url
stringlengths 90
342
| html
stringlengths 602
98.8k
| text_length
int64 602
98.8k
| __index_level_0__
int64 0
5.02k
|
|---|---|---|---|
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/01%3A_Introduction_to_Organic_Chemistry/1.02%3A_A_Bit_of_History |
You may not be much interested in the way that organic chemistry developed, but if you skip to the next section without reading further, you will miss some of the flavor of a truly great achievement - of how a few highly creative chemists were able, with the aid of a few simple tools, to determine the structures of molecules, far too small and too elusive to be seen individually with the finest optical microscope, manifesting themselves only by the collective behavior of at least millions of millions at once. Try to visualize the problems confronting the organic chemist of 100 years ago. You will have no more than reasonably pure samples of organic compounds, the common laboratory chemicals of today, glassware, balances, thermometers, means of measuring densities, and a few optical instruments. You also will have a relatively embryonic theory that there are molecules in those bottles and that one compound differs from another because its molecules have different members or kinds of atoms and different arrangements of bonds. Your task will be to determine what kinds and what numbers of atoms they contain, that is, to determine their . Obviously, a compound with formula \(C_2H_6O\) and one with \(C_2H_6O_2\) are not the same compound. But suppose two compounds from different sources both are \(C_2H_6O\). To decide whether these are the you could smell them (far better to than to inhale), taste them (emphatically not recommended), see if they have the same appearance and viscosity (if liquids), or use more sophisticated criteria: boiling point, melting point, density, or refractive index. Other possibilities would be to see if they both have the same solubility in water or other solvents and whether they give the same reaction products with various reagents. Of course, all this gets a bit tough when the compounds are not pure and no good ways are available to purify them, but that is part of the job. Think about how you might proceed. In retrospect it is surprising that in less than fifty years an enormous, even if incomplete, edifice of structural organic chemistry was constructed on the basis of the results of chemical reactions without determination of a single bond distance, and with no electronic theory as a guide. Interestingly, all of the subsequent developments of the quantum mechanical theory of chemical bonds has not altered this edifice in significant ways. Indeed, for a long time, a goal of molecular quantum mechanics was simply to be able to corroborate that when an organic chemist draws a single line between two carbon atoms to show that they are bonded, he in fact knows what he is doing. And that when he draws two (or three) bonds between the carbons to indicate a double (or triple) bond, quantum mechanics supports this also as a valid idea. Furthermore, when modern tools for determining organic structures that involve actually measuring the distances between the atoms became available, these provided great convenience, but no great surprises. To be sure, a few structures turned out to be incorrect because they were based on faulty or inadequate experimental evidence. But, on the whole, the modern three-dimensional representations of molecules that accord with actual measurements of bond distances and angles are in no important respect different from the widely used three-dimensional ball-and-stick models of organic molecules, and these, in essentially their present form, date from at least as far back as E. Paterno, in 1869. How was all of this achieved? Not by any very simple process. The essence of some of the important ideas follow, but it should be clear that what actually took place was far from straightforward. A diverse group of people was involved; many firmly committed to, if not having a vested interest in, earlier working hypotheses or that had served as useful bases for earlier experimentation, but were coming apart at the seams because they could not accommodate the new facts that kept emerging. As is usual in human endeavors, espousal of new and better ideas did not come equally quickly to all those used to thinking in particular ways. To illustrate, at least one famous chemist, Berthelot, still used \(HO\) as the formula for water twenty-five years after it seemed clear that \(H_2O\) was a better choice. Before structures of molecules could be established, there had to be a means of establishing molecular formulas and for this purpose the key concept was Avogadro's hypothesis, which can be stated in the form "equal volumes of gases at the same temperature and pressure contain the same number of molecules." Avogadro's hypothesis allowed assignment of molecular weights from measurements of gas densities. Then, with analytical techniques that permit determination of the weight percentages of the various elements in a compound, it became possible to set up a self-consistent set of relative atomic weights.\(^1\) From these and the relative molecular weights, one can assign molecular formulas. For example, if one finds that a compound contains \(22.0 \%\) carbon (atomic weight \(= 12.00\)), \(4.6 \%\) hydrogen (atomic weight \(= 1.008\)), and \(73.4 \%\) bromine (atomic weight \(= 79.90\)), then the ratios of the numbers of atoms are \(\left( 22.0/12.00 \right) : \left( 4.6/1.008 \right) : \left( 73.4/79.90 \right) = 1.83:4.56:0.92\). Dividing each of the last set of numbers by the smallest (\(0.92\)) gives \(1.99:4.96:1 \cong 2:5:1\), which suggests a molecular formula of \(C_2H_5Br\) or a multiple thereof. If we know that hydrogen gas is \(H_2\) and has a molecular weight of \(2 \times 1.008 = 2.016\), we can compare the weight of a given volume of hydrogen with the weight of the same volume of our unknown in the gas phase at the same temperature and pressure. If the experimental ratio of these weights turns out to be \(54\), then the molecular weight of the unknown would be \(2.016 \times 54 = 109\) and the formula \(C_2H_5Br\) would be correct. If we assume that the molecule is held together by chemical bonds, without knowing more, we could write numerous structures such as \(H-H-H-H-H-C-C-Br, H-C-Br-H-H-C-H-H\), and so on. However, if we also know of the existence of stable \(H_2\), but not \(H_3\); of stable \(Br_2\), but not of \(Br_3\); and of stable \(CH_3Br\), \(CH_2Br_2\), \(CHBr_3\), and \(CBr_4\), but not of \(CH_4Br\), \(CHBr\), \(CBr\), and so on, a pattern of what is called emerges. It will be seen that the above formulas all are consistent if hydrogen atoms and bromine atoms form just bond (are univalent) while carbon atoms form bonds (are tetravalent). This may seem almost naively simple today, but a considerable period of doubt and uncertainty preceded the acceptance of the idea of definite valences for the elements that emerged about 1852. If we accept hydrogen and bromine as being univalent and carbon as tetravalent, we can write as a structural formula for \(C_2H_5Br\).\(^2\) However, we also might have written There is a serious problem as to whether these formulas represent the or compounds. All that was known in the early days was that every purified sample of \(C_2H_5Br\), no matter how prepared, had a boiling point of \(38^\text{o}C\) and density of \(1.460 \: \text{g} \: \text{ml}^{-1}\). Furthermore, all looked the same, all smelled the same, and all underwent the same chemical reactions. There was no evidence that \(C_2H_5Br\) was a mixture or that more than one compound of this formula could be prepared. One might conclude, therefore, that all of the structural formulas above represent a single substance even though they superficially, at least, look different. Indeed, because \(H-Br\) and \(Br-H\) are two different ways of a formula for the same substance, we suspect that the same is true for There are, though, two of these structures that could be different from one another, namely In the first of these, \(CH_3-\) is located opposite the \(Br-\) and the \(H-\)'s on the carbon with the \(Br\) also are opposite one another. In the second formula, \(CH_3-\) and \(Br-\) are located to each other as are the \(H-\)'s on the same carbon. We therefore have a problem as to whether these two different formulas also represent different compounds. A brilliant solution to the problem posed in the preceding section came in 1874 when J. H. van't Hoff proposed that all four valences of carbon are equivalent and directed to the corners of a regular tetrahedron.\(^3\) If we redraw the structures for \(C_2H_5Br\) as \(1\), we see that there is only possible arrangement and, contrary to the impression we got from our earlier structural formulas, the bromine is located with respect to each of the hydrogens on the same carbon. A convenient way of representing organic molecules in three dimensions, which shows the tetrahedral relationships of the atoms very clearly, uses the so-called ball-and-stick models. The sticks that represent the bonds or valences form the tetrahedral angles of \(109.47^\text{o}\). The tetrahedral carbon does not solve all problems without additional postulates. For example, there are two different compounds known with the formula \(C_2H_4Br_2\). These substances, which we call , can be reasonably written as However, ball-and-stick models suggest further possibilities for the second structure, for example \(3\), \(4\), and \(5\): This is a problem apparently first clearly recognized by Paterno, in 1869. We call these rotational (or conformational) isomers, because one is converted to another by rotation of the halves of the molecule with respect to one another, with the \(C-C\) bond acting as an axle. If this is not clear, you should make a ball-and-stick model and see what rotation around the \(C-C\) bond does to the relationships between the atoms on the carbons. The difficulty presented by these possibilities finally was circumvented by a brilliant suggestion by van't Hoff of "free rotation," which holds that isomers corresponding to different rotational angles, such as \(3\), \(4\), and \(5\), do not have separate stable existence, but are interconverted by rotation around the \(C-C\) bond so rapidly that they are indistinguishable from one another. Thus there is only isomer corresponding to the different possible rotational angles and a total of only isomers of formula \(C_2H_4Br_2\). As we shall see, the idea of free rotation required extensive modification some 50 years after it was first proposed, but it was an extremely important paradigm, which, as often happens, became so deeply rooted as to become essentially an article of faith for later organic chemists. Free rotation will be discussed in more detail in Chapters 5 and 27. The problem of determining whether a particular isomer of \(C_2H_4Br_2\) is could be solved today in a few minutes by spectroscopic means, as will be explained in Chapter 9. However, at the time structure theory was being developed, the structure had to be deduced on the basis of chemical reactions, which could include either how the compound was formed or what it could be converted to. A virtually unassailable proof of structure, where it is applicable, is to determine how many different products each of a given group of isomers can give. For the \(C_2H_4Br_2\) pair of isomers, will be seen to give only possibility with one compound and with the other: Therefore, if we have two bottles, one containing one \(C_2H_4Br_2\) isomer and one the other and run the substitution test, the compound that gives only one product is \(6\) and the one that gives a mixture of two products is \(7\). Further, it will be seen that the test, besides telling which isomer is \(6\) and which is \(7\), establishes the structures of the two possible \(C_2H_3Br_3\) isomers, \(8\) and \(9\). Thus only \(8\) can be formed from both of the different \(C_2H_4Br_2\) isomers whereas \(9\) is formed from only one of them. There were already many interconversion reactions of organic compounds known at the time that valence theory, structural formulas, and the concept of the tetrahedral carbon came into general use. As a result, it did not take long before much of organic chemistry could be fitted into a concordant whole. One difficult problem was posed by the structures of a group of substitution products of benzene, \(C_6H_6\), called "aromatic compounds," which for a long time defied explanation. Benzene itself had been prepared first by Michael Faraday, in 1825. An ingenious solution for the benzene structure was provided by A. Kekule, in 1866, wherein he suggested (apparently as the result of a hallucinatory perception) that the six carbons were connected in a hexagonal ring with alternating single and double carbon-to-carbon bonds, and with each carbon connected to a single hydrogen, \(10\): This concept was controversial, to say the least, mainly on two counts. Benzene did not behave as expected, as judged by the behavior of other compounds with carbon-to-carbon double bonds and also because there should be two different dibromo substitution products of benzene with the bromine on adjacent carbons (\(11\) and \(12\)) but only one such compound could be isolated. Kekule explained the second objection away by maintaining that \(11\) and \(12\) were in rapid equilibrium through concerted bond shifts, in something like the same manner as the free-rotation hypothesis mentioned previously: However, the first objection could not be dismissed so easily and quite a number of alternative structures were proposed over the ensuing years. The controversy was not really resolved until it was established that benzene is a regular planar hexagon, which means that all of its \(C-C\) bonds have the same length, in best accord with a structure written not with double, not with single, but with 1.5 bonds between the carbons, as in \(13\): This. in turn, generated a massive further theoretical controversy over just how \(13\) should be interpreted, which, for a time, even became a part of "Cold-War" politics!\(^4\) We shall examine experimental and theoretical aspects of the benzene structure in some detail later. It is interesting that more than 100 years after Kekule's proposal the final story on the benzene structure is yet to be told.\(^5\) The combination of valence theory and the substitution method as described in gives, for many compounds, quite unequivocal proofs of structure. Use of chemical transformations for proofs of structure depends on the applicability of a simple guiding principle, often called the " ." As we shall see later, many exceptions are known and care is required to keep from making serious errors. With this caution, let us see how the principle may be applied. The compound \(C_2H_5Br\) discussed in reacts slowly with water to give a product of formula \(C_2H_6O\). The normal valence of oxygen is two, and we can write two, and only two, different structures, \(19\) and \(20\), for \(C_2H_6O\): The principle of least structural change favors \(19\) as the product, because the reaction to form it is a simple replacement of bromine bonded to carbon by \(-OH\), whereas formation of \(20\) would entail a much more drastic rearrangement of bonds. The argument is really a subtle one, involving an assessment of the reasonableness of various possible reactions. On the whole, however, it works rather well and, in the specific case of the \(C_2H_6O\) isomers, is strongly supported by the fact that treatment of \(19\) with strong hydrobromic acid (\(HBr\)) converts it back to \(C_2H_5Br\). In contrast, the isomer of structure \(20\) reacts with \(HBr\) to form two molecules of \(CH_3Br\): In each case, \(C-O\) bonds are broken and \(C-Br\) bonds are formed. We could conceive of many other possible reactions of \(C_2H_6O\) with \(HBr\), for example which, as indicated by \(\nrightarrow\), does occur, but hardly can be ruled out by the principle of least structural change itself. Showing how the probability of such alternative reactions can be evaluated will be a very large part of our later discussions. The substitution method and the interconversion reactions discussed for proof of structure possibly may give you erroneous ideas about the reactions and reactivity of organic compounds. We certainly do not wish to imply that it is a simple, straightforward process to make all of the possible substitution products of a compound such as In fact, as will be shown later, direct substitution of bromine for hydrogen with compounds such as this does not occur readily, and when it does occur, the four possible substitution products indeed are formed, but in far from equal amounts because there are for substitution at the different positions. Actually, some of the substitution products are formed only in very small quantities. Fortunately, this does not destroy the validity of the substitution method but does make it more difficult to apply. If direct substitution fails, some (or all) of the possible substitution products may have to be produced by indirect means. Nonetheless, you must understand that the success of the substitution method depends on determination of the total number of possible isomers - it does depend on how the isomers are prepared. Later, you will hear a lot about compounds or reagents being "reactive" and "unreactive." You may be exasperated by the loose way that these terms are used by organic chemists to characterize how fast various chemical changes occur. Many familiar inorganic reactions, such as the neutralization of hydrochloric acid with sodium hydroxide solution, are extremely fast at ordinary temperatures. But the same is not often true of reactions of organic compounds. For example, \(C_2H_5Br\) treated in two different ways is converted to gaseous compounds, one having the formula \(C_2H_6\) and the other \(C_2H_4\). The \(C_2H_4\) compound, , reacts with bromine to give \(C_2H_4Br_2\), but the \(C_2H_6\) compound, , does not react with bromine except at high temperatures or when exposed to sunlight (or similar intense light). The reaction products then are \(HBr\) and \(C_2H_5Br\), and later, \(HBr\) and \(C_2H_4Br_2\), \(C_2H_3Br_3\), and so on. We clearly can characterize \(C_2H_4\) as "reactive" and \(C_2H_6\) as "unreactive" toward bromine. The early organic chemists also used the terms "unsaturated" and "saturated" for this behavior, and these terms are still in wide use today. But we need to distinguish between "unsaturated" and "reactive," and between "saturated" and "unreactive," because these pairs of terms are not synonymous. The equations for the reactions of ethene and ethane with bromine are different in that ethene bromine, \(C_2H_4 + Br_2 \rightarrow C_2H_4Br_2\), whereas ethane bromine, \(C_2H_6 + Br_2 \rightarrow C_2H_5Br + HBr\). You should reserve the term "unsaturated" for compounds that can, at least potentially, react by , and "saturated' for compounds that can only be expected to react by . The difference between addition and substitution became much clearer with the development of the structure theory that called for carbon to be tetravalent and hydrogen univalent. Ethene then was assigned a structure with a carbon-to-carbon bond, and ethane a structure with a carbon-to-carbon bond: Addition of bromine to ethene subsequently was formulated as breaking one of the carbon-carbon bonds of the double bond and attaching bromine to these valences. Substitution was written similarly but here bromine and a \(C-H\) bond are involved: We will see later that the way in which these reactions actually occur is much more complicated than these simple equations indicate. In fact, such equations are regarded best as chemical accounting operations. The number of bonds is shown correctly for both the reactants and the products, and there is an indication of which bonds break and which bonds are formed in the overall process. However, do not make the mistake of assuming that no other bonds are broken or made in intermediate stages of the reaction. Much of what comes later in this book will be concerned with what we know, or can find out, about the of such reactions - a reaction mechanism being the actual sequence of events by which the reactants become converted to the products. Such information is of extraordinary value in defining and understanding the range of applicability of given reactions for practical preparations of desired compounds. The distinction we have made between "unsaturated" and "reactive" is best illustrated by a definite example. Ethene is "unsaturated" (and "reactive") toward bromine, but tetrachloroethene, \(C_2Cl_4\), will not add bromine at all under the same conditions and is clearly "unreactive." But is it also "saturated"? The answer is definitely no, because if we add a small amount of aluminum bromide, \(AlBr_3\), to a mixture of tetrachloroethene and bromine, addition does occur, although sluggishly: Obviously, tetrachloroethene is "unsaturated" in the sense it can undergo addition, even if it is unreactive to bromine in the absence of aluminum bromide. The aluminum bromide functions in the addition of bromine to tetrachloroethene as a , which is something that facilitates the conversion of reactants to products. The study of the nature and uses of catalysts will concern us throughout this book. Catalysis is our principal means of controlling organic reactions to help form the product we want in the shortest possible time. \(^1\)We will finesse here the long and important struggle of getting a truly self-consistent table of atomic weights. If you are interested in the complex history of this problem and the clear solution to it proposed by S. Cannizzaro in 1860, there are many accounts available in books on the history of chemistry. One example is J. R. Partington, , Vol. IV, Macmillan, London, 1964. Relative atomic weights now are based on \(^{12}C = 12\) (exactly). \(^2\)Formulas such as this appear to have been used first by Crum Brown, in 1864, after the originators of structural formulas, A. Kekule and A. Couper (1858), came up with rather awkward, impractical representations. It seems incredible today that even the drawing of these formulas was severely criticized for many years. The pot was kept boiling mainly by H. Kolbe, a productive German chemist with a gift for colorful invective and the advantage of a podium provided by being editor of an influential chemical journal. \(^3\)The name of J. A. Le Bel also is associated with this particular idea, but the record shows that Le Bel actually opposed the tetrahedral formulations, although, simultaneously with van't Hoff, he made a related very important contribution, as will be discussed in Chapter 5. \(^4\)The "resonance theory," to be discussed in detail in Chapters 6 and 21, was characterized in 1949 as a physically and ideologically inadmissible theory formulated by "decadent bourgeois scientists." See L. R. Graham, , Vintage Books, New York, 1973, Chapter VIII, for an interesting account of this controversy. \(^5\)Modern organic chemistry should not be regarded at all as a settled science, free of controversy. To be sure, personal attacks of the kind indulged in by Kolbe and others often are not published, but profound and indeed acrimonious differences of scientific interpretation exist and can persist for many years. and (1977) | 23,442 | 1,554 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Reactions_of_Main_Group_Elements_with_Carbonates |
Carbonates ions are formed by the reaction of carbonic acid with metals or organic compounds. In this case, the reaction of carbonates with the main group elements and its products, salts, is explained. In general, the most common main group elements used as carbonates are the Alkali and Alkaline metals. All main group carbonates, except Na, K, Rb and Cs are unstable to heat and insoluble in water. Usually all carbonates are soluble in acid, because of the formation of bicarbonate ion. Its regular physical appearance is of a white powder. The main uses of carbonates is as raw materials in different industrial processes such as drug development, glass making, pulp and paper industry, sodium chemicals (silicates), soap and detergent production, paper industry, water softener, clay and concrete production, among others. Other carbonates such as Beryllium Carbonate (BeCO ) and Tallium Carbonate (Tl CO ) are consider toxic and are used in fungicides and poison manufacture. From main group elements Sodium Carbonate (\(Na_2CO_3\)) and Calcium Carbonate (\(CaCO_3\)) are the most used. , known as soda ash, is a very important industrial chemical. It is mainly obtained by a method named by the chemical reaction of limestone (CaCO3) and sodium chloride (NaCl). 2 NaCl + CaCO → Na CO + CaCl Na CO common uses are in glass making, pulp and paper industry, sodium chemicals (silicates), soap and detergent production, paper industry and water softener. is the principal constituent of limestone (a sedimentary rock) and its pure state is obtained in three steps by the calcination of limestone and subsequent reaction with water and carbon dioxide. Ca(OH) (s) + CO (aq) → CaCO (s) + H O(l) Calcium Carbonate common uses are in glass, textile, paint, paper and plastic production, caulks industry, to produce ink and sealant. It is also used as a food additive (non toxic), as a drug development and chalk production. Hard water is the term used in relation with high amount of inorganic compounds such as carbonates, bicarbonates, sulfates or chlorides in water. The presence of high levels of carbonates and bicarbonates in water is denominated as temporary hardness of water. It is considered hard water when this inorganic compound exceed the 100 mg/L approximated, and it is usually expressed as the amount of calcium carbonate present in water. This is important when water is used for industrial process or cleaning purposes. The way precipitation is formed is when inorganic compounds in presence of high molecular organic compounds (such as soaps) produce undesirable insoluble precipitation. These precipitations are the responsible of a dirty or grayish white and low efficiency in cleaners. Other way to form precipitation is when hard water attains high temperatures; its inorganic compounds such as calcium carbonate precipitates leaving a deposited coating in water pipes or over boilers. Here is an example of what happens: Ca (aq)+2HCO (aq) → CaCO (s) + H O + CO One way to reduce temporary hardness such as calcium ions is by boiling or by the addition of calcium hydroxide (lime), but is not often used. The best way of soften water is by the addition of Soda Ash to the water or using an ion exchanged column. Some carbonates such as sodium carbonate or potassium carbonates are used to elaborate some detergents. Soda Ash (other name for sodium carbonate) is used since in one way it is cheap and also it helps soften water by precipitation of calcium and magnesium carbonates. Potassium carbonates are used because of its solubility. Some inorganic compounds such as calcium carbonate, sodium carbonate and potassium carbonate are used as unprocessed material to elaborate glass since their chemical characteristics, high quality and low prices. Calcium carbonates and sodium carbonates are used as raw material in the production of paper due to their low cost by using them instead of pulp and to improve the white and gloss of the paper. is the reaction between Li, Na, K, Rb and Cs with CO . All except Lithium are soluble in water and stable to heat. Is the reaction between Be, Mg, Ca, Sr and Ba with CO3. All are insoluble in water and unstable to heat. Is the reaction between Al and Tl with CO . Both are insoluble in water and unstable to heat. Is the reaction between Pb with CO . It is insoluble in water and unstable to heat Calcium Carbonate and Sodium Carbonate All main group carbonates, except Na, K, Rb and Cs are insoluble in water. Most communly Calcium Carbonate Sodium Carbonate ( Soda Ash) | 4,568 | 1,555 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/05%3A_Structure_Determination_Part_II_-_Nuclear_Magnetic_Resonance_Spectroscopy |
In the previous chapter, we learned about three important analytical techniques which allow us to deduce information about the structure of an organic molecule. In IR spectroscopy, transitions in the vibrational states of covalent bonds lead to the absorbance of specific infrared frequencies, telling us about the presence or absence of functional groups in the molecule of interest. In UV-Vis spectroscopy, transitions in the energy levels of electrons in pi bonds lead to the absorbance of ultraviolet and visible light, providing us with information about the arrangement of double bonds in a molecule. And in mass spectrometry, we are usually able to learn the molecular weight of a sample molecule, in addition to other kinds of information from analysis of the masses of molecular fragments. Although all three of these techniques provide us with valuable information about a molecule of interest, in most cases they do not – even in combination – tell us what we, as organic chemists, most want to know about our molecule. Specifically, these techniques do not allow us to determine its overall molecular structure – the framework, in other words, of its carbon-carbon and carbon-hydrogen bonds. It is this information that we need to have in order to be able to draw a Lewis structure of our molecule, and it is this information that is provided by an immensely powerful analytical technique called nuclear magnetic resonance (NMR) spectroscopy. In NMR, the nuclei of hydrogen, carbon, and other important elements undergo transitions in their magnetic states, leading to the absorbance of radiation in the radio frequency range of the electromagnetic spectrum. By analyzing the signals from these transitions, we learn about the chemical environment that each atom inhabits, including information about the presence of neighboring atoms. In this chapter, we will see how information from NMR, especially when combined with data from IR, UV-Vis, and MS experiments, can make it possible for us to form a complete picture of the atom-to-atom framework of an organic molecule.
| 2,098 | 1,556 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/07%3A_Obtaining_and_Preparing_Samples_for_Analysis/7.04%3A_Separating_the_Analyte_From_Interferents |
When an analytical method is selective for the analyte, analyzing a sample is a relatively simple task. For example, a quantitative analysis for glucose in honey is relatively easy to accomplish if the method is selective for glucose, even in the presence of other reducing sugars, such as fructose. Unfortunately, few analytical methods are selective toward a single species. In the absence of an interferent, the relationship between the sample’s signal, , and the analyte’s concentration, , is \[S_{samp}=k_{A} C_{A} \label{7.1}\] where is the analyte’s sensitivity. In Equation \ref{7.1}, and the equations that follow, you can replace the analyte’s concentration, , with the moles of analyte, , when working with methods, such as gravimetry, that respond to the absolute amount of analyte in a sample. In this case the interferent also is expressed in terms of moles. If an interferent, is present, then Equation \ref{7.1} becomes \[S_{samp}=k_{A} C_{A}+k_{I} C_{I} \label{7.2}\] where and are, respectively, the interferent’s sensitivity and concentration. A method’s selectivity for the analyte is determined by the relative difference in its sensitivity toward the analyte and the interferent. If is greater than , then the method is more selective for the analyte. The method is more selective for the interferent if is greater than . Even if a method is more selective for an interferent, we can use it to determine if the interferent’s contribution to is insignificant. The , , which we introduced in , provides a way to characterize a method’s selectivity. \[K_{A, I}=\frac{k_{I}}{k_{A}} \label{7.3}\] Solving Equation \ref{7.3} for , substituting into Equation \ref{7.2}, and simplifying, gives \[S_{samp}=k_{A}\left(C_{A}+K_{A, I} \times C_{I}\right) \label{7.4}\] An interferent, therefore, does not pose a problem as long as the product of its concentration and its selectivity coefficient is significantly smaller than the analyte’s concentration. \[K_{A, I} \times C_{I}<<C_{A} \nonumber\] If we cannot ignore an interferent’s contribution to the signal, then we must begin our analysis by separating the analyte and the interferent. | 2,180 | 1,557 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.08%3A_Introduction_to_Biochemistry |
Biochemistry is the study of chemical processes in living organisms, including, but not limited to, living matter. Biochemistry governs all living organisms and living processes. By controlling information flow through biochemical signaling and the flow of chemical energy through metabolism, biochemical processes give rise to the incredible complexity of life. Over the last decades of the 20th century, biochemistry become so successful at explaining living processes that now almost all areas of the life sciences from botany to medicine to genetics are engaged in biochemical research. Today, the main focus of pure biochemistry is in understanding how biological molecules give rise to the processes that occur within living cells, which in turn relates greatly to the study and understanding of whole organisms. Biochemistry is closely related to molecular biology, the study of the molecular mechanisms by which genetic information encoded in is able to result in the processes of life. Depending on the exact definition of the terms used, molecular biology can be thought of as a branch of biochemistry, or biochemistry as a tool with which to investigate and study molecular biology. Much of biochemistry deals with the structures, functions and interactions of biological macromolecules, such as proteins, nucleic acids, carbohydrates and lipids, which provide the structure of cells and perform many of the functions associated with life. The chemistry of the cell also depends on the reactions of smaller molecules and ions. These can be inorganic, for example water and metal ions, or organic, for example the amino acids which are used to synthesize proteins. The mechanisms by which cells harness energy from their environment via chemical reactions are known as metabolism. The findings of biochemistry are applied primarily in medicine, nutrition, and agriculture. In medicine, biochemists investigate the causes and cures of disease. In nutrition, they study how to maintain health and study the effects of nutritional deficiencies. In agriculture, biochemists investigate soil and fertilizers, and try to discover ways to improve crop cultivation, crop storage and pest control. Much of biochemistry deals with the structures and functions of cellular components such as proteins, carbohydrates, lipids, nucleic acids and other biomolecules—although increasingly processes rather than individual molecules are the main focus. | 2,465 | 1,558 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Catalysis/Examples/Examples_of_Catalysis/4._Examples_of_Acid_Catalysis_in_Organic_Chemistry |
This page looks at the use of acid catalysts in some organic reactions. It covers the nitration of benzene, the hydration of ethene to manufacture ethanol, and the reactions both to produce esters and to hydrolyse them under acidic conditions. You will find links to the full mechanisms for each of these reactions if you want them. Benzene is treated with a mixture of concentrated nitric acid and concentrated sulphuric acid at a temperature not exceeding 50°C. As the temperature increases there is a greater chance of getting more than one nitro group, -NO , substituted onto the ring. Nitrobenzene is formed. or: The concentrated sulfuric acid is acting as a catalyst. Because everything is present in the same liquid phase, this is a good example of homogeneous catalysis. Ethene is mixed with steam and passed over a catalyst consisting of solid silicon dioxide coated with phosphoric(V) acid. The temperature used is 300°C and the pressure is about 60 to 70 atmospheres. Because the catalyst is in a different phase from the reactants, this is an example of heterogeneous catalysis. This is a reversible reaction and only about 5% of the ethene reacts on each pass over the catalyst. When the reaction mixture is cooled, the ethanol and any excess steam condense, and the gaseous ethene can be recycled through the process. A conversion rate of about 95% is achieved by continual recycling in this way. are what is formed when an organic acid reacts with an alcohol in the presence of concentrated sulfuric acid as the catalyst. Everything is present in a single liquid phase, and so this is an example of homogeneous catalysis. For example, ethanoic acid reacts with ethanol to produce ethyl ethanoate. The ethyl ethanoate has the lowest boiling point of anything in the mixture, and so is distilled off as soon as it is formed. This helps to reduce the reverse reaction. In principle, this is the reverse of the esterification reaction but, in practice, it has to be done slightly differently. The ester is heated under reflux with a dilute acid such as dilute hydrochloric acid or dilute sulfuric acid. The equation for the reaction is simply the esterification equation written backwards. The dilute acid used as the catalyst also provides the water for the reaction. You need a large excess of water in order to increase the chances of the forward reaction happening and the ester hydrolyzing. You would normally hydrolyse esters quite differently by heating them with sodium hydroxide solution (alkaline hydrolysis). This an example of a catalytic reaction because the hydroxide ions are used up during the reaction. The main advantage of doing it like this is that it is a one-way reaction. The ester can be completely hydrolysed rather than only partially if the reaction is reversible. Jim Clark ( ) | 2,872 | 1,559 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/12%3A_Chromatographic_and_Electrophoretic_Methods/12.02%3A_General_Theory_of_Column_Chromatography |
Of the two methods for bringing the stationary phase and the mobile phases into contact, the most important is column chromatography. In this section we develop a general theory that we may apply to any form of column chromatography. Figure 12.2.1
provides a simple view of a liquid–solid column chromatography experiment. The sample is introduced as a narrow band at the top of the column. Ideally, the solute’s initial concentration profile is rectangular (Figure 12.2.2
a). As the sample moves down the column, the solutes begin to separate (Figure 12.2.1
b,c) and the individual solute bands begin to broaden and develop a Gaussian profile (Figure 12.2.2
b,c). If the strength of each solute’s interaction with the stationary phase is sufficiently different, then the solutes separate into individual bands (Figure 12.2.1
d and Figure 12.2.2
d).
. An alternative view of the separation in Figure 12.2.1
showing the concentration of each solute as a function of distance down the column. We can follow the progress of the separation by collecting fractions as they elute from the column (Figure 12.2.1
e,f), or by placing a suitable detector at the end of the column. A plot of the detector’s response as a function of elution time, or as a function of the volume of mobile phase, is known as a (Figure 12.2.3
), and consists of a peak for each solute. There are many possible detectors that we can use to monitor the separation. Later sections of this chapter describe some of the most popular. We can characterize a chromatographic peak’s properties in several ways, two of which are shown in Figure 12.2.4
. , , is the time between the sample’s injection and the maximum response for the solute’s peak. A chromatographic peak’s , , as shown in Figure 12.2.4
, is determined by extending tangent lines from the inflection points on either side of the peak through the baseline. Although usually we report and using units of time, we can report them using units of volume by multiplying each by the mobile phase’s velocity, or report them in linear units by measuring distances with a ruler. For example, a solute’s retention volume, , is \(t_\text{r} \times u\) where is the mobile phase’s velocity through the column. In addition to the solute’s peak, Figure 12.2.4
also shows a small peak that elutes shortly after the sample is injected into the mobile phase. This peak contains all , which move through the column at the same rate as the mobile phase. The time required to elute the nonretained solutes is called the column’s , . The goal of chromatography is to separate a mixture into a series of chromatographic peaks, each of which constitutes a single component of the mixture. The between two chromatographic peaks, , is a quantitative measure of their separation, and is defined as \[R_{A B}=\frac{t_{t, B}-t_{t,A}}{0.5\left(w_{B}+w_{A}\right)}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}} \label{12.1}\] where is the later eluting of the two solutes. As shown in Figure 12.2.5
, the separation of two chromatographic peaks improves with an increase in . If the areas under the two peaks are identical—as is the case in Figure 12.2.5
—then a resolution of 1.50 corresponds to an overlap of only 0.13% for the two elution profiles. Because resolution is a quantitative measure of a separation’s success, it is a useful way to determine if a change in experimental conditions leads to a better separation. In a chromatographic analysis of lemon oil a peak for limonene has a retention time of 8.36 min with a baseline width of 0.96 min. \(\gamma\)-Terpinene elutes at 9.54 min with a baseline width of 0.64 min. What is the resolution between the two peaks? Using Equation \ref{12.1} we find that the resolution is \[R_{A B}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}}=\frac{2(9.54 \text{ min}-8.36 \text{ min})}{0.64 \text{ min}+0.96 \text{ min}}=1.48 \nonumber\] Figure 12.2.6
shows the separation of a two-component mixture. What is the resolution between the two components? Use a ruler to measure \(\Delta t_\text{r}\), , and in millimeters. Because the relationship between elution time and distance is proportional, we can measure \(\Delta t_\text{r}\), , and using a ruler. My measurements are 8.5 mm for \(\Delta t_\text{r}\), and 12.0 mm each for and . Using these values, the resolution is \[R_{A B}=\frac{2 \Delta t_{t}}{w_{A}+w_{B}}=\frac{2(8.5 \text{ mm})}{12.0 \text{ mm}+12.0 \text{ mm}}=0.70 \nonumber\] Your measurements for \(\Delta t_\text{r}\), , and will depend on the relative size of your monitor or printout; however, your value for the resolution should be similar to the answer above. Equation \ref{12.1} suggests that we can improve resolution by increasing \(\Delta t_\text{r}\), or by decreasing and (Figure 12.2.7
). To increase \(\Delta t_\text{r}\) we can use one of two strategies. One approach is to adjust the separation conditions so that both solutes spend less time in the mobile phase—that is, we increase each —which provides more time to effect a separation. A second approach is to increase by adjusting conditions so that only one solute experiences a significant change in its retention time. The baseline width of a solute’s peak depends on the solutes movement within and between the mobile phase and the stationary phase, and is governed by several factors that collectively we call . We will consider each of these approaches for improving resolution in more detail, but first we must define some terms. Let’s assume we can describe a solute’s distribution between the mobile phase and stationary phase using the following equilibrium reaction \[S_{\text{m}} \rightleftharpoons S_{\text{s}} \nonumber\] where is the solute in the mobile phase and is the solute in the stationary phase. Following the same approach we used in for liquid–liquid extractions, the equilibrium constant for this reaction is an equilibrium partition coefficient, . \[K_{D}=\frac{\left[S_{\mathrm{s}}\right]}{\left[S_\text{m}\right]} \nonumber\] This is not a trivial assumption. In this section we are, in effect, treating the solute’s equilibrium between the mobile phase and the stationary phase as if it is identical to the equilibrium in a liquid–liquid extraction. You might question whether this is a reasonable assumption. There is an important difference between the two experiments that we need to consider. In a liquid–liquid extraction, which takes place in a separatory funnel, the two phases remain in contact with each other at all times, allowing for a true equilibrium. In chromatography, however, the mobile phase is in constant motion. A solute that moves into the stationary phase from the mobile phase will equilibrate back into a different portion of the mobile phase; this does not describe a true equilibrium. So, we ask again: Can we treat a solute’s distribution between the mobile phase and the stationary phase as an equilibrium process? The answer is yes, if the mobile phase velocity is slow relative to the kinetics of the solute’s movement back and forth between the two phase. In general, this is a reasonable assumption. In the absence of any additional equilibrium reactions in the mobile phase or the stationary phase, is equivalent to the distribution ratio, , \[D=\frac{\left[S_{0}\right]}{\left[S_\text{m}\right]}=\frac{(\operatorname{mol} \text{S})_\text{s} / V_\text{s}}{(\operatorname{mol} \text{S})_\text{m} / V_\text{m}}=K_{D} \label{12.2}\] where and are the volumes of the stationary phase and the mobile phase, respectively. A conservation of mass requires that the total moles of solute remain constant throughout the separation; thus, we know that the following equation is true. \[(\operatorname{mol} \text{S})_{\operatorname{tot}}=(\operatorname{mol} \text{S})_{\mathrm{m}}+(\operatorname{mol} \text{S})_\text{s} \label{12.3}\] Solving Equation \ref{12.3} for the moles of solute in the stationary phase and substituting into Equation \ref{12.2} leaves us with \[D = \frac{\left\{(\text{mol S})_{\text{tot}} - (\text{mol S})_\text{m}\right\} / V_{\mathrm{s}}}{(\text{mol S})_{\mathrm{m}} / V_{\mathrm{m}}} \nonumber\] Rearranging this equation and solving for the fraction of solute in the mobile phase, , gives \[f_\text{m} = \frac {(\text{mol S})_\text{m}} {(\text{mol S})_\text{tot}} = \frac {V_\text{m}} {DV_\text{s} + V_\text{m}} \label{12.4}\] which is identical to the result for a liquid-liquid extraction (see ). Because we may not know the exact volumes of the stationary phase and the mobile phase, we simplify Equation \ref{12.4} by dividing both the numerator and the denominator by ; thus \[f_\text{m} = \frac {V_\text{m}/V_\text{m}} {DV_\text{s}/V_\text{m} + V_\text{m}/V_\text{m}} = \frac {1} {DV_\text{s}/V_\text{m} + 1} = \frac {1} {1+k} \label{12.5}\] where \[k=D \times \frac{V_\text{s}}{V_\text{m}} \label{12.6}\] is the solute’s . Note that the larger the retention factor, the more the distribution ratio favors the stationary phase, leading to a more strongly retained solute and a longer retention time. Other (older) names for the retention factor are capacity factor, capacity ratio, and partition ratio, and it sometimes is given the symbol \(k^{\prime}\). Keep this in mind if you are using other resources. Retention factor is the approved name from the IUPAC Gold Book. We can determine a solute’s retention factor from a chromatogram by measuring the column’s void time, , and the solute’s retention time, (see ). Solving Equation \ref{12.5} for , we find that \[k=\frac{1-f_\text{m}}{f_\text{m}} \label{12.7}\] Earlier we defined as the fraction of solute in the mobile phase. Assuming a constant mobile phase velocity, we also can define as \[f_\text{m}=\frac{\text { time spent in the mobile phase }}{\text { time spent in the stationary phase }}=\frac{t_\text{m}}{t_\text{r}} \nonumber\] Substituting back into Equation \ref{12.7} and rearranging leaves us with \[k=\frac{1-\frac{t_{m}}{t_{t}}}{\frac{t_{\mathrm{m}}}{t_{\mathrm{r}}}}=\frac{t_{\mathrm{t}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{t_{\mathrm{r}}^{\prime}}{t_{\mathrm{m}}} \label{12.8}\] where \(t_\text{r}^{\prime}\) is the . In a chromatographic analysis of low molecular weight acids, butyric acid elutes with a retention time of 7.63 min. The column’s void time is 0.31 min. Calculate the retention factor for butyric acid. \[k_{\mathrm{but}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{7.63 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=23.6 \nonumber\] Figure 12.2.8
is the chromatogram for a two-component mixture. Determine the retention factor for each solute assuming the sample was injected at time = 0. Because the relationship between elution time and distance is proportional, we can measure , , and using a ruler. My measurements are 7.8 mm, 40.2 mm, and 51.5 mm, respectively. Using these values, the retention factors for solute A and solute B are \[k_{1}=\frac{t_{\mathrm{r} 1}-t_\text{m}}{t_\text{m}}=\frac{40.2 \text{ mm}-7.8 \text{ mm}}{7.8 \text{ mm}}=4.15 \nonumber\] \[k_{2}=\frac{t_{\mathrm{r} 2}-t_\text{m}}{t_\text{m}}=\frac{51.5 \text{ mm}-7.8 \text{ mm}}{7.8 \text{ mm}}=5.60 \nonumber\] Your measurements for , , and will depend on the relative size of your monitor or printout; however, your value for the resolution should be similar to the answer above. Selectivity is a relative measure of the retention of two solutes, which we define using a selectivity factor, \(\alpha\) \[\alpha=\frac{k_{B}}{k_{A}}=\frac{t_{r, B}-t_{\mathrm{m}}}{t_{r, A}-t_{\mathrm{m}}} \label{12.9}\] where solute has the smaller retention time. When two solutes elute with identical retention time, \(\alpha = 1.00\); for all other conditions \(\alpha > 1.00\). In the chromatographic analysis for low molecular weight acids described in , the retention time for isobutyric acid is 5.98 min. What is the selectivity factor for isobutyric acid and butyric acid? First we must calculate the retention factor for isobutyric acid. Using the void time from we have \[k_{\mathrm{iso}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{5.98 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=18.3 \nonumber\] The selectivity factor, therefore, is \[\alpha=\frac{k_{\text {but }}}{k_{\text {iso }}}=\frac{23.6}{18.3}=1.29 \nonumber\] Determine the selectivity factor for the chromatogram in . Using the results from , the selectivity factor is \[\alpha=\frac{k_{2}}{k_{1}}=\frac{5.60}{4.15}=1.35 \nonumber\] Your answer may differ slightly due to differences in your values for the two retention factors. Suppose we inject a sample that has a single component. At the moment we inject the sample it is a narrow band of finite width. As the sample passes through the column, the width of this band continually increases in a process we call . Column efficiency is a quantitative measure of the extent of band broadening. See and . When we inject the sample it has a uniform, or rectangular concentration profile with respect to distance down the column. As it passes through the column, the band broadens and takes on a Gaussian concentration profile. In their original theoretical model of chromatography, Martin and Synge divided the chromatographic column into discrete sections, which they called theoretical plates. Within each theoretical plate there is an equilibrium between the solute present in the stationary phase and the solute present in the mobile phase [Martin, A. J. P.; Synge, R. L. M. . , , 1358–1366]. They described column efficiency in terms of the number of , , \[N=\frac{L}{H} \label{12.10}\] where is the column’s length and is the height of a theoretical plate. For any given column, the column efficiency improves—and chromatographic peaks become narrower—when there are more theoretical plates. If we assume that a chromatographic peak has a Gaussian profile, then the extent of band broadening is given by the peak’s variance or standard deviation. The height of a theoretical plate is the peak’s variance per unit length of the column \[H=\frac{\sigma^{2}}{L} \label{12.11}\] where the standard deviation, \(\sigma\), has units of distance. Because retention times and peak widths usually are measured in seconds or minutes, it is more convenient to express the standard deviation in units of time, \(\tau\), by dividing \(\sigma\) by the solute’s average linear velocity, \(\overline{u}\), which is equivalent to dividing the distance it travels, , by its retention time, . \[\tau=\frac{\sigma}{\overline{u}}=\frac{\sigma t_{r}}{L} \label{12.12}\] For a Gaussian peak shape, the width at the baseline, , is four times its standard deviation, \(\tau\). \[w = 4 \tau \label{12.13}\] Combining Equation \ref{12.11}, Equation \ref{12.12}, and Equation \ref{12.13} defines the height of a theoretical plate in terms of the easily measured chromatographic parameters and . \[H=\frac{L w^{2}}{16 t_\text{r}^{2}} \label{12.14}\] Combing Equation \ref{12.14} and Equation \ref{12.10} gives the number of theoretical plates. \[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16\left(\frac{t_{\mathrm{r}}}{w}\right)^{2} \label{12.15}\] A chromatographic analysis for the chlorinated pesticide Dieldrin gives a peak with a retention time of 8.68 min and a baseline width of 0.29 min. Calculate the number of theoretical plates? Given that the column is 2.0 m long, what is the height of a theoretical plate in mm? Using Equation \ref{12.15}, the number of theoretical plates is \[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16 \times \frac{(8.68 \text{ min})^{2}}{(0.29 \text{ min})^{2}}=14300 \text{ plates} \nonumber\] Solving Equation \ref{12.10} for gives the average height of a theoretical plate as \[H=\frac{L}{N}=\frac{2.00 \text{ m}}{14300 \text{ plates}} \times \frac{1000 \text{ mm}}{\mathrm{m}}=0.14 \text{ mm} / \mathrm{plate} \nonumber\] For each solute in the chromatogram for , calculate the number of theoretical plates and the average height of a theoretical plate. The column is 0.5 m long. Because the relationship between elution time and distance is proportional, we can measure , , , and using a ruler. My measurements are 40.2 mm, 51.5 mm, 8.0 mm, and 13.5 mm, respectively. Using these values, the number of theoretical plates for each solute is \[N_{1}=16 \frac{t_{r,1}^{2}}{w_{1}^{2}}=16 \times \frac{(40.2 \text{ mm})^{2}}{(8.0 \text{ mm})^{2}}=400 \text { theoretical plates } \nonumber\] \[N_{2}=16 \frac{t_{r,2}^{2}}{w_{2}^{2}}=16 \times \frac{(51.5 \text{ mm})^{2}}{(13.5 \text{ mm})^{2}}=233 \text { theoretical plates } \nonumber\] The height of a theoretical plate for each solute is \[H_{1}=\frac{L}{N_{1}}=\frac{0.500 \text{ m}}{400 \text { plates }} \times \frac{1000 \text{ mm}}{\mathrm{m}}=1.2 \text{ mm} / \mathrm{plate} \nonumber\] \[H_{2}=\frac{L}{N_{2}}=\frac{0.500 \text{ m}}{233 \text { plates }} \times \frac{1000 \text{ mm}}{\mathrm{m}}=2.15 \text{ mm} / \mathrm{plate} \nonumber\] Your measurements for , , , and will depend on the relative size of your monitor or printout; however, your values for and for should be similar to the answer above. It is important to remember that a theoretical plate is an artificial construct and that a chromatographic column does not contain physical plates. In fact, the number of theoretical plates depends on both the properties of the column and the solute. As a result, the number of theoretical plates for a column may vary from solute to solute. One advantage of improving column efficiency is that we can separate more solutes with baseline resolution. One estimate of the number of solutes that we can separate is \[n_{c}=1+\frac{\sqrt{N}}{4} \ln \frac{V_{\max }}{V_{\min }} \label{12.16}\] where is the column’s , and and are the smallest and the largest volumes of mobile phase in which we can elute and detect a solute [Giddings, J. C. , Wiley-Interscience: New York, 1991]. A column with 10 000 theoretical plates, for example, can resolve no more than \[n_{c}=1+\frac{\sqrt{10000}}{4} \ln \frac{30 \mathrm{mL}}{1 \mathrm{mL}}=86 \text { solutes } \nonumber\] if and are 1 mL and 30 mL, respectively. This estimate provides an upper bound on the number of solutes and may help us exclude from consideration a column that does not have enough theoretical plates to separate a complex mixture. Just because a column’s theoretical peak capacity is larger than the number of solutes, however, does not mean that a separation is feasible. In most situations the practical peak capacity is less than the theoretical peak capacity because the retention characteristics of some solutes are so similar that a separation is impossible. Nevertheless, columns with more theoretical plates, or with a greater range of possible elution volumes, are more likely to separate a complex mixture. The smallest volume we can use is the column’s void volume. The largest volume is determined either by our patience—the maximum analysis time we can tolerate—or by our inability to detect solutes because there is too much band broadening. Our treatment of chromatography in this section assumes that a solute elutes as a symmetrical Gaussian peak, such as that shown in Figure 12.2.4
. This ideal behavior occurs when the solute’s partition coefficient, \[K_{\mathrm{D}}=\frac{[S_\text{s}]}{\left[S_\text{m}\right]} \nonumber\] is the same for all concentrations of solute. If this is not the case, then the chromatographic peak has an asymmetric peak shape similar to those shown in Figure 12.2.9
. The chromatographic peak in Figure 12.2.9
a is an example of , which occurs when some sites on the stationary phase retain the solute more strongly than other sites. Figure 12.2.9
b, which is an example of most often is the result of overloading the column with sample. As shown in Figure 12.2.9
a, we can report a peak’s asymmetry by drawing a horizontal line at 10% of the peak’s maximum height and measuring the distance from each side of the peak to a line drawn vertically through the peak’s maximum. The asymmetry factor, , is defined as \[T=\frac{b}{a} \nonumber\] The number of theoretical plates for an asymmetric peak shape is approximately \[N \approx \frac{41.7 \times \frac{t_{r}^{2}}{\left(w_{0.1}\right)^{2}}}{T+1.25}=\frac{41.7 \times \frac{t_{r}^{2}}{(a+b)^{2}}}{T+1.25} \nonumber\] where is the width at 10% of the peak’s height [Foley, J. P.; Dorsey, J. G. , , 730–737]. Asymmetric peaks have fewer theoretical plates, and the more asymmetric the peak the smaller the number of theoretical plates. For example, the following table gives values for for a solute eluting with a retention time of 10.0 min and a peak width of 1.00 min. 0.5 0.5 0.6 0.4 0.7 0.3 | 20,705 | 1,560 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.02%3A_Introduction_to_Hydrocarbons |
Hydrocarbons are organic compounds that contain carbon and hydrogen. The four general classes of hydrocarbons are: alkanes, alkenes, alkynes and arenes. Aromatic compounds derive their names from the fact that many of these compounds in the early days of discovery were grouped because they were oils with fragrant odors. The classifications for hydrocarbons, defined by nomenclature of organic chemistry are as follows: Because of differences in molecular structure, the empirical formula remains different between hydrocarbons; in linear, or "straight-run" alkanes, alkenes and alkynes, the amount of bonded hydrogen lessens in alkenes and alkynes due to the "self-bonding" or catenation of carbon preventing entire saturation of the hydrocarbon by the formation of double or triple bonds. The inherent ability of hydrocarbons to bond to themselves is known as catenation, and allows hydrocarbon to form more complex molecules, such as cyclohexane, and in rarer cases, arenes such as benzene. This ability comes from the fact that the bond character between carbon atoms is entirely non-polar, in that the distribution of electrons between the two elements is somewhat even due to the same electronegativity values of the elements (~0.30). | 1,262 | 1,561 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/22%3A_Biophysical_Reaction_Dynamics/22.03%3A_Representations_of_Dynamics |
We will survey different representation of time-dependent processes using examples from one-dimension. Watch the continuous time-dependent behavior of one or more particles/molecules in the system. Time‐dependent structural configurations A molecular dynamics trajectory will give you the position of all atoms as a function of time { ,t}. Although there is an enormous amount of information in such a trajectory, the raw data is often overwhelming and not of particularly high value itself. However, it is possible to project this high dimensional information in structural coordinates onto one or more collective variables ξ that forms a more meaningful representation of the dynamics, ξ(t). Alternatively, single molecule experiments can provide a chronological sequence of the states visited by molecule. State trajectories: Time‐dependent occupation of states A discretized representation of which state of the system the particle occupies. Requires that you define the boundaries of a state. Example: A two state trajectory for an equilibrium \(A \rightleftharpoons B\), where the time-dependent probability of being in state A is: \[P_{A}(t)=\left\{\begin{array}{l}1 \text { if } \xi(t)<\xi^{‡} \\0 \text { if } \xi(t)>\xi^{‡}\end{array}\right.\] With sufficient sampling, one can average over trajectories in order to develop a time-dependent probability distribution \(P(ξ,t)\) for the non-equilibrium evolution of an initial state. State Populations: Kinetics \[ \int_{\text {state A}}P(ξ,t)dξ=P_A(t) \] Time-correlation functions are commonly used to characterize trajectories of a fluctuating observable. These are described next | 1,653 | 1,563 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.03%3A_Chain_Reactions_I |
usually consist of many repeating elementary steps, each of which has a chain carrier. Once started, chain reactions continue until the reactants are exhausted. Fire and explosions are some of the phenomena associated with chain reactions. The are some intermediates that appear in the repeating elementary steps. These are usually . Once initiated, repeating elementary steps continue until the reactants are exhausted. When repeating steps generate more chain carriers, they are called , which leads to explosions. If the repeating elementary steps do not lead to the formation of new product, they are called . Addition of other materials in the reaction mixture can lead to the inhibition reaction to prevent the chain propagation reaction. When chain carriers react with one another forming stable product, the elementary steps are called . Explosions, polymerizations, and food spoilage often involve chain reactions. The chain reaction mechanism is involved in nuclear reactors; in this case the chain carriers are neutrons. The mechanisms describing chain reactions are useful models for describing chemical reactions. Most chemical chain reactions have very reactive intermediates called . The intermediate that maintains the chain reaction is called a These atoms or fragments are usually derived from stable molecules due to photo- or heat-dissociation. Usually, a free radical is marked by a dot beside the symbol (\(\ce{*}\)), which represents an odd electron exists on the species. This odd electron makes the intermediate very reactive. For example, the oxygen, chlorine and ethyl radicals are represented by \(\ce{O*}\), \(\ce{Cl*}\), and \(\ce{C2H5*}\), respectively. The \(\ce{Cl*}\) radicals can be formed by the homolytic photodissociation reaction: \[\ce{Cl2 + h\nu \rightarrow Cl* + *Cl} \nonumber \] The elementary steps used for mechanisms of chain reactions can be grouped into the following categories: For example, the chlorination of ethane is a chain reaction, and its mechanism is explained in the following way. If we mix chlorine, \(\ce{Cl2}\), and ethane, \(\ce{CH3CH3}\), together at room temperature, there is no detectable reaction. However, when the mixture is exposed to light, the reaction suddenly initiates, and explodes. To explain this, the following mechanism is proposed. Light (\(\ce{h\nu}\)) can often be used to initiate chain reactions since they can generate free radical intermediates via a photodissociation reaction. The initiation step can be written as: \[\ce{Cl2 + h\nu \rightarrow Cl* + *Cl} \nonumber \] Elementary steps in which the number of free radicals consumed is equal to the number of free radicals generated are called Once initiated, the following chain propagation steps repeat indefinitely or until the reactants are exhausted: \[\ce{Cl* +\; H3CCH3 \rightarrow ClH2CCH3 +\; H*} \nonumber \] \[\ce{Cl* +\; H3CCH3 \rightarrow H3CCH2* +\; HCl} \nonumber \] \[\ce{H* +\; Cl_2 \rightarrow HCl + Cl*} \nonumber \] and many other possibilities. In each of these steps, a radical is consumed, and another radical is generated. Thus, the chain reactions continue, releasing heat and . The heat and cause more radicals to form. Thus, the chain propagation steps cause . are elementary steps that generate more free radicals than they consume. Branching reactions result in an explosion. For example, in the reaction between hydrogen and oxygen, the following reaction may take place: \[\ce{H* +\; O2 \rightarrow HO* + *O*} \nonumber \] where \(\ce{*O*}\) is a di-radical, because the \(\ce{O}\) atom has an electronic configuration 2 2 2 2 . In this elementary step, three radicals are generated, whereas only one is consumed. The di-radical may react with a \(\ce{H2}\) molecule to form two radicals. \[\ce{*O* +\, H2 \rightarrow HO* +\, H*} \nonumber \] Thus, together chain branching reactions increase the number of chain carriers. Branching reactions contribute to the rapid explosion of hydrogen-oxygen mixtures, especially if the mixtures have proper proportions. The steps not leading to the formation of products are called or steps. For example, the following steps are inhibition reactions. \[\ce{Cl* +\; ClH2CCH3 \rightarrow H3CCH2* +\; Cl2} \nonumber \] \[\ce{Cl* +\; HCl \rightarrow H* +\; Cl2} \nonumber \] \[\ce{H* +\; ClH2CCH3 \rightarrow H3CCH3 + Cl*} \nonumber \] Furthermore, sometimes another reactive substance \(\ce{*A}\) may be added to the system to reduce the chain carriers to inhibit the chain reactions. \(\ce{Cl* + *A \rightarrow ClA\: (not\: reactive)}\) The species \(\ce{A*}\) is often called a In food industry, radical scavengers are added to prevent spoilage due to oxidation; these are called biological oxidants. The mechanisms in chain reactions are often quite complicated. When intermediates are detected, a reasonable mechanism can be proposed. Adding radical scavenger to prevent food spoilage is an important application in food chemistry. This application came from the application of the chain reaction model to natural phenomena. are elementary steps that consume radicals. When reactants are exhausted, free radicals combine with one another to give stable molecules (since unpaired electrons become paired). These elementary steps are responsible for the chain reactions' termination: \[\ce{Cl* + *Cl \rightarrow Cl-Cl} \nonumber \] \[\ce{H* + *H \rightarrow H-H} \nonumber \] \[\ce{H* + *Cl \rightarrow H-Cl} \nonumber \] \[\ce{H3CCH2* + *H2CCH3 \rightarrow CH3CH2-CH2CH3\: (forming\: a\: dimer)} \nonumber \] and other possibilities In chain reactions, many products are produced.
Identify steps for the names in the multiple choices.
Argon exists as a mono-atomic gas. All noble gases have mono-atomic molecules.
The reactant \(\ce{HCl}\) in the step is a product in the overall reaction. When \(\ce{HCl}\) reacts with \(\ce{Cl*}\), the reaction is retarded. \(\ce{Cl*}\) attacked one of the product molecule \(\ce{HCl}\) causing a reversal of the reaction. | 6,026 | 1,565 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_I_(Cortes)/05%3A_Orbital_Picture_of_Bonding-_Orbital_Combinations_Hybridization_Theory_and_Molecular_Orbitals/5.03%3A_Sigma_Bonding |
When atomic orbitals (pure or hybrid) of different atoms overlap to form covalent bonds, they may approach each other in two major ways: , or . Only head to head overlap is possible with orbitals because they are spherical. Hybrid orbitals also undergo mostly head to head overlap when forming covalent bonds. -orbitals, on the other hand, can approach each other either sideways or head to head. For now, however, we are concerned only with head to head overlap because that’s the only type that occurs in alkanes. We’ll discuss sideways overlap later in connection with alkenes and alkynes, that is, hydrocarbons that have double and triple bonds respectively. As illustrations, consider the bonds that have already been studied. The bond between two hydrogen atoms is an example of sigma bonding. The bonds between the orbitals of hybridized carbon and the orbitals of hydrogen in methane are also example of sigma bonds. sigma bond between orbitals sigma bonds between and orbitals Two carbons can also overlap to form a C–C sigma bond where two orbitals overlap head to head, such as in the formation of the ethane molecule: It can be easily seen that , also loosely known as single bonds. LINE-ANGLE FORMULAS In alkanes of 3 carbon atoms or more, the main carbon chain acquires a zig-zag structure due to the 109.5 angle between C–C bonds, such as in propane: Two representations of propane, where the zig-zag structure of the carbon chain becomes apparent Writing Lewis formulas, or even condensed formulas, for alkanes of many carbon atoms can quickly become cumbersome. A short hand notation that uses zig-zag lines has been developed. The resulting representations are known as . The beginning and the end of the zig-zag line, as well as any breaks in direction represent carbon atoms. Line angle representation for propane equivalent to or CH3CH2CH3 The arrows point to the positions of the carbon atoms. Every carbon atom has to form 4 bonds. . Other examples are: | 2,009 | 1,566 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Physical_Equilibria/Non-Ideal_Mixtures_of_Liquids |
This page looks at the phase diagrams for non-ideal mixtures of liquids, and introduces the idea of an azeotropic mixture (also known as an azeotrope or constant boiling mixture). It goes on to explain how this complicates the process of such a mixture. Due to , a plot the vapor pressure of an of two liquids against their composition will result in a straight line: In this case, pure A has the higher vapor pressure and so is the more volatile component. Raoult's Law only works for ideal mixtures. In these, the forces between the particles in the mixture are exactly the same as those in the pure liquids. The tendency for the particles to escape is the same in the mixture and in the pure liquids. That's not true in non-ideal mixtures. In mixtures showing a positive deviation from Raoult's Law, the vapor pressure of the mixture is always higher than you would expect from an ideal mixture. The deviation can be small - in which case, the straight line in the last graph turns into a slight curve. Notice that the highest vapor pressure anywhere is still the vapor pressure of pure A. Cases like this, where the deviation is small, behave just like ideal mixtures as far as distillation is concerned, and we do not need to say anything more about them. But some liquid mixtures have very large positive deviations from Raoult's Law, and in these cases, the curve becomes very distorted. Notice that mixtures over a range of compositions have higher vapor pressures than either pure liquid. The maximum vapor pressure is no longer that of one of the pure liquids. This has important consequences when we look at boiling points and distillation further down the page. The fact that the vapor pressure is higher than ideal in these mixtures means that molecules are breaking away more easily than they do in the pure liquids. That is because the intermolecular forces between molecules of A and B are less than they are in the pure liquids. You can see this when you mix the liquids. Less heat is evolved when the new attractions are set up than was absorbed to break the original ones. Heat will therefore be absorbed when the liquids mix. The enthalpy change of mixing is endothermic. The classic example of a mixture of this kind is ethanol and water. This produces a highly distorted curve with a maximum vapor pressure for a mixture containing 95.6% of ethanol by mass. In exactly the same way, you can have mixtures with vapor pressures which are less than would be expected by Raoult's Law. In some cases, the deviations are small, but in others they are much greater giving a minimum value for vapor pressure lower than that of either pure component. These are cases where the molecules break away from the mixture less easily than they do from the pure liquids. New stronger forces must exist in the mixture than in the original liquids. You can recognize this happening because heat is evolved when you mix the liquids - more heat is given out when the new stronger bonds are made than was used in breaking the original weaker ones. Many (although not all) examples of this involve actual reaction between the two liquids. The example of a major negative deviation is a mixture of nitric acid and water. These two covalent molecules react to give hydroxonium ions and nitrate ions. \[ \ce{H2O(l) + HNO3(l) <=> H3O^{+} (aq) + NO3(aq)^{-}} \label{1}\] You now have strong ionic attractions involved which increases the intermolecular interactions. Jim Clark ( ) | 3,490 | 1,567 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Calorimetry/Differential_Scanning_Calorimetry |
Calorimetry involves the experimental quantification of heat released in a chemical process, either a reaction or a conformational alteration. It can be used to determine parameters such as the (\(Δ_{r}H\)), which is the change in enthalpy associated with the process of a chemical reaction. When \(Δ_{r}H\) is a negative value, the process is exothermic and releases heat; when \(Δ_{r}H\) is a positive value, the process is endothermic and requires heat input. Calorimetry uses a closed system, meaning there is a system separated from its surroundings by some boundary, through which heat and energy but not mass are able to flow. Calorimetry may be conducted at either constant pressure or volume and allows one to monitor the change in temperature as a result of the chemical process being investigated. Differential scanning calorimetry is a specific type of calorimetry including both a sample substance and a reference substance, residing in separate chambers. While the reference chamber contains only a solvent (such as water), the sample chamber contains an equal amount of the same solvent in addition to the substance of interest, of which the Δ H is being determined. The \(Δ_{r}H\) due to the solvent is constant in both chambers, so any difference between the two can be attributed to the presence of the substance of interest. Each chamber is heated by a separate source in a way that their temperatures are always equal. This is accomplished through the use of thermocouples; the temperature of each chamber is constantly monitored and if a temperature difference is detected, then heat will be added to the cooler chamber to compensate for the difference. The heating rate used to maintain equivalent temperatures is logged as a function with respect to the temperature. For example, if the experimental goal is to determine the \(Δ_{r}H\) of a protein denaturation process, the reference cell could contain 100 mL H O, and the sample cell could contain 1 mg of the protein in addition to the same 100 mL H O. Therefore, the contribution of the solvent (H O) to the heat capacity of each cell would be equal, and the only difference would be the presence of the protein in the sample chamber. The following equation relates the change in temperature to the change in enthalpy: \[ dH = \int^{T_f}_{T_i}nC_p dt \] where \(dH\) is the rate of change in enthalpy, \(C_P\) is the of the calorimeter, \(dT\) is the rate of change in temperature, \(n\) is the number of moles of material, and \(T_i\) and \(T_f\) are the initial and final temperatures, respectively. This equation can be integrated to yield \[\Delta H = nC_p \Delta T\] where is the total change in enthalpy and is the change in temperature. The output yielded by differential scanning calorimetry is called a differential thermogram, which plots the required heat flow against temperature. Data analysis is highly dependent on the assumption that both the reference and sample cells are constantly and accurately maintained at equal temperatures. This graph indicates the change in power (electrical heat) as the temperatures of the two cells are gradually increased. A change in specific heat results in a small change in power, and can be either positive or negative depending on the particular process. The advent of an endothermic reaction will cause an increase in power as temperature increases, since additional heat is required to drive the reaction and still maintain the reference temperature. When an exothermic reaction occurs, the opposite effect is observed; power decreases because heat is released by the reaction and less power is required to maintain equivalent temperatures in the chambers. Differential scanning calorimetry can be used to study many different fields including biopolymer energetics where it is used to find the enthalpy of the protein denaturation process. A protein can be changed from its native state, in which it has a specific conformation due to non-covalent intramolecular interactions, to a denatured state where this characteristic structure is altered. Analysis of proteins through DSC can provide both the enthalpy of and information about the cooperativity of the denaturation process. A sharper peak in the thermogram indicates a higher level of cooperativity, meaning that when one structural association is disturbed, the likelihood of disruption at other points of association will be enhanced. DSC is also used in conjunction with differential thermal analysis. Through the combination of these two techniques, thermal behavior of inorganic compounds can be studied while the melting, boiling and decomposition points of organic compounds and polymers are found. | 4,723 | 1,568 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Stereoselectivity_in_Addition_Reactions_to_Double_Bonds/Oxidations/Epoxidation |
Some oxidation reactions of alkenes give cyclic ethers in which both carbons of a double bond become bonded to the same oxygen atom. These products are called or . An important method for preparing epoxides is by reaction with peracids, RCO H. The oxygen-oxygen bond of such peroxide derivatives is not only weak (ca. 35 kcal/mole), but in this case is polarized so that the acyloxy group is negative and the hydroxyl group is positive (recall that the acidity of water is about ten powers of ten weaker than that of a carboxylic acid). If we assume electrophilic character for the OH moiety, the following equation may be written. It is unlikely that a dipolar intermediate, as shown above, is actually formed. The epoxidation reaction is believed to occur in a single step with a transition state incorporating all of the bonding events shown in the equation. Consequently, epoxidations by peracids always have syn-stereoselectivity, and seldom give structural rearrangement. You may see the transition state by clicking the . Presumably the electron shifts indicated by the blue arrows induce a charge separation that is immediately neutralized by the green arrow electron shifts. The previous few reactions have been classified as reductions or oxidations, depending on the change in oxidation state of the functional carbons. It is important to remember that whenever an atom or group is reduced, some other atom or group is oxidized, and a balanced equation must balance the electron gain in the reduced species with the electron loss in the oxidized moiety, as well as numbers and kinds of atoms. Starting from an alkene (drawn in the box), the following diagram shows a hydrogenation reaction on the left (the catalyst is not shown) and an epoxidation reaction on the right. Examine these reactions, and for each identify which atoms are reduced and which are oxidized. Epoxides may be cleaved by aqueous acid to give glycols that are often diastereomeric with those prepared by the described above. Proton transfer from the acid catalyst generates the conjugate acid of the epoxide, which is attacked by nucleophiles such as water in the same way that the cyclic bromonium ion described above undergoes reaction. The result is of the double bond, in contrast to the syn-stereoselectivity of the earlier method. In the following equation this procedure is illustrated for a cis-disubstituted epoxide, which, of course, could be prepared from the corresponding cis-alkene. This hydration of an epoxide does not change the oxidation state of any atoms or groups. | 2,587 | 1,569 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Heterogeneous_Equilibria/Heterogeneous_Equilibria |
The 's are equilibrium constants in hetergeneous equilibria. If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship: Such a tendency is called the of the phase. As long as such a phase exists, its tendency or activity remains constant. However, the activities of substances in a gas phase are proportional to their partial pressures or concentrations. For a solution, their activities are proportional to their concentrations. Thus, their partial pressures or concentrations are indicators of their tendency to change. At equilibrium, these tendencies of changes reach certain proportions such that the forward and reverse changes are balanced. Similar to the equilibrium conditions of homogeneous systems, heterogeneous systems also tend to reach equilibrium conditions. can also be assigned to describe equilibrium conditions of heterogeneous systems. We shall look at several types of heterogeneous systems to illustrate how we deal with their behavior or change. Saturated solutions are typical heterogeneous equilibria. We all experience that when solid sugar is present in a sugar solution, putting more sugar in it will not increase its solubility. The sugar solution will not get any sweeter, that is if we really can taste the sweetness as proportional to the concentration. The solution mentioned above is called a saturated solution. The main criterion for a saturated solution is that the amount in the solid phase remains constant, or the concentration remains constant. A saturated solution is the result of equilibrium between the solute and its solution. For this type of equilibria, the equilibrium constant is the concentration of the saturated solution. We write the dissolution in an equation: \(\ce{C12H22O_{11\large{(s)}} \rightleftharpoons C12H22O_{11\large{(aq)}}}, \hspace{10px} K_{\ce c} = \ce{[C12H22O11]\: (saturated)}\) and \(\ce{[C12H22O11]}\) represents the saturated concentration of the solution, \(\ce{C12H22O_{11\large{(aq)}}}\). We simply treat the activity of the solid as 1 (unity). When a salt dissolves, the solution contains ions rather than molecules. The equilibrium constant is the product of the ion concentrations. This is illustrated next. When a salt dissolves in water, ions rather than molecules are present in the solution. For example, when silver chloride dissolves in water, \(\ce{Ag+}\) and \(\ce{Cl-}\) or more precisely \(\ce{Ag+(H2O)6}\) and \(\ce{Cl- (H2O)6}\) are present in the solution. We write the dissolving process and the equilibrium constant this way: \(\ce{AgCl_{\large{(s)}} \rightleftharpoons Ag+ + Cl- }, \hspace{10px} K = \ce{[Ag+] [Cl- ]}\) Since the solubility of \(\ce{AgCl}\) is small, the concentrations of \(\ce{Ag+}\) and \(\ce{Cl-}\) are very small. As we shall see later, the equilibrium constant of sparingly soluble salts is often designated as . More examples of dissolved salts and equilibrium constants are: \(\ce{CaCO3 \rightleftharpoons Ca^2+ + CO3^2-}, \hspace{10px} K_{\ce{sp}} = \ce{[Ca^2+] [CO3^2- ]}\);
\(\ce{Al3SO4 \rightleftharpoons 2 Al^3+ + 3 SO4^2-}, \hspace{10px} K_{\ce{sp}} = \ce{[Al^3+]^2 [SO4^2- ]^3}\) We shall deal with this type of equilibrium more extensively on other pages. Dissolving of a gas in a liquid involves changes of two phases. These types of changes are examples of heterogeneous equilibrium. For this type of equilibrium, the equilibrium constant is expressed by the partial pressure rather than by the ratio of pressure and concentration. For example, the dissolution of oxygen in water and the equilibrium constant are usually written in this way: \(\ce{O_{3\large{(g)}} \rightleftharpoons O_{3\large{(aq)}}}, \hspace{10px} K_{\ce p} = \ce{\dfrac{1}{P(O3)}}\); or \(\ce{O_{3\large{(aq)}} \rightleftharpoons O_{3\large{(g)}}}, \hspace{10px} K'_{\ce p} = \ce{P(O3)}\); The concentration is not 1 (unity), but we chose to express the equilibrium constant in terms of the partial pressure of oxygen. Ideally, in a closed system, the partial pressure of oxygen changes as it dissolves in water, and eventually reaches an equilibrium. But due to the small solubility of \(\ce{O3}\), the changes in partial pressure are not noticeable. Furthermore, this type of system has been investigated earlier by Henry, and he noticed that the solubility (concentration) of a gas in a liquid is proportional to the partial pressure. This is now known as Henry's law. For heterogeneous equilibria, the equilibrium constants, , should be expressed as a function of the concentrations of reactants and products of solution or gases. For convenience, the ACTIVITY of a SOLID or LIQUID is given as 1 (unity). For example, when limestone or shell (of shell fish), \(\ce{CaCO_{3\large{(s)}}}\), is heated, \(\ce{CO3}\) gas is released leaving the \(\ce{CaO}\) as a solid. We write the reaction and equilibrium constant in this form: \(\ce{CaCO_{3\large{(s)}} \rightleftharpoons CaO_{\large{(s)}} + CO_{3\large{(g)}}}, \hspace{10px} K_{\ce p} = \ce{P(CO3)}\) since the activities of the solids are considered unity. The is used to mean that the equilibrium constant is expressed in terms of partial pressures. In this example, is the saturated partial pressure of \(\ce{CO3}\) when \(\ce{CaCO3}\) and \(\ce{CaO}\) solids are present, and no net change will take place. We are used to the idea that water vapor pressure is a constant at a definite temperature. We seldom think of it being an equilibrium constant, but it is. The reaction can be represented by, \(\ce{H2O_{\large{(l)}} \rightleftharpoons H2O_{\large{(g)}}}, \hspace{10px} K_{\ce p} = \ce{P(H2O)}\)
\(\ce{H2O_{\large{(s)}} \rightleftharpoons H2O_{\large{(g)}}}, \hspace{10px} K_{\ce p} = \ce{P(H2O)}\) On the phase diagram of water, the equilibrium conditions between various phases are marked by curves. The sublimation, evaporation, and melting curves show the dependence of equilibrium on pressure and temperature. A change in temperature will shift the equilibrium along the paths on these curves. For your information, the vapor pressure of ice and water is listed in the data section. Here are some values: What do you expect the vapor pressure of ice at 5 deg C is? Is it higher than or lower than 6.343 mmHg? Well, find out! increase
The equilibrium constant, \(K = \ce{P(H2O)}\), varies with temperature. 4.579 mmHg
If the vapor pressures are different, the three phases cannot coexist. lower
The lowering of vapor pressure can be applied to explain the fact that salt solution freezes at lower temperature. | 6,890 | 1,570 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/13%3A_Intermolecular_Forces/13.01%3A_Intermolecular_Interactions |
Forces binding atoms in a molecule are due to chemical bonding. The energy required to break a bond is called the bond-energy. For example, the average bond-energy for \(\ce{O-H}\) bonds in water is 463 kJ/mol. On average, 463 kJ is required to break 6.023x10 \(\ce{O-H}\) bonds, or 926 kJ to convert 1.0 mole of water into 1.0 mol of \(\ce{O}\) and 2.0 mol of \(\ce{H}\) atoms. The forces holding molecules together are generally called intermolecular forces. The energy required to break molecules apart is much smaller than a typical bond-energy, but intermolecular forces play important roles in determining the properties of a substance. Intermolecular forces are particularly important in terms of how molecules interact and form biological organisms or even life. This link gives an excellent introduction to the interactions between molecules. In general, intermolecular forces can be divided into several categories. The four prominent types are: The division into types is for convenience in their discussion. Of course all types can be present simultaneously for many substances. Usually, intermolecular forces are discussed together with The States of Matter. Intermolecular forces also play important roles in solutions, a discussion of which is given in Hydration, solvation in water. A summary of the interactions is illustrated in the following diagram: See if you can answer the following questions. If you are looking for specific information, your study will be efficient. Some answers can be found in the Confidence Building Questions. Consider carefully the purpose of each question, and figure out what there is to be learned in it. Hint: a e f Discussion -
\(\ce{CO2}\), \(\ce{CH4}\), and \(\ce{N2}\) are symmetric, and hence they have no permanent dipole moments. A molecule with polar bonds unsymmetrically arranged will possess a permanent dipole. Hint: \(\ce{ICl}\) Discussion -
They have similar molecular weights: \(\mathrm{Br_2 = 160}\); \(\mathrm{ICl = 162}\). Their boiling points are 332 K and 370 K respectively. Hint: d. Discussion -
Induced dipoles are responsible for the London dispersion forces. The heavier the molecule, the larger the induced dipole will be. Thus, London dispersion forces are strong for heavy molecules. Hint: iodine, \(\ce{I2}\). Discussion -
Atomic weights for \(\ce{Br}\) and \(\ce{I}\) are 80 and 127 respectively. The higher the molecular weight, the stronger the London dispersion forces. Hint: water, \(\ce{H2O}\) Discussion -
The b.p. for \(\ce{H2O}\) is 100 deg C, and that of \(\ce{H2S}\) is -70 deg C. Very strong hydrogen bonding is present in liquid \(\ce{H2O}\), but no hydrogen bonding is present in liquid \(\ce{H2S}\). Hint: d. Discussion -
As more hydrogen bonds form when the temperature decreases, the volume expands, causing a decrease in density. Above 4 deg C, the thermal expansion is more prominent than the effect of hydrogen bonds. Hint: c. Discussion -
A hydrogen atom between two small, electronegative atoms (such as \(\ce{F}\), \(\ce{O}\), \(\ce{N}\)) causes a strong intermolecular interaction known as the hydrogen bond. The strength of a hydrogen bond depends upon the electronegativities and sizes of the two atoms. Hint: Ethanol has a higher boiling point. Discussion -
\(\ce{R-OH}\) group is both proton donor and acceptor for hydrogen bonding. Methyl groups have very weak hydrogen bonding, if any. | 3,433 | 1,571 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/07._Angular_Momentum/6%3A_Angular_Momentum |
Consider a variant of the one-dimensional particle in a box problem in which the x-axis is bent into a ring of radius R. We can write the same Schrödinger equation There are no boundary conditions in this case since the x-axis closes upon itself. A more appropriate independent variable for this problem is the angular position on the ring given by, \( \phi = x {/} R \) Schrödinger \[ -\dfrac{\hbar^2}{2mR^2} \dfrac{d^2 \psi (\phi)} {d (\phi)^2} = E \psi (\phi) \label{2}\] The kinetic energy of a body rotating in the xy-plane can be expressed as \[ E = \dfrac{L_z^2}{2I} \label{3}\] where \[I = mR^2\] is the moment of inertia and \( L_z\), the -component of angular momentum. (Since \( L = r \times p\), if and lie in the -plane, points in the -direction.) The structure of Equation \(\ref{2}\) suggests that this angular-momentum operator is given by \[ \hat{L_z} = -{i} \hbar \dfrac{\partial}{\partial \phi} \label{4}\] This result will follow from a more general derivation in the following Section. The Schrödinger equation (Equation \(\ref{2}\)) can now be written more compactly as \[ \psi \prime \prime \ (\phi) + m^2 \psi (\phi) = 0 \label{5}\] where \[m^2 \equiv 2IE/ \hbar^2\label{6}\] (Please do not confuse this variable m with the mass of the particle!) Possible solutions to (Equation \(\ref{5}\)) are \[ \psi (\phi) = \text{const}\, e^{\pm{i}m\phi} \label{7}\] For this wavefunction to be physically acceptable, it must be . Since \(\phi \) increased by any multiple of \(2\pi \) represents the same point on the ring, we must have \[ \psi (\phi + 2\pi ) = \psi (\phi) \label{8}\] and therefore \[e^{{i}m (\phi + 2\pi)} = e^{{i}m \phi} \label{9}\] This requires that \[e^{2\pi {i}m} = 1 \label{10}\] which is true only if \(m\) is an integer: \[ m = 0, \pm 1, \pm 2... \label{11} \] Using Equation \(\ref{6}\), this gives the quantized \[E_m = \dfrac{\hbar^2}{2I} m^2 \label{12}\] In contrast to the particle in a box, the eigenfunctions corresponding to \(+m \) and \(-m \) (Equation \(\ref{7}\)) are linearly independent, so both must be accepted. Therefore all eigenvalues, except \(E_0 \), are two-fold (or doubly) degenerate. The eigenfunctions can all be written in the form const \( e^{{i}m \phi} \), with allowed to take either positive and negative values (or 0), as in Equation \(\ref{10}\). The normalized eigenfunctions are \[ {\psi _{m}} (\phi) = \dfrac{1}{\sqrt{2 \pi}} e^{im\phi} \label{13}\] and can be verified to satisfy the normalization condition containing the complex conjugate \[ \int\limits_{0}^{2\pi} {\psi_{m}^*} (\phi) {\psi _{m}} (\phi) d\phi = 1 \] where we have noted that \( {\psi_{m}^*} (\phi) = (2\pi)^{-1/2} e^{-{i}m\phi} \). The mutual orthogonality of the functions (Equation \(\ref{13}\)) also follows easily, for \[\int\limits_{0}^{2\pi} {\psi_{m^\prime}^*} {\psi _{m}} (\phi) d\phi = \dfrac{1}{2\pi} \int\limits_{0}^{2\pi} e^{{i}(m-m^\prime) \phi} d\phi \] \[ = \dfrac{1}{2\pi} \int\limits_{0}^{2\pi} [cos(m-m^\prime)\phi + {i} sin(m-m^\prime)\phi] d\phi =0 \] for \( m^\prime \neq m \). The solutions in Equation \(\ref{12}\) are also eigenfunctions of the angular momentum operator (Equation \(\ref{4}\)), with \[ \hat{L_z} \psi_{m} (\phi) = m\hbar \psi_{m} (\phi), m = 0, \pm 1, \pm 2...\] This is a instance of a fundamental result in quantum mechanics, that any measured component of orbital angular momentum is restricted to integral multiples of \( \hbar \). The of the hydrogen atom, to be discussed in the next Chapter, can be derived from this principle alone. The benzene molecule consists of a ring of six carbon atoms around which six delocalized -electrons can circulate. A variant of the FEM for rings predicts the ground-state electron configuration which we can write as \( 1\pi^{2} 2\pi^{4} \), as shown here: The enhanced stability the benzene molecule can be attributed to the complete shells of \(\pi \)-electron orbitals, analogous to the way that noble gas electron configurations configuration fulfill Hückel's 4 cyclobutadiene cyclooctatetraene they contain partially-filled \[ \dfrac{hc}{\lambda} = E_2 - E_1 = \dfrac{\hbar^2}{2mR^2} {(2^2 -1^2)} \] The ring radius can be approximated by the C-C distance in benzene, 1.39 Å. We predict \( \lambda \approx \) 210 nm, whereas the experimental absorption has \( \lambda_{max} \approx \) 268 nm. The motion of a free particle on the surface of a sphere will involve components of angular momentum in three-dimensional space. Spherical polar coordinates provide the most convenient description for this and related problems with spherical symmetry. The position of an arbitrary point is described by three coordinates \( r , \theta, \phi \) as shown in Figure \(\Page {2}\). These are connected to Cartesian \[ x = r \sin\theta \cos \phi \] \[ y = r \sin \theta \sin \phi \] \[ z = r \cos \theta \] The radial variable represents the distance from to the origin, or the length of the vector : \[ r= \sqrt{x^2 +y^2 +z^2} \] The coordinate \( \theta \) is the angle between the vector and the -axis, similar to latitude in geography, but with = 0 and \( \theta = \pi \) corresponding to the North and South Poles, respectively. The angle describes the rotation of about the -axis, running from 0 to , similar to geographic longitude. The volume element in spherical polar coordinates is given by \[ d \tau = r^2 \sin \theta dr d \theta d\phi, \] \[ r \in \{0, \infty \} , \theta \in \{0, \pi\}, \phi \in \{0, 2\pi \} \] and represented graphically by the distorted cube in Figure \(\Page {1}\).
We also require the Laplacian operator \[ \nabla^{2} = \dfrac{1}{r^{2}} \dfrac{\partial}{\partial r} r^{2} \dfrac{\partial}{\partial r} + \dfrac{1}{r^2 sin \theta} \dfrac{\partial}{\partial \theta } sin \theta \dfrac{\partial}{\partial \theta } + \dfrac{1}{r^2 sin^2 \theta} \dfrac{\partial^2}{\partial \phi^2} \] A detailed derivation is given in Supplement 6. A particle of mass , free to move on the surface of a sphere of radius , can be located by the two angular variables \( \theta, \phi \). The Schrödinger equation therefore has the form \[ -\dfrac{\hbar^2}{2M} \nabla^{2} Y ({\theta , \phi}) = E Y ({\theta , \phi}) \] with the wavefunction conventionally written as \(Y ({\theta , \phi}) \). These functions are known as and have been used in applied mathematics long before quantum mechanics. Since \( = , a constant, the first term in the Laplacian does not contribute. The Schrödinger equation reduces to \[ \left\{ \dfrac{1}{sin \theta} \dfrac{\partial}{\partial \theta } sin \theta \dfrac{\partial}{\partial \theta } + \dfrac{1}{sin^2 \theta} \dfrac{\partial^2}{\partial \phi^2} + \lambda \right\} Y ({ \theta , \phi}) = 0 \] where \[ \lambda = \dfrac{2MR^2E}{\hbar^2} = \dfrac{2IE}{\hbar^2} \] again introducing the moment of inertia \( I = MR^2 \). The variables \( \theta \) and \( \phi \) can be separated in Equation \(\ref{22}\) after multiplying through by \( sin^2 \theta \). If we write \[ Y ({\theta , \phi }) = \Theta ({\theta }) \Phi ({\phi }) \] and follow the procedure used for the three-dimensional box, we find that dependence on \(\phi \) alone occurs in the term \[ \dfrac{\Phi^{\prime \prime} ({\phi)} }{\Phi ({\phi}) } = const \] This is identical in form to Equation \(\ref{5}\), with the constant equal to \( -m^2 \), and we can write down the analogous solutions \[ \Phi_{m}({\phi}) = \sqrt{\dfrac{1}{2 \pi}} e^{im \phi}, m=0, \pm 1, \pm 2 ... \] Substituting Equation \(\ref{24}\) into Equation \(\ref{22}\) and cancelling the functions \( \Phi ({\phi }) \), we obtain an ordinary differential equation for \( \Theta ({\theta }) \) \[ \left \{ \dfrac{1}{sin \theta} \dfrac{d}{d \theta} sin \theta \dfrac{d}{d \theta} - \dfrac{m^2}{sin^2 \theta} + \lambda \right \} \Theta ({\theta}) = 0 \] Consulting our friendly neighborhood mathematician, we learn that the single-valued, finite solutions to (Equation \(\ref{27}\)) are known as . The parameters \( \lambda \) and \(m\) are restricted to the values \[ \lambda = \ell ({ \ell + 1}) , \ell = 0, 1, 2 ... \] while \[ m = 0, \pm 1, \pm 2 ... \pm \ell ({2 \ell +1 values}) \] Putting Equation \(\ref{28}\) into Equation \(\ref{23}\), the allowed energy levels for a particle on a sphere are found to be \[ E_{\ell} = \dfrac{\hbar^2}{2I} \ell ({ \ell + 1}) \] Since the energy is independent of the second quantum number , the levels (Equation \(\ref{30}\)) are \( ({2 \ell+1}) \)-fold degenerate. The spherical harmonics constitute an orthonormal set satisfying the integral relations \[ \int_0^{\pi} \int_0^{2 \pi} Y_{\ell^{\prime} m^{\prime}}^* ({ \theta , \phi }) Y_{\ell m} ({\theta , \phi}) sin \theta d \theta d \phi = \delta_{\ell \ell^{\prime}} \delta_{mm^{\prime}} \] The following table lists the spherical harmonics through \( \ell \) = 2, which will be sufficient for our purposes. \[ Spherical Harmonics Y_{\ell m} ({\theta , \phi}) \] \[ Y_{00} = \left({\dfrac{1}{4 \pi}} \right)^{1/2} \] \[ Y_{10} = \left({\dfrac{3}{4 \pi}} \right)^{1/2} cos \theta \] \[ Y_{1 \pm 1} = \mp \left({\dfrac{3}{4 \pi}} \right)^{1/2} sin \theta e^{\pm i \phi} \] \[ Y_{20} = \left({\dfrac{5}{16 \pi}} \right)^{1/2} ({ 3 cos^2 \theta - 1}) \] \[ Y_{2 \pm 1} = \mp \left({\dfrac{15}{8 \pi}} \right)^{1/2} cos \theta sin \theta e^{\pm i \phi} \] \[ Y_{2 \pm 2} = \left({\dfrac{15}{32 \pi}} \right)^{1/2} sin^2 \theta e^{\pm 2i \phi} \] A graphical representation of these functions is given in Figure \(\Page {4}\). Surfaces of constant absolute value are drawn, positive where green and negative where red. Generalization of the energy-angular momentum relation in Equation \(\ref{3}\) to three dimensions gives \[ E = \dfrac{L^2}{2I} \] Thus from Equation \(\ref{21}\)-\(\ref{23}\) we can identify the operator for the square of total angular momentum \[\hat{L^2} = -\hbar^2 \left\{ \dfrac{1}{sin \theta} \dfrac{\partial}{\partial \theta} sin \theta \dfrac{\partial}{\partial \theta} + \dfrac{1}{sin^2 \theta} \dfrac{\partial^{2}}{\partial \phi^{2}} \right\} \] By Equations \(\ref{28}\) and \(\ref{29}\), the functions \( Y ({ \theta , \phi}) \) are simultaneous eigenfunctions of \( \hat{L^2} \) and \( \hat{L}_z \) such that and \[ \hat{L}_z Y_{\ell m} ({\theta, \phi}) = m \hbar Y_{\ell m} ({\theta , \phi}) \] But the \( Y_{\ell m} ({\theta , \phi}) \) are not eigenfunctions of either \( L_x \) and \(L_y \) (unless \(\ell \) = 0). Note that the magnitude of the total angular momentum \( \sqrt{\ell ({\ell +1}) } \hbar \) is greater than its maximum observable component in any direction, namely \(\ell \hbar \). The quantum-mechanical behavior of the angular momentum and its components can be represented by a vector model, illustrated in Figure 5. The angular momentum vector , with magnitude \( \sqrt{\ell ({\ell +1}) } \hbar \), can be pictured as precessing about the z-axis, with its -component \( L_z \) constant. The components \( L_x \) and \( L_y \) fluctuate in the course of precession, corresponding to the fact that the system is not in an eigenstate of either. There are 2\(\ell \) + 1 different allowed values for \( L_z \), with eigenvalues \( m \hbar ({ m = 0, \pm 1, \pm 2 ... \pm \ell }) \) equally spaced between \( + \ell \hbar \) and \( - \ell \hbar \). This discreteness in the allowed directions of the angular momentum vector is called . The existence of simultaneous eigenstates of \( \hat{L^2} \) and any one c omponent, conventionally \( \hat{L}_z \), is consistent with the commutation relations derived in Chap. 4: \[ \left[ \hat{L}_x , \hat{L}_y \right] = i \hbar \hat{L}_z et cyc\] and \[ \left[ \hat{L^2} , \hat{L}_z \right] = 0 \] The electron, as well as certain other fundamental particles, possesses an intrinsic angular momentum or , in addition to its orbital angular momentum. These two types of angular momentum are analogous to the daily and annual motions, respectively, of the Earth around the Sun. To distinguish the spin angular momentum from the orbital, we designate the quantum numbers as and \( m_s \), in place of \( \ell \) and . For the electron, the quantum number always has the value \( \dfrac{1}{2} \), while \(m_s \) can have one of two values,\( \pm \dfrac{1}{2} \). The electron is said to be an elementary particle of spin \( \dfrac{1}{2} \). The proton and neutron also have spin \( \dfrac{1}{2} \) and belong to the classification of particles called , which are governed by the Pauli exclusion principle. Other particles, including the photon, have integer values of spin and are classified as . These do obey the , so that an arbitrary number can occupy the same quantum state. A complete theory of spin requires relativistic quantum mechanics. For our purposes, it is sufficient to recognize the two possible internal states of the electron, which can be called `spin up' and `spin down.' These are designated, respectively, by \( \alpha \) and \( \beta \) as factors in the electron wavefunction. Spins play an essential role in determining the possible electronic states of atoms and molecules. | 13,052 | 1,572 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/05%3A_Energy_Changes_in_Chemical_Reactions/5.05%3A_Enthalpies_of_Solution |
Physical changes, such as melting or vaporization, and chemical reactions, in which one substance is converted to another, are accompanied by changes in enthalpy. Two other kinds of changes that are accompanied by changes in enthalpy are the dissolution of solids and the dilution of concentrated solutions. The dissolution of a solid can be described as follows:
\[ solute\left ( s \right ) + solvent\left ( l \right )\rightarrow soulution\left ( l \right ) \] The values of Δ for some common substances are given in Table \(\Page {1}\). The sign and the magnitude of Δ depend on specific attractive and repulsive interactions between the solute and the solvent; these factors will be discussed in Chapter 13. When substances dissolve, the process can be either exothermic (Δ < 0) or endothermic (Δ > 0), as you can see from the data in Table \(\Page {1}\). Substances with large positive or negative enthalpies of solution have commercial applications as instant cold or hot packs. Single-use versions of these products are based on the dissolution of either calcium chloride (CaCl , Δ = −81.3 kJ/mol) or ammonium nitrate (NH NO , Δ = +25.7 kJ/mol). Both types consist of a plastic bag that contains about 100 mL of water plus a dry chemical (40 g of CaCl or 30 g of NH NO ) in a separate plastic pouch. When the pack is twisted or struck sharply, the inner plastic bag of water ruptures, and the salt dissolves in the water. If the salt is CaCl , heat is released to produce a solution with a temperature of about 90°C; hence the product is an “instant hot compress.” If the salt is NH NO , heat is absorbed when it dissolves, and the temperature drops to about 0° for an “instant cold pack.” A similar product based on the of sodium acetate, not its dissolution, is marketed as a reusable hand warmer (Figure \(\Page {1}\) ). At high temperatures, sodium acetate forms a highly concentrated aqueous solution. With cooling, an unstable solution containing excess solute is formed. When the pack is agitated, sodium acetate trihydrate [CH CO Na·3H O] crystallizes, and heat is evolved:
\[ Na^{+}\left ( aq \right )+ CH_{3}CO_{2}^{-}\left ( aq \right ) + H_{2}O\left ( l \right ) \rightarrow CH_{3}CO_{2}Na\cdot \bullet H_{2}O\left ( s \right ) \quad \quad \Delta H = - \Delta H_{soln} = - 19.7 \; kJ/mol \] A bag of concentrated sodium acetate solution can be carried until heat is needed, at which time vigorous agitation induces crystallization and heat is released. The pack can be reused after it is immersed in hot water until the sodium acetate redissolves. \(\Page {1}\) The amount of heat released or absorbed when a substance is dissolved is not a constant; it depends on the final concentration of the solute. The Δ values given previously and in Table \(\Page {1}\) for example, were obtained by measuring the enthalpy changes at various concentrations and extrapolating the data to infinite dilution. Because Δ depends on the concentration of the solute, diluting a solution can produce a change in enthalpy. If the initial dissolution process is exothermic (Δ < 0), then the dilution process is also exothermic. This phenomenon is particularly relevant for strong acids and bases, which are often sold or stored as concentrated aqueous solutions. If water is added to a concentrated solution of sulfuric acid (which is 98% H SO and 2% H O) or sodium hydroxide, the heat released by the large negative Δ can cause the solution to boil. Dangerous spattering of strong acid or base can be avoided if the concentrated acid or base is slowly added to water, so that the heat liberated is largely dissipated by the water. Thus you should ; a useful way to avoid the danger is to remember: The is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure. Describe the distinction between Δ and Δ . Does adding water to concentrated acid result in an endothermic or an exothermic process? The following table lists ΔH values for some ionic compounds. If 1 mol of each solute is dissolved in 500 mL of water, rank the resulting solutions from warmest to coldest. | 4,191 | 1,573 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Addition_Reactions_of_Alkenes/Addition_of_Lewis_Acids_(Electrophilic_Reagents) |
The proton is not the only electrophilic species that initiates addition reactions to the double bond. Lewis acids like the halogens, boron hydrides and certain transition metal ions are able to bond to the alkene pi-electrons, and the resulting complexes rearrange or are attacked by nucleophiles to give addition products. The electrophilic character of the halogens is well known. Although fluorine is uncontrollably reactive, chlorine, bromine and to a lesser degree iodine react selectively with the double bond of alkenes. The addition of chlorine and bromine to alkenes, as shown in the following general equation, proceeds by an initial electrophilic attack on the pi-electrons of the double bond. Iodine adds reversibly to double bonds, but the equilibrium does not normally favor the addition product, so it is not a useful preparative method. Dihalo-compounds in which the halogens are juxtaposed in the manner shown are called , from the Latin , meaning neighboring. R C=CR + → R C -CR Other halogen containing reagents which add to double bonds include hypohalous acids, HOX, and sulfenyl chlorides, RSCl. These reagents are unsymmetrical, so their addition to unsymmetrical double bonds may in principle take place in two ways. In practice, these addition reactions are regioselective, with one of the two possible constitutionally isomeric products being favored. The electrophilic moiety of these reagents is the halogen. (CH ) C=CH + → (CH ) C -CH
(CH ) C=CH + C H → (CH ) C -CH C H The regioselectivity of the above reactions may be explained by the same mechanism we used to rationalize the Markovnikov rule. Thus, bonding of an electrophilic species to the double bond of an alkene should result in preferential formation of the more stable (more highly substituted) carbocation, and this intermediate should then combine rapidly with a nucleophilic species to produce the addition product. This is illustrated by the following equation. To apply this mechanism we need to determine the electrophilic moiety in each of the reagents. By using electronegativity differences we can dissect common addition reagents into electrophilic and nucleophilic moieties, as shown on the right. In the case of hypochlorous and hypobromous acids (HOX), these weak Brønsted acids (pK 's ca. 8) do not react as proton donors; and since oxygen is more electronegative than chlorine or bromine, the electrophile will be a halide cation. The nucleophilic species that bonds to the intermediate carbocation is then hydroxide ion, or more likely water (the usual solvent for these reagents), and the products are called halohydrins. Sulfenyl chlorides add in the opposite manner because the electrophile is a sulfur cation, RS , whereas the nucleophilic moiety is chloride anion (chlorine is more electronegative than sulfur). The addition products formed in reactions of alkenes with mercuric acetate and boron hydrides (compounds shown at the bottom of of the reagent list) are normally not isolated, but instead are converted to alcohols by a substitution reaction. These important synthetic transformations are illustrated for 2-methylpropene by the following equations, in which the electrophilic moiety is colored red and the nucleophile blue. The top reaction sequence illustrates the procedure and the bottom is an example of . The light blue vertical line separates the addition reaction on the left from the substitution on the right. The atoms or groups that have been added to the original double bond are colored orange in the final product. In both cases the overall reaction is the addition of water to the double bond, but the regioselectivity is reversed. The oxymercuration reaction gives the product predicted by Markovnikov's rule; hydroboration on the other hand gives the "anti-Markovnikov" product. Complementary reactions such as these are important because they allow us to direct a molecular transformation whichever way is desired. Mercury and boron are removed from the organic substrate in the second step of oxymercuration and hydroboration respectively. These reactions are seldom discussed in detail; however, it is worth noting that the mercury moiety is reduced to metallic mercury by borohydride (probably by way of radical intermediates), and boron is oxidized to borate by the alkaline peroxide. Addition of hydroperoxide anion to the electrophilic borane generates a tetra-coordinate boron peroxide, having the general formula R B-O-OH . This undergoes successive intramolecular shifts of alkyl groups from boron to oxygen, accompanied in each event by additional peroxide addition to electron deficient boron. The retention of configuration of the migrating alkyl group is attributed to the intramolecular nature of the rearrangement. Since the oxymercuration sequence gives the same hydration product as acid-catalyzed addition of water , we might question why this two-step procedure is used at all. The reason lies in the milder reaction conditions used for oxymercuration. The strong acid used for direct hydration may not be tolerated by other functional groups, and in some cases may cause molecular rearrangement . The addition of borane, BH , requires additional comment. In pure form this reagent is a dimeric gas B H , called diborane, but in ether or THF solution it is dissociated into a solvent coordinated monomer, R O-BH . Although diborane itself does not react easily with alkene double bonds, H.C. Brown (Purdue, Nobel Prize 1979) discovered that the solvated monomer adds rapidly under mild conditions. Boron and hydrogen have rather similar electronegativities, with hydrogen being slightly greater, so it is not likely there is significant dipolar character to the B-H bond. Since boron is electron deficient (it does not have a valence shell electron octet) the reagent itself is a Lewis acid and can bond to the pi-electrons of a double bond by displacement of the ether moiety from the solvated monomer. As shown in the following equation, this bonding might generate a dipolar intermediate consisting of a negatively-charged boron and a carbocation. Such a species would not be stable and would rearrange to a neutral product by the shift of a hydride to the carbocation center. Indeed, this hydride shift is believed to occur concurrently with the initial bonding to boron, as shown by the transition state drawn below the equation, so the discrete intermediate shown in the equation is not actually formed. Nevertheless, the carbocation stability rule cited above remains a useful way to predict the products from hydroboration reactions. . Note that this addition is unique among those we have discussed, in that it is a single-step process. Also, all three hydrogens in borane are potentially reactive, so that the alkyl borane product from the first addition may serve as the hydroboration reagent for two additional alkene molecules. | 6,930 | 1,574 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/Gen_Chem_Quantum_Theory/The_Periodic_Table |
The higher the electronegativity, the more attraction the element has towards bonding electrons. Bonding between two elements with a large electronegativity difference tends to be ionic. It is interesting to note that inert gases \(\ce{Kr}\) and \(\ce{Xe}\) have elenegativities similar to those of nitrogen and carbon. On the periodic table, there is a general trend. An element close to \(\ce{F}\) has a large electronegativity, whereas an element close to \(\ce{Cs}\) on the opposit corner of the periodic table from \(\ce{F}\) has the least electronegativity. The periodic table is a convenient way to correlate chemical properties. For example, from their position on the periodic table, we easily recognize them as metals, semimetals (metalloids), or nonmetals. Most elements are metals (M); the rest are metalloids (o), nonmetals (-), and inert gases (i), on the top right hand of the long periodic table. Look at some metals, and see if you can describe their characteristics. If not, please check some resource books to see if they give a description you like? are characterized by having a high melting point, high electric and heat conductivity, metallic bonding, being malleable and ductile, forming positive ions \(\ce{Cu^2+}\), \(\ce{Fe^3+}\), and forming alloys with one another. Check out a metal, and you'll be amazed how close it has come to your life. have low melting points, form molecules, and atoms in their solids are covalently bonded. They are poor conductors of heat and electricity, and they form negative ions or molecular compounds. Here are some non-metals. Note, however, that diamond is an exceptionally good heat conductor, but it does not conduct electricity. Noble gases: \(\ce{He}\), \(\ce{Ne}\), \(\ce{Ar}\), \(\ce{Kr}\), \(\ce{Xe}\), \(\ce{Rn}\)
Oxygen group: \(\ce{O}\), \(\ce{S}\), \(\ce{Se}\)
Nitrogen group: \(\ce{N}\), \(\ce{P}\), \(\ce{As}\)
Carbon group: \(\ce{C}\), \(\ce{Si}\)
Boron Between metals and nonmetals lie the metalloids as indicated above. The chemistry of these elements is best discussed in GROUPS such as alkali metals, carbon group, halogens, inert gases, etc. Hint: True, but paramagnetism and ferromagnetism may overcome diamagnetism.
Explain what is and how this type of material behaves in a magnetic field. Hint: Chromium is a paramagnetic element.
Paramagnetism is due to unpaired electrons in the material. Hint: \(\ce{Li}\)
Explain what is, and describe the trend for a group of elements.
It takes the most energy to remove an electron from \(\ce{Li}\) in this group. Which of the inert gases has the highest IE: \(\ce{He}\), \(\ce{Ne}\), \(\ce{Ar}\), \(\ce{Kr}\), or \(\ce{Rn}\)? Hint: \(\ce{Cl}\)
The electron affinities are given below: Hint: \(\ce{F}\)
You should know that \(\ce{F}\) is the most electronegative element, and \(\ce{Cs}\) is the least electronegative. Francium should be, but there is no natural occuring francium because its isotopes are all radioactive. Hint: The Pauling scale electronegativities are: oxygen, 3.4; chlorine, 3.2.
Due to the large electronegativity and the small size of oxygen atoms, oxygen compounds such as water and ethanol form hydrogen bonds. | 3,217 | 1,575 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/07%3A_Acids_bases_and_ions_in_aqueous_solution/7.01%3A_Introduction/7.1A%3A_Acid-Base_Theories_and_Concepts |
There are three primary theories of acid-base chemistry that are often taught together: Arrhenius theory, Brønsted-Lowry theory, and Lewis acid-base theory. Each theory is introduced below. The Swedish chemist Svante Arrhenius attributed the properties of acidity to hydrogen ions (\(\ce{H^{+}}\)) in 1884. An Arrhenius acid is a substance that, when added to water, increases the concentration of \(\ce{H^{+}}\) ions in the water. Note that chemists often write \(\ce{H^{+}(aq)}\) and refer to the hydrogen ion when describing acid-base reactions, but the free hydrogen nucleus does not exist alone in water. It exists in a hydrated form which for simplicity is often written as the hydronium (hydroxonium) ion, \(\ce{H3O^{+}}\). Thus, an Arrhenius acid can also be described as a substance that increases the concentration of hydronium ions when added to water. This definition stems from the equilibrium dissociation (self-ionization) of water into hydronium and hydroxide (\(\ce{OH^{-}}\)) ions: \[\ce{H_2O(l) + H_2O(l) ⇌ H_3O^{+}(aq) + OH^{-}(aq)} \nonumber\] with \(K_w\) defined as \(\ce{[H^{+},OH^{-}]}\). The value of \(K_w\) varies with temperature, as shown in the table below where at 25 °C \(K_w\) is approximately \(1.0 \times 10^{-14}\), i.e. \(pK_w= 14\). In pure water the majority of molecules are \(\ce{H2O}\), but the molecules are constantly dissociating and re-associating, and at any time a small number of the molecules (about 1 in 10 ) are hydronium and an equal number are hydroxide. Because the numbers are equal, pure water is neutral (not acidic or basic) and has an electrical conductivity of 5.5 microSiemen, μS/m. For comparison, sea water's conductivity is about one million times higher, 5 S/m (due to the dissolved salt). Although the term proton is often used for \(\ce{H^{+}}\), this should really be reserved for \(\ce{H}\) (protium) not \(\ce{D}\) (deuterium) or \(\ce{T}\) (tritium). The more general term, hydron covers all isotopes of hydrogen. An Arrhenius base, on the other hand, is a substance which increases the concentration of hydroxide ions when dissolved in water, hence decreasing the concentration of hydronium ions. To qualify as an Arrhenius acid, upon the introduction to water, the chemical must either cause, directly or otherwise: Conversely, to qualify as an Arrhenius base, upon the introduction to water, the chemical must either cause, directly or otherwise: The definition is expressed in terms of an equilibrium expression: \[\text{acid} + \text{base} ⇌ \text{conjugate base} + \text{conjugate acid}. \nonumber\] With an acid, \(\ce{HA}\), the equation can be written symbolically as: \[\ce{HA + B ⇌ A^{-} + HB^{+}} \nonumber\] The double harpoons sign, \(\ce{<=>}\), is used because the reaction can occur in both forward and backward directions. The acid, \(\ce{HA}\), can lose a hydron to become its conjugate base, \(\ce{A^{-}}\). The base, \(\ce{B}\), can accept a hydron to become its conjugate acid, \(\ce{HB^{+}}\). Most acid-base reactions are fast so that the components of the reaction are usually in dynamic equilibrium with each other. While the Arrhenius concept is useful for describing many reactions, it has limitations. In 1923, chemists Johannes Nicolaus Brønsted and Thomas Martin Lowry independently recognized that acid-base reactions involve the transfer of a hydron. A Brønsted-Lowry acid (or simply Brønsted acid) is a species that donates a hydron to a Brønsted-Lowry base. The Brønsted-Lowry acid-base theory has several advantages over the Arrhenius theory. Consider the following reactions of acetic acid (\(\ce{CH3COOH}\)): \[\ce{CH_3COOH + H_2O ⇌ CH_3COO^{-} + H_3O^{+}} \nonumber\] \[\ce{CH_3COOH + NH_3 ⇌ CH_3COO^{-} + NH_4^{+}} \nonumber\] Both theories easily describe the first reaction: \(\ce{CH3COOH}\) acts as an Arrhenius acid because it acts as a source of \(\ce{H3O^{+}}\) when dissolved in water, and it acts as a Brønsted acid by donating a hydron to water. In the second example \(\ce{CH3COOH}\) undergoes the same transformation, in this case donating a hydron to ammonia (\(\ce{NH3}\)), but it cannot be described using the Arrhenius definition of an acid because the reaction does not produce hydronium ions. To qualify as an Brønsted-Lowry acid, the chemical must either cause, directly or otherwise: Conversely, to qualify as an Brønsted-Lowry base, the chemical must either cause, directly or otherwise: A third concept was proposed in 1923 by Gilbert N. Lewis which includes reactions with acid-base characteristics that do not involve a hydron transfer. A Lewis acid is a species that reacts with a Lewis base to form a Lewis adduct. The Lewis acid accepts a pair of electrons from another species; in other words, it is an electron pair acceptor. Brønsted acid-base reactions involve hydron transfer reactions while Lewis acid-base reactions involve electron pair transfers. All Brønsted acids are Lewis acids, but not all Lewis acids are Brønsted acids. \[\ce{BF3 + F^{-} <=> BF4^{-}} \nonumber\] \[\ce{NH3 + H^{+} <=> NH4^{+}} \nonumber\] In the first example \(\ce{BF3}\) is a Lewis acid since it accepts an electron pair from the fluoride ion. This reaction cannot be described in terms of the Brønsted theory because there is no hydron transfer. The second reaction can be described using either theory. A hydron is transferred from an unspecified Brønsted acid to ammonia, a Brønsted base; alternatively, ammonia acts as a Lewis base and transfers a lone pair of electrons to form a bond with a hydrogen ion. To qualify as an Lewis acid, the chemical must Conversely, to qualify as an Lewis base, the chemical must: This acid-base theory was a revival of the oxygen theory of acids and bases, proposed by German chemist Hermann Lux in 1939 and further improved by Håkon Flood circa 1947. It is still used in modern geochemistry and for the electrochemistry of molten salts. This definition describes an acid as an oxide ion (\(\ce{O^{2-}}\)) acceptor and a base as an oxide ion donor. For example: \[\ce{MgO (base) + CO2 (acid) <=> MgCO3} \nonumber\] \[\ce{CaO (base) + SiO2 (acid) <=> CaSiO3}\nonumber\] \[\ce{NO3^{-} (base) + S2O7^{2-} (acid) <=> NO2+ + 2 SO4^{2-}}\nonumber\] Mikhail Usanovich developed a general theory that does not restrict acidity to hydrogen-containing compounds, and his approach, published in 1938, was even more general than the Lewis theory. Usanovich's theory can be summarized as defining an acid as anything that or , and a base as the reverse. This definition could even be applied to the concept of redox reactions (oxidation-reduction) as a special case of acid-base reactions. Some examples of Usanovich acid-base reactions include: A comparison of the above definitions of Acids and Bases shows that the Usanovich concept encompasses all of the others but some feel that because of this it is too general to be useful. | 6,903 | 1,576 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/05%3A_Energy_Changes_in_Chemical_Reactions/5.02%3A_Enthalpy_and_Reactions |
To study the flow of energy during a chemical reaction, we need to distinguish between a system , the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its surroundings , the rest of the universe, including the container in which the reaction is carried out (Figure \(\Page {1}\) ). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa. Three kinds of systems are important in chemistry. An open system can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A closed system can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An isolated system exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, , which must be true if . The state of a system is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A state function is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are . For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (Figure \(\Page {2}\) ). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose. The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel (see chapter opening image). The balanced chemical equation for the reaction is as follows:
\[ 2Al(s)+Fe_{2}O_{3}(s) \rightarrow 2Fe(s)+Al_{2}O_{3}(s) \] We can also write this chemical equation as
\[ 2Al(s)+Fe_{2}O_{3}(s) \rightarrow 2Fe(s)+Al_{2}O_{3}(s)+ heat \] to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called . In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings. A process in which heat ( ) is transferred a system its surroundings is described as exothermic . By convention, < 0 for an exothermic reaction. When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation:
\[ heat + H_{2}O\left (s \right ) \rightarrow H_{2}O\left (l \right ) \] When heat is transferred a system its surroundings, the process is endothermic . By convention, > 0 for an endothermic reaction. We have stated that the change in energy (Δ ) is equal to the sum of the heat produced and the work performed ( ). Work done by an expanding gas is called , also called . Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows:
\[ Cu\left (s \right ) + 4HNO_{3}\left (aq \right ) \rightarrow Cu(NO_{3})_{2}\left (aq \right ) + 2H_{2}O\left (l \right ) + 2NO_{3}\left (g \right ) \] If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure \(\Page {3}\) ). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., ). We find the amount of work done by multiplying the external pressure by the change in volume caused by movement of the piston (Δ ). At a constant external pressure (here, atmospheric pressure)
\[ w = -P\Delta V \] The negative sign associated with work done indicates that the system loses energy. If the volume increases at constant pressure (Δ > 0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (Δ < 0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy. The symbol in Equation \(\Page {3}\) represents the internal energy of a system, which is the sum of the kinetic energy and potential energy of all its components. Additionally, Δ = + , where is the heat produced by the system and w is the work performed by the system. It is the change in internal energy that produces heat plus work. Substituting Equation 8.1.5 we find that \[ \Delta E = q-P \Delta V \] Thus, if a reaction were carried out in a constant volume system (a pressure cooker for example) \[ \Delta E = q_{v} \] Where the subscipt v in q indicates that the process is carried out at constant volume. Thus, if the volume is held constant, the change in internal energy is equal to the heat flowing into or out of the system. However, we live in a constant pressure world, not a constant volume one. The atmospheric pressure at the surface of the earth where we live is roughly constant allowing for variation in barometric pressure due to weather or altitude. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy ( ) (from the Greek , meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy plus the product of its pressure and volume :
\[ H=E+PV \] Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function. If a chemical change occurs at constant pressure (for a given , Δ = 0), the change in enthalpy (Δ ) is
\[ \Delta H = \Delta (E + PV) = \Delta E + \Delta PV = \Delta E + P\Delta V \] Substituting + for Δ (Equation \(\Page {9}\) ) and − for Δ (Equation \(\Page {5}\) ), we obtain
\[ \Delta H = \Delta E + P\Delta V = q_{p} + w − w = q_{p} \] The subscript is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation \(\Page {10}\) we see that at constant pressure the change in enthalpy, Δ of the system, defined as − , is equal to the heat gained or lost.
\[\Delta H = H_{final} -H_{initial} = q_{p} \] Just as with Δ , because enthalpy is a state function, the magnitude of Δ depends on only the initial and final states of the system, not on the path taken. The enthalpy change is the same even if the process does occur at constant pressure. Most importantly, the change in enthalpy for a reaction can be determined by measuring the flow of heat into or out of the system. As we will see below, there are also cases where we want to know the amount of heat generated by a reaction, as for example for combustion of fuel. In that case knowing the molar change in enthalpy provides the information needed. Finally, looking at Equation 9.2.9 shows that for reactions where the volume does not change, or better put does not change much, ~ . As a practical matter, this includes all reactions where all the products and reactants are either in the solid, liquid or aqueous phases. This is not generally true when there are gases on either side of the equation, a question which we will discuss in the chapter on gases. To find Δ , measure . When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction (Δ ) , the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so Δ is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so Δ is positive. Thus Δ < 0 , and Δ > 0 . In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table: If Δ is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, (part (a) in Figure \(\Page {4}\) ). Conversely, if Δ is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, (part (b) in Figure \(\Page {4}\) ). Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion. Bond breaking requires an input of energy; bond making releases energy.
\[ \begin{matrix}
heat+ H_{2}O(s) \rightarrow H_{2}O(l) & \Delta H > 0
\end{matrix} \]
\[ \begin{matrix}
H_{2}O(l) \rightarrow H_{2}O(s) + heat & \Delta H < 0
\end{matrix} \] In both cases, the of the enthalpy change is the same; only the is different.
\[ 2Al\left (s \right )+Fe_{2}O_{3}\left (s \right ) \rightarrow 2Fe\left (s \right )+Al_{2}O_{3}\left (s \right )+ 815.5 \; kJ \] Thus Δ = −851.5 kJ/mol of Fe O . We can also describe Δ for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for Δ , in kilojoules rather than kilojoules per mole, is written after the reaction, as in , it is the value of Δ corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation:
\[ 2Al\left (s \right )+Fe_{2}O_{3}\left (s \right ) \rightarrow 2Fe\left (s \right )+Al_{2}O_{3}\left (s \right ) \;\;\;\; \Delta H_{rxn}= - 851.5 \; kJ \] If 4 mol of Al and 2 mol of Fe O react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows:
\[ - \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \] The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example \(\Page {1}\). Certain parts of the world, such as southern California and Saudi Arabia, are short of freshwater for drinking. One possible solution to the problem is to tow icebergs from Antarctica and then melt them as needed. If Δ is 6.01 kJ/mol for the reaction H O(s) → H O(l) at 0°C and constant pressure, how much energy would be required to melt a moderately large iceberg with a mass of 1.00 million metric tons (1.00 × 10 metric tons)? (A metric ton is 1000 kg.) energy per mole of ice and mass of iceberg energy required to melt iceberg Calculate the number of moles of ice contained in 1 million metric tons (1.00 × 10 metric tons) of ice. Calculate the energy needed to melt the ice by multiplying the number of moles of ice in the iceberg by the amount of energy required to melt 1 mol of ice. Because enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given Δ for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—so we need to calculate the number of moles of ice in the iceberg and multiply that number by Δ (+6.01 kJ/mol): \( \begin{matrix}
moles \; H_{2}O & = & 1.00\times 10^{6} \; metric \; tons H_{2}O \left ( \dfrac{1000 \; \cancel{kg}}{1 \; \cancel{metric \; ton}} \right ) \left ( \dfrac{1000 \; \cancel{g}}{1 \; \cancel{kg}} \right ) \left ( \dfrac{1 \; mol \; H_{2}O}{18.015 \; \cancel{g \; H_{2}O}} \right )\\
& = & 5.55\times 10^{10} \; mol H_{2}O
\end{matrix} \) The energy needed to melt the iceberg is thus \( \left ( \dfrac{6.01 \; kJ}{\cancel{mol \; H_{2}O}} \right )\left ( 5.55 \times 10^{10} \; \cancel{mol \; H_{2}O} \right )= 3.34 \times 10^{11} \; kJ \) Because so much energy is needed to melt the iceberg, this plan would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of different energy sources equivalent to the amount of energy needed to melt the iceberg are shown in the table below. If 17.3 g of powdered aluminum are allowed to react with excess Fe O , how much heat is produced? 273 kJ Equation \(\Page {6}\) : = + Equation \(\Page {5}\) : = − Δ Equation \(\Page {8}\) : Δ = Δ + Δ Equation \(\Page {9}\) : Δ = In chemistry, the small part of the universe that we are studying is the , and the rest of the universe is the . can exchange both matter and energy with their surroundings, can exchange energy but not matter with their surroundings, and can exchange neither matter nor energy with their surroundings. A is a property of a system that depends on only its present , not its history. A reaction or process in which heat is transferred from a system to its surroundings is . A reaction or process in which heat is transferred to a system from its surroundings is . is a state function used to measure the heat transferred from a system to its surroundings or vice versa at constant pressure. Only the can be measured. A negative Δ means that heat flows from a system to its surroundings; a positive Δ means that heat flows into a system from its surroundings. For a chemical reaction, the is the difference in enthalpy between products and reactants; the units of Δ are kilojoules per mole. Reversing a chemical reaction reverses the sign of Δ . The magnitude of Δ also depends on the physical state of the reactants and the products because processes such as melting solids or vaporizing liquids are also accompanied by enthalpy changes: the and the , respectively. The overall enthalpy change for a series of reactions is the sum of the enthalpy changes for the individual reactions, which is . The is the enthalpy change that occurs when a substance is burned in excess oxygen. The ) is the enthalpy change that accompanies the formation of a compound from its elements. ( ) are determined under : a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is defined to be zero. The ( ) can be calculated from the sum of the of the products (each multiplied by its stoichiometric coefficient) minus the sum of the standard enthalpies of formation of the reactants (each multiplied by its stoichiometric coefficient)—the “products minus reactants” rule. The is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure. Heat implies the flow of energy from one object to another. Describe the energy flow in an a. exothermic reaction. b. endothermic reaction. When a thermometer is suspended in an insulated thermos that contains a block of ice, the temperature recorded on the thermometer drops. Describe the direction of heat flow. In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. State whether each process is endothermic or exothermic. In each scenario, the system is defined as the mixture of chemical substances that undergoes a reaction. Determine whether each process is endothermic or exothermic. Is Earth’s environment an isolated system, an open system, or a closed system? Explain your answer. Why is it impossible to measure the absolute magnitude of the enthalpy of an object or a compound? Determine whether energy is consumed or released in each scenario. Explain your reasoning. The chapter states that enthalpy is an extensive property. Why? Describe a situation that illustrates this fact. The enthalpy of a system is affected by the physical states of the reactants and the products. Explain why. Is the distance a person travels on a trip a state function? Why or why not? | 17,942 | 1,578 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/18%3A_Carboxylic_Acids_and_Their_Derivatives/18.05%3A_Decarboxylation_of_Carboxylic_Acids |
The decarboxylation of \(\ce{RCO_2H}\) to give \(\ce{RH}\) and \(\ce{CO_2}\) can be calculated from bond energies and the stabilization energy of the carboxyl group to have \(\Delta H^0 = -7 \: \text{kcal mol}^{-1}\). This does not mean that the reaction goes easily. Special structural features are required. The simple aliphatic carboxylic acids do not lose carbon dioxide on heating, but when there are strongly electron-attracting groups attached to the \(\alpha\) carbon, decarboxylation often proceeds readily at \(100\)-\(150^\text{o}\). Examples include 3-Butenoic acid also undergoes decarboxylation but has to be heated to above \(200^\text{o}\): The mechanisms of thermal decarboxylation probably are not the same for all cases, but when the acid has a function such as \(\ce{O=C}\), \(\ce{N=C}\), \(\ce{O=N}\), or \(\ce{C=C}\) attached to the \(\alpha\) carbon then a cyclic elimination process appears to occur. For propanedioic acid the process is Carboxylate radicals can be generated in several ways. One is the thermal decomposition of diacyl peroxides, which are compounds with rather weak \(\ce{O-O}\) bonds: Another method involves of sodium or potassium carboxylate solutions, known as , in which carboxylate radicals are formed by transfer of an electron from the carboxylate ion to the anode. Decarboxylation may occur simultaneously with, or subsequent to, the formation of carboxylate radicals, leading to hydrocarbon radicals, which subsequently dimerize: An example is Decarboxylation of the silver salts of carboxylic acids in the presence of bromine or chlorine, the , often is useful for the synthesis of alkyl halides: The mechanism of this reaction seems to involve formation of carboxylate radicals through decomposition of an acyl hypobromic intermediate, \(12\): The Hunsdiecker reaction has certain disadvantages, mainly because it requires use of the pure silver salt, which is often difficult to prepare. With some acids, however, excellent results can be obtained using the acid itself and an excess of red mercuric oxide in place of the silver salt. or by heating the acid with lead tetraethanoate, \(\ce{Pb(O_2CCH_3)_4}\), and iodine, A somewhat similar decarboxylation reaction with formation of an alkene can be achieved by heating a carboxylic acid with lead tetraethanoate, \(\ce{Pb(O_2CCH_3)_4}\), in the presence of a catalytic amount of \(\ce{Cu(OCH_3)_2}\). A useful example is There is some competing decarboxylation of the ethanoic acid, but the conversions in this kind of reaction are usually good. The key steps in the reaction probably are exchange of carboxylic acid groups on tetravalent lead, cleavage of the \(\ce{Pb-O}\) bond to give the carboxylate radical, decarboxylation, oxidation of the alkyl radical by \(\ce{Cu}\) (II) to give the cation \(\left[ \ce{R} \cdot + \ce{Cu} (II) \rightarrow \ce{R}^\oplus + \ce{Cu} (I) \right]\), and finally loss of a proton to form the alkene. and (1977) | 2,980 | 1,579 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/07%3A_Further_Aspects_of_Covalent_Bonding/7.12%3A_Polarity_in_Polyatomic_Molecules/7.12.02%3A_Lecture_Demonstrations |
Add Magic Sand to water and observe "rabbit guts" like formations that suggest the importance of water in biogenesis. Add magic sand to isopropanol or hexane to demonstrate that it is the water, not the sand, that is "magical". A 2 cm x 2 cm square of index card is coated thoroughly on one side with graphite from a #2 pencil. About 100 mL each of water and carbon tetrachloride are added to a 250 mL beaker. If the square held in tweezers and is released at the interface of the water and carbon tetrachloride, it will invariable flip so that the bare paper side is toward the polar water, and the nonpolar graphite side is toward the nonpolar liquid. This is not a gravity effect, as proven by using hexane (floats on water) and water in the same experiment. Ed Vitz (Kutztown University), (University of | 836 | 1,580 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.06%3A_Controlling_the_Products_of_Reactions |
Whether in the synthetic laboratory or in industrial settings, one of the primary goals of modern chemistry is to control the identity and quantity of the products of chemical reactions. For example, a process aimed at synthesizing ammonia is designed to maximize the amount of ammonia produced using a given amount of energy. Alternatively, other processes may be designed to minimize the creation of undesired products, such as pollutants emitted from an internal combustion engine. To achieve these goals, chemists must consider the competing effects of the reaction conditions that they can control. One way to obtain a high yield of a desired compound is to make the reaction rate of the desired reaction much faster than the reaction rates of any other possible reactions that might occur in the system. Altering reaction conditions to control reaction rates, thereby obtaining a single product or set of products, is called kinetic control A second approach, called thermodynamic control , consists of adjusting conditions so that at equilibrium only the desired products are present in significant quantities. An example of thermodynamic control is the Haber-Bosch process used to synthesize ammonia via the following reaction:
\[N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \tag{15.6.1a}\] with \[ΔH_{rxn}=−91.8\; kJ/mol \tag{15.6.1b}\] Because the reaction converts 4 mol of gaseous reactants to only 2 mol of gaseous product, Le Chatelier’s principle predicts that the formation of NH will be favored when the pressure is increased. The reaction is exothermic, however (Δ = −91.8 kJ/mol), so the equilibrium constant decreases with increasing temperature, which causes an equilibrium mixture to contain only relatively small amounts of ammonia at high temperatures ( ). Taken together, these considerations suggest that the maximum yield of NH will be obtained if the reaction is carried out at as low a temperature and as high a pressure as possible. Unfortunately, at temperatures less than approximately 300°C, where the equilibrium yield of ammonia would be relatively high, the reaction is too to be of any commercial use. The industrial process therefore uses a mixed oxide (Fe O /K O) catalyst that enables the reaction to proceed at a significant rate at temperatures of 400°C–530°C, where the formation of ammonia is less unfavorable than at higher temperatures. Because of the low value of the equilibrium constant at high temperatures (e.g., = 0.039 at 800 K), there is no way to produce an equilibrium mixture that contains large proportions of ammonia at high temperatures. We can, however, control the temperature and the pressure while using a catalyst to convert a fraction of the N and H in the reaction mixture to NH , as is done in the Haber-Bosch process. This process also makes use of the fact that the product—ammonia—is less volatile than the reactants. Because NH is a liquid at room temperature at pressures greater than 10 atm, cooling the reaction mixture causes NH to condense from the vapor as liquid ammonia, which is easily separated from unreacted N and H . The unreacted gases are recycled until complete conversion of hydrogen and nitrogen to ammonia is eventually achieved. is a simplified layout of a Haber-Bosch process plant. The Sohio acrylonitrile process, in which propene and ammonia react with oxygen to form acrylonitrile, is an example of a kinetically controlled reaction:
\[\ce{CH_2=CHCH_{3(g)} + NH_{3(g)} + 3/2 O_{2(g)} <=> CH_2=CHC≡N_{(g)} + 3H_2O_{(g)}} \tag{15.6.2}\] Like most oxidation reactions of organic compounds, this reaction is highly exothermic (Δ ° = −519 kJ/mol) and has a very large equilibrium constant ( = 1.2 × 10 ). Nonetheless, the reaction shown in is not the reaction a chemist would expect to occur when propene or ammonia is heated in the presence of oxygen. Competing combustion reactions that produce CO and N from the reactants, such as those shown in and , are even more exothermic and have even larger equilibrium constants, thereby reducing the yield of the desired product, acrylonitrile:
\[ \ce{CH_2=CHCH_{3(g)} + 9/2 O_{2(g)} <=> 3CO_{2(g)} + 3H_2O_{(g)}} \tag{15.6.3a}\] with \[ΔH^°=−1926.1 \;kJ/mol \tag{15.6.3b}\] and \[K=4.5 \times 10^{338} \tag{15.6.3c}\]
\[\ce{2NH_{3(g)} + 3 O_{2(g)} <=> N_{2(g)} + 6H_2O_(g)} \tag{15.6.4a}\] with \[ΔH°=−1359.2 \;kJ/mol, \tag{15.6.4c}\] and \[K=4.4 \times 10^{234} \tag{15.6.4c}\] In fact, the formation of acrylonitrile in is accompanied by the release of approximately 760 kJ/mol of heat due to partial combustion of propene during the reaction. The Sohio process uses a catalyst that selectively accelerates the rate of formation of acrylonitrile without significantly affecting the reaction rates of competing combustion reactions. Consequently, acrylonitrile is formed more rapidly than CO and N under the optimized reaction conditions (approximately 1.5 atm and 450°C). The reaction mixture is rapidly cooled to prevent further oxidation or combustion of acrylonitrile, which is then washed out of the vapor with a liquid water spray. Thus controlling the kinetics of the reaction causes the desired product to be formed under conditions where equilibrium is not established. In industry, this reaction is carried out on an enormous scale. Acrylonitrile is the building block of the polymer called , found in all the products referred to collectively as , whose wide range of uses includes the synthesis of fibers woven into clothing and carpets. Controlling the amount of product formed requires that both thermodynamic and kinetic factors be considered. Recall that methanation is the reaction of hydrogen with carbon monoxide to form methane and water: \[CO_{(g)} +3H_{2(g)} \rightleftharpoons CH_{4(g)}+H_2O_{(g)}\] This reaction is the reverse of the steam reforming of methane described in Example 14. The reaction is exothermic (Δ ° = −206 kJ/mol), with an equilibrium constant at room temperature of = 7.6 × 10 . Unfortunately, however, CO and H do not react at an appreciable rate at room temperature. What conditions would you select to maximize the amount of methane formed per unit time by this reaction? balanced chemical equation and values of Δ ° and conditions to maximize yield of product Consider the effect of changes in temperature and pressure and the addition of an effective catalyst on the reaction rate and equilibrium of the reaction. Determine which combination of reaction conditions will result in the maximum production of methane. The products are highly favored at equilibrium, but the rate at which equilibrium is reached is too slow to be useful. You learned in that the reaction rate can often be increased dramatically by increasing the temperature of the reactants. Unfortunately, however, because the reaction is quite exothermic, an increase in temperature will shift the equilibrium to the left, causing more reactants to form and relieving the stress on the system by absorbing the added heat. If we increase the temperature too much, the equilibrium will no longer favor methane formation. (In fact, the equilibrium constant for this reaction is very temperature sensitive, decreasing to only 1.9 × 10 at 1000°C.) To increase the reaction rate, we can try to find a catalyst that will operate at lower temperatures where equilibrium favors the formation of products. Higher pressures will also favor the formation of products because 4 mol of gaseous reactant are converted to only 2 mol of gaseous product. Very high pressures should not be needed, however, because the equilibrium constant favors the formation of products. Thus optimal conditions for the reaction include carrying it out at temperatures greater than room temperature (but not too high), adding a catalyst, and using pressures greater than atmospheric pressure. Industrially, catalytic methanation is typically carried out at pressures of 1–100 atm and temperatures of 250°C–450°C in the presence of a nickel catalyst. (At 425°C°C, is 3.7 × 10 , so the formation of products is still favored.) The synthesis of methane can also be favored by removing either H O or CH from the reaction mixture by condensation as they form. Exercise As you learned in Example 10, the water–gas shift reaction is as follows: = 0.106 and Δ = 41.2 kJ/mol at 700 K. What reaction conditions would you use to maximize the yield of carbon monoxide? high temperatures to increase the reaction rate and favor product formation, a catalyst to increase the reaction rate, and atmospheric pressure because the equilibrium will not be greatly affected by pressure Changing conditions to affect the reaction rates to obtain a single product is called of the system. In contrast, is adjusting the conditions to ensure that only the desired product or products are present in significant concentrations at equilibrium. A reaction mixture will produce either product A or B depending on the reaction pathway. In the absence of a catalyst, product A is formed; in the presence of a catalyst, product B is formed. What conclusions can you draw about the forward and reverse rates of the reaction that produces A versus the reaction that produces B in (a) the absence of a catalyst and (b) the presence of a catalyst? Describe how you would design an experiment to determine the equilibrium constant for the synthesis of ammonia: \[N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \notag \] The forward reaction is exothermic (Δ ° = −91.8 kJ). What effect would an increase in temperature have on the equilibrium constant? What effect does a catalyst have on each of the following? How can the ratio / be used to determine in which direction a reaction will proceed to reach equilibrium? Industrial reactions are frequently run under conditions in which competing reactions can occur. Explain how a catalyst can be used to achieve reaction selectivity. Does the ratio / for the selected reaction change in the presence of a catalyst? The oxidation of acetylene via has Δ ° = −2600 kJ. What strategy would you use with regard to temperature, volume, and pressure to maximize the yield of product? The oxidation of carbon monoxide via has Δ ° = −283 kJ. If you were interested in maximizing the yield of CO , what general conditions would you select with regard to temperature, pressure, and volume? You are interested in maximizing the product yield of the system = 280 and Δ ° = −158 kJ. What general conditions would you select with regard to temperature, pressure, and volume? If SO has an initial concentration of 0.200 M and the amount of O is stoichiometric, what amount of SO is produced at equilibrium? Use low temperature and high pressure (small volume). | 10,767 | 1,583 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Calorimetry/00%3A_Front_Matter/02%3A_InfoPage |
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 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 integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 1,584 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/15%3A_Alcohols_and_Ethers/15.09%3A_Unsaturated_Alcohols_-_Alkenols |
The simplest unsaturated alcohols, ethenol (vinyl alcohol), is unstable with respect to ethanal and has never been isolated (see and ): Other simple unsaturated alkenols (enols) also rearrange to carbonyl compounds. However, ether and ester derivatives of enols are known and can be prepared by the addition of alcohols and carboxylic acids to alkynes. The esters are used to make many commercially important polymers ( ): The enol of 2-oxopropanoic acid (pyruvic acid) is of special biological interest because the phosphate ester of this compound is, like ATP ( ), a reservoir of chemical energy that can be utilized by coupling its hydrolysis \(\left( \Delta G^0 = -13 \: \text{kcal} \right)\) to thermodynamically less favorable reactions: In fact, the ester can be utilized to synthesize ATP from ADP; that is, it is a phosphorylating agent, and a more powerful one than ATP: Enols usually are unstable and are considerably more acidic than saturated alcohols. This means that the conjugate bases of the enols (the ) are more stable relative to the enols themselves than are alkoxide ions relative to alcohols. (Enolate anions are important reagents in the chemistry of carbonyl compounds and will be discussed in detail in .) The important factor here is delocalization of the negative charge on oxygen of enolate anions, as represented by the valence-bond structures \(14a\) and \(14b\): Because acidity depends on the in energy of the acid and its conjugate base, we must be sure that the stabilization of the enolate anion by electron delocalization represented by \(14a\) and \(14b\) is greater than the analogous stabilization of the neutral enol represented by \(15a\) and \(15b\): The rules for evaluating valence-bond structures ( ) tell us that the stabilization will be greatest when there are two or more nearly equivalent low-energy electron-pairing schemes. Inspection of \(14a\) and \(14b\) suggests that they will be more nearly equivalent than \(15a\) and \(15b\) because, although \(14b\) and \(15b\) have a negative charge on the carbon, in \(15b\) the oxygen has a positive charge. Another way of putting it is that \(15b\) represents an electron-pairing scheme with a , which intuitively is of higher energy than \(15a\) with no charge separation. Structures corresponding to \(14b\) and \(15b\) are not possible for saturated alkanols or their anions, hence we can see that enols should be more acidic than alcohols. Ascorbic acid (Vitamin C) is an example of a stable and quite acidic enol, or rather an enediol. It is a di-acid with p\(K_\text{a}\) values of 4.17 and 11.57: Other important examples of stable enol-type compounds are the aromatic alcohols, or . The \(K_\text{a}\)’s of these compounds are about \(10^{-10}\), some \(10^8\) times larger than the \(K_\text{a}\)’s for alcohols. The chemistry of these compounds, including their stability as enols, is discussed in . and (1977) | 2,949 | 1,585 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Electrophilic_Substitution_Reactions/E._An_Industrial_Alkylation_of_Benzene |
This page gives you the facts and simple, uncluttered mechanisms for the electrophilic substitution reaction between benzene and alkenes in the presence of a mixture of aluminium chloride and hydrogen chloride as the catalyst. Industrially, alkyl groups can be substituted into a benzene ring using a variant on Friedel-Crafts alkylation. One possibility is that instead of using a chloroalkane with an aluminium chloride catalyst, they use an alkene and a mixture of aluminium chloride and hydrogen chloride as the catalyst. This is a cheaper method because it saves having to make the chloroalkane first. To put an ethyl group on the ring (to make ethylbenzene), benzene is treated with a mixture of ethene, HCl and aluminium chloride. or better: The aluminium chloride and HCl aren't written into these equations because they are acting as catalysts. If you wanted to include them, you could write AlCl and HCl over the top of the arrow. The electrophile is CH CH . It is formed by reaction between the ethene and the HCl - exactly as if you were beginning to add the HCl to the ethene. The chloride ion is immediately picked up by the aluminium chloride to form an AlCl ion. That prevents the chloride ion from reacting with the CH CH ion to form chloroethane. The hydrogen is removed by the AlCl ion which was formed at the same time as the CH CH electrophile. The aluminium chloride and hydrogen chloride catalysts are re-generated in this second stage. The problem with more complicated alkenes like propene is that you have to be careful about the structure of the product. In each case, you can only really be sure of that structure if you work through the mechanism first. For example, the propyl group becomes attached to the ring via its middle carbon atom - and not its end one. You still need a mixture of aluminium chloride and hydrogen chloride as catalysts. When the propene reacts with the HCl, the hydrogen becomes attached to the end carbon atom. A secondary carbocation (carbonium ion) is formed because it is more stable than the primary one which would have been formed if the addition was the other way round. Because the positive charge is on the centre carbon atom, that is the one which will become attached to the ring. Again, the hydrogen is removed by the AlCl ion. The aluminium chloride and hydrogen chloride catalysts are re-generated in this second stage. | 2,407 | 1,586 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.08%3A_The_Collision_Model_of_Chemical_Kinetics |
In , you saw that it is possible to use kinetics studies of a chemical system, such as the effect of changes in reactant concentrations, to deduce events that occur on a microscopic scale, such as collisions between individual particles. Such studies have led to the , which is a useful tool for understanding the behavior of reacting chemical species. According to the collision model, a chemical reaction can occur only when the reactant molecules, atoms, or ions collide with more than a certain amount of kinetic energy and in the proper orientation. The collision model explains why, for example, most collisions between molecules do result in a chemical reaction. Nitrogen and oxygen molecules in a single liter of air at room temperature and 1 atm of pressure collide about 10 times per second. If every collision produced two molecules of NO, the atmosphere would have been converted to NO and then NO a long time ago. Instead, in most collisions, the molecules simply bounce off one another without reacting, much as marbles bounce off each other when they collide. The collision model also explains why such chemical reactions occur more rapidly at higher temperatures. For example, the reaction rates of many reactions that occur at room temperature approximately double with a temperature increase of only 10°C. In this section, we will use the collision model to analyze this relationship between temperature and reaction rates. In we discussed the kinetic molecular theory of gases, which showed that the average kinetic energy of the particles of a gas increases with increasing temperature. Because the speed of a particle is proportional to the square root of its kinetic energy, increasing the temperature will also increase the number of collisions between molecules per unit time. What the kinetic molecular theory of gases does not explain is why the reaction rate of most reactions approximately doubles with a 10°C temperature increase. This result is surprisingly large considering that a 10°C increase in the temperature of a gas from 300 K to 310 K increases the kinetic energy of the particles by only about 4%, leading to an increase in molecular speed of only about 2% and a correspondingly small increase in the number of bimolecular collisions per unit time. The collision model of chemical kinetics explains this behavior by introducing the concept of activation energy ( ) . We will define this concept using the reaction of NO with ozone, which plays an important role in the depletion of ozone in the ozone layer:
\( NO\left ( g \right ) + O_{3}\left ( g \right )\rightarrow NO_{2} \left ( g \right ) + O_{2}\left ( g \right ) \tag{14.7.1} \) Increasing the temperature from 200 K to 350 K causes the rate constant for this particular reaction to increase by a factor of more than 10, whereas the increase in the frequency of bimolecular collisions over this temperature range is only 30%. Thus something other than an increase in the collision rate must be affecting the reaction rate. The reaction rate, not the rate constant, will vary with concentration. The rate constant, however, does vary with temperature. shows a plot of the rate constant of the reaction of NO with O at various temperatures. The relationship is not linear but instead resembles the relationships seen in graphs of vapor pressure versus temperature ( ) and of conductivity versus temperature ( ). In all three cases, the shape of the plots results from a distribution of kinetic energy over a population of particles (electrons in the case of conductivity; molecules in the case of vapor pressure; and molecules, atoms, or ions in the case of reaction rates). Only a fraction of the particles have sufficient energy to overcome an energy barrier. The nonlinear shape of the curve is caused by a distribution of kinetic energy over a population of molecules. Only a fraction of the particles have enough energy to overcome an energy barrier, but as the temperature is increased, the size of that fraction increases. In the case of vapor pressure, particles must overcome an energy barrier to escape from the liquid phase to the gas phase. This barrier corresponds to the energy of the intermolecular forces that hold the molecules together in the liquid. In conductivity, the barrier is the energy gap between the filled and empty bands. In chemical reactions, the energy barrier corresponds to the amount of energy the particles must have to react when they collide. This energy threshold, called the activation energy, was first postulated in 1888 by the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry 1903). It is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. Molecules that collide with less than the threshold energy bounce off one another chemically unchanged, with only their direction of travel and their speed altered by the collision. Molecules that are able to overcome the energy barrier are able to react and form an arrangement of atoms called the activated complex or the transition state of the reaction. The activated complex is a reaction intermediate; it does not last long enough to be detected readily. Any phenomenon that depends on the distribution of thermal energy in a population of particles has a nonlinear temperature dependence. We can graph the energy of a reaction by plotting the potential energy of the system as the reaction progresses. shows a plot for the NO–O system, in which the vertical axis is potential energy and the horizontal axis is the , which indicates the progress of the reaction with time. The activated complex is shown in brackets with an asterisk. The overall change in potential energy for the reaction (Δ ) is negative, which means that the reaction releases energy. (In this case, Δ is −200.8 kJ/mol.) To react, however, the molecules must overcome the energy barrier to reaction ( is 9.6 kJ/mol). That is, 9.6 kJ/mol must be put into the system as the activation energy. Below this threshold, the particles do not have enough energy for the reaction to occur. Part (a) in illustrates the general situation in which the products have a lower potential energy than the reactants. In contrast, part (b) in illustrates the case in which the products have a higher potential energy than the reactants, so the overall reaction requires an input of energy; that is, it is energetically uphill, and Δ > 0. Although the energy changes that result from a reaction can be positive, negative, or even zero, in all cases an energy barrier must be overcome before a reaction can occur. This means that the activation energy is always positive. For similar reactions under comparable conditions, the one with the smallest will occur most rapidly. Whereas Δ is related to the tendency of a reaction to occur spontaneously, gives us information about the reaction rate and how rapidly the reaction rate changes with temperature. (For more information on spontaneous reactions, see .) Even when the energy of collisions between two reactant species is greater than , however, most collisions do not produce a reaction. The probability of a reaction occurring depends not only on the collision energy but also on the spatial orientation of the molecules when they collide. For NO and O to produce NO and O , a terminal oxygen atom of O must collide with the nitrogen atom of NO at an angle that allows O to transfer an oxygen atom to NO to produce NO ( . All other collisions produce no reaction. Because fewer than 1% of all possible orientations of NO and O result in a reaction at kinetic energies greater than , most collisions of NO and O are unproductive. The fraction of orientations that result in a reaction is called the steric factor ( ) , and, in general, its value can range from 0 (no orientations of molecules result in reaction) to 1 (all orientations result in reaction). shows both the kinetic energy distributions and a potential energy diagram for a reaction. The shaded areas show that at the lower temperature (300 K), only a small fraction of molecules collide with kinetic energy greater than ; however, at the higher temperature (500 K) a much larger fraction of molecules collide with kinetic energy greater than . Consequently, the reaction rate is much slower at the lower temperature because only a relatively few molecules collide with enough energy to overcome the potential energy barrier. For an A + B elementary reaction, all the factors that affect the reaction rate can be summarized in a single series of relationships: where
\( rate = k\left [ A \right ]\left [ B \right ] \tag{14.7.2} \) Arrhenius used these relationships to arrive at an equation that relates the magnitude of the rate constant for a reaction to the temperature, the activation energy, and the constant, , called the frequency factor :
\( k = A e^{-E_{a}/RT} \tag{14.7.3} \) The frequency factor is used to convert concentrations to collisions per second. is known as the Arrhenius equation and summarizes the collision model of chemical kinetics, where is the absolute temperature (in K) and is the ideal gas constant [8.314 J/(K·mol)]. indicates the sensitivity of the reaction to changes in temperature. The reaction rate with a large increases rapidly with increasing temperature, whereas the reaction rate with a smaller increases much more slowly with increasing temperature. If we know the reaction rate at various temperatures, we can use the Arrhenius equation to calculate the activation energy. Taking the natural logarithm of both sides of .3,
\( ln\;k = ln\;A - \dfrac{E_{a}}{RT} = ln\;A + \left [\left (-\dfrac{E_{a}}{R} \right ) \left ( \dfrac{1}{T} \right ) \right ] \tag{14.7.4} \) is the equation of a straight line, = + , where = ln and = 1/ . This means that a plot of ln versus 1/ is a straight line with a slope of − / and an intercept of ln . In fact, we need to measure the reaction rate at only two temperatures to estimate . Knowing the at one temperature allows us to predict the reaction rate at other temperatures. This is important in cooking and food preservation, for example, as well as in controlling industrial reactions to prevent potential disasters. The procedure for determining from reaction rates measured at several temperatures is illustrated in Example 13. Many people believe that the rate of a tree cricket’s chirping is related to temperature. To see whether this is true, biologists have carried out accurate measurements of the rate of tree cricket chirping ( ) as a function of temperature ( ). Use the data in the following table, along with the graph of ln[chirping rate] versus 1/ in to calculate for the biochemical reaction that controls cricket chirping. Then predict the chirping rate on a very hot evening, when the temperature is 308 K (35°C, or 95°F). chirping rate at various temperatures activation energy and chirping rate at specified temperature From the plot of ln versus 1/ in , calculate the slope of the line (− / ) and then solve for the activation energy. Express in terms of and and then in terms of and . Subtract the two equations; rearrange the result to describe / in terms of and . Using measured data from the table, solve the equation to obtain the ratio / . Using the value listed in the table for , solve for . If cricket chirping is controlled by a reaction that obeys the Arrhenius equation, then a plot of ln versus 1/ should give a straight line ( ). Also, the slope of the plot of ln versus 1/ should be equal to − / . We can use the two endpoints in to estimate the slope: \( slope = \dfrac{\Delta \;ln\; f}{\Delta \left ( 1/T \right )} = \dfrac{5.30 - 4.37}{3.34\times 10^{-3} \; K^{-1} - 3.48\times 10^{-3} \; K^{-1}} \) A computer best-fit line through all the points has a slope of −6.67 × 10 K, so our estimate is very close. We now use it to solve for the activation energy: \( E_{a} = \dfrac{0.93}{-0.041\times 10^{-3} \; K^{-1}} = 6.6 \times 10^{3} \; K \) If the activation energy of a reaction and the rate constant at one temperature are known, then we can calculate the reaction rate at any other temperature. We can use to express the known rate constant ( ) at the first temperature ( ) as follows: \( ln\;k_{1}= ln\; A -\dfrac{E_{a}}{RT_{1}} \) Similarly, we can express the unknown rate constant ( ) at the second temperature ( ) as follows: \( ln\;k_{2}= ln\; A -\dfrac{E_{a}}{RT_{2}} \) These two equations contain four known quantities ( , , , and ) and two unknowns ( and ). We can eliminate by subtracting the first equation from the second: \( ln\;k_{2} - ln\;k_{1}= \left (ln\; A -\dfrac{E_{a}}{RT_{2}} \right ) - \left (ln\; A -\dfrac{E_{a}}{RT_{1}} \right ) = -\dfrac{E_{a}}{RT_{2}} + \dfrac{E_{a}}{RT_{1}} \) Then \( ln\; \dfrac {k_{2}}{k_{1}}= -\dfrac{E_{a}}{R} \left (\dfrac{1}{T_{1}} - \dfrac{1}{T_{2}} \right ) \) To obtain the best prediction of chirping rate at 308 K ( ), we try to choose for and the measured rate constant and corresponding temperature in the data table that is closest to the best-fit line in the graph. Choosing data for = 296 K, where = 158, and using the calculated previously, \( ln\; \dfrac {k_{2}}{k_{1}}= -\dfrac{E_{a}}{R} \left (\dfrac{1}{T_{1}} - \dfrac{1}{T_{2}} \right ) = \dfrac{55\; kJ/mol}{8.314\; J/\left ( K\cdot mol \right )} \left ( \dfrac{1000 \;J}{1 \; kJ} \right ) \left ( \dfrac{1}{296 \; K}-\dfrac{1}{308 \; K} \right ) = 0.87 \) Thus / = 2.4 and = (2.4)(158) = 380, and the chirping rate on a night when the temperature is 308 K is predicted to be 380 chirps per minute. Exercise The equation for the decomposition of NO to NO and O is second order in NO : \( 2NO_{2}\left ( g \right ) \rightarrow 2NO\left ( g \right ) + O_{2}\left ( g \right ) \) Data for the reaction rate as a function of temperature are listed in the following table. Calculate for the reaction and the rate constant at 700 K. = 114 kJ/mol; = 18,600 M ·s = 1.86 × 10 M ·s . What results in a doubling of the reaction rate with a 10°C increase in temperature from 20° to 30°C? about 51 kJ/mol A minimum energy ( ) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the , or , of the reaction. At a given temperature, the higher the , the slower the reaction. The fraction of orientations that result in a reaction is the . The , steric factor, and activation energy are related to the rate constant in the : A plot of the natural logarithm of versus 1/ is a straight line with a slope of − / .
\( k = A e^{-E_{a}/RT} \tag{14.7.3} \) Although an increase in temperature results in an increase in kinetic energy, this increase in kinetic energy is not sufficient to explain the relationship between temperature and reaction rates. How does the activation energy relate to the chemical kinetics of a reaction? Why does an increase in temperature increase the reaction rate despite the fact that the average kinetic energy is still less than the activation energy? For any given reaction, what is the relationship between the activation energy and each of the following? If you are concerned with whether a reaction will occur rapidly, why would you be more interested in knowing the magnitude of the activation energy than the change in potential energy for the reaction? The product C in the reaction A + B → C + D can be separated easily from the reaction mixture. You have been given pure A and pure B and are told to determine the activation energy for this reaction to determine whether the reaction is suitable for the industrial synthesis of C. How would you do this? Why do you need to know the magnitude of the activation energy to make a decision about feasibility? Above , molecules collide with enough energy to overcome the energy barrier for a reaction. Is it possible for a reaction to occur at a temperature less than that needed to reach ? Explain your answer. What is the relationship between , , and ? How does an increase in affect the reaction rate? Of two highly exothermic reactions with different values of , which would need to be monitored more carefully: the one with the smaller value or the one with the higher value? Why? What happens to the approximate rate of a reaction when the temperature of the reaction is increased from 20°C to 30°C? What happens to the reaction rate when the temperature is raised to 70°C? For a given reaction at room temperature (20°C), what is the shape of a plot of reaction rate versus temperature as the temperature is increased to 70°C? Acetaldehyde, used in silvering mirrors and some perfumes, undergoes a second-order decomposition between 700 and 840 K. From the data in the following table, would you say that acetaldehyde follows the general rule that each 10 K increase in temperature doubles the reaction rate? Bromoethane reacts with hydroxide ion in water to produce ethanol. The activation energy for this reaction is 90 kJ/mol. If the reaction rate is 3.6 × 10 M/s at 25°C, what would the reaction rate be at the following temperatures? An enzyme-catalyzed reaction has an activation energy of 15 kcal/mol. How would the value of the rate constant differ between 20°C and 30°C? If the enzyme reduced the from 25 kcal/mol to 15 kcal/mol, by what factor has the enzyme increased the reaction rate at each temperature? The data in the following table are the rate constants as a function of temperature for the dimerization of 1,3-butadiene. What is the activation energy for this reaction? The reaction rate at 25°C is 1.0 × 10 M/s. Increasing the temperature to 75°C causes the reaction rate to increase to 7.0 × 10 M/s. Estimate for this process. If were 25 kJ/mol and the reaction rate at 25°C is 1.0 × 10 M/s, what would be the reaction rate at 75°C? The reaction rate will approximately double: 20°C to 30°C, the reaction rate increases by about 2 = 2; 20°C to 70°C, the reaction rate increases by about 2 = 32-fold. A plot of reaction rate versus temperature will give an exponential increase: rate ∝ 2 . 100 kJ/mol Thumbnail from | 18,570 | 1,587 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.01%3A_Prelude_to_Organic_Compounds/8.1.02%3A_Geology-_Earth's_Oldest_Fossils |
Finding the evidence for the earliest life on earth is a challenge that grips many geologists. Earth formed about 4.5 billion years ago, throughout the Hadean eon, until about 4.0 billion years ago, it was too hot to be inhabitable by known plants and animals. But chemical evidence from the oldest rocks, from formations 3.5 to 3.8 billion years old from the Archean (4.0 to 2.5 Billion years ago) eon from Greenland, South Africa, and Australia, suggest that living things might have emerged by that time. Paleontologists Claimed that microfossils proving the existence of life were found in the 3.465 billion year old Apex Chert from Western Australia. chert In our discussion of ionic substances, we described a number of macroscopic properties, such as electrical conductivity, crystal shape and cleavage, and characteristic chemical behavior of ions. These were understandable in terms of the microscopic picture of individual ions packed into a crystal lattice in a solid ionic compound or able to move past one another in a liquid or solution. The macroscopic properties of covalent and polar covalent substances can likewise be attributed to microscopic structure. We will see how the nature of the molecules in a covalently bonded substance influences its behavior. The number of covalent substances is far larger than the number of ionic compounds, largely because of the ability of one element, carbon, to form strong bonds with itself. Hydrogen, oxygen, nitrogen, and a number of other elements also bond strongly to carbon, and a tremendous variety of compounds can result. In the early days of chemistry such compounds were obtained from plants or animals rather than being synthesized by chemists, and so they came to be known as organic compounds. This distinguished them from the inorganic compounds available from nonliving portions of the earth’s surface. Today literally millions of carbon compounds can be synthesized in laboratories, and so this historical distinction is no longer valid. Nevertheless, the study of carbon compounds is still referred to as organic chemistry. Since organic compounds all involve covalent bonds, we will describe a number of them as we discuss covalent substances. Many are of considerable commercial importance, and you probably encounter them, perhaps without knowing it, every day. Macroscopic physical properties such as melting and boiling points depend on the strengths of the forces which hold microscopic particles together. In the case of molecules whose atoms are connected by covalent bonds, such intermolecular forces may be of three types. All molecules are attracted together by weak London forces. These depend on instantaneous polarization and increase in strength with the size of the molecular electron cloud. When a molecule contains atoms whose electronegativities differ significantly and the resulting bond dipoles do not cancel each other’s effects, dipole forces occur. This results in higher melting and boiling points than for nonpolar substances. The third type of intermolecular force, the hydrogen bond, occurs when one molecule contains a hydrogen atom connected to a highly electronegative partner. The other molecule must contain an electronegative atom, like fluorine, oxygen, or nitrogen, which has a lone pair. Although each hydrogen bond is weak compared with a covalent bond, large numbers of hydrogen bonds can have very significant effects. One example of this is in the properties of water. This highly unusual liquid plays a major role in making living systems and the earth’s environment behave as they do. Carbon normally forms four bonds, and carbon-carbon bonds are quite strong, allowing formation of long chains to which side branches and a variety of functional groups may be attached. Hence the number of molecular structures which can be adopted by organic compounds is extremely large. Functional groups containing oxygen atoms, nitrogen atoms, and multiple bonds often determine the chemical and physical properties of carbon compounds. Therefore organic chemistry may be systematized by studying related groups of compounds such as alkanes, cycloalkanes, aromatic compounds, alkenes and alkynes, alcohols, ethers, aldehydes and ketones, carboxylic acids, esters, amines, and so on. Within each of these categories chemical and physical behaviors are closely related to molecular structures. Some covalently bonded substances do not consist of individual small molecules. Instead, giant macromolecules form. Examples include most of the rocks in the crust of the earth as well as modern plastics. Properties of such substances depend on whether the macromolecules are three dimensional (like diamond), two dimensional (like graphite or sheets of mica) or one dimensional (like polyethylene). In the latter two cases the strengths of forces between adjacent macromolecules have a significant effect on the properties of the substances. | 4,953 | 1,588 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.19%3A_Structures_Relevant_to_Liganded_Hemoglobins |
Before the advent of techniques that enabled the preparation and stabilization of oxyhemoglobin crystals, key information on the probable structure of oxyhemoglobin and thence on the mechanism of cooperativity was extrapolated from structures of methemoglobin derivatives and from various five- and six-coordinate cobalt-porphyrinato complexes. The structures of these met derivatives have proved to be similar to that of oxyhemoglobin, at least in the stereochemistry of the metalloporphyrinato species and for the protein tertiary and quaternary structure as well. Two synthetic iron-dioxygen adducts built from the picket-fence porphyrins have been structurally characterized. The high, effectively fourfold symmetry of the binding pocket in these systems results in fourfold disorder of the angularly coordinated dioxygen molecule, and precludes the precise and accurate measurements of the Fe—O—O angle and O—O separation that are grist to the theoretical mills.* Figure 4.28B illustrates the stereochemistry for one conformer. Subsequently, the structures of several dioxygen adducts of biological oxygen carriers have been determined. Although the dioxygen moiety is usually ordered, the precision is tantalizingly just less than that needed to decide whether the apparently more-linear geometry seen for oxyerythrocruorin and oxyhemoglobin is significantly different from that for oxymyoglobin and therefore attributable to the water molecule or imidazole that is hydrogenbonded to the coordinated dioxygen ligand. Nonetheless, several interesting differences emerge. The axial base in oxymyoglobin and oxyhemoglobin is almost eclipsed; that is, \(\phi_{1}\) ≈ 0°. The axial base has moved from a tilted position in deoxyhemoglobin to a symmetric one in oxyhemoglobin. In the absence of steric constraints, the iron atom is essentially in the center of the porphyrin plane for Fe(PF)(1-MeIm)(O ), oxymyoglobin, and oxyhemoglobin. For the 2-methyl analogue, Fe(PF)(2-MeIm)(O ), the iron remains significantly out of the plane, as also appears to occur for oxyerythrocruorin. In the structure of Fe(TPP)(Py)(CO), Figure 4.30, a model for carbonyl hemoglobins, the iron atom is in the plane and the Fe—C\(\equiv\)O bond is linear and perpendicular, as expected. Not so for carbonyl hemoglobins, where the blob of electron density that is identified with the coordinated carbon monoxide lies substantially off the normal to the porphyrin. We return to this point shortly. In general, with the exception of the coordinated ligand, the structures of sixcoordinate low-spin hemoglobins, whether Fe or Fe , are similar. Indeed, the refined structures of oxy- and carbonmonoxyhemoglobin are superimposable within experimental uncertainties, except in the immediate vicinity of the diatomic ligand. Some metrical details are given in Table 4.8. Resol. Å Fe-N Å Fe-porph Å Doming Å Fe-L Å Fe-L Å Fe-XY deg. \(\varnothing_{1}\) deg. \(\varnothing_{2}\) deg. Tilt deg. 123(4) 148(8) 122(4) 143(8) 127(4) 145(8) 138(6) 180(11) 1.990(7) 1.988(13) 0.01 0.02 2.043(6) 2.041(5) 1.742(7) 1.748(7) 172.9(6) 175.9(6) Without exception to date (but see footnote 1 in Reference 168), in structurally characterized oxyhemoglobins, the coordinated dioxygen ligand is hydrogen-bonded to the distal histidine or to a water molecule—even though theoretical calculations show that hydrogen bonding would destabilize M—O moieties. This universal observation of hydrogen bonding in these biological systems is consistent with notions that electron density accumulates on the dioxygen molecule upon coordination. Given the errors associated with atomic positions (at best, ±0.20 Å) the x-ray crystallographic evidence could be equivocal, since hydrogen atoms on the distal imidazole are not observed. There are at least three lines of evidence that support the existence of a specific O • • • HN interaction. First, the EPR spectrum of cobalt oxyhemoglobin indicates that the coordinated dioxygen is hydrogen-bonded to something. Second, and more directly, in the neutron-diffraction structure of oxymyoglobin, where hydrogen and especially deuterium nuclides scatter strongly, the imino hydrogen or deuteron was located on the nitrogen atom closest to the coordinated dioxygen, as illustrated in Figure 4.31A. In contrast, in the neutron-diffraction structure of carbonmonoxymyoglobin, the alternative imidazole tautomer was observed (Figure 4.3IB). The absence of hydrogen bonding of the distal imidazole residue with the coordinated CO molecule is consistent with other lines of evidence that there is little accumulation of electron density on the carbonyl ligand. Third, but less directly, genetically engineered mutants have been produced in which the distal histidine has been replaced by glycine—sperm whale Mb E7His → Gly, and Hb\(\alpha\)E7His → Gly and HbA \(\beta\)E7His → Gly. For the myoglobin mutant, the O binding rate constant at room temperature increases by an order of magnitude, but the dissociation rate constant increases by two orders of magnitude, leading to a decrease in affinity of more than an order of magnitude, as derived from k /k . This leads to an estimate of the free energy associated with hydrogen bonding of \[\Delta G = -RTlog \bigg[\frac{P_{1/2}(O_{2})\cdotp Native}{P_{1/2}(O_{2})\cdotp Mutant} \bigg] = 1.5 kcal/mol \ldotp\] In addition, this mutant myoglobin autooxidizes rapidly compared to the native one. On the other hand, the affinity for CO is greatly increased, leading to a value of M for the mutant of 1300, compared to 16 for the native. Thus the distal histidine stabilizes a coordinated O ligand by hydrogen bonding and destabilizes a coordinated CO ligand by steric clash. A similar discrimination is seen for the \(\alpha\) chains of the hemoglobin mutant in the binding of the fourth O or CO molecule. For the \(\beta\) chains little difference is seen relative to the native protein: hydrogen bonding between the distal histidine and the coordinated dioxygen ligand appears to be much weaker in \(\beta\) chains, as evidenced by longer N(H) • • • O separations than those seen in the \(\alpha\) chains. Comparison of the crystal structures of the native and mutant \(\alpha\) (\(\beta\)E7His → Gly) structures reveals negligible changes in the distal environment, except for that occasioned by the replacement of —CH —C N H (histidine side chain) by —H (glycine side chain). Studies of hemoglobin mutants where the nonpolar distal residue \(\beta\)ValE11 (—CH(CH ) ) is replaced by alanine (—CH ), isoleucine (—CH(CH )CH CH ), and leucine (—CH CH(CH ) ) reveal that this valine offers steric hindrance to oxygen binding in the T state. Whereas the angularly coordinated O ligand fits comfortably around the distal histidine, a perpendicular and linear CO moiety cannot. Either the distal histidine rotates out of the way, or the CO tilts off axis, or the Fe—C\(\equiv\)O group bends, or some combination of these occurs. Notwithstanding the absence of bent M—CO moieties in the inorganic literature, reports of strongly bent M—CO groups appear in the biochemical literature. The controversy is illustrative of the synergistic interplay of data from models and proteins, and the importance of examining a problem with a miscellany of techniques. The molecular orbital model of ligand-metal interactions presented in Figure 4.22 does not preclude a bent M—C\(\equiv\)O moiety on symmetry grounds. Groups related to CO can bend; the normally linearly coordinated SCN moiety has been observed to become strongly bent under severe steric stress, with an Fe—N—CS of 140°. Unfortunately, the resolution in protein crystal structures is not sufficient to distinguish unequivocally a linear tilted stereochemistry from a bent one or from a combination of tilt and bend. Studies by the XANES technique have been interpreted in terms of a bent Fe—C\(\equiv\)O moiety (150°) \(\tag{4.50}\) both in MbCO and in the CO adduct of a chelated heme in micelles, the latter being an especially surprising result. From EXAFS data on a number of carbonyl adducts, interpretations were offered: linear or moderately bent (150°) FeCO moieties for unhindered model systems, and moderately bent or strongly bent (130°) FeCO moieties for hindered synthetic and biological systems. In the crystal structure of MbCO at 1.5 Å resolution, the CO group is disordered, and Fe—C\(\equiv\)O angles of 120° and 140° were proposed, although the alternative model of tilted, nearly linear Fe—C\(\equiv\)O stereochemistry could not be eliminated, and is indeed far more likely to account for the off-axis nature of the oxygen position. Vibrational spectroscopy confirms the existence of two major configurations, and indicates a third minor configuration of the Fe—C\(\equiv\)O moiety in MbCO. An elegant infrared study of the polarization of reattached carbon-monoxide molecules following photolysis of MbCO by linearly polarized light at 10 K gave tilt angles of the CO vector with respect to the heme normal of 15(3)°, 28(2)°, and 33(4)° for the three conformational substates; the former two values were confirmed in a similar study at room temperature. Note that these studies do not yield the tilt of the Fe—C bond to the heme normal. In three synthetic compounds with severe steric hindrance, the extent of bending and tilting of the Fe—CO moiety is small. In one nonporphyrinic system the Fe—C\(\equiv\)O group is bent by 9.4(5)° and tilted by 4.2°. In Fe(PocPF)(1,2-Me Im)(CO) the Fe—C\(\equiv\)O angle is 172.5(6)° and modest tilting of the Fe—CO group and substantial buckling of the porphyrin ring are apparent. In Fe(C Cap)(1-MeIm)(CO) the two independent Fe—C\(\equiv\)O angles are 172.9(6)° and 175.9(6)° and modest tilting of the Fe—CO group is again apparent. From a detailed analysis of the force constants describing the vibrational spectroscopy for the Fe—CO moiety, values of 171° for the Fe—CO angle, 9.5° for the tilt, and 11° for porphyrin buckling were calculated for MbCO. These results are particularly important, for in a model complex very closely related to Fe(Poc-PF)(1,2-Me Im)(CO), just mentioned, an EXAFS study suggested an Fe—C\(\equiv\)O bond angle of 127(4)°; that same study ascribed an Fe—C\(\equiv\)O bond angle of around 130° to MbCO. The structure of carbonmonoxyhemoglobin, Hb(CO) , now is interpreted in terms of a nearly linear tilted geometry. Clearly the geometry of attachment of CO to hemoglobins is perturbed by the surroundings of the ligand-binding site and hence the affinity of hemoglobins for CO is also perturbed. Unfortunately, a clear resolution of the geometry of the Fe—CO moiety in MbCO does not exist yet. | 10,698 | 1,589 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.E%3A_Properties_of_Gases_(Exercises) |
There are some gases in existence, such as \(\ce{NS2}\) and \(\ce{NCl2}\), which do not follow Boyle's Law, regardless of the pressure conditions they're under. Explain why this is the case. Boyle's Law assumes that the only factors changing are pressure (\(P\)) and volume (\(V\)), therefore, in order to apply Boyle's Law, both number of molecules (\(n\)) and temperature (\(T\)) must remain constant. However, both \(NS_2\) and \(NCl_2\) do not have constant \(n\) because they undergo association reactions when in contact with like molecules. \[ 2NH_2 \rightleftharpoons N_2S_2\] \[ 2NCl_2 \rightleftharpoons N_2Cl_2\] Therefore, may not be applied for these gases. An ideal gas originally at 0.85 atm and 66ºC was allowed to expand until its final volume, pressure, and temperature were 94 mL, 0.60 atm, and 45ºC, respectively. What was its initial volume? \[V_1=\dfrac{(P_2)(V_2)(T_1)}{(T_2)(P_1)}\] \[=\dfrac{(0.60\;atm)(94\;mL)(339\;K)}{(318\;K)(0.85\;atm)}=71\;mL\] Some ballpoint pens have a small hole in the main body of the pen. What is the purpose of the hole? The hole is there to allow the pressure to equalize and allow the ink to flow out of the pen. Starting with the ideal-gas equation, show how you ca calculate the molar mass of a gas from a knowledge of its density At STP (standard temperature and pressure), 0.280 /;L of a gas weighs 0.400/;g. Calculate the molar mass of the gas. First, we need to determine the moles of gas in the sample: \[n=\dfrac{(P)(V)}{(R)(T)}\] \[=\dfrac{(1.00\;atm)(0.280\;L)}{(0.08206\;L-atm/mol-K)(273\;K)}=0.0125\;moL\] \[=\dfrac{(0.400\;g)}{(0.0125\;mol)}=32.0\;g/moL\] Ozone molecules in the stratosphere absorb much of the harmful radiation from the sun. Typically, the temperature and partial pressure of ozone in the stratosphere are 250 K and 0.0010 atm, respectively. How many ozone molecules are present in 1.0 L of air under these conditions? Assume ideal-gas behavior. Dissolving 3.00 g of an impure sample of CaCO in an excess of HCl acid produced 0.656 L of CO (measured at 20ºC and 792 mmHg). Calculate the percent by mass of CaCO in the sample. A student breaks a thermometer and spills most of the mercury (Hg) onto the floor of a laboratory that measures 15.2 m long, 6.6 m wide, and 2.4 m high. The vapor pressure of mercury at 20ºC is 1.7 x 10 atm. a) The volume of the room is 15.2 m x 6.6 m x 2.4 m = 240 m . The pressure is 1.7 x 10 atm x 101325 Pa/atm = 0.17 Pa Ideal gas equation: \[n=\dfrac{PV}{RT}\] \[n=\dfrac{0.17\;Pa*(240)\;m^3}{\dfrac{8.314\;Pa-m^3}{K-mol}*(293K)}\] \[=0.017\;mole\] Mass of mercury vapor : m = n * M =(0.017 mole)(200.6 g/mole)(1000 mg/gram)} = 3400 mg b) The mass density of mercury vapor : [Hg] = m / V = \[\dfrac{3400\;mg}{240\;m^{-3}}\] =14 mg/m This value is indeed larger than the 0.050 mg Hg m allowed. c) U An oxide has the density of 1.75 g L at 1.12 atm and 100 C. Is this oxide O (g) or O \[M = \dfrac{\rho RT}{P} \] \[= \dfrac{(1.75 g L^{-1})(0.08206 L atm K^{-1}mol^{-1})(100+273)K)}{1.12 atm}=47.8 \dfrac{g}{mol}\] The molar mass of O (g) is 48 g mol , so the oxide must be O (g) You are given a gas at 750 torr with a volume of 0.62 Liters at 25°C. A while later, you measure the temperature and notice a 5°C temperature increase in the system. Assuming there is no pressure change and that this is an ideal gas, what is the new volume? First: List out what you have Vi=0.62 L T =25ºC +273.15= 298.15 We do not know the amount of moles so we should find that. PV=nRT So n =2.47x10 moles P =0.987 atm V =? T =298.15 K + 5 K= 303.15 K V =(nRT)/P =0.631 L The composition of air in your alveoli is 75% N 14% O , 5% CO , and 6% H O, by volume. (a) Calculate the average molar mass of this air sample. (b) Calculate the partial pressures of N , O , CO , and H O in your alveoli. assuming that alveolar pressure is 760 torr (At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of gas). The composition of air in your alveoli is 75% N 14% O , 5% CO , and 6% H O, by volume. (a) Calculate the average molar mass of this air sample. (b) Calculate the partial pressures of N , O , CO , and H O in your alveoli. (At constant pressure and temperature, the volume of a gas is directly proportional to the number of moles of gas.) In an expandable balloon we find a mixture of two gases only. The mass percent of the gases are: 65% 0 and 35 % H The mass of the gases combined is 4.6 grams. Calculate the total volume these two gases occupy at 298 K and 1.00 atm. You can find the Volume by using the Ideal Gas Law equation: \[V=\dfrac{nRT}{P}\] The mass of the two gases is given by: \( m_{H_{2}}= 4.6\times 0.35=1.6g\) \(m_{O_{2}}= 4.6\times 0.65=3.0g\) The number of moles of the mixture= \(n_{total}=n_{H_{2}}+n_{O_{2}}\) \( n_{total}=\dfrac{1.60g}{Q2.0g.mol^{-1}}+\dfrac{3.0g}{32g.mol^{-1}}=0.884mol\) The volume the two gases occupy = \(V=\dfrac{(0.884mol)(0.08206\dfrac{L.atm}{K.mol})(298K)}{1.0 atm}= 21.6 L\) The temperature in a small room in summer time is 31oC. The room (dimension 4mx4mx3m) is filled with 0.88 kg of water vapor. What is the partial pressure of water vapor in the air? A flask contains a mixture of two ideal gases, A and B. Show graphically how the total pressure of the system depends on the amount of A present. That is, plot the total pressure versus the mole fraction of A. Do the same for B on the same graph. The total number of moles of A and B is remained as constant. When you drink something from a straw, how does the liquid get from the glass up through the straw into your mouth? When you close your mouth around the straw, you close off the atmosphere from outside to inside. During suction, you increase the volume inside your mouth, thus lowering the internal pressure and the atmospheric pressure exerted onto the liquid over comes the internal pressure and so the liquid flows into your mouth. Consider the reaction of ferrous sulfide with a dilute acid: \[FeS_{(s)} + 2HCl_{(aq)} \rightarrow H_2S_{(g)} + FeCl_{2(l)}\] An amount of 5.10 grams of ferrous sulfide completely reacted in this reaction. The gas collected over the water at 27ºC has a volume 1.78 L. Find the pressure of the collected gas in atm. The vapor pressure of water at 27°C is 26.7 mmHg. Hint: Look at Base on the amount of reactant, use stoichiometry to find the number of moles of H2S gas produced and get partial pressure of hydrogen sulfide gas. Then use Dalton's law of partial pressure to find the total pressure of collected gas. \[ 5.10g FeS \times \dfrac{1mole FeS}{87.92g FeS} \times \dfrac{1mole H_{2}S}{1mole FeS} = 0.058mole H_{2}S\] \[ P_{H_{2}S} = \dfrac{nRT}{V} = \dfrac{0.058 mole H_{2}S\times0.08206 \dfrac{L\cdot atm}{mole\cdot K}\times(27+273)K}{1.78 L} = 0.802 atm\] \[ P_{total} = P_{H_{2}S} + P_{H_{2}O} = 0.802 atm + (26.7 mmHg \times \dfrac{1atm}{760mmHg}) = {\color{red} 0.837 atm} \] Tanks containing mixtures of oxygen and nitrogen are routinely lowered to the Aquarius underwater research lab in Florida. What is the percent composition of oxygen gas in the tank if the total pressure in the tank once it has reached Aquarius is 6.5 atm, and the partial pressure is 0.55 atm? For any ideal gas, \[PV=nRT\] At constant pressure and temperature, V is proportional to n. Thus: \[P_{O^{2}}=x_{O^{2}}P_{total}\] \[x_{O^{2}}=\dfrac{P_{O^{2}}}{P_{total}}=\dfrac{0.55atm}{6.5atm}=.085\] The percent composition is 8.5% A mixture of helium and oxygen gas is used to fill up an apparatus. If the total pressure within the apparatus is 10 atm and the partial pressure of helium is 2 atm within that apparatus, calculate the percent by volume of diatomic oxygen gas. \[ \chi_{He}=\frac{P_{He}}{P_{total} }, \ where \ \chi = Mole \ Fraction\] \[ \begin{align*} \chi_{He} &= \frac{2 atm}{10 atm} \\ \chi_{He} &= 0.2 \end{align*}\] \[ \begin{align*} \% \ by \ volume \ He &= \chi_{He} \cdot 100 \\ &= 0.2\ \cdot 100= 20\%\ \end{align*}\] \[ 100\% \ by \ volume \ total \ gas - 20\% \ by \ volume \ He= \ \underline{80\% \ by \ volume \ O_{2}}\] for more on mole fractions. Choose the more ideal system List the ideal conditions (the conditions that help allow a gas to behave ideally). Ideal conditions come from the kinetic theory of gases. This theory relies on three primary assumptions (as follows): Ideal conditions are at: This is because at these conditions is when there are relatively large distance between molecules. 1.00 mole of an unknown gas occupies a 0.865 L container at 0.90 atm and 25ºC. If the value of van der Waals constant .550, what is the value of for the gas? What does the value of indicate about the gas? \[P=4.0atm\\V=0.70L\\T=77.2K\\a=2.50\dfrac{atm\cdot L^2}{mol^2}\] \[(P+\dfrac{a n^2}{V^2})(V-nb)=nRT\rightarrow b=\dfrac{nRT}{P+\dfrac{an^2}{V^2}}-V(\dfrac{1}{-1})\] \[b=\dfrac{(1mol)(.08206\dfrac{L\cdot atm}{mol\cdot K})(77.2K)}{4.0atm+\dfrac{(Q2.5)1mol^2}{.70L^2}}-.70L(\dfrac{1}{-1mol})\\=4\times 10^-3\dfrac{L}{mol}\] The small value of \(b\) indicates that the gas molecules have a small molecular mass. The compressibility for compound 1 is 0.84 and 0.56 for compound. Which compound will occupy a smaller volume? Z is a prediction of the behavior of gases. A smaller value means that there are stronger intermolecular forces amongst the molecules, meaning a smaller volume occupied. Therefore, compound 2 will occupy a smaller volume than compound 1. True or false: The concept of temperature is a macroscopic. Explain your reasoning. True. The kinetic theory of gases hypothesizes that it deals with an extremely large amount of molecules. Therefore, temperature is a macroscopic concept since the average kinetic energy of the molecules within a given system is proportional to the temperature. A large amount of molecules must be included into this average in order for it to be properly utilized. What assumptions are made when applying the kinetic molecular theory to gases? Are all of these assumptions necessary? Why or why not? There are five assumptions made when applying the kinetic molecular theory to gases are: Most of these assumptions are in fact necessary to consider when applying the kinetic molecular theory to gases, however, assumption number 4 is not always crucial. This is due to the fact that it does not matter whether or not the collisions between the molecules and the walls are elastic if the walls of the container are at the same temperature as the molecules inside of the the container ( the gas) because although kinetic energy may be transferred from molecule to molecule, it will not be converted into any other forms of energy. A 3.50 L cube-shaped container container 3.5x10 molecules of \(N_2\) at 30ºC. What is pressure exerted on one wall? \[N=3.5\times 10^{23}\\V=3.50L\\T=303K\\k_B=1.38\times 10^{-23} \] \[ P=\dfrac{2N}{3V}\overline{E}_trans\rightarrow P=\dfrac{2N}{3V}(\dfrac{3}{2}k_BT)\] \[=\dfrac{2(3.5\times 10^{23})}{3(3.50L)}[\dfrac{3}{2}(1.38\times 10^{-23})(303K)]\] \[=4.18\times 10^2atm\] Imagine a container with walls 123 cm³. Argon (with a speed of 700 m/s) atoms at 300 K are colliding at right angles with the container's walls at a rate of \(5.0 \times 10^{20} m/s\). What would the force and pressure exerted on the container's walls be? Calculate the root-mean-square speed and the molar kinetic energy of N at 25 C The root-mean-square speed of \(N_2\) is: \[v_{rms} = \sqrt{\dfrac{3k_BT}{m}}\] or in terms of molar mass \(M\) instead of molecular mass (\(m\)) \[v_{rms} = \sqrt{\dfrac{3RT}{M}}\] \[ =\sqrt{\dfrac{(3)(8.3145 \, J\,\cancel{K} \cancel{mol^{-1}})(298.15\; \cancel{K})}{ 28.02 \; \cancel{g} \, \cancel{mol^{-1}} \frac{1\; kg}{1000\; \cancel;{g}} }} = 5.15\; m/s\] The molar kinetic energy of \(N_2\) at 25 C is: = \(\dfrac{3}{2}*\dfrac{8.3145J}{K mol}*298.15K*\dfrac{1kJ}{1000J}\) = 3.72 kJ/mol Find the \(v_{rms}\) of N_{2(g)} at 25ºC. What temperature must \(Cl_{2(g)}\) be to have the same \(v_{rms}\)? \[v{_{rms}} = \sqrt{\dfrac{3 R T}{M}}\] \[v_{rmsN_{2}} = \sqrt{\dfrac{3(8.314 J\cdot K^{-1}\cdot mol^{-1})(25+273)K}{28.02 x 10^{-3} kg\cdot mol^{-1}}}=515m \ s^{-1}\] \[ since \ v_{rmsN_{2}} = v_{rmsCl_{2}}, then \ v_{rmsCl_{2}} = 515m \ s^{-1}\] \[v_{rmsCl_{2}} = \sqrt{\dfrac{3(8.314 J\cdot K^{-1}\cdot mol^{-1})(T)}{70.9x10^{-3} g\cdot mol^{-1}}} = 515m \ s^{-1}\] T = 754 K Therefore, Cl (g) must be 754 K to have the same v as N (g) at 298 K a) The plot of the distribution of speeds for Cl at 100 K, 300 K, 600 K, and 1000 K b) The plot of the distribution of N Cl and CH at the same temperature (300K) Which gas would you expect to move faster at 255K, O or I ? And why? We'd expect O to move faster. This is because it is lighter. Think of the formula for average speed because it is divided by kg/mol the smaller the mass, the faster it goes. According to the Maxwell speed distribution, would \(O_{2(g)}\) have a wider speed distribution at 200 K or 1000 K? At lower temperatures, the speed distribution will be narrower and there will will be a smaller most probable speed while the higher temperature will have a wider speed distribution and a higher most probable speed. Therefore, \(O_{2(g)}\) will have a wider speed distribution at 1000 K. A CO molecule at unknown temperature at sea level is released to travel upward. Assuming that the temperature is constant and that the molecule doesn’t collide with another molecules and reaches a terminal height of 8.5 meters above sea level, what is the temperature? Do the same calculation for a H atom. [Hint: To calculate the altitude, h, the molecule will travel, equate its kinetic energy with the potential energy, mgh, where m is the mass and g is the acceleration due to gravity (9.81 m s ).] A CO molecule at unknown temperature at sea level is released to travel upward. Assuming that the temperature is constant and that the molecule doesn’t collide with another molecules and reaches a terminal height of 5000 meters above sea level, what is the temperature? Do the same calculation for a H atom. [Hint: To calculate the altitude, h, the molecule will travel, equate its kinetic energy with the potential energy, mgh, where m is the mass and g is the acceleration due to gravity (9.81 m s ).] Calculate the root-mean-square speed in cm/s for the following particles (hint, what is the average speed?): 0.010, 0.043, 0.027, 0.012, 0.041, 0.023, 0.011, 0.004, 0.007, and 0.009 (all in m/s). The root-mean-square speed of 6 particles is 2. The root-mean-square of the ensemble of particles is \[v_{rms}=\sqrt{\dfrac{\sum_{i=1}^{6} c^{2}_{i}}{m}}\] \[2.47=\sqrt{\dfrac{1.0^{2}+Q2.0^{2}+1.5^{2}+3.0^{2}+2.5^{2}+c_{6}}{6}}\] We solve for \(c_{6}\) and we get \[c^{2}_{6}=(Q2.47ms^{-1})^{2}(6)-(1.0^{2}+2.0^{2}+1.5^{2}+3.0^{2}+2.5^{2}) \; m^{2}s^{-2}\] \[c^{2}_{6}= 36.6m^{2}s^{-2}-22.5m^{2}s^{-2}\] \[c^{2}_{6}=14.1m^{2}s^{-2}\] \[c_{6}= 3.75ms^{-1}\] The average speed of the 6 particles is \[\bar{c}=\dfrac{\sum^{6}_{i=1}c_{i} }{N}\] \[\bar{c}=\dfrac{(1.0+Q2.0+1.5+3.0+Q2.5+3.75)ms^{-1}}{6}\] \[=2.30ms^{-1}\] The speeds of six gaseous molecules at a given temperature are 1.0m/s , 1.2m/s, 1.5m/s, 1.7m/s, 2.0m/s, and 2.3m/s. a) The root-mean-square-speed of those gases: \[v_{ = \sqrt{\dfrac{1}{N}*(v_1^{2}+v2^{2}+v_3^{2}+v_4^{2}+v_5^{2}+v_6^{2})}\] \[= \sqrt{\dfrac{1}{N}*(1^{2}+1.2^{2}+1.5^{2}+1.7^{2}+2^{2}+2.3^{2})}\] \[ = 1.68\; m/s\] The average speed is : \[v_{ = \dfrac{1+1.2+1.5+1.7+2+2.3}{6}\] \[ = 1.62\; m/s\] b) The average speed is smaller then the root-mean-square speed Calculate c for a sample of N (g) at 500K. \[c{_{mp}} = \sqrt{\dfrac{2RT}{M}}\] \[c{_{mp}} = \sqrt{\dfrac{2 \ (8.314 J \ K^{-1} \ mol^{-1}) (500K)}{.02802 kg \ mol^{-1}}} = 545 m \ s^{-1}\] Using the initial equation \(f(c)=4\pi c^2 \dfrac{m}{\pi K_BT} ^{3/2} e^{-\dfrac{mc^2}{2K_BT}}\) derive an expression for \(c_{mp}\) Step 1: $$f(c)=4\pi c^2 \dfrac{m}{\pi K_BT} ^{3/2} e^{-\dfrac{mc^2}{2K_BT}}\] $$=c^2 \dfrac{m}{\pi K_BT} ^{3/2} e^{-\dfrac{mc^2}{2K_BT}}\] Step 2: (Hint: Apply the to \(c^2\) and \(e^{-\dfrac{mc^2}{2K_BT}}\)) \[\dfrac{df(c)}{dc}=4\pi \dfrac{m}{2\pi K_BT}^{3/2} [2ce^{-\dfrac{mc^2}{2K_BT}} + c^2e^{-\dfrac{mc^2}{2K_BT}}-{\dfrac{mc}{2K_BT}}]\] Step 3: \[4\pi(\dfrac{m}{2\pi K_BT})^{3/2}ce^{-\dfrac{mc^2}{2K_BT}}(2-\dfrac{mc^2}{K_BT})\] \[\pi (\dfrac{m}{2\pi K_BT})^{3/2}c_{mp}e^{-\dfrac{mc_{mp}^2}{2K_BT}}(2-\dfrac{mc_{mp}^2}{K_BT})=0\] Step 5: \[2-\dfrac{mc^2 _{mp}}{K_BT}=0\] Step 6: \[c^2_{mp}=\dfrac{2K_BT}{m}\] Step 7: (Hint: Use Avogadro's number) \[c^2 _{mp}=\dfrac{2K_BT}{m}\dfrac{N_A}{N_A}=\dfrac{2RT}{M}\] Step 8: \[c_{mp}=\sqrt {\dfrac{2RT}{M}}\] Calculate the value of c for CH OH at 30°C. What is the ratio of the number of molecules of speed 400 m s to the number of molecules of speed c ? Calculate the value of the most probable speed, average speed, and the root-mean-square speed for \(C_5H_{12}\) at 25°C. \[c_{mp}=\sqrt{\dfrac{2RT}{M}}=\sqrt{\dfrac{2(8.314J/K*mol)(298.15K)}{72.15\times 10^{-3}kg/mol}}=26Q2.13m/s\] \[\begin{align*}\bar{c}&=\sqrt{\dfrac{8RT}{\pi *M}}=\sqrt{\dfrac{4}{\pi}}*\sqrt{\dfrac{2RT}{M}}=\sqrt{\dfrac{4}{\pi}}*c_{mp}\\&=\sqrt{\dfrac{4}{3.14159}}*262.13\;m/s=295.78m/s\end{align*}\] \[\begin{align*}c_{rmp}&=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3}{2}}*\sqrt{\dfrac{2RT}{M}}=\sqrt{\dfrac{3}{2}}*c_{mp}\\&=\sqrt{\dfrac{3}{2}}*262.13\; m/s=321.04\; m/s\end{align*}\] Calculate the value of c for CH OH at 30°C. What is the ratio of the number of molecules of speed 400 m s to the number of molecules of speed c ? How many times is the average distance between collision for a gas molecule larger than its molecular diameter, given that Pressure The mean free path of gas molecules is given as: \[ \gamma = \dfrac{1}{\sqrt{2}\pi d^{2}(N/V)} \] Hint: Look at concept of \[ PV = nRT \Rightarrow V = \dfrac{nRT}{P} \] \[ N/V = \dfrac{N}{n} \dfrac{P}{RT}\] \[ but (N/n) = N_{A} \] \[ \Rightarrow \dfrac{N}{V} = \dfrac{PN_{A}}{RT} \Rightarrow {\color{red} \gamma= \dfrac{RT}{\sqrt2 \pi d^{2} PN_{A}}} \] Gas A has a density of 0.835 g/cm3 and Gas B has a density of 14.67 g/cm3. From just this information, which do you expect to have a larger mean free path? Explain why. We would expect gas A to have a larger mean free path. A cylinder contains 25 molecules of an unknown ideal gas. What is the of the gas molecules? The cylinder has a volume of 20.0 cm and the gas has an atomic radius of 1.24 Å. (1 cm = 10 Å) For an ideal gas, the mean free path is calculated by: \[\lambda =\frac{\bar{c}}{Z_{1}}=\frac{1}{\sqrt{2}\pi d^{2}\frac{N}{V}}\] Thus, the mean free path of the unknown gas is: \[\lambda =\frac{1}{\sqrt{2}\pi (1.24\times 10^{8}\, cm)^{2}\, \frac{25}{20.0 cm^{2}}}=1.17\times 10^{-17}cm=1.17\times 10^{-19}m\] A rattle containing 50 spherical beads is shaken around by a toddler. What is the mean free path of the beads if the volume of the rattle is 500 cm and the diameter of each individual bead is 0.2 cm? Use the appropriate equation and plug in given values. \[ \begin{align} & \lambda = \frac{1}{\sqrt{2} \pi d^{2}(N/V)} \\ & \lambda = \frac{1}{\sqrt{2} \pi (\frac{0.2}{100} m)^{2}(50 beads/\frac{500}{100}m^{3})} \\ & \underline{ \lambda = 1.8 \cdot 10^{-4} \ \ m } \end{align}\] for more details about mean free path. Calculate the mean free path and the binary number of collisions per liter per second between Ar molecules at 298 K and 1.00 atm. Use 3.62 Å as the collision diameter of the Ar molecules. Assume ideal gas behavior. Calculate the mean free path and the binary number of collisions per liter per second between Ar molecules at 298 K and 1.00 atm. Use 3.62 Å as the collision diameter of the Ar molecules. Assume ideal gas behavior. \(CO_2\) has a collision diameter of 0.4 nanometers. At 25 oC and 1.2 atm pressure, what is the mean free path between binary collisions, and what is the rate of binary collisions per liter assuming the volume contains only pure carbon dioxide gas behaving ideally? For the molecules HCl at 350K and 1.21 atm, calculate the mean free path and the binary number of collisions per liter per second. The collision diameter of HCl molecules is 5.50 Å. Assume the HCl acts as an ideal gas. Calculate \(\dfrac{N}{V}\) to calculate the mean free path $$PV=NRT=\dfrac{N}{N_A}RT\] $$\dfrac{N}{V}=\dfrac{PN_A}{RT}=\dfrac{(1.21atm)(6.022\times 10^{23}mol^{-1})}{(0.08206L\,atm\,K^{-1}mol^{-1})(350K)}=Q2.537\] Turn L into \(m^{-3}\) $$Q2.537\times 10^{22}L^{-1}(\dfrac{1000L}{1m^{3}})=Q2.537\times 10^{22}m^{-3}\] $$\lambda=\dfrac{1}{\sqrt{2}\pi d^{2}(\dfrac{N}{V})}=\dfrac{1}{\sqrt{2}\pi (5.50\times 10^{-10}m)^{2}(Q2.537\times 10^{22}m^{-3})}=Q2.93\times 10^{-5}m\] Calculating the binary number of collisions : (Hint: Use the equation for average molecular speed.) $$\bar{C}=\sqrt{\dfrac{8RT}{\pi M}}=\sqrt{\dfrac{8(8.3145JK^{-1}mol^{-1}(300K)}{\pi (36.46\times 10^{-3}\dfrac{kg}{mol})}}\] $$z_{11}=\sqrt{\dfrac{2}{2}}\pi d^2\bar{C}\dfrac{N}{V}^{2}\] $$=1.80526\times 10^{29}m^{-3}s^{-1}\] (Hint: Turn into \(L^{-1}s^{-1}\)) $$z_{11}=\sqrt{\dfrac{2}{2}}\pi (5.50\times 10^{-10}m)^{2}(417.39ms^{-1})(\dfrac{1m^{3}}{1000L})=1.80527\times^{26} L^{-1}s^{-1}\] In a 4L container you have 5 grams of 0Q2. Assume the collision diameter of 0 is 3.55 Angstrongs . Calculate the mean free path of the gas. The mean free path is given by \[\lambda =\dfrac{1}{\sqrt{2}\pi d^{2}(\dfrac{N}{V})}\] V is given and we can find \(\left ( \dfrac{N}{V} \right)\) by calculating N. \[N=N_{A}.n=(6.022\times 10^{23}mol^{-1})(5gO_{2})(\dfrac{1mol}{32gO_{2}})\] \[ N=9.40\times 10^{22}\] \[\left ( \dfrac{N}{V} \right )=\dfrac{9.41\times 10^{22}}{4L}=Q2.35\times 10^{22}L^{-1}(1000Lm^{-3})\] \[\left ( \dfrac{N}{V}\right )=Q2.35\times 10^{25}m^{3}\] The mean free path is \[\lambda =\dfrac{1}{\sqrt{2}\pi d^{2}(\dfrac{N}{V})}\] \[\lambda =\dfrac{1}{\sqrt{2}\pi(3.55\times 10^{-10}m)^{2}(Q2.35\times 10^{25}m^{-3})}=7.60\times 10^{-8}m\] If a sealed container has of xenon molecules traveling at a speed of 24.6 m/s at 108 °C. How many moles of xenon gas do we have in the container? \(\bar{E}_{trans}=\dfrac{1}{2}mv^2=\dfrac{3}{2}k_BT\\ m=\dfrac{3k_BT}{v^2}=\dfrac{3(1.381\times 10^{-23}J/K)(381.15K)}{(24.6m/s)^2}*\dfrac{1kg*m^2/s^2}{1J}=Q2.609\times 10^{-23}kg\\ n=\dfrac{m}{molarmass}=Q2.609\times 10^{-23}kg*\dfrac{1000g}{1kg}*\dfrac{mol}{131.3g}=1.987\times 10^{-22}mol\) At room temperature (25oC) there are two flasks A and B containing gaseous Oxygen and Hydrogen, respectively, at their equilibrium states. Assuming there is no exchange of heat or work between the systems inside the flasks and the surroundings. Compare the initial velocities between an Oxygen and a Hydrogen molecule in each flask. In a insulated container of volume 2.50 L filled with nitrogen gas, the Z value is 6.75e collisions/second. The temperature measured is 55ºC. Assuming that the collision diameter of N is 3.75 Å, Hint: look at a) \[ Z_{1} = \sqrt{2} \pi d^{2} \bar{c} (\dfrac{N}{V}) \] \[ \Rightarrow N = \dfrac{VZ_{1}}{\sqrt2 \pi d^{2} \bar{c}}\] \[ \bar{c} = \sqrt{\dfrac{8RT}{\pi M}} = \sqrt{\dfrac{8(8.314 \dfrac{J}{mol \cdot K})(55+273)K}{\pi \cdot 0.014kg})} = 704.3 m/s \] \[ N = \dfrac{(2.50 L)(\dfrac{1m^{3}}{1000L}) (6.75 \times 10^{19} collision/s)}{\sqrt{2} \pi (3.75 \times 10^{-10}m)^{2} (704.3 m/s)} = 1.017 \times 10^{23} molecules\] \[ n = \dfrac{N}{N_A} = \dfrac{1.017 \times 10^{23} molecules}{6.022 \times 10^{23} \dfrac{molecules}{mole}} = {\color{red} 0.169 mole N_{2}} \] b) \[ Z_{11} = \dfrac{1}{2} Z_{1} (\dfrac{N}{V}) \] \[ Z_{11} = \dfrac{1}{2} (6.75 \times 10^{19}) \dfrac{1.017 \times 10^{23} molecules}{2.50L \times \dfrac{1m^{3}}{1000 L}} = {\color{red} 1.37 \times 10^{45} \dfrac{collisions}{m^{3} \cdot s}} \] c) Since N = nN and Z is proportional to N, therefore \[{\color{red} Z_{1} \propto n}\] \[{\color{red} Z_{11} \propto n^{2}} \] For molecular oxygen at \(56^{o}C\), calculate the number of collisions a single molecule makes in 1 second and the total number of binary collisions at P=1.0 atm and P=0.25 atm. How does pressure relate to these two quantities? The collision diameter of oxygen is 3.02 A. Step 1: (Hint: Calculate the density and average molecular speed) At P=1.0atm $$\dfrac{N}{V}=\dfrac{PN_A}{RT}=\dfrac{(1.0atm)(6.022\times 10^{23}mol^{-1})}{0.08206\,L\,atm\,K^{-1}mol^{-1})(273+56)}=Q2.23\times 10^{22}L^{-1}(\dfrac{1000L}{1m^{3}})=2.23\times 10^{22}m^{-3}\] Step 2: $$\bar{c}=\sqrt{\dfrac{8RT}{\pi M}}=\sqrt{\dfrac{8(8.3145JK^{-1}mol^{-1})(273+56K)}{\pi (31.998\times 10^{-3}kg\,mol^{-1})}}=466.578ms^{-1}\] Step 3: $$z_1=\sqrt{2}\pi d^2\bar{c}\dfrac{N}{V}=\sqrt{2}\pi(3.02\times 10^{-10}m)^{2}(466.578ms^{-1})(2.23\times 10^{22}m^{-3})=4.683\times 10^{28} collisions m^{-3}s^{-1}\] Step 4: $$z_{11}=\dfrac{\sqrt{2}}{2}\pi d\bar{c}(\dfrac{N}{V})^{2}=\dfrac{1}{2}(4.2\times 10^{6})(2.23\times 10^{22}m^{-3})=4.683\times 10^{28} collisionsm^{-3}s^{-1}\] For an ideal gas, \(z_1\) is proportional to P and \(z_{11}\) is proportional to \(P^{2}\). Reducing P to a fourth of its original value (from 1.0 atm to 0.25 atm) will also reduce 2, to one fourth its value at P=1.0atm. Therefore at P=0.25 atm, \[z_1=4.2\times 10^{5}collisions\,s^{-1}$$
$$z_{11}=4.683\times 10^{26}\,collisions\,m^{-3}s{-1}\] The relationship between the molar mass and root mean square speed of an ideal gas can be described as: \[\upsilon _{rms}=\sqrt{\frac{3RT}{\mathfrak{M}}}\] Use this equation to derive Grahm's Law of Effusion: \[\frac{r_{1}}{r_{2}}=\sqrt{\frac{\mathfrak{M_{2}}}{\mathfrak{M_{1}}}}\] For two gasses at standard temperature and pressure, \[\upsilon _{1rms}=\sqrt{\frac{3RT}{\mathfrak{M_{1}}}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \upsilon _{2rms}=\sqrt{\frac{3RT}{\mathfrak{M_{2}}}}\] For any of the two gases, \[\upsilon _{1rms}^{2}=\frac{3RT}{\mathfrak{M_{1}}}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \upsilon _{2rms}^{2}=\frac{3RT}{\mathfrak{M_{2}}}\] \[R=\frac{\upsilon _{1rms}^{2}\mathfrak{M_{1}}}{3T}=\frac{\upsilon _{2rms}^{2}\mathfrak{M_{2}}}{3T}\] \[\upsilon _{1rms}^{2}\mathfrak{M_{1}}=\upsilon _{2rms}^{2}\mathfrak{M_{2}}\] \[\frac{\upsilon _{1rms}^{2}}{\upsilon_{2rms}^{2}}=\frac{\mathfrak{M_{2}}}{\mathfrak{M_{1}}}\] \[\frac{\upsilon _{1rms}}{\upsilon_{2rms}}=\sqrt{\frac{\mathfrak{M_{2}}}{\mathfrak{M_{1}}}}\] Which can be rewritten as: \[\frac{r_{1}}{r_{2}}=\sqrt{\frac{\mathfrak{M_{2}}}{\mathfrak{M_{1}}}}\] Derive Graham's Law of Effusion (1) from the equation for calculating Kinetic Energy (2). \[ \begin{align} &\left(1 \right) \ \ \ \frac{v_{1}}{v_{2}}=\sqrt{\frac{m_2}{m_1}} \\ &\left(2 \right) \ \ \ KE =\frac{1}{2}mv^{2}\ \end{align}\] \[ \begin{align} & \frac{v_{1}}{v_{2}}=\sqrt{\frac{m_2}{m_1}} \ \ KE =\frac{1}{2}mv^{2}\ \\ & KE_1 = KE_2, \ for \ two\ substances \\ & \frac{1}{2}m_1 v_1^{2} = \frac{1}{2}m_2 v_2^{2}, \ multiply \ both \ sides \ by \ two \\ & m_1 v_1^{2} =m_2 v_2^{2} \\ & \frac{v_1^{2}}{v_2^{2}} = \frac{m_2}{m_1}, \ square \ root \ both \ sides \ of \ the \ equation \\ & \underline{ \frac{v_{1}}{v_{2}}=\sqrt{\frac{m_2}{m_1}} } \end{align}\] and for more information about effusion and Graham's law. How long would it take for oxygen to diffuse in and carbon dioxide to diffuse out of a cell that is 0.1µm thick given that the rate of diffusion of carbon dioxide is 0.0016 mm /s? \(\dfrac{t_{CO_2}}{t_{O_2}}=\dfrac{r_{O_2}}{r_{CO_2}}=\sqrt{\dfrac{M_{CO_2}}{M_{O_2}}}\\
r_{O_2}=(0.0016mm^2/s)\sqrt{\dfrac{44.01g/mol}{3Q2.0g/mol}}=0.0019mm^2/s\\
t_{CO_2}=0.1\mu m^2*\dfrac{(1mm)^2}{(100\mu m)^2}*\dfrac{s}{0.0016mm^2}=6.25\times 10^{-3}s\\
t_{O_2}=0.1\mu m^2*\dfrac{(1mm)^2}{(100\mu m)^2}*\dfrac{s}{0.0019mm^2}=5.26\times 10^{-3}s\) m = 28.01 g/mol t /t = (m /m ) m = (t /t ) * m = (14.1 min/11.0 min) * 28.01 g mol
m = 46.02 g mol The red gas is nitrogen dioxide, a common emission from pulp mills. We know because the diffusion rate corresponds with a molar mass of 46.01g, the molar mass of NO An amount of CO gas is effused through a small opening at the rate of 0.596 L/min. A homogenous mixture of O and N in the atmosphere is put under the same conditions for effusion and after 5 minutes, 5.03 L of the mixture was effused. Find the percent composition of the homogenous mixture (molar fraction). Hint: Look at concept of \[ \dfrac{r_{CO_{2}}}{r_{mixture}} = \sqrt{\dfrac{M_{mixture}}{M_{CO_{2}}}} \] \[M_{mixture} = (\dfrac{r_{CO^{2}}}{r_{mixture}})^{2} (M_{CO_{2}}) = (\dfrac{0.596 \dfrac{L}{min}}{\frac{5.03 L}{5 min}})^2 (44.01 \dfrac{g}{mol}) = 15.447 \dfrac{g}{mol}\] \[ M_{mixture} = M_{N_{2}}X_{N_{2}} + M_{O_{2}}X_{O_{2}} \] \[ M_{mixture} = M_{N_{2}}X_{N_{2}} + M_{O_{2}}(1-X_{N_{2}}) \] \[ 15.447 \dfrac{g}{mol} = (14.01 \dfrac{g}{mol})(X_{N_{2}}) + (16.00 \dfrac{g}{mol})(1-X_{N_{2}}) \] \[ 15.447 = 14.01 X_{N_{2}} + 16.00 - 16.00X_{N_{2}} = -1.99X_{N_{2}} +16.00 \] \[ X_{N_{2}} = 0.278 \] \[ X_{O_{2}} = 1-0.278 = 0.722 \] \[ {\color{red} \% of N_{2} = 27.8\%} \] \[ {\color{red} \% of O_{2} = 72.2\%} \] You are trying to isolate gaseous Nitrogen-14 via effusion in a 50:50 mixture of Nitrogen-14 and Nitrogen-15 respectively. What will be the percentage of enrichment of the mixture after a single stage of separation? Use Graham's Law of Diffusion and Effusion to solve this problem. \[ \begin{align} & \frac{r_1}{r_2} = \sqrt{\frac{M_{2}}{M_{1}}} \\ & \frac{ rate \ of \ ^{14}\textrm{N} }{ rate \ of \ ^{15}\textrm{N} } = \sqrt{ \frac{^{15}\textrm{N}}{^{14}\textrm{N}} } \end{align}\] \[ \begin{align} & \frac{rate \ of \ {}^{14}\textrm{N}}{rate \ of \ {}^{15}\textrm{N}} = \sqrt{\frac{15.0001 \ g \ mol^{-1}}{14.00307 \ g \ mol^{-1}}} \\ & \frac{rate \ of \ {}^{14}\textrm{N}}{rate \ of \ {}^{15}\textrm{N}} = 1.035 \end{align}\] Since the ratio of rates of effusion is 1.035, in a single step of separation 3.5% more nitrogen-14 will effuse out of the mixture than nitrogen-15. Thus the percent of enrichment would be 3.5% since there is 3.5% more of N-14 now. and b for more information about effusion and Graham's law. A mixture of Neon and Helium gas is effused at standard temperature and pressure. Assuming the mixture is equimolar, what is the composition of the gas mixture after effusion? For the effusion of a binary gas mixture, we can use : \[\frac{r_{1}}{r_{2}}=\sqrt{\frac{\mathfrak{M_{2}}}{\mathfrak{M_{1}}}}\] The ratio of the two gasses following effusion given r corresponds with neon and r corresponds with helium is: \[\frac{r_{1}}{r_{2}}=\sqrt{\frac{4.003\frac{g}{mol}}{20.180\frac{g}{mol}}}=0.4454\] Therefore, to calculate the mole fraction of each gas, \[x_{Ne}=\frac{0.4454}{0.4454+1}=0.3081\] \[x_{He}=1-x_{Ne}=1-0.3081=0.6919\] g mol /n = (m /m ) m = m /( /n ) 28.01 g mol /(1.18) = 20.11 g mol | 30,657 | 1,590 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/05%3A_Introduction_To_Reactions_In_Aqueous_Solutions/5.4%3A_Oxidation-Reduction%3A_Some_General_Principles |
The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, coherent, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O ). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction. Any oxidation must be accompanied by a reduction and vice versa. Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water). The reaction is as follows: \[ Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g) \label{5.4.1}\] Oxidation and reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants, which follows the mnemonic oxidation is loss reduction is gain, or oil rig. The oxidation state of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O or H , are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is \[ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s) \label{5.4.2} \] Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al O , electrons are transferred as follows (the superscript 0 emphasizes the oxidation state of the elements): \[ 4 Al^0 + 3 O_2^0 \rightarrow 4 Al^{3+} + 6 O^{2-} \label{5.4.3}\] and are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In , for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen: \[ electrons \, lost = 4 \, Al \, atoms \times {3 \, e^- \, lost \over Al \, atom } = 12 \, e^- \, lost \label{5.4.4a}\] \[ electrons \, gained = 6 \, O \, atoms \times {2 \, e^- \, gained \over O \, atom} = 12 \, e^- \, gained \label{5.4.4a}\] The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained. An additional example of a redox reaction, the reaction of sodium metal with oxygen in air, is illustrated in . In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained. Redox Reactions: Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. Previosuly, you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; ), magnesium oxide (MgO), and calcium chloride (CaCl ). In covalent compounds, in contrast, atoms share electrons. Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions. A set of rules for assigning oxidation states to atoms in chemical compounds follows. . In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound. Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in . Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H ion, whereas HCl forms H and Cl ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF but −½ in KO . Note that an oxidation state of −½ for O in KO is perfectly acceptable. The reduction of copper(I) oxide shown in Equation 5.4.5 demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H and Cu. From rule 4, hydrogen in H O has an oxidation state of +1, and from rule 5, oxygen in both Cu O and H O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows: \[ \overset {+1}{Cu_2} \underset {-2}{O} (s) + \overset {0}{H_2} (g) \rightarrow 2 \overset {0}{Cu} (s) + \overset {+1}{H_2} \underset {-2}{O} (g) \label{5.4.5} \] Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). So this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge: \[ electrons \, lost = 2 \, H \, atoms \times {1 \, e^- \, lost \over H \, atom } = 2 \, e^- \, lost \label{5.4.6a}\] \[ electrons \, gained = 2 \, Cu \, atoms \times {1 \, e^- \, gained \over Cu \, atom} = 2 \, e^- \, gained \label{5.4.6b}\] Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo. Assign oxidation states to all atoms in each compound. : molecular or empirical formula : oxidation states : Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1. : a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF ), the oxidation state of sulfur must be +6: [(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0 b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2: [(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0 c. Note that (NH ) SO is an ionic compound that consists of both a polyatomic cation (NH ) and a polyatomic anion (SO ) (see ). We assign oxidation states to the atoms in each polyatomic ion separately. For NH , hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3: [(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH ion For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6: [(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron: Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe O can be viewed as having two Fe ions and one Fe ion per formula unit, giving a net positive charge of +8 per formula unit. Fe O is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.” e. Initially, we assign oxidation states to the components of CH CO H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of [(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0 So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH ) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of [(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3 To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus \[ \underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \] Thus the sum of the oxidation states of the two carbon atoms is indeed zero. Assign oxidation states to all atoms in each compound. : A widely encountered class of oxidation–reduction reactions is the reaction of aqueous solutions of acids or metal salts with solid metals. An example is the corrosion of metal objects, such as the rusting of an automobile (Figure \(\Page {1}\)). Rust is formed from a complex oxidation–reduction reaction involving dilute acid solutions that contain Cl ions (effectively, dilute HCl), iron metal, and oxygen. When an object rusts, iron metal reacts with HCl(aq) to produce iron(II) chloride and hydrogen gas: \(Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g) \label{5.4.5}\) In subsequent steps, FeCl undergoes oxidation to form a reddish-brown precipitate of Fe(OH) . Many metals dissolve through reactions of this type, which have the general form \[metal + acid \rightarrow salt + hydrogen \label{5.4.6}\] Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves: \[ Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) \label{5.4.7}\] Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul! Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation 5.4.8) and the reduction of silver salts by copper (Equation 5.4.9 and Figure \(\Page {2}\)): \[ Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s) \label{5.4.8}\] \[ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s) \label{5.4.9}\] The reaction in Equation 5.4.8 is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object. By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing Zn . Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal: \[ Zn(s) + Mg^{2+}(aq) \cancel{\rightarrow} Zn^{2+}(aq) + Mg(s) \label{5.4.10}\] \[ Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s) \label{5.4.11}\] Magnesium has a greater tendency to be oxidized than zinc does. Pairwise reactions of this sort are the basis of the activity series (Figure \(\Page {4}\)), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the , , and . In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, which have the least tendency to be oxidized. (at the bottom of the series). When using the activity series to predict the outcome of a reaction, keep in mind that . Because magnesium is above zinc in Figure \(\Page {4}\), magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example \(\Page {2}\) demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation. reactants overall reaction and net ionic equation \[ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s) \] Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction. \[ Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) \] Lead(II) sulfate is the white solid that forms on corroded battery terminals. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation. In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the (Table \(\Page {1}\)), in which the overall reaction is separated into an oxidation equation and a reduction equation. are reactions of metals with either acids or another metal salt that result in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions can be predicted using the (Figure \(\Page {4}\)), which arranges metals and H in decreasing order of their tendency to be oxidized. Any metal will reduce metal ions below it in the activity series. lie at the top of the activity series, whereas are at the bottom of the activity series. | 17,851 | 1,591 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Exercises%3A_Fleming/6.E%3A_Putting_the_Second_Law_to_Work_(Exercises) |
Using , calculate the standard reaction Gibbs functions (\(\ G^o\)) for the following reactions at 298 K. Estimate \(\ G\) at 1000 K from its value at 298 K for the reaction \[C(s) + 2 H_2(g) \rightarrow CH_4(g)\] with \(\Delta G = -50.75\, kJ \,at\, 298\, K\)/ The standard Gibbs function for formation (\(\Delta G_f^o\)) of \(PbO_2(s)\) is -217.4 kJ/mol at 298 K. Assuming \(O_2\) is an ideal gas, find the standard Helmholtz function for formation (\(\Delta A_f^o\) for \(PbO_2\) at 298K. Calculate the entropy change for 1.00 mol of an ideal monatomic gas (C = 3/2 R) undergoing an expansion and simultaneous temperature increase from 10.0 L at 298 K to 205.0 L at 455 K. Consider a gas that obeys the equation of state \[ p =\dfrac{nRT}{V-nb}\] Show that \[\left( \dfrac{\partial C_p}{\partial p} \right)_T=0\] for an ideal gas. Derive the thermodynamic equation of state \[\left( \dfrac{\partial H}{\partial p} \right)_T = V( 1- T \alpha)\] Derive the thermodynamic equation of state \[\left( \dfrac{\partial U}{\partial V} \right)_T = T \dfrac{ \alpha}{\kappa_T} -p\] The “Joule Coefficient” is defined by \[ \mu_J = \left( \dfrac{\partial T}{\partial V} \right)_U \] Show that \[ \mu_J = \dfrac{1}{C_V} \left( p - \dfrac{T \alpha}{\kappa_T }\right)\] and evaluate the expression for an ideal gas. Derive expressions for the pressure derivatives \[ \left( \dfrac{\partial X}{\partial p} \right)_T\] where\(X\) is \(U\), \(H\), \(A\), \(G\), and \(S\) at constant temperature in terms of measurable properties. (The derivation of \( \left( \dfrac{\partial H}{\partial p} \right)_T\) was done in problem Q6.7). Evaluate the expressions for for a van der Waals gas. Derive expressions for the volume derivatives \[ \left( \dfrac{\partial X}{\partial V} \right)_T\] where \(X\) is \(U\), \(H\), \(A\), \(G\), and \(S\) at constant temperature in terms of measurable properties. (The derivation of \( \left( \dfrac{\partial U}{\partial V} \right)_T\) was done in problem Q8.8.) Evaluate the expressions for for a van der Waals gas. Evaluate the difference between \(C_p\) and \(C_V\) for a gas that obeys the equation of state \[ p =\dfrac{nRT}{V-nb}\] The adiabatic compressibility (\(k_S\)) is defined by \[ \kappa_S = \dfrac{1}{V} \left( \dfrac{\partial V}{\partial p} \right)_S\] Show that for an ideal gas, \[ \kappa_S = \dfrac{1}{p \gamma}\] | 2,371 | 1,592 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/08%3A_Substitution_Reactions_of_Alkyl_Halides/8.06%3A_More_About_the_Stereochemistry_of_(S_N2)_and_(S_N1)_Reactions |
A biomolecular nucleophilic substitution (S 2) reaction is a type of nucleophilic substitution whereby a lone pair of electrons on a nucleophile attacks an electron deficient electrophilic center and bonds to it, resulting in the expulsion of a leaving group. It is possible for the nucleophile to attack the electrophilic center in two ways. The following diagram illustrates these two types of nucleophilic attacks, where the frontside attack results in retention of configuration; that is, the product has the same configuration as the substrate. The backside attack results in inversion of configuration, where the product's configuration is opposite that of the substrate. S 2 Experimental observation shows that all S 2 reactions proceed with inversion of configuration; that is, the nucleophile will always attack from the backside in all S 2 reactions. To think about why this might be true, remember that the nucleophile has a lone pair of electrons to be shared with the electrophilic center, and the leaving group is going to take a lone pair of electrons with it upon leaving. Because like charges repel each other, the nucleophile will always proceed by a backside displacement mechanism. The S 2 reaction is stereospecific. A stereospecific reaction is one in which different stereoisomers react to give different stereoisomers of the product. For example, if the substrate is an R enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the R enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the S enantiomer. Conversely, if the substrate is an S enantiomer, a frontside nucleophilic attack results in retention of configuration, and the formation of the S enantiomer. A backside nucleophilic attack results in inversion of configuration, and the formation of the R enantiomer. In conclusion, S 2 reactions that begin with the R enantiomer as the substrate will form the S enantiomer as the product. Those that begin with the S enantiomer as the substrate will form the R enantiomer as the product. This concept also applies to substrates that are and substrates that are . If the configuration is the substrate, the resulting product will be . Conversely, if the configuration is the substrate, the resulting product will be . | 2,373 | 1,593 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/15%3A_Principles_of_Chemical_Equilibrium/15.6%3A_Altering_Equilibrium_Conditions_-_Le_Chateliers_Principle |
As we saw in the previous section, reactions proceed in both directions (reactants go to products and products go to reactants). We can tell a reaction is at equilibrium if the reaction quotient (\(Q\)) is equal to the equilibrium constant (\(K\)). We next address what happens when a system at equilibrium is disturbed so that \(Q\) is no longer equal to \(K\). If a system at equilibrium is subjected to a perturbance or (such as a change in concentration) the changes. Since this stress affects the concentrations of the reactants and the products, the value of \(Q\) will no longer equal the value of \(K\). To re-establish equilibrium, the system will either shift toward the products (if \((Q \leq K)\) or the reactants (if \((Q \geq K)\) until \(Q\) returns to the same value as \(K\). This process is described by When a chemical system at equilibrium is disturbed, it returns to equilibrium by counteracting the disturbance. As described in the previous paragraph, the disturbance causes a change in \(Q\); the reaction will shift to re-establish \(Q = K\). Le Chatelier's principle can be used to predict changes in equilibrium concentrations when a system that is at equilibrium is subjected to a stress. However, if we have a mixture of reactants and products that have not yet reached equilibrium, the changes necessary to reach equilibrium may not be so obvious. In such a case, we can compare the values of \(Q\) and \(K\) for the system to predict the changes. A chemical system at equilibrium can be shifted out of equilibrium by adding or removing one or more of the reactants or products. The concentrations of both reactants and products then undergo additional changes to return the system to equilibrium. The stress on the system in Figure \(\Page {1}\) is the reduction of the equilibrium concentration of (lowering the concentration of one of the reactants would cause \(Q\) to be larger than ). As a consequence, Le Chatelier's principle leads us to predict that the concentration of Fe(SCN) should decrease, increasing the concentration of SCN part way back to its original concentration, and increasing the concentration of Fe above its initial equilibrium concentration. The effect of a change in concentration on a system at equilibrium is illustrated further by the equilibrium of this chemical reaction: \[\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)} \label{15.7.1a} \] \[K_c=\mathrm{50.0 \; at\; 400°C} \label{15.7.1b} \] The numeric values for this example have been determined experimentally. A mixture of gases at 400 °C with \(\mathrm{[H_2] = [I_2]} = 0.221\; M\) and \(\ce{[HI]} = 1.563 \;M\) is at equilibrium; for this mixture, \(Q_c = K_c = 50.0\). If \(\ce{H_2}\) is introduced into the system so quickly that its concentration doubles before it begins to react (new \(\ce{[H_2]} = 0.442\; M\)), the reaction will shift so that a new equilibrium is reached, at which This gives: \[\begin{align*} Q_c &=\mathrm{\dfrac{[HI]^2}{[H_2,I_2]}} \\[4pt] &=\dfrac{(1.692)^2}{(0.374)(0.153)} \\[4pt] &= 50.0 =K_c \label{15.7.2} \end{align*} \] We have stressed this system by introducing additional \(\ce{H_2}\). The stress is relieved when the reaction shifts to the right, using up some (but not all) of the excess \(\ce{H_2}\), reducing the amount of uncombined \(\ce{I_2}\), and forming additional \(\ce{HI}\). A Discussing Le Chatelier’s Principle (Changing Concentrations): Sometimes we can change the position of equilibrium by changing the pressure of a system. However, changes in pressure have a measurable effect only in systems in which gases are involved, and then only when the chemical reaction produces a change in the total number of gas molecules in the system. An easy way to recognize such a system is to look for different numbers of moles of gas on the reactant and product sides of the equilibrium. While evaluating pressure (as well as related factors like volume), it is important to remember that equilibrium constants are defined with regard to concentration (for ) or partial pressure (for ). Some changes to total pressure, like adding an inert gas that is not part of the equilibrium, will change the total pressure but not the partial pressures of the gases in the equilibrium constant expression. Thus, addition of a gas not involved in the equilibrium will not perturb the equilibrium. As we increase the pressure of a gaseous system at equilibrium, either by decreasing the volume of the system or by adding more of one of the components of the equilibrium mixture, we introduce a stress by increasing the partial pressures of one or more of the components. In accordance with Le Chatelier's principle, a shift in the equilibrium that reduces the total number of molecules per unit of volume will be favored because this relieves the stress. The reverse reaction would be favored by a decrease in pressure. Consider what happens when we increase the pressure on a system in which \(\ce{NO}\), \(\ce{O_2}\), and \(\ce{NO_2}\) are at equilibrium: \[\ce{2NO(g) + O2(g) \rightleftharpoons 2NO2(g)} \label{15.7.3} \] The formation of additional amounts of \(\ce{NO2}\) decreases the total number of molecules in the system because each time two molecules of \(\ce{NO_2}\) form, a total of three molecules of \(\ce{NO}\) and \(\ce{O_2}\) are consumed. This reduces the total pressure exerted by the system and reduces, but does not completely relieve, the stress of the increased pressure. On the other hand, a decrease in the pressure on the system favors decomposition of \(\ce{NO_2}\) into \(\ce{NO}\) and \(\ce{O_2}\), which tends to restore the pressure. Now consider this reaction: \[\ce{N2(g) + O2(g) \rightleftharpoons 2NO(g)} \label{15.7.4} \] Because there is no change in the total number of molecules in the system during reaction, a change in pressure does not favor either formation or decomposition of gaseous nitrogen monoxide. Le Chatelier’s Principle (Changes in Pressure or Volume): Changing concentration or pressure perturbs an equilibrium because the reaction quotient is shifted away from the equilibrium value. Changing the temperature of a system at equilibrium has a different effect: A change in temperature actually changes the value of the equilibrium constant. However, we can qualitatively predict the effect of the temperature change by treating it as a stress on the system and applying Le Chatelier's principle. When hydrogen reacts with gaseous iodine, heat is evolved. \[\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)}\;\;\ ΔH=\mathrm{−9.4\;kJ\;(exothermic)} \label{15.7.5} \] Because this reaction is exothermic, we can write it with heat as a product. \[\ce{H2(g) + I2(g) \rightleftharpoons 2HI(g)} + \text{heat} \label{15.7.6} \] Increasing the temperature of the reaction increases the internal energy of the system. Thus, increasing the temperature has the effect of increasing the amount of one of the products of this reaction. The reaction shifts to the left to relieve the stress, and there is an increase in the concentration of H and I and a reduction in the concentration of HI. Lowering the temperature of this system reduces the amount of energy present, favors the production of heat, and favors the formation of hydrogen iodide. When we change the temperature of a system at equilibrium, the equilibrium constant for the reaction changes. Lowering the temperature in the HI system increases the equilibrium constant: At the new equilibrium the concentration of HI has increased and the concentrations of H and I decreased. Raising the temperature decreases the value of the equilibrium constant, from 67.5 at 357 °C to 50.0 at 400 °C. Temperature affects the equilibrium between \(\ce{NO_2}\) and \(\ce{N_2O_4}\) in this reaction \[\ce{N2O4(g) \rightleftharpoons 2NO2(g)}\;\;\; ΔH=\mathrm{57.20\; kJ} \label{15.7.7} \] The positive ΔH value tells us that the reaction is endothermic and could be written \[\ce{heat}+\ce{N_2O4(g) \rightleftharpoons 2NO2(g)} \label{15.7.8} \] At higher temperatures, the gas mixture has a deep brown color, indicative of a significant amount of brown \(\ce{NO_2}\) molecules. If, however, we put a stress on the system by cooling the mixture (withdrawing energy), the equilibrium shifts to the left to supply some of the energy lost by cooling. The concentration of colorless \(\ce{N_2O_4}\) increases, and the concentration of brown \(\ce{NO_2}\) decreases, causing the brown color to fade. The overview of how different disturbances affect the reaction equilibrium properties is tabulated in Table \(\Page {1}\). Write an equilibrium constant expression for each reaction and use this expression to predict what will happen to the concentration of the substance in bold when the indicated change is made if the system is to maintain equilibrium. : equilibrium systems and changes : equilibrium constant expressions and effects of changes : Write the equilibrium constant expression, remembering that pure liquids and solids do not appear in the expression. From this expression, predict the change that must occur to maintain equilibrium when the indicated changes are made. : Because \(HgO_{(s)}\) and \(Hg_{(l)}\) are pure substances, they do not appear in the equilibrium constant expression. Thus, for this reaction, \(K = [O_2]\). The equilibrium concentration of \(O_2\) is a constant and does not depend on the amount of \(HgO\) present. Hence adding more \(HgO\) will not affect the equilibrium concentration of \(O_2\), so no compensatory change is necessary. \(NH_4HS\) does not appear in the equilibrium constant expression because it is a solid. Thus \(K = [NH_3,H_2S]\), which means that the concentrations of the products are inversely proportional. If adding \(H_2S\) triples the \(H_2S\) concentration, for example, then the \(NH_3\) concentration must decrease by about a factor of 3 for the system to remain at equilibrium so that the product of the concentrations equals \(K\). For this reaction, \(K = \frac{[isobutane]}{[\textit{n-butane}]}\), so halving the concentration of isobutane means that the n-butane concentration must also decrease by about half if the system is to maintain equilibrium. Write an equilibrium constant expression for each reaction. What must happen to the concentration of the substance in bold when the indicated change occurs if the system is to maintain equilibrium? \(K = \dfrac{[H_2]}{[HBr]}\); \([H_2]\) must decrease by about a factor of 3. \(K = \dfrac{1}{[N_2]}\); solid lithium does not appear in the equilibrium constant expression, so no compensatory change is necessary. \(K = \dfrac{1}{[SO_2,Cl_2]}\); \([SO_2]\) must decrease by about half. Le Chatelier’s Principle (Changes in Temperature): As we learned during our study of kinetics, a catalyst can speed up the rate of a reaction. Though this increase in reaction rate may cause a system to reach equilibrium more quickly (by speeding up the forward and reverse reactions), a catalyst has no effect on the value of an equilibrium constant nor on equilibrium concentrations. The interplay of changes in concentration or pressure, temperature, and the lack of an influence of a catalyst on a chemical equilibrium is illustrated in the industrial synthesis of ammonia from nitrogen and hydrogen according to the equation \[\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g)} \label{15.7.9} \] A large quantity of ammonia is manufactured by this reaction. Each year, ammonia is among the top 10 chemicals, by mass, manufactured in the world. About 2 billion pounds are manufactured in the United States each year. Ammonia plays a vital role in our global economy. It is used in the production of fertilizers and is, itself, an important fertilizer for the growth of corn, cotton, and other crops. Large quantities of ammonia are converted to nitric acid, which plays an important role in the production of fertilizers, explosives, plastics, dyes, and fibers, and is also used in the steel industry. Haber was born in Breslau, Prussia (presently Wroclaw, Poland) in December 1868. He went on to study chemistry and, while at the University of Karlsruhe, he developed what would later be known as the Haber process: the catalytic formation of ammonia from hydrogen and atmospheric nitrogen under high temperatures and pressures. For this work, Haber was awarded the 1918 Nobel Prize in Chemistry for synthesis of ammonia from its elements (Equation \ref{15.7.9}). The Haber process was a boon to agriculture, as it allowed the production of fertilizers to no longer be dependent on mined feed stocks such as sodium nitrate. Currently, the annual production of synthetic nitrogen fertilizers exceeds 100 million tons and synthetic fertilizer production has increased the number of humans that arable land can support from 1.9 persons per hectare in 1908 to 4.3 in 2008. The availability of nitrogen is a strong limiting factor to the growth of plants. Despite accounting for 78% of air, diatomic nitrogen (\(\ce{N_2}\)) is nutritionally unavailable to a majority of plants due the tremendous stability of the nitrogen-nitrogen triple bond. Therefore, the nitrogen must be converted to a more bioavailable form (this conversion is called nitrogen fixation). Legumes achieve this conversion at ambient temperature by exploiting bacteria equipped with suitable enzymes. In addition to his work in ammonia production, Haber is also remembered by history as one of the fathers of chemical warfare. During World War I, he played a major role in the development of poisonous gases used for trench warfare. Regarding his role in these developments, Haber said, “During peace time a scientist belongs to the World, but during war time he belongs to his country.” Haber defended the use of gas warfare against accusations that it was inhumane, saying that death was death, by whatever means it was inflicted. He stands as an example of the ethical dilemmas that face scientists in times of war and the double-edged nature of the sword of science. Like Haber, the products made from ammonia can be multifaceted. In addition to their value for agriculture, nitrogen compounds can also be used to achieve destructive ends. Ammonium nitrate has also been used in explosives, including improvised explosive devices. Ammonium nitrate was one of the components of the bomb used in the attack on the Alfred P. Murrah Federal Building in downtown Oklahoma City on April 19, 1995. Systems at equilibrium can be disturbed by changes to temperature, concentration, and, in some cases, volume and pressure; volume and pressure changes will disturb equilibrium if the number of moles of gas is different on the reactant and product sides of the reaction. The system's response to these disturbances is described by Le Chatelier's principle: The system will respond in a way that counteracts the disturbance. Not all changes to the system result in a disturbance of the equilibrium. Adding a catalyst affects the rates of the reactions but does not alter the equilibrium, and changing pressure or volume will not significantly disturb systems with no gases or with equal numbers of moles of gas on the reactant and product side. ). | 15,217 | 1,594 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/01%3A_Introduction_to_Organic_Chemistry/1.02%3A_A_Bit_of_History |
You may not be much interested in the way that organic chemistry developed, but if you skip to the next section without reading further, you will miss some of the flavor of a truly great achievement - of how a few highly creative chemists were able, with the aid of a few simple tools, to determine the structures of molecules, far too small and too elusive to be seen individually with the finest optical microscope, manifesting themselves only by the collective behavior of at least millions of millions at once. Try to visualize the problems confronting the organic chemist of 100 years ago. You will have no more than reasonably pure samples of organic compounds, the common laboratory chemicals of today, glassware, balances, thermometers, means of measuring densities, and a few optical instruments. You also will have a relatively embryonic theory that there are molecules in those bottles and that one compound differs from another because its molecules have different members or kinds of atoms and different arrangements of bonds. Your task will be to determine what kinds and what numbers of atoms they contain, that is, to determine their . Obviously, a compound with formula \(C_2H_6O\) and one with \(C_2H_6O_2\) are not the same compound. But suppose two compounds from different sources both are \(C_2H_6O\). To decide whether these are the you could smell them (far better to than to inhale), taste them (emphatically not recommended), see if they have the same appearance and viscosity (if liquids), or use more sophisticated criteria: boiling point, melting point, density, or refractive index. Other possibilities would be to see if they both have the same solubility in water or other solvents and whether they give the same reaction products with various reagents. Of course, all this gets a bit tough when the compounds are not pure and no good ways are available to purify them, but that is part of the job. Think about how you might proceed. In retrospect it is surprising that in less than fifty years an enormous, even if incomplete, edifice of structural organic chemistry was constructed on the basis of the results of chemical reactions without determination of a single bond distance, and with no electronic theory as a guide. Interestingly, all of the subsequent developments of the quantum mechanical theory of chemical bonds has not altered this edifice in significant ways. Indeed, for a long time, a goal of molecular quantum mechanics was simply to be able to corroborate that when an organic chemist draws a single line between two carbon atoms to show that they are bonded, he in fact knows what he is doing. And that when he draws two (or three) bonds between the carbons to indicate a double (or triple) bond, quantum mechanics supports this also as a valid idea. Furthermore, when modern tools for determining organic structures that involve actually measuring the distances between the atoms became available, these provided great convenience, but no great surprises. To be sure, a few structures turned out to be incorrect because they were based on faulty or inadequate experimental evidence. But, on the whole, the modern three-dimensional representations of molecules that accord with actual measurements of bond distances and angles are in no important respect different from the widely used three-dimensional ball-and-stick models of organic molecules, and these, in essentially their present form, date from at least as far back as E. Paterno, in 1869. How was all of this achieved? Not by any very simple process. The essence of some of the important ideas follow, but it should be clear that what actually took place was far from straightforward. A diverse group of people was involved; many firmly committed to, if not having a vested interest in, earlier working hypotheses or that had served as useful bases for earlier experimentation, but were coming apart at the seams because they could not accommodate the new facts that kept emerging. As is usual in human endeavors, espousal of new and better ideas did not come equally quickly to all those used to thinking in particular ways. To illustrate, at least one famous chemist, Berthelot, still used \(HO\) as the formula for water twenty-five years after it seemed clear that \(H_2O\) was a better choice. Before structures of molecules could be established, there had to be a means of establishing molecular formulas and for this purpose the key concept was Avogadro's hypothesis, which can be stated in the form "equal volumes of gases at the same temperature and pressure contain the same number of molecules." Avogadro's hypothesis allowed assignment of molecular weights from measurements of gas densities. Then, with analytical techniques that permit determination of the weight percentages of the various elements in a compound, it became possible to set up a self-consistent set of relative atomic weights.\(^1\) From these and the relative molecular weights, one can assign molecular formulas. For example, if one finds that a compound contains \(22.0 \%\) carbon (atomic weight \(= 12.00\)), \(4.6 \%\) hydrogen (atomic weight \(= 1.008\)), and \(73.4 \%\) bromine (atomic weight \(= 79.90\)), then the ratios of the numbers of atoms are \(\left( 22.0/12.00 \right) : \left( 4.6/1.008 \right) : \left( 73.4/79.90 \right) = 1.83:4.56:0.92\). Dividing each of the last set of numbers by the smallest (\(0.92\)) gives \(1.99:4.96:1 \cong 2:5:1\), which suggests a molecular formula of \(C_2H_5Br\) or a multiple thereof. If we know that hydrogen gas is \(H_2\) and has a molecular weight of \(2 \times 1.008 = 2.016\), we can compare the weight of a given volume of hydrogen with the weight of the same volume of our unknown in the gas phase at the same temperature and pressure. If the experimental ratio of these weights turns out to be \(54\), then the molecular weight of the unknown would be \(2.016 \times 54 = 109\) and the formula \(C_2H_5Br\) would be correct. If we assume that the molecule is held together by chemical bonds, without knowing more, we could write numerous structures such as \(H-H-H-H-H-C-C-Br, H-C-Br-H-H-C-H-H\), and so on. However, if we also know of the existence of stable \(H_2\), but not \(H_3\); of stable \(Br_2\), but not of \(Br_3\); and of stable \(CH_3Br\), \(CH_2Br_2\), \(CHBr_3\), and \(CBr_4\), but not of \(CH_4Br\), \(CHBr\), \(CBr\), and so on, a pattern of what is called emerges. It will be seen that the above formulas all are consistent if hydrogen atoms and bromine atoms form just bond (are univalent) while carbon atoms form bonds (are tetravalent). This may seem almost naively simple today, but a considerable period of doubt and uncertainty preceded the acceptance of the idea of definite valences for the elements that emerged about 1852. If we accept hydrogen and bromine as being univalent and carbon as tetravalent, we can write as a structural formula for \(C_2H_5Br\).\(^2\) However, we also might have written There is a serious problem as to whether these formulas represent the or compounds. All that was known in the early days was that every purified sample of \(C_2H_5Br\), no matter how prepared, had a boiling point of \(38^\text{o}C\) and density of \(1.460 \: \text{g} \: \text{ml}^{-1}\). Furthermore, all looked the same, all smelled the same, and all underwent the same chemical reactions. There was no evidence that \(C_2H_5Br\) was a mixture or that more than one compound of this formula could be prepared. One might conclude, therefore, that all of the structural formulas above represent a single substance even though they superficially, at least, look different. Indeed, because \(H-Br\) and \(Br-H\) are two different ways of a formula for the same substance, we suspect that the same is true for There are, though, two of these structures that could be different from one another, namely In the first of these, \(CH_3-\) is located opposite the \(Br-\) and the \(H-\)'s on the carbon with the \(Br\) also are opposite one another. In the second formula, \(CH_3-\) and \(Br-\) are located to each other as are the \(H-\)'s on the same carbon. We therefore have a problem as to whether these two different formulas also represent different compounds. A brilliant solution to the problem posed in the preceding section came in 1874 when J. H. van't Hoff proposed that all four valences of carbon are equivalent and directed to the corners of a regular tetrahedron.\(^3\) If we redraw the structures for \(C_2H_5Br\) as \(1\), we see that there is only possible arrangement and, contrary to the impression we got from our earlier structural formulas, the bromine is located with respect to each of the hydrogens on the same carbon. A convenient way of representing organic molecules in three dimensions, which shows the tetrahedral relationships of the atoms very clearly, uses the so-called ball-and-stick models. The sticks that represent the bonds or valences form the tetrahedral angles of \(109.47^\text{o}\). The tetrahedral carbon does not solve all problems without additional postulates. For example, there are two different compounds known with the formula \(C_2H_4Br_2\). These substances, which we call , can be reasonably written as However, ball-and-stick models suggest further possibilities for the second structure, for example \(3\), \(4\), and \(5\): This is a problem apparently first clearly recognized by Paterno, in 1869. We call these rotational (or conformational) isomers, because one is converted to another by rotation of the halves of the molecule with respect to one another, with the \(C-C\) bond acting as an axle. If this is not clear, you should make a ball-and-stick model and see what rotation around the \(C-C\) bond does to the relationships between the atoms on the carbons. The difficulty presented by these possibilities finally was circumvented by a brilliant suggestion by van't Hoff of "free rotation," which holds that isomers corresponding to different rotational angles, such as \(3\), \(4\), and \(5\), do not have separate stable existence, but are interconverted by rotation around the \(C-C\) bond so rapidly that they are indistinguishable from one another. Thus there is only isomer corresponding to the different possible rotational angles and a total of only isomers of formula \(C_2H_4Br_2\). As we shall see, the idea of free rotation required extensive modification some 50 years after it was first proposed, but it was an extremely important paradigm, which, as often happens, became so deeply rooted as to become essentially an article of faith for later organic chemists. Free rotation will be discussed in more detail in Chapters 5 and 27. The problem of determining whether a particular isomer of \(C_2H_4Br_2\) is could be solved today in a few minutes by spectroscopic means, as will be explained in Chapter 9. However, at the time structure theory was being developed, the structure had to be deduced on the basis of chemical reactions, which could include either how the compound was formed or what it could be converted to. A virtually unassailable proof of structure, where it is applicable, is to determine how many different products each of a given group of isomers can give. For the \(C_2H_4Br_2\) pair of isomers, will be seen to give only possibility with one compound and with the other: Therefore, if we have two bottles, one containing one \(C_2H_4Br_2\) isomer and one the other and run the substitution test, the compound that gives only one product is \(6\) and the one that gives a mixture of two products is \(7\). Further, it will be seen that the test, besides telling which isomer is \(6\) and which is \(7\), establishes the structures of the two possible \(C_2H_3Br_3\) isomers, \(8\) and \(9\). Thus only \(8\) can be formed from both of the different \(C_2H_4Br_2\) isomers whereas \(9\) is formed from only one of them. There were already many interconversion reactions of organic compounds known at the time that valence theory, structural formulas, and the concept of the tetrahedral carbon came into general use. As a result, it did not take long before much of organic chemistry could be fitted into a concordant whole. One difficult problem was posed by the structures of a group of substitution products of benzene, \(C_6H_6\), called "aromatic compounds," which for a long time defied explanation. Benzene itself had been prepared first by Michael Faraday, in 1825. An ingenious solution for the benzene structure was provided by A. Kekule, in 1866, wherein he suggested (apparently as the result of a hallucinatory perception) that the six carbons were connected in a hexagonal ring with alternating single and double carbon-to-carbon bonds, and with each carbon connected to a single hydrogen, \(10\): This concept was controversial, to say the least, mainly on two counts. Benzene did not behave as expected, as judged by the behavior of other compounds with carbon-to-carbon double bonds and also because there should be two different dibromo substitution products of benzene with the bromine on adjacent carbons (\(11\) and \(12\)) but only one such compound could be isolated. Kekule explained the second objection away by maintaining that \(11\) and \(12\) were in rapid equilibrium through concerted bond shifts, in something like the same manner as the free-rotation hypothesis mentioned previously: However, the first objection could not be dismissed so easily and quite a number of alternative structures were proposed over the ensuing years. The controversy was not really resolved until it was established that benzene is a regular planar hexagon, which means that all of its \(C-C\) bonds have the same length, in best accord with a structure written not with double, not with single, but with 1.5 bonds between the carbons, as in \(13\): This. in turn, generated a massive further theoretical controversy over just how \(13\) should be interpreted, which, for a time, even became a part of "Cold-War" politics!\(^4\) We shall examine experimental and theoretical aspects of the benzene structure in some detail later. It is interesting that more than 100 years after Kekule's proposal the final story on the benzene structure is yet to be told.\(^5\) The combination of valence theory and the substitution method as described in gives, for many compounds, quite unequivocal proofs of structure. Use of chemical transformations for proofs of structure depends on the applicability of a simple guiding principle, often called the " ." As we shall see later, many exceptions are known and care is required to keep from making serious errors. With this caution, let us see how the principle may be applied. The compound \(C_2H_5Br\) discussed in reacts slowly with water to give a product of formula \(C_2H_6O\). The normal valence of oxygen is two, and we can write two, and only two, different structures, \(19\) and \(20\), for \(C_2H_6O\): The principle of least structural change favors \(19\) as the product, because the reaction to form it is a simple replacement of bromine bonded to carbon by \(-OH\), whereas formation of \(20\) would entail a much more drastic rearrangement of bonds. The argument is really a subtle one, involving an assessment of the reasonableness of various possible reactions. On the whole, however, it works rather well and, in the specific case of the \(C_2H_6O\) isomers, is strongly supported by the fact that treatment of \(19\) with strong hydrobromic acid (\(HBr\)) converts it back to \(C_2H_5Br\). In contrast, the isomer of structure \(20\) reacts with \(HBr\) to form two molecules of \(CH_3Br\): In each case, \(C-O\) bonds are broken and \(C-Br\) bonds are formed. We could conceive of many other possible reactions of \(C_2H_6O\) with \(HBr\), for example which, as indicated by \(\nrightarrow\), does occur, but hardly can be ruled out by the principle of least structural change itself. Showing how the probability of such alternative reactions can be evaluated will be a very large part of our later discussions. The substitution method and the interconversion reactions discussed for proof of structure possibly may give you erroneous ideas about the reactions and reactivity of organic compounds. We certainly do not wish to imply that it is a simple, straightforward process to make all of the possible substitution products of a compound such as In fact, as will be shown later, direct substitution of bromine for hydrogen with compounds such as this does not occur readily, and when it does occur, the four possible substitution products indeed are formed, but in far from equal amounts because there are for substitution at the different positions. Actually, some of the substitution products are formed only in very small quantities. Fortunately, this does not destroy the validity of the substitution method but does make it more difficult to apply. If direct substitution fails, some (or all) of the possible substitution products may have to be produced by indirect means. Nonetheless, you must understand that the success of the substitution method depends on determination of the total number of possible isomers - it does depend on how the isomers are prepared. Later, you will hear a lot about compounds or reagents being "reactive" and "unreactive." You may be exasperated by the loose way that these terms are used by organic chemists to characterize how fast various chemical changes occur. Many familiar inorganic reactions, such as the neutralization of hydrochloric acid with sodium hydroxide solution, are extremely fast at ordinary temperatures. But the same is not often true of reactions of organic compounds. For example, \(C_2H_5Br\) treated in two different ways is converted to gaseous compounds, one having the formula \(C_2H_6\) and the other \(C_2H_4\). The \(C_2H_4\) compound, , reacts with bromine to give \(C_2H_4Br_2\), but the \(C_2H_6\) compound, , does not react with bromine except at high temperatures or when exposed to sunlight (or similar intense light). The reaction products then are \(HBr\) and \(C_2H_5Br\), and later, \(HBr\) and \(C_2H_4Br_2\), \(C_2H_3Br_3\), and so on. We clearly can characterize \(C_2H_4\) as "reactive" and \(C_2H_6\) as "unreactive" toward bromine. The early organic chemists also used the terms "unsaturated" and "saturated" for this behavior, and these terms are still in wide use today. But we need to distinguish between "unsaturated" and "reactive," and between "saturated" and "unreactive," because these pairs of terms are not synonymous. The equations for the reactions of ethene and ethane with bromine are different in that ethene bromine, \(C_2H_4 + Br_2 \rightarrow C_2H_4Br_2\), whereas ethane bromine, \(C_2H_6 + Br_2 \rightarrow C_2H_5Br + HBr\). You should reserve the term "unsaturated" for compounds that can, at least potentially, react by , and "saturated' for compounds that can only be expected to react by . The difference between addition and substitution became much clearer with the development of the structure theory that called for carbon to be tetravalent and hydrogen univalent. Ethene then was assigned a structure with a carbon-to-carbon bond, and ethane a structure with a carbon-to-carbon bond: Addition of bromine to ethene subsequently was formulated as breaking one of the carbon-carbon bonds of the double bond and attaching bromine to these valences. Substitution was written similarly but here bromine and a \(C-H\) bond are involved: We will see later that the way in which these reactions actually occur is much more complicated than these simple equations indicate. In fact, such equations are regarded best as chemical accounting operations. The number of bonds is shown correctly for both the reactants and the products, and there is an indication of which bonds break and which bonds are formed in the overall process. However, do not make the mistake of assuming that no other bonds are broken or made in intermediate stages of the reaction. Much of what comes later in this book will be concerned with what we know, or can find out, about the of such reactions - a reaction mechanism being the actual sequence of events by which the reactants become converted to the products. Such information is of extraordinary value in defining and understanding the range of applicability of given reactions for practical preparations of desired compounds. The distinction we have made between "unsaturated" and "reactive" is best illustrated by a definite example. Ethene is "unsaturated" (and "reactive") toward bromine, but tetrachloroethene, \(C_2Cl_4\), will not add bromine at all under the same conditions and is clearly "unreactive." But is it also "saturated"? The answer is definitely no, because if we add a small amount of aluminum bromide, \(AlBr_3\), to a mixture of tetrachloroethene and bromine, addition does occur, although sluggishly: Obviously, tetrachloroethene is "unsaturated" in the sense it can undergo addition, even if it is unreactive to bromine in the absence of aluminum bromide. The aluminum bromide functions in the addition of bromine to tetrachloroethene as a , which is something that facilitates the conversion of reactants to products. The study of the nature and uses of catalysts will concern us throughout this book. Catalysis is our principal means of controlling organic reactions to help form the product we want in the shortest possible time. \(^1\)We will finesse here the long and important struggle of getting a truly self-consistent table of atomic weights. If you are interested in the complex history of this problem and the clear solution to it proposed by S. Cannizzaro in 1860, there are many accounts available in books on the history of chemistry. One example is J. R. Partington, , Vol. IV, Macmillan, London, 1964. Relative atomic weights now are based on \(^{12}C = 12\) (exactly). \(^2\)Formulas such as this appear to have been used first by Crum Brown, in 1864, after the originators of structural formulas, A. Kekule and A. Couper (1858), came up with rather awkward, impractical representations. It seems incredible today that even the drawing of these formulas was severely criticized for many years. The pot was kept boiling mainly by H. Kolbe, a productive German chemist with a gift for colorful invective and the advantage of a podium provided by being editor of an influential chemical journal. \(^3\)The name of J. A. Le Bel also is associated with this particular idea, but the record shows that Le Bel actually opposed the tetrahedral formulations, although, simultaneously with van't Hoff, he made a related very important contribution, as will be discussed in Chapter 5. \(^4\)The "resonance theory," to be discussed in detail in Chapters 6 and 21, was characterized in 1949 as a physically and ideologically inadmissible theory formulated by "decadent bourgeois scientists." See L. R. Graham, , Vintage Books, New York, 1973, Chapter VIII, for an interesting account of this controversy. \(^5\)Modern organic chemistry should not be regarded at all as a settled science, free of controversy. To be sure, personal attacks of the kind indulged in by Kolbe and others often are not published, but profound and indeed acrimonious differences of scientific interpretation exist and can persist for many years. and (1977) | 23,442 | 1,595 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/15%3A_Aromaticity_(Reactions_of_Benzene)/15.18%3A_It_Is_Important_to_Have_More_Than_One_Way_to_Carry_Out_a_Reaction |
The reaction of aldehydes and ketones with zinc amalgam (Zn/Hg alloy) in concentrated hydrochloric acid, which reduces the aldehyde or ketone to a hydrocarbon, is called Clemmensen reduction. This alternative reduction involves heating a carbonyl compound with finely divided, amalgamated zinc in a hydroxylic solvent (often an aqueous mixture) containing a mineral acid such as HCl. The mercury alloyed with the zinc does not participate in the reaction, it serves only to provide a clean active metal surface. The mechanism of Clemmensen reduction is not fully understood; intermediacy of radicals are implicated. Clemmensen reduction is complementary to , which also converts aldehydes and ketones to hydrocarbons, in that the former is carried out in strongly acidic conditions and the latter in strongly basic conditions. ) ), | 855 | 1,596 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.03%3A_First-Order_Reactions |
A is a reaction that proceeds at a rate that depends on only one reactant concentration. Differential rate laws are generally used to describe what is occurring on a molecular level during a reaction, whereas integrated rate laws are used for determining the reaction order and the value of the rate constant from experimental measurements. The differential equation describing first-order kinetics is given below: \[ Rate = - \dfrac{d[A]}{dt} = k[A]^1 = k[A] \label{1} \] is the reaction rate (in units of molar/time) and \(k\) is the reaction rate coefficient (in units of 1/time). However, the units of \(k\) vary for non-first-order reactions. These differential equations are , which simplifies the solutions as demonstrated below. First, write the differential form of the rate law. \[ Rate = - \dfrac{d[A]}{dt} = k[A] \nonumber \] Rearrange to give: \[ \dfrac{d[A]}{[A]} = - k\,dt \nonumber \] Second, integrate both sides of the equation. \[ \begin{align*} \int_{[A]_o}^{[A]} \dfrac{d[A]}{[A]} &= -\int_{t_o}^{t} k\, dt \label{4a} \\[4pt] \int_{[A]_{o}}^{[A]} \dfrac{1}{[A]} d[A] &= -\int_{t_o}^{t} k\, dt \label{4b} \end{align*} \] Recall from calculus that: \[ \int \dfrac{1}{x} = \ln(x) \nonumber \] Upon integration, \[ \ln[A] - \ln[A]_o = -kt \nonumber \] Rearrange to solve for one form of the rate law: \[ \ln[A] = \ln[A]_o - kt \nonumber \] This can be rearranged to: \[ \ln [A] = -kt + \ln [A]_o \nonumber \] This can further be arranged into y=mx +b form: \[ \ln [A] = -kt + \ln [A]_o \nonumber \] The equation is a straight line with slope m: \[mx=-kt \nonumber \] and y-intercept b: \[b=\ln [A]_o \nonumber \] Now, recall from that \[ \ln {\left(\dfrac{[A]_t}{ [A]_o}\right)}= -kt \nonumber \] where [A] is the concentration at time \(t\) and \([A]_o\) is the concentration at time 0, and \(k\) is the first-order rate constant. Because the logarithms of numbers do not have any units, the product \(-kt\) also lacks units. This concludes that unit of \(k\) in a first order of reaction must be time . Examples of time include s or min . Thus, the equation of a straight line is applicable: \[ \ln [A] = -kt + \ln [A]_o.\label{15} \] To test if it the reaction is a first-order reaction, plot the natural logarithm of a reactant concentration versus time and see whether the graph is linear. If the graph is linear and has a negative slope, the reaction must be a first-order reaction. To create another form of the rate law, raise each side of the previous equation to the exponent, \(e\): \[ \large e^{\ln[A]} = e^{\ln[A]_o - kt} \label{16} \] Simplifying gives the second form of the rate law: \[ [A] = [A]_{o}e^{- kt}\label{17} \] The integrated forms of the rate law can be used to find the population of reactant at any time after the start of the reaction. Plotting \(\ln[A]\) with respect to time for a first-order reaction gives a straight line with the slope of the line equal to \(-k\). More information can be found in the article on . This general relationship, in which a quantity changes at a rate that depends on its instantaneous value, is said to follow an exponential law. Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms. The reason that the exponential function \(y=e^x\) so efficiently describes such changes is that dy/dx = e ; that is, e is its own derivative, making the rate of change of \(y\) identical to its value at any point. The following graphs represents concentration of reactants versus time for a first-order reaction. Plotting \(\ln[A]\) with respect to time for a first-order reaction gives a straight line with the slope of the line equal to \(-k\). The half-life (\(t_{1/2}\)) is a timescale on which the initial population is decreased by half of its original value, represented by the following equation. After a period of one half-life, \(t = t_{1/2}\) and we can write \[\dfrac{[A]_{1/2}}{[A]_o} = \dfrac{1}{2}=e^{-k\,t_{1/2}} \label{18} \] Taking logarithms of both sides (remember that \(\ln e^x = x\)) yields \[ \ln 0.5 = -kt\label{19} \] Solving for the half-life, we obtain the simple relation \[ t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{k}\label{20} \] This indicates that the half-life of a first-order reaction is a constant. \[k = \dfrac{0.693}{600 \;s} = 0.00115 \;s^{-1} \nonumber \] As a check, can be used to confirm that this calculation generates the correct units of inverse time. Notice that, for first-order reactions, the half-life is of the initial concentration of reactant, which is a unique aspect to first-order reactions. The practical implication of this is that it takes as much time for [A] to decrease from 1 M to 0.5 M as it takes for [A] to decrease from 0.1 M to 0.05 M. In addition, the rate constant and the half life of a first-order process are inversely related. If 3.0 g of substance \(A\) decomposes for 36 minutes the mass of unreacted A remaining is found to be 0.375 g. What is the half life of this reaction if it follows first-order kinetics? There are two ways to approach this problem: The "simple inspection approach" and the "brute force approach" This approach is used when one can recognize that the final concentration of \(A\) is \(\frac{1}{8}\) of the initial concentration and hence three half lives \(\left(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\right)\) have elapsed during this reaction. \[t_{1/2} = \dfrac{36\, \text{min}}{3}= 12 \; \text{min} \nonumber \] This approach works only when the final concentration is \(\left(\frac{1}{2}\right)^n\) that of the initial concentration, then \(n\) is the number of half lives that have elapsed. If this is not the case, then approach #2 can be used. e \(t_{1/2}\) \[\begin{align*} \dfrac{[A]_t}{[A]_o} &= e^{-k\,t} \\[4pt] k &= -\dfrac{\ln \dfrac{[A]_t}{[A]_o}}{t} \\[4pt] &= -\dfrac{\ln \dfrac{0.375\, g}{3\, g}}{36\, \text{min}} \\[4pt] &= 0.0578 \, \text{min}^{-1} \end{align*} \] Therefore, via Equation \ref{20} \[t_{1/2}=\dfrac{\ln{2}}{k} \approx \dfrac{0.693}{0.0578 \, \text{min}^{-1}} \approx 12\, \text{min} \nonumber \] The first approach is considerably faster (if the number of half lives evolved is apparent). Calculate the half-life of the reactions below: Use the half life reaction that contains initial concentration and final concentration. Plug in the appropriate variables and solve to obtain: Determine the percent \(\ce{H2O2}\) that decomposes in the time using \(k=6.40 \times 10^{-5} s^{-1}\) ) | 6,787 | 1,597 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/12%3A_Chromatographic_and_Electrophoretic_Methods/12.03%3A_Optimizing_Chromatographic_Separations |
Now that we have defined the solute retention factor, selectivity, and column efficiency we are able to consider how they affect the resolution of two closely eluting peaks. Because the two peaks have similar retention times, it is reasonable to assume that their peak widths are nearly identical. If the number of theoretical plates is the same for all solutes—not strictly true, but not a bad assumption—then from equation , the ratio / is a constant. If two solutes have similar retention times, then their peak widths must be similar. Equation , therefore, becomes \[R_{A B}=\frac{t_{r, B}-t_{r, A}}{0.5\left(w_{B}+w_{A}\right)} \approx \frac{t_{r, B}-t_{r, A}}{0.5\left(2 w_{B}\right)}=\frac{t_{r, B}-t_{r, A}}{w_{B}} \label{12.1}\] where is the later eluting of the two solutes. Solving equation 12.2.15 for and substituting into Equation \ref{12.1} leaves us with the following result. \[R_{A B}=\frac{\sqrt{N_{B}}}{4} \times \frac{t_{r, B}-t_{r, A}}{t_{r, B}} \label{12.2}\] Rearranging equation provides us with the following equations for the retention times of solutes and . \[t_{r, A}=k_{A} t_{\mathrm{m}}+t_{\mathrm{m}} \quad \text { and } \quad t_{\mathrm{r}, B}=k_{B} t_{\mathrm{m}}+t_{\mathrm{m}} \nonumber\] After substituting these equations into Equation \ref{12.2} and simplifying, we have \[R_{A B}=\frac{\sqrt{N_{B}}}{4} \times \frac{k_{B}-k_{A}}{1+k_{B}} \nonumber\] Finally, we can eliminate solute ’s retention factor by substituting in equation . After rearranging, we end up with the following equation for the resolution between the chromatographic peaks for solutes A and B. \[R_{A B}=\frac{\sqrt{N_{B}}}{4} \times \frac{\alpha-1}{\alpha} \times \frac{k_{B}}{1+k_{B}} \label{12.3}\] In addition to resolution, another important factor in chromatography is the amount of time needed to elute a pair of solutes, which we can approximate using the retention time for solute . \[t_{r, s}=\frac{16 R_{AB}^{2} H}{u} \times\left(\frac{\alpha}{\alpha-1}\right)^{2} \times \frac{\left(1+k_{B}\right)^{3}}{k_{B}^{2}} \label{12.4}\] where is the mobile phase’s velocity. Although Equation \ref{12.3} is useful for considering how a change in , \(\alpha\), or qualitatively affects resolution—which suits our purpose here—it is less useful for making accurate quantitative predictions of resolution, particularly for smaller values of and for larger values of . For more accurate predictions use the equation \[R_{A B}=\frac{\sqrt{N}}{4} \times(\alpha-1) \times \frac{k_{B}}{1+k_{\mathrm{avg}}} \nonumber\] where is ( + /2. For a derivation of this equation and for a deeper discussion of resolution in column chromatography, see Foley, J. P. “Resolution Equations for Column Chromatography,” , , , 1275-1279. Equation \ref{12.3} and Equation \ref{12.4} contain terms that correspond to column efficiency, selectivity, and the solute retention factor. We can vary these terms, more or less independently, to improve resolution and analysis time. The first term, which is a function of the number of theoretical plates (for Equation \ref{12.3}) or the height of a theoretical plate (for Equation \ref{12.4}), accounts for the effect of column efficiency. The second term is a function of \(\alpha\) and accounts for the influence of column selectivity. Finally, the third term in both equations is a function of and accounts for the effect of solute ’s retention factor. A discussion of how we can use these parameters to improve resolution is the subject of the remainder of this section. One of the simplest ways to improve resolution is to adjust the retention factor for solute . If all other terms in Equation \ref{12.3} remain constant, an increase in will improve resolution. As shown by the curve in Figure 12.3.1
, however, the improvement is greatest if the initial value of is small. Once exceeds a value of approximately 10, a further increase produces only a marginal improvement in resolution. For example, if the original value of is 1, increasing its value to 10 gives an 82% improvement in resolution; a further increase to 15 provides a net improvement in resolution of only 87.5%. Any improvement in resolution from increasing the value of generally comes at the cost of a longer analysis time. The curve in Figure 12.3.1
shows the relative change in the retention time for solute as a function of its retention factor. Note that the minimum retention time is for = 2. Increasing from 2 to 10, for example, approximately doubles solute B’s retention time. The relationship between retention factor and analysis time in Figure 12.3.1
works to our advantage if a separation produces an acceptable resolution with a large . In this case we may be able to decrease with little loss in resolution and with a significantly shorter analysis time. To increase without changing selectivity, \(\alpha\), any change to the chromatographic conditions must result in a general, nonselective increase in the retention factor for both solutes. In gas chromatography, we can accomplish this by decreasing the column’s temperature. Because a solute’s vapor pressure is smaller at lower temperatures, it spends more time in the stationary phase and takes longer to elute. In liquid chromatography, the easiest way to increase a solute’s retention factor is to use a mobile phase that is a weaker solvent. When the mobile phase has a lower solvent strength, solutes spend proportionally more time in the stationary phase and take longer to elute. Adjusting the retention factor to improve the resolution between one pair of solutes may lead to unacceptably long retention times for other solutes. For example, suppose we need to analyze a four-component mixture with baseline resolution and with a run-time of less than 20 min. Our initial choice of conditions gives the chromatogram in Figure 12.3.2
a. Although we successfully separate components 3 and 4 within 15 min, we fail to separate components 1 and 2. Adjusting conditions to improve the resolution for the first two components by increasing provides a good separation of all four components, but the run-time is too long (Figure 12.3.2
b). This problem of finding a single set of acceptable operating conditions is known as the . One solution to the general elution problem is to make incremental adjustments to the retention factor as the separation takes place. At the beginning of the separation we set the initial chromatographic conditions to optimize the resolution for early eluting solutes. As the separation progresses, we adjust the chromatographic conditions to decrease the retention factor—and, therefore, to decrease the retention time—for each of the later eluting solutes (Figure 12.3.2
c). In gas chromatography this is accomplished by temperature programming. The column’s initial temperature is selected such that the first solutes to elute are resolved fully. The temperature is then increased, either continuously or in steps, to bring off later eluting components with both an acceptable resolution and a reasonable analysis time. In liquid chromatography the same effect is obtained by increasing the solvent’s eluting strength. This is known as a gradient elution. We will have more to say about each of these in later sections of this chapter. A second approach to improving resolution is to adjust the selectivity, \(\alpha\). In fact, for \(\alpha \approx 1\) usually it is not possible to improve resolution by adjusting the solute retention factor, , or the column efficiency, . A change in \(\alpha\) often has a more dramatic effect on resolution than a change in . For example, changing \(\alpha\) from 1.1 to 1.5, while holding constant all other terms, improves resolution by 267%. In gas chromatography, we adjust \(\alpha\) by changing the stationary phase; in liquid chromatography, we change the composition of the mobile phase to adjust \(\alpha\). To change \(\alpha\) we need to selectively adjust individual solute retention factors. Figure 12.3.3
shows one possible approach for the liquid chromatographic separation of a mixture of substituted benzoic acids. Because the retention time of a compound’s weak acid form and its weak base form are different, its retention time will vary with the pH of the mobile phase, as shown in Figure 12.3.3
a. The intersections of the curves in Figure 12.3.3
a show pH values where two solutes co-elute. For example, at a pH of 3.8 terephthalic acid and -hydroxybenzoic acid elute as a single chromatographic peak. Figure 12.3.3
a shows that there are many pH values where some separation is possible. To find the optimum separation, we plot a for each pair of solutes. The , , and curves in Figure 12.3.3
b show the variation in a with pH for the three pairs of solutes that are hardest to separate (for all other pairs of solutes, \(\alpha\) > 2 at all pH levels). The shading shows windows of pH values in which at least a partial separation is possible—this figure is sometimes called a window diagram—and the highest point in each window gives the optimum pH within that range. The best overall separation is the highest point in any window, which, for this example, is a pH of 3.5. Because the analysis time at this pH is more than 40 min (Figure 12.3.3
a), choosing a pH between 4.1–4.4 might produce an acceptable separation with a much shorter analysis time. Let’s use benzoic acid, C H COOH, to explain why pH can affect a solute’s retention time. The separation uses an aqueous mobile phase and a nonpolar stationary phase. At lower pHs, benzoic acid predominately is in its weak acid form, C H COOH, and partitions easily into the nonpolar stationary phase. At more basic pHs, however, benzoic acid is in its weak base form, C H COO . Because it now carries a charge, its solubility in the mobile phase increases and its solubility in the nonpolar stationary phase decreases. As a result, it spends more time in the mobile phase and has a shorter retention time. Although the usual way to adjust pH is to change the concentration of buffering agents, it also is possible to adjust pH by changing the column’s temperature because a solute’s p value is pH-dependent; for a review, see Gagliardi, L. G.; Tascon, M.; Castells, C. B. “Effect of Temperature on Acid–Base Equilibria in Separation Techniques: A Review,” , , , 35–57. A third approach to improving resolution is to adjust the column’s efficiency by increasing the number of theoretical plates, . If we have values for and \(\alpha\), then we can use Equation \ref{12.3} to calculate the number of theoretical plates for any resolution. Table 12.3.1
provides some representative values. For example, if \(\alpha\) = 1.05 and = 2.0, a resolution of 1.25 requires approximately 24 800 theoretical plates. If our column provides only 12 400 plates, half of what is needed, then a separation is not possible. How can we double the number of theoretical plates? The easiest way is to double the length of the column, although this also doubles the analysis time. A better approach is to cut the height of a theoretical plate, , in half, providing the desired resolution without changing the analysis time. Even better, if we can decrease by more than 50%, it may be possible to achieve the desired resolution with an even shorter analysis time by also decreasing or \(\alpha\). 0.5 1.5 3.0 To decrease the height of a theoretical plate we need to understand the experimental factors that affect band broadening. There are several theoretical treatments of band broadening. We will consider one approach that considers four contributions: variations in path lengths, longitudinal diffusion, mass transfer in the stationary phase, and mass transfer in the mobile phase. As solute molecules pass through the column they travel paths that differ in length. Because of this difference in path length, two solute molecules that enter the column at the same time will exit the column at different times. The result, as shown in Figure 12.3.4
, is a broadening of the solute’s profile on the column. The contribution of to the height of a theoretical plate, , is \[H_{p}=2 \lambda d_{p} \label{12.5}\] where is the average diameter of the particulate packing material and \(\lambda\) is a constant that accounts for the consistency of the packing. A smaller range of particle sizes and a more consistent packing produce a smaller value for \(\lambda\). For a column without packing material, is zero and there is no contribution to band broadening from multiple paths. An inconsistent packing creates channels that allow some solute molecules to travel quickly through the column. It also can creates pockets that temporarily trap some solute molecules, slowing their progress through the column. A more uniform packing minimizes these problems. The second contribution to band broadening is the result of the solute’s in the mobile phase. Solute molecules are in constant motion, diffusing from regions of higher solute concentration to regions where the concentration of solute is smaller. The result is an increase in the solute’s band width (Figure 12.3.5
). The contribution of longitudinal diffusion to the height of a theoretical plate, , is \[H_{d}=\frac{2 \gamma D_{m}}{u} \label{12.6}\] where is the solute’s diffusion coefficient in the mobile phase, is the mobile phase’s velocity, and \(\gamma\) is a constant related to the efficiency of column packing. Note that the effect of on band broadening is inversely proportional to the mobile phase velocity: a higher velocity provides less time for longitudinal diffusion. Because a solute’s diffusion coefficient is larger in the gas phase than in a liquid phase, longitudinal diffusion is a more serious problem in gas chromatography. As the solute passes through the column it moves between the mobile phase and the stationary phase. We call this movement between phases . As shown in Figure 12.3.6
, band broadening occurs if the solute’s movement within the mobile phase or within the stationary phase is not fast enough to maintain an equilibrium in its concentration between the two phases. On average, a solute molecule in the mobile phase moves down the column before it passes into the stationary phase. A solute molecule in the stationary phase, on the other hand, takes longer than expected to move back into the mobile phase. The contributions of mass transfer in the stationary phase, , and mass transfer in the mobile phase, , are given by the following equations \[H_{s}=\frac{q k d_{f}^{2}}{(1+k)^{2} D_{s}} u \label{12.7}\] \[H_{m}=\frac{f n\left(d_{p}^{2}, d_{c}^{2}\right)}{D_{m}} u \label{12.8}\] where is the thickness of the stationary phase, is the diameter of the column, and are the diffusion coefficients for the solute in the stationary phase and the mobile phase, is the solute’s retention factor, and is a constant related to the column packing material. Although the exact form of is not known, it is a function of particle size and column diameter. Note that the effect of and on band broadening is directly proportional to the mobile phase velocity because a smaller velocity provides more time for mass transfer. The abbreviation in Equation \ref{12.7} means “is a function of.” The height of a theoretical plate is a summation of the contributions from each of the terms affecting band broadening. \[H=H_{p}+H_{d}+H_{s}+H_{m} \label{12.9}\] An alternative form of this equation is the \[H=A+\frac{B}{u}+C u \label{12.10}\] which emphasizes the importance of the mobile phase’s velocity. In the van Deemter equation, accounts for the contribution of multiple paths ( ), / accounts for the contribution of longitudinal diffusion ( ), and accounts for the combined contribution of mass transfer in the stationary phase and in the mobile phase ( and ). There is some disagreement on the best equation for describing the relationship between plate height and mobile phase velocity [Hawkes, S. J. , , 393–398]. In addition to the van Deemter equation, other equations include \[H=\frac{B}{u}+\left(C_s+C_{m}\right) u \nonumber\] where and are the mass transfer terms for the stationary phase and the mobile phase and \[H=A u^{1 / 3}+\frac{B}{u}+C u \nonumber\] All three equations, and others, have been used to characterize chromatographic systems, with no single equation providing the best explanation in every case [Kennedy, R. T.; Jorgenson, J. W. , , 1128–1135]. To increase the number of theoretical plates without increasing the length of the column, we need to decrease one or more of the terms in Equation \ref{12.9}. The easiest way to decrease is to adjust the velocity of the mobile phase. For smaller mobile phase velocities, column efficiency is limited by longitudinal diffusion, and for higher mobile phase velocities efficiency is limited by the two mass transfer terms. As shown in Figure 12.3.7
—which uses the van Deemter equation—the optimum mobile phase velocity is the minimum in a plot of as a function of . The remaining parameters that affect the terms in Equation \ref{12.9} are functions of the column’s properties and suggest other possible approaches to improving column efficiency. For example, both and are a function of the size of the particles used to pack the column. Decreasing particle size, therefore, is another useful method for improving efficiency. For a more detailed discussion of ways to assess the quality of a column, see Desmet, G.; Caooter, D.; Broeckhaven, K. “Graphical Data Represenation Methods to Assess the Quality of LC Columns,” , , 8593–8602. Perhaps the most important advancement in chromatography columns is the development of open-tubular, or . These columns have very small diameters ( ≈ 50–500 μm) and contain no packing material ( = 0). Instead, the capillary column’s interior wall is coated with a thin film of the stationary phase. Plate height is reduced because the contribution to from (Equation \ref{12.5}) disappears and the contribution from (Equation \ref{12.8}) becomes smaller. Because the column does not contain any solid packing material, it takes less pressure to move the mobile phase through the column, which allows for longer columns. The combination of a longer column and a smaller height for a theoretical plate increases the number of theoretical plates by approximately \(100 \times\). Capillary columns are not without disadvantages. Because they are much narrower than packed columns, they require a significantly smaller amount of sample, which may be difficult to inject reproducibly. Another approach to improving resolution is to use thin films of stationary phase, which decreases the contribution to from (Equation \ref{12.7}). The smaller the particles, the more pressure is needed to push the mobile phase through the column. As a result, for any form of chromatography there is a practical limit to particle size. | 18,942 | 1,598 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Buffers/Henderson-Hasselbalch_Approximation |
The Henderson-Hasselbalch approximation allows us one method to approximate the pH of a buffer solution. The basic equation is as follows: \[ pH \approx pK_a + \log_{10} \dfrac{[A^-]}{[HA]} \nonumber \] We have straightforward calculations for strong acids and bases, but the computations behind buffers are rather complex and time consuming. By using the fact that weak acids and bases barely ionize, allowing us to approximate the pH of buffer solutions using initial concentrations. Though the approximation has a few restrictions, it simplifies a lengthy calculation into a simple equation derived from \(K\). (1878-1942) was a talented biochemist, among many other titles, who spent most of his career at Harvard. He was responsible for developing the components of the equation after studying equilibrium reactions that took place within blood as a result of respiration (specializing in "fatigue"). His equation was incomplete without the solid calculations into it provided by Hasselbalch. (1874-1962) was a chemist who studied pH closely. He also studied blood and reactions that took place with oxygen, to put in the simplest of terms. He eventually modified Henderson's equation by putting mathematical into it creating a solid relationship. For a weak acid \( HA \) and its conjugate base \(A^-\): \(HA + H_2O\leftrightharpoons H^+ + A^-\) which has an \(K_a\). The Henderson-Hasselbalch approximation is derived from this acid ionization constant. \[ \begin{align} K_a &= \dfrac{[H^+,A^-]}{[HA]} \\[4pt] -\log_{10} K_a &= -\log_{10} \dfrac{[H^+,A^-]}{[HA]} \\[4pt] - \log_{10} K_a &= - \log_{10} [H^+] - \log_{10} \dfrac{[A^-]}{[HA]} \\[4pt] pK_a &= pH - \log_{10} \dfrac{[A^-]}{[HA]} \\[4pt] pH &= pK_a + \log_{10} \dfrac{[A^-]}{[HA]} \label{HH1} \end{align} \] Equation \ref{HH1} is formulated in terms of equilibrium concentrations in solution. Since \( HA\) is a weak acid and weakly dissociates and we can introduce two approximations \[ [HA]\approx[HA]_i \nonumber \] and \[ [A^-]\approx[A^-]_i \nonumber \] Hence, we can use the initial concentrations rather than equilibrium concentrations because \[pK_a + \log_{10} \dfrac{[A^-]_i}{[HA]_i} \approx pK_a + \log_{10} \dfrac{[A^-]}{[HA]} = pH \nonumber \] Find \([H^+]\) in a solution of 1.0 M \(HNO_2\) and 0.225 M \(NaNO_2\). The \(K_a\) for \(HNO_2\) is \(5.6 \times 10^{−4}\) (Table E1). \[pK_a = -\log_{10} K_a = -\log_{10} (7.4 \times 10^{-4}) = 3.14 \nonumber \] \[pH = pK_a + \log_{10} \left( \dfrac{[NO_2^-]}{[HNO_2]}\right)\nonumber \] \[pH = 3.14 + \log_{10} \left( \dfrac{1}{0.225} \right)\nonumber \] \[pH = 3.14 + 0.648 = 3.788\nonumber \] \[[H^+] = 10^{-pH} = 10^{-3.788} = 1.6 \times 10^{-4} \nonumber \] What ratio \( \frac{[A^-]}{[HA]} \) will create an acetic acid buffer of pH 5.0? (\(K_a\) acetic acid is \(1.75 \times 10^{−5}\) (Table E1). \[pKa = -log Ka = -log(1.8 \times10^{-5}) = 4.74\nonumber \] \[pH = pK_a + \log_{10} \left( \dfrac{[A^-]}{[HA]}\right) \nonumber \] \[5.0 = 4.74 + \log_{10} \left( \dfrac{[A^-]}{[HA]}\right)\nonumber \] \[0.26 = \log_{10} \left( \dfrac{[A^-]}{[HA]}\right)\nonumber \] \[10^{0.26} = \dfrac{[A^-]}{[HA]}\nonumber \] \[1.8 = \dfrac{[A^-]}{[HA]}\nonumber \] Similarly, for a weak base \( B\) and its conjugate acid \(HB^+\): \[B + H_2O \leftrightharpoons OH^- + HB^+ \nonumber \] which has an base ionization constant \(K_b\). The Henderson-Hasselbalch approximation for basic buffers is derived from this base ionization constant. \[ \begin{align} K_b &= \dfrac{[OH^-,HB^+]}{[B]} \\[4pt] -\log_{10} K_b &= -\log_{10} \dfrac{[OH^-,HB^+]}{[B]} \\[4pt] -\log_{10} K_b &= - \log_{10} [OH^-] - \log_{10} \dfrac{[HB^+]}{[B]} \\[4pt] pK_b &= pOH - \log_{10} \dfrac{[HB^+]}{[B]} \\[4pt] pOH &= pK_b + \log_{10} \dfrac{[HB^+]}{[B]} \end{align} \nonumber \] Note that \( B\) is a weak base and does not dissociate completely and we can say \([B] \approx [B]_i\) and \( [HB^+] \approx [HB^+]_i\). Hence, we can use the initial concentrations because \[pK_b + \log_{10} \dfrac{[HB^+]_i}{[B]_i} \approx pK_b + \log_{10} \dfrac{[HB^+]}{[B]} = pOH \nonumber \] You prepare a buffer solution of 0.323 M \(NH_3\) and \((NH_4)_2SO_4\). What molarity of \((NH_4)_2SO_4\) is necessary to have a pH of 8.6? The \(pK_b\) for \(NH_3\) is 4.74 (Table E2). \[pK_a+ pK_b = 14\nonumber \] \[pK_a= 14 - 4.74 = 9.26\nonumber \] \[pH = pK_a + \log_{10} \left( \frac{[A^-]}{[HA]}\right)\nonumber \] \[8.6 = 9.26 + \log_{10} \left( \dfrac{0.323}{[NH_4^+]}\right)\nonumber \] \[-0.66 = \log_{10} \left( \dfrac{0.323}{[NH_4^+]}\right)\nonumber \] \[[NH_4^+] = 1.48 M\nonumber \] What is the pH of a buffer 0.500 moles acetic acid and 0.500 moles acetate ion and the total volume is 5 L when you add 0.350 moles HCl? The \(K_a\) for acetic acid is \(1.75 \times 10^{−5}\) (Table E1). \[pK_a = -\log_{10} K_a = -\log_{10} (1.75 \times 10^{−5}) = 4.756\nonumber \] \[pH = pK_a + \log_{10} \left( \dfrac{[acetate]}{[\text{acetic acid}] + [HCl]} \right)\nonumber \] note that mole ratio also works in place of concentrations as both the acid and base are in the same solution \[pH = 4.756 + \log_{10} (0.588)\nonumber \] \[pH = 4.756 - 0.230 = 4.52\nonumber \] What is the range of an acetic acid buffer described in problem \(\Page {4}\) without the added \(\ce{HCl}\)? The effective buffer range is of magnitude 2 pH units with the \(pK_a\) as the midpoint. \[pK_a = -\log_{10} K_a = -\log_{10} (1.75 \times 10^{−5}) = 4.756\nonumber \] Hence, the buffer range would be 3.75 to 5.75. | 5,498 | 1,599 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/22%3A_Biophysical_Reaction_Dynamics/22.05%3A_Time-Correlation_Functions |
Time‐Correlation Functions Time-correlation functions are commonly used to characterize the dynamics of a random (or stochastic) process. If we observe the behavior of an internal variable A describing the behavior of one molecule at thermal equilibrium, it may be subject to microscopic fluctuations. Although there may seem to be little information in this noisy trajectory, this dynamics is not entirely random, since they are a consequence of time-dependent interactions with the environment. We can provide a statistical description of the characteristic time scales and amplitudes to these changes by comparing the value of A at time t with the value of A at a later time t’. We define a time-correlation function as the product of these values averaged over an equilibrium ensemble: \[C_{AA}(t-t') \equiv \langle A(t)A(t')\rangle \] Correlation functions do not depend on the absolute point of observation (t and t’), but rather the time interval between observations (for stationary random processes). So, we can define the time interval \( \tau \equiv t-t'\), and express our function as \(C_{AA}(\tau)\). We can see that when we evaluate C at τ=0, we obtain the mean square value of \( A , \langle A^2 \rangle\). At long times, as thermal fluctuations act to randomize the system, the values of become uncorrelated: \(\lim_{\tau\to\infty} C_{AA}(\tau)=\langle A\rangle ^2\). It is therefore common to redefine the correlation function in terms of the deviation from average \[ \delta A=A-\langle A\rangle \] \[ C_{\delta A\delta A}(t)=\langle \delta A(t)\delta A(0) \rangle = C_{AA}(t)-\langle A\rangle ^2 \] Then \(C_{\delta A \delta A}(0)\) gives the variance for the random process, and the correlation function decays to zero as τ → ∞. The characteristic time scale for this relaxation is the correlation time, \(\tau_c\). which we can obtain from \[ \tau_c = \dfrac{1}{\langle \delta A^2 \rangle } \int_0^{\infty}dt \langle \delta A(t) \delta A(0)\rangle \] The classical correlation function can be obtained from an equilibrium probability distribution as \[ C_{AA}(t-t')=\int \mathrm{d}p \int \mathrm{d}q A(p,q;t) A(p,q;t') P_{eq}(p,q) \] In practice, correlation function are more commonly obtained from trajectories by calculating it as a time average \[C_{A A}(\tau)=\overline{A(\tau) A(0)}=\lim _{T \rightarrow \infty} \frac{1}{T} \int_{0}^{T} d t^{\prime} A_{i}\left(\tau+t^{\prime}\right) A_{i}\left(t^{\prime}\right)\] If the time-average value of C is to be equal to the equilibrium ensemble average value of C, we say thesystem is ergodic. Example: Velocity Autocorrelation Function for Gas A dilute gas of molecules has a Maxwell–Boltzmann distribution of velocities, for which we will focus on the velocity component along the \(\hat{x}\) direction, x . We know that the average velocity is \(\langle v_x \rangle=0\). The velocity correlation function is \[c_{v_xv_x}(\tau)=\langle v_x(\tau)v_x(0)\rangle \nonumber\] The average translational energy is \(\frac{1}{2}m\langle v_x^2 \rangle = k_BT/2 \), so \[C_{v_xv_x}(0)=\langle v_x^2(0) \rangle = \dfrac{k_BT}{m} \nonumber \] For time scales that are short compared to the average collision time between molecules, the velocity of any given molecule remains constant and unchanged, so the correlation function for the velocity is also unchanged at k T/m. This non-interacting regime corresponds to the behavior of an ideal gas. For any real gas, there will be collisions that randomize the direction and speed of the molecules, so that any molecule over a long enough time will sample the various velocities within the Maxwell–Boltzmann distribution. From the trajectory of x-velocities for a given molecule we can calculate \(C_{v_xv_x}(\tau)\) using time averaging. The correlation function will drop on with a correlation time τc, which is related to mean time between collisions. After enough collisions, the correlation with the initial velocity is lost and \(C_{v_xv_x}(\tau)\) approaches \(\langle v_x^2 \rangle = 0 \). Finally, we can determine the diffusion constant for the gas, which relates the time and mean square displacement of the molecules: \(\langle x^2(t)\rangle = 2D_xt\). From \(D_x= \int_0^{\infty}dt\langle v_x(t)v_x(0)\rangle \) we have \(D_x = k_BT\tau_c/m\). In viscous fluids \(\tau_c/m\) is called the mobility. Calculating a Correlation Function from a Trajectory We can evaluate eq. (22.5.6) for a discrete and finite trajectory in which we are given a series of observations of the dynamical variable at equally separated time points . The separation between time points is = δt, and the length of the trajectory is δt. Then we have \[C_{AA} = \dfrac{1}{T} \sum_{i,j=1}^{N} \delta t A(t_i) A(t_j) = \dfrac{1}{N}\sum_{i,j=1}^NA_iA_j \] where \(A_i=A(t_i)\). To make this more useful we want to express it as the time interval between points \(\tau = t_j-t_i = (j-i)\delta t\), and average over all possible pairwise products of separated by τ. Defining a new count integer , we can express the delay as \(\tau = n\delta t\). For a finite data set there are a different number of observations to average over at each time interval ( ). We have the most pairwise products— to be precise—when the time points are equal ( ). We only have one data pair for the maximum delay τ = . Therefore, the number of pairwise products for a given delay τ is . So we can write eq. (22.5.7) as \[C_{AA} (\tau) = C(n) = \dfrac{1}{N-n}\sum_{i-1}^{N-n}A_{i+n}A_i \] Note that this expression will only be calculated for positive values of , for which . As an example consider the following calculation for fluctuations in fluorescence intensity in an FCS experiment. This trajectory consists of 32000 consecutive measurements separated by 44 μs, and is plotted as a deviation from the mean δ ‒ The correlation function obtained from eq. (22.5.8) is We can see that the decay of the correlation function is observed for sub-ms time delays. From eq. (22.5.4) we find that the correlation time is τ = 890 μs. | 6,034 | 1,602 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.04%3A_Ea_and_Catalysts |
For reactions that follow the Arrhenius rate law a can be re-defined as a substance that lowers the energy of activation a by providing a pathway (reaction mechanism), or transition state. For example, it is well known that iodide ions catalyze the decomposition of hydrogen peroxide \(\ce{H2O2}\), \[\ce{2 H2O2 \rightarrow 2 H2O + O2} \nonumber\] Thus, by dissolving solid \(\ce{KI}\) in a solution of hydrogen peroxide, the formation of oxygen bubbles is accelerated. Of course, the reaction depends on concentrations of reactants and catalyst, but for a definite (or fixed) concentration, the relative can be compared as exemplified by the following examples. At 300 K, the activation energy, a, for the decomposition of \(\ce{H2O2}\) has been measured to be 75.3 kJ/mol. In the presence of a definite concentration of iodide ions, \(\ce{I-}\), the activation energy a has been estimated to be 56.5 kJ/mol. How much faster is the decomposition when the same concentration of iodide is present in the reaction? If and represent the of decomposition in the absence of iodide and in the presence of iodide ion (at a definite concentration) respectively, then \(rate = k\ce{f([H2O2])}\)
\(rate\,' = k\,'\ce{f([H2O2])}\) where \(\ce{f([H2O2])}\) is a function of the concentration of the reactant. Note that and are rates of the reactions in the absence of and in the presence of iodide ions. \(\begin{align*}
k &= \ce{A e}^{-75300/(8.3145\times300)}\\ \\
k\,' &= \ce{A e}^{-56500/(8.3145\times300)}\\ \\
\dfrac{k}{k\,'} &= \dfrac{\ce e^{-75300/(8.3145\times300)}}{\ce e^{-56500/(8.3145\times300)}}\\ \\
&= \ce e^{-(75300 - 56500)/(8.3145\times300)}\\ \\
&= \ce e^{-18800/(8.3145\times300)}\\ \\
&= 1877
\end{align*}\) Thus, = 1877 times , because the has increased 1877 times. Note that the presence of a catalyst allows the reaction to proceed at the same low temperature, but achieve a much faster rate of reaction. The activation energy of a reaction is 19.0 kJ/mol. When a catalyst is used at a particular concentration, the rate increases ten fold. Calculate the energy of activation when the catalyst is present. This example differs from example 1 in that we know how much faster the reaction is and want to evaluate the activation energy. Let the activation energy in the presence of the catalyst be a, then \(\begin{align*}
\dfrac{\ce e^{-E\ce a/(8.3145\times300)}}{\ce e^{-19000/(8.3145\times300)}} &= \dfrac{k_{\ce{catalyst}}}{k}\\ \\
&= 10\\ \\
\ce e^{-E\ce a/(8.3145\times300)} &= 10 \times \ce e^{-19000/(8.3145\times300)}\\ \\
&= 0.00492\\ \\
\dfrac{-E\ce a}{(8.3145\times300)} &= \ln0.00492\\ \\
-E\ce a &= (8.3145\times300)\times(-5.315)\\ \\
E\ce a &= \mathrm{13257\: J/mol}\\ \\
&= \mathrm{13.3\: kJ/mol}
\end{align*}\) The details of the calculation are given to illustrate the mathematic skills involved. Check out the units throughout the calculation please.
Calculate the time required to accomplish a certain task when the rate is different.
Calculate energy of activation when the rate of increase is known.
Calculate the time required to collect when no catalyst is used. | 3,169 | 1,603 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Solubilty/An_Introduction_to_Solubility_Products |
This page discusses how solubility products are defined, including their units units. It also explores the relationship between the solubility product of an ionic compound and its solubility. Barium sulfate is almost insoluble in water. It is not totally insoluble—very small amounts do dissolve. This is true of any "insoluble" ionic compound. If solid barium sulfate is shaken with water, a small amount of barium ions and sulfate ions break away from the surface of the solid and go into solution. Over time, some of these return from solution to stick onto the solid again. An equilibrium is established when the rate at which some i ons are breaking away from the solid lattice is matched by the rate at which others are returning. Consider the balanced equation for the barium sulfate reaction: \[ BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)\nonumber \] The position of this equilibrium lies very far to the left. The great majority of the barium sulfate is present as solid—there are no visible changes to the solid. However, the equilibrium does exist, and an equilibrium constant can be written. The equilibrium constant is called the solubility product, and is given the symbol K : \[ K_{sp} = [Ba^{2+} (aq)] [SO_4^{2-}(aq)]\nonumber \] For simplicity, solubility product expressions are often written without the state symbols: \[ K_{sp} = [Ba^{2+}] [SO_4^{2-}]\nonumber \] Notice that there is no solid barium sulfate term. For many simple equilibria, the equilibrium constant expression has terms for the right side of the equation divided by terms for the left side. But in this case, there is no term for the concentration of the solid barium sulfate. This is a heterogeneous equilibrium, one which contains substances in more than one state. In a heterogeneous equilibrium, concentration terms for solids are left out of the expression. Here is the corresponding equilibrium for calcium phosphate, Ca (PO ) : \[ Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)\nonumber \] Below is the solubility product expression: \[ K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2\nonumber \] As with any other equilibrium constant, the concentrations are raised to the power of their respective stoichiometric coefficients in the equilibrium equation. Solubility products only apply to sparingly soluble ionic compounds, not to normally soluble compounds such as sodium chloride. Interactions between the ions in the solution interfere with the simple equilibrium. The units for solubility products differ depending on the solubility product expression, and you need to be able to work them out each time. Below is the solubility product expression for barium sulfate (\ce{BaSO4}\)): \[ K_{sp} = [Ba^{2+}] [SO_4^{2-}]\nonumber \] Each concentration has the unit mol dm . In this case, the units for the solubility product in this case are the following: Recall the solubility product expression for calcium phosphate (\(\ce{CaPO4}\)) \[ K_{sp} = [Ca^{2+}]^3[PO_4^{3-}]^2\nonumber \] The units this time will be: Recall the barium sulfate equation: \[ BaSO_4(s) \rightleftharpoons Ba^{2+}(aq) + SO_4^{2-}(aq)\nonumber \] The corresponding solubility product expression is the following: \[ K_{sp}= [Ba^{2+},SO_4^{2-}]\nonumber \] K for barium sulfate at 298 K is 1.1 x 10 mol dm . In order for this equilibrium constant (the solubility product) to apply, solid barium sulfate must be present in a saturated solution of barium sulfate. This is indicated by the equilibrium equation. If barium ions and sulfate ions exist in solution in the presence of some solid barium sulfate at 298 K, and multiply the concentrations of the ions together, the answer will be 1.1 x 10 mol dm . It is possible to multiply ionic concentrations and obtain a value less than this solubility product; in such cases the solution is too dilute, and no equilibrium exists. If the concentrations are lowered enough, no precipitate can form. However, it is impossible to calculate a product greater than the solubility product. If solutions containing barium ions and sulfate ions are mixed such that the product of the concentrations would exceed K , a precipitate forms. Enough solid is produced to reduce the concentrations of the barium and sulfate ions down to the value of the solubility product. | 4,316 | 1,604 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Ionization_Constants/Weak_Acids_and_Bases_1 |
Weak acids and bases are only partially ionized in their solutions, whereas strong acids and bases are completely ionized when dissolved in water. Some common weak acids and bases are given here. Furthermore, weak acids and bases are very common, and we encounter them often both in the academic problems and in everyday life. The ionization of weak acids and bases is a chemical equilibrium phenomenon. The equilibrium principles are essential for the understanding of equilibria of weak acids and weak bases. In this connection, you probably realize that and . Acetic acid, \(\ce{CH3COOH}\), is a typical weak acid, and it is the ingredient of vinegar. It is partially ionized in its solution. \[\ce{CH3COOH \rightleftharpoons CH3COO- + H+} \nonumber \] The structure of the acetate ion, \(\ce{CH3COO-}\), is shown below. In a solution of acetic acid, the equilibrium concentrations are found to be \(\mathrm{[CH_3COOH] = 1.000}\); \(\mathrm{[CH_3COO^-] = 0.0042}\). Evaluate the pH of this solution and the equilibrium constant of ionization of acetic acid.
From the ionization of acetic acid, \(\begin{alignat}{3}
\ce{&CH3COOH \:\rightleftharpoons\: &&CH3COO- + &&\:\:\:\:H+}\\
&\:\:\:\:\:\:\:0.100 &&\:\:\:\:0.0042 &&0.0042
\end{alignat}\) we conclude that \(\begin{align}
\ce{[H+]} &= \ce{[CH3COO- ]}\\
&= 0.0042
\end{align}\) Thus, \(\mathrm{pH = -\log0.0042 = 2.376}\). The equilibrium constant of ionzation, \[K = \dfrac{(0.0042)^2}{1.000} = 1.78\times10^{-5} \nonumber \] The equilibrium constant of an acid is represented by ; and similar to the pH scale, a p scale is defined by \[\mathrm{p\mathit K_a = - \log \mathit K_a} \nonumber \] and for acetic acid, p = 4.75. Note that = 10 There are many weak acids, which do not completely dissociate in aqueous solution. As a general discussion of weak acids, let \(\ce{HA}\) represent a typical weak acid. Then its ionization can be written as: \[\ce{HA \rightleftharpoons H+ + A-} \nonumber \] In a solution whose label concentration is \(C = \ce{[HA]} + \ce{[A- ]}\), l et us assume that is the concentration that has undergone ionization. Thus, at equilibrium, the concentrations are \[\ce{[HA]} = C - x \nonumber \] \[\ce{[H+]} = \ce{[A- ]} = x \nonumber \] Make sure you understand why they are so, because you will have to set up these relationships in your problem solving. In summary, we formulate them as \(\begin{array}{cccccl}
\ce{HA &\rightleftharpoons &H+ &+ &A- &}\\
C & & & & &\leftarrow \textrm{ initial concentration, assume }\textit{x}\textrm{ M ionized}\\
C-x & &x\:\:\: & &x &\leftarrow \textrm{ equilibrium concentration}
\end{array}\) \(K_{\ce a} = \dfrac{x^2}{C - x}\) \[\mathrm{p\mathit K_a = - \log \mathit K_a} \nonumber \] \[\mathrm{\mathit K_a = 10^{-\large{p\mathit K}_{\Large a}}} \nonumber \] The p values of many weak acids are listed in table form in handbooks, and some of these values are given in Table E1. The p of acetic acid is 4.75. Find the pH of acetic acid solutions of labeled concentrations of 1.0 M, 0.010 M, and 0.00010 M. Assume the label concentration as and mole ionized, then the ionization and the equilibrium concentrations can be represented by the below. with \[K_{\ce a} = \dfrac{x^2}{C - x} \nonumber \] The equation is then \[x^2 + K_{\ce a} x = C K_{\ce a} = 0 \nonumber \] The solution of is then \[x = \dfrac{- K_{\ce a} + \sqrt{K_{\ce a}^2 + 4 C K_{\ce a}}}{2} \nonumber \] Recall that - 1.78e-5, the values of for various are given below: In the above calculations, the following cases may be considered: \(\begin{array}{cccccl}
\ce{HA &\rightleftharpoons &H+ &+ &A- &}\\
C-x & &x & &x &\\
\\
\ce{H2O &\rightleftharpoons &H+ &+ &OH- &}\\
55.6 &&y &&y &\leftarrow (\ce{[H2O]} = 55.6)
\end{array}\) Thus, \(\begin{align}
\ce{[H+]} &= (x+y), \\
\ce{[A- ]} &= x, \\
\ce{[OH- ]} &= y,
\end{align}\) and the two equilibria are \[K_{\ce a} = \dfrac{(x+y) x}{C - x}\label{1} \] and \[\begin{align}
K_{\ce w} &= (x+y)\, y, \label{2} \\
(K_{\ce w} &= \textrm{1E-14})
\end{align} \nonumber \] \[x^2 + (y + K_{\ce a}) x - C K_{\ce a} = 0 \nonumber \] \[x = \dfrac{ -(y+K_{\ce a}) + ((y+K_{\ce a})^2 + 4 C K_{\ce a})^{1/2}}{2} \nonumber \] One of the many methods to find a suitable solution for this problem is to use iterations, or successive approximations. The above procedure is actually a general method that always gives a satisfactory solution. This technique has to be used to calculate the pH of dilute weak acid solutions. Further discussion is given in the . The discussion on weak acids provides a paradigm for the discussion of weak bases. For weak base \(\ce{B}\), the ionization is \[\ce{B- + H2O \rightleftharpoons HB + OH-} \nonumber \] and \[K_{\ce b} = \ce{\dfrac{[HB] [OH- ]}{[B- ]}} \nonumber \] The pOH can be calculated for a basic solution if is given. In this case, the discussion is similar and parallel to that given above for the calculation of pH of weak acids when is known. A weak acid is a compound that b The pH of a solution depends on both the concentration and the degree of ionization, (or using as an indicator). In contrast, a strong acid is completely ionized in solution. The acidity constant, Ka, for a strong acid is b Infinity is a concept, it does not represent a definite value. Derive your answer from the definition of equilibrium constant. A strong acid is "completely" ionized in its solution, but the concentration of the conjugate acid is not zero. Thus, a very large is more realistic than infinity. Household vinegar is usually 5% acetic acid by volume. Calculate the molarity of this solution. Assume density of solution to be 1 g/mL. The formula weight of \(\ce{CH3COOH}\) is 60. 0.8 M
Assume 1 L solution. You have 50 mL acetic acid in 1 L vinegar. The density is 1 g/mL, thus, you have 50 g acetic acid. There is 50 mL vinegar in 1.0 L of vinegar, \(\mathrm{\dfrac{50\: g}{60\: g\: per\: mol} = 0.8\: mol/L}\) Acetic acid is a typical and familiar compound that provides a good example for numerical problems. Its acidity constant Ka is \(1.85 \times 10^{-5}\). What is the pH of a concentrated vinegar, which is a 1.0 M acetic acid solution? 4.3e-3
Use the approximation of \(\ce{[H+]} = \sqrt{K_{\ce a} \times C}\). \(\ce{[H+]} = (1.85\textrm{e-}5)^{1/2} = 4.3\textrm{e-}3\). The approximation is justified because 0.0043/1.0 = 0.4%. Note that most textbooks give = 1.75e-5, but we assume a slightly different value in this and the following problems. If you dilute the vinegar 100 times in a soup that you are cooking, the concentration of your soup is 0.010 M in acetic acid. Other ingredients are ignored. What is the pH of this solution? 3.4 Use the approximation of \(\ce{[H+]} = \sqrt{K_{\ce a} \times C}\). The concentration of \(\ce{H+}\) goes from 3.3E-3 in a 1 M solution down to 4.3E-4 M in a 0.01 M solution. The concentration of \(\ce{H+}\) decreases 10 times when the concentration of the acid decreases 100 times. What is the pH of a 0.010 M \(\ce{HCl}\) solution? 2 \(\ce{[H+]} = \mathrm{0.010\: M}\). The concentration of \(\ce{H+}\) is from \(\ce{HCl}\), which is a strong acid. The pH of a 0.01 M \(\ce{HCl}\) solution is lower than that of a 1 M acetic acid solution; compare with the previous problem. What is the pH of a \(1.0 \times 10^{-4}\) acetic acid solution (\(K_a = 1.85\times 10^{-5}\))? 4.4
Note that \(\ce{[H+]} = \textrm{4.3E-5}\) is 4% of \(\ce{[HAc]}\) (= 1.0E-4). Use the formula \(x = \dfrac{ -K_{\ce a} + (K_{\ce a}^2 + 4 C K_{\ce a})^{1/2}}{2}\) and see what you get. You should use the quadratic formula to calculate \(\ce{[H+]}\). The value using the quadratic formula is 4.5 rather than 4.4 from \(\sqrt{C\times K_{\ce a}}\). What is the pH of a \(1.0\times 10^{-3}\) M chloroacetic acid solution (\(K_a = 1.4\times 10^{-3}\))? This is an interesting numerical problem. Make a good effort to solve it. 3.2 Even a strong acid with concentration of 1.0E-3 M gives a pH of 3. When and are comparable, you have to use the quadratic formula. What is the pH of a \(1.0\times 10^{-6}\) M chloroacetic acid solution (\(K_a = 1.4\times 10^{-3}\))? 6.0 At this concentration, the acid is almost completely ionized. What is the pH of a \(1.0\times 10^{-7}\) M chloroacetic acid solution (\(K_a = 1.4\times 10^{-3}\))? 6.7 From the previous question, you know that the chloroacetic acid should have been completely ionized. Thus, \(\ce{[H+]}\) is about 2e-7; half of that is contributed by the self-ionization of water. This corresponds to a pH of 6.7. You have done a number of numerical problems involving various concentrations of some weak acids. These problems are inter- related. If you do not yet have the complete picture, you should review all these questions. Better yet, review the module. What is the pH of a \(1.0 \times 10^{-9}\) M solution of chloroacetic acid, \(K_a = 1.4 \times 10^{-3}\)? 7 Critical judgment is required. The pH is entirely due to the self-ionization of water at this concentration. | 9,067 | 1,605 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.05%3A_Elementary_Steps |
A for a reaction is a collection of elementary processes (also called elementary steps or elementary reactions) that explains how the overall reaction proceeds. A mechanism is a proposal from which you can work out a rate law that agrees with the observed rate laws. The fact that a mechanism explains the experimental results is not a proof that the mechanism is correct. A mechanism is our rationalization of a chemical reaction, and devising mechanisms is an excellent academic exercise. An is also called an or . It expresses how molecules or ions actually react with each other. The equation in an elementary step represents the reaction at the molecular level, not the overall reaction. Based on numbers of molecules involved in the elementary step, there are three kinds of elementary steps: unimolecular step (or process), bimolecular process, and trimolecular process. An elementary step is proposed to give the reaction rate expression. The rate of an elementary step is always written according to the proposed equation. This practice is very different from the derivation of rate laws for an overall reaction. When a molecule or ion decomposes by itself, such an elementary step is called a . A unimolecular step is always a first order reaction. The following examples are given to illustrate this point: \(\ce{O3 \rightarrow O2 + O}\), \(Rate = k \ce{[O3]}\) or in general \(\ce{A \rightarrow B + C + D}\), \(Rate = k \ce{[A]}\) \(\mathrm{A^* \rightarrow X + Y}\), \(Rate = k \mathrm{[A^*]}\) \(\mathrm{A^*}\) represents an excited molecule. A involves two reacting molecules or ions. The rates for these steps are 2nd order, and some examples are given to illustrate how you should give the rate expression. The simulation illustrates a bimolecular process. \(\ce{NO + O3 \rightarrow NO2 +O2}\), \(Rate = k \ce{[NO] [O3]}\) \(\ce{Cl + CH4 \rightarrow HCl + CH3}\), \(Rate = k \ce{[Cl] [CH4]}\) \(\mathrm{Ar + O_3 \rightarrow Ar + O_3^*}\), \(Rate = k \ce{[Ar] [O3]}\)
\(\ce{A + A \rightarrow B + C}\), \(Rate = k \ce{[A]^2}\) \(\ce{A + B \rightarrow X + Y}\), \(Rate = k \ce{[A] [B]}\) A involves the collision of three molecules. For example: \(\ce{O + O2 + N2 \rightarrow O3 + N2}\), \(Rate = k \ce{[O] [O2] [N2]}\) \(\ce{O + NO + N2 \rightarrow NO2 + N2}\), \(Rate = k \ce{[O] [NO] [N2]}\) The \(\ce{N2}\) molecules in the above trimolecular elementary steps are involved with energy transfer. They can not be canceled. They are written in the equation to give an expression for the s. In general, trimolecular steps may be, \(\ce{A + A + A \rightarrow products}\), \(Rate = k \ce{[A]^3}\) \(\ce{A + A + B \rightarrow products}\), \(Rate = k \ce{[A]^2 [B]}\) \(\ce{A + B + C \rightarrow products}\), \(Rate = k \ce{[A] [B] [C]}\) Three molecules colliding at an instant is rare, but occasionally these are some of the ways reactions take place. Elementary processes are written to show how a chemical reaction progresses leading to an overall reaction. Such a collection is called a . In a mechanism, elementary steps proceed at various speeds. , but the concentrations of reactants in that step must be expressed in terms of the concentrations of the reactants. The following example illustrates how elementary steps are used to represent a reaction mechanism. In particular, a slow step in a mechanism determines the rate of a reaction. If the reaction \(\ce{2 NO2 + F2 \rightarrow 2 NO2F}\) follows the mechanism \(\mathrm{i.}\:\; \ce{NO2 + F2 \rightarrow NO2F + F\: (slow)}\\
\mathrm{ii.}\: \ce{NO2 + F \rightarrow NO2F\: (fast)}\) Work out the rate law.
Since step i. is the rate-determining step, the rate law is \(-\dfrac{1}{2} \ce{\dfrac{d[NO2]}{dt}} = k \ce{[NO2] [F2]}\) Addition of i. and ii. gives the overall reaction.
This example illustrates that the overall reaction equation has nothing to do with the order of the reaction. The elementary process in the rate-determining step determines the order. Other possible elementary steps in this reaction are: \(\ce{F + F \rightarrow F2}\) \(\ce{F + F2 \rightarrow F2 + F}\) \(\ce{NO2F + F \rightarrow F + NO2F}\) but they do not lead to the formation of products. To propose a mechanism requires the knowledge of chemistry to give plausible elementary processes. A freshman in chemistry will not be asked to propose mechanisms, but you will be asked to give the rate laws from a given mechanism. The number of particles involved in an elementary step is called the , and in general, we consider only the molecularity of 1, 2, and 3. Types of elementary steps are summarized below. In the table, \(\ce{A}\), \(\ce{B}\), and \(\ce{C}\) represent reactants, intermediates, or products in the elementary process. Hint: b. \(\ce{A + A \rightarrow products}\)
Recognize and name all three elementary steps. Hint: Third order
Give the order of any elementary step. \(\mathrm{Hg^* + Ar \rightarrow Hg + Ar^*}\) Hint: First order for the light emission step.
Emission of light is first order, but use a number. | 5,032 | 1,606 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Atomic_Orbitals |
This page discusses atomic orbitals at an introductory level. It explores s and p orbitals in some detail, including their shapes and energies. d orbitals are described only in terms of their energy, and f orbitals are only mentioned in passing. When a planet moves around the sun, its definite path, called an orbit, can be plotted. A drastically simplified view of the atom looks similar, in which the electrons orbit around the nucleus. The truth is different; electrons, in fact, inhabit regions of space known as Orbits and orbitals sound similar, but they have quite different meanings. It is essential to understand the difference between them. To plot a path for something, the exact location and trajectory of the object must be known. This is imossible for electrons. The states that it is impossible to define with absolute precision, at the same time, both the position and the momentum of an electron. That makes it impossible to plot an orbit for an electron around a nucleus. Consider a single hydrogen atom: at a particular instant, the position of the electron is plotted. The position is plotted again soon afterward, and it is in a different position. There is no way to tell how it moved from the first place to the second. This process is repeated many times, eventually creating a 3D map of the places that the electron is likely to be found. In the hydrogen case, the electron can be found anywhere within a spherical space surrounding the nucleus. The figure above shows a of this spherical space. 95% of the time (or any arbitrary, high percentage), the electron is found within a fairly easily defined region of space quite close to the nucleus. Such a region of space is called an and it can be thought of as the region of space the electron inhabits. It is impossible to know what the electron is doing inside the orbital, so the electron's actions are ignored completely. All that can be said is that if an electron is in a particular orbital, it has a particular, definable energy. The orbital occupied by the hydrogen electron is called a The number represents the fact that the orbital is in the energy level closest to the nucleus. The letter indicates the shape of the orbital: s orbitals are spherically symmetric around the nucleus—they look like hollow balls made of chunky material with the nucleus at the center. The orbital shown above is a This is similar to a 1s orbital, except that the region where there is the greatest chance of finding the electron is further from the nucleus. This is an orbital at the second energy level. There is another region of slightly higher electron density (where the dots are thicker) nearer the nucleus ("electron density" is another way of describing the likelihood of an electron at a particular place). 2s (and 3s, 4s, etc.) electrons spend some of their time closer to the nucleus than might be expected. The effect of this is to slightly reduce the energy of electrons in s orbitals. The nearer the nucleus the electrons get, the lower their energy. 3s, 4s (etc.) orbitals are progressively further from the nucleus. Not all electrons inhabit s orbitals (in fact, very few electrons live occupy s orbitals). At the first energy level, the only orbital available to electrons is the 1s orbital, but at the second level, as well as a 2s orbital, there are A p orbital is shaped like 2 identical balloons tied together at the nucleus. The orbital shows where there is a 95% chance of finding a particular electron. At any one energy level it is possible to have three absolutely equivalent p orbitals pointing mutually at right angles to each other. These are arbitrarily given the symbols and . This is simply for convenience; the x, y, and z directions change constantly as the atom tumbles in space. The p orbitals at the second energy level are called 2p , 2p and 2p . There are similar orbitals at subsequent levels: 3p , 3p , 3p , 4p , 4p , 4p and so on. All levels except for the first level have p orbitals. At the higher levels the lobes are more elongated, with the most likely place to find the electron more distant from the nucleus. In addition to s and p orbitals, there are two other sets of orbitals that become available for electrons to inhabit at higher energy levels. At the third level, there is a set of five d orbitals (with complicated shapes and names) as well as the 3s and 3p orbitals (3p , 3p , 3p ). At the third level there are nine total orbitals. At the fourth level, as well the 4s and 4p and 4d orbitals there are an additional seven f orbitals, adding up to 16 orbitals in all. s, p, d and f orbitals are then available at all higher energy levels as well. An atom can be pictured as a very strange house (somewhat an inverted pyramid), with the nucleus living on the ground floor, and then various rooms (orbitals) on the higher floors occupied by the electrons. On the first floor there is only 1 room (the 1s orbital); on the second floor there are 4 rooms (the 2s, 2p , 2p and 2p orbitals); on the third floor there are 9 rooms (one 3s orbital, three 3p orbitals and five 3d orbitals); and so on. However, the rooms are not large: each orbital can only hold 2 electrons. A convenient way of showing the orbitals that the electrons live in is to draw "electrons-in-boxes". Orbitals can be represented as boxes with the electrons depicted with arrows. Often an up-arrow and a down-arrow are used to show that the electrons are different. The need for all electrons in an atom to be different originates from quantum theory. If the electrons inhabit different orbitals, they can have identical properties, but if they are both in the same orbital there must be a distinction between them. Quantum theory allocates them a property known as "spin," represented by the direction the arrow is pointing. A 1s orbital holding 2 electrons is drawn as shown on the right, but it can be written even more quickly as 1s . This is read as "one s two," not as "one s squared." Do not confuse the energy level with the number of electrons in this notation. is a German word meaning or . Moving from one atom to the next in the periodic table, the electronic structure of the next atom can be determined by fitting an extra electron into the next available orbital. Electrons fill low energy orbitals (closer to the nucleus) before they fill higher energy ones. If there is a choice between orbitals of equal energy, they fill the orbitals singly as far as possible before pairing up. This filling of orbitals singly where possible is known as . It only applies to orbitals with exactly the same energies (as with p orbitals, for example), and helps to minimize the repulsions between electrons and so makes the atom more stable. The diagram (not to scale) summarizes the energies of the orbitals up to the 4p level that you will need to know when you are using the . Notice that the s orbital always has a slightly lower energy than the p orbitals at the same energy level, so the s orbital always fills with electrons before the corresponding p orbitals do. The oddity is the position of the 3d orbitals. They are at a slightly higher level than the 4s, so the 4s orbital fills first, followed by all the 3d orbitals and then the 4p orbitals. Jim Clark ( ) | 7,300 | 1,607 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Ionization_Constants/Acid_and_Base_Strength |
All acids and bases do not ionize or dissociate to the same extent. This leads to the statement that acids and bases are not all of equal strength in producing H and OH ions in solution. The terms "strong" and "weak" give an indication of the strength of an acid or base. The terms strong and weak describe the ability of acid and base solutions to conduct electricity. If the acid or base conducts electricity strongly, it is a strong acid or base. If the acid or base conducts electricity weakly, it is a weak acid or base. The instructor will test the conductivity of various solutions with a light bulb apparatus. The light bulb circuit is incomplete. If the circuit is completed by a solution containing a large number of ions, the light bulb will glow brightly indicating a strong ability to conduct electricity as shown for HCl. If the circuit is completed by a solution containing large numbers of molecules and either no ions or few ions, the solution does not conduct or conducts very weakly as shown for acetic acid. An acid or base which strongly conducts electricity contains a large number of ions and is called a and an acid or base which conducts electricity only weakly contains only a few ions and is called a . The bond strengths of acids and bases are implied by the relative amounts of molecules and ions present in solution. The bonds are represented as: where A is a negative ion, and M is a positive ion Acids or bases with strong bonds exist predominately as molecules in solutions and are called "weak" acids or bases. Acids or bases with weak bonds easily dissociate into ions and are called "strong" acids or bases. Acids and bases behave differently in solution based on their strength. Acid or base "strength" is a measure of how readily the molecule ionizes in water. Some acids and bases ionize rapidly and almost completely in solution; these are called strong acids and strong bases. For example, hydrochloric acid (HCl) is a strong acid. When placed in water, virtually every HCl molecule splits into a H ion and a Cl ion in the reaction. \[\ce{HCl(aq) + H2O(l) <=> H3O^{+}(aq) + Cl^{-}(aq)} \nonumber\] For a strong acid like HCl, if you place 1 mole of HCl in a liter of water, you will get roughly 1 mole of H 0 ions and 1 mole of Cl ions. In a weak acid like hydrofluoric acid (HF), not all of the HF molecules split up, and although there will be some H and F ions released, there will still be HF molecules in solution . A similar concept applies to bases, except the reaction is different. A strong base like sodium hydroxide (NaOH) will also dissociate completely into water; if you put in 1 mole of NaOH into water, you will get 1 mole of hydroxide ions. \[\ce{NaOH(aq) + H2O(l) <=> Na^{+}(aq) + OH^{-}(aq) + H2O(l)} \nonumber\] The terms "strong" and "weak" in this context do not relate to how corrosive or caustic the substance is, but only its capability to ionize in water. The ability of a substance to eat through other materials or damage skin is more of a function of the properties of that acid, as well as its concentration. Although, strong acids are more directly dangerous at lower concentrations a strong acid is not necessarily more dangerous than a weak one. For example, hydrofluoric acid is a weak acid , but it is extremely dangerous and should be handled with great care. Hydrofluoric acid is particularly dangerous because it is capable of eating through glass, as seen in the video in the links section . The percent dissociation of an acid or base is mathematically indicated by the acid ionization constant (K ) or the base ionization constant (K ) . These terms refer to the ratio of reactants to products in equilibrium when the acid or base reacts with water. For acids the expression will be K = [H O ,A ]/[HA] where HA is the concentration of the acid at equilibrium, and A is the concentration of its conjugate base at equilibrium and for bases the expression will be \[K_b = \dfrac{[\ce{OH^{-}},\ce{HB^{+}}]}{\ce{B}}\] where B is the concentration of the base at equilibrium and HB is the concentration of its conjugate acid at equilibrium The stronger an acid is, the lower the pH it will produce in solution. pH is calculated by taking the negative logarithm of the concentration of hydronium ions. For strong acids, you can calculate the pH by simply taking the negative logarithm of its molarity as it completely dissociates into its conjugate base and hydronium. The same goes for strong bases, except the negative logarithm gives you the pOH as opposed to the pH. For weak acids and bases, the higher the K or K , the more acidic or basic the solution. To find the pH for a weak acid or base, you must use the K equation and a RICE table to determine the pH. All acids have a conjugate base that forms when they react with water, and similarly, all bases have a conjugate acid that reacts when they form with water. You can judge the relative strength of a conjugate by the \(K_a\) or \(K_b\) value of the substance because \(K_a \times K_b\) is equal to the ionization constant of water, K which is equal to \(1 \times 10^{-14}\) at room temperature. The higher the Ka, the stronger the acid is, and the weaker its conjugate base is. Similarly, the higher the K , the stronger the substance is as a base, and the more weakly acidic its conjugate acid is. For an acid that reacts with water in the reaction \[HA_{(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + A^-_{(aq)}\] \[K_a = \dfrac{[H_3O^+,A^-]}{[HA]}\] where each bracketed term represents the concentration of that substance in solution. Relation of K , K , K \[K_w = K_a \times K_b \nonumber\] Partial List of Strong Acids: Hydrochlroic acid (HCl), Nitric Acid (HNO ), Perchloric Acid (HClO ), Sulfuric Acid (H SO ) Partial List of Strong Bases: Sodium Hydroxide (NaOH), Barium Hydroxide (Ba(OH) ), Calcium Hydroxide (Ca(OH) ), Lithium Hydroxide (LiOH) (Hydroxides of Group I and II elements are generally strong bases) Partial List of Weak Acids: Acetic Acid (CH COOH), Carbonic Acid (H CO ), Phosphoric Acid (H PO ) Partial List of Weak Bases: Ammonia (NH ), Calcium Carbonate (CaCO ), Sodium Acetate (NaCH COO) Find the pH of 0.5 grams of HCl disolved into 100 ml of water: First find moles of acid: grams / molar mass = moles 0.5 grams / (36.5 g/mole) = 0.014 moles HCl Then find molarity: moles / volume = molarity 0.014 moles / 0.100 L = 0.14 M HCl is a strong acid and completely dissociates in water, therefore the pH will be equal to the negative logarithm of the concentration of HCl pH = -log(H O ) pH = -log(0.14) = 0.85 The K value for acetic acid is 1.76*10 , and the K value for benzoic acid is 6.46*10 , if two solutions are made, one from each acid, with equal concentrations, which one will have the lower pH? The K value is a measure of the ratio between reactants and products at equilibrium. For an acid, the reaction will be HA + H O --> A + H O . PH is based on the concentration of the hydronium ion (H O ) which is a product of the reaction of acid and water. A higher K value means a higher ratio of reactants to products, and so the acid with the higher K value will be producing more hydronium, and therefore have a lower pH. Therefore the solution of benzoic acid will have a lower pH. The K value of ammonium (NH ) is 5.6*10 , the K value of ammonia (NH 1.8*10 , is ammonium more strongly acidic than ammonia is basic? The relative strength of an acid or base depends on how high its K or K value is, in this case, the K value is far lower than the K value so the ammonia is more strongly basic than ammonium is acidic. | 7,645 | 1,608 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/11%3A_Reactions_and_Other_Chemical_Processes/11.06%3A_Adiabatic_Flame_Temperature |
\( \newcommand{\tx}[1]{\text{#1}} % text in math mode\)
\( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)
\( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)
\( \newcommand{\st}{^\circ} % standard state symbol\)
\( \newcommand{\id}{^{\text{id}}} % ideal\)
\( \newcommand{\rf}{^{\text{ref}}} % reference state\)
\( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)
\( \newcommand{\K}{\units{K}} % kelvins\)
\( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)
\( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\)
\( \newcommand{\Pa}{\units{Pa}}\)
\( \newcommand{\mol}{\units{mol}} % mole\)
\( \newcommand{\V}{\units{V}} % volts\)
\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) With a few simple approximations, we can estimate the temperature of a flame formed in a flowing gas mixture of oxygen or air and a fuel. We treat a moving segment of the gas mixture as a closed system in which the temperature increases as combustion takes place. We assume that the reaction occurs at a constant pressure equal to the standard pressure, and that the process is adiabatic and the gas is an ideal-gas mixture. The principle of the calculation is similar to that used for a constant-pressure calorimeter as explained by the paths shown in Fig. 11.11. When the combustion reaction in the segment of gas reaches reaction equilibrium, the advancement has changed by \(\Del\xi\) and the temperature has increased from \(T_1\) to \(T_2\). Because the reaction is assumed to be adiabatic at constant pressure, \(\Del H\expt\) is zero. Therefore, the sum of \(\Del H(\tx{rxn},T_1)\) and \(\Del H(\tx{P})\) is zero, and we can write \begin{equation} \Del\xi\Delsub{c}H\st(T_1) + \int_{T_1}^{T_2}\!C_p(\tx{P})\dif T = 0 \tag{11.6.1} \end{equation} where \(\Delsub{c}H\st(T_1)\) is the standard molar enthalpy of combustion at the initial temperature, and \(C_p(\tx{P})\) is the heat capacity at constant pressure of the product mixture. The value of \(T_2\) that satisfies Eq. 11.6.1 is the flame temperature. Problem 11.9 presents an application of this calculation. Several factors cause the actual temperature in a flame to be lower: the process is never completely adiabatic, and in the high temperature of the flame there may be product dissociation and other reactions in addition to the main combustion reaction. | 8,975 | 1,609 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Fission_and_Fusion/Fission_Chain_Reaction |
Nuclear Chain Reactions are a simple, yet powerful method which to produce both constructive and destructive forces. Only understood to a significant degree within the last century, nuclear chain reactions have many practical uses in the modern era. Chain reactions can be addressed into two categories: first, controlled (like a nuclear power plant) and uncontrolled (an atomic bomb). Both are motivated by fission reactions, which are elaborated in this section. Nuclear chain reactions are one of the modern applications of the fission process. Capitalizing upon the large amounts of energy, nuclear chain reactions can be used for constructive and destructive means. To understand how a chain reaction operates, it is best to search in depth of the nature of fission reactions. By understanding fission reactions, a chain reaction is then just an extrapolation of the excess particles produced by fission for a large quantity of an element. The neutron was discovered by Sir James Chadwick in 1932. In 1939 atomic fission had been discovered by Lise Meitner (1878 to 1968) with her nephew Otto Frisch and was documented in her paper where the term fission was coined. On December 2, 1942 Dr. Enrico Fermi, a Nobel Prize winner, produced the first sustained chain reaction at the University of Chicago. His assumption was that said reactions produce transuranium products. Otto Hahn and F. Strassmann reproduced Fermi's experiments and found contrary. By 1943 the first controlled chain reaction had taken place and by 1945, the first atomic weapon was produced. By 1951 nuclear chain reactions took a more practical route and became a means to produce electric power. Chain Reactions are basically fission reactions which through the products produce more chain reactions. One of the most well-known and useful examples of a chain reaction is of U which is used to harness nuclear energy. For U on average 2.5 neutrons are emitted, starting on average two more fission reactions. Below is a simple fission process: The elements which the daughter nuclei are composed of is based upon probability (compare the fragments of Figures 4 and 5). Understanding Fission is the key for chain reactions. Hence, below is a brief overview of the different aspects of fission. Below are scientific approaches to fission reactions, and the process which fission is produced. Fission Products can be produced by alpha, gamma, beta, charged particles, and through spontaneous reactions. A fission product is affected by the initial mass of nuclide, and excitation energy (expressed in MeV). Yield is the result is for any fission event, and is highly motivated by probability. This section covers experimental yields and charge yields. The yield can be expressed in two different experimental categories: which can be expressed by the equation: \(Y{(\%)} = \left(\dfrac{N}{P}\right)*{100}\) Where Y is the absolute yield, N is the number of atoms a given nucleus formed, and P is the number of fission events. Absolute Yield can be broken down into three different categories: \(A + n \rightarrow B + C\) \(A + n \rightarrow B + C\) \(A \rightarrow D + n\) \(D \rightarrow E + \beta\) \(E \rightarrow C + \alpha\) (this process is through natural decay, where Element A decomposes to Element C) Note: Sum of total yields for binary fission will be 200% Relative Yield \( \gamma_i\) is the ratio of absolute yields formed in fission, which can be expressed through the equation: \(\gamma_i = \dfrac{Y_i}{Y}\) where \(\gamma_i\) is the relative yield, Y is the absolute yield of specific nuclide, which is daughter to a specific nuclei Y also expressed as a yield. Relative yield also has respective independent, cumulative, total yields. An approximation of charge ratio (Z/A) is the same for both daughter nuclides as it was for the parent atom. The distribution of charge for daughter nuclides for a fixed mass A can be approximated with the equation: \(fragment\;charge = e^{-\dfrac{(Z-Z_p)^2}{c}}\) where Z is proton number, Z is approximately the charge ratio of the parent nuclide , and c is a number between 0.8 and 1.0 (depending on fission reaction). Fission products behave in specific trends, for example: fission is almost always a binary process, as ternary fission is on average 1000 less times probable (In most cases Ternary fission will appear for 1 to 3 events per 1000). Being able to predict the outcome of a fission event is vital for harvesting nuclear energy from power plants. Other trends that can be found is the nature of symmetric and asymmetric fission products. Fission will be symmetric or asymmetric depending on how many nucleons are required to form closed shells. Symmetric peaks can coexist in asymmetric peaks, such as Ra . Also, a symmetric mass distribution leads to an asymmetric fission process. Note that asymmetric and Symmetric fission are both present in heavy nuclides. : The preferred form of fission leading to two daughter nuclides of unequal masses. Asymmetric fission has an average yield 600 times greater than symmetric fission. Number of outer nucleons drastically affects shape of nucleus, which in turn may change the nature of fission symmetry. The probability of asymmetric fission is high for fission motivated by thermal neutrons (such as U , U , Pu ). Preferred for spontaneous reactions (such as Th , U , Cm , and Cf ), symmetric fission is caused by the preference in atomic nuclei to follow the 'magic numbers' most commonly of 50 proton 82 neutron nucleus. Symmetric Produces two daughter nuclides of equal or near equal masses. The range for atomic mass numbers that will be the product of symmetric fission are in the range of 127 to 139 amu. Symmetric fission also is preferred for high excitation energies (at or over 40 MeV). As a function of initial excitation energy, the probability of symmetric fission increases more rapidly than asymmetric fission. Note that multiple combinations are possible for fission chain reactions. An unstable nucleus by the liquid drop model can be expressed in the inequality: \(\dfrac{Z^2}{A} > {18}\) Which for any proton number Z and Atomic Mass A, holding this inequality to be true will release energy through symmetric fission. A notable exception is Zr which upon the border of this inequality is stable. Although bear in mind that the above inequality is limited because of binding energy and therefore is extremely slow if not impossible. A far more probable inequality would is: \(\dfrac{Z^2}{A} > {47}\) Which the second inequality holds for half-lives for super heavy atoms. Note that spontaneous Fission is extremely rare for even radioactive samples. Neutron binding energy is approximately 6 to 7 MeV. Neutron emission of atoms is favored over the fission process when energetically favorable. For the sake of uranium, the neutron emission and fission thresholds are only a few MeV apart. Thus odd-neutron atoms are 'slow neutron fission' and even neutron atoms are 'fast neutron fission'. With this implication it can be deduced that with enough initial excitement energy, multi neutron emission instead of fission is possible. One of the intriguing portions of operating fission according to chance is the objective of producing more than one neutron before the fission reaction takes place. Hence, is a competition between binding energies of the molecule and neutrons. For example of a second neutron binding energy is greater than the first neutron binding energy, then second chance fission thresholds occur before second neutron emission. The major energy of fission lies in the kinetics of the daughter nuclides. The magnitude of the kinetic energy is equal to the coulomb potential energy of fragments. The rest of the energy is released in excitation energy, which is usually neutron emission, and finally gamma radiation for the residual energy. Yield of neutrons that accompany fission is usually a linear relationship expressed as: 1 neutron to 8.5 MeV. A neutron on average carries 5.5 MeV in binding energy and 1.5 MeV in Kinetic energy. Hence, 7 MeV. Neutron yield of lighter fragments is three times greater than heavier fragments. Excitation energy is placed more into fragments and thus neutron emission than kinetics. One of the best examples of Chain reactions is within nuclear power plants. As a fact, nuclear power provides 17 percent of the world's electricity. There are many types of reactors, which all use realatively the same materials, and components to complete a nuclear fission chain reaction. The other spectrum of chain reactions, which produce enough energy to cause an explosion. Below are destructive chain reactions by accident. International Nuclear Event Scale (INES): The scale is written by degree of danger (lower in the list is greater danger) 3. A sample of Element Lol produced 15 fragment nuclides in 45 fission events, what is the absolute Yield Y(%)? 4. A new isotope of Xenon has been discovered with an atomic mass of 62 by the practical equation for spontaneous fission, will Xe undergo spontaneous fission? (Ignoring that this is absolutely impossible) 5. For a yield of a given mass, would the end link in a radioactive chain equal the total yield? | 9,236 | 1,610 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/03%3A_The_Langmuir_Isotherm/3.04%3A_Variation_of_Surface_Coverage_with_Temperature_and_Pressure |
Application of the assumptions of the Langmuir Isotherm leads to readily derivable expressions for the pressure dependence of the surface coverage (see Sections 3.2 and 3.3) - in the case of a simple, reversible molecular adsorption process the expression is \[ \theta = \dfrac{bP}{1+bP} \label{Eq1}\] where \(b = b(T)\). This is illustrated in the graph below which shows the characteristic Langmuir variation of coverage with pressure for molecular adsorption. Note two extremes in Equation \(\ref{Eq1}\): At a given pressure the extent of adsorption is determined by the value of \(b\), which is dependent upon both the temperature (T) and the enthalpy (heat) of adsorption. Remember that the magnitude of the adsorption enthalpy (a negative quantity itself) reflects the strength of binding of the adsorbate to the substrate. The value of \(b\) is increased by Therefore the set of curves shown below illustrates the effect of either (i) increasing the magnitude of the adsorption enthalpy at a fixed temperature, or (ii) decreasing the temperature for a given adsorption system. A given equilibrium surface coverage may be attainable at various combinations of pressure and temperature as highlighted below … note that as the temperature is lowered the pressure required to achieve a particular equilibrium surface coverage decreases. - this is often used as justification for one of the main ideologies of surface chemistry ; specifically, that it is possible to study technologically-important (high pressure / high temperature) surface processes within the low pressure environment of typical surface analysis systems by working at low temperatures. It must be recognized however that, at such low temperatures, kinetic restrictions that are not present at higher temperatures may become important. If you wish to see how the various factors relating to the adsorption and desorption of molecules influence the surface coverage then try out the Interactive Demonstration of the Langmuir Isotherm (note - this is based on the derivation discuss previously ). It has been shown in previous sections how the value of is dependent upon the enthalpy of adsorption. It has also just been demonstrated how the value of influences the pressure/temperature (P-T) dependence of the surface coverage. The implication of this is that it must be possible to determine the enthalpy of adsorption for a particular adsorbate/substrate system by studying the P-T dependence of the surface coverage. Various methods based upon this idea have been developed for the determination of adsorption enthalpies - one method is outlined below: Step 1: Involves determination of a number of adsorption isotherms, where a single isotherm is a coverage / pressure curve at a fixed temperature (Figure 3.4.1). Step 2: It is then possible to read off a number of pairs of values of pressure and temperature which yield the same surface coverage (Figure 3.4.2) Step 3: The Clausius-Clapeyron equation \[ \left( \dfrac{\partial \ln P}{\partial \frac{1}{T}} \right)_{\theta} = \dfrac{\Delta H_{abs}}{R}\] may then be applied to this set of (P-T) data and a plot of ( ln P ) v's (1/T) should give a straight line, the slope of which yields the adsorption enthalpy (Figure . The value obtained for the adsorption enthalpy is that pertaining at the surface coverage for which the P-T data was obtained, but steps 2 & 3 may be repeated for different surface coverages enabling the adsorption enthalpy to be determined over the whole range of coverages. This method is applicable only when the adsorption process is thermodynamically . | 3,628 | 1,611 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.7%3A_Polyprotic_Acids |
We can classify acids by the number of protons per molecule that they can give up in a reaction. Acids such as \(\ce{HCl}\), \(\ce{HNO3}\), and \(\ce{HCN}\) that contain one ionizable hydrogen atom in each molecule are called . Their reactions with water are: \[\ce{HCl}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{Cl-}(aq) \nonumber \] \[\ce{HNO3}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{NO3-}(aq) \nonumber \] \[\ce{HCN}(aq)+\ce{H2O}(l)⟶\ce{H3O+}(aq)+\ce{CN-}(aq) \nonumber \] Even though it contains four hydrogen atoms, acetic acid, \(\ce{CH3CO2H}\), is also monoprotic because only the hydrogen atom from the carboxyl group (\(\ce{-COOH}\)) reacts with bases: Similarly, monoprotic bases are bases that will accept a single proton. contain two ionizable hydrogen atoms per molecule; ionization of such acids occurs in two steps. The first ionization always takes place to a greater extent than the second ionization. For example, sulfuric acid, a strong acid, ionizes as follows: \[ \ce{H2SO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HSO4-}(aq) \nonumber \] with \(K_{\ce a1} > 10^2;\: {complete\: dissociation}\). \[ \ce{HSO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{SO4^{2−}}(aq) \nonumber \] with \( K_{\ce a2}=1.2×10^{−2}\). This stepwise ioniza process occurs for all polyprotic acids. When we make a solution of a weak diprotic acid, we get a solution that contains a mixture of acids. Carbonic acid, \(\ce{H2CO3}\), is an example of a weak diprotic acid. The first ionization of carbonic acid yields hydronium ions and bicarbonate ions in small amounts. \[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \nonumber \] with \[K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+,HCO3- ]}{[H2CO3]}}=4.3×10^{−7} \nonumber \] The bicarbonate ion can also act as an acid. It ionizes and forms hydronium ions and carbonate ions in even smaller quantities. \[\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber \] with \[ K_{\ce{HCO3-}}=\ce{\dfrac{[H3O+,CO3^2- ]}{[HCO3- ]}}=4.7×10^{−11} \nonumber \] \(K_{\ce{H2CO3}}\) is larger than \(K_{\ce{HCO3-}}\) by a factor of 10 , so H CO is the dominant producer of hydronium ion in the solution. This means that little of the \(\ce{HCO3-}\) formed by the ionization of H CO ionizes to give hydronium ions (and carbonate ions), and the concentrations of H O and \(\ce{HCO3-}\) are practically equal in a pure aqueous solution of H CO . If the first ionization constant of a weak diprotic acid is larger than the second by a factor of at least 20, it is appropriate to treat the first ionization separately and calculate concentrations resulting from it before calculating concentrations of species resulting from subsequent ionization. This can simplify our work considerably because we can determine the concentration of H O and the conjugate base from the first ionization, then determine the concentration of the conjugate base of the second ionization in a solution with concentrations determined by the first ionization. When we buy soda water (carbonated water), we are buying a solution of carbon dioxide in water. The solution is acidic because CO reacts with water to form carbonic acid, H CO . What are \(\ce{[H3O+]}\), \(\ce{[HCO3- ]}\), and \(\ce{[CO3^2- ]}\) in a saturated solution of CO with an initial [H CO ] = 0.033 ? \[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \label{step1} \tag{equilibrium step 1} \] \[\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \hspace{20px} K_{\ce a2}=4.7×10^{−11} \label{step2} \tag{equilibrium step 2} \] As indicated by the ionization constants, H CO is a much stronger acid than \(\ce{HCO3-}\), so \(\ce{H2CO3}\) is the dominant producer of hydronium ion in solution. Thus there are two parts in the solution of this problem: To summarize: \(\ce{H3O+}\) \(\ce{HCO3-}\). Since \ref{step1} is has a much bigger \(K_{a1}=4.3×10^{−7}\) than \(K_{a2}=4.7×10^{−11}\) for \ref{step2}, we can safely ignore the second ionization step and focus only on the first step (but address it in next part of problem). \[\ce{H2CO3}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HCO3-}(aq) \hspace{20px} K_{\ce a1}=4.3×10^{−7} \nonumber \] An abbreviated table of changes and concentrations shows: Substituting the equilibrium concentrations into the equilibrium constant gives us: \[K_{\ce{H2CO3}}=\ce{\dfrac{[H3O+,HCO3- ]}{[H2CO3]}}=\dfrac{(x)(x)}{0.033−x}=4.3×10^{−7} \nonumber \] Solving the preceding equation making our standard assumptions gives: \[x=1.2×10^{−4} \nonumber \] Thus: \[\ce{[H2CO3]}=0.033\:M \nonumber \] \[\ce{[H3O+]}=\ce{[HCO3- ]}=1.2×10^{−4}\:M \nonumber \] Since the \ref{step1} is has a much bigger \(K_a\) than \ref{step2}, we can the equilibrium conditions calculated from first part of example as the initial conditions for an Table for the \ref{step2}: \[\ce{HCO3-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{CO3^2-}(aq) \nonumber \] \[ \begin{align*} K_{\ce{HCO3-}}&=\ce{\dfrac{[H3O+,CO3^2- ]}{[HCO3- ]}} \\[4pt] &=\dfrac{(1.2×10^{−4}\:M + y) (y)}{(1.2×10^{−4}\:M - y)} \end{align*} \nonumber \] To avoid solving a quadratic equation, we can assume \(y \ll 1.2×10^{−4}\:M \) so \[K_{\ce{HCO3-}} = 4.7×10^{−11} \approx \dfrac{(1.2×10^{−4}\:M ) (y)}{(1.2×10^{−4}\:M)} \nonumber \] Rearranging to solve for \(y\) \[y \approx \dfrac{ (4.7×10^{−11})(1.2×10^{−4}\:M )}{ 1.2×10^{−4}\:M} \nonumber \] \[[\ce{CO3^2-}]=y \approx 4.7×10^{−11} \nonumber \] : In part 1 of this example, we found that the \(\ce{H2CO3}\) in a 0.033- solution ionizes slightly and at equilibrium \([\ce{H2CO3}] = 0.033\, M\), \([\ce{H3O^{+}}] = 1.2 × 10^{−4}\), and \(\ce{[HCO3- ]}=1.2×10^{−4}\:M\). In part 2, we determined that \(\ce{[CO3^2- ]}=5.6×10^{−11}\:M\). The concentration of \(H_2S\) in a saturated aqueous solution at room temperature is approximately 0.1 . Calculate \(\ce{[H3O+]}\), \(\ce{[HS^{−}]}\), and \(\ce{[S^{2−}]}\) in the solution: \[\ce{H2S}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HS-}(aq) \hspace{20px} K_{\ce a1}=8.9×10^{−8} \nonumber \] \[\ce{HS-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{S^2-}(aq) \hspace{20px} K_{\ce a2}=1.0×10^{−19} \nonumber \] \([\ce{H2S}] = 0.1 M\), \(\ce{[H3O+]} = [HS^{−}] = 0.0001\, M\), \([S^{2−}] = 1 × 10^{−19}\, M\) We note that the concentration of the sulfide ion is the same as . This is due to the fact that each subsequent dissociation occurs to a lesser degree (as acid gets weaker). A is an acid that has three dissociable protons that undergo stepwise ionization: Phosphoric acid is a typical example: \[\ce{H3PO4}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{H2PO4-}(aq) \nonumber \] with \(K_{\ce a1}=7.5×10^{−3} \). \[\ce{H2PO4-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{HPO4^2-}(aq) \nonumber \] with \( K_{\ce a2}=6.2×10^{−8} \). \[\ce{HPO4^2-}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{PO4^3-}(aq) \nonumber \] with \( K_{\ce a3}=4.2×10^{−13} \). As with the diprotic acids, the differences in the ionization constants of these reactions tell us that in each successive step the degree of ionization is significantly weaker. This is a general characteristic of polyprotic acids and successive ionization constants often differ by a factor of about 10 to 10 . This set of three dissociation reactions may appear to make calculations of equilibrium concentrations in a solution of H PO complicated. However, because the successive ionization constants differ by a factor of 10 to 10 , the calculations can be broken down into a series of parts similar to those for diprotic acids. Polyprotic bases can accept more than one hydrogen ion in solution. The carbonate ion is an example of a , since it can accept up to two protons. Solutions of alkali metal carbonates are quite alkaline, due to the reactions: \[\ce{H2O}(l)+\ce{CO3^2-}(aq)⇌\ce{HCO3-}(aq)+\ce{OH-}(aq) \nonumber \] and \[\ce{H2O}(l)+\ce{HCO3-}(aq)⇌\ce{H2CO3}(aq)+\ce{OH-}(aq) \nonumber \] An acid that contains more than one ionizable proton is a polyprotic acid. The protons of these acids ionize in steps. The differences in the acid ionization constants for the successive ionizations of the protons in a polyprotic acid usually vary by roughly five orders of magnitude. As long as the difference between the successive values of of the acid is greater than about a factor of 20, it is appropriate to break down the calculations of the concentrations of the ions in solution into a series of steps. | 8,340 | 1,613 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/04%3A_The_Second_Law_of_Thermodynamics/4.09%3A_Phase_Equilibria |
Nearly all of us have heated a pan of water with the lid in place and shortly thereafter heard the sounds of the lid rattling and hot water spilling onto the stovetop. When a liquid is heated, its molecules obtain sufficient kinetic energy to overcome the forces holding them in the liquid and they escape into the gaseous phase. By doing so, they generate a population of molecules in the vapor phase above the liquid that produces a pressure—the vapor pressure of the liquid. In the situation we described, enough pressure was generated to move the lid, which allowed the vapor to escape. If the vapor is contained in a sealed vessel, however, such as an unvented flask, and the vapor pressure becomes too high, the flask will explode (as many students have unfortunately discovered). In this section, we describe vapor pressure in more detail and explain how to quantitatively determine the vapor pressure of a liquid. Because the molecules of a liquid are in constant motion, we can plot the fraction of molecules with a given kinetic energy ( ) against their kinetic energy to obtain the kinetic energy distribution of the molecules in the liquid (Figure \(\Page {1}\)), just as we did for a gas. As for gases, increasing the temperature increases both the average kinetic energy of the particles in a liquid and the range of kinetic energy of the individual molecules. If we assume that a minimum amount of energy (\(E_0\)) is needed to overcome the intermolecular attractive forces that hold a liquid together, then some fraction of molecules in the liquid always has a kinetic energy greater than \(E_0\). The fraction of molecules with a kinetic energy greater than this minimum value increases with increasing temperature. Any molecule with a kinetic energy greater than \(E_0\) has enough energy to overcome the forces holding it in the liquid and escape into the vapor phase. Before it can do so, however, a molecule must also be at the surface of the liquid, where it is physically possible for it to leave the liquid surface; that is, only molecules at the surface can undergo evaporation (or vaporization), where molecules gain sufficient energy to enter a gaseous state above a liquid’s surface, thereby creating a vapor pressure. To understand the causes of vapor pressure, consider the apparatus shown in Figure \(\Page {2}\). When a liquid is introduced into an evacuated chamber (part (a) in Figure \(\Page {2}\)), the initial pressure above the liquid is approximately zero because there are as yet no molecules in the vapor phase. Some molecules at the surface, however, will have sufficient kinetic energy to escape from the liquid and form a vapor, thus increasing the pressure inside the container. As long as the temperature of the liquid is held constant, the fraction of molecules with \(KE > E_0\) will not change, and the rate at which molecules escape from the liquid into the vapor phase will depend only on the surface area of the liquid phase. As soon as some vapor has formed, a fraction of the molecules in the vapor phase will collide with the surface of the liquid and reenter the liquid phase in a process known as condensation (part (b) in Figure \(\Page {2}\)). As the number of molecules in the vapor phase increases, the number of collisions between vapor-phase molecules and the surface will also increase. Eventually, a will be reached in which exactly as many molecules per unit time leave the surface of the liquid (vaporize) as collide with it (condense). At this point, the pressure over the liquid stops increasing and remains constant at a particular value that is characteristic of the liquid at a given temperature. The rates of evaporation and condensation over time for a system such as this are shown graphically in Figure \(\Page {3}\). Two opposing processes (such as evaporation and condensation) that occur at the same rate and thus produce no change in a system, constitute a dynamic equilibrium. In the case of a liquid enclosed in a chamber, the molecules continuously evaporate and condense, but the amounts of liquid and vapor do not change with time. The pressure exerted by a vapor in dynamic equilibrium with a liquid is the equilibrium vapor pressure of the liquid. If a liquid is in an container, however, most of the molecules that escape into the vapor phase will collide with the surface of the liquid and return to the liquid phase. Instead, they will diffuse through the gas phase away from the container, and an equilibrium will never be established. Under these conditions, the liquid will continue to evaporate until it has “disappeared.” The speed with which this occurs depends on the vapor pressure of the liquid and the temperature. Volatile liquids have relatively high vapor pressures and tend to evaporate readily; nonvolatile liquids have low vapor pressures and evaporate more slowly. Although the dividing line between volatile and nonvolatile liquids is not clear-cut, as a general guideline, we can say that substances with vapor pressures greater than that of water (Figure \(\Page {4}\)) are relatively volatile, whereas those with vapor pressures less than that of water are relatively nonvolatile. Thus diethyl ether (ethyl ether), acetone, and gasoline are volatile, but mercury, ethylene glycol, and motor oil are nonvolatile. The equilibrium vapor pressure of a substance at a particular temperature is a characteristic of the material, like its molecular mass, melting point, and boiling point. It does depend on the amount of liquid as long as at least a tiny amount of liquid is present in equilibrium with the vapor. The equilibrium vapor pressure does, however, depend very strongly on the temperature and the intermolecular forces present, as shown for several substances in Figure \(\Page {4}\). Molecules that can hydrogen bond, such as ethylene glycol, have a much lower equilibrium vapor pressure than those that cannot, such as octane. The nonlinear increase in vapor pressure with increasing temperature is steeper than the increase in pressure expected for an ideal gas over the corresponding temperature range. The temperature dependence is so strong because the vapor pressure depends on the fraction of molecules that have a kinetic energy greater than that needed to escape from the liquid, and this fraction increases exponentially with temperature. As a result, sealed containers of volatile liquids are potential bombs if subjected to large increases in temperature. The gas tanks on automobiles are vented, for example, so that a car won’t explode when parked in the sun. Similarly, the small cans (1–5 gallons) used to transport gasoline are required by law to have a pop-off pressure release. Volatile substances have low boiling points and relatively weak intermolecular interactions; nonvolatile substances have high boiling points and relatively strong intermolecular interactions. A Discussing Vapor Pressure and Boiling Points. The exponential rise in vapor pressure with increasing temperature in Figure \(\Page {4}\) allows us to use natural logarithms to express the nonlinear relationship as a linear one. \[ \boxed{\ln P =\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T} \right) + C} \label{Eq1} \] where Plotting \(\ln P\) versus the inverse of the absolute temperature (\(1/T\)) is a straight line with a slope of −Δ / . Equation \(\ref{Eq1}\), called the , can be used to calculate the \(ΔH_{vap}\) of a liquid from its measured vapor pressure at two or more temperatures. The simplest way to determine \(ΔH_{vap}\) is to measure the vapor pressure of a liquid at temperatures and insert the values of \(P\) and \(T\) for these points into Equation \(\ref{Eq2}\), which is derived from the Clausius–Clapeyron equation: \[ \ln\left ( \dfrac{P_{1}}{P_{2}} \right)=\dfrac{-\Delta H_{vap}}{R}\left ( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \label{Eq2} \] Conversely, if we know Δ and the vapor pressure \(P_1\) at any temperature \(T_1\), we can use Equation \(\ref{Eq2}\) to calculate the vapor pressure \(P_2\) at any other temperature \(T_2\), as shown in Example \(\Page {1}\). A Discussing the Clausius-Clapeyron Equation. Link: The experimentally measured vapor pressures of liquid Hg at four temperatures are listed in the following table: From these data, calculate the enthalpy of vaporization (Δ ) of mercury and predict the vapor pressure of the liquid at 160°C. (Safety note: mercury is highly toxic; when it is spilled, its vapor pressure generates hazardous levels of mercury vapor.) vapor pressures at four temperatures Δ of mercury and vapor pressure at 160°C The table gives the measured vapor pressures of liquid Hg for four temperatures. Although one way to proceed would be to plot the data using Equation \(\ref{Eq1}\) and find the value of Δ from the slope of the line, an alternative approach is to use Equation \(\ref{Eq2}\) to obtain Δ directly from two pairs of values listed in the table, assuming no errors in our measurement. We therefore select two sets of values from the table and convert the temperatures from degrees Celsius to kelvin because the equation requires absolute temperatures. Substituting the values measured at 80.0°C ( ) and 120.0°C ( ) into Equation \(\ref{Eq2}\) gives \[\begin{align*} \ln \left ( \dfrac{0.7457 \; \cancel{Torr}}{0.0888 \; \cancel{Torr}} \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot K}\left ( \dfrac{1}{\left ( 120+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right)K} \right) \\[4pt] \ln\left ( 8.398 \right) &=\dfrac{-\Delta H_{vap}}{8.314 \; J/mol\cdot \cancel{K}}\left ( -2.88\times 10^{-4} \; \cancel{K^{-1}} \right) \\[4pt] 2.13 &=-\Delta H_{vap} \left ( -3.46 \times 10^{-4} \right) J^{-1}\cdot mol \\[4pt] \Delta H_{vap} &=61,400 \; J/mol = 61.4 \; kJ/mol \end{align*} \nonumber \] We can now use this value of Δ to calculate the vapor pressure of the liquid ( ) at 160.0°C ( ): \[ \ln\left ( \dfrac{P_{2} }{0.0888 \; torr} \right)=\dfrac{-61,400 \; \cancel{J/mol}}{8.314 \; \cancel{J/mol} \; K^{-1}}\left ( \dfrac{1}{\left ( 160+273 \right)K}-\dfrac{1}{\left ( 80.0+273 \right) K} \right) \nonumber \] Using the relationship \(e^{\ln x} = x\), we have \[\begin{align*} \ln \left ( \dfrac{P_{2} }{0.0888 \; Torr} \right) &=3.86 \\[4pt] \dfrac{P_{2} }{0.0888 \; Torr} &=e^{3.86} = 47.5 \\[4pt] P_{2} &= 4.21 Torr \end{align*} \nonumber \] At 160°C, liquid Hg has a vapor pressure of 4.21 torr, substantially greater than the pressure at 80.0°C, as we would expect. The vapor pressure of liquid nickel at 1606°C is 0.100 torr, whereas at 1805°C, its vapor pressure is 1.000 torr. At what temperature does the liquid have a vapor pressure of 2.500 torr? 1896°C As the temperature of a liquid increases, the vapor pressure of the liquid increases until it equals the external pressure, or the atmospheric pressure in the case of an open container. Bubbles of vapor begin to form throughout the liquid, and the liquid begins to boil. The temperature at which a liquid boils at exactly 1 atm pressure is the normal boiling point of the liquid. For water, the normal boiling point is exactly 100°C. The normal boiling points of the other liquids in Figure \(\Page {4}\) are represented by the points at which the vapor pressure curves cross the line corresponding to a pressure of 1 atm. Although we usually cite the normal boiling point of a liquid, the boiling point depends on the pressure. At a pressure greater than 1 atm, water boils at a temperature greater than 100°C because the increased pressure forces vapor molecules above the surface to condense. Hence the molecules must have greater kinetic energy to escape from the surface. Conversely, at pressures less than 1 atm, water boils below 100°C. Typical variations in atmospheric pressure at sea level are relatively small, causing only minor changes in the boiling point of water. For example, the highest recorded atmospheric pressure at sea level is 813 mmHg, recorded during a Siberian winter; the lowest sea-level pressure ever measured was 658 mmHg in a Pacific typhoon. At these pressures, the boiling point of water changes minimally, to 102°C and 96°C, respectively. At high altitudes, on the other hand, the dependence of the boiling point of water on pressure becomes significant. Table \(\Page {1}\) lists the boiling points of water at several locations with different altitudes. At an elevation of only 5000 ft, for example, the boiling point of water is already lower than the lowest ever recorded at sea level. The lower boiling point of water has major consequences for cooking everything from soft-boiled eggs (a “three-minute egg” may well take four or more minutes in the Rockies and even longer in the Himalayas) to cakes (cake mixes are often sold with separate high-altitude instructions). Conversely, pressure cookers, which have a seal that allows the pressure inside them to exceed 1 atm, are used to cook food more rapidly by raising the boiling point of water and thus the temperature at which the food is being cooked. As pressure increases, the boiling point of a liquid increases and vice versa. Use Figure \(\Page {4}\) to estimate the following. Data in Figure \(\Page {4}\), pressure, and boiling point corresponding boiling point and pressure Ethylene glycol is an organic compound primarily used as a raw material in the manufacture of polyester fibers and fabric industry, and polyethylene terephthalate resins (PET) used in bottling. Use the data in Figure \(\Page {4}\) to estimate the following. 200°C 450 mmHg Because the molecules of a liquid are in constant motion and possess a wide range of kinetic energies, at any moment some fraction of them has enough energy to escape from the surface of the liquid to enter the gas or vapor phase. This process, called or , generates a above the liquid. Molecules in the gas phase can collide with the liquid surface and reenter the liquid via . Eventually, a is reached in which the number of molecules evaporating and condensing per unit time is the same, and the system is in a state of . Under these conditions, a liquid exhibits a characteristic that depends only on the temperature. We can express the nonlinear relationship between vapor pressure and temperature as a linear relationship using the . This equation can be used to calculate the enthalpy of vaporization of a liquid from its measured vapor pressure at two or more temperatures. are liquids with high vapor pressures, which tend to evaporate readily from an open container; have low vapor pressures. When the vapor pressure equals the external pressure, bubbles of vapor form within the liquid, and it boils. The temperature at which a substance boils at a pressure of 1 atm is its . ( ) | 14,787 | 1,614 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Advanced_Thermodynamics/8._Phase_Transitions |
When chemical reactions occur, the system makes transition among multiple minima at the molecular level. The figure below illustrates the molecular connection between free energy \(G\), its derivative \(\Delta G\), and the free energy as a function of reaction coordinate, \(G(x)\). The free energy \(G(x)\) as a function of coordinate (e.g. bond distance) has two local minima, A’ being lower in free energy. When A converts to A’, \(\Delta G\) kJ/mole are released per infinitesimal amount of reaction \(d\xi\). Note that both x and \(\xi\) can vary between 0 and 1, but the meaning is very different: when all of the substance is in state A, \(x=\xi=0\), and when it is all in state A', \(x=\xi=1\). However, while the substance can have \(\xi=0.5\) when the reaction is half completed, almost none of it will ever be at x=0.5. Rather, half will be at x=1 and half at x=1. So far, we have treated macroscopic pure substances and mixtures as though they had a single minimum in the free energy as a function of reaction coordinate. However, thermodynamics does not forbid multiple minima even at the macroscopic level, and can be used to make comparative statements about the minima. A phase is a local minimum in the free energy surface. Unlike ordinary chemical reactions, transition between phases can occur even when only one pure substance is present: \[ A^{(1)} \rightarrow A^{(2)} \] For phase transitions, we call the reaction coordinate “order parameter.” The superscripts refer to the phases. An order parameter is a thermodynamic variable scaled to zero in one phase, nonzero in an(other) phase. Example: a gas-liquid transition order parameter \[ O = \rho - \rho_{gas}\] or \[ O = \dfrac{ \rho - \rho_{gas}}{ \rho_{liq} - \rho_{gas}}\] in general \[ O = X - X^{(2)}\] or \[ O = \dfrac{ X - X^{(2)}}{ X^{(2)} - X^{(1)}}\] Thermodynamics cannot make statements about the details of the barrier (e.g. its height), or how fast a transition can occur. The transition itself is a rather delicate matter – it violates P2 since temporarily \(\Delta G >0\) if the transition occurs at constant \(T\) and \(P\). The solution to this dilemma: if climbing the barrier were required of the entire macroscopic system, phase transitions could indeed occur. Rather, a small portion of phase (1), called a nucleus, fluctuates to look like phase (2). The nucleus is at the barrier top in fig. 8.1. From this nucleus, phase (2) grows downhill in chemical potential if it is at lower free energy (P2). Thus the transition itself relies on microscopic fluctuations, and microscopic information is required to determine the barrier height, which is rather small. We need statistical mechanics to compute rates. If we are interested only in equilibrium, not how we get there, we can treat the phase transition like any other chemical reaction: \(A^{(1)}\) and \(A^{(2)}\) interconvert to yield mole numbers \(n^{(1)}_{eq}\) & \(n^{(2)}_{eq}\) or concentrations or pressures that minimize \(G\): \[ A{(1)} \rightarrow A^{(2)} \] \[G(T,P,n^{(i)})=\mu^{(1)}n^{(1)} +\mu^{(2)}n^{(2)} = \mu^{(1)}n^{(1)} + \mu^{(2)}(n-n^{(1)}) \] \[ dG = 0\] \[ \mu^{(1)} n^{(1)} +\mu^{(2)} n^{(2)} = (\mu^{(1)}-\mu^{(2)}) dn^{(2)} \] \[ \mu^{(1)} = \mu^{(2)}\] : Chemical potential of two phases as a function of temperature and pressure (Gibbs ensemble). At high \(T\), phase 1 is more stable, at (3)-(5) both phases coexist, at low \(T\) phase 2 is more stable. In this diagram at high T and P, the two chemical potentials become degenerate and only one phase exists at (6). A critical phase transition occurs when the chemical potential barrier between two phases just vanishes. | 3,672 | 1,615 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_6%3A_Materials/21%3A_Structure_and_Bonding_in_Solids/21.4%3A_Defects_and_Amorphous_Solids |
Few, if any, crystals are in that all unit cells consist of the ideal arrangement of atoms or molecules and all cells line up in a three dimensional space with no distortion. Some cells may have one or more atoms less whereas others may have one or more atoms than the ideal unit cell. The imperfection of crystals are called . Crystal defects are results of thermodynamic equilibrium contributed also by the increase in entropy term of the Gibb's free energy: \[\Delta G = \Delta H - T \Delta S\] Only at the unattainable absolute zero K will a crystal be perfect, in other words, no crystals are absolutely perfect. However, the degree of imperfection vary from compound to compound. On the other hand, some solid-like structure called also exist in a liquid. For example, the density of water is the highest at 277 K. The flickering clusters increase as temperature drops below 277 K, and the water density decreases as a result. The missing and lacking of atoms or ions in an ideal or imaginary crystal structure or lattice and the misalignment of unit cells in real crystals are called crystal defects or solid defects. Crystal defects occur as points, along lines, or in the form of a surface, and they are called respectively. Point defects can be divided into Frenkel defects and , and these often occur in ionic crystals. The former are due to misplacement of ions and vacancies. Charges are balanced in the whole crystal despite the presence of interstitial or extra ions and vacancies. On the other hand, when only vacancies of cation and anions are present with no interstitial or misplaced ions, the defects are called Schottky defects. Point defects are common in crystals with large anions such as AgBr, AgI, RbAgI . Due to the defects, the ions have some freedom to move about in crystals, making them relatively good conductors. These are called ionic conductors, unlike metals in which electrons are responsible for electric conductivity. Recently, ionic conductors have attracted a lot of attention because the fuel cell and battery technologies require conducting solids to separate the electrodes. Line defects are mostly due to misalignment of ions or presence of vacancies along a line. When lines of ions are missing in an otherwise perfect array of ions, an appeared. Edge dislocation is responsible for the ductility and malleability. In fact the hammering and stretching of materials often involve the movement of edge dislocation. Movements of dislocations give rise to their plastic behavior. Line dislocations usually do not end inside the crystal, and they either form loops or end at the surface of a single crystal. A dislocation is characterized by its Burgers vector: If you imagine going around the dislocation line, and exactly going back as many atoms in each direction as you have gone forward, you will not come back to the same atom where you have started. The Burgers vector points from start atom to the end atom of your journey (This "journey" is called Burgers circuit in dislocation theory). In this electron microscope image of the surface of a crystal, you see point defects and a Burger journey around an edge dislocation. The dislocation line is in the crystal, and the image shows its ending at the surface. A Burger vector is approximately perpendicular to the dislocation line, and the missing line of atoms is somewhere within the block of the Buerger journey. If the misalignment shifts a block of ions gradually downwards or upwards causing the formation of a screw like deformation, a is formed. The diagram here shows the idealized screw dislocation. Line defects weakens the structure along a one-dimensional space, and the defects type and density affects the mechanical properties of the solids. Thus, formation and study of dislocations are particularly important for structural materials such as metals. This link gives some impressive images of dislocations. Chemical etching often reveal pits which are visible under small magnifications. The Table of X-ray Crystallographic Data of Minerals (The CRC Handbook of Chemistry and Physics) list the following for bunsenite (NiO): Crystal system: cubic, structure type: rock salt, a = 4.177*10-8 cm. In the table of Physical Constant of Inorganic Compounds, the density of bunsenite (NiO) is 6.67g / cm3. From these values, evaluate the cell volume (volume of the unit cell), sum of Ni and O radii (rNi + rO), 2 Molar volumes, (X-ray) density, and the Schottky defect vacancy rate.
Actually, most of the required values have been listed in the table, but their evaluations illustrate the methods. These values are evaluated below: Cell volume = a3
= (4.177*10-8)3
= 72.88*10-24 cm3 \[ r_{Ni}+r_O = \dfrac{a}{ 2}\]
\[ = 2.088 \times 10^{-8}\] X-ray density = 4*(58.69+16.00) / (6.023*10 *72.88*10 )
= 6.806 g / cm , Compared to the observed density = 6.67 g / cm . The molar volumes (58.69+16.00) / density are thus 74.69 / 6.806 = 10.97 cm ; and 74.69 / 6.67 = 11.20 cm The vacancy rate = 6.806 - 6.67 / 6.806
= 0.02 (or 2%) These methods are hints to assignments. From the given conditions, we cannot calculate individual radii of Ni and O, but their sum is calculable. Plane defects occur along a 2-dimensional surface. The of a crystal is an obvious imperfection, because these surface atoms are different from those deep in the crystals. When a solid is used as a catalyst, the catalytic activity depends very much on the surface area per unit mass of the sample. For these powdery material, methods have been developed for the determination of unit areas per unit mass. Another surface defects are along the . A grain is a single crystal. If many seeds are formed when a sample starts to crystallize, each seed grow until they meet at the boundaries. Properties along these boundaries are different from the grains. A third plane defects are the . For example, in the close packing arrangement, the adjancent layers always have the AB relationship. In a ccp (fcc) close packing sequence, ...ABCABC..., one of the layer may suddenly be out of sequence, and become ..ABABCABC.... Similarly, in the hcp sequence, there is a possibility that one of the layer accidentally startes in the C location and resulting in the formation of a grain boundary. You already know that to obtain a perfectly pure substance is almost impossible. Purification is a costly process. In general, analytical reagent-grade chemicals are of high purity, and yet few of them are better than 99.9% pure. This means that a foreign atom or molecule is present for every 1000 host atoms or molecules in the crystal. Perhaps the most demanding of purity is in the electronic industry. Silicon crystals of 99.999 (called 5 nines) or better are required for IC chips productions. These crystal are with nitrogen group elements of P and As or boron group elements B, Al etc to form n- aand p-type semiconductors. In these crystals, the impurity atom substitute atoms of the host crystals. Presence minute foreign atoms with one electron more or less than the valence four silicon and germanium host atoms is the key of making n- and p-type semiconductors. Having many semiconductors connected in a single chip makes the integrated circuit a very efficient information processor. The electronic properties change dramatically due to these impurities. This is further described in Inorganic Chemistry by Swaddle. In other bulk materials, the presence of impurity usually leads to a lowering of melting point. For example, Hall and Heroult tried to electrolyze natural aluminum compounds. They discovered that using a 5% mixture of Al O (melting point 273 K) in cryolite Na AlF (melting point 1273 K) reduced the melting point to 1223 K, and that enabled the production of aluminum in bulk. Recent modifications lowered melting temperatures below 933 K. Some types of glass are made by mixing silica (SiO ), alumina (Al O ), calcium oxide (CaO), and sodium oxide (Na O). They are softer, but due to lower melting points, they are cheaper to produce. Color centers are imperfections in crystals that cause color (defects that cause color by absorption of light). Due to defects, metal oxides may also act as semiconductors, because there are many different types of electron traps. Electrons in defect region only absorb light at certain range of wavelength. The color seen are due to lights not absorbed. For example, a diamond with C vacancies (missing carbon atoms) absorbs light, and these centers give green color as shown here. Replacement of Al for Si in quartz give rise to the color of smoky quartz. A high temperature phase of ZnO , (x < 1), has electrons in place of the O vacancies. These electrons are color centers, often referred to as -centers (from the German word meaning color). Similarly, heating of ZnS to 773 K causes a loss of sulfur, and these material fluoresces strongly in ultraviolet light. Some non-stoichiometric solids are engineered to be n-type or p-type semiconductors. Nickel oxide NiO gain oxygen on heating in air, resulting in having Ni sites acting as electron trap, a p-type semiconductor. On the other hand, ZnO lose oxygen on heating, and the excess Zn metal atoms in the sample are ready to give electrons. The solid is an n-type semiconductor.
Correlate properties of a material to its structure.
Explain Frendel and Schottky defects.
Correlate properties of a material to its structure.
Specify the desirable properties of a material. | 9,547 | 1,616 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_4%3A_Equilibrium_in_Chemical_Reactions/15%3A_AcidBase_Equilibria/15.5%3A_Buffer_Solutions |
Buffers are solutions that maintain a relatively constant pH when an acid or a base is added. They therefore protect, or “buffer,” other molecules in solution from the effects of the added acid or base. Buffers contain either a weak acid (\(HA\)) and its conjugate base \((A^−\)) or a weak base (\(B\)) and its conjugate acid (\(BH^+\)), and they are critically important for the proper functioning of biological systems. In fact, every biological fluid is buffered to maintain its physiological pH. To understand how buffers work, let’s look first at how the ionization equilibrium of a weak acid is affected by adding either the conjugate base of the acid or a strong acid (a source of \(\ce{H^{+}}\)). Le Chatelier’s principle can be used to predict the effect on the equilibrium position of the solution. A typical buffer used in biochemistry laboratories contains acetic acid and a salt such as sodium acetate. The dissociation reaction of acetic acid is as follows: \[\ce{CH3COOH (aq) <=> CH3COO^{−} (aq) + H^{+} (aq)} \label{Eq1} \] and the equilibrium constant expression is as follows: \[K_a=\dfrac{[\ce{H^{+}},\ce{CH3COO^{-}}]}{[\ce{CH3CO2H}]} \label{Eq2} \] Sodium acetate (\(\ce{CH_3CO_2Na}\)) is a strong electrolyte that ionizes completely in aqueous solution to produce \(\ce{Na^{+}}\) and \(\ce{CH3CO2^{−}}\) ions. If sodium acetate is added to a solution of acetic acid, Le Chatelier’s principle predicts that the equilibrium in Equation \ref{Eq1} will shift to the left, consuming some of the added \(\ce{CH_3COO^{−}}\) and some of the \(\ce{H^{+}}\) ions originally present in solution. Because \(\ce{Na^{+}}\) is a , it has no effect on the position of the equilibrium and can be ignored. The addition of sodium acetate produces a new equilibrium composition, in which \([\ce{H^{+}}]\) is less than the initial value. Because \([\ce{H^{+}}]\) has decreased, the pH will be higher. Thus adding a salt of the conjugate base to a solution of a weak acid increases the pH. This makes sense because sodium acetate is a base, and adding any base to a solution of a weak acid should increase the pH. If we instead add a strong acid such as \(\ce{HCl}\) to the system, \([\ce{H^{+}}]\) increases. Once again the equilibrium is temporarily disturbed, but the excess \(\ce{H^{+}}\) ions react with the conjugate base (\(\ce{CH_3CO_2^{−}}\)), whether from the parent acid or sodium acetate, to drive the equilibrium to the left. The net result is a new equilibrium composition that has a lower [\(\ce{CH_3CO_2^{−}}\)] than before. In both cases, only the equilibrium composition has changed; the ionization constant \(K_a\) for acetic acid remains the same. Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium, in this case \(\ce{CH3CO2^{−}}\), will therefore shift the equilibrium in the direction that reduces the concentration of the common ion. The shift in equilibrium is via the common ion effect. Adding a common ion to a system at equilibrium affects the equilibrium composition, but not the ionization constant. A 0.150 M solution of formic acid at 25°C (pKa = 3.75) has a pH of 2.28 and is 3.5% ionized. : solution concentration and pH, \(pK_a\), and percent ionization of acid; final concentration of conjugate base or strong acid added : pH and percent ionization of formic acid : : A Because sodium formate is a strong electrolyte, it ionizes completely in solution to give formate and sodium ions. The \(\ce{Na^{+}}\) ions are spectator ions, so they can be ignored in the equilibrium equation. Because water is both a much weaker acid than formic acid and a much weaker base than formate, the acid–base properties of the solution are determined solely by the formic acid ionization equilibrium: \[\ce{HCO2H (aq) <=> HCO^{−}2 (aq) + H^{+} (aq)} \nonumber \] The initial concentrations, the changes in concentration that occur as equilibrium is reached, and the final concentrations can be tabulated. We substitute the expressions for the final concentrations into the equilibrium constant expression and make our usual simplifying assumptions, so \[\begin{align*} K_a=\dfrac{[H^+,HCO_2^−]}{[HCO_2H]} &=\dfrac{(x)(0.100+x)}{0.150−x} \\[4pt] &\approx \dfrac{x(0.100)}{0.150} \\[4pt] &\approx 10^{−3.75} \\[4pt] &\approx 1.8 \times 10^{−4} \end{align*} \nonumber \] Rearranging and solving for \(x\), \[\begin{align*} x &=(1.8 \times 10^{−4}) \times \dfrac{0.150 \;M}{ 0.100 \;M} \\[4pt] &=2.7 \times 10^{−4}\\[4pt] &=[H^+] \end{align*} \nonumber \] The value of \(x\) is small compared with 0.150 or 0.100 M, so our assumption about the extent of ionization is justified. Moreover, \[K_aC_{HA} = (1.8 \times 10^{−4})(0.150) = 2.7 \times 10^{−5} \nonumber \] which is greater than \(1.0 \times 10^{−6}\), so again, our assumption is justified. The final pH is: \[pH= −\log(2.7 \times 10^{−4}) = 3.57 \nonumber \] compared with the initial value of 2.29. Thus adding a salt containing the conjugate base of the acid has increased the pH of the solution, as we expect based on Le Chatelier’s principle; the stress on the system has been relieved by the consumption of \(\ce{H^{+}}\) ions, driving the equilibrium to the left. Because \(HCl\) is a strong acid, it ionizes completely, and chloride is a spectator ion that can be neglected. Thus the only relevant acid–base equilibrium is again the dissociation of formic acid, and initially the concentration of formate is zero. We can construct a table of initial concentrations, changes in concentration, and final concentrations. \[HCO_2H (aq) \leftrightharpoons H^+ (aq) +HCO^−_2 (aq) \nonumber \] To calculate the percentage of formic acid that is ionized under these conditions, we have to determine the final \([\ce{HCO2^{-}}]\). We substitute final concentrations into the equilibrium constant expression and make the usual simplifying assumptions, so \[K_a=\dfrac{[H^+,HCO_2^−]}{[HCO_2H]}=\dfrac{(0.200+x)(x)}{0.150−x} \approx \dfrac{x(0.200)}{0.150}=1.80 \times 10^{−4} \nonumber \] Rearranging and solving for \(x\), \[\begin{align*} x &=(1.80 \times 10^{−4}) \times \dfrac{ 0.150\; M}{ 0.200\; M} \\[4pt] &=1.35 \times 10^{−4}=[HCO_2^−] \end{align*} \nonumber \] Once again, our simplifying assumptions are justified. The percent ionization of formic acid is as follows: \[\text{percent ionization}=\dfrac{1.35 \times 10^{−4} \;M} {0.150\; M} \times 100\%=0.0900\% \nonumber \] Adding the strong acid to the solution, as shown in the table, decreased the percent ionization of formic acid by a factor of approximately 38 (3.45%/0.0900%). Again, this is consistent with Le Chatelier’s principle: adding \(\ce{H^{+}}\) ions drives the dissociation equilibrium to the left. A 0.225 M solution of ethylamine (\(\ce{CH3CH2NH2}\) with \(pK_b = 3.19\)) has a pH of 12.08 and a percent ionization of 5.4% at 20°C. Calculate the following: 11.16 1.3% A Discussing the Common Ion Effect: Now let’s suppose we have a buffer solution that contains equimolar concentrations of a weak base (\(B\)) and its conjugate acid (\(BH^+\)). The general equation for the ionization of a weak base is as follows: \[B (aq) +H_2O (l) \leftrightharpoons BH^+ (aq) +OH^− (aq) \label{Eq3} \] If the equilibrium constant for the reaction as written in Equation \(\ref{Eq3}\) is small, for example \(K_b = 10^{−5}\), then the equilibrium constant for the reverse reaction is very large: \(K = \dfrac{1}{K_b} = 10^5\). Adding a strong base such as \(OH^-\) to the solution therefore causes the equilibrium in Equation \(\ref{Eq3}\) to shift to the left, consuming the added \(OH^-\). As a result, the \(OH^-\) ion concentration in solution remains relatively constant, and the pH of the solution changes very little. Le Chatelier’s principle predicts the same outcome: when the system is stressed by an increase in the \(OH^-\) ion concentration, the reaction will proceed to the left to counteract the stress. If the \(pK_b\) of the base is 5.0, the \(pK_a\) of its conjugate acid is \[pK_a = pK_w − pK_b = 14.0 – 5.0 = 9.0. \nonumber \] Thus the equilibrium constant for ionization of the conjugate acid is even smaller than that for ionization of the base. The ionization reaction for the conjugate acid of a weak base is written as follows: \[BH^+ (aq) +H_2O (l) \leftrightharpoons B (aq) +H_3O^+ (aq) \label{Eq4} \] Again, the equilibrium constant for the reverse of this reaction is very large: K = 1/K = 10 . If a strong acid is added, it is neutralized by reaction with the base as the reaction in Equation \(\ref{Eq4}\) shifts to the left. As a result, the \(H^+\) ion concentration does not increase very much, and the pH changes only slightly. In effect, a buffer solution behaves somewhat like a sponge that can absorb \(H^+\) and \(OH^-\) ions, thereby preventing large changes in pH when appreciable amounts of strong acid or base are added to a solution. Buffers are characterized by the pH range over which they can maintain a more or less constant pH and by their buffer capacity, the amount of strong acid or base that can be absorbed before the pH changes significantly. Although the useful pH range of a buffer depends strongly on the chemical properties of the weak acid and weak base used to prepare the buffer (i.e., on \(K\)), its buffer capacity depends solely on the concentrations of the species in the buffered solution. The more concentrated the buffer solution, the greater its buffer capacity. As illustrated in Figure \(\Page {1}\), when \(NaOH\) is added to solutions that contain different concentrations of an acetic acid/sodium acetate buffer, the observed change in the pH of the buffer is inversely proportional to the concentration of the buffer. If the buffer capacity is 10 times larger, then the buffer solution can absorb 10 times more strong acid or base before undergoing a significant change in pH. A buffer maintains a relatively constant pH when acid or base is added to a solution. The addition of even tiny volumes of 0.10 M \(NaOH\) to 100.0 mL of distilled water results in a very large change in pH. As the concentration of a 50:50 mixture of sodium acetate/acetic acid buffer in the solution is increased from 0.010 M to 1.00 M, the change in the pH produced by the addition of the same volume of \(NaOH\) solution decreases steadily. For buffer concentrations of at least 0.500 M, the addition of even 25 mL of the \(NaOH\) solution results in only a relatively small change in pH. The pH of a buffer can be calculated from the concentrations of the weak acid and the weak base used to prepare it, the concentration of the conjugate base and conjugate acid, and the \(pK_a\) or \(pK_b\) of the weak acid or weak base. The procedure is analogous to that used in Example \(\Page {1}\) to calculate the pH of a solution containing known concentrations of formic acid and formate. An alternative method frequently used to calculate the pH of a buffer solution is based on a rearrangement of the equilibrium equation for the dissociation of a weak acid. The simplified ionization reaction is \(HA \leftrightharpoons H^+ + A^−\), for which the equilibrium constant expression is as follows: \[K_a=\dfrac{[H^+,A^-]}{[HA]} \label{Eq5} \] This equation can be rearranged as follows: \[[H^+]=K_a\dfrac{[HA]}{[A^−]} \label{Eq6} \] Taking the logarithm of both sides and multiplying both sides by −1, \[ \begin{align} −\log[H^+] &=−\log K_a−\log\left(\dfrac{[HA]}{[A^−]}\right) \\[4pt] &=−\log{K_a}+\log\left(\dfrac{[A^−]}{[HA]}\right) \label{Eq7} \end{align} \] Replacing the negative logarithms in Equation \(\ref{Eq7}\), \[pH=pK_a+\log \left( \dfrac{[A^−]}{[HA]} \right) \label{Eq8} \] or, more generally, \[pH=pK_a+\log\left(\dfrac{[base]}{[acid]}\right) \label{Eq9} \] Equation \(\ref{Eq8}\) and Equation \(\ref{Eq9}\) are both forms of the Henderson-Hasselbalch approximation, named after the two early 20th-century chemists who first noticed that this rearranged version of the equilibrium constant expression provides an easy way to calculate the pH of a buffer solution. In general, the validity of the Henderson-Hasselbalch approximation may be limited to solutions whose concentrations are at least 100 times greater than their \(K_a\) values. There are three special cases where the Henderson-Hasselbalch approximation is easily interpreted without the need for calculations: Each time we increase the [base]/[acid] ratio by 10, the pH of the solution increases by 1 pH unit. Conversely, if the [base]/[acid] ratio is 0.1, then pH = \(pK_a\) − 1. Each additional factor-of-10 decrease in the [base]/[acid] ratio causes the pH to decrease by 1 pH unit. If [base] = [acid] for a buffer, then pH = \(pK_a\). Changing this ratio by a factor of 10 either way changes the pH by ±1 unit. What is the pH of a solution that contains : concentration of acid, conjugate base, and \(pK_a\); concentration of base, conjugate acid, and \(pK_b\) : pH : Substitute values into either form of the Henderson-Hasselbalch approximation (Equations \ref{Eq8} or \ref{Eq9}) to calculate the pH. : According to the Henderson-Hasselbalch approximation (Equation \ref{Eq8}), the pH of a solution that contains both a weak acid and its conjugate base is \[pH = pK_a + \log([A−]/[HA]). \nonumber \] Inserting the given values into the equation, \[\begin{align*} pH &=3.75+\log\left(\dfrac{0.215}{0.135}\right) \\[4pt] &=3.75+\log 1.593 \\[4pt] &=3.95 \end{align*} \nonumber \] This result makes sense because the \([A^−]/[HA]\) ratio is between 1 and 10, so the pH of the buffer must be between the \(pK_a\) (3.75) and \(pK_a + 1\), or 4.75. This is identical to part (a), except for the concentrations of the acid and the conjugate base, which are 10 times lower. Inserting the concentrations into the Henderson-Hasselbalch approximation, \[\begin{align*} pH &=3.75+\log\left(\dfrac{0.0215}{0.0135}\right) \\[4pt] &=3.75+\log 1.593 \\[4pt] &=3.95 \end{align*} \nonumber \] This result is identical to the result in part (a), which emphasizes the point that the pH of a buffer depends only on the ratio of the concentrations of the conjugate base and the acid, not on the magnitude of the concentrations. Because the [A ]/[HA] ratio is the same as in part (a), the pH of the buffer must also be the same (3.95). In this case, we have a weak base, pyridine (Py), and its conjugate acid, the pyridinium ion (\(HPy^+\)). We will therefore use Equation \(\ref{Eq9}\), the more general form of the Henderson-Hasselbalch approximation, in which “base” and “acid” refer to the appropriate species of the conjugate acid–base pair. We are given [base] = [Py] = 0.119 M and \([acid] = [HPy^{+}] = 0.234\, M\). We also are given \(pK_b = 8.77\) for pyridine, but we need \(pK_a\) for the pyridinium ion. Recall from Equation 16.23 that the \(pK_b\) of a weak base and the \(pK_a\) of its conjugate acid are related: \[pK_a + pK_b = pK_w. \nonumber \] Thus \(pK_a\) for the pyridinium ion is \(pK_w − pK_b = 14.00 − 8.77 = 5.23\). Substituting this \(pK_a\) value into the Henderson-Hasselbalch approximation, \[\begin{align*} pH=pK_a+\log \left(\dfrac{[base]}{[acid]}\right) \\[4pt] &=5.23+\log\left(\dfrac{0.119}{0.234}\right) \\[4pt] & =5.23 −0.294 \\[4pt] &=4.94 \end{align*} \nonumber \] Once again, this result makes sense: the \([B]/[BH^+]\) ratio is about 1/2, which is between 1 and 0.1, so the final pH must be between the \(pK_a\) (5.23) and \(pK_a − 1\), or 4.23. What is the pH of a solution that contains The \(pK_a\) of benzoic acid is 4.20, and the \(pK_b\) of trimethylamine is also 4.20. 4.08 9.68 A Discussing Using the Henderson Hasselbalch Equation: The Henderson-Hasselbalch approximation ((Equation \(\ref{Eq8}\)) can also be used to calculate the pH of a buffer solution after adding a given amount of strong acid or strong base, as demonstrated in Example \(\Page {3}\). The buffer solution in Example \(\Page {2}\) contained 0.135 M \(\ce{HCO2H}\) and 0.215 M \(\ce{HCO2Na}\) and had a pH of 3.95. : composition and pH of buffer; concentration and volume of added acid or base : final pH : The added \(\ce{HCl}\) (a strong acid) or \(\ce{NaOH}\) (a strong base) will react completely with formate (a weak base) or formic acid (a weak acid), respectively, to give formic acid or formate and water. We must therefore calculate the amounts of formic acid and formate present after the neutralization reaction. We begin by calculating the millimoles of formic acid and formate present in 100 mL of the initial pH 3.95 buffer: \[ 100 \, \cancel{mL} \left( \dfrac{0.135 \, mmol\; \ce{HCO2H}}{\cancel{mL}} \right) = 13.5\, mmol\, \ce{HCO2H} \nonumber \] \[ 100\, \cancel{mL } \left( \dfrac{0.215 \, mmol\; \ce{HCO2^{-}}}{\cancel{mL}} \right) = 21.5\, mmol\, \ce{HCO2^{-}} \nonumber \] The millimoles of \(\ce{H^{+}}\) in 5.00 mL of 1.00 M \(\ce{HCl}\) is as follows: \[ 5.00 \, \cancel{mL } \left( \dfrac{1.00 \,mmol\; \ce{H^{+}}}{\cancel{mL}} \right) = 5\, mmol\, \ce{H^{+}} \nonumber \] Next, we construct a table of initial amounts, changes in amounts, and final amounts: \[\ce{HCO^{2−}(aq) + H^{+} (aq) <=> HCO2H (aq)} \nonumber \] The final amount of \(H^+\) in solution is given as “∼0 mmol.” For the purposes of the stoichiometry calculation, this is essentially true, but remember that the point of the problem is to calculate the final \([H^+]\) and thus the pH. We now have all the information we need to calculate the pH. We can use either the lengthy procedure of Example \(\Page {1}\) or the Henderson–Hasselbach approximation. Because we have performed many equilibrium calculations in this chapter, we’ll take the latter approach. The Henderson-Hasselbalch approximation requires the concentrations of \(HCO_2^−\) and \(HCO_2H\), which can be calculated using the number of millimoles (\(n\)) of each and the total volume (\(VT\)). Substituting these values into the Henderson-Hasselbalch approximation (Equation \(\ref{Eq9}\)): \[\begin{align*} pH &=pK_a+\log \left( \dfrac{[HCO_2^−]}{[HCO_2H]} \right) \\[4pt] &=pK_a+\log\left(\dfrac{n_{HCO_2^−}/V_f}{n_{HCO_2H}/V_f}\right) \\[4pt] &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \end{align*} \nonumber \] Because the total volume appears in both the numerator and denominator, it cancels. We therefore need to use only the ratio of the number of millimoles of the conjugate base to the number of millimoles of the weak acid. So \[\begin{align*} pH &=pK_a+\log\left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \\[4pt] &=3.75+\log\left(\dfrac{16.5\; mmol}{18.5\; mmol}\right) \\[4pt] &=3.75 −0.050=3.70 \end{align*} \nonumber \] Once again, this result makes sense on two levels. First, the addition of \(HCl \)has decreased the pH from 3.95, as expected. Second, the ratio of \(HCO_2^−\) to \(HCO_2H\) is slightly less than 1, so the pH should be between the \(pK_a\) and \(pK_a\) − 1. The procedure for solving this part of the problem is exactly the same as that used in part (a). We have already calculated the numbers of millimoles of formic acid and formate in 100 mL of the initial pH 3.95 buffer: 13.5 mmol of \(HCO_2H\) and 21.5 mmol of \(HCO_2^−\). The number of millimoles of \(OH^-\) in 5.00 mL of 1.00 M \(NaOH\) is as follows: With this information, we can construct a table of initial amounts, changes in amounts, and final amounts. \[\ce{HCO2H (aq) + OH^{−} (aq) <=> HCO^{−}2 (aq) + H2O (l)} \nonumber \] The final amount of \(OH^-\) in solution is not actually zero; this is only approximately true based on the stoichiometric calculation. We can calculate the final pH by inserting the numbers of millimoles of both \(HCO_2^−\) and \(HCO_2H\) into the simplified Henderson-Hasselbalch expression used in part (a) because the volume cancels: \[\begin{align*} pH &=pK_a+\log \left(\dfrac{n_{HCO_2^−}}{n_{HCO_2H}}\right) \\[4pt] &=3.75+\log \left(\dfrac{26.5\; mmol}{8.5\; mmol} \right) \\[4pt] &=3.75+0.494 =4.24 \end{align*} \nonumber \] Once again, this result makes chemical sense: the pH has increased, as would be expected after adding a strong base, and the final pH is between the \(pK_a\) and \(pK_a\) + 1, as expected for a solution with a \(HCO_2^−/HCO_2H\) ratio between 1 and 10. The buffer solution from Example \(\Page {2}\) contained 0.119 M pyridine and 0.234 M pyridine hydrochloride and had a pH of 4.94. 5.30 4.42 Only the amounts (in moles or millimoles) of the acidic and basic components of the buffer are needed to use the Henderson-Hasselbalch approximation, not their concentrations. A Discussing the Change in pH with the Addition of a Strong Acid to a Buffer: The Change in pH with the Addition of a Strong Base to a Buffer: The results obtained in Example \(\Page {3}\) and its corresponding exercise demonstrate how little the pH of a well-chosen buffer solution changes despite the addition of a significant quantity of strong acid or strong base. Suppose we had added the same amount of \(HCl\) or \(NaOH\) solution to 100 mL of an unbuffered solution at pH 3.95 (corresponding to \(1.1 \times 10^{−4}\) M HCl). In this case, adding 5.00 mL of 1.00 M \(HCl\) would lower the final pH to 1.32 instead of 3.70, whereas adding 5.00 mL of 1.00 M \(NaOH\) would raise the final pH to 12.68 rather than 4.24. (Try verifying these values by doing the calculations yourself.) Thus the presence of a buffer significantly increases the ability of a solution to maintain an almost constant pH. The most effective buffers contain concentrations of an acid and its conjugate base. A buffer that contains approximately equal amounts of a weak acid and its conjugate base in solution is equally effective at neutralizing either added base or added acid. This is shown in Figure \(\Page {2}\) for an acetic acid/sodium acetate buffer. Adding a given amount of strong acid shifts the system along the horizontal axis to the left, whereas adding the same amount of strong base shifts the system the same distance to the right. In either case, the change in the ratio of \(CH_3CO_2^−\) to \(CH_3CO_2H\) from 1:1 reduces the buffer capacity of the solution. A Discussing The Buffer Region: There is a strong correlation between the effectiveness of a buffer solution and the titration curves discussed in Section 16.5. Consider the schematic titration curve of a weak acid with a strong base shown in Figure \(\Page {3}\). As indicated by the labels, the region around \(pK_a\) corresponds to the midpoint of the titration, when approximately half the weak acid has been neutralized. This portion of the titration curve corresponds to a buffer: it exhibits the smallest change in pH per increment of added strong base, as shown by the nearly horizontal nature of the curve in this region. The nearly flat portion of the curve extends only from approximately a pH value of 1 unit less than the \(pK_a\) to approximately a pH value of 1 unit greater than the \(pK_a\), which is why buffer solutions usually have a pH that is within ±1 pH units of the \(pK_a\) of the acid component of the buffer. This schematic plot of pH for the titration of a weak acid with a strong base shows the nearly flat region of the titration curve around the midpoint, which corresponds to the formation of a buffer. At the lower left, the pH of the solution is determined by the equilibrium for dissociation of the weak acid; at the upper right, the pH is determined by the equilibrium for reaction of the conjugate base with water. In the region of the titration curve at the lower left, before the midpoint, the acid–base properties of the solution are dominated by the equilibrium for dissociation of the weak acid, corresponding to \(K_a\). In the region of the titration curve at the upper right, after the midpoint, the acid–base properties of the solution are dominated by the equilibrium for reaction of the conjugate base of the weak acid with water, corresponding to \(K_b\). However, we can calculate either \(K_a\) or \(K_b\) from the other because they are related by \(K_w\). Metabolic processes produce large amounts of acids and bases, yet organisms are able to maintain an almost constant internal pH because their fluids contain buffers. This is not to say that the pH is uniform throughout all cells and tissues of a mammal. The internal pH of a red blood cell is about 7.2, but the pH of most other kinds of cells is lower, around 7.0. Even within a single cell, different compartments can have very different pH values. For example, one intracellular compartment in white blood cells has a pH of around 5.0. Because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately pH 5.0 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the \(\ce{CO2}/\ce{HCO3^{−}}\) system, which dominates the buffering action of blood plasma. The acid–base equilibrium in the \(\ce{CO2}/\ce{HCO3^{−}}\) buffer system is usually written as follows: \[\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{Eq10} \] with \(K_a = 4.5 \times 10^{−7}\) and \(pK_a = 6.35\) at 25°C. In fact, Equation \(\ref{Eq10}\) is a grossly oversimplified version of the \(\ce{CO2}/\ce{HCO3^{-}}\) system because a solution of \(\ce{CO2}\) in water contains only rather small amounts of \(H_2CO_3\). Thus Equation \(\ref{Eq10}\) does not allow us to understand how blood is actually buffered, particularly at a physiological temperature of 37°C. As shown in Equation \(\ref{Eq11}\), \(\ce{CO2}\) is in equilibrium with \(\ce{H2CO3}\), but the equilibrium lies far to the left, with an \(\ce{H2CO3}/\ce{CO2}\) ratio less than 0.01 under most conditions: \[\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{Eq11} \] with \(K′ = 4.0 \times 10^{−3}\) at 37°C. The true \(pK_a\) of carbonic acid at 37°C is therefore 3.70, not 6.35, corresponding to a \(K_a\) of \(2.0 \times 10^{−4}\), which makes it a much stronger acid than Equation \ref{Eq10} suggests. Adding Equation \ref{Eq10} and Equation \ref{Eq11} and canceling \(\ce{H2CO3}\) from both sides give the following overall equation for the reaction of \(\ce{CO2}\) with water to give a proton and the bicarbonate ion: \[\ce{CO2 (aq) + H2O (l) <=> H2CO3 (aq)} \label{16.65a} \] with \(K'=4.0 \times 10^{−3} (37°C)\) \[\ce{H2CO3 (aq) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65b} \] with \(K_a=2.0 \times 10^{−4} (37°C)\) \[\ce{CO2 (aq) + H2O (l) <=> H^{+} (aq) + HCO^{-}3 (aq)} \label{16.65c} \] with \(K=8.0 \times 10^{−7} (37°C)\) The \(K\) value for the reaction in Equation \ref{16.65c} is the product of the true ionization constant for carbonic acid (\(K_a\)) and the equilibrium constant (K) for the reaction of \(\ce{CO2 (aq)} \) with water to give carbonic acid. The equilibrium equation for the reaction of \(\ce{CO2}\) with water to give bicarbonate and a proton is therefore \[K=\dfrac{[\ce{H^{+}},\ce{HCO3^{-}}]}{[\ce{CO2}]}=8.0 \times 10^{−7} \label{eq13} \] The presence of a gas in the equilibrium constant expression for a buffer is unusual. According to , \[[\ce{CO2}]=k P_{\ce{CO2}} \nonumber \] where \(k\) is the Henry’s law constant for \(\ce{CO2}\), which is \(3.0 \times 10^{−5} \;M/mmHg\) at 37°C. Substituting this expression for \([\ce{CO2}]\) in Equation \ref{eq13}, \[K=\dfrac{[\ce{H^{+}},\ce{HCO3^{-}}]}{(3.0 \times 10^{−5}\; M/mmHg)(P_{\ce{CO2}})} \nonumber \] where \(P_{\ce{CO2}}\) is in mmHg. Taking the negative logarithm of both sides and rearranging, \[pH=6.10+\log \left( \dfrac{ [\ce{HCO3^{−}}]}{(3.0 \times 10^{−5} M/mm \;Hg)\; (P_{\ce{CO2}}) } \right) \label{Eq15} \] Thus the pH of the solution depends on both the \(\ce{CO2}\) pressure over the solution and \([\ce{HCO3^{−}}]\). Figure \(\Page {4}\) plots the relationship between pH and \([\ce{HCO3^{−}}]\) under physiological conditions for several different values of \(P_{\ce{CO2}}\), with normal pH and \([\ce{HCO3^{−}}]\) values indicated by the dashed lines. According to Equation \ref{Eq15}, adding a strong acid to the \(\ce{CO2}/\ce{HCO3^{−}}\) system causes \([\ce{HCO3^{−}}]\) to decrease as \(\ce{HCO3^{−}}\) is converted to \(\ce{CO2}\). Excess \(\ce{CO2}\) is released in the lungs and exhaled into the atmosphere, however, so there is essentially no change in \(P_{\ce{CO2}}\). Because the change in \([\ce{HCO3^{−}}]/P_{CO_2}\) is small, Equation \ref{Eq15} predicts that the change in pH will also be rather small. Conversely, if a strong base is added, the \(\ce{OH^{-}}\) reacts with \(\ce{CO2}\) to form \(\ce{HCO3^{−}}\), but \(\ce{CO2}\) is replenished by the body, again limiting the change in both \([\ce{HCO3^{−}}]/P_{\ce{CO2}}\) and pH. The \(\ce{CO2}/\ce{HCO3^{−}}\) buffer system is an example of an open system, in which the total concentration of the components of the buffer change to keep the pH at a nearly constant value. If a passenger steps out of an airplane in Denver, Colorado, for example, the lower \(P_{\ce{CO2}}\) at higher elevations (typically 31 mmHg at an elevation of 2000 m versus 40 mmHg at sea level) causes a shift to a new pH and \([\ce{HCO3^{-}}]\). The increase in pH and decrease in \([\ce{HCO3^{−}}]\) in response to the decrease in \(P_{\ce{CO2}}\) are responsible for the general malaise that many people experience at high altitudes. If their blood pH does not adjust rapidly, the condition can develop into the life-threatening phenomenon known as altitude sickness. A Summary of the pH Curve for a Strong Acid/Strong Base Titration: Buffers are solutions that resist a change in pH after adding an acid or a base. Buffers contain a weak acid (\(HA\)) and its conjugate weak base (\(A^−\)). Adding a strong electrolyte that contains one ion in common with a reaction system that is at equilibrium shifts the equilibrium in such a way as to reduce the concentration of the common ion. The shift in equilibrium is called the common ion effect. Buffers are characterized by their pH range and buffer capacity. The useful pH range of a buffer depends strongly on the chemical properties of the conjugate weak acid–base pair used to prepare the buffer (the \(K_a\) or \(K_b\)), whereas its buffer capacity depends solely on the concentrations of the species in the solution. The pH of a buffer can be calculated using the Henderson-Hasselbalch approximation, which is valid for solutions whose concentrations are at least 100 times greater than their \(K_a\) values. Because no single buffer system can effectively maintain a constant pH value over the physiological range of approximately 5 to 7.4, biochemical systems use a set of buffers with overlapping ranges. The most important of these is the \(CO_2/HCO_3^−\) system, which dominates the buffering action of blood plasma. | 30,633 | 1,617 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/06%3A_Structures_and_Energetics_of_Metallic_and_Ionic_solids/6.14%3A_Lattice_Energy_-_The_Born-Haber_Cycle |
Ionic solids tend to be very stable compounds. The enthalpies of formation of the ionic molecules cannot alone account for this stability. These compounds have an additional stability due to the lattice energy of the solid structure. However, lattice energy cannot be directly measured. The Born-Haber cycle allows us to understand and determine the lattice energies of ionic solids. is a type of potential energy that may be defined in two ways. In one definition, the lattice energy is the energy required to break apart an ionic solid and convert its component atoms into gaseous ions. This definition causes the value for the lattice energy to always be positive, since this will always be an endothermic reaction. The other definition says that lattice energy is the reverse process, meaning it is the energy released when gaseous ions bind to form an ionic solid. As implied in the definition, this process will always be exothermic, and thus the value for lattice energy will be negative. Its values are usually expressed with the units kJ/mol. Lattice Energy is used to explain the stability of ionic solids. Some might expect such an ordered structure to be less stable because the entropy of the system would be low. However, the crystalline structure allows each ion to interact with multiple oppositely charge ions, which causes a highly favorable change in the enthalpy of the system. A lot of energy is released as the oppositely charged ions interact. It is this that causes ionic solids to have such high melting and boiling points. Some require such high temperatures that they decompose before they can reach a melting and/or boiling point. There are several important concept to understand before the Born-Haber Cycle can be applied to determine the lattice energy of an ionic solid; ionization energy, electron affinity, dissociation energy, sublimation energy, heat of formation, and Hess's Law. The values used in the Born-Haber Cycle are all predetermined changes in enthalpy for the processes described in the section above. Hess' Law allows us to add or subtract these values, which allows us to determine the lattice energy. Determine the energy of the metal and nonmetal in their elemental forms. (Elements in their natural state have an energy level of zero.) Subtract from this the heat of formation of the ionic solid that would be formed from combining these elements in the appropriate ration. This is the energy of the ionic solid, and will be used at the end of the process to determine the lattice energy. The Born-Haber Cycle requires that the elements involved in the reaction are in their gaseous forms. Add the changes in enthalpy to turn one of the elements into its gaseous state, and then do the same for the other element. Metals exist in nature as single atoms and thus no dissociation energy needs to be added for this element. However, many nonmetals will exist as polyatomic species. For example, Cl exists as Cl in its elemental state. The energy required to change Cl into 2Cl atoms must be added to the value obtained in Step 2. Both the metal and nonmetal now need to be changed into their ionic forms, as they would exist in the ionic solid. To do this, the ionization energy of the metal will be added to the value from Step 3. Next, the of the nonmetal will be subtracted from the previous value. It is subtracted because it is a release of energy associated with the addition of an electron. *This is a common error due to confusion caused by the definition of electron affinity, so be careful when doing this calculation. Now the metal and nonmetal will be combined to form the ionic solid. This will cause a release of energy, which is called the lattice energy. The value for the lattice energy is the difference between the value from Step 1 and the value from Step 4. -------------------------------------------------------------------------------------------------------------------------------------------- The diagram below is another representation of the Born-Haber Cycle. The Born-Haber Cycle can be reduced to a single equation: Heat of formation= Heat of atomization+ Dissociation energy+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy *Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive. Rearrangement to solve for lattice energy gives the equation: Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities) | 4,794 | 1,618 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Fundamentals_of_Thermodynamics/Bond_Energies |
Bond energies are limited in their application for the reasons discussed earlier. They were: These are serious drawbacks if you want energy information about most reactions. Fortunately there is another way. Remember in our definition of enthalpy (\(H\)) we said it was a "state function". The net enthalpy change (\(ΔH\) --which is the only kind of enthalpy quantity we can measure) is independent of path. What does this mean? The diagram shows two possible pathways for one reaction: The direct reaction of the elements nitrogen and oxygen in a 1:2 molar ratio will produce 2 moles of nitrogen dioxide and absorb 68 kJ. But it is also possible to begin with the same elements under different conditions (like a 1:1 molar ratio) and produce 2 moles of nitrogen monoxide. This process absorbs 180 kJ. The nitrogen monoxide can then react with additional oxygen to form 2 moles of nitrogen dioxide. This process releases 112 kJ. The sum of the two reactions involving nitrogen monoxide gives the production of nitrogen dioxide: \[180 \;kJ + N_2 + O-2 \rightarrow 2 NO\] \[\underline{2 NO + O_2 \rightarrow 2 NO_2 + 112\; kJ}\] And because that is true, the enthalpy changes are also additive: \[180 \;kJ + (-112\; kJ) = 68\; kJ\] So with a sufficiently ambitious catalog of reaction enthalpy changes it is sometimes possible to calculate--rather than measure--the enthalpy change for a new reaction. Give the following data: \[ H_2 + ½ O_2 \rightarrow H_2O + 286\; kJ\] \[N_2O_5 + H_2O \rightarrow 2HNO_3 + 77 \; kJ\] \[N_2 + 3O_2 + H_2 \rightarrow 2HNO_3 + 348 \; kJ\] Calculate \(\Delta{H_{rxn}}\) for \[2N_2 + 5O_2 \rightarrow 2N_2O_5\] To begin these problems, concentrate on the items in the balanced equation you want. Where are they found in the data available? \[ H_2 + ½ O_2 \rightarrow H_2O + 286\; kJ\] \[\color{red} N_2O_5 \color{black} + H_2O \rightarrow 2HNO_3 + 77 \; kJ\] \[ \color{red} N_2 \color{black} + \color{red} 3O_2 \color{black} + H_2 \rightarrow 2HNO_3 + 348 \; kJ\] Once these are located, the equations needs to be adjusted so that the substances appear in the same amount as in the desired reaction and on the same sides. \[ H_2 + ½ O_2 \rightarrow H_2O + 286\; kJ\] \[2 \times (2 HNO_3 + 77\; kJ \rightarrow \color{red} N_2O_5 \color{black} + 2 H_2O)\] \[2 \times (\color{red}N_2 \color{black} + \color{red} 3 O_2 \color{black} + H_2 \rightarrow 2 HNO_3 + 348 \; kJ)\] Notice that this procedure did not fix the \(O_2\) entirely. However, there are also things that we need to get rid of so that the equations will add up to give the desired reaction of \(H_2\), \(H_2O\), and \(HNO_3\). If the first reaction is reversed and doubles, this will happen: \[ 2 \times ( H_2O + 572\; kJ \rightarrow 2 H_2 + O_2)\] \[2 \times (2 HNO_3 + 77\; kJ \rightarrow \color{red} N_2O_5 \color{black} + 2 H_2O\] \[2 \times (\color{red}N_2 \color{black} + \color{red} 3 O_2 \color{black} + H_2 \rightarrow 2 HNO_3 + 348 \; kJ)\] Multiply all this out to get \[ {2 H_2O} + 572\; kJ \rightarrow \cancel{2 H_2} + {O_2}\] \[{4 HNO_3} + 154\; kJ \rightarrow 2 \color{red} N_2O_5 \color{black} + 2 H_2O\] \[2 \color{red} N_2 \color{black} + \color{red} O_2 \color{black} + 2 H_2 \rightarrow 4 HNO_3 + 696 \; kJ\] Then we add and simplify, much like we do with a series of half reactions \[ \cancel{2 H_2O} + 572\; kJ \rightarrow \cancel{2 H_2} + \cancel{O_2}\] \[\cancel{4 HNO_3} + 154\; kJ \rightarrow 2 \color{red} N_2O_5 \color{black} + \cancel{2 H_2O}\] \[\underline {2 \color{red} N_2 \color{black} + \cancel{6} \color{red} O_2 \color{black} + \cancel{2 H_2} \rightarrow \cancel{4 HNO_3} + 696 \; kJ}\] \[ 30 \;kJ + 2 N_2 + 5 O_2 \rightarrow 2N_2O_5\] So what's the advantage? Theoretically every reaction can be "rewritten" as a series of processes involving elements forming individual compounds. It does not matter whether the reaction actually occurs that way, of course, because the enthalpies are additive (Hess's Law). The heats of formation represent those reactions and therefore can be used in their place to determine the overall enthalpy change in a reaction based on a mathematical statement. This additivity of heats of reaction (or reaction enthalpies) is generally known as Hess's Law. But it is also possible to state the equivalent of Hess's Law in purely mathematical terms with the introduction of an additional concept: If elements in their standard states (normal atmospheric pressure) and \(25^oC\) are defined as having no enthalpy of formation (i.e., it takes no energy to get an element the way it would normally be), then all compounds will have some enthalpy change associated with their formation from those elements. For example, when liquid water is formed from gaseous hydrogen and oxygen, we can write the following thermochemical equation: \[H_2 + ½ O_2 \rightarrow H_2O\; + \;285.8\; kJ\] The 285.8 kJ is the enthalpy of formation for liquid water: The fact that the value is on the products side of the reaction shows this is an exothermic process. The value can also be written separately (as in a table). By convention, a negative sign is then applied to show that there is a net loss of enthalpy in the system as the reactants become products. That means heat is released to the surroundings. So we can say \(ΔH_f^o = -285.8\; kJ/mol\) of water. All kinds of enthalpies of formation have been tabulated (Table T1). For an endothermic reaction the enthalpy of reaction would be written on the reactant side: \[177.8 kJ \;+ \;CaCO_3 \rightarrow CaO + CO_2\] And when written in a table, the value would be \(ΔH_{rxn} = +177.8\; kJ/mol\). All "heats of reaction" are molar and therefore proportional. Stoichiometric amounts of heat can be determined for a given amount of starting material just as any other stoichiometric calculation would be done. The thermochemical equation for the acid/base neutralization reaction of hydrochloric acid with barium hydroxide solution is \[ 2HCl + Ba(OH)_2 \rightarrow BaCl_2 + 2 H_2O + 118 \; kJ\] How much heat is produces if 34.5 g of HCl reacts with a stoichiometric amount of Barium hydroxide? Step 1: Balance the Equation Given and confirmed Step 2: Find number of moles of \[ (34.5\; g) \times \dfrac{1 \; mol}{36.5\; g} = 0.945 \; mol \; HCl\] Step 3: Use ration to find kJ \[ (0.945 \;mol\; HCl) \times \dfrac{118\; kJ}{2 \; mol \;HCl} = 55.8 \; kJ\] So what's the advantage? Theoretically every reaction can be "rewritten" as a series of processes involving elements forming individual compounds. It does not matter whether the reaction actually occurs that way, of course, because the enthalpies are additive (Hess's Law). The heats of formation represent those reactions and therefore can be used in their place to determine the overall enthalpy change in a reaction based on a mathematical statement. \[\sum{ΔH^o_{f\;products}} - \sum{ΔH^o_{f\;reactants}}= ΔH_{rxn}\] That looks fearsome but it simply says that the overall enthalpy change for a reaction is the difference between the heat content of the products and the heat content of the reactants. If the products end up with more stored energy than the reactants, the enthalpy change will be positive (endothermic). If the products end up with less stored energy than the reactants, the enthalpy change will be negative (exothermic). Thermite is a generic term for a mixture of metal and metal oxide used to generate tremendous heat. During the reaction one metal is reduced and the other is oxidized. The classic thermite mixture consists of iron(III) oxide and aluminum: How much energy is released in this reaction? \[ΔH_{rxn} = -1676\; kJ - (-826 \;kJ) = -850 \;\text{kJ/mol of either oxide}\] The notion of looking at the enthalpy change as reflective of "stored energy" is important and suggestive. Observation indicates that chemically "stable" compounds tend to have very negative heats of formation. Carbon dioxide (-393.5 kJ/mol) and water (-285.5 kJ/mol) would be examples, but by no means the most extreme. Oxides of metals like iron(III) oxide and aluminum oxide have very negative heats of formation. In contrast, chemically "unstable" compounds tend to have rather positive heats of formation. These very reactive substances have energy stored within their bonds. Silver fulminate, \(Ag_2C_2N_2O_2\), is a good example. The heat of formation is +180 kJ/mol. Silver fulminate is one of a series of compounds containing transition metals and the "fulminate" group. All of the compounds are very unstable. When silver fulminate decomposes it does so according to the following reaction: How much energy is released in this reaction? If 0.0009 g of silver fulminate decomposes (as in the "Whipper Snappers" fireworks), how much energy is released? \[ΔH_{rxn} = 2(-110.5 kJ) - 180 kJ = -401\; kJ/mol\] So \(0.0009\; g\) of silver fulminate is \(3 \times 10^{-6}\) moles. \[(3 \times 10^{-6} mol) \times (401\; kJ/mol) = 0.001 \;kJ\] Information like this seems to indicate that there might be a "preference" in Nature for reactions in which the enthalpy change is negative---loss of energy seems to breed chemical "stability". Substances which don't meet that criterion tend to react until they do. Or do they? | 9,194 | 1,619 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/02%3A_Atoms_Molecules_and_Ions/2.01%3A_The_Atomic_Theory_of_Matter |
Take some aluminum foil. Cut it in half. Now there are two smaller pieces of aluminum foil. Cut one of the pieces in half again. Cut one of those smaller pieces in half again. Continue cutting, making smaller and smaller pieces of aluminum foil. It should be obvious that the pieces are still aluminum foil; they are just becoming smaller and smaller. But how far can this exercise be taken, at least in theory? Can one continue cutting the aluminum foil into halves forever, making smaller and smaller pieces? Or is there some limit, some absolute smallest piece of aluminum foil? Thought experiments like this—and the conclusions based on them—were debated as far back as the fifth century . John Dalton (1766-1844) is the scientist credited for proposing the atomic theory. The theory explains several concepts that are relevant in the observable world: the composition of a pure gold necklace, what makes the pure gold necklace different than a pure silver necklace, and what occurs when pure gold is mixed with pure copper. This section explains the theories that Dalton used as a basis for his theory: (1) , (2) , and (3) "Nothing comes from nothing" is an important idea in ancient Greek philosophy that argues that what exists has always , since no new matter can come into existence where there was none before. Antoine Lavoisier (1743-1794) restated this principle for chemistry with the law of conservation of mass, which "means that the atoms of an object cannot be created or destroyed, but can be moved around and be changed into different particles." This law says that when a chemical reaction rearranges atoms into a new product, the mass of the reactants (chemicals before the chemical reaction) is the same as the mass of the products (the new chemicals made). More simply, whatever you do, you will still have the same amount of stuff (however, certain nuclear reactions like fusion and fission can convert a small part of the mass into energy. The law of conservation of mass states that the total mass present before a chemical reaction is the same as the total mass present after the chemical reaction; in other words, . The law of conservation of mass was formulated by Lavoisier as a result of his combustion experiment, in which he observed that the mass of his original substance—a glass vessel, tin, and air—was equal to the mass of the produced substance—the glass vessel, “tin calx”, and the remaining air. Historically, this was a difficult concept for scientists to grasp. If this law was true, then how could a large piece of wood be reduced to a small pile of ashes? The wood clearly has a greater mass than the ashes. From this observation scientists concluded that mass had been lost. However, Figure \(\Page {1}\) shows that the burning of word does follow the law of conservation of mass. Scientists did not account for the gases that play a critical role in this reaction. The law of conservation of mass states that the total mass present before a chemical reaction is the same as the total mass present after the chemical reaction. Joseph Proust (1754-1826) formulated the (also called the or ). This law states that if a compound is broken down into its constituent elements, the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. Joseph Proust based this law primarily on his experiments with basic copper carbonate. The illustration below depicts this law in action. Law of Definite Proportions states that in a given type of chemical substance, the elements are always combined in the same proportions by mass. The Law of Definite Proportions applies when elements are reacted together to form product. Therefore, while the Law of Definite Proportions can be used to compare two experiments in which hydrogen and oxygen react to form water, the Law of Definite Proportions can be used to compare one experiment in which hydrogen and oxygen react to form water, and another experiment in which hydrogen and oxygen react to form hydrogen peroxide (peroxide is another material that can be made from hydrogen and oxygen). Oxygen makes up 88.8% of the mass of any sample of pure water, while hydrogen makes up the remaining 11.2% of the mass. You can get water by melting ice or snow, by condensing steam, from river, sea, pond, etc. It can be from different places: , , Australia, or anywhere. It can be made by chemical reactions like burning hydrogen in oxygen. However, if the water is , it will consist of 88.8 % oxygen by mass and 11.2 % hydrogen by mass, irrespective of its source or method of preparation. Many combinations of elements can react to form more than one compound. In such cases, this law states that the weights of one element that combine with a fixed weight of another of these elements are integer multiples of one another. It's easy to say this, but please make sure that you understand how it works. Nitrogen forms a very large number of oxides, five of which are shown here. The law of multiple proportions states that if two elements form more than one compound between them, the masses of one element combined with a fixed mass of the second element form in ratios of small integers. Consider two separate compounds are formed by only carbon and oxygen. The first compound contains 42.9% carbon and 57.1% oxygen (by mass) and the second compound contains 27.3% carbon and 72.7% oxygen (again by mass). Is this consistent with the law of multiple proportions? The states that the masses of one element which combine with a fixed mass of the second element are in a ratio of numbers. Hence, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole-number ratio. Thus for every 1 g of the first compound there are 0.57 g of oxygen and 0.429 g of carbon. The mass of oxygen per gram carbon is: \[ \dfrac{0.571\; \text{g oxygen}}{0.429 \;\text{g carbon}} = 1.33\; \dfrac{\text{g oxygen}}{\text{g carbon}}\nonumber \] Similarly, for 1 g of the second compound, there are 0.727 g oxygen and 0.273 g of carbon. The ration of mass of oxygen per gram of carbon is \[ \dfrac{0.727\; \text{g oxygen}}{0.273 \;\text{g carbon}} = 2.66\; \dfrac{\text{g oxygen}}{\text{g carbon}}\nonumber \] Dividing the mass of oxygen per g of carbon of the second compound: \[\dfrac{2.66}{1.33} = 2\nonumber \] Hence the masses of oxygen combine with carbon in a 2:1 ratio which s consistent with the Law of Multiple Proportions since they are whole numbers. The modern atomic theory, proposed about 1803 by the English chemist John Dalton (Figure \(\Page {4}\)), is a fundamental concept that states that all elements are composed of atoms. Previously, an atom was defined as the smallest part of an element that maintains the identity of that element. Individual atoms are extremely small; even the largest atom has an approximate diameter of only 5.4 × 10 m. With that size, it takes over 18 million of these atoms, lined up side by side, to equal the width of the human pinkie (about 1 cm). Dalton’s ideas are called the atomic theory because the concept of atoms is very old. The Greek philosophers Leucippus and Democritus originally introduced atomic concepts in the fifth century BC. (The word comes from the Greek word , which means “indivisible” or “uncuttable.”) Dalton had something that the ancient Greek philosophers didn’t have, however; he had experimental evidence, such as the formulas of simple chemicals and the behavior of gases. In the 150 years or so before Dalton, natural philosophy had been maturing into modern science, and the scientific method was being used to study nature. When Dalton announced a modern atomic theory, he was proposing a fundamental theory to describe many previous observations of the natural world; he was not just participating in a philosophical discussion. Dalton's Theory was a powerful development as it explained the three laws of chemical combination (above) and recognized a workable distinction between the fundamental particle of an element (atom) and that of a compound (molecule). Six postulates are involved in Dalton's Atomic Theory: In light of the current state of knowledge in the field of Chemistry, Dalton’s theory had a few drawbacks. According to Dalton’s postulates, Despite these drawbacks, the importance of Dalton’s theory should not be underestimated. He displayed exceptional insight into the nature of matter. and his ideas provided a framework that was later modified and expanded by other. Consequentiually, John Dalton is often considered to be the father of modern atomic theory. Fundamental Experiments in Chemistry: This article explains the theories that Dalton used as a basis for his theory: (1) the Law of Conservation of Mass, (2) the Law of Constant Composition, (3) the Law of Multiple Proportions. | 8,904 | 1,620 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Fundamentals_of_Thermodynamics/Troutons_rule |
Trouton's rule says that for many (but not all) liquids, the is approximately the same at ~85 J mol K . The (partial) success of the rule is due to the fact that the entropy of a gas is considerably larger than that of any liquid. \[S_{gas} \gg S_{liquid}\] Therefore, the entropy of the initial state (e.g. the liquid) is negligible in determining the entropy of vaporization \[\Delta S_{vap}= S_{gas} - S_{liquid} \approx S_{gas} \label{approx}\] When a liquid vaporizes its entropy goes from a modest value to a significantly larger one. This is related to the ratio of the enthalpy of vaporization and the temperature of the transition: \[ΔS_{vap}= \dfrac{ΔH_{vap}}{T} \label{Eq1}\] \(ΔS_{vap}\) is found to be approximately constant at the boiling point (Figure \(\Page {1}\)): \[ΔS_{vap} \approx 85 \,J\, mol^{−1}K^{−1} \label{Trule}\] This is Trouton’s rule, which is valid for many liquids (e.g, the entropy of vaporization of toluene is 87.30 J K mol , that of benzene is 89.45 J K mol , and that of chloroform is 87.92 J K mol ). Because of its convenience, the rule is used to estimate the enthalpy of vaporization of liquids whose boiling points are known. Trouton’s rule can be used to estimate the enthalpy of vaporization of liquids whose boiling points are known. Experimental values vary rather more than this and for gases such as neon, nitrogen, oxygen and methane whose liquids all boil below 150 K, have values that are in the range 65−75, benzene, many 'normal' liquids and liquid sodium, lithium and iodine, in the range 80−90 and ethanol, water, hydrogen fluoride in the range 105−115 J mol K . Thus is nothing unusual about 150 K, but rather an influence from intermolecular interactions. The value of ~85 J mol K corresponds to a interaction energy of ~9.5kT per molecule and so the boiling point gives an indication of the strength of the cohesive energy holding molecules together in the condensed phase. When the cohesive energy exceeds this value, as in water, then the ratio \(ΔH_{vap}/T\) (Equation \(\ref{Eq1}\)) is larger and conversely the ratio is smaller when the cohesive energy is less as in Neon or methane. The ≈9.5kT minimum energy per molecule is quite a modest energy; if a molecule has six near neighbors this corresponds to about 3kT/2 per interaction between two molecules, roughly the average thermal energy. There is no universal rule for the entropy of melting since a similar approximate like that used for Trouton's law (Equation \(\ref{approx}\)) does not exist. However, if the mature of the interactions are consistent between a set of solids, then a crude correlation can be identified (Figure \(\Page {1}\); orange symbols). For example, the entropies of vaporization of water, ethanol, formic acid and hydrogen fluoride are far from the predicted values. However, if the liquid presents hydrogen bonding or any other kind of high ordered structure, its entropy will be particularly low and the entropy gain during vaporization will greater, too. The enthalpy of vaporization is greater for hydrogen-bonding molecules than for plain alkanes. For low-molecular weight alcohols, this effect is pronounced. The longer the alkane chain becomes, the more the compound behaves like a pure alkane. Keeping in mind the relative molecular weights of the compounds, you can see there is a decreasing effect of the hydrogen bonding (and other) effects on the n-alcohol series as we move to larger chains and become less alcohol-like (structured liquid) and more alkane-like (unstructure liquid). This is much more obvious when \(\mathrm{\Delta}H\) is normalized to a per-carbon basis. Table \(\Page {1}\) shows that two different processes control the enthalpy of vaporization, and similarly the saturation vapor concentration (also known as vapor pressure) or boiling point. At the low-molecular weight end, hydrogen bonding dominates, so we see the behavior common to polar, hydrogen-bonding compounds. At the high-molecular weight end, we see the pattern observed for alkanes. Trouton's rule hardly works for high ordered substances exhibiting hydrogen bonding. Other factors like the enthalpy of vaporization for a long chained organic molecule {strength of Van der Waals forces} may also play some significance role. The experimentally determined enthalpy of vaporization if water is \(40.7\, kJ\,mol^{-1}\). Does water follow Trouton's rule in predicting the enthalpy of vaporization? From Equations \(\ref{Eq1}\) and \(\ref{Trule}\), we get \[ΔS_{vap}= \dfrac{\Delta H_{vap}}{T} \approx 85 \, J \,K^{-1} mol^{-1}\nonumber\] This predicts that (since water boils at 373.15 K under atmospheric pressure) \[ΔH_{vap} \approx (85 \, J \,K^{-1} mol^{-1} )( 373.15\, K) = 31.7\, kJ \,K^{-1} mol^{-1} \nonumber\] This deviates from the observed enthalpy change of \(40.7\, J\, K^{-1} mol^{-1}\) by 23%, which is a sizable deviation from experiment. So, Trouton's law does not apply to water. This deviation is similarly observed by comparing the determined entropy of vaporization (Equation \(\ref{Eq1}\)) to the estimate from Trouton's law \[ΔS_{vap} = \dfrac{40.7\, kJ\, mol^{-1}}{373.15 K} = 109.1 J\, K^{-1} mol^{-1} > 85 \, J \,K^{-1} mol^{-1} \nonumber \] Example \(\Page {1}\) demostrates the general observation that the \(S_{liquid}\) for structured liquids is lower than for unstructured liquids, so the entropy gain during vaporization (i.e., \(\Delta_{vap}\) in Equation \(\ref{approx}\)) will be greater. In contrast to water, the entropy of vaporization of formic acid has negative deviance from Trouton's rule. This fact indicates the existence of an orderly structure in the gas phase; it is known that formic acid forms a dimer structure even in the gas phase (Figure \(\Page {2}\)). As with water, where hydrogen bonding results in structured phase and reduced entropy of the liquid (positive deviation from Trouton's law), the dimerization in formic acid reduces the entropy of the gas (negative deviation from Trouton's law). Negative deviance can also occur as a result of a reduced gas phase entropy owing to a low population of excited rotational states in the gas phase, particularly in small molecules such as methane. Trouton's rule validity can be increased by considering \[\Delta {\bar {S}}_{vap}=4.5R+R\ln T\] | 6,309 | 1,621 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_6%3A_Materials/21%3A_Structure_and_Bonding_in_Solids/21.5%3A_Lattice_Energies_of_Crystals |
Recall that the reaction of a metal with a nonmetal usually produces an ionic compound; that is, electrons are transferred from the metal (the ) to the nonmetal (the ). Ionic compounds are usually rigid, brittle, crystalline substances with flat surfaces that intersect at characteristic angles. They are not easily deformed, and they melt at relatively high temperatures. \(\ce{NaCl}\), for example, melts at 801°C. These properties result from the regular arrangement of the ions in the crystalline lattice and from the strong electrostatic attractive forces between ions with opposite charges. While formation of ion pairs from isolated ions releases large amounts of energy, even more energy is released when these ion pairs condense to form an ordered three-dimensional array. In such an arrangement each cation in the lattice is surrounded by more than one anion (typically four, six, or eight) and vice versa, so it is more stable than a system consisting of separate pairs of ions, in which there is only one cation–anion interaction in each pair. Note that \(r_0\) may differ between the gas-phase dimer and the lattice. An ionic lattice is more stable than a system consisting of separate ion pairs. The lattice energy of nearly any ionic solid can be calculated rather accurately using a modified form of : \[U=−\dfrac{k′Q_1Q_2}{r_0} \label{21.5.1}\] where \(U\), which is always a positive number, represents the amount of energy required to dissociate 1 mol of an ionic solid into the gaseous ions. The proportionality constant in Equation \(\ref{21.5.1}\) is expanded below, but it is worthwhile to discuss its general features first. If we assume that \(ΔV = 0\), then the lattice energy, \(U\), is approximately equal to the change in enthalpy, \(ΔH\): \[\ce{MX(s) \rightarrow M^{+n} (g) + X^{−n} (g)} \;\;\; ΔH \approx U \label{21.5.2}\] As before, \(Q_1\) and \(Q_2\) are the charges on the ions and \(r_0\) is the internuclear distance. We see from Equation \(\ref{21.5.1}\) that lattice energy is directly related to the product of the ion charges and inversely related to the internuclear distance. The value of the constant \(k′\) depends on the specific arrangement of ions in the solid lattice and their valence electron configurations. Representative values for calculated lattice energies, which range from about 600 to 10,000 kJ/mol, are listed in Table \(\Page {1}\). Energies of this magnitude can be decisive in determining the chemistry of the elements. Because the lattice energy depends on the of the charges of the ions, a salt having a metal cation with a +2 charge (M ) and a nonmetal anion with a −2 charge (X ) will have a lattice energy four times greater than one with \(\ce{M^{+}}\) and \(\ce{X^{−}}\), assuming the ions are of comparable size (and have similar internuclear distances). For example, the calculated value of \(U\) for \(\ce{NaF}\) is 910 kJ/mol, whereas \(U\) for \(\ce{MgO}\) (containing \(\ce{Mg^{2+}}\) and \(\ce{O^{2−}}\) ions) is 3795 kJ/mol. Because lattice energy is related to the internuclear distance, it is also inversely proportional to the size of the ions. This effect is illustrated in Figure \(\Page {1}\), which shows that lattice energy decreases for the series \(\ce{LiX}\), \(\ce{NaX}\), and \(\ce{KX}\) as the radius of \(\ce{X^{−}}\) increases. Because \(r_0\) in Equation \(\ref{21.5.1}\) is the sum of the ionic radii of the cation and the anion ( = + ), increases as the cation becomes larger in the series, so the magnitude of decreases. A similar effect is seen when the anion becomes larger in a series of compounds with the same cation. Because the ionic radii of the cations decrease in the order \(\ce{K^{+} > Na^{+} > Li^{+}}\) for a given halide ion, the lattice energy decreases smoothly from \(\ce{Li^{+}}\) to \(\ce{K^{+}}\). Conversely, for a given alkali metal ion, the fluoride salt always has the highest lattice energy and the iodide salt the lowest. Lattice energies are highest for substances with small, highly charged ions. Arrange GaP, BaS, CaO, and RbCl in order of increasing lattice energy. four compounds order of increasing lattice energy Using Equation \(\ref{21.5.1}\), predict the order of the lattice energies based on the charges on the ions. For compounds with ions with the same charge, use the relative sizes of the ions to make this prediction. The compound GaP, which is used in semiconductor electronics, contains Ga and P ions; the compound BaS contains Ba and S ions; the compound CaO contains Ca and O ions; and the compound RbCl has Rb and Cl ions. We know from Equation \(\ref{21.5.1}\) that lattice energy is directly proportional to the product of the ionic charges. Consequently, we expect RbCl, with a (−1)(+1) term in the numerator, to have the lowest lattice energy, and GaP, with a (+3)(−3) term, the highest. To decide whether BaS or CaO has the greater lattice energy, we need to consider the relative sizes of the ions because both compounds contain a +2 metal ion and a −2 chalcogenide ion. Because Ba lies below Ca in the periodic table, Ba is larger than Ca . Similarly, S is larger than O . Because the cation and the anion in BaS are both larger than the corresponding ions in CaO, the internuclear distance is greater in BaS and its lattice energy will be lower than that of CaO. The order of increasing lattice energy is RbCl < BaS < CaO < GaP. Arrange InAs, KBr, LiCl, SrSe, and ZnS in order of decreasing lattice energy. InAs > ZnS > SrSe > LiCl > KBr There are many other factors to be considered such as covalent character and electron-electron interactions in ionic solids. But for simplicity, let us consider the ionic solids as a collection of positive and negative ions. In this simple view, appropriate number of cations and anions come together to form a solid. The positive ions experience both attraction and repulsion from ions of opposite charge and ions of the same charge. As an example, let us consider the the \(\ce{NaCl}\) crystal. In the following discussion, assume be the distance between Na and Cl ions. The nearest neighbors of Na are 6 Cl ions at a distance 1 , 12 Na ions at a distance 2 , 8 Cl at 3 , 6 Na at 4 , 24 Na at 5 , and so on. Thus, the electrostatic potential of a single ion in a crystal by approximating the ions by point charges of the surrounding ions: \[ E_{ion-lattice} = \dfrac{Z^2e^2}{4\pi\epsilon_or} M \label{12.5.4}\] The \(M\) - named after Erwin Medelung - is a geometrical factor that depends on the arrangement of ions in the solid. For example, \(M\) for \(\ce{NaCl}\) is a poorly converging series of interaction energies: \[ M= \dfrac{6}{1} - \dfrac{12}{2} + \dfrac{8}{3} - \dfrac{6}{4} + \dfrac{24}{5} ... \label{21.5.5}\] with The Madelung constant depends on the structure type and Equation \(\ref{21.5.5}\) is applicable only for the sodium chloride (e.g., rock salt) lattice geometry. Other values for other structural types are given in Table \(\Page {2}\). There are other factors to consider for the evaluation of lattice energy and the treatment by Max Born and Alfred Landé led to the formula for the evaluation of lattice energy for a mole of . The Born–Landé equation (Equation \(\ref{21.5.6}\)) is a means of calculating the lattice energy of a crystalline ionic compound and derived from the electrostatic potential of the ionic lattice and a repulsive potential energy term \[ U= \dfrac{N_A M z^{+}z^{-} e^2}{4\pi \epsilon_o r_o} \left( 1 - \dfrac{1}{n} \right) \label{21.5.6}\] where Estimate the lattice energy for \(\ce{NaCl.}\) Using the values giving in the discussion above, the estimation is given by Equation \ref{21.5.6} \[ \begin{align*}U_{NaCl} &= \dfrac{(6.022 \times 10^{23} /mol) (1.74756) (1)(1) (1.6022 \times 10 ^{-19}\,C)^2}{ 4\pi \, (8.854 \times 10^{-12} C^2/m ) (282 \times 10^{-12}\; m)} \left( 1 - \dfrac{1}{9.1} \right) \\[4pt] &= - 756\, kJ/mol \end{align*}\] Much more should be considered in order to evaluate the lattice energy accurately, but the above calculation leads you to a good start. When methods to evaluate the energy of crystallization or lattice energy lead to reliable values, these values can be used in the Born-Hable cycle to evaluate other chemical properties, for example the electron affinity, which is really difficult to determine directly by experiment. The magnitude of the forces that hold an ionic substance together has a dramatic effect on many of its properties. The , for example, is the temperature at which the individual ions have enough kinetic energy to overcome the attractive forces that hold them in place. At the melting point, the ions can move freely, and the substance becomes a liquid. Thus melting points vary with lattice energies for ionic substances that have similar structures. The melting points of the sodium halides (Figure \(\Page {2}\)), for example, decrease smoothly from NaF to NaI, following the same trend as seen for their lattice energies (Figure \(\Page {1}\)). Similarly, the melting point of MgO is 2825°C, compared with 996°C for NaF, reflecting the higher lattice energies associated with higher charges on the ions. In fact, because of its high melting point, MgO is used as an electrical insulator in heating elements for electric stoves. The of ionic materials—that is, their resistance to scratching or abrasion—is also related to their lattice energies. Hardness is directly related to how tightly the ions are held together electrostatically, which, as we saw, is also reflected in the lattice energy. As an example, MgO is harder than NaF, which is consistent with its higher lattice energy. In addition to determining melting point and hardness, lattice energies affect the solubilities of ionic substances in water. In general, . For example, the solubility of NaF in water at 25°C is 4.13 g/100 mL, but under the same conditions, the solubility of MgO is only 0.65 mg/100 mL, meaning that it is essentially insoluble. High lattice energies lead to hard, insoluble compounds with high melting points. In principle, lattice energies could be measured by combining gaseous cations and anions to form an ionic solid and then measuring the heat evolved. Unfortunately, measurable quantities of gaseous ions have never been obtained under conditions where heat flow can be measured. Instead, lattice energies are found using the experimentally determined enthalpy changes for other chemical processes, Hess’s law, and a called the . Developed by Max Born and Fritz Haber in 1919, the Born–Haber cycle describes a process in which an ionic solid is conceptually formed from its component elements in a stepwise manner. Let’s use the Born–Haber cycle to determine the lattice energy of \(\ce{CsF(s)}\). \(\ce{CsF}\) is a nearly ideal ionic compound because \(\ce{Cs}\) is the least electronegative element that is not radioactive and F is the most electronegative element. To construct a thermochemical cycle for the formation of \(\ce{CsF}\), we need to know its enthalpy of formation, Δ , which is defined by the following chemical reaction: \[2Cs_{(s)}+F_{2(g)} \rightarrow 2CsF_{(s)} \label{21.5.7}\] Because enthalpy is a state function, the overall \(ΔH\) for a series of reactions is the sum of the values of \(ΔH\) for the individual reactions. We can therefore use a thermochemical cycle to determine the enthalpy change that accompanies the formation of solid CsF from the parent (not ions). The Born–Haber cycle for calculating the lattice energy of cesium fluoride is shown in Figure \(\Page {1}\). This particular cycle consists of six reactions, Equation \(\ref{21.5.7}\) plus the following five reactions: \[Cs_{(s)} \rightarrow Cs_{(g)}\;\;\; ΔH_1=ΔH_{sub}=76.5\; kJ/mol \label{21.5.8a}\] This equation describes the of elemental cesium, the conversion of the solid directly to a gas. The accompanying enthalpy change is called the (Table \(\Page {4}\)) and is positive because energy is required to sublime a solid. Data from (2004). \[\ce{Cs(g) -> Cs^{+}(g) + e^{-}} \;\;\; ΔH_2=I_1=375.7\; kJ/mol \label{21.5.8b}\] This equation describes the ionization of cesium, so the enthalpy change is the first ionization energy of cesium. Recall that energy is needed to ionize any neutral atom. Hence, regardless of the compound, the enthalpy change for this portion of the Born–Haber cycle is always positive. \[\ce{1/2 F2(g) -> F(g)} \;\;\; ΔH_3=\frac{1}{2}D=79.4\; kJ/mol \label{21.5.8c}\] This equation describes the dissociation of fluorine molecules into fluorine atoms, where \(D\) is the energy required for dissociation to occur (Table \(\Page {5}\)). We need to dissociate only \(\ce{1/2}\) mol of \(F_{2(g)}\) molecules to obtain 1 mol of \(F_{(g)}\) atoms. The ΔH for this reaction, too, is positive because energy is required to dissociate any stable diatomic molecule into the component atoms. Data from (2004). \[F_{(g)}+ e^- \rightarrow F^-_{(g)} \;\;\; ΔH_4= EA = –328.2\; kJ/mol \label{21.5.8d}\] This equation describes the formation of a gaseous fluoride ion from a fluorine atom; the enthalpy change is the electron affinity of fluorine. Recall that can be positive, negative, or zero. In this case, Δ is negative because of the highly negative electron affinity of fluorine. \[ Cs^+_{(g)} + F^–_{(g)}→CsF_{(s)} \;\;\; ΔH_5=–U \label{21.5.8e}\] This equation describes the formation of the ionic solid from the gaseous ions. Because Reaction 5 is the reverse of the equation used to define lattice energy and is defined to be a number, Δ is always , as it should be in a step that forms bonds. If the enthalpy of formation of CsF from the elements is known (Δ = −553.5 kJ/mol at 298 K), then the thermochemical cycle shown in Figure \(\Page {3}\) has only one unknown, the quantity Δ = − . From Hess’s law, we can write \[ΔH_f = ΔH_1 + ΔH_2 + ΔH_3 + ΔH_4 + ΔH_5 \label{21.5.9}\] We can rearrange Equation \(\ref{21.5.9}\) to give \[−ΔH_5 = ΔH_1 + ΔH_2 + ΔH_3 + ΔH_4 − ΔH_f \label{21.5.10}\] Substituting for the individual Δ s, we obtain Substituting the appropriate values into this equation gives \[ U = 76.5\; kJ/mol + 375.7 \;kJ/mol + 79.4\; kJ/mol + (−328.2\; kJ/mole) − (−553.5\; kJ/mol) = 756.9\; kJ/mol \label{21.5.11}\] is larger in magnitude than any of the other quantities in Equation \(\ref{21.5.1}\)1. The process we have used to arrive at this value is summarized in Table \(\Page {6}\). Equation \(\ref{21.5.9}\) may be used as a tool for predicting which ionic compounds are likely to form from particular elements. As we have noted, Δ (Δ ), Δ ( ), and Δ ( ) are always positive numbers, and Δ can be quite large. In contrast, Δ ( ) is comparatively small and can be positive, negative, or zero. Thus the first three terms in Equation \(\ref{21.5.9}\) make the formation of an ionic substance energetically unfavorable, and the fourth term contributes little either way. The formation of an ionic compound will be exothermic (Δ < 0) if and only if Δ (− ) is a large negative number. This means that . Another example is the formation of BaO: \[Ba_{(s)}+\frac{1}{2}O_{2(g)} \rightarrow BaO_{(s)} \label{21.5.11a}\] ) with a Born–Haber cycle is compared with that for the formation of \(\ce{CsF}\) in Figure \(\Page {4}\). The lattice energy of BaO, with a dipositive cation and a dinegative anion, dominates the Born–Haber cycle. Remember from Equations \(\ref{21.5.1}\) and \(\ref{21.5.6}\) that lattice energies are directly proportional to the product of the charges on the ions and inversely proportional to the internuclear distance. Although the internuclear distances are not significantly different for BaO and CsF (275 and 300 pm, respectively), the larger ionic charges in \(\ce{BaO}\) produce a much higher lattice energy. Substituting values for \(\ce{BaO}\) (Δ = −548.0 kJ/mol) into the equation and solving for gives: \[\begin{align*} U&=ΔH_{sub}(Ba)+[I_1(Ba)+I_2(Ba)]+\frac{1}{2}D(O_2)+[EA_1(O)+EA_2(O)]−ΔH_f(BaO)\;\;\; \label{21.5.17} \\[4pt] &=180.0\; kJ/mol + 1468.1 \; kJ/mol + 249.2\; kJ/mol + 603\; kJ/mol−(−548.0\; kJ/mol) \\[4pt] &= 3048\; kJ/mol \end{align*}\] If the formation of ionic lattices containing multiple charged ions is so energetically favorable, why does CsF contain Cs and F ions rather than Cs and F ions? If we assume that for a Cs F salt would be approximately the same as for BaO, the formation of a lattice containing Cs and F ions would release 2291 kJ/mol (3048 kJ/mol − 756.9 kJ/mol) more energy than one containing Cs and F ions. To form the Cs ion from Cs , however, would require removing a 5 electron from a filled inner shell, which calls for a great deal of energy: = 2234.4 kJ/mol for Cs. Furthermore, forming an F ion is expected to be even more energetically unfavorable than forming an O ion. Not only is an electron being added to an already negatively charged ion, but because the F ion has a filled 2 subshell, the added electron would have to occupy an empty high-energy 3 orbital. Cesium fluoride, therefore, is not Cs F because the energy cost of forming the doubly charged ions would be greater than the additional lattice energy that would be gained. Lattice energy is usually the most important energy factor in determining the stability of an ionic compound. Use the thermodynamics data in the reference tables to calculate the lattice energy of \(\ce{MgH2}\). chemical compound and data from figures and tables lattice energy Write a series of stepwise reactions for forming \(\ce{MgH2}\) from its elements via the gaseous ions. Use and data from the specified figures above and tables to calculate the lattice energy. Hess’s law allows us to use a thermochemical cycle (the Born–Haber cycle) to calculate the lattice energy for a given compound. We begin by writing reactions in which we form the component ions from the elements in a stepwise manner and then assemble the ionic solid: lists the first and second ionization energies for the period 3 elements [ (Mg) = 737.7 kJ/mol, (Mg) = 1450.7 kJ/mol]. First electron affinities for all elements are given in Figure \(\Page {1}\) [ (H) = −72.8 kJ/mol]. Table \(\Page {4}\) lists selected enthalpies of sublimation [Δ (Mg) = 147.1 kJ/mol]. Table \(\Page {5}\) lists selected bond dissociation energies [ (H ) = 436.0 kJ/mol]. Enthalpies of formation (Δ = −75.3 kJ/mol for MgH ) are listed in . From Hess’s law, Δ is equal to the sum of the enthalpy changes for Reactions 1–5: For MgH , = 2701.2 kJ/mol. Once again, lattice energy provides the driving force for forming this compound because Δ , Δ , Δ > 0. When solving this type of problem, be sure to write the chemical equation for each step and double-check that the enthalpy value used for each step has the correct sign . Use data from the reference tables to calculate the lattice energy of Li O. Remember that the second electron affinity for oxygen [O (g) + e → O (g)] is (+744 kJ/mol). 2809 kJ/mol Ionic compounds have strong electrostatic attractions between oppositely charged ions in a regular array. The lattice energy (\(U\)) of an ionic substance is defined as the energy required to dissociate the solid into gaseous ions; \(U\) can be calculated from the charges on the ions, the arrangement of the ions in the solid, and the internuclear distance. Because depends on the product of the ionic charges, substances with di- or tripositive cations and/or di- or trinegative anions tend to have higher lattice energies than their singly charged counterparts. Higher lattice energies typically result in higher and increased because more thermal energy is needed to overcome the forces that hold the ions together. Lattice energies cannot be measured directly but are obtained from a thermochemical cycle called the , in which Hess’s law is used to calculate the lattice energy from the measured enthalpy of formation of the ionic compound, along with other thermochemical data. The Born–Haber cycle can be used to predict which ionic compounds are likely to form. , the conversion of a solid directly to a gas, has an accompanying enthalpy change called the . | 20,152 | 1,622 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/12%3A_Chromatographic_and_Electrophoretic_Methods/12.02%3A_General_Theory_of_Column_Chromatography |
Of the two methods for bringing the stationary phase and the mobile phases into contact, the most important is column chromatography. In this section we develop a general theory that we may apply to any form of column chromatography. Figure 12.2.1
provides a simple view of a liquid–solid column chromatography experiment. The sample is introduced as a narrow band at the top of the column. Ideally, the solute’s initial concentration profile is rectangular (Figure 12.2.2
a). As the sample moves down the column, the solutes begin to separate (Figure 12.2.1
b,c) and the individual solute bands begin to broaden and develop a Gaussian profile (Figure 12.2.2
b,c). If the strength of each solute’s interaction with the stationary phase is sufficiently different, then the solutes separate into individual bands (Figure 12.2.1
d and Figure 12.2.2
d).
. An alternative view of the separation in Figure 12.2.1
showing the concentration of each solute as a function of distance down the column. We can follow the progress of the separation by collecting fractions as they elute from the column (Figure 12.2.1
e,f), or by placing a suitable detector at the end of the column. A plot of the detector’s response as a function of elution time, or as a function of the volume of mobile phase, is known as a (Figure 12.2.3
), and consists of a peak for each solute. There are many possible detectors that we can use to monitor the separation. Later sections of this chapter describe some of the most popular. We can characterize a chromatographic peak’s properties in several ways, two of which are shown in Figure 12.2.4
. , , is the time between the sample’s injection and the maximum response for the solute’s peak. A chromatographic peak’s , , as shown in Figure 12.2.4
, is determined by extending tangent lines from the inflection points on either side of the peak through the baseline. Although usually we report and using units of time, we can report them using units of volume by multiplying each by the mobile phase’s velocity, or report them in linear units by measuring distances with a ruler. For example, a solute’s retention volume, , is \(t_\text{r} \times u\) where is the mobile phase’s velocity through the column. In addition to the solute’s peak, Figure 12.2.4
also shows a small peak that elutes shortly after the sample is injected into the mobile phase. This peak contains all , which move through the column at the same rate as the mobile phase. The time required to elute the nonretained solutes is called the column’s , . The goal of chromatography is to separate a mixture into a series of chromatographic peaks, each of which constitutes a single component of the mixture. The between two chromatographic peaks, , is a quantitative measure of their separation, and is defined as \[R_{A B}=\frac{t_{t, B}-t_{t,A}}{0.5\left(w_{B}+w_{A}\right)}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}} \label{12.1}\] where is the later eluting of the two solutes. As shown in Figure 12.2.5
, the separation of two chromatographic peaks improves with an increase in . If the areas under the two peaks are identical—as is the case in Figure 12.2.5
—then a resolution of 1.50 corresponds to an overlap of only 0.13% for the two elution profiles. Because resolution is a quantitative measure of a separation’s success, it is a useful way to determine if a change in experimental conditions leads to a better separation. In a chromatographic analysis of lemon oil a peak for limonene has a retention time of 8.36 min with a baseline width of 0.96 min. \(\gamma\)-Terpinene elutes at 9.54 min with a baseline width of 0.64 min. What is the resolution between the two peaks? Using Equation \ref{12.1} we find that the resolution is \[R_{A B}=\frac{2 \Delta t_{r}}{w_{B}+w_{A}}=\frac{2(9.54 \text{ min}-8.36 \text{ min})}{0.64 \text{ min}+0.96 \text{ min}}=1.48 \nonumber\] Figure 12.2.6
shows the separation of a two-component mixture. What is the resolution between the two components? Use a ruler to measure \(\Delta t_\text{r}\), , and in millimeters. Because the relationship between elution time and distance is proportional, we can measure \(\Delta t_\text{r}\), , and using a ruler. My measurements are 8.5 mm for \(\Delta t_\text{r}\), and 12.0 mm each for and . Using these values, the resolution is \[R_{A B}=\frac{2 \Delta t_{t}}{w_{A}+w_{B}}=\frac{2(8.5 \text{ mm})}{12.0 \text{ mm}+12.0 \text{ mm}}=0.70 \nonumber\] Your measurements for \(\Delta t_\text{r}\), , and will depend on the relative size of your monitor or printout; however, your value for the resolution should be similar to the answer above. Equation \ref{12.1} suggests that we can improve resolution by increasing \(\Delta t_\text{r}\), or by decreasing and (Figure 12.2.7
). To increase \(\Delta t_\text{r}\) we can use one of two strategies. One approach is to adjust the separation conditions so that both solutes spend less time in the mobile phase—that is, we increase each —which provides more time to effect a separation. A second approach is to increase by adjusting conditions so that only one solute experiences a significant change in its retention time. The baseline width of a solute’s peak depends on the solutes movement within and between the mobile phase and the stationary phase, and is governed by several factors that collectively we call . We will consider each of these approaches for improving resolution in more detail, but first we must define some terms. Let’s assume we can describe a solute’s distribution between the mobile phase and stationary phase using the following equilibrium reaction \[S_{\text{m}} \rightleftharpoons S_{\text{s}} \nonumber\] where is the solute in the mobile phase and is the solute in the stationary phase. Following the same approach we used in for liquid–liquid extractions, the equilibrium constant for this reaction is an equilibrium partition coefficient, . \[K_{D}=\frac{\left[S_{\mathrm{s}}\right]}{\left[S_\text{m}\right]} \nonumber\] This is not a trivial assumption. In this section we are, in effect, treating the solute’s equilibrium between the mobile phase and the stationary phase as if it is identical to the equilibrium in a liquid–liquid extraction. You might question whether this is a reasonable assumption. There is an important difference between the two experiments that we need to consider. In a liquid–liquid extraction, which takes place in a separatory funnel, the two phases remain in contact with each other at all times, allowing for a true equilibrium. In chromatography, however, the mobile phase is in constant motion. A solute that moves into the stationary phase from the mobile phase will equilibrate back into a different portion of the mobile phase; this does not describe a true equilibrium. So, we ask again: Can we treat a solute’s distribution between the mobile phase and the stationary phase as an equilibrium process? The answer is yes, if the mobile phase velocity is slow relative to the kinetics of the solute’s movement back and forth between the two phase. In general, this is a reasonable assumption. In the absence of any additional equilibrium reactions in the mobile phase or the stationary phase, is equivalent to the distribution ratio, , \[D=\frac{\left[S_{0}\right]}{\left[S_\text{m}\right]}=\frac{(\operatorname{mol} \text{S})_\text{s} / V_\text{s}}{(\operatorname{mol} \text{S})_\text{m} / V_\text{m}}=K_{D} \label{12.2}\] where and are the volumes of the stationary phase and the mobile phase, respectively. A conservation of mass requires that the total moles of solute remain constant throughout the separation; thus, we know that the following equation is true. \[(\operatorname{mol} \text{S})_{\operatorname{tot}}=(\operatorname{mol} \text{S})_{\mathrm{m}}+(\operatorname{mol} \text{S})_\text{s} \label{12.3}\] Solving Equation \ref{12.3} for the moles of solute in the stationary phase and substituting into Equation \ref{12.2} leaves us with \[D = \frac{\left\{(\text{mol S})_{\text{tot}} - (\text{mol S})_\text{m}\right\} / V_{\mathrm{s}}}{(\text{mol S})_{\mathrm{m}} / V_{\mathrm{m}}} \nonumber\] Rearranging this equation and solving for the fraction of solute in the mobile phase, , gives \[f_\text{m} = \frac {(\text{mol S})_\text{m}} {(\text{mol S})_\text{tot}} = \frac {V_\text{m}} {DV_\text{s} + V_\text{m}} \label{12.4}\] which is identical to the result for a liquid-liquid extraction (see ). Because we may not know the exact volumes of the stationary phase and the mobile phase, we simplify Equation \ref{12.4} by dividing both the numerator and the denominator by ; thus \[f_\text{m} = \frac {V_\text{m}/V_\text{m}} {DV_\text{s}/V_\text{m} + V_\text{m}/V_\text{m}} = \frac {1} {DV_\text{s}/V_\text{m} + 1} = \frac {1} {1+k} \label{12.5}\] where \[k=D \times \frac{V_\text{s}}{V_\text{m}} \label{12.6}\] is the solute’s . Note that the larger the retention factor, the more the distribution ratio favors the stationary phase, leading to a more strongly retained solute and a longer retention time. Other (older) names for the retention factor are capacity factor, capacity ratio, and partition ratio, and it sometimes is given the symbol \(k^{\prime}\). Keep this in mind if you are using other resources. Retention factor is the approved name from the IUPAC Gold Book. We can determine a solute’s retention factor from a chromatogram by measuring the column’s void time, , and the solute’s retention time, (see ). Solving Equation \ref{12.5} for , we find that \[k=\frac{1-f_\text{m}}{f_\text{m}} \label{12.7}\] Earlier we defined as the fraction of solute in the mobile phase. Assuming a constant mobile phase velocity, we also can define as \[f_\text{m}=\frac{\text { time spent in the mobile phase }}{\text { time spent in the stationary phase }}=\frac{t_\text{m}}{t_\text{r}} \nonumber\] Substituting back into Equation \ref{12.7} and rearranging leaves us with \[k=\frac{1-\frac{t_{m}}{t_{t}}}{\frac{t_{\mathrm{m}}}{t_{\mathrm{r}}}}=\frac{t_{\mathrm{t}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{t_{\mathrm{r}}^{\prime}}{t_{\mathrm{m}}} \label{12.8}\] where \(t_\text{r}^{\prime}\) is the . In a chromatographic analysis of low molecular weight acids, butyric acid elutes with a retention time of 7.63 min. The column’s void time is 0.31 min. Calculate the retention factor for butyric acid. \[k_{\mathrm{but}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{7.63 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=23.6 \nonumber\] Figure 12.2.8
is the chromatogram for a two-component mixture. Determine the retention factor for each solute assuming the sample was injected at time = 0. Because the relationship between elution time and distance is proportional, we can measure , , and using a ruler. My measurements are 7.8 mm, 40.2 mm, and 51.5 mm, respectively. Using these values, the retention factors for solute A and solute B are \[k_{1}=\frac{t_{\mathrm{r} 1}-t_\text{m}}{t_\text{m}}=\frac{40.2 \text{ mm}-7.8 \text{ mm}}{7.8 \text{ mm}}=4.15 \nonumber\] \[k_{2}=\frac{t_{\mathrm{r} 2}-t_\text{m}}{t_\text{m}}=\frac{51.5 \text{ mm}-7.8 \text{ mm}}{7.8 \text{ mm}}=5.60 \nonumber\] Your measurements for , , and will depend on the relative size of your monitor or printout; however, your value for the resolution should be similar to the answer above. Selectivity is a relative measure of the retention of two solutes, which we define using a selectivity factor, \(\alpha\) \[\alpha=\frac{k_{B}}{k_{A}}=\frac{t_{r, B}-t_{\mathrm{m}}}{t_{r, A}-t_{\mathrm{m}}} \label{12.9}\] where solute has the smaller retention time. When two solutes elute with identical retention time, \(\alpha = 1.00\); for all other conditions \(\alpha > 1.00\). In the chromatographic analysis for low molecular weight acids described in , the retention time for isobutyric acid is 5.98 min. What is the selectivity factor for isobutyric acid and butyric acid? First we must calculate the retention factor for isobutyric acid. Using the void time from we have \[k_{\mathrm{iso}}=\frac{t_{\mathrm{r}}-t_{\mathrm{m}}}{t_{\mathrm{m}}}=\frac{5.98 \text{ min}-0.31 \text{ min}}{0.31 \text{ min}}=18.3 \nonumber\] The selectivity factor, therefore, is \[\alpha=\frac{k_{\text {but }}}{k_{\text {iso }}}=\frac{23.6}{18.3}=1.29 \nonumber\] Determine the selectivity factor for the chromatogram in . Using the results from , the selectivity factor is \[\alpha=\frac{k_{2}}{k_{1}}=\frac{5.60}{4.15}=1.35 \nonumber\] Your answer may differ slightly due to differences in your values for the two retention factors. Suppose we inject a sample that has a single component. At the moment we inject the sample it is a narrow band of finite width. As the sample passes through the column, the width of this band continually increases in a process we call . Column efficiency is a quantitative measure of the extent of band broadening. See and . When we inject the sample it has a uniform, or rectangular concentration profile with respect to distance down the column. As it passes through the column, the band broadens and takes on a Gaussian concentration profile. In their original theoretical model of chromatography, Martin and Synge divided the chromatographic column into discrete sections, which they called theoretical plates. Within each theoretical plate there is an equilibrium between the solute present in the stationary phase and the solute present in the mobile phase [Martin, A. J. P.; Synge, R. L. M. . , , 1358–1366]. They described column efficiency in terms of the number of , , \[N=\frac{L}{H} \label{12.10}\] where is the column’s length and is the height of a theoretical plate. For any given column, the column efficiency improves—and chromatographic peaks become narrower—when there are more theoretical plates. If we assume that a chromatographic peak has a Gaussian profile, then the extent of band broadening is given by the peak’s variance or standard deviation. The height of a theoretical plate is the peak’s variance per unit length of the column \[H=\frac{\sigma^{2}}{L} \label{12.11}\] where the standard deviation, \(\sigma\), has units of distance. Because retention times and peak widths usually are measured in seconds or minutes, it is more convenient to express the standard deviation in units of time, \(\tau\), by dividing \(\sigma\) by the solute’s average linear velocity, \(\overline{u}\), which is equivalent to dividing the distance it travels, , by its retention time, . \[\tau=\frac{\sigma}{\overline{u}}=\frac{\sigma t_{r}}{L} \label{12.12}\] For a Gaussian peak shape, the width at the baseline, , is four times its standard deviation, \(\tau\). \[w = 4 \tau \label{12.13}\] Combining Equation \ref{12.11}, Equation \ref{12.12}, and Equation \ref{12.13} defines the height of a theoretical plate in terms of the easily measured chromatographic parameters and . \[H=\frac{L w^{2}}{16 t_\text{r}^{2}} \label{12.14}\] Combing Equation \ref{12.14} and Equation \ref{12.10} gives the number of theoretical plates. \[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16\left(\frac{t_{\mathrm{r}}}{w}\right)^{2} \label{12.15}\] A chromatographic analysis for the chlorinated pesticide Dieldrin gives a peak with a retention time of 8.68 min and a baseline width of 0.29 min. Calculate the number of theoretical plates? Given that the column is 2.0 m long, what is the height of a theoretical plate in mm? Using Equation \ref{12.15}, the number of theoretical plates is \[N=16 \frac{t_{\mathrm{r}}^{2}}{w^{2}}=16 \times \frac{(8.68 \text{ min})^{2}}{(0.29 \text{ min})^{2}}=14300 \text{ plates} \nonumber\] Solving Equation \ref{12.10} for gives the average height of a theoretical plate as \[H=\frac{L}{N}=\frac{2.00 \text{ m}}{14300 \text{ plates}} \times \frac{1000 \text{ mm}}{\mathrm{m}}=0.14 \text{ mm} / \mathrm{plate} \nonumber\] For each solute in the chromatogram for , calculate the number of theoretical plates and the average height of a theoretical plate. The column is 0.5 m long. Because the relationship between elution time and distance is proportional, we can measure , , , and using a ruler. My measurements are 40.2 mm, 51.5 mm, 8.0 mm, and 13.5 mm, respectively. Using these values, the number of theoretical plates for each solute is \[N_{1}=16 \frac{t_{r,1}^{2}}{w_{1}^{2}}=16 \times \frac{(40.2 \text{ mm})^{2}}{(8.0 \text{ mm})^{2}}=400 \text { theoretical plates } \nonumber\] \[N_{2}=16 \frac{t_{r,2}^{2}}{w_{2}^{2}}=16 \times \frac{(51.5 \text{ mm})^{2}}{(13.5 \text{ mm})^{2}}=233 \text { theoretical plates } \nonumber\] The height of a theoretical plate for each solute is \[H_{1}=\frac{L}{N_{1}}=\frac{0.500 \text{ m}}{400 \text { plates }} \times \frac{1000 \text{ mm}}{\mathrm{m}}=1.2 \text{ mm} / \mathrm{plate} \nonumber\] \[H_{2}=\frac{L}{N_{2}}=\frac{0.500 \text{ m}}{233 \text { plates }} \times \frac{1000 \text{ mm}}{\mathrm{m}}=2.15 \text{ mm} / \mathrm{plate} \nonumber\] Your measurements for , , , and will depend on the relative size of your monitor or printout; however, your values for and for should be similar to the answer above. It is important to remember that a theoretical plate is an artificial construct and that a chromatographic column does not contain physical plates. In fact, the number of theoretical plates depends on both the properties of the column and the solute. As a result, the number of theoretical plates for a column may vary from solute to solute. One advantage of improving column efficiency is that we can separate more solutes with baseline resolution. One estimate of the number of solutes that we can separate is \[n_{c}=1+\frac{\sqrt{N}}{4} \ln \frac{V_{\max }}{V_{\min }} \label{12.16}\] where is the column’s , and and are the smallest and the largest volumes of mobile phase in which we can elute and detect a solute [Giddings, J. C. , Wiley-Interscience: New York, 1991]. A column with 10 000 theoretical plates, for example, can resolve no more than \[n_{c}=1+\frac{\sqrt{10000}}{4} \ln \frac{30 \mathrm{mL}}{1 \mathrm{mL}}=86 \text { solutes } \nonumber\] if and are 1 mL and 30 mL, respectively. This estimate provides an upper bound on the number of solutes and may help us exclude from consideration a column that does not have enough theoretical plates to separate a complex mixture. Just because a column’s theoretical peak capacity is larger than the number of solutes, however, does not mean that a separation is feasible. In most situations the practical peak capacity is less than the theoretical peak capacity because the retention characteristics of some solutes are so similar that a separation is impossible. Nevertheless, columns with more theoretical plates, or with a greater range of possible elution volumes, are more likely to separate a complex mixture. The smallest volume we can use is the column’s void volume. The largest volume is determined either by our patience—the maximum analysis time we can tolerate—or by our inability to detect solutes because there is too much band broadening. Our treatment of chromatography in this section assumes that a solute elutes as a symmetrical Gaussian peak, such as that shown in Figure 12.2.4
. This ideal behavior occurs when the solute’s partition coefficient, \[K_{\mathrm{D}}=\frac{[S_\text{s}]}{\left[S_\text{m}\right]} \nonumber\] is the same for all concentrations of solute. If this is not the case, then the chromatographic peak has an asymmetric peak shape similar to those shown in Figure 12.2.9
. The chromatographic peak in Figure 12.2.9
a is an example of , which occurs when some sites on the stationary phase retain the solute more strongly than other sites. Figure 12.2.9
b, which is an example of most often is the result of overloading the column with sample. As shown in Figure 12.2.9
a, we can report a peak’s asymmetry by drawing a horizontal line at 10% of the peak’s maximum height and measuring the distance from each side of the peak to a line drawn vertically through the peak’s maximum. The asymmetry factor, , is defined as \[T=\frac{b}{a} \nonumber\] The number of theoretical plates for an asymmetric peak shape is approximately \[N \approx \frac{41.7 \times \frac{t_{r}^{2}}{\left(w_{0.1}\right)^{2}}}{T+1.25}=\frac{41.7 \times \frac{t_{r}^{2}}{(a+b)^{2}}}{T+1.25} \nonumber\] where is the width at 10% of the peak’s height [Foley, J. P.; Dorsey, J. G. , , 730–737]. Asymmetric peaks have fewer theoretical plates, and the more asymmetric the peak the smaller the number of theoretical plates. For example, the following table gives values for for a solute eluting with a retention time of 10.0 min and a peak width of 1.00 min. 0.5 0.5 0.6 0.4 0.7 0.3 | 20,705 | 1,623 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.06%3A_General_Aspects_of_the_Chemistry_of_Copper |
The chemistry of copper in biological systems is limited to oxidation states I and II. The Cu state has electronic configuration d . Unless there are ligand bands or strong ligand-to-copper charge-transfer bands, diamagnetic Cu species are colorless. Complexes of Cu (d ) are often blue in color. The single unpaired electron makes Cu amenable to electron paramagnetic resonance (EPR) techniques, at least if the electron spins of Cu centers are independent of one another. In oxyhemocyanin the spins are so strongly coupled (-J > 600 cm ) that at room temperature and below the system is effectively diamagnetic and the pair of Cu ions is EPR silent. In aqueous solutions the Cu ion is unstable with respect to disproportionation to Cu metal and Cu ion: \[Cu^{II} + 2e^{-} \rightleftharpoons Cu \qquad E^{o} = 0.3402 V\] \[Cu^{+} + e^{-} \rightleftharpoons Cu \qquad E^{o} = 0.522 V \tag{4.35}\] \[2 Cu^{+} \rightleftharpoons Cu + Cu^{II-} \qquad E^{o} = 0.182 V\] The Cu state may be stabilized by ligands, especially sulfur-containing ones, or by immobilization as afforded by a protein matrix, or in nonaqueous solvents, such as acetonitrile, in the absence of dioxygen. Whereas Cu thiolate species are stable, Cu thiolate species usually are unstable with respect to the disproportionation: \[2 Cu^{II}—SR \rightarrow 2 Cu^{I} + R—S—S—R \tag{4.36}\] Again, immobilization may give kinetic stability to Cu thiolate species, as occurs in the blue-copper family of electron-transport proteins. Copper(l) complexes are often two-coordinate with a linear arrangement of ligands. Three-, four-, and possibly five-coordinate complexes are known. In the presence of O , nonbiological copper(l) [and iron(II)] complexes are often susceptible to ligand degradation, which may give the illusion of O binding. The mechanisms by which this reaction occurs remain essentially unknown. Iron-porphyrin systems are rather more robust. Nonetheless, there are now several well-characterized copper(l) systems that reversibly bind dioxygen, at least at low temperature. One that has been structurally characterized features a dicopper(II)-peroxo moiety, while a second, with more properties in common with oxyhemocyanin, features a moiety. | 2,253 | 1,624 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/02%3A_Properties_of_Gases/2.07%3A_The_Maxwell_Distribution_Laws |
In the context of the Kinetic Molecular Theory of Gases, a gas contains a large number of particles in rapid motions. Each particle has a different speed, and each collision between particles changes the speeds of the particles. An understanding of the properties of the gas requires an understanding of the distribution of particle speeds. At temperatures above absolute zero, all molecules are in motion. In the case of a gas, this motion consists of straight-line jumps whose lengths are quite great compared to the dimensions of the molecule. Although we can never predict the velocity of a particular molecule, the fact that we are usually dealing with a huge number of them allows us to know what fraction of the molecules have kinetic energies (and hence velocities) that lie within any given range. The trajectory of an individual gas molecule consists of a series of straight-line paths interrupted by collisions. What happens when two molecules collide depends on their relative kinetic energies; in general, a faster or heavier molecule will impart some of its kinetic energy to a slower or lighter one. Two molecules having identical masses and moving in opposite directions at the same speed will momentarily remain motionless after their collision. If we could measure the instantaneous velocities of all the molecules in a sample of a gas at some fixed temperature, we would obtain a wide range of values. A few would be zero, and a few would be very high velocities, but the majority would fall into a more or less well defined range. We might be tempted to define an average velocity for a collection of molecules, but here we would need to be careful: molecules moving in opposite directions have velocities of opposite signs. Because the molecules are in a gas are in random thermal motion, there will be just about as many molecules moving in one direction as in the opposite direction, so the velocity vectors of opposite signs would all cancel and the average velocity would come out to zero. Since this answer is not very useful, we need to do our averaging in a slightly different way. The proper treatment is to average the of the velocities, and then take the square root of this value. The resulting quantity is known as the velocity \[ v_{rms} = \sqrt{\dfrac{\sum \nu^2}{n}}\] where \(n\) is the number of molecules in the system. The formula relating the RMS velocity to the temperature and molar mass is surprisingly simple (derived below), considering the great complexity of the events it represents: \[ v_{rms} = \sqrt{\dfrac{3RT}{M}} \label{rms1}\] where Equation \(\ref{rms1}\) can also be expressed as \[ v_{rms} = \sqrt{\dfrac{3k_bT}{m}} \label{rms2}\] where Equation \(\ref{rms2}\) is just the per atom version of Equation \(\ref{rms1}\) which is expressed in terms of per mol. Either equation will work. What is the \(v_{rms}\) of a nitrogen molecule at 300 K? The molar mass of \(N_2\) is 28.01 g/mol. Substituting in the above equation and expressing in energy units, we obtain \[v^2 =\dfrac{(3)(8.31 \;J \; mol^{-1} \; K^{-1})(300\;K)}{28.01 \times 10^{-3}\; Kg \; mol^{-1}} = 2.67 \times 10^5 \; \dfrac{J}{Kg} \nonumber\] Recalling the definition of the joule (1 J = 1 kg m s ) and taking the square root, \[ \bar{v} = \sqrt{ \left(2.67 \times 10^5\; \dfrac{\cancel{J}}{\cancel{Kg}}\right) \left( \dfrac{ 1 \;\cancel{kg}\; m^2\; s^{–2}}{ 1 \;\cancel{J }} \right)} = 517 \;m /s \nonumber \] or \[ 517\, \dfrac{\cancel{m}}{s} \left(\dfrac{1\; km}{10^3\; \cancel{m}} \right) \left(\dfrac{3600\; \cancel{s}}{1\; hr} \right) = 1860\; km/hr \nonumber \] this is fast! The velocity of a rifle bullet is typically 300-500 m s ; convert to common units to see the comparison for yourself. A simpler formula for estimating average molecular velocities than \(\ref{rms1}\) is \[ v_{rms} =157 \sqrt{\dfrac{T}{M}} \nonumber \] in which \(v\) is in units of meters/sec, \(T\) is the and \(M\) the molar mass in grams. If we were to plot the number of molecules whose velocities fall within a series of narrow ranges, we would obtain a slightly asymmetric curve known as a . The peak of this curve would correspond to the velocity. This velocity distribution curve is known as the , but is frequently referred to only by Boltzmann's name. The was first worked out around 1850 by the great Scottish physicist, (left, 1831-1879), who is better known for discovering the laws of electromagnetic radiation. Later, the Austrian physicist (1844-1906) put the relation on a sounder theoretical basis and simplified the mathematics somewhat. Boltzmann pioneered the application of statistics to the physics and thermodynamics of matter, and was an ardent supporter of the atomic theory of matter at a time when it was still not accepted by many of his contemporaries. The Maxwell-Boltzmann distribution is used to determine how many molecules are moving between velocities \(v\) and \(v + dv\). Assuming that the one-dimensional distributions are independent of one another, that the velocity in the and directions does not affect the velocity, for example, the Maxwell-Boltzmann distribution is given by \[ \large \dfrac{dN}{N} = \left(\dfrac{m}{2\pi k_bT} \right)^{1/2}exp\left [\dfrac{-mv^2}{2k_b T}\right] dv \label{1}\] where Additionally, the function can be written in terms of the scalar quantity speed \(v\) instead of the vector quantity velocity. This form of the function defines the distribution of the gas molecules moving at different speeds, between \(v_1\) and \(v_2\), thus \[f(v)=4\pi v^2 \left (\dfrac{m}{2\pi k_b T} \right)^{3/2}exp\left [\frac{-mv^2}{2k_bT}\right] \label{2}\] Finally, the Maxwell-Boltzmann distribution can be used to determine the distribution of the kinetic energy of for a set of molecules. The distribution of the kinetic energy is identical to the distribution of the speeds for a certain gas at any temperature. The Maxwell-Boltzmann distribution is a probability distribution and just like any such distribution, can be characterized in a variety of ways including. Higher temperatures allow a larger fraction of molecules to acquire greater amounts of kinetic energy, causing the Boltzmann plots to spread out. Figure \(\Page {2}\) shows how the Maxwell-Boltzmann distribution is affected by temperature. At lower temperatures, the molecules have less energy. Therefore, the speeds of the molecules are lower and the distribution has a smaller range. As the temperature of the molecules increases, the distribution flattens out. Because the molecules have greater energy at higher temperature, the molecules are moving faster. Notice how the left ends of the plots are anchored at zero velocity (there will always be a few molecules that happen to be at rest.) As a consequence, the curves flatten out as the higher temperatures make additional higher-velocity states of motion more accessible. The area under each plot is the same for a constant number of molecules. All molecules have the same kinetic energy ( /2) at the same temperature, so the fraction of molecules with higher velocities will increase as , and thus the molecular weight, decreases. Figure \(\Page {3}\) shows the dependence of the Maxwell-Boltzmann distribution on molecule mass. On average, heavier molecules move more slowly than lighter molecules. Therefore, heavier molecules will have a smaller speed distribution, while lighter molecules will have a speed distribution that is more spread out. The Maxwell-Boltzmann equation, which forms the basis of the kinetic theory of gases, defines the distribution of speeds for a gas at a certain temperature. From this distribution function, the most probable speed, the average speed, and the root-mean-square speed can be derived. Usually, we are more interested in the speeds of molecules rather than their component velocities. The Maxwell–Boltzmann distribution for the speed follows immediately from the distribution of the velocity vector, above. Note that the speed of an individual gas particle is: \[v =\sqrt{v_x^2+ v_y^2 = v_z^2}\] Three speed expressions can be derived from the Maxwell-Boltzmann distribution: The is the maximum value on the distribution plot (Figure \(\Page {4}\)}. This is established by finding the velocity when the derivative of Equation \(\ref{2}\) is zero \[\dfrac{df(v)}{dv} = 0 \] which is \[\color{red} v_{mp}=\sqrt {\dfrac {2RT}{M}} \label{3a}\] The is the sum of the speeds of all the molecules divided by the number of molecules. \[\color{red} v_{avg}= \bar{v} = \int_0^{\infty} v f(v) dv = \sqrt {\dfrac{8RT}{\pi M}} \label{3b}\] The is square root of the average speed-squared. \[ \color{red} v_{rms}= \bar{v^2} = \sqrt {\dfrac {3RT}{M}} \label{3c}\] where It follows that for gases that follow the Maxwell-Boltzmann distribution: \[v_{mp}< v_{avg}< v_{rms} \label{4}\] ( ) ) | 8,868 | 1,626 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/04%3A_Biological_and_Synthetic_Dioxygen_Carriers/4.05%3A_General_Aspects_of_the_Chemistry_of_Cobalt |
Many parallels exist between the chemistry of Fe - and Co -porphyrinato systems. Dioxygen binds to many Co complexes to give mononuclear 1:1 Co:O complexes with a bent geometry \(\tag{4.37}\) and dinuclear 2:1 Co:O complexes, analogous to those described for Fe systems in Reactions (4.29a) and (4.29b). Indeed, these dinuclear systems were the first nonbiological oxygen carriers to be isolated. The geometry of the dioxygen moiety, spanning two metals, may be or : \(\tag{4.38}\) However, whereas these dinuclear cobalt species are invariably octahedral, dinuclear copper-peroxo species are tetrahedral or distorted square pyramidal. In the late 1960s, 1:1 Co:O species were first isolated by use of a combination of low temperatures and specific Schiff-base ligands. It was found that cobalt corrins, such as vitamin B , also formed 1:1 dioxygen adducts, although this chemistry is not known to be utilized by living systems. Cobalt(Il) porphyrins also form 1:1 adducts but with low O affinity, especially in nonpolar, aprotic solvents. Thus hemoglobin and myoglobin may be reconstituted from a cobaltoheme with preservation not only of dioxygen-binding capabilities but also of cooperativity. The synthetic 1:1 Co:O complexes have proven to be very useful in increasing our understanding of factors that determine oxygen affinity for cobalt systems and by extrapolation for iron systems. Two important differences make Co systems more accessible. First, in contrast to iron systems, the cleavage reaction (4.29c) and redimerization to a \(\mu\)-oxo species (Reaction 4.29d) do not occur (see Figure 4.15). Thus Co complexes of O are stable in solution at room temperature without the need for protection illustrated in Figure 4.14. Second, for Co -porphyrinato systems, the equilibrium constant for the addition of a second axial base, such as pyridine or 1-methylimidazole, is small. Thus the disproportionation to four-coordinate and six-coordinate species that occurs for corresponding Fe systems (Reaction 4.33) does not occur. This difference simplifies the interpretation of spectral changes that are used to obtain thermodynamic and kinetic parameters of which there are now voluminous examples. Moreover, the 1:1 Co-O complexes are paramagnetic. From the small Co hyperfine splitting, it is deduced that the single unpaired electron resides primarily on the dioxygen moiety. From other experiments it is apparent that net transfer of electron density from the metal onto the dioxygen varies considerably, from about 0.1e to about 0.8e . For example, it is found for a given Co Schiff base, Co(bzacen), that the redox potential of the cobalt-Schiffbase center LCo, measured by cyclic voltammetry, E , \[B_{2}LCo^{III} + e^{-} \rightleftharpoons B_{2}LCo^{II} \tag{4.39}\] \[B = substituted\; pyridine\] is a linear function of log K(O ) as the axial base B is varied. The more easily the Co center may be oxidized, the higher is the O affinity, as illustrated in Figure 4.18A. The dioxygen affinity also increases as the basicity of the axial nitrogenous ligand increases. This effect is illustrated in Figure 4.18B. Because of differing steric requirements, dimethylformamide (DMF), substituted imidazole, and piperidine (pip) ligands do not fall on the correlation defined by the series of substituted pyridine species. Note the synergistic nature of dioxygen binding: in general, the more electron density that is pumped onto the metal by the axial base, the more electron density is available for donation into the \(\pi\)* orbitals of the dioxygen ligand. E and log K(O ) are also correlated, although more weakly, for a number of hemoglobins (Figure 4.18C). Here the porphyrin and axial base remain constant, but presumably the surroundings of the heme group and O binding site vary in a manner that is less well-defined than in the model systems of Figure 4.18A and B. Notwithstanding these various perturbations to the metal center, the O—O stretch occurs at about 1140 cm , placing all 1:1 cobalt and iron-dioxygen complexes of nitrogenous and other hard ligands into the superoxo class.* Cobalt(II) porphyrins and their adducts with diamagnetic molecules invariably have spin S = \(\frac{1}{2}\). (See Figure 4.16, but add one electron.) Thus the structural changes are less pronounced than for corresponding iron(II) systems. From the similarities in geometries and differences in electronic structures between cobalt- substituted and native hemoglobins and their models, many insights have been gained about the factors that determine oxygen affinity as well as how cooperativity might, or might not, work at the molecular level. The mechanism of cooperativity has also been probed by the substitution of other metalloporphyrins into the globin: for example, zinc porphyrins have been used for their excited triplet-state properties, manganese porphyrins for their EPR activity, and ruthenium porphyrins as a member of the iron triad. * Because the O—O stretch may be coupled with other ligand modes, its value should not be used to estimate superoxo character, although in a series of \(\mu\)-superoxo and \(\mu\)-peroxo complexes of carefully controlled stereochemistry, small changes in (O—O) have been correlated with the pK of the suite of ligands. | 5,321 | 1,627 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/16%3A_Aqueous_AcidBase_Equilibriums/16.01%3A_The_Autoionization_of_Water |
As you learned in and , acids and bases can be defined in several different ways ( ). Recall that the Arrhenius definition of an acid is a substance that dissociates in water to produce H ions (protons), and an Arrhenius base is a substance that dissociates in water to produce OH (hydroxide) ions. According to this view, an acid–base reaction involves the reaction of a proton with a hydroxide ion to form water. Although Brønsted and Lowry defined an acid similarly to Arrhenius by describing an acid as any substance that can a proton, the Brønsted–Lowry definition of a base is much more general than the Arrhenius definition. In Brønsted–Lowry terms, a base is any substance that can a proton, so a base is not limited to just a hydroxide ion. This means that for every Brønsted–Lowry acid, there exists a corresponding conjugate base with one fewer proton, as we demonstrated in . Consequently, all Brønsted–Lowry acid–base reactions actually involve conjugate acid–base pairs and the transfer of a from one substance (the acid) to another (the base). In contrast, the Lewis definition of acids and bases, discussed in , focuses on accepting or donating rather than protons. A Lewis base is an electron-pair donor, and a Lewis acid is an electron-pair acceptor. Because this chapter deals with acid–base equilibriums in , our discussion will use primarily the Brønsted–Lowry definitions and nomenclature. Remember, however, that all three definitions are just different ways of looking at the same kind of reaction: a proton is an acid, and the hydroxide ion is a base—no matter which definition you use. In practice, chemists tend to use whichever definition is most helpful to make a particular point or understand a given system. If, for example, we refer to a base as having one or more lone pairs of electrons that can accept a proton, we are simply combining the Lewis and Brønsted–Lowry definitions to emphasize the characteristic properties of a base. In , we also introduced the acid–base properties of water, its , and the definition of pH. The purpose of this section is to review those concepts and describe them using the concepts of chemical equilibrium developed in . The structure of the water molecule, with its polar O–H bonds and two lone pairs of electrons on the oxygen atom, was described in and , and the structure of liquid water was discussed in . Recall that because of its highly polar structure, liquid water can act as either an acid (by donating a proton to a base) or a base (by using a lone pair of electrons to accept a proton). For example, when a strong acid such as HCl dissolves in water, it dissociates into chloride ions (Cl ) and protons (H ). As you learned in , the proton, in turn, reacts with a water molecule to form the (H O ):
In this reaction, HCl is the acid, and water acts as a base by accepting an H ion. The reaction in is often written in a simpler form by removing H O from each side:
\[ HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \tag{16.1.2}\] In , the hydronium ion is represented by H , although free H ions do not exist in liquid water. Water can also act as an acid, as shown in . In this equilibrium reaction, H O donates a proton to NH , which acts as a base:
\[\underset{aicd}{H_2O_{(aq)}} + \underset{base}{NH_{3(aq)}} \rightleftharpoons \underset{acid}{NH^+_{4 (aq)}} + \underset{base}{OH^-_{(aq)}} \tag{16.1.3}\] Thus water is amphiprotic , meaning that it can behave as either an acid or a base, depending on the nature of the other reactant. Notice that is an equilibrium reaction as indicated by the double arrow. Because water is amphiprotic, one water molecule can react with another to form an OH ion and an H O ion in an autoionization process:
\[2H_2O(l) \rightleftharpoons H_3O^+_{(aq)}+OH^−_{(aq)} \tag{16.1.4}\] The equilibrium constant for this reaction can be written as follows:
\[ K=\dfrac{[H_3O^+(aq),OH^−(aq)]}{[H_2O(l)]^2} \tag{16.1.5}\] When pure liquid water is in equilibrium with hydronium and hydroxide ions at 25°C, the concentrations of the hydronium ion and the hydroxide ion are equal: [H O ] = [OH ] = 1.003 × 10 M. Thus the number of dissociated water molecules is very small indeed, approximately 2 ppb. We can calculate [H O] at 25°C from the density of water at this temperature (0.997 g/mL):
\[[H_2O(l)]=mol/L=(0.997\; \cancel{g}/mL)\left(\dfrac{1 \;mol}{18.02\; \cancel{g}}\right)\left(\dfrac{1000\; \cancel{mL}}{L}\right)=55.3\; M \tag{16.1.6}\] With so few water molecules dissociated, the equilibrium of the autoionization reaction ( ) lies far to the left. Consequently, [H O] is essentially unchanged by the autoionization reaction and can be treated as a constant. Incorporating this constant into the equilibrium expression allows us to rearrange to define a new equilibrium constant, the ion-product constant of liquid water ( ) \[K=\dfrac{K_w}{[H_2O(l)]^2} \tag{16.1.7a}\] with \[K_w = [H_3O^+(aq),OH^−(aq)]=[H_3O^+(aq),OH^−(aq)] \tag{16.1.7b}\] Substituting the values for [H O ] and [OH ] at 25°C into this expression,
\[K_w=(1.003 \times10^{−7})(1.003 \times 10^{−7})=1.006 \times 10^{−14} \tag{16.1.8}\] Thus, to three significant figures, = 1.01 × 10 M. Like any other equilibrium constant, varies with temperature, ranging from 1.15 × 10 at 0°C to 4.99 × 10 at 100°C. In pure water, the concentrations of the hydronium ion and the hydroxide ion are equal, and the solution is therefore neutral. If [H O ] > [OH ], however, the solution is acidic, whereas if [H O ] < [OH ], the solution is basic. For an aqueous solution, the H O concentration is a quantitative measure of acidity: the higher the H O concentration, the more acidic the solution. Conversely, the higher the OH concentration, the more basic the solution. In most situations that you will encounter, the H O and OH concentrations from the dissociation of water are so small (1.003 × 10 M) that they can be ignored in calculating the H O or OH concentrations of solutions of acids and bases, but this is not always the case. The is a concise way of describing the H O concentration and hence the acidity or basicity of a solution. Recall from that pH and the H (H O ) concentration are related as follows:
\[pH=−log_{10}[H^{+}(aq)] \tag{16.1.9}\] \[[H^{+}(aq)]=10^{−pH} \tag{16.1.10}\] Because the scale is logarithmic, a pH difference of between two solutions corresponds to a difference of a factor of in their hydronium ion concentrations. (Refer to Essential Skills 3 in , if you need to refresh your memory about how to use logarithms.) Recall also that the pH of a neutral solution is 7.00 ([H O ] = 1.0 × 10 M), whereas acidic solutions have pH < 7.00 (corresponding to [H O ] > 1.0 × 10 ) and basic solutions have pH > 7.00 (corresponding to [H O ] < 1.0 × 10 ). Similar notation systems are used to describe many other chemical quantities that contain a large negative exponent. For example, chemists use an analogous to describe the hydroxide ion concentration of a solution. The pOH and [OH ] are related as follows:
\[pOH=−log_{10}[OH^{−}(aq)] \tag{16.1.11}\] The constant can also be expressed using this notation, where p = −log . Because a neutral solution has [OH ] = 1.0 × 10 , the pOH of a neutral solution is 7.00. Consequently, the sum of the pH and the pOH for a neutral solution at 25°C is 7.00 + 7.00 = 14.00. We can show that the sum of pH and pOH is equal to 14.00 for aqueous solution at 25°C by taking the negative logarithm of both sides of :
\[−logKw=−log([H_{3}O^{+}(aq),OH{−}(aq)])=(−log[H_{3}O^{+}(aq)])+(−log[OH^{−}(aq)])=pH+pOH \tag{16.1.13}\] Thus at any temperature, pH + pOH = p , so at 25°C, where = 1.0 × 10 , pH + pOH = 14.00. More generally, the pH of any neutral solution is half of the p at that temperature. The relationship among pH, pOH, and the acidity or basicity of a solution is summarized graphically in over the common pH range of 0 to 14. Notice the inverse relationship between the pH and pOH scales. For any neutral solution, pH + pOH = 14.00 (at 25°C) and and pH=12pKw. The for water at 100°C is 4.99 × 10 . Calculate p for water at this temperature and the pH and the pOH for a neutral aqueous solution at 100°C. Report pH and pOH values to two decimal places. p , pH, and pOH Calculate p by taking the negative logarithm of . For a neutral aqueous solution, [H O ] = [OH ]. Use this relationship and to calculate [H O ] and [OH ]. Then determine the pH and the pOH for the solution. Because p is the negative logarithm of , we can write \[pK_w = −\log K_w = −\log(4.99 \times 10^{−13}) = 12.302 \notag \] The answer is reasonable: is between 10 and 10 , so p must be between 12 and 13. shows that = [H O ,OH ]. Because [H O ] = [OH ] in a neutral solution, we can let = [H O ] = [OH ]: \[K_w =[H_3O^+,OH^−]=(x)(x)=x^2 \notag \] \[x=\sqrt{K_w} =\sqrt{4.99 \times 10^{−13}} =7.06 \times 10^{−7}\; M \notag \] Because is equal to both [H O ] and [OH ], We could obtain the same answer more easily (without using logarithms) by using the p . In this case, we know that p = 12.302, and from , we know that p = pH + pOH. Because pH = pOH in a neutral solution, we can use directly, setting pH = pOH = . Solving to two decimal places we obtain the following: \[pK_w pH + pOH = y + y = 2y \notag \] \[y=\dfrac{pK_w}{2}=\dfrac{12.302}{2}=6.15=pH=pOH \notag \] Exercise Humans maintain an internal temperature of about 37°C. At this temperature, = 3.55 × 10 . Calculate p and the pH and the pOH of a neutral solution at 37°C. Report pH and pOH values to two decimal places. p = 13.45 pH = pOH = 6.73 Water is : it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion (H O ). The of liquid water produces OH and H O ions. The equilibrium constant for this reaction is called the and is defined as = [H O ,OH ]. At 25°C, is 1.01 × 10 ; hence pH + pOH = p = 14.00. : = [H O ,OH ] : pH = −log [H ] : [H ] = 10 : pOH = −log [OH ] : [OH ] = 10 : p = pH + pOH What is the relationship between the value of the equilibrium constant for the autoionization of liquid water and the tabulated value of the ion-product constant of liquid water ( )? The density of liquid water decreases as the temperature increases from 25°C to 50°C. Will this effect cause to increase or decrease? Why? Show that water is amphiprotic by writing balanced chemical equations for the reactions of water with HNO and NH . In which reaction does water act as the acid? In which does it act as the base? Write a chemical equation for each of the following. Show that for the sum of the following reactions is equal to . \[K_{auto} = \dfrac{[H_3O^+,OH^−]}{[H_2O]^2} \notag \] \[K_w = [H_3O^+,OH^−] = K_{auto}[H_2O]^2 \notag \] The autoionization of sulfuric acid can be described by the following chemical equation: \[H_2SO_{4(l)}+H_2SO_{4(l)} \rightleftharpoons H_3SO^+_{4(soln)}+H_SO^−_{4(soln)} \notag \] At 25°C, = 3 × 10 . Write an equilibrium constant expression for K(H SO ) that is analogous to . The density of H SO is 1.8 g/cm at 25°C. What is the concentration of H SO ? What fraction of H SO is ionized? An aqueous solution of a substance is found to have [H O] = 2.48 × 10 M. Is the solution acidic, neutral, or basic? The pH of a solution is 5.63. What is its pOH? What is the [OH ]? Is the solution acidic or basic? State whether each solution is acidic, neutral, or basic. Calculate the pH and the pOH of each solution. Calculate the pH and the pOH of each solution. The pH of stomach acid is approximately 1.5. What is the [H ]? Given the pH values in parentheses, what is the [H ] of each solution? A reaction requires the addition of 250.0 mL of a solution with a pH of 3.50. What mass of HCl (in milligrams) must be dissolved in 250 mL of water to produce a solution with this pH? If you require 333 mL of a pH 12.50 solution, how would you prepare it using a 0.500 M sodium hydroxide stock solution? \[K_{H_2SO_4}=[H_3SO_4^+,HSO_4^−]=K[H_2SO_4]_2 \notag \] \[[H_3SO_4^+] = 0.3 M \notag \] [H SO ] = 0.3 M; the fraction ionized is 0.02. pOH = 8.37; [OH ] = 4.3 × 10 M; acidic 2.9 mg HCl | 12,296 | 1,628 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Nuclear_Kinetics/Radioactive_Decay_Rates |
Radioactive decay is the loss of elementary particles from an unstable nucleus, ultimately changing the unstable element into another more stable element. There are five types of radioactive decay: alpha emission, beta emission, positron emission, electron capture, and gamma emission. Each type of decay emits a specific particle which changes the type of product produced. The number of protons and neutrons found in the daughter nuclei (the nuclei produced from the decay) are determined by the type of decay or emission that the original element goes through. In terms of entropy, radioactive decay can be defined as the tendency for matter and energy to gain inert uniformity or stability. For elements, uniformity is produced by having an equal number of neutrons and protons which in turn dictates the desired nuclear forces to keep the nuclear particles inside the nucleus. However, any instance where one particle becomes more frequent than another creates a nucleus that becomes unstable. The unstable nucleus then releases radiation in order to gain stability. For example, the stable element Beryllium usually contains 4 protons and 5 neutrons in its nucleus (this is not considered a very large difference). However, there exists a lighter isotope of Beryllium which contains 4 protons and only 3 neutrons, which gives a total mass of 7 amu. This lighter isotope decays into Lithium-7 through electron capture. A proton from Beryllium-7 captures a single electron and becomes a neutron. This reaction produces a new isotope (Lithium-7) that has the same atomic mass unit as Beryllium-7 but one less proton which stabilizes the element. Another example is the element Uranium-238 which has 54 more neutrons than its protons (Atomic umber =92). This element gains stability by passing through various types of decays (19 steps-- also known as the Uranium series) and is converted into Pb-206 (atomic number 82).For further information about different types of decay that Uranium goes through, refer to Decay Pathways). Due to the smaller size of the nucleus compared to the atom and the enormity of electromagnetic forces, it is impossible to predict radioactive decay. The atomic nucleus which is in the center of the atom is buffered by surrounding electrons and external conditions. Because of this, the study of decay is independent of the element's environment. In other words, the decay rate is independent of an element's physical state such as surrounding temperature and pressure. For a given element, the decay or disintegration rate is proportional to the number of atoms and the activity measured in terms of atoms per unit time. If "A" represents the disintegration rate and "N" is number of radioactive atoms, then the direct relationship between them can be shown as below: \[ A \propto N \label{1A}\] or mathematically speaking \[ A= \lambda N \label{1B} \] where Since the decay rate is dependent upon the number of radioactive atoms, in terms of chemical kinetics, one can say that radioactive decay is a first order reaction process. Even though radioactive decay is a first order reaction, where the rate of the reaction depends upon the concentration of one reactant (r = k [A,B] = k [A}) , it is not affected by factors that alter a typical chemical reactions. In other words, the reaction rate does not depend upon the temperature, pressure, and other physical determinants. However, like a typical rate law equation, radioactive decay rate can be integrated to link the concentration of a reactant with time. Also, radioactive decay is an exponential decay function which means the larger the quantity of atoms, the more rapidly the element will decay. Mathematically speaking, the relationship between quantity and time for radioactive decay can be expressed in following way: \[\dfrac{dN}{dt} = - \lambda N \label{2A}\] or more specifically \[\dfrac{dN(t)}{dt} = - \lambda N \label{2B}\] or via rearranging the separable differential equation \[\dfrac{dN(t)}{N (t)} = - \lambda dt \label{3}\] by Integrating the equation \[\ln N(t) = - \lambda t + C \label{4}\] with One could derive equation 4 in following manner, too. The decay rate constant, \(\lambda\), is in the units time . For further information about first-order reactions, refer to First-Order Reactions. There are two ways to characterize the decay constant: mean-life and half-life. In both cases the unit of measurement is seconds. As indicated by the name, mean-life is the average of an element's lifetime and can be shown in terms of following expression \[ N_t=N_o e^{-\lambda t} \label{5} \] \[1 = \int^{\infty}_ 0 c \cdot N_0 e^{-\lambda t} dt = c \cdot \dfrac{N_0}{\lambda} \label{6}\] Rearranging the equation: \[ c= \dfrac{\lambda}{N_o}\] Half-life is the time period that is characterized by the time it takes for half of the substance to decay (both radioactive and non-radioactive elements).The rate of decay remains constant throughout the decay process. There are three ways to show the exponential nature of half-life. \[ N_t=N_o\left( \dfrac{1}{2} \right)^{t/t_{1/2}} \label{7} \] \[ N_t=N_o e^{-t/\tau} \label{8} \] By comparing Equations 1, 3 and 4, one will get following expressions \[ \ln {\left( \dfrac{1}{2} \right)^{t/t_{1/2}}}= \ln(e^{-t/\tau}) = \ln (e^{-\lambda t} ) \label{9}\] or with \(\ln(e) = 1\), then \[ \dfrac{t}{t_{1/2}} \ln \left( \frac{1}{2} \right) = \dfrac{-t}{\tau} = -\lambda t \label{10}\] By canceling \(t\) on both sides, one will get following equation (for half-life) \[t_{1/2}= \dfrac{\ln(2)}{\lambda} \approx \dfrac{0.693}{\lambda} \label{11} \] or combining equations 1B and 11 \[ A = \dfrac{0.693}{t_{1/2}}N \label{12}\] Equation 11 is a constant, meaning the half-life of radioactive decay is constant. Half-life and the radioactive decay rate constant λ are inversely proportional which means the shorter the half-life, the larger \(\lambda\) and the faster the decay. This is a hypothetical radioactive decay graph. If the half-life were shorter, then the exponential decay graph would be steeper and the line would be decreasing at a faster rate; therefore, the amount of the radioactive nuclei would decrease as well. Radioactive decay is not always a one step phenomenon. Often times the parent nuclei changes into a radioactive daughter nuclei which also decays. In such cases, it is possible that the half-life of the parent nuclei is longer or shorter than the half-life of the daughter nuclei. Depending upon the substance, it is possible that both parent and daughter nuclei have similar half lives. Ba-140 Parent has a longer half-life than the daughter nuclei (La and Ce) Po-218 has a smaller half-life than its daughter nuclei (different species of Pb and Bi). Almost equal half-life for both parents (\(^{135}I\)) and daughter nuclei (\(^{135}Xe\) and \(^{135}Cs\)) Since the decay rate is constant, one can use the radioactive decay law and the half-life formula to find the age of organic material, which is known as radioactive dating. One of the forms of radioactive dating is radiocarbon dating. Carbon 14 (C-14) is produced in the upper atmosphere through the collision of cosmic rays with atmospheric 14N. This radioactive carbon is incorporated in plants and respiration and eventually with animals that feed upon plants. The ratio of C-14 to C-12 is 1:10^12 within plants as well as in the atmosphere. This ratio, however, increases upon the death of an animal or when a plant decays because there is no new income of carbon 14. By knowing the half-life of carbon-14 (which is 5730 years) one can calculate the rate of disintegration of the nuclei within the organism or substance and thereby determine its age. It is possible to use other radioactive elements in order to determine the age of nonliving substances as well. Odd Problem Solutions: 1. Using Equation 11, we can set \(t_{1/2} = 573\, yrs\) and solve for \(\lambda\). \[\lambda=1.209 \times 10^{-4}\; yr^{-1}\] 3. To find the number of atoms in a Carbon-14 sample, we will use dimensional analysis. First we convert 1.00mg to 0.001 grams. From the name, we know the atomic mass of Carbon-14 to be 14 g/mol. We know Avogadro's number expresses \(6.022 \times 10^{23}\) atoms. Now we have the equation: \[ N= \dfrac{0.001\; g}{(1\; mol/14\;g)(6.022 \times 10^{23})} \;atoms/ 1\;mol\] \(N=4.301 \times 10^{19} \, \text{atoms}\) 5. Using Equation 1B and Equation 12, we can combine them and solve for \(N\). By rearranging Equation 11, \(\lambda=\ln\; 2/t_{1/2}\) we can insert that into Equation 1B. Now we have the formula \(A=\ln 2/t_{1/2} N\). Now we have to convert 5.3 years to hours because the activity is measured in disintegration (atoms) per hour. \[t_{1/2}=5.3\; \cancel{years} \times \left(\dfrac{365\; \cancel{days}}{1\; \cancel{year}}\right) \times \left(\dfrac{24\;hr}{1\;\cancel{day}} \right)=46,428\; hours\] From equation 12, \(N\) can be calculated \[N = A\; \dfrac{t_{1/2}}{\ln 2}\] or \[N = (6,800\; dis/hr)\; \dfrac{46,428\; hr}{\ln 2} \approx 4.56 \times 10^8\; \text{atoms}\] The Cobalt-60 sample has 456,000,000 atoms. 7. We want to determine the decay constant. By looking at the first and last given values, we can use Equation 2 to solve for λ. \[\ln(276\;cpm / 2000\;cpm)=-\lambda \times 1250\;hr\] \[\lambda = 0.00158 \;hr^{-1}\] Then using Equation 11, we can solve for half-life. \[t_{1/2}=\dfrac{\ln 2}{0.00158}\] The half-life of the sample is 438 hours. | 9,443 | 1,629 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/06%3A_Bonding_in_Organic_Molecules/6.07%3A_Advanced_Quantum_Theory_of_Organic_Molecules |
In recent years, great progress has been made in quantum-mechanical calculations of the properties of small organic molecules by so-called methods, which means calculations from basic physical theory using only fundamental constants, without calibration from known molecular constants. Calculations that are calibrated by one or more known properties and then used to compute other properties are called "semiempirical" calculations. It should be made clear that there is no single, unique method. Rather, there is a multitude of approaches, all directed toward obtaining useful approximations to mathematical problems for which no solution in closed form is known or foreseeable. The calculations are formidable, because account must be taken of several factors: the attractive forces between the electrons and the nuclei, the interelectronic repulsions between individual electrons, the internuclear repulsions, and the electron spins. The success of any given method usually is judged by how well it reproduces known molecular properties with considerable premium for use of tolerable amounts of computer time. Unfortunately, many calculations do not start from a readily visualized physical model and hence give numbers that, although agreeing well with experiment, cannot be used to enhance one's qualitative understanding of chemical bonding. To be sure, this should not be regarded as a necessary condition for making calculations. But it also must be recognized that the whole qualitative orbital and hybridization approach to chemical bonding presented in this chapter was evolved from mathematical models used as starting points for early and semiempirical calculations. The success of any given method is judged by how well it reproduces known molecular properties. The efforts of many chemical theorists now are being directed to making calculations that could lead to useful new qualitative concepts of bonding capable of increasing our ability to predict the properties of complex molecules. One very successful procedure, called the "generalized valence-bond" (GVB) method, avoids specific hybridization assignments for the orbitals and calculates an optimum set of orbitals to give the most stable possible electronic configuration for the specified positions of the atomic nuclei. Each chemical bond in the GVB method involves two electrons with paired spins in two more or less localized atomic orbitals, one on each atom. Thus the bonds correspond rather closely to the qualitative formulations used previously in this chapter, for example Figure 6-14. The "generalized valence-bond" (GVB) method, avoids specific hybridization assignments for the orbitals and calculates an optimum set of orbitals to give the most stable possible electronic configuration for the specified positions of the atomic nuclei. Electron-amplitude contour diagrams of the GVB orbitals for ethene are shown in Figure 6-22. Let us be clear about what these contour lines represent. They are lines of analogous to topological maps for which contour lines are equal-altitude lines. The electron amplitudes shown are those calculated whose positions are shown with crosses. In general, the amplitudes decrease with distance from the nucleus. The regions of for \(s\)-like orbitals (middle-right of Figure 6-22) surround the nuclei as a set of concentric shells corresponding to the surfaces of the layers of an onion (Figure 6-23). With the \(sp^2\)-like orbitals, the amplitude is zero at the nucleus of the atom to which the orbital belongs. The physical significance of is that its square corresponds to the , a matter that we will discuss further in Chapter 21. The amplitude can be either positive or negative, but its square (the electron density) is positive, and this is the physical property that can be measured by appropriate experiments. Looking down on ethene, we see at the top of Figure 6-22 two identical \(C-C\) \(\sigma\)-bonding orbitals, one on each carbon, directed toward each other. The long dashed lines divide the space around the atom into regions of opposite orbital phase (solid is positive and dotted is negative). The contours for one of the \(C-H\) bonding orbitals are in the middle of the figure, and you will see that the orbital centered on the hydrogen is very much like an \(s\) orbital, while the one on the carbon is a hybrid orbital with considerable \(p\) character. There are three other similar sets of orbitals for the other ethene \(C-H\) bonds. When we look at the molecule edgewise, perpendicular to the \(C-C\) \(\sigma\) bond, we see the contours of the individual, essentially \(p\)-type, orbitals for \(\pi\) bonding. Ethyne shows two sets of these orbitals, as expected. What is the difference between the GVB orbitals and the ordinary hybrid orbitals we have discussed previously in this chapter? Consider the \(sp^2\)-like orbitals (upper part of Figure 6-22) and the \(sp^2\) hybrids shown in Figure 6-9. The important point is that the \(sp^2\) hybrid in Figure 6-9 is an atomic orbital calculated for a alone in space. The GVB orbital is much more physically realistic, because it is an orbital derived for a molecule with all of the nuclei and other electrons present. Nonetheless, the general shape of the GVB \(sp^2\)-like orbitals will be seen to correspond rather closely to the simple \(sp^2\) orbital in Figure 6-9. This should give us confidence in the qualitative use of our simple atomic-orbital models. and (1977) | 5,516 | 1,630 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.02%3A_Glassware_and_Equipment/1.2.01%3A_Storing_Samples_(Parafilm_Teflon_Tape) |
When samples must be stored for a period of time, they are best stored upright in screw-capped vials. Samples may evaporate through the joint over time, and if a highly volatile sample is used, the joint should be wrapped in Teflon tape (semi-stretchy white film, Figures 1.14 a+b) or Parafilm, (stretchy plastic film, Figures 1.14 c-e) to create a better seal. Teflon tape is less permeable to solvents than Parafilm, and volatile samples wrapped in Parafilm may still evaporate over a period of weeks. If a sample is to be stored in a round bottomed flask for some time, it should be stoppered with a cork stopper or plastic cap. A rubber stopper should not be used as it will tend to swell when exposed to organic vapors, and a glass stopper may freeze. | 768 | 1,631 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Reactions_of_Main_Group_Elements_with_Hydrogen |
Hydrides are binary compounds of hydrogen. There are three possible hydrides that can be formed: ionic hydrides, covalent hydrides, and metallic hydrides. form when hydrogen reacts with transition metals, therefore they will not be introduced in this module. Ionic hydrides form when hydrogen reacts with s-block metals, not including Be and Mg. These s-block elements are found in Group 1 and Group 2 of the periodic table and are the most active metals. Group 1 metals are referred to as alkali metals and have a charge of +1 Group 2 metals are referred to as alkaline earth metals and have a charge of +2. Both Group 1 and Group 2 metals have low electronegativity values (less than 1.2). | 713 | 1,633 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Reactions_of_Main_Group_Elements_with_Halogens |
This section describes the chemistry of halogens with the main group elements such as the , , and and . The word halogen itself means "salt former" in Greek. Halogens such as chlorine, bromine and iodine have properties that enable them to react with other elements to form important salts such as sodium chloride, also known as table salt. From a standard reduction potential table, it is determined that iodine and bromine cannot oxidize water to oxygen because they have smaller reduction potentials than oxygen. Thus, iodine and bromine do not react with water. However, fluorine and chlorine have larger reduction potentials, and can oxidize water. Fluorine reacts with water vapor to form oxygen and ozone: \[2F_2(g) + 2H_2O(g) \rightarrow 4HF(g)+ O_2(g) \] \[3F_2(g) + 2H_2O(g) \rightarrow 6HF(g) + O_3(g)\] The reaction of water with chlorine, shown below, proceeds very slowly. \[Cl_2 + H_2O \rightarrow H^+ + Cl^- + HClO\] Chlorine and bromine are moderately soluble in water. These solutions form solid hydrates within an ice lattice. These solutions are good oxidizing agents. Chlorine reacts reversibly with water to produce acids as in the following example, in which chloric acid and hydrochloric acid are formed: \[Cl_2 + H_2O \rightleftharpoons HClO + HCl\] Iodine is slightly soluble in water. It has the lowest standard reduction potential of the halogens, and is therefore the least powerful oxidizing agent. Air and other reagents can oxidize acidified solution of iodide ions. All the halogens react directly with hydrogen, forming covalent bonds and—at sufficient levels of purity—colorless gases at room temperature. Hydrogen reacts with fluorine, chlorine, bromine, and iodine, forming HF, HCl, HBr, and HI, respectively. The bond strength of these molecules decreases down the group: \(HF > HCl > HBr > HI\). Iodine and hydrogen react non-spontaneously to produce hydrogen iodide \[H_2 + I_2 \rightarrow 2HI\] All the hydrogen halides are soluble in water, in which they form strong acids (with the exception of \(HF\)). The general equation of hydrogen halide for the acid reaction is given below: \[HX + H_2O \rightarrow H_3O^+ + X^-\] All the react vigorously with halogens to produce salts, the most industrially important of which are NaCl and KCl. \[ 2Na(s) + Cl_2(g) \rightarrow 2NaCl(s)\] Sodium Chloride is used as a preservative for meat and to melt the ice on the roads (via ). KCl is important for plant fertilizers because of the positive impact of potassium on plant growth. These metal halides form white ionic crystalline solids and are all soluble in water except LiF, because of its high lattice enthalpy attributed to strong electrostatic attraction between Li and F ions. The react to form hydrated halides. These halides are ionic except for those involving (the least metallic of the group). Because alkaline earth metals tend to lose electrons and halogen atoms tend to gain electrons ( ), the chemical reaction between these groups is the following: \[M + X_2 \rightarrow MX_2\] where Anhydrous calcium chloride has strong affinity for water, absorbing enough to dissolve its own crystal lattice. It can be produced directly from limestone, or as a by-product by . All the elements react with Halogens to form trihalides. Aluminum Fluoride, \(AlF_3\), is an ionic compound with a high melting point. However, most of the other aluminum halides form molecules with the formula \(Al_2X_6\) (\(X\) represents chlorine, bromine, or iodine). When two \(AlX_3\) units (or, more generally, any two identical units) join together, the resulting molecule is called a dimer. Aluminum halides are very reactive Lewis acids. They accept electrons and form acid-base compound called adducts, as in the following example: \[AlCl_3 + (C_2H_5)_2O \rightarrow Al(C_2H_5)_2OCl_3\] In this reaction, \(AlCl_3\) is the and \((C_2H_5)_2O\) is the . elements form halides with general formula MX (CCl , SiCl , GeCl , SnCl , PbCl ), although some elements such as Ge, Sn, Pb can also form dihalides (MX ). The tetrahalides of carbon, such as CCl , cannot be hydrolyzed due to non-availability of vacant valence d-orbitals, but other tetrahalides can be hydrolyzed. Silicon reacts with halogens to form compounds of the form SiX , where X represents any common halogen. At room temperature, SiF is a colorless gas, SiCl is a colorless liquid, SiBr is a colorless liquid, and SiI forms colorless crystals. SiF and SiCl can be completely hydrolyzed, but SiBr can be only partially hydrolyzed. Lead and tin are metals in Group 14. Tin occurs as both SnO and SnO . SnCl is a good reducing agent and is found in tinstone. SnF was once used as additive to toothpaste but now is replaced by NaF. Sulfur reacts directly with all the halogens except iodine. It spontaneously combines with fluorine to form sulfur hexafluoride, SF , a colorless and inert gas. It can also form SF which is a powerful fluorinating agent. Sulfur and chlorine form SCl , a red liquid, which is used in the production of the poisonous . This reaction is shown below: 3SF + 4BCl → 4BF + 3SCl + 3Cl Oxygen combines with fluoride to form the compounds OF and O F . The structures of these molecules resemble that of hydrogen peroxide, although they are much more reactive. Common halogen oxides include \(Cl_2O\), \(ClO_2\), \(Cl_2O_4\), and \(I_2O_5\). Chlorine monoxide, the anhydride of hypochlorous acid, reacts vigorously with water as shown below, giving off chlorine and oxygen as products. \[Cl_2O + H_2O \rightleftharpoons 2HOCl\] Chlorine dioxide and chlorine perchlorate form when sulfuric acid reacts with potassium chlorate. These compounds are similar to the nitrogen compounds \(NO_2\) and \(N_2O_4\). Iodine pentoxide forms iodic anhydride when reacted with water, as shown: \[I_2O_5 + H_2O \rightarrow 2HIO_3\] Compounds that are made up of both oxygen and hydrogen are considered to be oxygen acids, or oxoacids. Common oxoacids are shown in the table below. HClO HBrO HIO HClO HBrO HIO HClO HBrO HIO HClO The oxoanions of chlorine are following: All these compounds have common uses. For example, sodium chlorite is used as bleaching agent for textiles. Chlorate is a very good oxidizing agent and is very important in matches and fireworks. Halogens have the ability to form compounds with other halogens ( ). They are represented with the notation XY, in which the X and Y refer to two different halogens. Examples of this type of molecule include IBr and BrCl. They can also form polyatomic molecules such as XY , XY , XY , corresponding to molecules such as IF , BrF , and IF . Most interhalogen compounds such as CIF and BrF are very reactive. \(AlCl_3\) is a molecular compound (molecular formula) \(AlF_3\) is an ionic compound (formula compound) | 6,836 | 1,634 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/07%3A_Case_Studies-_Kinetics/7.02%3A_Chymotrypsin |
Chymotrypsin is a digestive enzyme belonging to a super family of enzymes called serine proteases. It uses an active serine residue to perform hydrolysis on the C-terminus of the aromatic amino acids of other proteins. Chymotrypsin is a protease enzyme that cleaves on the C-terminal phenylalanine (F), tryptophan (W), and tyrosine (Y) on peptide chains. It shows specificity for aromatic amino acids because of its hydrophobic pocket. Chymotrypsin is one of the most studied enzymes due to its two phase kinetics: pre-steady-state and steady state. The study of these two kinetic states gives evidence of the "Ping-Pong" mechanism, the formation of covalent complexes leading to covalent hydrolysis reactions, and the rate of the catalyzed reactions. Synthesis of chymotrypsin occurs primarily in the pancreas. Instead of the active form, however, it is produced as an inactive zymogen called chymotrypsinogen to prevent its protease activity from digesting the pancreas. Upon secretion into the lumen of the small intestine, it is converted to its active form by another enzyme called trypsin. This dependence of a different enzyme for the activation of a protease is a common way for the body to prevent the digestion of organs and other harmful enzymatic side-effects. Chymotrypsin operates through a general mechanism known as the ping-pong mechanism (Figure \(\Page {1}\)) whereby the enzyme reacts with a substrate to form an enzyme intermediate. This intermediate has different properties than the initial enzyme, so to regenerate the initial enzymatic activity, it must react with a secondary substrate. This process is illustrated below: More specifically, chymotrypsin operates through a particular type of ping-pong mechanism called covalent hydrolysis. This means that the enzyme first forms a covalent bond with the target substrate, displacing the more stable moiety into solution. This enzyme-substrate complex is called the enzyme intermediate. The intermediate then reacts with water, which displaces the remaining part of the initial substrate and reforms the initial enzyme.
Chymotrypsin, like most enzymes, is specific in the types of substrates with which it reacts. As a protease, it cleaves polypeptides, and its inherent specificity allows it to act only on the carboxy-terminal of aromatic residues. It is a somewhat complicated mechanism, and is best explained in a series of steps.
Step 1: The target enters the active site of chymotrypsin, and it is held there by hydrophobic interactions between exposed non-polar groups of enzyme residues and the non-polar aromatic side-chain of the substrate. It is important to note the hydrogen bond between the Schiff nitrogen on histidine-57 and the oxygen side-chain of serine-195. Step 2: Aided by the histidine-serine hydrogen bonding, the hydroxyl group on serine-195 performs a nucleophilic attack on the carbonyl carbon of an aromatic amino acid while simultaneously transferring the hydroxyl hydrogen to the histidine Schiff nitrogen. This attack pushes the pi carbonyl electrons onto the carbonyl oxygen, forming a short-lived intermediate consisting of a c-terminal carbon with four single bonds: an oxygen anion, the beta-carbon of the aromatic amino acid, the n-terminus of the subsequent amino acid of the substrate protein, and the serine-195 side-chain oxygen. Step 3: This intermediate is short-lived, as the oxyanion electrons reform the pi bond with the c-terminus of the aromatic amino acid. The bond between the carboxy-terminus of the aromatic amino acid and the n-terminus of the subsequent residue is cleaved, and its electrons are used to extract the hydrogen of the protonated Schiff nitrogen on histidine-57. The bonds between the carbonyl carbon and the serine-195 oxygen remain in an ester configuration. This is called the acyl-enzyme intermediate. The c-terminal side of the polypeptide is now free to dissociate from the active site of the enzyme.
Step 4: Water molecules are now able to enter and bind to active site through hydrogen bonding between the hydrogen atoms of water and the histidine-57 Schiff nitrogen.
Step 5: The water oxygen now makes a nucleophilic attack on the carbonyl carbon of the acyl-enzyme intermediate, pushing the carbonyl’s pi electrons onto the carbonyl carbon as histidine-57 extracts one proton from water. This forms another quaternary carbon covalently bonded with serine, a hydroxyl, an oxyanion, and the aromatic amino acid. The proton on the recently protonated histidine-57 is now able to make a hydrogen bond with the serine oxygen.
Step 6: The oxyanion electrons reform the carbonyl pi bond, cleaving the bond between the carbonyl carbon and the serine hydroxyl. The electrons in this bond are used by the serine oxygen to deprotonate the histidine Schiff nitrogen and reform the original enzyme. The substrate no longer has affinity for the active site, and it soon dissociates from the complex.
Experiments were conducted in 1953 by B.S. Hartley and B.A. Kilby to investigate the kinetics of chymotrypsin-catalyzed hydrolysis. Instead of using a poly-peptide chain as a substrate, they used a nitro-phenyl ester, p-nitrophenyl acetate, that resembles an aromatic amino acid. Hydrolysis of this compound by chymotrypsin at the carbonyl group yields acetate and nitrophenolate, the latter of which absorbs near 400 nm light and its concentration can thus be measured by spectrophotometry (Figure \(\Page {2}\)). Spectrophotometric analysis of chymotrypsin acting on nitrophenylacetate showed that nitrophenolate was produced at a rate independent of substrate concentration, proving that the only factor contributing to the rate of product formation is the concentration of enzyme; this is typical for enzyme-substrate kinetics. However, when the slope of the 0-order absorbance plot was traced back to the starting point (time = 0), it was found that the initial concentration of nitrophenolate was not 0. In fact, it showed a 1:1 stoichiometric ratio with the amount of chymotrypsin used in the assay. This can only be explained by the fact that hydrolysis by chymotrypsin is biphasic in nature, meaning that it proceeds in two distinct steps. To analytically determine the rate of catalysis, all substrates, products, and intermediates must be defined. Refer to the figure below:
Using these abbreviations, kinetic analysis becomes less cumbersome. 1. The initial amount of enzyme can be represented as the sum of the free enzyme, the bound enzyme, and the inactive intermediate. \[[E]_o = [ES] + [^*ES] + [E] \nonumber \] 2. Assuming the initial step is the faster than the subsequent steps, the rate of nitrophenolate production can be described as: \[\dfrac{d[P_1]}{dt}=k_2[ES] \nonumber \] 3. Likewise, the rate of acetate formation can be represented by the equation: \[\dfrac{d[P_2]}{dt}=k_3[^*ES] \nonumber \] 4. Therefore, the net change in concentration of the inactive intermediate can be deduced: \[\dfrac{d[*ES]}{dt}=k_2[ES]-k_3[^*ES] \nonumber \] 5. The last inference that can be made from analysis of the measured kinetics data (Figure \(\Page {2}\)) is that the first step of the reaction equilibrates rapidly, and thus the change in bound substrate can be described in the following equation. This is a principal tenet in analyzing catalysis. \[\dfrac{d[ES]}{dt}=k_1[E,S]-k_{-1}[ES]=0 \nonumber \] 6. Where: \[\dfrac{k_{-1}}{k_1} = K_s = \dfrac{[E,S]}{[ES]} \nonumber \] 7. Combining all of these quantities, we can deduce the catalytic rate constant as: \[ k_{cat} = \dfrac{k_2k_3}{k_2+k_3} \nonumber \] 8. In ester hydrolysis, \(k_3 >> k_2\), so the resultant catalytic rate constant simplifies to: \[k_{cat}=k_2 \nonumber \] which is in agreement with the observed zeroth-order kinetics of Figure \(\Page {2}\). Speculate on how the catalytic rate constant can be determined from the spectrophotogram. The catalytic rate constant can be deduced from the graph by simply determining the slope of the line where the reaction demonstrates 0-order kinetics (the linear part) How can product be consistently produced if the rate of change of the ES complex is 0? This is pre-equilibrium kinetics in action. The ES complex is formed from E and S at a faster rate than any other step in the reaction. As soon as ES is converted to *ES, another mole of ES is produced from an infinite supply of E + S. This means that the amount of ES and E + S is constantly at equilibrium, and thus the change of either with respect to time is 0. How would the rate of product formation change if: Explain the role of hydrogen bonding in protein hydrolysis catalyzed by chymotrypsin. Initially, hydrogen bonding between the enzymes histidine and serine side chains weakens the bond of serine’s O-H. This allows for a facilitated nucleophilic attack of the hydroxyl Oxygen on the substrates carbonyl group. Conversely, in the final step of the reaction, the bound serine oxygen forms a hydrogen bond with a protonated histidine, which allows for easier cleavage from the substrate. What would the spectrophotogram look like if the reaction proceeded via a steady-state mechanism instead of pre-equilibrium. The graph would show similar 0-order kinetics, but the line would intercept the Y-axis at an absorbance of 0 instead of the 1:1 mole ratio of nitrophenolate to enzyme. | 9,348 | 1,637 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/02%3A_Reaction_Rates/2.10%3A_Zero-Order_Reactions |
In some reactions, the rate is independent of the reactant concentration. The rates of these do not vary with increasing nor decreasing reactants concentrations. This means that the rate of the reaction is equal to the rate constant, \(k\), of that reaction. This property differs from both and . Zero-order kinetics is an artifact of the conditions under which the reaction is carried out. For this reason, reactions that follow zero-order kinetics are often referred to as pseudo-zero-order reactions. Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left. There are two general conditions that can give rise to zero-order rates: This situation commonly occurs when a reaction is catalyzed by attachment to a solid surface ( ) or to an enzyme. Nitrous oxide will decompose exothermically into nitrogen and oxygen, at a temperature of approximately 575 °C \[\ce{2N_2O ->[\Delta, \,Ni] 2N_2(g) + O_2(g)} \nonumber \] This reaction in the presence of a hot platinum wire (which acts as a catalyst) is zero-order, but it follows more conventional when carried out entirely in the gas phase. \[\ce{2N_2O -> 2N_2(g) + O_2(g)} \nonumber \] In this case, the \(N_2O\) molecules that react are limited to those that have attached themselves to the surface of the solid catalyst. Once all of the sites on the limited surface of the catalyst have been occupied, additional gas-phase molecules must wait until the decomposition of one of the adsorbed molecules frees up a surface site. Enzyme-catalyzed reactions in organisms begin with the attachment of the substrate to the active site on the enzyme, leading to the formation of an . If the number of enzyme molecules is limited in relation to substrate molecules, then the reaction may appear to be zero-order. This is most often seen when two or more reactants are involved. Thus if the reaction \[ A + B \rightarrow \text{products} \tag{1} \] is first-order in both reactants so that \[\text{rate} = k [A,B] \tag{2} \] If \(B\) is present in , then the reaction will appear to be zero order in \(B\) (and first order overall). This commonly happens when \(B\) is also the solvent that the reaction occurs in. \[Rate = - \dfrac{d[A]}{dt} = k[A]^0 = k = constant \tag{3} \] where \(Rate\) is the reaction rate and \(k\) is the reaction rate coefficient. In this example, the units of \(k\) are M/s. The units can vary with other types of reactions. For zero-order reactions, the units of the rate constants are always M/s. In higher order reactions, \(k\) will have different units. Integration of the differential rate law yields the concentration as a function of time. Start with the general rate law equations \[Rate = k[A]^n \tag{4} \] First, write the differential form of the rate law with \(n=0\) \[Rate = - \dfrac{d[A]^0}{dt} = k \tag{5} \] then rearrange \[{d}[A] = -kdt \tag{6} \] Second, integrate both sides of the equation. \[\int_{[A]_{0}}^{[A]} d[A] = - \int_{0}^{t} kdt \tag{7} \] Third, solve for \([A]\). This provides the integrated form of the rate law. \[[A] = [A]_0 -kt \tag{8} \] The integrated form of the rate law allows us to find the population of reactant at any time after the start of the reaction. \[[A] = -kt + [A]_0 \tag{9} \] is in the form y = mx+b where slope = m = -k and the y- intercept = b = \([A]_0\) Zero-order reactions are applicable for a very narrow region of time. Therefore, the linear graph shown below (Figure 2) is only realistic over a limited time range. If we were to extrapolate the line of this graph downward to represent all values of time for a given reaction, it would tell us that as time progresses, the concentration of our reactant becomes negative. We know that concentrations can never be negative, which is why zero-order reaction kinetics is applicable for describing a reaction for only brief window and must eventually transition into kinetics of a different order.
To understand where the above graph comes from, let us consider a catalyzed reaction. At the beginning of the reaction, and for small values of time, the rate of the reaction is constant; this is indicated by the blue line in Figures 2; right. This situation typically happens when a catalyst is saturated with reactants. With respect to , this point of catalyst saturation is related to the \(V_{max}\). As a reaction progresses through time, however, it is possible that less and less substrate will bind to the catalyst. As this occurs, the reaction slows and we see a tailing off of the graph (Figure 2; right). This portion of the reaction is represented by the dashed black line. In looking at this particular reaction, we can see that reactions are zero-order under all conditions. They are only zero-order for a limited amount of time. If we plot rate as a function of time, we obtain the graph below (Figure 3). Again, this only describes a narrow region of time. The slope of the graph is equal to k, the rate constant. Therefore, k is constant with time. In addition, we can see that the reaction rate is completely independent of how much reactant you put in.
The half-life. \(t_{1/2}\), is a timescale in which each half-life represents the reduction of the initial population to 50% of its original state. We can represent the relationship by the following equation. \[[A] = \dfrac{1}{2} [A]_o \tag{10} \] Using the integrated form of the rate law, we can develop a relationship between zero-order reactions and the half-life. \[[A] = [A]_o - kt \tag{11} \] Substitute \[\dfrac{1}{2}[A]_o = [A]_o - kt_{\dfrac{1}{2}} \tag{12} \] Solve for \(t_{1/2}\) Notice that, for zero-order reactions, the half-life on the initial concentration of reactant and the rate constant. Which reaction represents a zero-order reaction? The kinetics of any reaction depend on the reaction mechanism, or rate law, and the initial conditions. If we assume for the reaction A -> Products that there is an initial concentration of reactant of [A] at time t=0, and the rate law is an integral order in A, then we can summarize the kinetics of the zero-order reaction as follows: | 6,276 | 1,638 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/23%3A_Barrier_Crossing_and_Activated_Processes/23.02%3A_Kramers_Theory |
In our treatment the motion of the reactant over the transition state was treated as a free transitional degree of freedom. This ballistic or inertial motion is not representative of dynamics in soft matter at room temperature. is the leading approach to describe diffusive barrier crossing. It accounts for friction and thermal agitation that reduce the fraction of successful barrier crossings. Again, the rates are obtained from the flux over barrier along reaction coordinate, Equation (23.1). One approach is to treat diffusive crossing over the barrier in a potential using the . The diffusive flux under influence of potential has two contributions: \[ J(\xi)=-D \frac{d C(\xi)}{d \xi}-\frac{C(\xi)}{\zeta} \frac{d U(\xi)}{d \xi} \nonumber\] As discussed earlier \(ζ\) is the and in one dimension: \[ \zeta=\frac{k_{B} T}{D} \nonumber\] Written in terms of a probability density \(P\) \[\begin{align*} J &=D\left(-\frac{P}{k_{B} T} \frac{d U}{d \xi}-\frac{d P}{d \xi}\right) \\ &=-D e^{-U / k_{B} T} \frac{d}{d \xi}\left(P e^{U / k_{B} T}\right) \end{align*}\] or \[ J e^{U / k_{B} T}=-D \frac{d}{d \xi} P e^{U / k_{B} T} \] Here we have assumed that \(D\) and \(ζ\) are not functions of \(ξ\). The next important assumption of Kramers’ theory is that we can solve for the diffusive flux using the steady-state approximation. This allows us to set: \(J\) = constant. Integrate along \(ξ\) over barrier. \[ \begin{align*} J \int_{a}^{b} e^{U / k_{B} T} d \xi * &=-D \int_{a}^{b} d P e^{U / k_{B} T} \\[4pt] J \int_{a}^{b} e^{U(\xi) / k_{B} T} d \xi &=D\left\{P_{R} e^{U_{R} / k_{B} T}-P_{P} e^{U_{P} / k_{B} T}\right\} \end{align*}\] \(P_i\) are the probabilities of occupying the \(R\) or \(P\) state, and \(U_i\) are the energies of the \(R\) and \(P\) states. The right hand side of this equation describes net flux across barrier. Let’s consider only flux from \(R\longrightarrow P\): \( J_{R\longrightarrow P} \), which we do by setting \(P_P\longrightarrow 0\). This is just a barrier escape problem. Also as a reference point, we set \(U_R(ξ_R) = 0\). \[ J_{R \rightarrow P}=\frac{D P_{R}}{\int_{a}^{b} e^{U(\xi) / k_{B} T} d \xi} \label{23.2.2}\] The flux is linearly proportional to the diffusion coefficient and the probability of being in the reactant state. The flux is reduced by a factor that describes the energetic barrier to be overcome. Now let’s evaluate with a specific form of the potential. The simplest form is to model \(U(ξ)\) with parabolas. The reactant well is given by \[ U_{R}=\frac{1}{2} m \omega_{R}^{2}\left(\xi-\xi_{R}\right)^{2} \label{23.2.3}\] and we set \(\xi_R \longrightarrow 0\). The barrier is modeled by an inverted parabola centered at the transition state with a barrier height for the forward reaction \(E_f\) and a width given by the barrier frequency \(ω_{bar}\): \[ U_{\mathrm{bar}}=E_{f}-\frac{1}{2} m \omega_{\mathrm{bar}}^{2}\left(\xi-\xi^{‡}\right)^{2} \nonumber\] In essence this is treating the evolution of the probability distribution as the motion of a fictitious particle with mass \(m\). First we evaluate the denominator in Equation \ref{23.2.2}. \(e^{U_{bar}/k_BT}\) is a probability density that is peaked at \( \xi^‡\), so changing the limits on the integral does not affect things much. \[\int_{a}^{b} e^{U_{b a r} / k_{B} T} d \xi \approx \int_{-\infty}^{+\infty} d \xi \exp \left[-\frac{m \omega_{B}^{2}\left(\xi-\xi^{‡}\right)^{2}}{2 k_{B} T}\right]=\sqrt{\frac{2 \pi k_{B} T}{m \omega_{B}^{2}}} \nonumber\] Then Equation \ref{23.2.2} becomes \[ J_{R \rightarrow P}=\omega_{\text {bar }} D \sqrt{\frac{m}{2 \pi k_{B} T}} e^{-E_{f} / k_{B} T} P_{R} \label{23.2.4}\] Next, let’s evaluate \(P_R\). For a the Gaussian well in Equation \ref{23.2.3}, the probability density along \(ξ\) is \(P_R = e^{-U_{R} / k_{B} T}\): \[ P_{R}(\xi)=\exp \left[-\frac{1}{2} m \omega_{R}^{2}\left(\xi-\xi_{R}\right)^{2} / k_{B} T\right] \nonumber\] \[ P_{R} \approx \int_{-\infty}^{+\infty} P_{R}(\xi) d \xi=\omega_{R} \sqrt{\frac{m}{2 \pi k_{B} T}} \nonumber\] Substituting this into Equation \ref{23.2.4} we have \[ J_{R \rightarrow P}=\omega_{R} \omega_{b a r} D\left(\frac{m}{2 \pi k_{B} T}\right) e^{-E_{f} / k_{B} T} \nonumber\] Using the Einstein relation \( D = k_BT/\zeta \), we find that the forward flux scales inversely with friction (or viscosity). \[ J_{R \rightarrow P}=\frac{\omega_{R}}{2 \pi} \frac{\omega_{b a r}}{\zeta} e^{-E_{f} / k_{B} T} \] Also, the factor of \(m\) disappears when the problem is expressed in mass-weighted coordinates \(\omega_{\text {bar }}=\sqrt{m} \omega_{\text {bar }}\). Note the similarity of Equation (9) to transition state theory. If we associate the period of the particle in the reactant well with the barrier crossing frequency, \[ \frac{\omega_{R}}{2 \pi} \Rightarrow v=\frac{k_{B} T}{h} \nonumber\] then we can also find that we an expression for the transmission coefficient in this model: \[ k_{diff}=\kappa_{diff} k_{TST} \nonumber\] \[ \kappa_{diff}=\dfrac{\omega_{bar}}{\zeta}\ll 1 \nonumber\] This is the reaction rate in the strong damping, or diffusive, limit. Hendrik Kramers actually solved a more general problem based on the that described intermediate to strong damping. The reaction rate was described as \[ k_{Kr}=\kappa_{Kr} k_{TST} \nonumber\] \[\kappa_{K r}=\frac{1}{\omega_{b a r}}\left(-\frac{\zeta}{2}+\sqrt{\frac{\zeta^{2}}{4}+\omega_{b a r}^{2}}\right) \nonumber\] \[ \zeta=\frac{1}{m k_{B} T} \int_{0}^{\infty} d t\langle\xi(0) \xi(t)\rangle \nonumber\] This shows a crossover in behavior between the strong damping (or diffusive) behavior described above and an intermediate damping regime: In the weak friction limit, Kramers argued that the reaction rate scaled as \[k_{\text {weak }} \sim \zeta k_{T S T} \nonumber\] That is, if you had no friction at all, the particle would just move back and forth between the reactant and product state without committing to a particular well. You need some dissipation to relax irreversibly into the product well. On the basis of this we expect an optimal friction that maximizes \(κ\), which balances the need for some dissipation but without so much that barrier crossing is exceedingly rare. This “ ” is captured by the interpolation formula \[\kappa^{-1}=\kappa_{K r}^{-1}+\kappa_{\text {weak }}^{-1} \nonumber\] | 6,310 | 1,640 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/12%3A_Cycloalkanes_Cycloalkenes_and_Cycloalkynes/12.07%3A_Cycloalkenes_and_Cycloalkanes |
The \(\ce{C-C=C}\) angle in alkenes normally is about \(122^\text{o}\), which is \(10^\text{o}\) larger than the normal \(\ce{C-C-C}\) angle in cycloalkanes. This means that we would expect about \(20^\text{o}\) more angle strain in small-ring cycloalkenes than in the cycloalkanes with the same numbers of carbons in the ring. Comparison of the data for cycloalkenes in Table 12-5 and for cycloalkanes in Table 12-3 reveals that this expectation is realized for cyclopropene, but is less conspicuous for cyclobutene and cyclopentene. The reason for this is not clear, but may be connected in part with the \(\ce{C-H}\) bond strengths (see Section 12-4B). Cyclopropene has rather exceptional properties compared to the other cycloalkenes. It is quite unstable and the liquid polymerizes spontaneously although slowly, even at \(-80^\text{o}\). This substance, unlike other alkenes, reacts rapidly with iodine and behaves like an alkyne in that one of its double-bond hydrogens is replaced in silver-ammonia solution to yield an alkynide-like silver complex. The \(\ce{C-C=C}\) bond angles in alkynes normally are \(180^\text{o}\) and the angle strain involved in making a small-ring cycloalkyne, such as cyclopropyne, apparently is prohibitive. The smallest reasonably stable member of the series is cyclooctyne, and its properties, along with those of some higher homologs, are shown in Table 12-5. Strong evidence has been adduced for the existence of cyclopentyne, cyclohexyne, and cycloheptyne as unstable reaction intermediates. and (1977) | 1,567 | 1,641 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.01%3A_General_Principles_of_Catalysis |
As can be seen from the Arrhenius equation, the magnitude of the activation energy, \(E_a\), determines the value of the rate constant, \(k\), at a given temperature and thus the overall reaction rate. Catalysts provide a means of reducing \(E_a\) and increasing the reaction rate. Catalysts are defined as substances that participate in a chemical reaction but are not changed or consumed. Instead they provide a new mechanism for a reaction to occur which has a lower activation energy than that of the reaction without the catalyst. refers to reactions in which the catalyst is in solution with at least one of the reactants whereas refers to reactions in which the catalyst is present in a different phase, usually as a solid, than the reactants. Figure \(\Page {1}\) shows a comparison of energy profiles of a reaction in the absence and presence of a catalyst. Consider a non-catalyzed elementary reaction \[\text{A} \overset{k}{\longrightarrow} \text{P}\] which proceeds at rate \(k\) at a certain temperature. The reaction rate can be expressed as \[\dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right] \label{Eq1}\] In the presence of a catalyst \(\text{C}\), we can write the reaction as \[\text{A} + \text{C} \overset{k_\text{cat}}{\longrightarrow} \text{P} + \text{C}\] and the reaction rate as \[\dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right] - k_\text{cat} \left[ \text{A} \right] \left[ \text{C} \right] \label{Eq2}\] where the first term represents the uncatalyzed reaction and the second term represents the catalyzed reaction. Because the reaction rate of the catalyzed reaction is often magnitudes larger than that of the uncatalyzed reaction (i.e. \(k_\text{cat} \gg k\)), the first term can often be ignored. A common example of homogeneous catalysts are acids and bases. For example, given an overall reaction is \(\text{S} \rightarrow \text{P}\). If \(k\) is the rate, then \[\dfrac{d \left[ \text{P} \right]}{dt} = k \left[ \text{S} \right]\] The purpose of an enzyme is to enhance the rate of production of the product \(\text{P}\). The equations of the acid-catalyzed reaction are \[\begin{align} \text{S} + \text{A}H &\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{S}H^+ + \text{A}^- \\ \text{S}H^+ + H_2 O &\overset{k_2}{\rightarrow} \text{P} + H_3 O^+ \\ H_3 O^+ + \text{A}^- &\overset{k_3}{\underset{k_{-3}}{\rightleftharpoons}} \text{A}H + H_2 O \end{align}\] The full set of kinetic equations is \[\begin{align} \dfrac{d \left[ \text{S} \right]}{dt} &= -k_1 \left[ \text{S} \right] \left[ \text{A} H \right] + k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] \\ \dfrac{ d \left[ \text{A} H \right]}{dt} &= -k_1 \left[ \text{S} \right] \left[ \text{A} H \right] + k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] - k_{-3} \left[ \text{A} H \right] + k_3 \left[ H_3 O^+ \right] \left[ \text{A}^- \right] \\ \dfrac{d \left[ \text{S} H^+ \right]}{dt} &= k_1 \left[ \text{S} \right] \left[ \text{A} H \right] - k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] - k_2 \left[ \text{S} H^+ \right] \\ \dfrac{d \left[ \text{A}^- \right]}{dt} &= k_1 \left[ \text{S} \right] \left[ \text{A} H \right] - k_{-1} \left[ \text{S} H^+ \right] \left[ \text{A}^- \right] -k_2 \left[ \text{A}^- \right] \left[ H_3 O^+ \right] + k_{-3} \left[ \text{A} H \right] \\ \dfrac{d \left[ \text{P} \right]}{dt} &= k_2 \left[ \text{S} H^+ \right] \\ \dfrac{d \left[ H_3 O^+ \right]}{dt} &= -k_2 \left[ \text{S} H^+ \right] - k_3 \left[ H_3 O^+ \right] \left[ \text{A}^- \right] + k_{-3} \left[ \text{A} H \right] \end{align} \label{Eq3}\] We cannot easily solve these, as they are nonlinear. However, let us consider two cases \(k_2 \gg k_{-1} \left[ \text{A}^- \right]\) and \(k_2 \ll k_{-1} \left[ \text{A}^- \right]\). In both cases, \(\text{S} H^+\) is consumed quickly, and we can apply a steady-state approximation: \[\dfrac{d \left[ \text{S} H^+ \right]}{dt} = k_1 \left[ \text{S} \right] \left[ \text{A} H \right] - k_{-1} \left[ \text{A}^- \right] \left[ \text{S} H^+ \right] - k_2 \left[ \text{S} H^+ \right] = 0 \label{Eq4}\] Rearranging in terms of \(\text{S} H^+\) yields \[\left[ \text{S} H^+ \right] = \dfrac{k_1 \left[ \text{S} \right] \left[ \text{A} H \right]}{k_{-1} \left[ \text{A}^- \right] + k_2} \label{5}\] and the rate of production of \(\text{P}\) can be written as \[\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{S} H^+ \right] = \dfrac{k_1 k_2 \left[ \text{S} \right] \left[ \text{A} H \right]}{k_{-1} \left[ \text{A}^- \right] + k_2} \label{Eq6}\] In the case where \(k_2 \gg k_{-1} \left[ \text{A}^- \right]\), Equation \(\ref{Eq6}\) can be written as \[\dfrac{d \left[ \text{P} \right]}{dt} = k_1 \left[ \text{S} \right] \left[ \text{A} H \right] \label{Eq7}\] which is known as a general acid-catalyzed reaction. On the other hand, if \(k_2 \ll k_{-1} \left[ \text{A}^- \right]\), we can use an equilibrium approximation to write the rate of production of \(\text{P}\) as \[\dfrac{d \left[ \text{P} \right]}{dt} = \dfrac{k_1 k_2 \left[ \text{S} \right] \left[ \text{A} H \right]}{k_{-1} \left[ \text{A}^- \right]} = \dfrac{k_1 k_2}{k_{-1} K} \left[ \text{S} \right] \left[ H^+ \right] \label{Eq8}\] where \(K\) is the acid dissociation constant: \[K = \dfrac{ \left[ \text{A}^- \right] \left[ H^+ \right]}{\left[ \text{A} H \right]} \label{Eq9}\] In this case, the reaction is hydrogen ion-catalyzed. Enzymes speed up the rate of chemical reactions because they lower the energy of activation, the energy that must be supplied in order for molecules to react with one another (Figure \(\Page {3}\)). Like homogeneous catalysts discussed above, enzymes lower the energy of activation by forming an enzyme-substrate complex allowing products of the enzyme reaction to be formed and released (Figure \(\Page {2}\)). Enzymes are substances present in the cell in small amounts which speed up or catalyze chemical reactions. Enzymes speed up the rate of chemical reactions because they lower the energy of activation, the energy that must be supplied in order for molecules to react with one another. Enzymes lower the energy of activation by forming an enzyme-substrate complex. Figure \(\Page {4}\) Enzymes are generally globular proteins. (Some RNA molecules called ribozymes can also be enzymes. These are usually found in the nuclear region of cells and catalyze the splitting of RNA molecules.). Enzymes are catalysts that breakdown or synthesize more complex chemical compounds. They allow chemical reactions to occur fast enough to support life. Enzymes speed up the rate of chemical reactions because they lower the energy of activation, the energy that must be supplied in order for molecules to react with one another. Anything that an enzyme normally combines with is called a substrate. Enzymes are very efficient with a typically enzyme generally able to catalyze between 1 and 10,000 molecules of substrate per second. The means that enzymes are only have to be present in small amounts in the cell since. They are not altered during their reaction and are highly specific for their substrate, with generally one specific enzyme dedicated for each specific chemical reaction. Enzyme activity is affected by a number of factors including: Enzymes are essential to maintain homeostasis because any malfunction of an enzyme could lead to diseases. Therefore, pharmaceutical companies study enzyme to manipulate and synthesis new medicine. Besides their medicinal applications, enzymes in industry are important because enzymes help breaking down cellulose, wastes, etc. Enzymes are essential in the process of making new products in many industries such as pharmaceutical, food, paper, wine, etc. ( ) (COMMUNITY COLLEGE OF BALTIMORE COUNTY, CATONSVILLE CAMPUS) | 7,818 | 1,643 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Entropy/Microstates |
Dictionaries define “macro” as large and “micro” as very small but a macrostate and a microstate in thermodynamics aren't just definitions of big and little sizes of chemical systems. Instead, they are two very different ways of looking at a system. (Admittedly, a macrostate always has to involve an amount of matter large enough for us to measure its volume or pressure or temperature, i.e. in “bulk”. But in thermodynamics, a microstate isn't just about a smaller amount of matter', it is a detailed look at the energy that molecules or other particles have.) A microstate is one of the huge number of different accessible arrangements of the molecules' motional energy* for a particular macrostate. *Motional energy includes the translational, rotational, and vibrational modes of molecular motion. In calculations involving entropy, the ΔH of any phase change in a substance (“phase change energy”) is added to motional energy, but it is unaltered in ordinary entropy change (of heating, expansion, reaction, etc.) unless the phase itself is changed. A is the thermodynamic state of any system that is exactly characterized by measurement of the system's properties such as P, V, T, H and number of moles of each constituent. Thus, a macrostate does not change over time if its observable properties do not change. In contrast, a for a system is all about time and the energy of the molecules in that system. "In a system its energy is constantly being redistributed among its particles. In liquids and gases, the particles themselves are constantly redistributing in location as well as changing in the quanta (the individual amount of energy that each molecule has) due to their incessantly colliding, bouncing off each other with (usually) a different amount of energy for each molecule after the collision.. Each specific way, each arrangement of the energy of each molecule in the whole system at one instant is called a ." One microstate then is something like a theoretical "absolutely photo" of the location and momentum of molecule and atom in the whole macrostate. (This is talking in ‘classical mechanics’ language where molecules are assumed to have location and momentum. In quantum mechanics the behavior of molecules is only described in terms of their energies on particular energy levels. That is a more modern view that we will use.) In the next instant the system immediately changes to another microstate. (A molecule moving at an average speed of around a thousand miles an hour collides with others about seven times in a billionth of a second. Considering a mole of molecules (6 x 10 ) traveling at a very large number of different speeds, the collisions occur — and thus changes in energy of trillions of molecules occurs — in far less than a trillionth of a second. That's why it is wise to talk in terms of “an instant”!) To take a photo like that may seem impossible and it is. In the next instant — and that really means in an short time — at least a couple of moving molecules out of the 6 x 10 23 will hit one another.. But if only one molecule moves a bit slower because it had hit another and made that other one move an exactly equal amount faster — then that would be a different microstate. (The total energy hasn't changed when molecular movement changes one microstate into another. Every microstate for a particular system has exactly the total energy of the macrostate because a microstate is just an instantaneous quantum energy-photo of the whole system.) That's why, in an instant for any particular state, its motional energy* has been rearranged as to what molecule has what amount of energy. In other words, the system — the macrostate — rapidly and successively changes to be in a gigantic number of different microstates out of the “gazillions” of microstates, (In solids, the location of the particles is almost the same from instant to instant, but not exactly, because the particles are vibrating a tiny amount from a fixed point at enormous speeds.) N and O molecules are at 298 K are gases, of course, and have a very wide range of speeds, from zero to more than two thousand miles an hour with an average of roughly a thousand miles an hour. They go only about 200 times their diameter before colliding violently with another molecule and losing or gaining energy. Occasionally, two molecules colliding head on at exactly the same speed would stop completely before being hit by another molecule and regaining some speed.) In liquids, the distance between collisions is very small, but the speeds are about the same as in a gas at the same temperature. Now we know what a microstate is, but what good is something that we can just imagine as an impossible fast camera shot? The answer is loud and clear. We can calculate the numbers for a given macrostate and we find that microstates give us answers about the relation between molecular motion and entropy — i.e., between molecules (or atoms or ions) constantly energetically speeding, colliding with each other, moving distances in space (or, just vibrating rapidly in solids) and what we measure in a macrostate as its entropy. As you have read elsewhere, entropy is a (macro) measure of the spontaneous dispersal of energy, how widely spread out it becomes (at a specific temperature). Then, because the number of microstates that are for a system indicates all the different ways that energy can be arranged in that system, the larger the number of microstates accessible, the greater is a system's entropy at a given temperature. It is that the energy of a system is smeared or spread out over a greater number of microstates that it is more dispersed. That can't occur because all the energy of the macrostate is always in only one microstate at one instant. The macrostate's energy is more "spread out" when there are larger numbers of microstates for a system because at any instant all the energy that is in one microstate can be in any one of the now-larger total of microstates, a greatly increased number of choices, far less chance of being “localized” — i.e., just being able to jump around from one to only a dozen other microstates or'only' a few millions or so! More possibilities mean more chances for the system to be in one of MANY more different microstates — that is what is meant by "the system's total energy can be more dispersed or spread out”: more choices/chances. That might be fine, but how can we find out how many microstates are accessible for a macrostate? (Remember, a macrostate is just any system whose thermodynamic qualities of P, V, T, H, etc. have been measured so the system is exactly defined.) Fortunately, Ludwig Boltzmann gives us the answer in S = k ln W, where S is the value of entropy in joules/mole at T, k is Boltzmann's constant of 1.4 x 10 J/K and W is the number of microstates. Thus, if we look in “Standard State Tables” listing the entropy of a substance that has been determined experimentally by heating it from 0 K to 298 K, we find that ice at 273 K has been calculated to have an S of 41.3 J/K mol. Inserting that value in the Boltzmann equation gives us a result that should boggle one's mind because it is among the largest numbers in science. (The estimated number of atoms in our entire galaxy is around 10 while the number for the whole universe may be about 10 . A very large number in math is 10 and called "a googol" — Google!) Crystalline ice at 273 K has 10 accessible microstates. (Writing 5,000 zeroes per page, it would take not just reams of paper, not just reams piled miles high, but light years high of reams of paper to list all those microstates!) Entropy and entropy change are concerned with the energy dispersed in a system and its temperature, q /T. Thus, entropy is measured by the number of accessible microstates, in any one of which the system's total energy might be at one instant, not by the orderly patterns of the molecules aligned in a crystal. Anyone who discusses entropy and calls "orderly" the energy distribution among those humanly incomprehensible numbers of different microstates for a crystalline solid — such as we have just seen for ice — is looking at the wrong thing. Liquid water at the same temperature of ice, 273 K has an S of 63.3 J/K . Therefore, there are 10 accessible microstates for water. | 8,363 | 1,644 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/04%3A_Reaction_Mechanisms/4.02%3A_Chain_Reactions |
usually consist of many repeating elementary steps, each of which has a chain carrier. Once started, chain reactions continue until the reactants are exhausted. Fire and explosions are some of the phenomena associated with chain reactions. The are some intermediates that appear in the repeating elementary steps. These are usually . Once initiated, repeating elementary steps continue until the reactants are exhausted. When repeating steps generate more chain carriers, they are called , which leads to explosions. If the repeating elementary steps do not lead to the formation of new product, they are called . Addition of other materials in the reaction mixture can lead to the inhibition reaction to prevent the chain propagation reaction. When chain carriers react with one another forming stable product, the elementary steps are called . Explosions, polymerizations, and food spoilage often involve chain reactions. The chain reaction mechanism is involved in nuclear reactors; in this case the chain carriers are neutrons. The mechanisms describing chain reactions are useful models for describing chemical reactions. Most chemical chain reactions have very reactive intermediates called . The intermediate that maintains the chain reaction is called a These atoms or fragments are usually derived from stable molecules due to photo- or heat-dissociation. Usually, a free radical is marked by a dot beside the symbol (\(\ce{*}\)), which represents an odd electron exists on the species. This odd electron makes the intermediate very reactive. For example, the oxygen, chlorine and ethyl radicals are represented by \(\ce{O*}\), \(\ce{Cl*}\), and \(\ce{C2H5*}\), respectively. The \(\ce{Cl*}\) radicals can be formed by the homolytic photodissociation reaction: \[\ce{Cl2 + h\nu \rightarrow Cl* + *Cl} \nonumber \] The elementary steps used for mechanisms of chain reactions can be grouped into the following categories: For example, the chlorination of ethane is a chain reaction, and its mechanism is explained in the following way. If we mix chlorine, \(\ce{Cl2}\), and ethane, \(\ce{CH3CH3}\), together at room temperature, there is no detectable reaction. However, when the mixture is exposed to light, the reaction suddenly initiates, and explodes. To explain this, the following mechanism is proposed. Light (\(\ce{h\nu}\)) can often be used to initiate chain reactions since they can generate free radical intermediates via a photodissociation reaction. The initiation step can be written as: \[\ce{Cl2 + h\nu \rightarrow Cl* + *Cl} \nonumber \] Elementary steps in which the number of free radicals consumed is equal to the number of free radicals generated are called Once initiated, the following chain propagation steps repeat indefinitely or until the reactants are exhausted: \[\ce{Cl* +\; H3CCH3 \rightarrow ClH2CCH3 +\; H*} \nonumber \] \[\ce{Cl* +\; H3CCH3 \rightarrow H3CCH2* +\; HCl} \nonumber \] \[\ce{H* +\; Cl_2 \rightarrow HCl + Cl*} \nonumber \] and many other possibilities. In each of these steps, a radical is consumed, and another radical is generated. Thus, the chain reactions continue, releasing heat and . The heat and cause more radicals to form. Thus, the chain propagation steps cause . are elementary steps that generate more free radicals than they consume. Branching reactions result in an explosion. For example, in the reaction between hydrogen and oxygen, the following reaction may take place: \[\ce{H* +\; O2 \rightarrow HO* + *O*} \nonumber \] where \(\ce{*O*}\) is a di-radical, because the \(\ce{O}\) atom has an electronic configuration 2 2 2 2 . In this elementary step, three radicals are generated, whereas only one is consumed. The di-radical may react with a \(\ce{H2}\) molecule to form two radicals. \[\ce{*O* +\, H2 \rightarrow HO* +\, H*} \nonumber \] Thus, together chain branching reactions increase the number of chain carriers. Branching reactions contribute to the rapid explosion of hydrogen-oxygen mixtures, especially if the mixtures have proper proportions. The steps not leading to the formation of products are called or steps. For example, the following steps are inhibition reactions. \[\ce{Cl* +\; ClH2CCH3 \rightarrow H3CCH2* +\; Cl2} \nonumber \] \[\ce{Cl* +\; HCl \rightarrow H* +\; Cl2} \nonumber \] \[\ce{H* +\; ClH2CCH3 \rightarrow H3CCH3 + Cl*} \nonumber \] Furthermore, sometimes another reactive substance \(\ce{*A}\) may be added to the system to reduce the chain carriers to inhibit the chain reactions. \(\ce{Cl* + *A \rightarrow ClA\: (not\: reactive)}\) The species \(\ce{A*}\) is often called a In food industry, radical scavengers are added to prevent spoilage due to oxidation; these are called biological oxidants. The mechanisms in chain reactions are often quite complicated. When intermediates are detected, a reasonable mechanism can be proposed. Adding radical scavenger to prevent food spoilage is an important application in food chemistry. This application came from the application of the chain reaction model to natural phenomena. are elementary steps that consume radicals. When reactants are exhausted, free radicals combine with one another to give stable molecules (since unpaired electrons become paired). These elementary steps are responsible for the chain reactions' termination: \[\ce{Cl* + *Cl \rightarrow Cl-Cl} \nonumber \] \[\ce{H* + *H \rightarrow H-H} \nonumber \] \[\ce{H* + *Cl \rightarrow H-Cl} \nonumber \] \[\ce{H3CCH2* + *H2CCH3 \rightarrow CH3CH2-CH2CH3\: (forming\: a\: dimer)} \nonumber \] and other possibilities In chain reactions, many products are produced.
Identify steps for the names in the multiple choices.
Argon exists as a mono-atomic gas. All noble gases have mono-atomic molecules.
The reactant \(\ce{HCl}\) in the step is a product in the overall reaction. When \(\ce{HCl}\) reacts with \(\ce{Cl*}\), the reaction is retarded. \(\ce{Cl*}\) attacked one of the product molecule \(\ce{HCl}\) causing a reversal of the reaction. | 6,026 | 1,645 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Quantum_States_of_Atoms_and_Molecules_(Zielinksi_et_al)/05%3A_Translational_States/5.01%3A_The_Free_Particle |
We obtain the Schrödinger equation for the free particle using the following steps. First write \[\hat {H} \psi = E \psi \label {5-1}\] Next define the Hamiltonian, \[\hat {H} = \hat {T} + \hat {V} \label {5-2}\] and substitute the potential energy operator \[\hat {V} = 0 \label {5-3}\] and the kinetic energy operator \[ \hat {T} = -\dfrac {\hbar ^2}{2m} \dfrac {d^2}{dx^2} \label {5-4} \] to obtain the Schrödinger equation for a free particle \[-\dfrac {\hbar ^2}{2m} \dfrac {d^2 \psi (x)}{dx^2} = E \psi (x) \label {5-5}\] A major problem in Quantum Mechanics is finding solutions to differential equations, e.g. Equation \(\ref{5-5}\). Differential equations arise because the operator for kinetic energy includes a second derivative. We will solve the differential equations for some of the more basic cases, but since this is not a course in Mathematics, we will not go into all the details for other more complicated cases. The solutions that we consider in the greatest detail will illustrate the general procedures and show how the physical concept of quantization arises mathematically. We already encountered Equation \(\ref{5-5}\) in the last chapter . There, we used our knowledge of some basic functions to find the solution. Now we solve this equation by using some algebra and mathematical logic. First we rearrange Equation \(\ref{5-5}\) and make the substitution \[k^2 = \dfrac {2mE}{\hbar^2}. \label{sub}\] The substitution in Equation \ref{sub} is only one way of making a simplification. You could also use a different formulation for the substitution \[\alpha = \dfrac {2mE}{\hbar ^2} \nonumber\] but then you would find later that \((\alpha)^{1/2}\) corresponds to the wavevector \(k\) which equals \(\frac {2 \pi}{\lambda}\) and \(\frac {P}{\hbar}\). So choosing a squared variable like \(k^2\) in Equation \ref{sub} is a choice made with foresight. Trial-and-error is one method scientists use to solve problems, and the results often look sophisticated and insightful after they have been found, like choosing \(k^2\) rather than \(α\). Since \(E\) is the kinetic energy, \[ E = \dfrac {p^2}{2m} \label {5-6}\] and we saw in previous chapters that the momentum \(p\) and the wavevector \(k\) are related, \[p = \hbar k \label {5-7}\] we also could recognize that \(\dfrac {2mE}{\hbar ^2}\) is just \(k^2\) as shown here in Equation \(\ref{5-8}\). \[ \begin{align} \dfrac {2mE}{\hbar ^2} &= \left (\dfrac {2m}{\hbar ^2}\right ) \left ( \dfrac {p^2}{2m} \right ) \\[4pt] &= \left (\dfrac {\cancel{2m}}{\cancel{\hbar ^2}}\right ) \left ( \dfrac {\cancel{\hbar ^2} k^2}{\cancel{2m}} \right ) \\[4pt] &= k^2 \label {5-8} \end{align}\] The result for Equation \(\ref{5-5}\) after rearranging and substitution of result from Equation \(\ref{5-8}\) is \[\left ( \dfrac {d^2}{dx^2} + k^2 \right ) \psi (x) = 0 \label {5-9}\] This can be solved in the same way that a algebraic quadratic equation is solved. It is separated into two factors, and each is set equal to 0. This factorization produces two that can be integrated. The details are shown in the following equations. \[\left ( \dfrac {d^2}{dx^2} + k^2 \right ) \psi (x) = \left(\dfrac {d}{dx} + ik\right) \left(\dfrac {d}{dx} - ik \right) \psi (x) = 0 \label {5-10}\] Equation \(\ref{5-10}\) will be true if either \[ \left( \dfrac {d}{dx} + ik \right) \psi (x) = 0\] or \[ \left( \dfrac {d}{dx} - ik \right) \psi (x) = 0 \label {5-11}\] Rearranging and designating the two equations and the two solutions simultaneously by a + sign and a – sign produces \[\dfrac {d \psi _{ \pm} (x) }{\psi _{\pm} (x)} = \pm ik\,dx \label {5-12}\] which leads to \[\ln \psi _\pm (x) = \pm ikx + C _{\pm} \label {5-13}\] and finally \[\psi _{\pm} (x) = A_{\pm} e^{\pm ikx} \label {5-14}\] The constants \(A_+\) and \(A_-\) result from the constant of integration. The values of these constants are determined by some physical constraint that is imposed upon the solution. Such a constraint is called a . For the , discussed previously, the boundary condition is that the wavefunction must be zero at the boundaries where the potential energy is infinite. The free particle does not have such a boundary condition because the particle is not constrained to any one place. Another constraint is normalization, and here the integration constants serve to satisfy the normalization requirement. Show that the operator \(\left(\dfrac {d^2}{dx^2} + k^2\right)\) equals \(\left(\dfrac {d}{dx} + ik\right) \left(\dfrac {d}{dx} - ik\right)\) and that the two factors commute since \(k\) does not depend on \(x\). The answer is Equation \(\ref{5-10}\). Use the normalization constraint to evaluate \(A_{\pm}\) in Equation \ref{5-14}. Since the integral of \(|\psi |^2\) over all values of x from \(-∞\) to \(+∞\) is infinite, it appears that the wavefunction \(\psi \) cannot be normalized. We can circumvent this difficulty if we imagine the particle to be in a region of space ranging from \(-L\) to \(+L\) and consider \(L\) to approach infinity. The normalization then proceeds in the usual way as shown below. Notice that the normalization constants are even though the wavefunctions are complex. \[ \begin{align*} \int \limits _{-L}^{+L} \psi ^* (x) \psi (x) dx &= A_{\pm} ^* A_{\pm} \int \limits _{-L}^{L} e^{\mp ikx} e^{\pm ikx} dx = 1 \\[4pt] |A_{\pm}|^2 \int \limits _{-L}^{+L} dx &= |A_{\pm}|^2 2L = 1 \\[4pt] A_{\pm} &= [2L]^{-1/2} \end{align*} \] Write the wavefunctions, \(\psi ^+\) and \(\psi ^−\), for the free particle, explicitly including the normalization factors found in Example \(\Page {1}\). Find solutions to each of the following differential equations. \[ \dfrac {d^2 y(x)}{dx^2} + 25y(x) = 0 \nonumber\] \[\dfrac {d^2 y(x)}{dx^2} -3y(x) = 0 \nonumber \] A neat property of linear differential equations is that sums of solutions also are solutions, or more generally, linear combinations of solutions are solutions. A linear combination is a sum with constant coefficients where the coefficients can be positive, negative, or imaginary. For example \[\psi(x) = C_1\psi _+(x) + C_2\psi _−(x) \label {5-15}\] where \(C_1\) and \(C_2\) are the constant coefficients. Inserting the functions from Equation \(\ref{5-14}\), one gets \[\psi (x) = \dfrac {C_1}{\sqrt {2L}} e^{+ikx} + \dfrac {C_2}{\sqrt {2L}} e^{-ikx} \label {5-16}\] By using , \[e^{\pm ikx} = \cos (kx) \pm i\sin (kx) \label {5-17}\] Equation \(\ref{5-15}\) is transformed into \[\psi (x) = C\cos (kx) + D\sin (kx) \label {5-18}\] where we see that k is just the wavevector \(\dfrac{2\pi}{\lambda}\) in the trigonometric form of the solution to the Schrödinger equation. This result is consistent with our previous discussion regarding the choice of \(k^2\) to represent \(\dfrac {2mE}{ħ^2}\). Find expressions for \(C\) and \(D\) in Equation \(\ref{5-18}\) for two cases: when \(C_1 = C_2\) = +1 and when \(C_1\) = +1 and \(C_2\) = -1. Verify that Equations \(\ref{5-16}\) and \(\ref{5-18}\) are solutions to the Schrödinger Equation (Equation \(\ref{5-5}\)) with the eigenvalue \(E = \dfrac {\hbar ^2 k^2 }{2m}\). Demonstrate that the wavefunctions you wrote for Exercise \(\Page {2}\) are eigenfunctions of the momentum operator with eigenvalues \(\hbar k\) and \(-\hbar k\). Determine whether \(\psi (x)\) in Equation \(\ref{5-16}\) is an eigenfunction of the momentum operator. The probability density for finding the free particle at any point in the segment \(-L\) to \(+L\) can be seen by plotting \(\psi ^*\psi \) from -L to +L. Sketch these plots for the two wavefunctions, \(\psi _+\) and \(\psi _−\), that you wrote for Exercise \(\Page {2}\). Demonstrate that the area between \(\psi ^*\psi \) and the x-axis equals 1 for any value of L. Why must this area equal 1 even as L approaches infinity? Are all points in the space equally probable or are some positions favored by the particle? We found wavefunctions that describe the free particle, which could be an electron, an atom, or a molecule. Each wavefunction is identified by the wavevector \(k\). A wavefunction tells us three things about the free particle: the energy of the particle, the momentum of the particle, and the probability density of finding the particle at any point. You have demonstrated these properties in Exercises \(\Page {5}\), \(\Page {6}\), and \(\Page {8}\). These ideas are discussed further in the following paragraphs. We first find the momentum of a particle described by \(\psi _+(x)\). We also can say that the particle is in the state \(\psi _+(x)\). The value of the momentum is found by operating on the function with the momentum operator. Remember this problem is one-dimensional so vector quantities such as the wavevector or the momentum appear as scalars. The result is shown in Example \(\Page {1}\). Extract the momentum from the wavefunction for a free electron. First we write the momentum operator and wavefunction as shown by I and II. The momentum operator tells us the mathematical operation to perform on the function to obtain the momentum. Complete the operation shown in II to get III, which simplifies to IV. \[ \underset{I}{-i\hbar \dfrac {d}{dx} \psi _+ (x)} = \underset{II}{-i\hbar \dfrac {d}{dx} A _+ e^{ikx}} = \underset{III}{(-i\hbar) (ik) A_+ e^{ikx}} = \underset{IV}{\hbar k \psi _+ (x)} \nonumber \] Example \(\Page {2}\) is another way to conclude that the momentum of this particle is \[p = ħk.\] Here the Compton-de Broglie momentum-wavelength relation \(p = \hbar k\) appears from the solution to the Schrödinger equation and the definition of the momentum operator! For an electron in the state \(\psi _−(x)\), we similarly find \(p = -\hbar k\). This particle is moving in the minus x direction, opposite from the particle with momentum \(+ħk\). Since \(k = \dfrac {2 \pi}{\lambda}\), what then is the meaning of the wavelength for a particle, e.g. an electron? The wavelength is the wavelength of the wavefunction that describes the properties of the electron. We are not saying that an electron is a wave in the sense that an ocean wave is a wave; rather we are saying that a wavefunction is needed to describe the wave-like properties of the electron. Why the electron has these wave-like properties, remains a mystery. We find the energy of the particle by operating on the wavefunction with the Hamiltonian operator as shown next in Equation \(\ref{5-19}\). Examine each step and be sure you see how the eigenvalue is extracted from the wavefunction. \[ \begin{align} \hat {H} \psi _{\pm} (x) &= \dfrac {-\hbar ^2}{2m} \dfrac {d^2}{dx^2} A_{\pm} e^{\pm ikx} \\[4pt] &= \dfrac {-\hbar ^2}{2m} (\pm ik)^2 A_{\pm} e^{\pm ikx} \\[4pt] & = \dfrac {\hbar ^2 k^2}{2m} A_{\pm}e^{\pm ikx} \label {5-19} \end{align}\] Notice again how the operator works on the wavefunction to extract a property of the system from it. We conclude that the energy of the particle is \[ E = \dfrac { \hbar ^2 k^2}{2m} \label {5-20}\] Which is just the classical relation between energy and momentum of a free particle, \(E = \dfrac {p^2}{2m}\). Note that an electron with momentum +ħk has the same energy as an electron with momentum -ħk. When two or more states have the same energy, the states and the energy level are said to be degenerate. We have not found any restrictions on the momentum or the energy. These quantities are not quantized for the free particle because there are no boundary conditions. Any wave with any wavelength fits into an unbounded space. Quantization results from boundary conditions imposed on the wavefunction, as we saw for the particle-in-a-box. Describe how the wavelength of a free particle varies with the energy of the particle. Summarize how the energy and momentum information is contained in the wavefunction and how this information is extracted from the wavefunction. The probability density of a free particle at a position in space \(x_0\) is \[\psi _{\pm} ^* (x_0) \psi _{\pm} (x_0) = (2L)^{-1} e^{\mp ikx_0} e^{\pm ikx_0} = (2L)^{-1} \label {5-21}\] From this result we see that the probability density has units of 1/m; it is the probability per meter of finding the electron at the point \(x_0\). This probability is independent of \(x_0\), the electron can be found any place along the x axis with equal probability. Although we have no knowledge of the position of the electron, we do know the electron momentum exactly. This relationship between our knowledge of position and momentum is a manifestation of the , which says that as the uncertainty in one quantity is reduced, the uncertainty in another quantity increases. For this case, we know the momentum exactly and have no knowledge of the position of the particle. The uncertainty in the momentum is zero; the uncertainty in the position is infinite. | 12,732 | 1,646 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/17%3A_Solubility_and_Complexation_Equilibria/17.05%3A_Solubility_and_pH |
The solubility of many compounds depends strongly on the pH of the solution. For example, the anion in many sparingly soluble salts is the conjugate base of a weak acid that may become protonated in solution. In addition, the solubility of simple binary compounds such as oxides and sulfides, both strong bases, is often dependent on pH. In this section, we discuss the relationship between the solubility of these classes of compounds and pH. We begin our discussion by examining the effect of pH on the solubility of a representative salt, M A , where A is the conjugate base of the weak acid HA. When the salt dissolves in water, the following reaction occurs:
\[MA_{(s)} \rightleftharpoons M^+_{(aq)} + A^−_{(aq)} \tag{17.4.1a}\] with The anion can also react with water in a hydrolysis reaction:
\[A^−_{(aq)} + H_2O_{(l)} \rightleftharpoons OH^−_{(aq)} + HA_{(aq)} \tag{17.4.2}\] Because of the reaction described in , the predicted solubility of a sparingly soluble salt that has a basic anion such as S , PO , or CO is increased, as described in . If instead a strong acid is added to the solution, the added H will react essentially completely with A to form HA. This reaction decreases [A ], which decreases the magnitude of the ion product ( = [M ,A ]). According to Le Chatelier’s principle, more MA will dissolve until = . Hence . In contrast, pH has little to no effect on the solubility of salts whose anion is the conjugate base of a stronger weak acid or a acid, respectively (e.g., chlorides, bromides, iodides, and sulfates). For example, the hydroxide salt Mg(OH) is relatively insoluble in water:
\[Mg(OH)_{2(s)} \rightleftharpoons Mg^{2+}_{(aq)} + 2OH^−_{(aq)} \tag{17.4.3a}\] with \[K_{sp} = 5.61 \times 10^{−12} \tag{17.4.3b}\] When acid is added to a saturated solution that contains excess solid Mg(OH) , the following reaction occurs, removing OH from solution:
\[H^+_{(aq)} + OH^−_{(aq)} \rightarrow H_2O_{(l)} \tag{17.4.5}\] The overall equation for the reaction of Mg(OH) with acid is thus
\[Mg(OH)_{2(s)} + 2H^+_{(aq)} \rightleftharpoons Mg^{2+}_{(aq)} + 2H_2O_{(l)} \tag{17.4.6}\] As more acid is added to a suspension of Mg(OH) , the equilibrium shown in is driven to the right, so more Mg(OH) dissolves. Such pH-dependent solubility is not restricted to salts that contain anions derived from water. For example, CaF is a sparingly soluble salt:
\[CaF_{2(s)} \rightleftharpoons Ca^{2+}_{(aq)} + 2F^−_{(aq)} \tag{17.4.7a}\] with \[K_{sp} = 3.45 \times 10^{−11} \tag{17.4.7b}\] When strong acid is added to a saturated solution of CaF , the following reaction occurs:
\[H^+_{(aq)} + F^−_{(aq)} \rightleftharpoons HF_{(aq)} \tag{17.4.8}\] Because the forward reaction decreases the fluoride ion concentration, more CaF dissolves to relieve the stress on the system. The net reaction of CaF with strong acid is thus
\[CaF_{2(s)} + 2H^+_{(aq)} \rightarrow Ca^{2+}_{(aq)} + 2HF_{(aq)} \tag{17.4.9}\] Example 7 shows how to calculate the solubility effect of adding a strong acid to a solution of a sparingly soluble salt. Sparingly soluble salts derived from weak acids tend to be more soluble in an acidic solution. Lead oxalate (PbC O ), lead iodide (PbI ), and lead sulfate (PbSO ) are all rather insoluble, with values of 4.8 × 10 , 9.8 × 10 , and 2.53 × 10 , respectively. What effect does adding a strong acid, such as perchloric acid, have on their relative solubilities? values for three compounds relative solubilities in acid solution Write the balanced chemical equation for the dissolution of each salt. Because the strongest conjugate base will be most affected by the addition of strong acid, determine the relative solubilities from the relative basicity of the anions. The solubility equilibriums for the three salts are as follows: \[PbC_2O_{4(s)} \rightleftharpoons Pb^{2+}_{(aq)} + C_2O^{2−}_{4(aq) \notag }\] \[PbI_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2I^−_{(aq)} \notag \] \[PbSO_{4(s)} \rightleftharpoons Pb^{2+}_{(aq)} + SO^{2−}_{4(aq)} \notag \] The addition of a strong acid will have the greatest effect on the solubility of a salt that contains the conjugate base of a weak acid as the anion. Because HI is a strong acid, we predict that adding a strong acid to a saturated solution of PbI will not greatly affect its solubility; the acid will simply dissociate to form H (aq) and the corresponding anion. In contrast, oxalate is the fully deprotonated form of oxalic acid (HO CCO H), which is a weak diprotic acid (p = 1.23 and p = 4.19). Consequently, the oxalate ion has a significant affinity for one proton and a lower affinity for a second proton. Adding a strong acid to a saturated solution of lead oxalate will result in the following reactions: \[C_2O^{2−}_{4(aq)} + H^+_{(aq)} \rightarrow HO_2CCO^−_{2(aq)} \notag \] These reactions will decrease [C O ], causing more lead oxalate to dissolve to relieve the stress on the system.The p of HSO (1.99) is similar in magnitude to the p of oxalic acid, so adding a strong acid to a saturated solution of PbSO will result in the following reaction: \[ SO^{2-}_{4(aq)} + H^+_{(aq)} \rightleftharpoons HSO^-_{4(aq)} \notag \] Because HSO has a pKa of 1.99, this reaction will lie largely to the left as written. Consequently, we predict that the effect of added strong acid on the solubility of PbSO will be significantly less than for PbC O . Exercise Which of the following insoluble salts—AgCl, Ag CO , Ag PO , and/or AgBr—will be substantially more soluble in 1.0 M HNO than in pure water? Ag CO and Ag PO Caves and their associated pinnacles and spires of stone provide one of the most impressive examples of pH-dependent solubility equilibriums (part (a) in ). Perhaps the most familiar caves are formed from limestone, such as Carlsbad Caverns in New Mexico, Mammoth Cave in Kentucky, and Luray Caverns in Virginia. The primary reactions that are responsible for the formation of limestone caves are as follows: \[CO_{2(aq)} + H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + HCO^−_{3(aq)} \tag{17.4.10}\] \[HCO^−_{3(aq)} \rightleftharpoons H+(aq) + CO^{2−}_{3(aq)} \tag{17.4.11}\] \[Ca^{2+}_{(aq)} + CO^{2−}_{3(aq)} \rightleftharpoons CaCO_{3(s)} \tag{17.4.12}\] Limestone deposits that form caves consist primarily of CaCO from the remains of living creatures such as clams and corals, which used it for making structures such as shells. When a saturated solution of CaCO in CO -rich water rises toward Earth’s surface or is otherwise heated, CO gas is released as the water warms. CaCO then precipitates from the solution according to the following equation (part (b) in ):
The forward direction is the same reaction that produces the solid called in teapots, coffee makers, water heaters, boilers, and other places where hard water is repeatedly heated. When groundwater-containing atmospheric CO ( and ) finds its way into microscopic cracks in the limestone deposits, CaCO dissolves in the acidic solution in the reverse direction of . The cracks gradually enlarge from 10–50 µm to 5–10 mm, a process that can take as long as 10,000 yr. Eventually, after about another 10,000 yr, a cave forms. Groundwater from the surface seeps into the cave and clings to the ceiling, where the water evaporates and causes the equilibrium in to shift to the right. A circular layer of solid CaCO is deposited, which eventually produces a long, hollow spire of limestone called a that grows down from the ceiling. Below, where the droplets land when they fall from the ceiling, a similar process causes another spire, called a , to grow up. The same processes that carve out hollows below ground are also at work above ground, in some cases producing fantastically convoluted landscapes like that of Yunnan Province in China ( ). One of the earliest classifications of substances was based on their solubility in acidic versus basic solution, which led to the classification of oxides and hydroxides as being either basic or acidic. Basic oxides and hydroxides either react with water to produce a basic solution or dissolve readily in aqueous acid. Acidic oxides or hydroxides either react with water to produce an acidic solution or are soluble in aqueous base. As shown in , there is a clear correlation between the acidic or the basic character of an oxide and the position of the element combined with oxygen in the periodic table. Compare, for example, the reactions of a typical metal oxide, cesium oxide, and a typical nonmetal oxide, sulfur trioxide, with water:
\[Cs_2O_{(s)} + H_2O_{(l)} \rightarrow 2Cs^+_{(aq)} + 2OH^−_{(aq)} \tag{17.4.14}\] \[SO_{3(g)} + H_2O_{(l)} \rightarrow H_2SO_{4(aq)} \tag{17.4.15}\] Cesium oxide reacts with water to produce a basic solution of cesium hydroxide, whereas sulfur trioxide reacts with water to produce a solution of sulfuric acid—very different behaviors indeed! Metal oxides generally react with water to produce basic solutions, whereas nonmetal oxides produce acidic solutions. The difference in reactivity is due to the difference in bonding in the two kinds of oxides. Because of the low electronegativity of the metals at the far left in the periodic table, their oxides are best viewed as containing discrete M cations and O anions. At the other end of the spectrum are nonmetal oxides; due to their higher electronegativities, nonmetals form oxides with covalent bonds to oxygen. Because of the high electronegativity of oxygen, however, the covalent bond between oxygen and the other atom, E, is usually polarized: E –O . The atom E in these oxides acts as a Lewis acid that reacts with the oxygen atom of water to produce an oxoacid. Oxides of metals in high oxidation states also tend to be acidic oxides for the same reason: they contain covalent bonds to oxygen. An example of an acidic metal oxide is MoO , which is insoluble in both water and acid but dissolves in strong base to give solutions of the molybdate ion (MoO ):
\[MoO_{3(s)} + 2OH^−_{(aq)} \rightarrow MoO^{2−}_{4(aq)} + H_2O_{(l)} \tag{17.4.16}\] As shown in , there is a gradual transition from basic metal oxides to acidic nonmetal oxides as we go from the lower left to the upper right in the periodic table, with a broad diagonal band of oxides of intermediate character separating the two extremes. Many of the oxides of the elements in this diagonal region of the periodic table are soluble in both acidic basic solutions; consequently, they are called amphoteric oxides (from the Greek , meaning “both,” as in , which was defined in ). Amphoteric oxides either dissolve in acid to produce water or dissolve in base to produce a soluble complex. As shown in , for example, mixing the amphoteric oxide Cr(OH) (also written as Cr O ·3H O) with water gives a muddy, purple-brown suspension. Adding acid causes the Cr(OH) to dissolve to give a bright violet solution of Cr (aq), which contains the [Cr(H O) ] ion, whereas adding strong base gives a green solution of the [Cr(OH) ] ion. The chemical equations for the reactions are as follows:
\(\mathrm{Cr(OH)_3(s)}+\mathrm{3H^+(aq)}\rightarrow\underset{\textrm{violet}}{\mathrm{Cr^{3+}(aq)}}+\mathrm{3H_2O(l)} \tag{17.4.17}\) All three beakers originally contained a suspension of brownish purple Cr(OH) (s) (center). When concentrated acid (6 M H SO ) was added to the beaker on the left, Cr(OH) dissolved to produce violet [Cr(H O) ] ions and water. The addition of concentrated base (6 M NaOH) to the beaker on the right caused Cr(OH) to dissolve, producing green [Cr(OH) ] ions. Aluminum hydroxide, written as either Al(OH) or Al O ·3H O, is amphoteric. Write chemical equations to describe the dissolution of aluminum hydroxide in (a) acid and (b) base. amphoteric compound dissolution reactions in acid and base Using and as a guide, write the dissolution reactions in acid and base solutions. In aqueous solution, Al forms the complex ion [Al(H O) ] . Exercise Copper(II) hydroxide, written as either Cu(OH) or CuO·H O, is amphoteric. Write chemical equations that describe the dissolution of cupric hydroxide both in an acid and in a base. \[Cu(OH)_{2(s)} + 2H^+_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2H_2O_{(l)} \notag \] \[Cu(OH)_{2(s)} + 2OH^−_{(aq)} \rightarrow [Cu(OH)_4]^2_{−(aq)} \notag \] Many dissolved metal ions can be separated by the selective precipitation of the cations from solution under specific conditions. In this technique, pH is often used to control the concentration of the anion in solution, which controls which cations precipitate. The concentration of anions in solution can often be controlled by adjusting the pH, thereby allowing the selective precipitation of cations. Suppose, for example, we have a solution that contains 1.0 mM Zn and 1.0 mM Cd and want to separate the two metals by selective precipitation as the insoluble sulfide salts, ZnS and CdS. The relevant solubility equilibriums can be written as follows:
\[ZnS_{(s)} \rightleftharpoons Zn^{2+}_{(aq)} + S^{2−}_{(aq)} \tag{17.4.19a}\] with \[K_{sp}= 1.6 \times 10^{−24} \tag{17.4.19b}\]
and \[CdS_{(s)} \rightleftharpoons Cd^{2+}_{(aq)} + S^{2−}_{(aq)} \tag{17.4.20a}\] with Because the S ion is quite basic and reacts extensively with water to give HS and OH , the solubility equilibriums are more accurately written as MS(s) = M (aq) +HS (aq) + OH rather than MS(s) = M (aq) +S2 (aq) Here we use the simpler form involving S , which is justified because we take the reaction of S with water into account later in the solution, arriving at the same answer using either equilibrium equation. The sulfide concentrations needed to cause ZnS and CdS to precipitate are as follows:
\[K_{sp} = [Zn^{2+},S^{2−}] \tag{17.4.21a}\] \[1.6 \times 10^{−24} = (0.0010\; M)[S^{2−}]\tag{17.4.21b}\] \[1.6 \times 10^{−21}\; M = [S^{2−}]\tag{17.4.21c}\] and \[K_{sp} = [Cd^{2+},S^{2−}] \tag{17.4.22a}\] \[8.0 \times 10^{−27} = (0.0010\; M)[S^{2−}]\tag{17.4.22b}\] \[8.0 \times 10^{−24}\; M = [S^{2−}] \tag{17.4.22c}\] Thus sulfide concentrations between 1.6 × 10 M and 8.0 × 10 M will precipitate CdS from solution but not ZnS. How do we obtain such low concentrations of sulfide? A saturated aqueous solution of H S contains 0.10 M H S at 20°C. The p for H S is 6.97, and p corresponding to the formation of [S ] is 12.90. The equations for these reactions are as follows:
\[H_2S_{(aq)} \rightleftharpoons H^+_{(aq)} + HS^−_{(aq)} \tag{17.4.24a}\] with \[pK_{a1} = 6.97 \; \text{and hence} \; K_{a1} = 1.1 \times 10^{−7} \tag{17.4.24b}\] \[HS^−_{(aq)} \rightleftharpoons H^+_{(aq)} + S^{2−}_{(aq)} \tag{17.4.24c}\] with \[pK_{a2} = 12.90 \; \text{and hence} \; K_{a2} = 1.3 \times 10^{−13} \tag{17.4.24d}\] We can show that the concentration of S is 1.3 × 10 by comparing and and recognizing that the contribution to [H ] from the dissociation of HS is negligible compared with [H ] from the dissociation of H S. Thus substituting 0.10 M in the equation for for the concentration of H S, which is essentially constant regardless of the pH, gives the following:
Substituting this value for [H ] and [HS ] into the equation for , Although [S ] in an H S solution is very low (1.3 × 10 M), bubbling H S through the solution until it is saturated would precipitate both metal ions because the concentration of S would then be much greater than 1.6 × 10 M. Thus we must adjust [S ] to stay within the desired range. The most direct way to do this is to adjust [H ] by adding acid to the H S solution (recall Le Chatelier's principle), thereby driving the equilibrium in to the left. The overall equation for the dissociation of H S is as follows:
\[H_2S_{(aq)} \rightleftharpoons 2H^+_{(aq)} + S^{2−}_{(aq)} \tag{17.4.26}\] Now we can use the equilibrium constant for the overall reaction, which is the product of and , and the concentration of H S in a saturated solution to calculate the H concentration needed to produce [S ] of 1.6 × 10 M:
\[K=K_{\textrm{a1}}K_{\textrm{a2}}=(1.1\times10^{-7})(1.3\times10^{-13})=1.4\times10^{-20}=\dfrac{[\mathrm{H^+}]^2[\mathrm{S^{2-}}]}{[\mathrm{H_2S}]} \tag{17.4.27}\] Thus adding a strong acid such as HCl to make the solution 0.94 M in H will prevent the more soluble ZnS from precipitating while ensuring that the less soluble CdS will precipitate when the solution is saturated with H S. A solution contains 0.010 M Ca and 0.010 M La . What concentration of HCl is needed to precipitate La (C O ) ·9H O but not Ca(C O )·H O if the concentration of oxalic acid is 1.0 M? values are 2.32 × 10 for Ca(C O ) and 2.5 × 10 for La (C O ) ; p = 1.25 and p = 3.81 for oxalic acid. concentrations of cations, values, and concentration and p values for oxalic acid concentration of HCl needed for selective precipitation of La (C O ) Write each solubility product expression and calculate the oxalate concentration needed for precipitation to occur. Determine the concentration range needed for selective precipitation of La (C O ) ·9H O. Add the equations for the first and second dissociations of oxalic acid to get an overall equation for the dissociation of oxalic acid to oxalate. Substitute the [ox ] needed to precipitate La (C O ) ·9H O into the overall equation for the dissociation of oxalic acid to calculate the required [H ]. Because the salts have different stoichiometries, we cannot directly compare the magnitudes of the solubility products. Instead, we must use the equilibrium constant expression for each solubility product to calculate the concentration of oxalate needed for precipitation to occur. Using ox for oxalate, we write the solubility product expression for calcium oxalate as follows: \[K_{sp} = [Ca^{2+},ox^{2−}] = (0.010)[ox^{2−}] = 2.32 \times 10^{−9} \notag \] \[[ox^{2−}] = 2.32 \times 10^{−7}\; M \notag \] The expression for lanthanum oxalate is as follows: \[K_{sp} = [La^{3+}]^2[ox^{2−}]^3 = (0.010)^2[ox^{2−}]^3 = 2.5 \times 10^{−27} \notag \] \[[ox^{2−}] = 2.9 \times 10^{−8}\; M \notag \] Thus lanthanum oxalate is less soluble and will selectively precipitate when the oxalate concentration is between 2.9 × 10 M and 2.32 × 10 M. To prevent Ca from precipitating as calcium oxalate, we must add enough H to give a maximum oxalate concentration of 2.32 × 10 M. We can calculate the required [H ] by using the overall equation for the dissociation of oxalic acid to oxalate: Substituting the desired oxalate concentration into the equilibrium constant expression, Thus adding enough HCl to give [H ] = 6.1 M will cause only La (C O ) ·9H O to precipitate from the solution. Exercise A solution contains 0.015 M Fe and 0.015 M Pb . What concentration of acid is needed to ensure that Pb precipitates as PbS in a saturated solution of H S, but Fe does not precipitate as FeS? values are 6.3 × 10 for FeS and 8.0 × 10 for PbS. 0.018 M H The anion in many sparingly soluble salts is the conjugate base of a weak acid. At low pH, protonation of the anion can dramatically increase the solubility of the salt. Oxides can be classified as acidic oxides or basic oxides. either react with water to give an acidic solution or dissolve in strong base; most acidic oxides are nonmetal oxides or oxides of metals in high oxidation states. either react with water to give a basic solution or dissolve in strong acid; most basic oxides are oxides of metallic elements. Oxides or hydroxides that are soluble in both acidic and basic solutions are called . Most elements whose oxides exhibit amphoteric behavior are located along the diagonal line separating metals and nonmetals in the periodic table. In solutions that contain mixtures of dissolved metal ions, the pH can be used to control the anion concentration needed to selectively precipitate the desired cation. Which of the following will show the greatest increase in solubility if 1 M HNO is used instead of distilled water? Explain your reasoning. Of the compounds Sn(CH CO ) and SnS, one is soluble in dilute HCl and the other is soluble only in hot, concentrated HCl. Which is which? Provide a reasonable explanation. Where in the periodic table do you expect to find elements that form basic oxides? Where do you expect to find elements that form acidic oxides? Because water can autoionize, it reacts with oxides either as a base (as OH ) or as an acid (as H O ). Do you expect oxides of elements in high oxidation states to be more acidic (reacting with OH ) or more basic (reacting with H O ) than the corresponding oxides in low oxidation states? Why? Given solid samples of CrO, Cr O , and CrO , which would you expect to be the most acidic (reacts most readily with OH )? Which would be the most basic (reacts most readily with H O )? Why? Which of these elements—Be, B, Al, N, Se, In, Tl, Pb—do you expect to form an amphoteric oxide? Why? A 1.0 L solution contains 1.98 M Al(NO ) . What are [OH ] and [H ]? What pH is required to precipitate the cation as Al(OH) ? = 1.3 × 10 and = 1.05 × 10 for the hydrated Al ion. A 1.0 L solution contains 2.03 M CoCl . What is [H ]? What pH is required to precipitate the cation as Co(OH) ? = 5.92 × 10 and = 1.26 × 10 for the hydrated Co ion. Given 100 mL of a solution that contains 0.80 mM Ag and 0.80 mM Cu , can the two metals be separated by selective precipitation as the insoluble bromide salts by adding 10 mL of an 8.0 mM solution of KBr? values are 6.27 × 10 for CuBr and 5.35 × 10 for AgBr. What maximum [Br ] will separate the ions? Given 100 mL of a solution that is 1.5 mM in Tl , Zn , and Ni , which ions can be separated from solution by adding 5.0 mL of a 12.0 mM solution of Na C O ? How many milliliters of 12.0 mM Na C O should be added to separate Tl and Zn from Ni ? [H ] = 4.56 × 10 ; [OH ] = 2.19 × 10 ; pH = 2.94 No; both metal ions will precipitate; AgBr will precipitate as Br is added, and CuBr will begin to precipitate at [Br ] = 8.6 × 10 M. | 21,944 | 1,647 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Environmental_Toxicology_(van_Gestel_et_al.)/05%3A_Population_Community_and_Ecosystem_Ecotoxicology/5.03%3A_Wildlife_population_ecotoxicology |
Nico van den Brink : Ansje Löhr, John Elliott You should be able to Pharmaceuticals, uncertainty, population decline, retrospective monitoring Historically, vulture populations in India, Pakistan and Nepal were too numerous to be effectively counted. In the mid-1990s numbers in northern India started to decline catastrophically, which was evidenced in the Keoladeo National Park (figure 1, Prakash 1999). Further monitoring of population numbers indicated unprecedented declines of over 90-99% from the mid-1990s to the early 2000s for Oriental White-backed vultures ( ), Long-billed vultures ( ) and also Slender-billed vultures ( ) (Prakash 1999). In the following years, similar declines were observed in Pakistan and Nepal, indicating that the causative factor was not restricted to a specific country or area. Total losses of vultures were estimated to be in the order of tens of millions. The first ideas about potential causes of those declines focussed on known infectious diseases or the possibility of new diseases to which the vulture population had not been previously exposed. However, no diseases were identified that had shown similar rates of mortalities in other bird species. Vultures are also considered to have a highly developed immune response given their diet of scavenging dead and often decaying animals. To obtain insights, initial interdisciplinary ecological studies were performed to provide a basic understanding of background mortality in the species affected. These studies started in large colonies in Pakistan, but were literally races against time, as some populations had already decreased by 50%, while others were already extirpated, (Gilbert et al., 2006). Despite those difficulties it was determined that mortalities were occurring principally in adult birds and not at the nestling phase. More in depth studies were performed to discriminate between natural mortality of for instance juvenile fledglings, which may be high in summer, just after fledging. After scrutinising the data no seasonality was observed in the abnormal, high mortality, indicating that this was not related to breeding activities. The investigations also revealed another important factor that these vultures were predominantly feeding on domestic livestock, while telemetric observations, using transmitters to assess flight and activity patterns f the birds, showed that the individual birds could range over very long distances to reach carcasses of livestock (up to over 100 km). Since no apparent causes for mortality were obtained in the ecological studies, more diagnostic investigations were started, focussing on infectious diseases and carried out in Pakistan (Oaks, 2011). However, that was easier said than done. Since large numbers of birds died, it was deemed essential to establish the logistics necessary to perform the diagnostics, including post-mortems, on all birds found dead. Although high numbers of birds died, hardly any fresh carcasses were available, due to remoteness of some areas, the presence of other scavengers and often hot conditions which fostered rapid decay of carcasses. Post-mortems on a selection off birds revealed that birds suspect of abnormal mortality all suffered from visceral gout, which is a white pasty smear covering tissues in the body including liver and heart. In birds, this is indicative for kidney failure. Birds metabolise nitrogen into uric acid (mammals into urea) which is normally excreted with the faeces. However, in case of kidney failure the uric acid is not excreted but deposited in the body. Further inspections of more birds confirmed this, and the working hypothesis became that the increased mortality was caused by a factor inducing kidney failure in the birds. Based on the establishment of kidney failure as the causative factor, histological and pathological studies were performed on several birds found dead which revealed that in birds with visceral gout, kidney lesions were severe with acute renal tubular necrosis (Oaks et al., 2004), confirming the kidney failure hypothesis. However, no indications of inflammatory cell infiltrations were apparent, ruling out the possibilities of infectious diseases. Those observation shifted the focus to potential toxic effects, although no previous case was known with a chemical causing such severe and extremely acute effects. First the usual suspects for kidney failure were addressed, like trace metals (cadmium, lead) but also other acute toxic chemicals like organophosphorus and carbamate pesticides and organochlorine chemicals None of those chemicals occurred at levels of concern and were ruled out. That left the researchers without leads to any clear causative factor, even after years of study! Some essential pieces of information were available, however: 1) acute renal failure seemed associated with the mortality, 2) no infectious agent was likely to be causative pointing to chemical toxicity, 3) since exposure was likely to be via the diet the chemical exposure needed to be related to livestock (the predominant diet for the vultures), pointing to compounds present in livestock such as veterinarian products, 4) widespread use of veterinarian chemicals had started relatively recently. After a survey of veterinarians in the affected areas of Pakistan, a single veterinarian pharmaceutical matched the criteria, diclofenac. This is a non-steroid anti-inflammatory drug (NSAID) since long used in human medicine but only introduced since the 1990s as a veterinarian pharmaceutical in India, Pakistan and surrounding countries. NSAIDs are known nephrotoxic compounds, although no cases were known with such acute and sever impacts. Chemical analyses of kidneys of vultures confirmed that kidneys of birds with visceral gout contained diclofenac, birds without signs of visceral gout did not. Also kidneys from birds that showed visceral gout and that died in captivity while being studied, were positive for diclofenac, as was the meat they were fed with. This all indicated diclofenac toxicity as the cause of the mortality, which was validated in exposure studies, dosing captive vultures with diclofenac. The species of vultures appeared extremely sensitive to diclofenac, showing toxic effects at 1% of the therapeutic dose for livestock mammalian species. This underlying mechanism for that sensitivity has yet to be explained, but initially it was also unclear why the populations were impacted to such severe extent. That was found to be related to the feeding ecology of the vultures. They were shown to fly long ranges to search for carcasses, and as a result of that they show very aggregated feeding, i.e. a lot of birds on a single carcass (Green et al., 2004). Hence, a single contaminated carcass may expose an unexpected large part of the population to diclofenac. Hence, a combination of extreme sensitivity, foraging ecology and human chemical use caused the onset of extreme population declines of some Asian vulture species of the Gyps genus, or so called "Old World vultures". This case demonstrated the challenges involved in attempting to disentangle the stressors causing very apparent population effects even on imperative species like vultures. It took several years of different groups of excellent researcher to perform the necessary research and forensic studies (under sometimes difficult conditions). Lessons learned are that even for compounds that have been used for a long time and thought to be well understood, unexpected effects may become evident. There is consensus that such effects may not be covered in current risk assessments of chemicals prior to their use and application, but this also draws attention to the need for continued post-market monitoring of organisms for potential exposure and effects. It should be noted that even nowadays, although the use of diclofenac is prohibited in larger parts of Asia, continued use still occurs due to its effectiveness in treating livestock and its low costs making it available to the farmers. Nevertheless, populations of Gyps vultures have shown to recover slowly. Green, R.E., Newton, I.A., Shultz, S., Cunningham, A.A., Gilbert, M., Pain, D.J., Prakash, V. (2004). Diclofenac poisoning as a cause of vulture population declines across the Indian subcontinent. Journal of Applied Ecology 41, 793-800. Gilbert, M., Watson, R.T., Virani, M.Z., Oaks, J.L., Ahmed, S., Chaudhry, M.J.I., Arshad, M., Mahmood, S., Ali, A., Khan, A.A. (2006). Rapid population declines and mortality clusters in three Oriental whitebacked vulture colonies in Pakistan due to diclofenac poisoning. Oryx 40, 388-399. Oaks, J.L., Gilbert, M., Virani, M.Z., Watson, R.T., Meteyer, C.U., Rideout, B.A., Shivaprasad, H.L., Ahmed, S., Chaudhry, M.J.I., Arshad, M., Mahmood, S., Ali, A., Khan, A.A. (2004). Diclofenac residues as the cause of vulture population decline in Pakistan. Nature 427, 630-633. Oaks, J.L., Watson, R.T. (2011). South Asian vultures in crisis: Environmental contamination with a pharmaceutical. In: Elliott, J.E., Bishop, C.A., Morrissey, C.A. (Eds.) Wildlife Ecotoxicology. Springer, New York, NY. pp. 413-441. Prakash, V. (1999). Status of vultures in Keoladeo National Park, Bharatpur, Rajasthan, with special reference to population crash in species. Journal of the Bombay Natural History Society 96, 365-378. Which ecological traits make vulture populations extremely vulnerable to exposure to chemicals like diclofenac? What indirect effects do you expect from the chemical induced crashes of the Asian vultures? : Nico van den Brink : Ansje Löhr, Michiel Kraak, Pim Leonards, John Elliott You should be able to: : Threshold levels, read across, species specific sensitivity To re-establish a viable population of otters in the Netherlands, a program was established in the mid-1990s to re-introduce otters in the Netherlands, including monitoring of PCBs and other organic contaminants in the otters. Otters were captured in e.g. Belarus, Sweden and Czech Republic. Initial results showed that these otters already contained < 1 mg/kg PCBs based on wet weight (van den Brink & Jansman, 2006), which was considered to be below the threshold limits mentioned before. Individual otters were radio-tagged, and most were recovered later as victims of car incidences. Over time, PCB concentrations had changed, although not in the same direction for all specimen. Females with high initial concentrations showed declining concentrations, due to lactation, while in male specimens most concentrations increased over time, as you would expect. Nevertheless, concentrations were in the range of the threshold levels, hence risks on effects could not be excluded. Since the re-introduction program was established in a relatively low contaminated area in the Netherlands, questions were raised for re-introduction plans in more contaminated areas, like the Biesbosch where contaminants may still affect otters . To assess potential risks of PCB contamination in e.g. the Biesbosch for otter populations a modelling study was performed in which concentrations in fish from the Biesbosch were modelled into concentrations in otters. Concentrations of PCBs in the fish differed between species (lipid rich fish such as eel greater concentrations than lean white fish), size of the fish (larger fish greater concentrations than smaller fish) and between locations within the Biesbosch. Using Biomagnification Factors (BMFs) specific for each PCB-congener (see section on ), total PCB concentrations in lipids of otters were calculated based on fish concentrations and different compositions of fish diets of the otters (e.g. white fish versus eel, larger fish versus smaller fish, different locations). Different diets resulted in different modelled PCB concentrations in the otters, however all modelled concentrations were above the earlier mentioned threshold levels (van den Brink and Sluiter, 2015). This would indicate that risks of effects for otters could not be ruled out, and led to the notion that release of otters in the Biesbosch would not be the best option. However, a major issue related to such risk assessment is whether the threshold levels derived from mink are applicable to otter. The resulting threshold levels for otter are rather low and exceedance of these concentrations has been noticed in several studies. For instance, in well-thriving Scottish otter populations PCBs levels have been recorded greater than 50 mg/kg lipid weight in livers (Kruuk & Conroy, 1996). This is an order of magnitude higher than the threshold levels, which would indicate that even at higher concentrations, at which effects are to be expected based on mink studies, populations of free ranging otters do not seem to be affected adversely. Based on this, the applicability of mink-derived threshold levels for otters may be open to discussion. The case on otters showed that the derivation of ecological relevant toxicological threshold levels may be difficult due to the fact that otters are not regularly used in toxicity tests. Application of data from a related species, in this case the American mink, however, may be limited due to differences in sensitivity. In this case, this could result in too conservative assessments of the risks, although it should be noted that this may be different in other combinations of species. Therefore, the read across of information of closely related species should therefore be performed with great care. Basu, N., Scheuhammer, A.M., Bursian, S.J., Elliott, J., Rouvinen-Watt, K., Chan, H.M. (2007). Mink as a sentinel species in environmental health. Environmental Research 103, 130-144. Harding, L.E., Harris, M.L., Stephen, C.R., Elliott, J.E. (1999). Reproductive and morphological condition of wild mink ( ) and river otters ( ) in relation to chlorinated hydrocarbon contamination. Environmental Health Perspectives 107, 141-147. Folland, W.R., Newsted, J.L., Fitzgerald, S.D., Fuchsman, P.C., Bradley, P.W., Kern, J., Kannan, K., Zwiernik, M.J. (2016). Enzyme induction and histopathology elucidate aryl hydrocarbon receptor-mediated versus non-aryl receptor-mediated effects of Aroclor 1268 in American Mink ( ). Environmental Toxicology and Chemistry 35, 619-634. Kruuk, H., Conroy, J.W.H. (1996). Concentrations of some organochlorines in otters ( L) in Scotland: Implications for populations. Environmental Pollution 92, 165-171. Leonards, P.E.G., De Vries, T.H., Minnaard, W., Stuijfzand, S., Voogt, P.D., Cofino, W.P., Van Straalen, N.M., Van Hattum, B. (1995). Assessment of experimental data on PCB‐induced reproduction inhibition in mink, based on an isomer‐ and congener‐specific approach using 2,3,7,8‐tetrachlorodibenzo‐p‐dioxin toxic equivalency. Environmental Toxicology and Chemistry 14, 639-652. Murk, A.J., Leonards, P.E.G., Van Hattum, B., Luit, R., Van der Weiden, M.E.J., Smit, M. (1998). Application of biomarkers for exposure and effect of polyhalogenated aromatic hydrocarbons in naturally exposed European otters ( ). Environmental Toxicology and Pharmacology 6, 91-102. Van den Brink, N.W., Jansman, H.A.H. (2006). Applicability of spraints for monitoring organic contaminants in free-ranging otters ( ). Environmental Toxicology & Chemistry 25, 2821-2826. Name three reasons why assessment of risk of PCBs to otters is relatively complicated How is it possible that although toxicity threshold levels are exceeded in some otter populations, for example in Scotland, the population seem to thrive really well? | 15,553 | 1,648 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Voltaic_Cells/Electrochemical_Cells_under_Nonstandard_Conditions/Concentration_Cell |
A concentration cell is an that is comprised of two half-cells with the same electrodes, but differing in concentrations. A concentration cell acts to dilute the more concentrated solution and concentrate the more dilute solution, creating a voltage as the cell reaches an equilibrium. This is achieved by transferring the electrons from the cell with the lower concentration to the cell with the higher concentration. The , commonly written as E , of a concentration cell is equal to zero because the electrodes are identical. But, because the ion concentrations are different, there is a potential difference between the two half-cells. One can find this potential difference via the , \[ E_{cell} = E^\circ_{cell} - \dfrac{0.0592}{n}\log Q \] at 25 C. The E stands for the voltage that can be measured using a voltmeter (make sure if the voltmeter measures it in millivolts that you convert the number before using it in the equation). Note that the Nernst Equation indicates that cell potential is dependent on concentration, which results directly from the dependence of free energy on concentration. Remember that to find Q you use this equation: \[ \ce{$aA$ + $bB$ <=> $cC$ + $dD$} \] \[ Q = \frac{[C]^{c}[D]^{d}}{[A]^{a}[B]^{b}} \] When Q=1, meaning that the concentrations for the products and reactants are the same, then taking the log of this equals zero. When this occurs, the E is equal to the E . Another way to use the E , or to find it, is using the equation below. \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}\] Fig.1 An example of a concentration cell These concepts are useful for understanding the electron transfer and what occurs in half-cells. The two compartments of a cell must be separated so they do not mix, but cannot be completely separated with no way for ions to be transferred. A wire cannot be used to connect the two compartments because it would react with the ions that flow from one side to another. Because of this, a salt bridge is an important part of a concentration cell. It solves the major problem of electrons beginning to pile up too much in the right beaker. This buildup is due to electrons moving from the left side, or left beaker, to the right side, or right beaker. The salt bridge itself can be in a few different forms, such as a salt solution in a U-tube or a porous barrier (direct contact). It evens the charge by moving ions to the left side, or left beaker. In the written expression which shows what is occurring in specific reactions, the salt bridge is represented by the double lines. An example of this would be: \[ \text{Zn(s)} | \text{Zn}^{2+} (1~\text{M}) || \text{Cu}^{2+} (1~\text{M}) | \text{Cu} \] The double lines between the Zn (1 M) and the Cu (1 M) signify the salt bridge. The single lines, however, do not represent bridges; they represent the different phase changes, for example, from solid zinc to liquid zinc solution. If there is a comma where you would expect to see a single line, this is not incorrect. It simply means that no phase changes occurred. In this type of reaction, there are two electrodes which are involved. These are known as the anode and the cathode, or the left and right side, respectively. The anode is the side which is losing electrons (oxidation) while the cathode is the side which is gaining electrons (reduction). A voltmeter (not to be confused with a different kind of voltmeter which also measures a type of energy) is used to measure the cell potential that is passed between the two sides. It is typically located in between the two cells. This cell potential (also known as an electromotive force) occurs due to the flow of electrons. The value it shows can be negative or positive depending on the direction in which the electrons are flowing. If the potential is positive then the transfer of electrons is spontaneous, but the reverse reaction is NONspontaneous. Conversely, if the value of the potential is negative, the transfer of electrons is nonspontaneous and the reverse reaction. The voltmeter measures this potential in volts or millivolts. The tendency of electrons to flow from one chemical to another is known as electrochemistry. This is what occurs in a concentration cell. The electrons flow from the left side (or left beaker) to the right side (or right beaker). Because the left side is losing electrons and the right is gaining them, the left side is called the oxidation side and the right side is the reduction side. Although you could switch the two to be on the opposite sides, this is the general way in which the set up is done. The oxidation side is called the anode and the reduction side is the cathode. It is the flow of the electrons that cause one side to be oxidized and the other to be reduced. Corrosion can occur on a concentration cell when the metal being used is in contact with different concentrations, causing parts of the metal to have different electric potential than the other parts. One element that is often linked to this corrosion is oxygen. In the areas in which there is a low oxygen concentration corrosion occurs. This can be somewhat prevented through sealing off the cell and keeping it clean, but even this cannot prevent any corrosion from occurring at some point. Corrosion is most frequently a problem when the cell is in contact with soil. Because of the variations that occur within soil, which is much greater than the variations that occur within a fluid, contact with soil often causes corrosion for the cell. A pH meter is a specific type of concentration cell that uses the basic setup of a concentration cell to determine the pH, or the acidity/basicity, of a specific solution. It is comprised of two electrodes and a voltmeter. One of the electrodes, the glass one has two components: a metal (commonly silver chloride) wire and a separate semi-porous glass part filled with a potassium chloride solution with a pH of 7 surrounding the AgCl. The other electrode is called the reference electrode, which contains a potassium chloride solution surrounding a potassium chloride wire. The purpose of this second electrode is to act as a comparison for the solution being tested. When the glass electrode comes into contact with a solution of different pH, an electric potential is created due to the reaction of the hydrogen ions with the metal ions. This potential is then measured by the voltmeter, which is connected to the electrode. The higher the voltage, the more hydrogen ions the solution contains, which means the solution is more acidic. Fig.2 An example of a pH meter 1.) For the concentration cell bellow determine the flow of electrons. \[ \text{Fe} | \text{Fe}^{2+}(0.01~\text{M}) || \text{Fe}^{2+}(0.1~\text{M})| \text{Fe}\] The cells will reach equilibrium if . As a result, Fe will be formed in the left compartment and metal iron will be deposited on the right electrode. 2.) Calculate cell potential for a concentration cell with two silver electrodes with concentrations 0.2 M and 3.0 M. Reaction: \[ \ce{Ag^{2+} + 2e^- -> Ag(s)}\] Cell Diagram: \[ \text{Ag(s)}|\text{Ag}^{2+}(0.2~\text{M})||\text{Ag}^{2+}(3.0~\text{M})|\text{Ag(s)}\] Nernst Equation: \[ E = E^\circ - \dfrac{0.0592}{2}\log \dfrac{0.02}{3.0}\] **E = 0 for concentration cells 3.) Calculate the concentration of the unknown, given the equation below and a cell potential of 0.26&nsbp;V. \[ \text{Ag}|\text{Ag}^+(x~\text{M})||\text{Ag}^+(1.0~\text{M})|\text{Ag} \] \[ E = E^\circ - \dfrac{0.0592}{1} \log \dfrac{x}{1.0}\] \[0.26 = 0 - 0.0592 \log \dfrac{x}{1.0}\] \[4.362 = -\log(x) + \log(1.0)\] \[\log(x) = \log(1.0) - 4.362\] | 7,637 | 1,649 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Two-photon_absorption |
Two-photon absorption may seem very similar to non-resonance , and was in fact predicted due to this phenomena. In both cases, a non-resonant photon is used for excitation. However, in the absorption case a secondary non-resonant photon is used for excitation as well, while in Raman a second non-resonant photon is emitted. For Raman this results in occupation of an energy state at the difference of the frequencies of the absorbed and emitted photon. However, in two-photon absorption this results in the occupation of an energy state at the sum of the frequencies of the absorbed photons. The basic process is illustrated below in . Two-photon absorption is not a feature of a specific type of spectroscopy. It can be utilized in any type of spectroscopy including , , , and . However, it will result in a change of selection rules for many of these spectroscopies. This means that the peak shapes should remain the same, but the intensities should be significantly smaller due to the result of the second order perturbation. In her 1931 dissertation, Maria Goeppert-Mayer postulated the existence of two-photon absorption for the first time. She says that she based the derivation of this phenomena off of Kramer and Heisenberg's derivation of the probability of two-photon emission which uses Dirac's dispersion theory for calculation. The invention of the laser had a great impact on the field of two-photon spectroscopy because of the necessity of a high-intensity electromagnetic field to induce transitions. Although, lasers were not formally invented until 1969, Bell Labs was testing masers in 1958, which were only capable of short pulses of intense electromagnetic radiation. In 1961 Kaiser and Garret reported the first two-photon absorption of a compound. They used the new laser technology to excite CaF\(_{2}\)Eu\(^{2+}\) with both red and blue light to induce a two photon transition. The following constants, variables, and operators are used in this section: First we must take a look at how the atomic system interacts with any perturbation \[ \hat{H} = \hat{H}_{0} + \hat{H}_{1}\hspace{1cm}\label{1}\] If we are only going to look at the effects of the electric dipole then: \[ \hat{H} = \hat{H}_{0} + \hat{H}_{\mu_{e}}\hspace{1cm}\label{2}\] Now we must consider how our states will be evolving in time because the is time-dependent. According to the time-dependent Schrödinger equation: \[ \Psi_{\textrm{i}}(t) = \textrm{exp}\left (-\frac{i\hat{H}(t-t_{0})}{\hbar}\right )\Psi_{\textrm{i}}(t_{0})\hspace{1cm} \label{3} \] We have described our ground state as it is moving in time so now we can look at our final state which we can only describe as being at a set energy value: \[\hat{H}_{0}\Psi_{f} = \hbar\omega_{f}\Psi_{f}\hspace{1cm} \label{4} \] Now that we have described both the final and the initial states. We can find the probability of such a transition by finding the magnitude of the projection of the intial state upon the final state, otherwise known as the magnitude of the inner product. \[ |\left <\Psi_{f}| \Psi_{i}(t) \right >|^{2} =\] \[ \left | \left <\Psi_{f} \left | \textrm{exp}\left (-\frac{i\hat{H}(t-t_{0})}{\hbar}\right )\right | \Psi_{\textrm{i}}(t_{0}) \right > \right | ^{2}\hspace{1cm} \label{5}\] Now that we've found the general formula for the probability of our transitions in a time-dependent system, we must apply the perturbation from earlier. To do this we use the following identity: \[ \textrm{exp} \left (\frac{i\hat{H}_{0}t}{\hbar}\right )\hat{H}_{\mu_{e}}\textrm{exp}\left (-\frac{i\hat{H}t}{\hbar}\right ) = i\hbar\frac{\textrm{d}}{\textrm{d}t}\left [\textrm{exp}\left (\frac{i\hat{H}_{0}t}{\hbar}\right )\textrm{exp}\left (-\frac{i\hat{H}t}{\hbar}\right )\right ] \hspace{1cm}\label{6} \] The issue now is that the electric dipole Hamiltonian is not in the exponential form, this prevents us from simply solving for the total Hamiltonian exponential. To solve this we integrate with respect to time, which nullifies the derivative. \[ \int^{t}_{t_{0}} \textrm{exp}\left (\frac{i\hat{H}_{0}t}{\hbar}\right )\hat{H}_{\mu_{e}}\textrm{exp}\left (-\frac{i\hat{H}t}{\hbar}\right ) \textrm{d}t_{1} \] \[= i\hbar\left [\textrm{exp}\left (\frac{i\hat{H}_{0}t}{\hbar}\right )\textrm{exp}\left (-\frac{i\hat{H}t}{\hbar}\right ) -\textrm{exp}\left (\frac{i\hat{H}_{0}t_{0}}{\hbar}\right )\textrm{exp}\left (-\frac{i\hat{H}t_{0}}{\hbar}\right )\right ]\hspace{1cm} \label{7}\] This equation can be simplified and solved for using the approximation that the initial time was well before the incident of interaction with the perturbation. This allows us to assume that the electric dipole Hamiltonian would yield 0 at the initial time. This allows for the following iterative equation. \[ \textrm{exp}\left (\frac{-i\hat{H}t}{\hbar} \right ) = \textrm{exp} \left ( -\frac{i\hat{H}_{0}t}{\hbar}\right ) \times \left [ 1 - \frac{i}{\hbar}\int^{t}_{-\infty} \textrm{exp}\left (\frac{i\hat{H}_{0}t}{\hbar}\right )\hat{H}_{\mu_{e}}\textrm{exp}(\epsilon t_{1})\textrm{exp}\left (-\frac{i\hat{H}t}{\hbar}\right ) \textrm{d}t_{1} \right ]\hspace{1cm} \label{8} \] Now for the purposes of two-photon absorption we will need to find the 2nd order perturbation. Thus we must find the second iteration of this equation. Where epsilon is a very small number, we come to the following equation for both the first and second order contribution: \[ \left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > = \frac{1}{\hbar}\frac{\textrm{exp}(-i\omega_{i}t)}{\omega_{i} - \omega_{f}} \left ( \left< \Psi_{f} \left | \hat{H}_{\mu_{e}} \right | \Psi_{i} \right > + \frac{1}{\hbar}\sum_{l}\frac{ \left< \Psi_{f} \left | \hat{H}_{\mu_{e}} \right | \Psi_{l} \right > \left< \Psi_{l} \left | \hat{H}_{\mu_{e}} \right | \Psi_{i} \right >}{\omega_{i} - \omega_{l}} \right )\hspace{1cm} \label{9} \] The following constants, variables and operators are introduced in this section: For the perturbation method described in the previous section the use of a single iteration accurately describes the transitions between a two levels. This describes the results of Fermi's Golden Rule. However, in order to describe these two-photon processes which transgress through more than one transition, we must look at the second iteration within the perturbation. \[\hat{H}_{\mu_{e}} = e\hat{\mu}(\hat{E}_{1} + \hat{E}_{2}) \hspace{1cm} \label{10}\] Where \[\hat{E_{n}} = \xi_{n}\textrm{cos}(\omega_{n} t)\hspace{1cm} \label{11} \] Thus \[\hat{H}_{\mu_{e}} = e\hat{\mu}(\xi_{1}\textrm{cos}(\omega_{1} t) + \xi_{2}\textrm{cos}(\omega_{2} t))\hspace{1cm} \label{12} \] Substituting this into the second iteration gives us: \[ \left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > = \] \[\frac{1}{\hbar}\frac{\textrm{exp}(-i\omega_{i}t)}{\omega_{i} - \omega_{f}} \left ( \left< \Psi_{f} \left |e\hat{\mu}(\hat{E}_{1} + \hat{E}_{2}) \right | \Psi_{i} \right > + \frac{1}{\hbar}\sum_{l}\frac{ \left< \Psi_{f} \left | e\hat{\mu}(\hat{E}_{1} + \hat{E}_{2}) \right | \Psi_{l} \right > \left< \Psi_{l} \left | e\hat{\mu}(\hat{E}_{1} + \hat{E}_{2}) \right | \Psi_{i} \right >}{\omega_{i} - \omega_{l}} \right )\hspace{1cm} \label{13} \] Looking at this equation we see that the first term is that corresponding to single photon transitions. However, because we are looking for two-photon transitions we can ignore this term and focus on the second term. This second term has a summation to distinguish between the different permutations of arranging the perturbations and achieve the same final state but it includes the interaction of two photons from the same light source which we do not need to consider at the moment. If we eliminate the terms of the sum responsible for these guys we should find the expression for the two permutations of two-photon transitions from our two light sources. Now let us implement these changes. \[ \left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > = \] \[ \frac{1}{\hbar^{2}}\frac{\textrm{exp}(-i\omega_{i}t)}{\omega_{i} - \omega_{f}} \left (\frac{ \left< \Psi_{f} \left | e\hat{\mu}\xi_{1}\textrm{cos}(\omega_{1} t) \right | \Psi_{l} \right > \left< \Psi_{l} \left | e\hat{\mu}\xi_{2}\textrm{cos}(\omega_{2} t) \right | \Psi_{i} \right > +\left< \Psi_{f} \left | e\hat{\mu}\xi_{2}\textrm{cos}(\omega_{2} t) \right | \Psi_{l} \right > \left< \Psi_{l} \left | e\hat{\mu}\xi_{1}\textrm{cos}(\omega_{1} t) \right | \Psi_{i} \right >}{\omega_{i} - \omega_{l}} \right ) \hspace{1cm} \label{14} \] This equation describes the electric dipole contribution towards two-photon transitions. The following constants, variables and operators are introduced in this section: \(\hat{Z}\) In order to take a thorough look at the selection rules for the two-photon system induced by the electric field contribution we must look qualitatively at the results of the equation for the probability of a transition. The Hamiltonian presented in the previous section for the electric dipole contribution is the result of a of the original electric dipole contribution. For the purposes of finding the electric dipole selection rules we shall revert back to the original form. \[\hat{H}_{\mu_{e}}(t) = \frac{qE}{m\omega}\hat{P}_{z}\textrm{sin }\omega t \hspace{1cm}\label{15} \] We will consider the effect of the sinusoid part of this equation later in this section. For now let us track the selection rules due only to the operator \(\hat{P}_{z} \). \[\left [ \hat{Z}, \hat{H}_{0} \right ] = i\hbar\frac{\partial \hat{H}_{0}}{\partial \hat{P}_{z}} = i\hbar\frac{\hat{P}_{z}}{m} \hspace{1cm} \label{16} \] This can be shown to be true by simply working out the commutator and leads to the following statement \[\left < \Psi_{f} \left | \left [\hat{Z},\hat{H}_{0} \right ] \right | \Psi_{i} \right > = \left < \Psi_{f} \left | \hat{Z}\hat{H}_{0} -\hat{H}_{0}\hat{Z} \right | \Psi_{i} \right > \hspace{1cm} \label{17} \] \[ = -(E_{f} - E_{i})\left < \Psi_{f} \left | \hat{Z} \right | \Psi_{i} \right > = \frac{i\hbar}{m}\left < \Psi_{f} \left | \hat{P}_{z} \right | \Psi_{i} \right > \hspace{1cm} \label{18}\] This shows the result of an electric dipole induced transition for a two level system. However, we need to slightly change things in order to incorporate the three levels that will exist as a result of the two-photon transition as well as the fact that we will have two different perturbations. Thus the Hamiltonian for the two-photon system will be: \[ \hat{H}_{\mu_{e}}(t) = \frac{qE_{1}}{m\omega_{1}}\hat{P}_{z}\textrm{sin }\omega_{1} t + \frac{qE_{2}}{m\omega_{2}}\hat{P}_{z}\textrm{sin }\omega_{2} t \hspace{1cm}\label{19} \] Similar to before we can ignore the sinusoid part for later in this section. If we do this we instead come to the result: \[(E_{f}-E_{l})(E_{l}-E_{i})\left < \Psi_{f} \left | \hat{Z} \right | \Psi_{l} \right > \left < \Psi_{l} \left | \hat{Z} \right | \Psi_{i} \right > \hspace{1cm}\label{20}\] Whereas the is difficult to find the selection rules for, the position operator is fairly straight forward. This is because the variable \(z\), which is the result of the operator \(\hat{Z}\), is equal to the following wavefunction. \[z = \sqrt{\frac{4\pi}{3}}rY^{0}_{1}(\theta) \hspace{1cm}\label{21} \] This works under the assumption that the light is polarized in the \(z\) direction but we can also polarize the light in multiple directions as there are two photons. If this is the case we need to expand to define both \(x\) and \(y\) in terms of an angular function. We'll only examine the x case here because it gives the same results. \[x = \sqrt{\frac{2\pi}{3}}r\left ( Y^{-1}_{1} - Y^{1}_{1} \right ) \hspace{1cm} \label{22} \] While for our case we have two position operators acting on different kets, however, that has no effect on the ket because multiplication is commutative. This means the angular part of the inner product will result in the integral for the double \(z\) and double \(x\) case respectively. Because all of the results retain the same \(l\) values that will not be addressed until after we first look at the \(m\) values. \[\int d\Omega Y^{m_{f}\ast}_{l_{f}}(\theta,\phi) Y^{0}_{1}(\theta)Y^{m_{l}}_{l_{l}}(\theta,\phi)Y^{m_{l}\ast}_{l_{l}}(\theta,\phi)Y^{0}_{1}(\theta)Y^{m_{i}}_{l_{i}}(\theta,\phi)\hspace{1cm} \label{23}\] \[\int d\Omega Y^{m_{f}\ast}_{l_{f}}(\theta,\phi) \left ( Y^{-1}_{1} - Y^{1}_{1} \right )Y^{m_{l}}_{l_{l}}(\theta,\phi)Y^{m_{l}\ast}_{l_{l}}(\theta,\phi)\left ( Y^{-1}_{1} - Y^{1}_{1} \right )Y^{m_{i}}_{l_{i}}(\theta,\phi)\hspace{1cm} \label{24}\] The intermediate or virtual state angular functions have no effect since they are complex conjugates of each other, their \(l\) and \(m\) values will annihilate each other. Thus the integral we need to be concerned with is: \[\int d\Omega Y^{m_{f}\ast}_{l_{f}}(\theta,\phi) Y^{0}_{1}(\theta)Y^{0}_{1}(\theta)Y^{m_{i}}_{l_{i}}(\theta,\phi)\hspace{1cm} \label{25})\] \[\int d\Omega Y^{m_{f}\ast}_{l_{f}}(\theta,\phi) \left ( Y^{-1}_{1} - Y^{1}_{1} \right )\left ( Y^{-1}_{1} - Y^{1}_{1} \right )Y^{m_{i}}_{l_{i}}(\theta,\phi)\hspace{1cm} \label{26}\] Because the basis to the selection rule is that only a sum of \(m\) values that results in 0 will result in an even function to integrate over, we can now determine a range for \(m\) values. For purely \(Z\) polarized two-photon absorption the sum of the \(m\) values without the initial and final states is still 0, thus the \(\Delta m\) must remain 0. However, for the inclusion of \(Z\) and \(X\) or \(Y\). the sum without the initial and final states will be \(\pm 1 \). Thus in order to retain a total sum of 0 the \(\Delta m\) must be \(\pm 1\). Finally for the pure \(X\), pure \(Y\), and \(XY\) cases the sum from the operators can results in \(\pm 2\) or 0. Therefore, the \(\Delta m=\pm 2, 0 \) to negate the operators. Now to talk about the \(l\) values. The selection rule that governs them is that we need the sum of all the \(l\) values to be even. Complex conjugate \(l\) values are negative, the same holds true for the \(m\) values. Thus because each operator is has an \(l = 1\) and we now have two operators thus we can have the following possible changes to result in even sums. \[\Delta l = 0, \pm 2 \] Now the rationality behind the change in the selection rules will be discussed. Each photon has a unit of momentum with it. During a single photon excitation that momentum must be transferred due to conservation of momentum. Thus if it destructively interferes with the electron it can lower the total momentum by one unit and if it constructively interferes with the electron is can increase the total momentum by one unit. However, for the two photon case, we now have to consider not only the ways in which the photon and electron will interfere but also how the photons will interfere with each other. This results in a doubly constructive interference state with an increase of 2 units, one constructive and one destructive interference state that results in no change and a doubly destructive interference state that results in a decrease of 2 units. If we take a look at the equation presented at the end of the electric dipole section, it is a simple analysis to derive the basic selection rules for electric dipole induced two-photon transitions. However, we do not need to consider the constant values to determine selection rules as they only determine the intensity of the signal, not the existence of the peak. Where C is a constant: \[ \left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > = \] \[C \left ( \left< \Psi_{f} \left | e\hat{\mu}\xi_{1}\textrm{cos}(\omega_{1} t) \right | \Psi_{l} \right > \left< \Psi_{l} \left | e\hat{\mu}\xi_{2}\textrm{cos}(\omega_{2} t) \right | \Psi_{i} \right > +\left< \Psi_{f} \left | e\hat{\mu}\xi_{2}\textrm{cos}(\omega_{2} t) \right | \Psi_{l} \right > \left< \Psi_{l} \left | e\hat{\mu}\xi_{1}\textrm{cos}(\omega_{1} t) \right | \Psi_{i} \right > \right ) \hspace{1cm} \label{27} \] Although for one-photon transitions the dipole moment is of considerable importance. We notice that because both terms contain two dipole moments, regardless of whether they are odd or even, their combination will still become even. Furthermore, we can see that because the wave part of the electric field is an even function, that this too will preserve parity of the transition in both terms, even if it were an odd function as each term contains two of such functions. Thus the result is that as long as the final and initial state conserve parity (inversion symmetry or antisymmetry is maintained) the transition will be allowed. This is different from one-photon processes where parity conservation is forbidden and makes two-photon spectroscopy especially useful. The following constants, variables, and operators are introduced in this section: The magnetic dipole contribution is determined by the Magnetic dipole Hamiltonian: \[\hat{H}_{DM} = -\frac{q}{2m}(\hat{L_{x}} + 2\hat{S_{x}})B\textrm{cos}(\omega t) \hspace{1cm} \label{28} \] where B is the strength of the magnetic field. So for the two photon case we have: \[\hat{H}_{DM} = -\frac{q}{2m}\left [(\hat{L_{x}} + 2\hat{S_{x}})B_{1}\textrm{cos}(\omega_{1} t) +(\hat{L_{x}} + 2\hat{S_{x}})B_{2}\textrm{cos}(\omega_{2} t)\right ]\hspace{1cm} \label{29}\] Be aware that the photons do not need to be from the same polarization as they are here. One could be polarized in the X direction while the other is polarized in the Z direction. We will cover these situations in the derivation of the selection rules. If we similarly substitute this Hamiltonian into the second iteration for perturbation we get: \[ \left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > = \frac{1}{\hbar}\frac{\textrm{exp}(-i\omega_{i}t)}{\omega_{i} - \omega_{f}} \Bigg( \left< \Psi_{f} \left |-\frac{q}{2m}(\hat{L_{x}} + 2\hat{S_{x}})\left [B_{1}\textrm{cos}(\omega_{1} t) +B_{2}\textrm{cos}(\omega_{2} t)\right ] \right | \Psi_{i} \right > + \] \[\small{\sum_{l}\frac{ \left< \Psi_{f} \left |-\frac{q}{2m}(\hat{L}_{x} + 2\hat{S}_{x})\left [B_{1}\textrm{cos}(\omega_{1} t) +B_{2}\textrm{cos}(\omega_{2} t)\right ] \right | \Psi_{l} \right> \left<\Psi_{l} \left |-\frac{q}{2m}(\hat{L}_{x} + 2\hat{S}_{x})\left [B_{1}\textrm{cos}(\omega_{1} t) +B_{2}\textrm{cos}(\omega_{2} t)\right ] \right | \Psi_{i} \right >}{\hbar(\omega_{i} - \omega_{l})} \Bigg)} \hspace{1cm}\label{30}\] We can ignore the first term of this equation as it corresponds to the magnetic dipole contribution of the single photon transitions. From the summation we can ignore the squared terms as well because they represent the two photon transitions due to the beam interfering with itself. While that does exist and may be relevant for some spectroscopic work, here we are looking at the two-photon absorption from two light sources. \[ \left < \Psi_{f} \left | \hat{H} \right | \Psi_{i} \right > = \] \[ \small{\frac{q^{2}\textrm{cos }(\omega_{1}t)\textrm{cos }(\omega_{2} t)}{4m^{2}\hbar^{2}}\frac{\textrm{exp}(-i\omega_{i}t)}{\omega_{i} - \omega_{f}} \left ( \frac{ \left< \Psi_{f} \left | (\hat{L}_{x} +2\hat{S}_{x}) \right | \Psi_{l} \right > \left< \Psi_{l} \left | (\hat{L}_{x} +2\hat{S}_{x}) \right | \Psi_{i} \right > +\left< \Psi_{f} \left | (\hat{L}_{x} +2\hat{S}_{x}) \right | \Psi_{l} \right > \left< \Psi_{l} \left | (\hat{L}_{x} +2\hat{S}_{x}) \right | \Psi_{i} \right >}{\omega_{i} - \omega_{l}} \right )} \hspace{1cm} \label{31}\] Just as before, our selection rules stem from the magnetic dipole Hamiltonian: \[\hat{H}_{DM} = -\frac{q}{2m}(\hat{L_{x}} + 2\hat{S_{x}})B\textrm{cos}(\omega t) \hspace{1cm} \label{32} \] Here, as before, the sinusoid part is even and thus will have no effect on the selection rules. We must look at the operators to determine the selection rules. Because neither \(\hat{L}_{x}\) or \(\hat{S}_{x}\) have any effect on the value of \(l\) the selection rule for \(\Delta l\) is 0. However both \(\hat{L}_{x}\) and \(\hat{S}_{x}\) can change their corresponding \(m_{l}\) and \(m_{s}\) values by one unit. For \(m_{s}\) this corresponds to changing the spin of the electron. If the magnetic field is perpendicular to the spin of the system, then the \(\Delta m_{s}\) can also have a value of 0. If we extrapolate this to a two photon system then the same \(m_{s}\) values should still apply given that even if it is changed by 1 by the first photon, it can change back by -1 or remain in the \(+\frac{1}{2}\) state as those are the only two options. Therefore, we should expect the same selection rules for \(\Delta m_{s} = 0, \pm1\) in a two-photon system. Two-photon absorption has traditionally been a large part of the spectroscopy field. However, it was only within the past decade that biomedical applications have returned it to the spotlight. Two-photon absorption, and potentially all nonlinear absorptions, allow for unprecedented depth in medical imaging technology. Traditional single photon methods result in an enormous amount of scattering from the biological tissue samples. However, these nonlinear optics allow for the assignment of the scattered photons to their origins. Much of the research in this area is focused on maximizing the depth and clarity of the signal for this potentially non-intrusive medical procedure. Several parameters have been looked at including excitation wavelength, beam size, pulse width, and pulse frequency. As of now the largest depths achievable are around 1 mm. | 21,406 | 1,650 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/07._Angular_Momentum/Spherical_Harmonics |
are a group of functions used in math and the physical sciences to solve problems in disciplines including geometry, partial differential equations, and group theory. The general, normalized Spherical Harmonic is depicted below: \[ Y_{l}^{m}(\theta,\phi) = \sqrt{ \dfrac{(2l + 1)(l - |m|)!}{4\pi (l + |m|)!} } P_{l}^{|m|}(\cos\theta)e^{im\phi} \] One of the most prevalent applications for these functions is in the description of angular quantum mechanical systems. Utilized first by Laplace in 1782, these functions did not receive their name until nearly ninety years later by Lord Kelvin. Any is a function that satisfies : \[ \nabla^2 \psi = 0 \] These harmonics are classified as spherical due to being the solution to the angular portion of Laplace's equation in the coordinate system. Laplace's work involved the study of gravitational potentials and Kelvin used them in a collaboration with Peter Tait to write a textbook. In the 20th century, Erwin Schrödinger and Wolfgang Pauli both released papers in 1926 with details on how to solve the "simple" hydrogen atom system. Now, another ninety years later, the exact solutions to the hydrogen atom are still used to analyze multi-electron atoms and even entire molecules. Much of modern physical chemistry is based around framework that was established by these quantum mechanical treatments of nature. The 2p and 2p (angular) probability distributions depicted on the left and graphed on the right using "desmos". As Spherical Harmonics are unearthed by working with Laplace's equation in spherical coordinates, these functions are often products of trigonometric functions. These products are represented by the \( P_{l}^{|m|}(\cos\theta)\) term, which is called a . The details of where these polynomials come from are largely unnecessary here, lest we say that it is the set of solutions to a differential equation that forms from attempting to solve Laplace's equation. Unsurprisingly, that equation is called "Legendre's equation", and it features a transformation of \(\cos\theta = x\). As the general function shows above, for the spherical harmonic where \(l = m = 0\), the bracketed term turns into a simple constant. The exponential equals one and we say that: \[ Y_{0}^{0}(\theta,\phi) = \sqrt{ \dfrac{1}{4\pi} }\] What is not shown in full is what happens to the Legendre polynomial attached to our bracketed expression. In the simple \(l = m = 0\) case, it disappears. It is no coincidence that this article discusses both quantum mechanics and two variables, \(l\) and \(m\). These are exactly the angular momentum quantum number and magnetic quantum number, respectively, that are mentioned in General Chemistry classes. If we consider spectroscopic notation, an angular momentum quantum number of zero suggests that we have an if all of \(\psi(r,\theta,\phi)\) is present. This s orbital appears spherically symmetric on the boundary surface. In other words, the function looks like a ball. This is consistent with our constant-valued harmonic, for it would be constant-radius. Extending these functions to larger values of \(l\) leads to increasingly intricate Legendre polynomials and their associated \(m\) values. The \({Y_{1}^{0}}^{*}Y_{1}^{0}\) and \({Y_{1}^{1}}^{*}Y_{1}^{1}\) functions are plotted above. Recall that these functions are multiplied by their complex conjugate to properly represent the of "probability-density" (\(\psi^{*}\psi)\). It is also important to note that these functions alone are not referred to as orbitals, for this would imply that both the radial and angular components of the wavefunction are used. Identify the location(s) of all planar nodes of the following spherical harmonic: \[Y_{2}^{0}(\theta,\phi) = \sqrt{ \dfrac{5}{16\pi} }(3cos^2\theta - 1)\] Nodes are points at which our function equals zero, or in a more natural extension, they are locations in the probability-density where the electron will not be found (i.e. \(\psi^{*}\psi = 0)\). As this specific function is real, we could square it to find our probability-density. \[Y_{2}^{0} = [Y_{2}^{0}]^2 = 0\] As the non-squared function will be computationally easier to work with, and will give us an equivalent answer, we do not bother to square the function. The constant in front can be divided out of the expression, leaving: \[3cos^2\theta - 1 = 0\] \[\theta = cos^{-1}\bigg[\pm\dfrac{1}{\sqrt3}\bigg]\] \[\theta = 54.7^o \& 125.3^o\] We have described these functions as a set of solutions to a differential equation but we can also look at Spherical Harmonics from the standpoint of operators and the field of linear algebra. For the curious reader, a more in depth treatment of Laplace's equation and the methods used to solve it in the spherical domain are presented in this section of the text. For a brief review, partial differential equations are often simplified using a separation of variables technique that turns one PDE into several ordinary differential equations (which is easier, promise). This allows us to say \(\psi(r,\theta,\phi) = R_{nl}(r)Y_{l}^{m}(\theta,\phi)\), and to form a linear operator that can act on the Spherical Harmonics in an eigenvalue problem. The more important results from this analysis include (1) the recognition of an \(\hat{L}^2\) operator and (2) the fact that the Spherical Harmonics act as an eigenbasis for the given vector space. The \(\hat{L}^2\) operator is the operator associated with the square of angular momentum. It is directly related to the Hamiltonian operator (with zero potential) in the same way that kinetic energy and angular momentum are connected in classical physics. \[\hat{H} = \dfrac{\hat{L}^2}{2I}\] for \(I\) equal to the moment of inertia of the represented system. It is a linear operator (follows rules regarding additivity and homogeneity). More specifically, it is Hermitian. This means that when it is used in an eigenvalue problem, all eigenvalues will be real and the eigenfunctions will be orthogonal. In Dirac notation, orthogonality means that the inner product of any two different eigenfunctions will equal zero: \[\langle \psi_{i} | \psi_{j} \rangle = 0\] When we consider the fact that these functions are also often normalized, we can write the classic relationship between eigenfunctions of a quantum mechanical operator using a piecewise function: the Kronecker delta. \[\langle \psi_{i} | \psi_{j} \rangle = \delta_{ij} \, for \, \delta_{ij} = \begin{cases} 0 & i \neq j \ 1 & i = j \end{cases} \] This relationship also applies to the spherical harmonic set of solutions, and so we can write an orthonormality relationship for each quantum number: \[\langle Y_{l}^{m} | Y_{k}^{n} \rangle = \delta_{lk}\delta_{mn}\] The parity operator is sometimes denoted by "P", but will be referred to as \(\Pi\) here to not confuse it with the momentum operator. When this Hermitian operator is applied to a function, the signs of all variables within the function flip. This operator gives us a simple way to determine the symmetry of the function it acts on. Recall that even functions appear as \(f(x) = f(-x)\), and odd functions appear as \(f(-x) = -f(x)\). Combining this with \(\Pi\) gives the conditions: Using the parity operator and properties of integration, determine \(\langle Y_{l}^{m}| Y_{k}^{n} \rangle\) for any \( l\) an even number and \(k\) an odd number. As this question is for even and odd pairing, the task seems quite daunting, but analyzing the parity for a few simple cases will lead to a dramatic simplification of the problem. Start with acting the parity operator on the simplest spherical harmonic, \(l = m = 0\): \[\Pi Y_{0}^{0}(\theta,\phi) = \sqrt{\dfrac{1}{4\pi}} = Y_{0}^{0}(-\theta,-\phi)\] Now we can scale this up to the \(Y_{2}^{0}(\theta,\phi)\) case given in example one: \[\Pi Y_{2}^{0}(\theta,\phi) = \sqrt{ \dfrac{5}{16\pi} }(3cos^2(-\theta) - 1)\] but cosine is an even function, so again, we see: \[ Y_{2}^{0}(-\theta,-\phi) = Y_{2}^{0}(\theta,\phi)\] It appears that for every even, angular QM number, the spherical harmonic is even. As it turns out, every odd, angular QM number yields odd harmonics as well! If this is the case (verified after the next example), then we now have a simple task ahead of us. Note: Odd functions with symmetric integrals must be zero. \[\langle Y_{l}^{m}| Y_{k}^{n} \rangle = \int_{-\inf}^{\inf} (EVEN)(ODD)d\tau \] An even function multiplied by an odd function is an odd function (like even and odd numbers when multiplying them together). As such, this integral will be zero always, no matter what specific \(l\) and \(k\) are used. As one can imagine, this is a powerful tool. The impact is lessened slightly when coming off the heels off the idea that Hermitian operators like \(\hat{L}^2\) yield orthogonal eigenfunctions, but general parity of functions is useful! Consider the question of wanting to know the expectation value of our colatitudinal coordinate \(\theta\) for any given spherical harmonic with even-\(l\). \[\langle \theta \rangle = \langle Y_{l}^{m} | \theta | Y_{l}^{m} \rangle \] \[\langle \theta \rangle = \int_{-\inf}^{\inf} (EVEN)(ODD)(EVEN)d\tau \] Again, a complex sounding problem is reduced to a very straightforward analysis. Using integral properties, we see this is equal to zero, for any even-\(l\). A photo-set reminder of why an eigenvector (blue) is special. From . Lastly, the Spherical Harmonics form a complete set, and as such can act as a basis for the given (Hilbert) space. This means any spherical function can be written as a linear combination of these basis functions, (for the basis spans the space of continuous spherical functions by definition): \[f(\theta,\phi) = \sum_{l}\sum_{m} \alpha_{lm} Y_{l}^{m}(\theta,\phi) \] While any particular basis can act in this way, the fact that the Spherical Harmonics can do this shows a nice relationship between these functions and the Fourier Series, a basis set of sines and cosines. Spherical Harmonics are considered the higher-dimensional analogs of these Fourier combinations, and are incredibly useful in applications involving frequency domains. In the past few years, with the advancement of computer graphics and rendering, modeling of dynamic lighting systems have led to a new use for these functions. In order to do any serious computations with a large sum of Spherical Harmonics, we need to be able to generate them via computer in real-time (most specifically for real-time graphics systems). This requires the use of either recurrence relations or generating functions. While at the very top of this page is the general formula for our functions, the Legendre polynomials are still as of yet undefined. The two major statements required for this example are listed: \( P_{l}(x) = \dfrac{1}{2^{l}l!} \dfrac{d^{l}}{dx^{l}}[(x^{2} - 1)^{l}]\) \( P_{l}^{|m|}(x) = (1 - x^{2})^{\tiny\dfrac{|m|}{2}}\dfrac{d^{|m|}}{dx^{|m|}}P_{l}(x)\) Using these recurrence relations, write the spherical harmonic \(Y_{1}^{1}(\theta,\phi)\). To solve this problem, we can break up our process into four major parts. The first is determining our \(P_{l}(x)\) function. As \(l = 1\): \( P_{1}(x) = \dfrac{1}{2^{1}1!} \dfrac{d}{dx}[(x^{2} - 1)]\) \( P_{1}(x) = \dfrac{1}{2}(2x)\) \( P_{1}(x) = x\) Now that we have \(P_{l}(x)\), we can plug this into our Legendre recurrence relation to find the : \( P_{1}^{1}(x) = (1 - x^{2})^{\tiny\dfrac{1}{2}}\dfrac{d}{dx}x\) \( P_{1}^{1}(x) = (1 - x^{2})^{\tiny\dfrac{1}{2}}\) At the halfway point, we can use our general definition of Spherical Harmonics with the newly determined Legendre function. With \(m = l = 1\): \[ Y_{1}^{1}(\theta,\phi) = \sqrt{ \dfrac{(2(1) + 1)(1 - 1)!}{4\pi (1 + |1|)!} } (1 - x^{2})^{\tiny\dfrac{1}{2}}e^{i\phi} \] \[ Y_{1}^{1}(\theta,\phi) = \sqrt{ \dfrac{3}{8\pi} } (1 - x^{2})^{\tiny\dfrac{1}{2}}e^{i\phi} \] The last step is converting our Cartesian function into the proper coordinate system or making the switch from x to \(\cos\theta\). \[ Y_{1}^{1}(\theta,\phi) = \sqrt{ \dfrac{3}{8\pi} } (1 - (\cos\theta)^{2})^{\tiny\dfrac{1}{2}}e^{i\phi} \] \[ Y_{1}^{1}(\theta,\phi) = \sqrt{ \dfrac{3}{8\pi} } (sin^{2}\theta)^{\tiny\dfrac{1}{2}}e^{i\phi} \] \[ Y_{1}^{1}(\theta,\phi) = \sqrt{ \dfrac{3}{8\pi} }sin\theta e^{i\phi} \] As a side note, there are a number of different relations one can use to generate Spherical Harmonics or Legendre polynomials. Often times, efficient computer algorithms have much longer polynomial terms than the short, derivative-based statements from the beginning of this problem. As a final topic, we should take a closer look at the two recursive relations of Legendre polynomials together. As derivatives of even functions yield odd functions and vice versa, we note that for our first equation, an even \(l\) value implies an even number of derivatives, and this will yield another even function. When we plug this into our second relation, we now have to deal with \(|m|\) derivatives of our \(P_{l}\) function. We are in luck though, as in the spherical harmonic functions there is a separate component entirely dependent upon the sign of \(m\). As such, any changes in parity to the Legendre polynomial (to create the associated Legendre function) will be undone by the flip in sign of \(m\) in the azimuthal component. Parity only depends on \(l\)! This confirms our prediction from the second example that any Spherical Harmonic with even-\(l\) is also even, and any odd-\(l\) leads to odd \(Y_{l}^{m}\). | 13,483 | 1,651 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/13%3A_The_Phase_Rule_and_Phase_Diagrams/13.02%3A__Phase_Diagrams-_Binary_Systems |
\( \newcommand{\tx}[1]{\text{#1}} % text in math mode\)
\( \newcommand{\subs}[1]{_{\text{#1}}} % subscript text\)
\( \newcommand{\sups}[1]{^{\text{#1}}} % superscript text\)
\( \newcommand{\st}{^\circ} % standard state symbol\)
\( \newcommand{\id}{^{\text{id}}} % ideal\)
\( \newcommand{\rf}{^{\text{ref}}} % reference state\)
\( \newcommand{\units}[1]{\mbox{$\thinspace$#1}}\)
\( \newcommand{\K}{\units{K}} % kelvins\)
\( \newcommand{\degC}{^\circ\text{C}} % degrees Celsius\)
\( \newcommand{\br}{\units{bar}} % bar (\bar is already defined)\)
\( \newcommand{\Pa}{\units{Pa}}\)
\( \newcommand{\mol}{\units{mol}} % mole\)
\( \newcommand{\V}{\units{V}} % volts\)
\( \newcommand{\timesten}[1]{\mbox{$\,\times\,10^{#1}$}}\)
\( \newcommand{\per}{^{-1}} % minus one power\)
\( \newcommand{\m}{_{\text{m}}} % subscript m for molar quantity\)
\( \newcommand{\CVm}{C_{V,\text{m}}} % molar heat capacity at const.V\)
\( \newcommand{\Cpm}{C_{p,\text{m}}} % molar heat capacity at const.p\)
\( \newcommand{\kT}{\kappa_T} % isothermal compressibility\)
\( \newcommand{\A}{_{\text{A}}} % subscript A for solvent or state A\)
\( \newcommand{\B}{_{\text{B}}} % subscript B for solute or state B\)
\( \newcommand{\bd}{_{\text{b}}} % subscript b for boundary or boiling point\)
\( \newcommand{\C}{_{\text{C}}} % subscript C\)
\( \newcommand{\f}{_{\text{f}}} % subscript f for freezing point\)
\( \newcommand{\mA}{_{\text{m},\text{A}}} % subscript m,A (m=molar)\)
\( \newcommand{\mB}{_{\text{m},\text{B}}} % subscript m,B (m=molar)\)
\( \newcommand{\mi}{_{\text{m},i}} % subscript m,i (m=molar)\)
\( \newcommand{\fA}{_{\text{f},\text{A}}} % subscript f,A (for fr. pt.)\)
\( \newcommand{\fB}{_{\text{f},\text{B}}} % subscript f,B (for fr. pt.)\)
\( \newcommand{\xbB}{_{x,\text{B}}} % x basis, B\)
\( \newcommand{\xbC}{_{x,\text{C}}} % x basis, C\)
\( \newcommand{\cbB}{_{c,\text{B}}} % c basis, B\)
\( \newcommand{\mbB}{_{m,\text{B}}} % m basis, B\)
\( \newcommand{\kHi}{k_{\text{H},i}} % Henry's law constant, x basis, i\)
\( \newcommand{\kHB}{k_{\text{H,B}}} % Henry's law constant, x basis, B\)
\( \newcommand{\arrow}{\,\rightarrow\,} % right arrow with extra spaces\)
\( \newcommand{\arrows}{\,\rightleftharpoons\,} % double arrows with extra spaces\)
\( \newcommand{\ra}{\rightarrow} % right arrow (can be used in text mode)\)
\( \newcommand{\eq}{\subs{eq}} % equilibrium state\)
\( \newcommand{\onehalf}{\textstyle\frac{1}{2}\D} % small 1/2 for display equation\)
\( \newcommand{\sys}{\subs{sys}} % system property\)
\( \newcommand{\sur}{\sups{sur}} % surroundings\)
\( \renewcommand{\in}{\sups{int}} % internal\)
\( \newcommand{\lab}{\subs{lab}} % lab frame\)
\( \newcommand{\cm}{\subs{cm}} % center of mass\)
\( \newcommand{\rev}{\subs{rev}} % reversible\)
\( \newcommand{\irr}{\subs{irr}} % irreversible\)
\( \newcommand{\fric}{\subs{fric}} % friction\)
\( \newcommand{\diss}{\subs{diss}} % dissipation\)
\( \newcommand{\el}{\subs{el}} % electrical\)
\( \newcommand{\cell}{\subs{cell}} % cell\)
\( \newcommand{\As}{A\subs{s}} % surface area\)
\( \newcommand{\E}{^\mathsf{E}} % excess quantity (superscript)\)
\( \newcommand{\allni}{\{n_i \}} % set of all n_i\)
\( \newcommand{\sol}{\hspace{-.1em}\tx{(sol)}}\)
\( \newcommand{\solmB}{\tx{(sol,$\,$$m\B$)}}\)
\( \newcommand{\dil}{\tx{(dil)}}\)
\( \newcommand{\sln}{\tx{(sln)}}\)
\( \newcommand{\mix}{\tx{(mix)}}\)
\( \newcommand{\rxn}{\tx{(rxn)}}\)
\( \newcommand{\expt}{\tx{(expt)}}\)
\( \newcommand{\solid}{\tx{(s)}}\)
\( \newcommand{\liquid}{\tx{(l)}}\)
\( \newcommand{\gas}{\tx{(g)}}\)
\( \newcommand{\pha}{\alpha} % phase alpha\)
\( \newcommand{\phb}{\beta} % phase beta\)
\( \newcommand{\phg}{\gamma} % phase gamma\)
\( \newcommand{\aph}{^{\alpha}} % alpha phase superscript\)
\( \newcommand{\bph}{^{\beta}} % beta phase superscript\)
\( \newcommand{\gph}{^{\gamma}} % gamma phase superscript\)
\( \newcommand{\aphp}{^{\alpha'}} % alpha prime phase superscript\)
\( \newcommand{\bphp}{^{\beta'}} % beta prime phase superscript\)
\( \newcommand{\gphp}{^{\gamma'}} % gamma prime phase superscript\)
\( \newcommand{\apht}{\small\aph} % alpha phase tiny superscript\)
\( \newcommand{\bpht}{\small\bph} % beta phase tiny superscript\)
\( \newcommand{\gpht}{\small\gph} % gamma phase tiny superscript\) \( \newcommand{\upOmega}{\Omega}\) \( \newcommand{\dif}{\mathop{}\!\mathrm{d}} % roman d in math mode, preceded by space\)
\( \newcommand{\Dif}{\mathop{}\!\mathrm{D}} % roman D in math mode, preceded by space\)
\( \newcommand{\df}{\dif\hspace{0.05em} f} % df\) \(\newcommand{\dBar}{\mathop{}\!\mathrm{d}\hspace-.3em\raise1.05ex{\Rule{.8ex}{.125ex}{0ex}}} % inexact differential \)
\( \newcommand{\dq}{\dBar q} % heat differential\)
\( \newcommand{\dw}{\dBar w} % work differential\)
\( \newcommand{\dQ}{\dBar Q} % infinitesimal charge\)
\( \newcommand{\dx}{\dif\hspace{0.05em} x} % dx\)
\( \newcommand{\dt}{\dif\hspace{0.05em} t} % dt\)
\( \newcommand{\difp}{\dif\hspace{0.05em} p} % dp\)
\( \newcommand{\Del}{\Delta}\)
\( \newcommand{\Delsub}[1]{\Delta_{\text{#1}}}\)
\( \newcommand{\pd}[3]{(\partial #1 / \partial #2 )_{#3}} % \pd{}{}{} - partial derivative, one line\)
\( \newcommand{\Pd}[3]{\left( \dfrac {\partial #1} {\partial #2}\right)_{#3}} % Pd{}{}{} - Partial derivative, built-up\)
\( \newcommand{\bpd}[3]{[ \partial #1 / \partial #2 ]_{#3}}\)
\( \newcommand{\bPd}[3]{\left[ \dfrac {\partial #1} {\partial #2}\right]_{#3}}\)
\( \newcommand{\dotprod}{\small\bullet}\)
\( \newcommand{\fug}{f} % fugacity\)
\( \newcommand{\g}{\gamma} % solute activity coefficient, or gamma in general\)
\( \newcommand{\G}{\varGamma} % activity coefficient of a reference state (pressure factor)\)
\( \newcommand{\ecp}{\widetilde{\mu}} % electrochemical or total potential\)
\( \newcommand{\Eeq}{E\subs{cell, eq}} % equilibrium cell potential\)
\( \newcommand{\Ej}{E\subs{j}} % liquid junction potential\)
\( \newcommand{\mue}{\mu\subs{e}} % electron chemical potential\)
\( \newcommand{\defn}{\,\stackrel{\mathrm{def}}{=}\,} % "equal by definition" symbol\) \( \newcommand{\D}{\displaystyle} % for a line in built-up\)
\( \newcommand{\s}{\smash[b]} % use in equations with conditions of validity\)
\( \newcommand{\cond}[1]{\\[-2.5pt]{}\tag*{#1}}\)
\( \newcommand{\nextcond}[1]{\\[-5pt]{}\tag*{#1}}\)
\( \newcommand{\R}{8.3145\units{J$\,$K$\per\,$mol$\per$}} % gas constant value\)
\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) As explained in Sec. 8.2, a phase diagram is a kind of two-dimensional map that shows which phase or phases are stable under a given set of conditions. This section discusses some common kinds of binary systems, and Sec. 13.3 will describe some interesting ternary systems. A binary system has two components; \(C\) equals \(2\), and the number of degrees of freedom is \(F=4-P\). There must be at least one phase, so the maximum possible value of \(F\) is 3. Since \(F\) cannot be negative, the equilibrium system can have no more than four phases. We can independently vary the temperature, pressure, and composition of the system as a whole. Instead of using these variables as the coordinates of a three-dimensional phase diagram, we usually draw a two-dimensional phase diagram that is either a temperature–composition diagram at a fixed pressure or a pressure–composition diagram at a fixed temperature. The position of the system point on one of these diagrams then corresponds to a definite temperature, pressure, and overall composition. The composition variable usually varies along the horizontal axis and can be the mole fraction, mass fraction, or mass percent of one of the components, as will presently be illustrated by various examples. The way in which we interpret a two-dimensional phase diagram to obtain the compositions of individual phases depends on the number of phases present in the system. The examples that follow show some of the simpler kinds of phase diagrams known for binary systems. Temperature–composition phase diagram for a binary system exhibiting a eutectic point. Figure 13.1 is a temperature–composition phase diagram at a fixed pressure. The composition variable \(z\B\) is the mole fraction of component B in the system as a whole. The phases shown are a binary liquid mixture of A and B, pure solid A, and pure solid B. The one-phase liquid area is bounded by two curves, which we can think of either as freezing-point curves for the liquid or as solubility curves for the solids. These curves comprise the liquidus. As the mole fraction of either component in the liquid phase decreases from unity, the freezing point decreases. The curves meet at point a, which is a . At this point, both solid A and solid B can coexist in equilibrium with a binary liquid mixture. The composition at this point is the , and the temperature here (denoted \(T\subs{e}\)) is the . (“Eutectic” comes from the Greek for .) \(T\subs{e}\) is the lowest temperature for the given pressure at which the liquid phase is stable. Suppose we combine \(0.60\mol\) A and \(0.40\mol\) B (\(z\B=0.40\)) and adjust the temperature so as to put the system point at b. This point is in the one-phase liquid area, so the equilibrium system at this temperature has a single liquid phase. If we now place the system in thermal contact with a cold reservoir, heat is transferred out of the system and the system point moves down along the (path of constant overall composition) b–h. The cooling rate depends on the temperature gradient at the system boundary and the system’s heat capacity. At point c on the isopleth, the system point reaches the boundary of the one-phase area and is about to enter the two-phase area labeled A(s) + liquid. At this point in the cooling process, the liquid is saturated with respect to solid A, and solid A is about to freeze out from the liquid. There is an abrupt decrease (break) in the cooling rate at this point, because the freezing process involves an extra enthalpy decrease. At the still lower temperature at point d, the system point is within the two-phase solid–liquid area. The tie line through this point is line e–f. The compositions of the two phases are given by the values of \(z\B\) at the ends of the tie line: \(x\B\sups{s}=0\) for the solid and \(x\B\sups{l} =0.50\) for the liquid. From the general lever rule (Eq. 8.2.8), the ratio of the amounts in these phases is \begin{equation} \frac{n\sups{l} }{n\sups{s}} = \frac{z\B-x\B\sups{s}}{x\B\sups{l} -z\B} = \frac{0.40-0}{0.50-0.40} = 4.0 \tag{13.2.1} \end{equation} Since the total amount is \(n\sups{s}+n\sups{l} =1.00\mol\), the amounts of the two phases must be \(n\sups{s}=0.20\mol\) and \(n\sups{l} =0.80\mol\). When the system point reaches the eutectic temperature at point g, cooling halts until all of the liquid freezes. Solid B freezes out as well as solid A. During this , there are at first three phases: liquid with the eutectic composition, solid A, and solid B. As heat continues to be withdrawn from the system, the amount of liquid decreases and the amounts of the solids increase until finally only \(0.60\mol\) of solid A and \(0.40\mol\) of solid B are present. The temperature then begins to decrease again and the system point enters the two-phase area for solid A and solid B; tie lines in this area extend from \(z\B{=}0\) to \(z\B{=}1\). Temperature–composition phase diagrams such as this are often mapped out experimentally by observing the cooling curve (temperature as a function of time) along isopleths of various compositions. This procedure is . A break in the slope of a cooling curve at a particular temperature indicates the system point has moved from a one-phase liquid area to a two-phase area of liquid and solid. A temperature halt indicates the temperature is either the freezing point of the liquid to form a solid of the same composition, or else a eutectic temperature. Temperature–composition phase diagrams with single eutectics.
(a) Two pure solids and a liquid mixture (E. W. Washburn, , Vol. IV, McGraw-Hill, New York, 1928, p. 98).
(b) Two solid solutions and a liquid mixture. Figure 13.2 shows two temperature–composition phase diagrams with single eutectic points. The left-hand diagram is for the binary system of chloroform and carbon tetrachloride, two liquids that form nearly ideal mixtures. The solid phases are pure crystals, as in Fig. 13.1. The right-hand diagram is for the silver–copper system and involves solid phases that are solid solutions (substitutional alloys of variable composition). The area labeled s\(\aph\) is a solid solution that is mostly silver, and s\(\bph\) is a solid solution that is mostly copper. Tie lines in the two-phase areas do not end at a vertical line for a pure solid component as they do in the system shown in the left-hand diagram. The three phases that can coexist at the eutectic temperature of \(\tx{1,052}\K\) are the melt of the eutectic composition and the two solid solutions. Temperature–composition phase diagram for the binary system of \(\alpha\)-naphthylamine (A) and phenol (B) at \(1\br\) (J. C. Philip, , , 814–834, 1903). Section 12.5.4 discussed the possibility of the appearance of a when a binary liquid mixture is cooled. An example of this behavior is shown in Fig. 13.3, in which the solid compound contains equal amounts of the two components \(\alpha\)-naphthylamine and phenol. The possible solid phases are pure A, pure B, and the solid compound AB. Only one or two of these solids can be present simultaneously in an equilibrium state. The vertical line in the figure at \(z\B=0.5\) represents the solid compound. The temperature at the upper end of this line is the melting point of the solid compound, \(29\units{\(\degC\)}\). The solid melts to give a liquid of the same composition. A melting process with this behavior is called a . The cooling curve for liquid of this composition would display a halt at the melting point. The phase diagram in Fig. 13.3 has two eutectic points. It resembles two simple phase diagrams like Fig. 13.1 placed side by side. There is one important difference: the slope of the freezing-point curve (liquidus curve) is nonzero at the composition of a pure component, but is zero at the composition of a solid compound that is completely dissociated in the liquid (as derived theoretically in Sec. 12.5.4). Thus, the curve in Fig. 13.3 has a relative maximum at the composition of the solid compound (\(z\B=0.5\)) and is rounded there, instead of having a cusp—like a Romanesque arch rather than a Gothic arch. Temperature–composition phase diagram for the binary system of H\(_2\)O and NaCl at \(1\br\). (Data from Roger Cohen-Adad and John W. Lorimer, , Solubility Data Series, Vol. 47, Pergamon Press, Oxford, 1991; and E. W. Washburn, , Vol. III, McGraw-Hill, New York, 1928.) An example of a solid compound that does not melt congruently is shown in Fig. 13.4. The solid hydrate \(\ce{NaCl*2H2O}\) is \(61.9\%\) NaCl by mass. It decomposes at \(0\units{\(\degC\)}\) to form an aqueous solution of composition \(26.3\%\) NaCl by mass and a solid phase of anhydrous NaCl. These three phases can coexist at equilibrium at \(0\units{\(\degC\)}\). A phase transition like this, in which a solid compound changes into a liquid and a different solid, is called or melting, and the point on the phase diagram at this temperature at the composition of the liquid is a . Figure 13.4 shows there are two other temperatures at which three phases can be present simultaneously: \(-21\units{\(\degC\)}\), where the phases are ice, the solution at its eutectic point, and the solid hydrate; and \(109\units{\(\degC\)}\), where the phases are gaseous H\(_2\)O, a solution of composition \(28.3\%\) NaCl by mass, and solid NaCl. Note that both segments of the right-hand boundary of the one-phase solution area have positive slopes, meaning that the solubilities of the solid hydrate and the anhydrous salt both increase with increasing temperature. When two liquids that are partially miscible are combined in certain proportions, phase separation occurs (Sec. 11.1.6). Two liquid phases in equilibrium with one another are called . Obviously the two phases must have different compositions or they would be identical; the difference is called a . A binary system with two phases has two degrees of freedom, so that at a given temperature and pressure each conjugate phase has a fixed composition. Temperature–composition phase diagram for the binary system of methyl acetate (A) and carbon disulfide (B) at \(1\br\) (data from P. Ferloni and G. Spinolo, , 70, 1974). All phases are liquids. The open circle indicates the critical point. The typical dependence of a miscibility gap on temperature is shown in Fig. 13.5. The miscibility gap (the difference in compositions at the left and right boundaries of the two-phase area) decreases as the temperature increases until at the , also called the , the gap vanishes. The point at the maximum of the boundary curve of the two-phase area, where the temperature is the upper consolute temperature, is the or . At this point, the two liquid phases become identical, just as the liquid and gas phases become identical at the critical point of a pure substance. Critical opalescence (Sec. 8.2.3) is observed in the vicinity of this point, caused by large local composition fluctuations. At temperatures at and above the critical point, the system is a single binary liquid mixture. Suppose we combine \(6.0\mol\) of component A (methyl acetate) and \(4.0\mol\) of component B (carbon disulfide) in a cylindrical vessel and adjust the temperature to \(200\K\). The overall mole fraction of B is \(z\B=0.40\). The system point is at point a in the two-phase region. From the positions of points b and c at the ends of the tie line through point a, we find the two liquid layers have compositions \(x\B\aph=0.20\) and \(x\B\bph=0.92\). Since carbon disulfide is the more dense of the two pure liquids, the bottom layer is phase \(\phb\), the layer that is richer in carbon disulfide. According to the lever rule, the ratio of the amounts in the two phases is given by \begin{equation} \frac{n\bph}{n\aph} = \frac{z\B-x\B\aph}{x\B\bph-z\B} = \frac{0.40-0.20}{0.92-0.40} = 0.38 \tag{13.2.2} \end{equation} Combining this value with \(n\aph+n\bph=10.0\mol\) gives us \(n\aph=7.2\mol\) and \(n\bph=2.8\mol\). If we gradually add more carbon disulfide to the vessel while gently stirring and keeping the temperature constant, the system point moves to the right along the tie line. Since the ends of this tie line have fixed positions, neither phase changes its composition, but the amount of phase \(\phb\) increases at the expense of phase \(\pha\). The liquid–liquid interface moves up in the vessel toward the top of the liquid column until, at overall composition \(z\B=0.92\) (point c), there is only one liquid phase. Now suppose the system point is back at point a and we raise the temperature while keeping the overall composition constant at \(z\B=0.40\). The system point moves up the isopleth a–d. The phase diagram shows that the ratio \((z\B-x\B\aph)/(x\B\bph-z\B)\) decreases during this change. As a result, the amount of phase \(\pha\) increases, the amount of phase \(\phb\) decreases, and the liquid–liquid interface moves down toward the bottom of the vessel until at \(217\K\) (point d) there again is only one liquid phase. Phase diagrams for the binary system of toluene (A) and benzene (B). The curves are calculated from Eqs. 13.2.6 and 13.2.7 and the saturation vapor pressures of the pure liquids.
(a) Pressure–composition diagram at \(T=340\K\).
(b) Temperature–composition diagram at \(p=1\br\). Toluene and benzene form liquid mixtures that are practically ideal and closely obey Raoult’s law for partial pressure. For the binary system of these components, we can use the vapor pressures of the pure liquids to generate the liquidus and vaporus curves of the pressure–composition and temperature–composition phase diagram. The results are shown in Fig. 13.6. The composition variable \(z\A\) is the overall mole fraction of component A (toluene). The equations needed to generate the curves can be derived as follows. Consider a binary liquid mixture of components A and B and mole fraction composition \(x\A\) that obeys Raoult’s law for partial pressure (Eq. 9.4.2): \begin{equation} p\A = x\A p\A^* \qquad p\B = (1-x\A)p\B^* \tag{13.2.3} \end{equation} Strictly speaking, Raoult’s law applies to a liquid–gas system maintained at a constant pressure by means of a third gaseous component, and \(p\A^*\) and \(p\B^*\) are the vapor pressures of the pure liquid components at this pressure and the temperature of the system. However, when a liquid phase is equilibrated with a gas phase, the partial pressure of a constituent of the liquid is practically independent of the total pressure (Sec. 12.8.1), so that it is a good approximation to apply the equations to a liquid–gas system and treat \(p\A^*\) and \(p\B^*\) as functions only of \(T\). When the binary system contains a liquid phase and a gas phase in equilibrium, the pressure is the sum of \(p\A\) and \(p\B\), which from Eq. 13.2.3 is given by \begin{gather} \s {\begin{split} p & = x\A p\A^* + (1-x\A)p\B^* \cr & = p\B^* + (p\A^*-p\B^*)x\A \end{split} } \tag{13.2.4} \cond{(\(C{=}2\), ideal liquid mixture)} \end{gather} where \(x\A\) is the mole fraction of A in the liquid phase. Equation 13.2.4 shows that in the two-phase system, \(p\) has a value between \(p\A^*\) and \(p\B^*\), and that if \(T\) is constant, \(p\) is a linear function of \(x\A\). The mole fraction composition of the gas in the two-phase system is given by \begin{equation} y\A = \frac{p\A}{p} = \frac{x\A p\A^*}{p\B^* + (p\A^*-p\B^*)x\A } \tag{13.2.5} \end{equation} A binary two-phase system has two degrees of freedom. At a given \(T\) and \(p\), each phase must have a fixed composition. We can calculate the liquid composition by rearranging Eq. 13.2.4: \begin{gather} \s {x\A = \frac{p-p\B^*}{p\A^*-p\B^*}} \tag{13.2.6} \cond{(\(C{=}2\), ideal liquid mixture)} \end{gather} The gas composition is then given by \begin{gather} \s {\begin{split} y\A & = \frac{p\A}{p} = \frac{x\A p\A^*}{p} \cr & = \left( \frac{p-p\B^*}{p\A^*-p\B^*}\right) \frac{p\A^*}{p} \end{split} } \tag{13.2.7} \cond{(\(C{=}2\), ideal liquid mixture)} \end{gather} If we know \(p\A^*\) and \(p\B^*\) as functions of \(T\), we can use Eqs. 13.2.6 and 13.2.7 to calculate the compositions for any combination of \(T\) and \(p\) at which the liquid and gas phases can coexist, and thus construct a pressure–composition or temperature–composition phase diagram. In Fig. 13.6(a), the liquidus curve shows the relation between \(p\) and \(x\A\) for equilibrated liquid and gas phases at constant \(T\), and the vaporus curve shows the relation between \(p\) and \(y\A\) under these conditions. We see that \(p\) is a linear function of \(x\A\) but not of \(y\A\). In a similar fashion, the liquidus curve in Fig. 13.6(b) shows the relation between \(T\) and \(x\A\), and the vaporus curve shows the relation between \(T\) and \(y\A\), for equilibrated liquid and gas phases at constant \(p\). Neither curve is linear. Liquidus and vaporus surfaces for the binary system of toluene (A) and benzene. Cross-sections through the two-phase region are drawn at constant temperatures of \(340\K\) and \(370\K\) and at constant pressures of \(1\br\) and \(2\br\). Two of the cross-sections intersect at a tie line at \(T=370\K\) and \(p=1\br\), and the other cross-sections are hatched in the direction of the tie lines. A liquidus curve is also called a curve or a curve. Other names for a vaporus curve are curve and curve. These curves are actually cross-sections of liquidus and vaporus in a three-dimensional \(T\)–\(p\)–\(z\A\) phase diagram, as shown in Fig. 13.7. In this figure, the liquidus surface is in view at the front and the vaporus surface is hidden behind it. Most binary liquid mixtures do not behave ideally. The most common situation is deviations from Raoult’s law. (In the molecular model of Sec. 11.1.5, positive deviations correspond to a less negative value of \(k\subs{AB}\) than the average of \(k\subs{AA}\) and \(k\subs{BB}\).) Some mixtures, however, have specific A–B interactions, such as solvation or molecular association, that prevent random mixing of the molecules of A and B, and the result is then deviations from Raoult’s law. If the deviations from Raoult’s law, either positive or negative, are large enough, the constant-temperature liquidus curve exhibits a maximum or minimum and behavior results. Binary system of methanol (A) and benzene at \(45\units{\(\degC\)}\) (Hossein Toghiani, Rebecca K. Toghiani, and Dabir S. Viswanath, , , 63–67, 1994).
(a) Partial pressures and total pressure in the gas phase equilibrated with liquid mixtures. The dashed lines indicate Raoult’s law behavior.
(b) Pressure–composition phase diagram at \(45\units{\(\degC\)}\). Open circle: azeotropic point at \(z\A=0.59\) and \(p=60.5\units{kPa}\). Figure 13.8 shows the azeotropic behavior of the binary methanol-benzene system at constant temperature. In Fig. 13.8(a), the experimental partial pressures in a gas phase equilibrated with the nonideal liquid mixture are plotted as a function of the liquid composition. The partial pressures of both components exhibit positive deviations from Raoult’s law, consistent with the statement in Sec. 12.8.2 that if one constituent of a binary liquid mixture exhibits positive deviations from Raoult’s law, with only one inflection point in the curve of fugacity versus mole fraction, the other constituent also has positive deviations from Raoult’s law. The total pressure (equal to the sum of the partial pressures) has a maximum value greater than the vapor pressure of either pure component. The curve of \(p\) versus \(x\A\) becomes the liquidus curve of the pressure–composition phase diagram shown in Fig. 13.8(b). Points on the vaporus curve are calculated from \(p=p\A/y\A\). In practice, the data needed to generate the liquidus and vaporus curves of a nonideal binary system are usually obtained by allowing liquid mixtures of various compositions to boil in an equilibrium still at a fixed temperature or pressure. When the liquid and gas phases have become equilibrated, samples of each are withdrawn for analysis. The partial pressures shown in Fig. 13.8(a) were calculated from the experimental gas-phase compositions with the relations \(p\A=y\A p\) and \(p\B=p-p\A\). If the constant-temperature liquidus curve has a maximum pressure at a liquid composition not corresponding to one of the pure components, which is the case for the methanol–benzene system, then the liquid and gas phases are mixtures of identical compositions at this pressure. This behavior was deduced at the end of Sec. 12.8.3. On the pressure–composition phase diagram, the liquidus and vaporus curves both have maxima at this pressure, and the two curves coincide at an . A binary system with negative deviations from Raoult’s law can have an isothermal liquidus curve with a pressure at a particular mixture composition, in which case the liquidus and vaporus curves coincide at an azeotropic point at this minimum. The general phenomenon in which equilibrated liquid and gas mixtures have identical compositions is called , and the liquid with this composition is an azeotropic mixture or (Greek: ). An azeotropic mixture vaporizes as if it were a pure substance, undergoing an equilibrium phase transition to a gas of the same composition. Liquidus and vaporus surfaces for the binary system of methanol (A) and benzene (Hossein Toghiani, Rebecca K. Toghiani, and Dabir S. Viswanath, , , 63–67, 1994). Cross-sections are hatched in the direction of the tie lines. The dashed curve is the azeotrope vapor-pressure curve. If the liquidus and vaporus curves exhibit a on a pressure–composition phase diagram, then they exhibit a on a temperature–composition phase diagram. This relation is explained for the methanol–benzene system by the three-dimensional liquidus and vaporus surfaces drawn in Fig. 13.9. In this diagram, the vaporus surface is hidden behind the liquidus surface. The hatched cross-section at the front of the figure is the same as the pressure–composition diagram of Fig. 13.8(b), and the hatched cross-section at the top of the figure is a temperature–composition phase diagram in which the system exhibits a . A binary system containing an azeotropic mixture in equilibrium with its vapor has two species, two phases, and one relation among intensive variables: \(x\A =y\A\). The number of degrees of freedom is then \(F = 2+s-r-P = 2+2-1-2 = 1\); the system is univariant. At a given temperature, the azeotrope can exist at only one pressure and have only one composition. As \(T\) changes, so do \(p\) and \(z\A\) along an as illustrated by the dashed curve in Fig. 13.9. Temperature–composition phase diagrams of binary systems exhibiting (a) no azeotropy, (b) a minimum-boiling azeotrope, and (c) a maximum-boiling azeotrope. Only the one-phase areas are labeled; two-phase areas are hatched in the direction of the tie lines. Figure 13.10 summarizes the general appearance of some relatively simple temperature–composition phase diagrams of binary systems. If the system does not form an azeotrope ( behavior), the equilibrated gas phase is richer in one component than the liquid phase at all liquid compositions, and the liquid mixture can be separated into its two components by fractional distillation. The gas in equilibrium with an azeotropic mixture, however, is not enriched in either component. Fractional distillation of a system with an azeotrope leads to separation into one pure component and the azeotropic mixture. Temperature–composition phase diagrams of binary systems with partially-miscible liquids exhibiting (a) the ability to be separated into pure components by fractional distillation, (b) a minimum-boiling azeotrope, and (c) boiling at a lower temperature than the boiling point of either pure component. Only the one-phase areas are labeled; two-phase areas are hatched in the direction of the tie lines. More complicated behavior is shown in the phase diagrams of Fig. 13.11. These are binary systems with partially-miscible liquids in which the boiling point is reached before an upper consolute temperature can be observed. Pressure–composition phase diagram for the binary system of CuSO\(_4\) (A) and H\(_2\)O (B) at \(25\units{\(\degC\)}\) (Thomas S. Logan, , , 148–149, 1958; E. W. Washburn, , Vol. VII, McGraw-Hill, New York, 1930, p. 263). As an example of a two-component system with equilibrated solid and gas phases, consider the components \(\ce{CuSO4}\) and \(\ce{H2O}\), denoted A and B respectively. In the pressure–composition phase diagram shown in Fig. 13.12, the composition variable \(z\B\) is as usual the mole fraction of component B in the system as a whole. The anhydrous salt and its hydrates (solid compounds) form the series of solids \(\ce{CuSO4}\), \(\ce{CuSO4*H2O}\), \(\ce{CuSO4*3H2O}\), and \(\ce{CuSO4*5H2O}\). In the phase diagram these formulas are abbreviated A, AB, AB\(_3\), and AB\(_5\). The following dissociation equilibria (dehydration equilibria) are possible: \begin{align*} \ce{CuSO4*H2O}\tx{(s)} & \arrows \ce{CuSO4}\tx{(s)} + \ce{H2O}\tx{(g)}\cr \ce{1/2CuSO4*3H2O}\tx{(s)} & \arrows \ce{1/2CuSO4*H2O}\tx{(s)} + \ce{H2O}\tx{(g)}\cr \ce{1/2CuSO4*5H2O}\tx{(s)} & \arrows \ce{1/2CuSO4*3H2O}\tx{(s)} + \ce{H2O}\tx{(g)} \end{align*} The equilibria are written above with coefficients that make the coefficient of H\(_2\)O(g) unity. When one of these equilibria is established in the system, there are two components and three phases; the phase rule then tells us the system is univariant and the pressure has only one possible value at a given temperature. This pressure is called the of the higher hydrate. The dissociation pressures of the three hydrates are indicated by horizontal lines in Fig. 13.12. For instance, the dissociation pressure of \(\ce{CuSO4*5H2O}\) is \(1.05\timesten{-2}\units{\(\br\)}\). At the pressure of each horizontal line, the equilibrium system can have one, two, or three phases, with compositions given by the intersections of the line with vertical lines. A fourth three-phase equilibrium is shown at \(p=3.09\timesten{-2}\units{\(\br\)}\); this is the equilibrium between solid \(\ce{CuSO4*5H2O}\), the saturated aqueous solution of this hydrate, and water vapor. Consider the thermodynamic equilibrium constant of one of the dissociation reactions. At the low pressures shown in the phase diagram, the activities of the solids are practically unity and the fugacity of the water vapor is practically the same as the pressure, so the equilibrium constant is almost exactly equal to \(p\subs{d}/p\st\), where \(p\subs{d}\) is the dissociation pressure of the higher hydrate in the reaction. Thus, a hydrate cannot exist in equilibrium with water vapor at a pressure below the dissociation pressure of the hydrate because dissociation would be spontaneous under these conditions. Conversely, the salt formed by the dissociation of a hydrate cannot exist in equilibrium with water vapor at a pressure above the dissociation pressure because hydration would be spontaneous. If the system contains dry air as an additional gaseous component and one of the dissociation equilibria is established, the partial pressure \(p\subs{H\(_2\)O}\) of H\(_2\)O is equal (approximately) to the dissociation pressure \(p\subs{d}\) of the higher hydrate. The prior statements regarding dissociation and hydration now depend on the value of \(p\subs{H\(_2\)O}\). If a hydrate is placed in air in which \(p\subs{H\(_2\)O}\) is less than \(p\subs{d}\), dehydration is spontaneous; this phenomenon is called (Latin: ). If \(p\subs{H\(_2\)O}\) is greater than the vapor pressure of the saturated solution of the highest hydrate that can form in the system, the anhydrous salt and any of its hydrates will spontaneously absorb water and form the saturated solution; this is (Latin: ). If the two-component equilibrium system contains only two phases, it is bivariant corresponding to one of the areas in Fig. 13.12. Here both the temperature and the pressure can be varied. In the case of areas labeled with two phases, the pressure has to be applied to the solids by a fluid (other than H\(_2\)O) that is not considered part of the system. Pressure–temperature–composition behavior in the binary heptane–ethane system (W. B. Kay, , , 459–465, 1938). The open circles are critical points; the dashed curve is the critical curve. The dashed line a–b illustrates retrograde condensation at \(450\K\). Binary phase diagrams begin to look different when the pressure is greater than the critical pressure of either of the pure components. Various types of behavior have been observed in this region. One common type, that found in the binary system of heptane and ethane, is shown in Fig. 13.13. This figure shows sections of a three-dimensional phase diagram at five temperatures. Each section is a pressure–composition phase diagram at constant \(T\). The two-phase areas are hatched in the direction of the tie lines. At the left end of each tie line (at low \(z\A\)) is a vaporus curve, and at the right end is a liquidus curve. The vapor pressure curve of pure ethane (\(z\A{=}0\)) ends at the critical point of ethane at \(305.4\K\); between this point and the critical point of heptane at \(540.5\K\), there is a continuous , which is the locus of critical points at which gas and liquid mixtures become identical in composition and density. Consider what happens when the system point is at point a in Fig. 13.13 and the pressure is then increased by isothermal compression along line a–b. The system point moves from the area for a gas phase into the two-phase gas–liquid area and then out into the gas-phase area again. This curious phenomenon, condensation followed by vaporization, is called . Under some conditions, an isobaric increase of \(T\) can result in vaporization followed by condensation; this is . Pressure–temperature–composition behavior in the binary xenon–helium system (J. de Swann Arons and G. A. M. Diepen, , , 2322–2330, 1966). The open circles are critical points; the dashed curve is the critical curve. A different type of high-pressure behavior, that found in the xenon–helium system, is shown in Fig. 13.14. Here, the critical curve begins at the critical point of the less volatile component (xenon) and continues to temperatures and pressures than the critical temperature and pressure of either pure component. The two-phase region at pressures above this critical curve is sometimes said to represent , or , because we would not usually consider a liquid to exist beyond the critical points of the pure components. Of course, the coexisting phases in this two-phase region are not gases in the ordinary sense of being tenuous fluids, but are instead high-pressure fluids of liquid-like densities. If we want to call both phases gases, then we have to say that pure gaseous substances at high pressure do not necessarily mix spontaneously in all proportions as they do at ordinary pressures. If the pressure of a system is increased isothermally, eventually solid phases will appear; these are not shown in Figs. 13.13 and Fig. 13.14. | 38,443 | 1,652 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.01%3A_The_Concept_of_Chemical_Equilibrium |
Chemical equilibrium is a dynamic process that consists of a forward reaction, in which reactants are converted to products, and a reverse reaction, in which products are converted to reactants. At equilibrium, the forward and reverse reactions proceed at equal rates. Consider, for example, a simple system that contains only one reactant and one product, the reversible dissociation of dinitrogen tetroxide (N O ) to nitrogen dioxide (NO ). You may recall from that NO is responsible for the brown color we associate with smog. When a sealed tube containing solid N O (mp = −9.3°C; bp = 21.2°C) is heated from −78.4°C to 25°C, the red-brown color of NO appears ( ). The reaction can be followed visually because the product (NO ) is colored, whereas the reactant (N O ) is colorless:
\[N_2O_4{(g)} colorless \rightleftharpoons 2NO_{2(g)} \; red-brown \tag{15.1.1}\] The double arrow indicates that both the forward and reverse reactions are occurring simultaneously; it is read “is in equilibrium with.” shows how the composition of this system would vary as a function of time at a constant temperature. If the initial concentration of NO were zero, then it increases as the concentration of N O decreases. Eventually the composition of the system stops changing with time, and chemical equilibrium is achieved. Conversely, if we start with a sample that contains no N O but an initial NO concentration twice the initial concentration of N O in part (a) in , in accordance with the stoichiometry of the reaction, we reach exactly the same equilibrium composition, as shown in part (b) in . Thus equilibrium can be approached from in a chemical reaction. shows the forward and reverse reaction rates for a sample that initially contains pure NO . Because the initial concentration of N O is zero, the forward reaction rate (dissociation of N O ) is initially zero as well. In contrast, the reverse reaction rate (dimerization of NO ) is initially very high (2.0 × 10 M/s), but it decreases rapidly as the concentration of NO decreases. (Recall from that the reaction rate of the dimerization reaction is expected to decrease rapidly because the reaction is second order in NO : rate = [NO ] , where is the rate constant for the reverse reaction shown in .) As the concentration of N O increases, the rate of dissociation of N O increases—but more slowly than the dimerization of NO —because the reaction is only first order in N O (rate = [N O ], where is the rate constant for the forward reaction in ). Eventually, the forward and reverse reaction rates become identical, , and the system has reached chemical equilibrium. If the forward and reverse reactions occur at rates, then the system is at equilibrium. The rate of dimerization of NO (reverse reaction) decreases rapidly with time, as expected for a second-order reaction. Because the initial concentration of N O is zero, the rate of the dissociation reaction (forward reaction) at = 0 is also zero. As the dimerization reaction proceeds, the N O concentration increases, and its rate of dissociation also increases. Eventually the rates of the two reactions are equal: chemical equilibrium has been reached, and the concentrations of N O and NO no longer change. At equilibrium, the forward reaction rate is equal to the reverse reaction rate. The three reaction systems (1, 2, and 3) depicted in the accompanying illustration can all be described by the equation where the blue circles are A and the purple ovals are B. Each set of panels shows the changing composition of one of the three reaction mixtures as a function of time. Which system took the longest to reach chemical equilibrium? three reaction systems relative time to reach chemical equilibrium Compare the concentrations of A and B at different times. The system whose composition takes the longest to stabilize took the longest to reach chemical equilibrium. In systems 1 and 3, the concentration of A decreases from through but is the same at both and . Thus systems 1 and 3 are at equilibrium by . In system 2, the concentrations of A and B are still changing between and , so system 2 may not yet have reached equilibrium by . Thus system 2 took the longest to reach chemical equilibrium. Exercise In the following illustration, A is represented by blue circles, B by purple squares, and C by orange ovals; the equation for the reaction is A + B ⇌ C. The sets of panels represent the compositions of three reaction mixtures as a function of time. Which, if any, of the systems shown has reached equilibrium? system 2 is a dynamic process consisting of forward and reverse reactions that proceed at equal rates. At equilibrium, the composition of the system no longer changes with time. The composition of an equilibrium mixture is independent of the direction from which equilibrium is approached. What is meant when a reaction is described as “having reached equilibrium”? What does this statement mean regarding the forward and reverse reaction rates? What does this statement mean regarding the concentrations or amounts of the reactants and the products? Is it correct to say that the reaction has “stopped” when it has reached equilibrium? Explain your answer and support it with a specific example. Why is chemical equilibrium described as a dynamic process? Describe this process in the context of a saturated solution of NaCl in water. What is occurring on a microscopic level? What is happening on a macroscopic level? Which of these systems exists in a state of chemical equilibrium? Both forward and reverse reactions occur but at the same rate. Na and Cl ions continuously leave the surface of an NaCl crystal to enter solution, while at the same time Na and Cl ions in solution precipitate on the surface of the crystal. | 5,843 | 1,653 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Radioactivity/Discovery_of_Radioactivity |
The discovery of radioactivity took place over several years beginning with the discovery of x-rays in 1895 by Wilhelm Conrad Roentgen and continuing with such people as Henri Becquerel and the Curie family. The application of x-rays and radioactive materials is far reaching in medicine and industry. Radioactive material is used in everything from nuclear reactors to isotope infused saline solutions. These technologies allow us to utilize great amounts of energy and observe biological systems in ways which were unthinkable less than a century ago. What is the definition of radioactive? If you look up the meaning in the dictionary the convoluted answer that you will receive is: This definition begs the questions: What are ionizing radiation or particles? What exactly is meant by emission? Can you see or feel these particles? What makes something radioactive? Received the first Noble Prize in physics for his discovery of x-rays in 1901. On November 8, 1895, at the University of Wurzburg, Roentgen was working in the lab when he noticed a strange fluorescence coming from a nearby table. Upon further observation he found that it originated from a partially evacuated Hittof-Crookes tube, covered in opaque black paper which he was using to study cathode rays. He concluded that the fluorescence, which penetrated the opaque black paper, must have been caused by rays. This phenomenon was later coined x-rays and though the phenomenon of x-rays is not the same as radioactivity, Roentgen opened the door for radioactive discovery. Received the Noble Prize in physics for being the first to discover radioactivity as a phenomenon separate from that of x-rays and document the differences between the two. Henri Becquerel learned of Roentgen's discovery of x-rays through the fluorescence that some materials produce. Using a method similar to that of Roentgen, Becquerel surrounded several photographic plates with black paper and florescent salts. With the intention of further advancing the study of x-rays, Becquerel intended to place the concealed photographic paper in the sunlight and observe what transpired. Unfortunately, he had to delay his experiment because the skies over Paris were overcast. He placed the wrapped plates into a dark desk drawer. After a few days Becquerel returned to his experiment unwrapping the photographic paper and developing it, expecting only a light imprint from the salts. Instead, the salts left very distinct outlines in the photographic paper suggesting that the salts, regardless of lacking an energy source, continually fluoresced. What Becquerel had discovered was radioactivity. Pierre and Marie were award the Noble Prize in Physics in 1903 for their work on radioactivity. Marie Curie became the first woman to be awarded the nobel prize and the first person to obtain two nobel prizes when she won the prize for the discovery of Polonium and Radium in 1911. Though it was Henri Becquerel that discovered radioactivity, it was Marie Curie who coined the term. Using a device invented by her husband and his brother, that measured extremely low electrical currents, Curie was able to note that uranium electrified the air around it. Further investigation showed that the activity of uranium compounds depended upon the amount of uranium present and that radioactivity was not a result of the interactions between molecules, but rather came from the atom itself. Using Pitchblende and chalcolite Curie found that Thorium was radioactive as well. She later discovered two new radioactive elements: Radium and Polonium which took her several years since these elements are difficult to extract and extremely rare. Unfortunately, the Curies died young. Pierre Curie was killed in a street accident and Marie died of aplastic anemia, almost certainly a result of radiation exposure. Ernest Rutherford is considered the father of nuclear physics. With his gold foil experiment he was able to unlock the mysteries of the atomic structure. He received the noble prize in chemistry in 1908. In 1909 at the University of Manchester, Rutherford was bombarding a piece of gold foil with Alpha particles. Rutherford noted that although most of the particles went straight through the foil, one in every eight thousand was deflected back. "It was as if you fired a fifteen inch naval shell at a piece of tissue paper and the shell came right back and hit you," Rutherford said. He concluded that though an atom consists of mostly empty space, most of its mass is concentrated in a very small positively charged region known as the nucleus, while electrons buzz around on the outside. Rutherford was also able to observe that radioactive elements underwent a process of decay over time which varied from element to element. In 1919, Rutherford used alpha particles to transmutate one element (Oxygen) into another element (Nitrogen). Papers at the timed called it "splitting the atom." We now have the essentials to utilize radioactive elements. Roentgen gave us x-rays, Becquerel discovered radioactivity, the Curies were able to discover which elements were radioactive, and Rutherford brought about transmutation and the "splitting of the atom." All of these discoveries and curiosity came with a price. Time showed the damaging effects of radiation exposure and the incredible destruction that could be harnessed from these elements. Radioactive isotopes are presently used in many aspects of human life today. Most people recognize radioactivity's contributions to industry, research and war, but it is even used within many peoples homes. Here are a few examples of how radioactive isotopes are utilized today. Most people have radioactive material in their very own homes, or at least we would hope so. Why? Because in most every smoke detector unit today there is a very small amount of Americium-241. How does it work? Well Americium-241 is present in the detector in oxide form and it emits alpha particles and very low energy gamma rays. The alpha rays are absorbed in the detector, while the non-harmful gamma rays are able to escape. The alpha particles collide with oxygen and nitrogen in the air of the detector's ionization chamber producing charged particles, or ions. A small electric voltage runs across the chamber which is used to collect these ions and operate a small electric current between two electrodes. When smoke enters the chamber it absorbs the alpha particles disrupting the rate of ionization in the chamber, thereby turning off the electrical current, which sets off the alarm. For more information go to: On June 7th 1954 the the USSR produced the world's very first nuclear power plant. These plants, though clean burning, produce a great deal of toxic nuclear waste which is difficult to eliminate. To date, approximately 15% of the worlds electricity and 6% of the worlds power is produced in nuclear power plants. With the rise in gas prices many countries around the world considered increasing their use nuclear energy. The problem with nuclear energy is that although it is "clean" in the sense that only water vapor is emitted into the atmosphere, it has its share of problems. It must be kept constantly regulated, and is extremely hard to dispose of. In the past, poor regulation of nuclear power has caused major problems, such as the Chernobyl incident in 1986. Even when regulated properly, the waste can cause contamination which lasts for many years and destroys natural resources. For more information and a specific example go to: www.world-nuclear.org/info/ch...byl/inf07.html Large scale gamma irradiation is used to sterilize disposable medical supplies such as syringes, gloves and other instruments that would be damaged by heat sterilization. Large scale gamma irradiation is also used for killing parasites found in wool, wood and other widely distributed products. In the 1960's the irradiation of meat was allowed by the US, and it is now a commonly used food sterilization method. Small scale irradiates are also used for blood transfusions and other medical sterilization procedures. Gamma Rays can be used to determine the ash content of coal. By bombarding stable elements with radioactive rays one can cause a fluorescence, the energy of fluorescent x-rays can help identify if any elements are represented in a material. The intensity of the rays can indicate the quantity of that material. This process is commonly used in element processing plants. Radioisotopes are used as tracers in medical research. People ingest these isotopes which allow researchers to study processes like digestion and locate medical problems like cancers and obstructions within an individual's digestive tract. Radioactive elements are also used in clearing angioplasty obstructions and eliminating cancer. To date the only country to utilize and actually use them is the United States. On August 6th and 9th 1945, the US dropped nuclear weapons on Nagasaki and Hiroshima, Japan. These weapons were a part of a top secret project known today as the Manhattan project. Though those within the blast zone were instantly killed, the effects of these weapons would be felt for many years to come. Many more people died in the months following the bombing due to radiation poisoning, and years later, birth defects would prove the effects of radioactive bombardment upon DNA. A good resource on the industrial and medical uses of radioactive isotopes: www.world-nuclear.org/info/inf56.htm All of the radioactive elements are concentrated between atomic numbers 84 and 118 on the periodic table, though Tc and Pm are an exception. Also note that there is a break between 110 and 118 on the table, which are suspected radioactive elements that have yet to be discovered. 29 radioactive elements have been identified by scientists to date: | 9,853 | 1,654 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Exercises%3A_Physical_and_Theoretical_Chemistry/Data-Driven_Exercises/Analysis_of_the_IR_Spectrum_of_Carbon_Monoxide |
The infrared (IR) spectrum of a simple heteronuclear diatomic reveals a series of regularly spaced peaks. A typical spectrum is illustrated in Figure \(\Page {1}\). This exercise will involve an analysis of the IR spectrum of carbon monoxide, which looks much like the spectrum in Figure \(\Page {1}\). To understand the information contained in the \(\ce{CO}\) spectrum, we must take into account the vibrational and rotational energy levels of the molecule. To a first approximation, the vibration of the carbon-oxygen bond is well described by the harmonic oscillator model. The energy levels of a harmonic oscillator are given by \[E_{v} = \omega \left( v+ \dfrac{1}{2} \right) \label{1}\] where \(v\) is the vibrational quantum number (0, 1, 2 ….) and \(ω\) is the harmonic frequency of oscillator. This frequency can in turn be calculated from \[\omega = \sqrt{\dfrac{k}{\mu}} \label{2}\] where \(k\) is the spring constant of the bond and \(μ\) is the reduced mass of the diatomic, \[\mu = \dfrac{m_Cm_O}{m_C+m_O} \label{3}\] The rotational energy levels of \(\ce{CO}\) can be described using the rigid rotor model, which has energies \[ E_J = BJ (J+1) \label{4}\] where \(J\) is the rotational quantum number (0, 1, 2, 3, …) and \(B\) is the rotational constant, \[ B =\dfrac{\hbar^2}{2I} = \dfrac{\hbar^2}{2 \mu R^2} \label{5}\] In Equation \ref{5}, \(I\) represents the and \(R\) the equilibrium bond distance. The combined rotational-vibrational energy (ro-vibrational energy) can be obtained by adding Equations \ref{1} and \ref{4} which yields \[E_{v,J} = \omega \left( v+ \dfrac{1}{2} \right) + BJ (J+1) \label{6}\] When a diatomic absorbs a photon of IR light, there can be a simultaneous change in vibrational and rotational quantum number. However, not all transitions are permitted. The selection rules that apply to this system tell us that only transitions whereby \(Δυ = ±1\) and \(ΔJ = ±1\) are allowed. At room temperature, nearly all of the \(\ce{CO}\) molecules reside in their ground vibrational state (\(v = 0\)), so the vibrational transition that predominates is the \(v = 0 \to 1\) transition. But a wide range of rotational states are thermally occupied within the ground vibrational state, so several \(ΔJ = ±1\) rotational transitions can coincide with the \(v = 0 \to 1\) jump. Figure \(\Page {2}\) shows a number of ro-vibrational energy levels for \(\ce{CO}\). By convention, ground state parameters are identified with a double prime, “, and excited state parameters with a single prime,’. Of the transitions that can occur between the \(v”\) and \(v’\) level, some will involve a decrease in rotational quantum number (\(ΔJ = -1\)) and the others will involve an increase (\(ΔJ = +1\)). The peaks in the IR spectrum that arise from \(ΔJ = -1\) transitions are called transitions and the \(ΔJ = +1\) transitions are called R-branch transitions (three possible P-branch and three possible R-branch transitions are shown in Figure \(\Page {2}\)). Referring back to Figure \(\Page {1}\), the P-branch transitions give rise to a series of peaks that appear at lower wave numbers. The R-branch series appears at higher wave numbers. There appears to be a peak missing between these two series; no peak appears here because \(ΔJ = 0\) transitions are not allowed.
In this assignment, you will be given the peak positions of several \(v = 0 \to 1\) transitions in the carbon monoxide IR spectrum with a goal of determining the rotational constant of this diatomic. To accomplish this, you will first identify pairs of transition peaks that terminate in the same \(J’\) level (one peak each from the \(P\) and \(R\) branches). Once we have identified the appropriate pairs, we will subtract their wave number values to obtain an energy difference. As an example, consider the \(R-1\) and \(P-2\) transitions in Figure \(\Page {2}\), which both terminate in the \(J’=1\) level. By subtracting the wavenumber values for this pair of transitions we obtain the energy difference between the \(J”=0\) and \(J”=2\) levels (in the ground vibrational state). This can be represented symbolically by writing \[ \Delta E_{J''=1} = \tilde{\nu}_{P-2} - \tilde{\nu}_{R-1} = E_{0,2} - E_{0,0} \label{7}\] In a similar fashion, the subtraction of successive pairs of peak positions will yield the energy difference between the \(J”=1\) and \(J”=3\) levels, and the \(J”=2\) and \(J”=4\) levels, which can be represented by writing \[ \Delta E_{J''=2} = \tilde{\nu}_{P-3} - \tilde{\nu}_{R-2} = E_{0,3} - E_{0,1} \nonumber\] \[ \Delta E_{J''=3} = \tilde{\nu}_{P-4} - \tilde{\nu}_{R-3} = E_{0,4} - E_{0,2} \nonumber\] or more generally as \[ \Delta E_{J''=3} = E_{0,J''+1} - E_{0,J"-1} \label{8}\] Using Equation \ref{6}, we can derive a theoretical expression for this energy difference as follows: \[ \Delta E_{J''} = \left[ \omega \left(\dfrac{1}{2} \right) + B (J''+1)(J''+2) \right] - \left[ \omega \left(\dfrac{1}{2} \right) + B (J''-1)(J'') \right] \nonumber\] which simplifies to \[ \Delta E_{J''} = 2B (2J +1 ) \label{9}\] According to Equation \ref{9}, a plot of successive energy differences versus \((2J"+1)\) is linear and the slope equals \(2B\). on the following link and save the numerical information to an appropriate folder. The data corresponds to the fundamental gas phase IR spectrum of \(\ce{^{12}C^{16}O}\); the absorption peak wave number positions are listed from smallest to largest in one long column (the P-branch peaks are listed first, the word ‘center’ indicates the relative placement of the missing peak, and then the R-branch peaks are listed). All wave number values have been corrected for the index of refraction in air (yielding peak positions that would be observed in a vacuum). The data was obtained from Rao, K.N. and Mathews, C.W., editors, Molecular Spectroscopy: Modern Research, Academic Press, New York, 1972.
Wave number positions for the \(v = 0 \to 1\) infrared transitions in \(\ce{^{12}C^{16}O}\) | 5,995 | 1,655 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.04%3A_The_Change_of_Concentration_with_Time_(Integrated_Rate_Laws) |
Either the differential rate law or the integrated rate law can be used to determine the reaction order from experimental data. Often, zeroth o know how to determine the reaction order from experimental data. A is one whose rate is independent of concentration; its differential rate law is \[\text{rate} = k. \nonumber \] We refer to these reactions as zeroth order because we could also write their rate in a form such that the exponent of the reactant in the rate law is 0: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm{reactant}]^0=k(1)=k \label{14.4.1} \] Because rate is independent of reactant concentration, a graph of the concentration of any reactant as a function of time is a straight line with a slope of \(−k\). The value of \(k\) is negative because the concentration of the reactant decreases with time. Conversely, a graph of the concentration of any product as a function of time is a straight line with a slope of \(k\), a positive value. The integrated rate law for a zeroth-order reaction also produces a straight line and has the general form \[[A] = [A]_0 − kt \label{14.4.2} \] where \([A]_0\) is the initial concentration of reactant \(A\). has the form of the algebraic equation for a straight line, \[y = mx + b, \nonumber \] with \(y = [A]\), \(mx = −kt\), and \(b = [A]_0\).) In a zeroth-order reaction, the rate constant must have the same units as the reaction rate, typically moles per liter per second. Although it may seem counterintuitive for the reaction rate to be independent of the reactant concentration(s), such reactions are rather common. They occur most often when the reaction rate is determined by available surface area. An example is the decomposition of N O on a platinum (Pt) surface to produce N and O , which occurs at temperatures ranging from 200°C to 400°C: \[\mathrm{2N_2O(g)}\xrightarrow{\textrm{Pt}}\mathrm{2N_2(g)}+\mathrm{O_2(g)} \label{14.4.3} \] Without a platinum surface, the reaction requires temperatures greater than 700°C, but between 200°C and 400°C, the only factor that determines how rapidly N O decomposes is the amount of Pt surface available (not the amount of Pt). As long as there is enough N O to react with the entire Pt surface, doubling or quadrupling the N O concentration will have no effect on the reaction rate. At very low concentrations of N O, where there are not enough molecules present to occupy the entire available Pt surface, the reaction rate is dependent on the N O concentration. The reaction rate is as follows: \[\textrm{rate}=-\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2O}]}{\Delta t} \right )=\dfrac{1}{2}\left (\dfrac{\Delta[\mathrm{N_2}]}{\Delta t} \right )=\dfrac{\Delta[\mathrm{O_2}]}{\Delta t}=k[\mathrm{N_2O}]^0=k \label{14.4.4} \] Thus the rate at which N O is consumed and the rates at which N and O are produced are independent of concentration. As shown in , the change in the concentrations of all species with time is linear. Most important, the exponent (0) corresponding to the N O concentration in the experimentally derived rate law is not the same as the reactant’s stoichiometric coefficient in the balanced chemical equation (2). For this reaction, as for all others, the rate law must be determined experimentally. A zeroth-order reaction that takes place in the human liver is the oxidation of ethanol (from alcoholic beverages) to acetaldehyde, catalyzed by the alcohol dehydrogenase. At high ethanol concentrations, this reaction is also a zeroth-order reaction. The overall reaction equation is where NAD (nicotinamide adenine dinucleotide) and NADH (reduced nicotinamide adenine dinucleotide) are the oxidized and reduced forms, respectively, of a species used by all organisms to transport electrons. When an alcoholic beverage is consumed, the ethanol is rapidly absorbed into the blood. Its concentration then decreases at a constant rate until it reaches zero ( \(\Page {3a}\)). An average 70 kg person typically takes about 2.5 h to oxidize the 15 mL of ethanol contained in a single 12 oz can of beer, a 5 oz glass of wine, or a shot of distilled spirits (such as whiskey or brandy). The actual rate, however, varies a great deal from person to person, depending on body size and the amount of alcohol dehydrogenase in the liver. The reaction rate does not increase if a greater quantity of alcohol is consumed over the same period of time because the reaction rate is determined only by the amount of enzyme present in the liver. Contrary to popular belief, the caffeine in coffee is ineffective at catalyzing the oxidation of ethanol. When the ethanol has been completely oxidized and its concentration drops to essentially zero, the rate of oxidation also drops rapidly (part (b) in \(\Page {3}\)). These examples illustrate two important points: A Discussing Zero-Order Reactions. Link: In a , the reaction rate is directly proportional to the concentration of one of the reactants. First-order reactions often have the general form A → products. The differential rate for a first-order reaction is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=k[\textrm A] \label{14.4.5} \] If the concentration of A is doubled, the reaction rate doubles; if the concentration of A is increased by a factor of 10, the reaction rate increases by a factor of 10, and so forth. Because the units of the reaction rate are always moles per liter per second, the units of a first-order rate constant are reciprocal seconds (s ). The integrated rate law for a first-order reaction can be written in two different ways: one using exponents and one using logarithms. The exponential form is as follows: \[[A] = [A]_0e^{−kt} \label{14.4.6} \] where \([A]_0\) is the initial concentration of reactant \(A\) at \(t = 0\); \(k\) is the rate constant; and is the base of the natural logarithms, which has the value 2.718 to three decimal places. Recall that an integrated rate law gives the relationship between reactant concentration and time. predicts that the concentration of A will decrease in a smooth exponential curve over time. By taking the natural logarithm of each side of and rearranging, we obtain an alternative logarithmic expression of the relationship between the concentration of \(A\) and \(t\): \[\ln[A] = \ln[A]_0 − kt \label{14.4.7} \] Because has the form of the algebraic equation for a straight line, \[y = mx + b, \nonumber \] with \(y = \ln[A]\) and \(b = \ln[A]_0\), a plot of \(\ln[A]\) versus \(t\) for a first-order reaction should give a straight line with a slope of \(−k\) and an intercept of \(\ln[A]_0\). Either the differential rate law ( ) or the integrated rate law ( ) can be used to determine whether a particular reaction is first order. Discussing the The First-Order Integrated Rate Law Equation: First-order reactions are very common. One reaction that exhibits apparent first-order kinetics is the hydrolysis of the anticancer drug cisplatin. Cisplatin, the first “inorganic” anticancer drug to be discovered, is unique in its ability to cause complete remission of the relatively rare, but deadly cancers of the reproductive organs in young adults. The structures of cisplatin and its hydrolysis product are as follows: Both platinum compounds have four groups arranged in a square plane around a Pt(II) ion. The reaction shown in is important because cisplatin, the form in which the drug is administered, is not the form in which the drug is active. Instead, at least one chloride ion must be replaced by water to produce a species that reacts with deoxyribonucleic acid (DNA) to prevent cell division and tumor growth. Consequently, the kinetics of the reaction in \(\Page {4}\) have been studied extensively to find ways of maximizing the concentration of the active species. If a plot of reactant concentration versus time is not linear but a plot of the natural logarithm of reactant concentration versus time is linear, then the reaction is first order. The rate law and reaction order of the hydrolysis of cisplatin are determined from experimental data, such as those displayed in ble \(\Page {1}\). The table lists initial rate data for four experiments in which the reaction was run at pH 7.0 and 25°C but with different initial concentrations of cisplatin. Because the reaction rate increases with increasing cisplatin concentration, we know this cannot be a zeroth-order reaction. Comparing Experiments 1 and 2 in \(\Page {1}\) shows that the reaction rate doubles [(1.8 × 10 M/min) ÷ (9.0 × 10 M/min) = 2.0] when the concentration of cisplatin is doubled (from 0.0060 M to 0.012 M). Similarly, comparing Experiments 1 and 4 shows that the reaction rate increases by a factor of 5 [(4.5 × 10 M/min) ÷ (9.0 × 10 M/min) = 5.0] when the concentration of cisplatin is increased by a factor of 5 (from 0.0060 M to 0.030 M). Because the reaction rate is directly proportional to the concentration of the reactant, the exponent of the cisplatin concentration in the rate law must be 1, so the rate law is rate = [cisplatin] . Thus the reaction is first order. Knowing this, we can calculate the rate constant using the differential rate law for a first-order reaction and the data in any row of Table \(\Page {1}\). For example, substituting the values for Experiment 3 into , 3.6 × 10 M/min = (0.024 M) 1.5 × 10 min = Knowing the rate constant for the hydrolysis of cisplatin and the rate constants for subsequent reactions that produce species that are highly toxic enables hospital pharmacists to provide patients with solutions that contain only the desired form of the drug. At high temperatures, ethyl chloride produces HCl and ethylene by the following reaction: \[\ce{CH_3CH_2Cl(g) ->[\Delta] HCl(g) + C_2H_4(g)} \nonumber \] Using the rate data for the reaction at 650°C presented in the following table, calculate the reaction order with respect to the concentration of ethyl chloride and determine the rate constant for the reaction. balanced chemical equation, initial concentrations of reactant, and initial rates of reaction reaction order and rate constant Use measured concentrations and rate data from any of the experiments to find the rate constant. The reaction order with respect to ethyl chloride is determined by examining the effect of changes in the ethyl chloride concentration on the reaction rate. Comparing Experiments 2 and 3 shows that doubling the concentration doubles the reaction rate, so the reaction rate is proportional to [CH CH Cl]. Similarly, comparing Experiments 1 and 4 shows that quadrupling the concentration quadruples the reaction rate, again indicating that the reaction rate is directly proportional to [CH CH Cl]. This behavior is characteristic of a first-order reaction, for which the rate law is rate = [CH CH Cl]. We can calculate the rate constant ( ) using any row in the table. Selecting Experiment 1 gives the following: 1.60 × 10 M/s = (0.010 M) 1.6 × 10 s = Sulfuryl chloride (SO Cl ) decomposes to SO and Cl by the following reaction: \[SO_2Cl_2(g) → SO_2(g) + Cl_2(g) \nonumber \] Data for the reaction at 320°C are listed in the following table. Calculate the reaction order with regard to sulfuryl chloride and determine the rate constant for the reaction. We can also use the integrated rate law to determine the reaction rate for the hydrolysis of cisplatin. To do this, we examine the change in the concentration of the reactant or the product as a function of time at a single initial cisplatin concentration. shows plots for a solution that originally contained 0.0100 M cisplatin and was maintained at pH 7 and 25°C. The concentration of cisplatin decreases smoothly with time, and the concentration of chloride ion increases in a similar way. When we plot the natural logarithm of the concentration of cisplatin versus time, we obtain the plot shown in part (b) in . The straight line is consistent with the behavior of a system that obeys a first-order rate law. We can use any two points on the line to calculate the slope of the line, which gives us the rate constant for the reaction. Thus taking the points from part (a) in for = 100 min ([cisplatin] = 0.0086 M) and = 1000 min ([cisplatin] = 0.0022 M), \[\begin{align*}\textrm{slope}&=\dfrac{\ln [\textrm{cisplatin}]_{1000}-\ln [\textrm{cisplatin}]_{100}}{\mathrm{1000\;min-100\;min}}
\\[4pt] -k&=\dfrac{\ln 0.0022-\ln 0.0086}{\mathrm{1000\;min-100\;min}}=\dfrac{-6.12-(-4.76)}{\mathrm{900\;min}}=-1.51\times10^{-3}\;\mathrm{min^{-1}}
\\[4pt] k&=1.5\times10^{-3}\;\mathrm{min^{-1}}\end{align*} \nonumber \] The slope is negative because we are calculating the rate of disappearance of cisplatin. Also, the rate constant has units of min because the times plotted on the horizontal axes in parts (a) and (b) in are in minutes rather than seconds. The reaction order and the magnitude of the rate constant we obtain using the integrated rate law are exactly the same as those we calculated earlier using the differential rate law. This must be true if the experiments were carried out under the same conditions. Example Using the First-Order Integrated Rate Law Equation: If a sample of ethyl chloride with an initial concentration of 0.0200 M is heated at 650°C, what is the concentration of ethyl chloride after 10 h? How many hours at 650°C must elapse for the concentration to decrease to 0.0050 M ( = 1.6 × 10 s ) ? initial concentration, rate constant, and time interval concentration at specified time and time required to obtain particular concentration The exponential form of the integrated rate law for a first-order reaction ( ) is [A] = [A] . Having been given the initial concentration of ethyl chloride ([A] ) and having the rate constant of = 1.6 × 10 s , we can use the rate law to calculate the concentration of the reactant at a given time . Substituting the known values into the integrated rate law, \[\begin{align*}[\mathrm{CH_3CH_2Cl}]_{\mathrm{10\;h}}&=[\mathrm{CH_3CH_2Cl}]_0e^{-kt}
\\[4pt] &=\textrm{0.0200 M}(e^{\large{-(1.6\times10^{-6}\textrm{ s}^{-1})[(10\textrm{ h})(60\textrm{ min/h})(60\textrm{ s/min})]}})
\\[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber \] We could also have used the logarithmic form of the integrated rate law ( ): \[\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=\ln [\mathrm{CH_3CH_2Cl}]_0-kt
\\[4pt] &=\ln 0.0200-(1.6\times10^{-6}\textrm{ s}^{-1})[(\textrm{10 h})(\textrm{60 min/h})(\textrm{60 s/min})]
\\[4pt] &=-3.912-0.0576=-3.970 \nonumber
\\[4pt] [\mathrm{CH_3CH_2Cl}]_{\textrm{10 h}}&=e^{-3.970}\textrm{ M} \nonumber
\\[4pt] &=0.0189\textrm{ M} \nonumber\end{align*} \nonumber \] To calculate the amount of time required to reach a given concentration, we must solve the integrated rate law for \(t\). Eq gives the following: \[\begin{align*}\ln[\mathrm{CH_3CH_2Cl}]_t &=\ln[\mathrm{CH_3CH_2Cl}]_0-kt
\\[4pt] kt &=\ln[\mathrm{CH_3CH_2Cl}]_0-\ln[\mathrm{CH_3CH_2Cl}]_t=\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t}
\\[4pt] t &=\dfrac{1}{k}\left (\ln\dfrac{[\mathrm{CH_3CH_2Cl}]_0}{[\mathrm{CH_3CH_2Cl}]_t} \right )=\dfrac{1}{1.6\times10^{-6}\textrm{ s}^{-1}}\left(\ln \dfrac{0.0200\textrm{ M}}{0.0050\textrm{ M}}\right)
\\[4pt] &=\dfrac{\ln 4.0}{1.6\times10^{-6}\textrm{ s}^{-1}}=8.7\times10^5\textrm{ s}=240\textrm{ h}=2.4\times10^2\textrm{ h} \nonumber \end{align*} \nonumber \] In the exercise in Example \(\Page {1}\), you found that the decomposition of sulfuryl chloride (\(\ce{SO2Cl2}\)) is first order, and you calculated the rate constant at 320°C. The simplest kind of is one whose rate is proportional to the square of the concentration of one reactant. These generally have the form \[\ce{2A → products.}\nonumber \] A second kind of second-order reaction has a reaction rate that is proportional to the product of the concentrations of two reactants. Such reactions generally have the form A + B → products. An example of the former is a dimerization reaction, in which two smaller molecules, each called a monomer, combine to form a larger molecule (a dimer). The differential rate law for the simplest second-order reaction in which 2A → products is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{2\Delta t}=k[\textrm A]^2 \label{14.4.8} \] Consequently, doubling the concentration of A quadruples the reaction rate. For the units of the reaction rate to be moles per liter per second (M/s), the units of a second-order rate constant must be the inverse (M ·s ). Because the units of molarity are expressed as mol/L, the unit of the rate constant can also be written as L(mol·s). For the reaction 2A → products, the following integrated rate law describes the concentration of the reactant at a given time: \[\dfrac{1}{[\textrm A]}=\dfrac{1}{[\textrm A]_0}+kt \label{14.4.9} \] Because has the form of an algebraic equation for a straight line, = + , with = 1/[A] and = 1/[A] , a plot of 1/[A] versus for a simple second-order reaction is a straight line with a slope of and an intercept of 1/[A] . Second-order reactions generally have the form 2A → products or A + B → products. Discussing the Second-Order Integrated Rate Law Equation: Simple second-order reactions are common. In addition to dimerization reactions, two other examples are the decomposition of NO to NO and O and the decomposition of to I and H . Most examples involve simple inorganic molecules, but there are organic examples as well. We can follow the progress of the reaction described in the following paragraph by monitoring the decrease in the intensity of the red color of the reaction mixture. Many cyclic organic compounds that contain two carbon–carbon double bonds undergo a dimerization reaction to give complex structures. One example is as follows: Figure \(\Page {7}\) For simplicity, we will refer to this reactant and product as “monomer” and “dimer,” respectively. The systematic name of the monomer is 2,5-dimethyl-3,4-diphenylcyclopentadienone. The systematic name of the dimer is the name of the monomer followed by “dimer.” Because the monomers are the same, the general equation for this reaction is 2A → product. This reaction represents an important class of organic reactions used in the pharmaceutical industry to prepare complex carbon skeletons for the synthesis of drugs. Like the first-order reactions studied previously, it can be analyzed using either the differential rate law ( ) or the integrated rate law ( ). To determine the differential rate law for the reaction, we need data on how the reaction rate varies as a function of monomer concentrations, which are provided in . From the data, we see that the reaction rate is not independent of the monomer concentration, so this is not a zeroth-order reaction. We also see that the reaction rate is not proportional to the monomer concentration, so the reaction is not first order. Comparing the data in the second and fourth rows shows that the reaction rate decreases by a factor of 2.8 when the monomer concentration decreases by a factor of 1.7: \[\dfrac{5.0\times10^{-5}\textrm{ M/min}}{1.8\times10^{-5}\textrm{ M/min}}=2.8\hspace{5mm}\textrm{and}\hspace{5mm}\dfrac{3.4\times10^{-3}\textrm{ M}}{2.0\times10^{-3} \textrm{ M}}=1.7 \nonumber \] Because (1.7) = 2.9 ≈ 2.8, the reaction rate is approximately proportional to the square of the monomer concentration. rate ∝ [monomer] This means that the reaction is second order in the monomer. Using and the data from any row in , we can calculate the rate constant. Substituting values at time 10 min, for example, gives the following: \[\begin{align}\textrm{rate}&=k[\textrm A]^2 \\8.0\times10^{-5}\textrm{ M/min}&=k(4.4\times10^{-3}\textrm{ M})^2 \\4.1 \textrm{ M}^{-1}\cdot \textrm{min}^{-1}&=k\end{align} \nonumber \] We can also determine the reaction order using the integrated rate law. To do so, we use the decrease in the concentration of the monomer as a function of time for a single reaction, plotted in . The measurements show that the concentration of the monomer (initially 5.4 × 10 M) decreases with increasing time. This graph also shows that the reaction rate decreases smoothly with increasing time. According to the integrated rate law for a second-order reaction, a plot of 1/[monomer] versus should be a straight line, as shown in . Any pair of points on the line can be used to calculate the slope, which is the second-order rate constant. In this example, = 4.1 M ·min , which is consistent with the result obtained using the differential rate equation. Although in this example the stoichiometric coefficient is the same as the reaction order, this is not always the case. The reaction order must always be determined experimentally. For two or more reactions of the same order, the reaction with the largest rate constant is the fastest. Because the units of the rate constants for zeroth-, first-, and second-order reactions are different, however, we cannot compare the magnitudes of rate constants for reactions that have different orders. At high temperatures, nitrogen dioxide decomposes to nitric oxide and oxygen. \[\mathrm{2NO_2(g)}\xrightarrow{\Delta}\mathrm{2NO(g)}+\mathrm{O_2(g)} \nonumber \] Experimental data for the reaction at 300°C and four initial concentrations of NO are listed in the following table: Determine the reaction order and the rate constant. balanced chemical equation, initial concentrations, and initial rates reaction order and rate constant We can determine the reaction order with respect to nitrogen dioxide by comparing the changes in NO concentrations with the corresponding reaction rates. Comparing Experiments 2 and 4, for example, shows that doubling the concentration quadruples the reaction rate [(5.40 × 10 ) ÷ (1.35 × 10 ) = 4.0], which means that the reaction rate is proportional to [NO ] . Similarly, comparing Experiments 1 and 4 shows that tripling the concentration increases the reaction rate by a factor of 9, again indicating that the reaction rate is proportional to [NO ] . This behavior is characteristic of a second-order reaction. We have rate = [NO ] . We can calculate the rate constant ( ) using data from any experiment in the table. Selecting Experiment 2, for example, gives the following: \[\begin{align*}\textrm{rate}&=k[\mathrm{NO_2}]^2
\\5.40\times10^{-5}\textrm{ M/s}&=k(\mathrm{\mathrm{0.010\;M}})^2
\\0.54\mathrm{\;M^{-1}\cdot s^{-1}}&=k\end{align*} \nonumber \] When the highly reactive species HO forms in the atmosphere, one important reaction that then removes it from the atmosphere is as follows: \[2HO_{2(g)} \rightarrow H_2O_{2(g)} + O_{2(g)} \nonumber \] The kinetics of this reaction have been studied in the laboratory, and some initial rate data at 25°C are listed in the following table: Determine the reaction order and the rate constant. If a plot of reactant concentration versus time is linear, but a plot of concentration versus time is linear, then the reaction is second order. If a flask that initially contains 0.056 M NO is heated at 300°C, what will be the concentration of NO after 1.0 h? How long will it take for the concentration of NO to decrease to 10% of the initial concentration? Use the integrated rate law for a second-order reaction ( ) and the rate constant calculated above. balanced chemical equation, rate constant, time interval, and initial concentration final concentration and time required to reach specified concentration We know and [NO ] , and we are asked to determine [NO ] at = 1 h (3600 s). Substituting the appropriate values into , \[\begin{align*}\dfrac{1}{[\mathrm{NO_2}]_{3600}}&=\dfrac{1}{[\mathrm{NO_2}]_0}+kt \\[4pt] &=\dfrac{1}{0.056\textrm{ M}}+[(0.54 \mathrm{\;M^{-1}\cdot s^{-1}})(3600\textrm{ s})] \\[4pt] &=2.0\times10^3\textrm{ M}^{-1}\end{align*} \nonumber \] Thus [NO ] = 5.1 × 10 M. In this case, we know and [NO ] , and we are asked to calculate at what time [NO ] = 0.1[NO ] = 0.1(0.056 M) = 0.0056 M. To do this, we solve for , using the concentrations given. \[ \begin{align*} t &=\dfrac{(1/[\mathrm{NO_2}])-(1/[\mathrm{NO_2}]_0)}{k} \\[4pt] &=\dfrac{(1/0.0056 \textrm{ M})-(1/0.056\textrm{ M})}{0.54 \;\mathrm{M^{-1}\cdot s^{-1}}} \\[4pt] &=3.0\times10^2\textrm{ s}=5.0\textrm{ min} \end{align*} \nonumber \] NO decomposes very rapidly; under these conditions, the reaction is 90% complete in only 5.0 min. In the previous exercise, you calculated the rate constant for the decomposition of HO as = 1.4 × 10 M ·s . This high rate constant means that HO decomposes rapidly under the reaction conditions given in the problem. In fact, the HO molecule is so reactive that it is virtually impossible to obtain in high concentrations. Given a 0.0010 M sample of HO , calculate the concentration of HO that remains after 1.0 h at 25°C. How long will it take for 90% of the HO to decompose? Use the integrated rate law for a second-order reaction ( ) and the rate constant calculated in the exercise in Example \(\Page {3}\). In addition to the simple second-order reaction and rate law we have just described, another very common second-order reaction has the general form \(A + B \rightarrow products\), in which the reaction is first order in \(A\) and first order in \(B\). The differential rate law for this reaction is as follows: \[\textrm{rate}=-\dfrac{\Delta[\textrm A]}{\Delta t}=-\dfrac{\Delta[\textrm B]}{\Delta t}=k[\textrm A,\textrm B] \label{14.4.10} \] Because the reaction is first order both in A and in B, it has an overall reaction order of 2. (The integrated rate law for this reaction is rather complex, so we will not describe it.) We can recognize second-order reactions of this sort because the reaction rate is proportional to the concentrations of each reactant. The reaction rate of a zeroth-order reaction is independent of the concentration of the reactants. The reaction rate of a first-order reaction is directly proportional to the concentration of one reactant. The reaction rate of a simple second-order reaction is proportional to the square of the concentration of one reactant. Knowing the rate law of a reaction gives clues to the reaction mechanism. | 26,220 | 1,656 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_2%3A_Chemical_Bonding_and_Structure/7%3A_Bonding_in_Organic_Molecules/7.2%3A_The_Alkanes |
We begin our study of organic chemistry with the hydrocarbons, the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding. Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons). , in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules. The word has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C). We previously introduced the three simplest alkanes—methane (CH ), ethane (C H ), and propane (C H ) and they are shown again in Figure \(\Page {1}\). The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure \(\Page {2}\)). Methane (CH ), ethane (C H ), and propane (C H ) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH unit. The first 10 members of this series are given in Table \(\Page {1}\). Consider the series in Figure \(\Page {3}\). The sequence starts with C H , and a CH unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH group) is called a homologous series. The members of such a series, called , have properties that vary in a regular and predictable manner. The principle of gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series. The principle of homology allows us to write a general formula for alkanes: C H . Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C H = C H . | 2,717 | 1,658 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.02%3A_Equations_and_Mass_Relationships/3.2.04%3A_Food-_Let's_Cook |
What do and have to do with chemistry? We often use a calculation in everyday life without even realizing it, especially while cooking, preparing a grocery list or building something! You: "Stoichiometry? What is that?" Me: "It's using unit conversions in order to answer a question." You: "Oh, that's all? Besides not being able to pronounce the word very well, I thought it was some scary type of calculation." Me: "Believe it or not, you've been doing these types of calculations for a long time already. It is just a bunch of unit conversions!" Then once you recognize that you are in fact using the same calculations in your everyday life as in this chemistry course, hopefully this page will help you relate what you are "thinking" to what it looks like when you write it out as a math equation.
The thing that takes getting used to is how we can relate what we use every day to something that we are learning.
Read through these first three examples to get a feeling for how we actually use these calculations in our everyday life. Then move on to examples 4 and 5 to see how we relate these everyday calculations to chemistry.
If you get stumped on Examples 4 and 5, refer back to Examples 1, 2 and 3. Keep in mind that you will be using the EXACT same calculation, you are just replacing words like "eggs, ham, cheese, and cookies", with chemical names or chemical formulas. If you get confused, try replacing the words of a chemistry problem with more familiar words and see if that helps you learn these chemical calculations. For instance, you could replace NH with the word "egg" to help you feel more comfortable. I want to have friends over for lunch on Saturday and make grilled cheese sandwiches that require two slices of bread and one slice of cheese. I open the refrigerator to find that I have 5 slices of cheese. I look in the bread box to find that I have 10 slices of bread. How many sandwiches can I make? These three steps are the first steps to solving this type of question, in chemistry, it is called a "stoichiometry" problem. Step 1) Write out the recipe, also known as an + → 2 slices of bread + 1 slice of cheese → 1 grilled cheese sandwich Step 2) Find Quantity (moles) and Identify useful unit conversions and/or What we know:
1) We know we have 5 slices of cheese
2) We know we have 10 slices of bread. According to the equation written in step 1, ratios or unit conversions can be written by using the number in front of the ingredient, also known as a .
2 slices of bread is required for one slice of cheese.
Remember, the powerful thing about ratios is that we can also write them "upside-down".
The following ratios or unit conversions that can be written from this equation are: \(\begin{align} \frac{\text{1 slice of cheese}}{\text{2 slices of bread}}&=\frac{\text{2 slices of bread}}{\text{1 slice of cheese}} \\ & \\ \frac{\text{1 sandwich}}{\text{2 slices of bread}}&=\frac{\text{2 slices of bread}}{\text{1 sandwich}} \\ & \\ \frac{\text{1 sandwich}}{\text{1 slice of cheese}}&=\frac{\text{1 slice of cheese}}{\text{1 sandwich}} \\ \end{align}\) Before we start a calculation, let us visualize what we have to start with. You may even get an answer without even have to "think" about.
Let's re-organize this picture to represent the ratio that we determined in step 2. Is it easier to see how many sandwiches we can make now? Remember, the purpose of this page is to help you identify calculations you do in your everyday life that you probably don't even realize you're doing. If you did come to an answer in your head, this is what the equation looks like when written out. \(\begin{align} & mol_{\text{sandwiches}}=\text{5 slices of cheese}\times \frac{\text{1 sandwich}}{\text{1 slice of cheese}}=\text{5 sandwiches} \\ & \\ & mol_{\text{sandwiches}}=\text{10 slices of bread}\times \frac{\text{1 sandwich}}{\text{2 slices of bread}}=\text{5 sandwiches} \\ \end{align}\) In this case it doesn't matter which one starting material or that you start with. As long as you know the relationship or the between what you are looking for and what you know, then you will always be able to come to an answer. To re-cap the steps we used to solve this problem:
Step 1) Write equation
Step 2) Write or find moles
Step 3) Use molar ratio You just volunteered to provide cookies for your school's bake sale. They ask you to bring 100 cookies. You go home and pull out this recipe from your recipe box. Cookies Recipe - Ingredients You don't have any of the ingredients at home so we will need to go to the grocery store. We need to make a list so we make sure to get enough ingredients to make 100 cookies. Write an equation. We will use our recipe as our equation. Identify useful unit conversions. If we used our recipe we could relate any one ingredient on the recipe to any other ingredient on the recipe. Similar to making sandwiches. In chemistry terms this is called a . A is just a unit conversion. It allows us to get from what we know to what we want. Another helpful unit conversion that I will highlight is the following: 1 recipe = 20 cookies
If we need 100 cookies, how many batches of cookies do we need to make?
You might have already answered five? Let's set up an equation using the unit conversions from Step 1 and 2, \(batches_{\text{cookies}}=\text{100 cookies}\times \frac{\text{1 batch}}{\text{20 cookies}}=\text{5 batches}\) Now that I know how many batches of cookies I need, I can write my grocery list. To re-cap the steps we used to solve this problem: Step 1) Write equation
Step 2) Write or find moles
Step 3) Use molar ratio One of my favorite breakfast foods are cheese omelets in the morning. I am on a diet so I always measure the amount of each ingredient I use. I always make two in case someone else wants one. Below is the recipe for my 'perfect' omelet. Recipe: 6 Large eggs - 200. g per one egg
1 cups of shredded cheese - 50. g per one cup
I open the refrigerator this morning to find an excess of large eggs and 250. g of cheese that is about to spoil. 6 Large eggs + 1 cups of cheese → 2 omelets According to the equation written in step 1, these molar ratios can be written by using the number in front of the ingredient, also known as a .
Once again, when you are first starting out, it is helpful to write the ratios or the stoichiometric ratios connecting all of the components of the recipe.
It makes it easier to pick out which one ratio you will use as a unit conversion to help you calculate what the question is asking.
As you can see, there are more molar ratios this time because there are more starting materials. Additionally, the question gave us information about the mass per unit food item. Therefore there are 8 unit conversions in this question. We can determine the mass of 2 omelets by using the law of conservation of mass. Therefore, I will add the mass of 6 eggs and 1 cup of cheese to calculate the last unit conversion listed above. In order for us to be able to compare the starting materials and answer question 2 and 3, we need to convert all of the starting materials from into the associated with recipe or equation .
The actual quantity of cheese can be calculated using the mass per one unit of material. In this case, 1 cup of cheese is 50. grams.
In chemistry terms, this is called the . Molar mass can be used as a . Refer above for useful unit conversions
To relate the idea of molar mass to everyday life is something we do all of the time without even thinking about it.
For instance, you have an assortment of candy in a bucket. Three kids come up to you and ask you for some candy.
Do you weigh the candy and give the three kids equivalent masses? No. That would be unrealistic and probably a little messy.
Instead, we just decide that one candy bar should be given to each kid. Kid #1 receives one Snickers bar, kid #2 receives one Milky Way and kid #3 receives one Kit Kat bar. They are all happy because they received their OWN candy bar; they don't care that one might weigh a little bit more than the other candy bars.
To summarize, Snickers, Milky Way, and Kit Kat all represent ONE candy bar, however they have different masses...but that's okay. To relate it even more to chemistry, one mole of any compound is equivalent to one mole of any other compound, they just have a different mass depending on what substance you are talking about; just substitute these words and you are back to our everyday example; 'mole' = 'candy bar' and 'compound' = 'type of candy bar'.
You can use this idea in reverse, if you know the mass, then the amount of candy bars or 'moles' can be calculated.
\(\text{ }mol_{\text{cheese (actual)}}=\text{250. g of cheese}\times \frac{\text{1 cup of cheese}}{\text{50. g of cheese}}=\text{5.0 cups of cheese}\) Now that we know how much cheese we have, we can calculate how many eggs we need based on our recipe in Step 1. We can also calculate how many omelets we will make. We can represent it visually before writing an actual calculation.
Based on these calculations, we just answered question 1 and 2. We will use 30 eggs from the refrigerator in order to use up all of the cheese in order to make 10 omelets.
We will need one more step in order to answer question 3, how much will all of our omelets weigh? Let's find out how many 10 omelets will weigh using the useful unit conversion that we identified in Step 2. \(\text{ }mass_{\text{omelet}}=\text{10 omelets}\times \frac{\text{1250 g omelet}}{\text{2 omelets}}=\text{6250 g of omelet made}\)
The thing that takes getting used to is how we can relate what we use every day to something that we are learning. If you get stumped on the next two examples, refer back to Examples 1, 2 and 3. Keep in mind that you will be using the EXACT same calculation, you are just replacing words like "eggs, ham, cheese, and cookies", with chemical names or chemical formulas. If you get confused, try replacing the words of a chemistry problem with more familiar words and see if that helps you learn these chemical calculations. In the examples above you saw a lot of whole numbers. In chemistry you will be seeing more decimal numbers. Imagine that you could make 2.5 omelets and that is the same thing as producing or using 2.5 mol of a chemical substance. Find the amount of water produced when 3.68 mol NH is consumed according to Equation \(\ref{3}\). The amount of water produced must be in the stoichiometric ratio S(H O/NH ) to the amount of ammonia consumed: \(\large\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{\text{3}}} \right)=\Large\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}}\) Multiplying both sides , by we have \(\begin{align} \large n_{\text{H}_{\text{2}}\text{O produced}} &= \large n_{\text{NH}_{\text{3}}\text{ consumed}} \normalsize \times\text{S}\left( \frac{\ce{H2O}}{\ce{NH3}} \right) \\ { } \\ & =\text{3.68 mol NH}_3 \times \frac{\text{6 mol }\ce{H2O}}{\text{4 mol NH}_3} \\ & =\text{5.52 mol }\ce{H2O} \end{align}\) This is a typical illustration of the use of a as a . Example 4 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Step 3 when using the molar ratio. Simply remember that the coefficients in a balanced chemical equation give molar ratios, and that the proper choice results in cancellation of units. In road-map form \[\large \text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \\ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced} \nonumber \] or symbolically. \[\large n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}} \nonumber \] When using stoichiometric ratios, be sure you indicate moles . You can only cancel moles of the substance. In other words, 1 mol NH cancels 1 mol NH but does not cancel 1 mol H O. The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product. The chemical reaction in this example is of environmental interest. Iron pyrite (FeS ) is often an impurity in coal, and so burning this fuel in a power plant produces sulfur dioxide (SO ), a major air pollutant. Calculate the mass of sulfur dioxide (SO ) produced when 3.84 mol O is reacted with FeS according to the equation \[\ce{4 FeS2 (s) + 11 O2 (g) → 2 Fe2O3 (s) + 8 SO2 (g)} \nonumber\] The problem asks that we calculate the mass of SO produced. As we learned above in and in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of SO to the mass of SO . Therefore this problem in effect is asking that we calculate the amount of SO produced from the amount of O consumed. We will start this problem the same as in Example 4. It requires the molar ratio The equation was given to us in the question. The question is asking us about the amount of SO produced from the amount of O consumed so I want both of those substances in the molar ratio. \(\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}=\frac{\text{11 mol O}_{\text{2}}}{\text{8 mol SO}_{\text{2}}}\)
The question is also asking about the mass of SO , so I want to write the molar mass.
\(\frac{\text{1 mol SO}_{\text{2}}}{\text{64.06 g SO}_{\text{2}}}=\frac{\text{64.06 g SO}_{\text{2}}}{\text{1 mol SO}_{\text{2}}}\) The problem gives us the amount of oxygen gas in moles, so we do not need to use the molar mass equation like we did in Example 3 - I am on a Diet! The of SO produced is then \(\begin{align} n_{\text{SO}_{\text{2}}}\text{ produced}&=n_{\text{O}_{\text{2}}\text{ actual}}\text{ }\!\!\times\!\!\text{ conversion factor} \\n_{\text{SO}_{\text{2}}}\text{ produced}&=\text{3}\text{.84 mol O}_{\text{2}}\times \frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}=\text{2}\text{.79 mol SO}_{\text{2}}\text{ produced} \end{align}\) The question is asking about the mass of SO not the amount of SO . We can use the unit conversion or molar mass written in Step 2. The of SO is \(\begin{align} \text{mass}_{\text{SO}_{\text{2}}}&=\text{2}\text{.79 mol SO}_{\text{2}}\times \frac{\text{64}\text{.06 g SO}_{\text{2}}}{\text{1 mol SO}_{\text{2}}}=\text{179 g SO}_{\text{2}} \end{align}\) With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of O to moles of SO and the molar mass will convert moles of SO to grams of SO . A schematic road map for the one-step calculation can be written as \(mol_{\text{(O}_{\text{2}}\text{)}}\text{ }\xrightarrow{ratio\text{(SO}_{\text{2}}\text{/O}_{\text{2}}\text{)}}\text{ }mol_{\text{(SO}_{\text{2}}\text{)}}\text{ }\xrightarrow{M.M._{\text{SO}_{\text{2}}}}\text{ }mass_{\text{(SO}_{\text{2}}\text{)}}\) This is how the equation would look if it were all done in one step: \(\text{mass}_{\text{(SO}_{\text{2}}\text{)}}=\text{3}\text{.84 mol O}_{\text{2}}\times \text{ }\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}\text{ }\times \text{ }\frac{\text{64}\text{.06 g}}{\text{1 mol SO}_{\text{2}}}=\text{179 g}\) Notice that all of the units that are the same on the top and the bottom of the ratios can cancel, leaving just the units that you want at the end. | 15,742 | 1,659 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/12%3A_Chromatographic_and_Electrophoretic_Methods/12.03%3A_Optimizing_Chromatographic_Separations |
Now that we have defined the solute retention factor, selectivity, and column efficiency we are able to consider how they affect the resolution of two closely eluting peaks. Because the two peaks have similar retention times, it is reasonable to assume that their peak widths are nearly identical. If the number of theoretical plates is the same for all solutes—not strictly true, but not a bad assumption—then from equation , the ratio / is a constant. If two solutes have similar retention times, then their peak widths must be similar. Equation , therefore, becomes \[R_{A B}=\frac{t_{r, B}-t_{r, A}}{0.5\left(w_{B}+w_{A}\right)} \approx \frac{t_{r, B}-t_{r, A}}{0.5\left(2 w_{B}\right)}=\frac{t_{r, B}-t_{r, A}}{w_{B}} \label{12.1}\] where is the later eluting of the two solutes. Solving equation 12.2.15 for and substituting into Equation \ref{12.1} leaves us with the following result. \[R_{A B}=\frac{\sqrt{N_{B}}}{4} \times \frac{t_{r, B}-t_{r, A}}{t_{r, B}} \label{12.2}\] Rearranging equation provides us with the following equations for the retention times of solutes and . \[t_{r, A}=k_{A} t_{\mathrm{m}}+t_{\mathrm{m}} \quad \text { and } \quad t_{\mathrm{r}, B}=k_{B} t_{\mathrm{m}}+t_{\mathrm{m}} \nonumber\] After substituting these equations into Equation \ref{12.2} and simplifying, we have \[R_{A B}=\frac{\sqrt{N_{B}}}{4} \times \frac{k_{B}-k_{A}}{1+k_{B}} \nonumber\] Finally, we can eliminate solute ’s retention factor by substituting in equation . After rearranging, we end up with the following equation for the resolution between the chromatographic peaks for solutes A and B. \[R_{A B}=\frac{\sqrt{N_{B}}}{4} \times \frac{\alpha-1}{\alpha} \times \frac{k_{B}}{1+k_{B}} \label{12.3}\] In addition to resolution, another important factor in chromatography is the amount of time needed to elute a pair of solutes, which we can approximate using the retention time for solute . \[t_{r, s}=\frac{16 R_{AB}^{2} H}{u} \times\left(\frac{\alpha}{\alpha-1}\right)^{2} \times \frac{\left(1+k_{B}\right)^{3}}{k_{B}^{2}} \label{12.4}\] where is the mobile phase’s velocity. Although Equation \ref{12.3} is useful for considering how a change in , \(\alpha\), or qualitatively affects resolution—which suits our purpose here—it is less useful for making accurate quantitative predictions of resolution, particularly for smaller values of and for larger values of . For more accurate predictions use the equation \[R_{A B}=\frac{\sqrt{N}}{4} \times(\alpha-1) \times \frac{k_{B}}{1+k_{\mathrm{avg}}} \nonumber\] where is ( + /2. For a derivation of this equation and for a deeper discussion of resolution in column chromatography, see Foley, J. P. “Resolution Equations for Column Chromatography,” , , , 1275-1279. Equation \ref{12.3} and Equation \ref{12.4} contain terms that correspond to column efficiency, selectivity, and the solute retention factor. We can vary these terms, more or less independently, to improve resolution and analysis time. The first term, which is a function of the number of theoretical plates (for Equation \ref{12.3}) or the height of a theoretical plate (for Equation \ref{12.4}), accounts for the effect of column efficiency. The second term is a function of \(\alpha\) and accounts for the influence of column selectivity. Finally, the third term in both equations is a function of and accounts for the effect of solute ’s retention factor. A discussion of how we can use these parameters to improve resolution is the subject of the remainder of this section. One of the simplest ways to improve resolution is to adjust the retention factor for solute . If all other terms in Equation \ref{12.3} remain constant, an increase in will improve resolution. As shown by the curve in Figure 12.3.1
, however, the improvement is greatest if the initial value of is small. Once exceeds a value of approximately 10, a further increase produces only a marginal improvement in resolution. For example, if the original value of is 1, increasing its value to 10 gives an 82% improvement in resolution; a further increase to 15 provides a net improvement in resolution of only 87.5%. Any improvement in resolution from increasing the value of generally comes at the cost of a longer analysis time. The curve in Figure 12.3.1
shows the relative change in the retention time for solute as a function of its retention factor. Note that the minimum retention time is for = 2. Increasing from 2 to 10, for example, approximately doubles solute B’s retention time. The relationship between retention factor and analysis time in Figure 12.3.1
works to our advantage if a separation produces an acceptable resolution with a large . In this case we may be able to decrease with little loss in resolution and with a significantly shorter analysis time. To increase without changing selectivity, \(\alpha\), any change to the chromatographic conditions must result in a general, nonselective increase in the retention factor for both solutes. In gas chromatography, we can accomplish this by decreasing the column’s temperature. Because a solute’s vapor pressure is smaller at lower temperatures, it spends more time in the stationary phase and takes longer to elute. In liquid chromatography, the easiest way to increase a solute’s retention factor is to use a mobile phase that is a weaker solvent. When the mobile phase has a lower solvent strength, solutes spend proportionally more time in the stationary phase and take longer to elute. Adjusting the retention factor to improve the resolution between one pair of solutes may lead to unacceptably long retention times for other solutes. For example, suppose we need to analyze a four-component mixture with baseline resolution and with a run-time of less than 20 min. Our initial choice of conditions gives the chromatogram in Figure 12.3.2
a. Although we successfully separate components 3 and 4 within 15 min, we fail to separate components 1 and 2. Adjusting conditions to improve the resolution for the first two components by increasing provides a good separation of all four components, but the run-time is too long (Figure 12.3.2
b). This problem of finding a single set of acceptable operating conditions is known as the . One solution to the general elution problem is to make incremental adjustments to the retention factor as the separation takes place. At the beginning of the separation we set the initial chromatographic conditions to optimize the resolution for early eluting solutes. As the separation progresses, we adjust the chromatographic conditions to decrease the retention factor—and, therefore, to decrease the retention time—for each of the later eluting solutes (Figure 12.3.2
c). In gas chromatography this is accomplished by temperature programming. The column’s initial temperature is selected such that the first solutes to elute are resolved fully. The temperature is then increased, either continuously or in steps, to bring off later eluting components with both an acceptable resolution and a reasonable analysis time. In liquid chromatography the same effect is obtained by increasing the solvent’s eluting strength. This is known as a gradient elution. We will have more to say about each of these in later sections of this chapter. A second approach to improving resolution is to adjust the selectivity, \(\alpha\). In fact, for \(\alpha \approx 1\) usually it is not possible to improve resolution by adjusting the solute retention factor, , or the column efficiency, . A change in \(\alpha\) often has a more dramatic effect on resolution than a change in . For example, changing \(\alpha\) from 1.1 to 1.5, while holding constant all other terms, improves resolution by 267%. In gas chromatography, we adjust \(\alpha\) by changing the stationary phase; in liquid chromatography, we change the composition of the mobile phase to adjust \(\alpha\). To change \(\alpha\) we need to selectively adjust individual solute retention factors. Figure 12.3.3
shows one possible approach for the liquid chromatographic separation of a mixture of substituted benzoic acids. Because the retention time of a compound’s weak acid form and its weak base form are different, its retention time will vary with the pH of the mobile phase, as shown in Figure 12.3.3
a. The intersections of the curves in Figure 12.3.3
a show pH values where two solutes co-elute. For example, at a pH of 3.8 terephthalic acid and -hydroxybenzoic acid elute as a single chromatographic peak. Figure 12.3.3
a shows that there are many pH values where some separation is possible. To find the optimum separation, we plot a for each pair of solutes. The , , and curves in Figure 12.3.3
b show the variation in a with pH for the three pairs of solutes that are hardest to separate (for all other pairs of solutes, \(\alpha\) > 2 at all pH levels). The shading shows windows of pH values in which at least a partial separation is possible—this figure is sometimes called a window diagram—and the highest point in each window gives the optimum pH within that range. The best overall separation is the highest point in any window, which, for this example, is a pH of 3.5. Because the analysis time at this pH is more than 40 min (Figure 12.3.3
a), choosing a pH between 4.1–4.4 might produce an acceptable separation with a much shorter analysis time. Let’s use benzoic acid, C H COOH, to explain why pH can affect a solute’s retention time. The separation uses an aqueous mobile phase and a nonpolar stationary phase. At lower pHs, benzoic acid predominately is in its weak acid form, C H COOH, and partitions easily into the nonpolar stationary phase. At more basic pHs, however, benzoic acid is in its weak base form, C H COO . Because it now carries a charge, its solubility in the mobile phase increases and its solubility in the nonpolar stationary phase decreases. As a result, it spends more time in the mobile phase and has a shorter retention time. Although the usual way to adjust pH is to change the concentration of buffering agents, it also is possible to adjust pH by changing the column’s temperature because a solute’s p value is pH-dependent; for a review, see Gagliardi, L. G.; Tascon, M.; Castells, C. B. “Effect of Temperature on Acid–Base Equilibria in Separation Techniques: A Review,” , , , 35–57. A third approach to improving resolution is to adjust the column’s efficiency by increasing the number of theoretical plates, . If we have values for and \(\alpha\), then we can use Equation \ref{12.3} to calculate the number of theoretical plates for any resolution. Table 12.3.1
provides some representative values. For example, if \(\alpha\) = 1.05 and = 2.0, a resolution of 1.25 requires approximately 24 800 theoretical plates. If our column provides only 12 400 plates, half of what is needed, then a separation is not possible. How can we double the number of theoretical plates? The easiest way is to double the length of the column, although this also doubles the analysis time. A better approach is to cut the height of a theoretical plate, , in half, providing the desired resolution without changing the analysis time. Even better, if we can decrease by more than 50%, it may be possible to achieve the desired resolution with an even shorter analysis time by also decreasing or \(\alpha\). 0.5 1.5 3.0 To decrease the height of a theoretical plate we need to understand the experimental factors that affect band broadening. There are several theoretical treatments of band broadening. We will consider one approach that considers four contributions: variations in path lengths, longitudinal diffusion, mass transfer in the stationary phase, and mass transfer in the mobile phase. As solute molecules pass through the column they travel paths that differ in length. Because of this difference in path length, two solute molecules that enter the column at the same time will exit the column at different times. The result, as shown in Figure 12.3.4
, is a broadening of the solute’s profile on the column. The contribution of to the height of a theoretical plate, , is \[H_{p}=2 \lambda d_{p} \label{12.5}\] where is the average diameter of the particulate packing material and \(\lambda\) is a constant that accounts for the consistency of the packing. A smaller range of particle sizes and a more consistent packing produce a smaller value for \(\lambda\). For a column without packing material, is zero and there is no contribution to band broadening from multiple paths. An inconsistent packing creates channels that allow some solute molecules to travel quickly through the column. It also can creates pockets that temporarily trap some solute molecules, slowing their progress through the column. A more uniform packing minimizes these problems. The second contribution to band broadening is the result of the solute’s in the mobile phase. Solute molecules are in constant motion, diffusing from regions of higher solute concentration to regions where the concentration of solute is smaller. The result is an increase in the solute’s band width (Figure 12.3.5
). The contribution of longitudinal diffusion to the height of a theoretical plate, , is \[H_{d}=\frac{2 \gamma D_{m}}{u} \label{12.6}\] where is the solute’s diffusion coefficient in the mobile phase, is the mobile phase’s velocity, and \(\gamma\) is a constant related to the efficiency of column packing. Note that the effect of on band broadening is inversely proportional to the mobile phase velocity: a higher velocity provides less time for longitudinal diffusion. Because a solute’s diffusion coefficient is larger in the gas phase than in a liquid phase, longitudinal diffusion is a more serious problem in gas chromatography. As the solute passes through the column it moves between the mobile phase and the stationary phase. We call this movement between phases . As shown in Figure 12.3.6
, band broadening occurs if the solute’s movement within the mobile phase or within the stationary phase is not fast enough to maintain an equilibrium in its concentration between the two phases. On average, a solute molecule in the mobile phase moves down the column before it passes into the stationary phase. A solute molecule in the stationary phase, on the other hand, takes longer than expected to move back into the mobile phase. The contributions of mass transfer in the stationary phase, , and mass transfer in the mobile phase, , are given by the following equations \[H_{s}=\frac{q k d_{f}^{2}}{(1+k)^{2} D_{s}} u \label{12.7}\] \[H_{m}=\frac{f n\left(d_{p}^{2}, d_{c}^{2}\right)}{D_{m}} u \label{12.8}\] where is the thickness of the stationary phase, is the diameter of the column, and are the diffusion coefficients for the solute in the stationary phase and the mobile phase, is the solute’s retention factor, and is a constant related to the column packing material. Although the exact form of is not known, it is a function of particle size and column diameter. Note that the effect of and on band broadening is directly proportional to the mobile phase velocity because a smaller velocity provides more time for mass transfer. The abbreviation in Equation \ref{12.7} means “is a function of.” The height of a theoretical plate is a summation of the contributions from each of the terms affecting band broadening. \[H=H_{p}+H_{d}+H_{s}+H_{m} \label{12.9}\] An alternative form of this equation is the \[H=A+\frac{B}{u}+C u \label{12.10}\] which emphasizes the importance of the mobile phase’s velocity. In the van Deemter equation, accounts for the contribution of multiple paths ( ), / accounts for the contribution of longitudinal diffusion ( ), and accounts for the combined contribution of mass transfer in the stationary phase and in the mobile phase ( and ). There is some disagreement on the best equation for describing the relationship between plate height and mobile phase velocity [Hawkes, S. J. , , 393–398]. In addition to the van Deemter equation, other equations include \[H=\frac{B}{u}+\left(C_s+C_{m}\right) u \nonumber\] where and are the mass transfer terms for the stationary phase and the mobile phase and \[H=A u^{1 / 3}+\frac{B}{u}+C u \nonumber\] All three equations, and others, have been used to characterize chromatographic systems, with no single equation providing the best explanation in every case [Kennedy, R. T.; Jorgenson, J. W. , , 1128–1135]. To increase the number of theoretical plates without increasing the length of the column, we need to decrease one or more of the terms in Equation \ref{12.9}. The easiest way to decrease is to adjust the velocity of the mobile phase. For smaller mobile phase velocities, column efficiency is limited by longitudinal diffusion, and for higher mobile phase velocities efficiency is limited by the two mass transfer terms. As shown in Figure 12.3.7
—which uses the van Deemter equation—the optimum mobile phase velocity is the minimum in a plot of as a function of . The remaining parameters that affect the terms in Equation \ref{12.9} are functions of the column’s properties and suggest other possible approaches to improving column efficiency. For example, both and are a function of the size of the particles used to pack the column. Decreasing particle size, therefore, is another useful method for improving efficiency. For a more detailed discussion of ways to assess the quality of a column, see Desmet, G.; Caooter, D.; Broeckhaven, K. “Graphical Data Represenation Methods to Assess the Quality of LC Columns,” , , 8593–8602. Perhaps the most important advancement in chromatography columns is the development of open-tubular, or . These columns have very small diameters ( ≈ 50–500 μm) and contain no packing material ( = 0). Instead, the capillary column’s interior wall is coated with a thin film of the stationary phase. Plate height is reduced because the contribution to from (Equation \ref{12.5}) disappears and the contribution from (Equation \ref{12.8}) becomes smaller. Because the column does not contain any solid packing material, it takes less pressure to move the mobile phase through the column, which allows for longer columns. The combination of a longer column and a smaller height for a theoretical plate increases the number of theoretical plates by approximately \(100 \times\). Capillary columns are not without disadvantages. Because they are much narrower than packed columns, they require a significantly smaller amount of sample, which may be difficult to inject reproducibly. Another approach to improving resolution is to use thin films of stationary phase, which decreases the contribution to from (Equation \ref{12.7}). The smaller the particles, the more pressure is needed to push the mobile phase through the column. As a result, for any form of chromatography there is a practical limit to particle size. | 18,942 | 1,660 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/05%3A_Dioxygen_Reactions/5.03%3A_Copper-zinc_Superoxide_Dismutase |
Two families of metalloproteins are excellent catalysts for the disproportionation of superoxide (Reaction 5.95). \[2O_{2}^{-} + 2 H^{+} \xrightarrow{SOD} O_{2} + H_{2}O_{2} \tag{5.95}\] These are (1) the copper-zinc superoxide dismutases, CuZnSOD, found in almost all eukaryotic cells and a very few prokaryotes, and (2) the manganese and iron superoxide dismutases, MnSOD and FeSOD, the former found in the mitochondria of eukaryotic cells, and both found in many prokaryotes. Recent studies of bacterial and yeast mutants that were engineered to contain no superoxide dismutases demonstrated that the cells were unusually sensitive to dioxygen and that the sensitivity to dioxygen was relieved when an SOD gene was reintroduced into the cells. These results indicate that the superoxide dismutase enzymes playa critical role in dioxygen metabolism, but they do not define the chemical agent responsible for dioxygen toxicity (see Section III). Several transition-metal complexes have been observed to catalyze superoxide disproportionation; in fact, aqueous copper ion, Cu , is an excellent SOD catalyst, comparable in activity to CuZnSOD itself! Free aqueous Cu would not itself be suitable for use as an SOD , however, because it is too toxic (see Section III) and because it binds too strongly to a large variety of cellular components and thus would not be present as the free ion. (Most forms of complexed cupric ion show much less superoxide dismutase activity than the free ion.) Aside from aqueous copper ion, few other complexes are as effective as the SOD enzymes. Two mechanisms (Reactions 5.96 to 5.99) have been proposed for catalysis of superoxide disproportionation by metal complexes and metalloenzymes. Mechanism I: $$M^{n+} + O_{2}^{-} \rightarrow M^{(n-1)+} + O_{2} \tag{5.96}\] \[M^{(n-1)+} + O_{2}^{-} \rightarrow M^{n+}(O_{2}^{2-}) \xrightarrow{2H^{+}} M^{n+} + H_{2}O_{2} \tag{5.97}\] Mechanism II: $$M^{n+} + O_{2}^{-} \rightarrow M^{n+} (O_{2}^{-}) \tag{5.98}\] \[M^{n+} (O_{2}^{-}) + O_{2}^{-} \rightarrow M^{n+}(O_{2}^{2-}) \xrightarrow{2 H^{+}} M^{n} + H_{2}O_{2} \tag{5.99}\] \[+O_{2}\] In Mechanism I, which is favored for the SOD enzymes and most redox-active metal complexes with SOD activity, superoxide reduces the metal ion in the first step, and then the reduced metal ion is reoxidized by another superoxide, presumably via a metal-peroxo complex intermediate. In Mechanism II, which is proposed for nonredox metal complexes but may be operating in other situations as well, the metal ion is never reduced, but instead forms a superoxo complex, which is reduced to a peroxo complex by a second superoxide ion. In both mechanisms, the peroxo ligands are protonated and dissociate to give hydrogen peroxide. Analogues for each of the separate steps of Reactions (5.96) to (5.99) have been observed in reactions of superoxide with transition-metal complexes, thereby establishing the feasibility of both mechanisms. For example, superoxide was shown to reduce Cu phen) to give Cu (phen) (phen = 1,10-phenanthroline), a reaction analogous to Reaction (5.96). On the other hand, superoxide reacts with Cu (tet b) to form a superoxo complex (a reaction analogous to Reaction 5.98), presumably because Cu (tet b) is not easily reduced to the cuprous state, because the ligand cannot adjust to the tetrahedral geometry that Cu prefers. \(\tag{5.100}\) Reaction of superoxide with a reduced metal-ion complex to give oxidation of the complex and release of hydrogen peroxide (analogous to Reaction 5.97) has been observed in the reaction of Fe EDTA with superoxide. Reduction of a Co superoxo complex by free superoxide to give a peroxo complex (analogous to Reaction 5.99) has also been observed. If a metal complex can be reduced by superoxide and if its reduced form can be oxidized by superoxide, both at rates competitive with superoxide disproportionation, the complex can probably act as an SOD by Mechanism I. Mechanism II has been proposed to account for the apparent catalysis of superoxide disproportionation by Lewis acidic nonredox-active metal ions under certain conditions. However, this mechanism should probably be considered possible for redox metal ions and the SOD enzymes as well. It is difficult to distinguish the two mechanisms for redox-active metal ions and the SOD enzymes unless the reduced form of the catalyst is observed directly as an intermediate in the reaction. So far it has not been possible to observe this intermediate in the SOD enzymes or the metal complexes. The x-ray crystal structure of the oxidized form of CuZnSOD from bovine erythrocytes shows a protein consisting of two identical subunits held together almost entirely by hydrophobic interactions. Each subunit consists of a flattened cylindrical barrel of \(\beta\)-pleated sheet from which three external loops of irregular structure extend (Figure 5.15). The metal-binding region of the protein binds Cu and Zn in close proximity to each other, bridged by the imidazolate ring of a histidyI side chain. Figure 5.16 represents the metal-binding region. The Cu ion is coordinated to four histidyl imidazoles and a water in a highly distorted square-pyramidal geometry with water at the apical position. The Zn ion is coordinated to three histidyl imidazoles (including the one shared with copper) and an aspartyl carboxylate group, forming a distorted tetrahedral geometry around the metal ion. One of the most unusual aspects of the structure of this enzyme is the occurrence of the bridging imidazolate ligand, which holds the copper and zinc ions 6 Å apart. Such a configuration is not unusual for imidazole complexes of metal ions, which sometimes form long polymeric imidazolate-bridged structures. \(\tag{5.101}\) However, no other imidazolate-bridged bi- or polymetallic metalloprotein has yet been identified. The role of the zinc ion in CuZnSOD appears to be primarily structural. There is no evidence that water, anions, or other potential ligands can bind to the zinc, so it is highly unlikely that superoxide could interact with that site. Moreover, removal of zinc under conditions where the copper ion remains bound to the copper site does not significantly diminish the SOD activity of the enzyme. However, such removal does result in a diminished thermal stability, i.e., the zinc-depleted protein denatures at a lower temperature than the native protein, supporting the hypothesis that the role of the zinc is primarily structural in nature. The copper site is clearly the site of primary interaction of superoxide with the protein. The x-ray structure shows that the copper ion lies at the bottom of a narrow channel that is large enough to admit only water, small anions, and similarly small ligands (Figure 5.17). In the lining of the channel is the positively charged side chain of an arginine residue, 5 Å away from the copper ion and situated in such a position that it could interact with superoxide and other anions when they bind to copper. Near the mouth of the channel, at the surface of the protein, are two positively charged lysine residues, which are believed to play a role in attracting anions and guiding them into the channel. Chemical modification of these lysine or arginine residues substantially diminishes the SOD activity, supporting their role in the mechanism of reaction with superoxide. The x-ray structural results described above apply only to the oxidized form of the protein, i.e., the form containing Cu . The reduced form of the enzyme containing Cu is also stable and fully active as an SOD. If, as is likely, the mechanism of CuZnSOD-catalyzed superoxide disproportionation is Mechanism I (Reactions 5.96-5.97), the structure of the reduced form is of critical importance in understanding the enzymatic mechanism. Unfortunately, that structure is not yet available. The mechanism of superoxide disproportionation catalyzed by CuZnSOD is generally believed to go by Mechanism I (Reactions 5.96-5.97), i.e., reduction of Cu to Cu by superoxide with the release of dioxygen, followed by reoxidation of Cu to Cu by a second superoxide with the release of HO or H O . The protonation of peroxide dianion, O , prior to its release from the enzyme is required, because peroxide dianion is highly basic and thus too unstable to be released in its unprotonated form. The source of the proton that protonates peroxide in the enzymatic mechanism is the subject of some interest. Reduction of the oxidized protein has been shown to be accompanied by the uptake of one proton per subunit. That proton is believed to protonate the bridging imidazolate in association with the breaking of the bridge upon reduction of the copper. Derivatives with Co substituted for Zn at the native zinc site have been used to follow the process of reduction of the oxidized Cu form to the reduced Cu form. The Co in the zinc site does not change oxidation state, but acts instead as a spectroscopic probe of changes occurring at the native zinc-binding site. Upon reduction (Reaction 5.102), the visible absorption band due to Co shifts in a manner consistent with a change occurring in the ligand environment of Co . The resulting spectrum of the derivative containing Cu in the copper site and Co in the zinc site is very similar to the spectrum of the derivative in which the copper site is empty and the zinc site contains Co . This result suggests strongly that the imidazolate bridge is cleaved and protonated and that the resulting imidazole ligand is retained in the coordination sphere of Co (Reaction 5.102). \(\tag{5.102}\) The same proton is thus an attractive possibility for protonation of peroxide as it is formed in the enzymatic mechanism (Reactions 5.103 and 5.104). \(\tag{5.103}\) \(\tag{5.104}\) Attractive as this picture appears, there are several uncertainties about it. For example, the turnover of the enzyme may be too fast for protonation and deprotonation of the bridging histidine to occur. Moreover, the mechanism proposed would require the presence of a metal ion at the zinc site to hold the imidazole in place and to regulate the pK of the proton being transferred. The observation that removal of zinc gives a derivative with almost full SOD activity is thus surprising and may also cast some doubt on this mechanism. Other criticisms of this mechanism have been recently summarized. Studies of CuZnSOD derivatives prepared by site-directed mutagenesis are also providing interesting results concerning the SOD mechanism. For example, it has been shown that mutagenized derivatives of human CuZnSOD with major differences in copper-site geometry relative to the wild-type enzyme may nonetheless remain fully active. Studies of these and similar derivatives should provide considerable insight into the mechanism of reaction of CuZnSOD with superoxide. Studies of the interaction of CuZnSOD and its metal-substituted derivatives with anions have been useful in predicting the behavior of the protein in its reactions with its substrate, the superoxide anion, O . Cyanide, azide, cyanate, and thiocyanate bind to the copper ion, causing dissociation of a histidyl ligand and the water ligand from the copper. Phosphate also binds to the enzyme at a position close to the Cu center, but it apparently does not bind directly to it as a ligand. Chemical modification of Arg-141 with phenylglyoxal blocks the interaction of phosphate with the enzyme, suggesting that this positively charged residue is the site of interaction with phosphate. Electrostatic calculations of the charges on the CuZnSOD protein suggest that superoxide and other anions entering into the vicinity of the protein will be drawn toward and into the channel leading down to the copper site by the distribution of positive charges on the surface of the protein, the positively charged lysines at the mouth of the active-site cavity, and the positively charged arginine and copper ion within the active-site region. Some of the anions studied, e.g., CN , F , N ,and phosphate, have been shown to inhibit the SOD activity of the enzyme. The source of the inhibition is generally assumed to be competition with superoxide for binding to the copper, but it may sometimes result from a shift in the redox potential of copper, which is known to occur sometimes when an anion binds to copper. In the example described above, studies of a metal-substituted derivative helped in the evaluation of mechanistic possibilities for the enzymatic reaction. In addition, studies of such derivatives have provided useful information about the environment of the metal-ion binding sites. For example, metal-ion-substituted derivatives of CuZnSOD have been prepared with Cu , Cu , Zn , Ag , Ni , or Co bound to the native copper site, and with Zn , Cu , Cu , Co , Hg , Cd , Ni , or Ag bound to the native zinc site. The SOD activities of these derivatives are interesting; only those derivatives with copper in the copper site have a high degree of SOD activity, whereas the nature of the metal ion in the zinc site or even its absence has little or no effect. Derivatives of CuZnSOD are known with Cu ion bound either to the native copper site or to the native zinc site. The electronic absorption spectra of these derivatives indicate that the ligand environments of the two sites are very different. Copper(II) is a d transition-metal ion, and its d-d transitions are usually found in the visible and near-IR regions of the spectrum. Copper(II) complexes with coordinated nitrogen ligands are generally found to have an absorption band between 500 and 700 nm, with an extinction coefficient below 100 M cm . Bands in the absorption spectra of complexes with geometries that are distorted away from square planar tend to be red-shifted because of a smaller d-d splitting, and to have higher extinction coefficients because of the loss of centrosymmetry. Thus the optical spectrum of CuZnSOD with an absorption band with a maximum at 680 nm (14,700 cm ; see Figure 5.18A) and an extinction coefficient of 155 M cm per Cu is consistent with the crystal structural results that indicate that copper(II) is bound to four imidazole nitrogens and a water molecule in a distorted square-pyramidal geometry. Metal-substituted derivatives with Cu at the native copper site but with Co , Cd , Hg , or Ni substituted for Zn at the native zinc site all have a band at 680 nm, suggesting that the substitution of another metal ion for zinc perturbs the copper site very little, despite the proximity of the two metal sites. The absorption spectra of native CuZnSOD and these CuMSOD derivatives also have a shoulder at 417 nm (24,000 cm ; see Figure 5.18A), which is at lower energy than normal imidazole-to-Cu charge-transfer transitions, and has been assigned to an imidazolate-to-Cu charge transfer, indicating that the imidazolate bridge between Cu and the metal ion in the native zinc site is present, as observed in the crystal structure of CuZnSOD. Derivatives with the zinc site empty, which therefore cannot have an imidazolate bridge, are lacking this 417 nm shoulder. Small but significant changes in the absorption spectrum are seen when the metal ion is removed from the zinc site, e.g., in copper-only SOD (Figure 5.18B). The visible absorption band shifts to 700 nm (14,300 cm ), presumably due to a change in ligand field strength upon protonation of the bridging imidazolate. In addition, the shoulder at 417 nm has disappeared, again due to the absence of the imidazolate ligand. The spectroscopic properties due to copper in the native zinc site are best observed in the derivative Ag CuSOD, which has Ag in the copper site and Cu in the zinc site (see Figure 5.18C), since the d Ag ion is spectroscopically silent. In this derivative, the d-d transition is markedly red-shifted from the visible region of the spectrum into the near-IR, indicating that the ligand environment of Cu in that site is either tetrahedral or five coordinate. The EPR properties of Cu in this derivative are particularly interesting (as discussed below). The derivative with Cu bound at both sites, CuCuSOD, has a visible-near IR spectrum that is nearly a superposition of the spectra of CuZnSOD and Ag CuSOD (see Figure 5.19), indicating that the geometry of Cu in each of these sites is little affected by the nature of the metal ion in the other site. EPR spectroscopy has also proven to be particularly valuable in characterizing the metal environments in CuZnSOD and derivatives. The EPR spectrum of native CuZnSOD is shown in Figure 5.20A. The g resonance is split by the hyperfine coupling between the unpaired electron on Cu and the I = \(\frac{3}{2}\) nuclear spin of copper. The A value, 130 G, is intermediate between the larger A typical of square-planar Cu complexes with four nitrogen donor ligands and the lower A observed in blue copper proteins (see Chapter 6). The large linewidth seen in the g region indicates that the copper ion is in a rhombic (i.e., distorted) environment. Thus, the EPR spectrum is entirely consistent with the distorted square-pyramidal geometry observed in the x-ray structure. Removal of zinc from the native protein to give copper-only SOD results in a perturbed EPR spectrum, with a narrower g resonance and a larger A value (142 G) more nearly typical of Cu in an axial N environment (Figure 5.20B). Apparently the removal of zinc relaxes some constraints imposed on the geometry of the active-site ligands, allowing the copper to adopt to a geometry closer to its preferred tetragonal arrangement. The EPR spectrum due to Cu in the native Zn site in the Ag CuSOD derivative indicates that Cu is in a very different environment than when it is in the native copper site (Figure 5.20C). The spectrum is strongly rhombic, with a low value of A (97 G), supporting the conclusion based on the visible spectrum that copper is bound in a tetrahedral or five-coordinate environment. This type of site is unusual either for copper coordination complexes or for copper proteins in general, but does resemble the Cu EPR signal seen when either laccase or cytochrome c oxidase is partially reduced (see Figure 5.21). Partial reduction disrupts the magnetic coupling between these Cu centers that makes them EPR-silent in the fully oxidized protein. The EPR spectrum of CuCuSOD is very different from that of any of the other copper-containing derivatives (Figure 5.22) because the unpaired spins on the two copper centers interact and magnetically couple across the imidazolate bridge, resulting in a triplet EPR spectrum. This spectrum is virtually identical with that of model imidazolate-bridged binuclear copper complexes. Electronic absorption and EPR studies of derivatives of CuZnSOD containing Cu have provided useful information concerning the nature of the metal binding sites of those derivatives. H NMR spectra of those derivatives are generally not useful, however, because the relatively slowly relaxing paramagnetic Cu center causes the nearby proton resonances to be extremely broad. This difficulty has been overcome in two derivatives, CuCoSOD and CuNiSOD, in which the fast-relaxing paramagnetic Co and Ni centers at the zinc site interact across the imidazolate bridge and increase the relaxation rate of the Cu center, such that well-resolved paramagnetically shifted H NMR spectra of the region of the proteins near the two paramagnetic metal centers in the protein can be obtained and the resonances assigned. The use of H NMR to study CuCoSOD derivatives of CuZnSOD in combination with electronic absorption and EPR spectroscopies has enabled investigators to compare active-site structures of a variety of wild-type and mutant CuZnSOD proteins in order to find out if large changes in active-site structure have resulted from replacement of nearby amino-acid residues. | 19,934 | 1,661 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/11%3A_Liquids_and_Intermolecular_Forces/11.E%3A_Liquids_and_Intermolecular_Forces_(Exercises) |
. In addition to these individual basis; please contact A liquid, unlike a gas, is virtually . Explain what this means using macroscopic and microscopic descriptions. What general physical properties do liquids share with solids? What properties do liquids share with gases? Using a kinetic molecular approach, discuss the differences and similarities between liquids and gases with regard to How must the ideal gas law be altered to apply the kinetic molecular theory of gases to liquids? Explain. Why are the root mean square speeds of molecules in liquids less than the root mean square speeds of molecules in gases? What is the main difference between intramolecular interactions and intermolecular interactions? Which is typically stronger? How are changes of state affected by these different kinds of interactions? Describe the three major kinds of intermolecular interactions discussed in this chapter and their major features. The hydrogen bond is actually an example of one of the other two types of interaction. Identify the kind of interaction that includes hydrogen bonds and explain why hydrogen bonds fall into this category. Which are stronger—dipole–dipole interactions or London dispersion forces? Which are likely to be more important in a molecule with heavy atoms? Explain your answers. Explain why hydrogen bonds are unusually strong compared to other dipole–dipole interactions. How does the strength of hydrogen bonds compare with the strength of covalent bonds? Liquid water is essential for life as we know it, but based on its molecular mass, water should be a gas under standard conditions. Why is water a liquid rather than a gas under standard conditions? Describe the effect of polarity, molecular mass, and hydrogen bonding on the melting point and boiling point of a substance. Why are intermolecular interactions more important for liquids and solids than for gases? Under what conditions must these interactions be considered for gases? Using acetic acid as an example, illustrate both attractive and repulsive intermolecular interactions. How does the boiling point of a substance depend on the magnitude of the repulsive intermolecular interactions? In group 17, elemental fluorine and chlorine are gases, whereas bromine is a liquid and iodine is a solid. Why? The boiling points of the anhydrous hydrogen halides are as follows: HF, 19°C; HCl, −85°C; HBr, −67°C; and HI, −34°C. Explain any trends in the data, as well as any deviations from that trend. Identify the most important intermolecular interaction in each of the following. Identify the most important intermolecular interaction in each of the following. Would you expect London dispersion forces to be more important for Xe or Ne? Why? (The atomic radius of Ne is 38 pm, whereas that of Xe is 108 pm.) Arrange Kr, Cl , H , N , Ne, and O in order of increasing polarizability. Explain your reasoning. Both water and methanol have anomalously high boiling points due to hydrogen bonding, but the boiling point of water is greater than that of methanol despite its lower molecular mass. Why? Draw the structures of these two compounds, including any lone pairs, and indicate potential hydrogen bonds. The structures of ethanol, ethylene glycol, and glycerin are as follows: Arrange these compounds in order of increasing boiling point. Explain your rationale. Do you expect the boiling point of H S to be higher or lower than that of H O? Justify your answer. Ammonia (NH ), methylamine (CH NH ), and ethylamine (CH CH NH ) are gases at room temperature, while propylamine (CH CH CH NH ) is a liquid at room temperature. Explain these observations. Why is it not advisable to freeze a sealed glass bottle that is completely filled with water? Use both macroscopic and microscopic models to explain your answer. Is a similar consideration required for a bottle containing pure ethanol? Why or why not? Which compound in the following pairs will have the higher boiling point? Explain your reasoning. Some recipes call for vigorous boiling, while others call for gentle simmering. What is the difference in the temperature of the cooking liquid between boiling and simmering? What is the difference in energy input? Use the melting of a metal such as lead to explain the process of melting in terms of what is happening at the molecular level. As a piece of lead melts, the temperature of the metal remains constant, even though energy is being added continuously. Why? How does the O–H distance in a hydrogen bond in liquid water compare with the O–H distance in the covalent O–H bond in the H O molecule? What effect does this have on the structure and density of ice? Explain why the hydrogen bonds in liquid HF are stronger than the corresponding intermolecular H⋅⋅⋅I interactions in liquid HI. Water is a liquid under standard conditions because of its unique ability to form four strong hydrogen bonds per molecule. As the atomic mass of the halogens increases, so does the number of electrons and the average distance of those electrons from the nucleus. Larger atoms with more electrons are more easily polarized than smaller atoms, and the increase in polarizability with atomic number increases the strength of London dispersion forces. These intermolecular interactions are strong enough to favor the condensed states for bromine and iodine under normal conditions of temperature and pressure. Water has two polar O–H bonds with H atoms that can act as hydrogen bond donors, plus two lone pairs of electrons that can act as hydrogen bond acceptors, giving a net of hydrogen bonds per H O molecule. Although methanol also has two lone pairs of electrons on oxygen that can act as hydrogen bond acceptors, it only has one O–H bond with an H atom that can act as a hydrogen bond donor. Consequently, methanol can only form hydrogen bonds per molecule on average, versus four for water. Hydrogen bonding therefore has a much greater effect on the boiling point of water. Vigorous boiling causes more water molecule to escape into the vapor phase, but does not affect the temperature of the liquid. Vigorous boiling requires a higher energy input than does gentle simmering. Why is a water droplet round? How is the environment of molecules on the surface of a liquid droplet different from that of molecules in the interior of the droplet? How is this difference related to the concept of surface tension? Explain the role of intermolecular and intramolecular forces in surface tension. A mosquito is able to walk across water without sinking, but if a few drops of detergent are added to the water, the insect will sink. Why? Explain how soaps or surfactants decrease the surface tension of a liquid. How does the meniscus of an aqueous solution in a capillary change if a surfactant is added? Illustrate your answer with a diagram. Of CH Cl , hexane, and ethanol, which has the lowest viscosity? Which has the highest surface tension? Explain your reasoning in each case. At 25°C, cyclohexanol has a surface tension of 32.92 mN/m , whereas the surface tension of cyclohexanone, which is very similar chemically, is only 25.45 mN/m . Why is the surface tension of cyclohexanone so much less than that of cyclohexanol? What is the relationship between Explain your answers in terms of a microscopic picture. What two opposing forces are responsible for capillary action? How do these forces determine the shape of the meniscus? Which of the following liquids will have a concave meniscus in a glass capillary? Explain your reasoning. How does viscosity depend on molecular shape? What molecular features make liquids highly viscous? Adding a soap or a surfactant to water disrupts the attractive intermolecular interactions between water molecules, thereby decreasing the surface tension. Because water is a polar molecule, one would expect that a soap or a surfactant would also disrupt the attractive interactions responsible for adhesion of water to the surface of a glass capillary. As shown in the sketch, this would decrease the height of the water column inside the capillary, as well as making the meniscus less concave. As the structures indicate, cyclohexanol is a polar substance that can engage in hydrogen bonding, much like methanol or ethanol; consequently, it is expected to have a higher surface tension due to stronger intermolecular interactions. Cohesive forces are the intermolecular forces that hold the molecules of the liquid together, while adhesive forces are the attractive forces between the molecules of the liquid and the walls of the capillary. If the adhesive forces are stronger than the cohesive forces, the liquid is pulled up into the capillary and the meniscus is concave. Conversely, if the cohesive forces are stronger than the adhesive forces, the level of the liquid inside the capillary will be lower than the level outside the capillary, and the meniscus will be convex. Viscous substances often consist of molecules that are much longer than they are wide and whose structures are often rather flexible. As a result, the molecules tend to become tangled with one another (much like overcooked spaghetti), which decreases the rate at which they can move through the liquid. The viscosities of five liquids at 25°C are given in the following table. Explain the observed trends in viscosity. The following table gives values for the viscosity, boiling point, and surface tension of four substances. Examine these data carefully to see whether the data for each compound are internally consistent and point out any obvious errors or inconsistencies. Explain your reasoning. Surface tension data (in dyn/cm) for propanoic acid (C H O ), and 2-propanol (C H O), as a function of temperature, are given in the following table. Plot the data for each compound and explain the differences between the two graphs. Based on these data, which molecule is more polar? 3. The plots of surface tension versus temperature for propionic acid and isopropanol have essentially the same slope, but at all temperatures the surface tension of propionic acid is about 30% greater than for isopropanol. Because surface tension is a measure of the cohesive forces in a liquid, these data suggest that the cohesive forces for propionic acid are significantly greater than for isopropanol. Both substances consist of polar molecules with similar molecular masses, and the most important intermolecular interactions are likely to be dipole–dipole interactions. Consequently, these data suggest that propionic acid is more polar than isopropanol. In extremely cold climates, snow can disappear with no evidence of its melting. How can this happen? What change(s) in state are taking place? Would you expect this phenomenon to be more common at high or low altitudes? Explain your answer. Why do car manufacturers recommend that an automobile should not be left standing in subzero temperatures if its radiator contains only water? Car manufacturers also warn car owners that they should check the fluid level in a radiator only when the engine is cool. What is the basis for this warning? What is likely to happen if it is ignored? Use Hess’s law and a thermochemical cycle to show that, for any solid, the enthalpy of sublimation is equal to the sum of the enthalpy of fusion of the solid and the enthalpy of vaporization of the resulting liquid. Three distinct processes occur when an ice cube at −10°C is used to cool a glass of water at 20°C. What are they? Which causes the greatest temperature change in the water? When frost forms on a piece of glass, crystals of ice are deposited from water vapor in the air. How is this process related to sublimation? Describe the energy changes that take place as the water vapor is converted to frost. What phase changes are involved in each process? Which processes are exothermic, and which are endothermic? What phase changes are involved in each process? Which processes are exothermic, and which are endothermic? Why do substances with high enthalpies of fusion tend to have high melting points? Why is the enthalpy of vaporization of a compound invariably much larger than its enthalpy of fusion? What is the opposite of fusion, sublimation, and condensation? Describe the phase change in each pair of opposing processes and state whether each phase change is exothermic or endothermic. Draw a typical heating curve (temperature versus amount of heat added at a constant rate) for conversion of a solid to a liquid and then to a gas. What causes some regions of the plot to have a positive slope? What is happening in the regions of the plot where the curve is horizontal, meaning that the temperature does not change even though heat is being added? If you know the mass of a sample of a substance, how could you use a heating curve to calculate the specific heat of the substance, as well as the change in enthalpy associated with a phase change? Draw the heating curve for a liquid that has been superheated. How does this differ from a normal heating curve for a liquid? Draw the cooling curve for a liquid that has been supercooled. How does this differ from a normal cooling curve for a liquid? When snow disappears without melting, it must be subliming directly from the solid state to the vapor state. The rate at which this will occur depends solely on the partial pressure of water, not on the total pressure due to other gases. Consequently, altitude (and changes in atmospheric pressure) will not affect the rate of sublimation directly. 3 The general equations and enthalpy changes for the changes of state involved in converting a solid to a gas are: \[ \text{solid} \rightarrow \text{liquid}: \Delta H_{fus} \] \[ \text{liquid} \rightarrow \text{gas}: \Delta H_{vap} \] \[ \text{solid} \rightarrow \text{gas}: \Delta{H_{sub}}= \Delta{H_{fus}} + \Delta{H_{vap}}\] The relationship between these enthalpy changes is shown schematically in the thermochemical cycle below: The formation of frost on a surface is an example of deposition, which is the reverse of sublimation. The change in enthalpy for deposition is equal in magnitude, but opposite in sign, to Δ , which is a positive number: Δ = Δ + Δ . The enthalpy of vaporization is larger than the enthalpy of fusion because vaporization requires the addition of enough energy to disrupt all intermolecular interactions and create a gas in which the molecules move essentially independently. In contrast, fusion requires much less energy, because the intermolecular interactions in a liquid and a solid are similar in magnitude in all condensed phases. Fusion requires only enough energy to overcome the intermolecular interactions that lock molecules in place in a lattice, thereby allowing them to move more freely. The portions of the curve with a positive slope correspond to heating a single phase, while the horizontal portions of the curve correspond to phase changes. During a phase change, the temperature of the system does not change, because the added heat is melting the solid at its melting point or evaporating the liquid at its boiling point. A superheated liquid exists temporarily as liquid with a temperature above the normal boiling point of the liquid. When a supercooled liquid boils, the temperature drops as the liquid is converted to vapor. Conversely, a supercooled liquid exists temporarily as a liquid with a temperature lower than the normal melting point of the solid. As shown below, when a supercooled liquid crystallizes, the temperature increases as the liquid is converted to a solid. The density of oxygen at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of oxygen. The density of propane at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of propane. Draw the cooling curve for a sample of the vapor of a compound that has a melting point of 34°C and a boiling point of 77°C as it is cooled from 100°C to 0°C. Propionic acid has a melting point of −20.8°C and a boiling point of 141°C. Draw a heating curve showing the temperature versus time as heat is added at a constant rate to show the behavior of a sample of propionic acid as it is heated from −50°C to its boiling point. What happens above 141°C? A 0.542 g sample of I requires 96.1 J of energy to be converted to vapor. What is the enthalpy of sublimation of I ? A 2.0 L sample of gas at 210°C and 0.762 atm condenses to give 1.20 mL of liquid, and 476 J of heat is released during the process. What is the enthalpy of vaporization of the compound? One fuel used for jet engines and rockets is aluminum borohydride [Al(BH ) ], a liquid that readily reacts with water to produce hydrogen. The liquid has a boiling point of 44.5°C. How much energy is needed to vaporize 1.0 kg of aluminum borohydride at 20°C, given a Δ of 30 kJ/mol and a molar heat capacity ( ) of 194.6 J/(mol•K)? How much energy is released when freezing 100.0 g of dimethyl disulfide (C H S ) initially at 20°C? Use the following information: melting point = −84.7°C, Δ = 9.19 kJ/mol, = 118.1 J/(mol•K). Δ Δ How much heat is released in the conversion of 1.00 L of steam at 21.9 atm and 200°C to ice at −6.0°C and 1 atm? How much heat must be applied to convert a 1.00 g piece of ice at −10°C to steam at 120°C? How many grams of boiling water must be added to a glass with 25.0 g of ice at −3°C to obtain a liquid with a temperature of 45°C? How many grams of ice at −5.0°C must be added to 150.0 g of water at 22°C to give a final temperature of 15°C? The transition from a liquid to a gaseous phase is accompanied by a drastic decrease in density. According to the data in the table and the plot, the boiling point of liquid oxygen is between 90 and 100 K (actually 90.2 K). 45.0 kJ/mol 488 kJ 32.6 kJ 57 g What is the relationship between the boiling point, vapor pressure, and temperature of a substance and atmospheric pressure? What is the difference between a volatile liquid and a nonvolatile liquid? Suppose that two liquid substances have the same molecular mass, but one is volatile and the other is nonvolatile. What differences in the molecular structures of the two substances could account for the differences in volatility? An “old wives’ tale” states that applying ethanol to the wrists of a child with a very high fever will help to reduce the fever because blood vessels in the wrists are close to the skin. Is there a scientific basis for this recommendation? Would water be as effective as ethanol? Why is the air over a strip of grass significantly cooler than the air over a sandy beach only a few feet away? If gasoline is allowed to sit in an open container, it often feels much colder than the surrounding air. Explain this observation. Describe the flow of heat into or out of the system, as well as any transfer of mass that occurs. Would the temperature of a sealed can of gasoline be higher, lower, or the same as that of the open can? Explain your answer. What is the relationship between the vapor pressure of a liquid and At 25°C, benzene has a vapor pressure of 12.5 kPa, whereas the vapor pressure of acetic acid is 2.1 kPa. Which is more volatile? Based on the intermolecular interactions in the two liquids, explain why acetic acid has the lower vapor pressure. Acetylene (C H ), which is used for industrial welding, is transported in pressurized cylinders. Its vapor pressure at various temperatures is given in the following table. Plot the data and use your graph to estimate the vapor pressure of acetylene at 293 K. Then use your graph to determine the value of Δ for acetylene. How much energy is required to vaporize 2.00 g of acetylene at 250 K? The following table gives the vapor pressure of water at various temperatures. Plot the data and use your graph to estimate the vapor pressure of water at 25°C and at 75°C. What is the vapor pressure of water at 110°C? Use these data to determine the value of Δ for water. The Δ of carbon tetrachloride is 29.8 kJ/mol, and its normal boiling point is 76.8°C. What is its boiling point at 0.100 atm? The normal boiling point of sodium is 883°C. If Δ is 97.4 kJ/mol, what is the vapor pressure (in millimeters of mercury) of liquid sodium at 300°C? An unknown liquid has a vapor pressure of 0.860 atm at 63.7°C and a vapor pressure of 0.330 atm at 35.1°C. Use the data in Table 11.6 in Section 11.5 to identify the liquid. An unknown liquid has a boiling point of 75.8°C at 0.910 atm and a boiling point of 57.2°C at 0.430 atm. Use the data in Table 11.6 in Section 11.5 to identify the liquid. If the vapor pressure of a liquid is 0.850 atm at 20°C and 0.897 atm at 25°C, what is the normal boiling point of the liquid? If the vapor pressure of a liquid is 0.799 atm at 99.0°C and 0.842 atm at 111°C, what is the normal boiling point of the liquid? The vapor pressure of liquid SO is 33.4 torr at −63.4°C and 100.0 torr at −47.7 K. The vapor pressure of CO at various temperatures is given in the following table: vapor pressure at 273 K is 3050 mmHg; Δ = 18.7 kJ/mol, 1.44 kJ 12.5°C Δ = 28.9 kJ/mol, -hexane Δ = 7.81 kJ/mol, 36°C The lines in a phase diagram represent boundaries between different phases; at any combination of temperature and pressure that lies on a line, two phases are in equilibrium. It is physically impossible for more than three phases to coexist at any combination of temperature and pressure, but in principle there can be more than one triple point in a phase diagram. The slope of the line separating two phases depends upon their relative densities. For example, if the solid–liquid line slopes up and to the , the liquid is less dense than the solid, while if it slopes up and to the , the liquid is denser than the solid. 1. Compare the solid and liquid states in terms of a. rigidity of structure. b. long-range order. c. short-range order. 2. How do amorphous solids differ from crystalline solids in each characteristic? Which of the two types of solid is most similar to a liquid? a. rigidity of structure b. long-range order c. short-range order 3. Why is the arrangement of the constituent atoms or molecules more important in determining the properties of a solid than a liquid or a gas? 4. Why are the structures of solids usually described in terms of the positions of the constituent atoms rather than their motion? 5. What physical characteristics distinguish a crystalline solid from an amorphous solid? Describe at least two ways to determine experimentally whether a material is crystalline or amorphous. 6. Explain why each characteristic would or would not favor the formation of an amorphous solid. a. slow cooling of pure molten material b. impurities in the liquid from which the solid is formed c. weak intermolecular attractive forces 7. A student obtained a solid product in a laboratory synthesis. To verify the identity of the solid, she measured its melting point and found that the material melted over a 12°C range. After it had cooled, she measured the melting point of the same sample again and found that this time the solid had a sharp melting point at the temperature that is characteristic of the desired product. Why were the two melting points different? What was responsible for the change in the melting point? 3. The arrangement of the atoms or molecules is more important in determining the properties of a solid because of the greater persistent long-range order of solids. Gases and liquids cannot readily be described by the spatial arrangement of their components because rapid molecular motion and rearrangement defines many of the properties of liquids and gases. 7. The initial solid contained the desired compound in an amorphous state, as indicated by the wide temperature range over which melting occurred. Slow cooling of the liquid caused it to crystallize, as evidenced by the sharp second melting point observed at the expected temperature. 1. four 3. fcc 5. molybdenum 7. sodium, unit cell edge = 428 pm, r = 185 pm 9. d = 0.5335 g/cm , r =151.9 pm 1. Four vials labeled A–D contain sucrose, zinc, quartz, and sodium chloride, although not necessarily in that order. The following table summarizes the results of the series of analyses you have performed on the contents: Match each vial with its contents. 2. Do ionic solids generally have higher or lower melting points than molecular solids? Why? Do ionic solids generally have higher or lower melting points than covalent solids? Explain your reasoning. 3. The strength of London dispersion forces in molecular solids tends to increase with molecular mass, causing a smooth increase in melting points. Some molecular solids, however, have significantly lower melting points than predicted by their molecular masses. Why? 4. Suppose you want to synthesize a solid that is both heat resistant and a good electrical conductor. What specific types of bonding and molecular interactions would you want in your starting materials? 5. Explain the differences between an interstitial alloy and a substitutional alloy. Given an alloy in which the identity of one metallic element is known, how could you determine whether it is a substitutional alloy or an interstitial alloy? 6. How are intermetallic compounds different from interstitial alloys or substitutional alloys? 1. a. NaCl, ionic solid b. quartz, covalent solid c. zinc, metal d. sucrose, molecular solid 5. In a substitutional alloy, the impurity atoms are similar in size and chemical properties to the atoms of the host lattice; consequently, they simply replace some of the metal atoms in the normal lattice and do not greatly perturb the structure and physical properties. In an interstitial alloy, the impurity atoms are generally much smaller, have very different chemical properties, and occupy holes between the larger metal atoms. Because interstitial impurities form covalent bonds to the metal atoms in the host lattice, they tend to have a large effect on the mechanical properties of the metal, making it harder, less ductile, and more brittle. Comparing the mechanical properties of an alloy with those of the parent metal could be used to decide whether the alloy were a substitutional or interstitial alloy. 1. Will the melting point of lanthanum(III) oxide be higher or lower than that of ferrous bromide? The relevant ionic radii are as follows: La , 104 pm; O , 132 pm; Fe , 83 pm; and Br , 196 pm. Explain your reasoning. 2. Draw a graph showing the relationship between the electrical conductivity of metallic silver and temperature. 3. Which has the higher melting point? Explain your reasoning in each case. a. Os or Hf b. SnO or ZrO c. Al O or SiO 4. Draw a graph showing the relationship between the electrical conductivity of a typical semiconductor and temperature. 3. a. Osmium has a higher melting point, due to more valence electrons for metallic bonding. b. Zirconium oxide has a higher melting point, because it has greater ionic character. c. Aluminum oxide has a higher melting point, again because it has greater ionic character. | 27,217 | 1,662 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Reactions_of_the_Hexaaqua_Ions_with_Ammonia |
This page describes and explains the reactions between complex ions of the type [M(H O) ] and ammonia solution.Reactions of the hexaaqua ions with ammonia solution are complicated by the fact that the ammonia can have two quite different functions. It can act as a base (in the Brønsted-Lowry sense), but it is also a possible ligand which can replace water molecules around the central metal ion. When it acts as a ligand, it is acting as a Lewis base. We need to look at these two functions separately. This is what happens when you only add small amounts of dilute ammonia solution to any of the hexaaqua ions. The ligand effect only happens with an excess of ammonia or with concentrated ammonia - and with some metals you don't even see it then. We'll talk through what happens if you add a small amount of dilute ammonia solution to a solution containing a 2+ hexaaqua ion. These have the formula [M(H O) ] , and they are acidic. Their acidity is shown in the reaction of the hexaaqua ions with water molecules from the solution: \[ \ce{ [M(H2O)6]^{2+} (aq) + H2O <=> [M(H2O)5(OH)]^{+} (aq) + H3O^{+} (aq) }\] They are acting as acids by donating hydrogen ions to water molecules in the solution. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify this: \[ \ce{ [M(H2O)6]^{2+} (aq) <=> [M(H2O)5(OH)]^{+} (aq) + H^{+} (aq) }\] There are two possible reactions. According to Le Chatelier's Principle, the position of equilibrium will move to the right, producing more of the new complex ion. Statistically, there is far more chance of an ammonia molecule hitting a hexaaqua metal ion than of hitting a hydrogen ion. There are far more hexaaqua ions present. If that happens, you get exactly the same new complex ion formed as above. \[ \ce{ [M(H2O)6]^{2+} (aq) + NH3 <=> [M(H2O)5(OH)]^{+} (aq) + NH4^{+} (aq) }\] Notice that this is still a reversible change (unlike the corresponding change when you add hydroxide ions). Ammonia is only a weak base. Whichever of the above reactions happens, you end up with [M(H O) (OH)] ions in solution. These are also acidic, and can lose hydrogen ions from another of the water ligands. Taking the easier version of the equilibrium: \[ \ce{ [M(H2O)5(OH)]^{+} (aq) <=> [M(H2O)4(OH)2] (s) + H^{+} (aq) }\] Adding ammonia again tips the equilibrium to the right - either by reacting with the hydrogen ions, or by reacting directly with the complex on the left-hand side. When this happens, the new complex formed no longer has a charge - this is a "neutral complex". It is insoluble in water - and so a precipitate is formed. This precipitate is often written without including the remaining water ligands. In other words we write it as M(OH) . A precipitate of the metal hydroxide has been formed. You can also usefully write the complete change as an overall equilibrium reaction. This will be important for later on. \[ \ce{ [M(H2O)6]^{2+} (aq) + 2NH3 <=> [M(H2O)4(OH)2] (s) + 2NH4^{+} (aq) }\] If you did the same reaction with a 3+ ion, the only difference is that you would have to remove a total of 3 hydrogen ions in order to get to the neutral complex. That would give the overall equation: \[ \ce{ [M(H2O)6]^{3+} (aq) + 3NH3 <=> [M(H2O)3(OH)3] (s) + 4NH4^{+} (aq) }\] Remember that we are concentrating for the moment on the ammonia acting as a base - in other words, on the formation of hydroxide precipitates when you add small amounts of ammonia solution to solutions containing hexaaqua metal The diagrams, however, will show the complete change so I don't have to repeat them later on. Ignore the cases where the precipitate dissolves in excess ammonia for the moment. Iron is very easily oxidized under alkaline conditions. Oxygen in the air oxidizes the iron(II) hydroxide precipitate to iron(III) hydroxide especially around the top of the tube. The darkening of the precipitate comes from the same effect. This is NOT a ligand exchange reaction. I have shown the original solution as very pale pink (the palest I can produce!), but in fact it is virtually colorless. The pale brown precipitate is oxidized to darker brown manganese(III) oxide in contact with oxygen from the air. Again, this isn't a ligand exchange reaction. You start and finish with colourless solutions, producing a white precipitate on the way. Starting from a colorless solution, you get a white precipitate. In each case you get a precipitate of the neutral complex - the metal hydroxide. Apart from minor differences in the exact shade of color you get, these are almost all exactly the same as the precipitates you get when you add a little sodium hydroxide solution to the solutions of the hexaaqua ions. The only real difference lies in the color of the cobalt precipitate. In some cases, ammonia replaces water around the central metal ion to give another soluble complex. This is known as a ligand exchange reaction, and involves an equilibrium such as this one: \[ [Cu(H_2O)_6]^{2+} +4NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O\] The formation of this new soluble complex causes the precipitate to dissolve. The ammonia attaches to the central metal ion using the lone pair of electrons on the nitrogen atom. Because it is a lone pair donor, it is acting as a Lewis base. Almost all text books leave the argument at this point, assuming that it is obvious why the formation of the complex causes some precipitates to dissolve - it is not! Consider the copper case as typical of any of them. There are two equilibria involved in this. The first is the one in which ammonia is acting as a base and producing the precipitate: \[ [Cu(H_2O)_6]^{2+} + 2NH_3 \rightleftharpoons [Cu(H_2O)_4(OH)_2] + 2NH_4^+\] The other one is the ligand exchange reaction: \[ [Cu(H_2O)_6]^{2+} + 4NH_3 \rightleftharpoons [Cu(NH_3)_4(H_2O)_2]^{2+} + 4H_2O\] Notice that the hexaaqua ion appears in both of these. There is now an interaction between the two equilibria: Looking at it like this is helpful in explaining why some precipitates dissolve in excess ammonia while others don't. It depends on the positions of the equilibria. To get the precipitate to dissolve, you obviously need the ligand exchange equilibrium to lie well to the right, but you need the acid-base equilibrium to be easy to pull back to the left. Jim Clark ( ) | 6,357 | 1,664 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.11%3A_Bond_Enthalpies_and_Exothermic_or_Endothermic_Reactions |
We are in a position to appreciate the general principle which determines whether a gaseous reaction will be exothermic or endothermic. If less energy is needed to break up the reactant molecules into their constituent atoms than is released when these atoms are reconstituted into product molecules, then the reaction will be exothermic. Usually an exothermic reaction corresponds to the breaking of weak bonds (with small ) and the making of strong bonds (with large ). In the hydrogen-fluorine reaction ( ) one quite strong bond (D = 436 kJ mol ) and one very weak bond (D = 159 kJ mol ) are broken, while two very strong H―F bonds are made. (Note that with a bond enthalpy of 566 kJ mol , H―F is the strongest of all single bonds.) The combustion of methane discussed in is another example of the formation of stronger bonds at the expense of weaker ones. The bond enthalpy of the O―H bond is not much different in magnitude from those of the C―H and O=O bonds which it replaces: All lie between 400 and 500 kJ mol . The determining factor making this reaction exothermic is the exceedingly large bond enthalpy of the C=O bond which at 803 kJ mol is almost twice as great as for the other bonds involved in the reaction. Not only this reaction but reactions in which CO with its two very strong C=O bonds is produced are exothermic. In other cases the number of bonds broken or formed can be important. A nice example of this is the highly exothermic (Δ °(298 K) = –483.7 kJ mol ) reaction between hydrogen and oxygen to form water: All three types of bonds involved have comparable bond enthalpies: but the reason the reaction is exothermic becomes obvious if we rewrite it to make the bonds visible: While three bonds must be broken (two H―H and one O=O bond), a total of four bonds are made (four O―H bonds). Since all the bonds are similar in strength, making more bonds than are broken means the release of energy. In mathematical terms In summary, there are two factors which determine whether a gaseous reaction will be exothermic or not: (1) the relative of the bonds as measured by the bond enthalpies, and (2) the relative of bonds broken and formed. An exothermic reaction corresponds to the formation of bonds, bonds, or both. Since the strength of chemical bonds is a factor in determining whether a reaction will release energy or not, it is obviously important to know which kinds of bonds will be strong and which weak, and we can make some empirical generalizations about the magnitudes of bond enthalpies. The first and most obvious of these is that triple bonds are stronger than double bonds which in turn are stronger than single bonds. As can be seen from , triple bonds have bond enthalpies in the range of 800 to 1000 kJ mol . Double bonds range between 400 and 800 kJ mol and single bonds are in the range of 150 to 500 kJ mol . A second generalization is that the strengths of bonds usually increase with . The bond enthalpies of the hydrogen halides, for instance, increase in the order HI < HBr < HCl < HF, and a similar order can be noted for bonds between carbon and halogens. There are exceptions to this rule, though. One would expect the C―N bond to be intermediate in strength between the C―C bond and the C―O bond. As the table shows, it is actually weaker than either. A third factor affecting the strength of bonds is the size of the atoms. For the most part smaller atoms form stronger bonds. The smallest atom, hydrogen, forms four of the five strongest single bonds in the table. This is not entirely a matter of size since hydrogen is also the most electropositive element featured. Difference in , however, does not explain why the H―H bond enthalpy is so large. If we look at the halogens (VIIA elements), we find that the bond enthalpies of I―I, Br―Br, and Cl―Cl increase as expected with decreasing size. The F―F bond is an exception, though: D (158 kJ mol ) is significantly than D (242 kJ mol ). Other notable exceptions to this rule are the N―N and O―O bonds. The occurrence of these weak single bonds is of considerable importance to the chemistry of compounds which contain them.The value of a particular bond enthalpy in a given molecule can sometimes be very informative about the nature of the bonding in the molecule. This is especially true of molecules in which resonance is a possibility, as the following example shows. When benzene is burned in oxygen according to the equation Indicating the required bond enthalpy by the symbol D and carefully counting the bonds broken and bonds formed, we have \[–3169 \text{kJ mol}^{-1} = 6 (416 \text{kJ mol}^{-1}) + 6D_{CC} + \frac{15}{2} (498 \text{kJ mol}^{-1}) – 12 (803 \text{mol}^{-1}) – 6 (467 \text{kJ mol}^{-1}) = –6207 \text{kJ mol}^{-1} + 6D_{CC} \nonumber \] or \[D_{CC} = 506 \text{kJ mol}^{-1} \nonumber \] | 4,885 | 1,665 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Enthalpy/Hydration |
The formation of a solution involves the interaction of solute with solvent molecules. Many different liquids can be used as solvents for liquid solutions, and water is the most commonly used solvent. When water is used as the solvent, the dissolving process is called . The interaction between water molecules and sodium ion is illustrated as one of the diagram below. This is a typical . At the molecular level, the ions interact with water molecules from all directions in a 3-dimensional space. This diagram depicts the concept of interaction only. The above diagram also display , and interactions. In the absence of these interactions, solvation takes place due to . Definitions of these terms are obvious from the diagrams. The meaning of the words used in the term also hints the nature of the interactions. Enthalpy of hydration, \(\Delta H_{hyd}\), of an ion is the amount of heat released when a mole of the ion dissolves in a large amount of water forming an infinite dilute solution in the process, \[M^{z+}_{(g)} + mH_2O \rightarrow M^{z+}_{(aq)} \label{1}\] where M (aq) represents ions surrounded by water molecules and dispersed in the solution. The approximate hydration energies of some typical ions are listed here. Figure \(\Page {1}\) illustrates the point that as the atomic numbers increases, so do the ionic size, leading to a decrease in absolute values of enthalpy of hydration. From the above table, an estimate can be made for the hydration energy of sodium chloride. The hydration energy of an ionic compound consists of two inseparable parts. The first part is the energy released when the solvent forms a coordination compound with the ions. This energy released is called the , \(\Delta H_{lig}\). The processes related to these energies are shown below: \[M^{z+} + nL \rightarrow ML_n^{z+} \;\;\; \Delta H_{lig} \label{2}\] \[ML_n^{z+} + solvent \rightarrow ML^{z+}_{n(sol)} \;\;\; \Delta H_{disp} \label{3}\] The second step is to disperse the ions or hydrated ions into the solvent medium, which has a dielectric constant different from vacuum. This amount of energy is called , \(\Delta H_{disp}\). Therefore, \[\Delta H_{hyd} = \Delta H_{disp} + \Delta H_{lig} \label{4}\] This idea is brought up just to point out that the formation of complex ions is part of the hydration process, even though the two energies are not separable. When stronger coordination is made between the ions and other ligands, they replace the coordinated water molecules if they are present. In the presence of NH molecules, they replace the water of Cu(H O) : \[Cu(H_2O)_6^{2+} + 6NH_3 \rightarrow Cu(NH_3)_6^{2+} + 6H_2O \label{5}\] In the discussion of lattice energy, we consider the ions separated into a gas form whereas in the dissolution process, the ions are also separated, but this time into ions dispersed in a medium with solvent molecules between ions. The medium or solvent has a dielectric constant. The molar enthalpy of solution, \(\Delta H_{sol}\), is the energy released when one mole solid is dissolved in a solvent. This quantity, the enthalpy of crystallization, and energy of hydration forms a cycle. Taking the salt \(\ce{NaCl}\) as an example, the following relationship is obvious, \[\Delta H_{sol} =\Delta H_{lattice} + \Delta H_{hydration} \label{4A}\] from the following diagram. The term enthalpy of crystallization is used in this diagram instead of lattice energy so that all the arrows point downward. Note that enthalpy of crystallization \(H_{cryst}\), and energy of crystallization, \(E_{cryst}\) refer to the same quantity, and they are used interchangably. The enthalpies of solution for some salts can be positive values, in these cases the temperatures of the solution decrease as the substances dissolve; the dissolving is an endothermic reaction. The energy levels of solids and solutions reverse in order of height. The molar enthalpy of solution, \(\Delta H_\ce{sol}\) , is the energy released when one mole of solid is dissolved in a solvent. Sometimes the enthalpy of hydration is also (mis)understood as \(\Delta H_\ce{sol}\) . When apply these values, make sure you understand the process involved. These values indicates that when aluminum chloride and sulfuric acid are dissolved in water, much heat is released. Due to the very small value of enthalpies of solution, the temperature changes are hardly noticed when LiNO and NaCl are dissolving. The lattice energy of NaCl calculated using the of the NaCl structure type is +788 kJ/mol. The estimated enthalpy of hydration for sodium and chloride ions are -406 and -363 kJ/mol respectively. Estimate the enthalpy of solvation for NaCl. Using the cycle in Figure \(\Page {2}\) and Equation \ref{4A}, we have \[\Delta H_{hyd} = \Delta H_{lattice} + \Delta H_{sol} \nonumber\] \[\Delta H_{sol} = -769 - (788)\; kJ = -19\, kJ/mol \nonumber\] A positive value indicates an endothermic reaction. However, the value is small, and depending on the source of data, the estimated value may change. This value of 19 kJ/mol is too high compared to the value given earlier for NaCl of 3.88 kJ/mol, due to a high value of lattice energy used. The enthalpy of crystallization for KCl is -715 kJ/mol. The enthalpies of hydration for potassium and chloride are -322 and -363 kJ/mol respectively. From these values, estimate the enthalpy of solution for KCl. The enthalpy of hydration for KCl is estimated to be \[\Delta H_{hyd}= -322 + (-363) = -685 kJ/mol \nonumber\] Thus, the enthalpy of solution is \[\Delta H_{sol}= -685 - (-715) = 30 kJ/mol \nonumber\] The enthalpy of solution given above is -17.22. The two values here indicates that dissolving \(\ce{KCl}\) into water is an endothermic reaction or change. Should temperature decrease or increase when \(\ce{KCl}\) dissolves? | 5,811 | 1,666 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Group_Theory/Group_Theory_and_its_Application_to_Chemistry |
Group Theory is the mathematical application of symmetry to an object to obtain knowledge of its physical properties. What group theory brings to the table, is how the symmetry of a molecule is related to its physical properties and provides a quick simple method to determine the relevant physical information of the molecule. The symmetry of a molecule provides you with the information of what energy levels the orbitals will be, what the orbitals symmetries are, what transitions can occur between energy levels, even bond order to name a few can be found, all without rigorous calculations. The fact that so many important physical aspects can be derived from symmetry is a very profound statement and this is what makes group theory so powerful. To a fully understand the math behind group theory one needs to take a look at the theory portion of the Group Theory topic or refer to one of the reference text listed at the bottom of the page. Never the less as Chemist the object in question we are examining is usually a molecule. Though we live in the 21 century and much is known about the physical aspects that give rise to molecular and atomic properties. The number of high level calculations that need to be performed can be both time consuming and tedious. To most experimentalist this task is takes away time and is usually not the integral part of their work. When one thinks of group theory applications one doesn't necessarily associated it with everyday life or a simple toy like a Rubik's cube. A Rubik's cube is an a cube that has a \(3 \times 3\) array of different colored tiles on each of its six surfaces, for a total of 54 tiles. Since the cube exist in 3D space, the three axis are \(x\), \(y\), \(z\). Since the rubik's cube only allows rotation which are called operations, there are three such operations around each of the \(x\), \(y\), \(z\) axis. Of course the ultimate challenge of a Rubik's cube is to place all six colors on each of the six faces. By performing a series of such operations on the Rubik's cube one can arrive at a solution (A link of a person solving a Rubik's cube in 10.4s with operations performed noted, the operations performed will not translate to chemistry applications but it is a good example of how symmetry operations arrive at a solution). The operations shown in the Rubik's cube case are inherent to the make up of the cube, i.e., the only operations allowed are the rotations along the x, y, z axis. Therefore the Rubik's cube only has x,y,z rotation operations. Similarly the operations that are specific to a molecule are dependent on its symmetry. These operations are given in the top row of the character table. The character table contains a wealth of information, for a more detailed discussion of the character table can be found in Group Theory Theoretical portion of the chemWiki. All operations in the character table are contained in the first row of the character table, in this case \(E\), \(C_3\), & \(\sigma_v\), these are all of the operations that can be preformed on the molecule that return the original structure. The first column contains the three irreducible representations from now on denoted as \(\Gamma_{ir}\), here they are \(A_1\), \(A_2\) & \(E\). The value of the \(\Gamma_{ir}\) denotes what the operation does. A value of 1 represents no change, -1 opposite change and 0 is a combination of 1 & -1 (0’s are found in degenerate molecules. The final two columns Rotation and Translation represented by \(R_x\),\(R_y\), \(R_z\) & \(x\), \(y\), \(z\) respectively. Where R's refer to rotation about an axis and the \(x\), \(y\), \(z\) refers to a translation about an axis, the \(\Gamma_{ir}\) the each \(R_x\), \(R_y\), \(R_z\) & \(x\), \(y\), \(z\) term is the irreducible symmetry of a rotation or translation operation. Like wise the final column the orbital symmetries relates the orbital wavefunction to a irreducible representation. This is a quick rule to follow for calculating Direct Products of irreproducible representations, such a calculation will be necessary for working through transition moment integrals. Following the basic rules given by the table given below. One can easily work through symmetry calculations very quickly. All molecules vibrate. While these vibrations can originate from several events, which will be covered later, the most basic of these occurs when an electron is excited within the electronic state from one eigenstate to another. The Morse potential (electronic state) describes the energy of the eigenstate as a function of the interatomic distance. When an electron is excited form one eigenstate to another within the electronic state there is a change in interatomic distance, this result in a vibration occurring. Vibrational energies arise from the absorption of polarizing radiation. Each vibrational state is assigned a \(\Gamma_{ir}\). A vibration occurs when an electron remains within the electronic state but changes from one eigenstate to another (The vibrations for the moment are only IR active vibrations, there are also Raman vibrations which will be discussed later in electronic spectroscopy), in the case of the Morse diagram above the eigenstates are denoted as \(\nu\). As you can see from the diagram the eigenstate is a function of energy versus interatomic distance. To predicting whether or not a vibrational transition, or for that matter a transition of any kind, will occur we use the transition moment integral. \[\int \Psi_i*\mu \Psi_f d\tau=\langle \Psi_i | \mu| \Psi_f \rangle\] The transition moment integral is written here in standard integral format, but this is equivalent to Bra & Ket format which is standard in most chemistry quantum mechanical text (The \(\langle \Psi_i |\) is the Bra portion, \(| \Psi_f \rangle\) is the Ket portion). The transition moment operator \(\mu\) is the operator the couples the initial state \(\Psi_i\) to the final state \(\Psi_f\), which is derived from the time independent Schrödinger equation. However using group theory we can ignore the detailed mathematical methods. We can use the \(\Gamma_{ir}\) of the vibrational energy levels and the symmetry of the transition moment operator to find out if the transition is allowed by selection rules. The selection rules for vibrations or any transition is that is allowed, for it to by allowed by group theory the answer , which is always the first \(\Gamma_{ir}\) in the character table for the molecule in question. Let’s work through an example: Ammonia (\(NH_3\)) with a \(C_{3v}\) symmetry. Consequently, all of the properties contained in the \(C_{3v}\) character table above are pertinent to the ammonia molecule. The principle axis is the axis that the highest order rotation can be preformed. In this case the z-axis pass through the lone pairs (pink sphere), which contains a \(C_3\) axis. The ?’s or mirror planes (\(\sigma_v\) parallel to z-axis & \(\sigma_h\) perpendicular to the z-axis). In ammonia there is no \(\sigma_h\) only three \(\sigma_v\)’s. The combination of \(C_3\) & \(\sigma_v\) leads to \(C_{3v}\) point group, which leads to the C3v character table. The number of transitions is dictated by 3N-6 for non-linear molecules and 3N-5 for linear molecules, where N is the number of atoms. The 6 & the 5 derive from three translations in the x,y,z plan and three rotations also in the x,y,z plan. Where a linear molecule only has two rotations in the x & y plans since the z axis has infinite rotation. This leads to only 5 degrees of freedom in the rotation and translation operations. In the case of Ammonia there will be 3(4)-6=6 vibrational transitions. This can be confirmed by working through the vibrations of the molecule. This work is shown in the table below. The vibrations that are yielded 2A1 & 2E (where E is doubly degenerate, meaing two vibration modes each) which total 6 vibrations. This calculation was done by using the character table to find out the rotation and translation values and what atoms move during each operation. Using the character table we can characterize the A1 vibration as IR active along the z-axis and raman active as well. The E vibration is IR active along both the x & y axis and is Raman active as well. From the character table the IR symmetries correspond to the x, y & z translations. Where the Raman active vibrations correspond to the symmetries of the d-orbitals. Infrared Spectroscopy (IR) measures the vibrations that occur within a single electronic state, such as the one shown above. Because the transition occurs within a single electronic state there is a variation in interatomic distance. The dipole moment is dictate by the equation. Where the magnitude of dipole moment; is the polarizability constant (actually a tensor) & is the magnitude of the electric field which can be described as the electronegitivity. Therefore when a vibration occurs within a single electronic state there is a change in the dipole moment, which is the definition of an active IR transition. \[ \left ( \frac{\mathrm{d\mu} }{\mathrm{d} q} \right )_{eq} \neq 0 \] In terms of group theory a change in the dipole is a change from one vibrational state to another, as shwon by the equation above. A picture of the vibrational states with respect ot the rotational states and electronic states is given below. In IR spectroscopy the transition occurs only from on vibrational state to another all within the same electronic state, shown below as B. When an electron is excited from one electronic state to another, this is what is called an electronic transition. A clear example of this is part C in the energy level diagram shown above. Just as in a vibrational transition the selection rules for electronic transitions are dictated by the transition moment integral. However we now must consider both the electronic state symmetries and the vibration state symmetries since the electron will still be coupled between two vibrational states that are between two electronic states. This gives us this modified transition moment integral: Where you can see that the symmetry of the initial electronic state & vibrational state are in the Bra and the final electronic and vibrational states are in the Ket. Though this appears to be a modified version of the transition moment integral, the same equation holds true for a vibrational transition. The only difference would be the electronic state would be the same in both the initial and final states. Which the dot product of yields the totally symmetric representation, making the electronic state irrelevant for purely vibrational spectroscopy. In transition that occurs is the excitation from one electronic state to another and the selection rules are dictated by the transition moment integral discussed in the electronic spectroscopy segment. However mechanically Raman does produce a vibration like IR, but the selection rules for Raman state there must be a change in the polarization, that is the volume occupied by the molecule must change. But as far as group theory to determine whether or not a transition is allowed one can use the transition moment integral presented in the electronic transition portion. Where one enters the starting electronic state symmetry and vibrational symmetry and final electronic state symmetry and vibrational state, perform the direct product with the different M's or polarizing operators For more information about this topic please explore the Raman spectroscopy portion of the Chemwiki For the purposes of Group Theory Raman and Fluorescence are indistinguishable. They can be treated as the same process and in reality they are quantum mechanically but differ only in how Raman photons scatter versus those of fluorescence. is the same as fluorescence except upon excitation to a singlet state there is an interconversion step that converts the initial singlet state to a triplet state upon relaxation. This process is longer than fluorescence and can last microseconds to several minutes. However despite the singlet to triplet conversion the transition moment integral still holds true and the symmetry of ground state and final state still need to contain the totally symmetric representation. Molecular Orbitals also follow the symmetry rules and indeed have their own ?ir. Below are the pi molecular orbitals for trans-2-butene and the corresponding symmetry of each molecular orbital. The ?ir of the molecular orbitals are created by simply preforming the operations of that molecule's character table on that orbital. In the case of trans-2-butene the point group is C2h, the operations are: E, C2, i & ?h. Each operation will result in a change in phase (since were dealing with p-orbitals) or it will result in no change. The first molecular orbital results in the totally symmetric representation, working through all four operations E, C2, i, ?h will only result in 1's meaning there is no change, giving the Ag symmetry state. These molecular orbitals also represent different electronic states and can be arranged energetically. Putting the orbital that has the lowest energy, the orbital with the fewest nodes at the bottom of the energy diagram and like wise working up form lowest energy to highest energy. The highest energy orbital will have the most nodes. Once you've set up your MO diagram and place the four pi electrons in the orbitals you see that the first two orbitals listed (lowest energy) are HOMO orbitals and the bottom two (highest energy) and LUMO orbitals. With this information if you have a transition from the totally symmetric HOMO orbital to the totally symmetric LUMO orbital the transition moment operator would need to have Ag symmetry (using the C2h) to give a result containing the totally symmetric representation. These four molecular orbitals represent four different electronic states. So transitions from one MO into another would be something that is measured typically with UV-Vis spectrometer. | 14,004 | 1,667 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/10.3%3A_Polar_Covalent_Bonds_and_Electrostatic_Potential_Maps |
The electron pairs shared between two atoms For example, while the shared electron pairs is shared equally in the covalent bond in \(Cl_2\), in \(NaCl\) the 3s electron is stripped from the Na atom and is incorporated into the electronic structure of the Cl atom - and the compound is most accurately described as consisting of individual \(Na^+\) and \(Cl^-\) ions (ionic bonding). For most covalent substances, their bond character falls these two extremes. We demonstrated below, the is a useful concept for describing the sharing of electrons between atoms within a covalent bond: The elements with the highest ionization energies are generally those with the most negative electron affinities, which are located toward the upper right corner of the periodic table (compare Figure \(\Page {2}\) and Figure \(\Page {2}\)). Conversely, the elements with the lowest ionization energies are generally those with the least negative electron affinities and are located in the lower left corner of the periodic table. Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativity (represented by the Greek letter , χ, pronounced “ky” as in “sky”), defined as the ability of an atom to attract electrons to itself . Elements with high electronegativities tend to acquire electrons in chemical reactions and are found in the upper right corner of the periodic table. Elements with low electronegativities tend to lose electrons in chemical reactions and are found in the lower left corner of the periodic table. Unlike ionization energy or electron affinity, the electronegativity of an atom is not a simple, fixed property that can be directly measured in a single experiment. In fact, an atom’s electronegativity should depend to some extent on its chemical environment because the properties of an atom are influenced by its neighbors in a chemical compound. Nevertheless, when different methods for measuring the electronegativity of an atom are compared, they all tend to assign similar relative values to a given element. For example, all scales predict that fluorine has the highest electronegativity and cesium the lowest of the stable elements, which suggests that all the methods are measuring the same fundamental property. Electronegativity is defined as the ability of an atom in to attract electrons to itself. The the value, the the attractiveness for electrons. Electronegativity is a function of: (1) the atom's (how strongly the atom holds on to its own electrons) and (2) the atom's (how strongly the atom attracts other electrons). Both of these are properties of the atom. An element that is will be has: and will The original electronegativity scale, developed in the 1930s by Linus Pauling (1901– 1994) was based on measurements of the strengths of covalent bonds between different elements. Pauling arbitrarily set the electronegativity of fluorine at 4.0 (although today it has been refined to 3.98), thereby creating a scale in which all elements have values between 0 and 4.0. Periodic variations in Pauling’s electronegativity values are illustrated in Figure \(\Page {1}\) and Figure \(\Page {2}\). If we ignore the inert gases and elements for which no stable isotopes are known, we see that fluorine (\(\chi = 3.98\)) is the most electronegative element and cesium is the least electronegative nonradioactive element (\(\chi = 0.79\)). Because electronegativities generally increase diagonally from the lower left to the upper right of the periodic table, elements lying on diagonal lines running from upper left to lower right tend to have comparable values (e.g., O and Cl and N, S, and Br). Pauling won two Nobel Prizes, one for chemistry in 1954 and one for peace in 1962. When he was nine, Pauling’s father died, and his mother tried to convince him to quit school to support the family. He did not quit school but was denied a high school degree because of his refusal to take a civics class. Pauling’s method is limited by the fact that many elements do not form stable covalent compounds with other elements; hence their electronegativities cannot be measured by his method. Other definitions have since been developed that address this problem (e.g., the ). An element’s electronegativity provides us with a single value that we can use to characterize the chemistry of an element. Elements with a high electronegativity (χ ≥ 2.2) have very negative affinities and large ionization potentials, so they are generally nonmetals and electrical insulators that tend to gain electrons in chemical reactions (i.e., they are ). In contrast, elements with a low electronegativity (\(\chi \le 1.8\)) have electron affinities that have either positive or small negative values and small ionization potentials, so they are generally metals and good electrical conductors that tend to lose their valence electrons in chemical reactions (i.e., they are ). In between the metals and nonmetals, along the heavy diagonal line running from B to At is a group of elements with intermediate electronegativities (χ ~ 2.0). These are the (or metalloids), elements that have some of the chemical properties of both nonmetals and metals. The distinction between metals and nonmetals is one of the most fundamental we can make in categorizing the elements and predicting their chemical behavior. Figure \(\Page {3}\) shows the strong correlation between electronegativity values, metallic versus nonmetallic character, and location in the periodic table. Electronegativity values from lower left to upper right in the periodic table. The are based on the relative electronegativities of the elements; the more electronegative element in a binary compound is assigned a negative oxidation state. As we shall see, electronegativity values are also used to predict bond energies, bond polarities, and the kinds of reactions that compounds undergo. On the basis of their positions in the periodic table, arrange Cl, Se, Si, and Sr in order of increasing electronegativity and classify each as a metal, a nonmetal, or a semimetal. four elements order by increasing electronegativity and classification Electronegativity increases from lower left to upper right in the periodic table (Figure \(\Page {2}\)). Because Sr lies far to the left of the other elements given, we can predict that it will have the lowest electronegativity. Because Cl lies above and to the right of Se, we can predict that χ > χ . Because Si is located farther from the upper right corner than Se or Cl, its electronegativity should be lower than those of Se and Cl but greater than that of Sr. The overall order is therefore χ < χ < χ < χ . To classify the elements, we note that Sr lies well to the left of the diagonal belt of semimetals running from B to At; while Se and Cl lie to the right and Si lies in the middle. We can predict that Sr is a metal, Si is a semimetal, and Se and Cl are nonmetals. On the basis of their positions in the periodic table, arrange Ge, N, O, Rb, and Zr in order of increasing electronegativity and classify each as a metal, a nonmetal, or a semimetal. Rb < Zr < Ge < N < O; metals (Rb, Zr); semimetal (Ge); nonmetal (N, O) Electronegativity: The two idealized extremes of chemical bonding: (1) ionic bonding—in which one or more electrons are transferred completely from one atom to another, and the resulting ions are held together by purely electrostatic forces—and (2) covalent bonding, in which electrons are shared equally between two atoms. Most compounds, however, have polar covalent bonds, which means that electrons are shared unequally between the bonded atoms. Figure \(\Page {4}\) compares the electron distribution in a polar covalent bond with those in an ideally covalent and an ideally ionic bond. Recall that a lowercase Greek delta (\(\delta\)) is used to indicate that a bonded atom possesses a partial positive charge, indicated by \(\delta^+\), or a partial negative charge, indicated by \(\delta^-\), and a bond between two atoms that possess partial charges is a polar bond. The polarity of a bond—the extent to which it is polar—is determined largely by the relative electronegativities of the bonded atoms. Electronegativity (χ) was defined as the ability of an atom in a molecule or an ion to attract electrons to itself. Thus there is a direct correlation between electronegativity and bond polarity. A bond is if the bonded atoms have equal electronegativities. If the electronegativities of the bonded atoms are not equal, however, the bond is toward the more electronegative atom. A bond in which the electronegativity of B (χ ) is greater than the electronegativity of A (χ ), for example, is indicated with the partial negative charge on the more electronegative atom: \( \begin{matrix}
_{less\; electronegative}& & _{more\; electronegative}\\
A\; \; &-& B\; \; \; \; \\
^{\delta ^{+}} & & ^{\delta ^{-}}
\end{matrix} \label{10.3.1} \) One way of estimating the ionic character of a bond—that is, the magnitude of the charge separation in a polar covalent bond—is to calculate the difference in electronegativity between the two atoms: Δχ = χ − χ . To predict the polarity of the bonds in Cl , HCl, and NaCl, for example, we look at the electronegativities of the relevant atoms: χ = 3.16, χ = 2.20, and χ = 0.93 (see Figure \(\Page {2}\)). Cl must be nonpolar because the electronegativity difference (Δχ) is zero; hence the two chlorine atoms share the bonding electrons equally. In NaCl, Δχ is 2.23. This high value is typical of an ionic compound (Δχ ≥ ≈1.5) and means that the valence electron of sodium has been completely transferred to chlorine to form Na and Cl ions. In HCl, however, Δχ is only 0.96. The bonding electrons are more strongly attracted to the more electronegative chlorine atom, and so the charge distribution is \( \begin{matrix}
_{\delta ^{+}}& & _{\delta ^{-}}\\
H\; \; &-& Cl
\end{matrix} \) Remember that electronegativities are difficult to measure precisely and different definitions produce slightly different numbers. In practice, the polarity of a bond is usually estimated rather than calculated. Bond polarity and ionic character increase with an increasing difference in electronegativity. As with bond energies, the electronegativity of an atom depends to some extent on its chemical environment. It is therefore unlikely that the reported electronegativities of a chlorine atom in NaCl, Cl , ClF , and HClO would be exactly the same. The asymmetrical charge distribution in a polar substance such as HCl produces a \( Qr \) is abbreviated by the Greek letter mu (µ). The dipole moment is defined as the product of the partial charge on the bonded atoms and the distance between the partial charges: \[ \mu=Qr \label{10.3.2} \] where is measured in coulombs (C) and in meters. The unit for dipole moments is the debye (D): \[ 1\; D = 3.3356\times 10^{-30}\; C\cdot ·m \label{10.3.3} \] When a molecule with a dipole moment is placed in an electric field, it tends to orient itself with the electric field because of its asymmetrical charge distribution (Figure \(\Page {2}\)). We can measure the partial charges on the atoms in a molecule such as HCl using Equation 10.3.2 If the bonding in HCl were purely ionic, an electron would be transferred from H to Cl, so there would be a full +1 charge on the H atom and a full −1 charge on the Cl atom. The dipole moment of HCl is 1.109 D, as determined by measuring the extent of its alignment in an electric field, and the reported gas-phase H–Cl distance is 127.5 pm. Hence the charge on each atom is \[ Q=\dfrac{\mu }{r} =1.109\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{127.8\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right )=2.901\times 10^{-20}\;C \label{10.3.4} \] By dividing this calculated value by the charge on a single electron (1.6022 × 10 C), we find that the electron distribution in HCl is asymmetric and that effectively it appears that there is a net negative charge on the Cl of about −0.18, effectively corresponding to about 0.18 e . This certainly does not mean that there is a fraction of an electron on the Cl atom, but that the distribution of electron probability favors the Cl atom side of the molecule by about this amount. \[ \dfrac{2.901\times 10^{-20}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}}=0.1811\;e^{-} \label{10.3.5} \] To form a neutral compound, the charge on the H atom must be equal but opposite. Thus the measured dipole moment of HCl indicates that the H–Cl bond has approximately 18% ionic character (0.1811 × 100), or 82% covalent character. Instead of writing HCl as \( \begin{matrix}
_{\delta ^{+}}& & _{\delta ^{-}}\\
H\; \; &-& Cl
\end{matrix} \) we can therefore indicate the charge separation quantitatively as \( \begin{matrix}
_{0.18\delta ^{+}}& & _{0.18\delta ^{-}}\\
H\; \; &-& Cl
\end{matrix} \) Our calculated results are in agreement with the electronegativity difference between hydrogen and chlorine χ = 2.20; χ = 3.16, χ − χ = 0.96), a value well within the range for polar covalent bonds. We indicate the dipole moment by writing an arrow above the molecule. In HCl, for example, the dipole moment is indicated as follows: The arrow shows the direction of electron flow by pointing toward the more electronegative atom. The charge on the atoms of many substances in the gas phase can be calculated using measured dipole moments and bond distances. Figure \(\Page {6}\) shows a plot of the percent ionic character versus the difference in electronegativity of the bonded atoms for several substances. According to the graph, the bonding in species such as NaCl(g) and CsF(g) is substantially less than 100% ionic in character. As the gas condenses into a solid, however, dipole–dipole interactions between polarized species increase the charge separations. In the crystal, therefore, an electron is transferred from the metal to the nonmetal, and these substances behave like classic ionic compounds. The data in Figure \(\Page {6}\) show that diatomic species with an electronegativity difference of less than 1.5 are less than 50% ionic in character, which is consistent with our earlier description of these species as containing polar covalent bonds. The use of dipole moments to determine the ionic character of a polar bond is illustrated in Example 11. In the gas phase, NaCl has a dipole moment of 9.001 D and an Na–Cl distance of 236.1 pm. Calculate the percent ionic character in NaCl. chemical species, dipole moment, and internuclear distance percent ionic character Compute the charge on each atom using the information given and Equation 10.3.2. Find the percent ionic character from the ratio of the actual charge to the charge of a single electron. The charge on each atom is given by \[ Q=\dfrac{\mu }{r} =9.001\;\cancel{D}\left ( \dfrac{3.3356\times 10^{-30}\; C\cdot \cancel{m}}{1\; \cancel{D}} \right )\left ( \dfrac{1}{236.1\; \cancel{pm}} \right )\left ( \dfrac{1\; \cancel{pm}}{10^{-12\;} \cancel{m}} \right )=1.272\times 10^{-19}\;C \] Thus NaCl behaves as if it had charges of 1.272 × 10 C on each atom separated by 236.1 pm. The percent ionic character is given by the ratio of the actual charge to the charge of a single electron (the charge expected for the complete transfer of one electron): \[ \% \; ionic\; character=\left ( \dfrac{1.272\times 10^{-19}\; \cancel{C}}{1.6022\times 10^{-19}\; \cancel{C}} \right )\left ( 100 \right )=79.39\%\simeq 79\% \] In the gas phase, silver chloride (AgCl) has a dipole moment of 6.08 D and an Ag–Cl distance of 228.1 pm. What is the percent ionic character in silver chloride? 55.5% Dipole Intermolecular Force: Electrostatic potential maps convey information about the charge distribution of a molecule because of the properties of the nucleus and nature of electrostatic potential energy. A region of higher than average electrostatic potential energy indicates the presence of a stronger positive charge or a weaker negative charger. Given the positive charge of the nuclei, the higher potential energy value indicates the absence of negative charges (less screening of the nuclei), which would mean that there are fewer electrons in this region. The converse is also true with a low electrostatic potential indicateing an abundance of electrons. This property of electrostatic potentials can be extrapolated to molecules as well. The first step involved in creating an electrostatic potential map is collecting a very specific type of data: electrostatic potential energy. An advanced computer program calculates the electrostatic potential energy at a set distance from the nuclei of the molecule. Electrostatic potential energy is fundamentally a measure of the strength of the nearby charges, nuclei and electrons, at a particular position. To accurately analyze the charge distribution of a molecule, a very large quantity of electrostatic potential energy values must be calculated. The best way to convey this data is to visually represent it, as in an electrostatic potential map. A computer program then imposes the calculated data onto an electron density model of the molecule. To make the electrostatic potential energy data easy to interpret, a color spectrum, with red as the lowest electrostatic potential energy value and blue as the highest, is employed to convey the varying intensities of the electrostatic potential energy values. The most important thing to consider when analyzing an electrostatic potential map is the charge distribution. The relative distributions of electrons will allow you to deduce everything you need to know from these maps. Recall the relationship between electrostatic potential and charge distribution. Areas of low potential, red, are characterized by an abundance of electrons. Areas of high potential, blue, are characterized by a relative absence of electrons. Oxygen has a higher electronegativity value than sulfur ( ), hence. oxygen atoms would have a higher electron density around them than sulfur atoms. Thus the spherical region that corresponds to an oxygen atom would have a red portion on it. Now note that there are two oxygen atoms in sulfur dioxide (Figure \(\Page {7}\)). There are two sphere shaped objects that have red regions. These areas correspond to the location of the oxygen atoms. The blue tainted sphere at the top corresponds to the location of the sulfur atom. A high electrostatic potential indicates the relative of electrons and a low electrostatic potential indicates an of electrons Electrostatic potential maps can also be used to determine the nature of the molecules chemical bond. there is a great deal of intermediary potential energy, the non red or blue regions, in this diagram. This indicates that the electronegativity difference is not very great. In a molecule with a great electronegativity difference, charge is very polarized, and there are significant differences in electron density in different regions of the molecule. This great electronegativity difference leads to regions that are almost entirely red and almost entirely blue. Greater regions of intermediary potential, yellow and green, and smaller or no regions of extreme potential, red and blue, are key indicators of a smaller electronegativity difference. The following electrostatic potential map of phosphoric acid \(\H_3PO_4\). What regions correspond to atoms of oxygen, hydrogen, and phosphorous respectively? You do not need to know the molecular structure to answer this question. You do need to know the relative electronegative values of these atoms ( ). Simply by knowing this, you can deduce that oxygen would be affiliated with the red region or redish regions of the diagram, and hydrogen would be affiliated with the blue region. Phosphorous would fall in between these two extremes, in the green region. Here is the molecular diagram of phosphoric acid: \[ \mu = Qr \label{10.3.2}\] The of an element is the relative ability of an atom to attract electrons to itself in a chemical compound and increases diagonally from the lower left of the periodic table to the upper right. The Pauling electronegativity scale is based on measurements of the strengths of covalent bonds between different atoms, whereas the Mulliken electronegativity of an element is the average of its first ionization energy and the absolute value of its electron affinity. Elements with a high electronegativity are generally nonmetals and electrical insulators and tend to behave as oxidants in chemical reactions. Conversely, elements with a low electronegativity are generally metals and good electrical conductors and tend to behave as reductants in chemical reactions. Compounds with have electrons that are shared unequally between the bonded atoms. The polarity of such a bond is determined largely by the relative electronegativites of the bonded atoms. The asymmetrical charge distribution in a polar substance produces a , which is the product of the partial charges on the bonded atoms and the distance between them. | 21,374 | 1,668 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/22%3A_Biophysical_Reaction_Dynamics/22.01%3A_Concepts_and_Definitions |
Time-dependent problems in molecular biophysics: How do molecular systems change? How does a molecular system change its microscopic configuration? How are molecules transported? How does a system sample its thermodynamically accessible states? Two types of descriptions of time-dependent processes: There is no single way to describe biophysical kinetics and dynamics, so we will survey a few approaches. The emphasis here will be on the description and analysis of time-dependent phenomena, and not on the experimental or computational methods used to obtain the data. Two common classes of problems: Now let’s start with some basic definitions of terms we will use often: Refers to many types of variables that are used to describe the structure or configuration of a system. For instance, this may refer to the positions of atoms in a MD simulation as a function of time {r ,t}, or these Cartesian variables might be transformed onto a set of internal coordinates (such as bond lengths, bond angles, and torsion angles), or these positions may be projected onto a different collective coordinate. Unlike our simple lattice models, the transformation from atomic to collective coordinate is complex when the objective is to calculate a partition function, since the atomic degrees of freedom are all correlated. Collective coordinate Example: Solvent coordinate in electron transfer. In polar solvation, the position of the electron is governed by the stabilization by the configuration of solvent dipoles. An effective collective coordinate could be the difference in electrostatic potential between the donor and acceptor sites: \(q ~ Φ_A‒Φ_D\). Example: RMSD variation of structure with coordinates from a reference state. \[ R M S D=\sqrt{\frac{1}{n} \sum_{i=1}^{n}\left(\mathbf{r}_{i}-\mathbf{r}_{i}^{0}\right)^{2}} \] where \(r\) is the position of an atom in an n atom molecule. Reaction coordinate A structure is characterized by an energy of formation. There are many forms of energy that we will use, including free energy (G, A), internal energy or enthalpy (E, H), interaction potential (U, V), ... so we will have to be careful to define the energy for a problem. Most of the time, though, we are interested in free energy. The energy landscape is used to express the relative stability of different states, the position and magnitude of barriers between states, and possible configurational entropy of certain states. It is closely related to the free energy of the system, and is often used synonymously with the potential of mean force. The energy landscape expresses how the energy of a system (typically, but it is not limited to, free energy) depends on one or more coordinates of the system. It is often used as a free energy analog of a potential energy surface. For many-particle systems, they can be presented as a reduced dimensional surface by projecting onto one or a few degrees of freedom of interest, by integrating over the remaining degrees of freedom. “Energy landscapes” represent the free energy (or rather the negative of the logarithm of the probability) along a particular coordinate. Let’s remind ourselves of some definitions. The free energy of the system is calculated from . \[ A = -k_BT \ln{Z} \] where Z is the partition function. The free energy is a number that reflects the thermally weighted number of microstates available to the system. The free energy determines the relative probability of occupying two states of the system: \[ \dfrac{P_A}{P_B} = e^{-(A_A-A_B)/k_BT} \] The energy landscape is most closely related to a potential of mean force \[ F(x) = -k_BT\ln{P(x)} \] P(x) is the probability density that reflects the probability for observing the system at a position x. As such it is equivalent to decomposing the free energy as a function of the coordinate x. Whereas the partition function is evaluated by integrating a Boltzmann weighting over all degrees of freedom, P(x) is obtained by integrating over all degrees of freedom x. We will use the term “state” in the thermodynamic sense: a distinguishable or on free energy surface. States refer to a region of phase–space where you persist long compared to thermal fluctuations. The regions where there is a high probability of observing the system. One state is distinguished from another kinetically by a time-scale separation. The rate of evolving within a state is faster than the rate of transition between states. Configuration Transition State | 4,482 | 1,669 |
Subsets and Splits