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https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Brown_et_al./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 | 4,110 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Reactive_Intermediates/Carbanions_II |
A carbanion is an anion in which carbon has an unshared pair of electrons and bears a negative charge usually with three substituents for a total of eight valence electrons.[1] The carbanion exists in a trigonal pyramidal geometry. Formally, a carbanion is the conjugate base of a carbon acid. \[\ce{R_3C-H + B^- \rightarrow R_3C^- + H-B}\] where B stands for the base. A carbanion is one of several reactive intermediates in organic chemistry. A carbanion is a nucleophile, which stability and reactivity determined by several factors: A carbanion is a reactive intermediate and is encountered in organic chemistry for instance in the E1cB elimination reaction and in organometallic chemistry in for instance a Grignard reaction or in alkyl lithium chemistry. Stable carbanions do however exist. In 1984 Olmstead presented the lithium crown ether salt of the triphenylmethyl carbanion from triphenylmethane, n-butyllithium and 12-crown-4 at low temperatures:[2] Adding n-butyllithium to triphenylmethane in THF at low temperatures followed by 12-crown-4 results in a red solution and the salt complex precipitates at −20 °C. The central C-C bond lengths are 145 pm with the phenyl ring propelled at an average angle of 31.2°. This propeller shape is less pronounced with a tetramethylammonium counterion.[3] One tool for the detection of carbanions in solution is proton NMR.[4] A spectrum of cyclopentadiene in DMSO shows four vinylic protons at 6.5 ppm and two methylene bridge protons at 3 ppm whereas the cyclopentadienyl anion has a single resonance at 5.50 ppm. Any molecule containing a C-H can lose a proton forming the carbanion. Hence any hydrocarbon containing C-H bonds can be considered an acid with a corresponding pKa value. Methane is certainly not an acid in its classical meaning yet its estimated pKa is 56. Compare this to acetic acid with pKa 4.76. The same factors that determine the stability of the carbanion also determine the order in pKa in carbon acids. These values are determined for the compounds either in water in order to compare them to ordinary acids, indimethyl sulfoxide in which the majority of carbon acids and their anions are soluble or in the gas phase. With DMSO the acidity window for solutes is limited to its own pKa of 35.5. Starting from methane in Table 1, the acidity increases: With the molecular geometry for a carbanion described as a trigonal pyramid the question is whether or not carbanions can display chirality, because if the activation barrier for inversion of this geometry is too low any attempt at introducing chirality will end inracemization, similar to the nitrogen inversion. However, solid evidence exists that carbanions can indeed be chiral for example in research carried out with certain organolithium compounds. The first ever evidence for the existence of chiral organolithium compounds was obtained in 1950. Reaction of chiral 2-iodooctane with sec-butyllithium in petroleum ether at −70 °C followed by reaction with dry ice yielded mostly racemic 2-methylbutyric acid but also an amount of optically active 2-methyloctanoic acid which could only have formed from likewise optical active 2-methylheptyllithium with the carbon atom linked to lithium the carbanion:[6] On heating the reaction to 0 °C the optical activity is lost. More evidence followed in the 1960s. A reaction of the cis isomer of 2-methylcyclopropyl bromide with sec-butyllithium again followed by carboxylation with dry ice yielded cis-2-methylcyclopropylcarboxylic acid. The formation of the trans isomer would have indicated that the intermediate carbanion was unstable.[7] In the same manner the reaction of (+)-(S)-l-bromo-l-methyl-2,2-diphenylcyclopropane with n-butyllithium followed by quench with methanol resulted in product with retention of configuration: Of recent date are chiral methyllithium compounds: The phosphate contains a chiral group with a hydrogen and a deuterium substituent. The stannyl group is replaced by lithium to intermediate which undergoes a phosphate-phosphorane rearrangement to phosphorane which on reaction with acetic acid givesalcohol . Once again in the range of −78 °C to 0 °C the chirality is preserved in this reaction sequence.[10] A carbanionic structure first made an appearance in the reaction mechanism for the benzoin condensation as correctly proposed by Clarke and Lapworth in 1907.[11] In 1904 Schlenk prepared Ph3C-NMe4+ in a quest for pentavalent nitrogen (fromTetramethylammonium chloride and Ph3CNa) [12] and in 1914 he demonstrated how triarylmethyl radicals could be reduced to carbonions by alkali metals.[13] The phrase carbanion was introduced by Wallis and Adams in 1933 as the negatively charged counterpart of the carbonium ion. [14,15] | 4,772 | 4,112 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/14%3A_Green_Chemistry_-_Protection-Free_Organic_Synthesis |
During the past two centuries, the art organic synthesis has seen enormous progress in several disciplines of organic chemistry. Soon after the strides made by stalwarts like Kekule, Fischer and Robison during late 19th century and early 20th century, chemistry witnessed the dawn of mechanistic organic chemistry. The era of spectroscopy and reagents soon followed to compliment and advance mechanistic organic chemistry and synthesis to new heights. During the middle of twentieth century, advances in Separation Science took over the classical art of purification. The last three decades of the century saw the birth of Logic in Synthesis. All these developments augured well for new developments in drug discovery leading to large-scale synthesis of several complex organic molecules. However, this hectic pace of developments in industrial chemistry has taken place at the cost of environment that is so vital for survival of life on earth. In chemical industry, this realization came about only after disasters like the one witnessed at Bhopal, India jolted the chemical world. The onus of development of new molecules rests on us (chemists) and therefore the onus of is once again on our shoulders. Clean Chemistry (meaning Green Chemistry) depends on several factors – clean starting materials, clean reagents, clean solvents, clean product, clean energy and (not the least) clean processes. While all these subtitles are discussed under Green Chemistry, we would restrict our present discussions to one aspect of the Green Process viz., protection free syntheses because this falls under our present discussions ‘Logic of Synthesis’. The Logic of Synthesis opens up a systematic methodology for the development of Synthetic Trees. When properly devised, a synthetic tree contains all plausible routes for the synthesis of the compound. The route chosen depends on the need and constraints of the developer scientist. Hence, the choice of clean chemistry starts here. The routes should be analyzed from the windows for Green Chemistry. One such green window is the lookout for a Protection Free synthetic route. An industrially feasible route has to be very efficient to be economical. By imposing such a condition - Protection Free - to the choice of routes, it should possible to arrive at such routes. Each protection / deprotection sequence adds two steps to the scheme. Thus, more the protecting groups, less is the overall yield. In addition, each protection / deprotection step adds to the byproduct load released into the environment. Thus, protection based planning has a double-edged sword hidden at every such step. To minimize loss (of precious final product) protection / deprotection sequence should therefore be treated as an avoidable technique to be used only sparingly. Always remember that protection methodology provides such a sense of security, one tends to abuse and over use such a facility. Once a chemist settles in such a comfort zone, one tends to spend less time / effort to look for alternate approaches. of routes, it should possible to arrive at such routes. Each protection / deprotection sequence adds two steps to the scheme. Thus, more the protecting groups, less is the overall yield. In addition, each protection / deprotection step adds to the byproduct load released into the environment. Thus, protection based planning has a double-edged sword hidden at every such step. To minimize loss (of precious final product) protection / deprotection sequence should therefore be treated as an avoidable technique to be used only sparingly. Always remember that protection methodology provides such a sense of security, one tends to abuse and over use such a facility. Once a chemist settles in such a comfort zone, one tends to spend less time / effort to look for alternate approaches. It should be possible to deliberately select such routes from the Synthetic Tree or device such a tree based on this additional criteria. One could select only those routes that do not call for protection or has a minimum of such techniques. Once a conscious decision is taken that protection free syntheses are in addition to being , one is more likely to come up with suitable schemes. Since industrial scale syntheses are mainly guided by cost factors, it is not surprising to find several industrial processes that are protection free syntheses. Protection / deprotection should be treated as a case of mind set rather than limitations of chemistry. Having said this in such strong terms, let us also concede that it may not be feasible to completely avoid protection / deprotection strategies. In support of these statements, we would soon look at some successful protection free synthesis in this chapter. In 2007, Phil S. Baran reported the total synthesis of some marine natural products using protecting groups. While discussing the synthesis, the authors suggest that the following guidelines would be helpful while planning retro-analysis under this philosophy. In an effort to prove their point, they have reported an efficient route for the synthesis of (+)-Ambiguine H. This synthesis is shown below . The strategic bond disconnections envisaged by these workers were guided on the above guidelines. Nature 446, 404 The proposed strategic disconnections for ambiguine H are shown below . The first target was (-)-Hapalindole U for which the t-prenyl unit is not required. The top six membered ring unit came from the well-known terpene through four steps shown in The α- position to the ketone in A was linked the C3 of indole without protection, using a reaction developed in their laboratory (J. Am. Chem. Soc., 127, 7459 (2004)) . The second coupling was attempted with several catalysts. The free radical Heck conditions failed. They led mainly to cyclisation at C2 of indole unit. Probing further, the reaction succeeded by using the technique of slow addition of Herrman’s catalyst. In the absence of NH group on indole, such reactions failed to give this coupling. Now conversion to the natural product (-)-Hapalindole U (A) needed stereoselective transformation of the carbonyl unit to isonitrile unit. This was achieved by stereocontroled, microwave assisted reductive amination followed by formylation and dehydration reactions to give (-)-Hapalindole U (A). All these reactions could be scaled up to the level of several grams. (-)-Hapalindole U was made in sufficient quantities and stored. The next target (+)-Ambiguine H was a very unstable compound, which was made as and when needed, using the following reactions. (-)-Hapalindole U had several very reactive units in proximity. Introducing a prenyl unit into the crowded C2 region of the indole unit was very difficult . Treatment of t-BuOCl followed by prenyl-9-BBN led to a complex series of reaction, resulting in a crystalline product B. The formation of this product was rationalized by a cascade reaction mechanism shown in figure. Photolysis of B led to a Norrish-type cleavage, followed by rearrangement that got rid of all the extra groups on the skeleton to yield Ambiguine H. The reaction was not allowed to reach completion because the product was also sensitive to photolysis. The reported yield is based on recovered B. The insistence that no protection protocols would be permitted in the scheme gave not only a short route to the target compound, but also led to the use of Rearrangement Transforms, use of Cascade Reactions and led to new discoveries in chemical reactivities. In old chemistry, you would come across several such complex targets (for that time), whose synthesis were achieved without (or with very few) protection / deprotection sequences. There are other syntheses reported in recent years that have followed a ‘no protection’ barrier in planning the syntheses . We could visit two more such synthesis in this chapter. Kessane, a constituent of Japanese Valarian root is of interest due to its sedative and anxiolitic effect. In 2003, Booker-Milburn K. I. and co-workers reported an industrial synthesis of this compound in 8 steps following the theme ‘no protection / deprotection’. The outline of the synthesis is presented in The use of transition metal reagents could aid in the construction of C – C bonds without the need for protective groups, as demonstrated by Baldwin’s group in 2006 . This synthesis of Bureothin was achieved in 7 steps with no protection protocol. Some rare molecular engineering marvels like R.B. Woodward’s synthesis of PGF open up new concepts on ‘Protection’ in molecular synthesis. Note the way he incorporated a protecting group scaffold into the molecular mainframe. Unlike the usual way of thinking of protection / deprotection as an isolated operation, these workers subsequently incorporated this protecting unit into the molecular frame, thereby achieving ‘carbon economy’ well ahead to his time. These concepts of green chemistry came decades later. Also revisit some of the biogenetic type cyclisation reaction discussed earlier in the light of Green Chemistry. Any doubt on feasibility of protection free synthesis could be laid to rest. However, at the present state of developments in common synthetic methodologies, the protection free methodology has its limitations too. At present, it appears inconceivable to bring about a medium sized peptide or nucleotide with such a protection free protocols. This could also be true to several other syntheses of complex molecules. The main thesis of this essay is that it is practical to conceive such protection free synthetic protocols. This falls within the call of Green Chemistry. Though this has been the watchword in industrial synthesis, Baran’s 2007 paper coined the idiom “Protection-free synthesis’ and drew attention to such a possibility. Since then several papers have appeared with this idiom. One can be assured that such a basic green concept cannot be a passing fade. It is more likely to mark a corner stone in the logic of organic synthesis. | 10,038 | 4,113 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Enzymes/Case_Studies/Horseradish_Peroxidase |
Horseradish Peroxidase is a metalloenzyme that exists in the root of the horseradish plant. There are a large number of peroxidase isoenzymes of horseradish with the most common being the C type. This type will be discussed in the remaining report because of the extensive research that has been completed on it. Horseradish peroxidase uses hydrogen peroxide to oxidize both organic and inorganic compounds. Horseradish peroxidase along with other heme peroxidases are brightly colored especially under the near-ultraviolet light. This property of heme peroxidases make them useful for attaching to “transparent” proteins so that they can be seen under different wavelengths. The heme group that is in horseradish peroxidase is simpler than those in mammalians and therefore makes it an excellent starting point in the in-depth study of heme peroxidases and their functions. It has been found that horseradish peroxidase when combined with other compounds is highly reactive toward human tumor cells. A better understanding of horseradish peroxidase could lead to a new targeted cancer therapy. Horseradish peroxidase C has two metal centers, one of iron heme group and two calcium atoms.1 The structure is shown in Figure 1. The heme group has a planar structure with the iron atom held tightly in the middle of a porphyrin ring which is comprised of four pyrrole molecules.3 Iron has two open bonding sites, one above and one below the plane of the heme group. The heme group has a histidine (enzyme) attached in the proximal histidine residue (His170) which is located below the heme group in Figure 1. The second histidine residue in the distal side of the heme group, above the heme group, is vacant in the resting state. This site is open for hydrogen peroxide to attach during reduction-oxidation reactions. An oxygen atom will bond to this vacant site during activation. The iron atom’s sixth octahedral position is considered the active site of the enzyme. During the enzyme reaction, the bonding of the hydrogen peroxide to the iron atom creates an octahedral configuration around the iron atom. Other small molecules can also bond to the distal site, creating the same octahedral configuration. Figure 2 shows the three-dimensional structure of horseradish peroxidase. The iron heme is in the center of the enzyme shown in black with the iron atom as the red sphere. The two calcium atoms are black spheres and lie within the helical regions of the enzyme, with one in the distal region and one in the proximal region. The α-Helical and β-sheet regions of the enzyme are shown as the multicolored helical structures. According to Veitch, “Each calcium site is seven-coordinate with oxygen-donor ligands provided by a combination of amino acid side-chain carboxylates (Asp), hydroxyl groups (Ser, Thr), backbone carbonyls and a structural water molecule (distal site only).” Both the heme group and the calcium atoms are crucial to the enzyme working properly and the loss of one would result in instability. Pure iron, at room temperature, has a body-centered cubic structure. However, according to the RCSB Protein Data Bank, x-ray diffraction shows that horseradish peroxidase has an orthorhombic structure. An orthorhombic structure is where all of the angles in the unit cell equal 90o and each of the three sides do no equal each other. In the case of horseradish peroxidase the three side lengths are: a=40.28 Å, b=67.46 Å, and c=117.11 Å. X-ray diffraction also provided the data to create the three-dimensional structure in Figure 2. The immediate ligand environment of the iron atom in the resting state boasts a point group of D . The molecule is not linear nor does it have tetrahedral, icosahedral, or octahedral symmetry (at least in the rest state). The molecule has a C rotation axis and has 4 C axes perpendicular to the C axis.There is not a σh plane due to the fifth nitrogen atom which is part of the His170 residue that is bonded to the bottom of the iron atom.There are 4 σv planes that contain the C axis which leads to a point group symmetry of D . When the metalloenzyme is not in the resting state the sixth octahedral position of the iron atom is filled which will lead to a change in point group designation.Figure 2 shows the structure of the entire metalloenzyme.It can be seen that there is no symmetry within this structure and horseradish peroxidase as a whole has a point group of C . There is no principal axis, mirror plane, or center of inversion in the enzyme. techniques are difficult to use with horseradish peroxidase due to the reactivity and photolability of the enzyme. This causes the horseradish peroxidase enzyme to break down into a second phase (HRP-II), ferric, and ferrous species. Scientists W. Anthony Oertling and Gerald T. Babcock attempted to mitigate this problem by lowering to cryogenic temperatures, pulsing a near-UV laser, and mixing hydrogen peroxide in the enzyme very quickly. Although some of the old problems still arose, they were able to get a Raman spectrum of the horseradish peroxidase. Horseradish peroxidase has three intermediate compounds all of which have a different color under optical spectroscopy. Because of this, the different stages of the redox cycle can be identified and a better understand of what constitutes each part of the cycle can be found. Figure 3 shows the five oxidation states of horseradish peroxidase. During the redox reaction, the hydrogen peroxide bonds to the vacant octahedral position on the iron atom which initiates the reaction. There are three different intermediate horseradish peroxidase compounds that form during the reaction. They are created as shown in the figure, with either an addition of an electron or a reaction with hydrogen peroxide. The reduction of compound I to compound II and compound II back to the rest state is carried out by reduction substrates. They are usually phenols or aromatic amines. To fully understand how the structure is changing during these stages, x-ray diffraction data must be taken. This, however, has posed a problem because electrons that are stirred up by the x-rays will alter the reduction-oxidation stage of the active site. | 6,250 | 4,114 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.01%3A_Prelude_to_Equilibria |
Chemical equilibrium is an essential concept in chemistry, allowing for an accurate understanding of the molecular processes behind those symbols in chemical equations. However, chemical equilibrium is easily misunderstood. To dispel any misconceptions before we begin our investigation of chemical equilibrium, we will first define what chemical equilibrium is not. Chemical equilibrium is not when there are equal amounts or concentrations of reactants and products. Instead, chemical equilibrium is defined as the state at which the concentration of the reactants and products are constant, not necessarily equal. Chemical equilibrium is not when the reaction "stops". Rather chemical equilibrium is achieved when the forward and reverse rates of a reaction are equal. We will learn more about that further along in the chapter, but before then it is essential to understand that chemical equilibrium is not a reaction coming to a stop. Now that we have alerted you to the two biggest misconceptions that pop up when learning about chemical equilibrium, we can proceed to learning more about what exactly chemical equilibrium is and why it's important. For a preview to what we will be learning in the coming chapter, check out the video below or move on to the next section. | 1,293 | 4,115 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Dynamic_Light_Scattering |
determining the size distribution of nanoparticles in a suspension and detecting small amounts of high mass species in protein samples. Figure 1. A Typical Speckle Pattern The dark areas in the speckle pattern represent regions where the diffracted light from the particles arrives out of phase interfering destructively and the bright areas represent regions where the diffracted light arrives in phase interfering constructively. In practice, particle samples are typically not stationary because they are suspended in a solution and as a result they are due to collisions with solvent molecules. This type of motion is called and it is vital for DLS analysis because it allows the use of the Stokes-Einstein equation to relate the velocity of a particle in solution to its hydrodynamic radius. \[D=\frac{kT}{6\pi \eta a}\] In the Stokes-Einstein equation, is the diffusion velocity of the particle, is the Boltzmann constant, is the temperature, is the viscosity of the solution and is the hydrodynamic radius of the particle. In a sample of particles experiencing Brownian Motion, the distance between particles is constantly changing and this results in a Doppler Shift between the frequency of incoming light and the frequency of scattered light. Since the distance between particles affects the phase overlap of the diffracted light, the brightness of the spots on the speckle pattern will fluctuate in intensity as the particles change position with respect to each other. In a typical DLS experiment, a suspension of analyte such as nanoparticles or polymer molecules is irradiated with monochromatic light from a laser while intensity of the diffracted light is measured. The detector is typically a photomultiplier positioned at 90° to the light source and it is used to collect light diffracted from the sample. Collimating lenses are used to focus laser light to the center of the sample holder and to prevent saturation of the . Ideally, the sample itself should be free of unwanted particles that could contribute to light scattering. For this reason dispersions are often filtered or purified before being measured. Samples are also diluted to low concentrations in order to prevent the particles from interacting with each other and disrupting Brownian Motion. Since the fluctuating intensity data contains a wide spectrum of Doppler shifted frequencies it is not usually measured directly but instead it is compiled for processing using a device called a digital correlator. The function of the correlator in a DLS system is essentially to compare the intensity of two signals \[G_{1}(t)=\lim_{t\to\infty}\frac{1}{T}\int_{-T}^{T}I(t)I(t+\tau )dt\] And is related to the intensity correlation function (G ) by the Siegart relationship \[G_{2}(\tau )=1+\beta \left |G_{1}(\tau ) \right |^{2}\] where Consider the fluctuating intensities of the speckle pattern mentioned earlier. If the intensity signal at a location on a speckle pattern is compared to itself with no change in time (t) then the correlator will measure a perfect correlation and it will assign a value of 1. However, if the same intensity signal is compared with another signal a short time later (t+ Figure 2. Exponential decay of the correlation function. The correlation function for a system experiencing Brownian motion \(G(t)\) decays exponentially with decay constant \[G(t)=e^{-\Gamma t}\] \[\Gamma=-Dq^{^{2}} \] where \[q=\frac{4\pi n }{\lambda}\sin(\dfrac{\Theta }{2})\] | 3,500 | 4,117 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.02%3A_The_Equilibrium_State |
A simple and instructive example of a is provided by the interconversion of the cis and trans isomers of : \[{cis-}\text{C}_{2}\text{H}_{2}\text{F}{2}\rightleftharpoons {trans-}\text{C}_{2}\text{H}_{2}\text{F}_{2} \nonumber \] The two molecules involved in this equilibrium were illustrated in Figure \(\Page {1}\). The only difference between them is that in the cis isomer the two fluorine atoms are on the side of the molecule, while in the trans isomer they are on sides of the molecule. Although their molecules are so similar, these two isomers of difluoroethene are distinct chemical substances. They both condense to liquids at low temperatures, but these liquids have different boiling points. At room temperature both are gases, but they may be separated from each other and analyzed quantitatively by the technique of gas chromatography. that the barrier to free rotation about the bond prevents -C H F from changing rapidly into -C H F . The same applies to the reverse reaction, conversion of the trans isomer to the cis. These reactions occur very slowly at higher temperatures, and even at 700 K (427°C), several weeks are required before equilibrium is reached and the concentrations of cis and trans species no longer vary with time. To study the reaction conveniently, a catalyst, such as I ( ) is added, speeding the reaction so that equilibrium is reached in a few minutes. When this is done, we always end up with a mixture which is slightly richer in the cis isomer. Furthermore, at a given temperature, the ratio of concentrations of the two isomers is always the same. For example, at 623 K the ratio \[\frac{\text{Equilibrium concentration of trans}}{\text{Equilibrium concentration of cis}}=\frac{[trans\text{-C}_{2}\text{H}_{2}\text{F}_{2}]}{[cis\text{-C}_{2}\text{H}_{2}\text{F}_{\text{2}}]}=\text{0.50 }\label{2} \] [In the second ratio in Eq. \(\ref{2}\) square brackets are used to indicate the concentrations of C H F and -C H F once equilibrium has been reached.] Apart from a change in temperature, nothing will alter this equilibrium ratio from 0.50. Whether we start with the pure cis isomer, the pure trans isomer, or even a mixture of isomers, the same ratio is obtained (see Figure \(\Page {2}\) ). Other variations, such as starting with half the amount of either isomer, changing the volume of the container, or heating the mixture to 1000 K and then cooling it to 623 K, are likewise without effect. Even adding a catalyst has no effect on the equilibrium ratio. If we heat -C H F and -C H F to 623 K with iodine added as a catalyst, the only difference is that equilibrium is achieved in a few minutes instead of a few weeks. The final composition is the same as in the uncatalyzed case. We have described this equilibrium between the cis and trans isomers of difluoroethene because it demonstrates very clearly the four features which are characteristic of chemical situation in which appreciable concentrations of reactants and products are in equilibrium with each other. These four features are It is not always easy to tell when a chemical system is in a genuine equilibrium state. We often find mixtures of substances whose compositions do not change with time but which are not really in equilibrium with each other. A mixture of hydrogen and oxygen gas at room temperature is a good example. Although hydrogen and oxygen do react with each other to form water at room temperature, this reaction is so slow that no detectable change is apparent even after a few years. However, if an appropriate catalyst is added, the two gases react explosively and are converted completely to water according to the equation \[\text{2H}_{2}\text{ }({g}) + \text{O}_{2}\text{ } ({g})\rightarrow \text{2H}_{2}\text{O} \text{ }({l}) \nonumber \] | 3,816 | 4,118 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Ethers/Reactivity_of_Ethers/Reactions_of_Epoxides |
Epoxides (oxiranes) are three-membered cyclic ethers that are easily prepared from alkenes by reaction with peracids. Because of the large angle strain in this small ring, epoxides undergo acid and base-catalyzed C–O bond cleavage more easily than do larger ring ethers. Among the following examples, the first is unexceptional except for the fact that it occurs under milder conditions and more rapidly than other ether cleavages. The second and third examples clearly show the exceptional reactivity of epoxides, since unstrained ethers present in the same reactant or as solvent do not react. The aqueous acid used to work up the third reaction, following the cleavage of the ethylene oxide, simply neutralizes the magnesium salt of the alcohol product.
Sulfur is below oxygen in the periodic table. To see examples of organosulfur compounds and their chemistry ),
| 888 | 4,119 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.06%3A_Wave_Mechanics |
The paradox described by Heisenberg’s uncertainty principle and the wavelike nature of subatomic particles such as the electron made it impossible to use the equations of classical physics to describe the motion of electrons in atoms. Scientists needed a new approach that took the wave behavior of the electron into account. In 1926, an Austrian physicist, Erwin Schrödinger (1887–1961; Nobel Prize in Physics, 1933), developed , a mathematical technique that describes the relationship between the motion of a particle that exhibits wavelike properties (such as an electron) and its allowed energies. In doing so, Schrödinger developed the theory of quantum mechanics, which is used today to describe the energies and spatial distributions of electrons in atoms and molecules. Schrödinger’s unconventional approach to atomic theory was typical of his unconventional approach to life. He was notorious for his intense dislike of memorizing data and learning from books. When Hitler came to power in Germany, Schrödinger escaped to Italy. He then worked at Princeton University in the United States but eventually moved to the Institute for Advanced Studies in Dublin, Ireland, where he remained until his retirement in 1955. Although quantum mechanics uses sophisticated mathematics, you do not need to understand the mathematical details to follow our discussion of its general conclusions. We focus on the properties of the that are the solutions of Schrödinger’s equations. A wave function (Ψ) is a mathematical function that relates the location of an electron at a given point in space (identified by , , and coordinates) to the amplitude of its wave, which corresponds to its energy. Thus each wave function is associated with a particular energy . The properties of wave functions derived from quantum mechanics are summarized here: If you are the captain of a ship trying to intercept an enemy submarine, you need to deliver your depth charge to the right location at the right time. Schrödinger’s approach uses three quantum numbers ( , , and ) to specify any wave function. The quantum numbers provide information about the spatial distribution of an electron. Although can be any positive integer, only certain values of and are allowed for a given value of . Introduction to Quantum Numbers: The (n) tells the average relative distance of an electron from the nucleus: \[n = 1, 2, 3, 4,… \label{8.6.1}\] As increases for a given atom, so does the average distance of an electron from the nucleus. A negatively charged electron that is, on average, closer to the positively charged nucleus is attracted to the nucleus more strongly than an electron that is farther out in space. This means that electrons with higher values of are easier to remove from an atom. All wave functions that have the same value of are said to constitute a principal shell because those electrons have similar average distances from the nucleus. As you will see, the principal quantum number corresponds to the used by Bohr to describe electron orbits and by Rydberg to describe atomic energy levels. The second quantum number is often called the azimuthal quantum number ( ). The value of describes the of the region of space occupied by the electron. The allowed values of depend on the value of and can range from 0 to − 1: \[l = 0, 1, 2,…, n − 1 \label{8.6.2}\] For example, if = 1, can be only 0; if = 2, can be 0 or 1; and so forth. For a given atom, all wave functions that have the same values of both and form a subshell. The regions of space occupied by electrons in the same subshell usually have the same shape, but they are oriented differently in space. Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell: The third quantum number is the magnetic quantum number ( ). The value of \(m_l\) describes the of the region in space occupied by an electron with respect to an applied magnetic field. The allowed values of \(m_l\) depend on the value of : can range from − to in integral steps: \[m_l = −l, −l + 1,…, 0,…, l − 1, l \label{8.6.3}\] For example, if \(l = 0\), \(m_l\) can be only 0; if = 1, can be −1, 0, or +1; and if = 2, can be −2, −1, 0, +1, or +2. Each wave function with an allowed combination of , , and values describes an atomic orbital, a particular spatial distribution for an electron. For a given set of quantum numbers, each principal shell has a fixed number of subshells, and each subshell has a fixed number of orbitals. How many subshells and orbitals are contained within the principal shell with = 4? value of number of subshells and orbitals in the principal shell We know that can have all integral values from 0 to − 1. If = 4, then can equal 0, 1, 2, or 3. Because the shell has four values of , it has four subshells, each of which will contain a different number of orbitals, depending on the allowed values of . For = 0, can be only 0, and thus the = 0 subshell has only one orbital. For = 1, can be 0 or ±1; thus the = 1 subshell has three orbitals. For = 2, can be 0, ±1, or ±2, so there are five orbitals in the = 2 subshell. The last allowed value of is = 3, for which can be 0, ±1, ±2, or ±3, resulting in seven orbitals in the = 3 subshell. The total number of orbitals in the = 4 principal shell is the sum of the number of orbitals in each subshell and is equal to : How many subshells and orbitals are in the principal shell with = 3? three subshells; nine orbitals Rather than specifying all the values of and every time we refer to a subshell or an orbital, chemists use an abbreviated system with lowercase letters to denote the value of for a particular subshell or orbital: The principal quantum number is named first, followed by the letter , , , or as appropriate. These orbital designations are derived from corresponding spectroscopic characteristics: harp, rinciple, iffuse, and undamental. A 1 orbital has = 1 and = 0; a 2 subshell has = 2 and = 1 (and has three 2 orbitals, corresponding to = −1, 0, and +1); a 3 subshell has = 3 and = 2 (and has five 3 orbitals, corresponding to = −2, −1, 0, +1, and +2); and so forth. We can summarize the relationships between the quantum numbers and the number of subshells and orbitals as follows (Table \(\Page {1}\)): Each principal shell has subshells, and each subshell has 2 + 1 orbitals. Because of wave–particle duality, scientists must deal with the probability of an electron being at a particular point in space. To do so required the development of , which uses to describe the mathematical relationship between the motion of electrons in atoms and molecules and their energies. Wave functions have five important properties: Quantum numbers provide important information about the energy and spatial distribution of an electron. The can be any positive integer; as increases for an atom, the average distance of the electron from the nucleus also increases. All wave functions with the same value of constitute a in which the electrons have similar average distances from the nucleus. The can have integral values between 0 and − 1; it describes the shape of the electron distribution. Wave functions that have the same values of both and constitute a , corresponding to electron distributions that usually differ in orientation rather than in shape or average distance from the nucleus. The can have 2 + 1 integral values, ranging from − to + , and describes the orientation of the electron distribution. Each wave function with a given set of values of , , and describes a particular spatial distribution of an electron in an atom, an . | 7,788 | 4,122 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Phases_and_Intermolecular_Forces/Dipole-dipole_Forces |
are probably the simplest to understand. You probably already know that in an ionic solid like NaCl, the solid is held together by Coulomb attractions between the oppositely-charges ions. The Na and Cl ions alternate so the Coulomb forces are attractive. Dipole-dipole forces work the same way, except that the charges are smaller. A good example is HF (this is also an example of a special type of dipole-dipole force called a ). In HF, the bond is a very polar covalent bond. That means there is a partial negative (δ-) charge on F and partial positive (δ+) charge on H, and the molecule has a permanent (the electrons always spend more time on F). In the liquid or solid HF, the molecules arrange themselves so that the δ- and δ+ are close together. These partial charges attract each other, and this attraction is what we call dipole-dipole forces. Any molecule with a permanent dipole has dipole-dipole forces that hold the molecules next to each other as a solid or liquid. | 1,006 | 4,123 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.17%3A_Molecular_and_Ionic_Equations |
One of the unfortunate byproducts of industrialized society is acid rain. Sulfur dioxide from burning coal and nitrogen oxides from vehicle emissions both form acids. When these acids react with limestone (calcium carbonate), reactions occur that dissolve the limestone and release water and carbon dioxide. Over a period of time, serious damage is caused to the limestone structure. When ionic compounds are dissolved into water, the polar water molecules break apart the solid crystal lattice, resulting in the hydrated ions being evenly distributed through the water. This process is called dissociation and is the reason that all ionic compounds are strong electrolytes. When two different ionic compounds that have been dissolved in water are mixed, a chemical reaction may occur between certain pairs of the hydrated ions. Consider the double-replacement reaction that occurs when a solution of sodium chloride is mixed with a solution of silver nitrate: \[\ce{NaCl} \left( aq \right) + \ce{AgNO_3} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{AgCl} \left( s \right)\nonumber \] The driving force behind this reaction is the formation of the silver chloride precipitate (see figure below). This is called a molecular equation. A is an equation in which the formulas of the compounds are written as though all substances exist as molecules. However, there is a better way to show what is happening in this reaction. All of the aqueous compounds should be written as ions, because they are present in the water as separated ions—a result of their dissociation. \[\ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) + \ce{Ag^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{NO_3^-} \left( aq \right) + \ce{AgCl} \left( s \right)\nonumber \] This equation is called an , an equation in which dissolved ionic compounds are shown as free ions. Some other double-replacement reactions do not produce a precipitate as one of the products. The production of a gas and/or a molecular compound such as water may also drive the reaction. For example, consider the reaction of a solution of sodium carbonate with a solution of hydrochloric acid \(\left( \ce{HCl} \right)\). The products of the reaction are aqueous sodium chloride, carbon dioxide, and water. The balanced molecular equation is: \[\ce{Na_2CO_3} \left( aq \right) + 2 \ce{HCl} \left( aq \right) \rightarrow 2 \ce{NaCl} \left( aq \right) + \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\nonumber \] The ionic equation is: \[2 \ce{Na^+} \left( aq \right) + \ce{CO_3^{2-}} \left( aq \right) + 2 \ce{H^+} \left( aq \right) + 2 \ce{Cl^-} \left( aq \right) \rightarrow 2 \ce{Na^+} \left( aq \right) + 2 \ce{Cl^-} \left( aq \right) + \ce{CO_2} \left( g \right) + \ce{H_2O} \left( l \right)\nonumber \] A single-replacement reaction is one in which an element replaces another element in a compound. An element is in either the solid, liquid, or gas state and is not an ion. The example below shows the reaction of solid magnesium metal with aqueous silver nitrate to form aqueous magnesium nitrate and silver metal. Balanced molecular equation: \[\ce{Mg} \left( s \right) + 2 \ce{AgNO_3} \left( aq \right) \rightarrow \ce{Mg(NO_3)_2} \left( aq \right) + 2 \ce{Ag} \left( s \right)\nonumber \] Ionic equation: \[\ce{Mg} \left( s \right) + 2 \ce{Ag^+} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right) \rightarrow \ce{Mg^{2+}} \left( aq \right) + 2 \ce{NO_3^-} \left( aq \right) + \ce{Ag} \left( s \right)\nonumber \] This type of single-replacement reaction is called a metal replacement. Other common categories of single-replacement reactions are: hydrogen replacement and halogen replacement. | 3,757 | 4,124 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.10%3A_Gas_Stoichiometry |
The Haber cycle reaction of gaseous nitrogen and hydrogen to form ammonia is a critical step in the production of fertilizer from ammonia. It is important to have an excess of the starting materials so that a maximum yield of ammonia can be achieved. By knowing how much ammonia is needed for the manufacture of a batch of fertilizer, the proper amounts of nitrogen and hydrogen gases can be incorporated into the process. You have learned how to use molar volume to solve stoichiometry problems for chemical reactions involving one or more gases at . Now, we can use the ideal gas law to expand our treatment of chemical reactions to solve stoichiometry problems for reactions that occur at any temperature and pressure. What volume of carbon dioxide is produced by the combustion of \(25.21 \: \text{g}\) of ethanol \(\left( \ce{C_2H_5OH} \right)\) at \(54^\text{o} \text{C}\) and \(728 \: \text{mm} \: \ce{Hg}\)? Before using the ideal gas law, it is necessary to write and balance the chemical equation. Recall that in most combustion reactions, the given substance reacts with \(\ce{O_2}\) to form \(\ce{CO_2}\) and \(\ce{H_2O}\). Here is the balanced equation for the combustion of ethanol: \[\ce{C_2H_5OH} \left( l \right) + 3 \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right) + 3 \ce{H_2O} \left( l \right)\nonumber \] The number of moles of carbon dioxide gas is first calculated by stoichiometry. Then, the ideal gas law is used to calculate the volume of \(\ce{CO_2}\) produced. \[25.21 \: \text{g} \: \ce{C_2H_5OH} \times \frac{1 \: \text{mol} \: \ce{C_2H_5OH}}{46.08 \: \text{g} \: \ce{C_2H_5OH}} \times \frac{2 \: \text{mol} \: \ce{CO_2}}{1 \: \text{mol} \: \ce{C_2H_5OH}} = 1.094 \: \text{mol} \: \ce{CO_2}\nonumber \] The moles of carbon dioxide \(\left( n \right)\) is now substituted into \(PV = nRT\) to solve for the volume. \[V = \frac{nRT}{P} = \frac{1.094 \: \text{mol} \times 62.36 \: \text{L} \cdot \text{mm} \: \text{Hg}/\text{K} \cdot \text{mol} \times 327 \: \text{K}}{728 \: \text{mm} \: \ce{Hg}} = 30.6 \: \text{L}\nonumber \] The mass of ethanol is slightly more than one half mole, meaning that the mole ratio results in slightly more than one mole of carbon dioxide being produced. Because of the elevated temperature and reduced pressure compared to STP, the resulting volume is larger than \(22.4 \: \text{L}\). | 2,385 | 4,126 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.13%3A_Standard_Reduction_Potentials |
The following table provides and ́ values for selected reduction reactions. Values are from the following sources (primarily the first two): Solids, gases, and liquids are identified; all other species are aqueous. Reduction reactions in acidic solution are written using H in place of H O . You may rewrite a reaction by replacing H with H O and adding to the opposite side of the reaction one molecule of H O per H ; thus H AsO + 2H +2 \(\rightleftharpoons\) HAsO +2H O becomes H AsO + 2H O +2 \(\rightleftharpoons\) HAsO +4H O Conditions for formal potentials ( ́) are listed next to the potential. For most of the reduction half-reactions gathered here, there are minor differences in values provided by the references above. In most cases, these differences are small and will not affect calculations. In a few cases the differences are not insignificant and the user may find discrepancies in calculations. For example, Bard, Parsons, and Jordon report an value of –1.285 V for \[\text{Zn(OH)}_4^{2-} + 2e^- \rightleftharpoons \text{Zn}(s) + 4\text{OH}^-\nonumber\] while Milazzo, Caroli, and Sharma report the value as –1.214 V, Swift reports the value as –1.22, Bratsch reports the value as –1.199 V, and Latimer reports the value as –1.216 V. \(\text{Al}^{3+} + 3e^- \rightleftharpoons \text{Al}(s)\) \(\text{AlF}_6^{3-} + 3e^- \rightleftharpoons \text{Al}(s) + 6\text{F}^-\) \(\text{Ba}^{2+} + 2e^- \rightleftharpoons \text{Ba}(s)\) \(\text{BaO}(s) + 2\text{H}^+ + 2e^- \rightleftharpoons \text{Ba}(s) + \text{H}_2\text{O}(l)\) \(\text{Be}^{2+} + 2e^- \rightleftharpoons \text{Be}(s)\) \(\text{Bi}^{3+} + 3e^- \rightleftharpoons \text{Bi}(s)\) 0.317 0.199 \(\text{B(OH)}_3 + 3\text{H}^+ + 3e^- \rightleftharpoons \text{B}(s) + 3\text{H}_2\text{O}(l)\) –0.890 \(\text{B(OH)}_4^- + 3e^- \rightleftharpoons \text{B}(s) + 4\text{OH}^-\) –1.811 1.087 \(\text{HOBr} + \text{H}^+ + 2e^- \rightleftharpoons \text{Br}^- + \text{H}_2\text{O}(l)\) 1.341 \(\text{HOBr} + \text{H}^+ + e^- \rightleftharpoons \frac{1}{2} \text{Br}_2 + \text{H}_2\text{O}(l)\) 1.604 \(\text{BrO}^- + \text{H}_2\text{O}(l) + 2e^- \rightleftharpoons \text{Br}^- + 2\text{OH}^-\) 0.76 in 1 M NaOH \(\text{BrO}_3^- +6\text{H}^+ + 5e^- \rightleftharpoons \frac{1}{2} \text{Br}_2(l) + 3\text{H}_2\text{O}(l)\) 1.5 \(\text{BrO}_3^- + 6\text{H}^+ +6e^- \rightleftharpoons \text{Br}^- + 3\text{H}_2\text{O}(l)\) 1.478 \(\text{Cd}^{2+} + 2e^- \rightleftharpoons \text{Cd}(s)\) –0.4030 \(\text{Cd(CN)}_4^{2-} + 2e^- \rightleftharpoons \text{Cd}(s) + 4\text{CN}^-\) –0.943 \(\text{Cd(NH}_3)_4^{2+} + 2e^- \rightleftharpoons \text{Cd}(s) + 4\text{NH}_3\) –0.622 \(\text{Ca}^{2+} + 2e^- \rightleftharpoons \text{Ca}(s)\) –2.84 \(\text{CO}_2(g) + 2\text{H}^+ + 2e^- \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(l)\) –0.106 \(\text{CO}_2(g) + 2\text{H}^+ +2e^- \rightleftharpoons \text{HCO}_2\text{H}\) –0.20 \(2\text{CO}_2(g) + 2\text{H}^+ +2e^- \rightleftharpoons \text{H}_2\text{C}_2\text{O}_4\) –0.481 \(\text{HCHO} + 2\text{H}^+ + 2e^- \rightleftharpoons \text{CH}_3\text{OH}\) 0.2323 \(\text{Ce}^{3+} + 3e^- \rightleftharpoons \text{Ce}(s)\) –2.336 \(\text{Ce}^{4+} + e^- \rightleftharpoons \text{Ce}^{3+}\) 1.72 1.70 in 1 M HClO 1.44 in 1 M H SO 1.61 in 1 M HNO 1.28 in 1 M HCl \(\text{Cl}_2(g) + 2e^- \rightleftharpoons 2\text{Cl}^-\) 1.396 \(\text{ClO}^- + \text{H}_2\text{O}(l) + e^- \rightleftharpoons \frac{1}{2} \text{Cl}_2(g) + 2\text{OH}^-\) 0.421 in 1 M NaOH \(\text{ClO}^- + \text{H}_2\text{O}(l) + 2e^- \rightleftharpoons \text{Cl}^- + 2\text{OH}^-\) 0.890 in 1 M NaOH \(\text{HClO}_2 + 2\text{H}^+ + 2e^- \rightleftharpoons \text{HOCl} + \text{H}_2\text{O}(l)\) 1.64 \(\text{ClO}_3^- + 2\text{H}^+ + e^- \rightleftharpoons \text{ClO}_2(g) + \text{H}_2\text{O}(l)\) 1.175 \(\text{ClO}_3^- + 3\text{H}^+ + 2e^- \rightleftharpoons \text{HClO}_2 + \text{H}_2\text{O}(l)\) 1.181 \(\text{ClO}_4^- + 2\text{H}^+ +2e^- \rightleftharpoons \text{ClO}_3^- + \text{H}_2\text{O}(l)\) 1.201 \(\text{Cr}^{3+} + 3e^- \rightleftharpoons \text{Cr}(s)\) –0.424 \(\text{Cr}^{2+} + 2e^- \rightleftharpoons \text{Cr}(s)\) –0.90 \(\text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6e^- \rightleftharpoons 2\text{Cr}^{3+} + 7\text{H}_2\text{O}(l)\) 1.36 \(\text{CrO}_4^{2-} + 4\text{H}_2\text{O}(l) + 3e^- \rightleftharpoons \text{Cr(OH)}_4^- + 4\text{OH}^-\) \(\text{Co}^{2+} + 2e^- \rightleftharpoons \text{Co}(s)\) –0.277 \(\text{Co}^{3+} + 3e^- \rightleftharpoons \text{Co}(s)\) 1.92 \(\text{Co(NH}_3)_6^{3+} + e^- \rightleftharpoons \text{Co(NH}_3)_6^{2+}\) 0.1 \(\text{Co(OH)}_3(s) + e^- \rightleftharpoons \text{Co(OH)}_2(s) + \text{OH}^-\) 0.17 \(\text{Co(OH)}_2(s) + 2e^- \rightleftharpoons \text{Co}(s) + 2\text{OH}^-\) –0.746 \(\text{Cu}^+ + e^- \rightleftharpoons \text{Cu}(s)\) 0.520 \(\text{Cu}^{2+} + e^- \rightleftharpoons \text{Cu}^+\) 0.159 \(\text{Cu}^{2+} + 2e^- \rightleftharpoons \text{Cu}(s)\) 0.3419 \(\text{Cu}^{2+} + \text{I}^- + e^- \rightleftharpoons \text{CuI}(s)\) 0.86 \(\text{Cu}^{2+} + \text{Cl}^- + e^- \rightleftharpoons \text{CuCl}(s)\) 0.559 \(\text{F}_2(g) + 2\text{H}^+ + 2e^- \rightleftharpoons 2\text{HF}(g)\) 3.053 \(\text{F}_2(g) + 2e^- \rightleftharpoons 2\text{F}^-\) 2.87 \(\text{Ga}^{3+} + 3e^- \rightleftharpoons \text{Ga}(s)\) \(\text{Au}^+ + e^- \rightleftharpoons \text{Au}(s)\) 1.83 \(\text{Au}^{3+} + 2e^- \rightleftharpoons \text{Au}^+\) 1.36 \(\text{Au}^{3+} + 3e^- \rightleftharpoons \text{Au}(s)\) 1.52 \(\text{AuCl}_4^- + 3e^- \rightleftharpoons \text{Au}(s) + 4\text{Cl}^-\) 1.002 \(2\text{H}^+ + 2e^- \rightleftharpoons \text{H}_2 (g)\) 0.00000 \(\text{H}_2\text{O}(l) + e^- \rightleftharpoons \frac{1}{2} \text{H}_2(g) + \text{OH}^-\) –0.828 \(\text{I}_2(s) + 2e^- \rightleftharpoons 2\text{I}^-\) 0.5355 \(\text{I}_3^- + 2e^- \rightleftharpoons 3\text{I}^-\) 0.536 \(\text{HIO} + \text{H}^+ + 2e^- \rightleftharpoons \text{I}^- + \text{H}_2\text{O}(l)\) 0.985 \(\text{IO}_3^- + 6\text{H}^+ + 5e^- \rightleftharpoons \frac{1}{2} \text{I}_2(s) + 3\text{H}_2\text{O}(l)\) 1.195 \(\text{IO}_3^- + 3\text{H}_2\text{O}(l) + 6e^- \rightleftharpoons \text{I}^- +6\text{OH}^-\) 0.257 \(\text{Fe}^{2+} + 2e^- \rightleftharpoons \text{Fe}(s)\) –0.44 \(\text{Fe}^{3+} + 3e^- \rightleftharpoons \text{Fe}(s)\) –0.037 \(\text{Fe}^{3+} + e^- \rightleftharpoons \text{Fe}^{2+}\) 0.771 0.70 in 1 M HCl 0.767 in 1 M HClO 0.746 in 1 M HNO 0.68 in 1 M H SO 0.44 in 0.3 M H PO \(\text{Fe(CN)}_6^{3-} + e^- \rightleftharpoons \text{Fe(CN)}_6^{4-}\) 0.356 \(\text{Fe(phen)}_3^{3+} + e^- \rightleftharpoons \text{Fe(phen)}_3^{2+}\) 1.147 \(\text{La}^{3+} + 3e^- \rightleftharpoons \text{La}(s)\) –2.38 \(\text{Pb}^{2+} + 2e^- \rightleftharpoons \text{Pb}(s)\) –0.126 \(\text{PbO}_2(s) + 4\text{H}^+ + 2e^- \rightleftharpoons \text{Pb}^{2+} + 2\text{H}_2\text{O}(l)\) 1.46 \(\text{PbO}_2(s) + \text{SO}_4^{2-} + 4\text{H}^+ + 2e^- \rightleftharpoons \text{PbSO}_4(s) + 2\text{H}_2\text{O}(l)\) 1.690 \(\text{PbSO}_4(s) + 2e^- \rightleftharpoons \text{Pb}(s) + \text{SO}_4^{2-}\) –0.356 \(\text{Li}^+ + e^- \rightleftharpoons \text{Li}(s)\) –3.040 \(\text{Mg}^{2+} + 2e^- \rightleftharpoons \text{Mg}(s)\) –2.356 \(\text{Mg(OH)}_2(s) + 2e^- \rightleftharpoons \text{Mg}(s) + 2\text{OH}^-\) –2.687 \(\text{Mn}^{2+} + 2e^- \rightleftharpoons \text{Mn}(s)\) –1.17 \(\text{Mn}^{3+} + e^- \rightleftharpoons \text{Mn}^{2+}\) 1.5 \(\text{MnO}_2(s) + 4\text{H}^+ + 2e^- \rightleftharpoons \text{Mn}^{2+} + 2\text{H}_2\text{O}(l)\) 1.23 \(\text{MnO}_4^- + 4\text{H}^+ +3e^- \rightleftharpoons \text{MnO}_2(s) + 2\text{H}_2\text{O}(l)\) 1.70 \(\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightleftharpoons \text{Mn}^{2+} + 4\text{H}_2\text{O}(l)\) 1.51 \(\text{MnO}_4^- + 2\text{H}_2\text{O}(l) + 3e^- \rightleftharpoons \text{MnO}_2(s) + 4\text{OH}^-\) 0.60 \(\text{Hg}^{2+} + 2e^- \rightleftharpoons \text{Hg}(l)\) 0.8535 \(2\text{Hg}^{2+} +2e^- \rightleftharpoons \text{Hg}_2^{2+}\) 0.911 \(\text{Hg}_2^{2+} + 2e^- \rightleftharpoons 2\text{Hg}(l)\) 0.7960 \(\text{Hg}_2\text{Cl}_2(s) + 2e^- \rightleftharpoons 2\text{Hg}(l) + 2\text{Cl}^-\) 0.2682 \(\text{HgO}(s) + 2\text{H}^+ + 2e^- \rightleftharpoons \text{Hg}(l) + \text{H}_2\text{O}(l)\) 0.926 \(\text{Hg}_2\text{Br}_2(s) + 2e^- \rightleftharpoons 2\text{Hg}(l) + 2\text{Br}^-\) 1.392 \(\text{Hg}_2\text{I}_2(s) + 2e^- \rightleftharpoons 2\text{Hg}(l) + 2\text{I}^-\) –0.0405 \(\text{Mo}^{3+} + 3e^- \rightleftharpoons \text{Mo}(s)\) –0.2 \(\text{MoO}_2(s) + 4\text{H}^+ + 4e^- \rightleftharpoons \text{Mo}(s) + 2\text{H}_2\text{O}(l)\) –0.152 \(\text{MoO}_4^{2-} + 4\text{H}_2\text{O}(l) + 6e^- \rightleftharpoons \text{Mo}(s) + 8\text{OH}^-\) –0.913 \(\text{Ni}^{2+} + 2e^- \rightleftharpoons \text{Ni}(s)\) –0.257 \(\text{Ni(OH)}_2(s) + 2e^- \rightleftharpoons \text{Ni}(s) + 2\text{OH}^-\) –0.72 \(\text{Ni(NH}_3)_6^{2+} + 2e^- \rightleftharpoons \text{Ni}(s) + 6\text{NH}_3\) –0.49 \(\text{N}_2(g) + 5\text{H}^+ + 4e^- \rightleftharpoons \text{N}_2\text{H}_5^+\) –0.23 \(\text{N}_2\text{O}(g) + 2\text{H}^+ + 2e^- \rightleftharpoons \text{N}_2(g) + \text{H}_2\text{O}(l)\) 1.77 \(2\text{NO}(g) + 2\text{H}^+ + 2e^- \rightleftharpoons \text{N}_2\text{O}(g) + \text{H}_2\text{O}(l)\) 1.59 \(\text{HNO}_2 + \text{H}^+ + e^- \rightleftharpoons \text{NO}(g) + \text{H}_2\text{O}(l)\) 0.996 \(2\text{HNO}_2 + 4\text{H}^+ + 4e^- \rightleftharpoons \text{N}_2\text{O}(g) + 3\text{H}_2\text{O}(l)\) 1.297 \(\text{NO}_3^- + 3\text{H}^+ + 2e^- \rightleftharpoons \text{HNO}_2 + \text{H}_2\text{O}(l)\) 0.94 \(\text{O}_2(g) + 2\text{H}^+ + 2e^- \rightleftharpoons \text{H}_2\text{O}_2\) 0.695 \(\text{O}_2(g) + 4\text{H}^+ + 4e^- \rightleftharpoons 2\text{H}_2\text{O}(l)\) 1.229 \(\text{H}_2\text{O}_2 + 2\text{H}^+ + 2e^- \rightleftharpoons 2\text{H}_2\text{O}(l)\) 1.763 \(\text{O}_2(g) + 2\text{H}_2\text{O}(l) + 4e^- \rightleftharpoons 4\text{OH}^-\) 0.401 \(\text{O}_3(g) + 2\text{H}^+ + 2e^- \rightleftharpoons \text{O}_2(g) + \text{H}_2\text{O}(l)\) 2.07 \(\text{P}(s, white) + 3\text{H}^+ + 3e^- \rightleftharpoons \text{PH}_3(g)\) –0.063 \(\text{H}_3\text{PO}_3 + 2\text{H}^+ + 2e^- \rightleftharpoons \text{H}_3\text{PO}_2 + \text{H}_2\text{O}(l)\) –0.499 \(\text{H}_3\text{PO}_4 + 2\text{H}^+ + 2e^- \rightleftharpoons \text{H}_3\text{PO}_3 + \text{H}_2\text{O}(l)\) –0.276 \(\text{Pt}^{2+} + 2e^- \rightleftharpoons \text{Pt}(s)\) 1.188 \(\text{PtCl}_4^{2-} + 2e^- \rightleftharpoons \text{Pt}(s) + 4\text{Cl}^-\) 0.758 \(\text{K}^+ + e^- \rightleftharpoons \text{K}(s)\) –2.924 \(\text{Ru}^{3+} + 3e^- \rightleftharpoons \text{Ru}(s)\) 0.249 \(\text{RuO}_2(s) + 4\text{H}^+ + 4e^- \rightleftharpoons \text{Ru}(s) + 2\text{H}_2\text{O}(l)\) 0.68 \(\text{Ru(NH}_3)_6^{3+} + e^- \rightleftharpoons \text{Ru(NH}_3)_6^{2+}\) 0.10 \(\text{Ru(CN)}_6^{3-} + e^- \rightleftharpoons \text{Ru(CN)}_6^{4-}\) 0.86 \(\text{Se}(s) + 2e^- \rightleftharpoons \text{Se}^{2-}\) –0.67 in 1 M NaOH \(\text{Se}(s) + 2\text{H}^+ + 2e^- \rightleftharpoons \text{H}_2\text{Se}(g)\) –0.115 \(\text{H}_2\text{SeO}_3 + 4\text{H}^+ + 4e^- \rightleftharpoons \text{Se}(s) + 3\text{H}_2\text{O}(l)\) 0.74 \(\text{SeO}_4^{3-} + 4\text{H}^+ + e^- \rightleftharpoons \text{H}_2\text{SeO}_3 + \text{H}_2\text{O}(l)\) 1.151 \(\text{SiF}_6^{2-} + 4e^- \rightleftharpoons \text{Si}(s) + 6\text{F}^-\) –1.37 \(\text{SiO}_2(s) + 4\text{H}^+ + 4e^- \rightleftharpoons \text{Si}(s) + 2\text{H}_2\text{O}(l)\) –0.909 \(\text{SiO}_2(s) + 8\text{H}^+ + 8e^- \rightleftharpoons \text{SiH}_4(g) + 2\text{H}_2\text{O}(l)\) –0.516 \(\text{Ag}^+ + e^- \rightleftharpoons \text{Ag}(s)\) 0.7996 \(\text{AgBr}(s) + e^- \rightleftharpoons \text{Ag}(s) + \text{Br}^-\) 0.071 \(\text{Ag}_2\text{C}_2\text{O}_4(s) + 2e^- \rightleftharpoons 2\text{Ag}(s) + \text{C}_2\text{O}_4^{2-}\) 0.47 \(\text{AgCl}(s) + e^- \rightleftharpoons \text{Ag}(s) + \text{Cl}^-\) 0.2223 \(\text{AgI}(s) + e^- \rightleftharpoons \text{Ag}(s) + \text{I}^-\) –0.152 \(\text{Ag}_2\text{S}(s) + 2e^- \rightleftharpoons 2\text{Ag}(s) + \text{S}^{2-}\) –0.71 \(\text{Ag(NH}_3)_2^+ + e^- \rightleftharpoons \text{Ag}(s) + 2\text{NH}_3\) –0.373 \(\text{Na}^+ + e^- \rightleftharpoons \text{Na}(s)\) –2.713 \(\text{Sr}^{2+} + 2e^- \rightleftharpoons \text{Sr}(s)\) –2.89 \(\text{S}(s) + 2e^- \rightleftharpoons \text{S}^{2-}\) –0.407 \(\text{S}(s) + 2\text{H}^+ + 2e^- \rightleftharpoons \text{H}_2\text{S}(g)\) 0.144 \(\text{S}_2\text{O}_6^{2-} + 4\text{H}^+ + 2e^- \rightleftharpoons 2\text{H}_2\text{SO}_3\) 0.569 \(\text{S}_2\text{O}_8^{2-} + 2e^- \rightleftharpoons 2\text{SO}_4^{2-}\) 1.96 \(\text{S}_4\text{O}_6^{2-} + 2e^- \rightleftharpoons 2\text{S}_2\text{O}_3^{2-}\) 0.080 \(2\text{SO}_3^{2-} + 2\text{H}_2\text{O}(l) + 2e^- \rightleftharpoons \text{S}_2\text{O}_4^{2-} + 4\text{OH}^-\) –1.13 \(2\text{SO}_3^{2-} + 3\text{H}_2\text{O}(l) + 4e^- \rightleftharpoons \text{S}_2\text{O}_3^{2-} + 6\text{OH}^-\) –0.576 in 1 M NaOH \(2\text{SO}_4^{2-} + 4\text{H}^+ + 2e^- \rightleftharpoons \text{S}_2\text{O}_6^{2-} + 2\text{H}_2\text{O}(l)\) –0.25 \(\text{SO}_4^{2-} + \text{H}_2\text{O}(l) + 2e^- \rightleftharpoons \text{SO}_3^{2-} + 2\text{OH}^-\) –0.936 \(\text{SO}_4^{2-} + 4\text{H}^+ + 2e^- \rightleftharpoons \text{H}_2\text{SO}_3 + \text{H}_2\text{O}(l)\) 0.172 \(\text{Tl}^{3+} + 2e^- \rightleftharpoons \text{Tl}^+\) 1.25 in 1 M HClO 0.77 in 1 M HCl \(\text{Tl}^{3+} + 3e^- \rightleftharpoons \text{Tl}(s)\) \(\text{Sn}^{2+} + 2e^- \rightleftharpoons \text{Sn}(s)\) –0.19 in 1 M HCl \(\text{Sn}^{4+} + 2e^- \rightleftharpoons \text{Sn}^{2+}\) 0.154 0.139 in 1 M HCl \(\text{Ti}^{2+} + 2e^- \rightleftharpoons \text{Ti}(s)\) –0.163 \(\text{Ti}^{3+} + e^- \rightleftharpoons \text{Ti}^{2+}\) –0.37 \(\text{WO}_2(s) + 4\text{H}^+ + 4e^- \rightleftharpoons \text{W}(s) + 2\text{H}_2\text{O}(l)\) –0.119 \(\text{WO}_3(s) + 6\text{H}^+ + 6e^- \rightleftharpoons \text{W}(s) + 3\text{H}_2\text{O}(l)\) –0.090 \(\text{U}^{3+} + 3e^- \rightleftharpoons \text{U}(s)\) –1.66 \(\text{U}^{4+} + e^- \rightleftharpoons \text{U}^{3+}\) –0.52 \(\text{UO}_2^+ + 4\text{H}^+ + e^- \rightleftharpoons \text{U}^{4+} + 2\text{H}_2\text{O}(l)\) 0.27 \(\text{UO}_2^{2+} + e^- \rightleftharpoons \text{UO}_2^+\) 0.16 \(\text{UO}_2^{2+} + 4\text{H}^+ + 2e^- \rightleftharpoons \text{U}^{4+} + 2\text{H}_2\text{O}(l)\) 0.327 \(\text{V}^{2+} + 2e^- \rightleftharpoons \text{V}(s)\) –1.13 \(\text{V}^{3+} + e^- \rightleftharpoons \text{V}^{2+}\) –0.255 \(\text{VO}^{2+} + 2\text{H}^+ + e^- \rightleftharpoons \text{V}^{3+} + \text{H}_2\text{O}(l)\) 0.337 \(\text{VO}_2^{+} + 2\text{H}^+ + e^- \rightleftharpoons \text{VO}^{2+} + \text{H}_2\text{O}(l)\) 1.000 \(\text{Zn}^{2+} + 2e^- \rightleftharpoons \text{Zn}(s)\) –0.7618 \(\text{Zn(OH)}_4^{2-} + 2e^- \rightleftharpoons \text{Zn}(s) + 4\text{OH}^-\) –1.285 \(\text{Zn(NH}_3)_4^{2+} + 2e^- \rightleftharpoons \text{Zn}(s) + 4\text{NH}_3\) –1.04 \(\text{Zn(CN)}_4^{2-} + 2e^- \rightleftharpoons \text{Zn}(s) + 4\text{CN}^-\) –1.34 | 14,864 | 4,130 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.03%3A_Quantum_Theory |
By the late 19th century, many physicists thought their discipline was well on the way to explaining most natural phenomena. They could calculate the motions of material objects using Newton’s laws of classical mechanics, and they could describe the properties of radiant energy using mathematical relationships known as Maxwell’s equations, developed in 1873 by James Clerk Maxwell, a Scottish physicist. The universe appeared to be a simple and orderly place, containing matter, which consisted of particles that had mass and whose location and motion could be accurately described, and electromagnetic radiation, which was viewed as having no mass and whose exact position in space could not be fixed. Thus matter and energy were considered distinct and unrelated phenomena. Soon, however, scientists began to look more closely at a few inconvenient phenomena that could be explained by the theories available at the time. One phenomenon that seemed to contradict the theories of classical physics was blackbody radiation The wavelength of energy emitted by an object depends on only its temperature, not its surface or composition. Hence an electric stove burner or the filament of a space heater glows dull red or orange when heated, whereas the much hotter tungsten wire in an incandescent light bulb gives off a yellowish light. The of radiation is a measure of the energy emitted per unit area. A plot of the intensity of blackbody radiation as a function of wavelength for an object at various temperatures is shown in Figure \(\Page {2}\). One of the major assumptions of classical physics was that energy increased or decreased in a smooth, continuous manner. For example, classical physics predicted that as wavelength decreased, the intensity of the radiation an object emits should increase in a smooth curve without limit at temperatures, as shown by the broken line for 6000 K in Figure \(\Page {2}\). Thus classical physics could not explain the sharp in the intensity of radiation emitted at shorter wavelengths (primarily in the ultraviolet region of the spectrum), which was referred to as the “ultraviolet catastrophe.” In 1900, however, the German physicist Max Planck (1858–1947) explained the ultraviolet catastrophe by proposing that the energy of electromagnetic waves is rather than continuous. This means that for each temperature, there is a maximum intensity of radiation that is emitted in a blackbody object, corresponding to the peaks in Figure \(\Page {2}\), so the intensity does not follow a smooth curve as the temperature increases, as predicted by classical physics. Thus energy could be gained or lost only in integral multiples of some smallest unit of energy, a quantum (the smallest possible unit of energy). Energy can be gained or lost only in integral multiples of a quantum.. In addition to being a physicist, Planck was a gifted pianist, who at one time considered music as a career. During the 1930s, Planck felt it was his duty to remain in Germany, despite his open opposition to the policies of the Nazi government. One of his sons was executed in 1944 for his part in an unsuccessful attempt to assassinate Hitler, and bombing during the last weeks of World War II destroyed Planck’s home. After WWII, the major German scientific research organization was renamed . Although quantization may seem to be an unfamiliar concept, we encounter it frequently. For example, US money is integral multiples of pennies. Similarly, musical instruments like a piano or a trumpet can produce only certain musical notes, such as C or F sharp. Because these instruments cannot produce a continuous range of frequencies, their frequencies are quantized. Even electrical charge is quantized: an ion may have a charge of −1 or −2 but −1.33 electron charges. Planck postulated that the energy of a particular quantum of radiant energy could be described Max Planck (1858–1947) by the equation \[ E=h u \label{8.3.1}\] where the proportionality constant is called Planck’s constant, one of the most accurately known fundamental constants in science. For our purposes, its value to four significant figures is generally sufficient: As the frequency of electromagnetic radiation increases, the magnitude of the associated quantum of radiant energy increases. By assuming that energy can be emitted by an object only in integral multiples of ν, Planck devised an equation that fit the experimental data shown in Figure \(\Page {2}\). We can understand Planck’s explanation of the ultraviolet catastrophe qualitatively as follows: At low temperatures, radiation with only relatively low frequencies is emitted, corresponding to low-energy quanta. As the temperature of an object increases, there is an increased probability of emitting radiation with higher frequencies, corresponding to higher-energy quanta. At any temperature, however, it is simply more probable for an object to lose energy by emitting a large number of lower-energy quanta than a single very high-energy quantum that corresponds to ultraviolet radiation. The result is a maximum in the plot of intensity of emitted radiation versus wavelength, as shown in Figure \(\Page {2}\), and a shift in the position of the maximum to lower wavelength (higher frequency) with increasing temperature. At the time he proposed his radical hypothesis, Planck could not explain energies should be quantized. Initially, his hypothesis explained only one set of experimental data—blackbody radiation. If quantization were observed for a large number of different phenomena, then quantization would become a law. In time, a theory might be developed to explain that law. As things turned out, Planck’s hypothesis was the seed from which modern physics grew. Only five years after he proposed it, Planck’s quantization hypothesis was used to explain a second phenomenon that conflicted with the accepted laws of classical physics. When certain metals are exposed to light, electrons are ejected from their surface (Figure \(\Page {3}\)). Classical physics predicted that the number of electrons emitted and their kinetic energy should depend on only the intensity of the light, not its frequency. In fact, however, each metal was found to have a characteristic threshold frequency of light; below that frequency, no electrons are emitted regardless of the light’s intensity. Above the threshold frequency, the number of electrons emitted was found to be proportional to the intensity of the light, and their kinetic energy was proportional to the frequency. This phenomenon was called the photoelectric effect (A phenomenon in which electrons are ejected from the surface of a metal that has been exposed to light). Albert Einstein (1879–1955; Nobel Prize in Physics, 1921) quickly realized that Planck’s hypothesis about the quantization of radiant energy could also explain the photoelectric effect. The key feature of Einstein’s hypothesis was the assumption that radiant energy arrives at the metal surface in particles that we now call photons (a energy given by Equation 8.3.1 Einstein postulated that each metal has a particular electrostatic attraction for its electrons that must be overcome before an electron can be emitted from its surface ( = ν ). If photons of light with energy less than strike a metal surface, no single photon has enough energy to eject an electron, so no electrons are emitted regardless of the intensity of the light. If a photon with energy greater than strikes the metal, then part of its energy is used to overcome the forces that hold the electron to the metal surface, and the excess energy appears as the kinetic energy of the ejected electron: \[ kinetic\; energy\; of\; ejected\; electron=E-E_{o}=h u -h u _{o}=h\left ( u - u _{o} \right ) \label{8.3.2}\] When a metal is struck by light with energy above the threshold energy , the of emitted electrons is proportional to the of the light beam, which corresponds to the number of photons per square centimeter, but the of the emitted electrons is proportional to the of the light. Thus Einstein showed that the energy of the emitted electrons depended on the frequency of the light, contrary to the prediction of classical physics. In 1900, Einstein was working in the Swiss patent office in Bern. He was born in Germany and throughout his childhood his parents and teachers had worried that he might be developmentally disabled. The patent office job was a low-level civil service position that was not very demanding, but it did allow Einstein to spend a great deal of time reading and thinking about physics. In 1905, his "miracle year" he published four papers that revolutionized physics. One was on the special theory of relativity, a second on the equivalence of mass and energy, a third on Brownian motion, and the fourth on the photoelectric effect, for which he received the Nobel Prize in 1921, the theory of relativity and energy-matter equivalence being still controversial at the time Planck’s and Einstein’s postulate that energy is quantized is in many ways similar to Dalton’s description of atoms. Both theories are based on the existence of simple building blocks, atoms in one case and quanta of energy in the other. The work of Planck and Einstein thus suggested a connection between the quantized nature of energy and the properties of individual atoms. A ruby laser, a device that produces light in a narrow range of wavelengths emits red light at a wavelength of 694.3 nm (Figure \(\Page {4}\)). What is the energy in joules of a single photon? wavelength energy of single photon. The energy of a single photon is given by = ν = /λ. An x-ray generator, such as those used in hospitals, emits radiation with a wavelength of 1.544 Å. The Photoelectric Effect: The properties of , the radiation emitted by hot objects, could not be explained with classical physics. Max Planck postulated that energy was quantized and could be emitted or absorbed only in integral multiples of a small unit of energy, known as a . The energy of a quantum is proportional to the frequency of the radiation; the proportionality constant is a fundamental constant (Planck’s constant). Albert Einstein used Planck’s concept of the quantization of energy to explain the , the ejection of electrons from certain metals when exposed to light. Einstein postulated the existence of what today we call , particles of light with a particular energy, = ν. Both energy and matter have fundamental building blocks: quanta and atoms, respectively. ( ) | 10,588 | 4,131 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.12%3A_Boiling_Point |
When we heat a liquid until it boils, the bubbles that form inside the liquid consist of pure vapor. If the liquid is well stirred while boiling occurs, the vapor in the bubbles will be in equilibrium with the liquid and will have a pressure equal to the vapor pressure at the boiling temperature. However, the pressure inside the bubbles must also be equal to the external pressure above the liquid. If this were not so, the bubbles would either suddenly collapse or suddenly expand. It follows therefore that when a liquid boils, the . Normally when we boil a liquid, we do so at atmospheric pressure. If this pressure is the standard pressure of 1 atm (101.3 kPa), then the temperature at which the liquid boils is referred to as its . This is the boiling point which is usually quoted in chemical literature. Not everyone lives at sea level, though. Denver, Colorado, for example, is about a mile high, and the average atmospheric pressure there is only 630 mmHg (84 kPa). Liquids attain a vapor pressure of 630 mmHg at a somewhat lower temperature than is required to produce 760 mmHg (1 atm). Consequently liquids in Denver boil some 4 to 5°C lower than the normal boiling point. Since the boiling point is often used to identify a liquid, chemists living at high altitudes must be careful to allow for this difference. The dependence of the boiling point on the external pressure can often be very useful. Chemists often purify liquids by boiling them and collecting the vapor, a process known as . Some liquids have such high normal boiling points that they begin to decompose before distillation can be carried out. Such a liquid can often be distilled at reduced pressure. The temperature of boiling is then much lower, and the risk of decomposition considerably less. The reverse procedure is used in a pressure cooker. The pressure inside the sealed cooker builds up until it is larger than atmospheric, and so the water used for cooking boils at a temperature above its normal boiling point. Therefore the cooking proceeds more rapidly. The following video highlights the idea that boiling point is dependent upon both temperature and pressure. In the video, water is boiled in a flask, which is then stoppered and removed from the heat source. When cold water is poured over the top of the flask, it cools the gas above the liquid water. This decreases the vapor pressure above the water. The lower vapor pressure corresponds to a lower boiling point, and therefore the water boils again. Note that if cooling had been applied to the liquid on the bottom, these subsequent boilings would not occur. From the figure displaying boiling points of four alkanes, estimate the boiling points of the four alkanes when the pressure is reduced to 600 mmHg. Reading along the 600-mmHg line in the graph, we find that it meets the vapor-pressure curve for pentane at about 29°C. Accordingly this is the boiling point of pentane at 600 mmHg. Similarly we find the boiling point of hexane to be 61C, and of heptane to be 90°C. The boiling point of octane is above 100°C and cannot be estimated from the graph. | 3,125 | 4,132 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/02%3A_Gases/2.07%3A_Real_Gases |
While the ideal gas law is sufficient for the prediction of large numbers of properties and behaviors for gases, there are a number of times that deviations from ideality are extremely important. Several equations of state have been suggested to account for the deviations from ideality. One simple, but useful, expression is that proposed by Johannes Diderik van der Waals (1837 – 1923) (Johannes Diderik van der Waals - Biographical, 2014) van der Waals’ equation introduced corrections to the pressure and volume terms of the ideal gas law in order to account for intermolecular interactions and molecular size respectively. \[ \left ( p + \dfrac{a}{V_m^2} \right) (V_m - b) = RT \label{vdw} \] or \[ p =\dfrac{RT}{V_m-b} - \dfrac{a}{V_m^2} \label{vdw2} \] In this expression, \(a\) and \(b\) are variables of a given substance which can be measured and tabulated. In general, molecules with large intermolecular forces will have large values of \(a\), and large molecules will have large values of b. Some van der Waals constants are given in Table \(\Page {1}\). The van der Walls model is useful because it makes it so simple to interpret the parameters in terms of molecular size and intermolecular forces. But it does have limitations as well (as is the case of every scientific model!) Some other useful two-parameter and three-parameter (or more) equations of state include the Redlich-Kwong, Dieterici, and Clausius models (Table \(\Page {2}\)). These have the advantage that they allow for temperature dependence on some of the parameters, which as will be seen later, is necessary to model certain behaviors of real gases. A very handy expression that allows for deviations from ideal behavior is the of state. This is a simple power series expansion in which the higher-order terms contain all of the deviations from the ideal gas law. \[ p =\dfrac{RT}{V_m} \left(1+ \dfrac{B(T)}{V_m} +\dfrac{C(T)}{V_m} \dots \right ) \label{viral} \] In the limit that B(T) (the ) and C(T) are zero, the equation becomes the ideal gas law. Also, the molar volume of gases are small, the contributions from the third, fourth, etc. terms decrease in magnitude, allowing one to truncate the series at a convenient point. The second virial coefficient can be predicted from a theoretical intermolecular potential function by \[B(T) = N_a \int _{r=0}^{\infty} \left[ 1- \exp \left(\dfrac{U(r)}{k_BT} \right) \right] 2\pi r^2 \,dr \nonumber \] The quality of an intermolecular potential can be determined (partially) by the potential’s ability to predict the value of the second virial coefficient, \(B(T)\). An intermolecular potential function is used to describe the interactions between molecules. These interactions will have to include attractive forces, which will draw molecules together, and repulsive forces which will push them apart. If the molecules are hard spheres, lacking any attractive interactions, the potential function is fairly simple. \[ U(r) ={\begin{cases} \infty &{\text{for }}r\leq \sigma\\0&{\text{for }}r>\sigma.\end{cases}} \nonumber \] In this function, \(\sigma\) is determined by the size of the molecules. If two molecules come within a distance \(r\) of one another, they collide, bouncing off in a perfectly elastic collision. Real molecules, however, with have a range of intermolecular separations through which they will experience attractive forces (the so-called “soft wall” of the potential surface.) And then at very small separations, the repulsive forces will dominate, pushing the molecules apart (the so-called “hard wall” of the potential surface.) A commonly used intermolecular potential, \(U(r)\), is the . This function has the form \[ U(r) = 4 \epsilon \left[ \underbrace{\left(\dfrac{\sigma}{r}\right)^{12}}_{\text{repulsive term}} - \underbrace{\left(\dfrac{\sigma}{r}\right)^{6}}_{\text{attractive term}} \right] \nonumber \] where \(\sigma\) governs the width of the potential well, and \(\epsilon\) governs the depth. The distance between molecules is given by \(r\). The repulsive interactions between molecules are contained in the first terms and the attractive interactions are found in the second term. A commonly used method of creating a power series based on another equation is the . This is an expansion of a function about a useful reference point where each of the terms is generated by differentiating the original function. For a function \(f(x)\), the Taylor series \(F(x)\) can be generated from the expression \[F(x) = f(a) + \left.\dfrac{d}{dx} f(x) \right|_{x=a} (x-a) + \dfrac{1}{2!} \left. \dfrac{d^2}{dx^2} f(x) \right|_{x=a} (x-a)^2 + \dots \nonumber \] This can be applied to any equation of state to derive an expression for the virial coefficients in terms of the parameters of the equation of state. The van der Waals equation can be written in terms of molar volume (Equation \ref{vdw2}). When multiplying the right hand side by \(\frac{u}{u}\) (where \(u = 1/v\)) yields: \[ p =\dfrac{RTu}{1-bu} - au^2 \nonumber \] This expression can be "Talyor" expanded (to the first three terms) about \(u = 0\) (which corresponds to an infinite molar volume.) The coefficient terms that are needed for the expansion are \[p(u=0)=0 \nonumber \] \[ \dfrac{dp}{du} \big|_{u=0} = \left [ \dfrac{RT}{1-bu} + \dfrac{bRTu}{(1-bu)^2} - 2au \right]_{u=0} = RT \nonumber \] \[ \dfrac{d^2p}{du^2} \big|_{u=0} = \dfrac{1}{2} \left [ \dfrac{bRT}{(1-bu)^2} + \dfrac{bRT}{(1-bu)^2} - \dfrac{2b^2RTu}{(1-bu)^3} - 2au \right]_{u=0} = RT -a \nonumber \] \[ \dfrac{d^3p}{du^3} \big|_{u=0} = RTb^2 \nonumber \] And the virial equation can then be expressed in terms of the van der Waals parameters as \[ p = 0 +RT(u) + (bRT -a)(u)^2 + RTb^{2(u)^3} \nonumber \] Substituting \(u = 1/V\) and simplifying gives the desired result: \[ p= RT \left[ \dfrac{1}{V} + \dfrac{b-\frac{a}{RT}}{V^2} + \dfrac{b^2}{V^3} + \dots \right] \nonumber \] And the second virial coefficient is given by \[ B(T) = b-\dfrac{a}{RT} \nonumber \] A useful way in which deviations from ideality can be expressed is by defining the (\(Z\)) given by \[ Z = \dfrac{pV_m}{RT} \nonumber \] where \(V_m\) is the molar volume. For an ideal gas, \(Z = 1\) under all combinations of \(P\), \(V_m\), and \(T\). However, real gases will show some deviation (although all gases approach ideal behavior at low p, high V , and high T.) The compression factor for nitrogen at several temperatures is shown below over a range of pressures. As can be seen, the gas behaves closer to ideally over a longer range of pressure at the higher temperatures. In general, there is one temperature, the , at which a gas will approach ideal behavior as the pressure goes to zero asymptotically, and thus behave ideally over a broad range of lower pressures. The Boyle temperature is found by solving or Using the virial equation of state (Equation \ref{viral}), the Boyle temperature can be expressed in terms of the virial coefficients. Starting with the compression factor \[Z = 1 +\dfrac{B}{V_m} + \dots \nonumber \] and then differentiating with respect to \(1/V_m\) yields \[ So it can be concluded that at the Boyle temperature, the second virial coefficient \(B\) is equal to zero. This should make some sense given that the first virial coefficient provides most of the deviation from the ideal gas law, and so it must vanish as the gas behaves more ideally. The isotherms (lines of constant temperature) of CO reveal a very large deviation from ideal behavior. At high temperatures, CO behaves according to Boyle’s Law. However, at lower temperatures, the gas begins to condense to form a liquid at high pressures. At one specific temperature, the , the isotherm begins to display this critical behavior. The temperature, pressure, and molar volume (\(p_c\), \(T_c\), and \(V_c\)) at this point define the . In order to solve for expressions for the critical constants, one requires three equations. The equation of state provides one relationship. The second can be generated by recognizing that the slope of the isotherm at the critical point is zero. And finally, the third expression is derived by recognizing that the isotherm passes through an inflection point at the critical point. Using the van der Waals equation as an example, these three equations can be generated as follows: Solving these expressions for \(p_c\), \(T_c\), and \(V_c\) yields \[p_c = \dfrac{a}{27b^2} \nonumber \] \[T_c = \dfrac{8a}{27bR} \nonumber \] \[V_c = 3b \nonumber \] The critical variables can be used in this fashion to determine the values of the molecular parameters used in an equation of state (such as the van der Waals equation) for a given substance. The was proposed by van der Waals in 1913 (van der Waals J. D., 1913). He noted that the compression factor at the critical point \[Z_c = \dfrac{p_cV_c}{RT_c} \nonumber \] is very nearly the same for any substance. This is consistent with what is predicted by the van der Waals equation, which predicts \(Z_c = 0.375\) irrespective of substance. Further, it can be noted that based on defined by \[p_r= \dfrac{p}{p_c} \nonumber \] \[V_r= \dfrac{V}{V_c} \nonumber \] \[T_r= \dfrac{T}{T_c} \nonumber \] \[[ \dfrac{p_cV_c}{RT_c} \approx 0.292 \nonumber \] Also, the reduced compression factor can be plotted as a function of reduced pressure for several substances at several reduced isotherms with surprising consistency irrespective of the | 9,424 | 4,133 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.02%3A_Atomic_Spectra |
The photoelectric effect provided indisputable evidence for the existence of the photon and thus the particle-like behavior of electromagnetic radiation. The concept of the photon, however, emerged from experimentation with , electromagnetic radiation emitted as the result of a source’s temperature, which produces a continuous spectrum of energies. More direct evidence was needed to verify the quantized nature of electromagnetic radiation. In this section, we describe how experimentation with visible light provided this evidence. Although objects at high temperature emit a continuous spectrum of electromagnetic radiation, a different kind of spectrum is observed when pure samples of individual elements are heated. For example, when a high-voltage electrical discharge is passed through a sample of hydrogen gas at low pressure, the resulting individual isolated hydrogen atoms caused by the dissociation of H emit a red light. Unlike blackbody radiation, the color of the light emitted by the hydrogen atoms does not depend greatly on the temperature of the gas in the tube. When the emitted light is passed through a prism, only a few narrow lines, called a line spectrum, which is a spectrum in which light of only a certain wavelength is emitted or absorbed, rather than a continuous range of wavelengths (Figure \(\Page {1}\)), rather than a continuous range of colors. The light emitted by hydrogen atoms is red because, of its four characteristic lines, the most intense line in its spectrum is in the red portion of the visible spectrum, at 656 nm. With sodium, however, we observe a yellow color because the most intense lines in its spectrum are in the yellow portion of the spectrum, at about 589 nm. Such were observed for many other elements in the late 19th century, which presented a major challenge because classical physics was unable to explain them. Part of the explanation is provided by the observation of only a few values of λ (or ν) in the line spectrum meant that only a few values of were possible. Thus ; in other words, only states that had certain values of energy were possible, or . If a hydrogen atom could have value of energy, then a continuous spectrum would have been observed, similar to blackbody radiation. In 1885, a Swiss mathematics teacher, Johann Balmer (1825–1898), showed that the frequencies of the lines observed in the visible region of the spectrum of hydrogen fit a simple equation that can be expressed as follows: \[ u=constant\; \left ( \dfrac{1}{2^{2}}-\dfrac{1}{n^{^{2}}} \right ) \label{\(\Page {1}\)}\] where = 3, 4, 5, 6. As a result, these lines are known as the . The Swedish physicist Johannes Rydberg (1854–1919) subsequently restated and expanded Balmer’s result in the : \[ \dfrac{1}{\lambda }=\Re\; \left ( \dfrac{1}{n^{2}_{1}}-\dfrac{1}{n^{2}_{2}} \right ) \label{\(\Page {2}\)}\] where \(n_1\) and \(n_2\) are positive integers, \(n_2 > n_1\), and \( \Re \) the , has a value of 1.09737 × 10 m . A mathematics teacher at a secondary school for girls in Switzerland, Balmer was 60 years old when he wrote the paper on the spectral lines of hydrogen that made him famous. Balmer published only one other paper on the topic, which appeared when he was 72 years old. Like Balmer’s equation, Rydberg’s simple equation described the wavelengths of the visible lines in the emission spectrum of hydrogen (with = 2, = 3, 4, 5,…). More important, Rydberg’s equation also described the wavelengths of other series of lines that would be observed in the emission spectrum of hydrogen: one in the ultraviolet ( = 1, = 2, 3, 4,…) and one in the infrared ( = 3, = 4, 5, 6). Unfortunately, scientists had not yet developed any theoretical justification for an equation of this form. The Bohr Atom: There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Most light is polychromatic and contains light of many wavelengths. Light that has only a single wavelength is monochromatic and is produced by devices called lasers, which use transitions between two atomic energy levels to produce light in a very narrow range of wavelengths. Atoms can also absorb light of certain energies, resulting in a transition from the ground state or a lower-energy excited state to a higher-energy excited state. This produces an , which has dark lines in the same position as the bright lines in the of an element. Atoms of individual elements emit light at only specific wavelengths, producing a rather than the continuous spectrum of all wavelengths produced by a hot object. Niels Bohr explained the line spectrum of the hydrogen atom by assuming that the electron moved in circular orbits and that orbits with only certain radii were allowed. Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the of the atom and was most stable; orbits farther away were higher-energy . Transitions from an excited state to a lower-energy state resulted in the emission of light with only a limited number of wavelengths. Bohr’s model could not, however, explain the spectra of atoms heavier than hydrogen. ( ) | 5,311 | 4,137 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/14%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy |
After you have completed Chapter 14, you should be able to You have already studied the chemistry of compounds that contain one carbon-carbon double bond. In this chapter, you will focus your attention on compounds that contain two or more such bonds. In particular you will study the properties of those compounds that contain two carbon-carbon double bonds which are separated by one carbon-carbon single bond. These compounds are called “conjugated dienes.” To understand the properties exhibited by conjugated dienes, you must first examine their bonding in terms of the molecular orbital theory introduced in . Then, you must learn how the products of a reaction are dependent on both thermodynamic and kinetic considerations. Which of these two factors is the most important can sometimes determine which of two possible products will predominate when a reaction is carried out under specific conditions. Although we shall not make extensive use of ultraviolet spectroscopy, this technique can often provide important information when conjugated compounds are being investigated. In general, ultraviolet spectroscopy is less useful than the other spectroscopic techniques introduced earlier.
| 1,219 | 4,139 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Acid_Rain/Sources_of_Nitrogen_Oxides |
A natural source of nitrogen oxides occurs from a lightning stroke. The very high temperature in the vicinity of a lightning bolt causes the gases oxygen and nitrogen in the air to react to form \[\ce{N2 + O2 -> NO}\] The nitric oxide very quickly reacts with more oxygen to form \[\ce{NO + O2 -> NO2}\] Both of the nitrogen compounds are known collectively as nitrogen oxides or \(\ce{NO_{x}}\). At normal temperatures the oxygen and nitrogen gases do not react together. In the presence of very high temperatures nitrogen and oxygen do react together to form nitric oxide. These conditions are found in the combustion of coal and oil at electric power plants, and also during the Both of these sources contribute about equally to the formation of nitrogen oxides. In areas of high automobile traffic, such as in large cities, the amount of nitrogen oxides emitted into the atmosphere can be quite significant. In the Los Angeles area, the main source of acid rain is from automobiles. In certain national parks such as Yosemite and Sequoia, automobile traffic is banned to limit the amount of air pollution damage to the trees and plants. This also has the effect of reducing the visual smog in the air. | 1,227 | 4,143 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Vitamins_Cofactors_and_Coenzymes/Vitamin_A |
β-carotene is the molecule that gives carrots, sweet potatoes, squash, and other yellow or orange vegetables their orange color. It is part of a family of chemicals called the carotenoids, which are found in many fruit and vegetables, as well as some animal products such as egg yolks. Carotenoids were first isolated in the early 19th century, and have been synthesized for use as food colorings since the 1950s. Biologically, β-carotene is most important as the precursor of vitamin A in the human diet. It also has anti-oxidant properties and may help in preventing cancer and other diseases. The long chain of alternating double bonds (conjugated) is responsible for the orange color of beta-carotene. The conjugated chain in carotenoids means that they absorb in the visible region - green/blue part of the spectrum. So β-carotene appears orange, because the red/yellow colors are reflected back to us. Vitamin A has several functions in the body. The most well known is its role in - hence carrots "make you able to see in the dark". The retinol is oxidized to its aldehyde, retinal, which complexes with a molecule in the eye called opsin. When a photon of light hits the complex, the retinal changes from the 11-cis form to the all-trans form, initiating a chain of events which results in the transmission of an impulse up the optic nerve. A more detailed explanation is in . Other roles of vitamin A are much less well understood. It is known to be involved in the synthesis of certain glycoproteins, and that deficiency leads to abnormal bone development, disorders of the reproductive system, xerophthalmia (a drying condition of the cornea of the eye) and ultimately death. Vitamin A is required for healthy skin and mucus membranes, and for night vision. Its absence from diet leads to a loss in weight and failure of growth in young animals, to the eye diseases; xerophthalmia, and night blindness, and to a general susceptibility to infections. It is thought to help prevent the development of cancer. Good sources of carotene, such as green vegetables are good potential sources of vitamin A. Vitamin A is also synthetically manufactured by extraction from fish-liver oil and by synthesis from beta-ionone. β- β-Carotene is converted into vitamin A in the liver. Two molecules of vitamin A are formed from on molecule of beta carotene. If you compare the two molecules, it is clear that vitamin A (retinol) is very closely related to half of the beta-carotene molecule. One way in which beta-carotene can be converted to vitamin A is to break it apart at the center and is thought to be most important biologically. The breakdown of beta-carotene occurs in the walls of the small intestine (intestinal mucosa) and is catalyzed by the enzyme β-carotene dioxygenase to form retinal. The retinal reduced to retinol by retinaldehyde reductase in the intestines. This is the reduction of an aldehyde by the addition of hydrogen atoms to make the alcohol, retinol. The absorption of retinol from the alimentary tract is favored by the simultaneous absorption of fat or oil, especially if these are unsaturated. Retinol is esterified to palmitic acid and delivered to the blood via chylomicrons. Finally the retinol formed is stored in the liver as retinyl esters. This is why cod liver oil used to be taken as a vitamin A supplement. It is also why you should never eat polar bear liver if you run out of food in the Arctic; vitamin A is toxic in excess and a modest portion of polar bear liver contains more than ! Beta-carotene, on the other hand, is a safe source of vitamin A. The efficiency of conversion of beta-carotene to retinol depends on the level in the diet. If you eat more beta-carotene, less is converted, and the rest is stored in fat reserves in the body. So too much beta-carotene can make you turn yellow, but will not kill you with hypervitaminosis. | 3,907 | 4,144 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/SN2/Nucleophile |
Now that we have determined what will make a good leaving group, we will now consider . That is, the relative strength of the nucleophile. Nucleophilicity depends on many factors, including charge, basicity, solvent, polarizability, and the nature of the substituents. Nucleophiles can be neutral or negatively charged. In either case, it is important that the nucleophile be a good Lewis base, meaning it has electrons it wants to share. The following diagram is just a reminder of some of the nucleophiles that were presented in the section covering nucleophilic substitution. In looking at these two types of nucleophiles, you should notice that a reactive atom, such as oxygen, in a neutral species can also be a reactive atom in a negatively charged species. For example, the O in OH is negatively charged, but the O in H O is neutral. It has been experimentally shown that a nucleophile containing a negatively charged reactive atom is better than a nucleophile containing a reactive atom that is neutral. The next diagram illustrates this concept. Notice that when oxygen is part of the hydroxide ion, it bears a negative charge, and when it is part of a water molecule, it is neutral. The O of OH is a better nucleophile than the O of H O, and results in a faster reaction rate. Similarly, when nitrogen is part of NH , it bears a negative charge, and when it is part of NH , it is neutral. The N of NH is a better nucleophile than the N of NH , and results in a faster reaction rate. To say that nucleophilicity follows basicity across a row means that, as basicity increases from right to left on the periodic table, nucleophilicity also increases. As basicity decreases from left to right on the periodic table, nucleophilicity also decreases. When it comes to nucleophilicity, do not assign this same rule when making comparisons between the halogens located in a column. In this case of moving up and down a column, nucleophilicity does not always follow basicity. It depends on the type of solvent you are using. In the section Nucleophilic Substitution, we assigned a relationship to leaving groups containing C, N, O, and F, showing that the strength of the leaving group follows electronegativity. This is based on the fact that the best leaving groups are those that are weak bases that do not want to share their electrons. The best nucleophiles however, are good bases that want to share their electrons with the electrophilic carbon. The relationship shown below, therefore, is the exact opposite of that shown for the strength of a leaving group. In general chemistry, we classified solvents as being either polar or nonpolar. Polar solvents can be further subdivided into protic and and aprotic solvents. A is a solvent that has a hydrogen atom bound to an oxygen or nitrogen. A few examples of protic solvents include H O, ROH, RNH , and R NH, where water is an example of an inorganic protic solvent, and alcohols and amides are examples of organic solvents. The diagram below shows a few examples of protic solvents we will see. Since oxygen and nitrogen are highly electronegative atoms, the O-H and N-H bonds that are present in protic solvents result in a hydrogen that is positively polarized. When protic solvents are used in nucleophilic substitution reactions, the positively polarized hydrogen of the solvent molecule can interact with the negatively charged nucleophile. In solution, molecules or ions that are surrounded by these solvent molecules are said to be . Solvation is the process of attraction and association of solvent molecules with ions of a solute. The solute, in this case, is a negatively charged nucleophile. The following diagram depicts the interaction that can occur between a protic solvent and a negatively charged nucleophile. The interactions are called . A hydrogen bond results from a from a dipole-dipole force between between an electronegative atom, such as a halogen, and a hydrogen atom bonded to nitrogen, oxygen or fluorine. In the case below, we are using an alcohol (ROH) as an example of a protic solvent, but be aware that this interaction can occur with other solvents containing a positively polarized hydrogen atom, such as a molecule of water, or amides of the form RNH and R NH. Why is this important? Solvation weakens the nucleophile; that is, solvation decreases nucleophilicity. This is because the solvent forms a "shell" around the nucleophile, impeding the nucleophile's ability to attack an electrophilic carbon. Furthermore, because the charge on smaller anions is more concentrated, . The picture below illustrates this concept. Notice how the smaller fluoride anion is represented as being more heavily solvated than the larger iodide anion. This means that the fluoride anion will be a weaker nucleophile than the iodide anion. In fact, it is important to note that fluoride will not function as a nucleophile at all in protic solvents. It is so small that solvation creates a situation whereby fluoride's lone pair of electrons are no longer accessible, meaning it is unable to participate in a nucleophilic substitution reaction. Previously we learned how nucleophilicity follows basicity when moving a row. In our discussion on the effect of protic solvents on nucleophilicity, we learned that solvation weakens the nucleophile, having the greatest effect on smaller anions. In effect, when using protic solvents, nucleophilicity does follow basicity when moving up and down a column. In fact, it's the exact opposite: when basicity , nucleophilicity and when basicity , nucleophilicity . An is a solvent that lacks a positively polarized hydrogen. The next diagram illustrates several polar aprotic solvents that you should become familiar with. Aprotic solvents, like protic solvents, are polar but, because they lack a positively polarized hydrogen, they do not form hydrogen bonds with the anionic nucleophile. The result, with respect to solvation, is a relatively weak interaction between the aprotic solvent and the nucleophile. The consequence of this weakened interaction is two-fold. First, by using an aprotic solvent we can raise the reactivity of the nucleophile. This can sometimes have dramatic effects on the rate at which a nucleophilic substitution reaction can occur. For example, if we consider the reaction between bromoethane and potassium iodide, the reaction occurs 500 times faster in acetone than in methanol. A second consequence that results from the weak interaction that occurs between aprotic solvents and nucleophiles is that, under some conditions, there can be an inversion of the reactivity order. An inversion would result in nucleophilicity following basicity up and down a column, as shown in the following diagram. When we considered the effects of protic solvents, remember that the iodide anion was the strongest nucleophile. Now, in considering aprotic solvents under some conditions, the fluoride anion is the strongest nucelophile. Thus far, our discussion on nucleophilicity and solvent choice has been limited to negatively charged nucleophiles, such as F , Cl , Br , and I . With respect to these anions we learned that, when using protic solvents, nucleophilicity does not follow basicity, and when using aprotic solvents, the same relationship can occur, or there could be an inversion in the order of reactivity. What happens as we move up and down a column when considering ? It turns out that, in the case of uncharged nucleophiles, size dictates nucleophilicity. This is because larger elements have bigger, more diffuse, and more polarizable electron clouds. This cloud facilitates the formation of a more effective orbital overlap in the transition state of bimolecular nucleophilic substitution (SN ) reactions, resulting in a transition state that is lower in energy and a nucleophilic substitution that occurs at a faster rate. In the section , we learned that the SN transition state is very crowded. Recall that there are a total of 5 groups around the electrophilic center. For this reason, sterically hindered nucleophiles react more slowly than those lacking steric bulk. | 8,172 | 4,145 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Reactive_Intermediates/Carbocations |
A is an ion with a positively-charged carbon . Among the simplest examples are methenium , methanium , and ethanium . Some carbocations may have two or more positive charges, on the same carbon atom or on different atoms; such as the ethylene dication . Until the early 1970s, all carbocations were called carbonium ions.[2] In present-day chemistry, a carbocation is any positively charged carbon atom, classified in two main categories according to the valence of the charged carbon: This nomenclature was proposed by G. A. Olah.[3] University-level textbooks discuss carbocations only as if they are carbenium ions, or discuss carbocations with a fleeting reference to the older phrase of carbonium ion or carbenium and carbonium ions. The history of carbocations dates back to 1891 when G. Merling[8] reported that he added bromine to tropylidene (cycloheptatriene) and then heated the product to obtain a crystalline, water-soluble material, \(C_7H_7Br\). He did not suggest a structure for it; however, Doering and Knox[9] convincingly showed that it was tropylium (cycloheptatrienylium) bromide. This ion is predicted to be aromatic by Hückel's rule. In 1902, Norris and Kehrman independently discovered that colorless triphenylmethanol gives deep-yellow solutions in concentrated sulfuric acid. Triphenylmethyl chloride similarly formed orange complexes with aluminium and tin chlorides. In 1902, Adolf von Baeyer recognized the salt-like character of the compounds formed. He dubbed the relationship between color and salt formation halochromy, of which malachite green is a prime example. Carbocations are reactive intermediates in many organic reactions. This idea, first proposed by Julius Stieglitz in 1899,[10] was further developed by Hans Meerwein in his 1922 study[11,12] of the Wagner-Meerwein rearrangement. Carbocations were also found to be involved in the \(S_N1\) reaction, the \(E1\0 reaction, and in rearrangement reactions such as the Whitmore 1,2 shift. The chemical establishment was reluctant to accept the notion of a carbocation and for a long time the Journal of the American Chemical Society refused articles that mentioned them. The first NMR spectrum of a stable carbocation in solution was published by Doering et al.[13] in 1958. It was the heptamethylbenzenium ion, made by treating hexamethylbenzene with methyl chloride and aluminium chloride. The stable 7-norbornadienyl cation was prepared by Story et al. in 1960[14] by reacting norbornadienyl chloride with silver tetrafluoroborate in sulfur dioxide at −80 °C. The NMR spectrum established that it was non-classically bridged (the first stable non-classical ion observed). In 1962, Olah directly observed the tert-butyl carbocation by nuclear magnetic resonance as a stable species on dissolving tert-butyl fluoride in magic acid. The NMR of the norbornyl cation was first reported by Schleyer et al.[15] and it was shown to undergo proton-scrambling over a barrier by Saunders et al.[16] Order of stability of examples of tertiary (III), secondary (II), and primary (I) alkyl , as well as the methyl cation (far right). Carbocations are often the target of nucleophilic attack by like (OH ) ions or ions. Carbocations typically undergo from less stable structures to equally stable or more stable ones with in excess of 10 /sec. This fact complicates synthetic pathways to many compounds. For example, when 3-pentanol is heated with aqueous HCl, the initially formed 3-pentyl carbocation rearranges to a statistical mixture of the 3-pentyl and 2-pentyl. These cations react with chloride ion to produce about 1/3 3-chloropentane and 2/3 2-chloropentane. A carbocation may be stabilized by by a carbon-carbon double bond next to the ionized carbon. Such cations as cation CH =CH–CH and cation C H –CH are more stable than most other carbocations. Molecules that can form allyl or benzyl carbocations are especially reactive. These carbocations where the C+ is adjacent to another carbon atom that has a double or triple bond have extra stability because of the overlap of the empty p orbital of the carbocation with the p orbitals of the π bond. This overlap of the orbitals allows the charge to be shared between multiple atoms – delocalization of the charge - and, therefore, stabilizes the carbocation. Some carbocations such as the norbornyl cation exhibit more or less symmetrical three centre bonding. Cations of this sort have been referred to as non-classical ions. The energy difference between "classical" carbocations and "non-classical" isomers is often very small, and in general there is little, if any, activation energy involved in the transition between "classical" and "non-classical" structures. In essence, the "non-classical" form of the 2-butyl carbocation is 2-butene with a proton directly above the centre of what would be the carbon-carbon double bond. "Non-classical" carbocations were once the subject of great controversy. One of George Olah's greatest contributions to chemistry was resolving this controversy.[17] Cyclopropylcarbinyl cations can be studied by NMR:[18,19] In the NMR spectrum of a dimethyl derivative, two nonequivalent signals are found for the two methyl groups, indicating that the molecular conformation of this cation not perpendicular (as in A) but is bisected (as in B) with the empty p-orbital and the cyclopropyl ring system in the same plane: In terms of bent bond theory, this preference is explained by assuming favorable orbital overlap between the filled cyclopropane bent bonds and the empty p-orbital. | 5,603 | 4,146 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Protein_Structure/Protein_Folding |
Proteins have several layers of structure each of which is important in the process of protein folding. The first most basic level of this structure is the sequence of amino acids themselves. The sequencing is important because it will determine the types of interactions seen in the protein as it is folding. A novel sequence-based method based on the assumption that protein-protein interactions are more related to amino acids at the surface than those at the core. This study shows that not only is the amino acids that are in a protein important but also the order in which they are sequenced. The interactions of the amino acids will determine what the secondary and tertiary structure of the protein will be. The next layer in protein structure is the secondary structure. The secondary structure includes architectural structures that extend in one dimension. Secondary structure includes α-Helixes and β-sheets ( Figure \(\Page {1}\)). The α-helices, the most common secondary structure in proteins, the peptide –CO–NH–groups in the backbone form chains held together by NH ̄OC hydrogen bonds.” The α-helices form the backbone of proteins and help to aid in the folding process. The β-sheets form in two distinct ways. They are able to form in both parallel β-pleated sheets and anti parallel β-pleated sheets. When the α-helix or β-sheet is formed, the excluded volumes generated by the backbone and side chains overlap, leading to an increase in the total volume available to the translational displacement of water molecules. This is important because it leads to a more thermodynamically stable conformation and leads to less strain on the protein as a whole and thus are aided by the conformation. The tertiary structure is the next layer in protein structure. This takes the α-Helixes and β-sheets and allows them to fold into a three dimensional structure. Most proteins take on a globular structure once folded. The description of globular protein structures as an ensemble of contiguous closed loops or tightened end fragments reveals fold elements crucial for the formation of stable structures and for navigating the very process of protein folding. The globular proteins generally have a hydrophobic core surrounded by a hydrophilic outer layer. These interactions are important because they lead to the global structure and help create channels and binding sites for enzymes. The last layer of protein structure is the quaternary structure. The folding transition and the functional transitions between useful states are encoded in the linear sequence of amino acids, and a long- term goal of structural biology is to be able to predict both the structure and function of molecules from the information in the sequence. The Subunit organization is the last level of structure in protein molecules. The organization of the subunits is important because that determines the types of interactions that can form and dictates its use in the body. Proteins are folded and held together by several forms of molecular interactions. The molecular interactions include the thermodynamic stability of the complex, the hydrophobic interactions and the disulfide bonds formed in the proteins. The figure below (Figure \(\Page {2}\)) is an example of protein folding. The biggest factor in a proteins ability to fold is the thermodynamics of the structure. The interaction scheme includes the short-range propensity to form extended conformations, residue-dependent long-range contact potentials, and orientation-dependent hydrogen bonds. The thermodynamics are a main stabilizing force within a protein because if it is not in the lowest energy conformation it will continue to move and adjust until it finds its most stable state. The use of energy diagrams and maps are key in finding out when the protein is in the most stable form possible. The next type of interaction in protein folding is the hydrophobic interactions within the protein. The framework model and the hydrophobic collapse model represent two canonical descriptions of the protein folding process. The first places primary reliance on the short-range interactions of secondary structure and the second assigns greater importance to the long-range interactions of tertiary structure. These hydrophobic interactions have an impact not just on the primary structure but then lead to changes seen in the secondary and tertiary structure as well. Globular proteins acquire distinct compact native con- formations in water as a result of the hydrophobic effect. When a protein has been folded in the correct way it usually exists with the hydrophobic core as a result of being hydrated by waters in the system around it which is important because it creates a charged core to the protein and can lead to the creation of channels within the protein. The hydrophobic interactions are found to affect time correlation functions in the vicinity of the native state even though they have no impact on same time characteristics of the structure fluctuations around the native state. The hydrophobic interactions are shown to have an impact on the protein even after it has found the most stable conformation in how the proteins can interact with each other as well as folding themselves. Another type of interaction seen when the protein is folding is the disulfide linkages that form in the protein (Figure \(\Page {3}\)). The disulfide bond, a sulfur- sulfur chemical bond that results from an oxidative process that links nonadjacent (in most cases) cysteine’s of a protein. These are a major way that proteins get into their folded form. The types of disulfide bonds are cysteine-cysteine linkage is a stable part of their final folded structure and those in which pairs of cysteines alternate between the reduced and oxidized states. The more common is the linkages that cause the protein to fold together and link back on itself compared to the cysteines that are changing oxidation states because the bonds between cysteines once created are fairly stable. Proteins can miss function for several reasons. When a protein is miss folded it can lead to denaturation of the protein. Denaturation is the loss of protein structure and function. The miss folding does not always lead to complete lack of function but only partial loss of functionality. The miss functioning of proteins can sometimes lead to diseases in the human body. Alzheimer's Disease (AD) is a neurological degenerative disease that affects around 5 million Americans, including nearly half of those who are age 85 or older. The predominant risk factors of AD are age, family history, and heredity. Alzheimer’s disease typically results in memory loss, confusion of time and place, misplacing places, and changes in mood and behavior. AD results in dense plaques in the brain that are comprised of fibrillar β-amyloid proteins with a well-orders β-sheet secondary structure. These plaques visually look like voids in the brain matter (see figure 5) and are directly connected to the deterioration of thought processes. It has been determined that AD is a protein misfolding disease, where the misfolded protein is directly related to the formation of these plaques in the brain. It is yet to be fully understood what exactly causes this protein misfolding to begin, but several theories point to oxidative stress in the brain to be the initiating factor. This oxidation results in damage to the phospholipids in the brain, which has been found to result in a faster accumulation of amyloid β-proteins. Cystic Fibrosis (CF) is a chronic disease that affects 30,000 Americans. The typical affects of CF is a production of thick, sticky mucus that clogs the lungs and leads to life-threatening lung infection, and obstructs the pancreas preventing proper food processing. CF is caused by protein misfolding. This misfolding then results in some change in the protein known as cystic fibrosis transmembrane conductance regulator (CFTR), which can result in this potentially fatal disease. In approximately 70% of CF cases, a deletion of phenylalanine at position 508 in the CFTR is deleted. This deletion of Phe508 seems to be directly connected to the formation of CF. The protein misfolding that results in CF occurs prior to birth, but it is not entirely clear as to why. | 8,349 | 4,147 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/The_Wild_Ionists |
Faraday proved that pure water doesn't conduct electricity, but conductivity increases when some types of solutes are added. These solutes are called . For a long time, people thought that the ions that let water conduct electricity (to conduct, charged particles must move, such as electrons through a metal wire or ions through solution) were formed by the electricity. Svante Arrhenius was the son of a minor university employee in Sweden. For his doctoral research, he wanted to study whether molecular weights of compounds could be measured using the conductivity of their solutions (a subject his adviser told him not to work on―he often ignored his advisers' advice). To do this, he needed to understand how the conductivity of electrolyte solutions depends on concentration, but he found that his data were unexpectedly confusing. Wisely, he realized that explaining the conductivities would actually be more interesting and useful than his original plan to measure molecular weights. He found that the electrolytes fell into 2 types: strong and weak. (like salt and HCl) easily conduct electricity, and their conductivity is proportional to the concentration. (like vinegar and ammonia) conduct electricity much less, but their conductivity is not directly proportional to concentration. As concentration decreases, conductivity does too, but conductivity/mass of solute increases. He proposed that electrolytes had an active state (that conducts) and an inactive state (that doesn't conduct electricity); strong electrolytes were entirely in the active state, while weak electrolytes would have some molecules in the active state and some in the inactive state. As a weak electrolyte solution was diluted, the % of molecules in the active state would increase toward 100%. But this was vague, and his advisers didn't like it and gave him a very low grade. Because the grade was so low, he had to travel and do research without pay in other labs for many years to establish himself, instead of becoming a professor himself. Arrhenius travelled and met other scientists. A professor named Ostwald helped introduce him. He became friends with Nernst (who would later do some important work on thermodynamics and electrochemistry). Nernst was a wild young man who thought about becoming an actor, had scars from fighting, and often got very drunk. He also began to correspond (write letters) with van't Hoff, whom we mentioned . Van't Hoff had moved on from carbon chemistry and was studying osmosis. occurs when water moves across a membrane (like the membrane of a biological cell) to go from a dilute solution on one side to a concentrated solution on the other. This can create a pressure difference. This is how some antibiotics kill bacteria: the bacteria explode from the pressure because the solution outside the cell is more dilute than the solution inside the cell. Van't Hoff found that the osmotic pressure depends on the concentration of the solutions, but for electrolytes, there's an extra "fudge factor"; the concentration seems higher than expected. It appeared that the electrolytes were breaking into pieces, and these pieces increases the apparent concentration of the solution when determining osmotic pressure. Arrhenius noticed that the "fudge factor" in Van't Hoff's data was related to the fractions in Arrhenius' active states. At this time, electrons hadn't been discovered yet. People thought that ions form in solution only when electricity is passed through. However, Arrhenius realized that electrolytes must be splitting into charged pieces (ions) when they dissolved, even without electricity, because these charged pieces affected van't Hoff's osmotic pressure measurements. The idea that salt, which forms so eagerly from the elements, would split up completely in water, seemed crazy to most scientists. (Now, we know that splitting into ions vs splitting into atoms is very different!) But Ostwald, Nernst and van't Hoff agreed with Arrhenius, and they were called "the wild army of the Ionists". They convinced other chemists pretty quickly that this new theory of ionization was correct. This discussion was inspired by first chapter of the book by Patrick Coffey. | 4,238 | 4,151 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Atmospheric_Chemistry/Photosynthesis |
Photosynthesis is the process of converting light energy ( ) to chemical energy and storing it in the chemical bonds of sugar-like molecules. This process occurs in plants and some algae (Kingdom Protista). Plants need only light energy, CO , and H O to make sugar. The process of photosynthesis takes place in the chloroplasts (chloro = green; plasti = formed, molded), specifically using chlorophyll (phyll = leaf), the green pigment involved in photosynthesis. As early as 1640, people have demonstrated that photosynthesis was a means of converting carbon dioxide in the air into plant material. Thus, photosynthesis has been studied by many scientists, and there are many sites related to photosynthesis. Some of these sites describe their recent recent research activities, whereas others are for educational purposes. Even books and journals are introduced in this link. The photosynthesis is a very complicated process, and we can only give an introduction here. Photosynthesis reduces carbon dioxide into carbohydrates, \[\ce{6 CO2 + 12 H2O + hv -> C6H12O6 + 6 O2 + 6 H2O}\] Thus, both electrons and energy are required. The electrons come from water molecules, and the energy is first absorbed by pigments known as chlorophylls and carotenoids. The former absorb blue (wavelength 430 nm) and red (wavelength 670 nm) light and the later absorbe blue-green light (wavelenghts between 400 and 500 nm). Green and yellow light are not absorbed. Reflection of these types of light makes plants appear green. There are many varieties of pigments. They are bonded to proteins which provide pigment molecules with the appropriate orientation and position with respect to each other. After absorption by pigment, light energy is transferred to chlorophylls that are bonded to special proteins. Pigments and protein involved with this actual primary electron transfer event together are called the . A large number of pigment molecules (100-5000), collectively referred to as , "harvest" light and transfer the light energy to the same reaction center. The purpose is to maintain a high rate of electron transfer in the reaction center, even at lower light intensities. Photosynthesis is responsible for the production of oxygen and carbohydrates in plants. All living organisms respire, and so do plants. In the respiration, oxygen is consumed and carbon dioxide is produced. respiration takes place all the time, but respiration is masked by higher rate of photosynthesis when the light intensities is high. Working with the green algae chlorella, Melvin Calvin and Andy Benson, at the University of California at Berkeley, elucidated the following pathway for the conversion of carbon dioxide into carbohydrates: | 2,731 | 4,152 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Pharmaceuticals/Drug_Receptor_Interactions |
The vast majority of drugs show a remarkably high correlation of structure and specificity to produce pharmacological effects. Experimental evidence indicates that drugs interact with receptor sites localized in macromolecules which have protein-like properties and specific three dimensional shapes. A minimum three point attachment of a drug to a receptor site is required. In most cases a rather specific chemical structure is required for the receptor site and a complementary drug structure. Slight changes in the molecular structure of the drug may drastically change specificity. Several chemical forces may result in a temporary binding of the drug to the receptor. Essentially any bond could be involved with the drug-receptor interaction. Covalent bonds would be very tight and practically irreversible. Since by definition the drug-receptor interaction is reversible, covalent bond formation is rather rare except in a rather toxic situation. Since many drugs contain acid or amine functional groups which are ionized at physiological pH, ionic bonds are formed by the attraction of opposite charges in the receptor site. Polar-polar interactions as in hydrogen bonding are a further extension of the attraction of opposite charges. The drug-receptor reaction is essentially an exchange of the hydrogen bond between a drug molecule, surrounding water, and the receptor site. Finally hydrophobic bonds are formed between non-polar hydrocarbon groups on the drug and those in the receptor site. These bonds are not very specific but the interactions do occur to exclude water molecules. Repulsive forces which decrease the stability of the drug-receptor interaction include repulsion of like charges and steric hindrance. Steric hindrance refers to certain 3-dimensional features where repulsion occurs between electron clouds, inflexible chemical bonds, or bulky alkyl groups. | 1,904 | 4,153 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Synthesis_of_Esters/Preparation_of_Esters |
This page describes ways of making esters in the lab from alcohols and phenols using carboxylic acids, acyl chlorides (acid chlorides) or acid anhydrides as appropriate. This method can be used for converting alcohols into esters, but it doesn't work with phenols - compounds where the -OH group is attached directly to a benzene ring. Phenols react with carboxylic acids so slowly that the reaction is unusable for preparation purposes. Esters are produced when carboxylic acids are heated with alcohols in the presence of an acid catalyst. The catalyst is usually concentrated sulphuric acid. Dry hydrogen chloride gas is used in some cases, but these tend to involve aromatic esters (ones where the carboxylic acid contains a benzene ring). If you are a UK A level student you won't have to worry about these. The esterification reaction is both slow and reversible. The equation for the reaction between an acid RCOOH and an alcohol R'OH (where R and R' can be the same or different) is: So, for example, if you were making ethyl ethanoate from ethanoic acid and ethanol, the equation would be: Carboxylic acids and alcohols are often warmed together in the presence of a few drops of concentrated sulfuric acid to observe the smell of the esters formed. You would normally use small quantities of everything heated in a test tube stood in a hot water bath for a couple of minutes. Because the reactions are slow and reversible, you don't get a lot of ester produced in this time. The smell is often masked or distorted by the smell of the carboxylic acid. A simple way of detecting the smell of the ester is to pour the mixture into some water in a small beaker. Apart from the very small ones, esters are fairly insoluble in water and tend to form a thin layer on the surface. Excess acid and alcohol both dissolve and are tucked safely away under the ester layer. Small esters like ethyl ethanoate smell like typical organic solvents (ethyl ethanoate is a common solvent in, for example, glues). As the esters get bigger, the smells tend towards artificial fruit flavoring - "pear drops", for example. If you want to make a reasonably large sample of an ester, the method used depends to some extent on the size of the ester. Small esters are formed faster than bigger ones. To make a small ester like ethyl ethanoate, you can gently heat a mixture of ethanoic acid and ethanol in the presence of concentrated sulphuric acid, and distil off the ester as soon as it is formed. This prevents the reverse reaction happening. It works well because the ester has the lowest boiling point of anything present. The ester is the only thing in the mixture which doesn't form hydrogen bonds, and so it has the weakest intermolecular forces. Larger esters tend to form more slowly. In these cases, it may be necessary to heat the reaction mixture under reflux for some time to produce an equilibrium mixture. The ester can be separated from the carboxylic acid, alcohol, water and sulphuric acid in the mixture by fractional distillation. This method will work for alcohols and phenols. In the case of phenols, the reaction is sometimes improved by first converting the phenol into a more reactive form. If you add an acyl chloride to an alcohol, you get a vigorous (even violent) reaction at room temperature producing an ester and clouds of steamy acidic fumes of hydrogen chloride. For example, if you add the liquid ethanoyl chloride to ethanol, you get a burst of hydrogen chloride produced together with the liquid ester ethyl ethanoate. \[ CH_3OCl + CH_3CH_2OH \longrightarrow CH_3COOCH_2CH_3 + HCl \] The substance normally called "phenol" is the simplest of the family of phenols. Phenol has an -OH group attached to a benzene ring - and nothing else. The reaction between ethanoyl chloride and phenol is similar to the ethanol reaction although not so vigorous. Phenyl ethanoate is formed together with hydrogen chloride gas. Benzoyl chloride has the formula C H COCl. The -COCl group is attached directly to a benzene ring. It is much less reactive than simple acyl chlorides like ethanoyl chloride. The phenol is first converted into the ionic compound sodium phenoxide (sodium phenate) by dissolving it in sodium hydroxide solution. The phenoxide ion reacts more rapidly with benzoyl chloride than the original phenol does, but even so you have to shake it with benzoyl chloride for about 15 minutes. Solid phenyl benzoate is formed. This reaction can again be used to make esters from both alcohols and phenols. The reactions are slower than the corresponding reactions with acyl chlorides, and you usually need to warm the mixture. In the case of a phenol, you can react the phenol with sodium hydroxide solution first, producing the more reactive phenoxide ion. Taking ethanol reacting with ethanoic anhydride as a typical reaction involving an alcohol: There is a slow reaction at room temperature (or faster on warming). There is no visible change in the colorless liquids, but a mixture of ethyl ethanoate and ethanoic acid is formed. \[ (CH_3CO)_2O + CH_3CH_2OH \longrightarrow CH_3COOCH_2CH_3 + CH_3COOH \] The reaction with phenol is similar, but will be slower. Phenyl ethanoate is formed together with ethanoic acid. This reaction is not important itself, but a very similar reaction is involved in the manufacture of aspirin (covered in detail on another page - link below). If the phenol is first converted into sodium phenoxide by adding sodium hydroxide solution, the reaction is faster. Phenyl ethanoate is again formed, but this time the other product is sodium ethanoate rather than ethanoic acid. Jim Clark ( ) | 5,651 | 4,155 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Carbohydrates/Case_Studies/Sugar_and_Teeth |
Sugar, saliva, and bacteria lead to a formidable combination that may lead to tooth decay. After eating sugar, particularly sucrose, and even within minutes of brushing your teeth, sticky glycoproteins (combination of carbohydrate and protein molecule) adhere to the teeth to start the formation of plaque. At the same time millions of bacteria known as also adhere to the glycoprotein. Although, many oral bacteria also adhere, only the is able to cause cavities. In the next stage, the bacteria use the fructose in a metabolism process of glycolysis to get energy. The end product of glycolysis under anaerobic conditions is lactic acid. The lactic acid creates extra acidity to decrease the pH to the extent of dissolving the calcium phosphate in the tooth enamel leading to the start of a cavity. Preventative measures include frequent brushing and flossing to prevent plaque build up. A diet rich in calcium and fluoride in the water lead to stronger tooth enamel. A diet of more complex carbon hydrates that are low in sugar and no sucrose snacks between meals is also a good preventative measure. Only the S. mutans bacteria has an enzyme called glucosyl transferase on its surface that is able to cause the polymerization of glucose on the sucrose with the release of the fructose. The same enzyme continues to add many glucose molecules to each other to form dextran which is very similar in structure to amylose in starch. The dextran along with the bacteria adheres tightly to the tooth enamel and leads to the formation of plaque. This is just the first phase of cavity formation. The graphic below shows only a portion of this process which shows the release of the fructose. The glucose undergoes further polymerization as stated above. In the next stage, the bacteria use the fructose in a metabolism process of glycolysis to get energy. The end product of glycolysis under anaerobic conditions is lactic acid. The lactic acid creates extra acidity to decrease the pH to the extent of dissolving the calcium phosphate in the tooth enamel leading to the start of a cavity. Preventative measures include frequent brushing and flossing to prevent plaque build up. A diet rich in calcium and fluoride in the water lead to stronger tooth enamel. A diet of more complex carbon hydrates that are low in sugar and no sucrose snacks between meals is also a good preventative measure. | 2,412 | 4,156 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Metabolism/Catabolism/Biological_Oxidation |
Before the concept of biological oxidation can be understood and explored, the fundamental chemical process through which oxidation and reduction take place should be first established. All reactions which involve electron flow are considered oxidation-reduction reactions. The basic definition can be defined as: One reactant is oxidized (loses electrons), while another is reduced (gains electrons). A couple of basic oxidation-reduction or "redox" example's are given here. The reaction of magnesium metal with oxygen, involves the oxidation of magnesium \[ 2Mg(s) + O_2(g)→ 2MgO(s) \label{1} \] Since the magnesium solid is oxidized, we expect to see a loss of electrons. Similarly, since oxygen must therefore be reduced, we should see a gain of electrons. As the magnesium is oxidized there is a loss of 2 electrons while simultaneously, oxygen gains those two electrons. Another example of a redox reaction is with the two gasses CO and H . This redox reaction also demonstrates the importance of implementing "oxidation numbers" in the methodology of redox reactions, allowing for the determination of which reactant is being reduced and which reactant is being oxidized. The reaction of carbon dioxide gas with hydrogen gas, involving the oxidation of hydrogen \[CO_2 (g) + H_2 (g) → 2CO (g) + H_2O (g) \label{3}\] Since the hydrogen gas is being oxidized (reductant), we expect to see an overall loss of electrons for the resulting molecule. Similarly, we expect to see a gain in the overall number of electrons for the resulting molecule of the oxidant (CO ). Here it is possible to infer that the carbon of CO is being reduced by review of its unique oxidation number. Such that, C (of CO ) goes from an oxidation number of +4 to C (of CO) having an oxidation number of +2, representing a loss of two electrons. Similarly, H is noted as going from an oxidation number of 0 to +1, or gaining one electron in a reduction process. For more information on oxidation numbers, review the following link: The flow of electrons is a vital process that provides the necessary energy for the survival of all organisms. The primary source of energy that drives the electron flow in nearly all of these organisms is the radiant energy of the sun, in the form of electromagnetic radiation or Light. Through a series of nuclear reactions, the sun is able to generate thermal energy (which we can feel as warmth) from (which we perceive as light). However, the particular wavelength of the electromagnetic spectrum we are able to detect with the human eye is only between 400 and 700 nm in wavelength. It should therefore be noted that the visible part of the electromagnetic spectrum is actually a small percentage of the whole; where a much greater percentage remains undetectable for the human eye. In physics, the use of the term "light" refers to electromagnetic radiation of any wavelength, independent of its detectability for the human eye. For plants, the upper and lower ends of the visible spectrum are the wavelengths that help drive the process of splitting water (H O) during photosynthesis, to release its electrons for the biological reduction of carbon dioxide (CO ) and the release of diatomic oxygen (O ) to the atmosphere. It is through the process of photosynthesis that plants are able to use the energy from light to convert carbon dioxide and water into the chemical energy storage form called glucose. Plants represent one of the most basic examples of biological oxidation and reduction. The chemical conversion of carbon dioxide and water into sugar (glucose) and oxygen is a light-driven reduction process: \[ 6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2 \label{5} \] The process by which non photosynthetic organisms and cells obtain energy, is through the consumption of the energy rich products of photosynthesis. By oxidizing these products, electrons are passed along to make the products carbon dioxide, and water, in an environmental recycling process. The process of oxidizing glucose and atmospheric oxygen allowed energy to be captured for use by the organism that consumes these products of the plant. The following reaction represents this process: \[ C_6H_{12}O_6 + O_2 \rightarrow 6CO_2 + 6H_2O +Energy \label{6}\] It is therefore through this process that heterotrophs (most generally "animals" which consume other organisms obtain energy) and autotrophs (plants which are able to produce their own energy) participate in an environmental cycle of exchanging carbon dioxide and water to produce energy containing glucose for organismal oxidation and energy production, and subsequently allowing the regeneration of the byproducts carbon dioxide and water, to begin the cycle again. Therefore, these two groups of organisms have been allowed to diverge interdependently through this natural life cycle. Biological oxidation-reduction reactions, or simply utilize multiple stages or processes of oxidation to produce large amounts of , which is used to synthesize the energy unit called adenosine triphosphate or ATP. To efficiently produce ATP, the process of glycolysis must be near an abundance of oxygen. Since glycolysis by nature is not an efficient process, if it lacks sufficient amounts of oxygen the end product pyruvate, is reduced to lactate with NADH as the reducing agent. However, in a more favorable aerobic process, the degradation of glucose through glycolysis proceeds with two additional processes known as the and the terminal yielding the end products carbon dioxide and water, which we exhale with each breath. The products NADH and FADH formed during glycolysis and the citric acid cycle are able to reduce molecular oxygen (O ) thereby releasing large amounts of Gibbs energy used to make ATP. The process by which electrons are transferred from NADH or FADH to O by a series of electron transfer carriers, is known as . It is through this process that ATP is able to form as a result of the transfer of electrons. Thee specific examples of redox reactions that are used in biological processes, involving the transfer of electrons and hydrogen ions as follows. During some biological oxidation reactions, there is a simultaneous transfer of hydrogen ions with electrons (1). In other instances, hydrogen ions may be lost by the substance being oxidized while transferring only its electrons to the substance being reduced (2). A third type of biological oxidation might involve only a transfer of electrons (3). It should be noted that biological oxidation rarely proceeds in a direct manner, and generally involves complex mechanisms of several enzymes. The outline below recaps the three processes of biological oxidation stated above, in descending order. In the last stage of the metabolic process (the terminal respiratory chain), the sequence by which electrons are carried is determined by relative redox potentials. The carrier molecules used to transfer electrons in this stage are called cytochromes, which are an electron-carrying protein containing a heme group. The iron atom of each cytochrome molecule can exist either in the oxidized (Fe ) or reduced (Fe ) form. Within the terminal respiratory chain, each carrier molecule alternates between the reduced state and the oxidized state, with molecular oxygen as the final electron acceptor at the end. It is through the knowledge of redox potentials, that the knowledge of biological processes can be further expanded. The standard reduction potential is denoted as and is often based on the hydrogen electrode scale of pH 7, rather than pH 0, a common reference point for listed values. Moreover, the superscript symbol ( ) denotes standard-state conditions, while the adjacent superscript symbol ( ' ) denotes the pH scale of 7 for biochemical processes. It therefore becomes possible to trace the energy transfer in cells back to the fundamental flow of electrons from one particular molecule to another. Where this electron flow occurs via the physics principle of higher potential to lower potential; similar to a ball rolling down a hill, as opposed to the opposite direction. All of these reactions involving electron flow can be attributed to the basic definition of the oxidation-reduction pathway stated above. | 8,302 | 4,157 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Chirality/Stereoisomers/Substituted_Cyclohexanes |
Because it is so common among natural and synthetic compounds, and because its conformational features are rather well understood, we shall focus on the six-membered cyclohexane ring in this discussion. In a sample of cyclohexane, the two identical chair conformers are present in equal concentration, and the hydrogens are all equivalent (50% equatorial & 50% axial) due to rapid interconversion of the conformers. When the cyclohexane ring bears a substituent, the two chair conformers are not the same. In one conformer the substituent is axial, in the other it is equatorial. Due to steric hindrance in the axial location, substituent groups prefer to be equatorial and that chair conformer predominates in the equilibrium. We noted earlier that cycloalkanes having two or more substituents on different ring carbon atoms exist as a pair (sometimes more) of configurational stereoisomers. Now we must examine the way in which favorable ring conformations influence the properties of the configurational isomers. Remember, configurational stereoisomers are stable and do not easily interconvert, whereas, conformational isomers normally interconvert rapidly. In examining possible structures for substituted cyclohexanes, it is useful to follow two principles. The following equations and formulas illustrate how the presence of two or more substituents on a cyclohexane ring perturbs the interconversion of the two chair conformers in ways that can be predicted. In the case of 1,1-disubstituted cyclohexanes, one of the substituents must necessarily be axial and the other equatorial, regardless of which chair conformer is considered. Since the substituents are the same in 1,1-dimethylcyclohexane, the two conformers are identical and present in equal concentration. In 1-t-butyl-1-methylcyclohexane the t-butyl group is much larger than the methyl, and that chair conformer in which the larger group is equatorial will be favored in the equilibrium( > 99%). Consequently, the methyl group in this compound is almost exclusively axial in its orientation. In the cases of 1,2-, 1,3- and 1,4-disubstituted compounds the analysis is a bit more complex. It is always possible to have both groups equatorial, but whether this requires a cis-relationship or a trans-relationship depends on the relative location of the substituents. As we count around the ring from carbon #1 to #6, the uppermost bond on each carbon changes its orientation from equatorial (or axial) to axial (or equatorial) and back. It is important to remember that the . Therefore, it should be clear that for cis-1,2-disubstitution, one of the substituents must be equatorial and the other axial; in the trans-isomer both may be equatorial. Because of the alternating nature of equatorial and axial bonds, the opposite relationship is true for 1,3-disubstitution (cis is all equatorial, trans is equatorial/axial). Finally, 1,4-disubstitution reverts to the 1,2-pattern. . For additional information about six-membered ring conformations . | 3,026 | 4,160 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Alkanes_Background |
This is an introductory page about alkanes, such as methane, ethane, propane, butane and the remainder of the common alkanes. This page addresses their formulae and isomerism, their physical properties, and an introduction to their chemical reactivity. Alkanes are the simplest family of hydrocarbons - compounds containing carbon and hydrogen only. Alkanes only contain carbon-hydrogen bonds and carbon-carbon single bonds. The formula of any of the alkanes follow the general formula \(C_nH_{2n+2}\) The first six alkanes are tabulated below: All of the alkanes containing 4 or more carbon atoms show structural isomerism, meaning that there are two or more different structural formulae that you can draw for each molecular formula. C H could be either of these two different molecules: These are named butane and 2-methylpropane, respectively. Cycloalkanes also only contain carbon-hydrogen bonds and carbon-carbon single bonds, but the carbon atoms are joined in a ring. The smallest cycloalkane is cyclopropane. If you count the carbons and the hydrogens, you will see that they no longer adhere to the general formula C H . By joining the carbon atoms in a ring, two hydrogen atoms are lost. The general formula for a cycloalkane is C H and these are non-planar molecules (with the exception of cyclopentane) and exist as "puckered rings". Cyclohexane has a ring structure that looks like this: This is known as the "chair" form of cyclohexane because of its shape, which vaguely resembles a chair. Cyclohexane also has a boat configuration (not shown). The boiling points shown are for the "straight chain" isomers in which there are more than one (Figure 1). Notice that the first four alkanes are gases at room temperature, and solids do not start to appear until about C H . The temperatures cannot be more precise than those given in this chart because each isomer has a different melting and boiling point. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers! Cycloalkanes have boiling points that are approximately 10 - 20 K higher than the corresponding straight chain alkane. There difference between carbon and hydrogen (2.1 vs. 1.9) is small; therefore, there is only a slight bond polarity, meaning that the only attractions between one molecule and its neighbors will be Van der Waals dispersion forces. These forces will be very small for a molecule like methane but will increase as the size of the molecules increase. Therefore, the boiling points of the alkanes increase with the molecular size. Regarding isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, bulky molecules (with substantial amounts of branching) to lie close together (compact) compared with long, thin molecules. The boiling points of the three isomers of C H are as follows: The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less "wriggly"! Alkanes (both normal and cycloalkanes) are virtually insoluble in water but dissolve in organic solvents. The liquid alkanes are good solvents for many other covalent compounds. When a molecular substance dissolves in water, the following must occur: Breaking either of these attractions requires energy, although the amount of energy required to break the forces in a compound, such as methane, is relatively negligible; this is not true of the in water. To simplify, a substance will dissolve if sufficient energy is released when the new bonds are formed between the substance and the water to make up for the energy required to break the original attractions. The only new attractions between the alkane and the water molecules are the Van der Waals forces. These forces to do not release a sufficient amount of energy to compensate for the energy required to break the hydrogen bonds in water. Therefore, the alkane does not dissolve. In most organic solvents, the primary forces of attraction between the solvent molecules are the Van der Waals forces composed of either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus, there is no barrier to solubility. Alkanes contain strong carbon-carbon single bonds and strong carbon-hydrogen bonds. The carbon-hydrogen bonds are only very slightly polar. Therefore, there is no portion of the molecule that carries any significant amount of positive or negative charge, which is required for other molecules to be attracted to it. For example, many organic reactions start because an ion or a polar molecule is attracted to a portion of an organic molecule, which carries some positive or negative charge. This attraction does not occur with alkanes because alkane molecules do not have this separation of charge. The net effect is that alkanes have a fairly restricted set of reactions, including the following: These reactions are covered in separate pages. Cycloalkanes are very similar to the alkanes in reactivity, except for the very small cycloalkanes, especially cyclopropane. Cyclopropane is much more reactive than what is expected because of the bond angles in the ring. Normally, when carbon forms four single bonds, the bond angles are approximately 109.5°. In cyclopropane, the bond angles are 60°. With the electron pairs this close together, there is a significant amount of repulsion between the bonding pairs joining the carbon atoms, making the bonds easier to break. A is a study of the energetics of different spatial arrangements of atoms relative to rotations about bonds. Conformational analyses are assisted greatly by a representation of molecules in a manner different to skeletal structures. A representation of a molecule in which the atoms and bonds are viewed along the axis about which rotation occurs is called a (Figure 2). In a Newman projection, the molecule is viewed along an axis containing two atoms bonded to each other and the bond between them, about which the molecule can rotate. In a Newman projection, the "substituents" of each atom composing the bond, be they hydrogens or functional groups, can then be viewed both in front of and behind the carbon-carbon bond. Specifically, one can observe the angle between a substituent on the front atom and a substituent on the back atom in the Newman projection, which is called the or . In ethane specifically, we can imagine two possible "extreme" conformations. In one case, the dihedral angle is 0° and the hydrogens on the first carbon line up with or eclipse the hydrogens on the second carbon. When the dihedral angle is 0° and the hydrogens line up perfectly, ethane has adopted the conformation (Figure 3). The other extreme occurs when the hydrogens on the first carbon are as far away as possible from those on the second carbon; this occurs at a dihedral angle of 60° and is called the conformation (Figure 3). The staggered conformation of ethane is a more stable, lower energy conformation than the eclipsed conformation because the eclipsed conformation involves unfavorable interactions between hydrogen atoms. Specifically, the negatively charged electrons in the bonds repel each other most when the bonds line up. Thus, ethane spends most of its time in the more stable staggered conformation. In butane, two of the substituents, one on each carbon atom being viewed, is a methyl group. Methyl groups are much larger than hydrogen atoms. Thus, when eclipsed conformations occur in butane, the interactions are especially unfavorable. There are four possible "extreme" conformations of butane: 1) , when the methyl groups eclipse each other; 2) , when the methyl groups are staggered but next to each other; 3) Eclipsed, when the methyl groups eclipse hydrogen atoms; and 4) , when the methyl groups are staggered and as far away from each other as possible (Figure 4). The fully eclipsed conformation is clearly the highest in energy and least favorable since the largest groups are interacting directly with each other. As the molecule rotates, it adopts the relatively stable gauche conformation. As it continues to rotate, it encounters less favorable eclipsed conformation in which a methyl group eclipses a hydrogen. As rotation continues, the molecule comes to the anti conformation, which is the most stable since the substituents are staggered and the methyl groups are as far away from each other as possible. An is a graph which represents the energy of a molecule as a function of some changing parameter. An energy diagram can be made as a function of dihedral angle for butane (Figure 5). Clearly, the anti and gauche conformations are significantly more stable than the eclipsed and fully eclipsed conformations. Butane spends most of its time in the anti and gauche conformations. There is one final, very important point. At room temperature, approximately 84 kJ/mol of thermal energy is available to molecules. Thus, if the barrier to a rotation is less than 84 kJ/mol, the molecule will rotate. In ethane and butane, the barriers to rotation are significantly less than 84 kJ/mol. Therefore, even though the eclipsed conformations are unfavorable, the molecules are able to adopt them. In reality, since these conformations are not as stable, the molecules will quickly pass through them at room temperature and return a staggered conformation. Molecules are constantly converting between different staggered conformations all the time, quickly passing through eclipsed conformations in between. Thus, in alkanes, no single "true" conformation exists all the time; the molecule instead constantly converts between conformations, spending more time in those that are more stable. This constant conversion lies in stark contract to alkenes, which adopt the or ( or ) conformations and at room temperature; they do not interconvert because the barrier to rotation is too high. Jim Clark ( ) | 10,313 | 4,161 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.09%3A_Disaccharides |
The sugar molecules listed in Figure \(\Page {1}\) are usually referred to as . This distinguishes them from the which are made up by condensing two sugar units. A familiar example of a disaccharide is ordinary cane sugar, sucrose, which may be obtained by condensing a molecule of α-glucose with one of the cyclic forms of fructose called β-fructose. The structure of sucrose is shown in Figure \(\Page {2}\). Other, less familiar, examples of disaccharides are lactose, which occurs in milk, and maltose, which are shown in Figure \(\Page {3}\). In order to digest a disaccharide like sucrose or lactose, the human body must have an enzyme which can catalyze hydrolysis of the linkage between the two monosaccharide units. Many Asians, Africans, and American Indians are incapable of synthesizing lactase, the enzyme that speeds hydrolysis of lactose. If such persons drink milk, the undigested lactose makes them sick. | 942 | 4,162 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.07%3A_Carbohydrates |
Carbohydrates are sugars and sugar derivatives whose formulas can be written in the general form: C (H O) . (The subscripts and are whole numbers.) Some typical carbohydrates are sucrose (ordinary cane sugar), C H O ; glucose (dextrose), C H O ; fructose (fruit sugar), C H O ; and ribose, C H O .Since the atom ratio H/O is 2/1 in each formula, these compounds were originally thought to be hydrates of carbon, hence their general name. Glucose, by far the most-common simple sugar, is a primary source of energy for both animals and plants. Because they contain free glucose molecules, dextrose tablets or foods such as grapes and honey can provide a noticeable “lift” for persons tired by physical exertion. The individual glucose molecules pass rapidly into the bloodstream when such foods are eaten. Glucose is also the monomer from which the polymers cellulose and starch are built up. The structural material of plants, from the woody parts of trees to the cell walls of most algae, is cellulose. Plants store energy in starch, providing a source of glucose for all but the simplest organisms. The energy in starchy foods is not as rapidly available, however, since the polymeric structure must be broken down before glucose is released. As a consequence of the ubiquity of starch and especially cellulose, carbohydrates are by far the most plentiful organic compounds in the biosphere. | 1,409 | 4,163 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.22%3A_Distillation |
It is immediately apparent from the cartoon in the section on ( seen below ) that the composition of the vapor in equilibrium with a solution and the composition of the solution itself must be different. While only 50 percent of the molecules in the liquid phase are benzene (red) molecules, in the vapor phase they represent 80 percent of the molecules. This difference in composition between the two phases is the reason we can separate components of a liquid mixture by distillation. In the laboratory, distillation is usually carried out in an apparatus similar to that shown in figure 1. The liquid mixture to be distilled is heated in a round-bottom flask until it boils. The emerging vapor travels into the condenser where it is cooled sufficiently to return to the liquid state. The condensed liquid can then be collected in a suitable container. In practice this container is usually changed several times during the course of the distillation. Each sample collected in this way is called a , and the whole process is called . A thermometer may be used to determine when vapor reaches the condenser, and to determine the progression of the distillation. This process can also be viewed in the following video. Here distillation follows the same process already described for the figure. Suppose we use the distillation apparatus shown in the figure to distill an equimolar mixture of benzene and toluene. We would first have to raise the temperature of the mixture to 92.4°C, at which point the total vapor pressure would be 1 atm (101.3 kPa) and the liquid would boil. At 92.4°C the vapor in equilibrium with an equimolar mixture has a mole fraction of benzene equal to 0.71 (slightly different from that in the cartoon on the page discussing ). Accordingly, when our mixture is distilled, vapor of this composition will be condensed and the first drops in the collection vessel will contain 71 percent benzene molecules and 29 percent toluene molecules. We will now have a sample which is much richer in benzene than the original. In theory we could go on repeating this process indefinitely. If we redistilled a liquid mixture having = 0.71, it would boil at 87°C and yield a condensed vapor whose mole fraction of benzene would be 0.87. A third distillation would yield benzene with = 0.95, and a fourth would yield a product with = 0.99. If we went on long enough, we could obtain a sample of benzene of any desired purity. In practice repeated distillation takes up a lot of time, and we would lose part of the sample each time. Fortunately there is a way out of this difficulty. We insert a glass column filled with round beads, or with specially designed packing, between the distillation flask and the condenser. When a vapor reaches the bottom-most glass beads en route to the condenser, part of it will condense. This condensate will be richer in the less volatile component, toluene, and so the vapor which passes on will be richer in benzene than before. As it passes through the column, the vapor will undergo this process several times. In effect the fractionating column allows us to perform several distillations in one operation. A well-designed laboratory column makes it possible to effect several hundred distillations in one pass. By using such a column, it would be possible to distill virtually pure benzene from the equimolar mixture, leaving virtually pure toluene behind. Fractional distillation is used on a very large scale in the refining of petroleum. Some of the large towers seen in petroleum refineries are gigantic fractionating columns. These columns enable the oil company to separate the petroleum into five or six fractions called . The more-volatile cuts can be used directly for gasoline, kerosene, etc. The less-volatile cuts contain the very long-chain alkanes and are usually converted into shorter-chain alkanes by cracking, as described in the section on . While fractional distillation can be used to separate benzene and toluene and most mixtures of alkanes which show only small deviations from ideal behavior, it cannot be used to separate mixtures which exhibit a maximum or a minimum in the vapor-pressure versus mole-fraction curve. Thus any attempt to separate benzene and methanol (seen in the graph on the page discussing ) by fractional distillation does not result in methanol distilling over initially but rather results in a mixture with = 0.614. A mixture of this composition is not only more volatile than methanol or benzene but also has a higher vapor pressure than any other mixture, as can be seen from that graph. This is why it distills first. Such a mixture is called a . Conversely, for a mixture such as chloroform and acetone which exhibits large enough negative deviations from Raoult’s law, a of low volatility distills at the end of a fractional distillation, preventing complete separation of the pure components. | 4,927 | 4,165 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.04%3A_Aluminum_Production |
Aluminum is easily , and so its ore, Al O , is difficult to reduce. In fact water is reduced rather than Al ( ), and so must be carried out in a molten salt. Even this is difficult because the melting point of Al O is above 2000°C—a temperature which is very difficult to maintain. The first successful method for reducing Al O is the one still used today. It was developed in the United States in 1886 by Charles Hall (1863 to 1914), who was then 23 years old and fresh out of Oberlin College. Hall realized that if Al O were dissolved in another molten salt, the melting point of the mixture would be lower than for either pure substance. The substance Hall used was cryolite, Na AlF , in which the Al O can be dissolved at just over 1000°C. The electrolytic cell used for the Hall process. (Figure \(\Page {1}\)) consists of a steel box lined with graphite. This contains the molten Na AlF and Al O and also serves as the cathode. The anode is a large cylinder of carbon. Passage of electrical current maintains the high temperature of the cell and causes the following half-equations to occur: \[\text{Al}^{3+} + \text{3}e^{-} \rightarrow \text{Al}(l) \label{1} \] \[\text{2O}^{2-} + \text{C}(s) \rightarrow \text{CO}_2(g) + \text{4}e^{-}\label{2} \] Since the carbon anode is consumed by the oxidation half-equation, it must be replaced periodically. Aluminum production requires vast quantities of electrical energy, both to maintain the high temperature and to cause half-equations \(\ref{1}\) and \(\ref{2}\) to occur. Currently about 5 percent of the total electrical energy produced in the United States goes into the Hall process. Much of this energy comes from combustion of and hence consumes a valuable, nonrenewable resource. Since Al is protected from oxidation back to Al O by a surface coating of oxide, it is a prime candidate for recycling, as well as for applications such as house siding, where it is expected to remain for a long time. Throwing away aluminum beverage cans, on the other hand, is a tremendous waste of energy. Several other easily oxidized metals are currently produced by electrolysis, but not in such large quantities as Al. Mg is obtained by electrolyzing molten MgCl which is derived from seawater, and Na and Ca are produced together from a molten mixture of NaCl and CaCl . | 2,346 | 4,166 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Concept_Development_Studies_in_Chemistry_(Hutchinson)/15%3A_Acid-Base_Equilibrium |
We have developed an understanding of involving phase transitions and involving reactions entirely in the gas phase. We will assume an understanding of the principles of dynamic equilibrium, reaction equilibrium constants, and . To understand application of these principles to reactions in solution, we will now assume a definition of certain classes of substances as being either acids or bases. An acid is a substance whose molecules donate positive hydrogen ions (protons) to other molecules or ions. When dissolved in pure water, acid molecules will transfer a hydrogen ion to a water molecule or to a cluster of several water molecules. This increases the concentration of \(\ce{H^+}\) ions in the solution. A base is a substance whose molecules accept hydrogen ions from other molecules. When dissolved in pure water, base molecules will accept a hydrogen ion from a water molecule, leaving behind an increased concentration of \(\ce{OH^-}\) ions in the solution. To understand what determines acid-base behavior, we will assume an understanding of the bonding, structure, and properties of individual molecules. Acids and bases are very common substances whose properties vary greatly. Many acids are known to be quite corrosive, with the ability to dissolve solid metals or burn flesh. Many other acids, however, are not only benign but vital to the processes of life. Far from destroying biological molecules, they carry out reactions critical for organisms. Similarly, many bases are caustic cleansers while many others are medications to calm indigestion pains. In this concept study, we will develop an understanding of the characteristics of molecules which make them either acids or bases. We will examine measurements about the relative strengths of acids and bases, and we will use these to develop a quantitative understanding of the relative strengths of acids and bases. From this, we can develop a qualitative understanding of the properties of molecules which determine whether a molecule is a strong acid or a weak acid, a strong base or a weak base. This understanding is valuable in predicting the outcomes of reactions, based on the relative quantitative strengths of acids and bases. These reactions are commonly referred to as neutralization reactions. A surprisingly large number of reactions, particularly in organic chemistry, can be understood as transfer of hydrogen ions from acid molecules to base molecules. From the definition of an acid given in the Foundation, a typical acid can be written as \(\ce{HA}\), representing the hydrogen ion which will be donated and the rest of the molecule which will remain as a negative ion after the donation. The typical reaction of an acid in aqueous solution reacting with water can be written as \[\ce{HA} \left( aq \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_3O^+} \left( aq \right) + \ce{A^-} \left( aq \right)\] In this reaction, \(\ce{HA} \left( aq \right)\) represents an acid molecule dissolved in aqueous solution. \(\ce{H_3O^+} \left( aq \right)\) is a notation to indicate that the donate proton has been dissolved in solution. Observations indicate that the proton is associated with several water molecules in a cluster, rather than attached to a single molecule. \(\ce{H_3O^+}\) is a simplified notation to represent this result. Similarly, the \(\ce{A^-} \left( aq \right)\) ion is solvated by several water molecules. This equation is referred to as . The equation above implies that a \(0.1 \: \text{M}\) solution of the acid \(\ce{HA}\) in water should produce \(\ce{H_3O^+}\) ions in solution with a concentration of \(0.1 \: \text{M}\). In fact, the concentration of \(\ce{H_3O^+}\) ions, \(\left[ \ce{H_3O^+} \right]\), can be measured by a variety of techniques. Chemists commonly use a measure of the \(\ce{H_3O^+}\) ion concentration called the , defined by: \[\text{pH} = -\text{log} \: \left[ \ce{H_3O^+} \right]\] We now observe the concentration \(\left[ \ce{H_3O^+} \right]\) produced by dissolving a variety of acids in solution at a concentration of \(0.1 \: \text{M}\), and the results are tabulated in Table 15.1. Note that there are several acids listed for which \(\left[ \ce{H_3O^+} \right] = 0.1 \: \text{M}\), and pH = 1. This shows that, for these acids, the acid ionization is complete: essentially every acid molecule is ionized in the solution according to the equation above. However, there are other acids listed for which \(\left[ \ce{H_3O^+} \right]\) is considerably less than \(0.1 \: \text{M}\) and the pH is considerably greater than 1. For each of these acids, therefore, not all of the acid molecules ionize according to the equation above. In fact, it is clear in Table 15.1 that in these acids the vast majority of the acid molecules do not ionize, and only a small percentage does ionize. From these observations, we distinguish two classes of acids: and . Strong acids are those for which nearly \(100\%\) of the acid molecules ionize, whereas weak acids are those for which only a small percentage of molecules ionize. There are seven strong acids listed in Table 15.1. From many observations, it is possible to determine that these seven acids are the only commonly observed strong acids. The vast majority of all substances with acidic properties are weak acids. We seek to characterize weak acid ionization quantitatively and to determine what the differences in molecular properties are between strong acids and weak acids. Table 15.1 shows that the pH of \(0.1 \: \text{M}\) acid solutions varies from one weak acid to another. If we dissolve 0.1 moles of acid in a \(1.0 \: \text{L}\) solution, the fraction of those acid molecules which will ionize varies from weak acid to weak acid. For a few weak acids, using the data in Table 15.1 we calculate the percentage of ionized acid molecules in \(0.1 \: \text{M}\) acid solutions in Table 15.2. We might be tempted to conclude from Table 15.2 that we can characterize the strength of each acid by the percent ionization of acid molecules in solution. However, before doing so, we observe the pH of a single acid, nitrous acid, in solution as a function of the concentration of the acid. \[\ce{HNO_2} \left( aq \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_3O^+} \left( aq \right) + \ce{NO_2^-} \left( aq \right)\] In this case, "concentration of the acid" refers to the number of moles of acid that we dissolved per liter of water. Our observations are listed in Table 15.3, which gives \(\left[ \ce{H_3O^+} \right]\), pH, and percent ionization as a function of nitrous acid concentration. Surprisingly, perhaps, the percent ionization varies considerably as a function of the concentration of the nitrous acid. We recall that this means that the fraction of molecules which ionize, according to the acid ionization equation, depends on how many acid molecules there are per liter of solution. Since some but not all of the acid molecules are ionized, this means that nitrous acid molecules are present in solution at the same time as the negative nitrite ions and the positive hydrogen ions. Recalling our observation of equilibrium in gas phase reactions, we can conclude that the acid dissociation equation achieves equilibrium for each concentration of the nitrous acid. Since we know that gas phase reactions come to equilibrium under conditions determined by the equilibrium constant, we might speculate that the same is true of reactions in aqueous solution, including acid ionization. We therefore define an analogy to the gas phase reaction equilibrium constant. In this case, we would not be interested in the pressures of the components, since the reactants and products are all in solution. Instead, we try a function composed of the equilibrium concentrations: \[K = \frac{\left[ \ce{H_3O^+} \right] \left[ \ce{NO_2^-} \right]}{\left[ \ce{HNO_2} \right] \left[ \ce{H_2O} \right]}\] The concentrations at equilibrium can be calculated from the data in Table 15.3 for nitrous acid. \(\left[ \ce{H_3O^+} \right]\) is listed and \(\left[ \ce{NO_2^-} \right] = \left[ \ce{H_3O^+} \right]\). Furthermore, if \(c_0\) is the initial concentration of the acid defined by the number of moles of acid dissolved in solution per liter of solution, then \(\left[ \ce{HA} \right] = c_0 - \left[ \ce{H_3O^+} \right]\). Note that the contribution of \(\left[ \ce{H_2O} \left( l \right) \right]\) to the value of the function \(K\) is simply a constant. This is because the "concentration" of water in the solution is simply the molar density of water, \(\frac{n_{H_2O}}{V} = 55.5 \: \text{M}\), which is not affected by the presence or absence of solute. All of the relevant concentrations, along with the function in the equilibrium constant equation are calculated and tabulated in Table 15.4. We note that the function \(K\) in the equation above is approximately, though only approximately, the same for all conditions analyzed in Table 15.4. Variation of the concentration by a factor of 1000 produces a change in \(K\) of only \(10\%\) to \(15\%\). Hence, we can regard the function \(K\) as a constant which approximately describes the acid ionization equilibrium for nitrous acid. By convention, chemists omit the constant concentration of water from the equilibrium expression, resulting in the , , defined as \[K_a = \frac{\left[ \ce{H_3O^+} \right] \left[ \ce{NO_2^-} \right]}{\left[ \ce{HNO_2} \right]}\] From an average of the data in Table 15.4, we can calculate that, at \(25^\text{o} \text{C}\) for nitrous acid, \(K_a = 5 \times 10^{-4}\). Acid ionization constants for the other weak acids in Table 15.3 are listed in Table 15.5. We make two final notes about the results in Table 15.5. First, it is clear the larger the value of \(K_a\), the stronger the acid. That is, when \(K_a\) is a large number, the percent ionization of the acid is larger, and vice versa. Second, the values of \(K_a\) vary over many orders of magnitude. As such, it is often convenient to define the quantity p\(K_a\), analogous to pH, for purposes of comparing acid strengths: \[\text{p} K_a = -\text{log} \: K_a\] The value of p\(K_a\) for each acid is also listed in Table 15.5. Note that a small value of p\(K_a\) implies a large value of \(K_a\) and thus a stronger acid. Weaker acids have larger values of p\(K_a\). \(K_a\) and p\(K_a\) thus give a simple quantitative comparison of the strength of weak acids. Since we have the ability to measure pH for acid solutions, we can measure pH for pure water as well. It might seem that this would make no sense, as we would expect \(\left[ \ce{H_3O^+} \right]\) to equal zero exactly in pure water. Surprisingly, this is incorrect: a measurement on pure water at \(25^\text{o} \text{C}\) yields pH = 7, so that \(\left[ \ce{H_3O^+} \right] = 1.0 \times 10^{-7} \: \text{M}\). There can be only one possible source for these ions: water molecules. The process \[\ce{H_2O} \left( l \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{H_3O^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\] is referred to as the of water. Note that, in this reaction, some water molecules behave as acid, donating protons, while other water molecules behave as base, accepting protons. Since at equilibrium \(\left[ \ce{H_3O^+} \right] = 1.0 \times 10^{-7} \: \text{M}\), it must also be true that \(\left[ \ce{OH^-} \right] = 1.0 \times 10^{-7} \: \text{M}\). We can write the equilibrium constant for the above equation, following our previous convention of omitting the pure water from the expression, and we find that, at \(25^\text{o} \text{C}\), \[\begin{align} K_w &= \left[ \ce{H_3O^+} \right] \left[ \ce{OH^-} \right] \\ &= 1.0 \times 10^{-14} \: \text{M} \end{align}\] (In this case, the subscript "w" refers to "water".) Autoionization of water occurs in pure water but must also occur when ions are dissolved in aqueous solutions. This includes the presence of acids ionized in solution. For example, we consider a solution of \(0.1 \: \text{M}\) acetic acid. Measurements show that, in this solution \(\left[ \ce{H_3O^+} \right] = 1.3 \times 10^{-3} \: \text{M}\) and \(\left[ \ce{OH^-} \right] = 7.7 \times 10^{-12} \: \text{M}\). We note two things from this observation: first, the value of \(\left[ \ce{OH^-} \right]\) is considerably less than in pure water; second, the autoionization equilibrium constant remains the same at \(1.0 \times 10^{-14}\). From these notes, we can conclude that the autoionization equilibrium of water occurs in acid solution, but the extent of autoionization is suppressed by the presence of the acid in solution. We consider a final note on the autoionization of water. The pH of pure water is 7 at \(25^\text{o} \text{C}\). Adding any acid to pure water, no matter how weak the acid, must increase \(\left[ \ce{H_3O^+} \right]\), thus producing a pH below 7. As such, we can conclude that, for all acidic solutions, pH is less than 7, or on the other hand, any solution with pH less than 7 is acidic. We have not yet examined the behavior of base molecules in solution, nor have we compared the relative strengths of bases. We have defined a base molecule as one which accepts a positive hydrogen ion from another molecule. One of the most common examples is ammonia, \(\ce{NH_3}\). When ammonia is dissolved in aqueous solution, the following reaction occurs: \[\ce{NH_3} \left( aq \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{NH_4^+} \left( aq \right) + \ce{OH^-} \left( aq \right)\] Due to the lone pair of electrons on the highly electronegative \(\ce{N}\) atom, \(\ce{NH_3}\) molecules will readily attach a free hydrogen ion forming the ammonium ion \(\ce{NH_4^+}\). When we measure the concentration of \(\ce{OH^-}\) for various initial concentrations of \(\ce{NH_3}\) in water, we observe the results in Table 15.6. We should anticipate that a base ionization equilibrium constant might exist comparable to the acid ionization equilibrium constant, and in Table 15.6, we have also calculated the value of the function \(K_b\) defined as: \[K_b = \frac{\left[ \ce{NH_4^+} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{NH_3} \right]}\] Given that we have dissolved a base in pure water, we might be surprised to discover the presence of positive hydrogen ions, \(\ce{H_3O^+}\), in solution, but a measurement of the pH for each of the solutions reveals small amounts. The pH for each solution is also listed in Table 15.6. The source of these \(\ce{H_3O^+}\) ions must be the autoionization of water. Note, however, that in each case in basic solution, the concentration of \(\ce{H_3O^+}\) ions is less than that in pure water. Hence, the presence of the base in solution has suppressed the autoionization. Because of this, in each case the pH of a basic solution is greater than 7. Base ionization is therefore quite analogous to acid ionization observed earlier. We now consider a comparison of the strength of an acid to the strength of the base. To do so, we consider a class of reactions called "neutralization reactions" which occur when we mix an acid solution with a base solution. Since the acid donates protons and the base accepts protons, we might expect, when mixing acid and base, to achieve a solution which is no longer acidic or basic. For example, if we mix together equal volumes of \(0.1 \: \text{M}\) \(\ce{HCl} \left( aq \right)\) and \(0.1 \: \text{M}\) \(\ce{NaOH} \left( aq \right)\), the following reaction occurs: \[\ce{HCl} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{Cl^-} \left( aq \right) + \ce{H_2O} \left( l \right)\] The resultant solution is simply a salt solution with \(\ce{NaCl}\) dissolved in water. This solution has neither acidic nor basic properties, and the pH is 7; hence the acid and base have neutralized each other. In this case, we have mixed together a strong acid with a strong base. Since both are strong and since we mixed equal molar quantities of each, the neutralization reaction is essentially complete. We next consider mixing together a weak acid solution with a strong base solution, again with equal molar quantities of acid and base. As an example, we mix \(100 \: \text{mL}\) of \(0.1 \: \text{M}\) acetic acid \(\left( \ce{HA} \right)\) solution with \(100 \: \text{mL}\) of \(0.1 \: \text{M}\) sodium hydroxide. In this discussion, we will abbreviate the acetic acid molecular formula \(\ce{CH_3COOH}\) as \(\ce{HA}\) and the acetate ion \(\ce{CH_3COO^-}\) as \(\ce{A^-}\). The reaction of \(\ce{HA}\) and \(\ce{NaOH}\) is: \[\ce{HA} \left( aq \right) + \ce{NaOH} \left( aq \right) \rightarrow \ce{Na^+} \left( aq \right) + \ce{A^-} \left( aq \right) + \ce{H_2O} \left( l \right)\] \(\ce{A^-} \left( aq \right)\) is the acetate ion in solution, formed when an acetic acid molecule donates the positive hydrogen ion. We have thus created a salt solution again, in this case of sodium acetate in water. Note that the volume of the combined solution is \(200 \: \text{mL}\), so the concentration of sodium acetate \(\left( \ce{NaA} \right)\) in solution is \(0.050 \: \text{M}\). Unlike our previous \(\ce{NaCl}\) salt solution, a measurement in this case reveals that the pH of the product salt solution is 9.4, so the solution is basic. Thus, mixing equal molar quantities of strong base with weak acid produces a basic solution. In essence, the weak acid does not fully neutralize the strong base. To understand this, we examine the behavior of sodium acetate in solution. Since the pH is greater than 7, then there is an excess of \(\ce{OH^-}\) ions in solution relative to pure water. These ions must have come from the reaction of sodium acetate with the water. Therefore, the negative acetate ions in solution must behave as a base, accepting positive hydrogen ions: \[\ce{A^-} \left( aq \right) + \ce{H_2O} \left( l \right) \rightarrow \ce{HA} \left( aq \right) + \ce{OH^-} \left( aq \right)\] The reaction of an ion with water to form either an acid or a base solution is referred to as . From this example, the salt of a weak acid behaves as a base in water, resulting in a pH greater than 7. To understand the extent to which the hydrolysis of the negative ion occurs, we need to know the equilibrium constant for this reaction. This turns out to be determined by the acid ionization constant for \(\ce{HA}\). To see this, we write the equilibrium constant for the hydrolysis of \(\ce{A^-}\) as \[K_h = \frac{\left[ \ce{HA} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{A^-} \right]}\] Multiplying numerator and denominator by \(\left[ \ce{H_3O^+} \right]\), we find that \[\begin{align} K_h &= \frac{\left[ \ce{HA} \right] \left[ \ce{OH^-} \right]}{\left[ \ce{A^-} \right]} \frac{\left[ \ce{H_3O^+} \right]}{\left[ \ce{H_3O^+} \right]} \\ &= \frac{K_w}{K_a} \end{align}\] Therefore, for the hydrolysis of acetate ions in solution, \(K_h = 5.8 \times 10^{-10}\). This is fairly small, so the acetate ion is a very weak base. We now have a fairly complete quantitative description of acid-base equilibrium. To complete our understanding of acid-base equilibrium, we need a predictive model which relates acid strength or base strength to molecular properties. In general, we expect that the strength of an acid is related either to the relative ease by which it can donate a hydrogen ion or by the relative stability of the remaining negative ion formed after the departure of the hydrogen ion. To begin, we note that there are three basic categories of acids which we have examined in this study. First, there are simple : \(\ce{HF}\); \(\ce{HCl}\); \(\ce{HBr}\); \(\ce{HI}\). Second, there are acids formed from main group elements combined with one or more oxygen atoms, such as \(\ce{H_2SO_4}\) or \(\ce{HNO_3}\). These are called . Third, there are the , organic molecules which contain the carboxylic functional group in Figure 15.1. We consider first the simple binary acids. \(\ce{HCl}\), \(\ce{HBr}\), and \(\ce{HI}\) are all strong acids, whereas \(\ce{HF}\) is a weak acid. In comparing the experimental values of p\(K_a\) values in Table 15.7, we note that the acid strength increases in the order \(\ce{HF} < \ce{HCl} < \ce{HBr} < \ce{HI}\). This means that the hydrogen ion can more readily separate from the covalent bond with the halogen atom \(\left( \ce{X} \right)\) as we move down the periodic table, as shown in Table 15.7. The decreasing strength of the \(\ce{H-X}\) bond is primarily due to the increase in the size of the \(\ce{X}\) atom as we move down the periodic table. We conclude that one factor which influences acidity is the strength of the \(\ce{H-X}\) bond: a weaker bond produces a stronger acid, and vice versa. In the acids in the other two categories, the hydrogen atom which ionizes is attached directly to an oxygen atom. Thus, to understand acidity in these molecules, we must examine what the oxygen atom is in turn bonded to. It is very interesting to note that, in examining compounds like \(\ce{R-O-H}\), where \(\ce{R}\) is an atom or group of atoms, we can get either acidic or basic properties. For example, \(\ce{NaOH}\) is a strong base, whereas \(\ce{HOCl}\) is a weak acid. This means that, when \(\ce{NaOH}\) ionizes in solution, the \(\ce{Na-O}\) linkage ionizes, whereas when \(\ce{HOCl}\) ionizes in solution, the \(\ce{H-O}\) bond ionizes. To understand this behavior, we compare the strength of the simple oxyacids \(\ce{HOI}\), \(\ce{HOBr}\), and \(\ce{HOCl}\). The p\(K_a\)'s for these acids are found experimentally to be, respectively, 10.6, 8.6, and 7.5. The acid strength for \(\ce{HOX}\) increases as we move up the periodic table in the halogen group. This means that the \(\ce{H-O}\) bond ionizes more readily when the oxygen atom is bonded to a more electronegative atom. We can add to this observation by comparing the strengths of the acids \(\ce{HOCl}\), \(\ce{HOClO}\), \(\ce{HOClO_2}\), and \(\ce{HOClO_3}\). (Note that the molecular formulae are more commonly written as \(\ce{HClO}\), \(\ce{HClO_2}\), \(\ce{HClO_3}\), and \(\ce{HClO_4}\). We have written them instead to emphasize the molecular structure.) The p\(K_a\)'s of these acids are, respectively, 7.5, 2.0, -2.7, and -8.0. In each case, the molecule with more oxygen atoms on the central \(\ce{Cl}\) atom is the stronger acid: \(\ce{HOClO}\) is more acidic than \(\ce{HOCl}\), etc. A similar result is found in comparing the oxyacids of nitrogen. \(\ce{HONO_2}\), nitric acid, is one of the strong acids, whereas \(\ce{HONO}\), nitrous acid, is a weak acid. Since oxygen atoms are very strongly electronegative, these trends add to our observation that increasing electronegativity of the attached atoms increases the ionization of the \(\ce{O-H}\) bond. Why would electronegativity play a role in acid strength? There are two conclusions we might draw. First, a greater electronegativity of the atom or atoms attached to the \(\ce{H-O}\) in the oxyacid apparently results in a weaker \(\ce{H-O}\) bond, which is thus more readily ionized. We know that an electronegative atom polarizes bonds by drawing the electrons in the molecule towards it. In this case, the \(\ce{Cl}\) in \(\ce{HOCl}\) and the \(\ce{Br}\) in \(\ce{HOBr}\) must polarize the \(\ce{H-O}\) bond, weakening it and facilitating the ionization of the hydrogen. In comparing \(\ce{HOCl}\) to \(\ce{HOClO}\), the added oxygen atom must increase the polarization of the \(\ce{H-O}\) bond, thus weakening the bond further and increasing the extent of ionization. A second conclusion has to do with the ion created by the acid ionization. The negative ion produced has a surplus electron, and the relative energy of this ion will depend on how readily that extra electron is attracted to the atoms of the ion. The more electronegative those atoms are, the stronger is the attraction. Therefore, the \(\ce{OCl^-}\) ion can more readily accommodate the negative charge than can the \(\ce{OBr^-}\) ion. And the \(\ce{OClO^-}\)ion can more readily accommodate the negative charge than can the \(\ce{OCl^-}\) ion. We conclude that the presence of strongly electronegative atoms in an acid increases the polarization of the \(\ce{H-O}\) bond, thus facilitating ionization of the acid, and increases the attraction of the extra electron to the negative ion, thus stabilizing the negative ion. Both of these factors increase the acid strength. Chemists commonly use both of these conclusions in understanding and predicting relative acid strength. The relative acidity of carbon compounds is a major subject of organic chemistry, which we can only visit briefly here. In each of the carboxylic acids, the \(\ce{H-O}\) group is attached to a carbonyl \(\ce{C=O}\) group, which is in turn bonded to other atoms, the comparison we observe here is between carboxylic acid molecules, denoted as \(\ce{RCOOH}\), and other organic molecules containing the \(\ce{H-O}\) group, such as alcohols denoted as \(\ce{ROH}\). (\(\ce{R}\) is simply an atom or group of atoms attached to the functional group.) The former are obviously acids whereas the latter group contains molecules which are generally extremely weak acids. One interesting comparison is for the acid and alcohol when \(\ce{R}\) is the benzene ring, \(\ce{C_6H_5}\). Benzoic acid, \(\ce{C_6H_5COOH}\), has p\(K_a\) = 4.2, whereas phenol, \(\ce{C_6H_5OH}\), has p\(K_a\) = 9.9. Thus, the presence of the doubly bonded oxygen atom on the carbon atom adjacent to the \(\ce{O-H}\) clearly increases the acidity of the molecule, and thus increases ionization of the \(\ce{O-H}\) bond. This observation is quite reasonable in the context of our previous conclusion. Adding an electronegative oxygen atom in near proximity to the \(\ce{O-H}\) bond both increases the polarization of the \(\ce{O-H}\) bond and stabilizes the negative ion produced by the acid ionization. In addition to the electronegativity effect, carboxylate anions, \(\ce{RCOO^-}\), exhibit resonance stabilization, as seen in Figure 15.2. The resonance results in a sharing of the negative charge over several atoms, thus stabilizing the negative ion. This is a major contributing factor in the acidity of carboxylic acids versus alcohols. Strong acids have a higher percent ionization than do weak acids. Why don't we use percent ionization as a measure of acid strength, rather than \(K_a\)? Using the data in Table 15.4 for nitrous acid, plot \(\left[ \ce{H_3O^+} \right]\) versus \(c_0\), the initial concentration of the acid, and versus \(\left[ \ce{HNO_2} \right]\), the equilibrium concentration of the acid. On a second graph, plot \(\left[ \ce{H_3O^+} \right]^2\) versus \(c_0\), the initial concentration of the acid, and versus \(\left[ \ce{HNO_2} \right]\), the equilibrium concentration of the acid. Which of these results gives a straight line? Using the equilibrium constant expression, explain your answer. Using Le Chatelier's principle, explain why the concentration of \(\left[ \ce{OH^-} \right]\) is much lower in acidic solution than it is in neutral solution. We considered mixing a strong base with a weak acid, but we did not consider mixing a strong acid with a weak acid. Consider mixing \(0.1 \: \text{M} \: \ce{HNO_3}\) and \(0.1 \: \text{M} \ce{HNO_2}\). Predict the pH of the solution and the percent ionization of the nitrous acid. Rationalize your prediction using Le Chatelier's principle. Imagine taking a \(0.5 \: \text{M}\) solution of nitrous acid and slowly adding water to it. Looking at Table 15.3, we see that, as the concentration of nitrous acid decreases, the percent ionization increases. By contrast, \(\left[ \ce{H_3O^+} \right]\) decreases. Rationalize these results using Le Chatelier's principle. We observed that mixing a strong acid and a strong base, in equal amounts and concentrations, produces a neutral solution, and that mixing a strong base with a weak acid, in equal amounts and concentrations, produces a basic solution. Imagine mixing a weak acid and a weak base, in equal amounts and concentrations. Predict whether the resulting solution will be acidic, basic, or neutral, and explain your prediction. Using the electronegativity arguments presented above, explain why, in general, compounds like \(\ce{M-O-H}\) are bases rather than acids, when \(\ce{M}\) is a metal atom. Predict the relationship between the properties of the metal atom \(\ce{M}\) and the strength of the base \(\ce{MOH}\). Ionization of sulfuric acid \(\ce{H_2SO_4}\) produces \(\ce{HSO_4^-}\), which is also an acid. However, \(\ce{HSO_4^-}\) is a much weaker acid than \(\ce{H_2SO_4}\). Using the conclusions from above, explain why \(\ce{HSO_4^-}\) is a much weaker acid. Predict and explain the relative acid strengths of \(\ce{H_2S}\) and \(\ce{HCl}\). Predict and explain the relative acid strengths of \(\ce{H_3PO_4}\) and \(\ce{H_3AsO_4}\). Using arguments from above, predict and explain the relative acidity of phenol and methanol. (a) (b) . ; Chemistry) | 28,961 | 4,167 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.07%3A_Stress-Strain_Relationships |
If you think about materials made from polymers, a couple of things might readily come to mind. You might think about the ubiquitous use of plastics in structures around us, such as automobile bumpers. You might think about a rubber band that you use to fasten something more firmly, either at home or in the lab. In either case, the function of the material relies on its response to stimuli. Do we want it to be flexible, to change shape, but then snap back to where it came from? Exactly how much flexibility do we need? Where is the trade-off between flexibility and strength? These questions are important in polymer chemistry. Consequently, we often need to probe how materials behave under different conditions so that we know how they can be employed most effectively. Tensile testing is one of the simplest ways to probe how a material responds to stress. Remember, the in this context means the force exerted on the material per unit of cross-sectional area. As such, it has units of pressure, such as Pascals (Pa). To do the experiment, the two ends of a sample are attached to two clamps, one of which is movable. The movable clamp is then pulled so that the sample becomes stretched. The force required to pull the sample is recorded and, given the cross-sectional area of the sample perpendicular to the force, the force is converted into units of strain. The experiment also measures the distance that the sample is stretched (the strain). That distance is usually expressed as a ratio, comparing the change in length to the original length of the sample. It can be written either as a fraction or as a percent. In the rough sketch below, we can see what typically happens in such an experiment. The sample stretches (the strain increases), and it gets harder to stretch as seen by the increasing force (and therefore increasing stress) that is needed to keep stretching it. Eventually, the stress plummets, because of the sample breaks. Already in this graph, we can see a couple of important pieces of information we can learn from tensile testing. Number one, how far can we stretch this material before it breaks? That quantity is called "strain at break". In this sample the strain at break looks like about 1.1 or 110%; that means the sample was stretched to twice its original length, and then some. Number two, how much stress can the sample support without breaking? That quantity is called "ultimate tensile strength". In this sample, the ultimate tensile strength is just over 750 Pa. If we look more closely at the graph from another sample, we will get a third important quantity, and see some additional features. The important quantity can be derived from the first part of the curve (A), in which stress increases linearly with strain. In this linear region, the material is behaving as a "Hookean solid", meaning it obeys Hooke's Law. Hooke's Law says that stress and strain should be directly proportional. In his original words, the extension is proportional to the force: in which is the force, is the extension of the solid, and is the proportionality constant. Hooke's Law is commonly applied to the behavior of mechanical springs, but it also holds for other solid materials. The slope in the linear region of the graph (A) would equal that proportionality constant, , because the graph shows the ratio of . In materials science, this slope is more commonly called "Young's modulus". It is a measure of the inherent stiffness of the material. Young's modulus: E = σ/ε in which σ = stress and ε = extensional strain. The extensional strain is just the strain observed by stretching the material. In the graph shown above, Young's modulus is around: 250 Pa / 0.1 = 2,500 Pa. This initial region of the curve, in which Hooke's Law is obeyed, is sometimes called the "linear elastic region". The word "elastic" does have an immediate connotation in everyday English, bringing to mind a rubber band that can be stretched, so we think of the stretching part when we hear the word "elastic". However, the returning motion is an essential feature of elastic behavior. The rubber band always comes back to its original shape. Likewise, within the linear elastic region, any solid material returns to its original shape after it is deformed under stress. With most solids, such as aluminum or concrete, the linear elastic region spans a very narrow range of strains. Just by looking, we wouldn't notice these materials being deformed. However, this linear stress-strain relationship is typical of solids. In rheology, that behavior is described as elastic. Of course, with many polymers, a sample can be stretched so far that you can see the change with the naked eye, and it still returns to that original shape. What makes polymers different? The long-chain structure of polymers does make them behave differently from other materials. These chains can undergo conformational change: each bond along the chain can rotate, converting the polymer chain into a slightly different shape. That ability gives a "soft material" a great deal of flexibility. The conformations of chains can adapt to accommodate stress, moving the chains into new shapes that offer lower-energy packing between each other. When the stress is removed, the chains eventually slide back into their original conformations. They return to their equilibrium shapes. The presence of physical or chemical crosslinks help the material return to its original shape, functioning as anchor points so that the chains don't stray too far. Those interactions might be found in hard-phase interactions in a microphase separated material, as illustrated here, although they could also be found in a homogeneous material. The linear elastic region isn't all we see in the stress-strain curve above. At point (B), the linear relationship is suddenly lost. The stress might even drop, as seen in this particular case. This feature on the graph is called the "yield point". The stress being experienced by the material, and the resulting strain, has become sufficient to overcome the natural elastic behavior of the solid. As noted previously, physical crosslinks such as hydrogen bonds help to reinforce the elastic behavior of a polymer sample. If, at some point, those interactions are overcome, the chains will start to slide more readily past each other. As a result, the material loses its elasticity. When the stress is released, the material will still spring back as the chains settle into a new conformational equilibrium. However, that equilibrium will not be the same as the one before. New physical crosslinks will form as groups form intermolecular interactions with the nearest neighbors they encounter. It isn't likely that these will always be the same groups that they were interacting with before. As a result, the material will settle into a slightly different shape. You have probably seen this happen before when a rubber band has been stretched too far, too often, or for too long. Chains have dropped their old interactions and picked up new ones that formed more easily in the extended shape. At point (C) of the example graph, the slope of the curve starts to increase. The same change in stress results in smaller and smaller changes in strain; the material is getting stiffer. This phenomenon is called "strain hardening". This feature would not always be observed, but if it did happen, what would explain it? In polymers, one explanation may lay in the fact that the volume of the material should remain constant as it is stretched. If the sample is getting longer, that means it is also getting narrower. As a consequence of that narrowing cross-section, chains become compressed together. At some point, physical crosslinks begin occurring between neighboring chains. These crosslinks don't occur in equilibrium positions, with polymer chains coiled around each other like they were in the original sample. These crosslinks occur when chains are extended, lying parallel to each other, at closer contact distances than normal. We can see the opposite sort of thing happening at (D), when the slope of the curve is decreasing instead of increasing. This phenomenon is called "strain-softening". Again, it might not be observed in all samples. When it does happen, what causes it? In this case, the answer is simpler. Having overcome the interactions that held the chains together, there is nothing left to resist further deformation. As the chains begin to disentangle from each other, it becomes even easier to pull them apart, facilitating the extension or stretching of the sample. Eventually, at the breaking point (E), the chains start to lose contact with each other at some location in the sample, resulting in a catastrophic rupture of the sample. In each of the following curves, estimate ultimate tensile strength and strain at break. In each of the following curves, calculate Young's modulus. In each of the following curves, identify any diagnostic features. | 9,078 | 4,168 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.10%3A_Weak_Acids |
Not all molecules which contain hydrogen are capable of donating protons. For example, methane (CH ) and other show no acidic properties at all. Carbon is not highly , and so electron density is fairly evenly shared in a C―H bond, and the hydrogen atom is unlikely to depart without at least one electron. Even when it is bonded to highly electronegative atoms like oxygen or fluorine, a hydrogen atom is not strongly acidic. Acetic acid has the projection formula Write an equation for transfer of a proton from acetic acid to water. By the electronegativity rule given above, only the hydrogen attached to oxygen should be acidic. The equation is \[\text{CH}_{3}\text{COOH} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{CH}_{3}\text{COO}^{-} \label{1} \] : To emphasize that only one hydrogen atom is acidic, the formula acetic acid is often written HC H O . You may also find the formulas HA or HOA for acetic acid, where A or OA represents the acetate ion, CH COO . The equation for the proton transfer between acetic acid and water is written with a double arrow because it occurs to only a limited extent. Like HgCl , acetic acid is a weak electrolyte. According to , 0.001 HC H O conducts slightly more than one-tenth as much current as the same concentration of the HCl or HNO . Therefore we can conclude that at a given instant only a little over 10 percent of the acetic acid molecules have donated protons to water molecules according to Eq. \(\ref{1}\). Nearly 90 percent are in molecular form as CH COOH (or HC H O ) and make no contribution to the current. Because acetic acid is not a strong enough proton donor to be entirely converted to hydronium ions in aqueous solution, it is called a . A given concentration of a weak acid produces fewer hydronium ions per unit volume and therefore less acidity than the same concentration of a strong acid. For those of you in laboratory, you can identify a weak acid using litmus paper. Below is an image of litmus paper after the addition of the weak acid acetic acid (CH COOH). Notice the red hue, which indicates a weak acid. If HCl of the same molarity were tested, the color would be darker, more purple than red. There are a large number of weak acids, but fortunately they fall into a few well-defined categories: These compounds have the general formula RCOOH. All react with water in the same way as acetic acid [Eq. \(\ref{1}\)]. The strength of carboxylic acids is dependent on the electronegative strength of the atoms in the "R" group. Consider the compounds F COOH and H COOH. Fluorine is the most electronegative element, while hydrogen is comparable to carbon in electronegativity. Thus, the fluorines pull electron density away from the carboxyl group. This removes electron density from the acidic oxygen-hydrogen bond, which weakens it. This weaker bond means that the hydrogen can be removed more easily, which creates a stronger acid. This concept can be applied to any R group. The more electronegative the R group, the stronger the carboxylic acid will be. These have the same general formula H XO as strong oxyacids, but the number of hydrogens is equal to or one less than the number of oxygens. For a weak oxyacid, in other words, ≤ + 1. Some examples are: Hydrogen fluoride (HF) has a very strong bond and does not donate its proton as readily as other hydrogen halides. Other molecules in this category are hydrogen sulfide (H S) and hydrogen cyanide (HCN). In the latter case, even though H is bonded to C, the electronegative N atom pulls some electron density away, and the HCN molecule is a very weak proton donor. Cations, especially those of charge +3 or more or of the transition metals, are surrounded closely by four to six water molecules in aqueous solution. An example is Cr(H O) , shown in Figure \(\Page {1}\). The positive charge of the metal ion pulls electron density away from the surrounding water molecules, weakening the hold of the oxygen atoms for the hydrogen atoms. The latter can consequently be more easily donated as protons: \[\text{Cr}(\text{H}_{2}\text{O})_{6}^{3+} + \text{H}_{2}\text{O} \rightleftharpoons \text{Cr}(\text{H}_{2}\text{O})_{5}\text{OH}^{2+} + \text{H}_{3}\text{O}^{+} \nonumber \] Certain other ions can donate protons. One example is the ammonium ion, NH : \[\text{NH}_{4}^{+} + \text{H}_{2}\text{O} \rightleftharpoons \text{NH}_{3} + \text{H}_{3}\text{O}^{+} \nonumber \] Anions formed when some acids donate protons can lose yet another H . An example of this is the hydrogen sulfate ion formed when sulfuric acid donates a proton: \[\text{H}_{2}\text{SO}_{4} + \text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{O}^{+} + \text{HSO}_{4}^{-} \nonumber \] \[\text{HSO}_{4}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{SO}_{4}^{2-} \nonumber \] Although sulfuric acid is strong, the negative charge on the hydrogen sulfate ion holds the proton tighter, and so the ion is a considerably weaker acid. Acids such as H SO , H S, H SO , and H CO are called because they can donate two protons. Phosphoric acid, H PO , is —it can donate three protons. | 5,188 | 4,169 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Named_Reactions/Beckmann_Rearrangement |
The Beckmann rearrangement is a reaction employed in many sectors to convert oximes to amides. The reaction has greatly been improved since its discovery in the sense of safety and viability. This work focuses on the history of the Beckmann rearrangement and improvements applied to current syntheses of mass-produced, widely available compounds that previously utilized expensive, toxic, and difficult to synthesize or hard to obtain reagents. The Beckmann rearrangement is a reaction discovered in the mid-1880’s by the chemist Ernst Otto Beckmann. The reaction converts oximes into their corresponding amides allowing the insertion of the nitrogen atom from the C=N bond into the carbon chain forming a C–N bond. Depending on the starting material, it could also produce nitriles from aldehydes. Traditional methods for the rearrangement involve harsh reaction conditions, such as a strongly acidic medium and high temperatures that can lead to undesired side products and unsuitable for sensitive substrates. Further historical accounts on the reaction’s discovery shall be discussed shortly, along with the current advancements for reactions that employ the Beckmann rearrangement for mass-produced compounds in the chemical, pharmaceutical, and agricultural sectors In 1883, a chemist by the name of Alois Janny was working on the reaction of acetoxime with phosphorus pentachloride but could not identify the products. It was not until 1886, when Beckmann published his work identifying the products that were being produced by a ketoxime (benzophenone oxime) reacting with phosphorus pentachloride or phosphorus oxychloride, and determined that the product of his reaction was benzanilide. Originally, Beckmann did not intend to identify the products that Janny failed to. Beckmann was working on a method to distinguish between aldehydes and ketones. Just like many other discoveries, Beckmann’s was predominantly accidental. Beckmann determined as well that a ketoxime could be reacted with other reagents, such as sulfuric acid. Beckmann presented his so-called “Beckmann mixture” which consisted of hydrochloric acid, acetic anhydride, and acetic acid. His traditional method calls for a highly acidic medium and high temperatures. The highly acidic medium and high temperatures are what can be considered harsh, as these conditions are not suitable for sensitive substrates. Methods employed today vary from “one-pot” reactions to a few step reactions, low-to-mild temperatures, have low reaction times, and use readily available, less toxic, cost-effective reagents. Within the last 25 years, green chemistry studies have focused on employing these methods in effort to minimize the cost of production, waste, and use of toxic reagents. In 1900, Otto Wallach discovered the compound caprolactam, and the Parace Celanese Corp., USA patented their synthesis of paracetamol in 1985. Their work states that paracetamol is produced by reacting a hydroxy aromatic ketone (4-hydroxyacetophenone) with a hydroxylamine salt to form the ketoxime and exposing the ketoxime to a catalyst which induces the Beckmann rearrangement forming the N-acyl-hydroxy aromatic amine. For the Beckmann rearrangement to take place, 4-hydroxyacetophenone oxime was allowed to react in a mixture of Amberlyst 15 (catalytic resin) and acetic acid, which was then refluxed under N for 2 h, resulting in a good yield (66.7%). reacting 4-hydroxyacetophenone oxime with ammonium persulphate and dimethyl sulfoxide in 1,4-dioxane. This mixture was heated to 100ºC and allowed to react 45 min, and results in a good yield. [1] E. Beckmann, Ber. Dtsch. Chem. Ges. , 19, 988-993. [2] A. Martínez-Asencio, M. Yus, D. J.Ramon, Tetrahedron, , 68, 3948-3951. [3] S. Srivastava and K. Kaur, New J. Chem., 0, 3. [4] B. Waskow, et al., Tetrahedron Lett., , 57, 5575-5580. [5] Beckmann Rearrangement (accessed April 20, 2021) [6] A. Janny, [8] E. Beckmann, Ber. Dtsch. Ch Tetrahedron Lett., , 42, [14] [15] S. B. Mhaske, P. S. Mahajan, A Process For Synthesis Of Amides Via Radical-mediated Beckmann Rearrangement, , 1-22. | 4,130 | 4,170 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.11%3A_Weak_Bases |
By analogy with , are not strong enough proton acceptors to react completely with water. A typical example is ammonia, which reacts only to a limited extent: \[\text{NH}_{3} + \text{H}_{2}\text{O} \rightleftharpoons \text{NH}_{4}^{+} + \text{OH}^{-} \nonumber \] To verify that NH is a weak base, a litmus test can be done. The NH is added to litmus paper, which indicates it's pH based on the color of the litmus paper. In the image below you can see the litmus paper turns a greenish blue when exposed to NH (pH 9-10). For comparison, a strong base of the same molarity like NaOH or KOH would have a darker blue coloring. Weak bases fall into two main categories. Amines may be derived from ammonia by replacing one or more hydrogens with carbon chains. They react with water in the same way as ammonia. Trimethyl amine behaves as follows: The fact that the molecules of a are not entirely converted into hydronium ions and anions implies that those anions must have considerable affinity for protons. For example, the anion of acetic acid, acetate ion, can accept protons as follows: \[\text{C}_{2}\text{H}_{3}\text{O}_{2}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{HC}_{2}\text{H}_{3}\text{O}_{2} + \text{OH}^{-} \nonumber \] | 1,258 | 4,171 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alcohols/Reactivity_of_Alcohols/Thionyl_Chloride |
If There is one thing you learn how to do well in Org 1, it’s make alcohols. Let’s count the ways: hydroboration, acid-catalyzed hydration, for starters, and then substitution of alkyl halides with water or \(HO^-\). If you want to extend it even further, There is dihydroxylation (to make diols) using \(OsO_4\) or cold \(KMnO_4\), and even opening of epoxides under acidic or basic conditions to give alcohols. There is just one issue here and it comes up once you try to use alcohols in synthesis. Let’s say you want to use that alcohol in a subsequent substitution step, getting rid of the \(HO^-)\ and replacing it with something else. See any problems with that? Remember that good leaving groups are weak bases – and the hydroxide ion, being a strong base, tends to be a pretty bad leaving group. So what can we do? What you want to do is convert the alcohol into a better leaving group. One way is to convert the alcohol into a sulfonate ester – we talked about that with \( TsCl\) and \(MsCl\). Alternatively, alcohols can be converted into alkyl chlorides with thionyl chloride (\(SOCl_2\)). This is a useful reaction, because the resulting are versatile compounds that can be converted into many compounds that are not directly accessible from the alcohol itself. If you take an alcohol and add thionyl chloride, it will be converted into an alkyl chloride. The byproducts here are hydrochloric acid (\(HCl\)) and sulfur dioxide (\(SO_2\)). Note: there are significant differences in how this reaction is taught at different schools. Consult your instructor to be 100% sure that this applies to your course). There is one important thing to note here: see the stereochemistry? It’s been inverted.*(white lie alert – see below) That’s an important difference between \(SOCl_2\) and TsCl, which leaves the stereochemistry alone. We’ll get to the root cause of that in a moment, but in the meantime, can you think of a mechanism which results in inversion of configuration at carbon? As an extra bonus, thionyl chloride will also convert (“acyl chlorides”). Like alcohols, carboxylic acids have their limitations as reactants: the hydroxyl group interferes with many of the reactions we learn for nucleophilic acyl substitution (among others). Conversion of the OH into Cl solves this problem. As you might have guessed, conversion of alcohols to alkyl halides proceeds through a substitution reaction – specifically, an \(S_N2\) mechanism. The first step is attack of the oxygen upon the sulfur of \(SOCl_2\), which results in displacement of chloride ion. This has the side benefit of converting the alcohol into a good leaving group: in the next step, chloride ion attacks the carbon in \(S_N2\) fashion, resulting in cleavage of the C–O bond with inversion of configuration. The \(HOSCl\) breaks down into \(HCl\) and sulfur dioxide gas, which bubbles away. Since the reaction proceeds through a backside attack (\(S_N2\)), there is inversion of configuration at the carbon The mechanism for formation of acid chlorides from carboxylic acids is similar. The conversion of caboxylic acids to acid chlorides is similar, but proceeds through a [1,2]-addition of chloride ion to the carbonyl carbon followed by [1,2]-elimination to give the acid chloride, \(SO_2\) and \(HCl\) Like many sulfur-containing compounds, thionyl chloride is noseworthy for its pungent smell. Thionyl chloride has a nauseating sickly-sweet odor to it that imprints itself forever upon your memory. One accident that occurred during my time as a TA involved a student dropping a flask with 5 mL of thionyl chloride into a rotovap bath outside the fume hood. The cloud of \(SO_2\) and \(HCl\) that formed cleared the teaching lab for half an hour, so you can imagine what thionyl chloride would do if exposed to the moisture in your lungs. Treat with caution, just as you would if you were working with phosgene. *Here’s the white lie. Although it’s often taught that \(SOCl_2\) leads to 100% inversion of configuration, in reality it’s not always that simple. Inversion of configuration with \(SOCl_2\) is very solvent dependent. Depending on the choice of solvent, one can get either straight inversion, or a mixture of retention and inversion. For the purposes of beginning organic classes, most students can ignore this message. ( ) | 4,344 | 4,173 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Spectroscopy/Overview_of__molecular_spectroscopy |
Electromagnetic radiation, as you may recall from a previous chemistry or physics class, is composed of electrical and magnetic waves which oscillate on perpendicular planes. Visible light is electromagnetic radiation. So are the gamma rays that are emitted by spent nuclear fuel, the x-rays that a doctor uses to visualize your bones, the ultraviolet light that causes a painful sunburn when you forget to apply sun block, the infrared light that the army uses in night-vision goggles, the microwaves that you use to heat up your frozen burritos, and the radio-frequency waves that bring music to anybody who is old-fashioned enough to still listen to FM or AM radio. Just like ocean waves, electromagnetic waves travel in a defined direction. While the speed of ocean waves can vary, however, the speed of electromagnetic waves – commonly referred to as the speed of light – is essentially a constant, approximately 300 million meters per second. This is true whether we are talking about gamma radiation or visible light. Obviously, there is a big difference between these two types of waves – we are surrounded by the latter for more than half of our time on earth, whereas we hopefully never become exposed to the former to any significant degree. The different properties of the various types of electromagnetic radiation are due to differences in their wavelengths, and the corresponding differences in their energies: High-energy radiation (such as gamma- and x-rays) is composed of very short waves – as short as 10 meter from crest to crest. Longer waves are far less energetic, and thus are less dangerous to living things. Visible light waves are in the range of 400 – 700 nm (nanometers, or 10 m), while radio waves can be several hundred meters in length. The notion that electromagnetic radiation contains a quantifiable amount of energy can perhaps be better understood if we talk about light as a stream of , called , rather than as a wave. (Recall the concept known as ‘wave-particle duality’: at the quantum level, wave behavior and particle behavior become indistinguishable, and very small particles have an observable ‘wavelength’). If we describe light as a stream of photons, the energy of a particular wavelength can be expressed as: \[E = \dfrac{ hc}{\lambda}\] where E is energy in kcal/mol, (the Greek letter ) is wavelength in meters, is 3.00 x 10 m/s (the speed of light), and is 9.537 x 10 kcal s mol , a number known as Planck’s constant. Because electromagnetic radiation travels at a constant speed, each wavelength corresponds to a given frequency, which is the number of times per second that a crest passes a given point. Longer waves have lower frequencies, and shorter waves have higher frequencies. Frequency is commonly reported in hertz (Hz), meaning ‘cycles per second’, or ‘waves per second’. The standard unit for frequency is s . When talking about electromagnetic waves, we can refer either to wavelength or to frequency - the two values are interconverted using the simple expression: \[\lambda \nu = c\] where (the Greek letter ‘ ) is frequency in s . Visible red light with a wavelength of 700 nm, for example, has a frequency of 4.29 x 10 Hz, and an energy of 40.9 kcal per mole of photons. The full range of electromagnetic radiation wavelengths is referred to as the . Notice in the figure above that visible light takes up just a narrow band of the full spectrum. White light from the sun or a light bulb is a mixture of all of the visible wavelengths. You see the visible region of the electromagnetic spectrum divided into its different wavelengths every time you see a rainbow: violet light has the shortest wavelength, and red light has the longest. : Visible light has a wavelength range of about 400-700 nm. What is the corresponding frequency range? What is the corresponding energy range, in kcal/mol of photons? In a spectroscopy experiment, electromagnetic radiation of a specified range of wavelengths is allowed to pass through a sample containing a compound of interest. The sample molecules absorb energy from some of the wavelengths, and as a result jump from a low energy ‘ground state’ to some higher energy ‘excited state’. Other wavelengths are absorbed by the sample molecule, so they pass on through. A detector on the other side of the sample records which wavelengths were absorbed, and to what extent they were absorbed. Here is the key to molecular spectroscopy: Thus, if the transition involves the molecule jumping from ground state A to excited state B, with an energy difference of E, the molecule will specifically absorb radiation with wavelength that corresponds to E, while allowing other wavelengths to pass through unabsorbed. By observing which wavelengths a molecule absorbs, and to what extent it absorbs them, we can gain information about the nature of the energetic transitions that a molecule is able to undergo, and thus information about its structure. These generalized ideas may all sound quite confusing at this point, but things will become much clearer as we begin to discuss specific examples. | 5,138 | 4,174 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.04%3A_The_Bohr_Atom |
In 1913, a Danish physicist, Niels Bohr (1885–1962; Nobel Prize in Physics, 1922), proposed a theoretical model for the hydrogen atom that explained its emission spectrum. Bohr’s model required only one assumption: Rutherford’s earlier model of the atom had also assumed that electrons moved in circular orbits around the nucleus and that the atom was held together by the electrostatic attraction between the positively charged nucleus and the negatively charged electron. Although we now know that the assumption of circular orbits was incorrect, Bohr’s insight was to propose that . Using classical physics, Niels Bohr showed that the energy of an electron in a particular orbit is given by \[ E_{n}=\dfrac{-\Re hc}{n^{2}} \label{\(\Page {3}\)}\] where \( \Re \) is the Rydberg constant, is Planck’s constant, is the speed of light, and is a positive integer corresponding to the number assigned to the orbit, with = 1 corresponding to the orbit closest to the nucleus. During the Nazi occupation of Denmark in World War II, Bohr escaped to the United States, where he became associated with the Atomic Energy Project. In his final years, he devoted himself to the peaceful application of atomic physics and to resolving political problems arising from the development of atomic weapons. As n decreases, the energy holding the electron and the nucleus together becomes increasingly negative, the radius of the orbit shrinks and more energy is needed to ionize the atom. The orbit with n = 1 is the lowest lying and most tightly bound. The negative sign in Equation \(\Page {3}\) indicates that the electron-nucleus pair is more tightly bound when they are near each other than when they are far apart. Because a hydrogen atom with its one electron in this orbit has the lowest possible energy, this is the ground state (the most stable arrangement of electrons for an element or a compound), the most stable arrangement for a hydrogen atom. As n increases, the radius of the orbit increases; the electron is farther from the proton, which results in a less stable arrangement with higher potential energy (Figure 2.10). A hydrogen atom with an electron in an orbit with n > 1 is therefore in an excited state. Any arrangement of electrons that is higher in energy than the ground state.: its energy is higher than the energy of the ground state. When an atom in an excited state undergoes a transition to the ground state in a process called decay, it loses energy by emitting a photon whose energy corresponds to the difference in energy between the two states (Figure \(\Page {1}\) ). So the difference in energy (Δ ) between any two orbits or energy levels is given by \( \Delta E=E_{n_{1}}-E_{n_{2}} \) where is the final orbit and the initial orbit. Substituting from Bohr’s equation (Equation \(\Page {3}\)) for each energy value gives \[ \Delta E=E_{final}-E_{initial}=-\dfrac{\Re hc}{n_{2}^{2}}-\left ( -\dfrac{\Re hc}{n_{1}^{2}} \right )=-\Re hc\left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{\(\Page {4}\)}\] If > , the transition is from a higher energy state (larger-radius orbit) to a lower energy state (smaller-radius orbit), as shown by the dashed arrow in part (a) in Figure \(\Page {3}\). Substituting /λ for Δ gives \[ \Delta E = \dfrac{hc}{\lambda }=-\Re hc\left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{\(\Page {5}\)}\] Canceling on both sides gives \[ \dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \label{\(\Page {6}\)}\] Except for the negative sign, this is the same equation that Rydberg obtained experimentally. The negative sign in Equation \(\Page {5}\) and Equation \(\Page {6}\) indicates that energy is released as the electron moves from orbit to orbit because orbit is at a higher energy than orbit . Bohr calculated the value of \(\Re\) from fundamental constants such as the charge and mass of the electron and Planck's constant and obtained a value of 1.0974 × 10 m , the same number Rydberg had obtained by analyzing the emission spectra. We can now understand the physical basis for the Balmer series of lines in the emission spectrum of hydrogen (part (b) in Figure 2.9 ). As shown in part (b) in Figure \(\Page {3}\), the lines in this series correspond to transitions from higher-energy orbits (n > 2) to the second orbit (n = 2). Thus the hydrogen atoms in the sample have absorbed energy from the electrical discharge and decayed from a higher-energy excited state (n > 2) to a lower-energy state (n = 2) by emitting a photon of electromagnetic radiation whose energy corresponds exactly to the difference in energy between the two states (part (a) in Figure \(\Page {3}\) ). The n = 3 to n = 2 transition gives rise to the line at 656 nm (red), the n = 4 to n = 2 transition to the line at 486 nm (green), the n = 5 to n = 2 transition to the line at 434 nm (blue), and the n = 6 to n = 2 transition to the line at 410 nm (violet). Because a sample of hydrogen contains a large number of atoms, the intensity of the various lines in a line spectrum depends on the number of atoms in each excited state. At the temperature in the gas discharge tube, more atoms are in the n = 3 than the n ≥ 4 levels. Consequently, the n = 3 to n = 2 transition is the most intense line, producing the characteristic red color of a hydrogen discharge (part (a) in Figure \(\Page {1}\) ). Other families of lines are produced by transitions from excited states with n > 1 to the orbit with n = 1 or to orbits with n ≥ 3. These transitions are shown schematically in Figure \(\Page {4}\) In contemporary applications, electron transitions are used in timekeeping that needs to be exact. Telecommunications systems, such as cell phones, depend on timing signals that are accurate to within a millionth of a second per day, as are the devices that control the US power grid. Global positioning system (GPS) signals must be accurate to within a billionth of a second per day, which is equivalent to gaining or losing no more than one second in 1,400,000 years. Quantifying time requires finding an event with an interval that repeats on a regular basis. To achieve the accuracy required for modern purposes, physicists have turned to the atom. The current standard used to calibrate clocks is the cesium atom. Supercooled cesium atoms are placed in a vacuum chamber and bombarded with microwaves whose frequencies are carefully controlled. When the frequency is exactly right, the atoms absorb enough energy to undergo an electronic transition to a higher-energy state. Decay to a lower-energy state emits radiation. The microwave frequency is continually adjusted, serving as the clock’s pendulum. In 1967, the second was defined as the duration of 9,192,631,770 oscillations of the resonant frequency of a cesium atom, called the . Research is currently under way to develop the next generation of atomic clocks that promise to be even more accurate. Such devices would allow scientists to monitor vanishingly faint electromagnetic signals produced by nerve pathways in the brain and geologists to measure variations in gravitational fields, which cause fluctuations in time, that would aid in the discovery of oil or minerals. The so-called Lyman series of lines in the emission spectrum of hydrogen corresponds to transitions from various excited states to the = 1 orbit. Calculate the wavelength of the lowest-energy line in the Lyman series to three significant figures. In what region of the electromagnetic spectrum does it occur? lowest-energy orbit in the Lyman series wavelength of the lowest-energy Lyman line and corresponding region of the spectrum We can use the Rydberg equation to calculate the wavelength: \[ \dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right ) \] For the Lyman series, = 1. The lowest-energy line is due to a transition from the = 2 to = 1 orbit because they are the closest in energy. \[ \dfrac{1}{\lambda }=-\Re \left ( \dfrac{1}{n_{2}^{2}} - \dfrac{1}{n_{1}^{2}}\right )=1.097\times m^{-1}\left ( \dfrac{1}{1}-\dfrac{1}{4} \right )=8.228 \times 10^{6}\; m^{-1} \] It turns out that spectroscopists (the people who study spectroscopy) use cm rather than m as a common unit. Wavelength is inversely proportional to energy but frequency is directly proportional as shown by Planck's formula, E=h\( u \). Spectroscopists often talk about energy and frequency as equivalent. The cm unit is particularly convenient. The infrared range is roughly 200 - 5,000 cm , the visible from 11,000 to 25.000 cm and the UV between 25,000 and 100,000 cm . The units of cm are called wavenumbers, although people often verbalize it as inverse centimeters. We can convert the answer in part A to cm . \[ \varpi =\dfrac{1}{\lambda }=8.228\times 10^{6}\cancel{m^{-1}}\left (\dfrac{\cancel{m}}{100\;cm} \right )=82,280\: cm^{-1} \] and \[\lambda = 1.215 \times 10^{−7}\; m = 122\; nm \] This emission line is called Lyman alpha. It is the strongest atomic emission line from the sun and drives the chemistry of the upper atmosphere of all the planets producing ions by stripping electrons from atoms and molecules. It is completely absorbed by oxygen in the upper stratosphere, dissociating O molecules to O atoms which react with other O molecules to form stratospheric ozone This wavelength is in the ultraviolet region of the spectrum. The Pfund series of lines in the emission spectrum of hydrogen corresponds to transitions from higher excited states to the = 5 orbit. Calculate the wavelength of the line in the Pfund series to three significant figures. In which region of the spectrum does it lie? 4.65 × 10 nm; infrared Bohr’s model of the hydrogen atom gave an exact explanation for its observed emission spectrum. The following are his key contributions to our understanding of atomic structure: Unfortunately, Bohr could not explain the electron should be restricted to particular orbits. Also, despite a great deal of tinkering, such as assuming that orbits could be ellipses rather than circles, his model could not quantitatively explain the emission spectra of any element other than hydrogen (Figure \(\Page {5}\)). In fact, Bohr’s model worked only for species that contained just one electron: H, He , Li , and so forth. Scientists needed a fundamental change in their way of thinking about the electronic structure of atoms to advance beyond the Bohr model. Thus far we have explicitly considered only the emission of light by atoms in excited states, which produces an emission spectrum (a spectrum produced by the emission of light by atoms in excited states). The converse, absorption of light by ground-state atoms to produce an excited state, can also occur, producing an absorption spectrum (a spectrum produced by the absorption of light by ground-state atoms). Because each element has characteristic emission and absorption spectra, scientists can use such spectra to analyze the composition of matter. When an atom emits light, it decays to a lower energy state; when an atom absorbs light, it is excited to a higher energy state. If white light is passed through a sample of hydrogen, hydrogen atoms absorb energy as an electron is excited to higher energy levels (orbits with ≥ 2). If the light that emerges is passed through a prism, it forms a continuous spectrum with lines (corresponding to no light passing through the sample) at 656, 468, 434, and 410 nm. These wavelengths correspond to the = 2 to = 3, = 2 to = 4, = 2 to = 5, and = 2 to = 6 transitions. Any given element therefore has both a characteristic emission spectrum and a characteristic absorption spectrum, which are essentially complementary images. Emission and absorption spectra form the basis of , which uses spectra to provide information about the structure and the composition of a substance or an object. In particular, astronomers use emission and absorption spectra to determine the composition of stars and interstellar matter. As an example, consider the spectrum of sunlight shown in Figure \(\Page {7}\) Because the sun is very hot, the light it emits is in the form of a continuous emission spectrum. Superimposed on it, however, is a series of dark lines due primarily to the absorption of specific frequencies of light by cooler atoms in the outer atmosphere of the sun. By comparing these lines with the spectra of elements measured on Earth, we now know that the sun contains large amounts of hydrogen, iron, and carbon, along with smaller amounts of other elements. During the solar eclipse of 1868, the French astronomer Pierre Janssen (1824–1907) observed a set of lines that did not match those of any known element. He suggested that they were due to the presence of a new element, which he named , from the Greek , meaning “sun.” Helium was finally discovered in uranium ores on Earth in 1895. Alpha particles are helium nuclei. Alpha particles emitted by the radioactive uranium, pick up electrons from the rocks to form helium atoms. The familiar red color of “neon” signs used in advertising is due to the emission spectrum of neon shown in part (b) in Figure \(\Page {5}\). Similarly, the blue and yellow colors of certain street lights are caused, respectively, by mercury and sodium discharges. In all these cases, an electrical discharge excites neutral atoms to a higher energy state, and light is emitted when the atoms decay to the ground state. In the case of mercury, most of the emission lines are below 450 nm, which produces a blue light (part (c) in Figure \(\Page {5}\)). In the case of sodium, the most intense emission lines are at 589 nm, which produces an intense yellow light. The Bohr Atom: The colors of fireworks are also due to atomic emission spectra. As shown in part (a) in Figure \(\Page {9}\), a typical shell used in a fireworks display contains gunpowder to propel the shell into the air and a fuse to initiate a variety of reactions that produce heat and small explosions. Thermal energy excites the atoms to higher energy states; as they decay to lower energy states, the atoms emit light that gives the familiar colors. When oxidant/reductant mixtures listed in Table \(\Page {1}\) are ignited, a flash of white or yellow light is produced along with a loud bang. Achieving the colors shown in part (b) in Figure \(\Page {9}\) requires adding a small amount of a substance that has an emission spectrum in the desired portion of the visible spectrum. For example, sodium is used for yellow because of its 589 nm emission lines. The intense yellow color of sodium would mask most other colors, so potassium and ammonium salts, rather than sodium salts, are usually used as oxidants to produce other colors, which explains the preponderance of such salts in Table \(\Page {1}\). Strontium salts, which are also used in highway flares, emit red light, whereas barium gives a green color. Blue is one of the most difficult colors to achieve. Copper(II) salts emit a pale blue light, but copper is dangerous to use because it forms highly unstable explosive compounds with anions such as chlorate. As you might guess, preparing fireworks with the desired properties is a complex, challenging, and potentially hazardous process. If you have the time here is a NOVA program about how fireworks are made. There is an intimate connection between the atomic structure of an atom and its spectral characteristics. Atoms of individual elements emit light at only specific wavelengths, producing a rather than the continuous spectrum of all wavelengths produced by a hot object. Niels Bohr explained the line spectrum of the hydrogen atom by assuming that the electron moved in circular orbits and that orbits with only certain radii were allowed. Lines in the spectrum were due to transitions in which an electron moved from a higher-energy orbit with a larger radius to a lower-energy orbit with smaller radius. The orbit closest to the nucleus represented the of the atom and was most stable; orbits farther away were higher-energy . Transitions from an excited state to a lower-energy state resulted in the emission of light with only a limited number of wavelengths. Bohr’s model could not, however, explain the spectra of atoms heavier than hydrogen. ( ) | 16,367 | 4,176 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Properties_of_Arenes/Polycyclic_Aromatics |
Polycyclic aromatics are compounds containing two or more fused aromatic rings. These fused benzene rings share two carbon atoms between them. These are also called polycyclic benoid or polycyclic aromatic hydrocarbons (PAHs). Many PAHs are carcinogenic and highly toxic. The naming system for PAHs is quite complex. A series of fused benzene rings that occur in a linear manner are called acenes. Another form of fusion of benzene rings is angular fusion, or "annulation", which is, as its name implies, an angled fusion of the benzene rings. Naphthalene is the simplest polycyclic aromatic hydrocarbon since it is only a bicyclic molecule made up of two aromatic benzenes. The pi electrons in this molecule are even more delocalized than those in the simpler benzene molecule. Naphthalene is also planar and has 4n + 2 pi electrons (10) giving it the stabilizing and resonating aromatic properties shared with benzene. Aromaticity is a property that the majority of fused benzenoid hydrocarbons share. The extent to which each polycyclic molecule is aromatic varies. As shown above, the structural isomers of anthracene and phenanthrene are very similar molecules made up of three benzene rings. Anthracene is fused linearly, whereas phenanthrene is fused at an angle. This difference in fusions causes the phenanthrene to have five resonance structures which is one more than anthracene. This extra resonance makes the phenanthrene around 6 kcal per mol more stable. | 1,489 | 4,177 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/18%3A_Chemical_Kinetics/18.12%3A_Heterogeneous_Catalysis |
Our previous discussion has concentrated on catalysts which are in the same phase as the reaction being catalyzed. This kind of catalysis is called . Many important industrial processes rely on , in which the catalyst is in a different phase. Usually the catalyst is a solid and the reactants are gases, and so the rate-limiting step occurs at the solid surface. Thus heterogeneous catalysis is also referred to as . The detailed mechanisms of most heterogeneous reactions are not yet understood, but certain sites on the catalyst surface appear to be able to weaken or break bonds in reactant molecules. These are called . One example of heterogeneous catalysis is hydrogenation of an unsaturated organic compound such as ethane (C H ) by metal catalysts such as Pt or Ni: Heterogeneous catalysts are used extensively in the petroleum industry. One example is the combination of SiO and Al O used to speed up cracking of long-chain hydrocarbons into the smaller molecules needed for gasoline. Another is the Pt catalyst used to reform hydrocarbon chains into aromatic ring structures. This improves the octane rating of gasoline, making it more suitable for use in automobile engines. Other industries also make effective use of catalysts. SO , obtained by burning sulfur (or even from burning coal), can be oxidized to SO over vanadium pentoxide, V O . This is an important step in manufacturing H SO . Another important heterogeneous catalyst is used in the Haber process for synthesis of NH from N and H . As with most industrial catalysts, its exact composition is a trade secret, but it is mainly Fe with small amounts of Al O and K O added. The surface catalyst you are most likely to be familiar with is found in the exhaust systems of many automobiles constructed since 1976. Such a catalytic converter contains from 1 to 3 g Pt in a fine layer on the surface of a honeycomb-like structure or small beads made of Al O . The catalyst speeds up oxidation of unburned hydrocarbons and CO which would otherwise be emitted from the exhaust as air pollutants. It apparently does this by adsorbing and weakening the bond in the O molecule. Individual O atoms are then more readily transferred to CO or hydrocarbon molecules, producing CO and H O. This action of the catalytic surface can be inhibited or poisoned if lead atoms (from tetraethyllead in leaded gasoline) react with the surface. Hence the prohibition of use of leaded fuel in cars equipped with catalytic converters. | 2,505 | 4,178 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.05%3A_Gay-Lussac's_Law |
Propane tanks are widely used with barbeque grills. However, it's not fun to find out half-way through grilling that you've run out of gas. You can buy gauges that measure the pressure inside the tank to see how much is left. The gauge measures pressure and will register a higher pressure on a hot day than it will on a cold day. So, you need to take the air temperature into account when you decide whether or not to refill the tank before your next cook-out. When the temperature of a sample of gas in a rigid container is increased, the pressure of the gas increases as well. The increase in kinetic energy results in the molecules of gas striking the walls of the container with more force, resulting in a greater pressure. The French chemist Joseph Gay-Lussac (1778-1850) discovered the relationship between the pressure of a gas and its absolute temperature. states that the pressure of a given mass of gas varies directly with the absolute temperature of the gas, when the volume is kept constant. Gay-Lussac's Law is very similar to Charles's Law, with the only difference being the type of container. Whereas the container in a Charles's Law experiment is flexible, it is rigid in a Gay-Lussac's Law experiment. The mathematical expressions for Gay-Lussac's Law are likewise similar to those of Charles's Law: \[\frac{P}{T} \: \: \: \text{and} \: \: \: \frac{P_1}{T_1} = \frac{P_2}{T_2}\nonumber \] A graph of pressure vs. temperature also illustrates a direct relationship. As a gas is cooled at constant volume, its pressure continually decreases until the gas condenses to a liquid. The gas in an aerosol can is under a pressure of \(3.00 \: \text{atm}\) at a temperature of \(25^\text{o} \text{C}\). It is dangerous to dispose of an aerosol can by incineration. What would the pressure in the aerosol can be at a temperature of \(845^\text{o} \text{C}\)? Use Gay-Lussac's Law to solve for the unknown pressure \(\left( P_2 \right)\). The temperatures have first been converted to Kelvin. First, rearrange the equation algebraically to solve for \(P_2\). \[P_2 = \frac{P_1 \times T_2}{T_1}\nonumber \] Now substitute the known quantities into the equation and solve. \[P_2 = \frac{3.00 \: \text{atm} \times 1118 \: \text{K}}{298 \: \text{K}} = 11.3 \: \text{atm}\nonumber \] The pressure increases dramatically due to large increase in temperature. | 2,384 | 4,181 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/10%3A_Gases/10.S%3A_Gases_(Summary) |
\[P = F / A \nonumber \] \[PV = constant \nonumber \] where \(P\) = pressure, \(V\) = volume \[\dfrac{V}{T} = constant \nonumber \] where \(V\) = volume, \(T\) = Temperature \[V = constant \times n \nonumber \] where V = volume, = number of moles \[PV = nRT \nonumber \] where P = pressure, V = volume, n = number of moles, R = gas constant, T = Temperature (always expressed on absolute-temperature scale, usually Kelvin) The (\(R\)) is the constant of proportionality in the ideal-gas equation. Some values of R are given below : 0°C and 1 atm. 1 mol of gas at has a volume of 22.41 L ( ) Density of a gas (\*\rho\) = density, M = molar mass): \[\rho = \dfrac{PM}{ } \nonumber \] Molar mass of a gas: \[M = \dfrac{ \rho RT }{ P} \nonumber \] The partial pressure of a gas in a mixture is its mole fraction times the total pressure : set of assumptions about the nature of gases. These assumptions, when translated into mathematical form, yield the ideal-gas equation The pressure of a gas is caused by collisions of molecules with the walls of the container. The magnitude of the pressure is determined by how often and how "hard" the molecules strike the walls. If two different gases are at the same temperature, they have the same average kinetic energy. If the temperature of a gas is doubled, its kinetic energy also doubles. Hence, molecular motion increases with increasing temperature. The rms speed is important because the average kinetic energy of the gas molecules, \(ε\), is related directly to \(u^2\): \[ ε = \dfrac{1}{2}mu^2 \nonumber \] where \(m\) is the mass of the molecule Because mass doesn’t change with temperature, the rms speed (and also the average speed) of molecules must increase as temperature increases A gas composed of light gas particles will have the same average kinetic energy as one composed of much heavier particles, provided that the two gases are at the same temperature. The mass, \(m\), of the particles in the lighter gas is smaller that that in the heavier gas. Consequently, the particles of the lighter gas must have a higher rms speed, \(u\), than the heavier one: \[u =\sqrt{\dfrac{3RT}{ M}} \nonumber \] Since M is in the denominator, the less massive the gas molecules, the higher the rms speed \[\dfrac{r_1}{ r_2} = \sqrt{\dfrac{ M_2}{ M_1}} \nonumber \] where \(r\) is the rate of effusion The ideal gas equation may be rearranged as follows to understand deviations from ideal-gas behavior: \[\dfrac{PV}{RT} = n \nonumber \] According to the ideal gas equation: \[ \underbrace{P = \dfrac{nRT}{ V}}_{\text{ideal gas}} \nonumber \] According to Van der Waals: \[P = \dfrac{nRT}{ V – nb} – \dfrac{n^2a}{V^2} \nonumber \] Correction for volume of molecules – Correction for molecular attractions \[\left[ P + \left(\dfrac{n^2a}{ V^2}\right) \right] (V – nb) = nRT \nonumber \] The Van der Waals constants and are different for each gas. The values of these constants generally increase with an increase in mass of the molecule and with an increase in the complexity of its structures. | 3,066 | 4,182 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.3%3A_Multiple_Bonds |
The hybrid orbital model appears to account well for the geometry of molecules involving single covalent bonds. Is it also capable of describing molecules containing double and triple bonds? We have already discussed that multiple bonds consist of σ and π bonds. Next we can consider how we visualize these components and how they relate to hybrid orbitals. The Lewis structure of ethene, C H , shows us that each carbon atom is surrounded by one other carbon atom and two hydrogen atoms. The π bond in the C=C double bond results from the overlap of the third (remaining) 2 orbital on each carbon atom that is not involved in hybridization. This unhybridized orbital (lobes shown in red and blue in Figure \(\Page {2}\)) is perpendicular to the plane of the hybrid orbitals. Thus the unhybridized 2 orbitals overlap in a side-by-side fashion, above and below the internuclear axis and form a π bond In an ethene molecule, the four hydrogen atoms and the two carbon atoms are all in the same plane. If the two planes of hybrid orbitals tilted relative to each other, the orbitals would not be oriented to overlap efficiently to create the π bond. The planar configuration for the ethene molecule occurs because it is the most stable bonding arrangement. This is a significant difference between σ and π bonds; rotation around single (σ) bonds occurs easily because the end-to-end orbital overlap does not depend on the relative orientation of the orbitals on each atom in the bond. In other words, rotation around the internuclear axis does not change the extent to which the σ bonding orbitals overlap because the bonding electron density is symmetric about the axis. Rotation about the internuclear axis is much more difficult for multiple bonds; however, this would drastically alter the off-axis overlap of the π bonding orbitals, essentially breaking the π bond. In molecules with sp hybrid orbitals, two unhybridized p orbitals remain on the atom (Figure \(\Page {3}\)). We find this situation in acetylene, H−C≡C−H, which is a linear molecule. The sp hybrid orbitals of the two carbon atoms overlap end to end to form a σ bond between the carbon atoms (Figure \(\Page {4}\)). The remaining orbitals form σ bonds with hydrogen atoms. The two unhybridized orbitals per carbon are positioned such that they overlap side by side and, hence, form two π bonds. The two carbon atoms of acetylene are thus bound together by one σ bond and two π bonds, giving a triple bond. Hybridization involves only σ bonds, lone pairs of electrons, and single unpaired electrons (radicals). Structures that account for these features describe the correct hybridization of the atoms. However, many structures also include resonance forms. Remember that resonance forms occur when various arrangements of π bonds are possible. Since the arrangement of π bonds involves only the unhybridized orbitals, resonance does not influence the assignment of hybridization. For example, molecule benzene has two resonance forms (Figure \(\Page {5}\)). We can use either of these forms to determine that each of the carbon atoms is bonded to three other atoms with no lone pairs, so the correct hybridization is . The electrons in the unhybridized orbitals form π bonds. Neither resonance structure completely describes the electrons in the π bonds. They are not located in one position or the other, but in reality are delocalized throughout the ring. Valence bond theory does not easily address delocalization. Bonding in molecules with resonance forms is better described by molecular orbital theory. Some acid rain results from the reaction of sulfur dioxide with atmospheric water vapor, followed by the formation of sulfuric acid. Sulfur dioxide, \(\ce{SO2}\), is a major component of volcanic gases as well as a product of the combustion of sulfur-containing coal. What is the hybridization of the \(S\) atom in \(\ce{SO2}\)? The resonance structures of \(\ce{SO2}\) are The sulfur atom is surrounded by two bonds and one lone pair of electrons in either resonance structure. Therefore, the electron-pair geometry is trigonal planar, and the hybridization of the sulfur atom is . Another acid in acid rain is nitric acid, HNO , which is produced by the reaction of nitrogen dioxide, NO , with atmospheric water vapor. What is the hybridization of the nitrogen atom in NO ? (Note: the lone electron on nitrogen occupies a hybridized orbital just as a lone pair would.) Multiple bonds consist of a σ bond located along the axis between two atoms and one or two π bonds. The σ bonds are usually formed by the overlap of hybridized atomic orbitals, while the π bonds are formed by the side-by-side overlap of unhybridized orbitals. Resonance occurs when there are multiple unhybridized orbitals with the appropriate alignment to overlap, so the placement of π bonds can vary. | 4,882 | 4,183 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/06%3A_Gases/6.5%3A_Gases_in_Chemical_Reactions |
With the , we can use the relationship between the amounts of gases (in moles) and their volumes (in liters) to calculate the stoichiometry of reactions involving gases, if the pressure and temperature are known. This is important for several reasons. Many reactions that are carried out in the laboratory involve the formation or reaction of a gas, so chemists must be able to quantitatively treat gaseous products and reactants as readily as they quantitatively treat solids or solutions. Furthermore, many, if not most, industrially important reactions are carried out in the gas phase for practical reasons. Gases mix readily, are easily heated or cooled, and can be transferred from one place to another in a manufacturing facility via simple pumps and plumbing. As a chemical engineer said to one of the authors, “Gases always go where you want them to, liquids sometimes do, but solids almost never do.” Sulfuric acid, the industrial chemical produced in greatest quantity (almost 45 million tons per year in the United States alone), is prepared by the combustion of sulfur in air to give SO , followed by the reaction of SO with O in the presence of a catalyst to give SO , which reacts with water to give H SO . The overall chemical equation is as follows: \[\rm 2S_{(s)}+3O_{2(g)}+2H_2O_{(l)}\rightarrow 2H_2SO_{4(aq)}\] What volume of O (in liters) at 22°C and 745 mmHg pressure is required to produce 1.00 ton (907.18 kg) of H SO ? reaction, temperature, pressure, and mass of one product volume of gaseous reactant Calculate the number of moles of H SO in 1.00 ton. From the stoichiometric coefficients in the balanced chemical equation, calculate the number of moles of O required. Use the ideal gas law to determine the volume of O required under the given conditions. Be sure that all quantities are expressed in the appropriate units. We begin by calculating the number of moles of H SO in 1.00 ton: \[\rm\dfrac{907.18\times10^3\;g\;H_2SO_4}{(2\times1.008+32.06+4\times16.00)\;g/mol}=9250\;mol\;H_2SO_4\] We next calculate the number of moles of O required: \[\rm9250\;mol\;H_2SO_4\times\dfrac{3mol\; O_2}{2mol\;H_2SO_4}=1.389\times10^4\;mol\;O_2\] After converting all quantities to the appropriate units, we can use the ideal gas law to calculate the volume of O : \[V=\dfrac{nRT}{P}=\rm\dfrac{1.389\times10^4\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times(273+22)\;K}{745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}}=3.43\times10^5\;L\] The answer means that more than 300,000 L of oxygen gas are needed to produce 1 ton of sulfuric acid. These numbers may give you some appreciation for the magnitude of the engineering and plumbing problems faced in industrial chemistry. In Example 5, we saw that Charles used a balloon containing approximately 31,150 L of H for his initial flight in 1783. The hydrogen gas was produced by the reaction of metallic iron with dilute hydrochloric acid according to the following balanced chemical equation: \[Fe_{(s)} + 2 HCl_{(aq)} \rightarrow H_{2(g)} + FeCl_{2(aq)}\] How much iron (in kilograms) was needed to produce this volume of H if the temperature was 30°C and the atmospheric pressure was 745 mmHg? 68.6 kg of Fe (approximately 150 lb) Ideal Gas law Equation and Reaction Stoichiometry: As shown in Figure \(\Page {1}\), a common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. Because the gas is less dense than liquid water, it bubbles to the top of the bottle, displacing the water. Eventually, all the water is forced out and the bottle contains only gas. If a calibrated bottle is used (i.e., one with markings to indicate the volume of the gas) and the bottle is raised or lowered until the level of the water is the same both inside and outside, then the pressure within the bottle will exactly equal the atmospheric pressure measured separately with a barometer (\(P_{\rm bar.}\)). Remember, however, when calculating the amount of gas formed in the reaction, the gas collected inside the bottle is pure. Instead, it is a mixture of the product gas and water vapor. All liquids (including water) have a measurable amount of vapor in equilibrium with the liquid because molecules of the liquid are continuously escaping from the liquid’s surface, while other molecules from the vapor phase collide with the surface and return to the liquid. The vapor thus exerts a pressure above the liquid, which is called the liquid’s . In the case shown in Figure \(\Page {1}\), the bottle is therefore actually filled with a mixture of O and water vapor, and the total pressure is, by of partial pressures, the sum of the pressures of the two components: \[P_{\rm tot}=P_{\rm gas}+P_{\rm H_2O}=P_{\rm bar.} \label{6.6.1}\] If we want to know the pressure of the gas generated in the reaction to calculate the amount of gas formed, we must first subtract the pressure due to water vapor from the total pressure. This is done by referring to tabulated values of the vapor pressure of water as a function of temperature (Table \(\Page {1}\)). As shown in Figure \(\Page {2}\), the vapor pressure of water increases rapidly with increasing temperature, and at the normal boiling point (100°C), the vapor pressure is exactly 1 atm. The methodology is illustrated in Example \(\Page {2}\). The only gases that cannot be collected using this technique are those that readily dissolve in water (e.g., NH , H S, and CO ) and those that react rapidly with water (such as F and NO ). Sodium azide (\(NaN_3\)) decomposes to form sodium metal and nitrogen gas according to the following balanced chemical equation: \[ 2NaN_3 \rightarrow 2Na_{(s)} + 3N_{2\; (g)}\] This reaction is used to inflate the air bags that cushion passengers during automobile collisions. The reaction is initiated in air bags by an electrical impulse and results in the rapid evolution of gas. If the \(N_2\) gas that results from the decomposition of a 5.00 g sample of \(NaN_3\) could be collected by displacing water from an inverted flask, as in Figure \(\Page {1}\), what volume of gas would be produced at 21°C and 762 mmHg? reaction, mass of compound, temperature, and pressure volume of nitrogen gas produced Calculate the number of moles of N gas produced. From the data in Table \(\Page {1}\), determine the partial pressure of N gas in the flask. Use the ideal gas law to find the volume of N gas produced. Because we know the mass of the reactant and the stoichiometry of the reaction, our first step is to calculate the number of moles of N gas produced: \[\rm\dfrac{5.00\;g\;NaN_3}{(22.99+3\times14.01)\;g/mol}\times\dfrac{3mol\;N_2}{2mol\;NaN_3}=0.115\;mol\; N_2\] The pressure given (762 mmHg) is the pressure in the flask, which is the sum of the pressures due to the N gas and the water vapor present. Table \(\Page {1}\) tells us that the vapor pressure of water is 18.65 mmHg at 21°C (294 K), so the partial pressure of the N gas in the flask is only \[\rm(762 − 18.65)\;mmHg \times\dfrac{1\;atm}{760\;atm}= 743.4\; mmHg \times\dfrac{1\;atm}{760\;atm}= 0.978\; atm.\] Solving the ideal gas law for and substituting the other quantities (in the appropriate units), we get \[V=\dfrac{nRT}{P}=\rm\dfrac{0.115\;mol\times0.08206\dfrac{atm\cdot L}{mol\cdot K}\times294\;K}{0.978\;atm}=2.84\;L\] A 1.00 g sample of zinc metal is added to a solution of dilute hydrochloric acid. It dissolves to produce H gas according to the equation Zn(s) + 2 HCl(aq) → H (g) + ZnCl (aq). The resulting H gas is collected in a water-filled bottle at 30°C and an atmospheric pressure of 760 mmHg. What volume does it occupy? 0.397 L Collecting a Product Gas over Water: The relationship between the amounts of products and reactants in a chemical reaction can be expressed in units of moles or masses of pure substances, of volumes of solutions, or of volumes of gaseous substances. The ideal gas law can be used to calculate the volume of gaseous products or reactants as needed. In the laboratory, gases produced in a reaction are often collected by the displacement of water from filled vessels; the amount of gas can then be calculated from the volume of water displaced and the atmospheric pressure. A gas collected in such a way is not pure, however, but contains a significant amount of water vapor. The measured pressure must therefore be corrected for the vapor pressure of water, which depends strongly on the temperature. Why are so many industrially important reactions carried out in the gas phase? The volume of gas produced during a chemical reaction can be measured by collecting the gas in an inverted container filled with water. The gas forces water out of the container, and the volume of liquid displaced is a measure of the volume of gas. What additional information must be considered to determine the number of moles of gas produced? The volume of some gases cannot be measured using this method. What property of a gas precludes the use of this method? Equal masses of two solid compounds (A and B) are placed in separate sealed flasks filled with air at 1 atm and heated to 50°C for 10 hours. After cooling to room temperature, the pressure in the flask containing A was 1.5 atm. In contrast, the pressure in the flask containing B was 0.87 atm. Suggest an explanation for these observations. Would the masses of samples A and B still be equal after the experiment? Why or why not? During the smelting of iron, carbon reacts with oxygen to produce carbon monoxide, which then reacts with iron(III) oxide to produce iron metal and carbon dioxide. If 1.82 L of CO at STP is produced, Complete decomposition of a sample of potassium chlorate produced 1.34 g of potassium chloride and oxygen gas. The combustion of a 100.0 mg sample of an herbicide in excess oxygen produced 83.16 mL of CO and 72.9 mL of H O vapor at STP. A separate analysis showed that the sample contained 16.44 mg of chlorine. If the sample is known to contain only C, H, Cl, and N, determine the percent composition and the empirical formula of the herbicide. The combustion of a 300.0 mg sample of an antidepressant in excess oxygen produced 326 mL of CO and 164 mL of H O vapor at STP. A separate analysis showed that the sample contained 23.28% oxygen. If the sample is known to contain only C, H, O, and N, determine the percent composition and the empirical formula of the antidepressant. Percent composition: 58.3% C, 4.93% H, 23.28% O, and 13.5% N; empirical formula: C H O N | 10,617 | 4,185 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/25%3A_Nuclear_Chemistry/25.01%3A_Radioactivity |
is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of \(\ce{^1_1H}\), neutrons. Recall that the number of protons in the nucleus is called the atomic number (\(Z\)) of the element, and the sum of the number of protons and the number of neutrons is the mass number (\(A\)). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term and identify it by the notation: \[ \large \ce{^{A}_{Z}X} \label{Eq1a} \] where Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, \(\ce{^{14}_6C}\) is called “carbon-14.” Protons and neutrons, collectively called , are packed together tightly in a nucleus. With a radius of about 10 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10 meters. Nuclei are extremely dense compared to bulk matter, averaging \(1.8 \times 10^{14}\) grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm . If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately \(6.4 \times 10^6\) meters, 30,000 times larger). Changes of nuclei that result in changes in their atomic numbers, mass numbers, or energy states are . To describe a nuclear reaction, we use an equation that identifies the nuclides involved in the reaction, their mass numbers and atomic numbers, and the other particles involved in the reaction. A balanced chemical reaction equation reflects the fact that during a chemical reaction, bonds break and form, and atoms are rearranged, but the total numbers of atoms of each element are conserved and do not change. A balanced nuclear reaction equation indicates that there is a rearrangement during a nuclear reaction, but of subatomic particles rather than atoms. Nuclear reactions also follow conservation laws, and they are balanced in two ways: If the atomic number and the mass number of all but one of the particles in a nuclear reaction are known, we can identify the particle by balancing the reaction. For instance, we could determine that \(\ce{^{17}_8O}\) is a product of the nuclear reaction of \(\ce{^{14}_7N}\) and \(\ce{^4_2He}\) if we knew that a proton, \(\ce{^1_1H}\), was one of the two products. Example \(\Page {1}\) shows how we can identify a nuclide by balancing the nuclear reaction. The reaction of an α particle with magnesium-25 \( (\ce{^{25}_{12}Mg})\) produces a proton and a nuclide of another element. Identify the new nuclide produced. The nuclear reaction can be written as: \[\ce{^{25}_{12}Mg + ^4_2He \rightarrow ^1_1H + ^{A}_{Z}X} \nonumber \] where Because the sum of the mass numbers of the reactants must equal the sum of the mass numbers of the products: \[\mathrm{25+4=A+1} \nonumber \] so \[ \mathrm{A=28} \nonumber \] Similarly, the charges must balance, so: \[\mathrm{12+2=Z+1} \nonumber \] so \[\mathrm{Z=13} \nonumber \] Check the periodic table: The element with nuclear charge = +13 is aluminum. Thus, the product is \(\ce{^{28}_{13}Al}\). The nuclide \(\ce{^{125}_{53}I}\) combines with an electron and produces a new nucleus and no other massive particles. What is the equation for this reaction? \[\ce{^{125}_{53}I + ^0_{−1}e \rightarrow ^{125}_{52}Te} \nonumber \] The two general kinds of nuclear reactions are nuclear decay reactions and nuclear transmutation reactions In a , also called radioactive decay, an unstable nucleus emits radiation and is transformed into the nucleus of one or more other elements. The resulting daughter nuclei have a lower mass and are lower in energy (more stable) than the parent nucleus that decayed. In contrast, in a , a nucleus reacts with a subatomic particle or another nucleus to form a product nucleus that is than the starting material. As we shall see, nuclear decay reactions occur spontaneously under all conditions, but nuclear transmutation reactions occur only under very special conditions, such as the collision of a beam of highly energetic particles with a target nucleus or in the interior of stars. We begin this section by considering the different classes of radioactive nuclei, along with their characteristic nuclear decay reactions and the radiation they emit. Nuclear decay reactions occur spontaneously under all conditions, whereas nuclear transmutation reactions are induced. Just as we use the number and type of atoms present to balance a chemical equation, we can use the number and type of nucleons present to write a balanced nuclear equation for a nuclear decay reaction. This procedure also allows us to predict the identity of either the parent or the daughter nucleus if the identity of only one is known. Regardless of the mode of decay, the total number of nucleons is conserved in all nuclear reactions. To describe nuclear decay reactions, chemists have extended the \(^A _Z \textrm{X}\) notation for nuclides to include radioactive emissions. lists the name and symbol for each type of emitted radiation. The most notable addition is the , a particle that has the same mass as an electron but a positive charge rather than a negative charge. Like the notation used to indicate isotopes, the upper left superscript in the symbol for a particle gives the mass number, which is the total number of protons and neutrons. For a proton or a neutron, = 1. Because neither an electron nor a positron contains protons or neutrons, its mass number is 0. The numbers should not be taken literally, however, as meaning that these particles have zero mass; ejection of a beta particle (an electron) simply has a negligible effect on the mass of a nucleus. Similarly, the lower left subscript gives the charge of the particle. Because protons carry a positive charge, = +1 for a proton. In contrast, a neutron contains no protons and is electrically neutral, so = 0. In the case of an electron, = −1, and for a positron, = +1. Because γ rays are high-energy photons, both and are 0. In some cases, two different symbols are used for particles that are identical but produced in different ways. For example, the symbol \(^0_{-1}\textrm e\), which is usually simplified to e , represents a free electron or an electron associated with an atom, whereas the symbol \(^0_{-1}\beta\), which is often simplified to β , denotes an electron that originates from within the nucleus, which is a β particle. Similarly, \(^4_{2}\textrm{He}^{2+}\) refers to the nucleus of a helium atom, and \(^4_{2}\alpha\) denotes an identical particle that has been ejected from a heavier nucleus. There are six fundamentally different kinds of nuclear decay reactions, and each releases a different kind of particle or energy. The essential features of each reaction are shown in . The most common are alpha and beta decay and gamma emission, but the others are essential to an understanding of nuclear decay reactions. Many nuclei with mass numbers greater than 200 undergo , which results in the emission of a helium-4 nucleus as an , \(^4_{2}\alpha\). The general reaction is as follows: \[\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A-4}_{Z-2} \textrm X'}+\underset{\textrm{alpha}\\ \textrm{particle}}{^4_2 \alpha}\label{Eq1} \] The daughter nuclide contains two fewer protons and two fewer neutrons than the parent. Thus α-particle emission produces a daughter nucleus with a mass number − 4 and a nuclear charge − 2 compared to the parent nucleus. Radium-226, for example, undergoes alpha decay to form radon-222: \[^{226}_{88}\textrm{Ra}\rightarrow ^{222}_{86}\textrm{Rn}+^{4}_{2}\alpha\label{Eq2} \] Because nucleons are conserved in this and all other nuclear reactions, the sum of the mass numbers of the products, 222 + 4 = 226, equals the mass number of the parent. Similarly, the sum of the atomic numbers of the products, 86 + 2 = 88, equals the atomic number of the parent. Thus the nuclear equation is balanced. Just as the total number of atoms is conserved in a chemical reaction, the total number of nucleons is conserved in a nuclear reaction. Nuclei that contain too many neutrons often undergo , in which a neutron is converted to a proton and a high-energy electron that is ejected from the nucleus as a β particle: \[\underset{\textrm{unstable} \\ \textrm{neutron in} \\ \textrm{nucleus}}{^1_0 \textrm n}\rightarrow \underset{\textrm{proton} \\ \textrm{retained} \\ \textrm{by nucleus}}{^{1}_{1} \textrm p}+\underset{\textrm{beta particle} \\ \textrm{emitted by} \\ \textrm{nucleus}}{^0_{-1} \beta}\label{Eq3} \] The general reaction for beta decay is therefore \[\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z+1} \textrm X'}+\underset{\textrm{beta particle}}{^0_{-1} \beta}\label{Eq4} \] Although beta decay does not change the mass number of the nucleus, it does result in an increase of +1 in the atomic number because of the addition of a proton in the daughter nucleus. Thus beta decay decreases the neutron-to-proton ratio, moving the nucleus toward the band of stable nuclei. For example, carbon-14 undergoes beta decay to form nitrogen-14: \[^{14}_{6}\textrm{C}\rightarrow ^{14}_{7}\textrm{N}+\,^{0}_{-1}\beta \nonumber \] Once again, the number of nucleons is conserved, and the charges are balanced. The parent and the daughter nuclei have the same mass number, 14, and the sum of the atomic numbers of the products is 6, which is the same as the atomic number of the carbon-14 parent. Because a positron has the same mass as an electron but opposite charge, is the opposite of beta decay. Thus positron emission is characteristic of neutron-poor nuclei, which decay by transforming a proton to a neutron and emitting a high-energy positron: \[^{1}_{1}\textrm{p}^+\rightarrow ^{1}_{0}\textrm{n}+\,^{0}_{+1}\beta^+\label{Eq6} \] The general reaction for positron emission is therefore \[\underset{\textrm{parent}}{^A_Z \textrm X}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\underset{\textrm{positron}}{^0_{+1} \beta^+} \nonumber \] Like beta decay, positron emission does not change the mass number of the nucleus. In this case, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Thus the neutron-to-proton ratio has increased, again moving the nucleus closer to the band of stable nuclei. For example, carbon-11 undergoes positron emission to form boron-11: \[^{11}_{6}\textrm{C}\rightarrow ^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta^+ \nonumber \] Nucleons are conserved, and the charges balance. The mass number, 11, does not change, and the sum of the atomic numbers of the products is 6, the same as the atomic number of the parent carbon-11 nuclide. A neutron-poor nucleus can decay by either positron emission or , in which an electron in an inner shell reacts with a proton to produce a neutron: \[^{1}_{1}\textrm{p} +\; ^{0}_{-1}\textrm{e}\rightarrow \, ^{1}_{0}\textrm n\label{Eq9} \] When a second electron moves from an outer shell to take the place of the lower-energy electron that was absorbed by the nucleus, an x-ray is emitted. The overall reaction for electron capture is thus \[\underset{\textrm{parent}}{^A_Z \textrm X}+\underset{\textrm{electron}}{^0_{-1} \textrm e}\rightarrow \underset{\textrm{daughter}}{^{A}_{Z-1} \textrm X'}+\textrm{x-ray} \nonumber \] Electron capture does not change the mass number of the nucleus because both the proton that is lost and the neutron that is formed have a mass number of 1. As with positron emission, however, the atomic number of the daughter nucleus is lower by 1 than that of the parent. Once again, the neutron-to-proton ratio has increased, moving the nucleus toward the band of stable nuclei. For example, iron-55 decays by electron capture to form manganese-55, which is often written as follows: \[^{55}_{26}\textrm{Fe}\overset{\textrm{EC}}{\rightarrow}\, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber \] The atomic numbers of the parent and daughter nuclides differ in , although the mass numbers are the same. To write a balanced nuclear equation for this reaction, we must explicitly include the captured electron in the equation: \[^{55}_{26}\textrm{Fe}+\,^{0}_{-1}\textrm{e}\rightarrow \, ^{55}_{25}\textrm{Mn}+\textrm{x-ray} \nonumber \] Both positron emission and electron capture are usually observed for nuclides with low neutron-to-proton ratios, but the decay rates for the two processes can be very different. Many nuclear decay reactions produce daughter nuclei that are in a nuclear excited state, which is similar to an atom in which an electron has been excited to a higher-energy orbital to give an electronic excited state. Just as an electron in an electronic excited state emits energy in the form of a photon when it returns to the ground state, a nucleus in an excited state releases energy in the form of a photon when it returns to the ground state. These high-energy photons are γ rays. (\(\gamma\)) can occur virtually instantaneously, as it does in the alpha decay of uranium-238 to thorium-234, where the asterisk denotes an excited state: \[^{238}_{92}\textrm{U}\rightarrow \, \underset{\textrm{excited} \\ \textrm{nuclear} \\ \textrm{state}}{^{234}_{90}\textrm{Th*}} + ^{4}_{2}\alpha\xrightarrow {\textrm{relaxation}\,}\,^{234}_{90}\textrm{Th} + \ce{^0_0\gamma} \nonumber \] If we disregard the decay event that created the excited nucleus, then \[^{234}_{88}\textrm{Th*} \rightarrow\, ^{234}_{88}\textrm{Th} + ^{0}_{0}\gamma \nonumber \] or more generally, \[^{A}_{Z}\textrm{X*} \rightarrow\, ^{A}_{Z}\textrm{X} + ^{0}_{0}\gamma \nonumber \] Gamma emission can also occur after a significant delay. For example, technetium-99 has a half-life of about 6 hours before emitting a \(γ\) ray to form technetium-99 (the is for metastable). Because γ rays are energy, their emission does not affect either the mass number or the atomic number of the daughter nuclide. Gamma-ray emission is therefore the only kind of radiation that does not necessarily involve the conversion of one element to another, although it is almost always observed in conjunction with some other nuclear decay reaction. Only very massive nuclei with high neutron-to-proton ratios can undergo , in which the nucleus breaks into two pieces that have different atomic numbers and atomic masses. This process is most important for the transactinide elements, with ≥ 104. Spontaneous fission is invariably accompanied by the release of large amounts of energy, and it is usually accompanied by the emission of several neutrons as well. An example is the spontaneous fission of \(^{254}_{98}\textrm{Cf}\), which gives a distribution of fission products; one possible set of products is shown in the following equation: \[^{254}_{98}\textrm{Cf}\rightarrow \,^{118}_{46}\textrm{Pd}+\,^{132}_{52}\textrm{Te}+4^{1}_{0}\textrm{n} \nonumber \] Once again, the number of nucleons is conserved. Thus the sum of the mass numbers of the products (118 + 132 + 4 = 254) equals the mass number of the reactant. Similarly, the sum of the atomic numbers of the products [46 + 52 + (4 × 0) = 98] is the same as the atomic number of the parent nuclide. Write a balanced nuclear equation to describe each reaction. radioactive nuclide and mode of decay balanced nuclear equation Identify the reactants and the products from the information given. Use the values of and to identify any missing components needed to balance the equation. a. We know the identities of the reactant and one of the products (a β particle). We can therefore begin by writing an equation that shows the reactant and one of the products and indicates the unknown product as \(^{A}_{Z}\textrm{X}\): \[^{35}_{16}\textrm{S}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{-1}\beta \nonumber \] Because both protons and neutrons must be conserved in a nuclear reaction, the unknown product must have a mass number of = 35 − 0 = 35 and an atomic number of = 16 − (−1) = 17. The element with = 17 is chlorine, so the balanced nuclear equation is as follows: \[^{35}_{16}\textrm{S}\rightarrow\,^{35}_{17}\textrm{Cl}+\,^{0}_{-1}\beta \nonumber \] We know the identities of both reactants: \(^{201}_{80}\textrm{Hg}\) and an inner electron, \(^{0}_{-1}\textrm{e}\). The reaction is as follows: \[^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{A}_{Z}\textrm{X} \nonumber \] Both protons and neutrons are conserved, so the mass number of the product must be = 201 + 0 = 201, and the atomic number of the product must be = 80 + (−1) = 79, which corresponds to the element gold. The balanced nuclear equation is thus \[^{201}_{80}\textrm{Hg}+\,^{0}_{-1}\textrm e\rightarrow\,^{201}_{79}\textrm{Au} \nonumber \] As in part (a), we are given the identities of the reactant and one of the products—in this case, a positron. The unbalanced nuclear equation is therefore \[^{30}_{15}\textrm{P}\rightarrow\,^{A}_{Z}\textrm{X}+\,^{0}_{+1}\beta \nonumber \] The mass number of the second product is = 30 − 0 = 30, and its atomic number is = 15 − 1 = 14, which corresponds to silicon. The balanced nuclear equation for the reaction is as follows: \[^{30}_{15}\textrm{P}\rightarrow\,^{30}_{14}\textrm{Si}+\,^{0}_{+1}\beta \nonumber \] Write a balanced nuclear equation to describe each reaction. \(^{11}_{6}\textrm{C}\rightarrow\,^{11}_{5}\textrm{B}+\,^{0}_{+1}\beta\) \(^{99}_{42}\textrm{Mo}\rightarrow\,^{99m}_{43}\textrm{Tc}+\,^{0}_{-1}\beta\) \(^{185}_{74}\textrm{W}\rightarrow\,^{181}_{72}\textrm{Hf}+\,^{4}_{2}\alpha +\,^{0}_{0}\gamma\) Predict the kind of nuclear change each unstable nuclide undergoes when it decays. nuclide type of nuclear decay Based on the neutron-to-proton ratio and the value of , predict the type of nuclear decay reaction that will produce a more stable nuclide. Predict the kind of nuclear change each unstable nuclide undergoes when it decays. beta decay positron emission or electron capture alpha decay The nuclei of all elements with atomic numbers greater than 83 are unstable. Thus all isotopes of all elements beyond bismuth in the periodic table are radioactive. Because alpha decay decreases by only 2, and positron emission or electron capture decreases by only 1, it is impossible for any nuclide with > 85 to decay to a stable daughter nuclide in a single step, except via nuclear fission. Consequently, radioactive isotopes with > 85 usually decay to a daughter nucleus that is radiaoctive, which in turn decays to a second radioactive daughter nucleus, and so forth, until a stable nucleus finally results. This series of sequential alpha- and beta-decay reactions is called a . The most common is the uranium-238 decay series, which produces lead-206 in a series of 14 sequential alpha- and beta-decay reactions ( ). Although a radioactive decay series can be written for almost any isotope with > 85, only two others occur naturally: the decay of uranium-235 to lead-207 (in 11 steps) and thorium-232 to lead-208 (in 10 steps). A fourth series, the decay of neptunium-237 to bismuth-209 in 11 steps, is known to have occurred on the primitive Earth. With a half-life of “only” 2.14 million years, all the neptunium-237 present when Earth was formed decayed long ago, and today all the neptunium on Earth is synthetic. Due to these radioactive decay series, small amounts of very unstable isotopes are found in ores that contain uranium or thorium. These rare, unstable isotopes should have decayed long ago to stable nuclei with a lower atomic number, and they would no longer be found on Earth. Because they are generated continuously by the decay of uranium or thorium, however, their amounts have reached a steady state, in which their rate of formation is equal to their rate of decay. In some cases, the abundance of the daughter isotopes can be used to date a material or identify its origin. The discovery of radioactivity in the late 19th century showed that some nuclei spontaneously transform into nuclei with a different number of protons, thereby producing a different element. When scientists realized that these naturally occurring radioactive isotopes decayed by emitting subatomic particles, they realized that—in principle—it should be possible to carry out the reverse reaction, converting a stable nucleus to another more massive nucleus by bombarding it with subatomic particles in a nuclear transmutation reaction. The first successful nuclear transmutation reaction was carried out in 1919 by Ernest Rutherford, who showed that α particles emitted by radium could react with nitrogen nuclei to form oxygen nuclei. As shown in the following equation, a proton is emitted in the process: \(^{4}_{2}\alpha + \, ^{14}_{7}\textrm{N} \rightarrow \,^{17}_{8}\textrm{O}+\,^{1}_{1}\textrm{p}\label{Eq17}\) Rutherford’s nuclear transmutation experiments led to the discovery of the neutron. He found that bombarding the nucleus of a light target element with an α particle usually converted the target nucleus to a product that had an atomic number higher by 1 and a mass number higher by 3 than the target nucleus. Such behavior is consistent with the emission of a proton after reaction with the α particle. Very light targets such as Li, Be, and B reacted differently, however, emitting a new kind of highly penetrating radiation rather than a proton. Because neither a magnetic field nor an electrical field could deflect these high-energy particles, Rutherford concluded that they were electrically neutral. Other observations suggested that the mass of the neutral particle was similar to the mass of the proton. In 1932, James Chadwick (Nobel Prize in Physics, 1935), who was a student of Rutherford’s at the time, named these neutral particles neutrons and proposed that they were fundamental building blocks of the atom. The reaction that Chadwick initially used to explain the production of neutrons was as follows: \[^{4}_{2}\alpha + \, ^{9}_{4}\textrm{Be} \rightarrow \,^{12}_{6}\textrm{C}+\,^{1}_{0}\textrm{n}\label{Eq18} \] Because α particles and atomic nuclei are both positively charged, electrostatic forces cause them to repel each other. Only α particles with very high kinetic energy can overcome this repulsion and collide with a nucleus ( ). Neutrons have no electrical charge, however, so they are not repelled by the nucleus. Hence bombardment with neutrons is a much easier way to prepare new isotopes of the lighter elements. In fact, carbon-14 is formed naturally in the atmosphere by bombarding nitrogen-14 with neutrons generated by cosmic rays: \[^{1}_{0}\textrm{n} + \, ^{14}_{7}\textrm{N} \rightarrow \,^{14}_{6}\textrm{C}+\,^{1}_{1}\textrm{p}\label{Eq19} \] In 1933, Frédéric Joliot and Iréne Joliot-Curie (daughter of Marie and Pierre Curie) prepared the first artificial radioactive isotope by bombarding aluminum-27 with α particles. For each Al that reacted, one neutron was released. Identify the product nuclide and write a balanced nuclear equation for this transmutation reaction. reactants in a nuclear transmutation reaction product nuclide and balanced nuclear equation Based on the reactants and one product, identify the other product of the reaction. Use conservation of mass and charge to determine the values of and of the product nuclide and thus its identity. Write the balanced nuclear equation for the reaction. Bombarding an element with α particles usually produces an element with an atomic number that is 2 greater than the atomic number of the target nucleus. Thus we expect that aluminum ( = 13) will be converted to phosphorus ( = 15). With one neutron released, conservation of mass requires that the mass number of the other product be 3 greater than the mass number of the target. In this case, the mass number of the target is 27, so the mass number of the product will be 30. The second product is therefore phosphorus-30, \(^{30}_{15}\textrm{P}\). The balanced nuclear equation for the reaction is as follows: \[^{27}_{13}\textrm{Al} + \, ^{4}_{2}\alpha \rightarrow \,^{30}_{15}\textrm{P}+\,^{1}_{0}\textrm{n} \nonumber \] Because all isotopes of technetium are radioactive and have short half-lives, it does not exist in nature. Technetium can, however, be prepared by nuclear transmutation reactions. For example, bombarding a molybdenum-96 target with deuterium nuclei \((^{2}_{1}\textrm{H})\) produces technetium-97. Identify the other product of the reaction and write a balanced nuclear equation for this transmutation reaction. neutron, \(^{1}_{0}\textrm{n}\) ; \(^{96}_{42}\textrm{Mo} + \, ^{2}_{1}\textrm{H} \rightarrow \,^{97}_{43}\textrm{Tc}+\,^{1}_{0}\textrm{n}\) We noted earlier in this section that very heavy nuclides, corresponding to ≥ 104, tend to decay by spontaneous fission. Nuclides with slightly lower values of , such as the isotopes of uranium ( = 92) and plutonium ( = 94), do not undergo spontaneous fission at any significant rate. Some isotopes of these elements, however, such as \(^{235}_{92}\textrm{U}\) and \(^{239}_{94}\textrm{Pu}\) undergo induced nuclear fission when they are bombarded with relatively low-energy neutrons, as shown in the following equation for uranium-235 and in : \[^{235}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{236}_{92}\textrm{U}\rightarrow \,^{141}_{56}\textrm{Ba}+\,^{92}_{36}\textrm{Kr}+3^{1}_{0}\textrm{n}\label{Eq20} \] Any isotope that can undergo a nuclear fission reaction when bombarded with neutrons is called a . During nuclear fission, the nucleus usually divides asymmetrically rather than into two equal parts, as shown in . Moreover, every fission event of a given nuclide does not give the same products; more than 50 different fission modes have been identified for uranium-235, for example. Consequently, nuclear fission of a fissile nuclide can never be described by a single equation. Instead, as shown in , a distribution of many pairs of fission products with different yields is obtained, but the mass ratio of each pair of fission products produced by a single fission event is always roughly 3:2. Uranium ( = 92) is the heaviest naturally occurring element. Consequently, all the elements with > 92, the , are artificial and have been prepared by bombarding suitable target nuclei with smaller particles. The first of the transuranium elements to be prepared was neptunium ( = 93), which was synthesized in 1940 by bombarding a U target with neutrons. As shown in , this reaction occurs in two steps. Initially, a neutron combines with a U nucleus to form U, which is unstable and undergoes beta decay to produce Np: \(^{238}_{92}\textrm{U} + \, ^{1}_{0}\textrm{n} \rightarrow \,^{239}_{92}\textrm{U}\rightarrow \,^{239}_{93}\textrm{Np}+\,^{0}_{-1}\beta\label{Eq21}\) Subsequent beta decay of Np produces the second transuranium element, plutonium ( = 94): \(^{239}_{93}\textrm{Np} \rightarrow \,^{239}_{94}\textrm{Pu}+\,^{0}_{-1}\beta\label{Eq22}\) Bombarding the target with more massive nuclei creates elements that have atomic numbers significantly greater than that of the target nucleus ( ). Such techniques have resulted in the creation of the superheavy elements 114 and 116, both of which lie in or near the “island of stability." A device called a particle accelerator is used to accelerate positively charged particles to the speeds needed to overcome the electrostatic repulsions between them and the target nuclei by using electrical and magnetic fields. Operationally, the simplest particle accelerator is the linear accelerator ( ), in which a beam of particles is injected at one end of a long evacuated tube. Rapid alternation of the polarity of the electrodes along the tube causes the particles to be alternately accelerated toward a region of opposite charge and repelled by a region with the same charge, resulting in a tremendous acceleration as the particle travels down the tube. A modern linear accelerator such as the Stanford Linear Accelerator (SLAC) at Stanford University is about 2 miles long. To achieve the same outcome in less space, a particle accelerator called a cyclotron forces the charged particles to travel in a circular path rather than a linear one. The particles are injected into the center of a ring and accelerated by rapidly alternating the polarity of two large D-shaped electrodes above and below the ring, which accelerates the particles outward along a spiral path toward the target. The length of a linear accelerator and the size of the D-shaped electrodes in a cyclotron severely limit the kinetic energy that particles can attain in these devices. These limitations can be overcome by using a synchrotron, a hybrid of the two designs. A synchrotron contains an evacuated tube similar to that of a linear accelerator, but the tube is circular and can be more than a mile in diameter. Charged particles are accelerated around the circle by a series of magnets whose polarities rapidly alternate. In , the parent nucleus is converted to a more stable daughter nucleus. Nuclei with too many neutrons decay by converting a neutron to a proton, whereas nuclei with too few neutrons decay by converting a proton to a neutron. Very heavy nuclei (with ≥ 200 and > 83) are unstable and tend to decay by emitting an . When an unstable nuclide undergoes radioactive decay, the total number of nucleons is conserved, as is the total positive charge. Six different kinds of nuclear decay reactions are known. results in the emission of an α particle, \(^4 _2 \alpha\), and produces a daughter nucleus with a mass number that is lower by 4 and an atomic number that is lower by 2 than the parent nucleus. converts a neutron to a proton and emits a high-energy electron, producing a daughter nucleus with the same mass number as the parent and an atomic number that is higher by 1. is the opposite of beta decay and converts a proton to a neutron plus a positron. Positron emission does not change the mass number of the nucleus, but the atomic number of the daughter nucleus is lower by 1 than the parent. In , an electron in an inner shell reacts with a proton to produce a neutron, with emission of an x-ray. The mass number does not change, but the atomic number of the daughter is lower by 1 than the parent. In , a daughter nucleus in a nuclear excited state undergoes a transition to a lower-energy state by emitting a γ ray. Very heavy nuclei with high neutron-to-proton ratios can undergo , in which the nucleus breaks into two pieces that can have different atomic numbers and atomic masses with the release of neutrons. Many very heavy nuclei decay via a —a succession of some combination of alpha- and beta-decay reactions. In , a target nucleus is bombarded with energetic subatomic particles to give a product nucleus that is more massive than the original. All —elements with > 92—are artificial and must be prepared by nuclear transmutation reactions. These reactions are carried out in particle accelerators such as linear accelerators, cyclotrons, and synchrotrons. \[^A_Z \textrm X\rightarrow \, ^{A-4}_{Z-2} \textrm X'+\,^4_2 \alpha \nonumber \] \[^A_Z \textrm X\rightarrow \, ^{A}_{Z+1} \textrm X'+\,^0_{-1} \beta \nonumber \] \[^A_Z \textrm X\rightarrow \, ^{A}_{Z-1} \textrm X'+\,^0_{+1} \beta \nonumber \] \[^A_Z \textrm X+\,^{0}_{-1} \textrm e\rightarrow \, ^{A}_{Z-1} \textrm X'+\textrm{x-ray} \nonumber \] \[^A_Z \textrm{X*}\rightarrow \, ^{A}_{Z} \textrm X+\,^0_{0} \gamma \nonumber \] | 31,833 | 4,186 |
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Different instructors set out widely varying requirements for chemical nomenclature. The following are probably the most commonly expected: Chemical nomenclature is far too big a topic to treat comprehensively, and it would be a useless diversion to attempt to do so in a beginning course; most chemistry students pick up chemical names and the rules governing them as they go along. But we can hardly talk about chemistry without mentioning chemical substances, all of which do have names— and often, more than one! All we will try to do here is cover what you need to know to make sense of first-year chemistry. For those of you who plan to go on in chemistry, the really fun stuff comes later! There are more than 100 million named chemical substances. Who thinks up the names for all these chemicals? Are we in danger of running out of new names? The answer to the last question is "no", for the simple reason that the vast majority of the names are not "thought up"; there are elaborate rules for assigning names to chemical substances on the basis of their structures. These are called ; they may be a bit ponderous, but they uniquely identify a given substance. The rules for these names are defined by an international body. But in order to make indexing and identification easier, every known chemical substance has its own numeric "personal ID", known as a CAS registry number. For example, caffeine is uniquely identified by the registry number 58-08-2. About 15,000 new numbers are issued every day. Many chemicals are so much a part of our life that we know them by their familiar names, just like our other friends. A given substance may have several common or ; ordinary cane sugar, for example, is more formally known as "sucrose", but asking for it at the dinner table by that name will likely be a conversation-stopper, and I won't even venture to predict the outcome if you try using its systematic name in the same context: "please pass the α-D-glucopyranosyl-(1,2)- β-D-fructofuranoside!" But "sucrose" would be quite appropriate if you need to distinguish this particular sugar from the hundreds of other named sugars. The only place you would come across a systematic name like the rather unwieldy one mentioned here is when referring (in print or in a computer data base) to a sugar that has no common name. Chemical substances have been a part the fabric of civilization and culture for thousands of years, and present-day chemistry retains a lot of this ancient baggage in the form of terms whose hidden cultural and historic connections add color and interest to the subject. Many common chemical names have reached us only after remarkably long journeys through time and place, as the following two examples illustrate. Most people can associate the name (\(NH_3\)) with a gas having a pungent odor; the systematic name "nitrogen trihydride" (which is rarely used) will tell you its formula. What it will not tell you is that smoke from burning camel dung (the staple fuel of North Africa) condenses on cool surfaces to form a crystalline deposit. The ancient Romans first noticed this on the walls and ceiling of the temple that the Egyptians had built to the Sun-god Amun in Thebes, and they named the material sal ammoniac, meaning "salt of Amun". In 1774, Joseph Priestly (the discoverer of oxygen) found that heating sal ammoniac produced a gas with a pungent odor, which a T. Bergman named "ammonia" eight years later. Alcohol entered the English language in the 17th Century with the meaning of a "sublimated" substance, then became the "pure spirit" of anything, and only became associated with "spirit of wine" in 1753. Finally, in 1852, it become a part of chemical nomenclature that denoted a common class of organic compound. But it's still common practice to refer to the specific substance CH CH OH as "alcohol" rather then its systematic name . The general practice among chemists is to use the more common chemical names whenever it is practical to do so, especially in spoken or informal written communication. For many of the very simplest compounds (including most of those you will encounter in a first-year course), the systematic and common names are the same, but where there is a difference and if the context permits it, the common name is usually preferred. Many of the "common" names we refer to in this lesson are known and used mainly by the scientific community. Chemical substances that are employed in the home, the arts, or in industry have acquired traditional or "popular" names that are still in wide use. Many, like sal ammoniac mentioned above, have fascinating stories to tell. : Minerals are solid materials that occur in the earth which are classified and named according to their compositions (which often vary over a continuous range) and the arrangement of the atoms in their crystal lattices. There are about 4000 named minerals. Many are named after places, people, or properties, and most frequently end with -ite. Chemistry is a major industry, so it is not surprising that many substances are sold under trademarked names. This is especially common in the pharmaceutical industry, which uses computers to churn out names that they hope will distinguish a new product from those of its competitors. Perhaps the most famous of these is Aspirin, whose name was coined by the German company Bayer in 1899. This trade name was seized by the U.S. government following World War I, and is no longer a protected trade mark in that country. Naming of chemical substances begins with the names of the . The discoverer of an element has traditionally had the right to name it, and one can find some interesting human and cultural history in these names, many of which refer to the element's properties or to geographic locations. Only some of the more recently-discovered (and artificially produced) elements are named after people. Some elements were not really "discovered", but have been known since ancient times; many of these have symbols that are derived from the Latin names of the elements. \(\Page {2}\) There is a lot of history and tradition in many of these names. For example, the Latin name for mercury, , means "water silver", or quicksilver. The appellation "quack", as applied to an incompetent physician, is a corruption of the Flemish word for quicksilver, and derives from the use of mercury compounds in 17th century medicine. The name "mercury" is of alchemical origin and is of course derived from the name of the Greek god after whom the planet is named; the enigmatic properties of the element, at the same time metallic, fluid, and vaporizable, suggest the same messenger with the winged feet who circles through the heavens close to the sun. The system used for naming chemical substances depends on the nature of the molecular units making up the compound. These are usually either ions or molecules; different rules apply to each. In this section, we discuss the simplest binary (two-atom) molecules. It is often necessary to distinguish between compounds in which the same elements are present in different proportions; and are familiar to everyone. Chemists, perhaps hoping it will legitimize them as scholars, employ Greek (of sometimes Latin) prefixes to designate numbers within names; you will encounter these frequently, and you should know them: You will occasionally see names such as hydrogen and chlorine used to distinguish the common forms of these elements (H , Cl ) from the atoms that have the same name when it is required for clarity. Examples: It will be apparent from these examples that chemists are in the habit of taking a few liberties in applying the strict numeric prefixes to the more commonly known substances. These two-element compounds are usually quite easy to name because most of them follow the systematic rule of adding the suffix to the root name of the second element, which is normally the more "negative" one. Several such examples are shown above. But as noted above, there are some important exceptions in which common or names take precedence over systematic names: An is an electrically charged atom or molecule— that is, one in which the number of electrons differs from the number of nuclear protons. Many simple compounds can be regarded, at least in a formal way, as being made up of a pair of ions having opposite charge signs. The positive ions, also known as , are mostly those of metallic elements which simply take the name of the element itself. The only important non-metallic cations you need to know about are (Later on, when you study acids and bases, you will learn that the first two represent the same chemical species.) Some of the metallic ions are , meaning that they can exhibit more than one electric charge. For these there are systematic names that use Roman numerals, and the much older and less cumbersome common names that mostly employ the Latin names of the elements, using the endings - and - to denote the lower and higher charges, respectively (Table \(\Page {4}\)). (In cases where more than two charge values are possible, the systematic names are used.) The non-metallic elements generally form ions ( ). The names of the monatomic anions all end with the suffix: There are a number of important polyatomic anions which, for naming purposes, can be divided into several categories. A few follow the pattern for the monatomic anions: The most common oxygen-containing anions ( ) have names ending in , but if a variant containing a small number of oxygen atoms exists, it takes the suffix . The above ions (with the exception of nitrate) can also combine with H+ to produce "acid" forms having smaller negative charges. For rather obscure historic reasons, some of them have common names that begin with which, although officially discouraged, are still in wide use: Chlorine, and to a smaller extent bromine and iodine, form a more extensive series of oxyanions that requires a somewhat more intricate naming convention: These compounds are formally derived from positive ions ( ) and negative ions ( ) in a ratio that gives an electrically neutral unit. Salts, of which ordinary "salt" (sodium chloride) is the most common example, are all solids under ordinary conditions. A small number of these (such as NaCl) do retain their component ions and are properly called "ionic solids". In many cases, however, the ions lose their electrically charged character and form largely-non-ionic solids such as CuCl The term "ion-derived solids" encompasses both of these classes of compounds. Most of the cations and anions described above can combine to form solid compounds that are usually known as . The one overriding requirement is that the resulting compound must be electrically neutral: thus the ions Ca and Br combine only in a 1:2 ratio to form calcium bromide, CaBr . Because no other simplest formula is possible, there is no need to name it "calcium dibromide". Since some metallic elements form cations having different positive charges, the names of ionic compounds derived from these elements must contain some indication of the cation charge. The older method uses the suffixes and to denote the lower and higher charges, respectively. In the cases of iron and copper, the Latin names of the elements are used: , . This system is still widely used, although it has been officially supplanted by the more precise, if slightly cumbersome Stock system in which one indicates the cationic charge (actually, the oxidation number) by means of Roman numerals following the symbol for the cation. In both systems, the name of the anion ends in - . Most acids can be regarded as a combination of a hydrogen ion H with an anion; the name of the anion is reflected in the name of the acid. Notice, in the case of the oxyacids, how the anion suffixes and become and , respectively, in the acid name. Since organic (carbon) compounds constitute the vast majority of all known chemical substances, organic nomenclature is a huge subject in itself. We present here only the very basic part of it that you need to know in first-year chemistry— much more awaits those of you who are to experience the pleasures of an organic chemistry course later on. The simplest organic compounds are built of straight chains of carbon atoms which are named by means of prefixes that denote the number of carbons in the chain. Using the convention C to denote a straight chain of atoms (don't even about branched chains!), the prefixes for chain lengths from 1 through 10 are given here: As you can see, chains from C onward use Greek number prefixes, so you don't have a lot new to learn here. The simplest of these compounds are hydrocarbons having the general formula C H . They are known generically as , and their names all combine the appropriate numerical prefix with the ending : All carbon atoms must have four bonds attached to them; notice the common convention of not showing hydrogen atoms explicitly. | 13,038 | 4,187 |
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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. | 8,334 | 4,188 |
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Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic. What we call "classical" physics is based on our experience of what we perceive as the "real world". Even without knowing the details of Newton's laws of motion that describe the behavior of macroscopic bodies, we have all developed an intuitive understanding of this behavior; it is a part of everyone's personal view of the world. By extension, we tend to view atoms and molecules in much the same way, that is, simply as miniature versions of the macroscopic objects we know from everyday life. It turns out, however, that our everyday view of the macroscopic world is only a first approximation of the reality that becomes apparent at the atomic level. Many of those who first encounter this microscopic world of quantum weirdness find it so foreign to prior experience that their first reaction is to dismiss it as pure fantasy. The fact is, however, that it is not only for real, but serves as the key that unlocks even some of the simplest aspects of modern Chemistry. Our goal in this lesson is to introduce you to this new reality, and to provide you with a conceptual understanding of it that will make Chemistry a more meaningful part of your own personal world. Near the end of the nineteenth century, the enormous success of the recently developed kinetic molecular theory of gases had dispelled most doubts about the atomic nature of matter; the material world was seen to consist of particles that had distinct masses and sizes, and which moved in trajectories just as definite as those of billiard balls. In the 1890s, however, certain phenomena began to be noticed that seemed to be inconsistent with this dichotomy of particles and waves. This prompted further questions and further experiments which led eventually to the realization that classical physics, while it appears to be "the truth", is by no means the whole truth. In particular, it cannot accurately describe the behavior of objects that are extremely small or fast-moving. began as an entirely empirical, experimental science, dealing with the classification and properties of substances and with their transformations in chemical reactions. As this large body of facts developed into a science (one of whose functions is always to explain and correlate known facts and to predict new ones), it has become necessary to focus increasingly on the nature and behavior of individual atoms and of their own constituent parts, especially the . Owing to their extremely small masses, electrons behave as which do not obey the rules of classical physics. The purpose of this introductory unit is to summarize the major ideas of quantum theory that will be needed to treat atomic and molecular structure later on in the course. Quantum theory can be presented simply as a set of assumptions which are developed through mathematical treatment. This is in fact the best route to take if one is to use quantum mechanics as a working tool. More than this, however, quantum theory brings with it a set of that have far-reaching philosophical implications and which should be a part of the intellectual equipment of anyone who claims to have a general education in the sciences. A major objective of this chapter will be to introduce you to "the quantum way of thinking" and to show how this led to a profound break with the past, and a shift in our way of viewing the world that has no parallel in Western intellectual history. The development of our ideas about light and radiation was not quite as direct. In the 17th century, heat was regarded as a substance called whose invisible atoms could flow from one object to another, thus explaining thermal conduction. This view of heat as a material fluid seemed to be confirmed by the observation that heat can pass through a vacuum, a phenomenon that we now call . Isaac Newton, whose experiments with a prism in 1672 led to his famous textbook " ", noted that light seemed to react with green plants to produce growth, and must therefore be a "substance" having atoms of its own. By 1800, the was generally accepted. And yet there were questions. Count Rumford's observation that the drill bits employed in boring cannons produced more frictional heat when they were worn and dull led to the overthrow of the caloric theory. In 1812, Christiaan Huygens showed how a number of optical effects could be explained if light had a wavelike nature, and this led Fresnel to develop an elaborate wave theory of light. By 1818 the question of "particle or wave" had become so confused that the French Academy held a great debate intended to settle the matter once for all. The mathematician Poisson pointed out that Fresnel's wave theory had a ridiculous consequence: the shadow cast by a circular disk should have a bright spot of light at its center, where waves arriving in phase would reinforce each other. Fresnel performed the experiment and was entirely vindicated: if the light source is sufficiently point-like (an extended source such as the sun or an ordinary lamp will not work), this diffraction effect is indeed observed. By this time it was known that radiant heat and "cold" could be focused and transmitted by mirrors, and in 1800 William Herschel discovered that radiant heat could be sensed in the dark region just beyond the red light refracted by a prism. Light and radiant heat, which had formerly been considered separate, were now recognized as one, although the question of precisely what was doing the "waving" was something of an embarrassment. By 1890, physicists thought they had tidied up the world into the two realms of particulate matter and of wavelike radiant energy, which by then had been shown by James Clerk Maxwell to be forms of electromagnetic energy. No sooner had all this been accomplished, than the cracks began to appear; these quickly widened into chasms, and within twenty years the entire foundations of classical physics had disintegrated; it would not be until the 1920's that anyone with a serious interest in the nature of the microscopic world would find a steady place to stand. The atom was the first to go. It had been known for some time that when a high voltage is applied to two separated pieces of metal in an evacuated tube, "cathode rays" pass between them. These rays could be detected by their ability to cause certain materials to give off light, or , and were believed to be another form of electromagnetic radiation. Then, in the 1890s, J.J. Thompson and Jean Perrin showed that cathode rays are composed of particles having a measurable mass (less than 1/1000 of that of the hydrogen atom), they carry a fixed negative electric charge, and that they come from atoms. This last conclusion went so strongly against the prevailing view of atoms as the ultimate, un-cuttable stuff of the world that Thompson only reluctantly accepted it, and having done so, quickly became the object of widespread ridicule. But worse was soon to come; not only were atoms shown to be the smallest units of matter, but the work of the Curies established that atoms are not even immutable; atoms of high atomic weight such as uranium and radium give off penetrating beams of radiation and in the process change into other elements, disintegrating through a series of stages until they turn into lead. Among the various kinds of radiation that accompany radioactive disintegration are the very same cathode rays that had been produced artificially by Thompson, and which we now know as . The wave theory of radiation was also running into difficulties. Any object at a temperature above absolute zero gives off radiant energy; if the object is moderately warm, we sense this as . As the temperature is raised, a larger proportion of shorter-wavelength radiation is given off, so that at sufficiently high temperatures the object becomes luminous. The origin of this radiation was thought to lie in the thermally-induced oscillations of the atoms within the object, and on this basis the mathematical physicist James Rayleigh had worked out a formula that related the wavelengths given off to the temperature. Unfortunately, this formula did not work; it predicted that most of the radiation given off at any temperature would be of very short wavelength, which would place it in the ultraviolet region of the spectrum. What was most disconcerting is that no one could say Rayleigh's formula did not work, based as it was on sound classical physics; this puzzle became known as the "scandal of the ultraviolet". In 1899 the German physicist Max Planck pointed out that one simple change in Rayleigh's argument would produce a formula that accurately describes the radiation spectrum of a perfect radiator, which is known as a "black body". Rayleigh assumed that such an object would absorb and emit amounts of radiation in amounts of any magnitude, ranging from minute to very large. This is just what one would expect on the basis of the similar theory of mechanical physics which had long been well established. Planck's change, for which he could offer no physical justification other than that it works, was to discard this assumption, and to require that the absorption or emission of radiation occur only in discrete chunks, or . Max Planck had unlocked the door that would lead to the resurrection of the corpuscular theory of radiation. Only a few years later, Albert Einstein would kick the door open and walk through. By 1900 it was known that a beam of light, falling on a piece of metal, could cause electrons to be ejected from its surface. Evidently the energy associated with the light overcomes the binding energy of the electron in the metal; any energy the light supplies in excess of this binding energy appears as kinetic energy of the emitted electron. What seemed peculiar, however, was that the energy of the ejected electrons did not depend on the intensity of the light as classical physics would predict. Instead, the energy of the photoelectrons (as they are called) varies with the color, or of the light; the higher the frequency (the shorter the wavelength), the greater the energy of the ejected electrons. In 1905, Albert Einstein, then an unknown clerk in the Swiss Patent Office, published a remarkable paper in which he showed that if light were regarded as a collection of individual particles, a number of phenomena, including the photoelectric effect, could be explained. Each particle of light, which we now know as a , has associated with it a distinct energy that is proportional to the frequency of the light, and which corresponds to Planck's energy quanta. The energy of the photon is given by \[ E = h
u = \dfrac{hc}{\lambda}\] in which is Planck's constant, 6.63×10 J-s, ν (Greek ) is the frequency, λ ( ) is the wavelength, and is the velocity of light, 3.00×10 m s . The photoelectric effect is only seen if the photon energy exceeds the binding energy of the electron in the metal; it is clear from the above equation that as the wavelength increases, decreases, and eventually no electrons will be released. Einstein had in effect revived the corpuscular theory of light, although it would not be until about 1915 that sufficient experimental evidence would be at hand to convince most of the scientific world— but not all of it: Max Planck, whose work had led directly to the revival of the particle theory of light, remained one of the strongest doubters. The 1905 volume of is now an expensive collector's item, for in that year Einstein published three major papers, any one of which would have guaranteed him his place in posterity. The first, on the photoelectric effect, eventually won him the Nobel Prize. The second paper, on Brownian motion, amounted to the first direct confirmation of the atomic theory of matter. The third paper, his most famous, "On the electrodynamics of moving bodies", set forth the special theory of relativity. The appearance of his general theory of relativity in 1919 would finally make Einstein into a reluctant public celebrity and scientific superstar. This theory explained gravity as a consequence of the curvature of space-time. The concept of energy was slow to develop in science, partly because it was not adequately differentiated from the related quantities of force and motion. It was generally agreed that some agent of motion and change must exist; Descartes suggested, for example, that God, when creating the world, had filled it with "vortices" whose motions never ceased, but which could be transferred to other objects and thus give them motion. Gradually the concepts of and developed; these later became kinetic and potential energy. Later on, the cannon-boring experiments of Benjamin Thompson (Count Rumford) revealed the connections between heat and work. Finally, the invention of the steam engine forced the birth of the science of thermodynamics, whose founding law was that a quantity known as can be transferred from one object to another through the processes of heat and work, but that the energy itself is strictly conserved. If Einstein's first 1905 paper put him on the scientific map, the third one made him a scientific celebrity. In effect, Einstein merely asked a simple question about Faraday's law of electromagnetic induction, which says that a moving electric charge (such as is produced by an electric current flowing in a conductor) will create a magnetic field. Similarly, a moving magnetic field will induce an electric current. In either case, something has to be moving. Why, Einstein asked, does this motion have to be relative to that of the room in which the experiment is performed— that is, relative to the Earth? A stationary charge creates no field, but we know that there is really no such thing as a stationary charge, since the Earth itself is in motion; what, then, do motion and velocity ultimately relate to? The answer, Einstein suggested, is that the only constant and unchanging velocity in the universe is that of light. This being so, the beam emitted by the headlight of a moving vehicle, for example, can travel no faster than the light coming from a stationary one. This in turn suggested (through the Lorentz transformation - we are leaving out a few steps here!) that mass, as well as velocity (and thus also, time) are relative in that they depend entirely on the motion of the observer. Two observers, moving at different velocities relative to each other, will report different masses for the same object, and will age at different rates. Further, the faster an object moves with respect to an observer, the greater is its mass, and the harder it becomes to accelerate it to a still greater velocity. As the velocity of an object approaches the speed of light, its mass approaches infinity, making it impossible for an object to move as fast as light. According to Einstein, the speed of light is really the speed in the universe. If you are sitting still, you are moving through the dimension at the speed of light. If you are flying in an airplane, your motion along the three cartesian dimensions subtracts from that along the fourth (time) coordinate, with the result that time, for you, passes more slowly. Relativity comes into chemistry in two rather indirect ways: it is responsible for the magnetic moment ("spin") of the electron, and in high-atomic weight atoms in which the electrons have especially high effective velocities, their greater [relativistic] masses cause them to be bound more tightly to the nucleus— accounting, among other things, for the color of gold, and for the unusual physical and chemical properties of mercury. Where does the additional mass of a moving body come from? Simply from the kinetic energy of the object; this equivalence of mass and energy, expressed by the famous relation , is the most well known consequence of special relativity. The reason that photons alone can travel at the velocity of light is that these particles possess zero rest mass to start with. You can think of ordinary matter as "congealed energy", trapped by its possession of rest mass, whereas light is energy that has been liberated of its mass. | 16,365 | 4,189 |
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Given the tremendously important role of proteins in the functioning of any organism, the question arises of how cells synthesize appropriate polypeptides. Although more than 11400 enzymes are currently known, these constitute a very small fraction of the possible combinations of the 20 amino acids in chains of 100 or more. Clearly a great many of these combinations are worthless, in the sense that they are not adapted to any biological function. It would be pointless for any organism to construct them. Indeed, if they did occur, it would be useful for a living system to hydrolyze them to their individual amino acids and use these to build up proteins which did carry out necessary functions. Therefore it is reasonable to expect a living cell to contain some kind of “blueprint” which specifies the structures of those proteins which are essential for the cell’s normal functioning. Without such a guide an incredible amount of effort would be wasted in synthesizing unusable polypeptides just to get small amounts of those that worked. The blueprint just described is contained in the molecular architecture of deoxyribonucleic acid (DNA). The structure of DNA also provides an obvious mechanism by which the information necessary to specify protein structure can be reproduced and passed from a parent cell to its progeny. The complicated task of building protein structures from the DNA blueprint involves several types of ribonucleic acids (RNA’s) whose structures are closely related to that of DNA. Hence the storage, reproduction, and application of information about protein structure depends on . | 1,628 | 4,190 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures |
Over three hundred Topics in Thermodynamics (which can addressed individually) describe the thermodynamic properties of aqueous solutions and aqueous mixtures. The Gibbs energies of these systems are discussed leading through successive differential operations to enthalpies, volumes, heat capacities, compressibilities (isobaric and isentropic) and expansibilities. These properties are linked quantitatively to corresponding partial molar properties including chemical potentials and generally partial molar volumes, partial molar enthalpies, partial molar heat capacities, partial molar expansibilities and compressibilities. Key equations link experimentally determined variables (e.g. densities) and partial molar properties (e.g. partial molar volumes) of the components in an aqueous solution/mixture. Extensive references are given to published papers describing application of the equations.
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https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/16%3A_Appendix/16.11%3A__Acid_Dissociation_Constants |
The following table provides p and values for selected weak acids. All values are from Martell, A. E.; Smith, R. M. , Vols. 1–4. Plenum Press: New York, 1976. Unless otherwise stated, values are for 25 C and for zero ionic strength. Those values in brackets are considered less reliable. Weak acids are arranged alphabetically by the names of the neutral compounds from which they are derived. In some cases—such as acetic acid—the compound is the weak acid. In other cases—such as for the ammonium ion—the neutral compound is the conjugate base. Chemical formulas or structural formulas are shown for the fully protonated weak acid. Successive acid dissociation constants are provided for polyprotic weak acids; where there is ambiguity, the specific acidic proton is identified. To find the value for a conjugate weak base, recall that \[K_\text{a} \times K_\text{b} = K_\text{w} \nonumber\] for a conjugate weak acid, HA, and its conjugate weak base, A . 4.42 5.42 \(3.8 \times 10^{-5}\) \(3.8 \times 10^{-6}\) 2.348 (\(\ce{COOH}\)) 9.867 (\(\ce{NH3}\)) \(4.49 \times 10^{-3}\) \(1.36 \times 10^{-10}\) 2.08 (\(\ce{COOH}\)) 4.96 (\(\ce{NH3}\)) \(8.3 \times 10^{-3}\) \(1.1 \times 10^{-5}\) 4.78 (\(\ce{NH3}\)) 9.97 (OH) \(1.7 \times 10^{-5}\) \(1.05 \times 10^{-10}\) 1.823 (COOH) 8.991 (\(\ce{NH3}\)) [12.48] (\(\ce{NH2}\)) \(1.50 \times 10^{-2}\) \(1.02 \times 10^{-9}\) [\(3.3 \times 10^{-13}\)] 2.24 6.96 11.50 \(5.8 \times 10^{-3}\) \(1.1 \times 10^{-7}\) \(3.2 \times 10^{-12}\) 2.14 (COOH) 8.72 (\(\ce{NH3}\)) \(7.2 \times 10^{-3}\) \(1.9 \times 10^{-9}\) 1.990 (\(\alpha\)-COOH) 3.900 (\(\beta\)-COOH) 10.002 (\(\ce{NH3}\)) \(1.02 \times 10^{-2}\) \(1.26 \times 10^{-4}\) \(9.95 \times 10^{-11}\) 9.236 [12.74] [13.80] \(5.81 \times 10^{-10}\) [\(1.82 \times 10^{-13}\)] [\(1.58 \times 10^{-14}\)] 6.352 10.329 \(4.45 \times 10^{-7}\) \(4.69 \times 10^{-11}\) 9.40 12.8 \(4.0 \times 10^{-10}\) \(1.6 \times 10^{-13}\) –0.2 6.51 1.6 \(3.1 \times 10^{-7}\) 3.128 (COOH) 4.761 (COOH) 6.396 (COOH) \(7.45 \times 10^{-4}\) \(1.73 \times 10^{-5}\) \(4.02 \times 10^{-7}\) [1.71] (COOH) 8.36 (SH) 10.77 (\(\ce{NH3}\)) [\(1.9 \times 10^{-2}\)] \(4.4 \times 10^{-9}\) \(1.7 \times 10^{-11}\) 10.66 12.0 \(2.2 \times 10^{-11}\) \(1. \times 10^{-12}\) 6.848 9.928 \(1.42 \times 10^{-7}\) \(1.18 \times 10^{-10}\) ethylenediaminetetracetic acid (EDTA) (\(\mu = 0.1 \text{ M}\)) 0.0 (COOH) 1.5 (COOH) 2.0 (COOH) 2.66 (COOH) 6.16 (NH) 10.24 (NH) 1.0 \(3.2 \times 10^{-2}\) \(1.0 \times 10^{-2}\) \(2.2 \times 10^{-3}\) \(6.9 \times 10^{-7}\) \(5.8 \times 10^{-11}\) 3.053 4.494 \(8.85 \times 10^{-4}\) \(3.21 \times 10^{-5}\) 2.33 (\(\alpha\)-COOH) 4.42 (\(\lambda\)-COOH) 9.95 (\(\ce{NH3}\)) \(5.9 \times 10^{-3}\) \(3.8\times 10^{-5}\) \(1.12 \times 10^{-10}\) 2.17 (COOH) 9.01 (\(\ce{NH3}\)) \(6.8 \times 10^{-3}\) \(9.8 \times 10^{-10}\) 2.350 (COOH) 9.778 (\(\ce{NH3}\)) \(4.47 \times 10^{-3}\) \(1.67 \times 10^{-10}\) 3.881 (COOH) 1.7 (COOH) 6.02 (NH) 9.08 (\(\ce{NH3}\)) \(2. \times 10^{-2}\) \(9.5 \times 10^{-7}\) \(8.3 \times 10^{-10}\) 7.02 13.9 \(9.5 \times 10^{-8}\) \(1.3 \times 10^{-14}\) 4.9 (NH) 9.81 (OH) \(1.2 \times 10^{-5}\) \(1.6 \times 10^{-10}\) 2.319 (COOH) 9.754 (\(\ce{NH3}\)) \(4.8 \times 10^{-3}\) \(1.76 \times 10^{-10}\) 2.329 (COOH) 9.747 (\(\ce{NH3}\)) \(4.69 \times 10^{-3}\) \(1.79 \times 10^{-10}\) 2.04 (COOH) 9.08 (\(\alpha \text{-} \ce{NH3}\)) 10.69 (\(\epsilon \text{-} \ce{NH3}\)) \(9.1 \times 10^{-3}\) \(8.3 \times 10^{-10}\) \(2.0 \times 10^{-11}\) 1.910 6.332 \(1.23 \times 10^{-2}\) \(4.66 \times 10^{-7}\) 3.459 (COOH) 5.097 (COOH) \(3.48 \times 10^{-4}\) \(8.00 \times 10^{-6}\) 2.847 5.696 \(1.42 \times 10^{-3}\) \(2.01 \times 10^{-6}\) 2.20 (COOH) 9.05 (\(\ce{NH3}\)) \(6.3 \times 10^{-3}\) \(8.9 \times 10^{-10}\) nitrilotriacetic acid (\(T = 20 \text{°C}), pK_\text{a1: } \mu = 0.1 \text{ M}\)) 1.1 (COOH) 1.650 (COOH) 2.940 (COOH) 10.334 (\(\ce{NH3}\)) \(8. \times 10^{-2}\) \(2.24 \times 10^{-2}\) \(1.15 \times 10^{-3}\) \(4.63 \times 10^{-11}\) 1.252 4.266 \(5.60 \times 10^{-2}\) \(5.42 \times 10^{-5}\) 2.20 (COOH) 9.31 (\(\ce{NH3}\)) \(6.3 \times 10^{-3}\) \(4.9 \times 10^{-10}\) 2.148 7.199 12.35 \(7.11 \times 10^{-3}\) \(6.32 \times 10^{-8}\) \(4.5 \times 10^{-13}\) 2.950 5.408 \(1.12 \times 10^{-3}\) \(3.91 \times 10^{-6}\) 1.952 (COOH) 10.650 (NH) \(1.12 \times 10^{-2}\) \(2.29 \times 10^{-11}\) 4.874 9.30 11.06 \(5.0 \times 10^{-10}\) \(8.7 \times 10^{-12}\) 2.97 (COOH) 13.74 (OH) \(1.1 \times 10^{-3}\) \(1.8 \times 10^{-14}\) 2.187 (COOH) 9.209 (\(\ce{NH3}\)) \(6.50 \times 10^{-3}\) \(6.18 \times 10^{-10}\) 4.207 5.636 \(6.21 \times 10^{-5}\) \(2.31 \times 10^{-6}\) strong 1.99 — \(1.0 \times 10^{-2}\) 1.91 7.18 \(1.2 \times 10^{-2}\) \(6.6 \times 10^{-8}\) 3.036 (COOH) 4.366 (COOH) \(9.20 \times 10^{-4}\) \(4.31 \times 10^{-5}\) 2.088 (COOH) 9.100 (\(\ce{NH3}\)) \(8.17 \times 10^{-3}\) \(7.94 \times 10^{-10}\) 0.6 1.6 \(3. \times 10^{-1}\) \(3. \times 10^{-2}\) 2.35 (COOH) 9.33 (\(\ce{NH3}\)) \(4.5 \times 10^{-3}\) \(4.7 \times 10^{-10}\) 2.17 (COOH) 9.19 (\(\ce{NH3}\)) 10.47 (OH) \(6.8 \times 10^{-3}\) \(6.5 \times 10^{-10}\) \(3.4 \times 10^{-11}\) 2.286 (COOH) 9.718 (\(\ce{NH3}\)) \(5.18 \times 10^{-3}\) \(1.91 \times 10^{-10}\) | 5,243 | 4,194 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.06%3A_Combined_Gas_Law |
The modern refrigerator takes advantage of the gas laws to remove heat from a system. Compressed gas in the coils is allowed to expand. This expansion lowers the temperature of the gas and transfers heat energy from the material in the refrigerator to the gas. As the gas is pumped through the coils, the pressure on the gas compresses it and raises the gas temperature. This heat is then dissipated through the coils into the outside air. As the compressed gas is pumped through the system again, the process repeats itself. To this point, we have examined the relationships between any two of the variables of \(P\), \(V\), and \(T\), while the third variable is held constant. However, situations do arise where all three variables change. The expresses the relationship between the pressure, volume, and absolute temperature of a fixed amount of gas. For a combined gas law problem, only the amount of gas is held constant. \[\frac{P \times V}{T} = k \: \: \: \text{and} \: \: \: \frac{P_1 \times V_1}{T_1} = \frac{P_2 \times V_2}{T_2}\nonumber \] \(2.00 \: \text{L}\) of a gas at \(35^\text{o} \text{C}\) and \(0.833 \: \text{atm}\) is brought to standard temperature and pressure (STP). What will be the new gas volume? Use the combined gas law to solve for the unknown volume \(\left( V_2 \right)\). is \(273 \: \text{K}\) and \(1 \: \text{atm}\). The temperatures have been converted to Kelvin. . First, rearrange the equation algebraically to solve for \(V_2\). \[V_2 = \frac{P_1 \times V_1 \times T_2}{P_2 \times T_1}\nonumber \] Now substitute the known quantities into the equation and solve. \[V_2 = \frac{0.833 \: \text{atm} \times 2.00 \: \text{L} \times 273 \: \text{K}}{1.00 \: \text{atm} \times 308 \: \text{K}} = 1.48 \: \text{L}\nonumber \] Both the increase in pressure and the decrease in temperature cause the volume of the gas sample to decrease. Since both changes are relatively small, the volume does not decrease dramatically. It may seem challenging to remember all the different gas laws introduced so far. Fortunately, Boyle's, Charles's, and Gay-Lussac's laws can all be easily derived from the combined gas law. For example, consider a situation where a change occurs in the volume and pressure of a gas while the temperature is being held constant. In that case, it can be said that \(T_1 = T_2\). Look at the combined gas law and cancel the \(T\) variable out from both sides of the equation. What is left over is Boyle's Law: \(P_1 \times V_1 = P_2 \times V_2\). Likewise, if the pressure is constant, then \(P_1 = P_2\) and cancelling \(P\) out of the equation leaves Charles's Law. If the volume is constant, then \(V_1 = V_2\) and cancelling \(V\) out of the equation leaves Gay-Lussac's Law. \(\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}\) | 2,800 | 4,195 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Atmospheric_Chemistry/Carbon_Oxides |
Carbon forms two important gases with oxygen: carbon monoxide, CO, and carbon dioxide, CO . Carbon oxides are important components of the atmosphere, and they are parts of the carbon cycle. Carbon dioxide is naturally produced by respiration and metabolism, and consumed by plants in their photosynthesis. Since the industrial revolution, greater amount of carbon dioxide has been generated for over a hundred years due to increased industrial activities. Today, information on carbon oxides is important. Issues related to carbon oxides have no boundaries. The Carbon Dioxide Information Analysis Center (CADIAC) provides global datasets on carbon dioxide and other atmosphere gases and climate. These datasets are available to international researchers, policymakers, managers, and educators to help evaluate complex environmental issues associated with potential climate change. Carbon monoxide is also a national and global concern. The Consumer Product Safety Commission (CPSC) considers CO a senseless killer, and it provides information on CO poisoning and detection. The formation of carbon oxides is due to electronic configurations of carbon and oxygen. They have 4 and 6 valence electrons respectively. Using these valence electrons, we can give the Lewis dot structure for CO and three resonance structures for CO2 as follows: These formulas suggest very strong bonding between carbon and oxygen in these gaseous molecules: triple bond in C O, and double bonds in O=C=O. However, a formula containing a triple bond contribute to the resonance structure. The chemical bonding is more of an interpretation of the molecules in view of their properties. Using results from quantum mechanical approach, we may start by reviewing the electronic configurations of carbon and oxygen: C: 1s 2s 2p
O: 1s 2s 2p Thus, the carbon has 4 valence electrons and oxygen has 6 valence electrons. The s and p atomic orbitals are available for chemical bonding. The valence bond approach suggests that p orbitals of carbon and oxygen are used in these molecules. In CO, only one such atomic orbital from each atom of C and O are employed to form a sigma, , bond, and overlapping of two p orbitals leads to the formation of the two pi, , bond. Thus, the bond order is 3 between C and O in C O. A \(CO\) molecule has the same number of electrons as \(N_2\), and these molecules are said to be . The N molecule is also represented by N N. The molecular orbital (MO) approach for CO is describe in the lecture, and the MO energy level diagram has been given there. The plots of contours of equal electron densities has also been shown in earlier lectures, and the diagram for CO molecular orbitals is shown below: The valence bond approach for CO bonding is also very interesting. The two sp hybrid orbitals of central carbon overlaps with one p orbital from each of the oxygen atoms to from the two C-O bonds in O-C-O. The two remaining p orbitals of carbon overlap with a p orbital each of the two oxygen atoms forming two bonds, leading to the formation of O=C=O. Here is a challenge: find a suitable diagram for either valence bond approach or for the MO approach for carbon dioxide in the web. Molecules of \(CO\) and \(NN\) have two \(\pi\) bonds. Since the two atoms in \(CO\) are different, this made \(CO\) much more reactive than nitrogen. Indeed, \(CO\) forms many carbonyls with metal atoms or ions. For example, you have encountered some of the following carbonyls on the page of Heterogeneous catalysts The study of metal carbonyls started with the discover of ferrocene Fe(C H ) and now hundreds if not thousands of metal carbonyls have been synthesized.. A bond is formed between the carbon of CO and the metal atom. Such a bonding is very sensible if we consider the sp hybrid atomic orbitals of carbon being used in this case. Since there are two electrons in this orbitals of CO, the metal atom gains at least some fraction of electron due to the formation of this bond. The empty antibonding \(pi\) orbital \(\pi^*\) of \(CO\) has the right symmetry and orientation to receive the back-donated electron from the filled d-orbitals of the metal atom. The back donation reinforce the sigma bond, and vice versa. This type of bonding has been called the by Cotton and Wilkinson in their . A diagram showing this type of bonding scheme is shown on page 159 in by Swaddle. Carbon oxides are useful commodities. A gas containing CO and hydrogen is called synthetic gas, because it can be converted to methanol using a catalyst. During the past few decades, many metal carbonyls have been prepared. These carbonyls are potential catalysts. When the metal carbonyl is a gas, the purified metal carbonyl gas can be used for the production of extra-pure metals. Carbon dioxide is also a useful industrial gas. It is widely used in food and beverage industry. Here are some of its applications. The critical temperature of carbon dioxide is only 304 K at a critical pressure of 7.39 MPa (almost 7 atm). These conditions can easily be met to generate supercritical carbon dioxide, which is a powerful and descriminating solvent. The supercritical fluid penetrates porous solids, evaporates without leaving a trace. Thus, this fluid is widely used as an extracting solvent. This fluid is also very useful in the field of analytical chemsitry. On the high technology side, carbon dioxide lasers can provide a continuous laser beam from several milliwatts to several killowatts with a typical efficiency of 30%, one of the most efficient laser generation devices. This link also illustrates the basic theory of laser. Among many applications of laser, carbon dioxide laser has been used for as an art of cosmetic surgery. On the other hand, the heavier carbon dioxide usually stay in lower grounds, and when its concentration is very high, it can be a thread to living beings who will die of asphyxiation. When Henry Ford put people to work on the assembly line, he did not worry about the consequence of automobile exhaust. He probably did not foresee the change of the society. Now everyone wants a piece of the carbondioxide generating machine. You can imagine that when all countries use as much energy as Canadians do, the carbon dioxide level of the atmosphere will be much higher. We need good measurements about carbon dioxide level in order to know how it is changing. The National Oceanic and Atmospheric Administration (NOAA) of the U.S. is keeping track of it, and the measurements at Barrow, Alaska. The annual increase has been reported to be 1.49 ppm by volume per year. On the other hand, the atmospheric CO concentration was about 280 ppm by volume in the 1700s before the industrial revolution, and it was 360 ppm in 1994. If you want more details about Carbon Dioxide Emissions in the U.S., this link is full of data. Engineers, scientists, politician and the general public believe that increase level of CO will cause the world to warm up, because some scientists have demonstrated their findings and the experts agreed. Lengthy discussion is required to present scientific evidences for the so called of CO , and hopefully some day you will be able to judge the argument yourself. I have not found simple and convincing evidence to present at this time. However, experts have suggested a correlation with the increase level of CO and the average temperature of the globe. The green house effect of \(CO_2\) has attracted attention not only of expert and politicians, the public pressure (mostly the news media) have made the reduction of \(CO_2\) emission an international priority. The United Nations Panel on Climate Change made several recommendations regarding \(CO_2\). Well, what can be done at a personal level, as a community, and as a country requires the determination of individuals. This is a challenge for us all especially engineers, because they are at the forefront of many industries. We face many front for a solution to the problem of carbon dioxide emission. The standard enthalpy of formation of NO is 90.25 kJ / mol and the standard entropies of \(N_2\), \(O_2\), \(NO\) are 191.61, 205.138 and 210.761 J (K mol) respectively. Calculate the Gibb's energy for the reaction \[\ce{N_2 + O_2 \rightarrow 2 NO}\nonumber \] at standard conditions. The entropy of formation for the above reaction is \[\begin{align*} \Delta S^o &= 2 \times 210.761 - (205.138 + 191.61) \\[4pt] &= 24.77 \frac{J}{K mol} \end{align*}\] The Gibb's energy of formation is then: \[ \begin{align*} \Delta G^o &= \Delta H^o - T\Delta S^o \\[4pt] &= 180.5 - 298*0.02477 \\[4pt] & = 180.5 - 7.38 \,kJ \\[4pt] &= 173.12\, kJ \end{align*}\] This value shows that the reaction is endothermic. What is the equilibrium constant for the reaction as written and what is the implication of the result in the discussion of NO in air? | 8,868 | 4,197 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Index_of_Hydrogen_Deficiency_(IHD) |
There is no simple way of predicting how many isomers a given molecular formula will yield, (it can range from one to many). . Keep in mind that there is rotation about all single bonds not involved in a ring, but not about double bonds. Because all of the formulas that you will be dealing with are based on the \(C\) atom, it may be useful to review the ways that \(C\) can bond to itself and to other atoms. We will limit ourselves, for now, to the \(C\) atom with four bonds. Below are the possible combinations of C having a total of four bonds. In a hydrocarbon where all the C atoms have only single bonds and no rings are involved, the compound would have the maximum number of H atoms. If any of the bonds are replaced with double or triple bonds, or if rings are involved, there would be a “deficiency” of H atoms. By calculating the index of hydrogen deficiency (IHD), we can tell from the molecular formula whether and how many multiple bonds and rings are involved. IHD is also called the . This will help cut down the possibilities one has to consider in trying to come up with all the isomers of a given formula. Here is a summary of how the index of hydrogen deficiency (IHD) works. (\(C_xH_y\)): \[IHD = \dfrac{2x + 2 - y}{2} \] (where \(x\) and \(y\) stand for # of C and H respectively.) IHD \(C_2H_4\) is This means it can have either one double bond or one ring, but it cannot have a triple bond. Since you cannot form a ring with only two C’s, it must have a double bond. IHD \(C_4H_6\) \[\dfrac{2(4) + 2 - 6}{2} = 2\] This means it can have either one double bond and a ring such as or two double bonds such as CH =CH−CH=CH or CH =C=CH−CH or two rings , or one triple bond, such as CH C≡CCH . O and S atoms do not affect the IHD. Do not forget that when double bonds and rings are involved, geometric isomers are possible. Calculate the IHD for each of the following and see whether it corresponds to the structure shown. (Obviously it should!) Don’t peek until you’ve worked it out yourself, but answers are provided at the bottom. a)
b) CH CHCHCH CHCH
c)
d)
e) CH C≡CCOCH a) IHD = 3
b) IHD = 2
c) IHD = 5
d) IHD = 1
e) IHD = 3 | 2,191 | 4,205 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Chemical_Reactions_and_Interactions/Redox_Reactions |
reactions are reactions in which electrons shift allegiance. Allegiance means loyalty or commitment to a group, like your allegiance to your family or your country. If you decide to leave your home and become a citizen of a new country, you have shifted allegiance. Of course, you might be able to share your allegiance between your old country and your new country by being a citizen of both countries at the same time. This is similar to what electrons do. They have allegiance to a nucleus, and sometimes during a reaction they will shift allegiance to a different nucleus. They might shift allegiance completely, if the new nucleus is much better for them than the old one, or they might shift partially, and share their loyalty between the two. (Also, electrons, like people, can spend some time near nuclei they don't belong to, just like you can visit countries where you don't live, and then go back near their main nucleus, although electrons' trips are much shorter than ours!) All neutral atoms have a population of electrons equal to their number of protons. Atoms, just like countries, always have elite citizens that they treat really well (power, money...) and these elite citizens are way too comfortable to shift allegiance unless they get kidnapped or something. Unlike countries, atoms always have only a small number of non-elite citizens, between 0 and 8, roughly, but usually no more than 3 or 4. These non-elite electrons will easily shift allegiance if they get a better offer. What are the good parts of the periodic table? Fluorine is like heaven for electrons, they will basically never leave. If fluorine is accepting immigrants, electrons will leave anywhere else to move to F. Oxygen is second best. (Fluorine seems nasty to us because we don't want our citizen electrons to leave us for fluorine. We have jobs for them to do!) The noble gases are aloof nations, their populations of electrons hardly ever change. Their citizens don't want to leave, and they don't want any troublesome immigrants either. In general, the upper right corner of the periodic table are good comfortable nations to live in. The bottom left is such a bad place to live that electrons will destroy property in their rush to leave. (Watch this video for an example) shows "standard of living" across the periodic table. Red elements will tend to keep their own electrons and attract electrons from other nuclei; yellow elements will tend to lose their non-elite electrons. Ionic substances are like alliances between nations with really different standards of living. The non-elite electrons of the metal will try to shift allegiance to the non-metal. In molecular substances, also called covalent, the electrons will have dual citizenship, although they might have slightly greater loyalty to one nuclei than another, because these are alliances between similarly comfortable non-metal nuclei. With all these dual citizen electrons, with partial loyalty to several nuclei, it can get hard to count the population of electrons at each nucleus. is one procedure for counting. In this procedure, electrons are counted as belonging completely to the nuclei they feel the greatest loyalty to. The number of electrons that feel primary loyalty to each nucleus is counted, and subtracted the number of protons in that nucleus. But there are shortcuts to make this faster using valence rules. For example, in CO , oxygen has a normal valence of 2. This means that it can accept 2 extra citizen electrons per nucleus, so it has an oxidation number of -2. Therefore, because there are 2 O nuclei, carbon loses 4 citizen electrons, who transfer their primary loyalty to O. Thus, carbon has an oxidation number of +4, because it has 4 fewer truly loyal electrons than protons. How to count oxidation states: refers to a process in which something loses electrons, and has its oxidation number increase. This usually happens to compounds that react with oxygen gas, which is why it is called oxidation. refers to a process in which something gains electrons, and its oxidation number is reduced. Actually this is not quite where the word came from. When a metal oxide is reduced to the elemental metal and an elemental non-metal or non-metal compound, the mass of solid decreases because the non-metal usually leaves as a gas. This reduction in quantity is where the word comes from. When you think about it, for one thing to be oxidized, another thing must be reduced (because electrons can't appear out of nowhere), which is why we often use the combination word redox. Many of the reaction types we've already seen involve redox. For instance, combination of elements is a redox reaction. Decomposition reactions are often redox reactions. Combustion reactions are redox reactions. (Dissolution/precipitation and acid-base are not redox reactions.) The examples of redox you've already seen involve electrons shifting loyalty at the same time that new bonds are made between nuclei. It's also possible for electrons to abandon one nucleus for another separate nucleus that isn't bonded to the first one. For example, if an elemental metal is placed in a solution of salt or acid, the metal can lose some electrons to the cation in the solution. The metal forms cations, and the former cation usually leaves the solution, forming hydrogen gas or elemental metal. Here are some examples: \[Zn(s) + 2HBr(aq) \rightarrow ZnBr_{2}(aq) + H_{2}(g)\] \[Fe(s) + Pb(NO_{3})_{2}(aq) \rightarrow Fe(NO_{3})_{2}(aq) + Pb(s)\] All that happened in these reactions is that the disloyal electrons left a less-comfortable home country for a more comfortable one. These are called because one cation is displaced (replaced) by another. To predict what reactions will occur on this pattern, you can use the , which is like a ranking of "worst places to live". It's pretty good for making predictions, because in general electrons will move to places lower on the list, but it doesn't take into account all the specific circumstances, like current population pressures. You'll learn more reliable ways to make predictions later. The activity series is: Copper metal doesn't react with acid, because copper comes after hydrogen in the list. However, some acids have other oxidants in them that are stronger than hydrogen ions. For instance, nitrate ions are oxidizing, and Cu will react with nitric acid. Even gold reacts with a concentrated acid mixture, aqua regia, made of nitric and hydrochloric acids. | 6,512 | 4,206 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./13.E%3A_Solutions_(Exercises) |
" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact Problems marked with a ♦ involve multiple concepts. ♦ Scuba divers utilize high-pressure gas in their tanks to allow them to breathe underwater. At depths as shallow as 100 ft (30 m), the pressure exerted by water is 4.0 atm. At 25°C, the values of Henry’s law constants for N , O , and He in blood are as follows: N2 = 6.5 × 10 mol/(L·atm); O2 = 1.28 × 10 mol/(L·atm); and He = 3.7 × 10 mol/(L·atm). What would be the concentration of nitrogen and oxygen in blood at sea level where the air is 21% oxygen and 79% nitrogen? What would be the concentration of nitrogen and oxygen in blood at a depth of 30 m, assuming that the diver is breathing compressed air? ♦ Many modern batteries take advantage of lithium ions dissolved in suitable electrolytes. Typical batteries have lithium concentrations of 0.10 M. Which aqueous solution has the higher concentration of ion pairs: 0.08 M LiCl or 1.4 M LiCl? Why? Does an increase in the number of ion pairs correspond to a higher or lower van’t Hoff factor? Batteries rely on a high concentration of unpaired Li+ ions. Why is using a more concentrated solution not an ideal strategy in this case? Hydrogen sulfide, which is extremely toxic to humans, can be detected at a concentration of 2.0 ppb. At this level, headaches, dizziness, and nausea occur. At higher concentrations, however, the sense of smell is lost, and the lack of warning can result in coma and death can result. What is the concentration of H S in milligrams per liter at the detection level? The lethal dose of hydrogen sulfide by inhalation for rats is 7.13 × 10 g/L. What is this lethal dose in ppm? The density of air is 1.2929 g/L. One class of antibiotics consists of cyclic polyethers that can bind alkali metal cations in aqueous solution. Given the following antibiotics and cation selectivities, what conclusion can you draw regarding the relative sizes of the cavities? Phenylpropanolamine hydrochloride is a common nasal decongestant. An aqueous solution of phenylpropanolamine hydrochloride that is sold commercially as a children’s decongestant has a concentration of 6.67 × 10 M. If a common dose is 1.0 mL/12 lb of body weight, how many moles of the decongestant should be given to a 26 lb child? The “freeze-thaw” method is often used to remove dissolved oxygen from solvents in the laboratory. In this technique, a liquid is placed in a flask that is then sealed to the atmosphere, the liquid is frozen, and the flask is evacuated to remove any gas and solvent vapor in the flask. The connection to the vacuum pump is closed, the liquid is warmed to room temperature and then refrozen, and the process is repeated. Why is this technique effective for degassing a solvent? Q13.7 Suppose that, on a planet in a galaxy far, far away, a species has evolved whose biological processes require even more oxygen than we do. The partial pressure of oxygen on this planet, however, is much less than that on Earth. The chemical composition of the “blood” of this species is also different. Do you expect their “blood” to have a higher or lower value of the Henry’s law constant for oxygen at standard temperature and pressure? Justify your answer. A car owner who had never taken general chemistry decided that he needed to put some ethylene glycol antifreeze in his car’s radiator. After reading the directions on the container, however, he decided that “more must be better.” Instead of using the recommended mixture (30% ethylene glycol/70% water), he decided to reverse the amounts and used a 70% ethylene glycol/30% water mixture instead. Serious engine problems developed. Why? The ancient Greeks produced “Attic ware,” pottery with a characteristic black and red glaze. To separate smaller clay particles from larger ones, the powdered clay was suspended in water and allowed to settle. This process yielded clay fractions with coarse, medium, and fine particles, and one of these fractions was used for painting. Which size of clay particles forms a suspension, which forms a precipitate, and which forms a colloidal dispersion? Would the colloidal dispersion be better characterized as an emulsion? Why or why not? Which fraction of clay particles was used for painting? The Tyndall effect is often observed in movie theaters, where it makes the beam of light from the projector clearly visible. What conclusions can you draw about the quality of the air in a movie theater where you observe a large Tyndall effect? Aluminum sulfate is the active ingredient in styptic pencils, which can be used to stop bleeding from small cuts. The Al ions induce aggregation of colloids in the blood, which facilitates formation of a blood clot. How can Al ions induce aggregation of a colloid? What is the probable charge on the colloidal particles in blood? ♦ The liver secretes bile, which is essential for the digestion of fats. As discussed in ", fats are biomolecules with long hydrocarbon chains. The globules of fat released by partial digestion of food particles in the stomach and lower intestine are too large to be absorbed by the intestine unless they are emulsified by bile salts, such as glycocholate. Explain why a molecule like glycocholate is effective at creating an aqueous dispersion of fats in the digestive tract. Answers 1 atm: 2.7 × 10 M O and 5.1 × 10 M N 4 atm: 1.1 × 10 M O and 2.1 × 10 M N 2.6 × 10 mg/L, 550 ppm 1.4 × 10 mol To obtain the same concentration of dissolved oxygen in their “blood” at a lower partial pressure of oxygen, the value of the Henry’s law constant would have to be higher. The large, coarse particles would precipitate, the medium particles would form a suspension, and the fine ones would form a colloid. A colloid consists of solid particles in a liquid medium, so it is not an emulsion, which consists of small particles of one liquid suspended in another liquid. The finest particles would be used for painting. | 6,045 | 4,207 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Esters/Reactivity_of_Esters/Transesterification |
Transesterification is the conversion of a carboxylic acid ester into a different carboxylic acid ester. When in ester is placed in a large excess of an alcohol along with presence of either an acid or a base there can be an exchange of alkoxy groups. The large excess of alcohol is used to drive the reaction forward. The most common method of transesterification is the reaction of the ester with an alcohol in the presence of an acid catalyst eg: This reaction has the following mechanism: Since both the reactants and the products are an ester and an alcohol, the reaction is reversible and the equilibrium constant is close to one. Consequently, the has to be exploited to drive the reaction to completion. The simplest way to do so is to use the alcohol as the solvent as well. Nucleophilic attack by an alkoxide 2) Leaving group removal 1) Protonation of the carbonyl by the acid. The carbonyl is now activated toward nucleophilic attack. 2) Nucleophilic attack on the carbonyl 3) Proton transfer 4) Removal of the leaving group 5) Deprotonation ) | 1,070 | 4,208 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Nucleic_Acids/Nucleic_Acids/Nucleic_Acids |
The nucleic acids are informational molecules because their primary structure contains a code or set of directions by which they can duplicate themselves and guide the synthesis of proteins. The synthesis of proteins - most of which are enzymes - ultimately governs the metabolic activities of the cell. In 1953, Watson, an American biologist, and Crick, an English biologist, proposed the double helix structure for DNA. This development set the stage for a new and continuing era of chemical and biological investigation. The two main events in the life of a cell - dividing to make exact copies of themselves, and manufacturing proteins - both rely on blueprints coded in our genes. There are two types of nucleic acids which are polymers found in all living cells. is found mainly in the nucleus of the cell, while is found mainly in the cytoplasm of the cell although it is usually synthesized in the nucleus. DNA contains the genetic codes to make RNA and the RNA in turn then contains the codes for the primary sequence of amino acids to make proteins. The best way to understand the structures of DNA and RNA is to identify and examine individual parts of the structures first. The complete hydrolysis of nucleic acids yields three major classes of compounds: pentose sugars, phosphates, and heterocyclic amines (or bases). A major requirement of all living things is a suitable source of phosphorus. One of the major uses for phosphorus is as the phosphate ion which is incorporated into DNA and RNA. There are two types of pentose sugars found in nucleic acids. This difference is reflected in their names--deoxyribonucleic acid indicates the presence of deoxyribose; while ribonucleic acid indicates the presence of ribose. In the graphic below, the structures of both ribose and deoxyribose are shown. Note the red -OH on one and the red -H on the other are the only differences. The alpha and beta designations are interchangeable and are not a significant difference between the two. Heterocyclic amines are sometimes called nitrogen bases or simply . The heterocyclic amines are derived from two root structures: purines or pyrimidines. The purine root has both a six and a five member ring; the pyrimidine has a single six member ring. There are two major purines, adenine (A) and guanine (G), and three major pyrimidines, cytosine (C), uracil (U), and thymine (T). The structures are shown in the graphic on the left. As you can see, these structures are called "bases" because the amine groups as part of the ring or as a side chain have a basic property in water . A major difference between DNA and RNA is that DNA contains thymine, but not uracil, while RNA contains uracil but not thymine. The other three heterocyclic amines, adenine, guanine, and cytosine are found in both DNA and RNA. For convenience, you may remember, the list of heterocyclic amines in DNA by the words: The Amazing Gene Code (TAGC). | 2,951 | 4,209 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Conjugation/Polymerization_of_Conjugated_Dienes |
Conjugated dienes (alkenes with two double bonds and a single bond in between) can be polymerized to form important compounds like rubber. This takes place, in different forms, both in nature and in the laboratory. Interactions between double bonds on multiple chains leads to cross-linkage which creates elasticity within the compound. For rubber compounds to be synthesized, 1,3-butadiene must be polymerized. Below is a simple illustration of how this compound is formed into a chain. The 1,4 polymerization is much more useful to polymerization reactions. Above, the green structures represent the base units of the polymers that are synthesized and the red represents the bonds between these units which form these polymers. Whether the 1,3 product or the 1,4 product is formed depends on whether the reaction is thermally or kinetically controlled. The most important synthetic rubber is Neoprene which is produced by the polymerization of 2-chloro-1,3-butadiene. In this illustration, the dashed lines represent repetition of the same base units, so both the products and reactants are polymers. The reaction proceeds with a mechanism similar to the mechanism. Cross-linkage between the chlorine atom of one chain and the double bond of another contributes to the overall elasticity of neoprene. This cross-linkage occurs as the chains lie next to each other at random angles, and the attractions between double bonds prevent them from sliding back and forth. The synthesis of rubber in nature in somewhat similar the artificial synthesis of rubber except that it takes place within a plant. Instead of the 2-chloro-1,3-butadiene used in the synthesis of neoprene, natural rubber is synthesized from 2-methyl-1,3-butadiene. As an electrophile, the plant synthesizes the pyrophosphate 3-methyl-3-butenyl pyrophosphate is from phosphoric acid and 3-methyl-3-buten-1-ol. This pyrophosphate then catalyzes the reaction that leads to natural rubber. The 3-methyl-3-butenyl pyrophosphate (OPP) is then used in the polymerization of natural rubber as it pulls electrons off 2-methyl-1,3-butadiene (see questions section for this process.) Draw out the mechanism for the natural synthesis of rubber from 3-methyl-3-butenyl pyrophosphate and 2-methyl-1,3-butadiene. Show the movement of electrons with arrows. | 2,327 | 4,210 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.03%3A_Molecules_as_Energy_Carriers_and_Converters |
Make sure you thoroughly understand the following essential concepts: All molecules at temperatures above absolue zero possess — the randomized kinetic energy associated with the various motions the molecules as a whole, and also the atoms within them, can undergo. Polyatomic molecules also possess in the form of chemical bonds. Molecules are thus both vehicles for , and the means of when the formation, breaking, or rearrangement of the chemical bonds within them is accompanied by the uptake or release of . When you buy a liter of gasoline for your car, a cubic meter of natural gas to heat your home, or a small battery for your flashlight, you are purchasing energy in a chemical form. In each case, some kind of a chemical change will have to occur before this energy can be released and utilized: the fuel must be burned in the presence of oxygen, or the two poles of the battery must be connected through an external circuit (thereby initiating a chemical reaction inside the battery.) And eventually, when each of these reactions is complete, our source of energy will be exhausted; the fuel will be used up, or the battery will be “dead”. Where did the energy go? It could have gone to raise the temperature of the products, to perform work in expanding any gaseous products or to push electrons through a circuit. The remainder will reside in the chemical potential energy associated with the products of the reaction. Chemical substances are made of atoms, or more generally, of positively charged nuclei surrounded by negatively charged electrons. A molecule such as dihydrogen, H , is held together by electrostatic attractions mediated by the electrons shared between the two nuclei. The total potential energy of the molecule is the sum of the repulsions between like charges and the attractions between electrons and nuclei: \[PE_{total} = PE_{electron-electron} + PE_{nucleus-nucleus} + PE_{nucleus-electron} \label{1-1}\] In other words, . This dependence is expressed by the familiar which serves as an important description of the chemical bond between two atoms. refers to movement of an object as a complete unit. Translational motions of molecules in solids or liquids are restricted to very short distances comparable to the dimensions of the molecules themselves, whereas in gases the molecules typically travel hundreds of molecular diameters between collisions. In gaseous hydrogen, for example, the molecules will be moving freely from one location to another; this is called , and the molecules therefore possess \[KE_{trans} = \dfrac{mv^2}{2}\] in which \(v\) stands for the average velocity of the molecules; you may recall from your study of gases that \(v\), and therefore \(KE_{trans}\), depends on the temperature. In addition to translation, molecules composed of two or more atoms can possess other kinds of motion. Because a chemical bond acts as a kind of spring, the two atoms in H will have a natural . In more complicated molecules, many different modes of vibration become possible, and these all contribute a vibrational term KE to the total kinetic energy. Finally, a molecule can undergo motions which give rise to a third term \(KE_{rot}\). Thus the total kinetic energy of a molecule is the sum \[KE_{total} = KE_{trans} + KE_{vib} + KE_{rot} \label{1-2}\] The total energy of the molecule (its ) is just the sum \[U = KE_{total} + PE_{total} \label{1-3}\] Although this formula is simple and straightforward, it cannot take us very far in understanding and predicting the behavior of even one molecule, let alone a large number of them. The reason, of course, is the chaotic and unpredictable nature of molecular motion. Fortunately, the behavior of a large collection of molecules, like that of a large population of people, can be described by statistical methods. As noted above, the of a substance is a measure of how sensitively its temperature is affected by a change in heat content; the greater the heat capacity, the less effect a given flow of heat will have on the temperature. We also pointed out that is a measure of the average kinetic energy due to of molecules. If vibrational or rotational motions are also active, these will also accept thermal energy and reduce the amount that goes into translational motions. Because the temperature depends only on the latter, the effect of the other kinds of motions will be to reduce the dependence of the internal energy on the temperature, thus raising the of a substance. Whereas monatomic molecules can only possess translational thermal energy, two additional kinds of motions become possible in polyatomic molecules. A linear molecule has an axis that defines two perpendicular directions in which rotations can occur; each represents an additional degree of freedom, so the two together contribute a total of ½ to the heat capacity. Vibrational and rotational motions are not possible for monatomic species such as the noble gas elements, so these substances have the lowest heat capacities. Moreover, as you can see in the leftmost column of Table 1, their heat capacities are all the same. This reflects the fact that translational motions are the same for all particles; all such motions can be resolved into three directions in space, each contributing one to the molecule and ½ to its heat capacity. ( is the gas constant, 8.314 J K ). For a non-linear molecule, rotations are possible along all three directions of space, so these molecules have a rotational heat capacity of 3/2 . Finally, the individual atoms within a molecule can move relative to each other, producing a vibrational motion. A molecule consisting of atoms can vibrate in 3 –6 different ways or modes. Each vibrational mode contributes (rather than ½ ) to the total heat capacity. (These results come from advanced mechanics and will not be proven here.) Now we are in a position to understand why more complicated molecules have higher heat capacities. The total kinetic energy of a molecule is the sum of those due to the various kinds of motions: \[KE_{total} = KE_{trans} + KE_{rot} + KE_{vib} \label{2-1}\] When a monatomic gas absorbs heat, all of the energy ends up in translational motion, and thus goes to increase its temperature. In a polyatomic gas, by contrast, the absorbed energy is partitioned among the other kinds of motions; since only the translational motions contribute to the temperature, the temperature rise is smaller, and thus the heat capacity is larger. There is one very significant complication, however: classical mechanics predicts that the energy is always partitioned equally between all degrees of freedom. Experiments, however, show that this is observed only at quite high temperatures. The reason is that these motions are all . This means that only certain increments of energy are possible for each mode of motion, and unless a certain minimum amount of energy is available, a given mode will not be active at all and will contribute nothing to the heat capacity. The shading indicates the average thermal energy available at 300 K. Only those levels within this range will have significant occupancy as indicated by the thickness of the lines in the two rightmost columns. At 300 K, only the lowest vibrational state and the first few rotational states will be active. Most of the thermal energy will be confined to the translational levels whose minute spacing (10 J) causes them to appear as a continuum. This plot is typical of those for other polyatomic molecules, and shows the practical consequences of the spacings of the various forms of thermal energy. Thus translational motions are available at virtually all temperatures, but contributions to heat acapacity by rotational or vibrational motions can only develop at temperatures sufficiently large to excite these motions. It turns out that translational energy levels are spaced so closely that they these motions are active almost down to absolute zero, so . Rotational motions do not get started until intermediate temperatures, typically 300-500K, so within this range heat capacities begin to increase with temperature. Finally, at very high temperatures, vibrations begin to make a significant contribution to the heat capacity The strong intermolecular forces of liquids and many solids allow heat to be channeled into vibrational motions involving more than a single molecule, further increasing heat capacities. One of the well known “anomalous” properties of liquid water is its high heat capacity (75 J mol K ) due to intermolecular hydrogen bonding, which is directly responsible for the moderating influence of large bodies of water on coastal climates. Metallic solids are a rather special case. In metals, the atoms oscillate about their equilibrium positions in a rather uniform way which is essentially the same for all metals, so they should all have about the same heat capacity. That this is indeed the case is embodied in the . In the 19th century these workers discovered that the molar heat capacities of all the metallic elements they studied were around to 25 J mol K , which is close to what classical physics predicts for crystalline metals. This observation played an important role in characterizing new elements, for it provided a means of estimating their molar masses by a simple heat capacity measurement. Under the special conditions in which the pressure is 1 atm and the reactants and products are at a temperature of 298 K, Δ becomes the Δ . Chemists usually refer to the "enthalpy change of a reaction" as simply the "enthalpy of reaction", or even more simply as the " ". But students are allowed to employ this latter shortcut only if they are able to prove that they know the meaning of enthalpy. Since most changes that occur in the laboratory, on the surface of the earth, and in organisms are subjected to an approximately constant pressure of "one atmosphere" and reasonably salubrious temperatures, most reaction heats quoted in the literature refer to Δ . But the high pressures and extreme temperatures frequently encountered by chemical engineers, geochemists, and practicioners of chemical oceanography, often preclude the convenience of the "standard" values. The rearrangement of atoms that occurs in a chemical reaction is virtually always accompanied by the liberation or absorption of heat. If the purpose of the reaction is to serve as a source of heat, such as in the combustion of a fuel, then these heat effects are of direct and obvious interest. We will soon see, however, that a study of the energetics of chemical reactions in general can lead us to a deeper understanding of chemical equilibrium and the basis of chemical change itself. In chemical thermodynamics, we define the zero of the enthalpy and internal energy as that of the elements as they exist in their stable forms at 298K and 1 atm pressure. Thus the enthalpies of Xe , O and C are all zero, as are those of H and Cl in the reaction \[H_{2(g)} + Cl_{2(g)} → 2 HCl_{(g)}\] The enthalpy of two moles of HCl is smaller than that of the reactants, so the difference is released as heat. Such a reaction is said to be . The reverse of this reaction would absorb the same quantity of heat from the surroundings and be . In comparing the internal energies and enthalpies of different substances as we have been doing here, it is important to compare equal numbers of moles, because energy is an of matter. However, heats of reaction are commonly expressed on a molar basis and treated as . We can characterize any chemical reaction by the change in the internal energy or enthalpy: \[ΔH = H_{final} – H_{initial} \label{3-1}\] The significance of this can hardly be exaggerated because Δ , being a state function, is entirely independent of how the system gets from the initial state to the final state. In other words, the value of Δ or Δ for a given change in state is independent of the of the process. Consider, for example, the oxidation of a lump of sugar to carbon dioxide and water: \[\ce{C12H22O11 + 12 O2(g) → 12 CO2(g) + 11 H2O(l)}\] This process can be carried out in many ways, for example by burning the sugar in air, or by eating the sugar and letting your body carry out the oxidation. Although the mechanisms of the transformation are completely different for these two pathways, the overall change in the enthalpy of the system (the atoms of carbon, hydrogen and oxygen that were originally in the sugar) will be identical, and can be calculated simply by looking up the of the reactants and products and calculating the difference \[ΔH = [12 \times H(\ce{CO2})] + [11 \times H(\ce{H2O})] – H(\ce{C12H22O11}) = –5606\, kJ\] The same quantity of heat is released whether the sugar is burnt in the air or oxidized in a series of enzyme-catalyzed steps in your body. When the temperature of a substance is raised, it absorbs heat. The enthalpy of a system increases with the temperature by the amount \(ΔH = C_p ΔT\). The defining relation \[ΔH = ΔU + P ΔV\] tells us that this change is dominated by the internal energy, subject to a slight correction for the work associated with volume change. Heating a substance causes it to expand, making Δ positive and causing the enthalpy to increase slightly more than the internal energy. Physically, what this means is that if the temperature is increased while holding the pressure constant, some extra energy must be expended to push back the external atmosphere while the system expands. The difference between the dependence of and on temperature is only really significant for gases, since the coefficients of thermal expansion of liquids and solids are very small. A plot of the enthalpy of a system as a function of its temperature is called an . The slope of the line is given by . The enthalpy diagram of a pure substance such as water shows that this plot is not uniform, but is interrupted by sharp breaks at which the value of is apparently infinite, meaning that the substance can absorb or lose heat without undergoing any change in temperature at all. This, of course, is exactly what happens when a substance undergoes a phase change; you already know that the temperature the water boiling in a kettle can never exceed 100 until all the liquid has evaporated, at which point the temperature of the steam will rise as more heat flows into the system. provides a concise view of its thermal behavior. The slope of the line is given by the heat capacity . All -vs.- plots show sharp breaks at which the value of is apparently infinite, meaning that the substance can absorb or lose heat without undergoing any change in temperature at all. This, of course, is exactly what happens when a substance undergoes a ; you already know that the temperature of the water boiling in a kettle can never exceed 100°C until all the liquid has evaporated, at which point the temperature (of the steam) will rise as more heat flows into the system. The lowest-temperature discontinuity on the CCl diagram corresponds to a solid-solid phase transition associated with a rearrangement of molecules in the crystalline solid. | 15,197 | 4,211 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Nucleic_Acids/RNA/Protein-RNA_Recognition |
RNA-protein interactions are behind a number of vital processes in the cell. Without the ability of particular proteins to bind RNA, the RNA would no longer be able to carry out its important functions as a component of the ribosome and spliceosome. Other examples of important RNA-protein interactions include binding of tRNA to aminoacyl-tRNA synthetases, a process vital to translation of genetic information into proteins necessary for continued biological function4 and regulation of post-transcriptional control of gene expression via the binding of RNA to riobonucleoproteins, or RNPs. Although not as well characterized as the binding between DNA and proteins, RNA-protein binding has been a field that has seen a great deal of growth in recent years. Although it was originally expected that RNA-protein binding motifs might fall neatly into categories the way DNA motifs did, the wide range of secondary and tertiary RNA structures that can be recognized by proteins requires more variety in binding motifs of the proteins, and the rules used to categorize them become correspondingly more complex.6 At this time all major families of RNA-binding proteins have been structurally characterized and these characterizations have led to a much better understanding of RNA recognition. | 1,311 | 4,212 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/21%3A_Nuclear_Chemistry/21.4%3A_Transmutation_and_Nuclear_Energy |
After the discovery of radioactivity, the field of nuclear chemistry was created and developed rapidly during the early twentieth century. A slew of new discoveries in the 1930s and 1940s, along with World War , combined to usher in the Nuclear Age in the mid-twentieth century. Science learned how to create new substances, and certain isotopes of certain elements were found to possess the capacity to produce unprecedented amounts of energy, with the potential to cause tremendous damage during war, as well as produce enormous amounts of power for society’s needs during peace. is the conversion of one nuclide into another. It can occur by the radioactive decay of a nucleus, or the reaction of a nucleus with another particle. The first manmade nucleus was produced in Ernest Rutherford’s laboratory in 1919 by a reaction, the bombardment of one type of nuclei with other nuclei or with neutrons. Rutherford bombarded nitrogen atoms with high-speed α particles from a natural radioactive isotope of radium and observed protons resulting from the reaction: \[\ce{^{14}_7N + ^4_2He ⟶ ^{17}_8O + ^1_1H} \nonumber \] The \(\ce{^{17}_8O}\) and \(\ce{^1_1H}\) nuclei that are produced are stable, so no further (nuclear) changes occur. To reach the kinetic energies necessary to produce transmutation reactions, devices called are used. These devices use magnetic and electric fields to increase the speeds of nuclear particles. In all accelerators, the particles move in a vacuum to avoid collisions with gas molecules. When neutrons are required for transmutation reactions, they are usually obtained from radioactive decay reactions or from various nuclear reactions occurring in nuclear reactors. The Chemistry in Everyday Life feature that follows discusses a famous particle accelerator that made worldwide news. Located near Geneva, the CERN (“Conseil Européen pour la Recherche Nucléaire,” or European Council for Nuclear Research) Laboratory is the world’s premier center for the investigations of the fundamental particles that make up matter. It contains the 27-kilometer (17 mile) long, circular Large Hadron Collider (LHC), the largest particle accelerator in the world (Figure \(\Page {1}\)). In the , particles are boosted to high energies and are then made to collide with each other or with stationary targets at nearly the speed of light. Superconducting electromagnets are used to produce a strong magnetic field that guides the particles around the ring. Specialized, purpose-built detectors observe and record the results of these collisions, which are then analyzed by CERN scientists using powerful computers. In 2012, CERN announced that experiments at the LHC showed the first observations of the Higgs boson, an elementary particle that helps explain the origin of mass in fundamental particles. This long-anticipated discovery made worldwide news and resulted in the awarding of the 2103 Nobel Prize in Physics to François Englert and Peter Higgs, who had predicted the existence of this particle almost 50 years previously. Prior to 1940, the heaviest-known element was uranium, whose atomic number is 92. Now, many artificial elements have been synthesized and isolated, including several on such a large scale that they have had a profound effect on society. One of these—element 93, neptunium (Np)—was first made in 1940 by McMillan and Abelson by bombarding uranium-238 with neutrons. The reaction creates unstable uranium-239, with a half-life of 23.5 minutes, which then decays into neptunium-239. Neptunium-239 is also radioactive, with a half-life of 2.36 days, and it decays into plutonium-239. The nuclear reactions are: \[\begin{align*}
\ce{^{238}_{92}U + ^1_0n &⟶ ^{239}_{92}U} && \\[4pt]
&\ce{^{239}_{92}U &⟶ ^{239}_{93}Np + ^0_{−1}e \,\,\,\mathit{t}_{1/2}} &&\textrm{half-life}=\mathrm{23.5\: min} \\[4pt]
&\ce{^{239}_{93}Np &⟶ ^{239}_{94}Pu + ^0_{−1}e\,\,\, \mathit{t}_{1/2}} &&\textrm{half-life}=\mathrm{2.36\: days}
\end{align*} \nonumber \] Plutonium is now mostly formed in nuclear reactors as a byproduct during the decay of uranium. Some of the neutrons that are released during U-235 decay combine with U-238 nuclei to form uranium-239; this undergoes β decay to form neptunium-239, which in turn undergoes β decay to form plutonium-239 as illustrated in the preceding three equations. It is possible to summarize these equations as: \[\mathrm{\ce{^{238}_{92}U} + {^1_0n}⟶ \ce{^{239}_{92}U} \xrightarrow{β^-} \ce{^{239}_{93}Np} \xrightarrow{β^-} \ce{^{239}_{94}Pu}} \nonumber \] Heavier isotopes of plutonium—Pu-240, Pu-241, and Pu-242—are also produced when lighter plutonium nuclei capture neutrons. Some of this highly radioactive plutonium is used to produce military weapons, and the rest presents a serious storage problem because they have half-lives from thousands to hundreds of thousands of years. Although they have not been prepared in the same quantity as plutonium, many other synthetic nuclei have been produced. Nuclear medicine has developed from the ability to convert atoms of one type into other types of atoms. Radioactive isotopes of several dozen elements are currently used for medical applications. The radiation produced by their decay is used to image or treat various organs or portions of the body, among other uses. The elements beyond element 92 (uranium) are called . As of this writing, 22 transuranium elements have been produced and officially recognized by IUPAC; several other elements have formation claims that are waiting for approval. Some of these elements are shown in Table \(\Page {1}\). seaborgium Sg 106 \(\ce{^{206}_{82}Pb + ^{54}_{24}Cr ⟶ ^{257}_{106}Sg + 3 ^1_0n}\) \(\ce{^{249}_{98}Cf + ^{18}_8O ⟶ ^{263}_{106}Sg + 4 ^1_0n}\) Many heavier elements with smaller binding energies per nucleon can decompose into more stable elements that have intermediate mass numbers and larger binding energies per nucleon—that is, mass numbers and binding energies per nucleon that are closer to the “peak” of the binding energy graph near 56. Sometimes neutrons are also produced. This decomposition is called , the breaking of a large nucleus into smaller pieces. The breaking is rather random with the formation of a large number of different products. Fission usually does not occur naturally, but is induced by bombardment with neutrons. The first reported nuclear fission occurred in 1939 when three German scientists, Lise Meitner, Otto Hahn, and Fritz Strassman, bombarded uranium-235 atoms with slow-moving neutrons that split the U-238 nuclei into smaller fragments that consisted of several neutrons and elements near the middle of the periodic table. Since then, fission has been observed in many other isotopes, including most actinide isotopes that have an odd number of neutrons. A typical nuclear fission reaction is shown in Figure \(\Page {2}\). Among the products of Meitner, Hahn, and Strassman’s fission reaction were barium, krypton, lanthanum, and cerium, all of which have nuclei that are more stable than uranium-235. Since then, hundreds of different isotopes have been observed among the products of fissionable substances. A few of the many reactions that occur for U-235, and a graph showing the distribution of its fission products and their yields, are shown in Figure \(\Page {3}\). Similar fission reactions have been observed with other uranium isotopes, as well as with a variety of other isotopes such as those of plutonium. A tremendous amount of energy is produced by the fission of heavy elements. For instance, when one mole of U-235 undergoes fission, the products weigh about 0.2 grams less than the reactants; this “lost” mass is converted into a very large amount of energy, about 1.8 × 10 kJ per mole of U-235. Nuclear fission reactions produce incredibly large amounts of energy compared to chemical reactions. The fission of 1 kilogram of uranium-235, for example, produces about 2.5 million times as much energy as is produced by burning 1 kilogram of coal. As described earlier, when undergoing fission U-235 produces two “medium-sized” nuclei, and two or three neutrons. These neutrons may then cause the fission of other uranium-235 atoms, which in turn provide more neutrons that can cause fission of even more nuclei, and so on. If this occurs, we have a nuclear action (Figure \(\Page {4}\)). On the other hand, if too many neutrons escape the bulk material without interacting with a nucleus, then no chain reaction will occur. Material that can sustain a nuclear fission chain reaction is said to be or . (Technically, fissile material can undergo fission with neutrons of any energy, whereas fissionable material requires high-energy neutrons.) Nuclear fission becomes self-sustaining when the number of neutrons produced by fission equals or exceeds the number of neutrons absorbed by splitting nuclei plus the number that escape into the surroundings. The amount of a fissionable material that will support a self-sustaining chain reaction is a . An amount of fissionable material that cannot sustain a chain reaction is a . An amount of material in which there is an increasing rate of fission is known as a . The critical mass depends on the type of material: its purity, the temperature, the shape of the sample, and how the neutron reactions are controlled (Figure \(\Page {5}\)). An atomic bomb (Figure \(\Page {6}\)) contains several pounds of fissionable material, \(\ce{^{235}_{92}U}\) or \(\ce{^{239}_{94}Pu}\), a source of neutrons, and an explosive device for compressing it quickly into a small volume. When fissionable material is in small pieces, the proportion of neutrons that escape through the relatively large surface area is great, and a chain reaction does not take place. When the small pieces of fissionable material are brought together quickly to form a body with a mass larger than the critical mass, the relative number of escaping neutrons decreases, and a chain reaction and explosion result. Chain reactions of fissionable materials can be controlled and sustained without an explosion in a (Figure \(\Page {7}\)). Any nuclear reactor that produces power via the fission of uranium or plutonium by bombardment with neutrons must have at least five components: nuclear fuel consisting of fissionable material, a nuclear moderator, reactor coolant, control rods, and a shield and containment system. We will discuss these components in greater detail later in the section. The reactor works by separating the fissionable nuclear material such that a critical mass cannot be formed, controlling both the flux and absorption of neutrons to allow shutting down the fission reactions. In a nuclear reactor used for the production of electricity, the energy released by fission reactions is trapped as thermal energy and used to boil water and produce steam. The steam is used to turn a turbine, which powers a generator for the production of electricity. consists of a fissionable isotope, such as uranium-235, which must be present in sufficient quantity to provide a self-sustaining chain reaction. In the United States, uranium ores contain from 0.05–0.3% of the uranium oxide U O ; the uranium in the ore is about 99.3% nonfissionable U-238 with only 0.7% fissionable U-235. Nuclear reactors require a fuel with a higher concentration of U-235 than is found in nature; it is normally enriched to have about 5% of uranium mass as U-235. At this concentration, it is not possible to achieve the supercritical mass necessary for a nuclear explosion. Uranium can be enriched by gaseous diffusion (the only method currently used in the US), using a gas centrifuge, or by laser separation. In the gaseous diffusion enrichment plant where U-235 fuel is prepared, UF (uranium hexafluoride) gas at low pressure moves through barriers that have holes just barely large enough for UF to pass through. The slightly lighter UF molecules diffuse through the barrier slightly faster than the heavier UF molecules. This process is repeated through hundreds of barriers, gradually increasing the concentration of UF to the level needed by the nuclear reactor. The basis for this process, Graham’s law, is described in the chapter on gases. The enriched UF gas is collected, cooled until it solidifies, and then taken to a fabrication facility where it is made into fuel assemblies. Each fuel assembly consists of fuel rods that contain many thimble-sized, ceramic-encased, enriched uranium (usually UO ) fuel pellets. Modern nuclear reactors may contain as many as 10 million fuel pellets. The amount of energy in each of these pellets is equal to that in almost a ton of coal or 150 gallons of oil. Neutrons produced by nuclear reactions move too fast to cause fission (Figure 21.5.5). They must first be slowed to be absorbed by the fuel and produce additional nuclear reactions. A is a substance that slows the neutrons to a speed that is low enough to cause fission. Early reactors used high-purity graphite as a moderator. Modern reactors in the US exclusively use heavy water \(\ce{( ^2_1H2O)}\) or light water (ordinary H O), whereas some reactors in other countries use other materials, such as carbon dioxide, beryllium, or graphite. A nuclear is used to carry the heat produced by the fission reaction to an external boiler and turbine, where it is transformed into electricity. Two overlapping coolant loops are often used; this counteracts the transfer of radioactivity from the reactor to the primary coolant loop. All nuclear power plants in the US use water as a coolant. Other coolants include molten sodium, lead, a lead-bismuth mixture, or molten salts. Nuclear reactors use (Figure \(\Page {8}\)) to control the fission rate of the nuclear fuel by adjusting the number of slow neutrons present to keep the rate of the chain reaction at a safe level. Control rods are made of boron, cadmium, hafnium, or other elements that are able to absorb neutrons. Boron-10, for example, absorbs neutrons by a reaction that produces lithium-7 and alpha particles: \[\ce{^{10}_5B + ^1_0n⟶ ^7_3Li + ^4_2He} \nonumber \] When control rod assemblies are inserted into the fuel element in the reactor core, they absorb a larger fraction of the slow neutrons, thereby slowing the rate of the fission reaction and decreasing the power produced. Conversely, if the control rods are removed, fewer neutrons are absorbed, and the fission rate and energy production increase. In an emergency, the chain reaction can be shut down by fully inserting all of the control rods into the nuclear core between the fuel rods. During its operation, a nuclear reactor produces neutrons and other radiation. Even when shut down, the decay products are radioactive. In addition, an operating reactor is thermally very hot, and high pressures result from the circulation of water or another coolant through it. Thus, a reactor must withstand high temperatures and pressures, and must protect operating personnel from the radiation. Reactors are equipped with a (or shield) that consists of three parts: In addition, reactors are often covered with a steel or concrete dome that is designed to contain any radioactive materials might be released by a reactor accident. Nuclear power plants are designed in such a way that they cannot form a supercritical mass of fissionable material and therefore cannot create a nuclear explosion. But as history has shown, failures of systems and safeguards can cause catastrophic accidents, including chemical explosions and nuclear meltdowns (damage to the reactor core from overheating). The following Chemistry in Everyday Life feature explores three infamous meltdown incidents. The importance of cooling and containment are amply illustrated by three major accidents that occurred with the nuclear reactors at nuclear power generating stations in the United States (Three Mile Island), the former Soviet Union (Chernobyl), and Japan (Fukushima). In March 1979, the cooling system of the Unit 2 reactor at Three Mile Island Nuclear Generating Station in Pennsylvania failed, and the cooling water spilled from the reactor onto the floor of the containment building. After the pumps stopped, the reactors overheated due to the high radioactive decay heat produced in the first few days after the nuclear reactor shut down. The temperature of the core climbed to at least 2200 °C, and the upper portion of the core began to melt. In addition, the zirconium alloy cladding of the fuel rods began to react with steam and produced hydrogen: \[\ce{Zr}(s)+\ce{2H2O}(g)⟶\ce{ZrO2}(s)+\ce{2H2}(g) \nonumber \] The hydrogen accumulated in the confinement building, and it was feared that there was danger of an explosion of the mixture of hydrogen and air in the building. Consequently, hydrogen gas and radioactive gases (primarily krypton and xenon) were vented from the building. Within a week, cooling water circulation was restored and the core began to cool. The plant was closed for nearly 10 years during the cleanup process. Although zero discharge of radioactive material is desirable, the discharge of radioactive krypton and xenon, such as occurred at the Three Mile Island plant, is among the most tolerable. These gases readily disperse in the atmosphere and thus do not produce highly radioactive areas. Moreover, they are noble gases and are not incorporated into plant and animal matter in the food chain. Effectively none of the heavy elements of the core of the reactor were released into the environment, and no cleanup of the area outside of the containment building was necessary (Figure \(\Page {9}\)). Another major nuclear accident involving a reactor occurred in April 1986, at the Chernobyl Nuclear Power Plant in Ukraine, which was still a part of the former Soviet Union. While operating at low power during an unauthorized experiment with some of its safety devices shut off, one of the reactors at the plant became unstable. Its chain reaction became uncontrollable and increased to a level far beyond what the reactor was designed for. The steam pressure in the reactor rose to between 100 and 500 times the full power pressure and ruptured the reactor. Because the reactor was not enclosed in a containment building, a large amount of radioactive material spewed out, and additional fission products were released, as the graphite (carbon) moderator of the core ignited and burned. The fire was controlled, but over 200 plant workers and firefighters developed acute radiation sickness and at least 32 soon died from the effects of the radiation. It is predicted that about 4000 more deaths will occur among emergency workers and former Chernobyl residents from radiation-induced cancer and leukemia. The reactor has since been encapsulated in steel and concrete, a now-decaying structure known as the sarcophagus. Almost 30 years later, significant radiation problems still persist in the area, and Chernobyl largely remains a wasteland. In 2011, the Fukushima Daiichi Nuclear Power Plant in Japan was badly damaged by a 9.0-magnitude earthquake and resulting tsunami. Three reactors up and running at the time were shut down automatically, and emergency generators came online to power electronics and coolant systems. However, the tsunami quickly flooded the emergency generators and cut power to the pumps that circulated coolant water through the reactors. High-temperature steam in the reactors reacted with zirconium alloy to produce hydrogen gas. The gas escaped into the containment building, and the mixture of hydrogen and air exploded. Radioactive material was released from the containment vessels as the result of deliberate venting to reduce the hydrogen pressure, deliberate discharge of coolant water into the sea, and accidental or uncontrolled events. An evacuation zone around the damaged plant extended over 12.4 miles away, and an estimated 200,000 people were evacuated from the area. All 48 of Japan’s nuclear power plants were subsequently shut down, remaining shuttered as of December 2014. Since the disaster, public opinion has shifted from largely favoring to largely opposing increasing the use of nuclear power plants, and a restart of Japan’s atomic energy program is still stalled (Figure \(\Page {10}\)). The energy produced by a reactor fueled with enriched uranium results from the fission of uranium as well as from the fission of plutonium produced as the reactor operates. As discussed previously, the plutonium forms from the combination of neutrons and the uranium in the fuel. In any nuclear reactor, only about 0.1% of the mass of the fuel is converted into energy. The other 99.9% remains in the fuel rods as fission products and unused fuel. All of the fission products absorb neutrons, and after a period of several months to a few years, depending on the reactor, the fission products must be removed by changing the fuel rods. Otherwise, the concentration of these fission products would increase and absorb more neutrons until the reactor could no longer operate. Spent fuel rods contain a variety of products, consisting of unstable nuclei ranging in atomic number from 25 to 60, some transuranium elements, including plutonium and americium, and unreacted uranium isotopes. The unstable nuclei and the transuranium isotopes give the spent fuel a dangerously high level of radioactivity. The long-lived isotopes require thousands of years to decay to a safe level. The ultimate fate of the nuclear reactor as a significant source of energy in the United States probably rests on whether or not a politically and scientifically satisfactory technique for processing and storing the components of spent fuel rods can be developed. The process of converting very light nuclei into heavier nuclei is also accompanied by the conversion of mass into large amounts of energy, a process called . The principal source of energy in the sun is a net fusion reaction in which four hydrogen nuclei fuse and produce one helium nucleus and two positrons. This is a net reaction of a more complicated series of events: \[\ce{4^1_1H ⟶ ^4_2He + 2^0_{+1}} \nonumber \] A helium nucleus has a mass that is 0.7% less than that of four hydrogen nuclei; this lost mass is converted into energy during the fusion. This reaction produces about 3.6 × 10 kJ of energy per mole of \(\ce{^4_2He}\) produced. This is somewhat larger than the energy produced by the nuclear fission of one mole of U-235 (1.8 × 10 kJ), and over 3 million times larger than the energy produced by the (chemical) combustion of one mole of octane (5471 kJ). It has been determined that the nuclei of the heavy isotopes of hydrogen, a deuteron, \(^2_1\) and a triton, \(^3_1\), undergo fusion at extremely high temperatures (thermonuclear fusion). They form a helium nucleus and a neutron: \[\ce{^2_1H + ^3_1H ⟶ ^4_2He + 2^1_0n} \nonumber \] This change proceeds with a mass loss of 0.0188 amu, corresponding to the release of 1.69 × 10 kilojoules per mole of \(\ce{^4_2He}\) formed. The very high temperature is necessary to give the nuclei enough kinetic energy to overcome the very strong repulsive forces resulting from the positive charges on their nuclei so they can collide. Useful fusion reactions require very high temperatures for their initiation—about 15,000,000 K or more. At these temperatures, all molecules dissociate into atoms, and the atoms ionize, forming plasma. These conditions occur in an extremely large number of locations throughout the universe—stars are powered by fusion. Humans have already figured out how to create temperatures high enough to achieve fusion on a large scale in thermonuclear weapons. A thermonuclear weapon such as a hydrogen bomb contains a nuclear fission bomb that, when exploded, gives off enough energy to produce the extremely high temperatures necessary for fusion to occur. Another much more beneficial way to create fusion reactions is in a , a nuclear reactor in which fusion reactions of light nuclei are controlled. Because no solid materials are stable at such high temperatures, mechanical devices cannot contain the plasma in which fusion reactions occur. Two techniques to contain plasma at the density and temperature necessary for a fusion reaction are currently the focus of intensive research efforts: containment by a magnetic field and by the use of focused laser beams (Figure \(\Page {11}\)). A number of large projects are working to attain one of the biggest goals in science: getting hydrogen fuel to ignite and produce more energy than the amount supplied to achieve the extremely high temperatures and pressures that are required for fusion. At the time of this writing, there are no self-sustaining fusion reactors operating in the world, although small-scale controlled fusion reactions have been run for very brief periods. It is possible to produce new atoms by bombarding other atoms with nuclei or high-speed particles. The products of these transmutation reactions can be stable or radioactive. A number of artificial elements, including technetium, astatine, and the transuranium elements, have been produced in this way. Nuclear power as well as nuclear weapon detonations can be generated through fission (reactions in which a heavy nucleus is split into two or more lighter nuclei and several neutrons). Because the neutrons may induce additional fission reactions when they combine with other heavy nuclei, a chain reaction can result. Useful power is obtained if the fission process is carried out in a nuclear reactor. The conversion of light nuclei into heavier nuclei (fusion) also produces energy. At present, this energy has not been contained adequately and is too expensive to be feasible for commercial energy production. | 25,774 | 4,214 |
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A burning match and a bonfire may have the same temperature, yet you would not sit around a burning match on a fall evening to stay warm. Why not? The temperature of 1 gram of burning wood is approximately the same for both a match and a bonfire. This is an intensive property and depends on the material (wood). However, the overall amount of produced heat depends on the amount of material; this is an extensive property. The amount of wood in a bonfire is much greater than that in a match; the total amount of produced heat is also much greater, which is why we can sit around a bonfire to stay warm, but a match would not provide enough heat to keep us from getting cold. Prepare a table identifying several energy transitions that take place during the typical operation of an automobile. Explain the difference between heat capacity and specific heat of a substance. Heat capacity refers to the heat required to raise the temperature of the mass of the substance 1 degree; specific heat refers to the heat required to raise the temperature of 1 gram of the substance 1 degree. Thus, heat capacity is an extensive property, and specific heat is an intensive one. Calculate the heat capacity, in joules and in calories per degree, of the following: Calculate the heat capacity, in joules and in calories per degree, of the following: How much heat, in joules and in calories, must be added to a 75.0–g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C? \(q=mCΔ°T\) \(q=(75.0g)\times(\dfrac{0.449\:J}{g\:°C})\times(1,510°K) = 50,800J\) 50,800J ; 12,200cal How much heat, in joules and in calories, is required to heat a 28.4-g (1-oz) ice cube from −23.0 °C to −1.0 °C? 1310 J; 313 cal How much would the temperature of 275 g of water increase if 36.5 kJ of heat were added? ΔT = 31.7° C If 14.5 kJ of heat were added to 485 g of liquid water, how much would its temperature increase? 7.15 °C A piece of unknown substance weighs 44.7 g and requires 2110 J to increase its temperature from 23.2 °C to 89.6 °C. a.) Solve for the specific heat \(C\) and compare the values with the chart \(q=mCΔ°T\) \(2110J=(44.7\:g)(C)(66.4°C)\) \(C=\dfrac{2110\:J}{2970\:g\:°C}\) \(C=\dfrac{0.711\:J}{g\:°C}\) b.) Silicon A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C. An aluminum kettle weighs 1.05 kg. Most people find waterbeds uncomfortable unless the water temperature is maintained at about 85 °F. Unless it is heated, a waterbed that contains 892 L of water cools from 85 °F to 72 °F in 24 hours. Estimate the amount of electrical energy required over 24 hours, in kWh, to keep the bed from cooling. Note that 1 kilowatt-hour (kWh) = 3.6 × 10 J, and assume that the density of water is 1.0 g/mL (independent of temperature). What other assumptions did you make? How did they affect your calculated result (i.e., were they likely to yield “positive” or “negative” errors)? We assume that the density of water is 1.0 g/cm (1 g/mL) and that it takes as much energy to keep the water at 85 °F as to heat it from 72 °F to 85 °F. We also assume that only the water is going to be heated. Energy required = 7.47 kWh A 500-mL bottle of water at room temperature and a 2-L bottle of water at the same temperature were placed in a refrigerator. After 30 minutes, the 500-mL bottle of water had cooled to the temperature of the refrigerator. An hour later, the 2-L of water had cooled to the same temperature. When asked which sample of water lost the most heat, one student replied that both bottles lost the same amount of heat because they started at the same temperature and finished at the same temperature. A second student thought that the 2-L bottle of water lost more heat because there was more water. A third student believed that the 500-mL bottle of water lost more heat because it cooled more quickly. A fourth student thought that it was not possible to tell because we do not know the initial temperature and the final temperature of the water. Indicate which of these answers is correct and describe the error in each of the other answers. Would the amount of heat measured for the reaction in be greater, lesser, or remain the same if we used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer. lesser; more heat would be lost to the coffee cup and the environment and so Δ for the water would be lesser and the calculated would be lesser Would the amount of heat absorbed by the dissolution in appear greater, lesser, or remain the same if the experimenter used a calorimeter that was a poorer insulator than a coffee cup calorimeter? Explain your answer. Would the amount of heat absorbed by the dissolution in appear greater, lesser, or remain the same if the heat capacity of the calorimeter were taken into account? Explain your answer. greater, since taking the calorimeter’s heat capacity into account will compensate for the thermal energy transferred to the solution from the calorimeter; this approach includes the calorimeter itself, along with the solution, as “surroundings”: = −( + ); since both and are negative, including the latter term ( ) will yield a greater value for the heat of the dissolution How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat. How much will the temperature of a cup (180 g) of coffee at 95 °C be reduced when a 45 g silver spoon (specific heat 0.24 J/g °C) at 25 °C is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water. The temperature of the coffee will drop 1 degree. A 45-g aluminum spoon (specific heat 0.88 J/g °C) at 24 °C is placed in 180 mL (180 g) of coffee at 85 °C and the temperature of the two become equal. The temperature of the cooling water as it leaves the hot engine of an automobile is 240 °F. After it passes through the radiator it has a temperature of 175 °F. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of 4.184 J/g °C. \(5.7 \times 10^2\; kJ\) A 70.0-g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter like that shown in . The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water? What is the specific heat of the metal? If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5 °C in a calorimeter like that in , what is the resulting temperature of the water? 38.5 °C A 0.500-g sample of KCl is added to 50.0 g of water in a calorimeter ( ). If the temperature decreases by 1.05 °C, what is the approximate amount of heat involved in the dissolution of the KCl, assuming the heat capacity of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? Dissolving 3.0 g of CaCl ( ) in 150.0 g of water in a calorimeter ( ) at 22.4 °C causes the temperature to rise to 25.8 °C. What is the approximate amount of heat involved in the dissolution, assuming the heat capacity of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic? 2.2 kJ; The heat produced shows that the reaction is exothermic. When 50.0 g of 0.200 M NaCl( ) at 24.1 °C is added to 100.0 g of 0.100 M AgNO ( ) at 24.1 °C in a calorimeter, the temperature increases to 25.2 °C as AgCl( ) forms. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat in joules produced. The addition of 3.15 g of Ba(OH) •8H O to a solution of 1.52 g of NH SCN in 100 g of water in a calorimeter caused the temperature to fall by 3.1 °C. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat absorbed by the reaction, which can be represented by the following equation: \[Ba(OH)_2 \cdot 8H_2O_{(s)} + 2NH_4SCN_{(aq)} \rightarrow Ba(SCN)_{2(aq)} + 2NH_{3(aq)} + 10H_2O_{(l)}\] 1.4 kJ The reaction of 50 mL of acid and 50 mL of base described in increased the temperature of the solution by 6.9 degrees. How much would the temperature have increased if 100 mL of acid and 100 mL of base had been used in the same calorimeter starting at the same temperature of 22.0 °C? Explain your answer. If the 3.21 g of NH NO in were dissolved in 100.0 g of water under the same conditions, how much would the temperature change? Explain your answer. 22.6. Since the mass and the heat capacity of the solution is approximately equal to that of the water, the two-fold increase in the amount of water leads to a two-fold decrease of the temperature change. When 1.0 g of fructose, C H O ( ), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 °C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/°C, what is for this combustion? When a 0.740-g sample of trinitrotoluene (TNT), C H N O , is burned in a bomb calorimeter, the temperature increases from 23.4 °C to 26.9 °C. The heat capacity of the calorimeter is 534 J/°C, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample? 11.7 kJ One method of generating electricity is by burning coal to heat water, which produces steam that drives an electric generator. To determine the rate at which coal is to be fed into the burner in this type of plant, the heat of combustion per ton of coal must be determined using a bomb calorimeter. When 1.00 g of coal is burned in a bomb calorimeter, the temperature increases by 1.48 °C. If the heat capacity of the calorimeter is 21.6 kJ/°C, determine the heat produced by combustion of a ton of coal (2.000 × 10 pounds). The amount of fat recommended for someone with a daily diet of 2000 Calories is 65 g. What percent of the calories in this diet would be supplied by this amount of fat if the average number of Calories for fat is 9.1 Calories/g? 30% A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories (16 kcal). What is the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g? What is the maximum mass of carbohydrate in a 6-oz serving of diet soda that contains less than 1 Calorie per can if the average number of Calories for carbohydrates is 4.1 Calories/g? 0.24 g A pint of premium ice cream can contain 1100 Calories. What mass of fat, in grams and pounds, must be produced in the body to store an extra 1.1 × 10 Calories if the average number of Calories for fat is 9.1 Calories/g? A serving of a breakfast cereal contains 3 g of protein, 18 g of carbohydrates, and 6 g of fat. What is the Calorie content of a serving of this cereal if the average number of Calories for fat is 9.1 Calories/g, for carbohydrates is 4.1 Calories/g, and for protein is 4.1 Calories/g? 1.4 × 10 Calories Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs $4.23, or a liter of isooctane (density, 0.6919 g/mL) that costs $0.45? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories. Explain how the heat measured in differs from the enthalpy change for the exothermic reaction described by the following equation: \[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l)\] The enthalpy change of the indicated reaction is for exactly 1 mol HCL and 1 mol NaOH; the heat in the example is produced by 0.0500 mol HCl and 0.0500 mol NaOH. Using the data in the check your learning section of , calculate Δ in kJ/mol of AgNO ( ) for the reaction: \[\ce{NaCl}(aq)+\ce{AgNO3}(aq)⟶\ce{AgCl}(s)+\ce{NaNO3}(aq)\] Calculate the enthalpy of solution (Δ for the dissolution) per mole of NH NO under the conditions described in . 25 kJ mol Calculate Δ for the reaction described by the equation. \(\ce{Ba(OH)2⋅8H2O}(s)+\ce{2NH4SCN}(aq)⟶\ce{Ba(SCN)2}(aq)+\ce{2NH3}(aq)+\ce{10H2O}(l)\) Calculate the enthalpy of solution (Δ for the dissolution) per mole of CaCl . 81 kJ mol Although the gas used in an oxyacetylene torch is essentially pure acetylene, the heat produced by combustion of one mole of acetylene in such a torch is likely not equal to the enthalpy of combustion of acetylene listed in . Considering the conditions for which the tabulated data are reported, suggest an explanation. How much heat is produced by burning 4.00 moles of acetylene under standard state conditions? 5204.4 kJ How much heat is produced by combustion of 125 g of methanol under standard state conditions? How many moles of isooctane must be burned to produce 100 kJ of heat under standard state conditions? 1.83 × 10 mol What mass of carbon monoxide must be burned to produce 175 kJ of heat under standard state conditions? When 2.50 g of methane burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion per mole of methane under these conditions? 802 kJ mol How much heat is produced when 100 mL of 0.250 M HCl (density, 1.00 g/mL) and 200 mL of 0.150 M NaOH (density, 1.00 g/mL) are mixed? \[\ce{HCl}(aq)+\ce{NaOH}(aq)⟶\ce{NaCl}(aq)+\ce{H2O}(l)\hspace{20px}ΔH^\circ_{298}=\mathrm{−58\:kJ}\] If both solutions are at the same temperature and the heat capacity of the products is 4.19 J/g °C, how much will the temperature increase? What assumption did you make in your calculation? A sample of 0.562 g of carbon is burned in oxygen in a bomb calorimeter, producing carbon dioxide. Assume both the reactants and products are under standard state conditions, and that the heat released is directly proportional to the enthalpy of combustion of graphite. The temperature of the calorimeter increases from 26.74 °C to 27.93 °C. What is the heat capacity of the calorimeter and its contents? 15.5 kJ/ºC Before the introduction of chlorofluorocarbons, sulfur dioxide (enthalpy of vaporization, 6.00 kcal/mol) was used in household refrigerators. What mass of SO must be evaporated to remove as much heat as evaporation of 1.00 kg of CCl F (enthalpy of vaporization is 17.4 kJ/mol)? The vaporization reactions for SO and CCl F are \(\ce{SO2}(l)⟶\ce{SO2}(g)\) and \(\ce{CCl2F}(l)⟶\ce{CCl2F2}(g)\), respectively. Homes may be heated by pumping hot water through radiators. What mass of water will provide the same amount of heat when cooled from 95.0 to 35.0 °C, as the heat provided when 100 g of steam is cooled from 110 °C to 100 °C. 7.43 g Which of the enthalpies of combustion in the table are also standard enthalpies of formation? Does the standard enthalpy of formation of H O( ) differ from Δ ° for the reaction \(\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(g)\)? No. Joseph Priestly prepared oxygen in 1774 by heating red mercury(II) oxide with sunlight focused through a lens. How much heat is required to decompose exactly 1 mole of red HgO( ) to Hg( ) and O ( ) under standard conditions? How many kilojoules of heat will be released when exactly 1 mole of manganese, Mn, is burned to form Mn O ( ) at standard state conditions? 459.6 kJ How many kilojoules of heat will be released when exactly 1 mole of iron, Fe, is burned to form Fe O ( ) at standard state conditions? The following sequence of reactions occurs in the commercial production of aqueous nitric acid: \(\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(l)\hspace{20px}ΔH=\mathrm{−907\:kJ}\) \(\ce{2NO}(g)+\ce{O2}(g)⟶\ce{2NO2}(g)\hspace{20px}ΔH=\mathrm{−113\:kJ}\) \(\ce{3NO2}+\ce{H2O}(l)⟶\ce{2HNO2}(aq)+\ce{NO}(g)\hspace{20px}ΔH=\mathrm{−139\:kJ}\) Determine the total energy change for the production of one mole of aqueous nitric acid by this process. 495 kJ/mol Both graphite and diamond burn. \(\ce{C}(s,\:\ce{diamond})+\ce{O2}(g)⟶\ce{CO2}(g)\) For the conversion of graphite to diamond: \(\ce{C}(s,\:\ce{graphite})⟶\ce{C}(s,\:\ce{diamond})\hspace{20px}ΔH^\circ_{298}=\mathrm{1.90\:kJ}\) Which produces more heat, the combustion of graphite or the combustion of diamond? From the molar heats of formation in , determine how much heat is required to evaporate one mole of water: \(\ce{H2O}(l)⟶\ce{H2O}(g)\) 44.01 kJ/mol Which produces more heat? \(\ce{Os}(s)⟶\ce{2O2}(g)⟶\ce{OsO4}(s)\) or \(\ce{Os}(s)⟶\ce{2O2}(g)⟶\ce{OsO4}(g)\) for the phase change \(\ce{OsO4}(s)⟶\ce{OsO4}(g)\hspace{20px}ΔH=\mathrm{56.4\:kJ}\) Calculate \(ΔH^\circ_{298}\) for the process \(\ce{Sb}(s)+\dfrac{5}{2}\ce{Cl2}(g)⟶\ce{SbCl5}(g)\) from the following information: \(\ce{Sb}(s)+\dfrac{3}{2}\ce{Cl2}(g)⟶\ce{SbCl3}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−314\:kJ}\) \(\ce{SbCl3}(s)+\ce{Cl2}(g)⟶\ce{SbCl5}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−80\:kJ}\) 394 kJ Calculate \(ΔH^\circ_{298}\) for the process \(\ce{Zn}(s)+\ce{S}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)\) from the following information: \(\ce{Zn}(s)+\ce{S}(s)⟶\ce{ZnS}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−206.0\:kJ}\) \(\ce{ZnS}(s)+\ce{2O2}(g)⟶\ce{ZnSO4}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−776.8\:kJ}\) Calculate Δ for the process \(\ce{Hg2Cl2}(s)⟶\ce{2Hg}(l)+\ce{Cl2}(g)\) from the following information: \(\ce{Hg}(l)+\ce{Cl2}(g)⟶\ce{HgCl2}(s)\hspace{20px}ΔH=\mathrm{−224\:kJ}\) \(\ce{Hg}(l)+\ce{HgCl2}(s)⟶\ce{Hg2Cl2}(s)\hspace{20px}ΔH=\mathrm{−41.2\:kJ}\) 265 kJ Calculate \(ΔH^\circ_{298}\) for the process \(\ce{Co3O4}(s)⟶\ce{3Co}(s)+\ce{2O2}(g)\) from the following information: \(\ce{Co}(s)+\dfrac{1}{2}\ce{O2}(g)⟶\ce{CoO}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−237.9\:kJ}\) \(\ce{3Co}(s)+\ce{O2}(g)⟶\ce{Co3O4}(s)\hspace{20px}ΔH^\circ_{298}=\mathrm{−177.5\:kJ}\) Calculate the standard molar enthalpy of formation of NO( ) from the following data: \(\ce{N2}(g)+\ce{2O2}⟶\ce{2NO2}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{66.4\:kJ}\) \(\ce{2NO}(g)+\ce{O2}⟶\ce{2NO2}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−114.1\:kJ}\) 90.3 mol of NO Using the data in , calculate the standard enthalpy change for each of the following reactions: Using the data in , calculate the standard enthalpy change for each of the following reactions: The following reactions can be used to prepare samples of metals. Determine the enthalpy change under standard state conditions for each. The decomposition of hydrogen peroxide, H O , has been used to provide thrust in the control jets of various space vehicles. Using the data in , determine how much heat is produced by the decomposition of exactly 1 mole of H O under standard conditions. \(\ce{2H2O2}(l)⟶\ce{2H2O}(g)+\ce{O2}(g)\) −54.04 kJ mol Calculate the enthalpy of combustion of propane, C H ( ), for the formation of H O( ) and CO ( ). The enthalpy of formation of propane is −104 kJ/mol. Calculate the enthalpy of combustion of butane, C H ( ) for the formation of H O( ) and CO ( ). The enthalpy of formation of butane is −126 kJ/mol. 2660 kJ mol Both propane and butane are used as gaseous fuels. Which compound produces more heat per gram when burned? The white pigment TiO is prepared by the reaction of titanium tetrachloride, TiCl , with water vapor in the gas phase: \(\ce{TiCl4}(g)+\ce{2H2O}(g)⟶\ce{TiO2}(s)+\ce{4HCl}(g)\). How much heat is evolved in the production of exactly 1 mole of TiO ( ) under standard state conditions? 67.1 kJ Water gas, a mixture of H and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon: \(\ce{C}(s)+\ce{H2O}(g)⟶\ce{CO}(g)+\ce{H2}(g)\). In the early days of automobiles, illumination at night was provided by burning acetylene, C H . Though no longer used as auto headlamps, acetylene is still used as a source of light by some cave explorers. The acetylene is (was) prepared in the lamp by the reaction of water with calcium carbide, CaC : \(\ce{CaC2}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s)+\ce{C2H2}(g)\). Calculate the standard enthalpy of the reaction. The \(ΔH^\circ_\ce{f}\) of CaC is −15.14 kcal/mol. 122.8 kJ From the data in , determine which of the following fuels produces the greatest amount of heat per gram when burned under standard conditions: CO( ), CH ( ), or C H ( ). The enthalpy of combustion of hard coal averages −35 kJ/g, that of gasoline, 1.28 × 10 kJ/gal. How many kilograms of hard coal provide the same amount of heat as is available from 1.0 gallon of gasoline? Assume that the density of gasoline is 0.692 g/mL (the same as the density of isooctane). 3.7 kg Ethanol, C H OH, is used as a fuel for motor vehicles, particularly in Brazil. Among the substances that react with oxygen and that have been considered as potential rocket fuels are diborane [B H , produces B O ( ) and H O( )], methane [CH , produces CO ( ) and H O( )], and hydrazine [N H , produces N ( ) and H O( )]. On the basis of the heat released by 1.00 g of each substance in its reaction with oxygen, which of these compounds offers the best possibility as a rocket fuel? The \(ΔH^\circ_\ce{f}\) of B H ( ), CH ( ), and N H ( ) may be found in . On the assumption that the best rocket fuel is the one that gives off the most heat, B H is the prime candidate. How much heat is produced when 1.25 g of chromium metal reacts with oxygen gas under standard conditions? Ethylene, C H , a byproduct from the fractional distillation of petroleum, is fourth among the 50 chemical compounds produced commercially in the largest quantities. About 80% of synthetic ethanol is manufactured from ethylene by its reaction with water in the presence of a suitable catalyst. \(\ce{C2H4}(g)+\ce{H2O}(g)⟶\ce{C2H5OH}(l)\) Using the data in the table in , calculate Δ ° for the reaction. 88.2 kJ The oxidation of the sugar glucose, C H O , is described by the following equation: \(\ce{C6H12O6}(s)+\ce{6O2}(g)⟶\ce{6CO2}(g)+\ce{6H2O}(l)\hspace{20px}ΔH=\mathrm{−2816\:kJ}\) The metabolism of glucose gives the same products, although the glucose reacts with oxygen in a series of steps in the body. Propane, C H , is a hydrocarbon that is commonly used as a fuel. During a recent winter month in Sheboygan, Wisconsin, it was necessary to obtain 3500 kWh of heat provided by a natural gas furnace with 89% efficiency to keep a small house warm (the efficiency of a gas furnace is the percent of the heat produced by combustion that is transferred into the house). | 22,520 | 4,215 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Lipids/Non-glyceride_Lipids/Sphingolipids |
Sphingolipids are named after the sphinx in Greek mythology, part woman and part lion, who devoured all who could not answer her riddles. Spingolipids appeared to Johann Thudichum in 1874 as part of the dangerous riddle of the brain. Sphingolipids are a second type of lipid found in cell membranes, particularly nerve cells and brain tissues. They do not contain glycerol, but retain the two alcohols with the middle position occupied by an amine. Sphingosine has three parts, a three carbon chain with two alcohols and amine attached and a long hydrocarbon chain. In sphingomyelin, the base sphingosine has several other groups attached as shown in the graphic on the left. A fatty acid is attached to the amine through amide bond. Phosphate is attached through a phosphate ester bond, and again through a phosphate ester bond to choline. The human brain and spinal cord is made up of gray and white regions. The white region is made of nerve axons wrapped in a white lipid coating, the myelin sheath, which provides insulation to allow rapid conduction of electrical signals. Multiple sclerosis caused by a gradual degradation of the myelin sheath. Sphingomyleins are located throughout the body in nerve cell membranes. They make up about 25 % of the lipids in the myelin sheath that surrounds and insulates cells of the central nervous system. Niemann-Pick disease is caused by a deficiency of an enzyme that breaks down excessive sphingomyelin, which then builds up on the liver, spleen, brain, and bone marrow. An effected child usually dies within several years. Glycolipids are complex lipids that contain carbohydrates. Cerebrosides are an example which contain the sphingosine backbone attached to a fatty acid and a carbohydrate. The carbohydrates are most often glucose or galactose. Those that contain several carbohydrates are called gangliosides. The example on the left is shown with glucose. Glucocerebroside has the specific function to be in the cell membranes of macrophages, (cells that protect the body by destroying foreign microorganisms. Galactocerebroside is found almost exclusively in the membranes of brain cells. There are several genetic diseases resulting from the absence of specific enzymes which breakdown the glycolipids. Tay-Sachs, which mainly effects Jewish children, results in a build up of gangliosides and result in death in several years. Gaucher's disease results in the excessive build up of glucocerebroside resulting in severe anemia and enlarged liver and spleen. | 2,535 | 4,216 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.06%3A_Polar_Lipids |
As was true of most nonpolar lipids, the structures of polar lipids are based on condensation of fatty acids with glycerol. The main difference is that only two of the three OH groups on glycerol are involved. The third is combined with a highly polar molecule: In one sense the polar lipids are like the anions of fatty acids, only more so. They contain two hydrophobic hydrocarbon tails and a head which may have several electrically charged sites. As in the case of soap and detergent molecules, the tails of polar lipids tend to avoid water and other polar substances, but the heads are quite compatible with such environments. The polar lipids are most commonly found as components of cell walls and other membranes. Nearly all hypotheses regarding membrane structure take as a fundamental component a lipid bilayer (Figure \(\Page {2}\) ). Bilayers made in the laboratory have many properties in common with membranes. Ions such as Na , K , and Cl cannot pass through them, but water molecules can. The hydrocarbon core of such a bilayer should have large electrical resistance, as does a membrane. Certain carrier molecules can transport K and other ions across a bilayer, apparently by wrapping a hydrophobic cloak around them to disguise their charges. Membrane proteins in a bilayer also allow for transport of ions and other molecules across the bilayer which could not cross otherwise. | 1,414 | 4,217 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/04%3A_The_Structure_of_Atoms/4.03%3A_The_Periodic_Table |
The similarities among macroscopic properties within each of the chemical families just described lead one to expect microscopic similarities as well. Atoms of sodium ought to be similar in some way to atoms of lithium, potassium, and the other alkali metals. This could account for the related chemical reactivities and analogous compounds of these elements. According to Dalton’s atomic theory, different kinds of atoms may be distinguished by their relative masses (atomic weights). Therefore it seems reasonable to expect some correlation between this microscopic property and macroscopic chemical behavior. You can see that such a relationship exists by listing symbols for the first dozen elements in order of increasing relative mass. Obtaining , we have Elements which belong to families we have already discussed are indicated by shading around their symbols. The second, third, and forth elements on the list (He, Li, and Be) are a noble gas, an alkali metal, and an alkaline-earth metal, respectively. Exactly the same sequence is repeated eight elements later (Ne, Na, and Mg), but this time a halogen (F) precedes the noble gas. If a list were made of all elements, we would find the sequence halogen, noble gas, alkali metal, and alkaline-earth metal several more times. In 1871 the Russian chemist Dmitri Ivanovich Mendeleev (1834 to 1907) proposed the . This law states that . That is, similar elements do not have similar atomic weights. Rather, as we go down a list of elements in order of atomic weights, corresponding properties are observed at regular intervals. To emphasize this periodic repetition of similar properties, Mendeleev arranged the symbols and atomic weights of the elements in the table shown below. Each vertical column of this contains a or of related elements. The alkali metals are in group I ( I), alkaline earths in group II, chalcogens in group VI, and halogens in group VII. Mendeleev was not quite sure where to put the coinage metals, and so they appear twice. Each time, however, copper, silver, and gold are arranged in a vertical column. The noble gases were discovered nearly a quarter century after Mendeleev’s first periodic table was published, but they, too, fit the periodic arrangement. In constructing his table, Mendeleev found that sometimes there were not enough elements to fill all the available spaces in each horizontal row or . When this was true, he assumed that eventually someone would discover the element or elements needed to complete a period. Mendeleev therefore left blank spaces for undiscovered elements and predicted their properties by averaging the characteristics of other elements in the same group. As an example of this process, look at the fourth numbered row ( ). Scandium (Sc) was unknown in 1872; so titanium (Ti) followed calcium (Ca) in order of atomic weights. This would have placed titanium below boron (B) in group III, but Mendeleev knew that the most common oxide of titanium, TiO , had a formula similar to an oxide of carbon CO , rather than of boron, B O . Therefore he placed titanium below carbon in group IV. He proposed that an undiscovered element, ekaboron, would eventually be found to fit below boron. (The prefix means “below.”) Properties predicted for ekaboron are shown in the following table. They agreed remarkably with those measured experimentally for scandium when it was discovered 7 years later. This agreement was convincing evidence that a periodic table is a good way to summarize a great many macroscopic, experimental facts. \(\Page {1}\) Comparison of Mendeleev’s Predictions with the Observed Properties of the Element Scandium. * Mendeleev used the name ''eka''boron because the blank space into which the element should fit was ''below'' boron in his periodic table. † The modern value of the atomic weight of scandium is 44.96. The modern periodic table differs in some ways from Mendeleev’s original version. It contains more than 40 additional elements, and its rows are longer instead of being squeezed under one another in staggered columns. For example, Mendeleev’s fourth and fifth rows are both contained in the fourth period of the modern table. This ends up placing gallium, not scandium underneath boron in the periodic table. This rearrangement is due to theory on the electronic structure of atoms, in particular ideas about and the . The extremely important idea of vertical groups of related elements is still retained, as are Mendeleev’s group numbers. The latter appear as roman numerals at the top of each column in the modern table. To end on a very practical note, there are a variety of sites with interactive periodic tables listed below. These tables can be helpful for understanding and learning cool facts about the elements themselves. | 4,819 | 4,218 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Solids/Solids |
This pattern or is generated by using a unit marked by the centers of any four @ or # signs. The choice is up to you in this case. Such a unit is called a primitive unit. The pattern has a square (or rectangular on some screens) appearance, and to preserve the square, we may use a square unit of as our . Such choices result in having two @ and # per unit cell, and these are called . Thus, if we know the arrangement of a unit cell, we can use our imagination to build a crystal structure, or use symbols or models to represent a crystal structure. Since each pattern has features shared by many structures, often such a pattern is called a . For example, the diamond, zinc blende, wurtzite, and \(\ce{NaCl}\) structures have been called lattices; however, the word lattice has a more formal definition by crystal physics and chemists. The above site gives a gallery of lattices. Inorganic Chemistry by Swaddle defined crystals as packed regular arrays of atoms, ions, or molecules in a pattern repeated periodically . | 1,047 | 4,219 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/General_Chemistry_Supplement_(Eames)/Thermochemistry/Hess'_Law_and_Enthalpy_of_Formation |
is an early statement of the law of conservation of energy (1840). It says that the heat liberated by a process doesn't depend on how the process happens (only on the starting and ending states: in other words, it's a state function). Now we know we should really use enthalpy for this, not heat, because enthalpy is a state function, so this is true, while heat is a process. Hess' Law lets us break a reaction or process into a series of small, easily measured steps, and then we can add up the ΔH of the steps to find the change in enthalpy of the whole thing. Hess' law is a great way to think about chemical processes and make predictions. We'll see lots of applications of Hess' law, but right now let's start with finding reaction enthalpies using standard enthalpies of formation. help us predict reaction enthalpies for many reactions if the products and reactants are well-studied, even if the specific reaction is new. To do this, we imagine that we take the reactants and separate them into their pure elements in a . The standard state is the element in its most stable form at room temperature and atmospheric pressure. Then we take the elements and recombine them to make the products. The reaction enthalpy is equal to the difference in the enthalpies of these processes. Let's look at a specific example. Here are some enthalpies of formation (in kJ/mol of reaction): \[C(s) + O_{2}(g) \rightarrow CO_{2}(g)\; \Delta H_{f} = -394\; kJ$$
$$C(s) + 2H_{2}(g) \rightarrow CH_{4}(g)\; \Delta H_{f} = -75\; kJ$$
$$2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)\; \Delta H_{f} = -572\; kJ\] Let's use these enthalpies of formation to calculate the enthalpy of combustion for 1 mol of methane. The reaction we want is
$$CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)$$
If we reverse a reaction, we change the sign on ΔH, and if we multiply the reaction by a constant coefficient, we multiply ΔH by the same coefficient. Let's combine the formation constant equations so they add up to the reaction we want: That's almost right but we're missing the state of the water: \[H_{2}O(l)\rightarrow H_{2}O(g)\; \Delta H=44\; kJ\] The full set is Note that everything but the desired reaction cancels. Now we just add up the enthalpies of each step, and we find that the enthalpy of combustion of 1 mole of methane is -803 kJ. Determine the equation for the desired process (the process for which you want to know the enthalpy change). Break it into steps for which you can look up the enthalpy changes. This probably means steps like formation from elements, and changes of state. (Later, we'll include other processes like ionization, etc) Arrange the steps so that everything cancels out leaving just the desired reaction. Make sure the coefficients on equations are correct (multiply the equation and ΔH by a constant if needed) and that all the components are in the correct state (like the example above, we had to convert from liquid water to gaseous water). Then just add it up! | 3,022 | 4,222 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.05%3A_Refining_of_Copper |
Unrefined or “blister” copper is about 99 percent pure when obtained from the ore, but it is desirable to increase this to 99.95 percent if the copper is to be used in electrical wiring. Even small concentrations of impurities noticeably lower copper’s electrical conductivity. Such a high degree of purity can be obtained by in a cell similar to that shown in Figure \(\Page {1}\). In such a cell a thin sheet of high-purity Cu serves as the cathode, and the anode is the impure Cu which is to be refined. The electrolyte is a solution of copper(II) sulfate. Some of the impurities are metals such as Fe and Zn which are more easily oxidized than Cu. When current passes through the cell, these impurities go into solution from the anode, along with Cu: \[\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + \text{2}e^{-} \nonumber \] \[\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + \text{2}e^{-} \nonumber \] \[\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + \text{2}e^{-} \nonumber \] These ions all migrate toward the cathode, but Cu ( ) is more readily reduced than Fe ( ) or Zn ( ) and so it is the only one that plates out. The impurity ions remain in solution. Other impurities, such as Ag, Au, and Pt, are less easily oxidized than Cu. These remain in metallic form and fall to the bottom of the cell, forming “anode sludge” from which they can later be recovered. The great value of Ag, Au, and Pt helps to offset the cost of refining. | 1,452 | 4,223 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.01%3A_Compressibility |
When we pack to go on vacation, there seems to always be "one more" thing that we need to get in our suitcase. Maybe it's another bathing suit, pair of shoes, book—whatever the item, we need to get it in. Fortunately, we can usually squeeze things together somehow. Perhaps there is a little space between folds of clothing, or we can rearrange the shoes; somehow we can get that last item in and close the suitcase. Scuba diving is a form of underwater diving in which a diver carries his own breathing gas, usually in the form of a tank of compressed air. The pressure in most commonly used scuba tanks ranges from 200 to 300 atmospheres. Gases are unlike other states of matter in that a gas expands to fill the shape and volume of its container. For this reason, gases can also be compressed so that a relatively large amount of gas can be forced into a small container. If the air in a typical scuba tank were transferred to a container at the standard pressure of \(1 \: \text{atm}\), the volume of that container would need to be about 2500 liters. is the measure of how much a given volume of matter decreases when placed under pressure. If we put pressure on a solid or a liquid, there is essentially no change in volume. The atoms, ions, or molecules that make up the solid or liquid are very close together. There is no space between the individual particles, so they cannot pack together. The kinetic-molecular theory explains why gases are more compressible than either liquids or solids. Gases are compressible because most of the volume of a gas is composed of the large amounts of empty space between the gas particles. At room temperature and standard pressure, the average distance between gas molecules is about ten times the diameter of the molecules themselves. When a gas is compressed, as when the scuba tank is being filled, the gas particles are forced closer together. Compressed gases are used in many situations. In hospitals, oxygen is often used for patients who have damaged lungs to help them breathe better. If a patient is having a major operation, the anesthesia that is administered will frequently be a compressed gas. Welding requires very hot flames produced by compressed acetylene and oxygen mixtures. Many summer barbeque grills are fueled by compressed propane. | 2,326 | 4,227 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/16%3A_Solutions/16.01%3A_Solute-Solvent_Combinations |
The focus of "Chapter 15: Water" was water's role in the formation of aqueous solutions. We examined the primary characteristics of a solution and how water is able to dissolve solid solutes; we differentiated between a solution, a suspension, and a colloid. There are many examples of solutions that do not involve water at all, or that involve solutes that are not solids. The table below summarizes the possible combinations of solute-solvent states, along with examples of each. Our air is a homogenous mixture of many different gases and therefore qualifies as a solution. Approximately \(78\%\) of the atmosphere is nitrogen, making it the solvent for this solution. The next major constituent is oxygen (about \(21\%\)), followed by the inert gas argon \(\left( 0.9\% \right)\), carbon dioxide \(\left( 0.03\% \right)\), and trace amounts of neon, methane, helium, and other gases. Solid-solid solutions such as brass, bronze, and sterling silver are called alloys. Bronze (composed mainly of copper with added tin) was widely used in making weapons in times past, dating back to at least 2400 B.C. This metal alloy was hard and tough, but was eventually replaced by iron. Perhaps the most familiar liquid-solid solution is dental amalgam, used to fill teeth when there is a cavity. Approximately \(50\%\) of the amalgam material is liquid mercury to which a powdered alloy of silver, tin, and copper is added. Mercury is used because it binds well with the solid metal alloy. However, the use of mercury-based dental amalgam has gone under question in recent years, because of concerns regarding the toxicity of mercury. | 1,649 | 4,228 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/06%3A_Equilibrium_Chemistry/6.06%3A_Ladder_Diagrams |
When we develop or evaluate an analytical method, we often need to understand how the chemistry that takes place affects our results. Suppose we wish to isolate Ag by precipitating it as AgCl. If we also need to control pH, then we must use a reagent that does not adversely affect the solubility of AgCl. It is a mistake to use NH to adjust the pH, for example, because it increases the solubility of AgCl (see ). One of the primary sources of determinate errors in many analytical methods is failing to account for potential chemical interferences. In this section we introduce the ladder diagram as a simple graphical tool for visualizing equilibrium chemistry. We will use ladder diagrams to determine what reactions occur when we combine several reagents, to estimate the approximate composition of a system at equilibrium, and to evaluate how a change to solution conditions might affect an analytical method. Although not specifically on the topic of ladder diagrams as developed in this section, the following papers provide appropriate background information: (a) Runo, J. R.; Peters, D. G. , , 708–713; (b) Vale, J.; Fernández‐Pereira, C.; Alcalde, M. , , 790–795; (c) Fernández‐Pereira, C.; Vale, J. , , 1–18; (d) Fernández‐ Pereira, C.; Vale, J.; Alcalde, M. , , 15–21; (e) Fernández‐Pereira, C.; Alcalde, M.; Villegas, R.; Vale, J. , , 520–525. Ladder diagrams are a great tool for helping you to think intuitively about analytical chemistry. We will make frequent use of them in the chapters to follow. Let’s use acetic acid, CH COOH, to illustrate the process we will use to draw and to interpret an acid–base ladder diagram. Before we draw the diagram, however, let’s consider the equilibrium reaction in more detail. Acetic acid's acid dissociation reaction and equilibrium constant expression are \[\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CH}_{3} \mathrm{COO}^{-}(a q) \nonumber\] \[K_{\mathrm{a}}=\frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}=1.75 \times 10^{-5} \nonumber\] First, let’s take the logarithm of each term in this equation and multiply through by –1 \[-\log K_{a}=4.76=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]-\log \frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \nonumber\] Now, let’s replace –log[H O ] with pH and rearrange the equation to obtain the result shown here. \[\mathrm{pH}=4.76+\log \frac{\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]} \label{6.1}\] Equation \ref{6.1} tells us a great deal about the relationship between pH and the relative amounts of acetic acid and acetate at equilibrium. If the concentrations of CH COOH and CH COO are equal, then Equation \ref{6.1} reduces to \[\mathrm{pH}=4.76+\log (1)=4.76+0=4.76 \nonumber\] If the concentration of CH COO is greater than that of CH COOH, then the log term in Equation \ref{6.1} is positive and the pH is greater than 4.76. This is a reasonable result because we expect the concentration of the conjugate base, CH COO , to increase as the pH increases. Similar reasoning will convince you that the pH is less than 4.76 when the concentration of CH COOH exceeds that of CH COO . Now we are ready to construct acetic acid’s ladder diagram (Figure 6.6.1
). First, we draw a vertical arrow that represents the solution’s pH, with smaller (more acidic) pH levels at the bottom and larger (more basic) pH levels at the top. Second, we draw a horizontal line at a pH equal to acetic acid’s p value. This line, or step on the ladder, divides the pH axis into regions where either CH COOH or CH COO is the predominate species. This completes the ladder diagram. Using the ladder diagram, it is easy to identify the predominate form of acetic acid at any pH. At a pH of 3.5, for example, acetic acid exists primarily as CH COOH. If we add sufficient base to the solution such that the pH increases to 6.5, the predominate form of acetic acid is CH COO . Draw a ladder diagram for the weak base ‐nitrophenolate and identify its predominate form at a pH of 6.00. To draw a ladder diagram for a weak base, we simply draw the ladder diagram for its conjugate weak acid. From , the p for ‐nitrophenol is 7.15. The resulting ladder diagram is shown in Figure 6.6.2
. At a pH of 6.00, ‐nitrophenolate is present primarily in its weak acid form. Draw a ladder diagram for carbonic acid, H CO . Because H CO is a diprotic weak acid, your ladder diagram will have two steps. What is the predominate form of carbonic acid when the pH is 7.00? Relevant equilibrium constants are in . From , the p values for H CO are 6.352 and 10.329. The ladder diagram for H CO is shown below. The predominate form at a pH of 7.00 is \(\text{HCO}_3^-\). A ladder diagram is particularly useful for evaluating the reactivity between a weak acid and a weak base. Figure 6.6.3
, for example, shows a single ladder diagram for acetic acid/acetate and for ‐nitrophenol/ ‐nitrophenolate. An acid and a base can not co‐exist if their respective areas of predominance do not overlap. If we mix together solutions of acetic acid and sodium ‐nitrophenolate, the reaction \[\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{NO}_{2}^{-}(a q)+\mathrm{CH}_{3} \mathrm{COOH}(a q)\rightleftharpoons \text{CH}_3\text{COO}^-(aq) + \text{C}_6\text{H}_4\text{NO}_2\text{H}(aq) \label{6.2}\] occurs because the areas of predominance for acetic acid and ‐nitrophenolate do not overlap. The solution’s final composition depends on which species is the limiting reagent. The following example shows how we can use the ladder diagram in Figure 6.6.3
to evaluate the result of mixing together solutions of acetic acid and ‐nitrophenolate. Predict the approximate pH and the final composition after mixing together 0.090 moles of acetic acid and 0.040 moles of ‐nitrophenolate. The ladder diagram in Figure 6.6.3
indicates that the reaction between acetic acid and ‐nitrophenolate is favorable. Because acetic acid is in excess, we assume the reaction of ‐nitrophenolate to ‐nitrophenol is complete. At equilibrium essentially no ‐nitrophenolate remains and there are 0.040 mol of ‐nitrophenol. Converting ‐nitrophenolate to ‐nitrophenol consumes 0.040 moles of acetic acid; thus \[\begin{array}{c}{\text { moles } \mathrm{CH}_{3} \mathrm{COOH}=0.090-0.040=0.050 \ \mathrm{mol}} \\ {\text { moles } \mathrm{CH}_{3} \mathrm{COO}^{-}=0.040 \ \mathrm{mol}}\end{array} \nonumber\] According to the ladder diagram, the pH is 4.74 when there are equal amounts of CH COOH and CH COO . Because we have slightly more CH COOH than CH COO , the pH is slightly less than 4.74. Using Figure 6.6.3
, predict the approximate pH and the composition of the solution formed by mixing together 0.090 moles of ‐nitrophenolate and 0.040 moles of acetic acid. The ladder diagram in Figure 6.6.3
indicates that the reaction between acetic acid and ‐nitrophenolate is favorable. Because ‐nitrophenolate is in excess, we assume the reaction of acetic acid to acetate is complete. At equilibrium essentially no acetic acid remains and there are 0.040 moles of acetate. Converting acetic acid to acetate consumes 0.040 moles of ‐nitrophenolate; thus \[\text { moles } p \text {-nitrophenolate }=0.090-0.040=0.050 \text { mol } \nonumber\] \[\text { moles } p\text{-nitrophenol }=0.040 \ \mathrm{mol} \nonumber\] According to the ladder diagram for this system, the pH is 7.15 when there are equal concentrations of ‐nitrophenol and ‐nitrophenolate. Because we have slightly more ‐nitrophenolate than we have ‐nitrophenol, the pH is slightly greater than 7.15. If the areas of predominance for an acid and a base overlap, then we do not expect that much of a reaction will occur. For example, if we mix together solutions of CH COO and ‐nitrophenol, we do not expect a significant change in the moles of either reagent. Furthermore, the pH of the mixture must be between 4.76 and 7.15, with the exact pH depending upon the relative amounts of CH COO and ‐nitrophenol. We also can use an acid–base ladder diagram to evaluate the effect of pH on other equilibria. For example, the solubility of CaF \[\mathrm{CaF}_{2}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+2 \mathrm{F}^{-}(a q) \nonumber\] is affected by pH because F is a weak base. From Le Châtelier’s principle, we know that converting F to HF will increase the solubility of CaF . To minimize the solubility of CaF we need to maintain the solution’s pH so that F is the predominate species. The ladder diagram for HF (Figure 6.6.4
) shows us that maintaining a pH of more than 3.17 will minimize solubility losses. We can apply the same principles for constructing and interpreting an acid–base ladder diagram to equilibria that involve metal–ligand complexes. For a complexation reaction we define the ladder diagram’s scale using the concentration of uncomplexed, or free ligand, pL. Using the formation of \(\text{Cd(NH}_3)^{2+}\) as an example \[\mathrm{Cd}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)^{2+}(a q) \nonumber\] we can show that log is the dividing line between the areas of predominance for Cd and for \(\text{Cd(NH}_3)^{2+}\). \[K_{1}=3.55 \times 10^{2}=\frac{\left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)^{2+}\right]}{\left[\mathrm{Cd}^{2+}\right]\left[\mathrm{NH}_{3}\right]} \nonumber\] \[\log K_{1}=\log \left(3.55 \times 10^{2}\right)=\log \frac{\left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)^{2+}\right]}{\left[\mathrm{Cd}^{2+}\right]}-\log \left[\mathrm{NH}_{3}\right] \nonumber\] \[\log K_{1}=2.55=\log \frac{\left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)^{2+}\right]}{\left[\mathrm{Cd}^{2+}\right]}+\mathrm{p} \mathrm{NH}_{3} \nonumber\] \[\mathrm{p} \mathrm{NH}_{3}=\log K_{1}+\log \frac{\left[\mathrm{Cd}^{2+}\right]}{\left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)^{2+}\right]}=2.55+\log \frac{\left[\mathrm{Cd}^{2+}\right]}{\left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)^{2+}\right]} \nonumber\] Thus, Cd is the predominate species when pNH is greater than 2.55 (a concentration of NH smaller than \(2.82 \times 10^{-3}\) M) and for a pNH value less than 2.55, \(\text{Cd(NH}_3)^{2+}\) is the predominate species. Figure 6.6.5
shows a complete metal–ligand ladder diagram for Cd and NH that includes additional Cd–NH complexes. Draw a single ladder diagram for the Ca(EDTA) and the Mg(EDTA) metal–ligand complexes. Use your ladder diagram to predict the result of adding 0.080 moles of Ca to 0.060 moles of Mg(EDTA) . EDTA is an abbreviation for the ligand ethylenediaminetetraacetic acid. Figure 6.6.6
shows the ladder diagram for this system of metal–ligand complexes. Because the predominance regions for Ca and Mg(EDTA) do not overlap, the reaction \[\mathrm{Ca}^{2+}(a q)+\mathrm{Mg}(\mathrm{EDTA})^{2-}(a q) \rightleftharpoons \mathrm{Ca}(\mathrm{EDTA})^{2-}(a q)+\mathrm{Mg}^{2+}(a q) \nonumber\] proceeds essentially to completion. Because Ca is the excess reagent, the composition of the final solution is approximately \[\text { moles } \mathrm{Ca}^{2+}=0.080-0.060=0.020 \ \mathrm{mol} \nonumber\] \[\text { moles } \mathrm{Ca}(\mathrm{EDTA})^{2-}=0.060 \ \mathrm{mol} \nonumber\] \[\text { moles } \mathrm{Mg}^{2+}=0.060 \ \mathrm{mol} \nonumber\] \[\text { moles } \mathrm{Mg}(\mathrm{EDTA})^{2-}=0 \ \mathrm{mol} \nonumber\] The metal–ligand ladder diagram in Figure 6.6.5
uses stepwise formation constants. We also can construct a ladder diagram using cumulative formation constants. For example, the first three stepwise formation constants for the reaction of Zn with NH \[\mathrm{Zn}^{2+}(a q)+\mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)^{2+}(a q) \quad K_{1}=1.6 \times 10^{2} \nonumber\] \[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)^{2+}(a q)+\mathrm{NH}_{3}(a q)\rightleftharpoons\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q) \quad K_{2}=1.95 \times 10^{2} \nonumber\] \[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{2}^{2+}(a q)+\mathrm{NH}_{3}(a q)=\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q) \quad K_{3}=2.3 \times 10^{2} \nonumber\] suggests that the formation of \(\text{Zn(NH}_3)_3^{2+}\) is more favorable than the formation of \(\text{Zn(NH}_3)^{2+}\) or \(\text{Zn(NH}_3)_2^{2+}\). For this reason, the equilibrium is best represented by the cumulative formation reaction shown here. \[\mathrm{Zn}^{2+}(a q)+3 \mathrm{NH}_{3}(a q)\rightleftharpoons \mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{3}^{2+}(a q) \quad \beta_{3}=7.2 \times 10^{6} \nonumber\] Because is greater than , which is greater than , the formation of the metal‐ligand complex \(\text{Zn(NH}_3)_3^{2+}\) is more favorable than the formation of the other metal ligand complexes. For this reason, at lower values of pNH the concentration of \(\text{Zn(NH}_3)_3^{2+}\) is larger than the concentrations of \(\text{Zn(NH}_3)^{2+}\) or \(\text{Zn(NH}_3)_2^{2+}\). The value of \(\beta_3\) is \[\beta_{3}=K_{1} \times K_{2} \times K_{3} \nonumber\] To see how we incorporate this cumulative formation constant into a ladder diagram, we begin with the reaction’s equilibrium constant expression. \[\beta_{3}=\frac{\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{3}^{2+}\right]}{\left[\mathrm{Zn}^{2+}\right]\left[\mathrm{NH}_{3}\right]^{3}} \nonumber\] Taking the log of each side \[\log \beta_{3}=\log \frac{\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{3}^{2+}\right]}{\left[\mathrm{Zn}^{2+}\right]}-3 \log \left[\mathrm{NH}_{3}\right] \nonumber\] and rearranging gives \[\mathrm{pNH}_{3}=\frac{1}{3} \log \beta_{3}+\frac{1}{3} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{3}^{2+}\right]} \nonumber\] When the concentrations of Zn and \(\text{Zn(NH}_3)_3^{2+}\) are equal, then \[\mathrm{p} \mathrm{NH}_{3}=\frac{1}{3} \log \beta_{3}=2.29 \nonumber\] In general, for the metal–ligand complex ML , the step for a cumulative formation constant is \[\mathrm{pL}=\frac{1}{n} \log \beta_{n} \nonumber\] Figure 6.6.7
shows the complete ladder diagram for the Zn –NH system. We also can construct ladder diagrams to help us evaluate redox equilibria. Figure 6.6.8
shows a typical ladder diagram for two half‐reactions in which the scale is the potential, . The Nernst equation defines the areas of predominance. Using the Fe /Fe half‐reaction as an example, we write \[E=E^{\circ}-\frac{R T}{n F} \ln \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]}=0.771-0.05916 \log \frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]} \nonumber\] At a potential more positive than the standard state potential, the predominate species is Fe , whereas Fe predominates at potentials more negative than . When coupled with the step for the Sn /Sn half‐reaction we see that Sn is a useful reducing agent for Fe . If Sn is in excess, the potential of the resulting solution is near +0.154 V. Because the steps on a redox ladder diagram are standard state potentials, a complication arises if solutes other than the oxidizing agent and reducing agent are present at non‐standard state concentrations. For example, the potential for the half‐reaction \[\mathrm{UO}_{2}^{2+}(a q)+4 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+2 e^{-} \rightleftharpoons \mathrm{U}^{4+}(a q)+6 \mathrm{H}_{2} \mathrm{O}(l) \nonumber\] depends on the solution’s pH. To define areas of predominance in this case we begin with the Nernst equation \[E=+0.327-\frac{0.05916}{2} \log \frac{\left[\mathrm{U}^{4+}\right]}{\left[\mathrm{UO}_{2}^{2+}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{4}} \nonumber\] and factor out the concentration of H O . \[E=+0.327+\frac{0.05916}{2} \log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{4}-\frac{0.05916}{2} \log \frac{\left[\mathrm{U}^{4+}\right]}{\left[\mathrm{UO}_{2}^{2+}\right]}\nonumber\] From this equation we see that the area of predominance for \(\text{UO}_2^{2+}\) and U is defined by a step at a potential where [U ] = [\(\text{UO}_2^{2+}\)]. \[E=+0.327+\frac{0.05916}{2} \log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{4}=+0.327-0.1183 \mathrm{pH} \nonumber\] Figure 6.6.9
shows how pH affects the step for the \(\text{UO}_2^{2+}\) /U half‐reaction. | 16,386 | 4,232 |
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When you have completed Chapter 26, you should be able to Amino acids are important biochemicals, as they are the building blocks from which proteins and polypeptides are assembled. We begin this chapter with an examination of some of the fundamental chemistry of amino acids: their structures, stereochemistry and synthesis. We then discuss the nature of peptides and of the peptide bond, and present the complex issue of determining the order in which the various amino‑acid residues occur in a given peptide. Once a chemist knows the exact order the of the residues in a given peptide, the next challenge is to determine a method by which the same peptide can be prepared in the laboratory. Thus, two sections are devoted to the problem of protein synthesis. The final sections in the chapter deal with the classification, overall structure and denaturation of proteins.
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Sulfur analogs of alcohols are called or , and ether analogs are called . The chemical behavior of thiols and sulfides contrasts with that of alcohols and ethers in some important ways. Since hydrogen sulfide (H S) is a much stronger acid than water (by more than ten million fold), we expect, and find, thiols to be stronger acids than equivalent alcohols and phenols. Thiolate conjugate bases are easily formed, and have proven to be excellent nucleophiles in S 2 reactions of alkyl halides and tosylates. R–S Na + (CH ) CH–Br (CH ) CH–S–R + Na Br Although the basicity of ethers is roughly a hundred times greater than that of equivalent sulfides, the nucleophilicity of sulfur is much greater than that of oxygen, leading to a number of interesting and useful electrophilic substitutions of sulfur that are not normally observed for oxygen. Sulfides, for example, react with alkyl halides to give ternary sulfonium salts (equation # 1) in the same manner that 3º-amines are alkylated to . Although equivalent oxonium salts of ethers are known, they are only prepared under extreme conditions, and are exceptionally reactive. Remarkably, sulfoxides (equation # 2), sulfinate salts (# 3) and sulfite anion (# 4) also alkylate on sulfur, despite the partial negative formal charge on oxygen and partial positive charge on sulfur. Oxygen assumes only two oxidation states in its organic compounds (–1 in peroxides and –2 in other compounds). Sulfur, on the other hand, is found in oxidation states ranging from –2 to +6, as shown in the following table (some simple inorganic compounds are displayed in orange). Try drawing Lewis-structures for the sulfur atoms in these compounds. If you restrict your formulas to valence shell electron octets, most of the higher oxidation states will have formal charge separation, as in equation 2 above. The formulas written here neutralize this charge separation by double bonding that expands the valence octet of sulfur. Indeed, the S=O double bonds do not consist of the customary σ & π-orbitals found in carbon double bonds. As a third row element, sulfur has that may be used for p-d bonding in a fashion similar to p-p (π) bonding. In this way sulfur may expand an argon-like valence shell octet by two (e.g. sulfoxides) or four (e.g. sulfones) electrons. Sulfoxides have a fixed pyramidal shape (the sulfur non-bonding electron pair occupies one corner of a tetrahedron with sulfur at the center). Consequently, sulfoxides having two different alkyl or aryl substituents are chiral. Enantiomeric sulfoxides are stable and may be isolated. Thiols also differ dramatically from alcohols in their oxidation chemistry. Oxidation of 1º and 2º-alcohols to aldehydes and ketones changes the oxidation state of carbon but not oxygen. Oxidation of thiols and other sulfur compounds changes the oxidation state of sulfur rather than carbon. We see some representative sulfur oxidations in the following examples. In the first case, mild oxidation converts thiols to disufides. An equivalent oxidation of alcohols to peroxides is not normally observed. The reasons for this different behavior are not hard to identify. The S–S single bond is nearly twice as strong as the O–O bond in peroxides, and the O–H bond is more than 25 kcal/mole stronger than an S–H bond. Thus, thermodynamics favors disulfide formation over peroxide.
Mild oxidation of disufides with chlorine gives alkylsulfenyl chlorides, but more vigorous oxidation forms sulfonic acids (2nd example). Finally, oxidation of sulfides with hydrogen peroxide (or peracids) leads first to sulfoxides and then to sulfones. The nomenclature of sulfur compounds is generally straightforward. The prefix denotes replacement of a functional oxygen by sulfur. Thus, -SH is a thiol and C=S a thione. The prefix denotes replacement of a carbon atom in a chain or ring by sulfur, although a single ether-like sulfur is usually named as a sulfide. For example, C H SC H is ethyl propyl sulfide and C H SCH SC H may be named 3,5-dithiaoctane. Sulfonates are sulfonate acid esters and sultones are the equivalent of lactones. Other names are noted in the table above. | 4,199 | 4,240 |
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A reaction’s equilibrium position defines the extent to which the reaction can occur. For example, we expect a reaction with a large equilibrium constant, such as the dissociation of HCl in water \[\ce{HCl}(aq) + \ce{H2O}(l) \ce{->} \ce{H3O+}(aq) + \ce{Cl-}(aq) \nonumber\] to proceed nearly to completion. A large equilibrium constant, however, does not guarantee that a reaction will reach its equilibrium position. Many reactions with large equilibrium constants, such as the reduction of \(\ce{MnO4-}\) by \(\ce{H2O}\) \[\ce{4 MnO4-}(aq) + \ce{2 H2O}(l) \ce{->} \ce{4 MnO2}(s) + \ce{3 O2}(g) + \ce{4 OH-}(aq) \nonumber\] do not occur to an appreciable extent. The study of the rate at which a chemical reaction approaches its equilibrium position is called kinetics. A study of a reaction’s kinetics begins with the measurement of its reaction rate. Consider, for example, the general reaction shown below, involving the aqueous solutes A, B, C, and D, with stoichiometries of , , , and . \[a \ce{A} + b \ce{B} \ce{<=>} c \ce{C} + d \ce{D} \label{16.1}\] The rate, or velocity, at which this reaction approaches its equilibrium position is determined by following the change in concentration of one reactant or one product as a function of time. For example, if we monitor the concentration of reactant A, we express the rate as \[R = - \frac {d[\ce{A}]} {dt} \label{16.2}\] where is the measured rate expressed as a change in concentration of A as a function of time. Because a reactant’s concentration decreases with time, we include a negative sign so that the rate has a positive value. We also can determine the rate by following the change in concentration of a product as a function of time, which we express as \[R^{\prime} = + \frac {d[\ce{C}]} {dt} \label{16.3}\] Rates determined by monitoring different species do not necessarily have the same value. The rate in Equation \ref{16.2} and the rate \(R^{\prime}\) in Equation \ref{16.3} have the same value only if the stoichiometric coefficients of A and C in reaction \ref{16.1} are identical. In general, the relationship between the rates and \(R^{\prime}\) is \[R = \frac {a} {c} \times R^{\prime} \nonumber\] A rate law describes how a reaction’s rate is affected by the concentration of each species in the reaction mixture. The rate law for Reaction \ref{16.1} takes the general form of \[R = k[\ce{A}]^{\alpha} [\ce{B}]^{\beta} [\ce{C}]^{\gamma} [\ce{D}]^{\delta} [\ce{E}]^{\epsilon} ... \label{16.4}\] where is the rate constant, and \(\alpha\), \(\beta\), \(\gamma\), \(\delta\), and \(\epsilon\) are the reaction orders of the reaction for each species present in the reaction. There are several important points about the rate law in Equation \ref{16.4}. First, a reaction’s rate may depend on the concentrations of both reactants and products, as well as the concentration of a species that does not appear in the reaction’s overall stoichiometry. Species E in Equation \ref{16.4}, for example, may be a catalyst that does not appear in the reaction’s overall stoichiometry, but which increases the reaction’s rate. Second, the reaction order for a given species is not necessarily the same as its stoichiometry in the chemical reaction. Reaction orders may be positive, negative, or zero, and may take integer or non-integer values. Finally, the reaction’s overall reaction order is the sum of the individual reaction orders for each species. Thus, the overall reaction order for Equation \ref{16.4} is \(\alpha + \beta +\gamma + \delta + \epsilon\). In this section we review the application of kinetics to several simple chemical reactions, focusing on how we can use the integrated form of the rate law to determine reaction orders. In addition, we consider how we can determine the rate law for a more complex system. The simplest case we can treat is a first-order reaction in which the reaction’s rate depends on the concentration of only one species. The simplest example of a first-order reaction is an irreversible thermal decomposition of a single reactant, which we represent as \[\ce{A} \ce{->} \text{products} \label{16.5}\] with a rate law of \[R = - \frac {d[\ce{A}]} {dt} = k[\ce{A}] \label{16.6}\] The simplest way to demonstrate that a reaction is first-order in A, is to double the concentration of A and note the effect on the reaction’s rate. If the observed rate doubles, then the reaction is first-order in A. Alternatively, we can derive a relationship between the concentration of A and time by rearranging Equation \ref{16.6} and integrating. \[\frac {d[\ce{A}]} {[\ce{A}]} = -kdt \nonumber\] \[\int_{[{A}]_0}^{[{A}]_t}\frac{1}{[A]}d[A] = - k \int_{o}^{t}dt \label{16.7}\] Evaluating the integrals in Equation \ref{16.7} and rearranging \[\ln \frac {[\ce{A}]_t} {[\ce{A}]_0} = -kt \label{16.8}\] \[\ln [\ce{A}]_t = \ln [\ce{A}]_0 - kt \label{16.9}\] shows that for a first-order reaction, a plot of \(\ln[\ce{A}]_t\) versus time is linear with a slope of – and a -intercept of \(\ln[\ce{A}]_0\). Equation \ref{16.8} and Equation \ref{16.9} are known as integrated forms of the rate law. Reaction \ref{16.5} is not the only possible form of a first-order reaction. For example, the reaction \[\ce{A} + \ce{B} \ce{->} \text{products} \label{16.10}\] will follow first-order kinetics if the reaction is first-order in A and if the concentration of B does not affect the reaction’s rate, which may happen if the reaction’s mechanism involves at least two steps. Imagine that in the first step, A slowly converts to an intermediate species, C, which reacts rapidly with the remaining reactant, B, in one or more steps, to form the products. \[\ce{A} \ce{->} \ce{B} (\text{slow}) \nonumber\] \[\ce{B} + \ce{C} \ce{->} \text{products} \nonumber\] Because a reaction’s rate depends only on those species in the slowest step—usually called the rate-determining step—and any preceding steps, species B will not appear in the rate law. The simplest reaction demonstrating second-order behavior is \[\ce{2 A} \ce{->} \text{products} \nonumber\] for which the rate law is \[R = - \frac {d[\ce{A}]} {dt} = k[\ce{A}]^2 \nonumber\] Proceeding as we did earlier for a first-order reaction, we can easily derive the integrated form of the rate law. \[\frac {d[\ce{A}]} {[\ce{A}]^2} = -kdt \nonumber\] \[\int_{[\ce{A}]_0}^{[\ce{A}]_t} = -k \int_0^t dt \nonumber\] \[\frac {1} {[\ce{A}]_t} = kt + \frac {1} {[\ce{A}]_0} \nonumber\] For a second-order reaction, therefore, a plot of ([A] ) versus is linear with a slope of and a -intercept of ([A] ) . Alternatively, we can show that a reaction is second-order in A by observing the effect on the rate when we change the concentration of A. In this case, doubling the concentration of A produces a four-fold increase in the reaction’s rate. The following data were obtained during a kinetic study of the hydration of -methoxyphenylacetylene by measuring the relative amounts of reactants and products by NMR [data from Kaufman, D,; Sterner, C.; Masek, B.; Svenningsen, R.; Samuelson, G. , , 885–886]. To determine the reaction’s order we plot ln(%p-methoxyphenylacetylene) versus time for a first-order reaction, and (%p-methoxyphenylacetylene) versus time for a second-order reaction (see below). Because a straight-line for the first-order plot fits the data nicely, we conclude that the reaction is first-order in -meth- oxyphenylacetylene. Note that when we plot the data using the equation for a second-order reaction, the data show curvature that does not fit the straight-line model. Unfortunately, most reactions of importance in analytical chemistry do not follow the simple first-order or second-order rate laws discussed above. We are more likely to encounter the second-order rate law given in Equation \ref{16.11} than that in Equation \ref{16.10}. \[R = k [\ce{A}] [\ce{B}] \label{16.11}\] Demonstrating that a reaction obeys the rate law in Equation \ref{16.11} is complicated by the lack of a simple integrated form of the rate law. Often we can simplify the kinetics by carrying out the analysis under conditions where the concentrations of all species but one are so large that their concentrations effectively remain constant during the reaction. For example, if the concentration of B is selected such that \([\ce{B}] >> [\ce{A}]\), then Equation \ref{16.11} simplifies to \[R = k^{\prime} [\ce{A}] \nonumber\] where the rate constant ́ is equal to [B]. Under these conditions, the reaction appears to follow first-order kinetics in A; for this reason we identify the reaction as pseudo-first-order in A. We can verify the reaction order for A using either the integrated rate law or by observing the effect on the reaction’s rate of changing the concentration of A. To find the reaction order for B, we repeat the process under conditions where \([\ce{A}] >> [\ce{B}]\). A variation on the use of pseudo-ordered reactions is the initial rate method. In this approach we run a series of experiments in which we change one-at-a-time the concentration of each species that might affect the reaction’s rate and measure the resulting initial rate. Comparing the reaction’s initial rate for two experiments in which only the concentration of one species is different allows us to determine the reaction order for that species. The application of this method is outlined in the following example. The following data was collected during a kinetic study of the iodation of acetone by measuring the concentration of unreacted I in solution [data from Birk, J. P.; Walters, D. L. , , 585–587]. \(6.65 \times 10^{-3}\) The order of the rate law with respect to the three reactants is determined by comparing the rates of two experiments in which there is a change in concentration for only one of the reactants. For example, in Experiments 1 and 2, only the \([\ce{H3O+}]\) changes; as doubling the \([\ce{H3O+}]\) doubles the rate, we know that the reaction is first-order in \(\ce{H3O+}\). Working in the same manner, Experiments 6 and 7 show that the reaction is also first order with respect to \([\ce{C3H6O}]\), and Experiments 6 and 8 show that the rate of the reaction is independentof the \([\ce{I2}]\). Thus, the rate law is \[R = k [\ce{C3H6O}] [\ce{H3O+}] \nonumber\] To determine the value of the rate constant, we substitute the rate, the \([\ce{H3O+}]\), and the \([\ce{H3O+}]\) for each experiment into the rate law and solve for . Using the data from Experiment 1, for example, gives a rate constant of \(3.31 \times 10^{-5} \text{ M}^{-1} \text{ s}^{-1}\). The average rate constant for the eight experiments is \(3.49 \times 10^{-5} \text{ M}^{-1} \text{ s}^{-1}\). | 10,663 | 4,241 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Chemistry_of_Cooking_(Rodriguez-Velazquez)/09%3A_Spices/9.02%3A_Introduction_to_Salt |
Historically, salt was a prestigious commodity. “The salt of the earth” describes an outstanding person. The word salary comes from the Latin salaria, which was the payment made to Roman soldiers for the purchase of salt. In Arabic, the phrase translated as “there is salt between us” expresses the covenant between humans and the divine. Though no longer a valuable commodity in the monetary sense, salt is still valuable in the sense of being crucial to human health. Common salt (sodium chloride) is 40% sodium and 60% chloride. An average adult consumes about 7 kg (15 lb.) per year. Salt can be found deposited in Earth’s layers in rock salt deposits. These deposits formed when the water in the oceans that covered Earth many millions of years ago evaporated. The salt was then covered by various types of rocks.Today, we have three basic methods of obtaining salt from natural sources: | 904 | 4,242 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Logic_of_Organic_Synthesis_(Rao)/02%3A_Rules_and_Guidelines_Governing_Organic_Synthesis |
There are a few rules that provide guidelines for planning strategies in organic synthesis. These rules and guidelines have come from the keen observations of chemists after looking into several examples from their own research and other published work in the literature. Nonetheless, they serve as guidelines to avoid pit-falls in planning. Since all these rules are governed by the underlying principles of and these basic mechanistic principles are the touchstones against which the conclusions reached are to be tested. Most of the rules that are useful in planning synthesis are collected here for convenience.
| 643 | 4,244 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Atmospheric_Chemistry/Carbon_Cycle |
A scientist and an engineer may be called upon to solve a particular problem involving coal (carbon), gasoline (hydrocarbon), combustion of carbon or carbon containing fuel, lime stone, sea shells, carbon monoxide, or carbon dioxide. When we formulate a solution, we should be aware of the impact not only of the problem, but also of the solution for such a problem. Otherwise, the solution may result in a problem that is more expensive to solve later. Thus, it is important to know how carbon evolve at a global scale. The carbon cycle? is part of the Earth cycle. The diagram from this link is shown here, because it illustrate the global cycle of carbon without including respiration and metabolism. It illustrates the geological processes. The carbon atoms undergo a complicated chemistry forming what is known as the global carbon cycle, as do oxygen, nitrogen, and other elements; but the carbon cycle is the most widely recognized. An animal produces carbon dioxide and consumes oxygen in its metabolism of food. Glucose is a typical food and a metabolic reaction can be represented by: \[C_6H_{12}O_6 + 6 O_2 \rightarrow 6 CO_2+ 6 H_2O\] A plant and green bacteria, on the other hand, produces oxygen and consumes carbon dioxide in its photo synthesis. Energy in the form of electromagnetic radiation (or photons) is supplied so that the low-energy-content carbon dioxide can be converted to high-energy-content glucose. An overall reaction for the complicated multi-step photosynthesis reaction can be represented by: \[6 CO_2+ 12 H_2O -- h\nu \rightarrow C_6H_{12}O_6 + 6 O_2 + 6 H_2O\] At a glance, animals and plants make food for each other. The plants convert solar energy into high-energy food for the animals. Water is a reactant and a product in the photosynthesis. Radioactive labeled studies showed that the oxygen in the water produced comes from those in carbon dioxide. You may be thinking about plants with leaves that give beautiful flowers. In fact, primitive plants in the ocean play a more important role in in the photosynthesis process, because of the large number of them. The solubility of carbon dioxide depends on its partial pressure. As we know, carbon dioxide dissolves in water to form carbonic acid: \[CO_2+ H_2O \rightarrow H_2CO_3\] \[H_2CO_3 \rightarrow H^+ + HCO_3^- K_{a1} = 4.2 \times 10^{-7}\] \[HCO_3^- \rightarrow H^+ + CO_3^{2-} K_{a2} = 4.8 \times 10^{-11}\] The dissolved carbon dioxide further reacts with metal ions in the water forming calcium and magnesium carbonates. The values for CaCO and MgCO are \(5\times 10^{-9}\) and \(3\times 10^{-3}\) respectively. Extensive limestone (CaCO ) and dolomite (mixture of CaCO and MgCO ) have been formed this way. \[CaCO_3 \rightarrow Ca^{2+} + CO_3^{2-} \;\;\ K_{sp} = 5 \times 10^{-9}\] \[MgCO_3 \rightarrow Mg^{2+} + CO_3^{2-} \;\;\; K_{sp} = 3 \times 10^{-3}\] Some believed that this is how lime stone produced. Lime stone is soluble in acidic solutions, which may be formed by dissolving large amount of carbon dioxide. \[CaCO_{3(s)} + 2 H^+_{(aq)} \rightarrow Ca^{2+}{(aq)} + H_2CO_{3(aq)}\] or, \[CO2(aq) + H2O(l) + CaCO3(s) \rightarrow Ca2+(aq) + 2 HCO3-(aq)\] When the concentration of carbon dioxide is reduced, the acidity decreases and the reverse reaction takes place forming a solid, CaCO (s). Thus, metabolism, photosynthesis, mineralization and geological process are the major chemical processes in the global carbon cycle. In general, it is known that rain water saturated with carbon dioxide has a pH of 5.6. Lower than 5.6 is called acid rain due to the presence of sulfur oxides and nitrogen oxides. Assume the water to be otherwise pure than the dissolved carbon dioxide, estimate the solubility of carbon dioxide in water? Since pH = 5.6, \[ [H+] = 10-5.6 M \notag \]
\[= 2.5\times 10^{-6} M \notag \] Thus, the contribution of hydrogen ions from self ionization of water (pH = 7) is negligible. We have \([H^+] = 2.5 \times 10^{-6}\; M\) = \([HCO_3^-]\) The ionization of dissolved carbon dioxide is represented by these reactions, \[CO_2 + H_2O \rightarrow H_2CO_3\notag \] \[H_2CO_3 \rightarrow H^+ + HCO_3^- \;\; \; K_{a1} = 4.2 \times 10^{-7}\notag \] \[HCO_3^- \rightarrow H^+ + CO_3^{2-} \;\;\; K_{a2} = 4.8 \times 10^{-11}\notag \] The major contribution to the production of hydrogen ion comes from the first ionization of H CO , and other contributions are almost negligible. If we assume the concentration of H CO to be M in its ionization, \[H_2CO_3 \rightarrow H^+ + HCO_3^- \;\;\; K_{a1} = 4.2 \times 10^{-7}\notag \] then, by definition of we have \[\dfrac{(2.5\times 10^{-6)} }{x} = 4.2\times 10^{-7}\notag \] Thus, \[x = 1.5 \times 10^{-5}\; M\notag \]
= 0.65 mg / L Thus, the solubility is about 0.65 ppm by weight.
Many assumptions have been made here, other wise the solution will be more complicated. Make sure you understand the assumptions. From the solubility products of \(CaCO_3\), estimate its molar solubility in natural water saturated with carbon dioxide at pH 5.6 and 298 K. From the estimates given in Example 1, we still have to consider these equilibria: \[H_2CO_3 \rightarrow H^+ + HCO_3^- \;\;\; K_{a1} = 4.2\times 10^{-7}\notag \] \[HCO_3^- \rightarrow H^+ + CO_3^{2-} \;\;\; K_{a2} = 4.8\times 10^{-11}\notag \] Since the second ionization constant << , it is safe to assume the following: \[[H+] = [HCO3-] = 2.5\times 10^{-6} M\notag \] in the above equilibria. The following formulation is derived by the definition of : [H ] [CO ] 2.5\times 10^{ [CO ]
-------------- = -------------------- = 4.8\times 10^{-11}
[HCO ] 2.5\times 10^{ Thus, [CO ] = 4.8\times 10^{ . By the definition of the solubility product, \[CaCO_3 \rightarrow Ca^{2+} + CO_3^{2-} \;\;\; K_{sp} = 5\times 10^{-9}\] we have [Ca2+] = 5\times 10^{-9/4.8\times 10^{-11
= 100 M
This value is obviously too high and unreasonable. The result is certainly incorrect. Thus, we should re-examine the last assumption. As CaCO dissolves, the concentration of carbonate ion also increase. If this concentration is high, then the contribution due to dissolved carbon dioxide is negligible. Effectively, we have [Ca ] = [CO ] = , and \[y^2 = 5 \times 10^{-9}\notag \] Thus, y = 7\times 10^{-5 M = 0.007 g CaCO3 / L. This results is obtained by ignoring the dissolved carbon dioxide. The true value is probably somewhere in between, because as calcium carbonates dissolves, the pH of the solution changes. | 6,459 | 4,245 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/SN2/Substrate |
Now that we have discussed the effects that the leaving group, nucleophile, and solvent have on biomolecular nucleophilic substitution (S 2) reactions, it's time to turn our attention to how the substrate affects the reaction. Although the substrate, in the case of nucleophilic substitution of haloalkanes, is considered to be the entire molecule circled below, we will be paying particular attention to the alkyl portion of the substrate. In other words, we are most interested in the electrophilic center that bears the leaving group. In the section , we learned that the SN transition state is very crowded. Recall that there are a total of five groups around the electrophilic center, the nucleophile, the leaving group, and three substituents. If each of the three substituents in this transition state were small hydrogen atoms, as illustrated in the first example below, there would be little steric repulsion between the incoming nucleophile and the electrophilic center, thereby increasing the ease at which the nucleophilic substitution reaction can occur. Remember, for the SN reaction to occur, the nucleophile must be able to attack the electrophilic center, resulting in the expulsion of the leaving group. If one of the hydrogens, however, were replaced with an R group, such as a methyl or ethyl group, there would be an increase in steric repulsion with the incoming nucleophile. If two of the hydrogens were replaced by R groups, there would be an even greater increase in steric repulsion with the incoming nucleophile. How does steric hindrance affect the rate at which an SN reaction will occur? As each hydrogen is replaced by an R group, the rate of reaction is significantly diminished. This is because the addition of one or two R groups shields the backside of the electrophilic carbon, impeding nucleophilic attack. The diagram below illustrates this concept, showing that electrophilic carbons attached to three hydrogen atoms results in faster nucleophilic substitution reactions, in comparison to primary and secondary haloalkanes, which result in nucleophilic substitution reactions that occur at slower or much slower rates, respectively. Notice that a tertiary haloalkane, that which has three R groups attached, does not undergo nucleophilic substitution reactions at all. The addition of a third R group to this molecule creates a carbon that is entirely blocked. Previously we learned that adding R groups to the electrophilic carbon results in nucleophilic substitution reactions that occur at a slower rate. What if R groups are added to neighboring carbons? It turns out that the addition of substitutes on neighboring carbons will slow nucleophilic substitution reactions as well. In the example below, 2-methyl-1-bromopropane differs from 1-bromopropane in that it has a methyl group attached to the carbon that neighbors the electrophilic carbon. The addition of this methyl group results in a significant decrease in the rate of a nucleophilic substitution reaction. If R groups were added to carbons farther away from the electrophilic carbon, we would still see a decrease in the reaction rate. However, branching at carbons farther away from the electrophilic carbon would have a much smaller effect. | 3,269 | 4,248 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Book3A_Medicines_by_Design/04%3A_Molecules_to_Medicines/4.02%3A_21st-Century_Science |
While strategies such as chemical genetics can quicken the pace of drug discovery, other approaches may help expand the number of molecular targets from several hundred to several thousand. Many of these new avenues of research hinge on biology. Relatively new brands of research that are stepping onto center stage in 21st-century science include genomics (the study of all of an organism's genetic material), proteomics (the study of all of an organism's proteins), and bioinformatics (using computers to sift through large amounts of biological data). The "omics" revolution in biomedicine stems from biology's gradual transition from a gathering, descriptive enterprise to a science that will someday be able to model and predict biology. If you think 25,000 genes is a lot (the number of genes in the human genome), realize that each gene can give rise to different molecular job. Scientists estimate that humans have hundreds of thousands of protein variants. Clearly, there's lots of work to be done, which will undoubtedly keep researchers busy for years to come. Doctors use the drug Gleevec to treat a form of leukemia, a disease in which abnormally high numbers of immune cells (larger, purple circles in photo) populate the blood. Recently, researchers made an exciting step forward in the treatment of cancer. Years of basic research investigating circuits of cellular communication led scientists to tailor-make a new kind of cancer medicine. In May 2001, the drug Gleevec™ was approved to treat a rare cancer of the blood called chronic myelogenous leukemia (CML). The Food and Drug Administration described Gleevec's approval as " …a testament to the groundbreaking scientific research taking place in labs throughout America." Researchers designed this drug to halt a cell-communication pathway that is always "on" in CML. Their success was founded on years of experiments in the basis biology of how cancer cells grow. The discovery of Gleevec in an example of the success of so-called molecular targeting: understanding how diseases arise at the level of cells, then figuring out ways to treat them. Scores of drugs, some to treat cancer but also many other health conditions, are in the research pipeline as a result of scientists' eavesdropping on how cells communicate. | 2,309 | 4,249 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Geochemistry_(Lower)/02%3A_The_Hydrosphere/2.02%3A_The_hydrosphere_and_the_oceans |
Arthur C. Clarke Water is the most abundant substance at the earth’s surface. Almost all of it is in the oceans, which cover 70% of the surface area of the earth. However, the amounts of water present in the atmosphere and on land (as surface runoff, lakes and streams) is great enough to make it a significant agent in transporting substances between the lithosphere and the oceans. Water interacts with both the atmosphere and the lithosphere, acquiring solutes from each, and thus provides the major chemical link between these two realms. The various transformations undergone by water through the different stages of the hydrologic cycle act to transport both dissolved and particulate substances between different geographic locations. Reservoir Volume / 10 , km Percent of total Oceans Ice caps and glaciers Deep groundwater Shallow groundwater It appears to have been bound up in the silica-based materials such as micas and amphiboles which accreted to form the Earth. The heat released during this process would have been sufficient to drive off this water, which amounted to about 0.01% by mass of the primoridal material. The refers to the steady state that exists between evaporation, condensation, percolation, runoff, and circulation of water. The cycle is driven by solar energy, mainly through direct vaporization, but also by convective motion induced by uneven heating. The major interphase transport process of the hydrologic cycle is evaporation of water from the ocean. However, 90% of this vapor falls directly back into the ocean as rain, while 10% is transported over the land. Of the latter, about two-thirds evaporates again and one third runs off to the ocean. Figure \(\Page {1}\): The climatic hydrological cycle at global scale (Diagram and text from a at the of the Institute for Global Environment and Society) The movement of water on the earth's surface and through the atmosphere is known as the hydrologic cycle. Water is taken up by the atmosphere from the earth's surface in vapor form through evaporation. It may then be moved from place to place by the wind until it is condensed back to its liquid phase to form clouds. Water then returns to the surface of the earth in the form of either liquid (rain) or solid (snow, sleet, etc.) precipitation. Water transport can also take place on or below the earth's surface by flowing glaciers, rivers, and ground water flow. The amounts of water precipitated onto the land and oceans are in approximate proportion to the relative surface areas, but evaporation from the ocean exceeds that from the land by about 37,400 km yr . This difference is the amount of water transported to the oceans by river runoff. When water condenses from the atmosphere in the form of rain, it is slightly enriched in H O . During epochs of glacial buildup the fraction of H O in the oceans consequently decreases. Observation of H O /H O ratios in marine sediments is one way of studying the timing and extent of past glaciations. Since the degree of heavy isotope enrichment of condensed water is temperature dependent, this same method can be used to estimate mean world temperatures in the distant past. The hydrologic cycle also has important effects on the energy budget of the earth. Atmospheric water vapor (along with carbon dioxide and methane) tends to absorb the long-wavelength infrared radiation emitted by the earth’s surface, partially trapping the incoming shorter-wavelength energy and thus maintaining the mean surface temperature about 30° higher than would be the case in the absence of water vapor. Of the 51% of the solar radiation incident on the atmosphere that reaches the earth’s surface, about half of this (23%) is used to evaporate water. During the ten days that an average molecule resides in the atmosphere, it will travel about 1000 km. The atmospheric transport of water from equatorial to subtropical regions serves as an important mechanism for the transport of thermal energy; at latitudes of about 40°, as much as one-third of the energy input comes from release of latent heat from water vapor formed in equatorial regions. About 97% of the earth’s water is contained in the two reservoirs which comprise the oceans. The upper mixed layer contains about 5% of the total; it is separated from the deeper and colder layer by the thermocline. Mixing between these two stratified layers is very slow; of the total ocean volume of 6.8 10 m , only about 0.71 10 m , or about 0.01%, moves between the two layers per year. The mean residence time of a water molecule in the deep layer is about 1600 years. The large-scale motions of ocean water are the primary means by which chemical substances, especially those taken up and excreted by organisms, are transported within the ocean. An understanding of the general patterns of this circulation is essential in order to analyze the observed distribution of many of the chemical elements in different parts of the ocean and in the oceanic sediments. The circulation of the surface waters of the ocean is driven by the prevailing winds. The latter arise from uneven heating of the earth’s surface, and are arranged in bands that parallel the equator. Although the motion of the waters at the surface of the ocean are driven by the winds, they do not follow them in a simple manner. The reasons are threefold: the Coriolis effect, the presence of land masses, and uneveness in the sea level due to regional differences in temperature and atmospheric pressure. Figure \(\Page {2}\): Atmospheric winds The most intense heat input into the atmosphere occurs near the equator, where the heated air rises and cools, producing intense local precipitation but little surface wind. After cooling and losing moisture, this air moves north and south and descends at a latitude of about 30°. As it descends, it warms (largely by adiabatic compression) and its relative humidity decreases. The extreme dryness of this air gives rise to the subtropical desert regions between about 15° and 30°. Part of this air flows back toward the equator, giving rise to the northeast and southwest trade winds; the deflection to the east or west is caused by the Coriolis effect. Another part of the descending air travels poleward, producing the prevailing westerlies. Eventually these collide with cold air masses moving away from the polar regions, producing a region of unstable air and storm activity known as a polar front. Some of this polar air picks up enough heat to rise and enter into polar cell circulation patterns. The flow of air in the prevailing westerlies is subject to considerable turbulence which gives rise to planetary waves. These are moving regions in which warm surface air is lifted to higher levels, producing lines of storms that travel from west to east, and exchanging more air between the polar and temperate regions. Figure \(\Page {3}\): In the Northern hemisphere, the Coriolis effect not only deflects south-moving objects to the east but it also causes currents flowing parallel to the equator to veer to the right of the direction of flow, i.e. to the north or south. In addition, prevailing westerly winds and the eastward rotation of the earth cause water to pile up by a few centimeters at the western edges of the oceans. The resultant downhill flow, interacting with Coriolis forces, produces a that runs south-to-north in the northern hemisphere. A similar but opposite effect gives rise to a south-flowing on continental east coasts. In contrast to the upper levels of the ocean, the deep ocean is ; the density increases with depth so as to inhibit the vertical transport of water. This stratification divides the deep oceans into several distinct water masses which undergo movement in a more or less horizonal plane, with adjacent masses sometimes moving in opposite directions. Figure \(\Page {4}\): As the cold water fills up the deepest regions and spills over ridges into other deep basins, it creates huge undersea cascades which rival the greatest terrestrial waterfalls in height and the largest rivers in volume. For more on this, see the Feb. 1989 The winds and atmospheric effects outlined above affect only the upper part of the ocean. Below 100 meters or so, oceanic circulation is driven by the of the seawater, which is determined by its temperature and its salinity. Variations in these two quantities give rise to the of the deep currents of the ocean. It all starts when seasonal ice begins to form in the polar regions. Because the salts dissolved in seawater cannot be accommodated within the ice structure, they are largely excluded from the new ice and remain in solution. This increases the density of the surrounding unfrozen water, causing it to sink into the bottom ocean. There are two major locations at which surface waters enter the deep ocean. The northern entry point is in the Norwegian Sea off Greenland; this water forms a mass known as the North Atlantic Deep Water (NADW) which flows southward across the equator. Figure \(\Page {5}\): From Don Reed, at San Jose State U. Most of the transport into the deep ocean takes place in the Weddell Sea off the coast of Antarctica. The highly saline water flows down the submerged Antarctic Slope to begin a 5000-year trip to the north across the bottom of the ocean. This is the major route by which dissolved CO and O (which are more soluble in this cold water) are transported into the deep ocean where it forms a water mass known as the Antarctic Bottom Water (AABW) which can be traced into all three oceans. In the south, a flow from the antarctic region forms a water mass known as the Antarctic Bottom Water (AABW) which can be traced into all three oceans. The Pacific Ocean lacks any major identifiable direct source of cold water, so it is less differentiated and its deep circulation is sluggish and poorly defined. Figure \(\Page {6}\): Temperature and salinity profiles in a north-south subsection of the Atlantic ocean. As is apparent from the figure above, the vertical profiles of temperature and especially of the salinity are not uniform. To some extent, these two parameters have opposite effects: in equatorial regions, temperatures are higher (leading to lower density) but evaporation rates are also higher (leading to higher density). In polar regions, the formation of sea ice raises the density of the seawater (because only a small proportion of salt is incorporated into the ice). The nature and extent of the deep ocean currents differ in the Atlantic, Pacific, and Indian oceans. These currents are much slower than the surface currents, and in fact have not been measured directly; their existence is however clearly implied by the chemical composition and temperature of water samples taken from various parts of the ocean. Estimated rates are of the order of kilometers per month, in contrast to the few kilometers per hour of surface waters. The deep currents are the indirect results of processes occuring at the surface in which cold water of high salinity is produced as sea ice forms in the arctic and antarctic regions. This water is so dense that it sinks to the bottom, displacing warmer or less saline water as it moves. Recirculation of deep water to the surface occurs to a very small extent in many regions, but it is especially pronounced where water entering the Antarctic Bottom mass displaces other bottom water, and where water piles up at the western edges of continents. This latter water flows downhill (forming the western boundary currents mentioned above) and is replaced by colder water from the deep ocean. The deep ocean contains few organisms to deplete the water of the nutrients it receives from the remains of the dead organisms floating down from above; this upwelled water is therefor exceptionally rich in nutrients, and strongly encourages the growth of new organisms that extend up the food chain to fish. Thus the wind-driven upwelling that occurs off the west coast of South America is responsible for the Peruvian fishing and guano fertilizer industry. About every seven years these prevailing winds disappear for a while, allowing warm equatorial waters to move in. This phenomenon is known as and it results in massive kills of plankton and fish. Decomposition of the dead organisms reduces the oxygen content of the water, causing the death of still more fish, and allowing reduced compounds such as hydrogen sulfide to accumulate. Matthias Tomczak's is an excellent source of more information on the topics covered above. Page last modified: 21.01.2008 For information about this Web site or to contact the author,
please see the | 12,703 | 4,250 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Supplemental_Modules_(Environmental_Chemistry)/Acid_Rain/Sources_of_Sulfur_Oxides |
It has been estimated that on a global basis, natural sources, such as volcanoes, contribute about as the same amount of sulfur oxides to the atmosphere as human industrial activities. This amounts to 75-100 million tons from each source per year. However, in industrial countries such as in Europe and North America, human activities contribute 95 % of the sulfur oxides and natural sources only 5 %. In the Western States, natural sources of sulfur oxides may be more important. In 1980, emissions of sulfur dioxide totaled 24.1 million tons in the United States. Of this total 66 % came from electric power companies. Electric power companies that burn coal are a major source of sulfur oxides. Other industrial plants contributed about 22 %. Smelting of metals such as copper, zinc, lead, and nickel can produce large amounts of sulfur dioxide. In Canada, 45% of the emissions are from smelting operations, compared to only 6 % in the United States. Coal contains mainly carbon with some hydrogen. When coal is burned it reacts with oxygen in the air to produce carbon dioxide and water and large amounts of heat. \[C + O_2 \rightarrow CO_2\] In addition, coal may contain from 1-4 % of the element, sulfur. When the coal is burned with oxygen in the air, the sulfur is reacted to form . \[S + O_2 \rightarrow SO_2\] In certain resort towns, a significant source of visible smog conditions results from the burning of large quantities of wood in fireplaces and stoves. The smoke contains solid particles which may provide the initial bit of solid or catalyst that initiates the reactions to produce sulfuric acid or nitric acid in the water droplets. This is a well recognized problem in Aspen and Vail Colorado. Steps are being taken to reduce the burning of wood. | 1,788 | 4,251 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Pharmaceuticals/Narcotic_Analgesic_Drugs |
Narcotic agents are potent analgesics which are effective for the relief of severe pain. Analgesics are selective central nervous system depressants used to relieve pain. The term analgesic means "without pain". Even in therapeutic doses, narcotic analgesics can cause respiratory depression, nausea, and drowsiness. Long term administration produces tolerance, psychic, and physical dependence called addiction. Narcotic agents may be classified into four categories: The main pharmacological action of analgesics is on the cerebrum and medulla of the central nervous system. Another effect is on the smooth muscle and glandular secretions of the respiratory and gastro-intestinal tract. The precise mechanism of action is unknown although the narcotics appear to interact with specific receptor sites to interfere with pain impulses. A schematic for an analgesic receptor site may look as shown in the graphic below with morphine. Three areas are needed: a flat areas to accommodate a flat nonpolar aromatic ring, a cavity to accept another series of rings perpendicular, and an anionic site for polar interaction of the amine group. Recently investigators have discovered two compounds in the brain called enkephalins which resemble morphine in structure. Each one is a peptide composed of 5 amino acids and differ only in the last amino acid. The peptide sequences are: tyr-gly-gly-phe-leu and tyr-gly-gly-phe-met. Molecular models show that the structures of the enkephalins has some similarities with morphine. The main feature in common appears to be the aromatic ring with the -OH group attached (tyr). Methadone and other similar analgesics have 2 aromatic rings which would be similar to the enkephalins (tyr and phe). Analgesics may relieve pain by preventing the release of acetylcholine. Enkephalin molecules are released from a nerve cell and bind to analgesic receptor sites on the nerve cell sending the impulse. The binding of enkephalin or morphine-like drugs changes the shape of the nerve sending the impulse in such a fashion as to prevent the cell from releasing acetylcholine. As a result, the pain impulse cannot be transmitted and the brain does not preceive pain. Morphine exerts a narcotic action manifested by analgesia, drowsiness, changes in mood, and mental clouding. The major medical action of morphine sought in the CNS is analgesia. Opiates suppress the "cough center" which is also located in the brain stem, the medulla. Such an action is thought to underlie the use of opiate narcotics as cough suppressants. Codeine appears to be particularly effective in this action and is widely used for this purpose. Narcotic analgesics cause an addictive physical dependence. If the drug is discontinued, withdrawal symptoms are experienced. Although the reasons for addiction and withdrawal symptoms are not completely known, recent experiments have provided some information. A nucleotide known as cyclicadenosine monophosphate (cAMP) is synthesized with the aid of the enzyme adenylate cyclase. Enkephalin and morphine-like drugs inhibit this enzyme and thus decrease the amount of cAMP in the cells. In order to compensate for the decreased cAMP, the cells synthesize more enzyme in an attempt to produce more cAMP. Since more enzyme has been produced, more morphine is required as an inhibitor to keep the cAMP at a low level. This cycle repeats itself causing an increase in the tolerance level and increasing the amounts of morphine required. If morphine is suddenly withheld, withdrawal symptoms are probably caused by a high concentration of cAMP since the synthesizing enzyme, adenylate cyclase, is no longer being inhibited. Morphine and codeine are contained in opium from the poppy ( ) plant found in Turkey, Mexico, Southeast Asia, China, and India. This plant is 3-4 feet tall with 5-8 egg shaped capsules on top. Ten days after the poppy blooms in June, incisions are made in the capsules permitting a milky fluid to ooze out. The following day the gummy mass (now brown) is carefully scraped off and pressed into cakes of raw opium to dry. Opium contains over 20 compounds but only morphine (10%) and codeine (0.5%) are of any importance. Morphine is extracted from the opium and isolated in a relatively pure form. Since codeine is in such low concentration, it is synthesized from morphine by an ether-type methylation of an alcohol group. Codeine has only a fraction of the potency compared to morphine. It is used with aspirin and as a cough suppressant. Heroin is synthesized from morphine by a relatively simple esterification reaction of two alcohol (phenol) groups with acetic anhydride (equivalent to acetic acid). Heroin is much more potent than morphine but without the respiratory depression effect. A possible reason may be that heroin passes the blood-brain barrier much more rapidly than morphine. Once in the brain, the heroin is hydrolyzed to morphine which is responsible for its activity. include meperidine and methadone. Meperidine is the most common subsitute for morphine. It exerts several pharmacological effects: analgesic, local anesthetic, and mild antihistamine. This multiple activity may be explained by its structural resemblance to morphine, atropine, and histamine. Methadone is more active and more toxic than morphine. It can be used for the relief of may types of pain. In addition it is used as a narcotic substitute in addiction treatment because it prevents morphine abstinence syndrome. Methadone was synthesized by German chemists during Wold War II when the United States and our allies cut off their opium supply. And it is difficult to fight a war without analgesics so the Germans went to work and synthesized a number of medications in use today, including demerol and darvon which is structurally simular to methadone. And before we go further lets clear up another myth. Methadone, or dolophine was not named after Adolf Hitler. The "dol" in dolophine comes from the latin root "dolor." The female name Dolores is derived from it and the term dol is used in pain research to measure pain e.g., one dol is 1 unit of pain. Even methadone, which looks strikingly different from other opioid agonists, has steric forces which produce a configuration that closely resembles that of other opiates. See the graphic on the left and the top graphic on this page. In other words, steric forces bend the molecule of methadone into the correct configuration to fit into the opiate receptor. When you take methadone it first must be metabolized in the liver to a product that your body can use. Excess methadone is also stored in the liver and blood stream and this is how methadone works its 'time release trick' and last for 24 hours or more. Once in the blood stream metabolized methadone is slowly passed to the brain when it is needed to fill opiate receptors. Methadone is the only effective treatment for heroin addiction. It works to smooth the ups and down of heroin craving and allows the person to function normally. Narcotic Antagonists prevent or abolish excessive respiratory depression caused by the administration of morphine or related compounds. They act by competing for the same analgesic receptor sites. They are structurally related to morphine with the exception of the group attached to nitrogen. Nalorphine precipitates withdrawal symptoms and produces behavioral disturbances in addition to the antogonism action. Naloxane is a pure antagonist with no morphine like effects. It blocks the euphoric effect of heroin when given before heroin. Naltrexone became clinically available in 1985 as a new narcotic antagonist. Its actions resemble those of naloxone, but naltrexone is well is well absorbed orally and is long acting, necessitating only a dose of 50 to 100 mg. Therefore, it is useful in narcotic treatment programs where it is desired to maintain an individual on chronic therapy with a narcotic antagonist. In individuals taking naltrexone, subsequent injection of an opiate will produce little or no effect. Naltrexone appears to be particularly effective for the treatment of narcotic dependence in addicts who have more to gain by being drug-free rather than drug dependant | 8,217 | 4,252 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Chemistry_of_Cooking_(Rodriguez-Velazquez)/09%3A_Spices/9.11%3A_Flavorings_in_Baking |
Flavors cannot be considered a truly basic ingredient in bakery products but are important in producing the most desirable products. Flavoring materials consist of: Note: Salt may also be classed as a flavoring material because it intensifies other flavors. These and others (such as ) enable the baker to produce a wide variety of attractively flavored pastries, cakes, and other bakery products. Flavor extracts, essences, emulsions, and aromas are all solutions of flavor mixed with a solvent, often ethyl alcohol. The flavors used to make extracts and essences are the extracted essential oils from fruits, herbs, and vegetables, or an imitation of the same. Many fruit flavors are obtained from the natural parts (e.g., rind of lemons and oranges or the exterior fruit pulp of apricots and peaches). In some cases, artificial flavor is added to enhance the taste, and artificial coloring may be added for eye appeal. Both the Canadian and U.S. departments that regulate food restrict these and other additives. The flavors are sometimes encapsulated in corn syrup and emulsifiers. They may also be coated with gum to preserve the flavor compounds and give longer shelf life to the product. Some of the most popular essences are compounded from both natural and artificial sources. These essences have the true taste of the natural flavors. Aromas are flavors that have an oil extract base. They are usually much more expensive than alcoholic extracts, but purer and finer in their aromatic composition. Aromas are used for flavoring delicate creams, sauces, and ice creams. Emulsions are homogenized mixtures of aromatic oils and water plus a stabilizing agent (e.g., vegetable gum). Emulsions are more concentrated than extracts and are less susceptible to losing their flavor in the oven. They can therefore be used more sparingly. | 1,851 | 4,253 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Nucleic_Acids/DNA/The_Replication_of_DNA |
This page takes a very simplified look at how DNA replicates (copies) itself. It gives only a brief over-view of the process, with no attempt to describe the mechanism. We'll explain exactly what "semi-conservative" means when we have got some diagrams to look at. First imagine what happens if the two individual strands in the DNA double helix start to unzip. The diagram shows this happening in the middle of the DNA double helix - you mustn't assume that the top of the diagram is the end of the chain. It isn't. Further up the double helix, the two strands will still be joined together. In fact, this is happening lots of times along the very long DNA molecule. Lengths of chain become separated to form what are known as "bubbles". If you feel the need to see this in more detail, read the rest of this page, and then have a quick look at the links above. Some of the hydrogen bonds get broken and the two strands become partly separated. The red dotted lines on the diagram just point out the original base pairs. These are bonds in any form. These base pairs are now much too far apart for any sort of bonding between them. Now suppose that you have a source of nucleotides - phosphate joined to deoxyribose joined to a base, including all the four sorts of bases needed for DNA. The next diagram shows what would happen if a nucleotide containing guanine (G) and one containing cytosine (C) were attracted to the top two bases on the left-hand strand of the unzipped DNA - and then joined together. How did they end up joined together? This is all under the control of a number of enzymes, one of which (DNA polymerase) is responsible for joining up nucleotides along the chain in this way. Now suppose the same sort of thing happened at the top of the right-hand strand. You would end up with . . . Now compare the double strands that you are forming on the left- and right-hand sides. They are exactly the same . . . and if you were to continue this process, they would continue to be the same. And if you compare the patterns of bases in the new DNA being formed with what was in the original DNA before it started to unzip, everything is the same. This is inevitable because of the way the bases pair together. Let's simplify the last diagram, and assume that the whole copying process is complete. The next diagram focusses on the short bit of the total DNA molecule that we have been looking at. A typical human DNA molecule is around 150 million base pairs long - you will have to imagine the rest of it! You have also got to remember that in reality the whole thing would have coiled into its double helix. Trying to draw that just makes everything look messy and complicated! The original DNA is shown all in blue. The red strands in the daughter DNA are the ones which have been built on the original blue strands during the replication process. You can see that each of the daughter molecules is made of half of the original DNA plus a new strand. That's all "semi-conservative replication" means. Half of the original DNA is conserved (kept) in each of the daughter molecules. The red and blue, of course, have no physical significance apart from as a way of making the diagrams clearer. All three of these DNA molecules will be identical in every way. Jim Clark ( ) | 3,302 | 4,255 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reaction_Fundamentals/Pushing_Arrows |
In organic chemistry, it is important to understand the concept of . In polar reaction mechanisms, such as the , electron flow will be designated by arrows indicating the movement of electrons from electron rich regions to electron poor regions. In considering this concept, we must look at the two types of arrows provided in the mechanisms shown below. The curved arrows indicate the movement of electrons. The first type of arrow, shown in pink, originates from the electron pair of the nucleophile and extends to the electrophilic carbon of the haloalkane. This type of movement does not indicate that electrons leave the nucleophile; rather, it means that electrons become shared between the nucleophile and the electrophilic atom. The second type of curved arrow, also shown in pink, originates from the R-X bond and extends to the halogen. This indicates cleavage of the bond, whereby the electron pair becomes separated from R, the electrophilic carbon, and ends up on the halogen atom. While we are using the concept of nucleophilic substitution mechanisms to explain electron flow, it is very important to understand that this concept will be applied in nearly all the mechanisms you learn throughout your course of study. The simplest way to think about this in any mechanism you learn is that electrons will be pushed from an electron rich species or site to an electron poor species or site, and the direction of the curved arrow will indicate this. | 1,482 | 4,256 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Enzymes/Case_Studies/Fe-only_Hydrogenase |
Hydrogenase enzymes catalyze the reversible oxidation of molecular hydrogen to protons and electrons [H2 ↔ 2H+ + 2e ] [1].The two most studied classes are Ni-Fe hydrogenases and Fe-only hydrogenases. In general, Ni-Fe hydrogenase enzymes consume molecular hydrogen as a fuel source, and Fe-only hydrogense enzymes produce molecular hydrogen [2]. Crystal structure of the hydrogenase was obtained from a sulfate-reducing microorganism to 1.6 Å resolution (fig 1). Desulfovibrio desulfuricans (DdH) is a 53 kDa protein located in the periplasm [3]. Crystal structure of the enzyme was obtained from Clostridium pasteurianum an anaerobic soil microorganism to 1.8 Å resolution (fig 2). This enzyme, CpI, is a 61 kDa protein found in the cytoplasm [4]. DdH is a dimer with three 4Fe-4S cubanes and a H-cluster (fig 3)[5]. CpI is a monomer with three 4Fe-4S cubanes, one 2Fe-2S cluster, and a H-cluster (fig 4)[5]. When enzyme is isolated in air it changes its state from Fe(II)-Fe(I)(OH) to Fe(II)-Fe(I)(vacant). Actually, an exchangeable water molecule is bound to the active site labeled “vacant” in the figure. The enzyme then accepts an electron and a proton so that Fe(II)-Fe(I) is oxidized to protonated Fe(I)-Fe(I) Complex.The enzyme changes configuration, and a hydride is transferred into the active site of Fe(II) by reduction. The subsequent addition of an electron and a proton allows for the formation of molecular hydrogen, and the loss of H returns the system back to its native active form [6]. | 1,526 | 4,257 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Macromolecules/Macromolecules |
A macromolecule is a very large molecule having a polymeric chain structure. Proteins, polysaccharides, genes, ruber, and synthetic polymers consist of macromolecules. For synthetic polymers, here are the abbreviations for some common polymers: There are only a few known inorganic macromolecules. For example, when liquid sulfur is poured into cold water, long chains of ...-S-S-S-S-S-... are formed. These molecules are present in a phase known as elastic sulfur. Inorganic macromolecules can be divided into several categories: solids formed mainly due to covalent bonds, organosilanes, siloxanes and organosiloxanes. We have mentioned diamond, graphite, silicon, germanium, etc as large molecules. In this section, some more examples are given: Zinc sulfide has two forms (or phases): the and the . These are typical structures, because many other common compounds have the same structure. Knowing the bonding, geometry, symmetry of these structures is important, because scientific discussion and applications are based on their structures. The difference between wurtzite and zinc blende illustrates some fundamental geometry and symmetry difference. Wurtzite is a typical mineral, often involving iron and zinc sulfide, and formulated as (Fe,Zn)S For example: ZnO, SiC, AlN, CaSe, BN, C(Hexagonal Diamond) all have the same crystal structure as wurtzite in terms of bonding, symmetry, packing sequence etc. The zinc blende is cubic. It can be perceived as a f.c.c. lattice of S with half of the tetrahedral sites occupied by Zn. A unit cell of the crystal structure is shown in Zinc blende, and you can see this structure from various perspective. The diagram on the page can be manipulated by moving the mouse. The same group has also got a wurtzite structure that you can manipulate. By the way, the zinc blende structure has the same bonding skeleton as the diamond structure. Thus, the wurtzite and zinc blende structures are two typical structural types for inorganic macromolecules. Since Si-Si is much weaker than C-C bonds, silanes Si H , are not as stable as alkanes. Methane is a stable gas, but silane SiH explode as soon as it comes in contact with air. Silane is made by reacting Mg Si with dilute HCl, but the silane produced burns as soon as it comes in contact with air at the surface of the solution: \[Mg_2Si + 4 HCl \rightarrow 2 MgCl_2 + SiH_4\] \[SiH_4 + 2 O_2 \rightarrow SiO_2 + 2 H_2O\] When an reacts with water, the Si-Si bonds break: \[R_3Si-SiR_3 + 2 H_2O \rightarrow 2 R_3SiOH + H_2\] where \(R\) is an alkyl or aryl group. Organosilanes have some unique applications, and silane coupling agents, R SiX (X being a halogen) are useful. Siloxanes and organosiloxanes have Si-O-Si-O-Si linkage, and these are stable polymers due to the strong Si-O-Si bonds. Since Si atoms tend to form four bonds, these polymers form two- and three-dimensional networks, making them excellent sealants. Silicates are based on Si-O-Si linkages. Quartz for example is based on three-dimensional frame work of these linkages. Sulfur is in the same group as oxygen, and its valence electrons have the electron configuration: 3s 3p . These six electrons usually occupy the four sp hybrid orbitals, two of which have a pair of electrons each, and the other two have only one electron each. Thus, sulfur usually form two bonds such as H S, its structure similar to H O. Sulfur atoms bond to each other forming the cyclic molecules such as S and S . The two bonds and the two lone pairs of electrons around the sulfur point to the direction of a tetrahedron, making the -S- angle approximately 100 degrees. Sulfur has three allotropes: rhombic, monoclinic, and plastic sulfur. At room temperature, monoclinic sulfur is the the stable form. When heated, monoclinic sulfur melts to form a viscous liquid at 119 degree C at the atmosphere pressure. At higher pressure, the monoclinic sulfur transforms into the rhombic sulfur. Both crystalline forms have the S crown shaped molecule and the plastic sulfur has a chain structure of unspecified number of atoms S (n is a very large unspecified number). Since we are talking about the element sulfur, we might include some information on it. Sulfur occur as minerals: pyrite, FeS (also known as fool's gold); gypsum, CaSO ºH O, (when dehydrated it is called paster of paris). Sulfur also occur as an element in nature because some bacterial converts sulfur oxides and other compounds to elemental sulfur. Elemental sulfur is extrated from under ground by hot water and air in a process called the Frasch process. Sulfur is easily oxidized to sulfur dioxide \[S_8 + 8 O_2 \rightarrow 8 SO_2\] The SO is a bent molecule with an O-S-O angle of 120 degrees. It reacts further with water and with oxygen as examplified by the following reaction: SO + H O = H SO (sulfurous acid)
2 SO + O = 2 SO (catalyst required)
SO + H O = H SO (sulfuric acid)
Na SO + S = Na S(=S)O (known as hypo) has only 5 valence elections, one less electron than sulfur. Thus it tends to form three bonds around a P atom. The analogous of ammonia is phosphine, PH , which is air stable with a melting point of -88 C. White phosphorus can be obtained by reducing phosphorus oxide with carbon: \[P_4O_{10} + 10 C \rightarrow P_4 + 10 CO\] The crystals consist of P molecules as shown here, and it undergoes natural combustion if not stored under water. The combustion leads to the formation of P O . White phosphorus converts to stable red, black, violet and scarlet phosphorus, which have complicated network of macromolecules. The bonding in P can be explained in the same manner as that described for sulfur, but that is left as an exercise. As most other non-metallic elements, phosphorus also form a complicated covalent solid. | 5,795 | 4,258 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_OpenStax/07%3A_Chemical_Bonding_(Exercises) |
Does a cation gain protons to form a positive charge or does it lose electrons? The protons in the nucleus do not change during normal chemical reactions. Only the outer electrons move. Positive charges form when electrons are lost. Iron(III) sulfate [Fe (SO ) ] is composed of Fe and \(\ce{SO4^2-}\) ions. Explain why a sample of iron(III) sulfate is uncharged. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: P, I, Mg, Cl, In, Cs, O, Pb, Co? P, I, Cl, and O would form anions because they are nonmetals. Mg, In, Cs, Pb, and Co would form cations because they are metals. Which of the following atoms would be expected to form negative ions in binary ionic compounds and which would be expected to form positive ions: Br, Ca, Na, N, F, Al, Sn, S, Cd? Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds: P ; Mg ; Al ; O ; Cl ; Cs Predict the charge on the monatomic ions formed from the following atoms in binary ionic compounds: Write the electron configuration for each of the following ions: [Ar]4 3 4 ; [Kr]4 5 5 1 [Kr]4 ; [He]2 2 ; [Ar]3 ; 1 (h) [He]2 2 (i) [Kr]4 5 (j) [Ar]3 (k) [Ar]3 , (l) [Ar]3 4 Write the electron configuration for the monatomic ions formed from the following elements (which form the greatest concentration of monatomic ions in seawater): Write out the full electron configuration for each of the following atoms and for the monatomic ion found in binary ionic compounds containing the element: 1 2 2 3 3 ; Al : 1 2 2 ; 1 2 2 3 3 3 4 4 ; 1 2 2 3 3 3 4 4 ; 1 2 2 3 3 3 4 4 5 ; Sr : 1 2 2 3 3 3 4 4 ; 1 2 ; Li : 1 ; 1 2 2 3 3 3 4 4 ; 1 2 2 3 3 3 4 4 ; 1 2 2 3 3 ; 1 2 2 3 3 From the labels of several commercial products, prepare a list of six ionic compounds in the products. For each compound, write the formula. (You may need to look up some formulas in a suitable reference.) Why is it incorrect to speak of a molecule of solid NaCl? NaCl consists of discrete ions arranged in a crystal lattice, not covalently bonded molecules. What information can you use to predict whether a bond between two atoms is covalent or ionic? Predict which of the following compounds are ionic and which are covalent, based on the location of their constituent atoms in the periodic table: ionic: (b), (d), (e), (g), and (i); covalent: (a), (c), (f), (h), (j), and (k) Explain the difference between a nonpolar covalent bond, a polar covalent bond, and an ionic bond. From its position in the periodic table, determine which atom in each pair is more electronegative: Cl; O; O; S; N; P; N From its position in the periodic table, determine which atom in each pair is more electronegative: From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: H, C, N, O, F; H, I, Br, Cl, F; H, P, S, O, F; Na, Al, H, P, O; Ba, H, As, N, O From their positions in the periodic table, arrange the atoms in each of the following series in order of increasing electronegativity: Which atoms can bond to sulfur so as to produce a positive partial charge on the sulfur atom? N, O, F, and Cl Which is the most polar bond? Identify the more polar bond in each of the following pairs of bonds: HF; CO; OH; PCl; NH; PO; CN Which of the following molecules or ions contain polar bonds? Write the Lewis symbols for each of the following ions: eight electrons: eight electrons: no electrons Be ; eight electrons: no electrons Ga ; no electrons Li ; eight electrons: Many monatomic ions are found in seawater, including the ions formed from the following list of elements. Write the Lewis symbols for the monatomic ions formed from the following elements: Write the Lewis symbols of the ions in each of the following ionic compounds and the Lewis symbols of the atom from which they are formed: (a) ; (b) ; (c) ; (d) ; (e) ; (f) In the Lewis structures listed here, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element: (a) (b) (c) (d) Write the Lewis structure for the diatomic molecule P , an unstable form of phosphorus found in high-temperature phosphorus vapor. Write Lewis structures for the following: Write Lewis structures for the following: (a) In this case, the Lewis structure is inadequate to depict the fact that experimental studies have shown two unpaired electrons in each oxygen molecule (b) ; (c) ; (d) ; (e) ; (f) ; (g) ; (h) ; (i) ; (j) ; (k) Write Lewis structures for the following: Write Lewis structures for the following: SeF : ; XeF : ; \(\ce{SeCl3+}\): ; Cl BBCl : Write Lewis structures for: Correct the following statement: “The bonds in solid PbCl are ionic; the bond in a HCl molecule is covalent. Thus, all of the valence electrons in PbCl are located on the Cl ions, and all of the valence electrons in a HCl molecule are shared between the H and Cl atoms.” Two valence electrons per Pb atom are transferred to Cl atoms; the resulting Pb ion has a 6 valence shell configuration. Two of the valence electrons in the HCl molecule are shared, and the other six are located on the Cl atom as lone pairs of electrons. Write Lewis structures for the following molecules or ions: Methanol, H COH, is used as the fuel in some race cars. Ethanol, C H OH, is used extensively as motor fuel in Brazil. Both methanol and ethanol produce CO and H O when they burn. Write the chemical equations for these combustion reactions using Lewis structures instead of chemical formulas. Many planets in our solar system contain organic chemicals including methane (CH ) and traces of ethylene (C H ), ethane (C H ), propyne (H CCCH), and diacetylene (HCCCCH). Write the Lewis structures for each of these molecules. Carbon tetrachloride was formerly used in fire extinguishers for electrical fires. It is no longer used for this purpose because of the formation of the toxic gas phosgene, Cl CO. Write the Lewis structures for carbon tetrachloride and phosgene. Identify the atoms that correspond to each of the following electron configurations. Then, write the Lewis symbol for the common ion formed from each atom: The arrangement of atoms in several biologically important molecules is given here. Complete the Lewis structures of these molecules by adding multiple bonds and lone pairs. Do not add any more atoms. the amino acid serine: urea: pyruvic acid: uracil: carbonic acid: (a) ; (b) ; (c) ; (d) ; (e) A compound with a molar mass of about 28 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound. A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen by mass. Write the Lewis structure for a molecule of the compound. Two arrangements of atoms are possible for a compound with a molar mass of about 45 g/mol that contains 52.2% C, 13.1% H, and 34.7% O by mass. Write the Lewis structures for the two molecules. How are single, double, and triple bonds similar? How do they differ? Each bond includes a sharing of electrons between atoms. Two electrons are shared in a single bond; four electrons are shared in a double bond; and six electrons are shared in a triple bond. Write resonance forms that describe the distribution of electrons in each of these molecules or ions. the formate ion: Write resonance forms that describe the distribution of electrons in each of these molecules or ions. the allyl ion: (a) ; (b) ; (c) ; (d) ; (e) Write the resonance forms of ozone, O , the component of the upper atmosphere that protects the Earth from ultraviolet radiation. Sodium nitrite, which has been used to preserve bacon and other meats, is an ionic compound. Write the resonance forms of the nitrite ion, \(\ce{NO2-}\). In terms of the bonds present, explain why acetic acid, CH CO H, contains two distinct types of carbon-oxygen bonds, whereas the acetate ion, formed by loss of a hydrogen ion from acetic acid, only contains one type of carbon-oxygen bond. The skeleton structures of these species are shown: Write the Lewis structures for the following, and include resonance structures where appropriate. Indicate which has the strongest carbon-oxygen bond. (a) (b) CO has the strongest carbon-oxygen bond because there is a triple bond joining C and O. CO has double bonds. Toothpastes containing sodium hydrogen carbonate (sodium bicarbonate) and hydrogen peroxide are widely used. Write Lewis structures for the hydrogen carbonate ion and hydrogen peroxide molecule, with resonance forms where appropriate. Determine the formal charge of each element in the following: H: 0, Cl: 0; C: 0, F: 0; P: 0, Cl 0; P: 0, F: 0 Determine the formal charge of each element in the following: Calculate the formal charge of chlorine in the molecules Cl , BeCl , and ClF . Cl in Cl : 0; Cl in BeCl : 0; Cl in ClF : 0 Calculate the formal charge of each element in the following compounds and ions: Draw all possible resonance structures for each of these compounds. Determine the formal charge on each atom in each of the resonance structures: ; (b) ; (c) ; (d) Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in nitrosyl chloride: ClNO or ClON? Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in hypochlorous acid: HOCl or OClH? HOCl Based on formal charge considerations, which of the following would likely be the correct arrangement of atoms in sulfur dioxide: OSO or SOO? Draw the structure of hydroxylamine, H NO, and assign formal charges; look up the structure. Is the actual structure consistent with the formal charges? The structure that gives zero formal charges is consistent with the actual structure: Iodine forms a series of fluorides (listed here). Write Lewis structures for each of the four compounds and determine the formal charge of the iodine atom in each molecule: Write the Lewis structure and chemical formula of the compound with a molar mass of about 70 g/mol that contains 19.7% nitrogen and 80.3% fluorine by mass, and determine the formal charge of the atoms in this compound. NF ; Which of the following structures would we expect for nitrous acid? Determine the formal charges: Sulfuric acid is the industrial chemical produced in greatest quantity worldwide. About 90 billion pounds are produced each year in the United States alone. Write the Lewis structure for sulfuric acid, H SO , which has two oxygen atoms and two OH groups bonded to the sulfur. Which bond in each of the following pairs of bonds is the strongest? Using the bond energies in , determine the approximate enthalpy change for each of the following reactions: Using the bond energies in , determine the approximate enthalpy change for each of the following reactions: When a molecule can form two different structures, the structure with the stronger bonds is usually the more stable form. Use bond energies to predict the correct structure of the hydroxylamine molecule: The greater bond energy is in the figure on the left. It is the more stable form. How does the bond energy of HCldiffer from the standard enthalpy of formation of HCl( )? Using the standard enthalpy of formation data in , show how the standard enthalpy of formation of HCl( ) can be used to determine the bond energy. \(\ce{HCl}(g)⟶\dfrac{1}{2}\ce{H2}(g)+\dfrac{1}{2}\ce{Cl2}(g)\hspace{20px}ΔH^\circ_1=−ΔH^\circ_{\ce f[\ce{HCl}(g)]}\\
\dfrac{1}{2}\ce{H2}(g)⟶\ce{H}(g)\hspace{105px}ΔH^\circ_2=ΔH^\circ_{\ce f[\ce H(g)]}\\
\underline{\dfrac{1}{2}\ce{Cl2}(g)⟶\ce{Cl}(g)\hspace{99px}ΔH^\circ_3=ΔH^\circ_{\ce f[\ce{Cl}(g)]}}\\
\ce{HCl}(g)⟶\ce{H}(g)+\ce{Cl}(g)\hspace{58px}ΔH^\circ_{298}=ΔH^\circ_1+ΔH^\circ_2+ΔH^\circ_3\) \(\begin{align}
D_\ce{HCl}=ΔH^\circ_{298}&=ΔH^\circ_{\ce f[\ce{HCl}(g)]}+ΔH^\circ_{\ce f[\ce H(g)]}+ΔH^\circ_{\ce f[\ce{Cl}(g)]}\\
&=\mathrm{−(−92.307\:kJ)+217.97\:kJ+121.3\:kJ}\\
&=\mathrm{431.6\:kJ}
\end{align}\) Using the standard enthalpy of formation data in , calculate the bond energy of the carbon-sulfur double bond in CS . Using the standard enthalpy of formation data in , determine which bond is stronger: the S–F bond in SF ( ) or in SF ( )? The S–F bond in SF is stronger. Using the standard enthalpy of formation data in , determine which bond is stronger: the P–Cl bond in PCl ( ) or in PCl ( )? Complete the following Lewis structure by adding bonds (not atoms), and then indicate the longest bond: The C–C single bonds are longest. Use the bond energy to calculate an approximate value of Δ for the following reaction. Which is the more stable form of FNO ? Use principles of atomic structure to answer each of the following: The first ionization energy of Mg is 738 kJ/mol and that of Al is 578 kJ/mol. Account for this difference. When two electrons are removed from the valence shell, the Ca radius loses the outermost energy level and reverts to the lower = 3 level, which is much smaller in radius. The +2 charge on calcium pulls the oxygen much closer compared with K, thereby increasing the lattice energy relative to a less charged ion. (c) Removal of the 4 electron in Ca requires more energy than removal of the 4 electron in K because of the stronger attraction of the nucleus and the extra energy required to break the pairing of the electrons. The second ionization energy for K requires that an electron be removed from a lower energy level, where the attraction is much stronger from the nucleus for the electron. In addition, energy is required to unpair two electrons in a full orbital. For Ca, the second ionization potential requires removing only a lone electron in the exposed outer energy level. In Al, the removed electron is relatively unprotected and unpaired in a orbital. The higher energy for Mg mainly reflects the unpairing of the 2 electron. The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 200.8 pm. NaF crystallizes in the same structure as LiF but with a Na–F distance of 231 pm. Which of the following values most closely approximates the lattice energy of NaF: 510, 890, 1023, 1175, or 4090 kJ/mol? Explain your choice. For which of the following substances is the least energy required to convert one mole of the solid into separate ions? (d) The reaction of a metal, M, with a halogen, X , proceeds by an exothermic reaction as indicated by this equation: \(\ce{M}(s)+\ce{X2}(g)⟶\ce{MX2}(s)\). For each of the following, indicate which option will make the reaction more exothermic. Explain your answers. The lattice energy of LiF is 1023 kJ/mol, and the Li–F distance is 201 pm. MgO crystallizes in the same structure as LiF but with a Mg–O distance of 205 pm. Which of the following values most closely approximates the lattice energy of MgO: 256 kJ/mol, 512 kJ/mol, 1023 kJ/mol, 2046 kJ/mol, or 4008 kJ/mol? Explain your choice. 4008 kJ/mol; both ions in MgO have twice the charge of the ions in LiF; the bond length is very similar and both have the same structure; a quadrupling of the energy is expected based on the equation for lattice energy Which compound in each of the following pairs has the larger lattice energy? Note: Mg and Li have similar radii; O and F have similar radii. Explain your choices. Which compound in each of the following pairs has the larger lattice energy? Note: Ba and K have similar radii; S and Cl have similar radii. Explain your choices. Na O; Na has a smaller radius than K ; BaS; Ba has a larger charge than K; (c) BaS; Ba and S have larger charges; BaS; S has a larger charge Which of the following compounds requires the most energy to convert one mole of the solid into separate ions? Which of the following compounds requires the most energy to convert one mole of the solid into separate ions? (e) The lattice energy of KF is 794 kJ/mol, and the interionic distance is 269 pm. The Na–F distance in NaF, which has the same structure as KF, is 231 pm. Which of the following values is the closest approximation of the lattice energy of NaF: 682 kJ/mol, 794 kJ/mol, 924 kJ/mol, 1588 kJ/mol, or 3175 kJ/mol? Explain your answer. Explain why the HOH molecule is bent, whereas the HBeH molecule is linear. The placement of the two sets of unpaired electrons in water forces the bonds to assume a tetrahedral arrangement, and the resulting HOH molecule is bent. The HBeH molecule (in which Be has only two electrons to bond with the two electrons from the hydrogens) must have the electron pairs as far from one another as possible and is therefore linear. What feature of a Lewis structure can be used to tell if a molecule’s (or ion’s) electron-pair geometry and molecular structure will be identical? Explain the difference between electron-pair geometry and molecular structure. Space must be provided for each pair of electrons whether they are in a bond or are present as lone pairs. Electron-pair geometry considers the placement of all electrons. Molecular structure considers only the bonding-pair geometry. Why is the H–N–H angle in NH smaller than the H–C–H bond angle in CH ? Why is the H–N–H angle in \(\ce{NH4+}\) identical to the H–C–H bond angle in CH ? Explain how a molecule that contains polar bonds can be nonpolar. As long as the polar bonds are compensated (for example. two identical atoms are found directly across the central atom from one another), the molecule can be nonpolar. As a general rule, MX molecules (where M represents a central atom and X represents terminal atoms; n = 2 – 5) are polar if there is one or more lone pairs of electrons on M. NH (M = N, X = H, n = 3) is an example. There are two molecular structures with lone pairs that are exceptions to this rule. What are they? Predict the electron pair geometry and the molecular structure of each of the following molecules or ions: Identify the electron pair geometry and the molecular structure of each of the following molecules or ions: What are the electron-pair geometry and the molecular structure of each of the following molecules or ions? electron-pair geometry: octahedral, molecular structure: square pyramidal; electron-pair geometry: tetrahedral, molecular structure: bent; (c) electron-pair geometry: octahedral, molecular structure: square planar; electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; electron-pair geometry: trigonal bypyramidal, molecular structure: seesaw; electron-pair geometry: tetrahedral, molecular structure: bent (109°) Predict the electron pair geometry and the molecular structure of each of the following ions: Identify the electron pair geometry and the molecular structure of each of the following molecules: electron-pair geometry: trigonal planar, molecular structure: bent (120°); electron-pair geometry: linear, molecular structure: linear; (c) electron-pair geometry: trigonal planar, molecular structure: trigonal planar; electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal; electron-pair geometry: tetrahedral, molecular structure: tetrahedral; electron-pair geometry: trigonal bipyramidal, molecular structure: seesaw; (g) electron-pair geometry: tetrahedral, molecular structure: trigonal pyramidal Predict the electron pair geometry and the molecular structure of each of the following: Which of the following molecules and ions contain polar bonds? Which of these molecules and ions have dipole moments? All of these molecules and ions contain polar bonds. Only ClF , \(\ce{ClO2-}\), PCl , SeF , and \(\ce{PH2-}\) have dipole moments. Which of the molecules and ions in contain polar bonds? Which of these molecules and ions have dipole moments? Which of the following molecules have dipole moments? SeS , CCl F , PCl , and ClNO all have dipole moments. Identify the molecules with a dipole moment: The molecule XF has a dipole moment. Is X boron or phosphorus? P The molecule XCl has a dipole moment. Is X beryllium or sulfur? Is the Cl BBCl molecule polar or nonpolar? nonpolar There are three possible structures for PCl F with phosphorus as the central atom. Draw them and discuss how measurements of dipole moments could help distinguish among them. Describe the molecular structure around the indicated atom or atoms: tetrahedral; trigonal pyramidal; (c) bent (109°); trigonal planar; bent (109°); bent (109°); (g) CH CCH tetrahedral, CH CCH linear; (h) tetrahedral; (i) H CCCH linear; H CCCH trigonal planar Draw the Lewis structures and predict the shape of each compound or ion: A molecule with the formula AB , in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion for each shape. A molecule with the formula AB , in which A and B represent different atoms, could have one of three different shapes. Sketch and name the three different shapes that this molecule might have. Give an example of a molecule or ion that has each shape. Draw the Lewis electron dot structures for these molecules, including resonance structures where appropriate: predict the molecular shapes for \(\ce{CS3^2-}\) and CS and explain how you arrived at your predictions (a) ; (b) ; (c) ; \(\ce{CS3^2-}\) includes three regions of electron density (all are bonds with no lone pairs); the shape is trigonal planar; CS has only two regions of electron density (all bonds with no lone pairs); the shape is linear What is the molecular structure of the stable form of FNO ? (N is the central atom.) A compound with a molar mass of about 42 g/mol contains 85.7% carbon and 14.3% hydrogen. What is its molecular structure? The Lewis structure is made from three units, but the atoms must be rearranged: Use the to perform the following exercises for a two-atom molecule: Use the to perform the following exercises for a real molecule. You may need to rotate the molecules in three dimensions to see certain dipoles. The molecular dipole points away from the hydrogen atoms. Use the to build a molecule. Starting with the central atom, click on the double bond to add one double bond. Then add one single bond and one lone pair. Rotate the molecule to observe the complete geometry. Name the electron group geometry and molecular structure and predict the bond angle. Then click the check boxes at the bottom and right of the simulator to check your answers. Use the to explore real molecules. On the Real Molecules tab, select H O. Switch between the “real” and “model” modes. Explain the difference observed. The structures are very similar. In the model mode, each electron group occupies the same amount of space, so the bond angle is shown as 109.5°. In the “real” mode, the lone pairs are larger, causing the hydrogens to be compressed. This leads to the smaller angle of 104.5°. Use the to explore real molecules. On the Real Molecules tab, select “model” mode and S O. What is the model bond angle? Explain whether the “real” bond angle should be larger or smaller than the ideal model angle. | 23,156 | 4,259 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Medicinal_Chemistry/Cardiovascular_Drugs |
Cardiovascular disease constitutes the largest single cause of death in the industrialized countries. As with cancer, which is a distant second in terms of mortality, cardiovascular disease morbidity increases with age, accounting for about two-thirds of all deaths in persons over 75. Even though some diseases affect primarily the heart and other diseases effect the vascular system, they cannot be divorced from each other. This obvious interdependence makes a unified imperative. One of the major diseases, atherosclerosis, affects and ultimately damages the heart, kidneys, and other organs. Hypertension is generally defined as an elevation of systolic and/or arterial blood pressure and a value of 140/90 torr is generally accepted as the upper limit of normotension. Certain risk factors (e.g., hypercholesterolemia, diabetes, smoking, and a family history of vascular disease) in conjunction with hypertension predispose to arteriosclerosis and consequent cardiovascular morbidity. Patient populations with sustained diastolic blood pressures in the range of 105 to 129 torr are unequivocally benefited by effective reduction of blood pressure. The benefits of antihypertensive treatment are the avoidance of accelerated or malignant hypertension, a lower incidence of hypertensive renal failure, and a decrease in the incidence of hemorrhagic stroke and cardiac failure. Only recently has it been demonstrated that aggressive care of patients with mild diastolic hypertension can apparently reduce the incidence of myocardial infarction. Since they where introduced in the 1960s, -adrenergic blocking agents have become the most commonly used drugs for cardiovascular diseases. Propranolol was the first -blocker to come into wide clinical use, and it remains the most important of these compounds. It is a highly potent, nonselective adrenergic blocking agent with not intrinsic sympathomimetic activity. However, because of its ability to block receptors in bronchial smooth muscle and skeletal muscle, propranolol interferes with bronchodiation produced by epinephrine and with glycogenolysis, which ordinarily occurs during hypoglycemia. Thus, the drug is usually not used in individuals with asthma and must be used cautiously in diabetics who are receiving insulin. As a consequence, there has been a search for -blockers that are cardioselective and a number of drugs now have been developed that exhibit some degree of specificity for 1-adrenergic receptors. Metaprolol is a prototype for these more specific drugs. Much of the pharmacology of -blockers can be deduced from a knowledge of the functions subserved by the involved receptors and the physiological conditions under which they are activated. Thus, b-receptor blockade has little effect on the normal heart with the subject at complete rest, but may have profound effects when sympathetic control of the heart is high, as during exercise. -blockers decrease heart rate and cardiac output, prolongs mechanical systole, and slightly decreases blood pressure in resting subjects. Peripheral resistance is increased as a result of compensatory sympathetic reflexes, and blood flow to all tissues except the brain is reduced. Some of these drugs also have direct actions on cell membranes, which are commonly described as membrane stabilizing and local anesthetic. The local anesthetic potency of propranolol is about equal to that of lidocaine. -blockers are effective antihypertensive agents. Chronic treatment of hypertensive patients with b-adrenergic blocking agents results in a slowly developing reduction in blood pressure. Several mechanisms have been proposed for the efficacy of b-blockers in the management of hypertension. Reduction in cardiac output occurs rather promptly after administration of b-blockers. However, the hypotensive effects of -blockers does not appear as rapidly. The release of renin from the juxtaglumarular apparatus is stimulated by adrenergic agonists, and this effect is blocked by drugs such as propranolol. Also, -adrenergic agonists are known to increase modestly the release of norepinephrine from adrenergic nerve terminals, which causes vasoconstriction. -blockers block this effect, and such impairment of the release of norepinephrine following sympathetic nerve stimulation might contribute to the antihypertensive effects of the drugs. a-Adrenergic blocking agents exist in most blood vessels, particularly in coetaneous resistance vessels. Since their stimulation leads to constriction and therefore blood pressure elevation, it stands to reason that blocking such stimulation would lead to a diminution of blood pressure. Phenoxybenzamine binds covalently to the a receptor and produces an irreversible type of blockade. Phentolamine and tolazoline bind reversibly and antagonize the actions of sympathomimetic amines competitively. is a climbing shrub indigenous of India and neighboring countries. In 1954, it was reported that rauwolfia or reserpine was helpful in the treatment of psychotic patients because it acts centrally to produce characteristic sedation and a state of indifference to environmental stimuli; effects which resemble phenothiazines (discussed in Psychoactive Drugs). Subsequent discovery of the ability of rauwolfia alkaloids and related compounds to deplete biogenic amines from storage sites in the body initiated a great number of investigations directed at elucidating the interaction between these amines and reserpine. There are a number of rauwolfia alkaloids with complex structures. The structure of reserpine is as follows: It is clear that reserpine interferes with intracellular storage of catecholamines, but the amounts of reserpine in tissues are much too small to assume a stoichiometric displacement. Reserpine antagonizes the uptake of norepinephrine by isolated chromaffin granules by inhibiting the ATP-Mg -dependent uptake mechanism of the granule membrane. The drug also binds to the vesicular membranes for days, accounting for the irreversibility of the process. Reserpine depletes stores of catecholamines in many organs, including the brain and adrenal medulla, and most of its pharmacological effects have been attributed to this action. Since it is the reuptake and not the release of catecholamines that is inhibited, the existing pool must be fully depleted before antihypertensive effects become apparent. Also,, after administration of relatively large doses, reserpine causes a transient sympathomimetic effect followed by a slowly developing fall in blood pressure frequently associated with bradycardia. For normal doses, reduced concentrations of catecholamines can be measured within a hour after administration of reserpine, and depletion is maximal at 24 hours. Most of the catecholamine is deaminated intraneuronally, and pharmacological effects of the released mediator are minimal unless MAO has been inhibited. The decrease in norepinephrine synthesis induced by reserpine is the block of dopamine uptake into storage granules that contain the enzyme dopamine -hydroxylase ( ). Furthermore, the increased concentration of free catecholamine presumably feeds back to inhibit tyrosine hydroxylase, since norepinephrine competes with the pterin cofactor for the enzyme. Supersensitivity to catecholamines is observed following chronic administration of reserpine. The site of change is presumably postjunctional and may by due to alterations of the adrenergic receptors. Such adaptive change is usual following chronic deprivation of transmitter. Reserpine is used as an effective antihypertensive agent, particularly when used with other agents such as diuretics. Its low cost, once-daily administration, and minimal change in effect when compliance is erratic make it useful as an agent for long-term treatment of patients with uncomplicated mild hypertension. However, reserpine causes mental depression in 25% of patients. A diuretic is a substance that increases the rate of urine volume output, as the name implies. Most diuretics also increase urinary excretion of solutes, especially sodium and chloride. In fact, most diuretics that are used clinically act by decreasing the rate of sodium reabsorbtion from the nephron tubules, which in turn causes naturesis and this in turn causes diuresis. The most common clinical use of diuretics is to reduce extracellular fluid volume, especially in diseases associated with edema and hypertension. Soon after the introduction of the sulfonamides as antibacterial agents in the 1930s, changes in the electrolyte balance of patients were noted as was systemic acidosis accompanied by an alkalization of the urine due to increased rate of HCO excretion. Proposals by several researchers established that inhibition of the enzyme carbonic anhydrase (CA) accounted for the electrolytic imbalances produced. Since the antibacterial sulfonamides were relatively weak inhibitors, a successful search for more potent CA inhibition ensued. Acetazolamide became the first successful drug introduced into clinical use. Among the enormous number of sulfonamides that have been synthesized and tested, acetazolamide has been studied the most extensively as an inhibitor of carbonic anhydrase. The other drugs of this class that are available in the U.S. are dichlorphenamide and methazolamide. There structural formulas are as follows: The kidneys control acid-base balance of the body by excreting either an acidic or a basic urine. The overall mechanism by which the kidneys accomplish this is as follows: Large numbers of bicarbonate ions are filtered continuously into the nephron tubules of the kidneys, and if they are excreted into the urine, this removes base from the blood. On the other hand, large numbers of hydrogen ions are also secreted into the tubular lumen by the tubular epithelial cells, thus removing acid from the blood. Bicarbonate ions enter the tubular lumen of the kidney nephron with the glumerular filtrate. Bicarbonate ions do not readily permeate the luminal membranes of the renal tubular cells; therefore, bicarbonate ions that are filtered by the glomerulus cannot be directly reabsorbed. Instead, bicarbonate is reabsorbed by a special process in which it first combines with hydrogen ions to from H CO , which eventually becomes CO and H O. \[\ce{ H^{+} + HCO3 <=> H2CO3 <=> CO2 + H2O}\] This reabsorbption of bicarbonate ions is initiated by a reaction in the tubules between bicarbonate ions filtered at the glomerulus and hydrogen ions secreted by the tubular cells. The H CO formed then dissociates into CO and H O. The CO can move easily across the tubular membrane; therefore, it instantly diffuses into the tubular cell, where it recombines with H O, under the influence of carbonic anhydrase, to generate new a H CO molecule. This H CO in turn dissociates to form bicarbonate ion and hydrogen ion. The hydrogen ions are secreted from the cell into the tubular lumen, by sodium-hydrogen pump, in exchange for sodium. The bicarbonate ions together with the exchanged Na+, then enter the peritubular blood supply. The hydrogen ion, now in the tubular lumen, combines with another bicarbonate ion to form H CO , which then again dehydrates to CO , which reenters the tubular cell. The net result is reabsorbtion of most of the bicarbonate. Some of these concepts can be seen here in an animated nephron (you must have the shockwave plugin to view this). A hydrogen-bonding mechanism that acts competitively explains the action of acetazolamide-like carbonic anhydrase inhibitors that have diuretic properties. Carbonic acid is the normal substrate that fits into a cavity of and complexes with the enzyme carbonic anhydrase. This complex is strongly stabilized by four hydrogen bonds (see figure below). The acetazolamide-like drugs fit into the cavity of the enzyme also effectively bond, presumably to the same four areas by hydrogen bonds (see figure below). Thus these sulfonamide agents competitively prevent the carbonic acid from binding at this site which inhibits the reabsorption of NaHCO and H 0. More than 99% of the enzyme activity in the kidney must be inhibited before physiological effects become apparent. Following the administration of acetazolamide, the urine volume promptly increases. The urinary excretion of bicarbonate and fixed cation, mostly sodium (also potassium) and the normally acidic pH becomes alkaline. As a result, the concentration of bicarbonate in the extracellular fluid decreases and metabolic acidosis results. The urinary concentration of chloride falls. The presence of carbonic anhydrase in a number of intraocular structure, including the cilliary processes, and the high concentration of bicarbonate in the aqueous humor have focused attention on the role that the enzyme might play in the secretion of aqueous humor. acetazolamide reduces the rate of aqueous humor formation; intaocular pressure in patients with glaucoma is correspondingly reduced. Thiazines and related compounds comprise the most frequently used antihypertensive agents in the United States. They were synthesized as an out growth of studies on carbonic anhydrase inhibitors. Thiazides act directly of the kidney to increase the excretion of sodium chloride and an accompanying volume of water. Also many of these drugs act as corbonic anhydrase inhibitors. At the molecular level, it is unknown how benzothiadiazines inhibit the reuptake of sodium chloride. Angina pectoris, or ischemic heart disease, is the name given to the symptomatic oppressive pain resulting from myocardial ischemia. In simplest terms it results when the oxygen demand of myocardial tissues exceeds the circulatory supply. Once a local anemia due to an obstruction exists, biochemical changes are inevitable. Metabolic products will accumulate in the area, contractility declines, and NE release occurs from sympathetic neurons. The end result is pain. The underlying pathology of typical angina is usually advanced atherosclerosis and stenosis of the coronary vasculature which causes localized oxygen starvation. Episodes are be precipitated by emotional stress or exercise, but they usually cease rapidly with rest or nitroglycerin. In contrast, variant angina is caused by vasospasm of the coronary vessel and may or may not be associated with severe atherosclerosis. Patients with variant angina develop chest pain while at rest. Until recently, many of the drugs used to prevent anginal attacks were no more effective than a placebo. In fact. the use of placebos has been reported to relieve symptoms in as many as 50% of patients with angina pectoris. For over a century, however, nitroglycerin has been known to be useful to prevent or relieve acute anginal attacks. More recently, the efficacy of -adrenergic antagonists has been established for the long-term prophylaxis of typical angina. In addition, the calcium channel blockers appear to be effective for the treatment of vasospastic angina. While antianginal agents provide only symptomatic treatment, administration of these drugs do appear to decrease the incidence of sudden death associated with myocardial ischemia and infarction. The organic nitrates and nitrites are dilators of arterial and venous smooth muscle. The venodilation results in decreased left and right ventricular end-diastolic pressure, which are greater on a percentage basis than is the decrease in systemic arterial pressure. Net systemic vascular resistance is usually relatively unaffected; heart rate is unchanged or slightly increased; and pulmonary vascular resistance is consistently reduced. These drugs correct the inadequacy of myocardial oxygenation by increasing the supply of oxygen to ischemic myocardium by direct dilatation of the coronary vasculature and by decreasing the oxygen demand for oxygen by a reduction in cardiac work. The latter results from the decrease in vascular pressure enabling the heart to pump blood easier. Glyceryl trinitrate, or nitroglycerine, is a dense sweet-smelling oil that is highly explosive. The most utilized dosage form of nitroglycerin has been the sublingual tablet. Buccal absorption is rapid, offering almost instantaneous relief of sufficient duration (<30 min) for the emergency. Nevertheless, because of the valuable properties of nitroglycerin is now known to have in angina pectoralis, continuous blood levels of the drug are desirable. Therefore, different and innovative dosage forms are being developed. Because nitroglycerin is efficiently absorbed through the skin, this has led to the introduction of nitroglycerin skin patches. These patches contain the drug in a form which results in its continuous release.
Nitric oxide (NO) has been shown to be an important messenger in many signal transduction processes. This free radical gas is naturally produced endogenously from arginine in a complete reaction that is catalyzed by nitric oxide synthetase (NOS). Nitroglycerin, nitrites, and organic nitrates also lead to the formation of the reactive free radical NO and is the basis of their mechanism of action. However, it is not known the exact enzyme that converts these drugs into NO. Nitric oxide diffuses freely across membranes but has a short life, less than a few seconds, because it is highly reactive. Nitric oxide then activates a heme prosthetic group on the enzyme guanylate cyclase in the cell membrane. The heme molecule is, in effect, functioning as an extremely sensitive detector of NO. A portion of the enzyme protrudes to the interior of the cell and causes the formation of cyclic guanosine monophosphate (cGMP), a so-called second messenger. The cGMP in turn has many effects, one of which is to change the degree of phosphorylation of several enzymes that indirectly inhibit contraction. Especially, the pump that pumps calcium ions from the sarcoplasm into the sarcoplasmic reticulum in activated as well as the cell membrane pump that pumps calcium ions out of the cell itself; these effects reduce the intracellular calcium ion concentration, thereby inhibiting contraction. , an experimental heart medication with a great side effect.More on the Digitalis is the dried leaf of the foxglove plant, . Seeds and leaves of a number of other species also contain active cardiac principles. The two molecules in digitalis responsible for its pharmacological activity are digoxigenin and digitoxigenin. As you can see from the structures below, these genins are chemically related to sex and adrenocortical hormones. Digitalis has a powerful action of the myocardium that is unrivaled in value for the treatment of heart failure. It is also used to slow the ventricular rate in the presence of atrial fibrillation and flutter. The main pharmacodynamic property of digitalis is its ability to increase the force of myocardial contraction. The beneficial effects of the drug in patients with heart failure - increased cardiac output; decreased heart size. venous pressure, and blood volume; diuresis and relief of edema - are all explained on the basis of increased contractile force, a positive inotropic action. The most important component to the positive inotropic effect of digitalis direct inhibition of the membrane-bound Na+, K+-activated ATPase, which leads to an increase in intracellular [Ca+]. It seems clear that digitalis, in therapeutic concentrations, exerts no direct effect on the contractile proteins or on the interactions between them. Also, it seems most unlikely that the positive inotropic effect of digitalis is due to any action of the cellular mechanisms that provide the chemical energy for contraction. The hydrolysis of ATP by the Na+, K+-ATPase provides the energy for the so-called sodium-potassium pump - the system in the sarcolemma of cardiac fibers that actively extrudes sodium and transports potassium into the fibers. Digitalis drugs bind specifically to the Na+, K+-ATPase, inhibit its enzymatic activity, and impair the active transport of these two monovalent cations. As a result, there is a gradual increase in intracellular [Na+] and a gradual decrease in internal [K+]. These changes are small at therapeutic concentrations of the drug. It is judged to be crucially related to the positive inotropic effect of digitalis. This is so because in cardiac fibers, intracellular Ca is exchanged for extracellular Na+ by a different transmembrane pump. When internal [Na+] is increased because of inhibition of the Na+, K+ pump by digitalis, the exchange of extracellular Na+ for intracellular Ca is diminished, and internal [Ca ] increases. The probable consequence of this is an increased store of Ca in the sarcoplasmic reticulum and, with each action potential, a greater release of Ca to activate the contractible apparatus. Interactions of Na+, K+-ATPase with its various substrates are complex. Thus binding affinities with ATP, cofactor MG2+, Na+, K+, and a digitalis glycoside are all important to the overall effect. It is now accepted that the digitalis receptor is one or more of the conformations of Na+, K+-ATPase that occur during the ion pump's operation, possibly during a state in which the drug helps to stabilize one of the intermediate states of the enzyme. Evidence suggests that the entire glycoside molecule participates in the proposed drug-receptor interaction. The steric relationship of the lactone ring to the steroid nucleus is absolute. The double bond is also critical since saturation results in an almost total loss of activity. The required stereochemical positioning of rings C and D in relation to each other (cis) and of A and B (cis), and the configuration of the OH at C-14 (between rings C and D) have all been established. The figure below is a highly simplified version of a proposed interaction of the lactone side chain with such a digitalis receptor. It is believed that the polarized carbonyl group, with its electron-rich oxygen, hydrogen bonds to a hydroxyl group on the enzyme's surface, probably to a serine residue, and the carbon atom bonded to the steroid nucleus is attracted to an electron dense site at a secondary location. From any viewpoint blood is chemically the most complex tissue in the body. In addition to the multitude of cells and platelets, it contains inorganic ions, various plasma proteins, hormones, lipids, vitamins, a large variety of enzymes, nucleic acid breakdown products, a large number of unknown types of environmentally ingested chemical at varying stages of metabolic conversion, gases, and water. Among these is a group of more than a dozen chemical factors that will cause the blood to coagulate when properly triggered. Hemostasis is the spontaneous arrest of bleeding from damaged vessels. Precapillary vessels contract immediately when cut. Within seconds, thrombocytes are bound to the exposed collagen of the injured vessel. Platelets also stick to each other and a viscous mass is formed. This platelet plug can stop bleeding quickly, but it must be reinforced by fibrin for long-term effectiveness. This reinforcement is initiated by the local stimulation of the coagulation process by the expose collagen of the cut vessel and the released contents and membranes of platelets. The two pathways of blood coagulation are shown in the figure below. The circled clotting factors are dependent on vitamin K for their activation. Thrombogenesis is an altered state of hemostasis. An intravascular thrombus results from a pathological disturbance of hemostasis. The arterial thrombus is initiated by the adhesion and the release of circulating platelets to a vessel wall. This initial adhesion and the release of adenosine diphosphate (ADP) from platelets is followed by platelet-platelet aggregation. The thrombus grows to occlusive proportions in the areas of slower arterial blood flow. A platelet plug formed solely by platelet interaction is unstable. After the initial aggregation and viscous metamorphosis of platelets, fibrin becomes an important constituent of a thrombus. Production of thrombin occurs by activation of the reactions of blood coagulation at the site of the platelet mass. This thrombin stimulates further platelet aggregation not only by inducting the release of mote ADP from the platelets but also by stimulating the synthesis of prostaglandins, which are more powerful than ADP. The first orally active anticoagulant, dicoumerol, which is a molecule consisting of two 4-hydroxycoumarin moieties bonded at their 3-position via a CH2 bridge, was isolated from decomposed yellow sweet clover. It was discovered and identified because of hemorrhagic death of cattle ingesting this improperly stored feed in the early 1920s (sweet clover disease). This followed demonstration that oral use of this compound increased clotting time and decreased the incidence of post surgical intravascular thrombus formation. The oral anticoagulants are antagonists of vitamin K. Their administration to man or other animals leads to the appearance of precursors of the four vitamin K-dependent clotting factors in plasma and liver. These precursor proteins are biologically inactive in tests of coagulation. The precursor protein to prothrombin can be activated to thrombin nonphysiologically by snake venoms, demonstrating that the portion of the molecule necessary for this activity is intact. However, the prothrombin and other precursor proteins cannot bind divalent cations such as calcium and thus cannot interact with phospholipid-containing membranes, which are their normal sites of activation. This was puzzling for some time because abnormal prothrombin has the same number to amino acid residues and gives the same amino acid analysis after acid hydrolysis as does normal prothrombin. NMR studies revealed that normal prothrombin contains -carboxyglutamate (see figure below), a previously unknown residue. The vitamin K-sensitive step in the synthesis of clotting factors is the carboxylation of ten or more glutamic acid residues at the amino-terminal end of the precursor protein to form -carboxyglutamate. These amino acid residues are much stronger chelators of calcium than glutamate. The binding of Ca2+ by prothrombin anchors it to phospholipid membranes derived from blood platelets following injury. Th functional significance of the binding of prothrombin to phospholipid surfaces is that it brings prothrombin into close proximity with two proteins that mediate its conversion into thrombin-Factor Xa and Factor V. The amino-terminal fragment of prothrombin, which contains the Ca2+-binding sites, is released in this activation step. Thrombin free in this way from the phospholid surface can then activate fibrinogen in the plasma. In hepatic microsomes, the reduction of vitamin K to its hydroquinone form (vitamin KH ) precedes the bicarbonate-dependent carboxylation of precursor prothrombin, descarboxyprothrombin, to prothrombin (see figure below). This carboxylase activity for the synthesis of prothrombin is linked to an epoxidase activity for vitamin KH , which oxidizes the vitamin to vitamin K epoxide (KO). An epoxide reductase, which requires NADH, converts vitamin K epoxide back to vitamin KH . This reaction is the site of action of the coumarins and the site of genetic resistance to the coumarins, which is also characterized by an increase requirement for vitamin K. Thus it is the recycling of the vitamin to its reduced active cofactor that results ultimately in decreased thrombin levels. The vitamin K analog chloro-K1 (in which the 2-methyl group of vitamin K1 is replaced by a chloro group) directly inhibits the carboxylase and epoxidase reaction, which are not sensitive to coumarins. Dicoumerol, the prototype of coumarin drugs, is of relatively low potency, with a slow onset of up to 5 days for peak activity and hypoprothrombinemia. The anticoagulant effect may persist for more than 1 week after stopping the drug. Even though over doses can be antidoted with vitamin K1, clinical adjustment of anticoagulation, particularly downward, is difficult. Warfarin has become the most widely used of the coumarin drugs. It is the most potent, with many patient being maintained on only 5 mg/day Warfarin was initially introduced as a rodenticide because it was thought too dangerous for human use. It is still used in pest control. The drug, as sweetened pellets, causes rats to die from internal bleeding. In the early 1960s rats resistant to warfarin were noted in London. They were nicknamed "super rats." Several years later this phenomenon appeared in humans. It has been shown to be an inherited autosomal dominant trait. Person with this trait require a 20-fold increase in the drug to achieve anticoagulation-easily fatal to normal patients. Explanations for this unusual phenomenon have been proposed. One is that a tissue protein regulating the synthesis of one or more of the clotting factors became genetically altered. Another is a mutation in the enzyme daphorase that makes it less susceptible to coumarin drug inhibition. | 29,061 | 4,260 |
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In terms of their bulk properties, how do liquids and solids differ? How are they similar? Liquids and solids are similar in that they are matter composed of atoms, ions, or molecules. They are incompressible and have similar densities that are both much larger than those of gases. They are different in that liquids have no fixed shape, and solids are rigid. In terms of the kinetic molecular theory, in what ways are liquids similar to solids? In what ways are liquids different from solids? In terms of the kinetic molecular theory, in what ways are liquids similar to gases? In what ways are liquids different from gases? They are similar in that the atoms or molecules are free to move from one position to another. They differ in that the particles of a liquid are confined to the shape of the vessel in which they are placed. In contrast, a gas will expand without limit to fill the space into which it is placed. Explain why liquids assume the shape of any container into which they are poured, whereas solids are rigid and retain their shape. What is the evidence that all neutral atoms and molecules exert attractive forces on each other? All atoms and molecules will condense into a liquid or solid in which the attractive forces exceed the kinetic energy of the molecules, at sufficiently low temperature. Open the to answer the following questions: Define the following and give an example of each: The types of intermolecular forces in a substance are identical whether it is a solid, a liquid, or a gas. Why then does a substance change phase from a gas to a liquid or to a solid? Why do the boiling points of the noble gases increase in the order He < Ne < Ar < Kr < Xe? The London forces typically increase as the number of electrons increase. Neon and HF have approximately the same molecular masses. Arrange each of the following sets of compounds in order of increasing boiling point temperature: (a) SiH < HCl < H O; (b) F < Cl < Br ; (c) CH < C H < C H ; (d) N < O < NO The molecular mass of butanol, C H OH, is 74.14; that of ethylene glycol, CH (OH)CH OH, is 62.08, yet their boiling points are 117.2 °C and 174 °C, respectively. Explain the reason for the difference. On the basis of intermolecular attractions, explain the differences in the boiling points of –butane (−1 °C) and chloroethane (12 °C), which have similar molar masses. Only rather small dipole-dipole interactions from C-H bonds are available to hold -butane in the liquid state. Chloroethane, however, has rather large dipole interactions because of the Cl-C bond; the interaction is therefore stronger, leading to a higher boiling point. On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of acetone (56.2 °C) and 1-propanol (97.4 °C), which have similar molar masses. The melting point of H O( ) is 0 °C. Would you expect the melting point of H S( ) to be −85 °C, 0 °C, or 185 °C? Explain your answer. −85 °C. Water has stronger hydrogen bonds so it melts at a higher temperature. Silane (SiH ), phosphine (PH ), and hydrogen sulfide (H S) melt at −185 °C, −133 °C, and −85 °C, respectively. What does this suggest about the polar character and intermolecular attractions of the three compounds? Explain why a hydrogen bond between two water molecules is weaker than a hydrogen bond between two hydrogen fluoride molecules. The hydrogen bond between two hydrogen fluoride molecules is stronger than that between two water molecules because the electronegativity of F is greater than that of O. Consequently, the partial negative charge on F is greater than that on O. The hydrogen bond between the partially positive H and the larger partially negative F will be stronger than that formed between H and O. Under certain conditions, molecules of acetic acid, CH COOH, form “dimers,” pairs of acetic acid molecules held together by strong intermolecular attractions: Draw a dimer of acetic acid, showing how two CH COOH molecules are held together, and stating the type of IMF that is responsible. Proteins are chains of amino acids that can form in a variety of arrangements, one of which is a helix. What kind of IMF is responsible for holding the protein strand in this shape? On the protein image, show the locations of the IMFs that hold the protein together: H-bonding is the principle IMF holding the DNA strands together. The H-bonding is between the \(\mathrm{N−H}\) and \(\mathrm{C=O}\). The density of liquid NH is 0.64 g/mL; the density of gaseous NH at STP is 0.0007 g/mL. Explain the difference between the densities of these two phases. Identify the intermolecular forces present in the following solids: (a) hydrogen bonding and dispersion forces; (b) dispersion forces; (c) dipole-dipole attraction and dispersion forces The test tubes shown here contain equal amounts of the specified motor oils. Identical metal spheres were dropped at the same time into each of the tubes, and a brief moment later, the spheres had fallen to the heights indicated in the illustration. Rank the motor oils in order of increasing viscosity, and explain your reasoning: Although steel is denser than water, a steel needle or paper clip placed carefully lengthwise on the surface of still water can be made to float. Explain at a molecular level how this is possible: (credit: Cory Zanker) The water molecules have strong intermolecular forces of hydrogen bonding. The water molecules are thus attracted strongly to one another and exhibit a relatively large surface tension, forming a type of “skin” at its surface. This skin can support a bug or paper clip if gently placed on the water. The surface tension and viscosity values for diethyl ether, acetone, ethanol, and ethylene glycol are shown here. You may have heard someone use the figure of speech “slower than molasses in winter” to describe a process that occurs slowly. Explain why this is an apt idiom, using concepts of molecular size and shape, molecular interactions, and the effect of changing temperature. Temperature has an effect on intermolecular forces: the higher the temperature, the greater the kinetic energies of the molecules and the greater the extent to which their intermolecular forces are overcome, and so the more fluid (less viscous) the liquid; the lower the temperature, the lesser the intermolecular forces are overcome, and so the less viscous the liquid. It is often recommended that you let your car engine run idle to warm up before driving, especially on cold winter days. While the benefit of prolonged idling is dubious, it is certainly true that a warm engine is more fuel efficient than a cold one. Explain the reason for this. The surface tension and viscosity of water at several different temperatures are given in this table. (a) As the water reaches higher temperatures, the increased kinetic energies of its molecules are more effective in overcoming hydrogen bonding, and so its surface tension decreases. Surface tension and intermolecular forces are directly related. (b) The same trend in viscosity is seen as in surface tension, and for the same reason. At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.63 mm? Refer to for the required information. Water rises in a glass capillary tube to a height of 17 cm. What is the diameter of the capillary tube? 9.5 × 10 m Heat is added to boiling water. Explain why the temperature of the boiling water does not change. What does change? Heat is added to ice at 0 °C. Explain why the temperature of the ice does not change. What does change? The heat is absorbed by the ice, providing the energy required to partially overcome intermolecular attractive forces in the solid and causing a phase transition to liquid water. The solution remains at 0 °C until all the ice is melted. Only the amount of water existing as ice changes until the ice disappears. Then the temperature of the water can rise. What feature characterizes the dynamic equilibrium between a liquid and its vapor in a closed container? Identify two common observations indicating some liquids have sufficient vapor pressures to noticeably evaporate? We can see the amount of liquid in an open container decrease and we can smell the vapor of some liquids. Identify two common observations indicating some solids, such as dry ice and mothballs, have vapor pressures sufficient to sublime? What is the relationship between the intermolecular forces in a liquid and its vapor pressure? The vapor pressure of a liquid decreases as the strength of its intermolecular forces increases. What is the relationship between the intermolecular forces in a solid and its melting temperature? Why does spilled gasoline evaporate more rapidly on a hot day than on a cold day? As the temperature increases, the average kinetic energy of the molecules of gasoline increases and so a greater fraction of molecules have sufficient energy to escape from the liquid than at lower temperatures. Carbon tetrachloride, CCl , was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °C, the vapor pressure of CCl is 54.0 kPa, and its enthalpy of vaporization is 33.05 kJ/mol. Use this information to estimate the normal boiling point for CCl . When is the boiling point of a liquid equal to its normal boiling point? When the pressure of gas above the liquid is exactly 1 atm How does the boiling of a liquid differ from its evaporation? Use the information in to estimate the boiling point of water in Denver when the atmospheric pressure is 83.3 kPa. approximately 95 °C A syringe at a temperature of 20 °C is filled with liquid ether in such a way that there is no space for any vapor. If the temperature is kept constant and the plunger is withdrawn to create a volume that can be occupied by vapor, what would be the approximate pressure of the vapor produced? Explain the following observations: (a) At 5000 feet, the atmospheric pressure is lower than at sea level, and water will therefore boil at a lower temperature. This lower temperature will cause the physical and chemical changes involved in cooking the egg to proceed more slowly, and a longer time is required to fully cook the egg. (b) As long as the air surrounding the body contains less water vapor than the maximum that air can hold at that temperature, perspiration will evaporate, thereby cooling the body by removing the heat of vaporization required to vaporize the water. The enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why. Explain why the molar enthalpies of vaporization of the following substances increase in the order CH < C H < C H , even though all three substances experience the same dispersion forces when in the liquid state. Dispersion forces increase with molecular mass or size. As the number of atoms composing the molecules in this homologous series increases, so does the extent of intermolecular attraction via dispersion forces and, consequently, the energy required to overcome these forces and vaporize the liquids. Explain why the enthalpies of vaporization of the following substances increase in the order CH < NH < H O, even though all three substances have approximately the same molar mass. The enthalpy of vaporization of CO ( ) is 9.8 kJ/mol. Would you expect the enthalpy of vaporization of CS ( ) to be 28 kJ/mol, 9.8 kJ/mol, or −8.4 kJ/mol? Discuss the plausibility of each of these answers. The boiling point of CS is higher than that of CO partially because of the higher molecular weight of CS ; consequently, the attractive forces are stronger in CS . It would be expected, therefore, that the heat of vaporization would be greater than that of 9.8 kJ/mol for CO . A value of 28 kJ/mol would seem reasonable. A value of −8.4 kJ/mol would indicate a release of energy upon vaporization, which is clearly implausible. The hydrogen fluoride molecule, HF, is more polar than a water molecule, H O (for example, has a greater dipole moment), yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water. Explain. Ethyl chloride (boiling point, 13 °C) is used as a local anesthetic. When the liquid is sprayed on the skin, it cools the skin enough to freeze and numb it. Explain the cooling effect of liquid ethyl chloride. The thermal energy (heat) needed to evaporate the liquid is removed from the skin. Which contains the compounds listed correctly in order of increasing boiling points? How much heat is required to convert 422 g of liquid H O at 23.5 °C into steam at 150 °C? 1130 kJ Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20-ounce bottle of water that had been in the refrigerator at 3.8 °C, how much heat is needed to convert all of that water into sweat and then to vapor?? (Note: Your body temperature is 36.6 °C. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.) Titanium tetrachloride, TiCl , has a melting point of −23.2 °C and has a Δ = 9.37 kJ/mol. (a) 13.0 kJ; (b) It is likely that the heat of vaporization will have a larger magnitude since in the case of vaporization the intermolecular interactions have to be completely overcome, while melting weakens or destroys only some of them. From the phase diagram for water, determine the state of water at: What phase changes will take place when water is subjected to varying pressure at a constant temperature of 0.005 °C? At 40 °C? At −40 °C? At low pressures and 0.005 °C, the water is a gas. As the pressure increases to 4.6 torr, the water becomes a solid; as the pressure increases still more, it becomes a liquid. At 40 °C, water at low pressure is a vapor; at pressures higher than about 75 torr, it converts into a liquid. At −40 °C, water goes from a gas to a solid as the pressure increases above very low values. Pressure cookers allow food to cook faster because the higher pressure inside the pressure cooker increases the boiling temperature of water. A particular pressure cooker has a safety valve that is set to vent steam if the pressure exceeds 3.4 atm. What is the approximate maximum temperature that can be reached inside this pressure cooker? Explain your reasoning. From the phase diagram for carbon dioxide, determine the state of CO at: (a) liquid; (b) solid; (c) gas; (d) gas; (e) gas; (f) gas Determine the phase changes that carbon dioxide undergoes as the pressure changes if the temperature is held at −50 °C? If the temperature is held at −40 °C? At 20 °C? Consider a cylinder containing a mixture of liquid carbon dioxide in equilibrium with gaseous carbon dioxide at an initial pressure of 65 atm and a temperature of 20 °C. Sketch a plot depicting the change in the cylinder pressure with time as gaseous carbon dioxide is released at constant temperature. Dry ice, CO ( ), does not melt at atmospheric pressure. It sublimes at a temperature of −78 °C. What is the lowest pressure at which CO ( ) will melt to give CO ( )? At approximately what temperature will this occur? (See for the phase diagram.) If a severe storm results in the loss of electricity, it may be necessary to use a clothesline to dry laundry. In many parts of the country in the dead of winter, the clothes will quickly freeze when they are hung on the line. If it does not snow, will they dry anyway? Explain your answer. Yes, ice will sublime, although it may take it several days. Ice has a small vapor pressure, and some ice molecules form gas and escape from the ice crystals. As time passes, more and more solid converts to gas until eventually the clothes are dry. Is it possible to liquefy nitrogen at room temperature (about 25 °C)? Is it possible to liquefy sulfur dioxide at room temperature? Explain your answers. Elemental carbon has one gas phase, one liquid phase, and three different solid phases, as shown in the phase diagram: What types of liquids typically form amorphous solids? Amorphous solids lack an ordered internal structure. Liquid materials that contain large, cumbersome molecules that cannot move readily into ordered positions generally form such solids. At very low temperatures oxygen, O , freezes and forms a crystalline solid. Which best describes these crystals? (e) molecular crystals As it cools, olive oil slowly solidifies and forms a solid over a range of temperatures. Which best describes the solid? (d) amorphous Explain why ice, which is a crystalline solid, has a melting temperature of 0 °C, whereas butter, which is an amorphous solid, softens over a range of temperatures. Ice has a crystalline structure stabilized by hydrogen bonding. These intermolecular forces are of comparable strength and thus require the same amount of energy to overcome. As a result, ice melts at a single temperature and not over a range of temperatures. The various, very large molecules that compose butter experience varied van der Waals attractions of various strengths that are overcome at various temperatures, and so the melting process occurs over a wide temperature range. Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances: (a) SiO , covalent network; (b) KCl, ionic; (c) Cu, metallic; (d) CO, molecular; (e) C (diamond), covalent network; (f) BaSO , ionic; (g) NH , molecular; (h) NH F, ionic; (i) C H OH, molecular Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances: (a) CaCl , ionic; (b) SiC, covalent network; (c) N , molecular; (d) Fe, metallic; (e) C (graphite), covalent network; (f) CH CH CH CH , molecular; (g) HCl, molecular; (h) NH NO , ionic; (i) K PO , ionic Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid: X = metallic; Y = covalent network; Z = ionic Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid: X = ionic; Y = metallic; Z = covalent network Identify the following substances as ionic, metallic, covalent network, or molecular solids: Substance A is malleable, ductile, conducts electricity well, and has a melting point of 1135 °C. Substance B is brittle, does not conduct electricity as a solid but does when molten, and has a melting point of 2072 °C. Substance C is very hard, does not conduct electricity, and has a melting point of 3440 °C. Substance D is soft, does not conduct electricity, and has a melting point of 185 °C. A = metallic; B = ionic; C = covalent network; D = molecular Substance A is shiny, conducts electricity well, and melts at 975 °C. Substance A is likely a(n): (b) metallic solid Substance B is hard, does not conduct electricity, and melts at 1200 °C. Substance B is likely a(n): (d) covalent network solid Describe the crystal structure of iron, which crystallizes with two equivalent metal atoms in a cubic unit cell. The structure of this low-temperature form of iron (below 910 °C) is body-centered cubic. There is one-eighth atom at each of the eight corners of the cube and one atom in the center of the cube. Describe the crystal structure of Pt, which crystallizes with four equivalent metal atoms in a cubic unit cell. What is the coordination number of a chromium atom in the body-centered cubic structure of chromium? eight What is the coordination number of an aluminum atom in the face-centered cubic structure of aluminum? Cobalt metal crystallizes in a hexagonal closest packed structure. What is the coordination number of a cobalt atom? 12 Nickel metal crystallizes in a cubic closest packed structure. What is the coordination number of a nickel atom? Tungsten crystallizes in a body-centered cubic unit cell with an edge length of 3.165 Å. (a) 1.370 Å; (b) 19.26 g/cm Platinum (atomic radius = 1.38 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of platinum. Barium crystallizes in a body-centered cubic unit cell with an edge length of 5.025 Å (a) 2.176 Å; (b) 3.595 g/cm Aluminum (atomic radius = 1.43 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of aluminum. The density of aluminum is 2.7 g/cm ; that of silicon is 2.3 g/cm . Explain why Si has the lower density even though it has heavier atoms. The crystal structure of Si shows that it is less tightly packed (coordination number 4) in the solid than Al (coordination number 12). The free space in a metal may be found by subtracting the volume of the atoms in a unit cell from the volume of the cell. Calculate the percentage of free space in each of the three cubic lattices if all atoms in each are of equal size and touch their nearest neighbors. Which of these structures represents the most efficient packing? That is, which packs with the least amount of unused space? Cadmium sulfide, sometimes used as a yellow pigment by artists, crystallizes with cadmium, occupying one-half of the tetrahedral holes in a closest packed array of sulfide ions. What is the formula of cadmium sulfide? Explain your answer. In a closest-packed array, two tetrahedral holes exist for each anion. If only half the tetrahedral holes are occupied, the numbers of anions and cations are equal. The formula for cadmium sulfide is CdS. A compound of cadmium, tin, and phosphorus is used in the fabrication of some semiconductors. It crystallizes with cadmium occupying one-fourth of the tetrahedral holes and tin occupying one-fourth of the tetrahedral holes in a closest packed array of phosphide ions. What is the formula of the compound? Explain your answer. What is the formula of the magnetic oxide of cobalt, used in recording tapes, that crystallizes with cobalt atoms occupying one-eighth of the tetrahedral holes and one-half of the octahedral holes in a closely packed array of oxide ions? Co O A compound containing zinc, aluminum, and sulfur crystallizes with a closest-packed array of sulfide ions. Zinc ions are found in one-eighth of the tetrahedral holes and aluminum ions in one-half of the octahedral holes. What is the empirical formula of the compound? A compound of thallium and iodine crystallizes in a simple cubic array of iodide ions with thallium ions in all of the cubic holes. What is the formula of this iodide? Explain your answer. In a simple cubic array, only one cubic hole can be occupied be a cation for each anion in the array. The ratio of thallium to iodide must be 1:1; therefore, the formula for thallium is TlI. Which of the following elements reacts with sulfur to form a solid in which the sulfur atoms form a closest-packed array with all of the octahedral holes occupied: Li, Na, Be, Ca, or Al? What is the percent by mass of titanium in rutile, a mineral that contains titanium and oxygen, if structure can be described as a closest packed array of oxide ions with titanium ions in one-half of the octahedral holes? What is the oxidation number of titanium? 59.95%; The oxidation number of titanium is +4. Explain why the chemically similar alkali metal chlorides NaCl and CsCl have different structures, whereas the chemically different NaCl and MnS have the same structure. As minerals were formed from the molten magma, different ions occupied the same cites in the crystals. Lithium often occurs along with magnesium in minerals despite the difference in the charge on their ions. Suggest an explanation. Both ions are close in size: Mg, 0.65; Li, 0.60. This similarity allows the two to interchange rather easily. The difference in charge is generally compensated by the switch of Si for Al . Rubidium iodide crystallizes with a cubic unit cell that contains iodide ions at the corners and a rubidium ion in the center. What is the formula of the compound? One of the various manganese oxides crystallizes with a cubic unit cell that contains manganese ions at the corners and in the center. Oxide ions are located at the center of each edge of the unit cell. What is the formula of the compound? Mn O NaH crystallizes with the same crystal structure as NaCl. The edge length of the cubic unit cell of NaH is 4.880 Å. Thallium(I) iodide crystallizes with the same structure as CsCl. The edge length of the unit cell of TlI is 4.20 Å. Calculate the ionic radius of TI . (The ionic radius of I is 2.16 Å.) 1.48 Å A cubic unit cell contains manganese ions at the corners and fluoride ions at the center of each edge. What is the spacing between crystal planes that diffract X-rays with a wavelength of 1.541 nm at an angle of 15.55° (first order reflection)? 2.874 Å A diffractometer using X-rays with a wavelength of 0.2287 nm produced first-order diffraction peak for a crystal angle = 16.21°. Determine the spacing between the diffracting planes in this crystal. A metal with spacing between planes equal to 0.4164 nm diffracts X-rays with a wavelength of 0.2879 nm. What is the diffraction angle for the first order diffraction peak? 20.2° Gold crystallizes in a face-centered cubic unit cell. The second-order reflection (n = 2) of X-rays for the planes that make up the tops and bottoms of the unit cells is at = 22.20°. The wavelength of the X-rays is 1.54 Å. What is the density of metallic gold? When an electron in an excited molybdenum atom falls from the L to the K shell, an X-ray is emitted. These X-rays are diffracted at an angle of 7.75° by planes with a separation of 2.64 Å. What is the difference in energy between the K shell and the L shell in molybdenum assuming a first-order diffraction? 1.74 × 10 eV | 25,819 | 4,261 |
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What is a spontaneous reaction? A reaction has a natural tendency to occur and takes place without the continual input of energy from an external source. What is a nonspontaneous reaction? Indicate whether the following processes are spontaneous or nonspontaneous. spontaneous; nonspontaneous; spontaneous; nonspontaneous; spontaneous; spontaneous A helium-filled balloon spontaneously deflates overnight as He atoms diffuse through the wall of the balloon. Describe the redistribution of matter and/or energy that accompanies this process. Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain. Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time. In the below Figure all possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, Δ , if the particles are initially evenly distributed between the two boxes, but upon redistribution all end up in Box (b). In Figure all of the possible distributions and microstates are shown for four different particles shared between two boxes. Determine the entropy change, Δ , for the system when it is converted from distribution to distribution (d). There are four initial microstates and four final microstates. \[ΔS=k\ln\dfrac{W_\ce{f}}{W_\ce{i}}=\mathrm{1.38×10^{−23}\:J/K×\ln\dfrac{4}{4}=0}\] How does the process described in the previous item relate to the system shown in ? Consider a system similar to the one below, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to \(\dfrac{1}{8}\). What does this comparison tell us about even larger systems? The probability for all the particles to be on one side is \(\dfrac{1}{32}\). This probability is noticeably lower than the \(\dfrac{1}{8}\) result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large. Consider the system shown in Figure. What is the change in entropy for the process where the energy is initially associated only with particle A, but in the final state the energy is distributed between two different particles? Consider the system shown in . What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)? There is only one initial state. For the final state, the energy can be contained in pairs A-C, A-D, B-C, or B-D. Thus, there are four final possible states. \[ΔS=k\ln\left(\dfrac{W_\ce{f}}{W_\ce{i}}\right)=\mathrm{1.38×10^{−23}\:J/K×\ln\left(\dfrac{4}{1}\right)=1.91×10^{−23}\:J/K}\] Arrange the following sets of systems in order of increasing entropy. Assume one mole of each substance and the same temperature for each member of a set. At room temperature, the entropy of the halogens increases from I to Br to Cl . Explain. The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, I is a solid, Br is a liquid, and Cl is a gas. Consider two processes: sublimation of I ( ) and melting of I ( ) (Note: the latter process can occur at the same temperature but somewhat higher pressure). \[\ce{I2}(s)⟶\ce{I2}(g)\] \[\ce{I2}(s)⟶\ce{I2}(l)\] Is Δ positive or negative in these processes? In which of the processes will the magnitude of the entropy change be greater? Indicate which substance in the given pairs has the higher entropy value. Explain your choices. C H OH( ) as it is a larger molecule (more complex and more massive), and so more microstates describing its motions are available at any given temperature. C H OH( ) as it is in the gaseous state. 2H( ), since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms). Predict the sign of the entropy change for the following processes: Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction. Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution. Negative. There is a net loss of three moles of gas from reactants to products. Positive. There is a net increase of seven moles of gas from reactants to products. Write the balanced chemical equation for the combustion of methane, CH ( ), to give carbon dioxide and water vapor. Explain why it is difficult to predict whether Δ is positive or negative for this chemical reaction. Write the balanced chemical equation for the combustion of benzene, C H ( ), to give carbon dioxide and water vapor. Would you expect Δ to be positive or negative in this process? \[\ce{C6H6}(l)+7.5\ce{O2}(g)⟶\ce{3H2O}(g)+\ce{6CO2}(g)\] There are 7.5 moles of gas initially, and 3 + 6 = 9 moles of gas in the end. Therefore, it is likely that the entropy increases as a result of this reaction, and Δ is positive. What is the difference between Δ , Δ °, and \(ΔS^\circ_{298}\) for a chemical change? Calculate \(ΔS^\circ_{298}\) for the following changes. 107 J/K; −86.4 J/K; 133.2 J/K; 118.8 J/K; −326.6 J/K; −171.9 J/K; (g) −7.2 J/K Determine the entropy change for the combustion of liquid ethanol, C H OH, under standard state conditions to give gaseous carbon dioxide and liquid water. Determine the entropy change for the combustion of gaseous propane, C H , under standard state conditions to give gaseous carbon dioxide and water. 100.6 J/K “Thermite” reactions have been used for welding metal parts such as railway rails and in metal refining. One such thermite reaction is: \[\ce{Fe2O3}(s)+\ce{2Al}(s)⟶\ce{Al2O3}(s)+\ce{2Fe}(s)\] Is the reaction spontaneous at room temperature under standard conditions? During the reaction, the surroundings absorb 851.8 kJ/mol of heat. Using the relevant \(S^\circ_{298}\) values listed in , calculate \(S^\circ_{298}\) for the following changes: −198.1 J/K; −348.9 J/K From the following information, determine \(ΔS^\circ_{298}\) for the following: By calculating Δ at each temperature, determine if the melting of 1 mole of NaCl( ) is spontaneous at 500 °C and at 700 °C. \[S^\circ_{\ce{NaCl}(s)}=\mathrm{72.11\:\dfrac{J}{mol⋅K}}\hspace{40px} S^\circ_{\ce{NaCl}(l)}=\mathrm{95.06\:\dfrac{J}{mol⋅K}}\hspace{40px ΔH^\circ_\ce{fusion}=\mathrm{27.95\: kJ/mol}\] What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem? As Δ < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K. It is assumed that these do not change significantly at the higher temperatures used in the problem. Use the standard entropy data in to determine the change in entropy for each of the reactions listed in . All are run under standard state conditions and 25 °C. 2.86 J/K; 24.8 J/K; −113.2 J/K; −24.7 J/K; 15.5 J/K; 290.0 J/K What is the difference between Δ , Δ °, and \(ΔG^\circ_{298}\) for a chemical change? A reactions has \(ΔH^\circ_{298}\) = 100 kJ/mol and \(ΔS^\circ_{298}=\textrm{250 J/mol⋅K}\). Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous? The reaction is nonspontaneous at room temperature. Above 400 K, Δ will become negative, and the reaction will become spontaneous. Explain what happens as a reaction starts with Δ < 0 (negative) and reaches the point where Δ = 0. Use the standard free energy of formation data in to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions. 465.1 kJ nonspontaneous; −106.86 kJ spontaneous; −53.6 kJ spontaneous; −83.4 kJ spontaneous; −406.7 kJ spontaneous; −30.0 kJ spontaneous Use the standard free energy data in to determine the free energy change for each of the following reactions, which are run under standard state conditions and 25 °C. Identify each as either spontaneous or nonspontaneous at these conditions. Given: \[\ce{P4}(s)+\ce{5O2}(g)⟶\ce{P4O10}(s) \hspace{20px} ΔG^\circ_{298}=\mathrm{−2697.0\: kJ/mol}\] \[\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(g) \hspace{20px} ΔG^\circ_{298}=\mathrm{−457.18\: kJ/mol}\] \[\ce{6H2O}(g)+\ce{P4O10}(g)⟶\ce{4H3PO4}(l) \hspace{20px} ΔG^\circ_{298}=\mathrm{−428.66\: kJ/mol}\] −1124.3 kJ/mol for the standard free energy change. The calculation agrees with the value in because free energy is a state function (just like the enthalpy and entropy), so its change depends only on the initial and final states, not the path between them. Is the formation of ozone (O ( )) from oxygen (O ( )) spontaneous at room temperature under standard state conditions? Consider the decomposition of red mercury(II) oxide under standard state conditions. \[\ce{2HgO}(s,\,\ce{red})⟶\ce{2Hg}(l)+\ce{O2}(g)\] The reaction is nonspontaneous; Above 566 °C the process is spontaneous. Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels. Calculate Δ ° for each of the following reactions from the equilibrium constant at the temperature given. 1.5 × 10 kJ; −21.9 kJ; −5.34 kJ; −0.383 kJ; 18 kJ; 71 kJ Calculate Δ ° for each of the following reactions from the equilibrium constant at the temperature given. Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of Δ ° given. = 41; = 0.053; = 6.9 × 10 ; = 1.9; = 0.04 Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of Δ ° given. Calculate the equilibrium constant at the temperature given. In each of the following, the value of Δ is not given at the temperature of the reaction. Therefore, we must calculate Δ ° from the values Δ ° and Δ ° and then calculate Δ from the relation ΔG° = ΔH° − TΔS°. Calculate the equilibrium constant at the temperature given. Consider the following reaction at 298 K: \[\ce{N2O4}(g)⇌\ce{2NO2}(g) \hspace{20px} K_P=0.142\] What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium. The standard free energy change is \(ΔG^\circ_{298}=−RT\ln K=\mathrm{4.84\: kJ/mol}\). When reactants and products are in their standard states (1 bar or 1 atm), = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): < 1, and \(ΔG_{298}\) becomes less positive as it approaches zero. At equilibrium, = , and Δ = 0. Determine the normal boiling point (in kelvin) of dichloroethane, CH Cl . Find the actual boiling point using the Internet or some other source, and calculate the percent error in the temperature. Explain the differences, if any, between the two values. Under what conditions is \(\ce{N2O3}(g)⟶\ce{NO}(g)+\ce{NO2}(g)\) spontaneous? The reaction will be spontaneous at temperatures greater than 287 K. At room temperature, the equilibrium constant ( ) for the self-ionization of water is 1.00 × 10 . Using this information, calculate the standard free energy change for the aqueous reaction of hydrogen ion with hydroxide ion to produce water. (Hint: The reaction is the reverse of the self-ionization reaction.) Hydrogen sulfide is a pollutant found in natural gas. Following its removal, it is converted to sulfur by the reaction \(\ce{2H2S}(g)+\ce{SO2}(g)⇌\dfrac{3}{8}\ce{S8}(s,\,\ce{rhombic})+\ce{2H2O}(l)\). What is the equilibrium constant for this reaction? Is the reaction endothermic or exothermic? = 5.35 × 10 The process is exothermic. Consider the decomposition of CaCO ( ) into CaO( ) and CO ( ). What is the equilibrium partial pressure of CO at room temperature? In the laboratory, hydrogen chloride (HCl( )) and ammonia (NH ( )) often escape from bottles of their solutions and react to form the ammonium chloride (NH Cl( )), the white glaze often seen on glassware. Assuming that the number of moles of each gas that escapes into the room is the same, what is the maximum partial pressure of HCl and NH in the laboratory at room temperature? (Hint: The partial pressures will be equal and are at their maximum value when at equilibrium.) 1.0 × 10 atm. This is the maximum pressure of the gases under the stated conditions. Benzene can be prepared from acetylene. \(\ce{3C2H2}(g)⇌\ce{C6H6}(g)\). Determine the equilibrium constant at 25 °C and at 850 °C. Is the reaction spontaneous at either of these temperatures? Why is all acetylene not found as benzene? Carbon dioxide decomposes into CO and O at elevated temperatures. What is the equilibrium partial pressure of oxygen in a sample at 1000 °C for which the initial pressure of CO was 1.15 atm? \[x=\mathrm{1.29×10^{−5}\:atm}=P_{\ce{O2}}\] Carbon tetrachloride, an important industrial solvent, is prepared by the chlorination of methane at 850 K. \[\ce{CH4}(g)+\ce{4Cl2}(g)⟶\ce{CCl4}(g)+\ce{4HCl}(g)\] What is the equilibrium constant for the reaction at 850 K? Would the reaction vessel need to be heated or cooled to keep the temperature of the reaction constant? Acetic acid, CH CO H, can form a dimer, (CH CO H) , in the gas phase. \[\ce{2CH3CO2H}(g)⟶\ce{(CH3CO2H)2}(g)\] The dimer is held together by two hydrogen bonds with a total strength of 66.5 kJ per mole of dimer. At 25 °C, the equilibrium constant for the dimerization is 1.3 × 10 (pressure in atm). What is Δ ° for the reaction? −0.16 kJ Nitric acid, HNO , can be prepared by the following sequence of reactions: \[\ce{4NH3}(g)+\ce{5O2}(g)⟶\ce{4NO}(g)+\ce{6H2O}(g)\] \[\ce{2NO}(g)+\ce{O2}(g)⟶\ce{2NO2}(g)\] \[\ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(l)+\ce{NO}(g)\] How much heat is evolved when 1 mol of NH ( ) is converted to HNO ( )? Assume standard states at 25 °C. Determine Δ for the following reactions. (a) Antimony pentachloride decomposes at 448 °C. The reaction is: \[\ce{SbCl5}(g)⟶\ce{SbCl3}(g)+\ce{Cl2}(g)\] An equilibrium mixture in a 5.00 L flask at 448 °C contains 3.85 g of SbCl , 9.14 g of SbCl , and 2.84 g of Cl . Chlorine molecules dissociate according to this reaction: \[\ce{Cl2}(g)⟶\ce{2Cl}(g)\] 1.00% of Cl molecules dissociate at 975 K and a pressure of 1.00 atm. Given that the \(ΔG^\circ_\ce{f}\) for Pb ( ) and Cl ( ) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, , for PbCl ( ). Determine the standard free energy change, \(ΔG^\circ_\ce{f}\), for the formation of S ( ) given that the \(ΔG^\circ_\ce{f}\) for Ag ( ) and Ag S( ) are 77.1 k/mole and −39.5 kJ/mole respectively, and the solubility product for Ag S( ) is 8 × 10 . 90 kJ/mol Determine the standard enthalpy change, entropy change, and free energy change for the conversion of diamond to graphite. Discuss the spontaneity of the conversion with respect to the enthalpy and entropy changes. Explain why diamond spontaneously changing into graphite is not observed. The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ. \[\ce{H2O}(l)⇌\ce{H2O}(g) \hspace{20px} ΔG^\circ_{298}=\mathrm{8.58\: kJ}\] (a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; = 0.031; The evaporation of water is spontaneous; \(P_{\ce{H2O}}\) must always be less than or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity. In glycolysis, the reaction of glucose (Glu) to form glucose-6-phosphate (G6P) requires ATP to be present as described by the following equation: \[\mathrm{Glu + ATP ⟶ G6P + ADP} \hspace{20px} ΔG^\circ_{298}=\mathrm{−17\: kJ}\] In this process, ATP becomes ADP summarized by the following equation: \[\mathrm{ATP⟶ADP} \hspace{20px} ΔG^\circ_{298}=\mathrm{−30\: kJ}\] Determine the standard free energy change for the following reaction, and explain why ATP is necessary to drive this process: \[\mathrm{Glu⟶G6P} \hspace{20px} ΔG^\circ_{298}=\:?\] One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P): \[\mathrm{G6P⇌F6P} \hspace{20px} ΔG^\circ_{298}=\mathrm{1.7\: kJ}\] (a) Nonspontaneous as \(ΔG^\circ_{298}>0\); \(ΔG^\circ_{298}=−RT\ln K,\) \(ΔG = 1.7×10^3 + \left(8.314 × 335 × \ln\dfrac{28}{128}\right) = \mathrm{−2.5\: kJ}\). The forward reaction to produce F6P is spontaneous under these conditions. Without doing a numerical calculation, determine which of the following will reduce the free energy change for the reaction, that is, make it less positive or more negative, when the temperature is increased. Explain. When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of Δ , Δ , and Δ for this process, and justify your choices. Δ is negative as the process is spontaneous. Δ is positive as with the solution becoming cold, the dissolving must be endothermic. Δ must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound. An important source of copper is from the copper ore, chalcocite, a form of copper(I) sulfide. When heated, the Cu S decomposes to form copper and sulfur described by the following equation: \[\ce{Cu2S}(s)⟶\ce{Cu}(s)+\ce{S}(s)\] What happens to \(ΔG^\circ_{298}\) (becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased? | 18,585 | 4,262 |
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Molecules of alcohols contain one or more hydroxyl groups (OH groups) substituted for hydrogen atoms along the carbon chain. The structure of the simplest alcohol, methanol (methyl alcohol), can be derived from that of methane by putting an OH in place of one of the H’s: The name, too, is derived from the name methane by replacing the final with (for alcoh ). The general formula for an alcohol may be written as R—OH, where R represents the hydrocarbon (alkane) portion of the molecule and is called an . In methanol, R is the methyl group CH . Methanol is also called wood alcohol because it can be obtained by heating wood in the absence of air, a process called . Methanol vapor given off when the wood is heated can be condensed to a liquid by cooling below its boiling point of 65°C. The effect of polarity and especially hydrogen bonding due to the OH group is evident when this is compared with the temperature of –85°C at which ethane, C H , boils. Both molecules contain 18 electrons and are nearly the same size, and so London forces should be about the same, but the OH group in one methanol molecule can form strong hydrogen bonds with an OH in another molecule. Methanol is an important industrial chemical—nearly 3 × 10 kg was produced worldwide in 2003 . Some was made by destructive distillation, but most was synthesized from hydrogen and carbon monoxide: \[\ce{2H_{2} (g) + CO (g) \longrightarrow CH_{3}OH(l)} \nonumber \] This reaction is carried out at pressures several hundred times normal atmospheric pressure, using metal oxides as catalysts. Methanol is mainly used to make other compounds from which plastics are manufactured, but some is consumed as fuel in jet engines and racing cars. Methanol is also a component of nonpermanent antifreeze and automobile windshield-washer solvent. The second member of the alcohol family is ethanol (ethyl alcohol)― the substance we commonly call . Ethanol is also known as grain alcohol because it is obtained when grain or sugar ferments. refers to a chemical reaction which is speeded up by enzymes and occurs in the absence of air. Ethanol can also be synthesized by adding H O to ethene, obtained during petroleum refining: This is a typical example of an . The H and OH from H O are added to the ethene molecule and held there by electrons made available when one-half of the double bond breaks. Ethanol is used as a solvent, in some special fuels, in antifreeze, and to manufacture a number of other chemicals. You are probably most familiar with it as a component of alcoholic beverages. Ethanol makes up 3 to 6 percent of beer, 12 to 15 percent of most wines, and 49 to 59 percent of distilled liquor. (The “proof” of an alcoholic beverage is just twice the percentage of ethanol.) Alcohol’s intoxicating effects are well known, and it is a mild depressant. Prolonged overuse can lead to liver damage. Methanol also produces intoxication but is much more poisonous than ethanol—it can cause blindness and death. Denatured alcohol is ethanol to which methanol or some other poison has been added, making it unfit for human consumption. Most of the ethanol not used in alcoholic beverages is denatured because in that form its sale is taxed at a much lower rate. Two isomers are possible for alcohols containing three carbon atoms: The and the in the names of these compounds indicate the position of the OH group along the carbon chain. The propanols are much less important commercially than methanol and ethanol, although 2-propanol is commonly found in rubbing alcohol. A carbon atom typically forms four bonds. Therefore, in an alcohol where carbon is bonded to an -OH group, there can be up to three carbon atoms bonded to the carbon atom bonded to the oxygen in -OH. If no carbon atom or one carbon atom is bonded directly, the compound is a alcohol. If two are bonded directly, it is a alcohol; with three it is a alcohol, as illustrated below. All alcohols can be completely oxidized to carbon dioxide and water by oxygen in the air; that is, all alcohols are combustible. Like hydrocarbons, combustion is an important reaction of alcohols, but more controlled oxidation is even more important, because it can convert alcohols into other compounds that are useful to society. The ease with which an alcohol can be oxidized and the extent of the oxidation depends on whether the alcohol is primary, secondary, or tertiary. For primary alcohols, controlled, stepwise oxidation first yields compounds called aldehydes; if more of the oxidizing agent is available, then aldehydes can be further oxidized to carboxylic acids. Schematically Primary alcohol → aldehyde → carboxylic acid Oxidation of an organic compound can usually be recognized because either an oxygen atom is added to a molecule or two hydrogen atoms are lost from a molecule. For example, stepwise oxidation of ethanol first produces the aldehyde ethanal (commonly called acetaldehyde); further oxidation produces the carboxylic acid, ethanoic acid (commonly called acetic acid). An aldehyde has the functional group –CHO, where the carbon atom is double-bonded to an oxygen atom. A carboxylic acid has the functional group –COOH, in which the carbon atom is double-bonded to an oxygen atom and single-bonded to an oxygen atom in an OH group. The structures below show the differences between ethanol, ethanal, and ethanoic acid. Note that acetaldehyde differs from ethanol by loss of one H atom from the oxygen atom and one H atom from the carbon on the right. Acetic acid differs from acetaldehyde by having an addition O atom on the right-hand carbon atom. Controlled oxidation can be carried out in the laboratory using an aqueous solution of potassium permanganate or an aqueous solution of potassium dichromate. When a similar controlled oxidation is applied to a secondary alcohol, such as 2-propanol (see below), the oxidized molecule contains a C=O group that has two other carbon atoms attached to the C atom. This >C=O group with two carbon atoms attached to the C is called a ketone. Again, note that in the ketone the number of hydrogen atoms is fewer by two than the number in the secondary alcohol. The oxidation corresponds with loss of two hydrogen atoms. Ketones are difficult to oxidize further, because there is no way to add another oxygen atom to the carbon atom in the >C=O group, nor is there a way to remove hydrogen atoms from the C and O atoms in the >C=O group. Tertiary alcohols, which have no hydrogen atoms attached to the carbon that is bonded to the –OH group, are difficult to oxidize. If a primary alcohol, a secondary alcohol, and a tertiary alcohol are dissolved in water in three beakers and then treated with either potassium permanganate or potassium dichromate, only the primary and secondary alcohols will react. (The reaction can be observed because both permanganate ions and dichromate ions are colored (purple and orange, respectively). Thus for the primary and secondary alcohols, the color will disappear, but for tertiary alcohols there will be no color change. The structures of two more alcohols which are of commercial importance and are familiar to many persons are shown below:
Each ethylene glycol molecule has two hydrogens which can participate in hydrogen bonding, and each glycerin molecule has three. Both substances have rather high boiling points (198°C for ethylene glycol and 290°C for glycerin) and are syrupy, viscous liquids at room temperature. Their resistance to flowing freely is due to the network of hydrogen bonds that links each molecule to several of its fellows, making it more difficult for them to slide past one another. This highlight again the effect of hydrogen bonding on intermolecular forces and physical properties. In fact, in our ethylene glycol has the highest boiling point compared with other compounds containing the same number of electrons. The two propanol isomers are also on this table, only exceeded in boiling point by ethylene glycol, and acetic acid. Ethylene glycol is the principal component of engine coolant for automobiles and is also used to manufacture polyester fibers. In 2005, nearly 1.8 × 10 kg was produced worldwide. Glycerin is used as a lubricant and in the manufacture of explosives: When nitroglycerin is mixed with a solid material such as nitrocellulose (which is made by treating cotton or wood pulp with nitric acid), the product is a form of dynamite. | 8,465 | 4,263 |
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The increased use of radioisotopes has led to increased concerns over the effects of these materials on biological systems (such as humans). All radioactive nuclides emit high-energy particles or electromagnetic waves. When this radiation encounters living cells, it can cause heating, break chemical bonds, or ionize molecules. The most serious biological damage results when these radioactive emissions fragment or ionize molecules. For example, alpha and beta particles emitted from nuclear decay reactions possess much higher energies than ordinary chemical bond energies. When these particles strike and penetrate matter, they produce ions and molecular fragments that are extremely reactive. The damage this does to biomolecules in living organisms can cause serious malfunctions in normal cell processes, taxing the organism’s repair mechanisms and possibly causing illness or even death (Figure \(\Page {1}\)). There is a large difference in the magnitude of the biological effects of (for example, light and microwaves) and , emissions energetic enough to knock electrons out of molecules (for example, α and β particles, γ rays, X-rays, and high-energy ultraviolet radiation) (Figure \(\Page {2}\)). Energy absorbed from nonionizing radiation speeds up the movement of atoms and molecules, which is equivalent to heating the sample. Although biological systems are sensitive to heat (as we might know from touching a hot stove or spending a day at the beach in the sun), a large amount of nonionizing radiation is necessary before dangerous levels are reached. Ionizing radiation, however, may cause much more severe damage by breaking bonds or removing electrons in biological molecules, disrupting their structure and function. The damage can also be done indirectly, by first ionizing H O (the most abundant molecule in living organisms), which forms a H O ion that reacts with water, forming a hydronium ion and a hydroxyl radical: Radiation can harm either the whole body (somatic damage) or eggs and sperm (genetic damage). Its effects are more pronounced in cells that reproduce rapidly, such as the stomach lining, hair follicles, bone marrow, and embryos. This is why patients undergoing radiation therapy often feel nauseous or sick to their stomach, lose hair, have bone aches, and so on, and why particular care must be taken when undergoing radiation therapy during pregnancy. Different types of radiation have differing abilities to pass through material (Figure \(\Page {4}\)). A very thin barrier, such as a sheet or two of paper, or the top layer of skin cells, usually stops alpha particles. Because of this, alpha particle sources are usually not dangerous if outside the body, but are quite hazardous if ingested or inhaled (see the Chemistry in Everyday Life feature on Radon Exposure). Beta particles will pass through a hand, or a thin layer of material like paper or wood, but are stopped by a thin layer of metal. Gamma radiation is very penetrating and can pass through a thick layer of most materials. Some high-energy gamma radiation is able to pass through a few feet of concrete. Certain dense, high atomic number elements (such as lead) can effectively attenuate gamma radiation with thinner material and are used for shielding. The ability of various kinds of emissions to cause ionization varies greatly, and some particles have almost no tendency to produce ionization. Alpha particles have about twice the ionizing power of fast-moving neutrons, about 10 times that of β particles, and about 20 times that of γ rays and X-rays. For many people, one of the largest sources of exposure to radiation is from radon gas (Rn-222). Radon-222 is an α emitter with a half–life of 3.82 days. It is one of the products of the radioactive decay series of U-238, which is found in trace amounts in soil and rocks. The radon gas that is produced slowly escapes from the ground and gradually seeps into homes and other structures above. Since it is about eight times more dense than air, radon gas accumulates in basements and lower floors, and slowly diffuses throughout buildings (Figure \(\Page {5}\)). Several different devices are used to detect and measure radiation, including Geiger counters, scintillation counters (scintillators), and radiation dosimeters (Figure \(\Page {6}\)). Probably the best-known radiation instrument, the (also called the Geiger-Müller counter) detects and measures radiation. Radiation causes the ionization of the gas in a Geiger-Müller tube. The rate of ionization is proportional to the amount of radiation. A contains a scintillator—a material that emits light (luminesces) when excited by ionizing radiation—and a sensor that converts the light into an electric signal. also measure ionizing radiation and are often used to determine personal radiation exposure. Commonly used types are electronic, film badge, thermoluminescent, and quartz fiber dosimeters. A variety of units are used to measure various aspects of radiation (Table \(\Page {1}\)). The unit for rate of radioactive decay is the , with 1 Bq = 1 disintegration per second. The and are much larger units and are frequently used in medicine (1 curie = 1 Ci = \(3.7 \times 10^{10}\) disintegrations per second). The SI unit for measuring radiation dose is the , with 1 Gy = 1 J of energy absorbed per kilogram of tissue. In medical applications, the is more often used (1 rad = 0.01 Gy; 1 rad results in the absorption of 0.01 J/kg of tissue). The SI unit measuring tissue damage caused by radiation is the . This takes into account both the energy and the biological effects of the type of radiation involved in the radiation dose. The roentgen equivalent for man (rem) is the unit for radiation damage that is used most frequently in medicine (1 rem = 1 Sv). Note that the tissue damage units (rem or Sv) includes the energy of the radiation dose (rad or Gy), along with a biological factor referred to as the RBE (for relative biological effectiveness), that is an approximate measure of the relative damage done by the radiation. These are related by: \[ \text{number of rems}=\text{RBE} \times \text{number of rads} \label{Eq2} \] with RBE approximately 10 for α radiation, 2(+) for protons and neutrons, and 1 for β and γ radiation. Cobalt-60 ( = 5.26 y) is used in cancer therapy since the \(\gamma\) rays it emits can be focused in small areas where the cancer is located. A 5.00-g sample of Co-60 is available for cancer treatment. The activity is given by: \[\textrm{Activity}=λN=\left( \dfrac{\ln 2}{t_{1/2} } \right) N=\mathrm{\left( \dfrac{\ln 2}{5.26\ y} \right) \times 5.00 \ g=0.659\ \dfrac{g}{y} \ of\ \ce{^{60}Co} \text{ that decay}} \nonumber \] And to convert this to decays per second: \[\mathrm{0.659\; \frac{g}{y} \times \dfrac{y}{365 \;day} \times \dfrac{1\; day}{ 24\; hours} \times \dfrac{1\; h}{3,600 \;s} \times \dfrac{1\; mol}{59.9\; g} \times \dfrac{6.02 \times 10^{23} \;atoms}{1 \;mol} \times \dfrac{1\; decay}{1\; atom}} \nonumber \] \[\mathrm{=2.10 \times 10^{14} \; \frac{decay}{s}} \nonumber \] (a) Since \(\mathrm{1\; Bq = 1\; \frac{ decay}{s}}\), the activity in Becquerel (Bq) is: \[\mathrm{2.10 \times 10^{14} \dfrac{decay}{s} \times \left(\dfrac{1\ Bq}{1 \; \frac{decay}{s}} \right)=2.10 \times 10^{14} \; Bq} \nonumber \] (b) Since \(\mathrm{1\ Ci = 3.7 \times 10^{11}\; \frac{decay}{s}}\), the activity in curie (Ci) is: \[\mathrm{2.10 \times 10^{14} \frac{decay}{s} \times \left( \dfrac{1\ Ci}{3.7 \times 10^{11} \frac{decay}{s}} \right) =5.7 \times 10^2\;Ci} \nonumber \] Tritium is a radioactive isotope of hydrogen (\(t_{1/2} = \mathrm{12.32\; years}\)) that has several uses, including self-powered lighting, in which electrons emitted in tritium radioactive decay cause phosphorus to glow. Its nucleus contains one proton and two neutrons, and the atomic mass of tritium is 3.016 amu. What is the activity of a sample containing 1.00mg of tritium (a) in Bq and (b) in Ci? \(\mathrm{3.56 \times 10^{11} Bq}\) \(\mathrm{0.962\; Ci}\) The effects of radiation depend on the type, energy, and location of the radiation source, and the length of exposure. As shown in Figure \(\Page {8}\), the average person is exposed to background radiation, including cosmic rays from the sun and radon from uranium in the ground (see the Chemistry in Everyday Life feature on Radon Exposure); radiation from medical exposure, including scans, radioisotope tests, X-rays, and so on; and small amounts of radiation from other human activities, such as airplane flights (which are bombarded by increased numbers of cosmic rays in the upper atmosphere), radioactivity from consumer products, and a variety of radionuclides that enter our bodies when we breathe (for example, carbon-14) or through the food chain (for example, potassium-40, strontium-90, and iodine-131). A short-term, sudden dose of a large amount of radiation can cause a wide range of health effects, from changes in blood chemistry to death. Short-term exposure to tens of rems of radiation will likely cause very noticeable symptoms or illness; a dose of about 500 rems is estimated to have a 50% probability of causing the death of the victim within 30 days of exposure. Exposure to radioactive emissions has a cumulative effect on the body during a person’s lifetime, which is another reason why it is important to avoid any unnecessary exposure to radiation. Health effects of short-term exposure to radiation are shown in Table \(\Page {2}\). It is impossible to avoid some exposure to ionizing radiation. We are constantly exposed to background radiation from a variety of natural sources, including cosmic radiation, rocks, medical procedures, consumer products, and even our own atoms. We can minimize our exposure by blocking or shielding the radiation, moving farther from the source, and limiting the time of exposure. We are constantly exposed to radiation from a variety of naturally occurring and human-produced sources. This radiation can affect living organisms. Ionizing radiation is the most harmful because it can ionize molecules or break chemical bonds, which damages the molecule and causes malfunctions in cell processes. It can also create reactive hydroxyl radicals that damage biological molecules and disrupt physiological processes. Radiation can cause somatic or genetic damage, and is most harmful to rapidly reproducing cells. Types of radiation differ in their ability to penetrate material and damage tissue, with alpha particles the least penetrating, but potentially most damaging, and gamma rays the most penetrating. Various devices, including Geiger counters, scintillators, and dosimeters, are used to detect and measure radiation, and monitor radiation exposure. We use several units to measure radiation: becquerels or curies for rates of radioactive decay; gray or rads for energy absorbed; and rems or sieverts for biological effects of radiation. Exposure to radiation can cause a wide range of health effects, from minor to severe, including death. We can minimize the effects of radiation by shielding with dense materials such as lead, moving away from the source of radiation, and limiting time of exposure. | 11,223 | 4,264 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Polymer_Chemistry_(Schaller)/04%3A_Polymer_Properties/4.05%3A_Crystallinity_in_Polymers |
We often think of polymers in the form of plastics: solid materials that serve some structural function, like a water bottle or some Venetian blinds. Polymers are different from many other such solids. Metals, for instance, have crystalline structures, in which atoms form a regularly repeating pattern, row upon row. Polymers, in contrast, are generally somewhat . Think of a strainer filled with cooked spaghetti, the long chains of pasta looping over each other randomly. That's how the molecules of amorphous or "shapeless" polymers tend to distribute themselves. Polymers are capable of forming more ordered structures. If they aligned instead with their long chains parallel to each other, they would be able to get much closer together, and intermolecular attractions between the chains would be much stronger. The energy of the system would decrease, so this more crystalline structure should be inherently favored. So why doesn't that always happen? Think about a polymer cooling from a completely melted state, in which case the structure is certainly amorphous. As it cools, the material becomes less like a liquid, and then less soft. With small molecules, that transition is relatively abrupt, as the molecules slide into place, guided by their strengthening intermolecular attractions. The molecules just have to rotate a little to face the correct direction, or maybe budge a little to the side. With polymers, the transition is much more gradual, because those great, long chains have to slide over each other and uncoil in order to lie parallel. At some point, there just isn't enough energy for them to keep working their way into the optimal alignment. As a result, many polymers are semi-crystalline, with regions called where portions of chains have aligned parallel to each other, but also with large amorphous areas that are much more randomly oriented. As a result, a polymer sample might be 80% amorphous with only 20% of its chain lengths aligned in crystalline lamellae. Even so, those crystalline domains exert a strong influence on the properties of the polymer. Because of the stronger intermolecular attractions between these closer chains, the chains are much less able to slide past each other. The result is a material that is stronger and more rigid, and that can be very important for reliable structural materials. How do we know whether a polymer sample contains crystalline domains? We can use differential scanning calorimetry to observe the transition between an ordered crystalline phase and a disordered melt phase with added heat. As we saw when we were looking at , a melting point shows up as a peak on a DSC trace. You can look for it at a temperature above , which you will recall looks more like a step in the baseline. Below , the chains are not mobile enough to move out of alignment with each other. In the example below, you can see at about 128 °C as well as a melting point, , when attractions between chains in the lamellae are overcome, at 155-160 °C. If this were a regular molecular or elemental solid, the melting point would be the same as the fusion point; the material would melt at the same temperature during warming as it froze during cooling. In polymers, it usually doesn't work out that way, and a is observed. In hysteresis, a sample has changed because of previous events, leading to results that don't seem to be reproducible in the way that we would normally expect. In general, the crystallization temperature, , is lower than the melting temperature, . That's because, as the material melts, the chains move out of alignment with each other and, because of chain entanglement, they are hindered from getting back into alignment with each other again, leading to a delay in crystallization. Describe what is observed in each of the following DSC traces. Crystallinity can also be probed in other ways. The classic approach is through X-ray diffraction. When X-rays pass through ordered materials, they give rise to diffraction patterns, bright spots in space where ricocheting X-rays have constructively interfered with each other, shining starkly amid the blackness where the X-rays have undergone destructive interference. The drawing below is a cartoon of an X-ray diffraction pattern. It might be captured by a digital camera or, in earlier days, a sheet of photographic film. The bright spots result from X-rays scattering out from the middle, where they have encountered the sample. In the middle, the black circle blocks the original X-ray beam, which would otherwise be too bright to allow observation of those spots around it. The data from an X-ray diffraction experiment is fed into a program that can mathematically deconvolute why the spots showed up where they did. The process has been compared to throwing a stone over your shoulder into a pond, then watching the ripples to decide where the stone must be. With one stone, that should be pretty easy. You would find the stone at the center of a circle of ripples. With two or more stones, interference patterns make the ripples more complex, and so it may take more work to determine where each stone lies in the pond. The picture below illustrates the results of a relatively simple X-ray diffraction experiment commonly used with crystalline polymer samples. The technique is called wide-angle X-ray scattering, or WAXS. It simply reports the angle at which scattering intensity is observed relative to the initial X-ray beam. That angle corresponds to a distance between atoms in the sample, which in this case usually corresponds to a distance between parallel polymer chains. Sometimes there can be more than one peak, but this example is a simple one to start with. The relationship between this scattering angle and an interatomic distance is given by Bragg's Law, a fundamental starting point in X-ray crystallography first expressed by and his son, Sir Lawrence Bragg. The law states that: \[nλ = 2d\sin θ \label{bragg1}\] in which is an integer (typically assumed to be 1), lambda is the wavelength of the X-ray used (commonly 1.541 Å, although others are possible), is the distance between regularly repeating atoms and is the scattering angle. Rearranging Equation \ref{bragg1}, we find: \[d = \dfrac{1.541}{2\sin θ}\] In other words, the distance between atoms is inversely proportional to the scattering angle. The greater the angle, the smaller the distance. In each of the following cases, determine which sample (A or B) has a higher crystalline content. | 6,543 | 4,265 |
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