the point group . The polarization of the IR transition is determined by which transition integral is non-zero . This is very much like the criterion for an electronically allowed transition. All \(|v=0\rangle\) are a 1 (totally symmetric) and \(|v=1\rangle\) has the same symmetry as the singly excited normal mode, \(q\). Since \[A_{1} \otimes \Gamma_{x, y, z} \otimes \Gamma_{q} \supset A_{1} \nonumber \] The transition will be IR active if the normal mode symmetry belongs to the same irreducible representation as x, y, or z for the point group of the molecule. This can be applied immediately to H 2 O. Example \(\PageIndex{1}\): Water Water has three vibrational modes. Three normal modes of water Whether any of these modes is active in the infrared region of the spectrum is the question. This is determined by evaluating the quantum mechanical transition probability integral, \[ P = e \int \Psi^f (vib) \hat{q} \Psi^i (vib) d \tau \nonumber \] where q is the spatial coordinate and is either x, y, or z. If this integral is non-zero the transition is allowed. Group theory enables us to determine if the integral is non-zero as follows. We evaluate the direct product (note the similarity to the quantum mechanical transition probability integral) Γ vib f Γ q Γ vib i and if it turns out to be equal to, or contain, the irreducible representation A 1 then the transition is allowed. If the direct product isn't equal to or contain A 1 then the transition is forbidden. Γ vib i , the representation for the ground state is always equal to A 1 and Γ vib f is always equal to A 1 and Γ vib f has the symmetry of the vibrational mode being excited, which in our case is either A 1 or B 2 . Thus we can see that a vibrational mode will be infrared active if it belongs to the same symmetry species as one of the Cartesian coordinates. You should verify that this is correct and also be able to show that all three vibrational modes of the water molecule are infrared active. Selection Rule for Raman Recall, for Raman activity, there must be a change in the polarizability with respect to the normal mode. The components of the polarizability transform as the quadratic functions of x,y,z. i.e. x 2 , y 2 , z 2 , xy, xz, yz or combinations thereof (x 2 -y 2 ). Thus we require that for the polarizability operator to transform as one of the quadratic terms e.g. with Thus since has the same symmetry as the singly excited normal mode, q . The transition integrals we know and love will be: for at least one representation of polarization. At least one quadratic form transforms as each one of the irreducible representations. Therefore, all possible vibrations of any C 2v molecule will be Raman allowed. But what about other point groups? Centrosymmetric molecules – exclusion rules There are important examples for which the IR allowed and Raman allowed transitions are mutually exclusive. There are molecules having a center of inversion. In this case, there normal coordinates, \(q\), must be either \(g\) or \(u\), depending on their behavior toward the \(\hat{i}\) operation. Thus \(|v=1\rangle\) must be either \(g\) or \(u\), while \(|v=0\rangle\) will always be \(g\). In centrosymmetric molecules, the x,y,z operators are u -operators, while all components of \(\hat{P}\) are g -operators (go look it up on the O h point group character table!) i.e. \[\hat{x}^{2}=(-x)^{2} \quad \therefore \text { g symmetry } \nonumber \] \[\hat{x}=-(-x) \quad \therefore \text { u symmetry } \nonumber \] This symmetry behavior leads to the rule for centrosymmetirc molecules. Excitation of g -normal modes are forbidden in the IR, while excitation of u -normal modes are forbidden in Raman. Example \(\PageIndex{1}\) Carbon Dioxide is a linear triatomic molecule with 3N-5=4 normal modes: e. The point group is D ∞ h and the irreducible representations are \(\Sigma_{g}^{+}\), \Sigma_{u}^{+}, and \(\Pi_{u}\) respectively. Thus only asymmetric stretch and bends are found in the IR, while only the symmetric stretch is found in Raman spectra. The IR spectrum of \(\ce{CO2}\) (4.0 kPa total pressure) is shown below. Note these extra bands: v 3 +v 1 =3610 cm -1 and v 3 +2*v 2 =3710 cm -1 Vibrational frequency nomenclature The \(3N-5\) (5) fundamental frequencies are usually numbered sequentially, \(\nu_1\), \(\nu_2\), \(\nu_3\),.. \(\nu_{3N-6}\). The convention is that the highest frequencies of totally symmetric vibrations in \(\nu_1\) follow all other totally symmetric vibrations in order of decreasing frequency. Once all totally symmetric vibrations are numbers, the next number goes to the highest frequency non-symmetric vibration. For H 2 O: ν 1 is the “symmetric stretch” ν 2 is the “asymmetric stretch” ν 3 is the “symmetric bend” Exception, the bending mode of a linear molecule is ν 2 For CO 2 : ν 1 is the “symmetric stretch” ν 2 is the “symmetric bend” ν 3 is the “asymmetric stretch” If IR and Raman spectra consisted only of fundamental and overtones, spectra would be easier to assign, but combinations bands, ν 1 ± ν 2 and overtones 2 ν 1 also appear, complicating the assignment of the fundamental frequencies. |
Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Properties_of_Carboxylic_Acids/Carboxyl_Derivatives/Carboxylic_Derivatives_-_Reduction_(Catalytic_Reduction) | Reductions of carboxylic acid derivatives might be expected to lead either to aldehydes or alcohols, functional groups having a lower oxidation state of the carboxyl carbon. Indeed, carboxylic acids themselves are reduced to alcohols by lithium aluminum hydride. At this point it will be useful to consider three kinds of reductions: catalytic hydrogenation complex metal hydride reductions diborane reduction. Catalytic Hydrogenation As a rule, the carbonyl group does not add hydrogen as readily as do the carbon-carbon double and triple bonds. Thus, it is fairly easy to reduce an alkene or alkyne function without affecting any carbonyl functions in the same molecule. By using a platinum catalyst and increased temperature and pressure, it is possible to reduce aldehydes and ketones to alcohols, but carboxylic acids, esters and amides are comparatively unreactive. The exceptional reactivity of acyl halides, on the other hand, facilitates their reduction under mild conditions, by using a poisoned palladium catalyst similar to that used for the partial reduction of alkynes to alkenes. This reduction stops at the aldehyde stage, providing us with a useful two-step procedure for converting carboxylic acids to aldehydes, as reaction #1 below demonstrates. Equivalent reductions of anhydrides have not been reported, but we might speculate that they would be reduced more easily than esters. The only other reduction of a carboxylic acid derivative that is widely used is that of nitriles to 1º-amines. Examples of these reductions are provided in the following diagram. The second and third equations illustrate the extreme difference in hydrogenation reactivity between esters and nitriles. This is further demonstrated by the last reaction, in which a nitrile is preferentially reduced in the presence of a carbonyl group and two benzene rings. The resulting 1º-amine immediately reacts with the carbonyl function to give a cyclic enamine product (colored light blue). In most nitrile reductions ammonia is added to inhibit the formation of a 2º-amine by-product. This may occur by way of an intermediate aldehyde imine created by addition of the first equivalent of hydrogen. The following equations show how such an imine species might react with the 1º-amine product to give a substituted imine (2nd equation), which would then add hydrogen to generate a 2º-amine. Excess ammonia shifts the imine equilibrium to the left, as written below. 0 1 2 3 4 5 (1) R–C≡N + H2 catalyst RCH=NH imine H2 RCH2NH2 1º-amine 0 1 2 3 4 5 (2) RCH=NH + RCH2NH2 imine 1º-amine NaN RCH=NCH2R + NH3 substituted imine H2 & catalyst RCH2NHCH2R 2º-amine Contributors William Reusch, Professor Emeritus ( Michigan State U. ), Virtual Textbook of Organic Chemistry |
Courses/Indiana_Tech/EWC%3A_CHEM_1000_-_Introductory_Chemistry_(Budhi)/07%3A_Chemical_Reactions/7.11%3A_The_Activity_Series%3A_Predicting_Spontaneous_Redox_Reactions | Learning Objectives Use the activity series to predict if a reaction will occur. We see below two metals that can be exposed to water. The picture on the left is of sodium, which has a violent reaction when it comes in contact with water. The picture on the right is of silver, a metal so unreactive with water that it can be made into drinking vessels. Both metals have a single \(s\) electron in their outer shell, so you would predict a similar reactivity from each. However, we have a tool that allows us to make better predictions about how certain elements will react with others. The Activity Series Single-replacement reactions only occur when the element that is doing the replacing is more reactive than the element that is being replaced. Therefore, it is useful to have a list of elements in order of their relative reactivity. The activity series is a list of elements in decreasing order of their reactivity. Since metals replace other metals, while nonmetals replace other nonmetals, they each have a separate activity series. The table \(\PageIndex{1}\) below is an activity series of most common metals, and the table \(\PageIndex{2}\) is an activity series of the halogens. Elements, from most to least reactive Reaction Occurring \(\ce{Li}\) \(\ce{K}\) \(\ce{Ba}\) \(\ce{Sr}\) \(\ce{Ca}\) \(\ce{Na}\) React with cold water, replacing hydrogen. \(\ce{Mg}\) \(\ce{Al}\) \(\ce{Zn}\) \(\ce{Cr}\) \(\ce{Fe}\) \(\ce{Cd}\) React with steam, but not cold water, replacing hydrogen. \(\ce{Co}\) \(\ce{Ni}\) \(\ce{Sn}\) \(\ce{Pb}\) Do not react with water. React with acids, replacing hydrogen. \(\ce{H_2}\) NaN \(\ce{Cu}\) \(\ce{Hg}\) \(\ce{Ag}\) \(\ce{Pt}\) \(\ce{Au}\) Unreactive with water or acids. Elements, from most to least reactive \(\ce{F_2}\) \(\ce{Cl_2}\) \(\ce{Br_2}\) \(\ce{I_2}\) For a single-replacement reaction, a given element is capable of replacing an element that is below it in the activity series. This can be used to predict if a reaction will occur. Suppose that small pieces of the metal nickel were placed into two separate aqueous solutions: one of iron (III) nitrate and one of lead (II) nitrate. Looking at the activity series, we see that nickel is below iron, but above lead. Therefore, the nickel metal will be capable of replacing the lead in a reaction, but will not be capable of replacing iron. \[\ce{Ni} \left( s \right) + \ce{Pb(NO_3)_2} \left( aq \right) \rightarrow \ce{Ni(NO_3)_2} \left( aq \right) + \ce{Pb} \left( s \right) \nonumber \] \[\ce{Ni} \left( s \right) + \ce{Fe(NO_3)_3} \left( aq \right) \rightarrow \text{NR (no reaction)} \nonumber \] In the descriptions that accompany the activity series of metals, a given metal is also capable of undergoing the reactions described below that section. For example, lithium will react with cold water, replacing hydrogen. It will also react with steam and with acids, since that requires a lower degree of reactivity. Examples \(\PageIndex{1}\) Use the activity series to predict if the following reactions will occur. If not, write \(\text{NR}\). If the reaction does occur, write the products of the reaction and balance the equation. \(\ce{Al} \left( s \right) + \ce{Zn(NO_3)_2} \left( aq \right) \rightarrow\) \(\ce{Ag} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow\) Solution Steps Example \(\PageIndex{1A}\) \(\ce{Al} \left( s \right) + \ce{Zn(NO_3)_2} \left( aq \right) \rightarrow\) Example \(\PageIndex{1B}\) \(\ce{Ag} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow\) Plan the problem. Compare the placements of aluminum and zinc on the activity series (Table \(\PageIndex{1}\)) Compare the placements of silver and hydrogen (Table \(\PageIndex{1}\)) Solve. Since aluminum is above zinc, it is capable of replacing it and a reaction will occur. The products of the reaction will be aqueous aluminum nitrate and solid zinc. Take care to write the correct formulas for the products before balancing the equation. Aluminum adopts a \(+3\) charge in an ionic compound, so the formula for aluminum nitrate is \(\ce{Al(NO_3)_3}\). The balanced equation is: \(2 \ce{Al} \left( s \right) + 3 \ce{Zn(NO_3)_2} \left( aq \right) \rightarrow 2 \ce{Al(NO_3)_3} \left( aq \right) + 3 \ce{Zn} \left( s \right)\) Since silver is below hydrogen, it is not capable of replacing hydrogen in a reaction with an acid. \(\ce{Ag} \left( s \right) + \ce{HCl} \left( aq \right) \rightarrow \text{NR}\) Exercise \(\PageIndex{1}\) Use the activity series to predict the products, if any, of each equation. \(\ce{FeCl2 + Zn →}\) \(\ce{HNO3 + Au →}\) Answer a The products are ZnCl 2 + Fe. Answer b No reaction. Summary Metals and halogens are ranked according to their ability to displace other metals or halogens below them in the activity series. |
Courses/Eastern_Mennonite_University/EMU%3A_Chemistry_for_the_Life_Sciences_(Cessna)/12%3A_Organic_Chemistry%3A_Alkanes_and_Halogenated_Hydrocarbons/12.E%3A_Exercises | 12.1: Organic Chemistry Concept Review Exercises Classify each compound as organic or inorganic. C 3 H 8 O CaCl 2 Cr(NH 3 ) 3 Cl 3 C 30 H 48 O 3 N Which compound is likely organic and which is likely inorganic? a flammable compound that boils at 80°C and is insoluble in water a compound that does not burn, melts at 630°C, and is soluble in water Answers organic inorganic inorganic organic organic inorganic Classify each compound as organic or inorganic. C 6 H 10 CoCl 2 C 12 H 22 O 11 Classify each compound as organic or inorganic. CH 3 NH 2 NaNH 2 Cu(NH 3 ) 6 Cl 2 Which member of each pair has a higher melting point? CH 3 OH and NaOH CH 3 Cl and KCl Which member of each pair has a higher melting point? C 2 H 6 and CoCl 2 CH 4 and LiH Answers organic inorganic organic NaOH KCl 12.2: Structures and Names of Alkanes Concept Review Exercises In the homologous series of alkanes, what is the molecular formula for the member just above C 8 H 18 ? Use the general formula for alkanes to write the molecular formula of the alkane with 12 carbon atoms. Answers C 9 H 20 C 12 H 26 Exercises What compounds contain fewer carbon atoms than C 3 H 8 and are its homologs? What compounds contain five to eight carbon atoms and are homologs of C 4 H 10 ? Answer CH 4 and C 2 H 6 12.3: Branched-Chain Alkanes Concept Review Exercises In alkanes, can there be a two-carbon branch off the second carbon atom of a four-carbon chain? Explain. A student is asked to write structural formulas for two different hydrocarbons having the molecular formula C 5 H 12 . She writes one formula with all five carbon atoms in a horizontal line and the other with four carbon atoms in a line, with a CH 3 group extending down from the first attached to the third carbon atom. Do these structural formulas represent different molecular formulas? Explain why or why not. Answers No; the branch would make the longest continuous chain of five carbon atoms. No; both are five-carbon continuous chains. Key Takeaway Alkanes with four or more carbon atoms can exist in isomeric forms. Exercises Briefly identify the important distinctions between a straight-chain alkane and a branched-chain alkane. How are butane and isobutane related? How do they differ? Name each compound. Write the structural formula for each compound. hexane octane Indicate whether the structures in each set represent the same compound or isomers. CH 3 CH 2 CH 2 CH 3 and CH 3 CH 2 CH 2 CH 2 CH 3 and Answers Straight-chain alkanes and branched-chain alkanes have different properties as well as different structures. pentane heptane no yes 12.4: Condensed Structural and Line-Angle Formulas Exercises Write the condensed structural formula for each structural formula. A condensed structural formula for isohexane can be written as (CH 3 ) 2 CHCH 2 CH 2 CH 3 . Draw the line-angle formula for isohexane. Draw a line-angle formula for the compound CH 3 CH 2 CH(CH 3 )CH 2 CH 2 CH 3 . Give the structural formula for the compound represented by this line-angle formula: Answers CH 3 CH 3 CH 3 CH 2 CH 3 CH 3 CH 2 CH 2 CH 2 CH 3 12.5: IUPAC Nomenclature Concept Review Exercises What is a CH 3 group called when it is attached to a chain of carbon atoms—a substituent or a functional group? Which type of name uses numbers to locate substituents—common names or IUPAC names? Answers substituent IUPAC names Exercises Briefly identify the important distinctions between an alkane and an alkyl group. How many carbon atoms are present in each molecule? 2-methylbutane 3-ethylpentane How many carbon atoms are present in each molecule? 2,3-dimethylbutane 3-ethyl-2-methylheptane Draw the structure for each compound. 3-methylpentane 2,2,5-trimethylhexane 4-ethyl-3-methyloctane Draw the structure for each compound. 2-methylpentane 4-ethyl-2-methylhexane 2,2,3,3-tetramethylbutane Name each compound according to the IUPAC system. Name each compound according to the IUPAC system. What is a substituent? How is the location of a substituent indicated in the IUPAC system? Briefly identify the important distinctions between a common name and an IUPAC name. Answers An alkane is a molecule; an alkyl group is not an independent molecule but rather a part of a molecule that we consider as a unit. 6 10 2,2,4,4-tetramethylpentane 3-ethylhexane Common names are widely used but not very systematic; IUPAC names identify a parent compound and name other groups as substituents. 12.6: Physical Properties of Alkanes Concept Review Exercises Without referring to a table, predict which has a higher boiling point—hexane or octane. Explain. If 25 mL of hexane were added to 100 mL of water in a beaker, which of the following would you expect to happen? Explain. Hexane would dissolve in water. Hexane would not dissolve in water and would float on top. Hexane would not dissolve in water and would sink to the bottom of the container. Answers octane because of its greater molar mass b; hexane is insoluble in water and less dense than water. Exercises Without referring to a table or other reference, predict which member of each pair has the higher boiling point. pentane or butane heptane or nonane For which member of each pair is hexane a good solvent? pentane or water sodium chloride or soybean oil Answer pentane nonane 12.7: Chemical Properties of Alkanes Concept Review Exercises Why are alkanes sometimes called paraffins? Which halogen reacts most readily with alkanes? Which reacts least readily? Answers Alkanes do not react with many common chemicals. They are sometimes called paraffins, from the Latin parum affinis, meaning “little affinity.” most readily: \(F_2\); least readily: \(I_2\) Exercises Why do alkanes usually not react with ionic compounds such as most laboratory acids, bases, oxidizing agents, or reducing agents? Write an equation for the complete combustion of methane (\(CH_4\)), the main component of natural gas). What is the most important reaction of alkanes? Name some substances other than oxygen that react readily with alkanes. Answers Alkanes are nonpolar; they do not attract ions. 12.8: Halogenated Hydrocarbons Concept Review Exercises What is the IUPAC name for the HFC that has the formula CH 2 FCF 3 ? (Hint: you must use a number to indicate the location of each substituent F atom.) What is the IUPAC name for the HCFC that has the formula CHCl 2 CF 3 ? Answers 1,1,1,2-tetrafluoroethane 1,1,1-trifluoro-2,2-dichloroethane Exercises Write the condensed structural formula for each compound. methyl chloride chloroform Write the condensed structural formula for each compound. ethyl bromide carbon tetrachloride Write the condensed structural formulas for the two isomers that have the molecular formula C 3 H 7 Br. Give the common name and the IUPAC name of each. Write the condensed structural formulas for the four isomers that have the molecular formula C 4 H 9 Br. Give the IUPAC name of each. What is a CFC? How are CFCs involved in the destruction of the ozone layer? Explain why each compound is less destructive to the ozone layer than are CFCs. fluorocarbons HCFCs Answers CH 3 Cl CHCl 3 CH 3 CH 2 CH 2 Br, propyl bromide, 1-bromopropane; CH 3 CHBrCH 3 , isopropyl bromide, 2-bromopropane compounds containing Cl, F, and C; by releasing Cl atoms in the stratosphere 12.9: Cycloalkanes Concept Review Exercises What is the molecular formula of cyclooctane? What is the IUPAC name for this compound? Answers C 8 H 16 ethylcyclopropane Exercises Draw the structure for each compound. ethylcyclobutane propylcyclopropane Draw the structure for each compound. methylcyclohexane butylcyclobutane Cycloalkyl groups can be derived from cycloalkanes in the same way that alkyl groups are derived from alkanes. These groups are named as cyclopropyl, cyclobutyl, and so on. Name each cycloalkyl halide. Halogenated cycloalkanes can be named by the IUPAC system. As with alkyl derivatives, monosubstituted derivatives need no number to indicate the position of the halogen. To name disubstituted derivatives, the carbon atoms are numbered starting at the position of one substituent (C1) and proceeding to the second substituted atom by the shortest route. Name each compound. Answers cyclopentyl bromide cyclohexyl chloride 12.10: Chapter Summary Additional Exercises You find an unlabeled jar containing a solid that melts at 48°C. It ignites readily and burns readily. The substance is insoluble in water and floats on the surface. Is the substance likely to be organic or inorganic? Give the molecular formulas for methylcyclopentane, 2-methylpentane, and cyclohexane. Which are isomers? What is wrong with each name? (Hint: first write the structure as if it were correct.) Give the correct name for each compound. 2-dimethylpropane 2,3,3-trimethylbutane 2,4-diethylpentane 3,4-dimethyl-5-propylhexane What is the danger in swallowing a liquid alkane? Distinguish between lighter and heavier liquid alkanes in terms of their effects on the skin. Following is the line formula for an alkane. Draw its structure and give its name. Write equations for the complete combustion of each compound. propane (a bottled gas fuel) octane (a typical hydrocarbon in gasoline). The density of a gasoline sample is 0.690 g/mL. On the basis of the complete combustion of octane, calculate the amount in grams of carbon dioxide (CO 2 ) and water (H 2 O) formed per gallon (3.78 L) of the gasoline when used in an automobile. Draw the structures for the five isomeric hexanes (C 6 H 14 ). Name each by the IUPAC system. Indicate whether the structures in each set represent the same compound or isomers. Consider the line-angle formulas shown here and answer the questions. Which pair of formulas represents isomers? Draw each structure. Which formula represents an alkyl halide? Name the compound and write its condensed structural formula. Which formula represents a cyclic alkane? Name the compound and draw its structure. What is the molecular formula of the compound represented by (i)? Answers organic Two numbers are needed to indicate two substituents; 2,2-dimethylpropane. The lowest possible numbers were not used; 2,2,3-trimethylbutane. An ethyl substituent is not possible on the second carbon atom; 3,5-dimethylheptane. A propyl substituent is not possible on the fifth carbon atom; 3,4,5-trimethyloctane. Lighter alkanes wash away protective skin oils; heavier alkanes form a protective layer. C 3 H 8 + 5O 2 → 3CO 2 + 4H 2 O 2C 8 H 18 + 25O 2 → 16CO 2 + 18H 2 O CH 3 CH 2 CH 2 CH 2 CH 2 CH 3 ; hexane ii and iii; CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 and iv; 3-chloropentane; CH 3 CH 2 CHClCH 2 CH 3 i; ethylcyclopentane; C 7 H 14 |
Courses/Los_Angeles_Trade_Technical_College/Foundations_of_Introductory_Chemistry-1/00%3A_Front_Matter/02%3A_InfoPage | This text is disseminated via the Open Education Resource (OER) LibreTexts Project ( https://LibreTexts.org ) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact [email protected] . More information on our activities can be found via Facebook ( https://facebook.com/Libretexts ), Twitter ( https://twitter.com/libretexts ), or our blog ( http://Blog.Libretexts.org ). This text was compiled on 04/21/2025 |
Courses/SUNY_Adirondack/CHM_103%3A_Principles_of_Chemistry/07%3A_Introduction_to_Chemical_Reactions/7.2%3A_The_Law_of_Conservation_of_Matter | Learning Objectives Correctly define a law as it pertains to science. State the law of conservation of matter. In science, a law is a general statement that explains a large number of observations. Before being accepted, a law must be verified many times under many conditions. Laws are therefore considered the highest form of scientific knowledge and are generally thought to be inviolable. Scientific laws form the core of scientific knowledge. One scientific law that provides the foundation for understanding in chemistry is the law of conservation of matter. It states that in any given system that is closed to the transfer of matter (in and out), the amount of matter in the system stays constant. A concise way of expressing this law is to say that the amount of matter in a system is conserved . With the development of more precise ideas on elements, compounds and mixtures, scientists began to investigate how and why substances react. French chemist A. Lavoisier laid the foundation to the scientific investigation of matter by describing that substances react by following certain laws. These laws are called the laws of chemical combination. These eventually formed the basis of Dalton's Atomic Theory of Matter. Law of Conservation of Mass According to this law, during any physical or chemical change, the total mass of the products remains equal to the total mass of the reactants. \[ \overbrace{\underbrace{\ce{HgO (s)}}_{\text{100 g}}}^{\text{Mercuric oxide}} \rightarrow \underbrace{\overbrace{\ce{Hg (l) }}^{\text{Mercury}}}_{\text{92.6 g}} + \underbrace{\overbrace{\ce{O2 (g)}}^{\text{Oxygen}}}_{\text{7.4 g}} \nonumber \] Another way of stating this is, "In a chemical reaction, matter is neither created nor destroyed." The law of conservation of mass is also known as the "law of indestructibility of matter." Example \(\PageIndex{1}\) If heating 10 grams of \(\ce{CaCO3}\) produces 4.4 g of \(\ce{CO2}\) and 5.6 g of \(\ce{CaO}\), show that these observations are in agreement with the law of conservation of mass. Solution Mass of the reactants, \(\ce{CaCO3}\) : \(10 \,g\) Mass of the products, \(\ce{CO2}\) and \(\ce{CaO}\): \(4.4 \,g+ 5.6\, g = 10\, g\). Because the mass of the reactants = the mass of the products, the observations are in agreement with the law of conservation of mass. What does this mean for chemistry? In any chemical change, one or more initial substances change into a different substance or substances. Both the initial and final substances are composed of atoms because all matter is composed of atoms. According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and kind of atoms after the chemical change as were present before the chemical change. It may seem as though burning destroys matter, but the same amount, or mass, of matter still exists after a campfire as before. Figure 5.1.1 shows that when wood burns, it combines with oxygen and changes not only to ashes, but also to carbon dioxide and water vapor. The gases float off into the air, leaving behind just the ashes. Suppose we had measured the mass of the wood before it burned and the mass of the ashes after it burned. Also suppose we had been able to measure the oxygen used by the fire and the gases produced by the fire. What would we find? The total mass of matter after the fire would be the same as the total mass of matter before the fire. Exercise \(\PageIndex{1}\) What is the law of conservation of matter? How does the law of conservation of matter apply to chemistry? Answer a: The law of conservation of matter states that in any given system that is closed to the transfer of matter, the amount of matter in the system stays constant Answer b: The law of conservation of matter says that in chemical reactions, the total mass of the products must equal the total mass of the reactants. Key Takeaway The amount of matter in a closed system is conserved. |
Courses/University_of_Kentucky/UK%3A_General_Chemistry/09%3A_Gases/9.1%3A_Gas_Pressure | Skills to Develop Define the property of pressure Define and convert among the units of pressure measurements Describe the operation of common tools for measuring gas pressure Calculate pressure from manometer data The earth’s atmosphere exerts a pressure, as does any other gas. Although we do not normally notice atmospheric pressure, we are sensitive to pressure changes—for example, when your ears “pop” during take-off and landing while flying, or when you dive underwater. Gas pressure is caused by the force exerted by gas molecules colliding with the surfaces of objects (Figure \(\PageIndex{1}\)). Although the force of each collision is very small, any surface of appreciable area experiences a large number of collisions in a short time, which can result in a high pressure. In fact, normal air pressure is strong enough to crush a metal container when not balanced by equal pressure from inside the container. Figure \(\PageIndex{1}\) : The atmosphere above us exerts a large pressure on objects at the surface of the earth, roughly equal to the weight of a bowling ball pressing on an area the size of a human thumbnail. Atmospheric pressure is caused by the weight of the column of air molecules in the atmosphere above an object, such as the tanker car. At sea level, this pressure is roughly the same as that exerted by a full-grown African elephant standing on a doormat, or a typical bowling ball resting on your thumbnail. These may seem like huge amounts, and they are, but life on earth has evolved under such atmospheric pressure. If you actually perch a bowling ball on your thumbnail, the pressure experienced is twice the usual pressure, and the sensation is unpleasant. A dramatic illustration of atmospheric pressure is provided in this brief video, which shows a railway tanker car imploding when its internal pressure is decreased. Pressure is defined as the force exerted on a given area: \[P=\dfrac{F}{A} \label{9.2.1}\] Since pressure is directly proportional to force and inversely proportional to area (Equation \ref{9.2.1}), pressure can be increased either by either increasing the amount of force or by decreasing the area over which it is applied. Correspondingly, pressure can be decreased by either decreasing the force or increasing the area. Let’s apply the definition of pressure (Equation \ref{9.2.1}) to determine which would be more likely to fall through thin ice in Figure \(\PageIndex{2}\).—the elephant or the figure skater? Figure \(\PageIndex{2}\) : Although (a) an elephant’s weight is large, creating a very large force on the ground, (b) the figure skater exerts a much higher pressure on the ice due to the small surface area of her skates. (credit a: modification of work by Guido da Rozze; credit b: modification of work by Ryosuke Yagi). A large African elephant can weigh 7 tons, supported on four feet, each with a diameter of about 1.5 ft (footprint area of 250 in 2 ), so the pressure exerted by each foot is about 14 lb/in 2 : \[\mathrm{pressure\: per\: elephant\: foot=14,000\dfrac{lb}{elephant}×\dfrac{1\: elephant}{4\: feet}×\dfrac{1\: foot}{250\:in^2}=14\:lb/in^2} \label{9.2.2}\] The figure skater weighs about 120 lbs, supported on two skate blades, each with an area of about 2 in 2 , so the pressure exerted by each blade is about 30 lb/in 2 : \[\mathrm{pressure\: per\: skate\: blade=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: blades}×\dfrac{1\: blade}{2\:in^2}=30\:lb/in^2} \label{9.2.3}\] Even though the elephant is more than one hundred-times heavier than the skater, it exerts less than one-half of the pressure and would therefore be less likely to fall though thin ice. On the other hand, if the skater removes her skates and stands with bare feet (or regular footwear) on the ice, the larger area over which her weight is applied greatly reduces the pressure exerted: \[\mathrm{pressure\: per\: human\: foot=120\dfrac{lb}{skater}×\dfrac{1\: skater}{2\: feet}×\dfrac{1\: foot}{30\:in^2}=2\:lb/in^2} \label{9.2.4}\] The SI unit of pressure is the pascal (Pa) , with 1 Pa = 1 N/m 2 , where N is the newton, a unit of force defined as 1 kg m/s 2 . One pascal is a small pressure; in many cases, it is more convenient to use units of kilopascal (1 kPa = 1000 Pa) or bar (1 bar = 100,000 Pa). In the United States, pressure is often measured in pounds of force on an area of one square inch— pounds per square inch (psi) —for example, in car tires. Pressure can also be measured using the unit atmosphere (atm) , which originally represented the average sea level air pressure at the approximate latitude of Paris (45°). Table \(\PageIndex{1}\) provides some information on these and a few other common units for pressure measurements Unit Name and Abbreviation Definition or Relation to Other Unit Comment pascal (Pa) 1 Pa = 1 N/m2 recommended IUPAC unit kilopascal (kPa) 1 kPa = 1000 Pa NaN pounds per square inch (psi) air pressure at sea level is ~14.7 psi NaN atmosphere (atm) 1 atm = 101,325 Pa air pressure at sea level is ~1 atm bar (bar, or b) 1 bar = 100,000 Pa (exactly) commonly used in meteorology millibar (mbar, or mb) 1000 mbar = 1 bar NaN inches of mercury (in. Hg) 1 in. Hg = 3386 Pa used by aviation industry, also some weather reports torr \(\mathrm{1\: torr=\dfrac{1}{760}\:atm}\) named after Evangelista Torricelli, inventor of the barometer millimeters of mercury (mm Hg) 1 mm Hg ~1 torr NaN Example \(\PageIndex{1}\): Conversion of Pressure Units The United States National Weather Service reports pressure in both inches of Hg and millibars. Convert a pressure of 29.2 in. Hg into: torr atm kPa mbar Solution This is a unit conversion problem. The relationships between the various pressure units are given in Table 9.2.1. \(\mathrm{29.2\cancel{in\: Hg}×\dfrac{25.4\cancel{mm}}{1\cancel{in}} ×\dfrac{1\: torr}{1\cancel{mm\: Hg}} =742\: torr}\) \(\mathrm{742\cancel{torr}×\dfrac{1\: atm}{760\cancel{torr}}=0.976\: atm}\) \(\mathrm{742\cancel{torr}×\dfrac{101.325\: kPa}{760\cancel{torr}}=98.9\: kPa}\) \(\mathrm{98.9\cancel{kPa}×\dfrac{1000\cancel{Pa}}{1\cancel{kPa}} \times \dfrac{1\cancel{bar}}{100,000\cancel{Pa}} \times\dfrac{1000\: mbar}{1\cancel{bar}}=989\: mbar}\) Exercise \(\PageIndex{1}\) A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, in kilopascals, and in bar? Answer 0.974 atm; 740 mm Hg; 98.7 kPa; 0.987 bar We can measure atmospheric pressure, the force exerted by the atmosphere on the earth’s surface, with a barometer (Figure \(\PageIndex{3}\)). A barometer is a glass tube that is closed at one end, filled with a nonvolatile liquid such as mercury, and then inverted and immersed in a container of that liquid. The atmosphere exerts pressure on the liquid outside the tube, the column of liquid exerts pressure inside the tube, and the pressure at the liquid surface is the same inside and outside the tube. The height of the liquid in the tube is therefore proportional to the pressure exerted by the atmosphere. Figure \(\PageIndex{3}\) : In a barometer, the height, h, of the column of liquid is used as a measurement of the air pressure. Using very dense liquid mercury (left) permits the construction of reasonably sized barometers, whereas using water (right) would require a barometer more than 30 feet tall. If the liquid is water, normal atmospheric pressure will support a column of water over 10 meters high, which is rather inconvenient for making (and reading) a barometer. Because mercury (Hg) is about 13.6-times denser than water, a mercury barometer only needs to be \(\dfrac{1}{13.6}\) as tall as a water barometer—a more suitable size. Standard atmospheric pressure of 1 atm at sea level (101,325 Pa) corresponds to a column of mercury that is about 760 mm (29.92 in.) high. The torr was originally intended to be a unit equal to one millimeter of mercury, but it no longer corresponds exactly. The pressure exerted by a fluid due to gravity is known as hydrostatic pressure , p : \[p=hρg \label{9.2.5}\] where \(h\) is the height of the fluid, \(ρ\) is the density of the fluid, and \(g\) is acceleration due to gravity. Example \(\PageIndex{2}\): Calculation of Barometric Pressure Show the calculation supporting the claim that atmospheric pressure near sea level corresponds to the pressure exerted by a column of mercury that is about 760 mm high. The density of mercury = \(13.6 \,g/cm^3\). Solution The hydrostatic pressure is given by Equation \ref{9.2.5}, with \(h = 760 \,mm\), \(ρ = 13.6\, g/cm^3\), and \(g = 9.81 \,m/s^2\). Plugging these values into the Equation \ref{9.2.5} and doing the necessary unit conversions will give us the value we seek. (Note: We are expecting to find a pressure of ~101,325 Pa:) \[\mathrm{101,325\:\mathit{N}/m^2=101,325\:\dfrac{kg·m/s^2}{m^2}=101,325\:\dfrac{kg}{m·s^2}}\] \[\begin {align*} p&\mathrm{=\left(760\: mm×\dfrac{1\: m}{1000\: mm}\right)×\left(\dfrac{13.6\: g}{1\:cm^3}×\dfrac{1\: kg}{1000\: g}×\dfrac{( 100\: cm )^3}{( 1\: m )^3}\right)×\left(\dfrac{9.81\: m}{1\:s^2}\right)}\\ &\mathrm{=(0.760\: m)(13,600\:kg/m^3)(9.81\:m/s^2)=1.01 \times 10^5\:kg/ms^2=1.01×10^5\mathit{N}/m^2} \\ & \mathrm{=1.01×10^5\:Pa} \end {align*}\] Exercise \(\PageIndex{2}\) Calculate the height of a column of water at 25 °C that corresponds to normal atmospheric pressure. The density of water at this temperature is 1.0 g/cm 3 . A manometer is a device similar to a barometer that can be used to measure the pressure of a gas trapped in a container. A closed-end manometer is a U-shaped tube with one closed arm, one arm that connects to the gas to be measured, and a nonvolatile liquid (usually mercury) in between. As with a barometer, the distance between the liquid levels in the two arms of the tube ( h in the diagram) is proportional to the pressure of the gas in the container. An open-end manometer (Figure \(\PageIndex{3}\)) is the same as a closed-end manometer, but one of its arms is open to the atmosphere. In this case, the distance between the liquid levels corresponds to the difference in pressure between the gas in the container and the atmosphere. Figure \(\PageIndex{4}\): A manometer can be used to measure the pressure of a gas. The (difference in) height between the liquid levels (h) is a measure of the pressure. Mercury is usually used because of its large density. Example \(\PageIndex{3}\): Calculation of Pressure Using an Open-End Manometer The pressure of a sample of gas is measured at sea level with an open-end Hg (mercury) manometer, as shown below. Determine the pressure of the gas in: mm Hg atm kPa Solution The pressure of the gas equals the hydrostatic pressure due to a column of mercury of height 13.7 cm plus the pressure of the atmosphere at sea level. (The pressure at the bottom horizontal line is equal on both sides of the tube. The pressure on the left is due to the gas and the pressure on the right is due to 13.7 cm of Hg plus atmospheric pressure.) In mm Hg, this is: 137 mm Hg + 760 mm Hg = 897 mm Hg \(\mathrm{897\cancel{mm Hg}×\dfrac{1\: atm}{760\cancel{mm Hg}}=1.18\: atm}\) \(\mathrm{1.18\cancel{atm}×\dfrac{101.325\: kPa}{1\cancel{atm}}=1.20×10^2\:kPa}\) Exercise \(\PageIndex{3}\) The pressure of a sample of gas is measured at sea level with an open-end Hg manometer, as shown below Determine the pressure of the gas in: mm Hg atm kPa Answer a 642 mm Hg Answer b 0.845 atm Answer c 85.6 kPa Application: Measuring Blood Pressure Blood pressure is measured using a device called a sphygmomanometer (Greek sphygmos = “pulse”). It consists of an inflatable cuff to restrict blood flow, a manometer to measure the pressure, and a method of determining when blood flow begins and when it becomes impeded (Figure \(\PageIndex{5}\)). Since its invention in 1881, it has been an essential medical device. There are many types of sphygmomanometers: manual ones that require a stethoscope and are used by medical professionals; mercury ones, used when the most accuracy is required; less accurate mechanical ones; and digital ones that can be used with little training but that have limitations. When using a sphygmomanometer, the cuff is placed around the upper arm and inflated until blood flow is completely blocked, then slowly released. As the heart beats, blood forced through the arteries causes a rise in pressure. This rise in pressure at which blood flow begins is the systolic pressure— the peak pressure in the cardiac cycle. When the cuff’s pressure equals the arterial systolic pressure, blood flows past the cuff, creating audible sounds that can be heard using a stethoscope. This is followed by a decrease in pressure as the heart’s ventricles prepare for another beat. As cuff pressure continues to decrease, eventually sound is no longer heard; this is the diastolic pressure— the lowest pressure (resting phase) in the cardiac cycle. Blood pressure units from a sphygmomanometer are in terms of millimeters of mercury (mm Hg). Figure \(\PageIndex{5}\): (a) A medical technician prepares to measure a patient’s blood pressure with a sphygmomanometer. (b) A typical sphygmomanometer uses a valved rubber bulb to inflate the cuff and a diaphragm gauge to measure pressure. (credit a: modification of work by Master Sgt. Jeffrey Allen) Throughout the ages, people have observed clouds, winds, and precipitation, trying to discern patterns and make predictions: when it is best to plant and harvest; whether it is safe to set out on a sea voyage; and much more. We now face complex weather and atmosphere-related challenges that will have a major impact on our civilization and the ecosystem. Several different scientific disciplines use chemical principles to help us better understand weather, the atmosphere, and climate. These are meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric effects on earth’s weather. Meteorologists seek to understand and predict the weather in the short term, which can save lives and benefit the economy. Weather forecasts (Figure \(\PageIndex{5}\)) are the result of thousands of measurements of air pressure, temperature, and the like, which are compiled, modeled, and analyzed in weather centers worldwide. Figure \(\PageIndex{6}\): Meteorologists use weather maps to describe and predict weather. Regions of high (H) and low (L) pressure have large effects on weather conditions. The gray lines represent locations of constant pressure known as isobars. (credit: modification of work by National Oceanic and Atmospheric Administration) In terms of weather, low-pressure systems occur when the earth’s surface atmospheric pressure is lower than the surrounding environment: Moist air rises and condenses, producing clouds. Movement of moisture and air within various weather fronts instigates most weather events. The atmosphere is the gaseous layer that surrounds a planet. Earth’s atmosphere, which is roughly 100–125 km thick, consists of roughly 78.1% nitrogen and 21.0% oxygen, and can be subdivided further into the regions shown in Figure \(\PageIndex{7}\): the exosphere (furthest from earth, > 700 km above sea level), the thermosphere (80–700 km), the mesosphere (50–80 km), the stratosphere (second lowest level of our atmosphere, 12–50 km above sea level), and the troposphere (up to 12 km above sea level, roughly 80% of the earth’s atmosphere by mass and the layer where most weather events originate). As you go higher in the troposphere, air density and temperature both decrease. Figure \(\PageIndex{7}\): Earth’s atmosphere has five layers: the troposphere, the stratosphere, the mesosphere, the thermosphere, and the exosphere. Climatology is the study of the climate, averaged weather conditions over long time periods, using atmospheric data. However, climatologists study patterns and effects that occur over decades, centuries, and millennia, rather than shorter time frames of hours, days, and weeks like meteorologists. Atmospheric science is an even broader field, combining meteorology, climatology, and other scientific disciplines that study the atmosphere. Summary Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar. Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers. Key Equations \(P=\dfrac{F}{A}\) p = hρg Glossary atmosphere (atm) unit of pressure; 1 atm = 101,325 Pa bar (bar or b) unit of pressure; 1 bar = 100,000 Pa barometer device used to measure atmospheric pressure hydrostatic pressure pressure exerted by a fluid due to gravity manometer device used to measure the pressure of a gas trapped in a container pascal (Pa) SI unit of pressure; 1 Pa = 1 N/m 2 pounds per square inch (psi) unit of pressure common in the US pressure force exerted per unit area torr unit of pressure; \(\mathrm{1\: torr=\dfrac{1}{760}\,atm}\) Contributors Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ). |
Courses/Los_Angeles_Trade_Technical_College/Foundations_of_Introductory_Chemistry-1/21%3A_Biochemistry/21.12%3A_Phospholipids | If you were to go to the dentist to get a tooth pulled, you would not want to feel any pain. The dentist would inject an anesthetic into your gum to numb it. One theory as to why anesthetics work deals with the movement of ions across the cell membrane. The anesthetic gets into the membrane structure and causes shifts in how ions move across the membrane. If ion movement is disrupted, nerve impulses will not be transmitted and you will not sense pain—at least, not until the anesthetic wears off. Phospholipids A phospholipid is a lipid that contains a phosphate group and is a major component of cell membranes. A phospholipid consists of a hydrophilic (water-loving) head and hydrophobic (water-fearing) tail (see figure below). The phospholipid is essentially a triglyceride in which a fatty acid has been replaced by a phosphate group of some sort. Following the rule of "like dissolves like", the hydrophilic head of the phospholipid molecule dissolves readily in water. The long fatty acid chains of a phospholipid are nonpolar, and thus avoid water because of their insolubility. In water, phospholipids spontaneously form a double layer called a lipid bilayer, in which the hydrophobic tails of phospholipid molecules are sandwiched between two layers of hydrophilic heads (see figure below). In this way, only the heads of the molecules are exposed to the water, while the hydrophobic tails interact only with each other. Phospholipid bilayers are critical components of cell membranes. The lipid bilayer acts as a barrier to the passage of molecules and ions into and out of the cell. However, an important function of the cell membrane is to allow selective passage of certain substances into and out of cells. This is accomplished by the embedding of various protein molecules in and through the lipid bilayer (see figure below). These proteins form channels through which certain specific ions and molecules are able to move. Many membrane proteins also contain attached carbohydrates on the outside of the lipid bilayer, allowing it to form hydrogen bonds with water. Summary A phospholipid is a lipid that contains a phosphate group. A phospholipid consists of a hydrophilic (water-loving) head and hydrophobic (water-fearing) tail. In water, phospholipids spontaneously form a double layer called a lipid bilayer, in which the hydrophobic tails of phospholipid molecules are sandwiched between two layers of hydrophilic heads. Phospholipid bilayers are critical components of cell membranes. |
Courses/Madera_Community_College/Concepts_of_Physical_Science/00%3A_Front_Matter/02%3A_InfoPage | This text is disseminated via the Open Education Resource (OER) LibreTexts Project ( https://LibreTexts.org ) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact [email protected] . More information on our activities can be found via Facebook ( https://facebook.com/Libretexts ), Twitter ( https://twitter.com/libretexts ), or our blog ( http://Blog.Libretexts.org ). This text was compiled on 04/21/2025 |
Courses/Lumen_Learning/Book%3A_US_History_I_(OS_Collection)_(Lumen)/06%3A_Rule_Britannia!_The_English_Empire%2C_1660-1763/06.1%3A_Introduction | Isaac Royall and his family, seen here in a 1741 portrait by Robert Feke, moved to Medford, Massachusetts, from the West Indian island of Antigua, bringing their slaves with them. They were an affluent British colonial family, proud of their success and the success of the British Empire. The eighteenth century witnessed the birth of Great Britain (after the union of England and Scotland in 1707) and the expansion of the British Empire. By the mid-1700s, Great Britain had developed into a commercial and military powerhouse; its economic sway ranged from India, where the British East India Company had gained control over both trade and territory, to the West African coast, where British slave traders predominated, and to the British West Indies, whose lucrative sugar plantations, especially in Barbados and Jamaica, provided windfall profits for British planters. Meanwhile, the population rose dramatically in Britain’s North American colonies. In the early 1700s the population in the colonies had reached 250,000. By 1750, however, over a million British migrants and African slaves had established a near-continuous zone of settlement on the Atlantic coast from Maine to Georgia. During this period, the ties between Great Britain and the American colonies only grew stronger. Anglo-American colonists considered themselves part of the British Empire in all ways: politically, militarily, religiously (as Protestants), intellectually, and racially. The portrait of the Royall family exemplifies the colonial American gentry of the eighteenth century. Successful and well-to-do, they display fashions, hairstyles, and furnishings that all speak to their identity as proud and loyal British subjects. CC licensed content, Shared previously US History. Authored by : P. Scott Corbett, Volker Janssen, John M. Lund, Todd Pfannestiel, Paul Vickery, and Sylvie Waskiewicz. Provided by : OpenStax College. Located at : http://openstaxcollege.org/textbooks/us-history . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/content/col11740/latest/ |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/01%3A_Modules/1.10%3A_Gibbs_Energies/1.10.32%3A_Gibbs_Energies-_Liquid_Mixtures-_Typically_Aqueous_(TA) | For many binary aqueous liquid mixtures, the pattern shown by the molar excess thermodynamic parameters is \(\mathrm{G}_{\mathrm{m}}^{\mathrm{E}}>0\); \(\left|\mathrm{T} \, \mathrm{S}_{\mathrm{m}}^{\mathrm{E}}\right|>\left|\mathrm{H}_{\mathrm{m}}^{\mathrm{E}}\right|\). This pattern of excess molar properties defines TA mixtures. \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) is positive because the excess molar entropy of mixing is large in magnitude and negative in sign. In these terms mixing is dominated by the entropy change. The excess molar enthalpy of mixing is smaller in magnitude than either \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) or \(\mathrm{T} \, {\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) but exothermic in water-rich mixtures. The word ‘Typically’ in the description stems from observation that this pattern in thermodynamic variables is rarely shown by non-aqueous systems. At the time the classification was proposed [1], most binary aqueous liquid mixtures seemed to follow this pattern. Among the many examples of this class of system are aqueous mixtures formed by ethanol, 2-methyl propan-2-ol and cyclic ethers including tetrahydrofuran[2]. In water-rich mixtures, a large in magnitude but negative in sign \(\mathrm{T} \, {\mathrm{S}_{\mathrm{m}}}^{\mathrm{E}}\) produces a large (positive) \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\). For mixtures rich in the apolar component m endothermic mixing produces a positive \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\). The key point is that in m water-rich mixtures a positive \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) emerges from a negative \(\mathrm{T} \, {\mathrm{S}_{\mathrm{m}}}^{\mathrm{E}}\). With reference to volumetric properties of these system, the partial molar volume \(\mathrm{V}(\mathrm{ROH})\) for monohydric alcohols can be extrapolated to infinite dilution; i.e. \(\operatorname{limit}\left(\mathrm{x}_{2} \rightarrow 0\right) \mathrm{V}(\mathrm{ROH})=\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})^{\infty}\) where \(x_{2}\) is the mole fraction of alcohol. The difference, \(\left[\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})^{\infty}-\mathrm{V}^{*}(\mathrm{ROH} ; \ell)\right]\) is negative. In fact this pattern is observed for both TA and Typical Non-aqueous binary aqueous mixtures. Examples where this pattern is observed included aqueous mixtures formed by DMSO, \(\mathrm{H}_{2}\mathrm{O}_{2}\) and \(\mathrm{CH}_{3}\mathrm{CN}\). Significantly \(\left[\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})-\mathrm{V}^{*}(\mathrm{ROH} ; \ell)\right]\) is negative for TA mixtures, decreasing from \(\left[\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})^{\infty}-\mathrm{V}^{*}(\mathrm{ROH} ; \ell)\right]\) with increase in mole fraction of \(\mathrm{ROH}\), accompanying by a tendency to immiscibility. The initial decrease in \(\left[\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})-\mathrm{V}^{*}(\mathrm{ROH} ; \ell)\right]\) with increase in \(x_{2}\) is more dramatic the more hydrophobic the non-aqueous component; for 2-methyl propan-2-ol aqueous mixtures at \(298.2 \mathrm{~K}\) and ambient pressure, the minimum occurs at an alcohol mole fraction 0.04 (at \(298.2 \mathrm{~K}\)). Many explanations have been offered for the complicated patters shown by the dependence of \(\left[\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})-\mathrm{V}^{*}(\mathrm{ROH} ; \ell)\right]\) on mole fraction composition. In one model, the negative \(\left[\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})-\mathrm{V}^{*}(\mathrm{ROH} ; \ell)\right]\) at low mole fractions of \(\mathrm{ROH}\) is accounted for in terms of a liquid clathrate in which part of the hydrophobic R-group ‘occupies’ a guest site in the water lattice. The decrease in \(\left[\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})-\mathrm{V}^{*}(\mathrm{ROH} ; \ell)\right]\) is accounted for in terms of an increasing tendency towards a clathrate structure. But with increase in \(x_{2}\) there comes a point where there is insufficient water to construct a liquid clathrate water host. Hence \(\left[\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})-\mathrm{V}^{*}(\mathrm{ROH} ; \ell)\right]\) increases. An important characteristic of TA mixtures is a tendency towards and in some cases actual decrease in liquid miscibility with increase in temperature. At ambient \(\mathrm{T}\) and \(\mathrm{p}\), the mixture 2-methyl propan-2-ol + water is miscible (but only just!) in all molar proportions. The corresponding mixtures prepared using butan-1-ol and butan-2-ol are partially miscible. TA systems are therefore often characterised by a Lower Critical Solution Temperature LCST. In fact nearly all examples quoted in the literature of systems having an LCST involve water as one component; e.g. \(\mathrm{LCST} = 322 \mathrm{~K}\) for 2-butoxyethanol + water [3]. This tendency to partial miscibility is often signalled by the properties of the completely miscible systems. Returning to the patterns shown by relative partial molar volumes, \(\left[\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})-\mathrm{V}^{*}(\mathrm{ROH} ; \ell)\right]\), a stage is reached whereby with increase in mole fraction of the non-aqueous component, this property increases after a minimum. Other properties of the mixtures also change dramatically including a marked increase in \(\left(\alpha_{a} / v^{2}\right)\) where \(\alpha_{\mathrm{a}}\) is the amplitude attenuation constant and \(ν\) is the frequency of the sound wave in the MHz range; e.g. \(70 \mathrm{~MHz}\). Actually the pattern is complicated. Over the range of mixture mole fractions \(x_{2}\) where \(\left[\mathrm{V}(\mathrm{ROH} ; \mathrm{aq})-\mathrm{V}^{*}(\mathrm{ROH} ; \ell)\right]\) decreases with increase in \(x_{2}\), the ratio \(\left(\alpha_{a} / v^{2}\right)\) hardly changes although the speed of sound increases. At a mole fraction \(x_{2}\) characteristic of the temperature and the non-aqueous component, \(\left(\alpha_{a} / v^{2}\right)\) increases sharply, reaching a maximum where the mixture has a strong tendency to immiscibility. This interplay between in-phase and out-of-phase components of the complex isentropic compressibility when the mole fraction composition of the mixture is changed highlights the molecular complexity of these systems. By way of contrast the ratio \(\left(\alpha_{a} / v^{2}\right)\) for DMSO + water mixtures (a TNAN system) changes gradually when the mole fraction of DMSO is changed. For TA mixtures where \(\left(\alpha_{a} / v^{2}\right)\) is a maximum [4], other evidence points to the fact these mixtures are micro-heterogeneous; cf. excess molar isobaric heat capacities. Phase separation of the mixture 2-methyl propan-2-ol is observed when butane gas is dissolved in the liquid mixture. The miscibility curve shows an LCST near \(282 \mathrm{~K}\) [5]. Footnotes [1] F. Franks in Hydrogen-Bonded Solvent Systems, ed. A. K. Covington and P. Jones, Taylor and Francis, London,1968, pp.31-47. [2] The following references refer to properties of TA binary liquid mixtures. alcohol + water mixtures. Isobaric heat capacities. H. Ogawa and S. Murakami, Thermochim. Acta, 1986, 109 ,145. G. I. Makhatadze and P. L. Privalov, J. Solution Chem.,1989, 18 ,927. R. Arnaud, L. Avedikian and J.-P. Morel,J. Chim. Phys., 1972,45. fluoroalkanol + water mixtures. \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) R. Jadot and M.Fralha,J. Chem.Eng. Data,1988,33,237. J. Murto and A. Kiveninen, Suomen Kemist. Ser. B, 1967,40,258. Enthalpies M. Denda, H. Touhara and K. Nakanishi, J. Chem. Thermodyn., 1987, 19 ,539. A. Kivinen, J. Murto and A.Vhtala, Suomen Kemist.1967, 40 ,298. Volumes; J. Murto, A. Kivinen, S. Kivimaa and R. Laakso, Suomen Kemist., Ser. B, 1967, 40 ,250. amine + water mixtures \({\mathrm{X}_{\mathrm{m}}}^{\mathrm{E}}\) and miscibility; J. L. Copp and D. H. Everett, m Discuss. Faraday Soc.,1953, 15 ,174. Et3N + water; miscibility; A. Bellemans, J.Chem.Phys.,1953, 21 , 368. THF + water \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) J. Matous, J. P. Novak, J. Sobr and J. Pick, Collect. Czech. Chem.Commun.,1972, 37 ,2653. C. Treiner, J.-F. Bocquet and M. Chemla, J. Chim..Phys., 1973, 70 , 72. \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) for \(\mathrm{THF} + \mathrm{~D}_{2}\mathrm{O}\); J. Lejcek, J. Matous, J. P. Novak and J. Pick, J. Chem. Thermodyn., 1975, 7 ,927. Vapour composition; W. Hayduk, H. Laudie and O. H. Smith, J. Chem.Eng Data,1973, 18 ,373. \({\mathrm{H}_{\mathrm{m}}}^{\mathrm{E}}\) H. Nakayama and K. Shinoda, J. Chem. Thermodyn., 1971, 3 ,401. Activity of water K. L. Pinder, J. Chem. Eng. Data, 1973, 18 ,275. Thermal expansivities O. Kiyohara, P. J. D’Arcy and G. C. Benson, Can. J. Chem., 1978, 56 ,2803. \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) and \({\mathrm{V}_{\mathrm{m}}}^{\mathrm{E}}\). Signer, H. Arm and H. Daeniker, Helvetica Chimica Acta, 1969, 52 , 2347. 2-Methyl propan-2-ol + water In the chemical literature, 2-methyl propan-2-ol is often called t-butanol but as Prof. David J. G. Ives often pointed out, there is no organic compound, t-butane. \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\) and \({\mathrm{H}_{\mathrm{m}}}^{\mathrm{E}}\) Y. Koga, W. W. Y. Siu and T. Y. H. Wong, J. Phys. Chem., 1990, 94 ,7700. Enthalpies Y. Koga, Can. J.Chem.,1986, 64 ,206;1988, 66 ,1187,3171. Volumes A. Hvidt, R. Moss and G. Nielsen, Acta Chem. Scand.,Sect. B, 1978, B32 , 274. M. Sakurai, Bull. Chem. Soc. Jpn.,1987, 160 ,1. \(\mathrm{C}_{\mathrm{p}}\) and \(\mathrm{V}\) B. de Visser, G. Perron, and J. E. Desnoyers, Can J. Chem.,1977, 55 ,856. X-ray scattering K. Nishikawa, Y. Kodera and T. Iijima, J. Phys. Chem., 1987, 91 ,3694. Sound velocity H. Endo and O. Nomoto,Bull. Chem. Soc. Jpn., 1973, 46 , 3004. Isentropic compressibilities; J. Lara and J. E. Desnoyers, J. Solution Chem., 1981, 10 ,465. Propan-1-ol + water \({\mathrm{V}_{\mathrm{m}}}^{\mathrm{E}}\) G. C. Benson and O. Kiyohara, J. Solution Chem.,1980, 9 ,791. M. I. Davis, Thermochim. Acta, 1990, 157 ,295. C. De Visser, G. Perron and J. E. Desnoyers, Can. J.Chem.,1977, 55 ,856. Propanone + water \({\mathrm{X}_{\mathrm{m}}}^{\mathrm{E}}\) references; M. J. Blandamer, N. J. Blundell, J. Burgess, H. J. Cowles and I. M. Horn, J. Chem. Soc. Faraday Trans.,1990, 86 ,283. Methyl vinyl ketone + water LCST = \(301 \mathrm{~K}\); UCST = \(356 \mathrm{~K}\); J. Vojtko and M.Cihova, J. Chem. Eng. Data, 1972, 17 ,337. [3] F. Elizalde, J Gracia and M. Costas, J Phys.Chem.,1988, 93 ,3565. [4] M. J Blandamer and D. Waddington, Adv. Mol. Relax.Processes,1970, 2 ,1. [5] R. W. Cargill and D. E. MacPhee, J. Chem. Soc. Faraday Trans.1, 1989, 85 , 2665; an excellent observation! [5] \({\mathrm{G}_{\mathrm{m}}}^{\mathrm{E}}\); M. J. Blandamer, J. Burgess, A. Cooney, H. J. Cowles, I. M. Horn, K. M. Martin, K.W. Morcom and P. Warrick, J. Chem. Soc. Faraday Trans., 1990, 86 , 2209. |
Courses/Grand_Rapids_Community_College/CHM_120_-_Survey_of_General_Chemistry(Neils)/7%3A_Equilibrium_and_Thermodynamics/7.10%3A_Gibbs_Free_Energy | Skills to Develop To understand the relationship between the standard Gibbs free energy change, the extent of reaction, and equilibrium To understand the relationship between Gibbs free energy change and the direction in which a reaction will proceed to reach equilibrium To understand the relationship between Gibbs free energy and work. Gibbs Free-Energy The Gibbs free energy (\(G\)), often called simply free energy, was named in honor of J. Willard Gibbs (1838–1903), an American physicist who first developed the concept. It is defined in terms of three other state functions with which you are already familiar: enthalpy, temperature, and entropy: \[ G = H − TS \label{Eq1}\] Because it is a combination of state functions, \(G\) is also a state function. The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. Standard Gibbs Free-Energy Change We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. This set of conditions required for determining ΔG° is rarely, if ever, encountered when carrying out chemical reactions, so the literal meaning of ΔG° is of questionable value. However, the sign and magnitude of ΔG° are extremely helpful in describing whether the reaction is product-favored or reactant-favored when it has reached equilibrium, which is sometimes known as the extent of a chemical reaction. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation \(\ref{Eq2}\): \[ΔG° = ΔH° − TΔS° \label{Eq2}\] ΔG° < 0 means that the reaction is product-favored at equilibrium; the odds are in favor of making a significant amount of product at equilibrium. ΔG° > 0 means that the reaction is reactant-favored at equilibrium; the odds are against making a significant amount of product at equilibrium. It is important to recognize that a positive value of ΔG° for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that, at equilibrium, the concentrations of the products will be less than the concentrations of the reactants. ΔG° = 0 means that the reaction is neither product-favored nor reactant-favored at equilibrium; there are 50:50 odds that a significant amount of product will be made at equilibrium. This situation is rare for chemical reactions, although it can happen. This situation describes all phase changes when they occur at the exact phase change temperature. Exercise \(\PageIndex{1}\) Calculate the standard free-energy change (\(ΔG^o\)) at 25°C for the reaction \[2H_2(g)+N_2(g) \rightleftharpoons N_2H_4(l) \nonumber . \] Is the reaction product-favored at equilibrium as written at 25°C? Hint At 25°C, the standard enthalpy change (\(ΔH^o\)) is 50.6 kJ/mol, and the standard entropy change (\(ΔS^o\)) is -0.332 kJ/(mol•K) Answer 149.5 kJ/mol no, not product-favored, but reactant-favored, at equilibrium Video Solution Calculated values of ΔG° are used to predict whether a reaction will be product-favored at equilibrium when the reactants and products are mixed under standard conditions. Very few reactions are actually carried out under standard conditions, and calculated values of ΔG° may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG, notice the missing nought!) under the actual experimental conditions, which are usually quite different from ΔG°. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate ΔG° at various temperatures, as long as no phase change occurs over the temperature range being considered. We can also use the signs of ΔH° and ΔS° to determine the sign of ΔG° for a given reaction at various temperatures. Using Equation \(\ref{Eq2}\), the sign of ΔG° can be predicted for a reaction, using the signs of ΔH° and ΔS°: \[ΔG° = ΔH° − TΔS° \] 0 1 2 3 sign of ΔH° sign of ΔS° sign of ΔG° Meaning _ + _ The reaction is product-favored at equilibrium at all temperatures; always likely; odds always favor it + _ + The reaction is reactant-favored at equilibrium at all temperatures; never likely; odds are always against it + + -/+ The sign of ΔG° is temperature dependent. The reaction will be product-favored at relatively high temperatures. _ _ -/+ The sign of ΔG° is temperature dependent. The reaction will be product-favored at relatively low temperatures. NOTE: If ΔS° and ΔH° for a reaction have the same sign, then the sign of ΔG° depends on the temperature and thus the relative magnitudes of the ΔH° and TΔS° terms. If ΔS° and ΔH° for a reaction have the same sign, then there is a specific temperature at which the sign of ΔG° must equal 0 as the reaction shifts from being reactant-favored at equilibrium to being product-favored at equilibrium, and vice versa. The temperature at which this flip over occurs, at which a given reaction is neither product-favored nor reactant-favored, can be calculated by setting ΔG° = 0 in Equation \(\ref{Eq2}\), as illustrated in Example \(\PageIndex{2}\). Example \(\PageIndex{2}\) The reaction of nitrogen and hydrogen gas to produce ammonia \[3H_2(g)+N_2(g) \rightleftharpoons 2NH_3(l) \nonumber . \] is one in which ΔH° is −91.8 kJ and ΔS° is −198.1 J/K (both are negative). Such reactions are predicted to be product-favored at low temperatures but reactant-favored at high temperatures. Calculate the temperature at which this reaction changes from product-favored to reactant-favored, assuming that ΔH° and ΔS° are independent of temperature. Given : ΔH° and ΔS° Asked for : temperature at which reaction changes from product-favored to reactant-favored Strategy : Set ΔG° equal to zero in Equation \(\ref{Eq2}\) and solve for T, the temperature at which the reaction becomes reactant-favored. SOLUTION The temperature at which the reaction becomes reactant-favored at equilibrium is found by setting ΔG° equal to zero and rearranging Equation \(\ref{Eq2}\) to solve for T: \[\begin{align*}\Delta G^\circ &=\Delta H^\circ - T\Delta S^\circ=0 \\[5pt] \Delta H^\circ &=T\Delta S^\circ \\[5pt] T=\dfrac{\Delta H^\circ}{\Delta S^\circ}&=\dfrac{(-\textrm{91.8 kJ})(\textrm{1000 J/kJ})}{-\textrm{198.1 J/K}}=\textrm{463 K}\end{align*}\] This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K. Exercise \(\PageIndex{2}\) ΔH°= -116.2 kJ and ΔS° = - 146.6 J/K for the reaction of nitric oxide and oxygen to form nitrogen dioxide. \[2NO(g)+O_2(g) \rightleftharpoons 2NO_2(l) \nonumber . \] Use those data to calculate the temperature at which this reaction changes from product-favored to reactant-favored. Answer 792.6 K Video Solution Gibbs Free Energy and the Direction of Spontaneous Reactions A second major goal of chemical thermodynamics is to establish criteria for predicting whether a particular reaction or process will occur spontaneously, that is, if it will proceed in the forward direction, as written, to reach equilibrium. We have developed one such criterion, the change in entropy of the universe: if ΔS univ > 0 for a process or a reaction, then the process will go in the forward direction as written to reach equilibrium. Conversely, if ΔS univ < 0, a process will go in the reverse direction as written to reach equilibrium; if ΔS univ = 0, the system is already at equilibrium. The sign of ΔS univ is a universally applicable and infallible indicator of the direction of a reaction as it proceeds towards equilibrium. Unfortunately, using ΔS univ requires that we calculate ΔS for both a system and its surroundings. This calculation is not particularly useful for two reasons: we are normally much more interested in the system than in the surroundings, and it is difficult to make quantitative measurements of the surroundings (i.e., the rest of the universe). A criterion of spontaneity that is based solely on the state functions of a system would be much more convenient, and is provided by the Gibbs free energy. The criterion for predicting the likely direction for the reaction to occur to reach equilibrium is based on (\(ΔG\)), the change in \(G\), at constant temperature and pressure. Although very few chemical reactions actually occur under conditions of constant temperature and pressure, most systems can be brought back to the initial temperature and pressure without significantly affecting the value of thermodynamic state functions such as \(G\). At constant temperature and pressure, \[ ΔG = ΔH − TΔS \label{Eq3}\] where all thermodynamic quantities are those of the system. The relationship between the entropy change of the surroundings and the heat gained or lost by the system provides the key connection between the thermodynamic properties of the system and the change in entropy of the universe. The relationship shown in Equation \(\ref{Eq3}\) allows us to predict spontaneity by focusing exclusively on the thermodynamic properties and temperature of the system. We predict that highly exothermic processes (\(ΔH \ll 0\)) that increase the disorder of a system (\(ΔS_{sys} \gg 0\)) would therefore occur spontaneously. An example of such a process is the decomposition of ammonium nitrate fertilizer. Ammonium nitrate was also used to destroy the Murrah Federal Building in Oklahoma City, Oklahoma, in 1995. For a system at constant temperature and pressure, we can summarize the following results: If \(ΔG < 0\), the process occurs spontaneously, that is, it proceeds in the forward direction, as written, until it reaches equilibrium. If \(ΔG = 0\), the system is at equilibrium. If \(ΔG > 0\), the process is not spontaneous as written but occurs spontaneously in the reverse direction, until it reaches equilibrium. To further understand how the various components of ΔG dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor. If this process is carried out at 1 atm and the normal boiling point of 100.00°C (373.15 K), we can calculate ΔG from the experimentally measured value of ΔH vap (40.657 kJ/mol). For vaporizing 1 mol of water, \(ΔH = 40,657; J\), so the process is highly endothermic. The value of ΔS is \(\textrm{108.96 J/K} \) Hence there is an increase in the disorder of the system. At the normal boiling point of water, \[\begin{align*}\Delta G_{100^\circ\textrm C}&=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C} \\[5pt] &=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})] \\[5pt] &=\textrm{0 J}\end{align*}\] The energy required for vaporization offsets the increase in disorder of the system. Thus ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions. Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of ΔG for the vaporization of 1 mol of water at 110°C, assuming that ΔH and ΔS do not change significantly with temperature, becomes \[\begin{align*}\Delta G_{110^\circ\textrm C}&=\Delta H-T\Delta S \\[5pt] &=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})] \\[5pt] &=-\textrm{1091 J}\end{align*}\] At 110°C, \(ΔG < 0\), and vaporization is predicted to occur spontaneously and irreversibly. We can also calculate \(ΔG\) for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions: \[\begin{align*}\Delta G_{90^\circ\textrm C}&=\Delta H-T\Delta S \\[5pt] &=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})] \\[5pt] &=\textrm{1088 J}\end{align*}\] At 90°C, ΔG > 0, and water does not spontaneously convert to water vapor. When using all the digits in the calculator display in carrying out our calculations, ΔG 110°C = 1090 J = −ΔG 90°C , as we would predict. Relating Enthalpy and Entropy changes under Equilibrium Conditions \(ΔG = 0\) only if \(ΔH = TΔS\). We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of ΔH and ΔS into the definition of ΔG (Equation \(\ref{Eq2}\)), setting \(ΔG = 0\), and solving for \(T\), 0 J =40,657 J−T(108.96 J/K) T=373.15 K Thus \(ΔG = 0\) at T = 373.15 K and 1 atm, which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water. At temperatures greater than 373.15 K, \(ΔG\) is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, \(ΔG\) is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure \(\PageIndex{1}\) shows how the \(ΔH\) and \(TΔS\) terms vary with temperature for the vaporization of water. When the two lines cross, \(ΔG = 0\), and \(ΔH = TΔS\). Figure \(\PageIndex{1}\): Temperature Dependence of ΔH and TΔS for the Vaporization of Water. Both ΔH and TΔS are temperature dependent, but the lines have opposite slopes and cross at 373.15 K at 1 atm, where ΔH = TΔS. Because ΔG = ΔH − TΔS, at this temperature ΔG = 0, indicating that the liquid and vapor phases are in equilibrium. The normal boiling point of water is therefore 373.15 K. Above the normal boiling point, the TΔS term is greater than ΔH, making ΔG < 0; hence, liquid water evaporates spontaneously. Below the normal boiling point, the ΔH term is greater than TΔS, making ΔG > 0. Thus liquid water does not evaporate spontaneously, but water vapor spontaneously condenses to liquid. The Relationship between ΔG and Work In the previous subsection, we learned that the value of ΔG allows us to predict the spontaneity of a physical or a chemical change. In addition, the magnitude of ΔG for a process provides other important information. The change in free energy (ΔG) is equal to the maximum amount of work that a system can perform on the surroundings while undergoing a spontaneous change (at constant temperature and pressure): ΔG = w max . To see why this is true, let’s look again at the relationships among free energy, enthalpy, and entropy expressed in Equation \(\ref{Eq2}\). We can rearrange this equation as follows: \[ ΔH = ΔG + TΔS \label{Eq4}\] This equation tells us that when energy is released during an exothermic process (ΔH < 0), such as during the combustion of a fuel, some of that energy can be used to do work (ΔG < 0), while some is used to increase the entropy of the universe (TΔS > 0). Only if the process occurs infinitely slowly in a perfectly reversible manner will the entropy of the universe be unchanged. (For more information on entropy and reversibility, see the previous section). Because no real system is perfectly reversible, the entropy of the universe increases during all processes that produce energy. As a result, no process that uses stored energy can ever be 100% efficient; that is, ΔH will never equal ΔG because ΔS has a positive value. One of the major challenges facing engineers is to maximize the efficiency of converting stored energy to useful work or converting one form of energy to another. As indicated in Table \(\PageIndex{1}\), the efficiencies of various energy-converting devices vary widely. For example, an internal combustion engine typically uses only 25%–30% of the energy stored in the hydrocarbon fuel to perform work; the rest of the stored energy is released in an unusable form as heat. In contrast, gas–electric hybrid engines, now used in several models of automobiles, deliver approximately 50% greater fuel efficiency. A large electrical generator is highly efficient (approximately 99%) in converting mechanical to electrical energy, but a typical incandescent light bulb is one of the least efficient devices known (only approximately 5% of the electrical energy is converted to light). In contrast, a mammalian liver cell is a relatively efficient machine and can use fuels such as glucose with an efficiency of 30%–50%. Device Energy Conversion Approximate Efficiency (%) large electrical generator mechanical → electrical 99 chemical battery chemical → electrical 90 home furnace chemical → heat 65 small electric tool electrical → mechanical 60 space shuttle engine chemical → mechanical 50 mammalian liver cell chemical → chemical 30–50 spinach leaf cell light → chemical 30 internal combustion engine chemical → mechanical 25–30 fluorescent light electrical → light 20 solar cell light → electricity 10-20 incandescent light bulb electricity → light 5 yeast cell chemical → chemical 2–4 Summary The change in Gibbs free energy, which is based solely on changes in state functions, is the criterion for describing two aspects of a reaction. Standard free-energy change: \[ΔG° = ΔH° − TΔS° \nonumber\] The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The sign of ΔG° tells us if a reaction will be reactant-favored (+ ΔG°) or product-favored ( - ΔG°) at equilibrium. The magnitude of ΔG° tells us the extent of the reaction. A large +ΔG° tells us the reaction makes very little product at equilibrium. A large - ΔG° tells us the reaction makes a great deal of product at equilibrium. Free-energy change: \[ΔG = ΔH − TΔS \nonumber\] The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written, but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. Contributors Modified by Tom Neils (Grand Rapids Community College) |
Courses/Chabot_College/Introduction_to_General_Organic_and_Biochemistry/07%3A_Chemical_Reactions_and_Quantities/7.07%3A_Gram-Mole_Conversions | Learning Objectives To convert between mass units and mole units. As we just discussed, molar mass is defined as the mass (in grams) of 1 mole of substance (or Avogadro's number of molecules or formula units). The simplest type of manipulation using molar mass as a conversion factor is a mole-gram conversion (or its reverse, a gram-mole conversion). We also established that 1 mol of Al has a mass of 26.98 g (Example \(\PageIndex{1}\)). Stated mathematically, 1 mol Al = 26.98 g Al We can divide both sides of this expression by either side to get one of two possible conversion factors: \[\mathrm{\dfrac{1\: mol\: Al}{26.98\: g\: Al}\quad and \quad \dfrac{26.98\: g\: Al}{1\: mol\: Al}} \nonumber\] The first conversion factor can be used to convert from mass to moles, and the second converts from moles to mass. Both can be used to solve problems that would be hard to do “by eye.” Example \(\PageIndex{1}\) What is the mass of 3.987 mol of Al? Solution The first step in a conversion problem is to decide what conversion factor to use. Because we are starting with mole units, we want a conversion factor that will cancel the mole unit and introduce the unit for mass in the numerator. Therefore, we should use the \(\mathrm{\dfrac{26.98\: g\: Al}{1\: mol\: Al}}\) conversion factor. We start with the given quantity and multiply by the conversion factor: \(\mathrm{3.987\: mol\: Al\times\dfrac{26.98\: g\: Al}{1\: mol\: Al}}\) Note that the mol units cancel algebraically. (The quantity 3.987 mol is understood to be in the numerator of a fraction that has 1 in the unwritten denominator.) Canceling and solving gives \(\mathrm{3.987\: mol\: Al\times \dfrac{26.98\: g\: Al}{1\: mol\: Al}=107.6\: g\: Al}\) Our final answer is expressed to four significant figures. Exercise \(\PageIndex{1}\) How many moles are present in 100.0 g of Al? (Hint: you will have to use the other conversion factor we obtained for aluminum.) Answer \(\mathrm{100.0\: g\: Al\times \dfrac{1\: mol\: Al}{26.98\: g\: Al}=3.706\: mol\: Al}\) Conversions like this are possible for any substance, as long as the proper atomic mass, formula mass, or molar mass is known (or can be determined) and expressed in grams per mole. Figure \(\PageIndex{1}\) is a chart for determining what conversion factor is needed, and Figure \(\PageIndex{2}\) is a flow diagram for the steps needed to perform a conversion. Example \(\PageIndex{2}\) A biochemist needs 0.00655 mol of bilirubin (C 33 H 36 N 4 O 6 ) for an experiment. How many grams of bilirubin will that be? Solution To convert from moles to mass, we need the molar mass of bilirubin, which we can determine from its chemical formula: 0 1 2 33 C molar mass: 33 × 12.01 g = 396.33 g 36 H molar mass: 36 × 1.01 g = 36.36 g 4 N molar mass: 4 × 14.01 g = 56.04 g 6 O molar mass: 6 × 16.00 g = 96.00 g Total: NaN 584.73 g The molar mass of bilirubin is 584.73 g. Using the relationship 1 mol bilirubin = 584.73 g bilirubin we can construct the appropriate conversion factor for determining how many grams there are in 0.00655 mol. Following the steps from Figure \(\PageIndex{2}\): \(\mathrm{0.00655\: mol\: bilirubin \times \dfrac{584.73\: g\: bilirubin}{mol\: bilirubin}=3.83\: g\: bilirubin}\) The mol bilirubin unit cancels. The biochemist needs 3.83 g of bilirubin. Exercise \(\PageIndex{2}\) A chemist needs 457.8 g of KMnO 4 to make a solution. How many moles of KMnO 4 is that? Answer \(\mathrm{457.8\: g\: KMnO_4\times \dfrac{1\: mol\: KMnO_4}{158.04\: g\: KMnO_4}=2.897\: mol\: KMnO_4}\) To Your Health: Minerals For our bodies to function properly, we need to ingest certain substances from our diets. Among our dietary needs are minerals, the noncarbon elements our body uses for a variety of functions, such developing bone or ensuring proper nerve transmission. The US Department of Agriculture has established some recommendations for the RDIs of various minerals. The accompanying table lists the RDIs for minerals, both in mass and moles, assuming a 2,000-calorie daily diet. Mineral Male (age 19–30 y) Male (age 19–30 y).1 Female (age 19–30 y) Female (age 19–30 y).1 Ca 1,000 mg 0.025 mol 1,000 mg 0.025 mol Cr 35 µg 6.7 × 10−7 mol 25 µg 4.8 × 10−7 mol Cu 900 µg 1.4 × 10−5 mol 900 µg 1.4 × 10−5 mol F 4 mg 2.1 × 10−4 mol 3 mg 1.5 × 10−4 mol I 150 µg 1.2 × 10−6 mol 150 µg 1.2 × 10−6 mol Fe 8 mg 1.4 × 10−4 mol 18 mg 3.2 × 10−4 mol K 3,500 mg 9.0 × 10−2 mol 3,500 mg 9.0 × 10−2 mol Mg 400 mg 1.6 × 10−2 mol 310 mg 1.3 × 10−2 mol Mn 2.3 mg 4.2 × 10−5 mol 1.8 mg 3.3 × 10−5 mol Mo 45 mg 4.7 × 10−7 mol 45 mg 4.7 × 10−7 mol Na 2,400 mg 1.0 × 10−1 mol 2,400 mg 1.0 × 10−1 mol P 700 mg 2.3 × 10−2 mol 700 mg 2.3 × 10−2 mol Se 55 µg 7.0 × 10−7 mol 55 µg 7.0 × 10−7 mol Zn 11 mg 1.7 × 10−4 mol 8 mg 1.2 × 10−4 mol Table \(\PageIndex{1}\) illustrates several things. First, the needs of men and women for some minerals are different. The extreme case is for iron; women need over twice as much as men do. In all other cases where there is a different RDI, men need more than women. Second, the amounts of the various minerals needed on a daily basis vary widely—both on a mass scale and a molar scale. The average person needs 0.1 mol of Na a day, which is about 2.5 g. On the other hand, a person needs only about 25–35 µg of Cr per day, which is under one millionth of a mole. As small as this amount is, a deficiency of chromium in the diet can lead to diabetes-like symptoms or neurological problems, especially in the extremities (hands and feet). For some minerals, the body does not require much to keep itself operating properly. Although a properly balanced diet will provide all the necessary minerals, some people take dietary supplements. However, too much of a good thing, even minerals, is not good. Exposure to too much chromium, for example, causes a skin irritation, and certain forms of chromium are known to cause cancer (as presented in the movie Erin Brockovich ). |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Chemistry_with_Applications_in_Spectroscopy_(Fleming)/09%3A_Molecules/9.07%3A_Vibronic_Transitions | Just as rotational motion is important in understanding vibrational spectra, vibrational (as well as rotational) motion(s) are important in understanding electronic transition is molecules. Electronic transitions in which vibrational structure is resolved are sometimes referred to a vibronic transition . When rotation is thrown in to the mix, the term “ rovibronic transitions ” is sometimes used. Vibronic transitions can be discussed in terms of the transition moment. Keeping in mind that the wavefunction for a vibronic state can be expressed as a product \[\Psi _{tot} = \psi _{elec} \psi _{vib}\nonumber \] and that the transition moment is given by \[\int \Psi _{tot}^{*} \vec{\mu }\, \Psi _{tot} d\tau\nonumber \] Substitution yields \[\int \left(\psi _{elec} \psi _{vib} \right)^{*} \vec{\mu }\, \left(\psi _{elec} \psi _{vib} \right) d\tau\nonumber \] Since the dipole moment operator is a derivative operator, the chain rule must be employed, which yields \[\int \psi _{elec}^{*} \psi _{elec} d\tau \int \psi _{vib}^{*} \vec{\mu }\psi _{vib} d\tau +\int \psi _{elec}^{*} \vec{\mu }\psi _{elec} d\tau \int \psi _{vib}^{*} \psi _{vib} d\tau\nonumber \] Since the electronic wavefunction must be orthogonal, the first term will vanish for transitions between two different electronic states. The second term however, does not vanish. In face, the magnitude of the \(\int \psi _{vib}^{*} \psi _{vib} d\tau\) will be determined by the overlap of the two vibrational levels. (Note that since these represent vibrational wavefunctions in different electronic state, there is no reason for the wavefunctions to be orthogonal.) Franck-Condon Factors The intensity of a band in a vibronic transition will be governed by the magnitude of the Frank-Condon Factor for the band. The Franck-Condon factor (FCF) is defined by \[FCF = \left[\int \psi _{vib}^{'} \psi _{vib}^{"} d\tau \right]^{2}\nonumber \] which is governed purely by the degree of overlap between the upper state vibrational wavefunction and that in the lower state. The overlap will be large for \(\Delta v = 0\) if the potential energy functions of the upper and lower states are similar (similar \(\omega _{e}\), \(\omega _{e} x _{e}\), \(r _{e}\), etc.) and strong sequences will be observed in the spectrum. If, however, the equilibrium bond length changes significantly, the maximum Franck-Condon overlap will occur for combinations of v’ and v” for which \(\Delta v \neq 0\). In these cases, strong progressions will be observed. The Franck-Condon principle is closely associated with the Born-Oppenheimer approximation . In cases where the Born-Oppenheimer breaks down, the Franck-Condon principle is compromised as well. |
Courses/Brevard_College/CHE_103_Principles_of_Chemistry_I/08%3A_Chemical_Reactions/8.09%3A_Limiting_Reagents | Learning Objectives Identify a limiting reagent from a set of reactants. Calculate how much product will be produced from the limiting reagent. Calculate how much reactant(s) remains when the reaction is complete. In addition to the assumption that reactions proceed all the way to completion, one additional assumption we have made about chemical reactions is that all the reactants are present in the proper quantities to react to products; this is not always the case. In Figure \(\PageIndex{2}\) we are taking hydrogen atoms and oxygen atoms (left) to make water molecules (right). However, there are not enough oxygen atoms to use up all the hydrogen atoms. We run out of oxygen atoms and cannot make any more water molecules, so the process stops when we run out of oxygen atoms. A similar situation exists for many chemical reactions: you usually run out of one reactant before all of the other reactant has reacted. The reactant you run out of is called the limiting reagent; the other reactant or reactants are considered to be in excess . A crucial skill in evaluating the conditions of a chemical process is to determine which reactant is the limiting reagent and which is in excess. The key to recognizing which reactant is the limiting reagent is based on a mole-mass or mass-mass calculation: whichever reactant gives the lesser amount of product is the limiting reagent. What we need to do is determine an amount of one product (either moles or mass), assuming all of each reactant reacts. Whichever reactant gives the least amount of that particular product is the limiting reagent. It does not matter which product we use, as long as we use the same one each time. It does not matter whether we determine the number of moles or grams of that product; however, we will see shortly that knowing the final mass of product can be useful. For example, consider this reaction: \[4As(s) + 3O_2(g) → 2As_2O_3(s)\nonumber \] Suppose we start a reaction with 50.0 g of As and 50.0 g of O 2 . Which one is the limiting reagent? We need to perform two mole-mass calculations, each assuming that each reactant reacts completely. Then we compare the amount of the product produced by each and determine which is less. The calculations are as follows: \[50.0\cancel{g\, As}\times \frac{1\cancel{mol\, As}}{74.92\cancel{g\, As}}\times \frac{2\, mol\, As_{2}O_{3}}{4\cancel{mol\, As}}=0.334\, mol\, As_{2}O_{3}\nonumber \] \[50.0\cancel{g\, O_{2}}\times \frac{1\cancel{mol\, O_{2}}}{32.00\cancel{g\, O_{2}}}\times \frac{2\, mol\, As_{2}O_{3}}{3\cancel{mol\, O_{2}}}=1.04\, mol\, As_{2}O_{3}\nonumber \] Comparing these two answers, it is clear that 0.334 mol of As 2 O 3 is less than 1.04 mol of As 2 O 3 , so arsenic is the limiting reagent. If this reaction is performed under these initial conditions, the arsenic will run out before the oxygen runs out. We say that the oxygen is "in excess." Identifying the limiting reagent, then, is straightforward. However, there are usually two associated questions: (1) what mass of product (or products) is then actually formed? and (2) what mass of what reactant is left over? The first question is straightforward to answer: simply perform a conversion from the number of moles of product formed to its mass, using its molar mass. For As 2 O 3 , the molar mass is 197.84 g/mol; knowing that we will form 0.334 mol of As 2 O 3 under the given conditions, we will get \[0.334\cancel{mol\, As_{2}O_{3}}\times \frac{197.84\, g\, As_{2}}{\cancel{1\, mol\, As_{2}O_{3}}}=66.1\, g\, As_{2}O_{3}\nonumber \] The second question is somewhat more convoluted to answer. First, we must do a mass-mass calculation relating the limiting reagent (here, As) to the other reagent (O 2 ). Once we determine the mass of O 2 that reacted, we subtract that from the original amount to determine the amount left over. According to the mass-mass calculation, \[50.0\cancel{g\, As}\times \frac{1\cancel{mol\, As}}{74.92\cancel{g\, As}}\times \frac{3\cancel{mol\, O_{2}}}{4\cancel{mol\, As}}\times \frac{32.00\, g\, O_{2}}{\cancel{1\, mol\, O_{2}}}=16.0\, g\, O_{2}\; reacted\nonumber \] Because we reacted 16.0 g of our original O 2 , we subtract that from the original amount, 50.0 g, to get the mass of O 2 remaining: 50.0 g O 2 − 16.0 g O 2 reacted = 34.0 g O 2 left over You must remember to perform this final subtraction to determine the amount remaining; a common error is to report the 16.0 g as the amount remaining. Example \(\PageIndex{1}\) A 5.00 g quantity of Rb is combined with 3.44 g of MgCl 2 according to this chemical reaction: \[2R b(s) + MgCl_2(s) → Mg(s) + 2RbCl(s) \nonumber \nonumber \] What mass of Mg is formed, and what mass of what reactant is left over? Solution Because the question asks what mass of magnesium is formed, we can perform two mass-mass calculations and determine which amount is less. \[5.00\cancel{g\, Rb}\times \frac{1\cancel{mol\, Rb}}{85.47\cancel{g\, Rb}}\times \frac{1\cancel{mol\, Mg}}{2\cancel{mol\, Rb}}\times \frac{24.31\, g\, Mg}{\cancel{1\, mol\, Mg}}=0.711\, g\, Mg \nonumber \] \[3.44\cancel{g\, MgCl_{2}}\times \frac{1\cancel{mol\, MgCl_{2}}}{95.21\cancel{g\, MgCl_{2}}}\times \frac{1\cancel{mol\, Mg}}{1\cancel{mol\, MgCl_{2}}}\times \frac{24.31\, g\, Mg}{\cancel{1\, mol\, Mg}}=0.878\, g\, Mg \nonumber \] The 0.711 g of Mg is the lesser quantity, so the associated reactant—5.00 g of Rb—is the limiting reagent. To determine how much of the other reactant is left, we have to do one more mass-mass calculation to determine what mass of MgCl2 reacted with the 5.00 g of Rb, and then subtract the amount reacted from the original amount. \[5.00\cancel{g\, Rb}\times \frac{1\cancel{mol\, Rb}}{85.47\cancel{g\, Rb}}\times \frac{1\cancel{mol\, MgCl_{2}}}{2\cancel{mol\, Rb}}\times \frac{95.21\, g\, Mg}{\cancel{1\, mol\, MgCl_{2}}}=2.78\, g\, MgCl_{2}\: \: reacted \nonumber \] Because we started with 3.44 g of MgCl2, we have 3.44 g MgCl2 − 2.78 g MgCl2 reacted = 0.66 g MgCl2 left Exercise \(\PageIndex{1}\) Given the initial amounts listed, what is the limiting reagent, and what is the mass of the leftover reagent? \[\underbrace{22.7\, g}_{MgO(s)}+\underbrace{17.9\, g}_{H_2S}\rightarrow MgS(s)+H_{2}O(l) \nonumber \] Answer H 2 S is the limiting reagent; 1.5 g of MgO are left over. Summary The limiting reagent is the reactant that produces the least amount of product. Mass-mass calculations can determine how much product is produced and how much of the other reactants remain. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/09%3A_Mixtures/9.01%3A_Composition_Variables | \( \newcommand{\tx}[1]{\text{#1}} % text in math mode\) \( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\) \( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\) \( \newcommand{\st}{^\circ} % standard state symbol\) \( \newcommand{\id}{^{\text{id}}} % ideal\) \( \newcommand{\rf}{^{\text{ref}}} % reference state\) \( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\) \( \newcommand{\K}{\units{K}} % kelvins\) \( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\) \( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\) \( \newcommand{\Pa}{\units{Pa}}\) \( \newcommand{\mol}{\units{mol}} % mole\) \( \newcommand{\V}{\units{V}} % volts\) \( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\) \( \newcommand{\per}{^{-1}} % minus one power\) \( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\) \( 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superscript\) \( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\) \( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\) \( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\) \( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\) \( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\) \( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\) \( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\) \( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \) \( \newcommand{\dq}{\dBar q} % heat differential\) \( \newcommand{\dw}{\dBar w} % work differential\) \( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\) \( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\) \( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\) \( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\) \( \newcommand{\Del}{\Delta}\) \( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\) \( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\) \( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\) \( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\) \( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\) \( \newcommand{\dotprod}{\small\bullet}\) \( \newcommand{\fug}{f} % fugacity\) \( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\) \( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\) \( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\) \( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\) \( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\) \( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\) \( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\) \( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\) \( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\) \( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\) \( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\) \( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \) \( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \) \( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) A composition variable is an intensive property that indicates the relative amount of a particular species or substance in a phase. 9.1.1 Species and substances We sometimes need to make a distinction between a species and a substance. A species is any entity of definite elemental composition and charge and can be described by a chemical formula, such as H\(_2\)O, H\(_3\)O\(^+\), NaCl, or Na\(^+\). A substance is a species that can be prepared in a pure state (e.g., N\(_2\) and NaCl). Since we cannot prepare a macroscopic amount of a single kind of ion by itself, a charged species such as H\(_3\)O\(^+\) or Na\(^+\) is not a substance. Chap. 10 will discuss the special features of mixtures containing charged species. 9.1.2 Mixtures in general The mole fraction of species \(i\) is defined by \begin{gather} \s{ x_i \defn \frac{n_i}{\sum_j n_j} \qquad \tx{or} \qquad y_i \defn \frac{n_i}{\sum_j n_j} } \tag{9.1.1} \cond{(\(P{=}1\))} \end{gather} where \(n_i\) is the amount of species \(i\) and the sum is taken over all species in the mixture. The symbol \(x_i\) is used for a mixture in general, and \(y_i\) is used when the mixture is a gas. The mass fraction , or weight fraction, of species \(i\) is defined by \begin{gather} \s{ w_i \defn \frac{m(i)}{m} = \frac{n_i M_i}{\sum_j n_j M_j} } \tag{9.1.2} \cond{(\(P{=}1\))} \end{gather} where \(m(i)\) is the mass of species \(i\) and \(m\) is the total mass. The concentration , or molarity, of species \(i\) in a mixture is defined by \begin{gather} \s{ c_i \defn \frac{n_i}{V} } \tag{9.1.3} \cond{(\(P{=}1\))} \end{gather} The symbol M is often used to stand for units of mol L\(^{-1}\), or mol dm\(^{-3}\). Thus, a concentration of \(0.5\units{M}\) is \(0.5\) moles per liter, or \(0.5\) molar. Concentration is sometimes called “amount concentration” or “molar concentration” to avoid confusion with number concentration (the number of particles per unit volume). An alternative notation for \(c\A\) is [A]. A binary mixture is a mixture of two substances. 9.1.3 Solutions A solution , strictly speaking, is a mixture in which one substance, the solvent , is treated in a special way. Each of the other species comprising the mixture is then a solute . The solvent is denoted by A and the solute species by B, C, and so on. (Some chemists denote the solvent by subscript \(1\) and use \(2\), \(3\), and so on for solutes.) Although in principle a solution can be a gas mixture, in this section we will consider only liquid and solid solutions. We can prepare a solution of varying composition by gradually mixing one or more solutes with the solvent so as to continuously increase the solute mole fractions. During this mixing process, the physical state (liquid or solid) of the solution remains the same as that of the pure solvent. When the sum of the solute mole fractions is small compared to \(x\A\) (i.e., \(x\A\) is close to unity), the solution is called dilute . As the solute mole fractions increase, we say the solution becomes more concentrated . Mole fraction, mass fraction, and concentration can be used as composition variables for both solvent and solute, just as they are for mixtures in general. A fourth composition variable, molality, is often used for a solute. The molality of solute species B is defined by \begin{gather} \s{ m\B \defn \frac{n\B}{m(\tx{A})} } \tag{9.1.4} \cond{(solution)} \end{gather} where \(m(\tx{A})=n\A M\A\) is the mass of solvent. The symbol m is sometimes used to stand for units of mol kg\(^{-1}\), although this should be discouraged because m is also the symbol for meter. For example, a solute molality of \(0.6\units{m}\) is \(0.6\) moles of solute per kilogram of solvent, or \(0.6\) molal. 9.1.4 Binary solutions We may write simplified equations for a binary solution of two substances, solvent A and solute B. Equations 9.1.1–9.1.4 become \begin{gather} \s{ x\B = \frac{n\B}{n\A + n\B} } \tag{9.1.5} \cond{(binary solution)} \end{gather} \begin{gather} \s{ w\B = \frac{n\B M\B}{n\A M\A + n\B M\B} } \tag{9.1.6} \cond{(binary solution)} \end{gather} \begin{gather} \s{ c\B = \frac{n\B}{V} = \frac{n\B \rho}{n\A M\A + n\B M\B} } \tag{9.1.7} \cond{(binary solution)} \end{gather} \begin{gather} \s{ m\B = \frac{n\B}{n\A M\A} } \tag{9.1.8} \cond{(binary solution)} \end{gather} The right sides of Eqs. 9.1.5–9.1.8 express the solute composition variables in terms of the amounts and molar masses of the solvent and solute and the density \(\rho\) of the solution. To be able to relate the values of these composition variables to one another, we solve each equation for \(n\B\) and divide by \(n\A\) to obtain an expression for the mole ratio \(n\B/n\A\): \begin{gather} \s{\tx{from Eq. 9.1.5}} \tag{9.1.9} \qquad \s{\frac{n\B}{n\A} = \frac{x\B}{1-x\B}} \cond{(binary solution)} \end{gather} \begin{gather} \s{\tx{from Eq. 9.1.6}} \tag{9.1.10} \qquad \s{\frac{n\B}{n\A} = \frac{M\A w\B}{M\B(1-w\B)}} \cond{(binary solution)} \end{gather} \begin{gather} \s{\tx{from Eq. 9.1.7}} \tag{9.1.11} \qquad \s{\frac{n\B}{n\A} = \frac{M\A c\B}{\rho - M\B c\B}} \cond{(binary solution)} \end{gather} \begin{gather} \s{\tx{from Eq. 9.1.8}} \tag{9.1.12} \qquad \s{\frac{n\B}{n\A} = M\A m\B} \cond{(binary solution)} \end{gather} These expressions for \(n\B/n\A\) allow us to find one composition variable as a function of another. For example, to find molality as a function of concentration, we equate the expressions for \(n\B/n\A\) on the right sides of Eqs. 9.1.11 and 9.1.12 and solve for \(m\B\) to obtain \begin{equation} m\B = \frac{c\B}{\rho - M\B c\B} \tag{9.1.13} \end{equation} A binary solution becomes more dilute as any of the solute composition variables becomes smaller. In the limit of infinite dilution, the expressions for \(n\B/n\A\) become: \begin{gather} \s{\begin{split} \frac{n\B}{n\A} & = x\B \cr & = \frac{M\A}{M\B}w\B \cr & = \frac{M\A}{\rho\A^*}c\B = V\mA^* c\B \cr & = \s{ M\A m\B } \end{split}} \tag{9.1.14} \cond{(binary solution at} \nextcond{infinite dilution)} \end{gather} where a superscript asterisk (\({}^*\)) denotes a pure phase. We see that, in the limit of infinite dilution, the composition variables \(x\B\), \(w\B\), \(c\B\), and \(m\B\) are proportional to one another. These expressions are also valid for solute B in a multi solute solution in which each solute is very dilute; that is, in the limit \(x\A\ra 1\). The rule of thumb that the molarity and molality values of a dilute aqueous solution are approximately equal is explained by the relation \(M\A c\B/\rho\A^*=M\A m\B\) (from Eq. 9.1.14), or \(c\B/\rho\A^* = m\B\), and the fact that the density \(\rho\A^*\) of water is approximately \(1\units{kg L\(^{-1}\)}\). Hence, if the solvent is water and the solution is dilute, the numerical value of \(c\B\) expressed in mol L\(^{-1}\) is approximately equal to the numerical value of \(m\B\) expressed in mol kg\(^{-1}\). 9.1.5 The composition of a mixture We can describe the composition of a phase with the amounts of each species, or with any of the composition variables defined earlier: mole fraction, mass fraction, concentration, or molality. If we use mole fractions or mass fractions to describe the composition, we need the values for all but one of the species, since the sum of all fractions is unity. Other composition variables are sometimes used, such as volume fraction, mole ratio, and mole percent. To describe the composition of a gas mixture, partial pressures can be used (Sec. 9.3.1). When the composition of a mixture is said to be fixed or constant during changes of temperature, pressure, or volume, this means there is no change in the relative amounts or masses of the various species. A mixture of fixed composition has fixed values of mole fractions, mass fractions, and molalities, but not necessarily of concentrations and partial pressures. Concentrations will change if the volume changes, and partial pressures in a gas mixture will change if the pressure changes. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/01%3A_Modules/1.08%3A_Enthalpy/1.8.13%3A_Enthalpies-_Liquid_Mixtures | For an ideal binary liquid mixture the Gibbs energy at temperature T is given by equation (a). \[\mathrm{G}(\operatorname{mix} ; \mathrm{id})=\mathrm{n}_{1} \,\left[\mu_{1}^{*}(\ell)+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{x}_{1}\right)\right]+\mathrm{n}_{2} \,\left[\mu_{2}^{*}(\ell)+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{x}_{2}\right)\right] \nonumber \] From the Gibbs-Helmholtz equation, \[\mathrm{H}(\operatorname{mix} ; \mathrm{id})=\mathrm{n}_{1} \, \mathrm{H}_{1}^{*}(\ell)+\mathrm{n}_{2} \, \mathrm{H}_{2}^{*}(\ell) \nonumber \] Hence for an ideal binary liquid mixture, \[\mathrm{H}_{1}(\operatorname{mix} ; \mathrm{id})=\mathrm{H}_{1}^{*}(\ell) \text { and } \mathrm{H}_{2}(\operatorname{mix} ; \mathrm{id})=\mathrm{H}_{2}^{*}(\ell) \nonumber \] The molar enthalpy of a real binary liquid mixture is given by equation (d). \[\mathrm{H}_{\mathrm{m}}=\mathrm{x}_{1} \, \mathrm{H}_{1}(\operatorname{mix})+\mathrm{x}_{2} \, \mathrm{H}_{2}(\operatorname{mix}) \nonumber \] Therefore the molar enthalpy of mixing for a real binary liquid mixture is given by equation (e). \[\Delta_{\text {mix }} H_{m}=x_{1} \,\left[H_{1}(\operatorname{mix})-H_{1}^{*}(\ell)\right]+x_{2} \,\left[H_{2}(\operatorname{mix})-H_{2}^{*}(\ell]\right. \nonumber \] Significantly equations (b) and (e) show that the molar enthalpy of mixing of an ideal binary liquid mixture, \(\Delta_{\text {mix }} H_{m}(\mathrm{id})\) is zero. The latter condition offers an important point of reference for isobaric calorimetry. [1] If we discover that the mixing of two liquids (at constant pressure) is not zero, the measured molar heat of mixing [\(=\Delta_{\text {mix }} \mathrm{H}_{\mathrm{m}}\)] is an immediate indicator of the extent to which the properties of a given mixture are not ideal. Nevertheless it is important to set down a link between the measured enthalpies of mixing with the activity coefficients of two liquid components. To this end we start with the equation for the chemical potentials of liquid component 1 in a liquid mixture at temperature \(\mathrm{T}\) and pressure \(\mathrm{p}\) (which is close to ambient); equation (f). \[\mu_{1}(\mathrm{mix})=\mu_{1}^{*}(\ell)+\mathrm{R} \, \mathrm{T} \, \ln \left(\mathrm{x}_{1} \, \mathrm{f}_{1}\right) \nonumber \] where \[\operatorname{limit}\left(x_{1} \rightarrow 1\right) f_{1}=1 \text { at all } T \text { and } p \text {. } \nonumber \] The Gibbs - Helmholtz equation yields an equation for the partial molar enthalpy of component 1 in the liquid mixture. Thus \[\mathrm{H}_{1}(\operatorname{mix})=\mathrm{H}_{1}^{*}(\ell)-\mathrm{R} \, \mathrm{T}^{2} \,\left[\partial \ln \left(\mathrm{f}_{1}\right) / \partial \mathrm{T}\right]_{\mathrm{p}} \nonumber \] Similarly, \[\mathrm{H}_{2}(\operatorname{mix})=\mathrm{H}_{2}^{*}(\ell)-\mathrm{R} \, \mathrm{T}^{2} \,\left[\partial \ln \left(\mathrm{f}_{2}\right) / \partial \mathrm{T}\right]_{\mathrm{p}} \nonumber \] Hence, \[\begin{aligned} &\mathrm{H}_{\mathrm{m}}(\operatorname{mix})= \\ &\quad \mathrm{H}_{\mathrm{m}}(\operatorname{mix} ; \mathrm{id})-\mathrm{R} \, \mathrm{T}^{2} \,\left\{\mathrm{x}_{1} \,\left[\partial \ln \left(\mathrm{f}_{1}\right) / \partial \mathrm{T}\right]_{\mathrm{p}}+\mathrm{x}_{2} \,\left[\partial \ln \left(\mathrm{f}_{2}\right) / \partial \mathrm{T}\right]_{\mathrm{p}}\right\} \end{aligned} \nonumber \] We also obtain equations for the excess molar enthalpies of the two components (at defined \(\mathrm{T}\) and \(\mathrm{p}\)). \[\mathrm{H}_{1}^{\mathrm{E}}(\mathrm{mix})=-\mathrm{R} \, \mathrm{T}^{2} \,\left[\partial \ln \left(\mathrm{f}_{1}\right) / \partial \mathrm{T}\right]_{\mathrm{p}} \nonumber \] and \[\mathrm{H}_{2}^{\mathrm{E}}(\mathrm{mix})=-\mathrm{R} \, \mathrm{T}^{2} \,\left[\partial \ln \left(\mathrm{f}_{2}\right) / \partial \mathrm{T}\right]_{\mathrm{p}} \nonumber \] The excess molar enthalpy, \[\mathrm{H}_{\mathrm{m}}^{\mathrm{E}}(\text { mix })=-\mathrm{R} \, \mathrm{T}^{2} \,\left\{\mathrm{x}_{1} \,\left[\partial \ln \left(\mathrm{f}_{1}\right) / \partial \mathrm{T}\right]_{\mathrm{p}}+\mathrm{x}_{2} \,\left[\partial \ln \left(\mathrm{f}_{2}\right) / \partial \mathrm{T}\right]_{\mathrm{p}}\right\} \nonumber \] At fixed pressure, the differential dependence of \(\mathrm{H}_{\mathrm{m}}^{\mathrm{E}}(\operatorname{mix})\) on temperature yields the corresponding excess isobaric heat capacity of mixing. Footnotes [1] J. B. Ott and C. J. Wormald, Experimental Thermodynamics, IUPAC Chemical Data Series, No. 39, ed. K. N. Marsh and P. A. G. O’Hara, Blackwell, Oxford, 1994, chapter 8. |
Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/02%3A_Chromatography/2.03%3A_Thin_Layer_Chromatography_(TLC)/2.3C%3A_The_Retention_Factor | A convenient way for chemists to report the results of a TLC plate in lab notebooks is through a " retention factor ",\(^2\) or \(R_f\) value , which quantitates a compound's movement (Equation \ref{2}). \[R_f = \dfrac{\text{distance traveled by the compound}}{\text{distance traveled by the solvent front}} \label{2}\] To measure how far a compound traveled, the distance is measured from the compound's original location (the baseline marked with pencil) to the compound's location after elution (the approximate middle of the spot, Figure 2.14a). Due to the approximate nature of this measurement, ruler values should be recorded only to the nearest millimeter. To measure how far the solvent traveled, the distance is measured from the baseline to the solvent front. The solvent front (Figure 2.14b) is essential to this \(R_f\) calculation. When removing a TLC plate from its chamber, the solvent front needs to be marked immediately with pencil, as the solvent will often evaporate rapidly. The \(R_f\) value is a ratio, and it represents the relative distance the spot traveled compared to the distance it could have traveled if it moved with the solvent front. An \(R_f\) of 0.55 means the spot moved \(55\%\) as far as the solvent front, or a little more than halfway. Since an \(R_f\) is essentially a percentage, it is not particularly important to let a TLC run to any particular height on the TLC plate. In Figure 2.15, a sample of acetophenone was eluted to different heights, and the \(R_f\) was calculated in each case to be similar, although not identical . Slight variations in \(R_f\) arise from error associated with ruler measurements, but also different quantities of adsorbed water on the TLC plates that alter the properties of the adsorbent. \(R_f\) values should always be regarded as approximate. Although in theory a TLC can be run to any height, it's customary to let the solvent run approximately \(0.5 \: \text{cm}\) from the top of the plate to minimize error in the \(R_f\) calculations, and to achieve the best separation of mixtures. A TLC plate should not be allowed to run completely to the top of the plate as it may affect the results. However, if using a saturated, sealed TLC chamber, the \(R_f\) can still be calculated. \(^2\)Sometimes the \(R_f\) is called the retardation factor, as it is a measurement of how the movement of the spots is slowed, or retarded. |
Courses/University_of_North_Carolina_Charlotte/CHEM_2141%3A__Survey_of_Physical_Chemistry/08%3A_Optional-_Special_topics/8.04%3A_Toxicology_MSDT/8.4.06%3A_Principles_of_Toxicology/8.4.6.01%3A_Introduction_to_Toxicology/8.4.6.1.02%3A_Basic_Terminology | Toxicology Defined Toxicology is an evolving medical science and toxicology terminology is evolving with it. Most terms are very specific and will be defined as they appear in the tutorial. However, some terms are more general and used throughout the various sections. The most commonly used terms are introduced in this section. Toxicology is the study of the adverse effects of chemicals or physical agents on living organisms. A toxicologist is a scientist who determines the harmful effects of agents and the cellular, biochemical, and molecular mechanisms responsible for the effects. Toxinology , a specialized area of study, looks at microbial, plant and animal venoms, poisons, and toxins. Terminology and definitions for materials that cause toxic effects are not always consistently used in the literature. The most common terms are toxicant , toxin , poison , toxic agent , toxic substance , and toxic chemical. Toxicant , toxin , and poison are often used interchangeably in the literature but there are subtle differences as shown below: Toxicants : Substances producing adverse biological effects of any kind. May be chemical or physical in nature. Effects may be acute or chronic. Figure \(\PageIndex{1}\): Pesticide chemicals are toxicants (Image Source: iStock Photos, ©) Toxins: Peptides or proteins produced by living organisms. Venoms are toxins injected by a bite or sting. Figure \(\PageIndex{2}\): Amanita muscaria mushroom contains a neurotoxin (Image Source: iStock Photos, ©) Poisons: Toxins produced by organisms. Figure \(\PageIndex{3}\): Black Widow spiders produce a poison that is a toxin (Image Source: Texas Parks & Wildlife, ©) A toxic agent is anything that can produce an adverse biological effect. It may be chemical, physical, or biological in form. For example, toxic agents may be: Chemical (such as cyanide) Physical (such as radiation) Biological (such as snake venom) The toxicity of the agent is dependent on the dose . A distinction is made for diseases people get from living organisms. Organisms that invade and multiply within another organism and produce their effects by biological activity are not classified as toxic agents but as biological agents . An example of this is a virus that damages cell membranes resulting in cell death. If the invading organisms excrete chemicals which are the basis for their toxicity , the excreted substances are known as biological toxins . In that case, the organisms are called toxic organisms . A specific example is tetanus. Tetanus is caused by a bacterium, Clostridium tetani . The bacteria C. tetani itself does not cause disease by invading and destroying cells . Rather, a toxin ( neurotoxin ) that the bacteria excrete travels to the nervous system and produces the disease (Figure 8). Toxic Substances A toxic substance is simply a material that has toxic properties. It may be a discrete toxic chemical or a mixture of toxic chemicals . For example, lead chromate, asbestos, and gasoline are all toxic substances . More specifically: Lead chromate is a discrete toxic chemical . Asbestos is a toxic material that does not have an exact chemical composition but comprises a variety of fibers and minerals. Gasoline s a toxic substance rather than a toxic chemical in that it contains a mixture of many chemicals . Toxic substances may not always have a constant composition. The composition of gasoline varies with octane level, manufacturer, time of season, and other factors. Figure \(\PageIndex{4}\): Examples of toxic substances : lead chromate (left), asbestos (center), and gasoline (right) (Image Source: iStock Photos, ©) Systemic Toxicants and Organ Toxicants Toxic substances may be systemic toxicants or organ toxicants . A systemic toxicant affects the entire body or many organs rather than a specific site. For example, potassium cyanide is a systemic toxicant in that it affects virtually every cell and organ in the body by interfering with the cells ’ ability to use oxygen. Toxicants may also affect only specific tissues or organs while not producing damage to the body as a whole. These specific sites are known as the target organs or target tissues . Benzene is a specific organ toxicant in that it is primarily toxic to the blood-forming tissues . Lead is also a specific organ toxicant ; however, it has three target organs : the central nervous system , the kidneys, and the hematopoietic system. A toxicant may affect a specific type of tissue (such as connective tissue ) that is present in several organs . The toxic site is then considered the target tissue . Figure \(\PageIndex{5}\): Systemic toxicant and organ toxicant (Image Source: iStock Photos, ©) Types of Cells The body is composed of many types of cells , which can be classified in several ways. Table 1 shows examples of one classification of one type of cells . Cell Types Examples Basic structure cuboidal cells Tissue type hepatocytes of the liver Germ cells ova and sperm Somatic cells non-reproductive cells of the body Germ cells are involved in reproduction and can give rise to a new organism. They have only a single set of chromosomes peculiar to a specific sex. Male germ cells give rise to sperm and female germ cells develop into ova. Toxicity to germ cells can cause effects in a developing fetus that lead to outcomes such as birth defects or miscarriage. Somatic cells are all body cells except the reproductive germ cells . ( Somatic cells include the "basic structure" and " tissue type" cells listed in Table 1). They have two sets (or pairs) of chromosomes. In an exposed individual, toxicity to somatic cells causes a variety of toxic effects, such as dermatitis, death, and cancer . Figure 6 illustrates the differences between germ cells and somatic cells . Natural and Man-Made Chemicals Often, people mistakenly assume that all man-made chemicals are harmful and natural chemicals are beneficial. In reality, natural chemicals can be just as harmful to human health as man-made chemicals , and in many cases, more harmful. Figure 12 compares the toxicity of several natural and man-made chemicals . |
Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/24%3A_Organometallic_chemistry-_d-block_elements/24.02%3A_Common_Types_of_Ligand_-_Bonding_and_Spectroscopy | Learning Objectives In this lecture you will learn the following Understand the role lead by ligands in stabilizing organometallic transition metal complexes. Know about various synthetic methods available for preparing the organometallic transition metal complexes. Understand the various factors like β-elimination and other bimolecular decomposition pathways that contribute to the observed instability of the organometallic transition metal complexes. Obtain insight about making stable organometallic transition metal complexes by suppression of the destabilizing factors mentioned. Ligands play a vital role in stabilizing transition metal complexes. The stability as well as the reactivity of a metal in its complex form thus depend upon the number and the type of ligands it is bound to. In this regard, the organometallic carbon based ligands come in diverse varieties displaying a wide range of binding modes to a metal. In general, the binding modes of the carbon-derived ligands depend upon the hybridization state of the metal bound carbon atom. These ligands can thus bind to a metal in many different ways as depicted below. Lastly, these ligands can either be of (a) purely σ−donor type, or depending upon the capability of the ligand to form the multiple bonds may also be of (b) a σ−donor/π−acceptor type, in which the σ−interaction is supplemented by a varying degree of π−interaction. Preparation of transition metal-alkyl and transition metal-aryl complexes The transition metal−alkyl and transition−metal aryl complexes are usually prepared by the following routes discussed below, Metathesis This involves the reactions of metal halides with organolithium, organomagnesium, organoaluminium, organotin and organozinc reagents. Of the different organoalkyl compounds listed above, the organolithium and organomagnesium compounds are strongly carbanionic while the remaining main group organometallics like the organoalkyl, organozinc and organotin reagents are relatively less carbanionic in nature. Thus, the main group organometallic reagents have attenuated alkylating power, that can be productively used in partial exchange of halide ligands. Alkene insertion or Hydrometallation As the name implies, this category of reaction involves an insertion reaction between metal hydride and alkene as shown below. These type reactions are relevant to certain homogeneous catalytic processes in which insertion of an olefin to M−H bond is often observed. Carbene insertion This category represents the reaction of metal hydrides with carbenes. Metallate alkylation reaction This category represents the reaction of carbonylate anions with alkyl halides as shown below. Preparation of transition metal-alkyl and transition metal-aryl complexes (contd..) Metallate acylation reaction This category involves the reaction of carbonylate anions with acyl halides. Oxidative addition reaction Many unsaturated 16 VE transition metal complexes having d 8 or d 10 configuration undergo oxidative addition reactions with alkyl halides. The oxidative addition reactions proceed with the oxidation state as well as coordination number of the metal increasing by +2. Addition reaction This category involves the reaction of an activated metal bound olefin complex with a nucleophile as shown below. Thermodynamic Stability and Kinetic Lability The transition metal organometallic compounds are often difficult to synthesize under ordinary laboratory conditions and require stringent experimental protocols involving the exclusion of air and moisture for doing so. As a consequence, many homoleptic binary transition metal−alkyl and transition metal−aryl compounds like, Et 2 Fe or Me 2 Ni cannot be made under normal laboratory conditions. More interestingly, most of the examples of transition metal−aryl and transition metal−alkyl compounds, known in the literature, invariably contain additional ligands like η 5 -C 5 H 5 , CO, PR 3 or halides. For example, Transition metal−carbon (TM−C) bond energy values are important for understanding the instability of transition metal organometallic compounds. In general, the TM−C bonds are weaker than the transition metal−main group element (TM−MGE) bonds (MGE = F, O, Cl, and N) and more interestingly so, unlike the TM−MGE bond energies, the TM−C bond energy values increase with increasing atomic number. The steric effects of the ligands also play a crucial role in influencing the TM−C bond energies and thus have to be given due consideration. Contrary to the popular belief, the difficulty in obtaining transition metal−aryl and transition metal−alkyl complexes does primarily arise from the thermodynamic reasons but rather the kinetic ones. β−elimination is by far the most general decomposition mechanism that contribute to the instability of transition metal organometallic compounds. β−elimination results in the formation of metal hydrides and olefin as shown below. β−elimination can also be reversible as shown below. The instability of transition metal organometallic compounds can arise out of kinetic lability like in the case of the β−elimination reactions that trigger decomposition of these complexes. Thus, the suppression of the decomposition reactions provides a viable option for the stabilization of the transition metal organometallic complexes. The β−elimination reactions in transition metal organometallic complexes may be suppressed under any of the following three conditions. Formation of the leaving olefin becomes sterically or energetically unfavorable In the course of β−elimination, this situation arises when the olefinic bond is formed at a bridgehead carbon atom or when a double bond is formed with the elements of higher periods. For instance, the norbornyl group is less prone to decomposition by β−elimination because that would require the formation of olefinic double bond at a bridgehead carbon atom in the subsequent olefin, i.e . norbornene, and which is energetically unfavorable. Absence of β−hydrogen atom in organic ligands Transition metal bound ligands that do not possess β−hydrogen cannot decompose by β−elimination pathway and hence such complexes are generally more stable than the ones containing β−hydrogen atoms. For example, the neopentyl complex, Ti[CH 2 C(CH 3 ) 3 ] 4 (m.p 90 °C), and the benzyl complex, Zr(CH 2 Ph) 4 (m.p. 132 °C), exhibit higher thermal stability as both of the neopentyl and benzyl ligands lack β−hydrogens. Central metal atom is coordinatively saturated Transition metal organometallic complexes in which the central metal atom is coordinatively saturated tend to be more stable due to the lack of coordination space available around the metal center to facilitate β−elimination reaction or other decomposition reactions. Thus, the absence of free coordination sites at the metal is crucial towards enhancing the stability of the transition metal organometallic complexes. For example, Ti(Me) 4 , which is coordinatively unsaturated can undergo a bimolecular decomposition reaction via a binuclear intermediate ( A ), is unstable and exhibits a decomposition temperature of –40 °C. On the contrary, Pb(Me) 4 , that cannot undergo decomposition by such bimolecular pathway, is more stable and distills at 110 °C at 1 bar atmospheric pressure. The Ti(Me) 4 decomposes by dimerization involving the formation of Ti−C ( 3c−2e ) bonds. For Pb(Me) 4 , such bimolecular decomposition pathway is not feasible, as being a main group element it has higher outer d orbital for extending the coordination number. If the free coordination site of Ti(Me) 4 is blocked by another ligand, as in [(bipy)Ti(Me) 4 ], then the thermal stability of the complex, [(bipy)Ti(Me) 4 ], increased significantly. Other bidentate chelating ligands like bis (dimethylphosphano)ethane (dmpe) also serve the same purpose. Coordinative saturation thus brings in kinetic stabilization in complexes. For example, Ti(Me) 4 is extremely reactive as it is coordinatively unsaturated, while W(Me) 4 is relatively inert for reasons of being sterically shielded and hence, coordinatively saturated. Thus, if all of the above discussed criteria for the suppression of β-elimination are taken care of, then extremely stable organometallic complexes can be obtained like the one shown below. Problems 1. Arrange the following compounds in the order of their stability. ( a ). Ti(Et) 4 ( b ). Ti(Me) 4 and ( c ). Ti(6-norbornyl) 4 Ans: Ti(Et) 4 < Ti(Me) 4 < Ti(6-norbornyl) 4 2. Predict the product of the reaction given below. Ans: Equi molar amounts of (Bu 3 P)CuD and CH 2 =CDC 2 H 5 3. Will the compound β-eliminate, (a). readily, (b). slowly and (c). not at all. Explain your answer with proper reasoning. Ans: Not at all as the ß-hydrogens are pointing away from the metal and cannot participate in ß-elimination recation. Self Assessment test 1. Write the product(s) of the reactions. Ans: Summary Ligands assume a pivotal role in the stabilization of the organometallic transition metal complexes. There are several methods available for the preparation of the organometallic transition metal complexes. The observed instability of the organometallic transition metal complexes can be attributed to two main phenomena namely β-elimination and bimolecular decomposition reaction that severely undermine the instability of these complexes. The suppression of these decomposition pathway thus pave way for obtaining highly stable organometallic transition metal complexes. |
Courses/Grand_Rapids_Community_College/CHM_120%3A_Survey_of_General_Chemistry_(Crandell)/09%3A_Equilibrium_and_Acid_Base_Chemistry/9.02%3A_Equilibrium_Constants | Learning Objectives By the end of this section, you will be able to: Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures Relate the magnitude of an equilibrium constant to properties of the chemical system The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient ( Q ) . For a reversible reaction described by \[m A +n B +\rightleftharpoons x C +y D \nonumber \] the reaction quotient is derived directly from the stoichiometry of the balanced equation as \[Q_c=\frac{[ C ]^x[ D ]^y}{[ A ]^m[ B ]^n} \nonumber \] where the subscript c denotes the use of molar concentrations in the expression. If the reactants and products are gaseous, a reaction quotient may be similarly derived using partial pressures: \[Q_p=\frac{P_{ C }^x P_{ D }^y}{P_{ A }{ }^m P_{ B }{ }^n} \nonumber \] Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing Q . In most cases, this will introduce only modest errors in calculations involving reaction quotients. Example \(\PageIndex{1}\): Writing Reaction Quotient Expressions Write the concentration-based reaction quotient expression for each of the following reactions: \(\ce{3 O2(g) \rightleftharpoons 2 O3(g)}\) \(\ce{N2(g) + 3 H2(g) \rightleftharpoons 2 NH3(g)}\) \(\ce{4 NH3(g) + 7 O2(g) \rightleftharpoons 4 NO2(g) + 6 H2O(g)}\) Solution \[Q_c=\frac{\left[ \ce{O3} \right]^2}{\left[ \ce{O2} \right]^3} \nonumber \] \[Q_c=\frac{\left[ \ce{NH3} \right]^2}{\left[ \ce{N2} \right]\left[ \ce{H2} \right]^3} \nonumber \] \[Q_c=\frac{\left[ \ce{NO2} \right]^4\left[ \ce{H2O} \right]^6}{\left[ \ce{NH3} \right]^4\left[ \ce{O2} \right]^7} \nonumber \] Exercise \(\PageIndex{1}\) Write the concentration-based reaction quotient expression for each of the following reactions: \(\ce{2SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)}\) \(\ce{C4H8(g) \rightleftharpoons 2 C2H4(g)}\) \(\ce{2 C4H10(g) + 13 O2(g) \rightleftharpoons 8 CO2(g) + 10 H2O (g)}\) Answer \[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]} \nonumber \] \[Q_c=\frac{\left[ \ce{C2H4} \right]^2}{\left[ \ce{C4H8} \right]} \nonumber \] \[Q_c=\frac{\left[ \ce{CO2} \right]^8\left[ \ce{H2O} \right]^{10}}{\left[ \ce{C4H10} \right]^2\left[ \ce{O2} \right]^{13}} \nonumber \] The numerical value of \(Q\) varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. To illustrate this point, consider the oxidation of sulfur dioxide: \[\ce{2 SO2(g) + O2(g) \rightleftharpoons 2 SO3(g)} \nonumber \] Two different experimental scenarios are depicted in Figure \(\PageIndex{1}\), one in which this reaction is initiated with a mixture of reactants only, SO 2 and O 2 , and another that begins with only product, SO 3 . For the reaction that begins with a mixture of reactants only, Q is initially equal to zero: \[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{0^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=0 \nonumber \] As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of Q c ), product concentration increases (as does the numerator of Q c ), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of Q c . If the reaction begins with only product present, the value of Q c is initially undefined (immeasurably large, or infinite): \[Q_c=\frac{\left[ \ce{SO3} \right]^2}{\left[ \ce{SO2} \right]^2\left[ \ce{O2} \right]}=\frac{\left[ \ce{SO3} \right]^2}{0} \rightarrow \infty \nonumber \] In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of Q c decrease with time, the reactant concentrations and the denominator of Q c increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium. The constant value of Q exhibited by a system at equilibrium is called the equilibrium con stant, \(K\): \[K \equiv Q \text { at equilibrium } \nonumber \] Comparison of the data plots in Figure \(\PageIndex{1}\) shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action . Definition: Law of Mass Action At a given temperature, the reaction quotient for a system at equilibrium is constant. Example \(\PageIndex{2}\): Evaluating a Reaction Quotient Gaseous nitrogen dioxide forms dinitrogen tetroxide according to this equation: \[\ce{2 NO2(g) <=> N2O4(g)} \nonumber \] When 0.10 mol NO 2 is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO 2 ] = 0.016 M and [N 2 O 4 ] = 0.042 M . What is the value of the reaction quotient before any reaction occurs? What is the value of the equilibrium constant for the reaction? Solution As for all equilibrium calculations in this text, use the simplified equations for \(Q\) and \(K\) and disregard any concentration or pressure units, as noted previously in this section. (a) Before any product is formed \[\left[ \ce{NO2} \right]=\frac{0.10~\text{mol} }{1.0~\text{L} }=0.10~\text{M} \nonumber \] \[[\ce{N2O4}] = 0~\text{M} \nonumber \] Thus \[Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0}{0.10^2}=0 \nonumber \] (b) At equilibrium, \[K_c=Q_c=\frac{\left[ \ce{N2O4} \right]}{\left[ \ce{NO2} \right]^2}=\frac{0.042}{0.016^2}=1.6 \times 10^2. \nonumber \] The equilibrium constant is \(1.6 \times 10^{2}\). Exercise \(\PageIndex{2}\) For the reaction \[\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)} \nonumber \] the equilibrium concentrations are [SO 2 ] = 0.90 M , [O 2 ] = 0.35 M , and [SO 3 ] = 1.1 M . What is the value of the equilibrium constant, K c ? Answer K c = 4.3 By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large K will reach equilibrium when most of the reactant has been converted to product, whereas a small K indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of K does not indicate how rapidly or slowly equilibrium will be reached. Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer. The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing Q to K for an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur. To further illustrate this important point, consider the reversible reaction shown below: \[\ce{CO(g) + H2O(g) \rightleftharpoons CO2(g) + H2(g)} \quad K_c=0.640 \quad T =800{ }^{\circ} C \nonumber \] The bar charts in Figure \(\PageIndex{2}\) represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction’s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established. Example \(\PageIndex{3}\): Predicting the Direction of Reaction Given here are the starting concentrations of reactants and products for three experiments involving this reaction: \[\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \quad K_c=0.64 \nonumber \] Determine in which direction the reaction proceeds as it goes to equilibrium in each of the three experiments shown. Reactants/Products Experiment 1 Experiment 2 Experiment 3 [CO]i 0.020 M 0.011 M 0.0094 M [H2O]i 0.020 M 0.0011 M 0.0025 M [CO2]i 0.0040 M 0.037 M 0.0015 M [H2]i 0.0040 M 0.046 M 0.0076 M Solution Experiment 1: \[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0040)(0.0040)}{(0.020)(0.020)}=0.040 \nonumber \] Q c < K c (0.040 < 0.64) The reaction will proceed in the forward direction. Experiment 2: \[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.037)(0.046)}{(0.011)(0.0011)}=1.4 \times 10^2 \nonumber \] Q c > K c (140 > 0.64) The reaction will proceed in the reverse direction. Experiment 3: \[Q_c=\frac{\left[ \ce{CO2} \right]\left[ \ce{H2} \right]}{[ \ce{CO} ]\left[ \ce{H2O} \right]}=\frac{(0.0015)(0.0076)}{(0.0094)(0.0025)}=0.48 \nonumber \] Q c < K c (0.48 < 0.64) The reaction will proceed in the forward direction. Exercise \(\PageIndex{3}\) Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium. A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl 2 (g), and 0.500 mol of NOCl: \[\ce{2 NO(g) + Cl2(g) <=> 2 NOCl(g)} \quad K_c=4.6 \times 10^4 \nonumber \] A 5.0-L flask containing 17 g of NH 3 , 14 g of N 2 , and 12 g of H 2 : \[\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)} \quad K_c=0.060 \nonumber \] A 2.00-L flask containing 230 g of SO 3 (g): \[\ce{2 SO3(g) <=> 2 SO2(g) + O2(g)} \quad K_c=0.230 \nonumber \] Answer (a) Q c = 6.45 \times 10^{3} DELMAR, forward. (b) Q c = 0.23, reverse. (c) Q c = 0, forward. Homogeneous Equilibria A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions . These solutions are most commonly either liquid or gaseous phases, as shown by the examples below: \[\begin{aligned} \ce{C2H2(aq) + 2 Br2(aq) & \rightleftharpoons C2H2Br4(aq)} & K_c & =\frac{\left[ C_2 H_2 Br_4\right]}{\left[ C_2 H_2\right]\left[ Br_2\right]^2} \\[4pt] \ce{I2(aq) + I^{-}(aq) & \rightleftharpoons I_3^{-}(aq)} & K_c & =\frac{\left[ I_3-\right.}{\left[ I_2\right]\left[ I^{-}\right]} \\[4pt] \ce{HF(aq) + H2O(l) & \rightleftharpoons H3O^{+}(aq) + F^{-}(aq)} & K_c & =\frac{\left[ H_3 O^{+}\right]\left[ F^{-}\right]}{[ HF ]} \\[4pt] \ce{NH3(aq) + H2O(l) & \rightleftharpoons NH4^{+}(aq) + OH^{-}(aq)} & K_c & =\frac{\left[ NH_4^{+}\right]\left[ OH^{-}\right]}{\left[ NH_3\right]} \end{aligned} \nonumber \] These examples all involve aqueous solutions, those in which water functions as the solvent. In the last two examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K ) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn’t be included . Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species. Note It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that K a = K eq [H 2 O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation K a = K eq (\(\textit{a}_{H_2O}\)). Because \(\textit{a}_{H_2O}\) = 1 for a dilute solution, K a = K eq (1), or K a = K eq . |
Bookshelves/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/05%3A_Properties_of_Compounds/5.02%3A_Carbohydrate_Structures | Learning Outcomes Describe the structure and function of carbohydrates. Identify functional groups of carbohydrates. Give general name for a carbohydrate molecule (i.e. aldotetrose, ketopentose, etc) Label carbohydrates as either D- or L-enantiomers. Draw the mirror image of a carbohydrate molecule. Distinguish between monosaccharides, disaccharides, and polysaccharides. Describe the structure of complex carbohydrates. Recognize how carbohydrates determine blood type. The brain is a marvelous organ. And it's a hungry one, too. The major fuel for the brain is the carbohydrate glucose. The average adult brain represents about \(2\%\) of our body's weight, but uses \(25\%\) of the glucose in the body. Moreover, specific areas of the brain use glucose at different rates. If you are concentrating hard, (taking a test, for example) certain parts of the brain need a lot of extra glucose while other parts of the brain only use their normal amount. Something to think about. As a child, you may have been told that sugar is bad for you. Well, that's not exactly true. Essentially, carbohydrates are made of sugar, from a single sugar molecule to thousands of sugar molecules all attached together. Why? One reason is to store energy. But that does not mean you should eat it by the spoonful. Carbohydrates Carbohydrates are organic compounds that contain only carbon \(\left( \ce{C} \right)\), hydrogen \(\left( \ce{H} \right)\), and oxygen \(\left( \ce{O} \right)\). They contain a chain of carbons, an aldehyde or a ketone, and hydroxyl groups. Every carbon atom is attached to one oxygen atom. There are thousands of different carbohydrates, but they all consist of one or more smaller units called monosaccharides. Monosaccharides The general formula for a monosaccharide is \(\left( \ce{CH_2O} \right) _n \), where \(n\) can be any number greater than two. For example, if \(n\) is 6, then the formula can be written \(\ce{C_6H_{12}O_6}\). This is the formula for the monosaccharide glucose. Another monosaccharide, fructose, has the same chemical formula as glucose, but the atoms are arranged differently. Carbohydrates have many isomers because of the arrangement of the \(\ce{-OH}\) groups in their structures. Compare the glucose and fructose molecules in the figure below. Can you identify their differences? The only differences are the positions of some of the atoms. These differences affect the properties of the two monosaccharides. Monosaccharides can be classified by the number of carbon atoms they contain: diose (2), triose (3), tetrose (4), pentose (5), hexose (6), heptose (7), and so on. They can also be classified based on whether or not they contain an aldehyde (aldose) or ketone (ketose). We can also combine these two designations to refer to classes of carbohydrates. For example, an aldohexose is a carbohydrate (indicated by the -ose ending) with six carbons ( hex ) and an aldehyde group ( aldo ). A ketopentose is a carbohydrate with a ketone and 5 carbons. Both glucose and fructose are hexoses because they contain six carbons but glucose is an aldohexose while fructose (also known as "fruit sugar") is a ketohexose. Other common monosaccharides include galactose (part of lactose), xylose ("wood sugar"), ribose (in RNA), and deoxyribose (in DNA). Fischer Projections There are several ways to draw the structure of carbohydrate molecules. The Fischer projection (straight chain) makes it appear that the molecule is flat but it is a three-dimensional molecule. Although we will not be concerned with the 3D orientation, know that the arrangement in the Fischer projection does provide information about the orientation of atoms around each carbon atom. These projections simplify the drawing of molecules yet retain important information about the arrangement of atoms within the structure. The figure below shows the Fischer projections for the enantiomers (non-superimposable mirror images) of ephedrine and pseudoephedrine. While it may appear that the molecules are the same, they are not because the Fischer projection does not explicitly show the three-dimensional geometry of the molecule. Fischer projections provide an easy way to distinguish among the many, similar carbohydrate molecules that exist. For example, there are sixteen aldohexoses (see figure below). Note the different patterns of the \(\ce{-OH}\) bonds on the left and right sides of the Fischer projection for each. Changing the orientation of one or more of the \(\ce{-OH}\) groups changes the identity of the molecule. Each carbohydrate molecule also has an enantiomer and the two are designated as the D- and L- versions of the compound. The designation is based on the orientation of the \(\ce{-OH}\) group on the chiral carbon farthest from the aldehyde or ketone. The structures of D-glucose and L-glucose are shown in the figure below. The orientation of all \(\ce{-OH}\) groups are reversed but only the arrangement of at the carbon indicated by the arrow determines whether the sugar is a D-sugar with the \(\ce{-OH}\) group on the right or an L-sugar with the \(\ce{-OH}\) group on the left. Haworth Structures Like Fischer projections, the Haworth structures provide information about a molecule's three-dimensional structure without explicitly showing it in the drawing. Carbohydrates are present in the body in both the chain and ring forms with the latter being more common. Haworth projections provide a simple way to display the ring structures and may or may not show the hydrogen atoms attached to each carbon. Remember, every carbon has four bonds so hydrogens are implied when the structure does not show all four bonds. When the cyclic monosaccharide forms, there are two versions that can form, called \(\alpha\) (alpha\) and \(\beta\) (beta) (see figure below). The arrow in the figure indicates the anomeric carbon which it the location where the ring forms and where the orientation of the \(\ce{-OH}\) group can change. The orientation of the other \(\ce{-OH}\) groups are fixed because they are determined by the orientation of the \(\ce{-OH}\) groups in the particular monosaccharide (compare to the orientation of the \(\ce{-OH}\) groups on the left and right sides of the Fischer projections). Each monosaccharide can exist in either \(\alpha\) or \(\beta\) form and the two forms will interconvert as the ring opens and closes. The \(\alpha\) form occurs when the \(\ce{-OH}\) group on the anomeric carbon is pointing down and the \(\beta\) version exists when the \(\ce{-OH}\) group on the anomeric carbon is pointing up. As a result of these different orientations, we can have four forms of each monosaccharide. For example, glucose can exist as \(\alpha\)-D-glucose, \(\alpha\)-L-glucose, \(\beta\)-D-glucose, or \(\beta\)-L-glucose. While the \(\alpha\) and \(\beta\) forms can interconvert, the same cannot be said for D and L versions. Naturally occurring monosaccharides are in the D version, called "D sugars". The arrangement within the D or L form is fixed and they cannot interconvert. Disaccharides If two monosaccharides bond together, they form a carbohydrate called a disaccharide . Two monosaccharides will bond together through a dehydration reaction, in which a water molecule is lost. A dehydration reaction is a condensation reaction , a chemical reaction in which two molecules combine to form one single molecule, losing a small molecule in the process. In the dehydration reaction, this small molecule is water. The bond between two monosaccharides is known as a glycosidic bond . An example of a disaccharide is sucrose (table sugar), which consists of the monosaccharides glucose and fructose (see figure below). Other common disaccharides include lactose ("milk sugar") and maltose. Monosaccharides and disaccharides are also called simple sugars . They provide the major source of energy to living cells. Got milk? Milk is one of the basic foods needed for good nutrition, especially for growing children. It contains vitamins and minerals necessary for healthy development. Unfortunately, milk and other dairy products also contain lactose, a carbohydrate that can make some people very ill. Lactose intolerance is a condition in which the lactose in milk cannot be digested well in the small intestine. The undigested lactose then moves into the large intestine where bacteria attack it, forming large amounts of gas. Symptoms of lactose intolerance include bloating, cramps, nausea, and vomiting. Avoidance of foods containing lactose is recommended for people who show signs of lactose intolerance. Since dairy products can provide many vital nutrients, tablets can be taken that provide the needed digestive materials in the small intestine. Lactose-free milk is also readily available. Oligosaccharides An oligosaccharide is a saccharide polymer containing a small number (typically two to ten) of monosaccharides. Oligosaccharides can have many functions; for example, they are commonly found on the plasma membrane of animal cells where they can play a role in cell-cell recognition. In general, they are found attached to compatible amino acid side-chains in proteins or to lipids. Oligosaccharides are often found as a component of glycoproteins or glycolipids . They are often used as chemical markers on the outside of cells, often for cell recognition. Oligosaccharides are also responsible for determining blood type. Blood Type Carbohydrates attached to red blood cells also determine blood type (see figure below). Of the four blood types, type O has the fewest types of saccharides attached to it while type AB has the most. As a result, type O blood is considered the universal donor because it doesn't have any saccharides present that will appear as foreign when transfused into blood of another type. The reverse is not true. For example, if type A blood is given to a patient with type O blood, it will be rejected by the body because there is an unknown species being introduced to the body. Type A blood cells contain N-acetyl-galactosamine which is not present in type O blood. A person with type O blood would undergo rejection upon receiving type A blood. The Rhesus factor (Rh) in blood also affects donor and acceptor properties but it does not depend on carbohydrates. The Rh factor is determined by the presence (Rh+) or absence (Rh-) of a specific protein on the surface of red blood cells. Polysaccharides Polysaccharides are long carbohydrate molecules of repeated monomer units joined together by glycosidic bonds. A polysaccharide may contain anywhere from a few monosaccharides to several thousand monosaccharides. Polysaccharides are also called complex carbohydrates . Polysaccharides have a general formula of \(\ce{C_x(H_2O)_y}\), where \(x\) is usually a large number between 200 and 2500. Starches are one of the more common polysaccharides. Starch is made up of a mixture of amylose \(\left( 15 \right.\)-\(\left. 20\% \right)\) and amylopectin \(\left( 80 \right.\)-\(\left. 85\% \right)\). Amylose consists of a linear chain of several hundred glucose molecules and amylopectin is a branched molecules made of several thousand glucose units. Starches can be digested by hydrolysis reactions , catalyzed by enzymes called amylases , which can break the glycosidic bonds. Humans and other animals have amylases, so they can digest starches. Potato, rice, wheat, and maize are major sources of starch in the human diet. The formations of starches are the ways that plants store glucose. Glycogen is sometimes referred to as animal starch . Glycogen is used for long-term energy storage in animal cells. Glycogen is made primarily by the liver and the muscles. Are we there yet? As the weather warms up, the runners come out. Not just the casual joggers, but those really serious ones who actually enjoy running all 26.2 miles of a marathon. Prior to these races (and a lot of shorter ones), you hear a lot about carbo-loading. This practice involves eating a lot of starch in the days prior to the race. The starch is converted to glucose, which is normally used for biochemical energy. Excess glucose is stored as glycogen in liver and muscle tissue to be used when needed. If there is a lot of glycogen available, the muscles will have more biochemical energy to draw on when needed for the long run. The rest of us will just sit at the sidewalk restaurant eating our spaghetti and enjoying watching other people work hard. The main functions of polysaccharides are to store energy and form structural tissues. Examples of several other polysaccharides and their roles are listed in the table below. These complex carbohydrates play important roles in living organisms. Complex Carbohydrate Function Organism Starch Stores energy Plants Amylose Stores energy Plants Glycogen Stores energy Animals Cellulose Forms cell walls Plants Chitin Forms an exoskeleton Some animals |
Courses/University_of_Kentucky/UK%3A_General_Chemistry/19%3A_Transition_Metals_and_Coordination_Chemistry/19.3%3A_Optical_and_Magnetic_Properties_of_Coordination_Compounds | Skills to Develop Outline the basic premise of crystal field theory (CFT) Identify molecular geometries associated with various d-orbital splitting patterns Predict electron configurations of split d orbitals for selected transition metal atoms or ions Explain spectral and magnetic properties in terms of CFT concepts The behavior of coordination compounds cannot be adequately explained by the same theories used for main group element chemistry. The observed geometries of coordination complexes are not consistent with hybridized orbitals on the central metal overlapping with ligand orbitals, as would be predicted by valence bond theory. The observed colors indicate that the d orbitals often occur at different energy levels rather than all being degenerate, that is, of equal energy, as are the three p orbitals. To explain the stabilities, structures, colors, and magnetic properties of transition metal complexes, a different bonding model has been developed. Just as valence bond theory explains many aspects of bonding in main group chemistry, crystal field theory is useful in understanding and predicting the behavior of transition metal complexes. Crystal Field Theory To explain the observed behavior of transition metal complexes (such as how colors arise), a model involving electrostatic interactions between the electrons from the ligands and the electrons in the unhybridized d orbitals of the central metal atom has been developed. This electrostatic model is crystal field theory (CFT). It allows us to understand, interpret, and predict the colors, magnetic behavior, and some structures of coordination compounds of transition metals. CFT focuses on the nonbonding electrons on the central metal ion in coordination complexes not on the metal-ligand bonds. Like valence bond theory, CFT tells only part of the story of the behavior of complexes. However, it tells the part that valence bond theory does not. In its pure form, CFT ignores any covalent bonding between ligands and metal ions. Both the ligand and the metal are treated as infinitesimally small point charges. All electrons are negative, so the electrons donated from the ligands will repel the electrons of the central metal. Let us consider the behavior of the electrons in the unhybridized d orbitals in an octahedral complex. The five d orbitals consist of lobe-shaped regions and are arranged in space, as shown in Figure \(\PageIndex{1}\). In an octahedral complex, the six ligands coordinate along the axes. Figure \(\PageIndex{1}\) : The directional characteristics of the five d orbitals are shown here. The shaded portions indicate the phase of the orbitals. The ligands (L) coordinate along the axes. For clarity, the ligands have been omitted from the \(d_{x^2−y^2}\) orbital so that the axis labels could be shown. In an uncomplexed metal ion in the gas phase, the electrons are distributed among the five d orbitals in accord with Hund's rule because the orbitals all have the same energy. However, when ligands coordinate to a metal ion, the energies of the d orbitals are no longer the same. In octahedral complexes, the lobes in two of the five d orbitals, the \(d_{z^2}\) and \(d_{x^2−y^2}\) orbitals, point toward the ligands (Figure \(\PageIndex{1}\)). These two orbitals are called the e g orbitals (the symbol actually refers to the symmetry of the orbitals, but we will use it as a convenient name for these two orbitals in an octahedral complex). The other three orbitals, the d xy , d xz , and d yz orbitals, have lobes that point between the ligands and are called the t 2 g orbitals (again, the symbol really refers to the symmetry of the orbitals). As six ligands approach the metal ion along the axes of the octahedron, their point charges repel the electrons in the d orbitals of the metal ion. However, the repulsions between the electrons in the e g orbitals (the \(d_{z^2}\) and \(d_{x^2−y^2}\) orbitals) and the ligands are greater than the repulsions between the electrons in the t 2 g orbitals (the d zy , d xz , and d yz orbitals) and the ligands. This is because the lobes of the e g orbitals point directly at the ligands, whereas the lobes of the t 2 g orbitals point between them. Thus, electrons in the e g orbitals of the metal ion in an octahedral complex have higher potential energies than those of electrons in the t 2 g orbitals. The difference in energy may be represented as shown in Figure \(\PageIndex{2}\). Figure \(\PageIndex{2}\) : In octahedral complexes, the e g orbitals are destabilized (higher in energy) compared to the t 2g orbitals because the ligands interact more strongly with the d orbitals at which they are pointed directly. The difference in energy between the e g and the t 2 g orbitals is called the crystal field splitting and is symbolized by Δoct , where oct stands for octahedral. The magnitude of Δ oct depends on many factors, including the nature of the six ligands located around the central metal ion, the charge on the metal, and whether the metal is using 3 d , 4 d , or 5 d orbitals. Different ligands produce different crystal field splittings. The increasing crystal field splitting produced by ligands is expressed in the spectrochemical series , a short version of which is given here: \[\large \underset{\textrm{a few ligands of the spectrochemical series, in order of increasing field strength of the ligand}}{\xrightarrow{\ce{I- <Br- <Cl- <F- <H2O<C2O4^2- <NH3<\mathit{en}<NO2- <CN-}}}\] In this series, ligands on the left cause small crystal field splittings and are weak-field ligands , whereas those on the right cause larger splittings and are strong-field ligands . Thus, the Δ oct value for an octahedral complex with iodide ligands (I − ) is much smaller than the Δ oct value for the same metal with cyanide ligands (CN − ). Electrons in the d orbitals follow the aufbau (“filling up”) principle, which says that the orbitals will be filled to give the lowest total energy, just as in main group chemistry. When two electrons occupy the same orbital, the like charges repel each other. The energy needed to pair up two electrons in a single orbital is called the pairing energy (P) . Electrons will always singly occupy each orbital in a degenerate set before pairing. P is similar in magnitude to Δ oct . When electrons fill the d orbitals, the relative magnitudes of Δ oct and P determine which orbitals will be occupied. In [Fe(CN) 6 ] 4− , the strong field of six cyanide ligands produces a large Δ oct . Under these conditions, the electrons require less energy to pair than they require to be excited to the e g orbitals (Δ oct > P). The six 3 d electrons of the Fe 2+ ion pair in the three t 2 g orbitals (Figure \(\PageIndex{3}\)). Complexes in which the electrons are paired because of the large crystal field splitting are called low-spin complexes because the number of unpaired electrons (spins) is minimized. Figure \(\PageIndex{3}\) : Iron(II) complexes have six electrons in the 5d orbitals. In the absence of a crystal field, the orbitals are degenerate. For coordination complexes with strong-field ligands such as [Fe(CN) 6 ] 4− , Δ oct is greater than P, and the electrons pair in the lower energy t 2g orbitals before occupying the eg orbitals. With weak-field ligands such as H 2 O, the ligand field splitting is less than the pairing energy, Δ oct less than P, so the electrons occupy all d orbitals singly before any pairing occurs. In [Fe(H 2 O) 6 ] 2+ , on the other hand, the weak field of the water molecules produces only a small crystal field splitting (Δ oct < P). Because it requires less energy for the electrons to occupy the e g orbitals than to pair together, there will be an electron in each of the five 3 d orbitals before pairing occurs. For the six d electrons on the iron(II) center in [Fe(H 2 O) 6 ] 2+ , there will be one pair of electrons and four unpaired electrons (Figure \(\PageIndex{3}\)). Complexes such as the [Fe(H 2 O) 6 ] 2+ ion, in which the electrons are unpaired because the crystal field splitting is not large enough to cause them to pair, are called high-spin complexes because the number of unpaired electrons (spins) is maximized. A similar line of reasoning shows why the [Fe(CN) 6 ] 3− ion is a low-spin complex with only one unpaired electron, whereas both the [Fe(H 2 O) 6 ] 3+ and [FeF 6 ] 3− ions are high-spin complexes with five unpaired electrons. Example \(\PageIndex{1}\): High- and Low-Spin Complexes Predict the number of unpaired electrons. K 3 [CrI 6 ] [Cu(en) 2 (H 2 O) 2 ]Cl 2 Na 3 [Co(NO 2 ) 6 ] Solution The complexes are octahedral. Cr 3+ has a d 3 configuration. These electrons will all be unpaired. Cu 2+ is d 9 , so there will be one unpaired electron. Co 3+ has d 6 valence electrons, so the crystal field splitting will determine how many are paired. Nitrite is a strong-field ligand, so the complex will be low spin. Six electrons will go in the t 2 g orbitals, leaving 0 unpaired. Exercise \(\PageIndex{1}\) The size of the crystal field splitting only influences the arrangement of electrons when there is a choice between pairing electrons and filling the higher-energy orbitals. For which d -electron configurations will there be a difference between high- and low-spin configurations in octahedral complexes? Answer d 4 , d 5 , d 6 , and d 7 Example \(\PageIndex{2}\): CFT for Other Geometries CFT is applicable to molecules in geometries other than octahedral. In octahedral complexes, remember that the lobes of the e g set point directly at the ligands. For tetrahedral complexes, the d orbitals remain in place, but now we have only four ligands located between the axes (Figure \(\PageIndex{4}\)). None of the orbitals points directly at the tetrahedral ligands. However, the e g set (along the Cartesian axes) overlaps with the ligands less than does the t 2 g set. By analogy with the octahedral case, predict the energy diagram for the d orbitals in a tetrahedral crystal field. To avoid confusion, the octahedral e g set becomes a tetrahedral e set, and the octahedral t 2 g set becomes a t 2 set. Figure \(\PageIndex{4}\) : This diagram shows the orientation of the tetrahedral ligands with respect to the axis system for the orbitals. Solution Since CFT is based on electrostatic repulsion, the orbitals closer to the ligands will be destabilized and raised in energy relative to the other set of orbitals. The splitting is less than for octahedral complexes because the overlap is less, so Δ tet is usually small \(\left(Δ_\ce{tet}=\dfrac{4}{9}Δ_\ce{oct}\right)\): Exercise \(\PageIndex{2}\) Explain how many unpaired electrons a tetrahedral d 4 ion will have. Answer 4; because Δ tet is small, all tetrahedral complexes are high spin and the electrons go into the t 2 orbitals before pairing The other common geometry is square planar. It is possible to consider a square planar geometry as an octahedral structure with a pair of trans ligands removed. The removed ligands are assumed to be on the z -axis. This changes the distribution of the d orbitals, as orbitals on or near the z -axis become more stable, and those on or near the x- or y -axes become less stable. This results in the octahedral t 2 g and the e g sets splitting and gives a more complicated pattern with no simple Δ oct . The basic pattern is: Magnetic Moments of Molecules and Ions Experimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as O 2 that contain unpaired electrons are paramagnetic. Paramagnetic substances are attracted to magnetic fields. Many transition metal complexes have unpaired electrons and hence are paramagnetic. Molecules such as N 2 and ions such as Na + and [Fe(CN) 6 ] 4− that contain no unpaired electrons are diamagnetic. Diamagnetic substances have a slight tendency to be repelled by magnetic fields. When an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic. The size of the magnetic moment of a system containing unpaired electrons is related directly to the number of such electrons: the greater the number of unpaired electrons, the larger the magnetic moment. Therefore, the observed magnetic moment is used to determine the number of unpaired electrons present. The measured magnetic moment of low-spin d 6 [Fe(CN) 6 ] 4− confirms that iron is diamagnetic, whereas high-spin d 6 [Fe(H 2 O) 6 ] 2+ has four unpaired electrons with a magnetic moment that confirms this arrangement. Colors of Transition Metal Complexes When atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the d orbitals often allows photons in the visible range to be absorbed. The human eye perceives a mixture of all the colors, in the proportions present in sunlight, as white light. Complementary colors, those located across from each other on a color wheel, are also used in color vision. The eye perceives a mixture of two complementary colors, in the proper proportions, as white light. Likewise, when a color is missing from white light, the eye sees its complement. For example, when red photons are absorbed from white light, the eyes see the color green. When violet photons are removed from white light, the eyes see lemon yellow. The blue color of the [Cu(NH 3 ) 4 ] 2+ ion results because this ion absorbs orange and red light, leaving the complementary colors of blue and green (Figure \(\PageIndex{5}\)). Figure \(\PageIndex{5}\): (a) An object is black if it absorbs all colors of light. If it reflects all colors of light, it is white. An object has a color if it absorbs all colors except one, such as this yellow strip. The strip also appears yellow if it absorbs the complementary color from white light (in this case, indigo). (b) Complementary colors are located directly across from one another on the color wheel. (c) A solution of [Cu(NH 3 ) 4 ] 2+ ions absorbs red and orange light, so the transmitted light appears as the complementary color, blue. Example \(\PageIndex{3}\): Colors of Complexes The octahedral complex [Ti(H 2 O) 6 ] 3+ has a single d electron. To excite this electron from the ground state t 2 g orbital to the e g orbital, this complex absorbs light from 450 to 600 nm. The maximum absorbance corresponds to Δ oct and occurs at 499 nm. Calculate the value of Δ oct in Joules and predict what color the solution will appear. Solution Using Planck's equation (refer to the section on electromagnetic energy), we calculate: \[v=\dfrac{c}{λ}\mathrm{\:so\:\dfrac{3.00×10^8\: m/s}{\dfrac{499\: nm×1\: m}{10^9\:nm}}=6.01×10^{14}\:Hz} \nonumber\] \[E=hnu\mathrm{\:so\:6.63×10^{−34}\:\textrm{J⋅s}×6.01×10^{14}\:Hz=3.99×10^{−19}\:Joules/ion} \nonumber\] Because the complex absorbs 600 nm (orange) through 450 (blue), the indigo, violet, and red wavelengths will be transmitted, and the complex will appear purple. Exercise \(\PageIndex{3}\) A complex that appears green, absorbs photons of what wavelengths? Answer red, 620–800 nm Small changes in the relative energies of the orbitals that electrons are transitioning between can lead to drastic shifts in the color of light absorbed. Therefore, the colors of coordination compounds depend on many factors. As shown in Figure \(\PageIndex{6}\), different aqueous metal ions can have different colors. In addition, different oxidation states of one metal can produce different colors, as shown for the vanadium complexes in the link below. Figure \(\PageIndex{6}\): The partially filled d orbitals of the stable ions Cr 3+ (aq), Fe 3+ (aq), and Co 2+ (aq) (left, center and right, respectively) give rise to various colors. (credit: Sahar Atwa) The specific ligands coordinated to the metal center also influence the color of coordination complexes. For example, the iron(II) complex [Fe(H 2 O) 6 ]SO 4 appears blue-green because the high-spin complex absorbs photons in the red wavelengths (Figure \(\PageIndex{7}\)). In contrast, the low-spin iron(II) complex K 4 [Fe(CN) 6 ] appears pale yellow because it absorbs higher-energy violet photons. Figure \(\PageIndex{7}\): Both (a) hexaaquairon(II) sulfate and (b) potassium hexacyanoferrate(II) contain d 6 iron(II) octahedral metal centers, but they absorb photons in different ranges of the visible spectrum. In general, strong-field ligands cause a large split in the energies of d orbitals of the central metal atom (large Δ oct ). Transition metal coordination compounds with these ligands are yellow, orange, or red because they absorb higher-energy violet or blue light. On the other hand, coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light. Video \(\PageIndex{8}\): Watch this video of the reduction of vanadium complexes to observe the colorful effect of changing oxidation states. A coordination compound of the Cu + ion has a d 10 configuration, and all the e g orbitals are filled. To excite an electron to a higher level, such as the 4 p orbital, photons of very high energy are necessary. This energy corresponds to very short wavelengths in the ultraviolet region of the spectrum. No visible light is absorbed, so the eye sees no change, and the compound appears white or colorless. A solution containing [Cu(CN) 2 ] − , for example, is colorless. On the other hand, octahedral Cu 2+ complexes have a vacancy in the e g orbitals, and electrons can be excited to this level. The wavelength (energy) of the light absorbed corresponds to the visible part of the spectrum, and Cu 2+ complexes are almost always colored—blue, blue-green violet, or yellow (Figure \(\PageIndex{8}\)). Although CFT successfully describes many properties of coordination complexes, molecular orbital explanations (beyond the introductory scope provided here) are required to understand fully the behavior of coordination complexes. Figure \(\PageIndex{8}\): (a) Copper(I) complexes with d 10 configurations such as CuI tend to be colorless, whereas (b) d 9 copper(II) complexes such as Cu(NO 3 ) 2 ·5H 2 O are brightly colored. Summary Crystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal d orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting (Δ oct ) depends on the nature of the ligands bonded to the metal. Strong-field ligands produce large splitting and favor low-spin complexes, in which the t 2 g orbitals are completely filled before any electrons occupy the e g orbitals. Weak-field ligands favor formation of high-spin complexes. The t 2 g and the e g orbitals are singly occupied before any are doubly occupied. Glossary crystal field splitting (Δ oct ) difference in energy between the t 2 g and e g sets or t and e sets of orbitals crystal field theory model that explains the energies of the orbitals in transition metals in terms of electrostatic interactions with the ligands but does not include metal ligand bonding e g orbitals set of two d orbitals that are oriented on the Cartesian axes for coordination complexes; in octahedral complexes, they are higher in energy than the t 2 g orbitals geometric isomers isomers that differ in the way in which atoms are oriented in space relative to each other, leading to different physical and chemical properties high-spin complex complex in which the electrons maximize the total electron spin by singly populating all of the orbitals before pairing two electrons into the lower-energy orbitals low-spin complex complex in which the electrons minimize the total electron spin by pairing in the lower-energy orbitals before populating the higher-energy orbitals pairing energy (P) energy required to place two electrons with opposite spins into a single orbital spectrochemical series ranking of ligands according to the magnitude of the crystal field splitting they induce strong-field ligand ligand that causes larger crystal field splittings t 2 g orbitals set of three d orbitals aligned between the Cartesian axes for coordination complexes; in octahedral complexes, they are lowered in energy compared to the e g orbitals according to CFT weak-field ligand ligand that causes small crystal field splittings Contributors Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/[email protected] ). |
Courses/University_of_Wisconsin_Oshkosh/Chem_370%3A_Physical_Chemistry_1_-_Thermodynamics_(Gutow)/06%3A_Reaction_Kinetics/6.01%3A_Reaction_Rates_and_Rate_Laws | The Reaction Rate The rate of a chemical reaction (or the reaction rate ) can be defined by the time needed for a change in concentration to occur. But there is a problem in that this allows for the definition to be made based on concentration changes for either the reactants or the products. Plus, due to stoichiometric concerns, the rates at which the concentrations are generally different! Toward this end, the following convention is used. For a general reaction \[a A + b B \rightarrow c C + d D \nonumber \] the reaction rate can be defined by any of the ratios \[\text{rate} = - \dfrac{1}{a} \dfrac{\Delta [A]}{dt} = - \dfrac{1}{b} \dfrac{\Delta[B]}{dt} = + \dfrac{1}{c} \dfrac{\Delta [C]}{dt} = + \dfrac{1}{d} \dfrac{ \Delta [D]}{dt} \nonumber \] Or for infinitesimal time intervals \[\text{rate} = - \dfrac{1}{a} \dfrac{d[A]}{dt} = - \dfrac{1}{b} \dfrac{d[C]}{dt} = + \dfrac{1}{c} \dfrac{d[C]}{dt} = + \dfrac{1}{d} \dfrac{d[D]}{dt} \nonumber \] Example \(\PageIndex{1}\): Under a certain set of conditions, the rate of the reaction \[\ce{N_2 + 3 H_2 \rightarrow 2 NH_3} \nonumber \] the reaction rate is \(6.0 \times 10^{-4}\, M/s\). Calculate the time-rate of change for the concentrations of N 2 , H 2 , and NH 3 . Solution : Due to the stoichiometry of the reaction, \[\text{rate} = - \dfrac{d[N_2]}{dt} = - \dfrac{1}{3} \dfrac{d[H_2]}{dt} = + \dfrac{1}{2} \dfrac{d[NH_3]}{dt} \nonumber \] so \[\begin{align*} \dfrac{d[N_2]}{dt} &= -6.0 \times 10^{-4} \,M/s \\[4pt] \dfrac{d[H_2]}{dt} &= -2.0 \times 10^{-4} \,M/s \\[4pt] \dfrac{d[NH_3]}{dt} &= 3.0 \times 10^{-4} \,M/s \end{align*} \] Note: The time derivatives for the reactants are negative because the reactant concentrations are decreasing, and those of products are positive since the concentrations of products increase as the reaction progresses. The Rate Law As shown above, the rate of the reaction can be followed experimentally by measuring the rate of the loss of a reactant or the rate of the production of a product. The rate of the reaction is often related to the concentration of some or all of the chemical species present at a given time. An equation called the rate law is used to show this relationship. The rate law cannot be predicted by looking at the balanced chemical reaction but must be determined by experiment. For example, the rate law for the reaction \[\ce{Cl2 (g) + CO (g) → Cl2CO (g) } \nonumber \] was experimentally determined to be \[ \text{rate} = k[Cl_2]^{3/2}[CO] \nonumber \] In this equation, \( k \) is the rate constant, and \( [\ce{Cl_2}] \) and \( [\ce{CO}] \) are the molar concentrations of Cl 2 and of CO. Each exponent is called the order of the given species. Thus, the rate law is 3/2 order in Cl 2 and first order in CO. The sum of the individual reactant orders is called the reaction order . This reaction has a reaction order of 5/2. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/07%3A_Approximation_Methods/7.03%3A_Trial_Functions_Can_Be_Linear_Combinations_of_Functions_That_Also_Contain_Variational_Parameters | Demonstrate that variational problems can include changing parameters within the elements (normal variational method) and changing coefficients of a basis set (linear variational method) An alternative approach to the general problem of introducing variational parameters into wavefunctions is the construction of a wavefunction as a linear combination of other functions each with one or multiple parameters that can be varied For hydrogen, the radial function decays, or decreases in amplitude, exponentially as the distance from the nucleus increases. For helium and other multi-electron atoms, the radial dependence of the total probability density does not fall off as a simple exponential with increasing distance from the nucleus as it does for hydrogen. More complex single-electron functions therefore are needed in order to model the effects of electron-electron interactions on the total radial distribution function. One way to obtain more appropriate single-electron functions is to use a sum of exponential functions in place of the hydrogenic spin-orbitals. An example of such a wavefunction created from a sum or linear combination of exponential functions is written as \[ \varphi _{1s} (r_1) = \sum _j c_j e^{-\zeta _j r_j / a_o} \label{9-37} \] The linear combination permits weighing of the different exponentials through the adjustable coefficients (\(c_j\)) for each term in the sum. Each exponential term has a different rate of decay through the zeta-parameter \(\zeta _j\). The exponential functions in Equation \(\ref{9-37}\) are called basis functions . Basis functions are the functions used in linear combinations to produce the single-electron orbitals that in turn combine to create the product multi-electron wavefunctions. Originally the most popular basis functions used were the STO’s, but today STO’s are not used in most quantum chemistry calculations. However, they are often the functions to which more computationally efficient basis functions are fitted. Physically, the \(\zeta _j\) parameters account for the effective nuclear charge (often denoted with \(Z_{eff}\)). The use of several zeta values in the linear combination essentially allows the effective nuclear charge to vary with the distance of an electron from the nucleus. This variation makes sense physically. When an electron is close to the nucleus, the effective nuclear charge should be close to the actual nuclear charge. When the electron is far from the nucleus, the effective nuclear charge should be much smaller. See Slater's rules for a rule-of-thumb approach to evaluate \(Z_{eff}\) values. A term in Equation \(\ref{9-37}\) with a small \(\zeta\) will decay slowly with distance from the nucleus. A term with a large \(\zeta\) will decay rapidly with distance and not contribute at large distances. The need for such a linear combination of exponentials is a consequence of the electron-electron repulsion and its effect of screening the nucleus for each electron due to the presence of the other electrons. Make plots of \(\varphi\) in Equation \(\ref{9-37}\) using three equally weighted terms with \(\zeta\) = 1.0, 2.0, and 5.0. Also plot each term separately. Computational procedures in which an exponential parameter like \(\zeta\) is varied are more precisely called the Nonlinear Variational Method because the variational parameter is part of the wavefunction and the change in the function and energy caused by a change in the parameter is not linear. The optimum values for the zeta parameters in any particular calculation are determined by doing a variational calculation for each orbital to minimize the ground-state energy. When this calculation involves a nonlinear variational calculation for the zetas, it requires a large amount of computer time. The use of the variational method to find values for the coefficients, \(\{c_j\}\), in the linear combination given by Equation \(\ref{9-37}\) above is called the Linear Variational Method because the single-electron function whose energy is to be minimized (in this case \(\varphi _{1s}\)) depends linearly on the coefficients. Although the idea is the same, it usually is much easier to implement the linear variational method in practice. Nonlinear variational calculations are extremely costly in terms of computer time because each time a zeta parameter is changed, all of the integrals need to be recalculated. In the linear variation, where only the coefficients in a linear combination are varied, the basis functions and the integrals do not change. Consequently, an optimum set of zeta parameters were chosen from variational calculations on many small multi-electron systems, and these values, which are given in Table 7.3.1 , generally can be used in the STOs for other and larger systems. Atom \(\zeta _{1s}\) \(\zeta _{2s,2p}\) H 1.24 - He 1.69 - Li 2.69 0.80 Be 3.68 1.15 B 4.68 1.50 C 5.67 1.72 N 6.67 1.95 O 7.66 2.25 F 8.56 2.55 Ne 9.64 2.88 Compare the value \(\zeta _{1s}\) = 1.24 in Table 7.3.1 for hydrogen with the value you obtained in Exercise 7.3.1 . and comment on possible reasons for any difference. Why are the zeta values larger for 1s than for 2s and 2p orbitals? Why do the \(\zeta _{1s}\) values increase by essentially one unit for each element from He to Ne while the increase for the \(\zeta _{2s, 2p}\) values is much smaller? Answer ζ values represent the rate of decay in the radial function of an orbital. ζ values are larger for 1s than 2s and 2p orbitals because 1s orbitals have a smaller radial function. As a result, 1s orbitals decrease faster in radial function as you move further from the nucleus, and have a larger ζ value to represent this faster decay. The ζ values for 1s increase essentially by one unit for each element from He to Ne because the 1s orbital is closest to the nucleus, and experiences the greatest effects from change in electronegativity as nuclear density increases from He to Ne. This increase in electronegativity causes the radial function to decay more and more rapidly as atomic number/nucleus density increase. The 2s and 2p orbitals don't experience as great a change in radial function decay rate because they are shielded by the 1s orbital. The discussion above gives us some new ideas about how to write flexible, useful single-electron wavefunctions that can be used to construct multi-electron wavefunctions for variational calculations. Single-electron functions built from the basis function approach are flexible because they have several adjustable parameters, and useful because the adjustable parameters still have clear physical interpretations. Such functions will be needed in the Hartree-Fock method discussed elsewhere. |
Courses/Modesto_Junior_College/Chemistry_143%3A_Introductory_College_Chemistry_(Brzezinski)/CHEM_143%3A_Text_(Brzezinski)/06%3A_Chemical_Compounds/6.06%3A_Lewis_Structures_of_Ionic_Compounds-_Electrons_Transferred/6.6.01%3A_Covalent_Lewis_Structures-_Electrons_Shared | Learning Objectives Define covalent bond . Illustrate covalent bond formation with Lewis electron dot diagrams. Ionic bonding typically occurs when it is easy for one atom to lose one or more electrons and another atom to gain one or more electrons. However, some atoms won’t give up or gain electrons easily. Yet they still participate in compound formation. How? There is another mechanism for obtaining a complete valence shell: sharing electrons. When electrons are shared between two atoms, they make a bond called a covalent bond . Let us illustrate a covalent bond by using H atoms, with the understanding that H atoms need only two electrons to fill the 1 s subshell. Each H atom starts with a single electron in its valence shell: \[\mathbf{H\, \cdot }\; \; \; \; \; \mathbf{\cdot \: H} \nonumber \] The two H atoms can share their electrons: \[\mathbf{H}\: \mathbf{: H} \nonumber \] We can use circles to show that each H atom has two electrons around the nucleus, completely filling each atom’s valence shell: Because each H atom has a filled valence shell, this bond is stable, and we have made a diatomic hydrogen molecule. (This explains why hydrogen is one of the diatomic elements.) For simplicity’s sake, it is not unusual to represent the covalent bond with a dash, instead of with two dots: H–H Because two atoms are sharing one pair of electrons, this covalent bond is called a single bond . As another example, consider fluorine. F atoms have seven electrons in their valence shell: These two atoms can do the same thing that the H atoms did; they share their unpaired electrons to make a covalent bond. Note that each F atom has a complete octet around it now: We can also write this using a dash to represent the shared electron pair: There are two different types of electrons in the fluorine diatomic molecule. The bonding electron pair makes the covalent bond. Each F atom has three other pairs of electrons that do not participate in the bonding; they are called lone pair electrons . Each F atom has one bonding pair and three lone pairs of electrons. Covalent bonds can be made between different elements as well. One example is HF . Each atom starts out with an odd number of electrons in its valence shell: The two atoms can share their unpaired electrons to make a covalent bond: We note that the H atom has a full valence shell with two electrons, while the F atom has a complete octet of electrons. Example \(\PageIndex{1}\): Use Lewis electron dot diagrams to illustrate the covalent bond formation in HBr. Solution HBr is very similar to HF, except that it has Br instead of F. The atoms are as follows: The two atoms can share their unpaired electron: Exercise \(\PageIndex{1}\) Use Lewis electron dot diagrams to illustrate the covalent bond formation in Cl 2 . Answer When working with covalent structures, it sometimes looks like you have leftover electrons. You apply the rules you learned so far, and there are still some electrons that remain unattached. You can't just leave them there. So where do you put them? Multiple Covalent Bonds Some molecules are not able to satisfy the octet rule by making only single covalent bonds between the atoms. Consider the compound ethene, which has a molecular formula of \(\ce{C_2H_4}\). The carbon atoms are bonded together, with each carbon also bonded to two hydrogen atoms. two \(\ce{C}\) atoms \(= 2 \times 4 = 8\) valence electrons four \(\ce{H}\) atoms \(= 4 \times 1 = 4\) valence electrons total of 12 valence electrons in the molecule If the Lewis electron dot structure was drawn with a single bond between the carbon atoms and with the octet rule followed, it would look like this: This Lewis structure is incorrect because it contains a total of 14 electrons. However, the Lewis structure can be changed by eliminating the lone pairs on the carbon atoms and having to share two pairs instead of only one pair. A double covalent bond is a covalent bond formed by atoms that share two pairs of electrons. The double covalent bond that occurs between the two carbon atoms in ethane can also be represented by a structural formula and with a molecular model as shown in the figure below. A triple covalent bond is a covalent bond formed by atoms that share three pairs of electrons. The element nitrogen is a gas that composes the majority of Earth's atmosphere. A nitrogen atom has five valence electrons, which can be shown as one pair and three single electrons. When combining with another nitrogen atom to form a diatomic molecule, the three single electrons on each atom combine to form three shared pairs of electrons. Each nitrogen atom follows the octet rule with one lone pair of electrons, and six electrons that are shared between the atoms. Summary Covalent bonds are formed when atoms share electrons. Lewis electron dot diagrams can be drawn to illustrate covalent bond formation. Double bonds or triple bonds between atoms may be necessary to properly illustrate the bonding in some molecules. Contributions & Attributions Anonymous by request |
Courses/Clackamas_Community_College/CH_112%3A_Chemistry_for_Health_Sciences/08%3A_Solids_Liquids_and_Gases/8.05%3A_Gas_Laws | Learning Objectives To predict the properties of gases using the gas laws. Experience has shown that several properties of a gas can be related to each other under certain conditions. The properties are pressure ( P ), volume ( V ), temperature ( T , in kelvins), and amount of material expressed in moles ( n ). What we find is that a sample of gas cannot have any random values for these properties. Instead, only certain values, dictated by some simple mathematical relationships, will occur. Boyle’s Law The first simple relationship, referred to as a gas law, is between the pressure of a gas and its volume. If the amount of gas in a sample and its temperature are kept constant, then as the pressure of a gas is increased, the volume of the gas decreases proportionately. Mathematically, this is written as \[\mathrm{P \propto \dfrac{1}{V}} \nonumber \] where the “∝” symbol means “is proportional to.” This is one form of Boyle’s law, which relates the pressure of a gas to its volume. A more useful form of Boyle’s law involves a change in conditions of a gas. For a given amount of gas at a constant temperature, if we know the initial pressure and volume of a gas sample and the pressure or volume changes, we can calculate what the new volume or pressure will be. That form of Boyle’s law is written \[P_iV_i = P_fV_f \label{Eq1} \] where the subscript \(i\) refers to initial conditions and the subscript \(f\) refers to final conditions. To use \(\ref{Eq1}\), you need to know any three of the variables so that you can algebraically calculate the fourth variable. Also, the pressure quantities must have the same units, as must the two volume quantities. If the two similar variables don’t have the same variables, one value must be converted to the other value’s unit. Example \(\PageIndex{1}\): Increasing Pressure in a Gas What happens to the volume of a gas if its pressure is increased? Assume all other conditions remain the same. Solution If the pressure of a gas is increased, the volume decreases in response. Exercise \(\PageIndex{1}\): Increasing Volume in a Gas What happens to the pressure of a gas if its volume is increased? Assume all other conditions remain the same. Answer If the volume of a gas is increased, the pressure decreases. Example \(\PageIndex{2}\): Gas Compression If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is reduced to 0.987 atm? Assume that the amount and the temperature of the gas remain constant. Solution The key in problems like this is to be able to identify which quantities represent which variables from the relevant equation. The way the question is worded, you should be able to tell that 1.56 atm is P i , 7.02 L is V i , and 0.987 atm is P f . What we are looking for is the final volume— V f . Therefore, substituting these values into P i V i = P f V f : (1.56 atm)(7.02 L) = (0.987 atm) × V f The expression has atmospheres on both sides of the equation, so they cancel algebraically: (1.56)(7.02 L) = (0.987) × V f Now we divide both sides of the expression by 0.987 to isolate V f , the quantity we are seeking: \(\mathrm{\dfrac{(1.56)(7.02\: L)}{0.987}=V_f}\) Performing the multiplication and division, we get the value of V f , which is 11.1 L. The volume increases. This should make sense because the pressure decreases, so pressure and volume are inversely related. Exercise \(\PageIndex{2}\) If a sample of gas has an initial pressure of 3.66 atm and an initial volume of 11.8 L, what is the final pressure if the volume is reduced to 5.09 L? Assume that the amount and the temperature of the gas remain constant. Answer 8.48 atm If the units of similar quantities are not the same, one of them must be converted to the other quantity’s units for the calculation to work out properly. It does not matter which quantity is converted to a different unit; the only thing that matters is that the conversion and subsequent algebra are performed properly. The following example illustrates this process. Example \(\PageIndex{3}\) If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is changed to 1,775 torr? Does the answer make sense? Assume that the amount and the temperature of the gas remain constant. Solution This example is similar to Example \(\PageIndex{2}\), except now the final pressure is expressed in torr. For the math to work out properly, one of the pressure values must be converted to the other unit. Let us change the initial pressure to torr: \(\mathrm{1.56\: atm\times\dfrac{760\: torr}{1\: atm}=1,190\: torr}\) Now we can use Boyle’s law: (1,190 torr)(7.02 L) = (1,775 torr) × V f Torr cancels algebraically from both sides of the equation, leaving (1,190)(7.02 L) = (1,775) × V f Now we divide both sides of the equation by 1,775 to isolate V f on one side. Solving for the final volume, \(\mathrm{V_f=\dfrac{(1,190)(7.02\: L)}{1,775}=4.71\: L}\) Because the pressure increases, it makes sense that the volume decreases. The answer for the final volume is essentially the same if we converted the 1,775 torr to atmospheres: \(\mathrm{1,775\: torr\times\dfrac{1\: atm}{760\: torr}=2.336\: atm}\). Using Boyle’s law: (1.56 atm)(7.02 L) = (2.335 atm) × V f ; \(\mathrm{V_f=\dfrac{(1.56\: atm)(7.02\: L)}{2.336\: atm}=4.69\: L}\). Exercise \(\PageIndex{3}\) If a sample of gas has an initial pressure of 375 torr and an initial volume of 7.02 L, what is the final pressure if the volume is changed to 4,577 mL? Does the answer make sense? Assume that amount and the temperature of the gas remain constant. Answer 575 torr To Your Health: Breathing Breathing certainly is a major contribution to your health! Without breathing, we could not survive. Curiously, the act of breathing itself is little more than an application of Boyle’s law. The lungs are a series of ever-narrowing tubes that end in a myriad of tiny sacs called alveoli. It is in the alveoli that oxygen from the air transfers to the bloodstream and carbon dioxide from the bloodstream transfers to the lungs for exhalation. For air to move in and out of the lungs, the pressure inside the lungs must change, forcing the lungs to change volume—just as predicted by Boyle’s law. The pressure change is caused by the diaphragm, a muscle that covers the bottom of the lungs. When the diaphragm moves down, it expands the size of our lungs. When this happens, the air pressure inside our lungs decreases slightly. This causes new air to rush in, and we inhale. The pressure decrease is slight—only 3 torr, or about 0.4% of an atmosphere. We inhale only 0.5–1.0 L of air per normal breath. Exhaling air requires that we relax the diaphragm, which pushes against the lungs and slightly decreases the volume of the lungs. This slightly increases the pressure of the air in the lungs, and air is forced out; we exhale. Only 1–2 torr of extra pressure is needed to exhale. So with every breath, our own bodies are performing an experimental test of Boyle’s law. Charles’s Law Another simple gas law relates the volume of a gas to its temperature. Experiments indicate that as the temperature of a gas sample is increased, its volume increases as long as the pressure and the amount of gas remain constant. The way to write this mathematically is \[\mathrm V \propto T \nonumber \] At this point, the concept of temperature must be clarified. Although the Kelvin scale is the preferred temperature scale, the Celsius scale is also a common temperature scale used in science. The Celsius scale is based on the melting and boiling points of water and is actually the common temperature scale used by most countries around the world (except for the United States, which still uses the Fahrenheit scale). The value of a Celsius temperature is directly related to its Kelvin value by a simple expression: Kelvin temperature = Celsius temperature + 273 Thus, it is easy to convert from one temperature scale to another. The Kelvin scale is sometimes referred to as the absolute scale because the zero point on the Kelvin scale is at absolute zero, the coldest possible temperature. On the other temperature scales, absolute zero is −260°C or −459°F. The expression relating a gas volume to its temperature begs the following question: to which temperature scale is the volume of a gas related? The answer is that gas volumes are directly related to the Kelvin temperature . Therefore, the temperature of a gas sample should always be expressed in (or converted to) a Kelvin temperature. Example \(\PageIndex{4}\): Increasing Temperature What happens to the volume of a gas if its temperature is decreased? Assume that all other conditions remain constant. Solution If the temperature of a gas sample is decreased, the volume decreases as well. Exercise \(\PageIndex{4}\) What happens to the temperature of a gas if its volume is increased? Assume that all other conditions remain constant. Answer The temperature increases. As with Boyle’s law, the relationship between volume and temperature can be expressed in terms of initial and final values of volume and temperature, as follows: \[\mathrm{\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}} \nonumber \] where V i and T i are the initial volume and temperature, and V f and T f are the final volume and temperature. This is Charles’s law. The restriction on its use is that the pressure of the gas and the amount of gas must remain constant. (Charles’s law is sometimes referred to as Gay-Lussac’s law, after the scientist who promoted Charles’s work.) Example \(\PageIndex{5}\) A gas sample at 20°C has an initial volume of 20.0 L. What is its volume if the temperature is changed to 60°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant. Solution Although the temperatures are given in degrees Celsius, we must convert them to the kelvins before we can use Charles’s law. Thus, 20°C + 273 = 293 K = T i 60°C + 273 = 333 K = T f Now we can substitute these values into Charles’s law, along with the initial volume of 20.0 L: \(\mathrm{\dfrac{20.0\: L}{293\: K}=\dfrac{V_f}{333\: K}}\) Multiplying the 333 K to the other side of the equation, we see that our temperature units will cancel: \(\mathrm{\dfrac{(333\: K)(20.0\: L)}{293\: K}=V_f}\) Solving for the final volume, V f = 22.7 L. So, as the temperature is increased, the volume increases. This makes sense because volume is directly proportional to the absolute temperature (as long as the pressure and the amount of the remain constant). Exercise \(\PageIndex{5}\) A gas sample at 35°C has an initial volume of 5.06 L. What is its volume if the temperature is changed to −35°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant. Answer 3.91 L Combined Gas Law Other gas laws can be constructed, but we will focus on only two more. The combined gas law brings Boyle’s and Charles’s laws together to relate pressure, volume, and temperature changes of a gas sample: \[\mathrm{\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}} \nonumber \] To apply this gas law, the amount of gas should remain constant. As with the other gas laws, the temperature must be expressed in kelvins, and the units on the similar quantities should be the same. Because of the dependence on three quantities at the same time, it is difficult to tell in advance what will happen to one property of a gas sample as two other properties change. The best way to know is to work it out mathematically. Example \(\PageIndex{6}\) A sample of gas has P i = 1.50 atm, V i = 10.5 L, and T i = 300 K. What is the final volume if P f = 0.750 atm and T f = 350 K? Solution Using the combined gas law, substitute for five of the quantities: \(\mathrm{\dfrac{(1.50\: atm)(10.5\: L)}{300\: K}=\dfrac{(0.750\: atm)(V_f)}{350\: K}}\) We algebraically rearrange this expression to isolate V f on one side of the equation: \(\mathrm{V_f=\dfrac{(1.50\: atm)(10.5\: L)(350\: K)}{(300\: K)(0.750\: atm)}=24.5\: L}\) Note how all the units cancel except the unit for volume. A sample of gas has P i = 0.768 atm, V i = 10.5 L, and T i = 300 K. What is the final pressure if V f = 7.85 L and T f = 250 K? Answer 0.856 atm Example \(\PageIndex{7}\) A balloon containing a sample of gas has a temperature of 22°C and a pressure of 1.09 atm in an airport in Cleveland. The balloon has a volume of 1,070 mL. The balloon is transported by plane to Denver, where the temperature is 11°C and the pressure is 655 torr. What is the new volume of the balloon? Solution The first task is to convert all quantities to the proper and consistent units. The temperatures must be expressed in kelvins, and the pressure units are different so one of the quantities must be converted. Let us convert the atmospheres to torr: 22°C + 273 = 295 K = T i 11°C + 273 = 284 K = T f \(\mathrm{1.09\: atm\times\dfrac{760\: torr}{1\: atm}=828\: torr = P_i}\) Now we can substitute the quantities into the combined has law: \(\mathrm{\dfrac{(828\: torr)(1,070\: mL)}{295\: K}=\dfrac{(655\: torr)\times V_f}{284\: K}}\) To solve for V f , we multiply the 284 K in the denominator of the right side into the numerator on the left, and we divide 655 torr in the numerator of the right side into the denominator on the left: \(\mathrm{\dfrac{(828\: torr)(1,070\: mL)(284\: K)}{(295\: K)(655\: torr)}=V_f}\) Notice that torr and kelvins cancel, as they are found in both the numerator and denominator. The only unit that remains is milliliters, which is a unit of volume. So V f = 1,300 mL. The overall change is that the volume of the balloon has increased by 230 mL. A balloon used to lift weather instruments into the atmosphere contains gas having a volume of 1,150 L on the ground, where the pressure is 0.977 atm and the temperature is 18°C. Aloft, this gas has a pressure of 6.88 torr and a temperature of −15°C. What is the new volume of the gas? Answer 110,038 L The Ideal Gas Law So far, the gas laws we have used have focused on changing one or more properties of the gas, such as its volume, pressure, or temperature. There is one gas law that relates all the independent properties of a gas under any particular condition, rather than a change in conditions. This gas law is called the ideal gas law. The formula of this law is as follows: PV = nRT In this equation, P is pressure, V is volume, n is amount of moles, and T is temperature. R is called the ideal gas law constant and is a proportionality constant that relates the values of pressure, volume, amount, and temperature of a gas sample. The variables in this equation do not have the subscripts i and f to indicate an initial condition and a final condition. The ideal gas law relates the four independent properties of a gas under any conditions. The value of R depends on what units are used to express the other quantities. If volume is expressed in liters and pressure in atmospheres, then the proper value of R is as follows: \[\mathrm{R=0.08205 \: \dfrac{L\cdot atm}{mol\cdot K}} \nonumber \] This may seem like a strange unit, but that is what is required for the units to work out algebraically. Example \(\PageIndex{8}\) What is the volume in liters of 1.45 mol of N 2 gas at 298 K and 3.995 atm? Solution Using the ideal gas law where P = 3.995 atm, n = 1.45, and T = 298, \(\mathrm{(3.995\: atm)\times V=(1.45\: mol)\left(0.08205\: \dfrac{L\cdot atm}{mol\cdot K}\right)(298\: K)}\) On the right side, the moles and kelvins cancel. Also, because atmospheres appear in the numerator on both sides of the equation, they also cancel. The only remaining unit is liters, a unit of volume. So 3.995 × V = (1.45)(0.08205)(298) L Dividing both sides of the equation by 3.995 and evaluating, we get V = 8.87 L. Note that the conditions of the gas are not changing. Rather, the ideal gas law allows us to determine what the fourth property of a gas (here, volume) must be if three other properties (here, amount, pressure, and temperature) are known. What is the pressure of a sample of CO 2 gas if 0.557 mol is held in a 20.0 L container at 451 K? Answer 1.03 atm For convenience, scientists have selected 273 K (0°C) and 1.00 atm pressure as a set of standard conditions for gases. This combination of conditions is called standard temperature and pressure (STP). Under these conditions, 1 mol of any gas has about the same volume. We can use the ideal gas law to determine the volume of 1 mol of gas at STP: \[\mathrm{(1.00\: atm)\times V=(1.00\: mol)\left(0.08205\: \dfrac{L\cdot atm}{mol\cdot K}\right)(273\: K)} \nonumber \] This volume is 22.4 L. Because this volume is independent of the identity of a gas, the idea that 1 mol of gas has a volume of 22.4 L at STP makes a convenient conversion factor: 1 mol gas = 22.4 L (at STP) Example \(\PageIndex{9}\) Cyclopropane (C 3 H 6 ) is a gas that formerly was used as an anesthetic. How many moles of gas are there in a 100.0 L sample if the gas is at STP? Solution We can set up a simple, one-step conversion that relates moles and liters: \(\mathrm{100.0\: L\: C_3H_6\times \dfrac{1\: mol}{22.4\: L}=4.46\: mol\: C_3H_6}\) There are almost 4.5 mol of gas in 100.0 L. Note: Because of its flammability, cyclopropane is no longer used as an anesthetic gas. Freon is a trade name for a series of fluorine- and chlorine-containing gases that formerly were used in refrigeration systems. What volume does 8.75 mol of Freon have at STP? Note: Many gases known as Freon are no longer used because their presence in the atmosphere destroys the ozone layer, which protects us from ultraviolet light from the sun. Answer 196 L Airbags Airbags (Figure \(\PageIndex{3}\)) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN 3 . When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN 3 to initiate its decomposition: \[\ce{2NaN3}(s)\rightarrow \ce{3N2}(g)+\ce{2Na}(s) \nonumber \] This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN 3 will generate approximately 50 L of N 2 . Dalton's Law of Partial Pressures The ideal gas equation of state applies to mixtures just as to pure gases. It was in fact with a gas mixture, ordinary air, that Boyle, Gay-Lussac and Charles did their early experiments. The only new concept we need in order to deal with gas mixtures is the partial pressure , a concept invented by the famous English chemist John Dalton (1766-1844). Dalton reasoned that the low density and high compressibility of gases indicates that they consist mostly of empty space; from this it follows that when two or more different gases occupy the same volume, they behave entirely independently. The contribution that each component of a gaseous mixture makes to the total pressure of the gas is known as the partial pressure of that gas. The definition of Dalton's Law of Partial Pressures that address this is: The total pressure of a gas is the sum of the partial pressures of its components which is expressed algebraically as \[P_{total}=P_1+P_2+P_3 ... = \sum_i P_i \nonumber \] or, equivalently \[ P_{total} = \dfrac{RT}{V} \sum_i n_i \nonumber \] There is also a similar relationship based on volume fractions , known as Amagat's law of partial volumes . It is exactly analogous to Dalton's law, in that it states that the total volume of a mixture is just the sum of the partial volumes of its components. But there are two important differences: Amagat's law holds only for ideal gases which must all be at the same temperature and pressure. Dalton's law has neither of these restrictions. Although Amagat's law seems intuitively obvious, it sometimes proves useful in chemical engineering applications. We will make no use of it in this course. Example \(\PageIndex{10}\) Three flasks having different volumes and containing different gases at various pressures are connected by stopcocks as shown. When the stopcocks are opened, What will be the pressure in the system? Which gas will be most abundant in the mixture? Assume that the temperature is uniform and that the volume of the connecting tubes is negligible. Solution The trick here is to note that the total number of moles n T and the temperature remain unchanged, so we can make use of Boyle's law PV = constant. We will work out the details for CO 2 only, denoted by subscripts a. For CO 2 , \[P_aV_a = (2.13\; atm)(1.50\; L) = 3.20\; L \cdot atm \nonumber \] Adding the PV products for each separate container, we obtain \[A\sum_i P_i V_i = 6.36\; L \cdot atm = n_T RT \nonumber \] We will call this sum P 1 V 1 . After the stopcocks have been opened and the gases mix, the new conditions are denoted by P 2 V 2 . From Boyle's law (\(\ref{Eq1}\), \[P_1V_1 = P_2V_2 = 6.36\; L \cdot atm \nonumber \] \[V_2 = \sum_i V_i = 4.50\; L \nonumber \] Solving for the final pressure P 2 we obtain (6.36 L-atm)/(4.50 L) = 1.41 atm . For part (b) , note that the number of moles of each gas is n = PV/RT . The mole fraction of any one gas is X i = n i / n T . For CO 2 , this works out to (3.19/ RT ) / (6.36/ RT ) = 0.501. Because this exceeds 0.5, we know that this is the most abundant gas in the final mixture. Dalton’s law states that in a gas mixture (\(P_{total}\)) each gas will exert a pressure independent of the other gases (\(P_n\)) and each gas will behave as if it alone occupies the total volume. By extension, the partial pressure of each gas can be calculated by multiplying the total pressure (\(P_{total}\)) by the gas percentage (%). \[P_{Total} = P_1 + P_2 + P_3 + P_4 + ... + P_n \nonumber \] or \[P_n = \dfrac{\text{% of individual gas}_n}{P_{Total}} \nonumber \] Gas Partial Pressure (mm Hg) Percentage (%) Nitrogen, (N_2\) \(P_{N_2}\) = 594 78.000 Oxygen, \(O_2\) \(P_{O_2}\)= 160 21.000 Carbon Dioxide, \(CO_2\) \(P_{CO_2}\) = 0.25 0.033 Water Vapor, \(H_2O\) \(P_{H_2O}\) = 5.7 0.750 Other trace gases \(P_{Other}\) = 0.05 0.220 Total air \(P_{Total}\) = 760 1.000 Application of Dalton's Law: Collecting Gases over Water A common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. This arrangement is called a pneumatic trough , and was widely used in the early days of chemistry. As the gas enters the bottle it displaces the water and becomes trapped in the upper part. The volume of the gas can be observed by means of a calibrated scale on the bottle, but what about its pressure? The total pressure confining the gas is just that of the atmosphere transmitting its force through the water. (An exact calculation would also have to take into account the height of the water column in the inverted tube.) But liquid water itself is always in equilibrium with its vapor, so the space in the top of the tube is a mixture of two gases: the gas being collected, and gaseous H 2 O. The partial pressure of H 2 O is known as the vapor pressure of water and it depends on the temperature. In order to determine the quantity of gas we have collected, we must use Dalton's Law to find the partial pressure of that gas. Example \(\PageIndex{11}\) Oxygen gas was collected over water as shown above. The atmospheric pressure was 754 torr, the temperature was 22°C, and the volume of the gas was 155 mL. The vapor pressure of water at 22°C is 19.8 torr. Use this information to estimate the number of moles of \(O_2\) produced. Solution From Dalton's law, \[P_{O_2} = P_{total} – P_{H_2O} = 754 – 19.8 = 734 \; torr = 0.966\; atm \nonumber \] Now use the Ideal Gas Law to convert to moles \[ n =\dfrac{PV}{RT} = \dfrac{(0.966\; atm)(0.155\;L)}{(0.082\; L atm mol^{-1} K^{-1})(295\; K)}= 0.00619 \; mol \nonumber \] Henry’s Law Henry's law is one of the gas laws formulated by William Henry in 1803. It states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. To explain this law, Henry derived the equation: \[ C =k P_{gas} \nonumber \] where Henry’s Law tells us that the greater the pressure of gas above the surface of a liquid, the higher the concentration of the gas in the liquid. Also, Henry’s law tells us that gases diffuse from areas of high gas concentration to areas of low gas concentration. Applicability of Henry's Law Henry's law only works if the molecules are at equilibrium. Henry's law does not work for gases at high pressures (e.g., \(N_{2\;(g)}\) at high pressure becomes very soluble and harmful when in the blood supply ). Henry's law does not work if there is a chemical reaction between the solute and solvent (e.g., \(HCl_{(g)}\) reacts with water by a dissociation reaction to generate \(H_3O^+\) and \(Cl^-\) ions). Application of Henry's Law: Scuba diving Our respiratory systems are designed to maintain the proper oxygen concentration in the blood when the partial pressure of O 2 is 0.21 atm, its normal sea-level value. Below the water surface, the pressure increases by 1 atm for each 10.3 m increase in depth; thus a scuba diver at 10.3 m experiences a total of 2 atm pressure pressing on the body. In order to prevent the lungs from collapsing, the air the diver breathes should also be at about the same pressure. But at a total pressure of 2 atm, the partial pressure of \(O_2\) in ordinary air would be 0.42 atm; at a depth of 100 ft (about 30 m), the \(O_2\) pressure of 0.8 atm would be far too high for health. For this reason, the air mixture in the pressurized tanks that scuba divers wear must contain a smaller fraction of \(O_2\). This can be achieved most simply by raising the nitrogen content, but high partial pressures of N 2 can also be dangerous, resulting in a condition known as nitrogen narcosis. The preferred diluting agent for sustained deep diving is helium, which has very little tendency to dissolve in the blood even at high pressures. Career Focus: Respiratory Therapist Certain diseases—such as emphysema, lung cancer, and severe asthma—primarily affect the lungs. Respiratory therapists help patients with breathing-related problems. They can evaluate, help diagnose, and treat breathing disorders and even help provide emergency assistance in acute illness where breathing is compromised. Most respiratory therapists must complete at least two years of college and earn an associate’s degree, although therapists can assume more responsibility if they have a college degree. Therapists must also pass state or national certification exams. Once certified, respiratory therapists can work in hospitals, doctor’s offices, nursing homes, or patient’s homes. Therapists work with equipment such as oxygen tanks and respirators, may sometimes dispense medication to aid in breathing, perform tests, and educate patients in breathing exercises and other therapy. Because respiratory therapists work directly with patients, the ability to work well with others is a must for this career. It is an important job because it deals with one of the most crucial functions of the body. Key Takeaway The physical properties of gases are predictable using mathematical formulas known as gas laws. \(C\) is the solubility of a gas at a fixed temperature in a particular solvent (in units of M or mL gas/L) \(k\) is Henry's law constant (often in units of M/atm) \(P_{gas}\) is the partial pressure of the gas (often in units of Atm) |
Courses/Ursinus_College/CHEM322%3A_Inorganic_Chemistry/08%3A_Electronic_Structure_of_Coordination_Complexes | Coordination compounds are important to all areas of chemistry, engineering, the life and environmental sciences, and beyond. In the synthetic laboratory catalytic amounts of coordination compounds enable organic chemists to synthesize new compounds selectively and in high yield under mild conditions. Applied industrially, coordination compound catalysts serve as vital catalysts that facilitate the conversion of raw petrochemical or bio-derived feedstocks into useful industrial and consumer products. Without them life as we know it would be impossible, as many biochemical systems are coordination complexes. Examples include the hemoglobin that transports oxygen around our bodies and the myoglobin that stores it, the photosystems that harvest light and use light energy in photosynthesis, the constituents of the respiratory chain, and many of the enzymes involved in the expression and transmission of genetic information. In studying coordination chemistry you are about to take your first steps into a vast and exciting world. 8.1: Introduction to Coordination Complexes 8.1.1: What are Coordination Complexes? 8.1.2: History of Coordination Complexes 8.1.3: Nomenclature and Ligands 8.1.4: Coordination Numbers and Structures 8.1.5: Isomerism 8.2: Crystal Field Theory 8.2.1: Crystal Field Theory 8.2.2: Crystal Field Stabilization Energy 8.2.3: Non-octahedral Complexes 8.3: Crystal Field Theory and Magnetism 8.3.1: Jahn-Teller Distortions 8.3.2: Magnetism 8.3.3: Magnetic Moments of Transition Metals 8.3.4: Ferro-, Ferri- and Antiferromagnetism 8.4: Ligand Field Theory 8.4.1: Ligand Field Theory 8.4.2: The Spectrochemical Series 8.4.3: Factors That Affect Ligand Field Splitting 8.4.4: Octahedral vs. Tetrahedral Geometries 8.5: Absorption Spectroscopy of Coordination Complexes 8.5.1: Absorption of Light 8.5.2: Colors of Coordination Complexes 8.5.3: Charge-Transfer Spectra 8.6: Tanabe Sugano Diagrams 8.6.1: Tanabe-Sugano Diagrams 8.6.2: Selection Rules 8.6.3: Applications of Tanabe-Sugano Diagrams 8.6.4: Tetrahedral Complexes |
Courses/Lebanon_Valley_College/CHM_311%3A_Physical_Chemistry_I_(Lebanon_Valley_College)/06%3A_Molecular_Structure_and_Bonding/6.19%3A_Molecular_Term_Symbols_Designate_Symmetry | The quantum numbers for diatomic molecules are similar from the atomic quantum numbers. Be cautious, because the rules for finding the possible combinations are different. The total orbital angular momentum quantum number For the molecular case, this number is called \(Λ\) instead of \(L\). It follows the same naming convention as \(L\), except that instead of using capital English letters, it uses capital Greek letters: \(Λ = 0 \rightarrow Σ \) \(Λ = 1 \rightarrow Π\) \(Λ = 2 \rightarrow Δ \) \(Λ = 3 \rightarrow Φ \) Unlike \(L\), there is not a general formula for finding the possible combinations of \(Λ\). You have to examine the individual microstates. This is easier than it sounds. The total magnetic quantum number \(M_L\): \(M_L\) works like \(M_l\) with atoms, except that there is no formula for finding the combinations. The total spin magnetic quantum number \(M_S\): \(M_S\) works exactly like \(M_s\). Electrons can either point with or against the z ‐ axis, and being in a molecular orbital versus an atomic orbital doesn’t change this. \(M_S\) can range from \(m_{s1} + m_{s2}\) to \(m_{s1} ‐ m_{s2}\). Two New Symmetries: Parity and Reflection Molecular orbitals are more complex than atomic ones and require more modifiers to completely define. Parity (sometimes called “inversion”) tells you if the orbital is symmetric or anti‐symmetric when an inversion operation is performed. The symmetry notation u and g are sometimes used when describing molecular orbitals. This refers to the operation of inversion, which requires starting at an arbitrary point in the orbital, traveling straight through the center, and then continuing outwards an equal distance from the center. The orbital is designated g (for gerade, even) if the phase is the same, and u (for ungerade, uneven) if the phase changes sign. To determine whether or not a given state is \(g\) or \(u\), find the parity of each individual open‐shell electron and uses these simple (Laporte rules): \(g + g \rightarrow g \) \(g + u \rightarrow u \) \(u + u \rightarrow g \) What is the parity of the state \(1σ_g^21σ_u^22σ_g^22σ_u^22π_u^12π_u^1\) ? Solution Since both open shell electrons are ungerade, the overall parity is g. Helpful hint: bonding sigma orbitals and anti ‐ bonding pi orbitals are always gerade. Anti ‐ bonding sigmas and bonding pis are always ungerade. Draw them and see for yourself. Reflection determines if a given orbital is symmetric or anti‐symmetric upon reflection through a plane that contains both nuclei. The choice of symmetry planes is arbitrary. As long as you pick a plane and stick with it, you will always get the right answer. When an orbital is symmetric, it is labeled +. When an orbital is anti ‐ symmetric, it is labeled ‐ . To find the overall reflection of a state, use these rules: (+)(+) \rightarrow + (+)(‐) \rightarrow ‐ (‐)(‐) \rightarrow + Reflection only applies to Σ states! For Λ > 0, there are no reflection labels! If you experiment with the rules, you will quickly realize why this is the case. What is the reflection of the state \(1σ_g^21σ_u^22σ_g^22σ_u^22π_u^12π_u^1\) ? Solution You need to know what the orbitals look like. Draw a picture and then pick a plane. For this example, the plane of the page is selcted, but the orthogonal plane would have worked just as well. The “vertical” orbital is + The “horizontal” orbital is ‐ Since one is + and one is ‐ , the overall reflection is ‐ . Try using the orthogonal plane and convince yourself that you still get the same answer. What are the term symbols for \(O_2\) ? Solution The molecular orbital diagram for \(O_2\) is Where I chose arbitrary configurations for the last two electrons. There are two open ‐ shell electrons occupying the anti‐bonding \(π_g\) orbitals. These are the only electrons that matter. It is easiest to simply draw all of the permutations and figure out the bounds on \(Λ\) and \(M_L\) by inspection. If we do this, it is easy to see that \(Λ = 2,0\) and that \(M_L = 2,0, ‐ 2\) 0 1 2 \(M_s= -1\) \(M_s= 0\) \(M_s= 2\) 0 1 0 1 2 1 0 1 0 The top row is a \(Λ=2\) \(M_S =0\) state, so it is \(^1Δ\). Both electrons are in the rightmost orbital. This orbital is gerade, and (g)(g) = g, so the parity label is g. We do not assign reflection labels to non Σ states, so the term symbol is \(^1Δ_g\). After removing the used up microstates, the chart becomes 0 1 2 \(M_s= -1\) \(M_s= 0\) \(M_s= 2\) 1 1 0 This is a Λ =0 state with three possible spin configurations, so it is \(^3Σ\). We know that the electrons are in different sub‐orbitals (if you cannot see this, try drawing all of the possible combinations that give \(Λ =0\)). Both of the orbitals are gerade, so the overall parity is gerade. One of the orbitals will be +, the other will be ‐ . The final answer is \(^1Δ_g \(^3Σ^+_g\) Write the term symbols for \(O_2^‐\) Solution First draw the electron configuration diagram. There are only two possible configurations. It should be easy to see that the term symbol is \(^2\Pi_g\). |
Courses/Rio_Hondo/Chemistry_110%3A_An_Introduction_to_General_Organic_and_Biological_Chemistry_(Garg)/11%3A_Alcohols_Thiols_Aldehydes_and_Ketones/11.06%3A_Reactions_of_Alcohols | Learning Objectives Give two major types of reactions of alcohols. Describe the result of the oxidation of a primary alcohol. Describe the result of the oxidation of a secondary alcohol. Chemical reactions in alcohols occur mainly at the functional group, but some involve hydrogen atoms attached to the OH -bearing carbon atom or to an adjacent carbon atom. Of the three major kinds of alcohol reactions, which are summarized in Figure \(\PageIndex{1}\), two—dehydration and oxidation—are considered here. The third reaction type— esterification —is covered elsewhere. Dehydration As noted in Figure \(\PageIndex{1}\), an alcohol undergoes dehydration in the presence of a catalyst to form an alkene and water. The reaction removes the OH group from the alcohol carbon atom and a hydrogen atom from an adjacent carbon atom in the same molecule: Under the proper conditions, it is possible for the dehydration to occur between two alcohol molecules. The entire OH group of one molecule and only the hydrogen atom of the OH group of the second molecule are removed. The two ethyl groups attached to an oxygen atom form an ether molecule. ( Ethers are discussed in elsewhere ) Thus, depending on conditions, one can prepare either alkenes or ethers by the dehydration of alcohols. Both dehydration and hydration reactions occur continuously in cellular metabolism, with enzymes serving as catalysts and at a temperature of about 37°C. The following reaction occurs in the "Embden–Meyerhof" pathway Although the participating compounds are complex, the reaction is the same: elimination of water from the starting material. The idea is that if you know the chemistry of a particular functional group, you know the chemistry of hundreds of different compounds. Oxidation Primary and secondary alcohols are readily oxidized. We saw earlier how methanol and ethanol are oxidized by liver enzymes to form aldehydes. Because a variety of oxidizing agents can bring about oxidation, we can indicate an oxidizing agent without specifying a particular one by writing an equation with the symbol [O] above the arrow. For example, we write the oxidation of ethanol—a primary alcohol—to form acetaldehyde—an aldehyde—as follows: We shall see that aldehydes are even more easily oxidized than alcohols and yield carboxylic acids. Secondary alcohols are oxidized to ketones . The oxidation of isopropyl alcohol by potassium dichromate (\(\ce{K2Cr2O7}\)) gives acetone, the simplest ketone: Unlike aldehydes, ketones are relatively resistant to further oxidation, so no special precautions are required to isolate them as they form. Note that in oxidation of both primary (RCH 2 OH) and secondary (R 2 CHOH) alcohols, two hydrogen atoms are removed from the alcohol molecule, one from the OH group and other from the carbon atom that bears the OH group. These reactions can also be carried out in the laboratory with chemical oxidizing agents. One such oxidizing agent is potassium dichromate. The balanced equation (showing only the species involved in the reaction) in this case is as follows: \[\ce{8H^{=} + Cr2O7^{2-} + 3CH3CH2OH -> 3CH3CHO + 2Cr^{3+} + 7H2O} \nonumber \] Alcohol oxidation is important in living organisms. Enzyme-controlled oxidation reactions provide the energy cells need to do useful work. One step in the metabolism of carbohydrates involves the oxidation of the secondary alcohol group in isocitric acid to a ketone group: The overall type of reaction is the same as that in the conversion of isopropyl alcohol to acetone. Tertiary alcohols (R 3 COH) are resistant to oxidation because the carbon atom that carries the OH group does not have a hydrogen atom attached but is instead bonded to other carbon atoms. The oxidation reactions we have described involve the formation of a carbon-to-oxygen double bond. Thus, the carbon atom bearing the OH group must be able to release one of its attached atoms to form the double bond. The carbon-to-hydrogen bonding is easily broken under oxidative conditions, but carbon-to-carbon bonds are not. Therefore tertiary alcohols are not easily oxidized. Example \(\PageIndex{1}\) Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow. CH 3 CH 2 CH 2 CH 2 CH 2 OH Solution The first step is to recognize the class of each alcohol as primary, secondary, or tertiary. This alcohol has the OH group on a carbon atom that is attached to only one other carbon atom, so it is a primary alcohol. Oxidation forms first an aldehyde and further oxidation forms a carboxylic acid. This alcohol has the OH group on a carbon atom that is attached to three other carbon atoms, so it is a tertiary alcohol. No reaction occurs. This alcohol has the OH group on a carbon atom that is attached to two other carbon atoms, so it is a secondary alcohol; oxidation gives a ketone. Exercise \(\PageIndex{1}\) Write an equation for the oxidation of each alcohol. Use [O] above the arrow to indicate an oxidizing agent. If no reaction occurs, write “no reaction” after the arrow. Summary Alcohols can be dehydrated to form either alkenes (higher temperature, excess acid) or ethers (lower temperature, excess alcohol). Primary alcohols are oxidized to form aldehydes. Secondary alcohols are oxidized to form ketones. Tertiary alcohols are not readily oxidized. |
Courses/Fullerton_College/Introductory_Biochemistry/10%3A_Amino_Acids_Proteins_and_Enzymes/10.13%3A_Basic_Principles_of_Catalysis | A printable version of this section is here: BiochemFFA_4_1.pdf . The entire textbook is available for free from the authors at http://biochem.science.oregonstate.edu/content/biochemistry-free-and-easy If there is a magical component to life, an argument can surely be made for it being catalysis. Thanks to catalysis, reactions that can take hundreds of years to complete in the uncatalyzed “real world,” occur in seconds in the presence of a catalyst. Chemical catalysts, such as platinum, can speed reactions, but enzymes (which are simply super-catalysts with a “twist,” as we shall see) put chemical catalysts to shame (Figure 4.1). To understand enzymatic catalysis, it is necessary first to understand energy. Chemical reactions follow the universal trend of moving towards lower energy, but they often have a barrier in place that must be overcome. The secret to catalytic action is reducing the magnitude of that barrier. Before discussing enzymes, it is appropriate to pause and discuss an important concept relating to chemical/biochemical reactions. That concept is equilibrium and it is very often misunderstood. The “equi" part of the word relates to equal, as one might expect, but it does not relate to absolute concentrations. What happens when a biochemical reaction is at equilibrium is that the concentrations of reactants and products do not change over time. This does not mean that the reactions have stopped. Remember that reactions are reversible, so there is a forward reaction and a reverse reaction: if you had 8 molecules of A, and 4 of B at the beginning, and 2 molecules of A were converted to B, while 2 molecules of B were simultaneously converted back to A, the number of molecules of A and B remain unchanged, i.e., the reaction is at equilibrium. However, you will notice that this does not mean that there are equal numbers of A and B molecules. Concentration Matters So, contrary to the perceptions of many students, the concentrations of products and reactants are not equal at equilibrium, unless the ΔG°’ for a reaction is zero, because when this is the case, \[ΔG = \ln \left(\dfrac{[\rm{Products}]}{[\rm{Reactants}]} \right)\] since the ΔG°’ is zero. Because ΔG itself is zero at equilibrium, then \[[Products] = [Reactants].\] This is the only circumstance where \[[Products] = [Reactants]\] at equilibrium. Reiterating, at equilibrium, the concentrations of reactant and product do not change over time. That is, for a reaction \(A \rightleftharpoons B [A]\) at time zero when equilibrium is reached, \([A]_{T_0}\), will be the same 5 minutes later (assuming A and B are chemically stable). Thus, \[[A]_{T_0} = [A]_{T+5}\] Similarly, \[[B]_{T_0} = [B]_{T+5}\] For that matter, at any amount of time X after equilibrium has been reached, \[[A]T0 = [A]T+5 = [A]TX\] and \[[B]T0 = [B] T+5 = [B]TX\] However, unless ΔG°’ = 0, it is wrong to say [A]T0 = [B]T0 As we study biochemical reactions and reaction rates, it is important to remember that 1) reactions do not generally start at equilibrium; 2) all reactions move in the direction of equilibrium; and 3) reactions in cells behave just like those in test tubes - they do not begin at equilibrium, but they move towards it. Dynamic reactions The reactions occurring in cells, though, are very dynamic and complex. In a test tube, they can be studied one at a time. In cells, the product of one reaction is often the substrate for another one. Reactions in cells are interconnected in this way, giving rise to what are called metabolic pathways. There are, in fact, thousands of different interconnected reactions going on continuously in cells. Attempts to study a single reaction in the chaos of a cell is daunting to say the least. For this reason, biochemists isolate enzymes from cells and study reactions individually. It is with this in mind that we begin our consideration of the phenomenon of catalysis by describing, first, the way in which enzymes work. Activation energy Figure 4.2 schematically depicts the energy changes that occur during the progression of a simple reaction. In order for the reaction to proceed, an activation energy must be overcome in order for the reaction to occur. In Figure 4.3, the activation energy for a catalyzed reaction is overlaid. As you can see, the reactants start at the same energy level for both catalyzed and uncatalyzed reactions and that the products end at the same energy for both as well. The catalyzed reaction, however, has a lower energy of activation (dotted line) than the uncatalyzed reaction. This is the secret to catalysis - overall ΔG for a reaction does NOT change with catalysis, but the activation energy is lowered. Figure 4.3 - Energy changes during the course of an uncatalyzed reaction (solid green line) and a catalyzed reaction (dotted green line). Image by Aleia Kim Reversibility The extent to which reactions will proceed forward is a function of the size of the energy difference between the product and reactant states. The lower the energy of the products compared to the reactants, the larger the percentage of molecules that will be present as products at equilibrium. It is worth noting that since an enzyme lowers the activation energy for a reaction that it can speed the reversal of a reaction just as it speeds a reaction in the forward direction. At equilibrium, of course, no change in concentration of reactants and products occurs. Thus, enzymes speed the time required to reach equilibrium, but do not affect the balance of products and reactants at equilibrium. Exceptions The reversibility of enzymatic reactions is an important consideration for equilibrium, the measurement of enzyme kinetics, for Gibbs free energy, for metabolic pathways, and for physiology. There are some minor exceptions to the reversibility of reactions, though. They are related to the disappearance of a substrate or product of a reaction. Consider the first reaction below which is catalyzed by the enzyme carbonic anhydrase: \[CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons HCO_3^- + H^+\] In the forward direction, carbonic acid is produced from water and carbon dioxide. It can either remain intact in the solution or ionize to produce bicarbonate ion and a proton. In the reverse direction, water and carbon dioxide are produced. Carbon dioxide, of course, is a gas and can leave the solution and escape. When reaction molecules are removed, as they would be if carbon dioxide escaped, the reaction is pulled in the direction of the molecule being lost and reversal cannot occur unless the missing molecule is replaced. In the second reaction occurring on the right, carbonic acid (H2CO3) is “removed” by ionization. This too would limit the reaction going back to carbon dioxide in water. This last type of “removal” is what occurs in metabolic pathways. In this case, the product of one reaction (carbonic acid) is the substrate for the next (formation of bicarbonate and a proton). In the metabolic pathway of glycolysis , ten reactions are connected in this manner and reversing the process is much more complicated than if just one reaction was being considered. General mechanisms of action As noted above, enzymatically catalyzed reactions are orders of magnitude faster than uncatalyzed and chemical-catalyzed reactions. The secret of their success lies in a fundamental difference in their mechanisms of action. Every chemistry student has been taught that a catalyst speeds a reaction without being consumed by it. In other words, the catalyst ends up after a reaction just the way it started so it can catalyze other reactions, as well. Enzymes share this property, but in the middle, during the catalytic action, an enzyme is transiently changed. In fact, it is the ability of an enzyme to change that leads to its incredible efficiency as a catalyst. Changes These changes may be subtle electronic ones, more significant covalent modifications, or structural changes arising from the flexibility inherent in enzymes, but not present in chemical catalysts. Flexibility allows movement and movement facilitates alteration of electronic environments necessary for catalysis. Enzymes are, thus, much more efficient than rigid chemical catalysts as a result of their abilities to facilitate the changes necessary to optimize the catalytic process. Substrate binding Another important difference between the mechanism of action of an enzyme and a chemical catalyst is that an enzyme has binding sites that not only ‘grab’ the substrate (molecule involved in the reaction being catalyzed), but also place it in a position to be electronically induced to react, either within itself or with another substrate. The enzyme itself may play a role in the electronic induction or the induction may occur as a result of substrates being placed in very close proximity to each other. Chemical catalysts have no such ability to bind substrates and are dependent upon them colliding in the right orientation at or near their surfaces. Active site Reactions in an enzyme are catalyzed at a specific location within it known as the ‘active site’. Substrates bind at the active site and are oriented to provide access for the relevant portion of the molecule to the electronic environment of the enzyme where catalysis occurs. Enzyme flexibility As mentioned earlier, a difference between an enzyme and a chemical catalyst is that an enzyme is flexible. Its slight changes in shape (often arising from the binding of the substrate itself) help to optimally position substrates for reaction after they bind. Figure 4.5 - Lysozyme with substrate binding site (blue), active site (red) and bound substrate (black). Wikipedia Induced fit These changes in shape are explained, in part, by Koshland’s Induced Fit Model of Catalysis (Figure 4.6), which illustrates that not only do enzymes change substrates, but that substrates also transiently change enzyme structure. At the end of the catalysis, the enzyme is returned to its original state. Koshland’s model is in contrast to the Fischer Lock and Key model, which says simply that an enzyme has a fixed shape that is perfectly matched for binding its substrate(s). Enzyme flexibility also is important for control of enzyme activity. Enzymes alternate between the T (tight) state, which is a lower activity state and the R (relaxed) state, which has greater activity. Induced Fit The Koshland Induced Fit model of catalysis postulates that enzymes are flexible and change shape on binding substrate. Changes in shape help to 1) aid binding of additional substrates in reactions involving more than one substrate and/or 2) facilitate formation of an electronic environment in the enzyme that favors catalysis. This model is in contrast to the Fischer Lock and Key Model of catalysis which considers enzymes as having pre-formed substrate binding sites. Ordered binding The Koshland model is consistent with multi-substrate binding enzymes that exhibit ordered binding of substrates. For these systems, binding of the first substrate induces structural changes in the enzyme necessary for binding the second substrate. There is considerable experimental evidence supporting the Koshland model. Hexokinase, for example, is one of many enzymes known to undergo significant structural alteration after binding of substrate. In this case, the two substrates are brought into very close proximity by the induced fit and catalysis is made possible as a result. Enzyme kinetics To understand how an enzyme enhances the rate of a reaction, we must understand enzyme kinetics. We present a model here proposed by Leonor Michaelis and Maud Menten. In order to understand the model, it is necessary to understand a few parameters. First, we describe a reaction in simple terms proceeding as follows E + S ⇄ ES -> E + P where E is enzyme, S is substrate, and P is product. In this scheme, ES is the Enzyme-Substrate complex, which is simply the enzyme bound to its substrate. We could define the ES state a bit further with E + S ⇄ ES -> ES* -> EP -> E + P where ES* is the activated state and EP is the enzyme-product complex before release of the product. The first consideration we have is velocity. The velocity of a reaction is the rate of creation of product over time, measured as the concentration of product per time. The time is a critical consideration when measuring velocity. In a closed system (in which an enzyme operates), all reactions will advance towards equilibrium. Enzymatically catalyzed reactions are no different in the end result from non-enzymatic reactions, except that they get to equilibrium faster. Equilibrium At equilibrium, the ratio of product to reactant does not change. That is a property of equilibrium. Since the system is closed, the concentration of product over time will not change. The velocity will thus be zero under these conditions and we will have learned nothing about the reaction if we wait too long to study it. Velocity Consequently, in Michaelis-Menten kinetics, velocity is measured as initial velocity (V 0 ). This is accomplished by measuring the rate of formation of product early in the reaction before equilibrium is established and under these conditions, there is very little if any of the reverse reaction occurring. The other two assumptions are related. First, we use conditions where there is much more substrate than enzyme. This makes sense. If the substrate is not in great excess, then the enzyme’s conversion of substrate to product will occur much faster than the enzyme can bind substrate. Experimental considerations Now we turn our attention to how studies of the kinetic properties of an enzyme are conducted. To perform an analysis, one would do the following experiment - 20 different tubes would be set up with enzyme buffer (to keep the enzyme stable), the same amount of enzyme, and then a different amount of substrate in each tube, ranging from tiny amounts in the first tubes to very large amounts in the last tubes. The reaction would be allowed to proceed for a fixed, short amount of time and then the reaction would be stopped and the amount of product contained in each tube would be determined. The initial velocity (V 0 ) of the reaction then would be the concentration of product found in each tube divided by the time that the reaction was allowed to run. Data from the experiment would be plotted on a graph using initial velocity (V 0 ) on the Y-axis and the concentration of substrate on the X-axis, each tube, of course having a unique reaction velocity corresponding to a unique substrate concentration. For an enzyme following Michaelis-Menten kinetics, a curve like that shown in Figure 4.18 or 4.19 would result. At low concentration of substrate, it is limiting and the enzyme converts it into product as soon as it can bind it. Consequently, at low concentrations of substrate, the rate of increase of [P] is almost linear with [S] (Figure 4.19). Figure 4.19 - Linear relationship between [P] and [S] at low [S] Non-linear increase As the substrate concentration increases, however, the velocity of the reaction in tubes with higher substrate concentration ceases to increase linearly and instead begins to flatten out, indicating that as the substrate concentration gets higher and higher, the enzyme has a harder time keeping up to convert the substrate to product. Saturation Not surprisingly, when the enzyme becomes completely saturated with substrate, it will not have to wait for substrate to diffuse to it and will therefore be operating at maximum velocity. For an enzyme following Michaelis-Menten kinetics will have its velocity (v) at any given substrate concentration given by the following equation: V max Two terms in the equation above require explanation. The first is V max . It refers to the maximum velocity of an enzymatic reaction. Maximum velocity for a reaction occurs when an enzyme is saturated with substrate. Saturation is important because it means (per the assumption above) that none of the enzyme molecules are “waiting” for substrate after a product is released. Saturation ensures that another substrate is always instantly available. The unit of V max is concentration of product per time = [P]/time. On a plot of initial velocity versus substrate concentration (V0 vs. [S]), V max is the value on the Y axis that the curve asymptotically approaches (dotted line in Figure 4.20). It should be noted that the value of V max depends on the amount of enzyme used in a reaction. If you double the amount of enzyme used, you will double the V max . If one wanted to compare the velocities of two different enzymes, it would be necessary to use the same amounts of enzyme in the reaction each one catalyzes. Km The second term is Km (also known as Ks). Referred to as the Michaelis constant, Km is the substrate concentration that causes the enzyme to work at half of maximum velocity (Vmax/2). What it measures, in simple terms, is the affinity an enzyme has for its substrate. The value of Km is inversely related to the affinity of the enzyme for its substrate. Enzymes with a high Km value will have a lower affinity for their substrate (will take more substrate to get to Vmax/2) whereas those with a low Km will have high affinity and take less substrate to get to Vmax/2. The unit of Km is concentration. Affinities of enzymes for substrates vary considerably, so knowing Km helps us to understand how well an enzyme is suited to the substrate being used. Measurement of Km depends on the measurement of Vmax. Common mistake A common mistake students make in describing Vmax is saying that Km = Vmax /2. This is, of course, not true. Km is a substrate concentration and is the amount of substrate it takes for an enzyme to reach Vmax /2. On the other hand Vmax /2 is a velocity and a velocity certainly cannot equal a concentration. Lineweaver-Burk plots The study of enzyme kinetics is typically the most math intensive component of biochemistry and one of the most daunting aspects of the subject for many students. Although attempts are made to simplify the mathematical considerations, sometimes they only serve to confuse or frustrate students. Such is the case with modified enzyme plots, such as Lineweaver-Burk (Figure 4.26). Indeed, when presented by professors as simply another thing to memorize, who can blame students? In reality, both of these plots are aimed at simplifying the determination of parameters, such as \(K_m\) and \(V_{max}\). In making either of these modified plots, it is important to recognize that the same data is used as in making a V0 vs. [S] plot. The data are simply manipulated to make the plotting easier. Figure 4.26 - A Lineweaver-Burk plot of \(1/V_0\) vs \(1/[S]\). Image by Aleia Kim Double reciprocal For a LineWeaver-Burk plot, the manipulation is using the reciprocal of the values of both the velocity and the substrate concentration. The inverted values are then plotted on a graph as 1/V 0 vs. 1/[S]. Because of these inversions, Lineweaver-Burk plots are commonly referred to as ‘double-reciprocal’ plots. As can be seen in Figure 4.26, the value of Km on a Lineweaver Burk plot is easily determined as the negative reciprocal of the x-intercept , whereas the Vmax is the inverse of the y-intercept. Coenzymes Organic molecules that assist enzymes and facilitate catalysis are co-factors called coenzymes. The term co-factor is a broad category usually subdivided into inorganic ions and coenzymes. If the coenzyme is very tightly or covalently bound to the enzyme, it is referred to as a prosthetic group. Enzymes without their co-factors are inactive and referred to as apoenzymes. Enzymes containing all of their co-factors are called holoenzymes. |
Courses/Arkansas_Northeastern_College/CH14133%3A_Chemistry_for_General_Education/09%3A_Electrons_in_Atoms_and_the_Periodic_Table/9.09%3A_Periodic_Trends-_Atomic_Size%2C_Ionization_Energy%2C_and_Metallic_Character | Learning Objectives Be able to state how certain properties of atoms vary based on their relative position on the periodic table. One of the reasons the periodic table is so useful is because its structure allows us to qualitatively determine how some properties of the elements vary versus their position on the periodic table. The variations of properties versus positions on the periodic table are called periodic trends . There is no other tool in science that allows us to judge relative properties of a class of objects like this, which makes the periodic table a very useful tool. Many periodic trends are general. There may be a few points where an opposite trend is seen, but there is an overall trend when considered across a whole row or down a whole column of the periodic table. The first periodic trend we will consider is atomic radius. The atomic radius is an indication of the size of an atom. Although the concept of a definite radius of an atom is a bit fuzzy, atoms behave as if they have a certain radius. Such radii can be estimated from various experimental techniques, such as the x-ray crystallography of crystals. As you go down a column of the periodic table, the atomic radii increase. This is because the valence electron shell is getting larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus. This trend can be summarized as follows: \[as\downarrow PT,atomic\; radius \uparrow \nonumber \] where PT stands for periodic table. Going across a row on the periodic table, left to right, the trend is different. Even though the valence shell maintains the same principal quantum number, the number of protons—and hence the nuclear charge—is increasing as you go across the row. The increasing positive charge casts a tighter grip on the valence electrons, so as you go across the periodic table, the atomic radii decrease. Again, we can summarize this trend as follows: \[as\rightarrow PT,atomic\; radius \downarrow \nonumber \] Figure \(\PageIndex{1}\) shows spheres representing the atoms of the s and p blocks from the periodic table to scale, showing the two trends for the atomic radius. Example \(\PageIndex{1}\): Atomic Radii Referring only to a periodic table and not to Figure \(\PageIndex{1}\), which atom is larger in each pair? Si or S S or Te Solution Si is to the left of S on the periodic table; it is larger because as you go across the row, the atoms get smaller. S is above Te on the periodic table; Te is larger because as you go down the column, the atoms get larger. Exercise \(\PageIndex{1}\): Atomic Radii Referring only to a periodic table and not to Figure \(\PageIndex{1}\), which atom is smaller, Ca or Br? Answer Br Ionization energy (IE) is the amount of energy required to remove an electron from an atom in the gas phase: \[A(g)\rightarrow A^{+}(g)+e^{-}\; \; \; \; \; \Delta H\equiv IE \nonumber \] IE is usually expressed in kJ/mol of atoms. It is always positive because the removal of an electron always requires that energy be put in (i.e., it is endothermic). IE also shows periodic trends. As you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus. Thus, \[as\downarrow PT,\; IE\downarrow \nonumber \] However, as you go across the periodic table and the electrons get drawn closer in, it takes more energy to remove an electron; as a result, IE increases: \[as\rightarrow PT,\; IE\uparrow \nonumber \] Figure \(\PageIndex{2}\) shows values of IE versus position on the periodic table. Again, the trend is not absolute, but the general trends going across and down the periodic table should be obvious. IE also shows an interesting trend within a given atom. This is because more than one IE can be defined by removing successive electrons (if the atom has them to begin with): First Ionization Energy (IE 1 ): \[A(g) → A^+(g) + e^- \nonumber \] Second Ionization Energy (IE 2 ): \[A^{+}(g) → A^{2+}(g) + e^- \nonumber \] Third Ionization Energy (IE 3 ): \[A^{2+}(g) → A^{3+}(g) + e^- \nonumber \] and so forth. Each successive IE is larger than the previous because an electron is being removed from an atom with a progressively larger positive charge. However, IE takes a large jump when a successive ionization goes down into a new shell. For example, the following are the first three IEs for Mg, whose electron configuration is 1 s 2 2 s 2 2 p 6 3 s 2 : First Ionization Energy (IE 1 ) = 738 kJ/mol: \[\ce{Mg (g) -> Mg+ (g) + e− }\nonumber\] Second Ionization Energy (IE 2 ) = 1,450 kJ/mol: \[\ce{Mg+ (g) -> Mg^2+ (g) + e− }\nonumber\] Third Ionization Energy (IE 3 ) = 7,734 kJ/mol: \[\ce{Mg^2+ (g) -> Mg^3+ (g) + e− }\nonumber\] The second IE is twice the first, which is not a surprise: the first IE involves removing an electron from a neutral atom, while the second one involves removing an electron from a positive ion. The third IE, however, is over five times the previous one. Why is it so much larger? Because the first two electrons are removed from the 3 s subshell, but the third electron has to be removed from the n = 2 shell (specifically, the 2 p subshell, which is lower in energy than the n = 3 shell). Thus, it takes much more energy than just overcoming a larger ionic charge would suggest. It is trends like this that demonstrate that electrons within atoms are organized in groups. Example \(\PageIndex{2}\): Ionization Energies Which atom in each pair has the larger first ionization energy? Ca or Sr K or K + Solution Because Sr is below Ca on the periodic table, it is easier to remove an electron from it; thus, Ca has the higher IE. Because K + has a positive charge, it will be harder to remove another electron from it, so its IE is larger than that of K. Indeed, it will be significantly larger because the next electron in K + to be removed comes from another shell. Exercise \(\PageIndex{2}\): Ionization Energies Which atom has the lower ionization energy, C or F? Answer C The opposite of IE is described by electron affinity (EA), which is the energy change when a gas-phase atom accepts an electron: \[\ce{A (g) + e- -> A- (g) } \; \; \; \; \; \Delta H\equiv EA \nonumber \] EA is also usually expressed in kJ/mol. EA also demonstrates some periodic trends, although they are less obvious than the other periodic trends discussed previously. Generally, as you go across the periodic table, EA increases its magnitude: \[as\rightarrow PT,\; EA\uparrow \nonumber \] There is not a definitive trend as you go down the periodic table; sometimes EA increases, sometimes it decreases. Figure \(\PageIndex{3}\) shows EA values versus position on the periodic table for the s - and p -block elements. The trend is not absolute, especially considering the large positive EA values for the second column. However, the general trend going across the periodic table should be obvious. Example \(\PageIndex{3}\): Electron Affinities Predict which atom in each pair will have the highest magnitude of Electron Affinity. C or F Na or S Solution C and F are in the same row on the periodic table, but F is farther to the right. Therefore, F should have the larger magnitude of EA. Na and S are in the same row on the periodic table, but S is farther to the right. Therefore, S should have the larger magnitude of EA. Exercise \(\PageIndex{3}\): Electron Affinities Predict which atom will have the highest magnitude of Electron Affinity: As or Br. Answer Br Metallic Character The metallic character is used to define the chemical properties that metallic elements present. Generally, metals tend to lose electrons to form cations. Nonmetals tend to gain electrons to form anions. They also have a high oxidation potential—therefore they are easily oxidized and are strong reducing agents. Metals also form basic oxides; the more basic the oxide, the higher the metallic character. As you move across the table from left to right, the metallic character decreases, because the elements easily accept electrons to fill their valance shells. Therefore, these elements take on the nonmetallic character of forming anions. As you move up the table, the metallic character decreases, due to the greater pull that the nucleus has on the outer electrons. This greater pull makes it harder for the atoms to lose electrons and form cations. Uses of the Periodic Properties of Elements Predict greater or smaller atomic size and radial distribution in neutral atoms and ions. Measure and compare ionization energies. Compare electron affinities and electronegativities. Predict redox potential. Compare metallic character with other elements; ability to form cations. Predict reactions that may or may not occur due to the trends. Determine greater cell potential (sum of oxidation and reduction potential) between reactions. Complete chemical reactions according to trends. Summary Certain properties—notably atomic radius, ionization energies, and electron affinities—can be qualitatively understood by the positions of the elements on the periodic table. The major trends are summarized in the figure below. There are three factors that help in the prediction of the trends in the Periodic Table: number of protons in the nucleus, number of shells, and shielding effect. |
Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/14%3A_Electrophilic_Reactions/14.02%3A_Electrophilic_Addition_to_Alkenes | Addition of \(HBr\) to alkenes The simplest type of electrophilic reaction to visualize is the addition of a haloacid such as \(HBr\) to an isolated alkene. It is not a biological reaction, but nonetheless can serve as a convenient model to introduce some of the most important ideas about electrophilic reactions. Electrophilic addition of \(HBr\) to an alkene: Step 1 is an acid-base reaction: the \(\pi \) electrons of the alkene act as a base and extract the acidic proton of \(HBr\). This leaves one of the carbons with a new bond to hydrogen, and the other with an incomplete octet and a positive formal charge. In step 2, the nucleophilic bromide anion attacks the electrophilic carbocation to form a new carbon-bromine bond. Overall, the \(HBr\) molecule - in the form of a proton and a bromide anion - has been added to the double bond. To understand how \(\pi \)-bonded electrons in an alkene could be basic, let's first review the bonding picture for alkenes. Recall ( section 2.1) that the both of the carbons in an alkene group are \(sp^2\) hybridized, meaning that each carbon has three \(sp^2\) hybrid orbitals extending out in the same plane at \(180^{\circ}\) angles (trigonal planar geometry), and a single, unhybridized \(p\) orbital oriented perpendicular to that plane - one lobe above the plane, one lobe below. The unhybridized \(p\) orbitals on the two alkene carbons overlap, in a side-by-side fashion, to form the \(pi \) bond, which extends above and below the plane formed by the \(s\) bonds. two electrons shared in this π bond are farther away from the carbon nuclei than the electrons in the carbon-carbon \(s\) bond, and thus are more accessible to the acidic proton. In addition, recall that molecular orbital (MO) theory tells us that \(p\) orbitals are higher in energy than s orbitals (section 2.2). As a consequence, it is easier to break the \(p\) bond of an alkene than it is to break the \(s\) bond: the \(p\) bond is more reactive. As the \(HBr\) molecule approaches the alkene, a new \(s\) bond is formed between one of the alkene carbons and the electron-poor proton from \(HBr\). The carbon, which was \(sp^2\) hybridized when it was part of the alkene, is now \(sp^3\) hybridized. The other alkene carbon is still \(sp^2\) hybridized, but it now bears a positive formal charge because it has only three bonds, and its \(p\) orbital is empty. But it won't stay empty for long: a carbocation is a very reactive, unstable intermediate. The bromide ion will rapidly act as a nucleophile, filling the orbital with a pair of electrons, and now with four \(s\) bonds the carbon is \(sp^3\)-hybridized. The first step in the electrophilic addition reaction is much slower than the second step, because the intermediate carbocation species is higher in energy than either the reactants or the products, and as a result the energy barrier for the first step is also higher than for the second step. The slower first step is the rate-determining step: a change in the rate of the slow step will effect the rate of the overall reaction, while a change in the rate of the fast step will not. It is important to recognize the inherent difference between an electrophilic addition to an alkene and a conjugate addition to an alkene in the \(\square \square \square\) position, the latter of which we studied earlier in section 13.4. In both reactions, a proton and a nucleophile add to the double bond of an alkene. In a conjugate addition, the nucleophilic attack takes place first, resulting in a negatively charged intermediate (an enolate). Protonation is the second step. Also, of course, the alkene must be conjugated to a carbonyl or imine. In an electrophilic addition, proton abstraction occurs first, generating a positively-charged intermediate. Nucleophilic attack is the second step. No conjugated carbonyl or imine group is required: in fact a nearby carbonyl group would actually slow down a hypothetical electrophilic addition reaction down because a carbonyl is an electron withdrawing, carbocation-destabilized group. The stereochemistry of electrophilic addition Depending on the structure of the starting alkene, electrophilic addition has the potential to create two new chiral centers. Addition of \(HBr\) to an alkene is not stereoselective: the reaction results in racemization at both of the alkene carbons. Consider the addition of \(HBr\) to cis-3,4-dimethyl-3-hexene. The initial proton abstraction step creates a new chiral center, and because the acidic proton could be added to either side of the planar alkene carbon with equal probability, the center could have either \(S\) or \(R\) configuration. Likewise, in the second step the nucleophilic bromide ion could attack from either side of the planar carbocation, leading to an equal mixture of \(S\) and \(R\) configuration at that carbon as well. Therefore, we expect the product mixture to consist of equal amounts of four different stereoisomers. Predict the product(s) of electrophilic addition of \(HBr\) to the following alkenes. Draw all possible stereoisomers that could form, and take care not to draw identical structures twice. trans-2-butene cis-3-hexene cyclopentene The regiochemistry of electrophilic addition In many cases of electrophilic addition to an alkene, regiochemistry comes into play: the reaction can result in the formation of two different constitutional isomers. Consider the electrophilic addition of \(HBr\) to 2-methylpropene: Note that carbon #1 and carbon #2 in the starting alkene are not the same - carbon #2 is bonded to two methyl groups, and carbon #1 to two hydrogen atoms. The initial protonation step could therefore go two different ways, resulting in two different carbocation intermediates. Notice how pathway ' a ' gives a tertiary carbocation intermediate (\(I_a\)), while pathway ' b ' gives a primary carbocation intermediate (\(I_b\)) We know from section 8.5 that the tertiary carbocation \(I_a\) is lower in energy. Consequently, the transition state \(TS(a)\) leading to \(I_a\) is lower in energy than \(TS(b)\), meaning that \(I_a\) forms faster than \(I_b\). Because the protonation step is the rate determining step for the reaction, tertiary alkyl bromide A will form faster than the primary alkyl bromide B, and thus A will be the predominant product of the reaction. The electrophilic addition of \(HBr\) to 2-methylpropene is regioselective: more than one constitutional isomer can potentially form, but one isomer is favored over the other. It is generally observed that in electrophilic addition of haloacids to alkenes, the more substituted carbon is the one that ends up bonded to the heteroatom of the acid, while the less substituted carbon is protonated. This 'rule of thumb' is known as Markovnikov's rule , after the Russian chemist Vladimir Markovnikov who proposed it in 1869. While it is useful in many cases, Markovikov's rule does not apply to all electrophilic addition reactions. It is better to use a more general principle: When an asymmetrical alkene undergoes electrophilic addition, the product that predominates is the one that results from the more stable of the two possible carbocation intermediates. How is this different from Markovnikov's original rule? Consider the following hypothetical reaction, in which the starting alkene incorporates two trifluoromethyl substituents: Now when \(HBr\) is added, it is the less substituted carbocation that forms faster in the rate-determining protonation step, because in this intermediate the carbon bearing the positive charge is located further away from the electron-withdrawing, cation-destabilizing fluorines. As a result, the predominant product is the secondary rather than the tertiary bromoalkane. This is referred to as an anti-Markovnikov addition product, because it 'breaks' Markovnikov's rule. If the two possible carbocation intermediates in an electrophilic addition reaction are of similar stability, the product will be a mixture of constitutional isomers. Electrophilic addition of water and alcohol The (non-biochemical) addition of water to an alkene is very similar mechanistically to the addition of a haloacid such as \(HBr\) or \(HCl\), and the same stereochemical and regiochemical principles apply. A catalytic amount of a strong acid such as phosphoric or sulfuric acid is required, so that the acidic species in solution is actually \(H_3O^+\). Note that \(H_3O^+\) is regenerated in the course of the reaction. Figure 14.2.10 If an alkene is treated with methanol and a catalytic amount of strong acid, the result is an ether: Figure 14.2.11 Draw a mechanism for the ether-forming reaction above. Addition to conjugated alkenes Electrophilic addition to conjugated alkenes presents additional regiochemical possibilities, due to resonance delocalization of the allylic carbocation intermediate. Addition of one molar equivalent of \(HBr\) to 1,3-butadiene, for example, leads to a mixture of three products, two of which are a pair of enantiomers due to the creation of a chiral center at carbon #2. Figure 14.2.12 Explain why 4-bromo-1-butene is not a significant product of the reaction above. Predict the major product(s) of the following reactions. Draw all possible stereoisomers , and take care not to draw the same structure twice. Hint - are the double bonds in an aromatic ring likely to undergo electrophilic addition? Biochemical electrophilic addition reactions Myrcene is an isoprenoid compound synthesized by many different kinds of plants and used in the preparation of perfumes. Recently an enzymatic pathway for the degradation of myrcene has been identified in bacteria (J. Biol. Chem 2010, 285, 30436). The first step of this pathway is electrophilic addition of water to a conjugated alkene system. Figure 14.2.13 Draw a mechanism for the above reaction, showing two resonance contributors of the carbocation intermediate. How would you characterize the intermediate? Although the hydration of myrcene above looks very familiar, many enzyme-catalyzed electrophilic addition reactions differ from what we have seen so far, in that the electron-poor species attacked by the p-bonded electrons in the initial step is a carbocation rather than an acidic proton: Figure 14.2.14 \(\square\)-terpineol, a major component in the sap of pine trees, is formed in an electrophilic addition reaction. The first thing that happens (which we will refer to below as 'step a', in order to keep the step numbering consistent what the addition mechanisms we have seen so far) is departure of a pyrophosphate leaving group, forming an allylic carbocation electrophile. Figure 14.2.15 The actual electrophilic addition stage of the reaction begins with step 1, as the π electrons an alkene are drawn toward one of the two carbons that share the positive charge, effectively closing a six-membered ring. A water molecule then attacks the second carbocation intermediate (step 2), which completes the addition process. Notice something important about the regiochemical course of the reaction: step 1 results in the formation of a six-membered ring and a tertiary carbocation. As we have stressed before, biochemical reactions tend to follow energetically favorable mechanistic pathways. An alternate regiochemical course to step 1 shown above could result in a seven-membered ring and a secondary carbocation , a much less energetically favorable intermediate in terms of both carbocation stability and ring size. Draw a mechanism for this hypothetical alternate reaction, and show the product that would result after the addition of water in a hypothetical 'step 2'. |
Courses/Georgia_Southern_University/CHEM_1152%3A_Survey_of_Chemistry_II_(Osborne)/07%3A_Lipids/7.03%3A_Phospholipids | Learning Objectives Identify the distinguishing characteristics of membrane lipids. Describe membrane components and how they are arranged. All living cells are surrounded by a cell membrane. Plant cells (Figure \(\PageIndex{1a}\)) and animal cells (Figure \(\PageIndex{1b}\)) contain a cell nucleus that is also surrounded by a membrane and holds the genetic information for the cell. Everything between the cell membrane and the nuclear membrane—including intracellular fluids and various subcellular components such as the mitochondria and ribosomes—is called the cytoplasm. The membranes of all cells have a fundamentally similar structure, but membrane function varies tremendously from one organism to another and even from one cell to another within a single organism. This diversity arises mainly from the presence of different proteins and lipids in the membrane. The lipids in cell membranes are highly polar but have dual characteristics: part of the lipid is ionic and therefore dissolves in water, whereas the rest has a hydrocarbon structure and therefore dissolves in nonpolar substances. Often, the ionic part is referred to as hydrophilic , meaning “water loving,” and the nonpolar part as hydrophobic , meaning “water fearing” (repelled by water). When allowed to float freely in water, polar lipids spontaneously cluster together in any one of three arrangements: micelles, monolayers, and bilayers (Figure \(\PageIndex{2}\)). Micelles are aggregations in which the lipids’ hydrocarbon tails—being hydrophobic—are directed toward the center of the assemblage and away from the surrounding water while the hydrophilic heads are directed outward, in contact with the water. Each micelle may contain thousands of lipid molecules. Polar lipids may also form a monolayer, a layer one molecule thick on the surface of the water. The polar heads face into water, and the nonpolar tails stick up into the air. Bilayers are double layers of lipids arranged so that the hydrophobic tails are sandwiched between an inner surface and an outer surface consisting of hydrophilic heads. The hydrophilic heads are in contact with water on either side of the bilayer, whereas the tails, sequestered inside the bilayer, are prevented from having contact with the water. Bilayers like this make up every cell membrane (Figure \(\PageIndex{3}\)). In the bilayer interior, the hydrophobic tails (that is, the fatty acid portions of lipid molecules) interact by means of dispersion forces. The interactions are weakened by the presence of unsaturated fatty acids. As a result, the membrane components are free to mill about to some extent, and the membrane is described as fluid. The lipids found in cell membranes can be categorized in various ways. Phospholipids are lipids containing phosphorus. Glycolipids are sugar-containing lipids. The latter are found exclusively on the outer surface of the cell membrane, acting as distinguishing surface markers for the cell and thus serving in cellular recognition and cell-to-cell communication. Sphingolipids are phospholipids or glycolipids that contain the unsaturated amino alcohol sphingosine rather than glycerol. Diagrammatic structures of representative membrane lipids are presented in Figure \(\PageIndex{4}\). Phosphoglycerides (also known as glycerophospholipids ) are the most abundant phospholipids in cell membranes. They consist of a glycerol unit with fatty acids attached to the first two carbon atoms, while a phosphoric acid unit, esterified with an alcohol molecule (usually an amino alcohol, as in Figure \(\PageIndex{5a}\)) is attached to the third carbon atom of glycerol (Figure \(\PageIndex{5b}\)). Notice that the phosphoglyceride molecule is identical to a triglyceride up to the phosphoric acid unit (Figure \(\PageIndex{5b}\)). There are two common types of phosphoglycerides. Phosphoglycerides containing ethanolamine as the amino alcohol are called phosphatidylethanolamines or cephalins . Cephalins are found in brain tissue and nerves and also have a role in blood clotting. Phosphoglycerides containing choline as the amino alcohol unit are called phosphatidylcholines or lecithins . Lecithins occur in all living organisms. Like cephalins, they are important constituents of nerve and brain tissue. Egg yolks are especially rich in lecithins. Commercial-grade lecithins isolated from soybeans are widely used in foods as emulsifying agents. An emulsifying agent is used to stabilize an emulsion—a dispersion of two liquids that do not normally mix, such as oil and water. Many foods are emulsions. Milk is an emulsion of butterfat in water. The emulsifying agent in milk is a protein called casein . Mayonnaise is an emulsion of salad oil in water, stabilized by lecithins present in egg yolk. Sphingomyelins , the simplest sphingolipids , each contain a fatty acid, a phosphoric acid, sphingosine , and choline (Figure \(\PageIndex{6}\)). Because they contain phosphoric acid, they are also classified as phospholipids. Sphingomyelins are important constituents of the myelin sheath surrounding the axon of a nerve cell. Multiple sclerosis is one of several diseases resulting from damage to the myelin sheath. Most animal cells contain sphingolipids called cerebrosides (Figure \(\PageIndex{7}\)). Cerebrosides are composed of sphingosine, a fatty acid, and galactose or glucose. They therefore resemble sphingomyelins but have a sugar unit in place of the choline phosphate group. Cerebrosides are important constituents of the membranes of nerve and brain cells. The sphingolipids called gangliosides are more complex, usually containing a branched chain of three to eight monosaccharides and/or substituted sugars. Because of considerable variation in their sugar components, about 130 varieties of gangliosides have been identified. Most cell-to-cell recognition and communication processes (e.g., blood group antigens) depend on differences in the sequences of sugars in these compounds. Gangliosides are most prevalent in the outer membranes of nerve cells, although they also occur in smaller quantities in the outer membranes of most other cells. Because cerebrosides and gangliosides contain sugar groups, they are also classified as glycolipids . Membrane Proteins If membranes were composed only of lipids, very few ions or polar molecules could pass through their hydrophobic “sandwich filling” to enter or leave any cell. However, certain charged and polar species do cross the membrane, aided by proteins that move about in the lipid bilayer. The two major classes of proteins in the cell membrane are integral proteins, which span the hydrophobic interior of the bilayer, and peripheral proteins, which are more loosely associated with the surface of the lipid bilayer (Figure \(\PageIndex{3}\)). Peripheral proteins may be attached to integral proteins, to the polar head groups of phospholipids, or to both by hydrogen bonding and electrostatic forces. Small ions and molecules soluble in water enter and leave the cell by way of channels through the integral proteins. Some proteins, called carrier proteins , facilitate the passage of certain molecules, such as hormones and neurotransmitters, by specific interactions between the protein and the molecule being transported. Example \(\PageIndex{1}\) Classify the following as a phospholipid, a glycolipid, and/or a sphingolipid. (Some lipids can be given more than one classification.) Solution phospholipid sphingolipid and glycolipid Exercise \(\PageIndex{1}\) Classify the following as a phospholipid, a glycolipid, and/or a sphingolipid. (Some lipids can be given more than one classification.) Summary Lipids are important components of biological membranes. These lipids have dual characteristics: part of the molecule is hydrophilic, and part of the molecule is hydrophobic. Membrane lipids may be classified as phospholipids, glycolipids, and/or sphingolipids. Proteins are another important component of biological membranes. Integral proteins span the lipid bilayer, while peripheral proteins are more loosely associated with the surface of the membrane. |
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_331_-_Organic_Chemistry_(Lund)/04%3A_Conformations_and_Stereochemistry/4.05%3A_Optical_Activity | Chiral molecules, as we learned in the introduction to this chapter, have an interesting optical property. You may know from studying physics that light waves are oscillating electric and magnetic fields. In ordinary light, the oscillation is randomly oriented in an infinite number of planes. When ordinary light is passed through a polarizer, all planes of oscillation are filtered out except one, resulting in plane-polarized light . A beam of plane-polarized light, when passed through a sample of a chiral compound, interacts with the compound in such a way that the angle of oscillation will rotate. This property is called optical activity . If a compound rotates plane polarized light in the clockwise (+) direction, it is said to be dextrorotatory , while if it rotates light in the counterclockwise (-) direction it is levorotatory . The magnitude of the observed optical activity is dependent on temperature, the wavelength of light used, solvent, concentration of the chiral sample, and the path length of the sample tube (path length is the length that the plane-polarized light travels through the chiral sample). Typically, optical activity measurements are made in a 1 decimeter (10 cm) path-length sample tube at 25 ° C, using as a light source the so-called “D-line” from a sodium lamp, which has a wavelength of 589 nm. The specific rotation of a pure chiral compound at 25 ° is expressed by the expression: \[[\alpha]_{\mathrm{D}}^{25}=\frac{\alpha_{\mathrm{obs}}}{l c} \nonumber\] where \(\alpha_{obs}\) is the observed rotation, \(l\) is path length in decimeters, and \(c\) is the concentration of the sample in grams per 100 mL. In other words, the specific rotation of a chiral compound is the optical rotation that is observed when 1 g of the compound is dissolved in enough of a given solvent to make 100 mL solution, and the rotation is measured in a 1 dm cuvette at 25 o C using light from a sodium lamp. Every chiral molecule has a characteristic specific rotation, which is recorded in the chemical literature as a physical property just like melting point or density. Different enantiomers of a compound will always rotate plane-polarized light with an equal but opposite magnitude. ( S )-ibuprofen, for example, has a specific rotation of +54.5 o (dextrorotatory) in methanol, while ( R )-ibuprofen has a specific rotation of -54.5 o . There is no relationship between chiral compound's R/S designation and the direction of its specific rotation. For example, the S enantiomer of ibuprofen is dextrorotatory, but the S enantiomer of glyceraldehyde is levorotatory. A 50:50 mixture of two enantiomers (a racemic mixture) will have no observable optical activity, because the two optical activities cancel each other out. In a structural drawing, a 'squigly' bond from a chiral center indicates a mixture of both R and S configurations. Chiral molecules are often labeled according to whether they are dextrorotatory or levorotatory as well as by their R/S designation. For example, the pure enantiomers of ibuprofen are labeled ( S )-(+)-ibuprofen and ( R )-(-)-ibuprofen, while ( ± )-ibuprofen refers to the racemic mixture, which is the form in which the drug is sold to consumers. Exercise 3.14 The specific rotation of ( R )-limonene is +11.5 o in ethanol. What is the expected observed rotation of a sample of 6.00 g ( S )-limonene dissolved in ethanol to a total volume of 80.0 mL in a 1.00 dm (10.0 cm) pathlength cuvette? Exercise 3.15 The specific rotation of ( S )-carvone is +61 ° , measured 'neat' (pure liquid sample, no solvent). The optical rotation of a mixture of R and S carvone is measured at -23 ° . Which enantiomer is in excess in the mixture? Solutions to exercises All of the twenty natural amino acids except glycine have a chiral center at their a lpha-carbon (recall that basic amino acid structure and terminology was introduced in section 1.3). Virtually all of the amino acids found in nature, both in the form of free amino acids or incorporated into peptides and proteins, have what is referred to in the biochemical literature as the ' L ' configuration: All but one of the 19 L -amino acids have S stereochemistry at the a- carbon, using the rules of the R/S naming system. D -amino acids are very rare in nature, but we will learn about an interesting example of a peptide containing one D -amino acid residue later in chapter 12. Exercise 3.16 Which of the 20 common L -amino acids found in nature has the R configuration? Refer to the amino acid table for structures. Solutions to exercises Khan Academy video tutorial on optical acitivity |
Bookshelves/Introductory_Chemistry/Chemistry_for_Allied_Health_(Soult)/02%3A_Elements_and_Ions | Chemistry is the study of matter and the changes it undergoes. Atoms are the building blocks of matter and are the smallest unit of an element. 2.1: Isotopes and Atomic Mass Atoms are the fundamental building blocks of all matter and are composed of protons, neutrons, and electrons. Isotopes are atoms of the same element with a different mass due to a different number of neutrons. 2.2: Matter Living things are made of matter. In fact, matter is the "stuff" of which all things are made (see figure below. Anything that occupies space and has mass is known as matter. Matter, in turn, consists of chemical substances. Chemistry is the study of matter and the changes it undergoes. 2.3: Mole and Molar Mass A mole is used as a measure of the amount of substance present. Atoms are very small so using more common counting numbers, such as a dozen, are impractical. 2.4: Electron Arrangements The number and arrangement of electrons in an atom control the types and number of bonds that an atom can form so understanding their arrangement helps explain the behavior of an atom in a compound or molecule. 2.5: Ion Formation Ions form from the gain or loss of electrons. The electron arrangements allows for the prediction of how many electrons will be gained or lost. 2.6: Ionic Compounds Ionic compounds form from at least one negatively-charged anion and one positively charged cation. The charges on the ions allow the prediction of the number of each that are combined to form a single formula unit of an ionic compound. 2.7: Elements and Ions (Exercises) These are homework exercises to accompany Chapter 2 of the University of Kentucky's LibreText for CHE 103 - Chemistry for Allied Health. Solutions are available below the questions. |
Courses/Stanford_Online_High_School/TEN2C-Carbon/01%3A_Carbon_Chains/1.06%3A_Naming_Alkanes/1.6.02%3A_Nomenclature_of_unsaturated_hydrocarbons | Alkenes The simplest alkenes are ethene (traditionally called ethylene) , C 2 H 4 or CH 2 =CH 2 , and propene (or propylene) , C 3 H 6 or CH 3 CH=CH 2 (part (a) in the f igure below ). The names of alkenes that have more than three carbon atoms use the same stems as the names of the alkanes (see t able above, “The First 10 Straight-Chain Alkanes” ) but end in – ene instead of – ane . As with alkanes, more than one structure is possible for alkenes with four or more carbon atoms. For example, an alkene with four carbon atoms has three possible structures. One is CH 2 =CHCH 2 CH 3 (1-butene), which has the double bond between the first and second carbon atoms in the chain. The other two structures have the double bond between the second and third carbon atoms and are forms of CH 3 CH=CHCH 3 ((but-2-ene, formerly called 2-butene). All four carbon atoms in but-2-ene lie in the same plane, so there are two possible structures (part (a) in Figure 3.7.2 ). If the two methyl groups are on the same side of the double bond, the compound is cis -but-2-ene or cis -2-butene (from the Latin cis , meaning “on the same side”). If the two methyl groups are on opposite sides of the double bond, the compound is trans -but-2-ene or trans -2-butene (from the Latin trans , meaning “across”). (NB: We will learn more complete naming for such isomers in section 4.3.) These are distinctly different molecules: cis -but-2-ene melts at −138.9°C, whereas trans -but-2-ene melts at −105.5°C. Some Simple (a) Alkenes, (b) Alkynes, and (c) Cyclic Hydrocarbons. The positions of the carbon atoms in the chain are indicated by C 1 or C2. Some names here are traditional names rather than modern IUPAC names. Just as a number indicates the positions of branches in an alkane, the number in the name of an alkene specifies the position of the first carbon atom of the double bond. The name is based on the lowest possible number starting from either end of the carbon chain, so CH 3 CH 2 CH=CH 2 is called but-1-ene (formerly 1-butene), not but-3-ene. Note that CH 2 =CHCH 2 CH 3 and CH 3 CH 2 CH=CH 2 are different ways of writing the same molecule (but-1-ene) in two different orientations. The name of a compound does not depend on its orientation. As illustrated for 1-butene, both condensed structural formulas and molecular models show different orientations of the same molecule. It is important to be able to recognize the same structure no matter what its orientation. Note The positions of groups or multiple bonds are always indicated by the lowest number possible. Alkynes The simplest alkyne is ethyne , traditionally called acetylene , C 2 H 2 or HC≡CH (part (b) in the figure above). Because a mixture of acetylene and oxygen burns with a flame that is hot enough (>3000°C) to cut metals such as hardened steel, acetylene is widely used in cutting and welding torches. The names of other alkynes are similar to those of the corresponding alkanes but end in – yne . For example, HC≡CCH 3 is propyne , and CH 3 C≡CCH 3 is but-2-yne because the multiple bond begins on the second carbon atom. Note The number of bonds between carbon atoms in a hydrocarbon is indicated in the suffix: alk ane : only carbon–carbon single bonds alk ene : at least one carbon–carbon double bond alk yne : at least one carbon–carbon triple bond Aromatic hydrocarbons Alkanes, alkenes, alkynes, and cyclic hydrocarbons are generally called aliphatic hydrocarbons. The name comes from the Greek aleiphar , meaning “oil,” because the first examples were extracted from animal fats. In contrast, the first examples of aromatic hydrocarbons , also called arenes , were obtained by the distillation and degradation of highly scented (thus aromatic ) resins from tropical trees. The simplest aromatic hydrocarbon is benzene (C 6 H 6 ), which was first obtained from a coal distillate. The word aromatic now refers to benzene and structurally similar compounds. As shown in part (a) in Figure 3.7.4 , it is possible to draw the structure of benzene in two different but equivalent ways, depending on which carbon atoms are connected by double bonds or single bonds. Toluene is similar to benzene, except that one hydrogen atom is replaced by a –CH 3 group; it has the formula C 7 H 8 (part (b) in Figure 3.7.4 ). The chemical behavior of aromatic compounds differs from the behavior of aliphatic compounds. Benzene and toluene are found in gasoline, and benzene is the starting material for preparing substances as diverse as aspirin and nylon. Figure 3.7.4: Two Aromatic Hydrocarbons: (a) Benzene and (b) Toluene Figure 3.7.5 illustrates two of the molecular structures possible for hydrocarbons that have six carbon atoms. As shown, compounds with the same molecular formula can have very different structures. Figure 3.7.5: Two Hydrocarbons with the Molecular Formula C6H12 Example 3.7.1 Write the condensed structural formula for each hydrocarbon. n-heptane 2-pentene 2-butyne cyclooctene (give the skeletal formula) Given : name of hydrocarbon Asked for : condensed structural formula Strategy : Use the prefix to determine the number of carbon atoms in the molecule and whether it is cyclic. From the suffix, determine whether multiple bonds are present. Identify the position of any multiple bonds from the number(s) in the name and then write the condensed structural formula. [reveal-answer q=”86147″]Show Solution[/reveal-answer] [hidden-answer a=”86147″] a. A The prefix hept- tells us that this hydrocarbon has seven carbon atoms, and n- indicates that the carbon atoms form a straight chain. The suffix -ane tells that it is an alkane, with no carbon–carbon double or triple bonds. B The condensed structural formula is CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 , which can also be written as $$CH_3(CH_2)_5CH_3$$. b. A The prefix pent- tells us that this hydrocarbon has five carbon atoms, and the suffix -ene indicates that it is an alkene, with a carbon–carbon double bond. B The 2- tells us that the double bond begins on the second carbon of the five-carbon atom chain. The condensed structural formula of the compound is therefore CH 3 CH=CHCH 2 CH 3 . c. A The prefix but- tells us that the compound has a chain of four carbon atoms, and the suffix -yne indicates that it has a carbon–carbon triple bond. B The 2- tells us that the triple bond begins on the second carbon of the four-carbon atom chain. So the condensed structural formula for the compound is CH3C≡CCH 3 . d. A The prefix cyclo- tells us that this hydrocarbon has a ring structure, and oct- indicates that it contains eight carbon atoms, which we can draw as The suffix -ene tells us that the compound contains a carbon–carbon double bond, but where in the ring do we place the double bond? B Because all eight carbon atoms are identical, it doesn’t matter. We can draw the structure of cyclooctene as [/hidden-answer] Exercise Write the condensed structural formula for each hydrocarbon. n-octane 2-hexene 1-heptyne cyclopentane (give the skeletal formula) [reveal-answer q=”895946″]Show Answer[/reveal-answer] [hidden-answer a=”895946″] CH 3 (CH 2 ) 6 CH 3 CH 3 CH=CHCH 2 CH 2 CH 3 HC≡C(CH 2 ) 4 CH 3 [/hidden-answer] The general name for a group of atoms derived from an alkane is an alkyl group . The name of an alkyl group is derived from the name of the alkane by adding the suffix – yl . Thus the –CH 3 fragment is a methyl group, the –CH 2 CH 3 fragment is an ethyl group, and so forth, where the dash represents a single bond to some other atom or group. Similarly, groups of atoms derived from aromatic hydrocarbons are aryl groups , which sometimes have unexpected names. For example, the –C 6 H 5 fragment is derived from benzene, but it is called a phenyl group. In general formulas and structures, alkyl and aryl groups are often abbreviated as R . Structures of alkyl and aryl groups. The methyl group is an example of an alkyl group, and the phenyl group is an example of an aryl group . Summary The simplest organic compounds are the hydrocarbons , which contain only carbon and hydrogen. Alkanes contain only carbon–hydrogen and carbon–carbon single bonds, alkenes contain at least one carbon–carbon double bond, and alkynes contain one or more carbon–carbon triple bonds. Hydrocarbons can also be cyclic , with the ends of the chain connected to form a ring. Collectively, alkanes, alkenes, and alkynes are called aliphatic hydrocarbons . Aromatic hydrocarbons , or arenes , are another important class of hydrocarbons that contain rings of carbon atoms related to the structure of benzene (C 6 H 6 ). A derivative of an alkane or an arene from which one hydrogen atom has been removed is called an alkyl group or an aryl group , respectively. CC licensed content, Shared previously 3.7: Names of Formulas of Organic Compounds. Located at : https://chem.libretexts.org/Textbook_Maps/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.7%3A__Names_of_Formulas_of_Organic_Compounds . Project : Chemistry LibreTexts. License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike |
Courses/Prince_Georges_Community_College/CHEM_2000%3A_General_Chemistry_for_Engineers_-_F21/10%3A_Chemical_Bonding_II-_Valance_Bond_Theory_and_Molecular_Orbital_Theory/10.04%3A_VSPER_Theory_-_Predicting_Molecular_Geometries | Valence shell electron-pair repulsion theory ( VSEPR theory) enables us to predict the molecular structure, including approximate bond angles around a central atom, of a molecule from an examination of the number of bonds and lone electron pairs in its Lewis structure. The VSEPR model assumes that electron pairs in the valence shell of a central atom will adopt an arrangement that minimizes repulsions between these electron pairs by maximizing the distance between them. The electrons in the valence shell of a central atom form either bonding pairs of electrons, located primarily between bonded atoms, or lone pairs. The electrostatic repulsion of these electrons is reduced when the various regions of high electron density assume positions as far from each other as possible. VSEPR theory predicts the arrangement of electron pairs around each central atom and, usually, the correct arrangement of atoms in a molecule. We should understand, however, that the theory only considers electron-pair repulsions. Other interactions, such as nuclear-nuclear repulsions and nuclear-electron attractions, are also involved in the final arrangement that atoms adopt in a particular molecular structure. As a simple example of VSEPR theory, let us predict the structure of a gaseous BeF 2 molecule. The Lewis structure of BeF 2 (Figure \(\PageIndex{2}\)) shows only two electron pairs around the central beryllium atom. With two bonds and no lone pairs of electrons on the central atom, the bonds are as far apart as possible, and the electrostatic repulsion between these regions of high electron density is reduced to a minimum when they are on opposite sides of the central atom. The bond angle is 180° (Figure \(\PageIndex{2}\)). Figure \(\PageIndex{2}\): The BeF 2 molecule adopts a linear structure in which the two bonds are as far apart as possible, on opposite sides of the Be atom. Figure \(\PageIndex{3}\) illustrates this and other electron-pair geometries that minimize the repulsions among regions of high electron density (bonds and/or lone pairs). Two regions of electron density around a central atom in a molecule form a linear geometry; three regions form a trigonal planar geometry; four regions form a tetrahedral geometry; five regions form a trigonal bipyramidal geometry; and six regions form an octahedral geometry. |
Courses/University_of_Kentucky/UK%3A_General_Chemistry/18%3A_Representative_Metals_Metalloids_and_Nonmetals/18.E%3A_Representative_Metals_Metalloids_and_Nonmetals_(Exercises) | 18.1: Periodicity How do alkali metals differ from alkaline earth metals in atomic structure and general properties? The alkali metals all have a single s electron in their outermost shell. In contrast, the alkaline earth metals have a completed s subshell in their outermost shell. In general, the alkali metals react faster and are more reactive than the corresponding alkaline earth metals in the same period. Why does the reactivity of the alkali metals decrease from cesium to lithium? Predict the formulas for the nine compounds that may form when each species in column 1 of Table reacts with each species in column 2. 1 2 Na I Sr Se Al O \[\ce{Na + I2 ⟶ 2NaI\\ 2Na + Se ⟶ Na2Se\\ 2Na + O2 ⟶ Na2O2}\] \[\ce{Sr + I2⟶SrI2\\ Sr + Se⟶SeSe\\ 2Sr + O2⟶2SrO}\] \[\ce{2Al + 3I2⟶2AlI3\\ 2Al + 3Se⟶Al2Se3\\ 4Al + 3O2⟶2Al2O3}\] Predict the best choice in each of the following. You may wish to review the chapter on electronic structure for relevant examples. (a) the most metallic of the elements Al, Be, and Ba (b) the most covalent of the compounds NaCl, CaCl 2 , and BeCl 2 (c) the lowest first ionization energy among the elements Rb, K, and Li (d) the smallest among Al, Al + , and Al 3+ (e) the largest among Cs + , Ba 2+ , and Xe Sodium chloride and strontium chloride are both white solids. How could you distinguish one from the other? The possible ways of distinguishing between the two include infrared spectroscopy by comparison of known compounds, a flame test that gives the characteristic yellow color for sodium (strontium has a red flame), or comparison of their solubilities in water. At 20 °C, NaCl dissolves to the extent of \(\mathrm{\dfrac{35.7\: g}{100\: mL}}\) compared with \(\mathrm{\dfrac{53.8\: g}{100\: mL}}\) for SrCl 2 . Heating to 100 °C provides an easy test, since the solubility of NaCl is \(\mathrm{\dfrac{39.12\: g}{100\: mL}}\), but that of SrCl 2 is \(\mathrm{\dfrac{100.8\: g}{100\: mL}}\). Density determination on a solid is sometimes difficult, but there is enough difference (2.165 g/mL NaCl and 3.052 g/mL SrCl 2 ) that this method would be viable and perhaps the easiest and least expensive test to perform. The reaction of quicklime, CaO, with water produces slaked lime, Ca(OH) 2 , which is widely used in the construction industry to make mortar and plaster. The reaction of quicklime and water is highly exothermic: \[\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s) \hspace{20px} ΔH=\mathrm{−350\: kJ\:mol^{−1}}\] (a) What is the enthalpy of reaction per gram of quicklime that reacts? (b) How much heat, in kilojoules, is associated with the production of 1 ton of slaked lime? Write a balanced equation for the reaction of elemental strontium with each of the following: (a) oxygen (b) hydrogen bromide (c) hydrogen (d) phosphorus (e) water (a) \(\ce{2Sr}(s)+\ce{O2}(g)⟶\ce{2SrO}(s)\); (b) \(\ce{Sr}(s)+\ce{2HBr}(g)⟶\ce{SrBr2}(s)+\ce{H2}(g)\); (c) \(\ce{Sr}(s)+\ce{H2}(g)⟶\ce{SrH2}(s)\); (d) \(\ce{6Sr}(s)+\ce{P4}(s)⟶\ce{2Sr3P2}(s)\); (e) \(\ce{Sr}(s)+\ce{2H2O}(l)⟶\ce{Sr(OH)2}(aq)+\ce{H2}(g)\) How many moles of ionic species are present in 1.0 L of a solution marked 1.0 M mercury(I) nitrate? What is the mass of fish, in kilograms, that one would have to consume to obtain a fatal dose of mercury, if the fish contains 30 parts per million of mercury by weight? (Assume that all the mercury from the fish ends up as mercury(II) chloride in the body and that a fatal dose is 0.20 g of HgCl 2 .) How many pounds of fish is this? 11 lb The elements sodium, aluminum, and chlorine are in the same period. (a) Which has the greatest electronegativity? (b) Which of the atoms is smallest? (c) Write the Lewis structure for the simplest covalent compound that can form between aluminum and chlorine. (d) Will the oxide of each element be acidic, basic, or amphoteric? Does metallic tin react with HCl? Yes, tin reacts with hydrochloric acid to produce hydrogen gas. What is tin pest, also known as tin disease? Compare the nature of the bonds in PbCl 2 to that of the bonds in PbCl 4 . In PbCl 2 , the bonding is ionic, as indicated by its melting point of 501 °C. In PbCl 4 , the bonding is covalent, as evidenced by it being an unstable liquid at room temperature. Is the reaction of rubidium with water more or less vigorous than that of sodium? How does the rate of reaction of magnesium compare? 18.2: Occurrence and Preparation of the Representative Metals Write an equation for the reduction of cesium chloride by elemental calcium at high temperature. \[\ce{2CsCl}(l)+\ce{Ca}(g)\:\mathrm{\overset{countercurrent \\ fractionating \\ tower}{\xrightarrow{\hspace{40px}}}}\:\ce{2Cs}(g)+\ce{CaCl2}(l)\] Why is it necessary to keep the chlorine and sodium, resulting from the electrolysis of sodium chloride, separate during the production of sodium metal? Give balanced equations for the overall reaction in the electrolysis of molten lithium chloride and for the reactions occurring at the electrodes. You may wish to review the chapter on electrochemistry for relevant examples. Cathode (reduction): \(\ce{2Li+} + \ce{2e-}⟶\ce{2Li}(l)\); Anode (oxidation): \(\ce{2Cl-}⟶\ce{Cl2}(g)+\ce{2e-}\); Overall reaction: \(\ce{2Li+}+\ce{2Cl-}⟶\ce{2Li}(l)+\ce{Cl2}(g)\) The electrolysis of molten sodium chloride or of aqueous sodium chloride produces chlorine. Calculate the mass of chlorine produced from 3.00 kg sodium chloride in each case. You may wish to review the chapter on electrochemistry for relevant examples. What mass, in grams, of hydrogen gas forms during the complete reaction of 10.01 g of calcium with water? 0.5035 g H 2 How many grams of oxygen gas are necessary to react completely with 3.01 × 10 21 atoms of magnesium to yield magnesium oxide? Magnesium is an active metal; it burns in the form of powder, ribbons, and filaments to provide flashes of brilliant light. Why is it possible to use magnesium in construction? Despite its reactivity, magnesium can be used in construction even when the magnesium is going to come in contact with a flame because a protective oxide coating is formed, preventing gross oxidation. Only if the metal is finely subdivided or present in a thin sheet will a high-intensity flame cause its rapid burning. Why is it possible for an active metal like aluminum to be useful as a structural metal? Describe the production of metallic aluminum by electrolytic reduction. Extract from ore: \(\ce{AlO(OH)}(s)+\ce{NaOH}(aq)+\ce{H2O}(l)⟶\ce{Na[Al(OH)4]}(aq)\) Recover: \(\ce{2Na[Al(OH)4]}(s)+\ce{H2SO4}(aq)⟶\ce{2Al(OH)3}(s)+\ce{Na2SO4}(aq)+\ce{2H2O}(l)\) Sinter: \(\ce{2Al(OH)3}(s)⟶\ce{Al2O3}(s)+\ce{3H2O}(g)\) Dissolve in Na 3 AlF 6 ( l ) and electrolyze: \(\ce{Al^3+}+\ce{3e-}⟶\ce{Al}(s)\) What is the common ore of tin and how is tin separated from it? A chemist dissolves a 1.497-g sample of a type of metal (an alloy of Sn, Pb, Sb, and Cu) in nitric acid, and metastannic acid, H 2 SnO 3 , is precipitated. She heats the precipitate to drive off the water, which leaves 0.4909 g of tin(IV) oxide. What was the percentage of tin in the original sample? 25.83% Consider the production of 100 kg of sodium metal using a current of 50,000 A, assuming a 100% yield. (a) How long will it take to produce the 100 kg of sodium metal? (b) What volume of chlorine at 25 °C and 1.00 atm forms? What mass of magnesium forms when 100,000 A is passed through a MgCl 2 melt for 1.00 h if the yield of magnesium is 85% of the theoretical yield? 39 kg 18.3: Structure and General Properties of the Metalloids Give the hybridization of the metalloid and the molecular geometry for each of the following compounds or ions. You may wish to review the chapters on chemical bonding and advanced covalent bonding for relevant examples. (a) GeH 4 (b) SbF 3 (c) Te(OH) 6 (d) H 2 Te (e) GeF 2 (f) TeCl 4 (g) \(\ce{SiF6^2-}\) (h) SbCl 5 (i) TeF 6 Write a Lewis structure for each of the following molecules or ions. You may wish to review the chapter on chemical bonding. (a) H 3 BPH 3 (b) \(\ce{BF4-}\) (c) BBr 3 (d) B(CH 3 ) 3 (e) B(OH) 3 (a) H 3 BPH 3 : ; (b) \(\ce{BF4-}\): ; (c) BBr 3 : ; (d) B(CH 3 ) 3 : ; (e) B(OH) 3 : Describe the hybridization of boron and the molecular structure about the boron in each of the following: (a) H 3 BPH 3 (b) \(\ce{BF4-}\) (c) BBr 3 (d) B(CH 3 ) 3 (e) B(OH) 3 Using only the periodic table, write the complete electron configuration for silicon, including any empty orbitals in the valence shell. You may wish to review the chapter on electronic structure. 1 s 2 2 s 2 2 p 6 3 s 2 3 p 2 3 d 0 . Write a Lewis structure for each of the following molecules and ions: (a) (CH 3 ) 3 SiH (b) \(\ce{SiO4^4-}\) (c) Si 2 H 6 (d) Si(OH) 4 (e) \(\ce{SiF6^2-}\) Describe the hybridization of silicon and the molecular structure of the following molecules and ions: (a) (CH 3 ) 3 SiH (b) \(\ce{SiO4^4-}\) (c) Si 2 H 6 (d) Si(OH) 4 (e) \(\ce{SiF6^2-}\) (a) (CH 3 ) 3 SiH: sp 3 bonding about Si; the structure is tetrahedral; (b) \(\ce{SiO4^4-}\): sp 3 bonding about Si; the structure is tetrahedral; (c) Si 2 H 6 : sp 3 bonding about each Si; the structure is linear along the Si-Si bond; (d) Si(OH) 4 : sp 3 bonding about Si; the structure is tetrahedral; (e) \(\ce{SiF6^2-}\): sp 3 d 2 bonding about Si; the structure is octahedral Describe the hybridization and the bonding of a silicon atom in elemental silicon. Classify each of the following molecules as polar or nonpolar. You may wish to review the chapter on chemical bonding. (a) SiH 4 (b) Si 2 H 6 (c) SiCl 3 H (d) SiF 4 (e) SiCl 2 F 2 (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar Silicon reacts with sulfur at elevated temperatures. If 0.0923 g of silicon reacts with sulfur to give 0.3030 g of silicon sulfide, determine the empirical formula of silicon sulfide. Name each of the following compounds: (a) TeO 2 (b) Sb 2 S 3 (c) GeF 4 (d) SiH 4 (e) GeH 4 (a) tellurium dioxide or tellurium(IV) oxide; (b) antimony(III) sulfide; (c) germanium(IV) fluoride; (d) silane or silicon(IV) hydride; (e) germanium(IV) hydride Write a balanced equation for the reaction of elemental boron with each of the following (most of these reactions require high temperature): (a) F 2 (b) O 2 (c) S (d) Se (e) Br 2 Why is boron limited to a maximum coordination number of four in its compounds? Boron has only s and p orbitals available, which can accommodate a maximum of four electron pairs. Unlike silicon, no d orbitals are available in boron. Write a formula for each of the following compounds: (a) silicon dioxide (b) silicon tetraiodide (c) silane (d) silicon carbide (e) magnesium silicide From the data given in Appendix I , determine the standard enthalpy change and the standard free energy change for each of the following reactions: (a) \(\ce{BF3}(g)+\ce{3H2O}(l)⟶\ce{B(OH)3}(s)+\ce{3HF}(g)\) (b) \(\ce{BCl3}(g)+\ce{3H2O}(l)⟶\ce{B(OH)3}(s)+\ce{3HCl}(g)\) (c) \(\ce{B2H6}(g)+\ce{6H2O}(l)⟶\ce{2B(OH)3}(s)+\ce{6H2}(g)\) (a) Δ H ° = 87 kJ; Δ G ° = 44 kJ; (b) Δ H ° = −109.9 kJ; Δ G° = −154.7 kJ; (c) Δ H ° = −510 kJ; Δ G ° = −601.5 kJ A hydride of silicon prepared by the reaction of Mg 2 Si with acid exerted a pressure of 306 torr at 26 °C in a bulb with a volume of 57.0 mL. If the mass of the hydride was 0.0861 g, what is its molecular mass? What is the molecular formula for the hydride? Suppose you discovered a diamond completely encased in a silicate rock. How would you chemically free the diamond without harming it? A mild solution of hydrofluoric acid would dissolve the silicate and would not harm the diamond. 18.4: Structure and General Properties of the Nonmetals Carbon forms a number of allotropes, two of which are graphite and diamond. Silicon has a diamond structure. Why is there no allotrope of silicon with a graphite structure? Nitrogen in the atmosphere exists as very stable diatomic molecules. Why does phosphorus form less stable P 4 molecules instead of P 2 molecules? In the N 2 molecule, the nitrogen atoms have an σ bond and two π bonds holding the two atoms together. The presence of three strong bonds makes N 2 a very stable molecule. Phosphorus is a third-period element, and as such, does not form π bonds efficiently; therefore, it must fulfill its bonding requirement by forming three σ bonds. Write balanced chemical equations for the reaction of the following acid anhydrides with water: (a) SO 3 (b) N 2 O 3 (c) Cl 2 O 7 (d) P 4 O 10 (e) NO 2 Determine the oxidation number of each element in each of the following compounds: (a) HCN (b) OF 2 (c) AsCl 3 (a) H = 1+, C = 2+, and N = 3−; (b) O = 2+ and F = 1−; (c) As = 3+ and Cl = 1− Determine the oxidation state of sulfur in each of the following: (a) SO 3 (b) SO 2 (c) \(\ce{SO3^2-}\) Arrange the following in order of increasing electronegativity: F; Cl; O; and S. S < Cl < O < F Why does white phosphorus consist of tetrahedral P 4 molecules while nitrogen consists of diatomic N 2 molecules? 18.5: Occurrence, Preparation, and Compounds of Hydrogen Why does hydrogen not exhibit an oxidation state of 1− when bonded to nonmetals? The electronegativity of the nonmetals is greater than that of hydrogen. Thus, the negative charge is better represented on the nonmetal, which has the greater tendency to attract electrons in the bond to itself. The reaction of calcium hydride, CaH 2 , with water can be characterized as a Lewis acid-base reaction: \[\ce{CaH2}(s)+\ce{2H2O}(l)⟶\ce{Ca(OH)2}(aq)+\ce{2H2}(g)\] Identify the Lewis acid and the Lewis base among the reactants. The reaction is also an oxidation-reduction reaction. Identify the oxidizing agent, the reducing agent, and the changes in oxidation number that occur in the reaction. In drawing Lewis structures, we learn that a hydrogen atom forms only one bond in a covalent compound. Why? Hydrogen has only one orbital with which to bond to other atoms. Consequently, only one two-electron bond can form. What mass of CaH 2 is necessary to react with water to provide enough hydrogen gas to fill a balloon at 20 °C and 0.8 atm pressure with a volume of 4.5 L? The balanced equation is: \[\ce{CaH2}(s)+\ce{2H2O}(l)⟶\ce{Ca(OH)2}(aq)+\ce{2H2}(g)\] What mass of hydrogen gas results from the reaction of 8.5 g of KH with water? \[\ce{KH + H2O ⟶ KOH + H2}\] 0.43 g H 2 18.6: Occurrence, Preparation, and Properties of Carbonates Carbon forms the \(\ce{CO3^2-}\) ion, yet silicon does not form an analogous \(\ce{SiO3^2-}\) ion. Why? Complete and balance the following chemical equations: (a) hardening of plaster containing slaked lime \[\ce{Ca(OH)2 + CO2 ⟶}\] (b) removal of sulfur dioxide from the flue gas of power plants \[\ce{CaO + SO2 ⟶}\] (c) the reaction of baking powder that produces carbon dioxide gas and causes bread to rise \[\ce{NaHCO3 + NaH2PO4 ⟶}\] (a) \(\ce{Ca(OH)2}(aq)+\ce{CO2}(g)⟶\ce{CaCO3}(s)+\ce{H2O}(l)\); (b) \(\ce{CaO}(s)+\ce{SO2}(g)⟶\ce{CaSO3}(s)\); (c) \(\ce{2NaHCO3}(s)+\ce{NaH2PO4}(aq)⟶\ce{Na3PO4}(aq)+\ce{2CO2}(g)+\ce{2H2O}(l)\) Heating a sample of Na 2 CO 3 ⋅ x H 2 O weighing 4.640 g until the removal of the water of hydration leaves 1.720 g of anhydrous Na 2 CO 3 . What is the formula of the hydrated compound? 18.7: Occurrence, Preparation, and Properties of Nitrogen Write the Lewis structures for each of the following: (a) NH 2− (b) N 2 F 4 (c) \(\ce{NH2-}\) (d) NF 3 (e) \(\ce{N3-}\) (a) NH 2− : ; (b) N 2 F 4 : ; (c) \(\ce{NH2-}\): ; (d) NF 3 : ; (e) \(\ce{N3-}\): For each of the following, indicate the hybridization of the nitrogen atom (for \(\ce{N3-}\), the central nitrogen). (a) N 2 F 4 (b) \(\ce{NH2-}\) (c) NF 3 (d) \(\ce{N3-}\) Explain how ammonia can function both as a Brønsted base and as a Lewis base. Ammonia acts as a Brønsted base because it readily accepts protons and as a Lewis base in that it has an electron pair to donate. Brønsted base: \(\ce{NH3 + H3O+ ⟶ NH4+ + H2O}\) Lewis base: \(\ce{2NH3 + Ag+ ⟶ [H3N−Ag−NH3]+}\) Determine the oxidation state of nitrogen in each of the following. You may wish to review the chapter on chemical bonding for relevant examples. (a) NCl 3 (b) ClNO (c) N 2 O 5 (d) N 2 O 3 (e) \(\ce{NO2-}\) (f) N 2 O 4 (g) N 2 O (h) \(\ce{NO3-}\) (i) HNO 2 (j) HNO 3 For each of the following, draw the Lewis structure, predict the ONO bond angle, and give the hybridization of the nitrogen. You may wish to review the chapters on chemical bonding and advanced theories of covalent bonding for relevant examples. (a) NO 2 (b) \(\ce{NO2-}\) (c) \(\ce{NO2+}\) (a) NO 2 : Nitrogen is sp 2 hybridized. The molecule has a bent geometry with an ONO bond angle of approximately 120°. (b) \(\ce{NO2-}\): Nitrogen is sp 2 hybridized. The molecule has a bent geometry with an ONO bond angle slightly less than 120°. (c) \(\ce{NO2+}\): Nitrogen is sp hybridized. The molecule has a linear geometry with an ONO bond angle of 180°. How many grams of gaseous ammonia will the reaction of 3.0 g hydrogen gas and 3.0 g of nitrogen gas produce? Although PF 5 and AsF 5 are stable, nitrogen does not form NF 5 molecules. Explain this difference among members of the same group. Nitrogen cannot form a NF 5 molecule because it does not have d orbitals to bond with the additional two fluorine atoms. The equivalence point for the titration of a 25.00-mL sample of CsOH solution with 0.1062 M HNO 3 is at 35.27 mL. What is the concentration of the CsOH solution? 18.8: Occurrence, Preparation, and Properties of Phosphorus Write the Lewis structure for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry. (a) PH 3 (b) \(\ce{PH4+}\) (c) P 2 H 4 (d) \(\ce{PO4^3-}\) (e) PF 5 (a) ; (b) ; (c) ; (d) ; (e) Describe the molecular structure of each of the following molecules or ions listed. You may wish to review the chapter on chemical bonding and molecular geometry. (a) PH 3 (b) \(\ce{PH4+}\) (c) P 2 H 4 (d) \(\ce{PO4^3-}\) Complete and balance each of the following chemical equations. (In some cases, there may be more than one correct answer.) (a) \(\ce{P4 + Al⟶}\) (b) \(\ce{P4 + Na⟶}\) (c) \(\ce{P4 + F2⟶}\) (d) \(\ce{P4 + Cl2⟶}\) (e) \(\ce{P4 + O2⟶}\) (f) \(\ce{P4O6 + O2⟶}\) (a) \(\ce{P4}(s)+\ce{4Al}(s)⟶\ce{4AlP}(s)\); (b) \(\ce{P4}(s)+\ce{12Na}(s)⟶\ce{4Na3P}(s)\); (c) \(\ce{P4}(s)+\ce{10F2}(g)⟶\ce{4PF5}(l)\); (d) \(\ce{P4}(s)+\ce{6Cl2}(g)⟶\ce{4PCl3}(l)\) or \(\ce{P4}(s)+\ce{10Cl2}(g)⟶\ce{4PCl5}(l)\); (e) \(\ce{P4}(s)+\ce{3O2}(g)⟶\ce{P4O6}(s)\) or \(\ce{P4}(s)+\ce{5O2}(g)⟶\ce{P4O10}(s)\); (f) \(\ce{P4O6}(s)+\ce{2O2}(g)⟶\ce{P4O10}(s)\) Describe the hybridization of phosphorus in each of the following compounds: P 4 O 10 , P 4 O 6 , PH 4 I (an ionic compound), PBr 3 , H 3 PO 4 , H 3 PO 3 , PH 3 , and P 2 H 4 . You may wish to review the chapter on advanced theories of covalent bonding. What volume of 0.200 M NaOH is necessary to neutralize the solution produced by dissolving 2.00 g of PCl 3 is an excess of water? Note that when H 3 PO 3 is titrated under these conditions, only one proton of the acid molecule reacts. 291 mL How much POCl 3 can form from 25.0 g of PCl 5 and the appropriate amount of H 2 O? How many tons of Ca 3 (PO 4 ) 2 are necessary to prepare 5.0 tons of phosphorus if the yield is 90%? 28 tons Write equations showing the stepwise ionization of phosphorous acid. Draw the Lewis structures and describe the geometry for the following: (a) \(\ce{PF4+}\) (b) PF 5 (c) \(\ce{PF6-}\) (d) POF 3 (a) ; (b) ; (c) ; (d) Why does phosphorous acid form only two series of salts, even though the molecule contains three hydrogen atoms? Assign an oxidation state to phosphorus in each of the following: (a) NaH 2 PO 3 (b) PF 5 (c) P 4 O 6 (d) K 3 PO 4 (e) Na 3 P (f) Na 4 P 2 O 7 (a) P = 3+; (b) P = 5+; (c) P = 3+; (d) P = 5+; (e) P = 3−; (f) P = 5+ Phosphoric acid, one of the acids used in some cola drinks, is produced by the reaction of phosphorus(V) oxide, an acidic oxide, with water. Phosphorus(V) oxide is prepared by the combustion of phosphorus. (a) Write the empirical formula of phosphorus(V) oxide. (b) What is the molecular formula of phosphorus(V) oxide if the molar mass is about 280. (c) Write balanced equations for the production of phosphorus(V) oxide and phosphoric acid. (d) Determine the mass of phosphorus required to make 1.00 × 10 4 kg of phosphoric acid, assuming a yield of 98.85%. 18.9: Occurrence, Preparation, and Compounds of Oxygen Predict the product of burning francium in air. FrO 2 Using equations, describe the reaction of water with potassium and with potassium oxide. Write balanced chemical equations for the following reactions: (a) zinc metal heated in a stream of oxygen gas (b) zinc carbonate heated until loss of mass stops (c) zinc carbonate added to a solution of acetic acid, CH 3 CO 2 H (d) zinc added to a solution of hydrobromic acid (a) \(\ce{2Zn}(s)+\ce{O2}(g)⟶\ce{2ZnO}(s)\); (b) \(\ce{ZnCO3}(s)⟶\ce{ZnO}(s)+\ce{CO2}(g)\); (c) \(\ce{ZnCO3}(s)+\ce{2CH3COOH}(aq)⟶\ce{Zn(CH3COO)2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\); (d) \(\ce{Zn}(s)+\ce{2HBr}(aq)⟶\ce{ZnBr2}(aq)+\ce{H2}(g)\) Write balanced chemical equations for the following reactions: (a) cadmium burned in air (b) elemental cadmium added to a solution of hydrochloric acid (c) cadmium hydroxide added to a solution of acetic acid, CH 3 CO 2 H Illustrate the amphoteric nature of aluminum hydroxide by citing suitable equations. \(\ce{Al(OH)3}(s)+\ce{3H+}(aq)⟶\ce{Al^3+}+\ce{3H2O}(l)\); \(\ce{Al(OH)3}(s)+\ce{OH-}⟶\ce{[Al(OH)4]-}(aq)\) Write balanced chemical equations for the following reactions: (a) metallic aluminum burned in air (b) elemental aluminum heated in an atmosphere of chlorine (c) aluminum heated in hydrogen bromide gas (d) aluminum hydroxide added to a solution of nitric acid Write balanced chemical equations for the following reactions: (a) sodium oxide added to water (b) cesium carbonate added to an excess of an aqueous solution of HF (c) aluminum oxide added to an aqueous solution of HClO 4 (d) a solution of sodium carbonate added to solution of barium nitrate (e) titanium metal produced from the reaction of titanium tetrachloride with elemental sodium (a) \(\ce{Na2O}(s)+\ce{H2O}(l)⟶\ce{2NaOH}(aq)\); (b) \(\ce{Cs2CO3}(s)+\ce{2HF}(aq)⟶\ce{2CsF}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\); (c) \(\ce{Al2O3}(s)+\ce{6HClO4}(aq)⟶\ce{2Al(ClO4)3}(aq)+\ce{3H2O}(l)\); (d) \(\ce{Na2CO3}(aq)+\ce{Ba(NO3)2}(aq)⟶\ce{2NaNO3}(aq)+\ce{BaCO3}(s)\); (e) \(\ce{TiCl4}(l)+\ce{4Na}(s)⟶\ce{Ti}(s)+\ce{4NaCl}(s)\) What volume of 0.250 M H 2 SO 4 solution is required to neutralize a solution that contains 5.00 g of CaCO 3 ? Which is the stronger acid, HClO 4 or HBrO 4 ? Why? HClO 4 is the stronger acid because, in a series of oxyacids with similar formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s). The stronger attraction of the oxygen electron results in a stronger attraction of oxygen for the electrons in the O-H bond, making the hydrogen more easily released. The weaker this bond, the stronger the acid. Write a balanced chemical equation for the reaction of an excess of oxygen with each of the following. Remember that oxygen is a strong oxidizing agent and tends to oxidize an element to its maximum oxidation state. (a) Mg (b) Rb (c) Ga (d) C 2 H 2 (e) CO Which is the stronger acid, H 2 SO 4 or H 2 SeO 4 ? Why? You may wish to review the chapter on acid-base equilibria. As H 2 SO 4 and H 2 SeO 4 are both oxyacids and their central atoms both have the same oxidation number, the acid strength depends on the relative electronegativity of the central atom. As sulfur is more electronegative than selenium, H 2 SO 4 is the stronger acid. 18.10: Occurrence, Preparation, and Properties of Sulfur Explain why hydrogen sulfide is a gas at room temperature, whereas water, which has a lower molecular mass, is a liquid. Give the hybridization and oxidation state for sulfur in SO 2 , in SO 3 , and in H 2 SO 4 . SO 2 , sp 2 4+; SO 3 , sp 2 , 6+; H 2 SO 4 , sp 3 , 6+ Which is the stronger acid, NaHSO 3 or NaHSO 4 ? Determine the oxidation state of sulfur in SF 6 , SO 2 F 2 , and KHS. SF 6 : S = 6+; SO 2 F 2 : S = 6+; KHS: S = 2− Which is a stronger acid, sulfurous acid or sulfuric acid? Why? Oxygen forms double bonds in O 2 , but sulfur forms single bonds in S 8 . Why? Sulfur is able to form double bonds only at high temperatures (substantially endothermic conditions), which is not the case for oxygen. Give the Lewis structure of each of the following: (a) SF 4 (b) K 2 SO 4 (c) SO 2 Cl 2 (d) H 2 SO 3 (e) SO 3 Write two balanced chemical equations in which sulfuric acid acts as an oxidizing agent. There are many possible answers including: \[\ce{Cu}(s)+\ce{2H2SO4}(l)⟶\ce{CuSO4}(aq)+\ce{SO2}(g)+\ce{2H2O}(l)\] \[\ce{C}(s)+\ce{2H2SO4}(l)⟶\ce{CO2}(g)+\ce{2SO2}(g)+\ce{2H2O}(l)\] Explain why sulfuric acid, H 2 SO 4 , which is a covalent molecule, dissolves in water and produces a solution that contains ions. How many grams of Epsom salts (MgSO 4 ⋅7H 2 O) will form from 5.0 kg of magnesium? 5.1 × 10 4 g 18.11: Occurrence, Preparation, and Properties of Halogens What does it mean to say that mercury(II) halides are weak electrolytes? Why is SnCl4 not classified as a salt? SnCl4 is not a salt because it is covalently bonded. A salt must have ionic bonds. The following reactions are all similar to those of the industrial chemicals. Complete and balance the equations for these reactions: (a) reaction of a weak base and a strong acid \[\ce{NH3 + HClO4⟶}\] (b) preparation of a soluble silver salt for silver plating \[\ce{Ag2CO3 + HNO3⟶}\] (c) preparation of strontium hydroxide by electrolysis of a solution of strontium chloride \[\ce{SrCl2}(aq)+\ce{H2O}(l)\xrightarrow{\ce{electrolysis}}\] Which is the stronger acid, HClO3 or HBrO3? Why? In oxyacids with similar formulas, the acid strength increases as the electronegativity of the central atom increases. HClO3 is stronger than HBrO3; Cl is more electronegative than Br. What is the hybridization of iodine in IF3 and IF5? Predict the molecular geometries and draw Lewis structures for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry. (a) IF 5 (b) \(\ce{I3-}\) (c) PCl 5 (d) SeF 4 (e) ClF 3 (a) ; (b) ; (c) ; (d) ; (e) Which halogen has the highest ionization energy? Is this what you would predict based on what you have learned about periodic properties? Name each of the following compounds: (a) BrF 3 (b) NaBrO 3 (c) PBr 5 (d) NaClO 4 (e) KClO (a) bromine trifluoride; (b) sodium bromate; (c) phosphorus pentabromide; (d) sodium perchlorate; (e) potassium hypochlorite Explain why, at room temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid. What is the oxidation state of the halogen in each of the following? (a) H 5 IO 6 (b) \(\ce{IO4-}\) (c) ClO 2 (d) ICl 3 (e) F 2 (a) I: 7+; (b) I: 7+; (c) Cl: 4+; (d) I: 3+; Cl: 1−; (e) F: 0 Physiological saline concentration—that is, the sodium chloride concentration in our bodies—is approximately 0.16 M . A saline solution for contact lenses is prepared to match the physiological concentration. If you purchase 25 mL of contact lens saline solution, how many grams of sodium chloride have you bought? 18.12: Occurrence, Preparation, and Properties of the Noble Gases Give the hybridization of xenon in each of the following. You may wish to review the chapter on the advanced theories of covalent bonding. (a) XeF 2 (b) XeF 4 (c) XeO 3 (d) XeO 4 (e) XeOF 4 (a) sp 3 d hybridized; (b) sp 3 d 2 hybridized; (c) sp 3 hybridized; (d) sp 3 hybridized; (e) sp 3 d 2 hybridized; What is the molecular structure of each of the following molecules? You may wish to review the chapter on chemical bonding and molecular geometry. (a) XeF 2 (b) XeF 4 (c) XeO 3 (d) XeO 4 (e) XeOF 4 Indicate whether each of the following molecules is polar or nonpolar. You may wish to review the chapter on chemical bonding and molecular geometry. (a) XeF 2 (b) XeF 4 (c) XeO 3 (d) XeO 4 (e) XeOF 4 (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar What is the oxidation state of the noble gas in each of the following? You may wish to review the chapter on chemical bonding and molecular geometry. (a) XeO 2 F 2 (b) KrF 2 (c) \(\ce{XeF3+}\) (d) \(\ce{XeO6^4-}\) (e) XeO 3 A mixture of xenon and fluorine was heated. A sample of the white solid that formed reacted with hydrogen to yield 81 mL of xenon (at STP) and hydrogen fluoride, which was collected in water, giving a solution of hydrofluoric acid. The hydrofluoric acid solution was titrated, and 68.43 mL of 0.3172 M sodium hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon. The empirical formula is XeF 6 , and the balanced reactions are: \[\ce{Xe}(g)+\ce{3F2}(g)\xrightarrow{Δ}\ce{XeF6}(s)\] \[\ce{XeF6}(s)+\ce{3H2}(g)⟶\ce{6HF}(g)+\ce{Xe}(g)\] Basic solutions of Na 4 XeO 6 are powerful oxidants. What mass of Mn(NO 3 ) 2 •6H 2 O reacts with 125.0 mL of a 0.1717 M basic solution of Na 4 XeO 6 that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate? |
Courses/Victor_Valley_College/CHEM100_Victor_Valley_College/12%3A_Organic_Compounds_of_Oxygen_and_Nitrogen | Ethanol and resveratrol, a phenol, are representatives of two of the families of oxygen-containing compounds that we consider in this chapter. Two other classes, aldehydes and ketones, are formed by the oxidation of alcohols. Ethers, another class, are made by the dehydration of alcohols. 12.1: Alcohols - Nomenclature and Classification In the IUPAC system, alcohols are named by changing the ending of the parent alkane name to -ol. Alcohols are classified according to the number of carbon atoms attached to the carbon atom that is attached to the OH group. 12.2: Physical Properties of Alcohols Alcohols have higher boiling points than do ethers and alkanes of similar molar masses because the OH group allows alcohol molecules to engage in hydrogen bonding. Alcohols of four or fewer carbon atoms are soluble in water because the alcohol molecules engage in hydrogen bonding with water molecules; comparable alkane molecules cannot engage in hydrogen bonding. 12.3: Reactions that Form Alcohols Many alcohols are made by the hydration of alkenes. Ethanol can be made by the fermentation of sugars or starch from various sources. 12.4: Reactions of Alcohols Alcohols can be dehydrated to form either alkenes (higher temperature, excess acid) or ethers (lower temperature, excess alcohol). Primary alcohols are oxidized to form aldehydes. Secondary alcohols are oxidized to form ketones. Tertiary alcohols are not readily oxidized. 12.5: Aldehydes and Ketones- Structure and Names The common names of aldehydes are taken from the names of the corresponding carboxylic acids: formaldehyde, acetaldehyde, and so on. The common names of ketones, like those of ethers, consist of the names of the groups attached to the carbonyl group, followed by the word ketone. Stem names of aldehydes and ketones are derived from those of the parent alkanes, using an -al ending for an aldehydes and an -one ending for a ketone. 12.6: Properties of Aldehydes and Ketones The polar carbon-to-oxygen double bond causes aldehydes and ketones to have higher boiling points than those of ethers and alkanes of similar molar masses but lower than those of comparable alcohols that engage in intermolecular hydrogen bonding. Aldehydes are readily oxidized to carboxylic acids, whereas ketones resist oxidation. 12.7: Ethers To give ethers common names, simply name the groups attached to the oxygen atom, followed by the generic name ether. If both groups are the same, the group name should be preceded by the prefix di-. Ether molecules have no OH group and thus no intermolecular hydrogen bonding. Ethers therefore have quite low boiling points for a given molar mass. Ether molecules have an oxygen atom and can engage in hydrogen bonding with water molecules. An ether molecule has about the same solubility in water as 12.8: Carboxylic Acids - Structures and Names Simple carboxylic acids are best known by common names based on Latin and Greek words that describe their source (e.g., formic acid, Latin formica, meaning “ant”). Greek letters, not numbers, designate the position of substituted acids in the common naming convention. IUPAC names are derived from the LCC of the parent hydrocarbon with the -e ending of the parent alkane replaced by the suffix -oic and the word acid. 12.9: Acidity of Carboxylic Acids 12.10: Esters - Structures and Names An ester has an OR group attached to the carbon atom of a carbonyl group. 12.11: Amines - Structures and Names An amine is a derivative of ammonia in which one, two, or all three hydrogen atoms are replaced by hydrocarbon groups. Amines are classified as primary, secondary, or tertiary by the number of hydrocarbon groups attached to the nitrogen atom. Amines are named by naming the alkyl groups attached to the nitrogen atom, followed by the suffix -amine. 12.12: Amines as Bases and Heterocylclic Amines Amines are bases; they react with acids to form salts. Salts of aniline are properly named as anilinium compounds, but an older system is used to name drugs: the salts of amine drugs and hydrochloric acid are called “hydrochlorides.” Heterocyclic amines are cyclic compounds with one or more nitrogen atoms in the ring. 12.13: Amides- Structures and Names Amides have a general structure in which a nitrogen atom is bonded to a carbonyl carbon atom. In names for amides, the -ic acid of the common name or the -oic ending of the IUPAC for the corresponding carboxylic acid is replaced by -amide. Template:HideTOC |
Bookshelves/Analytical_Chemistry/Chemometrics_Using_R_(Harvey)/03%3A_Visualizing_Data/3.03%3A_Creating_Plots_From_Scratch_in_R_Using_Base_Graphics | As we saw in the last section, the functions to create dot charts, stripcharts, boxplots, barplots, and histograms have arguments that we can use to alter the appearance of the function’s output. For example, here is the full list of arguments available when we use dotchart() that control what the plot shows. dotchart(x, labels = NULL, groups = NULL, gdata = NULL, cex = par("cex"), pt.cex = cex, pch = 21, gpch = 21, bg = par("bg"), color = par("fg"), gcolor = par("fg"), lcolor = "gray", xlim = range(x[is.finite(x)]), main = NULL, xlab = NULL, ylab = NULL, ...) Each of the arguments has a default value, which means we need not specify the value for an argument unless we wish to change its value, as we did when we set pch to 19. The final argument of ... indicates that we can change any of a long list of graphical parameters that control what we see when we use dotchart. Creating a Simple Scatterplot Using R One of the most common, and most important, visualizations in analytical chemistry is a scatterplot in which we are interested in the relationship, if any, between two measurement by plotting the values for one variable along the x -axis and the values for the other variable along y -axis. For this exercise, we will use some data from the Puget Sound Data Hoard that gives the mass and the diameter for 816 M&Ms obtained from a 14.0-oz bag of plain M&Ms, a 12.7-oz bag of peanut M&Ms, and a 12.7-oz bag of peanut butter M&Ms. Let’s read the data into R and store it in a data frame with the name psmm_data . You can download a copy of the data using this link saving it in your working directory. psmm_data = read.csv("data/PugetSoundM&MData.csv") We might expect that as the diameter of an M&M increases so will the mass of the M&M. We might also expect that the relationship between diameter and mass may depend on whether the M&Ms are plain, peanut, or peanut butter. So that we can access data for each type of M&M, let’s use the which() function to create vectors that designate the row numbers for each of the three types of M&Ms. pb_id = which(psmm_data$type == "peanut butter") plain_id = which(psmm_data$type == "plain") peanut_id = which(psmm_data$type == "peanut") Typically we are interested in how one variable affects the other variable. We call the former the independent variable and place it on the x -axis and we call the latter the dependent variable and place it on the y -axis. Here we will use diameter as the independent variable and mass as the dependent variable. To create a scatterplot for the plain M&Ms we use the function plot(x, y) where x is the data to plot on the x -axis and y is the data to plot on the y -axis. plot(x = psmm_data$diameter[plain_id], y = psmm_data$mass[plain_id]) Customizing a Plot Created Using R Although our scatterplot shows that the mass of a plain M&M increases as its diameter increases, it is not a particularly attractive plot. In addition to specifying x and y , the plot function allows us to pass additional arguments to customize our plot; here are some of these optional arguments: type = “ option ” . This argument specifies how points are displayed; there are a number of options, but the most useful are “p” for points (this is the default), “l” for lines without points, “b” for both points and lines that do not touch the points, “o” for points and lines that pass through the points, “h” for histogram-like vertical lines, and “s” for stair steps; use “n” if you wish to suppress the points. pch = number . This argument selects the symbol used to plot the data, with the number assigned to each symbol shown below. The default option is 1, or an open circle. Symbols 15–20 are filled using the color of the symbol’s boundary, and symbols 21–25 can take a background color that is different from the symbol’s boundary. See later in this document for more details about setting colors. The figure below shows the different options. # code from http://www.sthda.com/english/wiki/r-...available-in-r oldPar = par() par(font = 2, mar = c(0.5, 0, 0, 0)) y = rev(c(rep(1, 6),rep(2, 5), rep(3, 5), rep(4, 5), rep(5, 5))) x = c(rep(1:5, 5), 6) plot(x, y, pch = 0:25, cex = 1.5, ylim = c(1, 5.5), xlim = c(1, 6.5), axes = FALSE, xlab = "", ylab = "", bg = "blue") text(x, y, labels = 0:25, pos = 3) par(mar = oldPar$mar, font = oldPar$font) lty = number . This argument specifies the type of line to draw; the options are 1 for a solid line (this is the default), 2 for a dashed line, 3 for a dotted line, 4 for a dot-dash line, 5 for a long-dash line, and 6 for a two-dash line. lwd = number . This argument sets the width of the line. The default is 1 and any other entry simply scales the width relative to the default; thus lwd = 2 doubles the width and lwd = 0.5 cuts the width in half. bty = “ option ” . This argument specifies the type of box to draw around the plot; the options are “o” to draw all four sides (this is the default), “l” to draw on the left side and the bottom side only, “7” to draw on the top side and the right side only, “c” to draw all but the right side, “u” to draw all but the top side, “]” to draw all but the left side, and “n” to omit all four sides. axes = logical . This argument indicates whether the axes are drawn (TRUE) or not drawn (FALSE); the default is TRUE. xlim = c( begin , end ) . This argument sets the limits for the x -axis, overriding the default limits set by the plot() command. ylim = c( begin , end ) . This argument sets the limits for the y -axis, overriding the default limits set by the plot() command. xlab = “ text ” . This argument specifies the label for the x -axis, overriding the default label set by the plot() command. ylab = “ text ” . This argument specifies the label for the y -axis, overriding the default label set by the plot() command. main = “ text ” . This argument specifies the main title, which is placed above the plot, overriding the default title set by the plot() command. sub = “ text ” . This argument specifies the subtitle, which is placed below the plot, overriding the default subtitle set by the plot() command. cex = number . This argument controls the relative size of the symbols used to plot points. The default is 1 and any other entry simply scales the size relative to the default; thus cex = 2 doubles the size and cex = 0.5 cuts the size in half. cex.axis = number . This argument controls the relative size of the text used for the scale on both axes; see the entry above for cex for more details. cex.lab = number . This argument controls the relative size of the text used for the label on both axes; see the entry above for cex for more details. cex.main = number . This argument controls the relative size of the text used for the plot’s main title; see the entry above for cex for more details. cex.sub = number . This argument controls the relative size of the text used for the plot’s subtitle; see the entry above for cex for more details. col = number or “ string ” . This argument controls the color of the symbols used to plot points. There are 657 available colors, for which the default is “black” or 24. You can see a list of colors (number and text string) by typing colors() in the console. col.axis = number or “ string ” . This argument controls the color of the text used for the scale on both axes; see the entry above for col for more details. col.lab = number or “ string ” . This argument controls the color of the text used for the label on both axes; see the entry above for col for more details. col.main = number or “ string ” . This argument controls the color of the text used for the plot’s main title; see the entry above for col for more details. col.sub = number or “ string ” . This argument controls the color of the text used for the plot’s subtitle; see the entry above for col for more details. bg = number or “ string ” . This argument sets the background color for the plot symbols 21–25; see the entries above for pch and for col for more details. Let’s use some of these arguments to improve our scatterplot by adding some color to and adjusting the size of the symbols used to plot the data, and by adding a title and some more informative labels for the two axes. plot(x = psmm_data$diameter[plain_id], y = psmm_data$mass[plain_id], xlab = "diameter of M&Ms", ylab = "mass of M&Ms", main = "Diameter and Mass of Plain M&Ms", pch = 19, cex = 0.5, col = "blue") Modifying an Existing Plot Created Using R We can modify an existing plot in a number of useful ways, such as adding a new set of data, adding a reference line, adding a legend, adding text, and adding a set of grid lines; here are some of the things we can do: points(x, y, . . . ) . This command is identical to the plot() command, but overlays the new points on the current plot instead of first erasing the previous plot. Note: the points() command can not re-scale the axes; thus, you must ensure that your original plot—created using the plot() command—has x -axis and y -axis limits that meet your needs. abline(h = number , . . . ) . This command adds a horizontal line at y = number with the line’s color, type, and size set using the optional arguments. abline(v = number , . . . ) . This command adds a vertical line at x = number with the line’s color, type, and size set using the optional arguments. abline(b = number , a = number , . . . ) . This command adds a diagonal line defined by a slope ( b ) and a y -intercept ( a ); the line’s color, type, and size are set using the optional arguments. As we will see in Chapter 8, this is a useful command for displaying the results of a linear regression. legend( location , legend , . . . ) . This command adds a legend to the current plot. The location is specified in one of two ways: • by giving the x and y coordinates for the legend’s upper-left corner using x = number and y = number ) • by using location = “ keyword ” where the keyword is one of “topleft”, “top”, “topright”, “right”, “bottomright”, “bottom”, “bottomleft”, or “left”; the optional argument inset = number moves the legend in from the margin when using a keyword (it takes a value from 0 to 1 as a fraction of the plot’s area; the default is 0) The legend is added as a vector of character strings (one for each item in the legend), and any accompanying formatting, such as plot symbols, lines, or colors, are passed along as vectors of the same length; look carefully at the example at the end of this section to see how this command works. text( location , label , . . . ) . This command adds the text given by “ label ” to the current plot. The location is specified by providing values for x and y using x = number and y = number . By default, the text is centered at its location; to set the text so that it is left-justified (which is easier to work with), add the argument adj = c(0, NA) . grid( col , lty , lwd ) . This command adds a set of grid lines to the plot using the color, line type, and line width defined by “ col ”, “ lty ”, and “ lwd ”, respectively. Here is an example of a figure in which we show how the diameter and mass vary as a function of the type of M&Ms, add a legend, add a grid, and add some text that identifies the source of the data. Note the use of the functions max and min to identify the limits needed to display results for all of the data. # determine minimum and maximum values for diameter and mass so that we can # set limits for the x-axis and y-axis that will allow plotting of all data xmax = max(psmm_data$diameter) xmin = min(psmm_data$diameter) ymax = max(psmm_data$mass) ymin = min(psmm_data$mass) # create the initial plot using data for plain M&Ms, xlim and ylim values # ensure plot window will allow plotting of all data plot(x = psmm_data$diameter[plain_id], y = psmm_data$mass[plain_id], xlab = "diameter of M&Ms", ylab = "mass of M&Ms", main = "Diameter and Mass of M&Ms", pch = 19, cex = 0.65, col = "red", xlim = c(xmin, xmax), ylim = c(ymin, ymax)) # add the data for the peanut and peanut butter M&Ms using points() points(x = psmm_data$diameter[peanut_id], y = psmm_data$mass[peanut_id], pch = 18, col = "brown", cex = 0.65) points(x = psmm_data$diameter[pb_id], y = psmm_data$mass[pb_id], pch = 17, col = "blue", cex = 0.65) # add a legend, gird, and explanatory text legend(x = "topleft", legend = c("plain", "peanut", "peanut butter"), col = c("red", "brown", "blue"), pch = c(19, 18, 17), bty = "n") grid(col = "gray") text(x = 16.5, y = 1, label = "data from University of Puget Sound Data Hoard", cex = 0.5) Our new plot shows that the individual M&Ms are reasonably well separated from each other in the space created by the variables diameter and mass, although a few M&Ms encroach into the space occupied by other types of M&Ms. We also see that the distribution of plain M&Ms is much more compact than for peanut and peanut butter M&Ms, which makes sense given the likely variability in the size of individual peanuts and the softer consistency of peanut butter. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/19%3A_The_Distribution_of_Outcomes_for_Multiple_Trials/19.05%3A_Problems | 1. Leland got a train set for Christmas. It came with seven rail cars. (We say that all seven cars are “distinguishable.”) Four of the rail cars are box cars and three are tank cars. If we distinguish between permutations in which the box cars are coupled (lined up) differently but not between permutations in which tank cars are coupled differently, how many ways can the seven cars be coupled so that all of the tank cars are together? What are they? What formula can we use to compute this number? (Hint: We can represent one of the possibilities as \(b_1b_2b_3b_4T\). This is one of the possibilities in which the first four cars behind the engine are all box cars. There are \(4!\) such possibilities; that is, there are \(4!\) possible permutations for placing the four box cars.) 2. If we don’t care about the order in which the box cars are coupled, and we don’t care about the order in which the tank cars are coupled, how many ways can the rail cars in problem 1 be coupled so that all of the tank cars are together? What are they? What formula can we use to compute this number? 3. If we distinguish between permutations in which either the box cars or the tank cars in problem 1 are ordered differently, how many ways can the rail cars be coupled so that all of the tank cars are together? What formula can we use to compute this number? 4. How many ways can all seven rail cars in problem 1 be coupled if the tank cars need not be together? 5. If, as in the previous problem, we distinguish between permutations in which any of the rail cars are ordered differently, how many ways can the rail cars be coupled so that not all of the tank cars are together? 6. If we distinguish between box cars and tank cars, but we do not distinguish one box car from another box car, and we do not distinguish one tank car from another tank car, how many ways can the rail cars in problem 1 be coupled? 7. If Leland gets five flat cars for his birthday, he will have four box cars, three tank cars and five flat cars. How many ways will Leland be able to couple (permute) these twelve rail cars? 8. If we distinguish between box cars and tank cars, between box cars and flat cars, and between tank cars and flat cars, but we do not distinguish one box car from another box car, and we do not distinguish one tank car from another tank car, and we do not distinguish one flat car from another flat car, how many ways can the rail cars in problem seven be coupled? What formula can we use to compute this number? 9. We are given four distinguishable marbles, labeled \(A--D\), and two cups, labeled \(1\) and \(2\). We want to explore the number of ways we can put two marbles in cup \(1\) and two marbles in cup \(2\). This is the number of combinations, \(C\left(2,2\right)\), for the population set \(N_1=2\), \(N_2=2\). (a) One combination is \({\left[AB\right]}_1{\left[CD\right]}_2\). Find the remaining combinations. What is \(C\left(2,2\right)\)? (b) There are four permutations for the combination given in (a):\(\ {\left[AB\right]}_1{\left[CD\right]}_2\); \({\left[BA\right]}_1{\left[CD\right]}_2\); \({\left[AB\right]}_1{\left[DC\right]}_2\); \({\left[BA\right]}_1{\left[DC\right]}_2\). Find all of the permutations for each of the remaining combinations. (c) How many permutations are there for each combination? (d) Write down all of the possible permutations of marbles \(A--D\). Show that there is a one-to-one correspondence with the permutations in (b). (e) Show that the total number of permutations is equal to the number of combinations times the number of permutations possible for each combination. 10. We are given seven distinguishable marbles, labeled \(A--G\), and two cups, labeled \(1\) and \(2\). We want to find the number of ways we can put three marbles in cup \(1\) and four marbles in cup\(\ 2\). That is, we seek \(C\left(3,4\right)\), the number of combinations in which \(N_1=3\) and \(N_2=4\). \({\left[ABC\right]}_1{\left[DEFG\right]}_2\) is one such combination. (a) How many different ways can these marbles be placed in different orders without exchanging any marbles between cup \(1\) and cup \(2\)? (This is the number of permutations associated with this combination.) (b) Find a different combination with \(N_1=3\) and \(N_2=4\). (c) How many permutations are possible for the marbles in (b)? How many permutations are possible for any combination with \(N_1=3\) and \(N_2=4\)? (d) If \(C\left(3,4\right)\) is the number of combinations in which \(N_1=3\) and \(N_2=4\), and if \(P\) is the number of permutations for each such combination, what is the total number of permutations possible for 7 marbles? (e) How else can one express the number of permutations possible for 7 marbles? (f) Equate your conclusions in (d) and (e). Find \(C\left(3,4\right)\). 11. (a) Calculate the probabilities of 0, 1, 2, 3, and 4 heads in a series of four tosses of an unbiased coin. The event of 2 heads is\(\ 20\%\) of these five events. Note particularly the probability of the event: 2 heads in 4 tosses. (b) Calculate the probabilities of 0, 1, 2, 3, … , 8, and 9 heads in a series of nine tosses of an unbiased coin. The events of 4 heads and 5 heads comprise \(20\%\) of these ten cases. Calculate the probability of 4 heads or 5 heads; i.e., the probability of being in the middle \(20\%\) of the possible events. (c) Calculate the probabilities of 0, 1, 2, 3, … , 13, and 14 heads in a series of fourteen tosses of an unbiased coin. The events of 6 heads, 7 heads, and 8 heads comprise 20% of these fifteen cases. Calculate the probability of 6, 7, or 8 heads; i.e., the probability of being in the middle \(20\%\) of the possible events. (d) What happens to the probabilities for the middle \(20\%\) of possible events as the number of tosses becomes very large? How does this relate to the fraction heads in a series of tosses when the total number of tosses becomes very large? 12. Let the value of the outcome heads be one and the value of the outcome tails be zero. Let the “score” from a particular simultaneous toss of \(n\) coins be \[\mathrm{score}=1\times \left(\frac{number\ of\ heads}{number\ of\ coins}\right)\ +0\times \left(\frac{number\ of\ tails}{number\ of\ coins}\right) \nonumber \] Let us refer to the distribution of scores from tosses of \(n\) coins as the “\(S_n\) distribution.” (a) The \(S_1\) distribution comprises two outcomes: \(\mathrm{\{}\)1 head, 0 tail\(\mathrm{\}}\) and \(\mathrm{\{}\)0 head, 1 tail\(\mathrm{\}}\). What is the mean of the \(S_1\) distribution? (b) What is the variance of the \(S_1\) distribution? (c) What is the mean of the \(S_n\) distribution? (d) What is the variance of the \(S_n\) distribution? 13. Fifty unbiased coins are tossed simultaneously. (a) Calculate the probability of 25 heads and 25 tails. (b) Calculate the probability of 23 heads and 27 tails. (c) Calculate the probability of 3 heads and 47 tails. (d) Calculate the ratio of your results for parts (a) and (b). (e) Calculate the ratio of your results for parts (a) and (c). 14. For \(N=3,\ 6\) and \(10\), calculate\(\) (a) The exact value of \(N!\) (b) The value of \(N!\) according to the approximation \[N!\approx N^N \left(2\pi N\right)^{1/2}\mathrm{exp}\left(-N\right)\mathrm{exp}\left(\frac{1}{12N}\right) \nonumber \] (c) The value of N! according to the approximation \[N!\approx N^N \left(2\pi N\right)^{1/2}\mathrm{exp}\left(-N\right) \nonumber \] (d) The value of N! according to the approximation \[N!\approx N^N\mathrm{exp}\left(-N\right) \nonumber \] (e) The ratio of the value in (b) to the corresponding value in (a). (f) The ratio of the value in (c) to the corresponding value in (a). (g) The ratio of the value in (d) to the corresponding value in (a). (h) Comment. 15. Find , \(d ~ \ln N! /dN\) using each of the approximations \[N!\approx N^N \left(2\pi N\right)^{1/2} \mathrm{exp}\left(-N\right)\mathrm{exp}\left(\frac{1}{12N}\right)\approx N^N \left(2\pi N\right)^{1/2} \mathrm{exp}\left(-N\right)\approx N^N\mathrm{exp}\left(-N\right) \nonumber \] How do the resulting approximations for \(d ~ \ln N! /dN\) compare to one another as \(N\) becomes very large? 16. There are three energy levels available to any one molecule in a crystal of the substance. Consider a crystal containing \(1000\) molecules. These molecules are distinguishable because each occupies a unique site in the crystalline lattice. How many combinations (microstates) are associated with the population set \(N_1=800\), \(N_2=150\), \(N_3=50\)? |
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_333_-_Organic_Chemistry_III_(Lund)/15%3A_Oxidation_and_Reduction_Reactions/15.0%3A_Prelude_to_Oxidation_and_Reduction_Reactions | Introduction Theo Ross was not doing very well at his summer job, and he was frustrated. His boss had given him specific instructions, and yet Theo kept botching the job, over and over again. Theo was not used to failure – he had achieved almost perfect scores on both the ACT and SAT college entrance exams, and was headed to Stanford University in the fall. Why couldn't he get it right? It wasn't brain surgery, after all. Well, actually – it was brain surgery. An April 17, 2014 article in Sports Illustrated tells Theo's story. Wanting to do something interesting over the summer of 2010 before he started college, Theo had applied for a research internship at the National Institutes of Health in Bethesda, Maryland. This was an extremely competitive program normally reserved for outstanding college students, but somehow Theo had managed to win a coveted spot in the program, working with Dr. Dorian McGavern, a neurologist studying how meningitis effects the brain. Dr. McGavern assigned Theo the task of performing 'skull-thinning' surgery on mice, part of which involved using a special saw to shave down a small section of the bone in order to gain access to the brain. It was a delicate procedure, something that even some experienced neurosurgeons who had tried it had found challenging. Any small slip resulted in a concussion to the mouse's brain, rendering it useless for the study. Theo just couldn't get the hang of it, and ended up concussing one mouse after another. You have probably heard the old expression: “when life gives you lemons, make lemonade”. Theo made a lot of lemonade that summer. Theo and Dr. McGavern eventually realized that his failure at the procedure actually presented an opportunity to observe what happens to a brain right after a concussive injury. Theo started doing more skull-thinning surgeries, but now the goal was to cause concussion, rather than to avoid it. The concussed mice (who had been anaesthetized prior to the surgery) were immediately strapped under a microscope so that Theo could observe how their brains responded to the injury. This was new, and very exciting stuff: most of what neurologists knew about concussions up to that point had come from MRI (magnetic resonance imaging – see chapter 5) or autopsies. Nobody knew very much about what happens at the cellular level in a brain in the minutes and hours after a concussion has occurred. In addition, the problem of traumatic head injuries and the long-lasting effects they cause was becoming an increasingly hot topic in the news, critically relevant to thousands of veterans returning from Iraq and Afghanistan as well as to football players and other athletes in contact sports– including Theo, who had been a competitive wrestler in high school. (You might have been wondering why Theo's story appeared in Sports Illustrated – now you know.) Theo spent the rest of that summer, and every spring break and summer vacation over the next few years, working in McGavern's lab on the new project. He and McGavern found evidence that the 'hidden' damage to a concussed brain – that which went undetected in MRI scans but could come back to haunt the victim years later in the form of recurring headaches, memory loss, and depression – may be caused by a type of molecule referred to as 'reactive oxygen species', or ROS, leaking from damaged tissues into the brain. ROS are potentially harmful byproducts of respiration such as hydrogen peroxide (\(H_2O_2\)) that are constantly being produced in our cells. Although ROS can cause serious oxidative damage if they are allowed to build up, our bodies have evolved ways to deal with them, using so-called 'ROS scavengers' to convert them to something innocuous like water. With this new understanding, Theo and his mentor had another idea: what if they could prevent the ROS from causing further damage to a recently concussed brain by applying an scavenger to the injury? After some trial and error, they found that an ROS scavenger compound called glutathione, when applied directly to the skull of a concussed mouse within a few minutes to three hours after the injury, could permeate the bone and react with the ROS. Brain cells from these glutathione-treated mice appeared normal, with none of the signs of ROS damage Theo was used to seeing. The road from an initial scientific discovery to a safe and effective medical treatment is often a very long one, but Theo Roth and Dorian McGavern appear to have made a discovery that could eventually help prevent some of the most devastating and long-term damage caused by traumatic head injuries. In the end, it's a very good thing that Theo's hands were not cut out for brain surgery. The chemistry of oxidation and reduction - often called 'redox' chemistry - is central to Theo Roth's discovery about what happens to a concussed brain at the molecular level. This chapter is dedicated to redox chemistry. We'll begin with a reminder of what you learned in General Chemistry about the fundamentals of redox reactions in the context of inorganic elements such as iron, copper and zinc: reduction is a gain of electrons, and oxidation is a loss of electrons. Then, we'll expand our understanding to include bioorganic redox reactivity, examining among other things how alcohols are converted to ketones and aldehydes, aldehydes are converted to carboxylic acids, and amines are converted to imines. We will also talk about redox reactions in the broader context of metabolism in living things. A central player in some of the biochemical redox reactions we will see is the coenzyme called glutathione, Theo Roth's 'magic bullet' molecule that was able to rescue mouse brain cells from death by oxidation. We'll see how glutathione acts as a mediator in the formation and cleavage of disulfide bonds in proteins, and how it acts as an 'ROS scavenger' to turn hydrogen peroxide into water. |
Courses/University_of_Arkansas_Little_Rock/Chem_1402%3A_General_Chemistry_1_(Belford)/Text/5%3A_Energy_and_Chemical_Reactions/5.6%3A_Calorimetry | Learning Objectives Apply the First Law of Thermodynamics to calorimetry Compare heat flow from hot to cold objects in an ideal calorimeter versus a real calorimeter Calculate heat, temperature change, and specific heat after thermal equilibrium is reached between two substances in a calorimeter Calculate the molar heat of enthalpy for a reactions using coffee cup calorimetry Compare and contrast coffee cup calorimetry and bomb calorimetry Calorimetry Calorimetery is an application of the First Law of Thermodynamics to heat transfer, and allows us to measure the enthalpies of reaction or the heat capacities of substances. From the first law we can state \[\Delta E_{Universe} = ΔE_{System} + ΔE_{Surrounding} = 0\] therefore, \[ ΔE_{System} = - ΔE_{Surrounding}\] A calorimeter is an instrument that has a thermally isolated compartment that is designed to prevent heat flow to the surroundings. Technically, this is called an adiabatic surface in that no heat flows in and out of the calorimeter, which is in contrast to isothermal, which means constant temperature. In truth, there is no such thing as a thermally isolated system, but we can design systems that approach it on the time scale of a measurement. We will consider two types of calorimeters, ideal calorimeters, where the calorimeter does not absorb heat itself, and real calorimeters, where the calorimeter absorbs heat. We will apply calorimetry to two different thermodynamic processes, that of transference of heat across objects at different temperatures, and that released or absorbed by a chemical reaction. We will look at two ways of measuring this heat, either the temperature change of the calorimeter, or the mass of a substance that undergoes a phase change within the calorimeter at a constant temperature. If we consider the work in a calorimeter to be zero, and noting that Δ E = Δ Q + W, the above equation becomes \[\Delta Q_{system} = -\Delta Q_{Surrounding}\] If we can measure the heat absorbed or lost by the surroundings we can use the above equivalence statement to calculate the heat lost or absorbed the system, and this is the basis for calorimetry. Figure \(\PageIndex{1}\) shows the basic layout of an ideal calorimeter where the gray shaded box acts as a surface that prevent any heat entering or leaving the solution, and so the solution functions as the surroundings. In the first exothermic process heat is released by the system and the temperature rises, while in the second endothermic process heat is absorbed by the system and the temperature decreases. Heat Transfer Between Hot and Cold Objects in an Ideal Calorimeter When a hot object (T H ) is placed in thermal contact with a cold object (T C ) energy is transferred between the two, but the rate of heat transfer from the hot to the cold is greater, so it cools down while the cold object heats up, until they reach the same temperature (T F ). Once they are at the same temperature the system is at thermal equilibrium and the rate of transfer of energy from the hot to cold is the same as cold to hot. That is, energy is a vectoral flow, and even at the same temperature, there is a flow of energy across the boundary of two objects, but the flow is the same. Starting with the first law for two isolated objects and considering no work, we have \[q_{cold} + q_{hot}=0\] so \[q_{cold} =- q_{hot}=0\] The amount of heat lost by a warmer object equals the amount of heat gained by a cooler object. From section 5.2 we know that for a substance in a constant phase the heat relates to the temperature by equation 5.2.3 (q=mc\(\Delta|)T, which substituting into the above two equations give \[ m_Cc_C \Delta T _C + m_Hc_H \Delta T_H=0 \] or \[ m_Cc_C \Delta T _C =- m_Hc_H \Delta T_H \] where \[\Delta T_H= T_F-T_H \; \; and \; \; \Delta T_C= T_F-T_C\] In an ideal calorimeter we consider the calorimeter to absorb no heat itself, so if we have material of a uniform composition, we can relate the heat transferred from the hot object to the cold object by eq. 5.6.7. Example \(\PageIndex{1}\) Ideal calorimeter problem What is T F when 2.00g Au at 100.0°C is placed into 25.0 ml H 2 O at 30.0°C? You can obtain the specific heat capacities for gold and water from table 5.2.1 . Solution You are solving 5.6.7 for \(T_F\). \[T_F=\frac{m_Hc_HT_H+m_cc_cT_c}{m_Hc_H+m_cc_c} = 30.17^oC\] The solution to this problem is worked out in video \(\PageIndex{1}\) Heat Transfer Between Hot and Cold Objects in a Real Calorimeter In a real calorimeter the calorimeter itself will absorb some of the heat, even if it transfers no heat to the surroundings. That is, the calorimeter itself has a heat capacitance. This usually needs to be measured for each calorimeter, and following the conventions of section 4.2, we use a capital C to represent the capacitance of an object, or the calorimeter constant. That is, the calorimeter constant is the calorimeter's heat capacitance. Consider dropping a hot object into cold water that has been sitting in a calorimeter long enough to have the same temperature as the calorimeter. From eq. 5.6 we know that Q Cold = -Q Hot , But now there are two cold objects, the water and the calorimeter, so \[Q_c=m_cc_c \Delta T_c + C_{cal} \Delta T_c\] where C cal is the heat capacitance of the calorimeter. Exercise \(\PageIndex{1}\) Example \(\PageIndex{2}\) (below) is the same problem as Example \(\PageIndex{1}\) (above) but it is in a real calorimeter. Do you think T F in a real calorimeter will be hotter or colder than in the ideal? Answer It will be colder because in example \(\PageIndex{1}\) the 2 g Au was transferring its heat to 25 mL water, while in example \(\PageIndex{2}\) it is transferring the heat to 25 mL of water AND the calorimeter itself. is the same problem as Example \(\PageIndex{1}\) but it is in a real calorimeter. Before solving this or looking at the answer, Example \(\PageIndex{2}\) Real calorimeter problem What is T F when 2.00g Au at 100.0°C is placed into 25.0 ml H 2 O at 30.0°C? The calorimeter constant for the calorimeter is 36.0 J/°C and you can obtain the specific heat capacities for gold and water from table 5.2.1 . Solution You are solving eq. 5.6.10 for \(T_F\), but there are two entities absorbing the heat \[ Q_c +Q_H=0 \\ (m_cc_c \Delta T _c + C_{cal} \Delta T _C) + m_Hc_H \Delta T_H=0 \] \[ T_F=\frac{m_Hc_HT_H+\left ( m_cc_c +C_{cal}\right )T_c}{m_Hc_H+C_{cal} + m_cc_c} = 30.12^oC \] The solution to this problem is worked out in video \(\PageIndex{2}\) below. Exercise \(\PageIndex{2}\) Calculate the calorimeter constant if 150.0 g of lead at 100.0°C were placed in a calorimeter with 50.0 g of water at 22.0°C and the resulting temperature of the mixture was 26.2 °C , you can obtain the specific heat capacities for lead and water from table 5.2.1 . Answer As in example \(\PageIndex{2}\), we start with the first law (for a system that does no work). \[ Q_c +Q_H=0 \\ (m_cc_c \Delta T _c + C_{cal} \Delta T _C) + m_Hc_H \Delta T_H=0 \] solve for C cal . \[c_{cal}=\frac{m_cc_c \Delta T _c + m_Hc_H \Delta T_H}{\Delta T_c} \\ =\frac{50.0g\left ( 4.184\frac{J}{g^oC} \right )\left ( 4.2^oC \right )+ 150.0g\left (0.128 \frac{J}{g^oC} \right )\left ( -73.8^oC \right )}{4.2^oC}=550J\] Calorimetry and Enthalpies of Reaction Calorimeters can also measure the molar enthalpy of reaction, by simply equating the heat absorbed/lost by the calorimeter to the heat lost/absorbed by the reaction. That is, the system is the reaction. To come up with the heat of reaction, you need to know what the limiting reagent is, and calculate it from that. Starting with the first law and considering there to be no work \[Q_{system}=-Q_{surrounding} \\\text{which becomes} \\ Q_{reaction}=-Q_{calorimeter}\] Q calorimeter represents the calorimeter and everything in it. That is, all the heat exchanged with the reaction is absorbed or released by the calorimeter and its contents. Q reaction is the enthalpy of reaction (\(\Delta H_{r}\)) for a system at constant pressure. If we know the moles of the limiting reagent, we can divide by that and come up with the molar enthalpy of reaction The following example shows how to calculate the molar enthalpy of reaction for the reaction between sulfuric acid and sodium hydroxide. Example \(\PageIndex{3}\) Calculate \(\Delta H_r\) for the neutralization of sodium hydroxide by sulfuric acid if 100.0 mL of 1.00M H 2 SO 4 reacts with 100.0mL of 1.00M NaOH if all reactants were at 25.0 o C before reaction and reached a final temperature of 32.0 o C after the reaction. Treat the solutions as if they had the same specific heat capacity and density of pure water, Solution Step 1: Identify limiting reagent so we can identify moles reacted. \[\underset{V=100.0mL\\M=1.00M}{H_2SO_4}+ \underset{100.0mL \\M=1.00M}{2NaOH} \rightarrow Na_2SO_4 + 2H_2O\] Sodium hydroxide (NaOH) is the limiting reagents (calculate moles of each reactant and divide by stoichiometric coefficient, review section 4.2 in necessary) \[n_{NaOH}=1.00\frac{molNaOH}{L}\left ( 0.1000L \right )=0.100mol \;NaOH\] Step 2: Calculate heat of reaction based on the first law of thermodynamics We note the heat transferred from the reaction is abosorbed by the calorimeter and its contents, noting that the calorimeter itself is ideal and so its calorimeter constant is zero (C cal =0). \[\begin{align}Q_R & =-Q_{calorimeter} \nonumber \\ & = -m_{contents}c_{H_2O}\Delta T \nonumber \\ & = -200.0g\left ( 4.184\frac{J}{g^oC} \right )\left ( 32.0^oC-25.0^oC \right ) \nonumber \\ &= -5.9kJ \end{align}\] Step 3: Relate this to the moles of the limiting reagent You know know that when 0.1000 M of NaOH completely reacted that -59.kJ of heat was released, so use the stoichiometric equation to calculate the heat of reaction. \[\frac{\Delta H_{r}}{mol \; NaOH}= \frac{-5.9kJ}{0.100mol} = -59kJ/mol \; NaOH\] Step 4: Relate this to the balanced equation to get the enthalpy of reaction Remember, the enthalpy of reaction is the energy related to the balanced equation, not each reactant or product. So for the neutralization fo sulfuric acid by sodium hydroxide it is per mole sulfuric acid consumed, per 2 moles sodium hydroxide consumed, per mole sodium sulfate produced and per two moles sodium chloride produced. So \[\Delta H_{reaction} = -59 \frac{kJ}{mol \;NaOH}(2 \; mol \; NaOH)= -120kJ\] Lets do a quick review of what the enthalpy of reaction means. So for the reaction H 2 SO 4 (aq) + 2NaOH(aq) --> Na 2 SO 4 (aq) + 2 NaCl it means is that for each mole of H 2 SO 4 consumed or Na 2 SO 4 produced- 120kJ of energy is released, while for each mole of NaOH consumed or NaCl produced -59kJ of energy is released when sodium hydroxide reacts with sulfuric acid. Constant-Volume Calorimetry: Combustion Data Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeter (A device used to measure energy changes in chemical processes. shown schematically in Figure \(\PageIndex{3}\)). The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated. Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. The heat released by a reaction carried out at constant volume is identical to the change in internal energy (\(ΔU\)) rather than the enthalpy change (\(ΔH\)); \(ΔU\) is related to \(ΔH\) by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that \(ΔU < ΔH\), the relationship between the measured temperature change and \(ΔH_{comb}\) is given in Equation \(\ref{5.5.9}\), where C bomb is the total heat capacity of the steel bomb and the water surrounding it: \[ \Delta H_{comb} < q_{comb} = q_{calorimater} = C_{bomb} \Delta T \label{5.5.9}\] To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (\(\ce{C6H5CO2H}\)) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its \(ΔH_{comb} = −26.38\, kJ/g\)). This value and the measured increase in temperature of the calorimeter can be used in Equation \(\ref{5.5.9}\) to determine \(C_{bomb}\). The use of a bomb calorimeter to measure the \(ΔH_{comb}\) of a substance is illustrated in Example \(\PageIndex{4}\). Example \(\PageIndex{4}\): Combustion of Glucose The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08 °C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the \(ΔH_{comb}\) of glucose? Given: mass and \(ΔT\) for combustion of standard and sample Asked for: \(ΔH_{comb}\) of glucose Strategy: Calculate the value of \(q_{rxn}\) for benzoic acid by multiplying the mass of benzoic acid by its \(ΔH_{comb}\). Then use Equation 5.6.21 to determine the heat capacity of the calorimeter (\(C_{bomb}\)) from \(q_{comb}\) and \(ΔT\). Calculate the amount of heat released during the combustion of glucose by multiplying the heat capacity of the bomb by the temperature change. Determine the Δ H comb of glucose by multiplying the amount of heat released per gram by the molar mass of glucose. Solution: The first step is to use Equation \(\ref{5.5.9}\) and the information obtained from the combustion of benzoic acid to calculate C bomb . We are given \(ΔT\), and we can calculate q comb from the mass of benzoic acid: \[ \begin{align*} q_{comb} &= \left ( 0.579 \; \cancel{g} \right )\left ( -26.38 \; kJ/\cancel{g} \right ) \\[4pt] &= - 15.3 \; kJ \end{align*} \] From Equation \(\ref{5.5.9}\), \[ \begin{align*} -C_{bomb} &= \dfrac{q_{comb}}{\Delta T} \\[4pt] &= \dfrac{-15.3 \; kJ}{2.08 \; ^{o}C} \\[4pt] &=- 7.34 \; kJ/^{o}C \end{align*} \] B According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose: \[ \begin{align*} q_{comb} &=-C_{bomb}\Delta T \\[4pt] &= \left ( -7.34 \; kJ/^{o}C \right )\left ( 3.64 \; ^{o}C \right ) \\[4pt] &=- 26.7 \; kJ \end{align*} \] Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the Δ H comb of glucose is \[ \begin{align*} \Delta H_{comb} &=\left ( \dfrac{-26.7 \; kJ}{1.732 \; \cancel{g}} \right )\left ( \dfrac{180.16 \; \cancel{g}}{mol} \right ) \\[4pt] &= -2780 \; kJ/mol \\[4pt] &=2.78 \times 10^{3} \; J/mol \end{align*} \] This result is in good agreement (< 1% error) with the value of \(ΔH_{comb} = −2803\, kJ/mol\) that calculated using enthalpies of formation. Exercise \(\PageIndex{4}\): Combustion of Benzoic Acid When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75 °C is observed. When 1.932 g of methylhydrazine (CH 3 NHNH 2 ) is ignited in the same calorimeter, the temperature increase is 4.64 °C. Calculate the Δ H comb of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle. Answer −1.30 × 10 3 kJ/mol |
Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/16%3A_AcidBase_Equilibria/16.08%3A_Relationship_Between_Ka_and_Kb | Learning Objectives To know the relationship between acid or base strength and the magnitude of \(K_a\), \(K_b\), \(pK_a\), and \(pK_b\). The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For example, the general equation for the ionization of a weak acid in water, where HA is the parent acid and A− is its conjugate base, is as follows: \[HA_{(aq)}+H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)}+A^−_{(aq)} \label{16.5.1}\] The equilibrium constant for this dissociation is as follows: \[K=\dfrac{[H_3O^+][A^−]}{[HA]} \label{16.5.2}\] The equilibrium constant for this reaction is the acid ionization constant \(K_a\), also called the acid dissociation constant: \[K_a=\dfrac{[H_3O^+][A^−]}{[HA]} \label{16.5.3}\] Thus the numerical values of K and \(K_a\) differ by the concentration of water (55.3 M). Again, for simplicity, \(H_3O^+\) can be written as \(H^+\) in Equation \(\ref{16.5.3}\). Keep in mind, though, that free \(H^+\) does not exist in aqueous solutions and that a proton is transferred to \(H_2O\) in all acid ionization reactions to form \(H^3O^+\). The larger the \(K_a\), the stronger the acid and the higher the \(H^+\) concentration at equilibrium. Like all equilibrium constants, acid–base ionization constants are actually measured in terms of the activities of \(H^+\) or \(OH^−\), thus making them unitless . The values of \(K_a\) for a number of common acids are given in Table \(\PageIndex{1}\). Acid \(HA\) \(K_a\) \(pK_a\) \(A^−\) \(K_b\) \(pK_b\) hydroiodic acid \(HI\) \(2 \times 10^{9}\) −9.3 \(I^−\) \(5.5 \times 10^{−24}\) 23.26 sulfuric acid (1)* \(H_2SO_4\) \(1 \times 10^{2}\) −2.0 \(HSO_4^−\) \(1 \times 10^{−16}\) 16.0 nitric acid \(HNO_3\) \(2.3 \times 10^{1}\) −1.37 \(NO_3^−\) \(4.3 \times 10^{−16}\) 15.37 hydronium ion \(H_3O^+\) \(1.0\) 0.00 \(H_2O\) \(1.0 \times 10^{−14}\) 14.00 sulfuric acid (2)* \(HSO_4^−\) \(1.0 \times 10^{−2}\) 1.99 \(SO_4^{2−}\) \(9.8 \times 10^{−13}\) 12.01 hydrofluoric acid \(HF\) \(6.3 \times 10^{−4}\) 3.20 \(F^−\) \(1.6 \times 10^{−11}\) 10.80 nitrous acid \(HNO_2\) \(5.6 \times 10^{−4}\) 3.25 \(NO2^−\) \(1.8 \times 10^{−11}\) 10.75 formic acid \(HCO_2H\) \(1.78 \times 10^{−4}\) 3.750 \(HCO_2−\) \(5.6 \times 10^{−11}\) 10.25 benzoic acid \(C_6H_5CO_2H\) \(6.3 \times 10^{−5}\) 4.20 \(C_6H_5CO_2^−\) \(1.6 \times 10^{−10}\) 9.80 acetic acid \(CH_3CO_2H\) \(1.7 \times 10^{−5}\) 4.76 \(CH_3CO_2^−\) \(5.8 \times 10^{−10}\) 9.24 pyridinium ion \(C_5H_5NH^+\) \(5.9 \times 10^{−6}\) 5.23 \(C_5H_5N\) \(1.7 \times 10^{−9}\) 8.77 hypochlorous acid \(HOCl\) \(4.0 \times 10^{−8}\) 7.40 \(OCl^−\) \(2.5 \times 10^{−7}\) 6.60 hydrocyanic acid \(HCN\) \(6.2 \times 10^{−10}\) 9.21 \(CN^−\) \(1.6 \times 10^{−5}\) 4.79 ammonium ion \(NH_4^+\) \(5.6 \times 10^{−10}\) 9.25 \(NH_3\) \(1.8 \times 10^{−5}\) 4.75 water \(H_2O\) \(1.0 \times 10^{−14}\) 14.00 \(OH^−\) \(1.00\) 0.00 acetylene \(C_2H_2\) \(1 \times 10^{−26}\) 26.0 \(HC_2^−\) \(1 \times 10^{12}\) −12.0 ammonia \(NH_3\) \(1 \times 10^{−35}\) 35.0 \(NH_2^−\) \(1 \times 10^{21}\) −21.0 *The number in parentheses indicates the ionization step referred to for a polyprotic acid. *The number in parentheses indicates the ionization step referred to for a polyprotic acid. *The number in parentheses indicates the ionization step referred to for a polyprotic acid. *The number in parentheses indicates the ionization step referred to for a polyprotic acid. *The number in parentheses indicates the ionization step referred to for a polyprotic acid. *The number in parentheses indicates the ionization step referred to for a polyprotic acid. *The number in parentheses indicates the ionization step referred to for a polyprotic acid. Weak bases react with water to produce the hydroxide ion, as shown in the following general equation, where B is the parent base and BH+ is its conjugate acid: \[B_{(aq)}+H_2O_{(l)} \rightleftharpoons BH^+_{(aq)}+OH^−_{(aq)} \label{16.5.4}\] The equilibrium constant for this reaction is the base ionization constant (K b ), also called the base dissociation constant: \[K_b=\dfrac{[BH^+][OH^−]}{[B]} \label{16.5.5}\] Once again, the concentration does not appear in the equilibrium constant expression.. The larger the \(K_b\), the stronger the base and the higher the \(OH^−\) concentration at equilibrium. The values of \(K_b\) for a number of common weak bases are given in Table \(\PageIndex{2}\). Base \(B\) \(K_b\) \(pK_b\) \(BH^+\) \(K_a\) \(pK_a\) hydroxide ion \(OH^−\) \(1.0\) 0.00* \(H_2O\) \(1.0 \times 10^{−14}\) 14.00 phosphate ion \(PO_4^{3−}\) \(2.1 \times 10^{−2}\) 1.68 \(HPO_4^{2−}\) \(4.8 \times 10^{−13}\) 12.32 dimethylamine \((CH_3)_2NH\) \(5.4 \times 10^{−4}\) 3.27 \((CH_3)_2NH_2^+\) \(1.9 \times 10^{−11}\) 10.73 methylamine \(CH_3NH_2\) \(4.6 \times 10^{−4}\) 3.34 \(CH_3NH_3^+\) \(2.2 \times 10^{−11}\) 10.66 trimethylamine \((CH_3)_3N\) \(6.3 \times 10^{−5}\) 4.20 \((CH_3)_3NH^+\) \(1.6 \times 10^{−10}\) 9.80 ammonia \(NH_3\) \(1.8 \times 10^{−5}\) 4.75 \(NH_4^+\) \(5.6 \times 10^{−10}\) 9.25 pyridine \(C_5H_5N\) \(1.7 \times 10^{−9}\) 8.77 \(C_5H_5NH^+\) \(5.9 \times 10^{−6}\) 5.23 aniline \(C_6H_5NH_2\) \(7.4 \times 10^{−10}\) 9.13 \(C_6H_5NH_3^+\) \(1.3 \times 10^{−5}\) 4.87 water \(H_2O\) \(1.0 \times 10^{−14}\) 14.00 \(H_3O^+\) \(1.0^*\) 0.00 *As in Table \(\PageIndex{1}\). *As in Table \(\PageIndex{1}\). *As in Table \(\PageIndex{1}\). *As in Table \(\PageIndex{1}\). *As in Table \(\PageIndex{1}\). *As in Table \(\PageIndex{1}\). *As in Table \(\PageIndex{1}\). There is a simple relationship between the magnitude of \(K_a\) for an acid and \(K_b\) for its conjugate base. Consider, for example, the ionization of hydrocyanic acid (\(HCN\)) in water to produce an acidic solution, and the reaction of \(CN^−\) with water to produce a basic solution: \[HCN_{(aq)} \rightleftharpoons H^+_{(aq)}+CN^−_{(aq)} \label{16.5.6}\] \[CN^−_{(aq)}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+HCN_{(aq)} \label{16.5.7}\] The equilibrium constant expression for the ionization of HCN is as follows: \[K_a=\dfrac{[H^+][CN^−]}{[HCN]} \label{16.5.8}\] The corresponding expression for the reaction of cyanide with water is as follows: \[K_b=\dfrac{[OH^−][HCN]}{[CN^−]} \label{16.5.9}\] If we add Equations \(\ref{16.5.6}\) and \(\ref{16.5.7}\), we obtain the following (recall that the equilibrium constant for the sum of two reactions is the product of the equilibrium constants for the individual reactions): \[\cancel{HCN_{(aq)}} \rightleftharpoons H^+_{(aq)}+\cancel{CN^−_{(aq)}} \;\;\; K_a=[H^+]\cancel{[CN^−]}/\cancel{[HCN]}\] \[\cancel{CN^−_{(aq)}}+H_2O_{(l)} \rightleftharpoons OH^−_{(aq)}+\cancel{HCN_{(aq)}} \;\;\; K_b=[OH^−]\cancel{[HCN]}/\cancel{[CN^−]}\] \[H_2O_{(l)} \rightleftharpoons H^+_{(aq)}+OH^−_{(aq)} \;\;\; K=K_a \times K_b=[H^+][OH^−]\] In this case, the sum of the reactions described by \(K_a\) and \(K_b\) is the equation for the autoionization of water, and the product of the two equilibrium constants is \(K_w\): \[K_aK_b = K_w \label{16.5.10}\] Thus if we know either \(K_a\) for an acid or \(K_b\) for its conjugate base, we can calculate the other equilibrium constant for any conjugate acid–base pair. Just as with \(pH\), \(pOH\), and pKw, we can use negative logarithms to avoid exponential notation in writing acid and base ionization constants, by defining \(pK_a\) as follows: \[pKa = −\log_{10}K_a \label{16.5.11}\] \[K_a=10^{−pK_a} \label{16.5.12}\] and \(pK_b\) as \[pK_b = −\log_{10}K_b \label{16.5.13}\] \[K_b=10^{−pK_b} \label{16.5.14}\] Similarly, Equation 16.5.10, which expresses the relationship between \(K_a\) and \(K_b\), can be written in logarithmic form as follows: \[pK_a + pK_b = pK_w \label{16.5.15}\] At 25°C, this becomes \[pK_a + pK_b = 14.00 \label{16.5.16}\] The values of \(pK_a\) and \(pK_b\) are given for several common acids and bases in Table 16.5.1 and Table 16.5.2, respectively, and a more extensive set of data is provided in Tables E1 and E2. Because of the use of negative logarithms, smaller values of \(pK_a\) correspond to larger acid ionization constants and hence stronger acids. For example, nitrous acid (\(HNO_2\)), with a \(pK_a\) of 3.25, is about a 1000 times stronger acid than hydrocyanic acid (HCN), with a \(pK_a\) of 9.21. Conversely, smaller values of \(pK_b\) correspond to larger base ionization constants and hence stronger bases. The relative strengths of some common acids and their conjugate bases are shown graphically in Figure 16.5. The conjugate acid–base pairs are listed in order (from top to bottom) of increasing acid strength, which corresponds to decreasing values of \(pK_a\). This order corresponds to decreasing strength of the conjugate base or increasing values of \(pK_b\). At the bottom left of Figure 16.5.2 are the common strong acids; at the top right are the most common strong bases. Notice the inverse relationship between the strength of the parent acid and the strength of the conjugate base. Thus the conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base. The conjugate base of a strong acid is a weak base and vice versa. We can use the relative strengths of acids and bases to predict the direction of an acid–base reaction by following a single rule: an acid–base equilibrium always favors the side with the weaker acid and base, as indicated by these arrows: \[\text{stronger acid + stronger base} \ce{ <=>>} \text{weaker acid + weaker base} \] In an acid–base reaction, the proton always reacts with the stronger base. For example, hydrochloric acid is a strong acid that ionizes essentially completely in dilute aqueous solution to produce \(H_3O^+\) and \(Cl^−\); only negligible amounts of \(HCl\) molecules remain undissociated. Hence the ionization equilibrium lies virtually all the way to the right, as represented by a single arrow: \[HCl_{(aq)} + H_2O_{(l)} \rightarrow \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)} \label{16.5.17}\] In contrast, acetic acid is a weak acid, and water is a weak base. Consequently, aqueous solutions of acetic acid contain mostly acetic acid molecules in equilibrium with a small concentration of \(H_3O^+\) and acetate ions, and the ionization equilibrium lies far to the left, as represented by these arrows: \[ \ce{ CH_3CO_2H_{(aq)} + H_2O_{(l)} <<=> H_3O^+_{(aq)} + CH_3CO_{2(aq)}^- }\] Similarly, in the reaction of ammonia with water, the hydroxide ion is a strong base, and ammonia is a weak base, whereas the ammonium ion is a stronger acid than water. Hence this equilibrium also lies to the left: \[H_2O_{(l)} + NH_{3(aq)} \ce{ <<=>} NH^+_{4(aq)} + OH^-_{(aq)}\] All acid–base equilibria favor the side with the weaker acid and base. Thus the proton is bound to the stronger base. Example \(\PageIndex{1}\): Butyrate and Dimethylammonium Ions Calculate \(K_b\) and \(pK_b\) of the butyrate ion (\(CH_3CH_2CH_2CO_2^−\)). The \(pK_a\) of butyric acid at 25°C is 4.83. Butyric acid is responsible for the foul smell of rancid butter. Calculate \(K_a\) and \(pK_a\) of the dimethylammonium ion (\((CH_3)_2NH_2^+\)). The base ionization constant \(K_b\) of dimethylamine (\((CH_3)_2NH\)) is \(5.4 \times 10^{−4}\) at 25°C. Given : \(pK_a\) and \(K_b\) Asked for : corresponding \(K_b\) and \(pK_b\), \(K_a\) and \(pK_a\) Strategy : The constants \(K_a\) and \(K_b\) are related as shown in Equation 16.5.10. The \(pK_a\) and \(pK_b\) for an acid and its conjugate base are related as shown in Equation 16.5.15 and Equation 16.5.16. Use the relationships pK = −log K and K = 10−pK (Equation 16.5.11 and Equation 16.5.13) to convert between \(K_a\) and \(pK_a\) or \(K_b\) and \(pK_b\). Solution : We are given the \(pK_a\) for butyric acid and asked to calculate the \(K_b\) and the \(pK_b\) for its conjugate base, the butyrate ion. Because the \(pK_a\) value cited is for a temperature of 25°C, we can use Equation 16.5.16: \(pK_a\) + \(pK_b\) = pKw = 14.00. Substituting the \(pK_a\) and solving for the \(pK_b\), \[4.83+pK_b=14.00\] \[pK_b=14.00−4.83=9.17\] Because \(pK_b = −\log K_b\), \(K_b\) is \(10^{−9.17} = 6.8 \times 10^{−10}\). In this case, we are given \(K_b\) for a base (dimethylamine) and asked to calculate \(K_a\) and \(pK_a\) for its conjugate acid, the dimethylammonium ion. Because the initial quantity given is \(K_b\) rather than \(pK_b\), we can use Equation 16.5.10: \(K_aK_b = K_w\). Substituting the values of \(K_b\) and \(K_w\) at 25°C and solving for \(K_a\), \[K_a(5.4 \times 10^{−4})=1.01 \times 10^{−14}\] \[K_a=1.9 \times 10^{−11}\] Because \(pK_a\) = −log \(K_a\), we have \(pK_a = −\log(1.9 \times 10^{−11}) = 10.72\). We could also have converted \(K_b\) to \(pK_b\) to obtain the same answer: \[pK_b=−\log(5.4 \times 10^{−4})=3.27\] \[pKa+pK_b=14.00\] \[pK_a=10.73\] \[K_a=10^{−pK_a}=10^{−10.73}=1.9 \times 10^{−11}\] If we are given any one of these four quantities for an acid or a base (\(K_a\), \(pK_a\), \(K_b\), or \(pK_b\)), we can calculate the other three. Exercise \(\PageIndex{1}\): Lactic Acid Lactic acid (\(CH_3CH(OH)CO_2H\)) is responsible for the pungent taste and smell of sour milk; it is also thought to produce soreness in fatigued muscles. Its \(pK_a\) is 3.86 at 25°C. Calculate \(K_a\) for lactic acid and \(pK_b\) and \(K_b\) for the lactate ion. Answer \(K_a = 1.4 \times 10^{−4}\) for lactic acid; \(pK_b\) = 10.14 and \(K_b = 7.2 \times 10^{−11}\) for the lactate ion Summary Two species that differ by only a proton constitute a conjugate acid–base pair. The magnitude of the equilibrium constant for an ionization reaction can be used to determine the relative strengths of acids and bases. For an aqueous solution of a weak acid, the dissociation constant is called the acid ionization constant (Ka). Similarly, the equilibrium constant for the reaction of a weak base with water is the base ionization constant (Kb). For any conjugate acid–base pair, \(K_aK_b = K_w\). Smaller values of \(pK_a\) correspond to larger acid ionization constants and hence stronger acids. Conversely, smaller values of \(pK_b\) correspond to larger base ionization constants and hence stronger bases. At 25°C, \(pK_a + pK_b = 14.00\). Acid–base reactions always proceed in the direction that produces the weaker acid–base pair. Key Takeaways The Ka and Kb values for a conjugated acid–base pairs are related through the K w value: \[K_aK_b = K_w \] The conjugate base of a strong acid is a very weak base, and the conjugate base of a very weak acid is a strong base. Key Equations Acid ionization constant: \[K_a=\dfrac{[H_3O^+][A^−]}{[HA]}\] Base ionization constant: \[K_b=\dfrac{[BH^+][OH^−]}{[B]} \] Relationship between \(K_a\) and \(K_b\) of a conjugate acid–base pair: \[K_aK_b = K_w \] Definition of \(pK_a\): \[pKa = −\log_{10}K_a \nonumber\] \[K_a=10^{−pK_a}\] Definition of \(pK_b\): \[pK_b = −\log_{10}K_b \nonumber\] \[K_b=10^{−pK_b} \] Relationship between \(pK_a\) and \(pK_b\) of a conjugate acid–base pair: \[pK_a + pK_b = pK_w \] \[pK_a + pK_b = 14.00 \; \text{at 25°C} \] |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Chemistry_with_Applications_in_Spectroscopy_(Fleming)/04%3A_The_Harmonic_Oscillator_and_Vibrational_Spectroscopy/4.06%3A_References | Fleming, P. E., & Mathews, C. W. (1996). A Reanalysis of the A\(^1\Pi\) –X 1 \(^1\Sigma\) +Transition of AlBr. Journal of Molecular Spectroscopy, 175(1), 31-36. doi:10.1006/jmsp.1996.0005 Gaydon, A. G. (1946). The determination of dissociation energies by the birge-sponer extrapolation. Proceedings of the Physical Society, 58(5), 525-537. Meyer, C. F., & Levin, A. A. (1929). Physical Review A, 34, 44. Morse, P. M. (1929). Diatomic Molecules According to the Wave Mechanics. II. Vibrational Levels. Physical Review, 34(1), 57-64. doi:10.1103/PhysRev.34.57 |
Courses/Lumen_Learning/Book%3A_Writing_About_Texts_(Lumen)/08%3A_Module_6%3A_Incorporating_Materials_and_Drafting_the_Essay/08.1%3A_Introduction | Module Six: Drafting the Research Essay Module Introduction We’ve talked so much about the reader, but you can’t forget that the opening line is important to the writer, too. To the person who’s actually boots-on-the-ground. Because it’s not just the reader’s way in, it’s the writer’s way in also, and you’ve got to find a doorway that fits us both. ~ Stephen King Novelist Stephen King’s sentiment above rings true for all writers—including students composing research arguments. Effective persuasive writing, after all, engages with the audience from the opening passages. Writers should attempt to blend creativity with credible content and a logical approach in order to craft a solid term paper, and much of that important work takes place from the outset, when the author establishes a bond with his or her audience and outlines the parameters of the project. This module is designed to help you begin the drafting process on your research argument by presenting advice on organizing the different parts of your essay and effectively incorporating the resources you will use to support your main points. To summarize our work thus far in the course, we have explored various formal approaches to creating persuasive rhetorical arguments. We have studied and practiced critical reading and response skills in the investigation of articles and speeches. We have examined the components of basic logic and investigated a variety of approaches to using definition and creative phrasing to develop complex, persuasive arguments. And we now have a research plan and a collection of resources that we will incorporate in our term paper. The primary assignment for this learning module is the submission of a three-page “pilot paper,” which should include the following elements: The first three pages of your research argument, which should include your introduction, your thesis, and at least the early passages of your essay’s first subdivision. At least two resources should be incorporated in accordance with MLA formatting principles. This module will offer advice on capturing a reader’s attention from the outset of a piece of writing, using sources effectively in college writing, and creating powerful conclusions that leave readers with a clear understanding of the components of your argument. (1) Objectives Upon completion of this module, the student will be able to: Produce the introductory passages of a ten-page argumentative research essay, including the introduction, the thesis statement, the early passages of the essay’s first subdivision, and two resources correctly introduced using Modern Language Association (MLA) in-text documentation principles. (1) Readings Online Learning Unit CC licensed content, Original Authored by : Florida State College at Jacksonville. License : CC BY: Attribution |
Courses/CSU_San_Bernardino/CHEM_2100%3A_General_Chemistry_I_(Mink)/09%3A_Gases/9.05%3A_Effusion_and_Diffusion_of_Gases | Learning Objectives By the end of this section, you will be able to: Define and explain effusion and diffusion State Graham’s law and use it to compute relevant gas properties If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure \(\PageIndex{1}\)). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure \(\PageIndex{1}\). The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs). We are often interested in the rate of diffusion , the amount of gas passing through some area per unit time: \[\text { rate of diffusion }=\frac{\text { amount of gas passing through an area }}{\text { unit of time }} \nonumber \] The diffusion rate depends on several factors: temperature; the mass of the atoms or molecules; the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation. A process involving movement of gaseous species similar to diffusion is effusion , the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure \(\PageIndex{2}\)). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same. If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure \(\PageIndex{3}\)). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion : The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles : \[\text { rate of effusion } \propto \frac{1}{\sqrt{ M }} \nonumber \] This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles: \[\frac{\text { rate of effusion of } A }{\text { rate of effusion of } B }=\frac{\sqrt{ M_{ B }}}{\sqrt{ M_{ A }}} \nonumber \] Example \(\PageIndex{1}\): Applying Graham’s Law to Rates of Effusion Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen. Solution From Graham’s law, we have: \[\frac{\text { rate of effusion of hydrogen }}{\text { rate of effusion of oxygen }}=\frac{\sqrt{32 gmol^{-1}}}{\sqrt{2 gmol^{-1}}}=\frac{\sqrt{16}}{\sqrt{1}}=\frac{4}{1} \nonumber \] Hydrogen effuses four times as rapidly as oxygen. Exercise \(\PageIndex{1}\) At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Under the same conditions, at what rate will sulfur dioxide effuse? Answer 52 mL/s Example \(\PageIndex{2}\): Effusion Time Calculations It takes 243 s for 4.46 10 −5 mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 10 −5 mol Ne to effuse? Solution It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion: \[\text { rate of effusion }=\frac{\text { amount of gas transferred }}{\text { time }} \nonumber \] and combine it with Graham’s law: \[\frac{\text { rate of effusion of gas Xe }}{\text { rate of effusion of gas Ne}}=\frac{\sqrt{ M_{ Ne }}}{\sqrt{ M_{ Xe }}} \nonumber \] To get: \[\frac{\frac{\text { amount of Xe transferred }}{\text { time for Xe }}}{\frac{\text { amount of Ne transferred }}{\text { time for Ne }}}=\frac{\sqrt{ M_{ Ne }}}{\sqrt{ M_{ Xe }}} \nonumber \] Noting that amount of A = amount of B , and solving for time for Ne : \[\frac{\frac{\text { amount of } Xe }{\text { time for Xe }}}{\frac{\text { amount of } Ne }{\text { time for } Ne }}=\frac{\text { time for } Ne }{\text { time for Xe }}=\frac{\sqrt{ M_{ Ne }}}{\sqrt{ M_{ Xe }}}=\frac{\sqrt{ M_{ Ne }}}{\sqrt{ M_{ Xe }}} \nonumber \] and substitute values: \[\frac{\text { time for Ne }}{243 s }=\sqrt{\frac{20.2~ \cancel{\text{g mol}} }{131.3 ~ \cancel{\text{g mol}} }}=0.392 \nonumber \] Finally, solve for the desired quantity: \[\text { time for } Ne =0.392 \times 243~\text{s} =95.3~\text{s} \nonumber \] Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for Xe, which means the time of effusion for Ne will be smaller than that for Xe. Exercise \(\PageIndex{2}\) A party balloon filled with helium deflates to \(\frac{2}{3}\) of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air (ℳ = 28.2 g/mol) to deflate to\(\frac{1}{2}\) of its original volume? Answer 32 h Example \(\PageIndex{3}\): Determining Molar Mass Using Graham’s Law An unknown gas effuses 1.66 times more rapidly than CO 2 . What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity? Solution From Graham’s law, we have: \[ \frac{\text { rate of effusion of Unknown }}{\text { rate of effusion of } CO_2}=\frac{\sqrt{ M_{ CO_2}}}{\sqrt{ M_{\text {Unknown }}}} \nonumber \] Plug in known data: \[\frac{1.66}{1}=\frac{\sqrt{44.0 g / mol }}{\sqrt{ M_{\text {Unknown }}}} \nonumber \] Solve: \[M_{\text {Unknown }}=\frac{44.0 g / mol }{(1.66)^2}=16.0 g / mol \nonumber \] The gas could well be CH 4 , the only gas with this molar mass. Exercise \(\PageIndex{3}\) Hydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas. Answer 163 g/mol How Sciences Interconnect: Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of 235 U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% 235 U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF 6 , the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The 235 UF 6 molecules have a higher average speed and diffuse through the barrier a little faster than the heavier 238 UF 6 molecules. The gas that has passed through the barrier is slightly enriched in 235 UF 6 and the residual gas is slightly depleted. The small difference in molecular weights between 235 UF 6 and 238 UF 6 only about 0.4% enrichment, is achieved in one diffuser (Figure \(\PageIndex{4}\)). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained. The large scale separation of gaseous 235 UF 6 from 238 UF 6 was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10 –6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF 6 . Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons. |
Courses/College_of_the_Canyons/Chem_151%3A_Preparatory_General_Chemistry_OER/06%3A_Chemical_Composition/6.01%3A_Counting_Nails_by_the_Pound | Counting by Weighing The size of molecule is so small that it is physically difficult, if not impossible, to directly count out molecules (Figure \(\PageIndex{1}\)). However, we can count them indirectly by using a common trick of "counting by weighing". Consider the example of counting nails in a big box at a hardware store. You need to estimate the number of nails in a box. The weight of an empty box is \(213 \,g\) and the weight of the box plus a bunch of big nails is \(1340\, g\). Assume that we know that the weight of one big nail is 0.450 g. Hopefully it's not necessary to tear open the package and count the nails. We agree that \[\text{mass of big nails} = 1340\, g - 213\, g = 1227 \,g\] Therefore \[\text{Number of big nails in box} = 1227\, g \dfrac{\text{1 big nail}}{0.450\, g} = 2,726.6\, \text{big nails} = 2,730 \,\text{big nails}. \label{eq2}\] You have just counted the number of big nails in the box by weighing them (rather than by counting them individually). Now consider if the box of nails weighed the same, but the box were filled with small nails with an individual mass of \(0.23\, g/\text{small nail}\) instead? You would do the same math, but use a different denominator in Equation \ref{eq2}: \[\text{Number of small nails in box} = 1227\, g \: \dfrac{\text{1 small nail}}{0.230\, g} = 5,334.7\, \text{small nails} = 5,335 \, \text{small nails}. \label{eq3}\] The individual mass is the conversion factor used in the calculation and changes, based on the nature of the nail (big or small). Let's ask a different question: how many dozens of nails are there in the same box of small nails described above? If we know the information from Equation \ref{eq3}, we can just use the conversion of how many nails are in a dozen: \[\text{5,335 small nails} \: \dfrac{\text{1 dozen}}{12 \text{small nails}} = 444.6 \,\text{dozen small nails}. \label{eq4}\] If we want to get this value from weighing, we use the "dozen mass" instead of individual mass: \[12 \times 0.23 g = 2.76\, g/\text{dozen small nails}. \label{eq5}\] So following Equation \ref{eq3}, we get: \[\text{Number of dozens of small nails} = 1227\, g \: \dfrac{\text{1 dozen small nails}}{2.76\, g} = 444.6 \,\text{dozen small nails} \label{eq6}\] and this is the same result as Equation \ref{eq4}. These calculations demonstrate the difference between individual mass (i.e., per individual) and collective mass (e.g., per dozen or per gross). The collective mass of most importance to chemistry is molar mass (i.e., mass per mole or mass per \(6.022 \times 10^{23}\)), which will covered in more detail in the next section. |
Courses/Fresno_City_College/Introductory_Chemistry_Atoms_First_for_FCC/04%3A_Compounds_and_Chemical_Bonds/4.05%3A_Ionic_Compounds/4.5.02%3A_Writing_Formulas_for_Ionic_Compounds | Learning Objectives Write the correct formula for an ionic compound. Recognize polyatomic ions in chemical formulas. Ionic compounds do not exist as molecules. In the solid state, ionic compounds are in crystal lattice containing many ions each of the cation and anion. An ionic formula, like \(\ce{NaCl}\), is an empirical formula. This formula merely indicates that sodium chloride is made of an equal number of sodium and chloride ions. Sodium sulfide, another ionic compound, has the formula \(\ce{Na_2S}\). This formula indicates that this compound is made up of twice as many sodium ions as sulfide ions. This section will teach you how to find the correct ratio of ions, so that you can write a correct formula. If you know the name of a binary ionic compound, you can write its chemical formula . Start by writing the metal ion with its charge, followed by the nonmetal ion with its charge. Because the overall compound must be electrically neutral, decide how many of each ion is needed in order for the positive and negative charges to cancel each other out. Example \(\PageIndex{1}\): Aluminum Nitride and Lithium Oxide Write the formulas for aluminum nitride and lithium oxide. Solution Unnamed: 0 Write the formula for aluminum nitride Write the formula for lithium oxide 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Al^{3+}} \: \: \: \: \: \ce{N^{3-}}\) \(\ce{Li^+} \: \: \: \: \: \ce{O^{2-}}\) 2. Use a multiplier to make the total charge of the cations and anions equal to each other. total charge of cations = total charge of anions 1(3+) = 1(3-) +3 = -3 total charge of cations = total charge of anions 2(1+) = 1(2-) +2 = -2 3. Use the multipliers as subscript for each ion. \(\ce{Al_1N_1}\) \(\ce{Li_2O_1}\) 4. Write the final formula. Leave out all charges and all subscripts that are 1. \(\ce{AlN}\) \(\ce{Li_2O}\) An alternative way to writing a correct formula for an ionic compound is to use the crisscross method . In this method, the numerical value of each of the ion charges is crossed over to become the subscript of the other ion. Signs of the charges are dropped. Example \(\PageIndex{2}\): The Crisscross Method for Lead (IV) Oxide Write the formula for lead (IV) oxide. Solution Crisscross Method Write the formula for lead (IV) oxide 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Pb^{4+}} \: \: \: \: \: \ce{O^{2-}}\) 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. NaN 3. Reduce to the lowest ratio. \(\ce{Pb_2O_4}\) 4. Write the final formula. Leave out all subscripts that are 1. \(\ce{PbO_2}\) Exercise \(\PageIndex{2}\) Write the chemical formula for an ionic compound composed of each pair of ions. the calcium ion and the oxygen ion the 2+ copper ion and the sulfur ion the 1+ copper ion and the sulfur ion Answer a: CaO Answer b: CuS Answer c: Cu 2 S Be aware that ionic compounds are empirical formulas and so must be written as the lowest ratio of the ions. Example \(\PageIndex{3}\): Sulfur Compound Write the formula for sodium combined with sulfur. Solution Crisscross Method Write the formula for sodium combined with sulfur 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Na^{+}} \: \: \: \: \: \ce{S^{2-}}\) 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. NaN 3. Reduce to the lowest ratio. This step is not necessary. 4. Write the final formula. Leave out all subscripts that are 1. \(\ce{Na_2S}\) Exercise \(\PageIndex{3}\) Write the formula for each ionic compound. sodium bromide lithium chloride magnesium oxide Answer a: NaBr Answer b: LiCl Answer c: MgO Writing Formulas for Ionic Compounds Containing Polyatomic Ions Polyatomic ions were introduced in a previous subsection. It is important to understand the relationship between the name, formula, and charge of polyatomic ions before beginning to use them within ionic compounds. The rule for constructing formulas for ionic compounds containing polyatomic ions is the same as for formulas containing monatomic (single-atom) ions: the positive and negative charges must balance. If more than one of a particular polyatomic ion is needed to balance the charge, the entire formula for the polyatomic ion must be enclosed in parentheses, and the numerical subscript is placed outside the parentheses. This is to show that the subscript applies to the entire polyatomic ion. An example is Ba(NO 3 ) 2 . Writing a formula for ionic compounds containing polyatomic ions also involves the same steps as for a binary ionic compound. Write the symbol and charge of the cation followed by the symbol and charge of the anion. Example \(\PageIndex{4}\): Calcium Nitrate Write the formula for calcium nitrate. Solution Crisscross Method Write the formula for calcium nitrate 1. Write the symbol and charge of the cation (metal) first and the anion (nonmetal) second. \(\ce{Ca^{2+}} \: \: \: \: \: \ce{NO_3^-}\) 2. Transpose only the number of the positive charge to become the subscript of the anion and the number only of the negative charge to become the subscript of the cation. NaN 3. Reduce to the lowest ratio. \(\ce{Ca_1(NO_3)_2}\) 4. Write the final formula. Leave out all subscripts that are 1. If there is only 1 of the polyatomic ion, leave off parentheses. \(\ce{Ca(NO_3)_2}\) Example \(\PageIndex{5}\) Write the chemical formula for an ionic compound composed of the potassium ion and the sulfate ion. Solution Explanation Answer Potassium ions have a charge of 1+, while sulfate ions have a charge of 2−. We will need two potassium ions to balance the charge on the sulfate ion, so the proper chemical formula is K2SO4. \(K_2SO_4\) Exercise \(\PageIndex{5}\) Write the chemical formula for an ionic compound composed of each pair of ions. the magnesium ion and the carbonate ion the aluminum ion and the acetate ion Answer a: MgCO 3 Answer b: Al(CH 3 COO) 3 Recognizing Ionic Compounds There are two ways to recognize ionic compounds. First, compounds between metal and nonmetal elements are usually ionic. For example, CaBr 2 contains a metallic element (calcium, a group 2 [or 2A] metal) and a nonmetallic element (bromine, a group 17 [or 7A] nonmetal). Therefore, it is most likely an ionic compound. (In fact, it is ionic.) In contrast, the compound NO 2 contains two elements that are both nonmetals (nitrogen, from group 15 [or 5A] , and oxygen, from group 16 [or 6A] . It is not an ionic compound; it belongs to the category of covalent compounds discussed elsewhere. Also note that this combination of nitrogen and oxygen has no electric charge specified, so it is not the nitrite ion. Second, if you recognize the formula of a polyatomic ion in a compound, the compound is ionic. For example, if you see the formula Ba(NO 3 ) 2 , you may recognize the “NO 3 ” part as the nitrate ion, NO 3 − . (Remember that the convention for writing formulas for ionic compounds is not to include the ionic charge.) This is a clue that the other part of the formula, Ba, is actually the Ba 2 + ion, with the 2+ charge balancing the overall 2− charge from the two nitrate ions. Thus, this compound is also ionic. Example \(\PageIndex{6}\) Identify each compound as ionic or not ionic. Na 2 O PCl 3 NH 4 Cl OF 2 Solution Explanation Answer a. Sodium is a metal, and oxygen is a nonmetal. Therefore, Na2O is expected to be ionic. \(Na_2O\), ionic b. Both phosphorus and chlorine are nonmetals. Therefore, PCl3 is not ionic. \(PCl_3\), not ionic c. The NH4 in the formula represents the ammonium ion, NH4+, which indicates that this compound is ionic. \(NH_4Cl\), ionic d. Both oxygen and fluorine are nonmetals. Therefore, OF2 is not ionic. \(OF_2\), not ionic Exercise \(\PageIndex{6}\) Identify each compound as ionic or not ionic. N 2 O FeCl 3 (NH 4 ) 3 PO 4 SOCl 2 Answer a: not ionic Answer b: ionic Answer c: ionic Answer d: not ionic Summary Formulas for ionic compounds contain the symbols and number of each atom present in a compound in the lowest whole number ratio. Contributions & Attributions This page was constructed from content via the following contributor(s) and edited (topically or extensively) by the LibreTexts development team to meet platform style, presentation, and quality: Marisa Alviar-Agnew ( Sacramento City College ) Henry Agnew (UC Davis) |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Quantum_Tutorials_(Rioux)/02%3A_Atomic_Structure/2.61%3A_One-dimensional_H-atom_with_Delta_Function_Potential | The One‐dimensional Hydrogen Atom with a Delta Function Potential Energy Interaction Between the Proton and Electron The energy Hamiltonian and a normalized wave function for the hydrogen atom with delta function interaction between the electron and proton is given below. \[ \begin{matrix} \text{H} = - \frac{1}{2} \frac{d^2}{dx^2} \blacksquare - \Delta (x) \blacksquare & \Psi (x) = \text{exp} \left( - |x| \right) & \int_{- \infty}^{ \infty} \Psi (x)^2 dx \rightarrow 1 \end{matrix} \nonumber \] The wave function is not well‐behaved in coordinate space making the evaluation of kinetic energy tricky. However, the evaluation of potential energy is straight forward. So the strategy is to evaluate potential energy in coordinate space and kinetic energy in momentum space. The latter requires a Fourier transform of the coordinate space wave function. Calculation of Potential Energy in Coordinate Space \[ V = \int_{- \infty}^{ \infty} \Psi (x) - \Delta (x) \Psi (x) \text{dx} \rightarrow - 1 \nonumber \] Fourier Transform of the Coordinate Wave Function into Momentum Space \[ \Psi (p) = \int_{- \infty}^{ \infty} \frac{ \text{exp(-i p x)}}{ \sqrt{2 \pi}} \Psi (x) \text{dx simplify} \rightarrow \frac{ \sqrt{2}}{ \sqrt{ \pi} \left( p^2 + 1 \right)} \nonumber \] Display the Momentum Wave Function to Show it is Well-behaved Calculation of Kinetic Energy in Momentum Space \[ \text{T} = \int_{- \infty}^{ \infty} \Phi (p) \frac{p^2}{2} \Phi (p) \text{dp simplify} \rightarrow \frac{1}{2} \nonumber \] Calculation of Total Energy \[ \text{E = T + V} \rightarrow \text{E} = - \frac{1}{2} \nonumber \] The Wigner Function The Wigner phase‐space distribution function is calculated using the momentum wave function. \[ \text{W(x, p)} = \frac{1}{2 \pi} \int{- \infty}^{ \infty} \overline{ \Phi \left( \text{p} + \frac{ \text{s}}{2} \right)} \text{exp(-i s x)} \Phi \left( \text{p} - \frac{ \text{s}}{2} \right) \text{ds} \nonumber \] \[ \begin{matrix} N = 50 & i = 0 .. N & x_i = -4 + \frac{8i}{N} & j = 0 .. N & p_j = -7 + \frac{14j}{N} & \text{Wigner}_{i,~j} = \text{W} \left( x_i,~ p_j \right) \end{matrix} \nonumber \] Next we look at two variational calculations on the same system. The first involves a gaussian trial wave function, and the second a trigonmetric trial wave function. Gaussian Trial Wave Function \[ \begin{matrix} \Psi_1 (x,~ \beta ) = \left( \frac{2 \beta}{ \pi} \right)^{ \frac{1}{4}} \text{exp}\left( - \beta x^2 \right) & \int_{- \infty}^{ \infty} \Psi_1 (x,~ \beta)^2 \text{dx assume, } \beta >0 \rightarrow 1 \end{matrix} \nonumber \] Evaluation of Variational Energy Integral \[ \text{E} ( \beta ) = \int_{- \infty}^{ \infty} \Psi_1 (x,~ \beta ) - \frac{1}{2} \frac{d^2}{dx^2} \Psi (x,~ \beta ) \text{dx... assume, } \beta > 0 \rightarrow \frac{ \beta}{2} - \frac{ \sqrt{2} \sqrt{3}}{ \sqrt{ \pi}} + \int_{- \infty}^{ \infty} - \Delta (x) \Psi_1 (x,~ \beta)^2 dx \nonumber \] Energy Minimization With Respect to Variational Parameter \[ \begin{matrix} \frac{d}{d \beta} \text{E} ( \beta = 0 \text{ solve, } \beta \rightarrow \frac{2}{ \pi} & \begin{array}{c|c} \text{E} ( \beta ) & _{ \text{simplify}}^{ \text{substitute, } \beta = \frac{2}{ \pi}} \rightarrow - \frac{1}{ \pi} = -0.318 \end{array} \end{matrix} \nonumber \] Error \[ \left| \frac{-.318+ .5}{-.5} \right| = 36.4 \% \nonumber \] Trigonometric Trial Wave Function \[ \begin{matrix} \Psi_2 ( \text{x}, ~ \beta ) = \sqrt{ \frac{ \beta}{2}} \text{sech} ( \beta \text{x} ) & \end{matrix} \nonumber \] Evaluation of Variational Energy Integral \[ \begin{matrix} \text{E} ( \beta ) = \begin{array}{c|c} \int_{ - \infty}^{ \infty} \Psi_2 ( \text{x},~ \beta ) - \frac{1}{2} \frac{d^2}{dx^2} \Psi_2 ( \text{x, } \beta ) dx ... & _{ \text{simplify}}^{ \text{assume, } \beta > 0} \rightarrow \frac{ \beta ( \beta -3}{6} \\ + \int_{ - \infty}^{ \infty} - \Delta ( \text{x}) \Psi_2 ( \text{x, } \beta )^2 dx \end{array} \end{matrix} \nonumber \] Energy Minimization With Respect to Variational Parameter \[ \begin{matrix} \frac{d}{d \beta} \text{E} ( \beta ) = 0 \text{ solve, } \beta \rightarrow \frac{3}{2} & \begin{array}{c|c} \text{E} ( \beta ) & _{ \text{simplify}}^{ \text{substitute, } \beta = \frac{3}{2}} \rightarrow - \frac{3}{8} = -0.375 \end{array} \end{matrix} \nonumber \] Error \[ \left| \frac{-.375 + .5}{-.5} \right| = 25 \% \nonumber \] Graphical Comparison of Exact Wave Function with Trial Wave Function This graphical comparison is consistent with the variational results presented above. The trigonometric trial function more closely resembles the exact wave function. |
Courses/University_of_Arkansas_Little_Rock/00001%3AF23_Gen_Chem_2_Lab/0%3A_General_Information/4%3A_Instrumentation/4%3A_pH_Meter | Multiple labs this semester will be using a pH meter, which uses a probe to measure the pH of aqueous solutions. pH Probes In this experiment we will use a Ph probe, which is an electronic device that measures the pH. These are very common and they should always be checked against standard solutions of known pH and calibrated if they read incorrectly. The pH probe is an electrochemical cell and we will cover these in chapter 19 , sections 19.3-19.5 and 19.7. The following YouTube from Oxford Press does an excellent job of describing how a pH probe works. It is imperative that you test your probe in a buffer to be sure it is reading accurately and if it is not, you will need to calibrate it. Video \(\PageIndex{1}\) 2:30 YouTuve describing the operation of a pH probe developed by Oxford University Press ( https://youtu.be/aIn4D2QXUy4 ). The pH reading is not accurate until the probe stabilizes, so when you change the pH you need to wait until the reading becomes steady before recording the value. Vernier Probes General Instructions for use of a pH meter The pH probe is stored in an electrolyte and should never be allowed to dry out. You should have DI water in a squeeze bottle that you can rinse the probe tip over a 600 mL beaker and be careful to never touch the probe surface. Between measurements you can fill a keep the probe in a 150 mL Erlenmeyer flask that is half full of water (this is more stable than a beaker). The first thing you should do is test your probe with a small amount of buffer solution solution of known pH and if it reads the buffer pH accurately, you are good to go, and if not, you need to calibrate the pH meter before proceeding. Calibration |
Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Sciences_Digital_Library/In_Class_Activities/Biological_Mass_Spectrometry%3A_Proteomics/Instructors_Manual/Section_4%3A_MS-MS_and_De_Novo_Sequencing/Section_4A._Tandem_MS | While enzymes, such as trypsin, can be used to cleave proteins and peptides at specific amino acid linkages, we can also fragment peptides inside of a mass spectrometer to obtain additional information. These types of experiments are called tandem MS or MS-MS experiments. These experiments are particularly helpful for “shotgun” proteomics or bottom-up proteomics. In these experiments, protein mixtures are first digested with enzymes (such as trypsin), then separated by one or more chromatography steps, and then electrosprayed into a mass spectrometer. A mass analyzer is then used to select a precursor ion with a specific m / z value for fragmentation. Fragmentation requires that some energy be added to the system. The most common method of fragmentation in MS-MS experiments is collision-induced dissociation (CID). In CID, the precursor ion is accelerated into an interaction cell that contains a collision gas, such as helium or nitrogen. When the precursor ion collides with the collision gas, the ion can fragment into two fragments, an ion and a neutral. The fragment ions are then analyzed to produce the MS/MS spectrum. To better understand the operation of a commonly used triple quadruple or “triple quad” MS/MS, watch an animation , then answer the questions below. Video Questions 1. What is the purpose of the skimmer in the instrument? A. The skimmer helps to remove neutrals so that only ions enter the mass analyzer. 2. The animation uses color to indicate difference m / z value ions. Which “color” of ion is selected as the precursor ion? A. The green ions are the precursor. They reach the end of the first quadrupole and enter the collision cell, while other m/z value ions are removed by the mass filter effect of the quadrupole. 3. Although this style of instrument is commonly called a “triple quad,” the collision cell is not actually a quadrupole. What is it? A. It is a hexapole. 4. The last quadrupole selected fragment ions to be sent to the detector. Neutrals also pass through this quadrupole. Why don’t they produce a signal at the detector? A. The conversion dynode (detector) is at 90° relative to the quadrupole ion path. The electrical potential on the dynode draws ions down, where their impact produces electrons that become the amplified current at the detector. Neutrals are not affected by this electrical potential so they continue along the quadruple axis and do not result in current. MS/MS experiments are useful because the fragment m/z values give information about the analyte’s molecular structure. Low energy CID often produces small neutral losses, such as H 2 O, CH 3 OH, CO, CO 2 , NH 3 , and CN. Higher energy collisions can lead to retrosynthetic reactions, in which characteristic bonds in the precursor ion are broken. For example, CID of peptides often results in cleavage of peptide bonds along the backbone. Reading Question 1. Complete the table below to summarize the expected mass differences from common neutral losses. Round expected masses to the nearest amu. Table 1. Common neutral losses produced by CID of peptide ions. Neutral Loss Mass (amu) NH3 NaN H2O NaN CN NaN CO NaN CH3OH NaN CO2 NaN A. Table 1. Common neutral losses produced by CID of peptide ions. Neutral Loss Mass (amu) NH3 17 H2O 18 CN 26 CO 28 CH3OH 32 CO2 44 The identity of the precursor ion can be scanned through the mass range so that a mass spectrum of each precursor ion’s corresponding fragments is obtained. These experiments produced large quantities of data extremely rapidly. Interpreting these large data sets usually involves specialized software programs that identify peptides from their fragments and then identify proteins from their peptides; however, for simple mixtures, the data may be interpreted manually since peptide fragmentation in tandem MS experiments is well-characterized. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/04%3A_Postulates_and_Principles_of_Quantum_Mechanics | 4.1: The Wavefunction Specifies the State of a System This page details the first postulate of quantum mechanics, asserting that a system's state is represented by its wavefunction, \(\psi\). Unlike classical mechanics, quantum states cannot precisely define position and momentum due to the uncertainty principle. Valid wavefunctions must be square-integrable, single-valued, continuous, and finite. The text clarifies that while some wavefunctions may meet certain criteria, others, like one that approaches infinity, are unacceptable. 4.2: Quantum Operators Represent Classical Variables This page describes the correspondence principle in quantum mechanics, stating that every classical observable has a corresponding quantum operator. It discusses observables like position and momentum, represented by operators acting on the wavefunction. Key operators include those for kinetic energy, potential energy, and the Hamiltonian, which combines both energy types. Additionally, a table outlines various quantum operators alongside their observables, emphasizing their mathematical forms. 4.3: Observable Quantities Must Be Eigenvalues of Quantum Mechanical Operators This page explores eigenvalue equations in quantum mechanics, detailing the Hamiltonian operator, Schrödinger Equation, and how observables are derived as eigenvalues from eigenstates. It discusses expectation values for position, momentum, and kinetic energy in one-dimensional systems, noting that the average position is central, with zero momentum for the ground state. 4.4: The Time-Dependent Schrödinger Equation This page clarifies the differences between the time-dependent and time-independent Schrödinger equations, highlighting their effects on wavefunctions. It details how the time-dependent equation shows wavefunction evolution, while the time-independent equation indicates stationary states with constant energies. 4.5: Eigenfunctions of Operators are Orthogonal This page explains Hermitian operators in quantum mechanics, highlighting that they correspond to experimental observables with real eigenvalues and orthogonal eigenstates. It discusses the orthogonality of eigenfunctions, proving that wavefunctions from different eigenvalues are orthogonal, illustrated with particle-in-a-box examples. The text also addresses degenerate eigenstates and their potential non-orthogonality, which can be rectified through the Gram-Schmidt Orthogonalization process. 4.6: Commuting Operators Allow Infinite Precision This page explains the Heisenberg Uncertainty Principle and commutation relations in quantum mechanics, highlighting the significance of operators and their commutation. It details conditions for operator commutation and illustrates with examples, particularly regarding angular momentum. The text concludes that non-commuting operators impose limitations on the uncertainties of measurable physical quantities. 4.E: Postulates and Principles of Quantum Mechanics (Exercises) This page delves into quantum mechanics, covering key concepts such as wavefunctions, operators, and their commutation relations, particularly in the context of angular momentum. It emphasizes the uncertainty principle, the role of Hermitian operators, and the time-dependent Schrödinger equation. The text discusses the implications of non-commuting operators, the conditions for eigenvalues, and boundary conditions affecting wavefunctions. |
Courses/University_of_Kansas/General_Organic_and_Biological_Chemistry/06%3A_Quantities_in_Chemical_Reactions/6.05%3A_Limiting_Reagent_and_Percent_Yield | Learning Objectives Define and determine theoretical yields, actual yields, and percent yields. Identify a limiting reagent from a set of reactants. Calculate how much product will be produced from the limiting reagent. Calculate how much reactant(s) remains when the reaction is complete. Yield In all the previous calculations we have performed involving balanced chemical equations, we made two assumptions: The reaction goes exactly as written. The reaction proceeds completely. In reality, such things as side reactions occur that make some chemical reactions rather messy. For example, in the actual combustion of some carbon-containing compounds, such as methane, some CO is produced as well as CO 2 . However, we will continue to ignore side reactions, unless otherwise noted. The second assumption, that the reaction proceeds completely, is more troublesome. Many chemical reactions do not proceed to completion as written, for a variety of reasons. When we calculate an amount of product assuming that all the reactant reacts, we calculate the theoretical yield , an amount that is theoretically produced as calculated using the balanced chemical reaction. In many cases, however, this is not what really happens. In many cases, less—sometimes, much less—of a product is made during the course of a chemical reaction. The amount that is actually produced in a reaction is called the actual yield . By definition, the actual yield is less than or equal to the theoretical yield. If it is not, then an error has been made. Both theoretical yields and actual yields are expressed in units of moles or grams. It is also common to see something called a percent yield. The percent yield is a comparison between the actual yield and the theoretical yield and is defined as \[ \text{percent yield} = \dfrac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \label{yield}\] It does not matter whether the actual and theoretical yields are expressed in moles or grams, as long as they are expressed in the same units. However, the percent yield always has units of percent. Proper percent yields are between 0% and 100%. In the laboratory, a student will occasionally obtain a yield that appears to be greater than 100%. This usually happens when the product is impure or is wet with a solvent such as water. If this is not the case, then the student must have made an error in weighing either the reactants or the products. The law of conservation of mass applies even to undergraduate chemistry laboratory experiments. A 100% yield means that everything worked perfectly, and the chemist obtained all the product that could have been produced. Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows the unlikelihood of a 100% yield. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair. In part because of the problems and costs of waste disposal, industrial production facilities face considerable pressures to optimize the yields of products and make them as close to 100% as possible. Example \(\PageIndex{1}\): A worker reacts 30.5 g of Zn with nitric acid and evaporates the remaining water to obtain 65.2 g of Zn(NO 3 ) 2 . What are the theoretical yield, the actual yield, and the percent yield? \[\ce{Zn(s) + 2HNO_3(aq) → Zn(NO_3)_2(aq) + H_2(g)} \nonumber \] Solution A mass-mass calculation can be performed to determine the theoretical yield. We need the molar masses of Zn (65.39 g/mol) and Zn(NO 3 ) 2 (189.41 g/mol). In three steps, the mass-mass calculation is: \[30.5\cancel{g\, Zn}\times \frac{1\, \cancel{mol\, Zn}}{65.39\cancel{g\, Zn}}\times \frac{1\, \cancel{mol\, Zn(NO_{3})_{2}}}{1\cancel{mol\, Zn}}\times \frac{189.41\, g\, Zn(NO_{3})_{2}}{1\cancel{mol\,Zn(NO_{3})_{2}}}=88.3\, g\, Zn(NO_{3})_{2}\nonumber \] Thus, the theoretical yield is 88.3 g of Zn(NO 3 ) 2 . The actual yield is the amount that was actually made, which was 65.2 g of Zn(NO 3 ) 2 . To calculate the percent yield, we take the actual yield and divide it by the theoretical yield and multiply by 100 (Equation \ref{yield}): \[\frac{65.2\, g\, Zn(NO_{3})_{2}}{88.3\, g\,Zn(NO_{3})_{2}}\times 100\%=73.8\%\nonumber \] The worker achieved almost three-fourths of the possible yield. Exercise \(\PageIndex{1}\) A synthesis produced 2.05 g of NH 3 from 16.5 g of N 2 . What is the theoretical yield and the percent yield? \[N_2(g) + 3H_2(g) → 2NH_3(g)\nonumber \] *Technically, this is a reversible reaction (with double arrows), but for this exercise consider it irreversible (single arrow). Answer theoretical yield = 20.1 g; percent yield = 10.2% Chemistry is Everywhere: Actual Yields in Drug Synthesis and Purification Many drugs are the product of several steps of chemical synthesis. Each step typically occurs with less than 100% yield, so the overall percent yield might be very small. The general rule is that the overall percent yield is the product of the percent yields of the individual synthesis steps. For a drug synthesis that has many steps, the overall percent yield can be very tiny, which is one factor in the huge cost of some drugs. For example, if a 10-step synthesis has a percent yield of 90% for each step, the overall yield for the entire synthesis is only 35%. Many scientists work every day trying to improve percent yields of the steps in the synthesis to decrease costs, improve profits, and minimize waste. Even purifications of complex molecules into drug-quality purity are subject to percent yields. Consider the purification of impure albuterol. Albuterol (C 13 H 21 NO 2 ; accompanying figure) is an inhaled drug used to treat asthma, bronchitis, and other obstructive pulmonary diseases. It is synthesized from norepinephrine, a naturally occurring hormone and neurotransmitter. Its initial synthesis makes very impure albuterol that is purified in five chemical steps. The details of the steps do not concern us; only the percent yields do: 0 1 impure albuterol → intermediate A percent yield = 70% intermediate A → intermediate B percent yield = 100% intermediate B → intermediate C percent yield = 40% intermediate C → intermediate D percent yield = 72% intermediate D → purified albuterol percent yield = 35% overall percent yield = 70% × 100% × 40% × 72% × 35% = 7.5% overall percent yield = 70% × 100% × 40% × 72% × 35% = 7.5% That is, only about one-fourteenth of the original material was turned into the purified drug. This demonstrates one reason why some drugs are so expensive—a lot of material is lost in making a high-purity pharmaceutical. Limiting Reagent In all the examples discussed thus far, the reactants were assumed to be present in stoichiometric quantities. Consequently, none of the reactants was left over at the end of the reaction. This is often desirable, as in the case of a space shuttle, where excess oxygen or hydrogen was not only extra freight to be hauled into orbit but also an explosion hazard. More often, however, reactants are present in mole ratios that are not the same as the ratio of the coefficients in the balanced chemical equation. As a result, one or more of them will not be used up completely but will be left over when the reaction is completed. In this situation, the amount of product that can be obtained is limited by the amount of only one of the reactants. The reactant that restricts the amount of product obtained is called the limiting reactant. The reactant that remains after a reaction has gone to completion is in excess. Consider a nonchemical example. Assume you have invited some friends for dinner and want to bake brownies for dessert. You find two boxes of brownie mix in your pantry and see that each package requires two eggs. The balanced equation for brownie preparation is thus \[ 1 \,\text{box mix} + 2 \,\text{eggs} \rightarrow 1 \, \text{batch brownies} \label{3.7.1}\] If you have a dozen eggs, which ingredient will determine the number of batches of brownies that you can prepare? Because each box of brownie mix requires two eggs and you have two boxes, you need four eggs. Twelve eggs is eight more eggs than you need. Although the ratio of eggs to boxes in is 2:1, the ratio in your possession is 6:1. Hence the eggs are the ingredient (reactant) present in excess, and the brownie mix is the limiting reactant. Even if you had a refrigerator full of eggs, you could make only two batches of brownies. Figure \(\PageIndex{1}\) : The Concept of a Limiting Reactant in the Preparation of Brownies A similar situation exists for many chemical reactions: you usually run out of one reactant before all of the other reactant has reacted. The reactant you run out of is called the limiting reagent ; t he other reactant or reactants are considered to be in excess . A crucial skill in evaluating the conditions of a chemical process is to determine which reactant is the limiting reagent and which is in excess. The key to recognizing which reactant is the limiting reagent is based on a mole-mass or mass-mass calculation: whichever reactant gives the lesser amount of product is the limiting reagent. What we need to do is determine an amount of one product (either moles or mass), assuming all of each reactant reacts. Whichever reactant gives the least amount of that particular product is the limiting reagent. It does not matter which product we use, as long as we use the same one each time. It does not matter whether we determine the number of moles or grams of that product; however, we will see shortly that knowing the final mass of product can be useful. For example, consider this reaction: \[4As(s) + 3O_2(g) → 2As_2O_3(s)\nonumber \] Suppose we start a reaction with 50.0 g of As and 50.0 g of O 2 . Which one is the limiting reagent? We need to perform two mole-mass calculations, each assuming that each reactant reacts completely. Then we compare the amount of the product produced by each and determine which is less. The calculations are as follows: \[50.0\cancel{g\, As}\times \frac{1\cancel{mol\, As}}{74.92\cancel{g\, As}}\times \frac{2\, mol\, As_{2}O_{3}}{4\cancel{mol\, As}}=0.334\, mol\, As_{2}O_{3}\nonumber \] \[50.0\cancel{g\, O_{2}}\times \frac{1\cancel{mol\, O_{2}}}{32.00\cancel{g\, O_{2}}}\times \frac{2\, mol\, As_{2}O_{3}}{3\cancel{mol\, O_{2}}}=1.04\, mol\, As_{2}O_{3}\nonumber \] Comparing these two answers, it is clear that 0.334 mol of As 2 O 3 is less than 1.04 mol of As 2 O 3 , so arsenic is the limiting reagent. If this reaction is performed under these initial conditions, the arsenic will run out before the oxygen runs out. We say that the oxygen is "in excess." Identifying the limiting reagent, then, is straightforward. However, there are usually two associated questions: (1) what mass of product (or products) is then actually formed? and (2) what mass of what reactant is left over? The first question is straightforward to answer: simply perform a conversion from the number of moles of product formed to its mass, using its molar mass. For As 2 O 3 , the molar mass is 197.84 g/mol; knowing that we will form 0.334 mol of As 2 O 3 under the given conditions, we will get \[0.334\cancel{mol\, As_{2}O_{3}}\times \frac{197.84\, g\, As_{2}}{\cancel{1\, mol\, As_{2}O_{3}}}=66.1\, g\, As_{2}O_{3}\nonumber \] The second question is somewhat more convoluted to answer. First, we must do a mass-mass calculation relating the limiting reagent (here, As) to the other reagent (O 2 ). Once we determine the mass of O 2 that reacted, we subtract that from the original amount to determine the amount left over. According to the mass-mass calculation, \[50.0\cancel{g\, As}\times \frac{1\cancel{mol\, As}}{74.92\cancel{g\, As}}\times \frac{3\cancel{mol\, O_{2}}}{4\cancel{mol\, As}}\times \frac{32.00\, g\, O_{2}}{\cancel{1\, mol\, O_{2}}}=16.0\, g\, O_{2}\; reacted\nonumber \] Because we reacted 16.0 g of our original O 2 , we subtract that from the original amount, 50.0 g, to get the mass of O 2 remaining: 50.0 g O 2 − 16.0 g O 2 reacted = 34.0 g O 2 left over You must remember to perform this final subtraction to determine the amount remaining; a common error is to report the 16.0 g as the amount remaining. Example \(\PageIndex{1}\): A 5.00 g quantity of Rb is combined with 3.44 g of MgCl 2 according to this chemical reaction: \[2R b(s) + MgCl_2(s) → Mg(s) + 2RbCl(s) \nonumber\nonumber \] What mass of Mg is formed, and what mass of what reactant is left over? Solution Because the question asks what mass of magnesium is formed, we can perform two mass-mass calculations and determine which amount is less. \[5.00\cancel{g\, Rb}\times \frac{1\cancel{mol\, Rb}}{85.47\cancel{g\, Rb}}\times \frac{1\cancel{mol\, Mg}}{2\cancel{mol\, Rb}}\times \frac{24.31\, g\, Mg}{\cancel{1\, mol\, Mg}}=0.711\, g\, Mg \nonumber \] \[3.44\cancel{g\, MgCl_{2}}\times \frac{1\cancel{mol\, MgCl_{2}}}{95.21\cancel{g\, MgCl_{2}}}\times \frac{1\cancel{mol\, Mg}}{1\cancel{mol\, MgCl_{2}}}\times \frac{24.31\, g\, Mg}{\cancel{1\, mol\, Mg}}=0.878\, g\, Mg \nonumber \] The 0.711 g of Mg is the lesser quantity, so the associated reactant—5.00 g of Rb—is the limiting reagent. To determine how much of the other reactant is left, we have to do one more mass-mass calculation to determine what mass of MgCl2 reacted with the 5.00 g of Rb, and then subtract the amount reacted from the original amount. \[5.00\cancel{g\, Rb}\times \frac{1\cancel{mol\, Rb}}{85.47\cancel{g\, Rb}}\times \frac{1\cancel{mol\, MgCl_{2}}}{2\cancel{mol\, Rb}}\times \frac{95.21\, g\, Mg}{\cancel{1\, mol\, MgCl_{2}}}=2.78\, g\, MgCl_{2}\: \: reacted \nonumber \] Because we started with 3.44 g of MgCl2, we have 3.44 g MgCl2 − 2.78 g MgCl2 reacted = 0.66 g MgCl2 left Exercise \(\PageIndex{1}\) Given the initial amounts listed, what is the limiting reagent, and what is the mass of the leftover reagent? \[\underbrace{22.7\, g}_{MgO(s)}+\underbrace{17.9\, g}_{H_2S}\rightarrow MgS(s)+H_{2}O(l) \nonumber \] Answer H 2 S is the limiting reagent; 1.5 g of MgO are left over. Summary Theoretical yield is the calculated yield using the balanced chemical reaction. Actual yield is what is actually obtained in a chemical reaction. Percent yield is a comparison of the actual yield with the theoretical yield. The limiting reagent is the reactant that produces the least amount of product. Mass-mass calculations can determine how much product is produced and how much of the other reactants remain. |
Courses/Mount_Aloysius_College/CHEM_100%3A_General_Chemistry_(O'Connor)/19%3A_Nucleic_Acids/19.06%3A_Mutations_and_Genetic_Diseases | Learning Objectives To describe the causes of genetic mutations and how they lead to genetic diseases. We have seen that the sequence of nucleotides in a cell’s deoxyribonucleic acid (DNA) is what ultimately determines the sequence of amino acids in proteins made by the cell and thus is critical for the proper functioning of the cell. On rare occasions, however, the nucleotide sequence in DNA may be modified either spontaneously (by errors during replication, occurring approximately once for every 10 billion nucleotides) or from exposure to heat, radiation, or certain chemicals. Any chemical or physical change that alters the nucleotide sequence in DNA is called a mutation. When a mutation occurs in an egg or sperm cell that then produces a living organism, it will be inherited by all the offspring of that organism. Common types of mutations include substitution (a different nucleotide is substituted), insertion (the addition of a new nucleotide), and deletion (the loss of a nucleotide). These changes within DNA are called point mutations because only one nucleotide is substituted, added, or deleted (Figure \(\PageIndex{1}\)). Because an insertion or deletion results in a frame-shift that changes the reading of subsequent codons and, therefore, alters the entire amino acid sequence that follows the mutation, insertions and deletions are usually more harmful than a substitution in which only a single amino acid is altered. The chemical or physical agents that cause mutations are called mutagens. Examples of physical mutagens are ultraviolet (UV) and gamma radiation. Radiation exerts its mutagenic effect either directly or by creating free radicals that in turn have mutagenic effects. Radiation and free radicals can lead to the formation of bonds between nitrogenous bases in DNA. For example, exposure to UV light can result in the formation of a covalent bond between two adjacent thymines on a DNA strand, producing a thymine dimer (Figure \(\PageIndex{2}\)). If not repaired, the dimer prevents the formation of the double helix at the point where it occurs. The genetic disease xeroderma pigmentosum is caused by a lack of the enzyme that cuts out the thymine dimers in damaged DNA. Individuals affected by this condition are abnormally sensitive to light and are more prone to skin cancer than normal individuals. Sometimes gene mutations are beneficial, but most of them are detrimental. For example, if a point mutation occurs at a crucial position in a DNA sequence, the affected protein will lack biological activity, perhaps resulting in the death of a cell. In such cases the altered DNA sequence is lost and will not be copied into daughter cells. Nonlethal mutations in an egg or sperm cell may lead to metabolic abnormalities or hereditary diseases. Such diseases are called inborn errors of metabolism or genetic diseases. A partial listing of genetic diseases is presented in Figure \(\PageIndex{1}\), and two specific diseases are discussed in the following sections. In most cases, the defective gene results in a failure to synthesize a particular enzyme. Disease Responsible Protein or Enzyme alkaptonuria homogentisic acid oxidase galactosemia galactose 1-phosphate uridyl transferase, galactokinase, or UDP galactose epimerase Gaucher disease glucocerebrosidase gout and Lesch-Nyhan syndrome hypoxanthine-guanine phosphoribosyl transferase hemophilia antihemophilic factor (factor VIII) or Christmas factor (factor IX) homocystinuria cystathionine synthetase maple syrup urine disease branched chain α-keto acid dehydrogenase complex McArdle syndrome muscle phosphorylase Niemann-Pick disease sphingomyelinase phenylketonuria (PKU) phenylalanine hydroxylase sickle cell anemia hemoglobin Tay-Sachs disease hexosaminidase A tyrosinemia fumarylacetoacetate hydrolase or tyrosine aminotransferase von Gierke disease glucose 6-phosphatase Wilson disease Wilson disease protein PKU results from the absence of the enzyme phenylalanine hydroxylase. Without this enzyme, a person cannot convert phenylalanine to tyrosine, which is the precursor of the neurotransmitters dopamine and norepinephrine as well as the skin pigment melanin. When this reaction cannot occur, phenylalanine accumulates and is then converted to higher than normal quantities of phenylpyruvate. The disease acquired its name from the high levels of phenylpyruvate (a phenyl ketone) in urine. Excessive amounts of phenylpyruvate impair normal brain development, which causes severe mental retardation. PKU may be diagnosed by assaying a sample of blood or urine for phenylalanine or one of its metabolites. Medical authorities recommend testing every newborn’s blood for phenylalanine within 24 h to 3 weeks after birth. If the condition is detected, mental retardation can be prevented by immediately placing the infant on a diet containing little or no phenylalanine. Because phenylalanine is plentiful in naturally produced proteins, the low-phenylalanine diet depends on a synthetic protein substitute plus very small measured amounts of naturally produced foods. Before dietary treatment was introduced in the early 1960s, severe mental retardation was a common outcome for children with PKU. Prior to the 1960s, 85% of patients with PKU had an intelligence quotient (IQ) less than 40, and 37% had IQ scores below 10. Since the introduction of dietary treatments, however, over 95% of children with PKU have developed normal or near-normal intelligence. The incidence of PKU in newborns is about 1 in 12,000 in North America. Every state in the United States has mandated that screening for PKU be provided to all newborns. Several genetic diseases are collectively categorized as lipid-storage diseases . Lipids are constantly being synthesized and broken down in the body, so if the enzymes that catalyze lipid degradation are missing, the lipids tend to accumulate and cause a variety of medical problems. When a genetic mutation occurs in the gene for the enzyme hexosaminidase A, for example, gangliosides cannot be degraded but accumulate in brain tissue, causing the ganglion cells of the brain to become greatly enlarged and nonfunctional. This genetic disease, known as Tay-Sachs disease , leads to a regression in development, dementia, paralysis, and blindness, with death usually occurring before the age of three. There is currently no treatment, but Tay-Sachs disease can be diagnosed in a fetus by assaying the amniotic fluid (amniocentesis) for hexosaminidase A. A blood test can identify Tay-Sachs carriers—people who inherit a defective gene from only one rather than both parents—because they produce only half the normal amount of hexosaminidase A, although they do not exhibit symptoms of the disease. Looking Closer: Recombinant DNA Technology More than 3,000 human diseases have been shown to have a genetic component, caused or in some way modulated by the person’s genetic composition. Moreover, in the last decade or so, researchers have succeeded in identifying many of the genes and even mutations that are responsible for specific genetic diseases. Now scientists have found ways of identifying and isolating genes that have specific biological functions and placing those genes in another organism, such as a bacterium, which can be easily grown in culture. With these techniques, known as recombinant DNA technology , the ability to cure many serious genetic diseases appears to be within our grasp. Isolating the specific gene or genes that cause a particular genetic disease is a monumental task. One reason for the difficulty is the enormous amount of a cell’s DNA, only a minute portion of which contains the gene sequence. Thus, the first task is to obtain smaller pieces of DNA that can be more easily handled. Fortunately, researchers are able to use restriction enzymes (also known as restriction endonucleases), discovered in 1970, which are enzymes that cut DNA at specific, known nucleotide sequences, yielding DNA fragments of shorter length. For example, the restriction enzyme EcoRI recognizes the nucleotide sequence shown here and cuts both DNA strands as indicated: Once a DNA strand has been fragmented, it must be cloned; that is, multiple identical copies of each DNA fragment are produced to make sure there are sufficient amounts of each to detect and manipulate in the laboratory. Cloning is accomplished by inserting the individual DNA fragments into phages (bacterial viruses) that can enter bacterial cells and be replicated. When a bacterial cell infected by the modified phage is placed in an appropriate culture medium, it forms a colony of cells, all containing copies of the original DNA fragment. This technique is used to produce many bacterial colonies, each containing a different DNA fragment. The result is a DNA library , a collection of bacterial colonies that together contain the entire genome of a particular organism. The next task is to screen the DNA library to determine which bacterial colony (or colonies) has incorporated the DNA fragment containing the desired gene. A short piece of DNA, known as a hybridization probe , which has a nucleotide sequence complementary to a known sequence in the gene, is synthesized, and a radioactive phosphate group is added to it as a “tag.” You might be wondering how researchers are able to prepare such a probe if the gene has not yet been isolated. One way is to use a segment of the desired gene isolated from another organism. An alternative method depends on knowing all or part of the amino acid sequence of the protein produced by the gene of interest: the amino acid sequence is used to produce an approximate genetic code for the gene, and this nucleotide sequence is then produced synthetically. (The amino acid sequence used is carefully chosen to include, if possible, many amino acids such as methionine and tryptophan, which have only a single codon each.) After a probe identifies a colony containing the desired gene, the DNA fragment is clipped out, again using restriction enzymes, and spliced into another replicating entity, usually a plasmid. Plasmids are tiny mini-chromosomes found in many bacteria, such as Escherichia coli ( E. coli ). A recombined plasmid would then be inserted into the host organism (usually the bacterium E. coli ), where it would go to work to produce the desired protein. Proponents of recombinant DNA research are excited about its great potential benefits. An example is the production of human growth hormone, which is used to treat children who fail to grow properly. Formerly, human growth hormone was available only in tiny amounts obtained from cadavers. Now it is readily available through recombinant DNA technology. Another gene that has been cloned is the gene for epidermal growth factor, which stimulates the growth of skin cells and can be used to speed the healing of burns and other skin wounds. Recombinant techniques are also a powerful research tool, providing enormous aid to scientists as they map and sequence genes and determine the functions of different segments of an organism’s DNA. In addition to advancements in the ongoing treatment of genetic diseases, recombinant DNA technology may actually lead to cures. When appropriate genes are successfully inserted into E. coli , the bacteria can become miniature pharmaceutical factories, producing great quantities of insulin for people with diabetes, clotting factor for people with hemophilia, missing enzymes, hormones, vitamins, antibodies, vaccines, and so on. Recent accomplishments include the production in E. coli of recombinant DNA molecules containing synthetic genes for tissue plasminogen activator, a clot-dissolving enzyme that can rescue heart attack victims, as well as the production of vaccines against hepatitis B (humans) and hoof-and-mouth disease (cattle). Scientists have used other bacteria besides E. coli in gene-splicing experiments and also yeast and fungi. Plant molecular biologists use a bacterial plasmid to introduce genes for several foreign proteins (including animal proteins) into plants. The bacterium is Agrobacterium tumefaciens , which can cause tumors in many plants, but which can be treated so that its tumor-causing ability is eliminated. One practical application of its plasmids would be to enhance a plant’s nutritional value by transferring into it the gene necessary for the synthesis of an amino acid in which the plant is normally deficient (for example, transferring the gene for methionine synthesis into pinto beans, which normally do not synthesize high levels of methionine). Restriction enzymes have been isolated from a number of bacteria and are named after the bacterium of origin. EcoRI is a restriction enzyme obtained from the R strain of E. coli . The roman numeral I indicates that it was the first restriction enzyme obtained from this strain of bacteria. Summary The nucleotide sequence in DNA may be modified either spontaneously or from exposure to heat, radiation, or certain chemicals and can lead to mutations. Mutagens are the chemical or physical agents that cause mutations. Genetic diseases are hereditary diseases that occur because of a mutation in a critical gene. |
Courses/SUNY_Oneonta/Chem_221%3A_Organic_Chemistry_I_(Bennett)/1%3ALecture_Textbook/01%3A_Introduction_to_Organic_Structure_and_Bonding_I/1.06%3A_Problems_for_Chapter_1 | Solutions to selected Chapter 1 problems P1.1: The figure below illustrates a section of an intermediate compound that forms during the protein synthesis process in the cell. Lone pairs are not shown, as is typical in drawings of organic compounds. a) The structure as drawn is incomplete, because it is missing formal charges - fill them in. b) How many hydrogen atoms are on this structure? c) Identify the two important biomolecule classes (covered in section 1.3) in the structure. P1.2 : Find, in Table 6 ('Structures of common coenzymes', in the tables section at the back of this book), examples of the following: a) a thiol b) an amide c) a secondary alcohol d) an aldehyde e) a methyl substituent on a ring f) a primary ammonium ion g) a phosphate anhydride h) a phosphate ester P1.3 : Draw line structures corresponding to the following compounds. Show all lone pair electrons (and don't forget that non-zero formal charges are part of a correctly drawn structure!) a) 2,2,4-trimethylpentane b) 3-phenyl-2-propenal c) 6-methyl-2,5-cyclohexadienone d) 3-methylbutanenitrile e) 2,6-dimethyldecane f) 2,2,5,5-tetramethyl-3-hexanol g) methyl butanoate h) N -ethylhexanamide i) 7-fluoroheptanoate j) 1-ethyl-3,3-dimethylcyclohexene P1.4: Reaction A below is part of the biosynthetic pathway for the amino acid methionine, and reaction B is part of the pentose phosphate pathway of sugar metabolism. a) What is the functional group transformation that is taking place in each reaction? b) Keeping in mind that the 'R' abbreviation is often used to denote parts of a larger molecule which are not the focus of a particular process, which of the following abbreviated structures could be appropriate to use for aspartate semialdehyde when drawing out details of reaction A? c) Again using the 'R' convention, suggest an appropriate abbreviation for the reactant in reaction B. P1.5 : Find, in the table of amino acid structures (Table 5), examples of the following: a) a secondary alcohol b) an amide c) a thiol d) a sulfide e) a phenol f) a side chain primary ammonium g) a side chain carboxylate h) a secondary amine P1.6: Draw correct Lewis structures for ozone (O 3 ), azide ion, (N 3 - ), and bicarbonate ion, HCO 3 - . Include lone pair electrons and formal charges, and use your General Chemistry textbook to review VSEPR theory, which will enable you to draw correct bond geometries. P1.7: Draw one example each of compounds fitting the descriptions below, using line structures. Be sure to include all non-zero formal charges. All atoms should fit one of the common bonding patters discussed in this chapter. There are many possible correct answers - be sure to check your drawings with your instructor or tutor. a) an 8-carbon molecule with secondary alcohol, primary amine, amide, and cis -alkene groups b) a 12-carbon molecule with carboxylate, diphosphate, and lactone (cyclic ester) groups. c) a 9-carbon molecule with cyclopentane, alkene, ether, and aldehyde groups P1.8: Three of the four structures below are missing formal charges. a) Fill in all missing formal charges (assume all atoms have a complete octet of valence electrons). b) Identify the following functional groups or structural elements (there may be more than one of each): carboxylate, carboxylic acid, cyclopropyl, amide, ketone, secondary ammonium ion, tertiary alcohol. c) Determine the number of hydrogen atoms in each compound. P1.9: a) Draw four constitutional isomers with the molecular formula C 4 H 8 . ( b) Draw two open-chain (non-cyclic) constitutional isomers of cyclohexanol (there are more than two possible answers). P1.10: Draw structures of four different amides with molecular formula C 3 H 7 NO. Solutions to selected Chapter 1 problems Organic Chemistry With a Biological Emphasis by Tim Soderberg (University of Minnesota, Morris) |
Courses/Centre_College/CHE_332%3A_Inorganic_Chemistry/11%3A_Organometallic_Reactions_and_Catalysis | Introduction This chapter is a survey of common organometallic reactions and catalytic cycles. The reactions of organometallic complexes are divided into two main categories: (1) Reactions that involve gain or loss of ligands, and (2) Reactions involving modification of ligands. Section 11.1: Reactions Involving Gain or Loss of Ligands Section 11.1.1: Ligand Dissociation and Substitution Section 11.1.2: Oxidative Addition Section 11.1.3: Reductive Elimination Section 11.1.4: Sigma Bond Metathasis Section 11.1.5: Application of Pincer Ligands Section 11.2: Reactions Invloving Modification of Unsaturated Ligands Section 11.2.1: Introduction to Insertion Section 11.2.2: CO Insertions (Alkyl Migration) Section 11.2.3: Migratory Insertion-1,2-Insertions Section 11.2.4: β-Elimination Reactions Section 11.2.5: Abstraction and Addition Section 11.3: Organometallic Catalysts Section 11.3.1: Catalytic Deuteration Section 11.3.2: Hydroformylation Section 11.3.3: Monsanto Acetic Acid Process Section 11.3.4: Wacker (Smidt) Process Section 11.3.5: Hydrogenation by Wilkinson's Catalyst Section 11.3.6: Olefin Metathesis Section 11.4: Heterogeneous Catalysts Section 11.4.1: Ziegler-Natta Polymerizations Section 11.4.2: Water-Gas Shift Reaction Section 11.5: Problems Section 11.5.1: Concept Review Questions Chapter 12 Section 11.5.2: Homework Problems Chapter 12 |
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