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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/19%3A_Transition_Metals_and_Coordination_Chemistry/19.2%3A_Coordination_Chemistry_of_Transition_Metals |
The hemoglobin in your blood, the chlorophyll in green plants, vitamin \(B_{12}\), and the catalyst used in the manufacture of polyethylene all contain coordination compounds. Ions of the metals, especially the transition metals, are likely to form complexes. Many of these compounds are highly colored (Figure \(\Page {1}\)). In the remainder of this chapter, we will consider the structure and bonding of these remarkable compounds. Remember that in most main group element compounds, the valence electrons of the isolated atoms combine to form chemical bonds that satisfy the octet rule. For instance, the four valence electrons of carbon overlap with electrons from four hydrogen atoms to form CH . The one valence electron leaves sodium and adds to the seven valence electrons of chlorine to form the ionic formula unit NaCl (Figure \(\Page {2}\)). Transition metals do not normally bond in this fashion. They primarily form coordinate covalent bonds, a form of the Lewis acid-base interaction in which both of the electrons in the bond are contributed by a donor (Lewis base) to an electron acceptor (Lewis acid). The Lewis acid in coordination complexes, often called a ion (or atom), is often a transition metal or inner transition metal, although main group elements can also form . The Lewis base donors, called , can be a wide variety of chemicals—atoms, molecules, or ions. The only requirement is that they have one or more electron pairs, which can be donated to the central metal. Most often, this involves a with a lone pair of electrons that can form a coordinate bond to the metal. The consists of the central metal ion or atom plus its attached ligands. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. The of the central metal ion or atom is the number of donor atoms bonded to it. The coordination number for the silver ion in [Ag(NH ) ] is two (Figure \(\Page {3}\)). For the copper(II) ion in [CuCl ] , the coordination number is four, whereas for the cobalt(II) ion in [Co(H O) ] the coordination number is six. Each of these ligands is , from the Greek for “one toothed,” meaning that they connect with the central metal through only one atom. In this case, the number of ligands and the coordination number are equal. Many other ligands coordinate to the metal in more complex fashions. are those in which two atoms coordinate to the metal center. For example, ethylenediamine (en, H NCH CH NH ) contains two nitrogen atoms, each of which has a lone pair and can serve as a Lewis base (Figure \(\Page {4}\)). Both of the atoms can coordinate to a single metal center. In the complex [Co(en) ] , there are three bidentate en ligands, and the coordination number of the cobalt(III) ion is six. The most common coordination numbers are two, four, and six, but examples of all coordination numbers from 1 to 15 are known. Any ligand that bonds to a central metal ion by more than one donor atom is a (or “many teeth”) because it can bite into the metal center with more than one bond. The term (pronounced “KEY-late”) from the Greek for “claw” is also used to describe this type of interaction. Many polydentate ligands are , and a complex consisting of one or more of these ligands and a central metal is a chelate. A chelating ligand is also known as a chelating agent. A chelating ligand holds the metal ion rather like a crab’s claw would hold a marble. Figure \(\Page {4}\) showed one example of a chelate and the heme complex in hemoglobin is another important example (Figure \(\Page {5}\)). It contains a polydentate ligand with four donor atoms that coordinate to iron. Polydentate ligands are sometimes identified with prefixes that indicate the number of donor atoms in the ligand. As we have seen, ligands with one donor atom, such as NH , Cl , and H O, are monodentate ligands. Ligands with two donor groups are bidentate ligands. Ethylenediamine, H NCH CH NH , and the anion of the acid glycine, \(\ce{NH2CH2CO2-}\) (Figure \(\Page {6}\)) are examples of bidentate ligands. Tridentate ligands, tetradentate ligands, pentadentate ligands, and hexadentate ligands contain three, four, five, and six donor atoms, respectively. The heme ligand (Figure \(\Page {5}\)) is a tetradentate ligand. The nomenclature of the complexes is patterned after a system suggested by Alfred Werner, a Swiss chemist and Nobel laureate, whose outstanding work more than 100 years ago laid the foundation for a clearer understanding of these compounds. The following five rules are used for naming complexes: When the complex is either a cation or a neutral molecule, the name of the central metal atom is spelled exactly like the name of the element and is followed by a Roman numeral in parentheses to indicate its oxidation state (Tables \(\Page {2}\), \(\Page {3}\), and \(\Page {3}\)). When the complex is an anion, the suffix -ate is added to the stem of the name of the metal, followed by the Roman numeral designation of its oxidation state. Sometimes, the Latin name of the metal is used when the English name is clumsy. For example, is used instead of , instead , and instead of . The oxidation state of the metal is determined based on the charges of each ligand and the overall charge of the coordination compound. For example, in [Cr(H O) Cl ]Br, the coordination sphere (in brackets) has a charge of 1+ to balance the bromide ion. The water ligands are neutral, and the chloride ligands are anionic with a charge of 1− each. To determine the oxidation state of the metal, we set the overall charge equal to the sum of the ligands and the metal: +1 = −2 + , so the oxidation state ( ) is equal to 3+. Determine the name of the following complexes and give the coordination number of the central metal atom. The complex potassium dicyanoargenate(I) is used to make antiseptic compounds. Give the formula and coordination number. K[Ag(CN) ]; coordination number two The most common structures of the complexes in coordination compounds are octahedral, tetrahedral, and square planar (Figure \(\Page {7}\)). For transition metal complexes, the coordination number determines the geometry around the central metal ion. Table \(\Page {3}\) compares coordination numbers to the molecular geometry: Unlike main group atoms in which both the bonding and nonbonding electrons determine the molecular shape, the nonbonding -electrons do not change the arrangement of the ligands. Octahedral complexes have a coordination number of six, and the six donor atoms are arranged at the corners of an octahedron around the central metal ion. Examples are shown in Figure \(\Page {8}\). The chloride and nitrate anions in [Co(H O) ]Cl and [Cr(en) ](NO ) , and the potassium cations in K [PtCl ], are outside the brackets and are not bonded to the metal ion. For transition metals with a coordination number of four, two different geometries are possible: tetrahedral or square planar. Unlike main group elements, where these geometries can be predicted from VSEPR theory, a more detailed discussion of transition metal orbitals (discussed in the section on Crystal Field Theory) is required to predict which complexes will be tetrahedral and which will be square planar. In tetrahedral complexes such as [Zn(CN) ] (Figure \(\Page {9}\)), each of the ligand pairs forms an angle of 109.5°. In square planar complexes, such as [Pt(NH ) Cl ], each ligand has two other ligands at 90° angles (called the positions) and one additional ligand at an 180° angle, in the position. Isomers are different chemical species that have the same chemical formula. Transition metals often form , in which the same atoms are connected through the same types of bonds but with differences in their orientation in space. Coordination complexes with two different ligands in the and positions from a ligand of interest form isomers. For example, the octahedral [Co(NH ) Cl ] ion has two isomers. In the , the two chloride ligands are adjacent to each other (Figure \(\Page {1}\)). The other isomer, the , has the two chloride ligands directly across from one another. Different geometric isomers of a substance are different chemical compounds. They exhibit different properties, even though they have the same formula. For example, the two isomers of [Co(NH ) Cl ]NO differ in color; the form is violet, and the form is green. Furthermore, these isomers have different dipole moments, solubilities, and reactivities. As an example of how the arrangement in space can influence the molecular properties, consider the polarity of the two [Co(NH ) Cl ]NO isomers. Remember that the polarity of a molecule or ion is determined by the bond dipoles (which are due to the difference in electronegativity of the bonding atoms) and their arrangement in space. In one isomer, chloride ligands cause more electron density on one side of the molecule than on the other, making it polar. For the isomer, each ligand is directly across from an identical ligand, so the bond dipoles cancel out, and the molecule is nonpolar. Identify which geometric isomer of [Pt(NH ) Cl ] is shown in Figure \(\Page {9}\)b. Draw the other geometric isomer and give its full name. In the Figure \(\Page {9}\)b, the two chlorine ligands occupy positions. The other form is shown in below. When naming specific isomers, the descriptor is listed in front of the name. Therefore, this complex is -diaminedichloroplatinum(II). The isomer of [Pt(NH ) Cl ] has each ligand directly across from an adjacent ligand. Draw the ion -diaqua- -dibromo- -dichlorocobalt(II). Another important type of isomers are , or , in which two objects are exact mirror images of each other but cannot be lined up so that all parts match. This means that optical isomers are nonsuperimposable mirror images. A classic example of this is a pair of hands, in which the right and left hand are mirror images of one another but cannot be superimposed. Optical isomers are very important in organic and biochemistry because living systems often incorporate one specific optical isomer and not the other. Unlike geometric isomers, pairs of optical isomers have identical properties (boiling point, polarity, solubility, etc.). Optical isomers differ only in the way they affect polarized light and how they react with other optical isomers. For coordination complexes, many coordination compounds such as [M(en) ] [in which M is a central metal ion such as iron(III) or cobalt(II)] form enantiomers, as shown in Figure \(\Page {11}\). These two isomers will react differently with other optical isomers. For example, helices are optical isomers, and the form that occurs in nature (right-handed DNA) will bind to only one isomer of [M(en) ] and not the other. The [Co(en) Cl ] ion exhibits geometric isomerism ( / ), and its isomer exists as a pair of optical isomers (Figure \(\Page {12}\)). occur when the coordination compound contains a ligand that can bind to the transition metal center through two different atoms. For example, the CN ligand can bind through the carbon atom (cyano) or through the nitrogen atom (isocyano). Similarly, SCN− can be bound through the sulfur or nitrogen atom, affording two distinct compounds ([Co(NH ) SCN] or [Co(NH ) NCS] ). (or ) occur when one anionic ligand in the inner coordination sphere is replaced with the counter ion from the outer coordination sphere. A simple example of two ionization isomers are [CoCl ,Br] and [CoCl Br,Cl]. Chlorophyll, the green pigment in plants, is a complex that contains magnesium (Figure \(\Page {13}\)). This is an example of a main group element in a coordination complex. Plants appear green because chlorophyll absorbs red and purple light; the reflected light consequently appears green. The energy resulting from the absorption of light is used in photosynthesis. One of the most important applications of transition metals is as industrial catalysts. As you recall from the chapter on kinetics, a catalyst increases the rate of reaction by lowering the activation energy and is regenerated in the catalytic cycle. Over 90% of all manufactured products are made with the aid of one or more catalysts. The ability to bind ligands and change oxidation states makes transition metal catalysts well suited for catalytic applications. Vanadium oxide is used to produce 230,000,000 tons of sulfuric acid worldwide each year, which in turn is used to make everything from fertilizers to cans for food. Plastics are made with the aid of transition metal catalysts, along with detergents, fertilizers, paints, and more (Figure \(\Page {14}\)). Very complicated pharmaceuticals are manufactured with catalysts that are selective, reacting with one specific bond out of a large number of possibilities. Catalysts allow processes to be more economical and more environmentally friendly. Developing new catalysts and better understanding of existing systems are important areas of current research. Many other coordination complexes are also brightly colored. The square planar copper(II) complex phthalocyanine blue (from Figure \(\Page {13}\)) is one of many complexes used as pigments or dyes. This complex is used in blue ink, blue jeans, and certain blue paints. The structure of heme (Figure \(\Page {15}\)), the iron-containing complex in hemoglobin, is very similar to that in chlorophyll. In hemoglobin, the red heme complex is bonded to a large protein molecule (globin) by the attachment of the protein to the heme ligand. Oxygen molecules are transported by hemoglobin in the blood by being bound to the iron center. When the hemoglobin loses its oxygen, the color changes to a bluish red. Hemoglobin will only transport oxygen if the iron is Fe ; oxidation of the iron to Fe prevents oxygen transport. Complexing agents often are used for water softening because they tie up such ions as Ca , Mg , and Fe , which make water hard. Many metal ions are also undesirable in food products because these ions can catalyze reactions that change the color of food. Coordination complexes are useful as preservatives. For example, the ligand , (HO CCH ) NCH CH N(CH CO H) , coordinates to metal ions through six donor atoms and prevents the metals from reacting (Figur \(\Page {16}\)). This ligand also is used to sequester metal ions in paper production, textiles, and detergents, and has pharmaceutical uses. Complexing agents that tie up metal ions are also used as drugs. British Anti-Lewisite (BAL), HSCH CH(SH)CH OH, is a drug developed during World War I as an antidote for the arsenic-based war gas Lewisite. is now used to treat poisoning by heavy metals, such as arsenic, mercury, thallium, and chromium. The drug is a ligand and functions by making a water-soluble chelate of the metal; the kidneys eliminate this metal chelate (Figure \(\Page {17}\)). Another polydentate ligand, enterobactin, which is isolated from certain bacteria, is used to form complexes of iron and thereby to control the severe iron buildup found in patients suffering from blood diseases such as Cooley’s anemia, who require frequent transfusions. As the transfused blood breaks down, the usual metabolic processes that remove iron are overloaded, and excess iron can build up to fatal levels. Enterobactin forms a water-soluble complex with excess iron, and the body can safely eliminate this complex. Ligands like BAL and enterobactin are important in medical treatments for heavy metal poisoning. However, chelation therapies can disrupt the normal concentration of ions in the body, leading to serious side effects, so researchers are searching for new chelation drugs. One drug that has been developed is dimercaptosuccinic acid (DMSA), shown in Figure \(\Page {18}\). Identify which atoms in this molecule could act as donor atoms. All of the oxygen and sulfur atoms have lone pairs of electrons that can be used to coordinate to a metal center, so there are six possible donor atoms. Geometrically, only two of these atoms can be coordinated to a metal at once. The most common binding mode involves the coordination of one sulfur atom and one oxygen atom, forming a five-member ring with the metal. Some alternative medicine practitioners recommend chelation treatments for ailments that are not clearly related to heavy metals, such as cancer and autism, although the practice is discouraged by many scientific organizations. Identify at least two biologically important metals that could be disrupted by chelation therapy. Ca, Fe, Zn, and Cu Ligands are also used in the electroplating industry. When metal ions are reduced to produce thin metal coatings, metals can clump together to form clusters and nanoparticles. When metal coordination complexes are used, the ligands keep the metal atoms isolated from each other. It has been found that many metals plate out as a smoother, more uniform, better-looking, and more adherent surface when plated from a bath containing the metal as a complex ion. Thus, complexes such as [Ag(CN) ] and [Au(CN) ] are used extensively in the electroplating industry. In 1965, scientists at Michigan State University discovered that there was a platinum complex that inhibited cell division in certain microorganisms. Later work showed that the complex was -diaminedichloroplatinum(II), [Pt(NH ) (Cl) ], and that the isomer was not effective. The inhibition of cell division indicated that this square planar compound could be an anticancer agent. In 1978, the Food and Drug Administration approved this compound, known as cisplatin, for use in the treatment of certain forms of cancer. Since that time, many similar platinum compounds have been developed for the treatment of cancer. In all cases, these are the isomers and never the isomers. The diamine (NH ) portion is retained with other groups, replacing the dichloro [(Cl) ] portion. The newer drugs include carboplatin, oxaliplatin, and satraplatin. The transition elements and main group elements can form coordination compounds, or complexes, in which a central metal atom or ion is bonded to one or more ligands by coordinate covalent bonds. Ligands with more than one donor atom are called polydentate ligands and form chelates. The common geometries found in complexes are tetrahedral and square planar (both with a coordination number of four) and octahedral (with a coordination number of six). and configurations are possible in some octahedral and square planar complexes. In addition to these geometrical isomers, optical isomers (molecules or ions that are mirror images but not superimposable) are possible in certain octahedral complexes. Coordination complexes have a wide variety of uses including oxygen transport in blood, water purification, and pharmaceutical use. | 18,860 | 2,552 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.02%3A_Equations_and_Mass_Relationships |
Consider the balanced chemical equation (i.e., catalytic oxidation of ammonia) such as \[ 4 \text{ N} \text{H}_{3} (g) + 5 \text{O}_{2} (g) \rightarrow 4 \text{ N} \text{O} (g) + 6 \text{ H}_{2} \text{O} (g) \label{1} \] not only tells how many molecules of each kind are involved in a reaction, it also indicates the of each substance that is involved. Equation \(\ref{1}\) (represented molecularly by the image below it) says that 4 NH can react with 5 O to give 4 NO and 6 H O . It also says that 4 NH would react with 5 O yielding 4 NO and 6 H O. The balanced equation does more than this, though. It also tells us that \(2 \cdot4 = 8 \text{mol NH}_3\) will react with \(2 \cdot5 = 10 \text{mol O}_2\), and that \(\small\frac{1}{2} \cdot4 = 2 \text{mol NH}_3\) requires only \(\small\frac{1}{2} \cdot5 = 2.5 \text{mol O}_2\). In other words, the equation indicates that exactly 5 mol O must react 4 mol NH consumed. For the purpose of calculating how much O is required to react with a certain amount of NH therefore, the significant information contained in Equation \(\ref{1}\) is the \[\frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{\text{3}}}\label{2} \] We shall call such a ratio derived from a balanced chemical equation a and give it the symbol . Thus, for Equation \(\ref{1}\), \[\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{NH}_{\text{3}}} \right)=\frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{\text{3}}} \label{3} \] The word comes from the Greek words , “element,“ and , “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another. Derive all possible stoichiometric ratios from Equation \(\ref{1}\). Any ratio of amounts of substance given by coefficients in the equation may be used: \[\begin{align*} &\text{S}\left(\frac{\ce{NH3}}{\ce{O2}}\right) = \frac{\text{4 mol NH}_3}{\text{5 mol O}_2} &\text{S}\left(\frac{\ce{O2}}{\ce{NO}}\right) &= \frac{\text{5 mol O}_2}{\text{4 mol NO}} \\ { } \\ &\text{S}\left(\frac{\ce{NH3}}{\ce{NO}}\right) = \frac{\text{4 mol NH}_3}{\text{4 mol NO}} &\space\text{S}\left(\frac{\ce{O2}}{\ce{H2O}}\right) &= \frac{\text{5 mol O}_2}{\text{6 mol }\ce{H2O}} \\ { } \\ &\text{S}\left(\frac{\ce{NH3}}{\ce{H2O}}\right) = \frac{\text{4 mol NH}_3}{\text{6 mol }\ce{H2O}} &\space\text{S}\left(\frac{\ce{NO}}{\ce{H2O}}\right) &= \frac{\text{4 mol NO}}{\text{6 mol }\ce{H2O}} \end{align*} \nonumber \] There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Equation \(\ref{3}\) gives one of them.] . Using Equation \(\ref{2}\) as an example, this means that the ratio of the amount of O consumed to the amount of NH consumed must be the stoichiometric ratio S(O /NH ): \[\frac{n_{\text{O}_{\text{2}}\text{ consumed}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}} =\text{S} \left(\frac{\text{O}_2}{\text{NH}_3}\right) = \frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{3}}\label{9} \] Similarly, the ratio of the amount of H O produced to the amount of NH consumed must be S(H O/NH ): \[\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}} =\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{3}} \right) = \frac{\text{6 mol H}_{\text{2}}\text{O}}{\text{4 mol NH}_{3}} \label{10} \] In general we can say that \[\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}} \label{11} \] or, in symbols, \[\text{S}\left( \frac{\text{X}}{\text{Y}} \right)= \frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}} \label{12} \] Note that in the word Equation \(\ref{11}\) and the symbolic Equation \(\ref{12}\), \(X\) and \(Y\) may represent reactant or product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants and the amounts of products will be in appropriate stoichiometric ratios. Find the amount of water produced when 3.68 mol NH is consumed according to Equation \(\ref{10}\). The amount of water produced must be in the stoichiometric ratio S(H O/NH ) to the amount of ammonia consumed: \[\text{S}\left( \dfrac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{\text{3}}} \right)=\dfrac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}} \nonumber \] Multiplying both sides consumed, by we have \[\begin{align} n_{\text{H}_{\text{2}}\text{O produced}} &= n_{\text{NH}_{\text{3}}\text{ consumed}} \normalsize \cdot\text{S}\left( \frac{\ce{H2O}}{\ce{NH3}} \right) \\ { } \\ & =\text{3.68 mol NH}_3 \cdot\frac{\text{6 mol }\ce{H2O}}{\text{4 mol NH}_3} \\ & =\text{5.52 mol }\ce{H2O} \end{align} \nonumber \] This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example \(\Page {2}\) is analogous to , where density was employed as a conversion factor between mass and volume. Example \(\Page {2}\) is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Equation \(\ref{9}\) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form \[ \text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \\ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced} \nonumber \] or symbolically. \[ n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}} \nonumber \] When using stoichiometric ratios, be sure you indicate moles . You can only cancel moles of the same substance. In other words, 1 mol NH cancels 1 mol NH but does not cancel 1 mol H O. The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product. Calculate the mass of sulfur dioxide (SO ) produced when 3.84 mol O is reacted with FeS according to the equation \[\ce{4FeS2 + 11O2 -> 2Fe2O3 + 8SO2} \nonumber \] The problem asks that we calculate the mass of SO produced. As we learned in , the molar mass can be used to convert from the amount of SO to the mass of SO . Therefore this problem in effect is asking that we calculate the amount of SO produced from the amount of O consumed. This is the same problem as in Example 2. It requires the stoichiometric ratio \(\text{S}\left( \frac{\text{SO}_{\text{2}}}{\text{O}_{\text{2}}} \right)=\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}\) The of SO produced is then \[\begin{align*} n_{\ce{SO2}\text{ produced}} & = n_{\ce{O2}\text{ consumed}}\text{ }\normalsize\cdot\text{ conversion factor} \\ & =\text{3.84 mol O}_2\cdot\frac{\text{8 mol SO}_2}{\text{11 mol O}_2} \\ & =\text{2.79 mol SO}_2 \end{align*} \nonumber \] The of SO is \[\begin{align*}\text{m}_{\text{SO}_{\text{2}}} & =\text{2.79 mol SO}_2\cdot\frac{\text{64.06 g SO}_2}{\text{1 mol SO}_2} \\& =\text{179 g SO}_2 \end{align*} \nonumber \] With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of O to moles of SO and the molar mass will convert moles of SO to grams of SO . A schematic road map for the one-step calculation can be written as \[ n_{\text{O}_{\text{2}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/O}_{\text{2}}\text{)}}\text{ }n_{\text{SO}_{\text{2}}}\text{ }\xrightarrow{M_{\text{SO}_{\text{2}}}}\text{ }m_{\text{SO}_{\text{2}}} \nonumber \] Thus \[ \text{m}_{\text{SO}_{\text{2}}}=\text{3}\text{.84 mol O}_{\text{2}}\cdot\text{ }\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}\normalsize\text{ }\cdot\text{ }\frac{\text{64}\text{.06 g}}{\text{1 mol SO}_{\text{2}}}=\normalsize\text{179 g} \nonumber \] These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations. The chemical reaction in this example is of environmental interest. Iron pyrite (FeS ) is often an impurity in coal, and so burning this fuel in a power plant produces sulfur dioxide (SO ), a major air pollutant. Our next example also involves burning a fuel and its effect on the atmosphere. What mass of oxygen would be consumed when 3.3 × 10 g, 3.3 Pg (petagrams), of octane (C H ) is burned to produce CO and H O? First, write a balanced equation \[\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O} \nonumber \] The problem gives the mass of C H burned and asks for the mass of O required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O consumed. Finally, the molar mass of O permits calculation of the mass of O . Symbolically \[ m_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{M_{\text{C}_{\text{8}}\text{H}_{\text{18}}}}\text{ }n_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/C}_{\text{8}}\text{H}_{\text{18}}\text{)}}\text{ }n_{\text{O}_{\text{2}}}\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}} \nonumber \] \[\begin{align} m_{\text{O}_{\text{2}}} & =\text{3}\text{.3 }\cdot\text{ 10}^{\text{15}}\text{ g }\cdot\text{ }\frac{\text{1 mol C}_{\text{8}}\text{H}_{\text{18}}}{\text{114 g}}\text{ }\cdot\text{ }\frac{\text{25 mol O}_{\text{2}}}{\text{2 mol C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\cdot \text{ }\frac{\text{32}\text{.00 g}}{\text{1 mol O}_{\text{2}}} \\ & =\text{1}\text{.2 }\cdot\text{ 10}^{\text{16}}\text{ g } \end{align*} \nonumber \] Thus 12 Pg (petagrams) of O would be needed. The large mass of oxygen obtained in this example is an estimate of how much O is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 10 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O . Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen. | 10,611 | 2,553 |
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We take advantage of changes between the gas, liquid, and solid states to cool a drink with ice cubes (solid to liquid), cool our bodies by perspiration (liquid to gas), and cool food inside a refrigerator (gas to liquid and vice versa). We use dry ice, which is solid CO , as a refrigerant (solid to gas), and we make artificial snow for skiing and snowboarding by transforming a liquid to a solid. In this section, we examine what happens when any of the three forms of matter is converted to either of the other two. These changes of state are often called phase changes. The six most common phase changes are shown in Figure \(\Page {1}\). Phase changes are accompanied by a change in the energy of a system. For example, converting a liquid, in which the molecules are close together, to a gas, in which the molecules are, on average, far apart, requires an input of energy (heat) to give the molecules enough kinetic energy to allow them to overcome the intermolecular attractive forces. The stronger the attractive forces, the more energy is needed to overcome them. Solids, which are highly ordered, have the strongest intermolecular interactions, whereas gases, which are very disordered, have the weakest. Thus any transition from a more ordered to a less ordered state (solid to liquid, liquid to gas, or solid to gas) requires an input of energy; it is . Conversely, any transition from a less ordered to a more ordered state (liquid to solid, gas to liquid, or gas to solid) releases energy; it is . The energy change associated with each common phase change is shown in Figure \(\Page {1}\). Previously, we defined the enthalpy changes associated with various chemical and physical processes. The melting points and molar , the energy required to convert from a solid to a liquid, a process known as fusion (or melting), as well as the normal boiling points and enthalpies of vaporization (Δ ) of selected compounds are listed in Table \(\Page {1}\). The substances with the highest melting points usually have the highest enthalpies of fusion; they tend to be ionic compounds that are held together by very strong electrostatic interactions. Substances with high boiling points are those with strong intermolecular interactions that must be overcome to convert a liquid to a gas, resulting in high enthalpies of vaporization. The enthalpy of vaporization of a given substance is much greater than its enthalpy of fusion because it takes more energy to completely separate molecules (conversion from a liquid to a gas) than to enable them only to move past one another freely (conversion from a solid to a liquid). Δ is positive for any transition from a more ordered to a less ordered state and negative for a transition from a less ordered to a more ordered state. The direct conversion of a solid to a gas, without an intervening liquid phase, is called sublimation. The amount of energy required to sublime 1 mol of a pure solid is the enthalpy of sublimation (Δ ). Common substances that sublime at standard temperature and pressure (STP; 0°C, 1 atm) include CO (dry ice); iodine (Figure \(\Page {2}\)); naphthalene, a substance used to protect woolen clothing against moths; and 1,4-dichlorobenzene. As shown in Figure \(\Page {1}\), the enthalpy of sublimation of a substance is the sum of its enthalpies of fusion and vaporization provided all values are at the same ; this is an application of . \[ΔH_{sub} =ΔH_{fus} +ΔH_{vap} \label{Eq1}\] Less energy is needed to allow molecules to move past each other than to separate them totally Fusion, vaporization, and sublimation are endothermic processes; they occur only with the absorption of heat. Anyone who has ever stepped out of a swimming pool on a cool, breezy day has felt the heat loss that accompanies the evaporation of water from the skin. Our bodies use this same phenomenon to maintain a constant temperature: we perspire continuously, even when at rest, losing about 600 mL of water daily by evaporation from the skin. We also lose about 400 mL of water as water vapor in the air we exhale, which also contributes to cooling. Refrigerators and air-conditioners operate on a similar principle: heat is absorbed from the object or area to be cooled and used to vaporize a low-boiling-point liquid, such as ammonia or the chlorofluorocarbons (CFCs) and the hydrofluorocarbons (HCFCs). The vapor is then transported to a different location and compressed, thus releasing and dissipating the heat. Likewise, ice cubes efficiently cool a drink not because of their low temperature but because heat is required to convert ice at 0°C to liquid water at 0°C, as demonstrated later in Example 8. The Thermodynamics of Phase Changes: The processes on the right side of Figure \(\Page {1}\)—freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve. Figure \(\Page {3}\) shows a heating curve, a plot of temperature versus heating time, for a 75 g sample of water. The sample is initially ice at 1 atm and −23°C; as heat is added, the temperature of the ice increases linearly with time. The slope of the line depends on both the mass of the ice and the specific heat ( ) of ice, which is the number of joules required to raise the temperature of 1 g of ice by 1°C. As the temperature of the ice increases, the water molecules in the ice crystal absorb more and more energy and vibrate more vigorously. At the melting point, they have enough kinetic energy to overcome attractive forces and move with respect to one another. As more heat is added, the temperature of the system does increase further but remains constant at 0°C until all the ice has melted. Once all the ice has been converted to liquid water, the temperature of the water again begins to increase. Now, however, the temperature increases more slowly than before because the specific heat capacity of water is than that of ice. When the temperature of the water reaches 100°C, the water begins to boil. Here, too, the temperature remains constant at 100°C until all the water has been converted to steam. At this point, the temperature again begins to rise, but at a rate than seen in the other phases because the heat capacity of steam is than that of ice or water. Thus . In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process, and as long as even a small amount of liquid water is present, the temperature of the system remains at 100°C during the boiling process. The rate at which heat is added does affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils. The temperature of a sample does not change during a phase change. If heat is added at a constant rate, as in Figure \(\Page {3}\), then the length of the horizontal lines, which represents the time during which the temperature does not change, is directly proportional to the magnitude of the enthalpies associated with the phase changes. In Figure \(\Page {3}\), the horizontal line at 100°C is much longer than the line at 0°C because the enthalpy of vaporization of water is several times greater than the enthalpy of fusion. A superheated liquid is a sample of a liquid at the temperature and pressure at which it should be a gas. Superheated liquids are not stable; the liquid will eventually boil, sometimes violently. The phenomenon of superheating causes “bumping” when a liquid is heated in the laboratory. When a test tube containing water is heated over a Bunsen burner, for example, one portion of the liquid can easily become too hot. When the superheated liquid converts to a gas, it can push or “bump” the rest of the liquid out of the test tube. Placing a stirring rod or a small piece of ceramic (a “boiling chip”) in the test tube allows bubbles of vapor to form on the surface of the object so the liquid boils instead of becoming superheated. Superheating is the reason a liquid heated in a smooth cup in a microwave oven may not boil until the cup is moved, when the motion of the cup allows bubbles to form. The cooling curve, a plot of temperature versus cooling time, in Figure \(\Page {4}\) plots temperature versus time as a 75 g sample of steam, initially at 1 atm and 200°C, is cooled. Although we might expect the cooling curve to be the mirror image of the heating curve in Figure \(\Page {3}\), the cooling curve is an identical mirror image. As heat is removed from the steam, the temperature falls until it reaches 100°C. At this temperature, the steam begins to condense to liquid water. No further temperature change occurs until all the steam is converted to the liquid; then the temperature again decreases as the water is cooled. We might expect to reach another plateau at 0°C, where the water is converted to ice; in reality, however, this does not always occur. Instead, the temperature often drops below the freezing point for some time, as shown by the little dip in the cooling curve below 0°C. This region corresponds to an unstable form of the liquid, a supercooled liquid. If the liquid is allowed to stand, if cooling is continued, or if a small crystal of the solid phase is added (a seed crystal), the supercooled liquid will convert to a solid, sometimes quite suddenly. As the water freezes, the temperature increases slightly due to the heat evolved during the freezing process and then holds constant at the melting point as the rest of the water freezes. Subsequently, the temperature of the ice decreases again as more heat is removed from the system. Supercooling effects have a huge impact on Earth’s climate. For example, supercooling of water droplets in clouds can prevent the clouds from releasing precipitation over regions that are persistently arid as a result. Clouds consist of tiny droplets of water, which in principle should be dense enough to fall as rain. In fact, however, the droplets must aggregate to reach a certain size before they can fall to the ground. Usually a small particle (a ) is required for the droplets to aggregate; the nucleus can be a dust particle, an ice crystal, or a particle of silver iodide dispersed in a cloud during (a method of inducing rain). Unfortunately, the small droplets of water generally remain as a supercooled liquid down to about −10°C, rather than freezing into ice crystals that are more suitable nuclei for raindrop formation. One approach to producing rainfall from an existing cloud is to cool the water droplets so that they crystallize to provide nuclei around which raindrops can grow. This is best done by dispersing small granules of solid CO (dry ice) into the cloud from an airplane. Solid CO sublimes directly to the gas at pressures of 1 atm or lower, and the enthalpy of sublimation is substantial (25.3 kJ/mol). As the CO sublimes, it absorbs heat from the cloud, often with the desired results. If a 50.0 g ice cube at 0.0°C is added to 500 mL of tea at 20.0°C, what is the temperature of the tea when the ice cube has just melted? Assume that no heat is transferred to or from the surroundings. The density of water (and iced tea) is 1.00 g/mL over the range 0°C–20°C, the specific heats of liquid water and ice are 4.184 J/(g•°C) and 2.062 J/(g•°C), respectively, and the enthalpy of fusion of ice is 6.01 kJ/mol. mass, volume, initial temperature, density, specific heats, and \(ΔH_{fus}\) final temperature Substitute the values given into the general equation relating heat gained to heat lost (Equation 5.39) to obtain the final temperature of the mixture. When two substances or objects at different temperatures are brought into contact, heat will flow from the warmer one to the cooler. The amount of heat that flows is given by \[q=mC_sΔT\] where is heat, is mass, is the specific heat, and Δ is the temperature change. Eventually, the temperatures of the two substances will become equal at a value somewhere between their initial temperatures. Calculating the temperature of iced tea after adding an ice cube is slightly more complicated. The general equation relating heat gained and heat lost is still valid, but in this case we also have to take into account the amount of heat required to melt the ice cube from ice at 0.0°C to liquid water at 0.0°C. Suppose you are overtaken by a blizzard while ski touring and you take refuge in a tent. You are thirsty, but you forgot to bring liquid water. You have a choice of eating a few handfuls of snow (say 400 g) at −5.0°C immediately to quench your thirst or setting up your propane stove, melting the snow, and heating the water to body temperature before drinking it. You recall that the survival guide you leafed through at the hotel said something about not eating snow, but you cannot remember why—after all, it’s just frozen water. To understand the guide’s recommendation, calculate the amount of heat that your body will have to supply to bring 400 g of snow at −5.0°C to your body’s internal temperature of 37°C. Use the data in Example \(\Page {1}\) 200 kJ (4.1 kJ to bring the ice from −5.0°C to 0.0°C, 133.6 kJ to melt the ice at 0.0°C, and 61.9 kJ to bring the water from 0.0°C to 37°C), which is energy that would not have been expended had you first melted the snow. Changes of state are examples of , or . All phase changes are accompanied by changes in the energy of a system. Changes from a more-ordered state to a less-ordered state (such as a liquid to a gas) are . Changes from a less-ordered state to a more-ordered state (such as a liquid to a solid) are always . The conversion of a solid to a liquid is called . The energy required to melt 1 mol of a substance is its enthalpy of fusion (Δ ). The energy change required to vaporize 1 mol of a substance is the enthalpy of vaporization (Δ ). The direct conversion of a solid to a gas is . The amount of energy needed to sublime 1 mol of a substance is its and is the sum of the enthalpies of fusion and vaporization. Plots of the temperature of a substance versus heat added or versus heating time at a constant rate of heating are called . Heating curves relate temperature changes to phase transitions. A , a liquid at a temperature and pressure at which it should be a gas, is not stable. A is not exactly the reverse of the heating curve because many liquids do not freeze at the expected temperature. Instead, they form a , a metastable liquid phase that exists below the normal melting point. Supercooled liquids usually crystallize on standing, or adding a of the same or another substance can induce crystallization. | 15,330 | 2,554 |
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The first law of thermodynamics governs changes in the state function we have called (U). Changes in the internal energy (ΔU) are closely related to changes in the enthalpy (ΔH), which is a measure of the heat flow between a system and its surroundings at constant pressure. You also learned previously that the enthalpy change for a chemical reaction can be calculated using tabulated values of enthalpies of formation. This information, however, does not tell us whether a particular process or reaction will occur spontaneously. Let’s consider a familiar example of spontaneous change. If a hot frying pan that has just been removed from the stove is allowed to come into contact with a cooler object, such as cold water in a sink, heat will flow from the hotter object to the cooler one, in this case usually releasing steam. Eventually both objects will reach the same temperature, at a value between the initial temperatures of the two objects. This transfer of heat from a hot object to a cooler one obeys the first law of thermodynamics: energy is conserved. Now consider the same process in reverse. Suppose that a hot frying pan in a sink of cold water were to become hotter while the water became cooler. As long as the same amount of thermal energy was gained by the frying pan and lost by the water, the first law of thermodynamics would be satisfied. Yet we all know that such a process cannot occur: heat always flows from a hot object to a cold one, never in the reverse direction. That is, by itself the magnitude of the heat flow associated with a process does not predict whether the process will occur spontaneously. For many years, chemists and physicists tried to identify a single measurable quantity that would enable them to predict whether a particular process or reaction would occur spontaneously. Initially, many of them focused on enthalpy changes and hypothesized that an exothermic process would always be spontaneous. But although it is true that many, if not most, spontaneous processes are exothermic, there are also many spontaneous processes that are not exothermic. For example, at a pressure of 1 atm, ice melts spontaneously at temperatures greater than 0°C, yet this is an endothermic process because heat is absorbed. Similarly, many salts (such as NH NO , NaCl, and KBr) dissolve spontaneously in water even though they absorb heat from the surroundings as they dissolve (i.e., ΔH > 0). Reactions can also be both spontaneous and highly endothermic, like the reaction of barium hydroxide with ammonium thiocyanate shown in Figure \(\Page {1}\). Thus enthalpy is not the only factor that determines whether a process is spontaneous. For example, after a cube of sugar has dissolved in a glass of water so that the sucrose molecules are uniformly dispersed in a dilute solution, they never spontaneously come back together in solution to form a sugar cube. Moreover, the molecules of a gas remain evenly distributed throughout the entire volume of a glass bulb and never spontaneously assemble in only one portion of the available volume. To help explain why these phenomena proceed spontaneously in only one direction requires an additional state function called entropy (S), a thermodynamic property of all substances that is proportional to their degree of disorder. In Chapter 13, we introduced the concept of entropy in relation to solution formation. Here we further explore the nature of this state function and define it mathematically. In 1824, at the age of 28, Nicolas Léonard Sadi (Figure \(\Page {2}\)) published the results of an extensive study regarding the efficiency of steam heat engines. In a later review of Carnot’s findings, Rudolf introduced a new thermodynamic property that relates the spontaneous heat flow accompanying a process to the temperature at which the process takes place. This new property was expressed as the ratio of the heat ( ) and the kelvin temperature ( ). The term refers to a process that takes place at such a slow rate that it is always at equilibrium and its direction can be changed (it can be “reversed”) by an infinitesimally small change is some condition. Note that the idea of a reversible process is a formalism required to support the development of various thermodynamic concepts; no real processes are truly reversible, rather they are classified as . Similar to other thermodynamic properties, this new quantity is a state function, and so its change depends only upon the initial and final states of a system. In 1865, Clausius named this property and defined its change for any process as the following: \[ \color{red} ΔS=\dfrac{q_\ce{rev}}{T} \label{Eq1}\] The entropy change for a real, irreversible process is then equal to that for the theoretical reversible process that involves the same initial and final states. Following the work of Carnot and Clausius, Ludwig developed a molecular-scale statistical model that related the entropy of a system to the number of possible for the system. A is a specific configuration of the locations and energies of the atoms or molecules that comprise a system like the following: \[S=k \ln W \label{Eq2}\] Here is the Boltzmann constant and has a value of 1.38 × 10 J/K. As for other state functions, the change in entropy for a process is the difference between its final ( ) and initial ( ) values: \[ΔS=S_\ce{f}−S_\ce{i}=k \ln W_\ce{f} − k \ln W_\ce{i}=k \ln\dfrac{W_\ce{f}}{W_\ce{i}} \label{Eq2a}\] For processes involving an increase in the number of microstates, > , the entropy of the system increases, Δ > 0. Conversely, processes that reduce the number of microstates, < , yield a decrease in system entropy, Δ < 0. This molecular-scale interpretation of entropy provides a link to the probability that a process will occur as illustrated in the next paragraphs. Consider the general case of a system comprised of particles distributed among boxes. The number of microstates possible for such a system is . For example, distributing four particles among two boxes will result in 2 = 16 different microstates as illustrated in Figure \(\Page {3}\). Microstates with equivalent particle arrangements (not considering individual particle identities) are grouped together and are called . The probability that a system will exist with its components in a given distribution is proportional to the number of microstates within the distribution. Since entropy increases logarithmically with the number of microstates, . For this system, the most probable configuration is one of the six microstates associated with distribution (c) where the particles are evenly distributed between the boxes, that is, a configuration of two particles in each box. The probability of finding the system in this configuration is \[\dfrac{6}{16}\) or \(\dfrac{3}{8}\] The least probable configuration of the system is one in which all four particles are in one box, corresponding to distributions (a) and (d), each with a probability of \[\dfrac{1}{16}\] The probability of finding all particles in only one box (either the left box or right box) is then \[\left(\dfrac{1}{16}+\dfrac{1}{16}\right)=\dfrac{2}{16}\) or \(\dfrac{1}{8}\] As you add more particles to the system, the number of possible microstates increases exponentially (2 ). A macroscopic (laboratory-sized) system would typically consist of moles of particles ( ~ 10 ), and the corresponding number of microstates would be staggeringly huge. Regardless of the number of particles in the system, however, the distributions in which roughly equal numbers of particles are found in each box are always the most probable configurations. is a macroscopic example of this particle-in-a-box model. For this system, the most probable distribution is confirmed to be the one in which the matter is most uniformly dispersed or distributed between the two flasks. The spontaneous process whereby the gas contained initially in one flask expands to fill both flasks equally therefore yields an increase in entropy for the system. A similar approach may be used to describe the spontaneous flow of heat. Consider a system consisting of two objects, each containing two particles, and two units of energy (represented as “*”) in Figure \(\Page {4}\). The hot object is comprised of particles and and initially contains both energy units. The cold object is comprised of particles and , which initially has no energy units. Distribution (a) shows the three microstates possible for the initial state of the system, with both units of energy contained within the hot object. If one of the two energy units is transferred, the result is distribution (b) consisting of four microstates. If both energy units are transferred, the result is distribution (c) consisting of three microstates. And so, we may describe this system by a total of ten microstates. The probability that the heat does not flow when the two objects are brought into contact, that is, that the system remains in distribution (a), is \(\dfrac{3}{10}\). More likely is the flow of heat to yield one of the other two distribution, the combined probability being \(\dfrac{7}{10}\). The most likely result is the flow of heat to yield the uniform dispersal of energy represented by distribution (b), the probability of this configuration being \(\dfrac{4}{10}\). As for the previous example of matter dispersal, extrapolating this treatment to macroscopic collections of particles dramatically increases the probability of the uniform distribution relative to the other distributions. This supports the common observation that placing hot and cold objects in contact results in spontaneous heat flow that ultimately equalizes the objects’ temperatures. And, again, this spontaneous process is also characterized by an increase in system entropy. Consider the system shown here. What is the change in entropy for a process that converts the system from distribution (a) to (c)? We are interested in the following change: The initial number of microstates is one, the final six: \[ΔS=k\ln\dfrac{W_\ce{c}}{W_\ce{a}}=\mathrm{1.38×10^{−23}\:J/K×\ln\dfrac{6}{1}=2.47×10^{−23}\:J/K}\] The sign of this result is consistent with expectation; since there are more microstates possible for the final state than for the initial state, the change in entropy should be positive. Consider the system shown in Figure \(\Page {3}\). What is the change in entropy for the process where the energy is transferred from the hot object ( ) to the cold object ( )? : 0 J/K Entropy: The relationships between entropy, microstates, and matter/energy dispersal described previously allow us to make generalizations regarding the relative entropies of substances and to predict the sign of entropy changes for chemical and physical processes. Consider the phase changes illustrated in Figure \(\Page {5}\). In the solid phase, the atoms or molecules are restricted to nearly fixed positions with respect to each other and are capable of only modest oscillations about these positions. With essentially fixed locations for the system’s component particles, the number of microstates is relatively small. In the liquid phase, the atoms or molecules are free to move over and around each other, though they remain in relatively close proximity to one another. This increased freedom of motion results in a greater variation in possible particle locations, so the number of microstates is correspondingly greater than for the solid. As a result, > and the process of converting a substance from solid to liquid (melting) is characterized by an increase in entropy, Δ > 0. By the same logic, the reciprocal process (freezing) exhibits a decrease in entropy, Δ < 0. Now consider the vapor or gas phase. The atoms or molecules occupy a greater volume than in the liquid phase; therefore each atom or molecule can be found in many more locations than in the liquid (or solid) phase. Consequently, for any substance, > > , and the processes of vaporization and sublimation likewise involve increases in entropy, Δ > 0. Likewise, the reciprocal phase transitions, condensation and deposition, involve decreases in entropy, Δ < 0. According to kinetic-molecular theory, the temperature of a substance is proportional to the average kinetic energy of its particles. Raising the temperature of a substance will result in more extensive vibrations of the particles in solids and more rapid translations of the particles in liquids and gases. At higher temperatures, the distribution of kinetic energies among the atoms or molecules of the substance is also broader (more dispersed) than at lower temperatures. Thus, the entropy for any substance increases with temperature (Figure \(\Page {6}\) ). The entropy of a substance is influenced by structure of the particles (atoms or molecules) that comprise the substance. With regard to atomic substances, heavier atoms possess greater entropy at a given temperature than lighter atoms, which is a consequence of the relation between a particle’s mass and the spacing of quantized translational energy levels (which is a topic beyond the scope of our treatment). For molecules, greater numbers of atoms (regardless of their masses) increase the ways in which the molecules can vibrate and thus the number of possible microstates and the system entropy. Finally, variations in the types of particles affects the entropy of a system. Compared to a pure substance, in which all particles are identical, the entropy of a mixture of two or more different particle types is greater. This is because of the additional orientations and interactions that are possible in a system comprised of nonidentical components. For example, when a solid dissolves in a liquid, the particles of the solid experience both a greater freedom of motion and additional interactions with the solvent particles. This corresponds to a more uniform dispersal of matter and energy and a greater number of microstates. The process of dissolution therefore involves an increase in entropy, Δ > 0. Considering the various factors that affect entropy allows us to make informed predictions of the sign of Δ for various chemical and physical processes as illustrated in Example . Predict the sign of the entropy change for the following processes. Indicate the reason for each of your predictions. Predict the sign of the enthalpy change for the following processes. Give a reason for your prediction. : (a) Positive; The solid dissolves to give an increase of mobile ions in solution. (b) Negative; The liquid becomes a more ordered solid. (c) Positive; The relatively ordered solid becomes a gas. (d) Positive; There is a net production of one mole of gas. Entropy (S) is a thermodynamic property of all substances. The greater the number of possible microstates for a system, the greater the disorder and the higher the entropy. Experiments show that the magnitude of ΔS is 80–90 J/(mol•K) for a wide variety of liquids with different boiling points. However, liquids that have highly ordered structures due to hydrogen bonding or other intermolecular interactions tend to have significantly higher values of ΔS . For instance, ΔS for water is 102 J/(mol•K). Another process that is accompanied by entropy changes is the formation of a solution. As illustrated in Figure \(\Page {4}\), the formation of a liquid solution from a crystalline solid (the solute) and a liquid solvent is expected to result in an increase in the number of available microstates of the system and hence its entropy. Indeed, dissolving a substance such as NaCl in water disrupts both the ordered crystal lattice of NaCl and the ordered hydrogen-bonded structure of water, leading to an increase in the entropy of the system. At the same time, however, each dissolved Na ion becomes hydrated by an ordered arrangement of at least six water molecules, and the Cl ions also cause the water to adopt a particular local structure. Both of these effects increase the order of the system, leading to a decrease in entropy. The overall entropy change for the formation of a solution therefore depends on the relative magnitudes of these opposing factors. In the case of an NaCl solution, disruption of the crystalline NaCl structure and the hydrogen-bonded interactions in water is quantitatively more important, so ΔS > 0. Dissolving NaCl in water results in an increase in the entropy of the system. Each hydrated ion, however, forms an ordered arrangement with water molecules, which decreases the entropy of the system. The magnitude of the increase is greater than the magnitude of the decrease, so the overall entropy change for the formation of an NaCl solution is positive. Predict which substance in each pair has the higher entropy and justify your answer. : amounts of substances and temperature : higher entropy : From the number of atoms present and the phase of each substance, predict which has the greater number of available microstates and hence the higher entropy. : Predict which substance in each pair has the higher entropy and justify your answer. Changes in entropy (ΔS), together with changes in enthalpy (ΔH), enable us to predict in which direction a chemical or physical change will occur spontaneously. Before discussing how to do so, however, we must understand the difference between a reversible process and an irreversible one. In a reversible process, every intermediate state between the extremes is an equilibrium state, regardless of the direction of the change. In contrast, an irreversible process is one in which the intermediate states are not equilibrium states, so change occurs spontaneously in only one direction. As a result, a reversible process can change direction at any time, whereas an irreversible process cannot. When a gas expands reversibly against an external pressure such as a piston, for example, the expansion can be reversed at any time by reversing the motion of the piston; once the gas is compressed, it can be allowed to expand again, and the process can continue indefinitely. In contrast, the expansion of a gas into a vacuum (P = 0) is irreversible because the external pressure is measurably less than the internal pressure of the gas. No equilibrium states exist, and the gas expands irreversibly. When gas escapes from a microscopic hole in a balloon into a vacuum, for example, the process is irreversible; the direction of airflow cannot change. Because work done during the expansion of a gas depends on the opposing external pressure (w = P ΔV), work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: w ≥ w . Whether a process is reversible or irreversible, ΔU = q + w. Because U is a state function, the magnitude of ΔU does not depend on reversibility and is independent of the path taken. So \[ΔU = q_{rev} + w_{rev} = q_{irrev} + w_{irrev} \label{Eq1a}\] Work done in a reversible process is always equal to or greater than work done in a corresponding irreversible process: w ≥ w . In other words, ΔU for a process is the same whether that process is carried out in a reversible manner or an irreversible one. We now return to our earlier definition of entropy, using the magnitude of the heat flow for a reversible process (q ) to define entropy quantitatively. At the atomic and molecular level, all energy is ; each particle possesses discrete states of kinetic energy and is able to accept thermal energy only in packets whose values correspond to the energies of one or more of these states. Polyatomic molecules can store energy in rotational and vibrational motions, and all molecules (even monatomic ones) will possess translational kinetic energy (thermal energy) at all temperatures above absolute zero. The energy difference between adjacent translational states is so minute that translational kinetic energy can be regarded as (non-quantized) for most practical purposes. The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an example here. Suppose that we have a system consisting of three molecules and three quanta of energy to share among them. We can give all the kinetic energy to any one molecule, leaving the others with none, we can give two units to one molecule and one unit to another, or we can share out the energy equally and give one unit to each molecule. All told, there are ten possible ways of distributing three units of energy among three identical molecules as shown here: Each of these ten possibilities represents a distinct that will describe the system at any instant in time. Those microstates that possess identical distributions of energy among the accessible quantum levels (and differ only in which particular molecules occupy the levels) are known as . Because all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates that can produce it. Thus in the system shown above, the configuration labeled will be observed 60% of the time, while will occur only 10% of the time. As the number of molecules and the number of quanta increases, the number of accessible microstates grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available microstates will be around 10 — a number that greatly exceeds the number of atoms in the observable universe! The number of possible configurations (as defined above) also increases, but in such a way as to greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large enough to be observable under normal conditions, only a single configuration (energy distribution amongst the quantum states) need be considered; even the second-most-probable configuration can be neglected. : any collection of molecules large enough in numbers to have chemical significance will have its therrmal energy distributed over an unimaginably large number of microstates. The number of microstates increases exponentially as more energy states ("configurations" as defined above) become accessible owing to \[\Delta S=\frac{q_{\textrm{rev}}}{T}\] Entropy (S) is a state function whose value increases with an increase in the number of available microstates. A reversible process is one for which all intermediate states between extremes are equilibrium states; it can change direction at any time. In contrast, an irreversible process occurs in one direction only. The change in entropy of the system or the surroundings is the quantity of heat transferred divided by the temperature. Entropy ( ) may be interpreted as a measure of the dispersal or distribution of matter and/or energy in a system, and it is often described as representing the “disorder” of the system. For a given substance, < < in a given physical state at a given temperature, entropy is typically greater for heavier atoms or more complex molecules. Entropy increases when a system is heated and when solutions form. Using these guidelines, the sign of entropy changes for some chemical reactions may be reliably predicted. ). | 23,313 | 2,556 |
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Acid–base reactions are essential in both biochemistry and industrial chemistry. Moreover, many of the substances we encounter in our homes, the supermarket, and the pharmacy are acids or bases. For example, aspirin is an acid (acetylsalicylic acid), and antacids are bases. In fact, every amateur chef who has prepared mayonnaise or squeezed a wedge of lemon to marinate a piece of fish has carried out an acid–base reaction. Before we discuss the characteristics of such reactions, let’s first describe some of the properties of acids and bases. We can define as substances that dissolve in water to produce H ions, whereas are defined as substances that dissolve in water to produce OH ions. In fact, this is only one possible set of definitions. Although the general properties of acids and bases have been known for more than a thousand years, the definitions of and have changed dramatically as scientists have learned more about them. In ancient times, an acid was any substance that had a sour taste (e.g., vinegar or lemon juice), caused consistent color changes in dyes derived from plants (e.g., turning blue litmus paper red), reacted with certain metals to produce hydrogen gas and a solution of a salt containing a metal cation, and dissolved carbonate salts such as limestone (CaCO ) with the evolution of carbon dioxide. In contrast, a base was any substance that had a bitter taste, felt slippery to the touch, and caused color changes in plant dyes that differed diametrically from the changes caused by acids (e.g., turning red litmus paper blue). Although these definitions were useful, they were entirely descriptive. The first person to define acids and bases in detail was the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry, 1903). According to the , an acid is a substance like hydrochloric acid that dissolves in water to produce H ions (protons; Equation \(\ref{4.3.1}\)), and a base is a substance like sodium hydroxide that dissolves in water to produce hydroxide (OH ) ions (Equation \(\ref{4.3.2}\)): \[ \underset{an\: Arrhenius\: acid}{HCl_{(g)}} \xrightarrow {H_2 O_{(l)}} H^+_{(aq)} + Cl^-_{(aq)} \label{4.3.1} \] \[ \underset{an\: Arrhenius\: base}{NaOH_{(s)}} \xrightarrow {H_2O_{(l)}} Na^+_{(aq)} + OH^-_{(aq)} \label{4.3.2} \] According to Arrhenius, the characteristic properties of acids and bases are due exclusively to the presence of H and OH ions, respectively, in solution. Although Arrhenius’s ideas were widely accepted, his definition of acids and bases had two major limitations: \[NH_{3\;(g)} + HCl_{(g)} \rightarrow NH_4Cl_{(s)} \label{4.3.3} \] Because of the limitations of the Arrhenius definition, a more general definition of acids and bases was needed. One was proposed independently in 1923 by the Danish chemist J. N. Brønsted (1879–1947) and the British chemist T. M. Lowry (1874–1936), who defined acid–base reactions in terms of the transfer of a proton (H ion) from one substance to another. According to Brønsted and Lowry, an acid ( is any substance that can donate a proton, and a base is any substance that can accept a proton. The Brønsted–Lowry definition of an acid is essentially the same as the Arrhenius definition, except that it is not restricted to aqueous solutions. The Brønsted–Lowry definition of a base, however, is far more general because the hydroxide ion is just one of many substances that can accept a proton. Ammonia, for example, reacts with a proton to form \(NH_4^+\), so in Equation \(\ref{4.3.3}\), \(NH_3\) is a Brønsted–Lowry base and \(HCl\) is a Brønsted–Lowry acid. Because of its more general nature, the Brønsted–Lowry definition is used throughout this text unless otherwise specified. Acids differ in the number of protons they can donate. For example, monoprotic acids are compounds that are capable of donating a single proton per molecule. Monoprotic acids include HF, HCl, HBr, HI, HNO , and HNO . All carboxylic acids that contain a single −CO H group, such as acetic acid (CH CO H), are monoprotic acids, dissociating to form RCO and H . . For example, H SO can donate two H ions in separate steps, so it is a diprotic acid (a and H PO , which is capable of donating three protons in successive steps, is a triprotic acid (Equation \(\ref{4.3.4}\), Equation \(\ref{4.3.5}\), and Equation \(\ref{4.3.6}\)): \[ H_3 PO_4 (l) \overset{H_2 O(l)}{\rightleftharpoons} H ^+ ( a q ) + H_2 PO_4 ^- (aq) \label{4.3.4} \] \[ H_2 PO_4 ^- (aq) \rightleftharpoons H ^+ (aq) + HPO_4^{2-} (aq) \label{4.3.5} \] \[ HPO_4^{2-} (aq) \rightleftharpoons H^+ (aq) + PO_4^{3-} (aq) \label{4.3.6} \] In chemical equations such as these, a double arrow is used to indicate that both the forward and reverse reactions occur simultaneously, so the forward reaction does not go to completion. Instead, the solution contains significant amounts of both reactants and products. Over time, the reaction reaches a state in which the concentration of each species in solution remains constant. The reaction is then said to be in We will not discuss the strengths of acids and bases quantitatively until next semester. Qualitatively, however, we can state that react essentially completely with water to give \(H^+\) and the corresponding anion. Similarly, dissociate essentially completely in water to give \(OH^−\) and the corresponding cation. Strong acids and strong bases are both strong electrolytes. In contrast, only a fraction of the molecules of weak acids and weak bases react with water to produce ions, so weak acids and weak bases are also weak electrolytes. Typically less than 5% of a weak electrolyte dissociates into ions in solution, whereas more than 95% is present in undissociated form. In practice, only a few strong acids are commonly encountered: HCl, HBr, HI, HNO , HClO , and H SO (H PO is only moderately strong). The most common strong bases are ionic compounds that contain the hydroxide ion as the anion; three examples are NaOH, KOH, and Ca(OH) . Common weak acids include HCN, H S, HF, oxoacids such as HNO and HClO, and carboxylic acids such as acetic acid. The ionization reaction of acetic acid is as follows: \[ CH_3 CO_2 H(l) \overset{H_2 O(l)}{\rightleftharpoons} H^+ (aq) + CH_3 CO_2^- (aq) \label{4.3.7} \] Although acetic acid is soluble in water, almost all of the acetic acid in solution exists in the form of neutral molecules (less than 1% dissociates). Sulfuric acid is unusual in that it is a strong acid when it donates its first proton (Equation \(\ref{4.3.8}\)) but a weak acid when it donates its second proton (Equation \(\ref{4.3.9}\)) as indicated by the single and double arrows, respectively: \[ \underset{strong\: acid}{H_2 SO_4 (l)} \xrightarrow {H_2 O(l)} H ^+ (aq) + HSO_4 ^- (aq) \label{4.3.8} \] \[ \underset{weak\: acid}{HSO_4^- (aq)} \rightleftharpoons H^+ (aq) + SO_4^{2-} (aq) \label{4.3.9} \] Consequently, an aqueous solution of sulfuric acid contains \(H^+_{(aq)}\) ions and a mixture of \(HSO^-_{4\;(aq)}\) and \(SO^{2−}_{4\;(aq)}\) ions, but no \(H_2SO_4\) molecules. All other polyprotic acids, such as H PO , are weak acids. The most common weak base is ammonia, which reacts with water to form small amounts of hydroxide ion: \[ NH_3 (g) + H_2 O(l) \rightleftharpoons NH_4^+ (aq) + OH^- (aq) \label{4.3.10} \] Most of the ammonia (>99%) is present in the form of NH (g). Amines, which are organic analogues of ammonia, are also weak bases, as are ionic compounds that contain anions derived from weak acids (such as S ). There is no correlation between the solubility of a substance and whether it is a strong electrolyte, a weak electrolyte, or a nonelectrolyte. Definition of Strong/Weak Acids & Bases: Table \(\Page {1}\) lists some common strong acids and bases. Acids other than the six common strong acids are almost invariably weak acids. The only common strong bases are the hydroxides of the alkali metals and the heavier alkaline earths (Ca, Sr, and Ba); any other bases you encounter are most likely weak. Remember that Many weak acids and bases are extremely soluble in water. Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these. compound acid or base strength Determine whether the compound is organic or inorganic. If inorganic, determine whether the compound is acidic or basic by the presence of dissociable H or OH ions, respectively. If organic, identify the compound as a weak base or a weak acid by the presence of an amine or a carboxylic acid group, respectively. Recall that all polyprotic acids except H SO are weak acids. Classify each compound as a strong acid, a weak acid, a strong base, a weak base, or none of these. strong base strong acid weak acid weak base none of these; formaldehyde is a neutral molecule | 8,819 | 2,558 |
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While the focus of this textbook is on organic reactions occuring in living cells, if you are a chemistry major, or are planning to take a standardized exam such as the MCAT, you will need to be familiar with a number of laboratory synthesis reactions. Here, we review the lab synthesis reactions covered in this text, which include most of the reactions typically covered in traditional organic texts. on the chapter/section number for direct links to the section where these reactions are introduced. : content below redirects to an older edition of the text, which differs from the current version in some content and organization. | 784 | 2,559 |
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The physical, chemical and spectroscopic properties of alcohols are relative to it’s chemical structures. Alcohols are compounds of the general formula ROH, where R is any alkyl or substituted alkyl group. The hydroxyl group (OH groups) is the characteristic functional group of alcohols and is one of the most important functional groups of naturally occurring organic molecules. All carbohydrates and their derivatives, including nucleic acids, have hydroxyl groups. Some amino acids, most steroids, many terpenes, and plant pigments have hydroxyl groups. These substances serve many diverse purposes for the support and maintenance of life. One extreme example is the potent toxin tetrodotoxin, which is isolated from puffer fish and has obvious use for defense against predators. This compound has special biochemical interest, having six different hydroxylic functions arranged on a cagelike structure: On the more practical side, vast quantities of simple alcohols —methanol, ethanol, 2-propanol, 1-butanol —and many ethers are made from petroleum-derived hydrocarbons. These alcohols are widely used as solvents and as intermediates for the synthesis of more complex substances. The reactions involving the hydrogens of alcoholic OH groups are expected to be similar to those of water, HOH, the simplest hydroxylic compound. Alcohols, ROH, can be regarded in this respect as substitution products of water. However, with alcohols we shall be interested not only in reactions that proceed at the O-H bond but also with processes that result in cleavage of the C-O bond, or changes in the organic group R. The simple ethers, ROR, do not have O-H bonds, and most of their reactions are limited to the substituent groups. The chemistry of ethers, therefore, is less varied than that of alcohols. This fact is turned to advantage in the widespread use of ethers as solvents for a variety of organic reactions, as we already have seen for . Nonetheless, cyclic ethers with small rings show enhanced reactivity because of ring strain and, for this reason, are valuable intermediates in organic synthesis. Before turning to the specific chemistry of alcohols and ethers, we remind you that the naming alcohols and ethers is summarized in naming alcohols, phenols and Naming Ethers. and (1977) | 2,313 | 2,561 |
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Prior to the early 1920's, chemists doubted the existence of molecules having molecular weights greater than a few thousand. This limiting view was challenged by , a German chemist with experience in studying natural compounds such as rubber and cellulose. In contrast to the prevailing rationalization of these substances as aggregates of small molecules, Staudinger proposed they were made up of composed of 10,000 or more atoms. He formulated a structure for rubber, based on a repeating isoprene unit (referred to as a monomer). For his contributions to chemistry, Staudinger received the 1953 Nobel Prize. The terms and were derived from the Greek roots (many), (one) and (part). Recognition that polymeric macromolecules make up many important natural materials was followed by the creation of synthetic analogs having a variety of properties. Indeed, applications of these materials as fibers, flexible films, adhesives, resistant paints and tough but light solids have transformed modern society. Some important examples of these substances are discussed in the following sections. There are two general types of polymerization reactions: addition polymerization and condensation polymerization. In addition polymerization, the monomers add to one another in such a way that the polymer contains all the atoms of the starting monomers. Ethylene molecules are joined together in long chains. Many natural materials—such as proteins, cellulose and starch, and complex silicate minerals—are polymers. Artificial fibers, films, plastics, semisolid resins, and rubbers are also polymers. More than half the compounds produced by the chemical industry are synthetic polymers. The polymerization can be represented by the reaction of a few monomer units: The bond lines extending at the ends in the formula of the product indicate that the structure extends for many units in each direction. Notice that all the atoms—two carbon atoms and four hydrogen atoms—of each monomer molecule are incorporated into the polymer structure. Because displays such as the one above are cumbersome, the polymerization is often abbreviated as follows: CH =CH → CH CH During the polymeriation of ethene, thousands of ethene molecules join together to make poly(ethene) - commonly called polythene. The reaction is done at high pressures in the presence of a trace of oxygen as an initiator. Some common addition polymers are listed in Table \(\Page {1}\). Note that all the monomers have carbon-to-carbon double bonds. Many polymers are mundane (e.g., plastic bags, food wrap, toys, and tableware), but there are also polymers that conduct electricity, have amazing adhesive properties, or are stronger than steel but much lighter in weight. The oxygen reacts with some of the ethene to give an organic peroxide. Organic peroxides are very reactive molecules containing oxygen-oxygen single bonds which are quite weak and which break easily to give free radicals. You can short-cut the process by adding other organic peroxides directly to the ethene instead of using oxygen if you want to. The type of the free radicals that start the reaction off vary depending on their source. For simplicity we give them a general formula: \(Ra ^{\bullet}\) In an ethene molecule, CH =CH , the two pairs of electrons which make up the double bond aren't the same. One pair is held securely on the line between the two carbon nuclei in a bond called a sigma bond. The other pair is more loosely held in an orbital above and below the plane of the molecule known as a \(\pi\) bond. It would be helpful - but not essential - if you read about the before you went on. If the diagram above is unfamiliar to you, then you certainly ought to read this background material. Imagine what happens if a free radical approaches the \(\pi\) bond in ethene. Don't worry that we've gone back to a simpler diagram. As long as you realise that the pair of electrons shown between the two carbon atoms is in a \(\pi\) bond - and therefore vulnerable - that's all that really matters for this mechanism. The sigma bond between the carbon atoms isn't affected by any of this. The free radical, Ra , uses one of the electrons in the \(\pi\) bond to help to form a new bond between itself and the left hand carbon atom. The other electron returns to the right hand carbon. You can show this using "curly arrow" notation if you want to: If you aren't sure about about you can follow this link. This is energetically worth doing because the new bond between the radical and the carbon is stronger than the \(\pi\) bond which is broken. You would get more energy out when the new bond is made than was used to break the old one. The more energy that is given out, the more stable the system becomes. What we've now got is a bigger free radical - lengthened by CH CH . That can react with another ethene molecule in the same way: So now the radical is even bigger. That can react with another ethene - and so on and so on. The polymer chain gets longer and longer. The chain does not, however, grow indefinitely. Sooner or later two free radicals will collide together. That immediately stops the growth of two chains and produces one of the final molecules in the poly(ethene). It is important to realise that the poly(ethene) is going to be a mixture of molecules of different sizes, made in this sort of random way. Because chain termination is a random process, poly(ethene) will be made up of chains of different lengths. A large number of important and useful polymeric materials are not formed by chain-growth processes involving reactive species such as radicals, but proceed instead by conventional functional group transformations of polyfunctional reactants. These polymerizations often (but not always) occur with loss of a small byproduct, such as water, and generally (but not always) combine two different components in an alternating structure. The polyester Dacron and the polyamide Nylon 66, shown here, are two examples of synthetic condensation polymers, also known as step-growth polymers. In contrast to chain-growth polymers, most of which grow by carbon-carbon bond formation, step-growth polymers generally grow by carbon-heteroatom bond formation (C-O & C-N in Dacron & Nylon respectively). Although polymers of this kind might be considered to be alternating copolymers, the repeating monomeric unit is usually defined as a combined moiety. Examples of naturally occurring condensation polymers are cellulose, the polypeptide chains of proteins, and poly(β-hydroxybutyric acid), a polyester synthesized in large quantity by certain soil and water bacteria. Formulas for these will be displayed below by clicking on the diagram. Condensation polymers form more slowly than addition polymers, often requiring heat, and they are generally lower in molecular weight. The terminal functional groups on a chain remain active, so that groups of shorter chains combine into longer chains in the late stages of polymerization. The presence of polar functional groups on the chains often enhances chain-chain attractions, particularly if these involve hydrogen bonding, and thereby crystallinity and tensile strength. The following examples of condensation polymers are illustrative. Note that for commercial synthesis the carboxylic acid components may actually be employed in the form of derivatives such as simple esters. Also, the polymerization reactions for Nylon 6 and Spandex do not proceed by elimination of water or other small molecules. Nevertheless, the polymer clearly forms by a step-growth process. Some Condensation Polymers The difference in Tg and Tm between the first polyester (completely aliphatic) and the two nylon polyamides (5th & 6th entries) shows the effect of intra-chain hydrogen bonding on crystallinity. The replacement of flexible alkylidene links with rigid benzene rings also stiffens the polymer chain, leading to increased crystalline character, as demonstrated for polyesters (entries 1, 2 &3) and polyamides (entries 5, 6, 7 & 8). The high Tg and Tm values for the amorphous polymer Lexan are consistent with its brilliant transparency and glass-like rigidity. Kevlar and Nomex are extremely tough and resistant materials, which find use in bullet-proof vests and fire resistant clothing. Many polymers, both addition and condensation, are used as fibers The chief methods of spinning synthetic polymers into fibers are from melts or viscous solutions. Polyesters, polyamides and polyolefins are usually spun from melts, provided the Tm is not too high. Polyacrylates suffer thermal degradation and are therefore spun from solution in a volatile solvent. Cold-drawing is an important physical treatment that improves the strength and appearance of these polymer fibers. At temperatures above T , a thicker than desired fiber can be forcibly stretched to many times its length; and in so doing the polymer chains become untangled, and tend to align in a parallel fashion. This cold-drawing procedure organizes randomly oriented crystalline domains, and also aligns amorphous domains so they become more crystalline. In these cases, the physically oriented morphology is stabilized and retained in the final product. This contrasts with elastomeric polymers, for which the stretched or aligned morphology is unstable relative to the amorphous random coil morphology. This cold-drawing treatment may also be used to treat polymer films (e.g. Mylar & Saran) as well as fibers. Step-growth polymerization is also used for preparing a class of adhesives and amorphous solids called epoxy resins. Here the covalent bonding occurs by an S 2 reaction between a nucleophile, usually an amine, and a terminal epoxide. In the following example, the same bisphenol A intermediate used as a monomer for Lexan serves as a difunctional scaffold to which the epoxide rings are attached. Bisphenol A is prepared by the acid-catalyzed condensation of acetone with phenol. ), Jim Clark ( ) | 10,017 | 2,562 |
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Implementing a sampling plan usually involves three steps: physically removing the sample from its target population, preserving the sample, and preparing the sample for analysis. Except for in situ sampling, we analyze a sample after we have removed it from its target population. Because sampling exposes the target population to potential contamination, our sampling device must be inert and clean. Once we remove a sample from its target population, there is a danger that it will undergo a chemical or physical change before we can complete its analysis. This is a serious problem because the sample’s properties will no longer e representative of the target population. To prevent this problem, we often preserve samples before we transport them to the laboratory for analysis. Even when we analyze a sample in the field, preservation may still be necessary. The initial sample is called the primary or , and it may be a single increment drawn from the target population or a composite of several increments. In many cases we cannot analyze the gross sample without first preparing the sample for analyze by reducing the sample’s particle size, by converting the sample into a more readily analyzable form, or by improving its homogeneity. Although you may never work with the specific samples highlighted in this section, the case studies presented here may help you in envisioning potential problems associated with your samples. There are many good examples of solution samples: commercial solvents; beverages, such as milk or fruit juice; natural waters, including lakes, streams, seawater, and rain; bodily fluids, such as blood and urine; and, suspensions, such as those found in many oral medications. Let’s use the sampling of natural waters and wastewaters as a case study in how to sample a solution. The chemical composition of a surface water—such as a stream, river, lake, estuary, or ocean—is influenced by flow rate and depth. Rapidly flowing shallow streams and rivers, and shallow (<5 m) lakes usually are well mixed and show little stratification with depth. To collect a grab sample we submerge a capped bottle below the surface, remove the cap and allow the bottle to fill completely, and replace the cap. Collecting a sample this way avoids the air–water interface, which may be enriched with heavy metals or contaminated with oil [Duce, R. A.; Quinn, J. G. Olney, C. E.; Piotrowicz, S. R.; Ray, S. J.; Wade, T. L. , , 161–163]. Slowly moving streams and rivers, lakes deeper than five meters, estuaries, and oceans may show substantial stratification with depth. Grab samples from near the surface are collected as described above, and samples at greater depths are collected using a sample bottle lowered to the desired depth (Figure 7.3.1
). Wells for sampling groundwater are purged before we collect samples because the chemical composition of water in a well-casing may differ significantly from that of the groundwater. These differences may result from contaminants introduced while drilling the well or by a change in the groundwater’s redox potential following its exposure to atmospheric oxygen. In general, a well is purged by pumping out a volume of water equivalent to several well-casing volumes or by pumping until the water’s temperature, pH, or specific conductance is constant. A municipal water supply, such as a residence or a business, is purged before sampling because the chemical composition of water standing in a pipe may differ significantly from the treated water supply. Samples are collected at faucets after flushing the pipes for 2-3 minutes. Samples from municipal wastewater treatment plants and industrial discharges often are collected as a 24-hour composite. An automatic sampler periodically removes an individual grab sample, adding it to those collected previously. The volume of each sample and the frequency of sampling may be constant, or may vary in response to changes in flow rate. Sample containers for collecting natural waters and wastewaters are made from glass or plastic. Kimax and Pyrex brand borosilicate glass have the advantage of being easy to sterilize, easy to clean, and inert to all solutions except those that are strongly alkaline. The disadvantages of glass containers are cost, weight, and the ease of breakage. Plastic containers are made from a variety of polymers, including polyethylene, polypropylene, polycarbonate, polyvinyl chloride, and Teflon. Plastic containers are light-weight, durable, and, except for those manufactured from Teflon, inexpensive. In most cases glass or plastic bottles are used interchangeably, although polyethylene bottles generally are preferred because of their lower cost. Glass containers are always used when collecting samples for the analysis of pesticides, oil and grease, and organics because these species often interact with plastic surfaces. Because glass surfaces easily adsorb metal ions, plastic bottles are preferred when collecting samples for the analysis of trace metals. In most cases the sample bottle has a wide mouth, which makes it easy to fill and to remove the sample. A narrow-mouth sample bottle is used if exposing the sample to the container’s cap or to the outside environment is a problem. Unless exposure to plastic is a problem, caps for sample bottles are manufactured from polyethylene. When polyethylene must be avoided, the container’s cap includes an inert interior liner of neoprene or Teflon. Here our concern is only with the need to prepare the gross sample by converting it into a form suitable for analysis. Some analytical methods require additional sample preparation steps, such as concentrating or diluting the analyte, or adjusting the analyte’s chemical form. We will consider these forms of sample preparation in later chapters that focus on specific analytical methods. After removing a sample from its target population, its chemical composition may change as a result of chemical, biological, or physical processes. To prevent a change in composition, samples are preserved by controlling the sample’s pH and temperature, by limiting its exposure to light or to the atmosphere, or by adding a chemical preservative. After preserving a sample, it is safely stored for later analysis. The maximum holding time between preservation and analysis depends on the analyte’s stability and the effectiveness of sample preservation. Table 7.3.1
summarizes preservation methods and maximum holding times for several analytes of importance in the analysis of natural waters and wastewaters. cool to 4 C; add H SO to pH < 2 HNO to pH < 2 HNO to pH < 2 none required 1 mL of 10 mg/mL HgCl or immediate extraction with a suitable non-aqueous solvent 7 days without extraction; 40 days with extraction none required Other than adding a preservative, solution samples generally do not need additional preparation before analysis. This is the case for samples of natural waters and wastewaters. Solution samples with particularly complex matricies—blood and milk are two common examples—may need addi- tional processing to separate analytes from interferents, a topic covered later in this chapter. Typical examples of gaseous samples include automobile exhaust, emissions from industrial smokestacks, atmospheric gases, and compressed gases. Also included in this category are aerosol particulates—the fine solid particles and liquid droplets that form smoke and smog. Let’s use the sampling of urban air as a case study in how to sample a gas. One approach for collecting a sample of urban air is to fill a stainless steel canister or a Tedlar/Teflon bag. A pump pulls the air into the container and, after purging, the container is sealed. This method has the advantage of being simple and of collecting a representative sample. Disadvantages include the tendency for some analytes to adsorb to the container’s walls, the presence of analytes at concentrations too low to detect with suitable accuracy and precision, and the presence of reactive analytes, such as ozone and nitrogen oxides, that may react with the container or that may otherwise alter the sample’s chemical composition during storage. When using a stainless steel canister, cryogenic cooling, which changes the sample from a gaseous state to a liquid state, may limit some of these disadvantages. Most urban air samples are collected by filtration or by using a trap that contains a solid sorbent. Solid sorbents are used for volatile gases (a vapor pressure more than 10 atm) and for semi-volatile gases (a vapor pressure between 10 atm and 10 atm). Filtration is used to collect aerosol particulates. Trapping and filtering allow for sampling larger volumes of gas—an important concern for an analyte with a small concentration—and stabilizes the sample between its collection and its analysis. In solid sorbent sampling, a pump pulls the urban air through a canister packed with sorbent particles. Typically 2–100 L of air are sampled when collecting a volatile compound and 2–500 m when collecting a semi-volatile gas. A variety of inorganic, organic polymer, and carbon sorbents have been used. Inorganic sorbents, such as silica gel, alumina, magnesium aluminum silicate, and molecular sieves, are efficient collectors for polar compounds. Their efficiency at absorbing water, however, limits their capacity for many organic analytes. 1 m is equivalent to 10 L. Organic polymeric sorbents include polymeric resins of 2,4-diphenyl- -phenylene oxide or styrene-divinylbenzene for volatile compounds, and polyurethane foam for semi-volatile compounds. These materials have a low affinity for water and are efficient for sampling all but the most highly volatile organic compounds and some lower molecular weight alcohols and ketones. Carbon sorbents are superior to organic polymer resins, which makes them useful for highly volatile organic compounds that will not absorb onto polymeric resins, although removing the compounds may be difficult. Non-volatile compounds normally are present either as solid particulates or are bound to solid particulates. Samples are collected by pulling a large volume of urban air through a filtering unit and collecting the particulates on glass fiber filters. The short term exposure of humans, animals, and plants to atmospheric pollutants is more severe than that for pollutants in other matrices. Because the composition of atmospheric gases can vary significantly over a time, the continuous monitoring of atmospheric gases such as O , CO, SO , NH , H O , and NO by in situ sampling is important [Tanner, R. L. in Keith, L. H., ed. , American Chemical Society: Washington, D. C., 1988, 275–286]. After collecting a gross sample of urban air, generally there is little need for sample preservation or preparation. The chemical composition of a gas sample usually is stable when it is collected using a solid sorbent, a filter, or by cryogenic cooling. When using a solid sorbent, gaseous compounds are released for analysis by thermal desorption or by extracting with a suitable solvent. If the sorbent is selective for a single analyte, the increase in the sorbent’s mass is used to determine the amount of analyte in the sample. Typical examples of solid samples include large particulates, such as those found in ores; smaller particulates, such as soils and sediments; tablets, pellets, and capsules used for dispensing pharmaceutical products and animal feeds; sheet materials, such as polymers and rolled metals; and tissue samples from biological specimens. Solids usually are heterogeneous and we must collect samples carefully if they are to be representative of the target population. Let’s use the sampling of sediments, soils, and ores as a case study in how to sample solids. Sediments from the bottom of streams, rivers, lakes, estuaries, and oceans are collected with a bottom grab sampler or with a corer. A bottom grab sampler (Figure 7.3.2
) is equipped with a pair of jaws that close when they contact the sediment, scooping up sediment in the process. Its principal advantages are ease of use and the ability to collect a large sample. Disadvantages include the tendency to lose finer grain sediment particles as water flows out of the sampler, and the loss of spatial information—both laterally and with depth—due to mixing of the sample. An alternative method for collecting sediments is the cylindrical coring device shown in Figure 7.3.3
). The corer is dropped into the sediment, collecting a column of sediment and the water in contact with the sediment. With the possible exception of sediment at the surface, which may experience mixing, samples collected with a corer maintain their vertical profile, which preserves information about how the sediment’s composition changes with depth. Collecting soil samples at depths of up to 30 cm is accomplished with a scoop or a shovel, although the sampling variance generally is high. A better tool for collecting soil samples near the surface is a soil punch, which is a thin-walled steel tube that retains a core sample after it is pushed into the soil and removed. Soil samples from depths greater than 30 cm are collected by digging a trench and collecting lateral samples with a soil punch. Alternatively, an auger is used to drill a hole to the desired depth and the sample collected with a soil punch. For particulate materials, particle size often determines the sampling method. Larger particulate solids, such as ores, are sampled using a riffle (Figure 7.3.4
), which is a trough with an even number of compartments. Because adjoining compartments empty onto opposite sides of the riffle, dumping a gross sample into the riffle divides it in half. By repeatedly passing half of the separated material back through the riffle, a sample of the desired size is collected. A sample thief (Figure 7.3.5
) is used for sampling smaller particulate materials, such as powders. A typical sample thief consists of two tubes that are nestled together. Each tube has one or more slots aligned down the length of the sample thief. Before inserting the sample thief into the material being sampled, the slots are closed by rotating the inner tube. When the sample thief is in place, rotating the inner tube opens the slots, which fill with individual samples. The inner tube is then rotated to the closed position and the sample thief withdrawn. Without preservation, a solid sample may undergo a change in composition due to the loss of volatile material, biodegradation, or chemical reactivity (particularly redox reactions). Storing samples at lower temperatures makes them less prone to biodegradation and to the loss of volatile material, but fracturing of solids and phase separations may present problems. To minimize the loss of volatile compounds, the sample container is filled completely, eliminating a headspace where gases collect. Samples that have not been exposed to O particularly are susceptible to oxidation reactions. For example, samples of anaerobic sediments must be prevented from coming into contact with air. Unlike gases and liquids, which generally require little sample preparation, a solid sample usually needs some processing before analysis. There are two reasons for this. First, as discussed in , the standard deviation for sampling, , is a function of the number of particles in the sample, not the combined mass of the particles. For a heterogeneous material that consists of large particulates, the gross sample may be too large to analyze. For example, a Ni-bearing ore with an average particle size of 5 mm may require a sample that weighs one ton to obtain a reasonable . Reducing the sample’s average particle size allows us to collect the same number of particles with a smaller, more manageable mass. Second, many analytical techniques require that the analyte be in solution. A reduction in particle size is accomplished by crushing and grinding the gross sample. The resulting particulates are then thoroughly mixed and divided into of smaller mass. This process seldom occurs in a single step. Instead, subsamples are cycled through the process several times until a final is obtained. Crushing and grinding uses mechanical force to break larger particles into smaller particles. A variety of tools are used depending on the particle’s size and hardness. Large particles are crushed using jaw crushers that can reduce particles to diameters of a few millimeters. Ball mills, disk mills, and mortars and pestles are used to further reduce particle size. A significant change in the gross sample’s composition may occur during crushing and grinding. Decreasing particle size increases the available surface area, which increases the risk of losing volatile components. This problem is made worse by the frictional heat that accompanies crushing and grinding. Increasing the surface area also exposes interior portions of the sample to the atmosphere where oxidation may alter the gross sample’s composition. Other problems include contamination from the materials used to crush and grind the sample, and differences in the ease with which particles are reduced in size. For example, softer particles are easier to reduce in size and may be lost as dust before the remaining sample is processed. This is a particular problem if the analyte’s distribution between different types of particles is not uniform. The gross sample is reduced to a uniform particle size by intermittently passing it through a sieve. Those particles not passing through the sieve receive additional processing until the entire sample is of uniform size. The resulting material is mixed thoroughly to ensure homogeneity and a subsample obtained with a riffle, or by c . As shown in Figure 7.3.6
, the gross sample is piled into a cone, flattened, and divided into four quarters. After discarding two diagonally opposed quarters, the remaining material is cycled through the process of coning and quartering until a suitable laboratory sample remains. If you are fortunate, your sample will dissolve easily in a suitable solvent, requiring no more effort than gently swirling and heating. Distilled water usually is the solvent of choice for inorganic salts, but organic solvents, such as methanol, chloroform, and toluene, are useful for organic materials. When a sample is difficult to dissolve, the next step is to try digesting it with an acid or a base. Table 7.3.2
lists several common acids and bases, and summarizes their use. Digestions are carried out in an open container, usually a beaker, using a hot-plate as a source of heat. The main advantage of an open-vessel digestion is cost because it requires no special equipment. Volatile reaction products, however, are lost, which results in a determinate error if they include the analyte. HCl (37% w/w) HNO (70% w/w) H SO (98% w/w) HClO (70% w/w) HCl:HNO (3:1 v/v) NaOH Many digestions now are carried out in a closed container using microwave radiation as the source of energy. Vessels for microwave digestion are manufactured using Teflon (or some other fluoropolymer) or fused silica. Both materials are thermally stable, chemically resistant, transparent to microwave radiation, and capable of withstanding elevated pressures. A typical microwave digestion vessel, as shown in Figure 7.3.7
, consists of an insulated vessel body and a cap with a pressure relief valve. The vessels are placed in a microwave oven (a typical oven can accommodate 6–14 vessels) and microwave energy is controlled by monitoring the temperature or pressure within one of the vessels. . Microwave digestion unit: on the left is a view of the unit’s interior showing the carousel that holds the digestion vessels; on the right is a close-up of a Teflon digestion vessel, which is encased in a thermal sleeve. The pressure relief value, which is part of the vessel’s blue cap, contains a membrane that ruptures if the internal pressure becomes too high. Inorganic samples that resist decomposition by digesting with acids or bases often are brought into solution by fusing with a large excess of an alkali metal salt, called a flux. After mixing the sample and the flux in a crucible, they are heated to a molten state and allowed to cool slowly to room temperature. The resulting melt usually dissolves readily in distilled water or dilute acid. Table 7.3.3
summarizes several common fluxes and their uses. Fusion works when other methods of decomposition do not because of the high temperature and the flux’s high concentration in the molten liquid. Disadvantages include contamination from the flux and the crucible, and the loss of volatile materials. Finally, we can decompose organic materials by dry ashing. In this method the sample is placed in a suitable crucible and heated over a flame or in a furnace. The carbon present in the sample oxidizes to CO , and hydrogen, sulfur, and nitrogen are volatilized as H O, SO , and N . These gases can be trapped and weighed to determine their concentration in the organic material. Often the goal of dry ashing is to remove the organic material, leaving behind an inorganic residue, or ash, that can be further analyzed. | 21,215 | 2,564 |
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A precipitation reaction is a reaction that yields an insoluble product—a precipitate—when two solutions are mixed. We described a precipitation reaction in which a colorless solution of silver nitrate was mixed with a yellow-orange solution of potassium dichromate to give a reddish precipitate of silver dichromate: \[AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + KNO_3(aq)\label{5.2.1A}\] This equation has the general form of an exchange reaction: \[ AC + BD \rightarrow \underset{insoluble}{AD} + BC \label{5.2.2B}\] Thus precipitation reactions are a subclass of exchange reactions that occur between ionic compounds when one of the products is insoluble. Because both components of each compound change partners, such reactions are sometimes called . Two important uses of precipitation reactions are to isolate metals that have been extracted from their ores and to recover precious metals for recycling. While chemical equations show the identities of the reactants and the products and gave the stoichiometries of the reactions, but they told us very little about what was occurring in solution. In contrast, equations that show only the hydrated species focus our attention on the chemistry that is taking place and allow us to see similarities between reactions that might not otherwise be apparent. Let’s consider the reaction of silver nitrate with potassium dichromate. When aqueous solutions of silver nitrate and potassium dichromate are mixed, silver dichromate forms as a red solid. The overall balanced chemical equation for the reaction shows each reactant and product as undissociated, electrically neutral compounds: \[2AgNO_3(aq) + K_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2KNO_3(aq) \label{5.2.1}\] Although Equation \(\ref{5.2.1}\) gives the identity of the reactants and the products, it does not show the identities of the actual species in solution. Because ionic substances such as AgNO and K Cr O are strong electrolytes, they dissociate completely in aqueous solution to form ions. In contrast, because Ag Cr O is not very soluble, it separates from the solution as a solid. To find out what is actually occurring in solution, it is more informative to write the reaction as a complete ionic equation showing which ions and molecules are hydrated and which are present in other forms and phases: \[2Ag^+(aq) + 2NO_3^-(aq) + 2K^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2K^+(aq) + 2NO_3^-(aq) \label{5.2.2}\] Note that K (aq) and NO (aq) ions are present on both sides of the equation, and their coefficients are the same on both sides. These ions are called spectator ions because they do not participate in the actual reaction. Canceling the spectator ions gives the net ionic equation, which shows only those species that participate in the chemical reaction: \[2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) \label{5.2.3}\] Both mass and charge must be conserved in chemical reactions because the numbers of electrons and protons do not change. For charge to be conserved, the sum of the charges of the ions multiplied by their coefficients must be the same on both sides of the equation. In Equation \(\ref{5.2.3}\), the charge on the left side is 2(+1) + 1(−2) = 0, which is the same as the charge of a neutral Ag Cr O formula unit. By eliminating the spectator ions, we can focus on the chemistry that takes place in a solution. For example, the overall chemical equation for the reaction between silver fluoride and ammonium dichromate is as follows: \[2AgF(aq) + (NH_4)_2Cr_2O_7(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4F(aq)\label{5.2.4}\] The complete ionic equation for this reaction is as follows: \[2Ag^+(aq) + 2F^-(aq) + 2NH_4^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s) + 2NH_4^+(aq) + 2F^-(aq)\label{5.2.5}\] Because two NH (aq) and two F (aq) ions appear on both sides of Equation \(\ref{5.2.5}\), they are spectator ions. They can therefore be canceled to give the net ionic equation (Equation \(\ref{5.2.6}\)), which is identical to Equation \(\ref{5.2.3}\): \[2Ag^+(aq) + Cr_2O_7^{2-}(aq) \rightarrow Ag_2Cr_2O_7(s)\label{5.2.6}\] If we look at net ionic equations, it becomes apparent that many different combinations of reactants can result in the same net chemical reaction. For example, we can predict that silver fluoride could be replaced by silver nitrate in the preceding reaction without affecting the outcome of the reaction. Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous barium nitrate with aqueous sodium phosphate to give solid barium phosphate and a solution of sodium nitrate. reactants and products overall, complete ionic, and net ionic equations Write and balance the overall chemical equation. Write all the soluble reactants and products in their dissociated form to give the complete ionic equation; then cancel species that appear on both sides of the complete ionic equation to give the net ionic equation. From the information given, we can write the unbalanced chemical equation for the reaction: \(Ba(NO_3)_2(aq) + Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + NaNO_3(aq)\) Because the product is Ba (PO ) , which contains three Ba ions and two PO ions per formula unit, we can balance the equation by inspection: \( 3Ba(NO_3)_2(aq) + 2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + 6NaNO_3(aq) \) This is the overall balanced chemical equation for the reaction, showing the reactants and products in their undissociated form. To obtain the complete ionic equation, we write each soluble reactant and product in dissociated form: \( 3Ba^{2+}(aq) + 6NO_3^-(aq) + 6Na^+(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s) + 6Na^+(aq) + 6NO_3^-(aq) \) The six NO (aq) ions and the six Na (aq) ions that appear on both sides of the equation are spectator ions that can be canceled to give the net ionic equation: \(3Ba^{2+}(aq) + 2PO_4^{3-}(aq) \rightarrow Ba_3(PO_4)_2(s)\) Write the overall chemical equation, the complete ionic equation, and the net ionic equation for the reaction of aqueous silver fluoride with aqueous sodium phosphate to give solid silver phosphate and a solution of sodium fluoride. overall chemical equation: \[3AgF(aq) + Na_3PO_4(aq) \rightarrow Ag_3PO_4(s) + 3NaF(aq) \nonumber\] complete ionic equation: \[3Ag^+(aq) + 3F^-(aq) + 3Na^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) + 3Na^+(aq) + 3F^-(aq) \nonumber\] net ionic equation: \[3Ag^+(aq) + PO_4^{3-}(aq) \rightarrow Ag_3PO_4(s) \nonumber\] So far, we have always indicated whether a reaction will occur when solutions are mixed and, if so, what products will form. As you advance in chemistry, however, you will need to predict the results of mixing solutions of compounds, anticipate what kind of reaction (if any) will occur, and predict the identities of the products. Students tend to think that this means they are supposed to “just know” what will happen when two substances are mixed. Nothing could be further from the truth: an infinite number of chemical reactions is possible, and neither you nor anyone else could possibly memorize them all. Instead, you must begin by identifying the various reactions that occur and then assessing which is the most probable (or least improbable) outcome. The most important step in analyzing an unknown reaction is to (not forgetting the solvent itself) so that you can assess which species are most likely to react with one another. The easiest way to make that kind of prediction is to attempt to place the reaction into one of several familiar classifications, refinements of the five general kinds of reactions (acid–base, exchange, condensation, cleavage, and oxidation–reduction reactions). In the sections that follow, we discuss three of the most important kinds of reactions that occur in aqueous solutions: precipitation reactions (also known as exchange reactions), acid–base reactions, and oxidation–reduction reactions. Determining the Products for Precipitation Reactions: Table \(\Page {2}\) gives guidelines for predicting the solubility of a wide variety of ionic compounds. To determine whether a precipitation reaction will occur, we identify each species in the solution and then refer to Table \(\Page {2}\) to see which, if any, combination(s) of cation and anion are likely to produce an insoluble salt. In doing so, it is important to recognize that and are relative terms that span a wide range of actual solubilities. We will discuss solubilities in more detail later, where you will learn that very small amounts of the constituent ions remain in solution even after precipitation of an “insoluble” salt. For our purposes, however, we will assume that precipitation of an insoluble salt is complete. Just as important as predicting the product of a reaction is knowing when a chemical reaction will occur. Simply mixing solutions of two different chemical substances does guarantee that a reaction will take place. For example, if 500 mL of a 1.0 M aqueous NaCl solution is mixed with 500 mL of a 1.0 M aqueous KBr solution, the final solution has a volume of 1.00 L and contains 0.50 M Na (aq), 0.50 M Cl (aq), 0.50 M K (aq), and 0.50 M Br (aq). As you will see in the following sections, none of these species reacts with any of the others. When these solutions are mixed, the only effect is to dilute each solution with the other (Figure \(\Page {1}\)). Using the information in Table \(\Page {2}\), predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs. reactants reaction and net ionic equation Because barium chloride and lithium sulfate are strong electrolytes, each dissociates completely in water to give a solution that contains the constituent anions and cations. Mixing the two solutions gives an aqueous solution that contains Ba , Cl , Li , and SO ions. The only possible exchange reaction is to form LiCl and BaSO : We now need to decide whether either of these products is insoluble. Table \(\Page {2}\) shows that LiCl is soluble in water (rules 1 and 4), but BaSO is not soluble in water (rule 5). Thus BaSO will precipitate according to the net ionic equation \[Ba^{2+}(aq) + SO_4^{2-}(aq) \rightarrow BaSO_4(s)\] Although soluble barium salts are toxic, BaSO is so insoluble that it can be used to diagnose stomach and intestinal problems without being absorbed into tissues. An outline of the digestive organs appears on x-rays of patients who have been given a “barium milkshake” or a “barium enema”—a suspension of very fine BaSO particles in water. According to Table \(\Page {2}\), RbCl is soluble (rules 1 and 4), but Co(OH) is not soluble (rule 5). Hence Co(OH) will precipitate according to the following net ionic equation: \(Co^{2+}(aq) + 2OH^-(aq) \rightarrow Co(OH)_2(s)\) According to Table \(\Page {2}\), both AlBr (rule 4) and Sr(NO ) (rule 2) are soluble. Thus no net reaction will occur. According to Table \(\Page {2}\), ammonium acetate is soluble (rules 1 and 3), but PbI is insoluble (rule 4). Thus Pb(C H O ) will dissolve, and PbI will precipitate. The net ionic equation is as follows: \(Pb^{2+} (aq) + 2I^-(aq) \rightarrow PbI_2(s) \) Using the information in Table \(\Page {2}\), predict what will happen in each case involving strong electrolytes. Write the net ionic equation for any reaction that occurs. Precipitation reactions can be used to recover silver from solutions used to develop conventional photographic film. Although largely supplanted by digital photography, conventional methods are often used for artistic purposes. Silver bromide is an off-white solid that turns black when exposed to light, which is due to the formation of small particles of silver metal. Black-and-white photography uses this reaction to capture images in shades of gray, with the darkest areas of the film corresponding to the areas that received the most light. The first step in film processing is to enhance the black/white contrast by using a developer to increase the amount of black. The developer is a reductant: because silver atoms catalyze the reduction reaction, grains of silver bromide that have already been partially reduced by exposure to light react with the reductant much more rapidly than unexposed grains. After the film is developed, any unexposed silver bromide must be removed by a process called “fixing”; otherwise, the entire film would turn black with additional exposure to light. Although silver bromide is insoluble in water, it is soluble in a dilute solution of sodium thiosulfate (Na S O ; ) because of the formation of [Ag(S O ) ] ions. Thus washing the film with thiosulfate solution dissolves unexposed silver bromide and leaves a pattern of metallic silver granules that constitutes the negative. This procedure is summarized in Figure \(\Page {2}\). The negative image is then projected onto paper coated with silver halides, and the developing and fixing processes are repeated to give a positive image. (Color photography works in much the same way, with a combination of silver halides and organic dyes superimposed in layers.) “Instant photo” operations can generate more than a hundred gallons of dilute silver waste solution per day. Recovery of silver from thiosulfate fixing solutions involves first removing the thiosulfate by oxidation and then precipitating Ag ions with excess chloride ions. A silver recovery unit can process 1500 L of photographic silver waste solution per day. Adding excess solid sodium chloride to a 500 mL sample of the waste (after removing the thiosulfate as described previously) gives a white precipitate that, after filtration and drying, consists of 3.73 g of AgCl. What mass of NaCl must be added to the 1500 L of silver waste to ensure that all the Ag ions precipitate? volume of solution of one reactant and mass of product from a sample of reactant solution mass of second reactant needed for complete reaction We can use the data provided to determine the concentration of Ag ions in the waste, from which the number of moles of Ag in the entire waste solution can be calculated. From the net ionic equation, we can determine how many moles of Cl are needed, which in turn will give us the mass of NaCl necessary. The first step is to write the net ionic equation for the reaction: \(Cl^-(aq) + Ag^+(aq) \rightarrow AgCl(s) \) We know that 500 mL of solution produced 3.73 g of AgCl. We can convert this value to the number of moles of AgCl as follows: \[ moles\: AgCl = \dfrac{grams\: AgCl} {molar\: mass\: AgCl} = 3 .73\: \cancel{g\: AgCl} \left( \dfrac{1\: mol\: AgCl} {143 .32\: \cancel{g\: AgCl}} \right) = 0 .0260\: mol\: AgCl \] Therefore, the 500 mL sample of the solution contained 0.0260 mol of Ag . The Ag concentration is determined as follows: \[ [Ag^+ ] = \dfrac{moles\: Ag^+} {liters\: soln} = \dfrac{0 .0260\: mol\: AgCl} {0 .500\: L} = 0 .0520\: M \] The total number of moles of Ag present in 1500 L of solution is as follows: \[ moles\: Ag^+ = 1500\: \cancel{L} \left( \dfrac{0 .520\: mol} {1\: \cancel{L}} \right) = 78 .1\: mol\: Ag^+ \] According to the net ionic equation, one Cl ion is required for each Ag ion. Thus 78.1 mol of NaCl are needed to precipitate the silver. The corresponding mass of NaCl is \[ mass\: NaCl = 78 .1 \: \cancel{mol\: NaCl} \left( \dfrac{58 .44\: g\: NaCl} {1\: \cancel{mol\: NaCl}} \right) = 4560\: g\: NaCl = 4 .56\: kg\: NaCl \] Note that 78.1 mol of AgCl correspond to 8.43 kg of metallic silver, which is worth about $7983 at 2011 prices ($32.84 per troy ounce). Silver recovery may be economically attractive as well as ecologically sound, although the procedure outlined is becoming nearly obsolete for all but artistic purposes with the growth of digital photography. Because of its toxicity, arsenic is the active ingredient in many pesticides. The arsenic content of a pesticide can be measured by oxidizing arsenic compounds to the arsenate ion (AsO ), which forms an insoluble silver salt (Ag AsO ). Suppose you are asked to assess the purity of technical grade sodium arsenite (NaAsO ), the active ingredient in a pesticide used against termites. You dissolve a 10.00 g sample in water, oxidize it to arsenate, and dilute it with water to a final volume of 500 mL. You then add excess AgNO solution to a 50.0 mL sample of the arsenate solution. The resulting precipitate of Ag AsO has a mass of 3.24 g after drying. What is the percentage by mass of NaAsO in the original sample? 91.0% Precipitation reactions are a subclass of double displacement reactions. Determining the Net Ionic Equation for a Precipitation Reaction: A complete ionic equation consists of the net ionic equation and spectator ions. Predicting the solubility of ionic compounds in water can give insight into whether or not a reaction will occur. The chemical equation for a reaction in solution can be written in three ways. The shows all the substances present in their undissociated forms; the shows all the substances present in the form in which they actually exist in solution; and the is derived from the complete ionic equation by omitting all , ions that occur on both sides of the equation with the same coefficients. Net ionic equations demonstrate that many different combinations of reactants can give the same net chemical reaction. In a , a subclass of exchange reactions, an insoluble material (a ) forms when solutions of two substances are mixed. To predict the product of a precipitation reaction, all species initially present in the solutions are identified, as are any combinations likely to produce an insoluble salt. ( ) | 17,704 | 2,565 |
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The process of obtaining quantitative information of a sample using a fast chemical reaction by reacting with a certain volume of reactant whose concentration is known is called titration. When an acid-base reaction is used, the process is called acid-base titration. When a redox reaction is used, the process is called a redox titration. Titration is also called volumetric analysis, which is a type of quantitative chemical analysis. In freshman chemistry, we treat titration this way. A titration is a technique where a solution of known concentration is used to determine the concentration of an unknown solution. Typically, the titrant (the known solution) is added from a buret to a known quantity of the analyte (the unknown solution) until the reaction is complete. Knowing the volume of titrant added allows the determination of the concentration of the unknown. Often, an indicator is used to usually signal the end of the reaction, the endpoint. For acid-base titration, a modern lab will usually monitor titration with a pH meter which is interfaced to a computer, so that you will be able to plot the pH or other physical quantities versus the volume that is added. In this module, we simulate this experiment graphically without using chemicals. A program that simulates titrations of strong acids and strong bases is very easy, because the calculation of pH in this experiment is very simple. An example of titration is the acetic acid and NaOH - strong base and weak acid - titration following the equation below. \[\ce{HC2H4O2(aq)+OH^{-}(aq) → C2H4O2(aq)+H2O(l) }\] The plot of pH as a function of titrant added is called a titration curve. Let us take a look at a titration process: Evaluate \(\ce{[H+]}\) and pH in the titration of 10.0 mL 1.0 M \(\ce{HCl}\) solution with 1.0 M \(\ce{NaOH}\) solution, and plot the titration curve. \(\begin{align}
\textrm{The amount of acid present} &= \mathrm{V_a\times C_a}\\
&= \mathrm{10.0\: mL \times \dfrac{1.0\: mol}{1000\: mL}}\\
&= \textrm{10 mmol (mili-mole)}
\end{align}\) \(\textrm{The amount of base NaOH added} = \mathrm{V_b\times C_b}\) \(\textrm{The amount of acid left} = \mathrm{V_a\times C_a - V_b\times C_b}\) \(\textrm{The concentration of acid and thus }\ce{[H+]}
= \mathrm{\dfrac{[V_a\times C_a - V_b\times C_b]}{V_a + V_b}}\) With the above formulation, we can build a table for various values as shown on the right. Plot the titration curve on a graph based on the data. At equivalence point, why is pH=7? What formula is used to calculate pH? Why does pH change rapidly at the equivalence point? Sketch titration curves when the concentrations of both acids and bases are 0.10, 0.0010 and 0.000010 M. What can you conclude from these sketches? What are \(\ce{[Na+]}\) and \(\ce{[Cl- ]}\) at the following points: initially (before any base is added), half-equivalence point; equivalence point, after 10.5 mL \(\ce{NaOH}\) is added, after 20.0 mL \(\ce{NaOH}\) is added? Well, when you have acquired the skill to calculate the pH at any point during titration, you may write a simulation program to plot the titration curve. Calculations for strong-acid_strong-base titration are simple, but when weak acid or base are involved, the calculations are somewhat more complicated. However, we are interested in this area and some simulation programs are available on the internet. What is the pH of a 0.5 M sodium acetate solution? | 3,430 | 2,566 |
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Asymmetric objects are chiral Symmetric objects are achiral Symmetric objects are superposable with their mirror images. They are one and the same. Asymmetric objects are nonsuperposable with their mirror images. They are different objects. In the case of molecules, chiral molecules and their mirror images are different molecules. Chiral molecules and their mirror images are a kind of stereoisomers called enantiomers. - Compounds that have the same molecular formula and the same connectivity, but different arrangement of the atoms in 3-dimensional space. Stereoisomers cannot be converted into each other without breaking bonds. - Nonsuperposable mirror images, or chiral molecules which are mirror images. - A tetrahedral carbon atom bearing four different substituents. - Asymmetrically substituted atoms in a molecular structure. The most common type encountered in this course will be the chiral carbon described above. - Stereoisomers which are not enantiomers (or mirror images). - Symmetric, or achiral molecules that contain stereocenters. Meso compounds and their mirror images are not stereoisomers, since they are identical. - The ability of chiral substances to rotate the plane of polarized light by a specific angle. - Ability of chiral substances to rotate the plane of polarized light to the right. - Ability of chiral substances to rotate the plane of polarized light to the left. - The measured angle of rotation of polarized light by a pure chiral sample under specified standard conditions (refer to textbook for a description of these). - A mixture consisting of equal amounts of enantiomers. A racemic mixture exhibits no optical activity because the activities of the individual enantiomers are equal and opposite in value, therby canceling each other out. - The difference in percent between two enantiomers present in a mixture in unequal amounts. For example, if a mixture contains 75% of one enantiomer and 25% of the other, the optical purity is 75-25 = 50%. - A description of the precise 3-dimensional topography of the molecule. - A description of the 3-dimensional topography of the molecule relative to an arbitrary standard. Absolute and relative configurations may or may not coincide - The term chiral center refers to an atom in the molecular structure. The term chiral molecule refers to the entire molecule. The presence of one chiral center renders the entire molecule chiral. The presence of two or more chiral centers may or may not result in the molecule being chiral. In the examples given below the chiral centers are indicated with an asterisk. The vertical broken line represents a plane of symmetry. - The primary criterion to determine molecular chirality is the absence of any symmetry elements. However, some achiral molecules have chiral conformations. For example the chair conformations of 1,2-disubstituted cyclohexanes are chiral, yet the molecule as a whole is considered achiral. On the whole, we can apply the following criteria. If the contributing conformations average out to an achiral conformation, then the molecule is considered achiral. Such molecules do not show optical activity. In the case of 1,2-disubstituted cyclohexanes the two most stable conformations are chiral. If we could freeze and isolate one of them, it would exhibit optical activity. But because they are mirror images in equilibrium, their optical activities cancel out and the sample is optically inactive. A similar example is illustrated by the conformations of (2R,3S)-1,2-dichlorobutane, which again is achiral, even though some of its conformations are chiral. (2R,3S)-2,3-dichlorobutane If a chiral conformation prevails over the others, then the molecule is considered chiral and it will show optical activity. The most common situations of this type involve molecules which are locked up into a chiral conformation due to steric interactions that impede free rotation around sigma bonds. In the example shown below, the two benzene rings cannot be coplanar because the steric interactions between the methyl and chlorine groups are too severe. The molecule is locked up in a conformation that has no symmetry, therefore it is chiral. Also notice that the molecule does not have any chiral centers. Its chirality is strictly due to a conformational effect. | 4,330 | 2,567 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.12%3A_Fossil_Fuels_and_the_Energy_Crisis |
A chemical fuel is any substance which will react with atmospheric oxygen, is available at reasonable cost and quantity, and produces environmentally acceptable reaction products. During the past century the most important sources of heat energy in the United States and other industrialized countries have been the : coal, petroleum, and natural gas. In 1971, for example, the United States obtained 43.5 percent of its energy from oil, 34.7 percent from natural gas, and 19.7 percent from coal. Only 1.5 percent was obtained from hydroelectric power and 0.6 percent from nuclear power. Other industrialized countries also obtain 95 percent or more of their energy from fossil fuels. Coal, petroleum, and natural gas consist primarily of , and so it is not hard to see why they make excellent fuels. When they burn in air, the principal products are water and carbon dioxide, compounds which contain the strongest double bond ( = 804 kJ mol ) and the third-strongest single bond ( = 463 kJ mol ) in . Thus more energy is liberated by bond formation than is needed to break the weaker C―C and C―H bonds in the fuel. Use the bond enthalpies given in to estimate the when 1 mol heptane, C H , is burned completely in oxygen: \[\ce{C_{7}H_{16}(g) + 11O_{2} \rightarrow 7CO_{2} + 8H_{2}O} \nonumber \] emembering that the projection formula for heptane is we can make up the following list of bonds broken and formed: Thus \[\triangle H = 14 174 \text{kJ mol}^{-1} – 18 664 \text{kJ mol}^{-1} = –4490 \text{kJ mol}^{-1} \nonumber \] Apart from the hydrocarbon compounds in fossil fuels, there are few substances which fulfill the criteria for a good fuel. One example is hydrogen gas: \[\ce{2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)} \nonumber \] with \(\triangle H^{o} (298 K) = –571.7 \text{kJ mol}^{-1}\). Hydrogen does not occur as the element at the surface of the earth, however, so it must be manufactured. Right now much of it is made as a by-product of petroleum refining, and so hydrogen will certainly not be an immediate panacea for our current petroleum shortage. Eventually, though, it may be possible to generate hydrogen economically by electrolysis of water with current provided by nuclear power plants, and so this fuel does merit consideration. | 2,285 | 2,569 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.02%3A_Simple_Distillation/5.2A%3A_Uses_of_Simple_Distillation |
Fermented grains can produce a maximum alcohol content of about \(16\%\), the upper limit of most beers and wines, as the yeast organism used in fermentation cannot survive in more concentrated alcoholic solutions. However, spirits such as gin, vodka, and rum usually have an alcohol content of roughly \(40\%\) by volume. Distillation is the method used to concentrate fermented solutions and produce hard liquors. A distillation of fermented grape juice is commonly done in academic laboratories (Figure 5.2). Figure 5.3 shows the distillation of red wine. Notice how the wine is noticeably purified in the process as the distillate is clear (Figure 5.3d). All concentrated alcoholic products at first distill clear, but some become colored during the aging process (from components leaching out of the barrels they are stored in, or from oxidation). The distillation of red wine can be proven to have concentrated the alcohol through a flame test, which is the origin of the term "proof". In the 16\(^\text{th}\) century when rum was a bartering item, it was regularly tested to ensure that it had not been watered down. An alcoholic sample that is \(57\%\) alcohol by volume is capable of catching on fire, while more dilute solutions cannot, enabling ignition to serve as "proof" of alcohol content. The red wine used in Figure 5.3 did not ignite, which is consistent with its label that states it to be only \(13\%\) alcohol (Figure 5.4a). The distillate easily caught on fire (Figure 5.4c+d), and density analysis showed it to be \(67\%\) alcohol. Access to clean, fresh water is a major problem facing the world today. In countries neighboring the ocean, seawater desalination is sometimes used to provide the country with drinkable water. Distillation is one of the main methods\(^1\) used to purify ocean water and works well since salt, microorganisms, and other components of seawater are non-volatile. The main disadvantage of distilling water is that the process requires a lot of energy, and unless engineered creatively, the economics can be a major deterrent to using the method. Heat released from a power plant is often used to provide the energy for the distillation of seawater, and Saudi Arabia and Israel mainly use coupled power plants and distilleries to obtain roughly half the fresh water needed for their countries.\(^2\) Along with drinking water, distilled water is necessary for scientific lab work as dissolved salts in tap water may interfere with some experiments. Many science buildings have their own distilled water generators (Figure 5.5a), enabling wash bottles and carboys to be filled with a turn of the spigot (Figure 5.5b). Distillation is an excellent purification tool for many liquids, and can be used to purify products from a chemical reaction. Figure 5.6 shows the distillation of a crude sample of isoamyl acetate, formed through a Fischer esterification reaction. The crude sample was originally yellow (Figure 5.6a), but the distillate was colorless (Figure 5.6d), making obvious the removal of some materials through the process. Distillation can also be used to purify reagents that have degraded over time. Figure 5.7a shows a bottle of benzaldehyde that has partially oxidized, as evidenced by the crystals of benzoic acid seen adhering to the inside of the glass (indicated with an arrow). A simple distillation purified the benzaldehyde sample, and its effectiveness was demonstrated through comparison of the infrared (IR) spectra of the original material (Figure 5.8a) and the distillate (Figure 5.8b). The broad region indicated by an arrow in Figure 5.8a represent the \(\ce{O-H}\) stretch of a carboxylic acid (benzoic acid), and the distillate's IR spectrum lacks this feature. \(^1\)More often, vacuum distillation is used for desalination, as the lowered pressure allows for boiling to occur at a lower temperature (which requires less energy). Vacuum distillation of seawater is still energy intensive. Reverse osmosis is also used for seawater desalination. \(^2\)Pyper, Julia, "Israel is creating a water surplus using desalination," , February 7, 2014. | 4,135 | 2,571 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/How_the_Mass_Spectrometer_Works |
This page describes how a mass spectrum is produced using a mass spectrometer. If something is moving and you subject it to a sideways force, instead of moving in a straight line, it will move in a curve - deflected out of its original path by the sideways force. Suppose you had a cannonball traveling past you and you wanted to deflect it as it went by you. All you've got is a jet of water from a hose-pipe that you can squirt at it. Frankly, its not going to make a lot of difference! Because the cannonball is so heavy, it will hardly be deflected at all from its original course. But suppose instead, you tried to deflect a table tennis ball traveling at the same speed as the cannonball using the same jet of water. Because this ball is so light, you will get a huge deflection. The amount of deflection you will get for a given sideways force depends on the mass of the ball. If you knew the speed of the ball and the size of the force, you could calculate the mass of the ball if you knew what sort of curved path it was deflected through. The less the deflection, the heavier the ball. You can apply exactly the same principle to atomic sized particles. Atoms can be deflected by magnetic fields - provided the atom is first turned into an ion. Electrically charged particles are affected by a magnetic field although electrically neutral ones aren't. The sequence is : It's important that the ions produced in the ionization chamber have a free run through the machine without hitting air molecules. The vaporized sample passes into the ionization chamber. The electrically heated metal coil gives off electrons which are attracted to the electron trap which is a positively charged plate. The particles in the sample (atoms or molecules) are therefore bombarded with a stream of electrons, and some of the collisions are energetic enough to knock one or more electrons out of the sample particles to make positive ions. Most of the positive ions formed will carry a charge of +1 because it is much more difficult to remove further electrons from an already positive ion. These positive ions are persuaded out into the rest of the machine by the ion repeller which is another metal plate carrying a slight positive charge. The positive ions are repelled away from the very positive ionization chamber and pass through three slits, the final one of which is at 0 volts. The middle slit carries some intermediate voltage. All the ions are accelerated into a finely focused beam. Different ions are deflected by the magnetic field by different amounts. The amount of deflection depends on: These two factors are combined into the mass/charge ratio. Mass/charge ratio is given the symbol m/z (or sometimes m/e). For example, if an ion had a mass of 28 and a charge of 1+, its mass/charge ratio would be 28. An ion with a mass of 56 and a charge of 2+ would also have a mass/charge ratio of 28. In the last diagram, ion stream A is most deflected - it will contain ions with the smallest mass/charge ratio. Ion stream C is the least deflected - it contains ions with the greatest mass/charge ratio. It makes it simpler to talk about this if we assume that the charge on all the ions is 1+. Most of the ions passing through the mass spectrometer will have a charge of 1+, so that the mass/charge ratio will be the same as the mass of the ion. Assuming 1+ ions, stream A has the lightest ions, stream B the next lightest and stream C the heaviest. Lighter ions are going to be more deflected than heavy ones. Only ion stream B makes it right through the machine to the ion detector. The other ions collide with the walls where they will pick up electrons and be neutralised. Eventually, they get removed from the mass spectrometer by the vacuum pump. When an ion hits the metal box, its charge is neutralized by an electron jumping from the metal on to the ion (right hand diagram). That leaves a space amongst the electrons in the metal, and the electrons in the wire shuffle along to fill it. A flow of electrons in the wire is detected as an electric current which can be amplified and recorded. The more ions arriving, the greater the current. How might the other ions be detected - those in streams A and C which have been lost in the machine? Remember that stream A was most deflected - it has the smallest value of m/z (the lightest ions if the charge is 1+). To bring them on to the detector, you would need to deflect them less - by using a smaller magnetic field (a smaller sideways force). To bring those with a larger m/z value (the heavier ions if the charge is +1) on to the detector you would have to deflect them more by using a larger magnetic field. If you vary the magnetic field, you can bring each ion stream in turn on to the detector to produce a current which is proportional to the number of ions arriving. The mass of each ion being detected is related to the size of the magnetic field used to bring it on to the detector. The machine can be calibrated to record current (which is a measure of the number of ions) against m/z directly. The mass is measured on the C scale. The output from the chart recorder is usually simplified into a "stick diagram". This shows the relative current produced by ions of varying mass/charge ratio. The stick diagram for molybdenum looks like this: You may find diagrams in which the vertical axis is labeled as either "relative abundance" or "relative intensity". Whichever is used, it means the same thing. The vertical scale is related to the current received by the chart recorder - and so to the number of ions arriving at the detector: the greater the current, the more abundant the ion. As you will see from the diagram, the commonest ion has a mass/charge ratio of 98. Other ions have mass/charge ratios of 92, 94, 95, 96, 97 and 100. That means that molybdenum consists of 7 different isotopes. Assuming that the ions all have a charge of 1+, that means that the masses of the 7 isotopes on the carbon-12 scale are 92, 94, 95, 96, 97, 98 and 100. Jim Clark ( ) | 6,047 | 2,572 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Fundamentals/Stirlings_Approximation |
Stirling's approximation is named after the Scottish mathematician James Stirling (1692-1770). In confronting statistical problems we often encounter factorials of very large numbers. The factorial \(N!\) is a product \(N(N-1)(N-2)...(2)(1)\). Therefore, \(\ln \,N!\) is a sum \[\left.\ln N!\right. = \ln 1 + \ln 2 + \ln 3 + ... + \ln N = \sum_{k=1}^N \ln k. \label{1} \] where we have used the property of logarithms that \(\log(abc) =\log(a) + \log(b) +\log(c)\). The sum is shown in figure below. Using one has \[\sum_{k=1}^N \ln k=\int_1^N \ln x\,dx+\sum_{k=1}^p\frac{B_{2k}}{2k(2k-1)}\left(\frac{1}{n^{2k-1}}-1\right)+R , \label{2} \] where = −1/2, = 1/6, = 0, = −1/30, = 0, = 1/42, = 0, = −1/30, ... are the , and \( is an error term which is normally small for suitable values of \( . Then, for large \(N\), \[\ln N! \sim \int_1^N \ln x\,dx \approx N \ln N -N . \label{3} \] after some further manipulation one arrives at (apparently Stirling's contribution was the prefactor of \(\sqrt{2\pi})\) \[N! = \sqrt{2 \pi N} \; N^{N} e^{-N} e^{\lambda_N} \label{4} \] where \[\dfrac{1}{12N+1} < \lambda_N < \frac{1}{12N}. \label{5} \] The sum of the area under the blue rectangles shown below up to \(N\) is \(\ln N!\). As you can see the rectangles begin to closely approximate the red curve as \(m\) gets larger. The area under the curve is given the integral of \(\ln x\). \[ \ln N! = \sum_{m=1}^N \ln m \approx \int_1^N \ln x\, dx \label{6} \] To solve the integral use \[ \int u\,dv=uv-\int v\,dy \label{7A} \] Here we let \(u = \ln x\) and \(dv = dx\). Then \(v = x\) and \(du = \frac{dx}{x}\). \[ \int_0^N \ln x \, dx = x \ln x|_0^N - \int_0^N x \dfrac{dx}{x} \label{7B} \] Notice that \(x/x = 1\) in the last integral and \(x \ln x\) is 0 when evaluated at zero, so we have \[ \int_0^N \ln x \, dx = N \ln N - \int_0^N dx \label{8} \] Which gives us Stirling’s approximation: \(\ln N! = N \ln N – N\). As is clear from the figure above Stirling’s approximation gets better as the number N gets larger (Table \(\Page {1}\)). Calculators often overheat at 200!, which is all right since clearly result are converging. In thermodynamics, we are often dealing very large N (i.e., of the order of Avagadro’s number) and for these values Stirling’s approximation is excellent. | 2,317 | 2,573 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/02%3A_Adsorption_of_Molecules_on_Surfaces/2.06%3A_The_Desorption_Process |
An adsorbed species present on a surface at low temperatures may remain almost indefinitely in that state. As the temperature of the substrate is increased, however, there will come a point at which the thermal energy of the adsorbed species is such that one of several things may occur: The last of these options is the desorption process. In the absence of decomposition the desorbing species will generally be the same as that originally adsorbed but this is not necessarily always the case. (An example where it is not is found in the adsorption of some alkali metals on metallic substrates exhibiting a high work function where, at low coverages, the desorbing species is the alkali metal ion as opposed to the neutral atom. Other examples would include certain isomerization reactions.) The rate of desorption, , of an adsorbate from a surface can be expressed in the general form: \[R_{des} = k N^x \label{1}\] with The order of desorption can usually be predicted because we are concerned with an of a "reaction": specifically, \[A_{(ads)} \rightarrow A_{(g)}\] \[M_{(ads)} \rightarrow M_{(g)}\] - will usually be a first order process (i.e. \(x = 1\) ). Examples include … \[2 A_{(ads)} \rightarrow A_{2 (g)}\] - will usually be a second order process (i.e. = 2\)). Examples include: The rate constant for the desorption process may be expressed in an , \[k_{des} = A \;\exp( -E_a^{des} / RT ) \label{2}\] with This then gives the following general expression for the rate of desorption \[ R_{des} = -\dfrac{dN}{dt} = \nu N^x \exp ( -E_a^{des} / RT) \label{3}\] In the particular case of simple molecular adsorption, the pre-exponential/frequency factor (\(\nu\)) may also be equated with the frequency of vibration of the bond between the molecule and substrate; this is because every time this bond is stretched during the course of a vibrational cycle can be considered an attempt to break the bond and hence an attempt at desorption. One property of an adsorbed molecule that is intimately related to the desorption kinetics is the - this is the average time that a molecule will spend on the surface under a given set of conditions (in particular, for a specified surface temperature) before it desorbs into the gas phase. For a first order process such as the desorption step of a molecularly adsorbed species: \[M_{(ads)} \rightarrow M_{(g)}\] the average time (\(\tau\)) prior to the process occurring is given by: \[\tau = \dfrac{1}{k_1} \label{4}\] where \(k_1\) is the first order rate constant (no proof of this will be given here). From equation 3 with \(x=1\), we know that \[k_1 =\nu \exp ( -E_a^{des} / RT) \label{5}\] and if we also substitute for \(E_a^{des}\) using the approximate relation \(E_a^{des} \approx -ΔH_{ads}\) discussed in , then we get the following expression for the surface residence time \[\tau = \tau_o \exp ( -ΔH_{ads} / RT ) \label{6}\] with | 2,911 | 2,575 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/05%3A_Cooperativity/19%3A_Self-Assembly/19.02%3A_Classical_Nucleation_Theory |
Let’s summarize the thermodynamic theory for the nucleation of a liquid droplet by the association of molecules from the vapor. The free energy for forming a droplet out of \(n\) molecules (which we refer to as monomers) has two contributions: a surface energy term that describes the energy needed to make droplet interface and a volume term that describes the cohesive energy of the monomers. \[ \Delta G_n = \gamma a - \Delta \epsilon V \] Note the similarity to our discussion of the , where \(γ\) was just the surface tension of water. \(Δε\) is the bulk cohesive energy—a positive number. Since this is a homogeneous cluster, we expect the cluster volume \(V\) to be proportional to \(n\) and, for a spherical droplet, the surface area a to be proportional to \(V^{2/3}\) and thus \(n^{2/3}\) (remember our discussion of hydrophobic collapse). To write this in terms of monomer units, we can express the total area in terms of the average surface area per molecule in the droplet: \[ \alpha = a/n \] and as the monomer volume \(V_0\). Then the free energy is \[ \Delta G_n = \gamma \alpha n^{2/3} - \Delta \epsilon V_0 n \label{2}\] and the chemical potential of the droplet as \[ \begin{align*} \Delta \mu_n &= \dfrac{\partial \Delta G_m}{\partial n} \\[4pt] &= \dfrac{2}{3}\gamma_0 \alpha n ^{-1/3} + \Delta \epsilon V_0 \label{3} \end{align*}\] These competing effects result in a maximum in \(ΔG\) versus \(n\), which is known as the \(n^{*}\). The free energy at \(n^{*}\) is positive and called the nucleation barrier ΔG . We find \(n^{*}\) by setting Equation \ref{3} equal to zero: \[ n^* = \left( \dfrac{2\gamma_0 \alpha}{3\Delta \epsilon V_0} \right)^3 \] and substituting into Equation \ref{2} \[ G^* = \dfrac{4}{27}\dfrac{(\gamma_0 \alpha )^3}{(\Delta \epsilon V_0)^2} \] For nucleation of a liquid droplet from vapor, if fewer than n monomers associate, there is not enough cohesive energy to allow the growth of a droplet and the nucleus will dissociate. If more than \(n^{*}\) monomers associate, the droplet is still unstable, but the direction of spontaneous change will increase the size of the droplet and a liquid phase will grow from the nucleus. The process of micelle formation requires a balance of attractive and repulsive forces that stabilize an aggregate, which can depend on surface and volume terms. Thus the ΔG has a similar form, but the signs of different factors may be positive or negative. _____________________________________ P. S. Richard, Nucleation: theory and applications to protein solutions and colloidal suspensions, J. Phys.: Condens. Matter 19 (3), 033101 (2007). | 2,633 | 2,576 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Free_Radical_Reactions_of_Alkenes/Allylic_Substitution |
We noted earlier that benzylic and allylic sites are exceptionally reactive in . Since carbon-carbon double bonds add chlorine and bromine in liquid phase solutions, radical substitution reactions by these halogens are often carried out at elevated temperature in the gas phase (first equation below). Formation of the ionic π-complexes that are intermediates in halogen addition is unfavorable in the absence of polar solvents, and generally favors substitution over addition. The brominating reagent, N-bromosuccinimide (NBS), has proven useful for achieving allylic or benzylic substitution in CCl solution at temperatures below its boiling point (77 ºC). One such application is shown in the second equation. The predominance of allylic substitution over addition in the NBS reaction is interesting. The N–Br bond is undoubtedly weak (probably less than 50 kcal/mol) so bromine atom abstraction by radicals should be very favorable. The resulting succinimyl radical might then establish a chain reaction by removing an allylic hydrogen from the alkene. One problem with this mechanism is that NBS is very insoluble in CCl , about 0.006 mole / liter at reflux. Although it is possible that the allylic bromination occurs at a solid-liquid interface, evidence for another pathway has been obtained. In the non-polar solvent used for these reactions, very low concentrations of bromine may be generated from NBS. This would serve as a source of bromine atoms, which would abstract allylic hydrogens irreversibly (an exothermic reaction) in competition with reversible addition to the double bond. The HBr produced in this way is known to react with NBS, giving a new bromine molecule and succinimide, as shown here. Ionic addition of bromine to the double bond would be very slow in these circumstances (CCl is a nonpolar solvent). HBr + (CH CO) NBr → Br + (CH CO) NH This mechanism is essentially the same as that for the of alkanes, with NBS serving as a source of very low concentrations of bromine. Unsymmetrical allylic radicals will react to give two regioisomers. Thus, 1-octene on bromination with NBS yields a mixture of 3-bromo-1-octene (ca. 18%) and 1-bromo-2-octene (82%) - both cis and trans isomers. RCH CH=CH + (CH CO) N → RCH CH=CH + RCH=CHCH + (CH CO) N | 2,295 | 2,577 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Elimination_Reactions/E1_Reactions/Electrophilic_Alkene_Addition_Mechanism |
The electrophilic Alkene Adddition reaction involves two sequential steps. The red arrows are curved or curly arrows showing electron movement. The alkene pi ( ) electrons react with a source of electron deficient H. The most stable carbocation is formed. Occasionally the first-formed carbocation undergoes a rearrangement to a more stable carbocation. Formation of the carbocation is the rate limiting (slowest) step in the mechanism.
The electron deficient carbocation reacts with an electron rich X- ion. Note the bromine attached to the C-2 position from the more stable secondary carbocation + + The first-formed carbocation is secondary, a methide migration converts this into the more stable tertiary carbocation. A small amount of the secondary carbocation reacts with Cl before rearranging, but the majority rearranges before reacting. | 865 | 2,579 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4J%3A_Cooling_Baths |
On occasion a solution may need to be cooled: to minimize evaporation of volatile liquids, induce crystallization, or to favor a certain reaction mechanism. Several cold baths are used for certain applications, with the simplest being an ice bath. When preparing an ice bath, it is important to use a mixture of ice water, as an ice-water slurry has better surface contact with a flask than ice alone. Sometimes salts are added to ice in order to create baths colder than \(0^\text{o} \text{C}\) ( ). It is also quite common to cool a solvent with dry ice (solid \(\ce{CO_2}\)) in order to achieve dramatically colder baths. In dry ice baths, dry ice is added to the solvent until a portion of dry ice remains. The most common dry ice bath is made with acetone and dry ice, and achieves a cold bath of \(-78^\text{o} \text{C}\). Other cold baths are shown in Table 1.8. Ice baths can be made in Tupperware containers, beakers, or almost any container (Figure 1.57a). Baths colder than \(-10^\text{o} \text{C}\) should be made in an insulating container or else they cannot be easily handled and will lose heat too quickly to the room. The most common cold bath container is a wide-mouthed Dewar (Figure 1.57b), which has a vacuum jacket between the bath and the exterior. An inexpensive homemade insulating bath can be made by nesting two crystallizing dishes, filling the gap with vermiculite (a packing material) and sealing the gap with silicone caulking (Figure 1.57c). Dry ice should not be handled with your bare hands or else you may get frostbite. Also avoid direct contact with baths colder than \(-10^\text{o} \text{C}\). \(^9\)J. A. Dean, , 15\(^\text{th}\) ed., McGraw-Hill, , Sect 11.1. | 1,718 | 2,581 |
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In many cases, the formation of a reactive intermediate (or even a longer lived intermediate) involves a reversible step. This is the case if the intermediate can decompose to reform reactants with a significant probability as well as moving on to form products. In many cases, this will lead to a pre-equilibrium condition in which the can be applied. An example of a reaction mechanism of this sort is \[ A + B \xrightleftharpoons [k_1]{k_{-1}} AB \nonumber \] \[ AB \xrightarrow{k_2} C \nonumber \] Given this mechanism, the application of the steady state approximation is cumbersome. However, if the initial step is assumed to achieve equilibrium, an expression can be found for \([AB]\). In order to derive this expression, one assumes that the rate of the forward reaction is equal to the rate of the reverse reaction for the initial step in the mechanism. \[ k_{1}[A,B] = k_{-1}[AB] \nonumber \] or \[\dfrac{ k_{1}[A,B]}{k_{-1}} = [AB] \nonumber \] This expression can be substituted into an expression for the rate of formation of the product \(C\): \[\dfrac{d[C]}{dt} = k_2[AB] \nonumber \] or \[\dfrac{d[C]}{dt} = \dfrac{ k_2 k_{1}}{k_{-1}}[A,B] \nonumber \] Which predicts a reaction rate law that is first order in \(A\), first order in \(B\), and second order overall. Given the following mechanism, apply the equilibrium approximation to the first step to predict the rate law suggested by the mechanism. \[ A + A \xrightleftharpoons [k_1]{k_{-1}} A_2 \nonumber \] \[ A_2+B \xrightarrow{k_2} C + A \nonumber \] If the equilibrium approximation is valid for the first step, \[ k_{1}[A]^2 = k_{-1}[A_2] \nonumber \] or \[\dfrac{ k_{1}[A]^2}{k_{-1}} \approx [A_2] \nonumber \] Plugging this into the rate equation for the second step \[\dfrac{d[C]}{dt} = k_2[A_2,B] \nonumber \] yields \[\dfrac{d[C]}{dt} = \dfrac{ k_2k_{1}}{k_{-1}} [A]^2[B] \nonumber \] Thus, the rate law has the form \[\text{rate} = k' [A]^2[B] \nonumber \] which is second order in \(A\), first order in \(B\) and third order over all, and in which the effective rate constant (\(k'\) is Sometimes, the equilibrium approximation can suggest rate laws that have negative orders with respect to certain species. For example, consider the following reaction \[A + 2B \rightarrow 2C \nonumber \] A proposed mechanism for which might be \[ A + B \xrightleftharpoons [k_1]{k_{-1}} I + C \nonumber \] \[ I+ B \xrightarrow{k_2} C \nonumber \] in which \(I\) is an intermediate. Applying the equilibrium approximation to the first step yields \[ k_{1}[A,B] = k_{-1}[I,C] \nonumber \] or \[\dfrac{ k_{1}[A,B]}{k_{-1}[C]} \approx [I] \nonumber \] Substituting this into an expression for the rate of formation of \(C\), one sees \[\dfrac{d[C]}{dt} = k_{2} [I] [B] \nonumber \] or \[\dfrac{d[C]}{dt} = \dfrac{ k_{1}[A,B]}{k_{-1}[C]} [B] = \dfrac{ k_{2} k_{1}[A,B]}{k_{-1}[C]} \nonumber \] The rate law is then of the form \[\text{rate} = k \dfrac{[A,B]^2}{[C]} \nonumber \] which is first order in \(A\), second order in \(B\), negative one order in \(C\), and second order overall. Also, In this case, the negative order in \(C\) means that a buildup of compound \(C\) will cause the reaction to slow. These sort of rate laws are not uncommon for reactions with a reversible initial step that forms some of the eventual reaction product. | 3,328 | 2,582 |
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The functions \(A\) and \(G\) are oftentimes referred to as functions. The reason for this is that they are a measure of the maximum work (in the case of \(\Delta A\)) or non p-V work (in the case of \(\Delta G\)) that is available from a process. To show this, consider the total differentials. First, consider the differential of \(A\). \[dA = dU -TdS - SdT \nonumber \] Substituting the combined first and second laws for \(dU\), but expressing the work term as \(dw\), yields \[dA = TdS -dw -TdS - SdT \nonumber \] And cancelling the \(TdS\) terms gives \[ dA = dw - SdT \nonumber \] or at constant temperature (\(dT = 0\)) \[dA = dw \nonumber \] Since the only assumption made here was that the change is reversible (allowing for the substitution of \(TdS\) for \(dq\)), and \(dw\) for a reversible change is the maximum amount of work, it follows that \(dA\) gives the maximum work that can be produced from a process at constant temperature. Similarly, a simple expression can be derived for \(dG\). Starting from the total differential of \(G\). \[dG = dU + pdV – pdV + Vdp – TdS – SdT \nonumber \] Using an expression for \(dU = dq + dw\), where \(dq = TdS\) and \(dw \) is split into two terms, one (\(dw_{pV}\)) describing the work of expansion and the other (\(dw_e\)) describing any other type of work (electrical, stretching, etc.) \[ dU - TdS + dW_{pV} + dW_e \nonumber \] \(dG\) can be expressed as \[dG = \cancel{TdS} - \cancel{pdV} +dw_e + \cancel{pdV} + Vdp – \cancel{TdS} – SdT \nonumber \] Cancelling the \(TdS\) and \(pdV\) terms leaves \[dG = +dw_e + Vdp – SdT \nonumber \] So at constant temperature (\(dT = 0\)) and pressure (\(dp = 0\)), \[dG = dw_e \nonumber \] This implies that \(dG\) gives the maximum amount of non p-V work that can be extracted from a process. This concept of \(dA\) and \(dG\) giving the maximum work (under the specified conditions) is where the term “free energy” comes from, as it is the energy that is to do work in the surroundings. If a system is to be optimized to do work in the soundings (for example a steam engine that may do work by moving a locomotive) the functions A and \(G\) will be important to understand. It will, therefore, be useful to understand how these functions change with changing conditions, such as volume, temperature, and pressure. | 2,335 | 2,584 |
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Before we consider the enzymatically controlled reactions of dioxygen in living systems, it is instructive to consider the uncontrolled and deleterious reactions that must also occur in aerobic organisms. Life originally appeared on Earth at a time when the atmosphere contained only low concentrations of dioxygen, and was reducing rather than oxidizing, as it is today. With the appearance of photosynthetic organisms approximately 2.5 billion years ago, however, the conversion to an aerobic, oxidizing atmosphere exposed the existing anaerobic organisms to a gradually increasing level of oxidative stress. Modern-day anaerobic bacteria, the descendants of the original primitive anaerobic organisms, evolved in ways that enabled them to avoid contact with normal atmospheric concentrations of dioxygen. Modern-day aerobic organisms, by contrast, evolved by developing aerobic metabolism to harness the oxidizing power of dioxygen and thus to obtain usable metabolic energy. This remarkably successful adaptation enabled life to survive and flourish as the atmosphere became aerobic, and also allowed larger, multicellular organisms to evolve. An important aspect of dioxygen chemistry that enabled the development of aerobic metabolism is the relatively slow rate of dioxygen reactions in the absence of catalysts. Thus, enzymes could be used to direct and control the oxidation of substrates either for energy generation or for biosynthesis. Nevertheless, the balance achieved between constructive and destructive oxidation is a delicate one, maintained in aerobic organisms by several means, e.g.: compartmentalization of oxidative reactions in mitochondria, peroxisomes, and chloroplasts; scavenging or detoxification of toxic byproducts of dioxygen reactions; repair of some types of oxidatively damaged species; and degradation and replacement of other species. The classification "anaerobic" actually includes organisms with varying degrees of tolerance for dioxygen: strict anaerobes, for which even small concentrations of O are toxic; moderate anaerobes, which can tolerate low levels of dioxygen; and microaerophiles, which require low concentrations of O for growth, but cannot tolerate normal atmospheric concentrations, i.e., 21 percent O , 1 atm pressure. Anaerobic organisms thrive in places protected from the atmosphere, for example, in rotting organic material, decaying teeth, the colon, and gangrenous wounds. Dioxygen appears to be toxic to anaerobic organisms largely because it depletes the reducing equivalents in the cell that are needed for normal biosynthetic reactions. Aerobic organisms can, of course, live in environments in which they are exposed to normal atmospheric concentrations of O . Nevertheless, there is much evidence that O is toxic to these organisms as well. For example, plants grown in varying concentrations of O have been observed to grow faster in lower than normal concentrations of O . grown under 5 atm of O ceased to grow unless the growth medium was supplemented with branched-chain amino acids or precursors. High concentrations of O damaged the enzyme dihydroxy acid dehydratase, an important component in the biosynthetic pathway for those amino acids. In mammals, elevated levels of O are clearly toxic, leading first to coughing and soreness of the throat, and then to convulsions when the level of 5 atm of 100 percent O is reached. Eventually, elevated concentrations of O lead to pulmonary edema and irreversible lung damage, with obvious damage to other tissues as well. The effects of high concentrations of O on humans is of some medical interest, since dioxygen is used therapeutically for patients experiencing difficulty breathing, or for those suffering from infection by anaerobic organisms. The major biochemical targets of O toxicity appear to be lipids, DNA, and proteins. The chemical reactions accounting for the damage to each type of target are probably different, not only because of the different reactivities of these three classes of molecules, but also because of the different environment for each one inside the cell. Lipids, for example, are essential components of membranes and are extremely hydrophobic. The oxidative damage that is observed is due to free-radical autoxidation (see Reactions 5.16 to 5.21), and the products observed are lipid hydroperoxides (see Reaction 5.23). The introduction of the hydroperoxide group into the interior of the lipid bilayer apparently causes that structure to be disrupted, as the configuration of the lipid rearranges in order to bring that polar group out of the hydrophobic membrane interior and up to the membrane-water interface. DNA, by contrast, is in the interior of the cell, and its exposed portions are surrounded by an aqueous medium. It is particularly vulnerable to oxidative attack at the base or at the sugar, and multiple products are formed when samples are exposed to oxidants . Since oxidation of DNA may lead to mutations, this type of damage is potentially very serious. Proteins also suffer oxidative damage, with amino-acid side chains, particularly the sulfur-containing residues cysteine and methionine, appearing to be the most vulnerable sites. The biological defense systems protecting against oxidative damage and its consequences are summarized below. Some examples of small-molecule antioxidants are \(\alpha\)-tocopherol (vitamin E; 5.24), which is found dissolved in cell membranes and protects them against lipid peroxidation, and ascorbate (vitamin C; 5.25) and glutathione (5.26), which are found in the cytosol of many cells. Several others are known as well. \(\tag{5.24}\) \(\tag{5.25}\) \(\tag{5.26}\) The enzymatic antioxidants are (a) catalase and the various peroxidases, whose presence lowers the concentration of hydrogen peroxide, thereby preventing it from entering into potentially damaging reactions with various cell components (see Section VI and Reactions 5.82 and 5.83), and (b) the superoxide dismutases, whose presence provides protection against dioxygen toxicity that is believed to be mediated by the superoxide anion, O (see Section VII and Reaction 5.95). Some of the enzymatic and nonenzymatic antioxidants in the cell are illustrated in Figure 5.1. Redox-active metal ions are present in the cell in their free, uncomplexed state only in extremely low concentrations. They are instead sequestered by metal-ion storage and transport proteins, such as ferritin and transferrin for iron (see Chapter 1) and ceruloplasmin for copper. This arrangement prevents such metal ions from catalyzing deleterious oxidative reactions, but makes them available for incorporation into metalloenzymes as they are needed. In vitro experiments have shown quite clearly that redox-active metal ions such as Fe or Cu are extremely good catalysts for oxidation of sulfhydryl groups by O (Reaction 5.27). \[4RSH + O_{2} \xrightarrow{M^{n+}} 2RSSR + 2H_{2}O \tag{5.27}\] In addition, in the reducing environment of the cell, redox-active metal ions catalyze a very efficient one-electron reduction of hydrogen peroxide to produce hydroxyl radical, one of the most potent and reactive oxidants known (Reactions 5.28 to 5.30). \[M^{n+} + Red^{-} \rightarrow M^{(n-1)+} + Red \tag{5.28}\] \[M^{(n-1)+} + H_{2}O_{2} \rightarrow M^{n+} + OH^{-} + HO \cdotp \tag{5.29}\] \[Red^{-} + H_{2}O_{2} \rightarrow Red + OH^{-} + HO \cdotp \tag{5.30}\] \[(Red^{-} = reducing\; agent)\] Binding those metal ions in a metalloprotein usually prevents them from entering into these types of reactions. For example, transferrin, the iron-transport enzyme in serum, is normally only 30 percent saturated with iron. Under conditions of increasing iron overload, the empty iron-binding sites on transferrin are observed to fill, and symptoms of iron poisoning are not observed until after transferrin has been totally saturated with iron. Ceruloplasmin and metallothionein may playa similar role in preventing copper toxicity. It is very likely that both iron and copper toxicity are largely due to catalysis of oxidation reactions by those metal ions. Repair of oxidative damage must go on constantly, even under normal conditions of aerobic metabolism. For lipids, repair of peroxidized fatty-acid chains is catalyzed by phospholipase A , which recognizes the structural changes at the lipid-water interface caused by the fatty-acid hydroperoxide, and catalyzes removal of the fatty acid at that site. The repair is then completed by enzymatic reacylation. Although some oxidatively damaged proteins are repaired, more commonly such proteins are recognized, degraded at accelerated rates, and then replaced. For DNA, several multi-enzyme systems exist whose function is to repair oxidatively damaged DNA. For example, one such system catalyzes recognition and removal of damaged bases, removal of the damaged part of the strand, synthesis of new DNA to fill in the gaps, and religation to restore the DNA to its original, undamaged state. Mutant organisms that lack these repair enzymes are found to be hypersensitive to O , H O , or other oxidants. One particularly interesting aspect of oxidant stress is that most aerobic organisms can survive in the presence of normally lethal levels of oxidants if they have first been exposed to lower, nontoxic levels of oxidants. This phenomenon has been observed in animals, plants, yeast, and bacteria, and suggests that low levels of oxidants cause antioxidant systems to be induced . In certain bacteria, the mechanism of this induction is at least partially understood. A DNA-binding regulatory protein named OxyR that exists in two redox states has been identified in these systems. Increased oxidant stress presumably increases concentration of the oxidized form, which then acts to turn on the transcription of the genes for some of the antioxidant enzymes. A related phenomenon may occur when bacteria and yeast switch from anaerobic to aerobic metabolism. When dioxygen is absent, these microorganisms live by fermentation, and do not waste energy by synthesizing the enzymes and other proteins needed for aerobic metabolism. However, when they are exposed to dioxygen, the synthesis of the respiratory apparatus is turned on. The details of this induction are not known completely, but some steps at least depend on the presence of heme, the prosthetic group of hemoglobin and other heme proteins, whose synthesis requires the presence of dioxygen. What has been left out of the preceding discussion is the identification of the species responsible for oxidative damage, i.e., the agents that directly attack the various vulnerable targets in the cell. They were left out because the details of the chemistry responsible for dioxygen toxicity are largely unknown. In 1954, Rebeca Gerschman formulated the "free-radical theory of oxygen toxicity" after noting that tissues subjected to ionizing radiation resemble those exposed to elevated levels of dioxygen. Fourteen years later, Irwin Fridovich proposed that the free radical responsible for dioxygen toxicity was superoxide, O , based on his identification of the first of the superoxide dismutase enzymes. Today it is still not known if superoxide is the principal agent of dioxygen toxicity, and, if so, what the chemistry responsible for that toxicity is. There is no question that superoxide is formed during the normal course of aerobic metabolism, although it is difficult to obtain estimates of the amount under varying conditions, because, even in the absence of a catalyst, superoxide disproportionates quite rapidly to dioxygen and hydrogen peroxide (Reaction 5.4) and therefore never accumulates to any great extent in the cell under normal conditions of pH. One major problem in this area is that a satisfactory chemical explanation for the purported toxicity of superoxide has never been found, despite much indirect evidence from experiments that the presence of superoxide can lead to undesirable oxidation of various cell components and that such oxidation can be inhibited by superoxide dismutase. The mechanism most commonly proposed is production of hydroxyl radicals via Reactions (5.28) to (5.30) with Red = O , which is referred to as the "Metal-Catalyzed Haber-Weiss Reaction". The role of superoxide in this mechanism is to reduce oxidized metal ions, such as Cu or Fe , present in the cell in trace amounts, to a lower oxidation state. Hydroxyl radical is an extremely powerful and indiscriminate oxidant. It can abstract hydrogen atoms from organic substrates, and oxidize most reducing agents very rapidly. It is also a very effective initiator of free-radical autoxidation reactions (see Section II.C above). Therefore, reactions that produce hydroxyl radical in a living cell will probably be very deleterious. The problem with this explanation for superoxide toxicity is that the only role played by superoxide here is that of a reducing agent of trace metal ions. The interior of a cell is a highly reducing environment, however, and other reducing agents naturally present in the cell such as, for example, ascorbate anion can also act as Red in Reaction (5.28), and the resulting oxidation reactions due to hydroxyl radical are therefore no longer inhibitable by SOD. Other possible explanations for superoxide toxicity exist, of course, but none has ever been demonstrated experimentally. Superoxide might bind to a specific enzyme and inhibit it, much as cytochrome oxidase is inhibited by cyanide or hemoglobin by carbon monoxide. Certain enzymes may be extraordinarily sensitive to direct oxidation by superoxide, as has been suggested for the enzyme aconitase, an iron-sulfur enzyme that contains an exposed iron atom. Another possibility is that the protonated and therefore neutral form of superoxide, HO , dissolves in membranes and acts as an initiator of lipid peroxidation. It has also been suggested that superoxide may react with nitric oxide, NO, in the cell producing peroxynitrite, a very potent oxidant. One particularly appealing mechanism for superoxide toxicity that has gained favor in recent years is the "Site-Specific Haber-Weiss Mechanism." The idea here is that traces of redox-active metal ions such as copper and iron are bound to macromolecules under normal conditions in the cell. Most reducing agents in the cell are too bulky to come into close proximity to these sequestered metal ions. Superoxide, however, in addition to being an excellent reducing agent, is very small, and could penetrate to these metal ions and reduce them. The reduced metal ions could then react with hydrogen peroxide, generating hydroxyl radical, which would immediately attack at a site near the location of the bound metal ion. This mechanism is very similar to that of the metal complexes that cause DNA cleavage; by reacting with hydrogen peroxide while bound to DNA, they generate powerful oxidants that react with DNA with high efficiency because of their proximity to it (see Chapter 8). Although we are unsure what specific chemical reactions superoxide might undergo inside of the cell, there nevertheless does exist strong evidence that the superoxide dismutases play an important role in protection against dioxygen-induced damage. Mutant strains of bacteria and yeast that lack superoxide dismutases are killed by elevated concentrations of dioxygen that have no effect on the wild-type cells. This extreme sensitivity to dioxygen is alleviated when the gene coding for a superoxide dismutase is reinserted into the cell, even if the new SOD is of another type and from a different organism. In summary, we know a great deal about the sites that are vulnerable to oxidative damage in biological systems, about the agents that protect against such damage, and about the mechanisms that repair such damage. Metal ions are involved in all this chemistry, both as catalysts of deleterious oxidative reactions and as cofactors in the enzymes that protect against and repair such damage. What we still do not know at this time, however, is how dioxygen initiates the sequence of chemical reactions that produce the agents that attack the vulnerable biological targets | 16,226 | 2,585 |
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The rate of adsorption, \(R_{ads}\), of a molecule onto a surface can be expressed in the same manner as any kinetic process. For example, when it is expressed in terms of the partial pressure of the molecule in the gas phase above the surface: \[R_{ads} = k' P^x\] where: If the rate constant is then expressed in an , then we obtain a kinetic equation of the form: \[R_{ads} = A e^{-E_a / RT} P^ x\] where \(E_a\) is the for adsorption and \(A\) the factor. It is much more informative, however, to consider the factors controlling this process at the molecular level. The rate of adsorption is governed by (1) the rate of arrival of molecules at the surface and (2) the proportion of incident molecules which undergo adsorption. Hence, we can express the rate of adsorption (per unit area of surface, i.e., molecules m s ) as a product of the incident molecular flux, \(F\), and a sticking probability, \(S\). \[R_{ads} = S\, F \label{flux-stick}\] The sticking probability varies from 0 (never sticking) to 1 (always sticking). The flux (in molecules m s of incident molecules is given by the \[F = \dfrac{P}{ \sqrt{2πmkT}}\] where The sticking probability is clearly a property of the adsorbate / substrate system under consideration but must lie in the range 0 < < 1; it may depend upon various factors - foremost amongst these being the existing coverage of adsorbed species (\(θ\)) and the presence of any activation barrier to adsorption. In general,therefore \[S = f (θ) e^{-E_a / RT} \] where, once again, \(E_a\) is the activation energy for adsorption and \(f(θ)\) is some, as yet undetermined, function of the existing surface coverage of adsorbed species. Combining the equations for \(S\) and \(F\) yields the following expression for the rate of adsorption: \[ R =\dfrac{f(θ) P}{\sqrt{2\pi m kT}} e^{-E_a/RT} \label{eq5}\] For a discussion of some of the factors which determine the magnitude of the activation energy of adsorption you should see which looks at the typical PE curve associated with various types of adsorption process. If a surface is initially clean and it is then exposed to a gas pressure under conditions where the rate of desorption is very slow, then the coverage of adsorbed molecules may initially be estimated simply by consideration of the kinetics of adsorption. As noted above, the rate of adsorption is given by Equation \(\ref{flux-stick}\), which can be written in term of a derivative \[\dfrac{dN_{ads}}{dt} = S \, F \label{diffEq}\] where \(N_{ads}\) is the number of adsorbed species per unit area of surface. Equation \(\ref{diffEq}\) must be integrated to obtain an expression for \(N_{ads}\), since the sticking probability is coverage (and hence also time) dependent. However, if it is assumed that the sticking probability is essentially constant (which may be a reasonable approximation for relatively low coverages), then this integration simply yields: \[ N_{ads} = SFt\] | 2,955 | 2,586 |
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In many chemical operations it is an advantage not only to be able to form a precipitate but to be able to redissolve it. Fortunately, there is a wide class of sparingly soluble salts which can almost always be redissolved by adding acid. These are precipitates in which the anion is basic; i.e., they are the salts of weak acids. An example of such a precipitate is calcium carbonate, whose solubility equilibrium is \[\text{CaCO}_{3}\text{ }({s}) \rightleftharpoons \text{Ca}^{2+} ({aq}) + \text{CO}_{3}^{2-} ({aq}) \nonumber \] If acid is now added to this solution, some of the carbonate ions become protonated and transformed into HCO ions. As a result, the concentration of the carbonate ion is reduced. In accord with Le Chatelier’s principle, the system will respond to this reduction by trying to produce more carbonate ions. Some solid CaCO will dissolve, and the equilibrium will be shifted to the right. If enough acid is added, the carbonate-ion concentration in the solution can be reduced so as to make the ion product ( = × ) smaller than the solubility product so that the precipitate dissolves. A similar behavior is shown by other precipitates involving basic anions. Virtually all the carbonates, sulfides, hydroxides, and phosphates which are sparingly soluble in water can be dissolved in acid. Thus, for instance, we can dissolve precipitates like ZnS, Mg(OH) , and Ca (PO ) because all the following equilibria \[\text{ZnS}\text{ }({s})\rightleftharpoons \text{Zn}^{2+} ({aq}) + \text{S}^{2-} ({aq}) \nonumber \] \[\text{Mg(OH)}_{2}\text{ }({s})\rightleftharpoons \text{Mg}^{2+} ({aq}) + \text{2OH}^{-} ({aq}) \nonumber \] \[\text{Ca}_{3}(\text{PO}_{4})_{2}\text{ }({s})\rightleftharpoons \text{3Ca}^{2+} ({aq}) + \text{2PO}_{4}^{3-} ({aq}) \nonumber \] can be shifted to the right by attacking the basic species S , OH , and PO with hydronium ions. Very occasionally we find an exception to this rule. Mercury(II) sulfide, HgS, is notorious for being insoluble. The solubility product for the equilibrium \[\text{HgS}\text{ }({s})\rightleftharpoons \text{Hg}^{2+} ({aq}) + \text{S}^{2-} ({aq} \nonumber \] is so minute that not even concentrated acid will reduce the sulfide ion sufficiently to make smaller than . Occasionally the shift in a solubility-product equilibrium caused by a decrease in pH may be undesirable. One example of this was mentioned in the . Acid rainfall can occur when oxides of sulfur and other acidic air pollutants are removed from the atmosphere. In some parts of the United States pH values as low as 4.0 have been observed. These acid solutions dissolve marble and limestone (CaCO ) causing considerable property damage. This is especially true in Europe, where some statues and other works of art have been almost completely destroyed over the last half century. | 2,845 | 2,588 |
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After completing this section, you should be able to Make certain that you can define, and use in context, the key terms below. An is a reaction in which a substrate is initially attacked by an electrophile, and the overall result is the addition of one or more relatively simple molecules across a multiple bond. The mechanism for the addition of hydrogen halide to propene shown in the reading is quite detailed. Normally, an organic chemist would write the reaction scheme as follows: However, the more detailed mechanism shown in the reading does allow you to see the exact fate of all the electrons involved in the reaction. In your previous chemistry course, you were probably taught the importance of balancing chemical equations. It may come as a surprise to you that organic chemists usually do not balance their equations, and often represent reactions using a format which is quite different from the carefully written, balanced equations encountered in general chemistry courses. In fact, organic chemists are rarely interested in the inorganic products of their reactions; furthermore, most organic reactions are non-quantitative in nature. In many of the reactions in this course, the percentage yield is indicated beneath the products: you are not expected to memorize these figures. The question of yield is very important in organic chemistry, where two, five, ten or even twenty reactions may be needed to synthesize a desired product. For example, if a chemist wishes to prepare compound D by the following reaction sequence: \[A → B → C → D \nonumber \] and each of the individual steps gives only a 50% yield, one mole of A would give only 1 mol 0.125 mol of D You will gain first-hand experience of such situations in the laboratory component of this course. One of the most important reactions for alkenes is called electrophilic addition. In this chapter several variations of the electrophilic addition reaction will be discussed. Each case will have aspects common among all electrophilic addition. In this section, the electrophilic addition reaction will be discussed in general to provide a better understanding of subsequent alkene reactions. As discussed in Section 6-5, the double bond in alkenes is electron rich due to the prescience of 4 electrons instead of the two in a single bond. Also, the pi electrons are positioned above and below the double bond making them more accessibly for reactions. Overall, double bonds can easily donate lone pair electrons to act like a nucleophile (nucleus-loving, electron rich, a Lewis acid). During an electrophilic addition reactions double bonds donate lone pair electrons to an electrophile (Electron-loving, electron poor, a Lewis base). There are many types of electrophilic addition, but this section will focus on the addition of hydrogen halides (HX). Many of the basic ideas discussed will aplicable to subsequent electrophilic addition reactions. Overall during this reaction the pi bond of the alkene is broken to form two single, sigma bonds. As shown in the reaction mechanism, one of these sigma bonds is connected to the H and the other to the X of the hydrogen halide. This reaction works well with HBr and HCl. HI can also but used but is is usually generated during the reaction by reacting potassium iodidie (KI) with phosphoric acid (H PO ). All alkenes undergo addition reactions with the hydrogen halides. A hydrogen atom joins to one of the carbon atoms originally in the double bond, and a halogen atom to the other. For example, with ethene and hydrogen chloride, you get chloroethane: With but-2-ene you get 2-chlorobutane: What happens if you add the hydrogen to the carbon atom at the right-hand end of the double bond, and the chlorine to the left-hand end? You would still have the same product. The chlorine would be on a carbon atom next to the end of the chain - you would simply have drawn the molecule flipped over in space. That would be different of the alkene was unsymmetrical - that's why we have to look at them separately. During the first step of the mechanism, the 2 pi electrons from the double bond attack the H in the HBr electrophile which is shown by a curved arrow. The two pi electrons form a C-H sigma bond between the hydrogen from HBr and a carbon from the double bond. Simultaneously the electrons from the H-X bond move onto the halogen to form a halide anion. The removal of pi electrons form the double bond makes one of the carbons become an electron deficient carbocation intermediate. This carbon is sp hybridized and the positive charge is contained in an unhybridized p orbital. The formed carbocation now can act as an electrophile and accept an electron pair from the nucleophilic halide anion. The electron pair becomes a X-C sigma bond to create the neutral alkyl halide product of electrophilic addition. All of the halides (HBr, HCl, HI, HF) can participate in this reaction and add on in the same manner. Although different halides do have different rates of reaction, due to the H-X bond getting weaker as X gets larger (poor overlap of orbitals)s. An energy diagram for the two-step electrophilic addition mechanism is shown below. The energy diagram has two peaks which represent the transition state for each mechanistic step. The peaks are separated by a valley which represents the high energy carbocation reaction intermediate. Because the energy of activation for the first step of the mechanism (ΔE ) is much larger than the second (ΔE ), the first step of the mechanism is the rate-determining step. Both the alkene and the hydrogen halide are reactants in the first step of the mechanism, this electrophilic addition is a second order reaction and the rate law expression can be written rate = k[Alkene,HX]. Also, any structural feature which can stabilize the transition state between the reactants the carbocation intermediate will lower ΔE and thereby increase the reaction rate. Overall, the alkyl halide product of this reaction more stable than the reactants making the reaction exothermic. Reaction rates increase in the order HF - HCl - HBr - HI. Hydrogen fluoride reacts much more slowly than the other three, and is normally ignored in talking about these reactions. When the hydrogen halides react with alkenes, the hydrogen-halogen bond has to be broken. The bond strength falls as you go from HF to HI, and the hydrogen-fluorine bond is particularly strong. Because it is difficult to break the bond between the hydrogen and the fluorine, the addition of HF is bound to be slow. This applies to unsymmetrical alkenes as well as to symmetrical ones. For simplicity the examples given below are all symmetrical ones- but they don't have to be. Reaction rates increase as the alkene gets more complicated - in the sense of the number of alkyl groups (such as methyl groups) attached to the carbon atoms at either end of the double bond. For example: There are two ways of looking at the reasons for this - both of which need you to know about the mechanism for the reactions. Alkenes react because the electrons in the pi bond attract things with any degree of positive charge. Anything which increases the electron density around the double bond will help this. Alkyl groups have a tendency to "push" electrons away from themselves towards the double bond. The more alkyl groups you have, the more negative the area around the double bonds becomes. The more negatively charged that region becomes, the more it will attract molecules like hydrogen chloride. The more important reason, though, lies in the stability of the intermediate ion formed during the reaction. The three examples given above produce these carbocations (carbonium ions) at the half-way stage of the reaction: The stability of the intermediate ions governs the activation energy for the reaction. As you go towards the more complicated alkenes, the activation energy for the reaction falls. That means that the reactions become faster. Organic reaction equations are often written in one of two ways. The reactant for the reaction is written to the left of the reaction arrow. The products are written to the right of the arrow. The reagent for the reaction is written above the arrow. Other reaction conditions such as the solvent or the temperature can be written above or below the reaction arrow. Alternativley the reactant and reagent can both be written to the left of the reaction arrow. This is typically done to highlight the importance of the reactant. The solvent and reaction temperature are still written above or below the reaction arrow. The reaction products are still written to the right of the reaction arrow. | 8,713 | 2,589 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/10%3A_Electrolyte_Solutions/10.01%3A_Single-ion_Quantities |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) Consider a solution of an electrolyte solute that dissociates completely into a cation species and an anion species. Subscripts \(+\) and \(-\) will be used to denote the cation and anion, respectively. The solute molality \(m\B\) is defined as the amount of solute formula unit divided by the mass of solvent. We first need to investigate the relation between the chemical potential of an ion species and the electric potential of the solution phase. The electric potential \(\phi\) in the interior of a phase is called the , or . It is defined as the work needed to reversibly move an infinitesimal test charge into the phase from a position infinitely far from other charges, divided by the value of the test charge. The electrical potential energy of a charge in the phase is the product of \(\phi\) and the charge. Consider a hypothetical process in which an infinitesimal amount \(\dif n_+\) of the cation is transferred into a solution phase at constant \(T\) and \(p\). The quantity of charge transferred is \(\delta Q=z_+F\dif n_+\), where \(z_+\) is the charge number (\(+1\), \(+2\), etc.) of the cation, and \(F\) is the Faraday constant. (The Faraday constant is the charge per amount of protons.) If the phase is at zero electric potential, the process causes no change in its electrical potential energy. However, if the phase has a finite electric potential \(\phi\), the transfer process changes its electrical potential energy by \(\phi \delta Q=z_+F\phi\dif n_+\). Consequently, the internal energy change depends on \(\phi\) according to \begin{equation} \dif U(\phi) = \dif U(0) + z_+F\phi\dif n_+ \tag{10.1.1} \end{equation} where the electric potential is indicated in parentheses. The change in the Gibbs energy of the phase is given by \(\dif G = \dif(U-TS+pV)\), where \(T\), \(S\), \(p\), and \(V\) are unaffected by the value of \(\phi\). The dependence of \(\dif G\) on \(\phi\) is therefore \begin{equation} \dif G(\phi) = \dif G(0) + z_+F\phi\dif n_+ \tag{10.1.2} \end{equation} The Gibbs fundamental equation for an open system, \(\dif G=-S\dif T+V\difp+\sum_i\mu_i\dif n_i\) (Eq. 9.2.34), assumes the electric potential is zero. From this equation and Eq. 10.1.2, the Gibbs energy change during the transfer process at constant \(T\) and \(p\) is found to depend on \(\phi\) according to \begin{equation} \dif G(\phi) = \left[ \mu_+(0) + z_+F\phi \right] \dif n_+ \tag{10.1.3} \end{equation} The chemical potential of the cation in a phase of electric potential \(\phi\), defined by the partial molar Gibbs energy \(\bpd{G(\phi)}{n_+}{T,p}\), is therefore given by \begin{equation} \mu_+(\phi)=\mu_+(0)+z_+F\phi \tag{10.1.4} \end{equation} The corresponding relation for an anion is \begin{equation} \mu_-(\phi)=\mu_-(0)+z_-F\phi \tag{10.1.5} \end{equation} where \(z_-\) is the charge number of the anion (\(-1\), \(-2\), etc.). For a charged species in general, we have \begin{equation} \mu_i(\phi)=\mu_i(0)+z_iF\phi \tag{10.1.6} \end{equation} We define the on a molality basis in the same way as for a nonelectrolyte solute, with the additional stipulation that the ion is in a phase of zero electric potential. Thus, the standard state is a hypothetical state in which the ion is at molality \(m\st\) with behavior extrapolated from infinite dilution on a molality basis, in a phase of pressure \(p=p\st\) and electric potential \(\phi{=}0\). The \(\mu_+\st\) or \(\mu_-\st\) of a cation or anion is the chemical potential of the ion in its standard state. Single-ion activities \(a_+\) and \(a_-\) in a phase of zero electric potential are defined by relations having the form of Eq. 9.7.8: \begin{equation} \mu_+(0)=\mu_+\st+RT\ln a_+ \qquad \mu_-(0)=\mu_-\st+RT\ln a_- \tag{10.1.7} \end{equation} As explained in Sec. 9.7, \(a_+\) and \(a_-\) should depend on the temperature, pressure, and composition of the phase, and not on the value of \(\phi\). From Eqs. 10.1.4, 10.1.5, and 10.1.7, the relations between the chemical potential of a cation or anion, its activity, and the electric potential of its phase, are found to be \begin{equation} \mu_+=\mu_+\st + RT\ln a_+ + z_+ F\phi \qquad \mu_-=\mu_-\st + RT\ln a_- + z_i F\phi \tag{10.1.8} \end{equation} These relations are definitions of single-ion activities in a phase of electric potential \(\phi\). For a charged species in general, we can write \begin{equation} \mu_i=\mu_i\st + RT\ln a_i + z_i F\phi \tag{10.1.9} \end{equation} Note that we can also apply this equation to an uncharged species, because the charge number \(z_i\) is then zero and Eq. 10.1.9 becomes the same as Eq. 9.7.2. Some thermodynamicists call the quantity \((\mu_i\st+RT\ln a_i)\), which depends only on \(T\), \(p\), and composition, the of ion \(i\), and the quantity \((\mu_i\st+RT\ln a_i+z_iF\phi)\) the with symbol \(\tilde{\mu}_i\). Of course there is no experimental way to evaluate either \(\mu_+\) or \(\mu_-\) relative to a reference state or standard state, because it is impossible to add cations or anions by themselves to a solution. We can nevertheless write some theoretical relations involving \(\mu_+\) and \(\mu_-\). For a given temperature and pressure, we can write the dependence of the chemical potentials of the ions on their molalities in the same form as that given by Eq. 9.5.18 for a nonelectrolyte solute: \begin{equation} \mu_+=\mu_+\rf + RT\ln\left(\g_+\frac{m_+}{m\st}\right) \qquad \mu_-=\mu_-\rf + RT\ln\left(\g_-\frac{m_-}{m\st}\right) \tag{10.1.10} \end{equation} Here \(\mu_+\rf\) and \(\mu_-\rf\) are the chemical potentials of the cation and anion in solute reference states. Each reference state is defined as a hypothetical solution with the same temperature, pressure, and electric potential as the solution under consideration; in this solution, the molality of the ion has the standard value \(m\st\), and the ion behaves according to Henry’s law based on molality. \(\g_+\) and \(\g_-\) are single-ion activity coefficients on a molality basis. The single-ion activity coefficients approach unity in the limit of infinite dilution: \begin{gather} \s{ \g_+ \ra 1 \quad \tx{and} \quad \g_- \ra 1 \quad \tx{as} \quad m\B \ra 0} \tag{10.1.11} \cond{(constant \(T\), \(p\), and \(\phi\))} \end{gather} In other words, we assume that in an extremely dilute electrolyte solution each individual ion behaves like a nonelectrolyte solute species in an ideal-dilute solution. At a finite solute molality, the values of \(\g_+\) and \(\g_-\) are the ones that allow Eq. 10.1.10 to give the correct values of the quantities \((\mu_+-\mu_+\rf)\) and \((\mu_- -\mu_-\rf)\). We have no way to actually measure these quantities experimentally, so we cannot evaluate either \(\g_+\) or \(\g_-\). We can define single-ion pressure factors \(\G_+\) and \(\G_-\) as follows: \begin{equation} \G_+\defn\exp\left(\frac{\mu_+\rf-\mu_+\st}{RT}\right) \approx \exp\left[ \frac{V_+^{\infty}(p-p\st)}{RT} \right] \tag{10.1.12} \end{equation} \begin{equation} \G_-\defn\exp\left(\frac{\mu_-\rf-\mu_-\st}{RT}\right) \approx \exp\left[ \frac{V_-^{\infty}(p-p\st)}{RT} \right] \tag{10.1.13} \end{equation} The approximations in these equations are like those in Table 9.6 for nonelectrolyte solutes; they are based on the assumption that the partial molar volumes \(V_+\) and \(V_-\) are independent of pressure. From Eqs. 10.1.7, 10.1.10, 10.1.12, and 10.1.13, the single-ion activities are related to the solution composition by \begin{equation} a_+=\G_+\g_+\frac{m_+}{m\st} \qquad a_-=\G_-\g_-\frac{m_-}{m\st} \tag{10.1.14} \end{equation} Then, from Eq. 10.1.9, we have the following relations between the chemical potentials and molalities of the ions: \begin{equation} \mu_+=\mu_+\st + RT\ln(\G_+\g_+m_+/m\st) + z_+F\phi \tag{10.1.15} \end{equation} \begin{equation} \mu_-=\mu_-\st + RT\ln(\G_-\g_-m_-/m\st) + z_-F\phi \tag{10.1.16} \end{equation} Like the values of \(\g_+\) and \(\g_-\), values of the single-ion quantities \(a_+\), \(a_-\), \(\G_+\), and \(\G_-\) cannot be determined by experiment. | 15,416 | 2,591 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/27%3A_Reactions_of_Organic_Compounds/27.01%3A_Organic_Reactions%3A_An_Introduction |
These are the four "prototypical" organic chemistry reactions, though several others which can be categorized as one of these are generally referred to by other names. Look at these reactions and ask yourself this question for each: At least 80% of the reactions students in organic chemistry fall into one of these four categories. The sooner you can get into the habit of recognizing bond formation and breakage the better off you will be. A fifth reaction is also discussed: rearrangement reactions. Another important category of organic reactions are straight-forward Brønsted–Lowry acid-base reactions (Figure \(\Page {2}\). You’ll notice that they might look very different on the surface – all those different structures! – . we are breaking an H-(atom) bond and forming an H-(atom) bond. At the same time we are also connecting the two “leftover” partners to form a , composed of two oppositely-charged ions. Let’s look at that last reaction in more detail. Here, we are breaking a C-H bond and an (ionic) Na-NH2 bond, and forming an N-H bond as well as an (ionic) C-Na bond. There are four “actors” in this reaction – as there are in acid-base reaction – and we have names for all of them. We can also draw the reverse of the previous reaction. Look at this carefully. we are still breaking a bond to H and forming a bond to H, but we’ve swapped everything. we are breaking N-H and C-Na, and forming N-Na and C-H. However, tells us that this reaction does not happen to any appreciable extent. In this acid/base reaction, which bonds are broken and with are formed, if any? \[\ce{(CH3-CH2)3N} + \ce{HCl} \rightarrow \ce{(CH3-CH2)3NH^+} + \ce{Cl^-}
\nonumber\] When you take an alkene (or alkyne) and add certain types of reagents to them, you get results like this. See if you can recognize the bonds formed and broken. Here’s the basic pattern: break a C-C multiple bond (also called a π bond) and form two new single bonds (“σ-bonds” to carbon). This reaction is called an “addition reaction” and it’s an extremely common type of reaction. Note that the reaction occurs at the carbons that are a part of a multiple bond – nothing else on the molecule is affected. In this addition reaction, which bonds are broken and with are formed, if any? \[\ce{CH3-C(=O)-CH3} \ce{->[\ce{ LiAlH4 },\ce{ H^+/H2O }]} \ce{CH3-CH(OH)-CH3}
\nonumber\] Note: This reaction could be described as a (nucleophilic) addition reaction or a reduction. Here are three examples of nucleophilic substitution reactions. In each case, we are This is an extremely common pattern for organic chemistry reactions. Here’s some interesting results that tell us. We don’t get this information by thinking about what happens and predicting – we have to nature in order to get her to give up her secrets. In this substitution reaction , which bonds are broken and with are formed, if any? \[\ce{CH3-C(=O)-OH + CH3-OH} \ce{->[\ce{H2SO4},\text{ Δ and reflux }]} \ce{CH3-C(=O)-O-CH3} + \ce{H2O}
\nonumber\] Note: This reaction would usually be called a condensation reaction. Let’s look at two simple cases. One important thing to keep in mind: although you see “Na OCH3” written here, the “Na+” (sodium) is not important for our purposes. It could alternatively be K+ (potassium) or Li+ (lithium). It’s just balancing the negative charge on the oxygen. When you take an alkyl halide and add a strong base (such as \(\ce{NaOCH3}\) or \(\ce{NaOCH2CH3}\)) a reaction occurs. See if you can recognize the bonds broken and formed. This reaction results in the of a new C-C double bond (π bond) and breaking two single bonds to carbon (in these cases, one of them is H and the other is a halide such as Cl or Br). We are also forming a new bond between H and the O. So in this respect, this reaction incorporates a pattern we have seen before – an acid-base reaction. Finally, we also form a salt in this reaction. Since we are primarily interested in the organic product (that is, the one containing carbon), you might find that the salt byproduct is not written in some reaction schemes, but that doesn’t mean that it’s not there. The three events in the elimination reactyio of Figure \(\Page {3}\) (formation of C-C π bond, breakage of two adjacent carbon sigma bonds) are the exact of the addition reactions discussed above. Another example of an elimination is when 2-bromopropane is heated under reflux with a concentrated solution of sodium or potassium hydroxide in ethanol. Heating under reflux involves heating with a condenser placed vertically in the flask to avoid loss of volatile liquids. Propene is formed and, because this is a gas, it passes through the condenser and can be collected. In this reaction, which bonds are broken and with are formed, if any? \[\ce{CH3-CH(OH)-CH3} \ce{->[H_2SO_4,K_2Cr_2O_7 ]} \ce{CH3-C(=O)-CH3}
\nonumber\] Note: This reaction could also be described an oxidation reaction. Rearrangement reactions can accompany many of the reactions we’ve previously covered such as substitution, addition, and elimination reactions. These reactions are comparatively rare. In fact, if you don’t look closely, sometimes you can miss the fact that a rearrangement reaction has occurred. Let’s look at a substitution reaction first. On the top is a “typical” substitution reaction: we are taking an alkyl halide and adding water. The C-Br bond is broken and a C-OH bond is formed. If you look at the table on the right you’ll see this follows the typical pattern of substitution reactions. However if we change thing about this alkyl halide – move the bromine to C-3 instead of C-2 – now when we run this reaction we see a different product emerge. It is also a substitution reaction (we are replacing Br with OH) but it’s on a That’s because if you look closely, you can see there are actually 3 bonds broken and 3 bonds formed. The C2-H bond broke and the C3-H bond formed. In this Rearrangement reaction, which bonds are broken and with are formed, if any? \[ \ce{CH3-CH2-CH2-C(OH)=CH2} → \ce{CH3-CH2-CH2-C(=O)-CH3}
\nonumber\] Note: This reaction could also be described an oxidation reaction. ( ) | 6,150 | 2,592 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/09%3A_Mixtures/9.06%3A_Evaluation_of_Activity_Coefficients |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) This section describes several methods by which activity coefficients of nonelectrolyte substances may be evaluated. Section 9.6.3 describes an osmotic coefficient method that is also suitable for electrolyte solutes, as will be explained in Sec. 10.6. Suppose we equilibrate a liquid mixture with a gas phase. If component \(i\) of the liquid mixture is a volatile nonelectrolyte, and we are able to evaluate its fugacity \(\fug_i\) in the gas phase, we have a convenient way to evaluate the activity coefficient \(\g_i\) in the liquid. The relation between \(\g_i\) and \(\fug_i\) will now be derived. When component \(i\) is in transfer equilibrium between two phases, its chemical potential is the same in both phases. Equating expressions for \(\mu_i\) in the liquid mixture and the equilibrated gas phase (from Eqs. 9.5.14 and 9.5.11, respectively), and then solving for \(\g_i\), we have \begin{equation} \mu_i^* + RT\ln\left(\g_i x_i\right) = \mu_i\rf\gas + RT\ln (\fug_i/p) \tag{9.6.1} \end{equation} \begin{equation} \g_i = \exp \left[ \frac{\mu_i\rf\gas -\mu_i^*}{RT} \right] \times \frac{\fug_i}{x_i p} \tag{9.6.2} \end{equation} On the right side of Eq. 9.6.2, only \(\fug_i\) and \(x_i\) depend on the liquid composition. We can therefore write \begin{equation} \g_i = C_i\frac{\fug_i}{x_i} \tag{9.6.3} \end{equation} where \(C_i\) is a factor whose value depends on \(T\) and \(p\), but not on the liquid composition. Solving Eq. 9.6.3 for \(C_i\) gives \(C_i=\g_i x_i/\fug_i\). Now consider Eq. 9.5.20. It says that as \(x_i\) approaches 1 at constant \(T\) and \(p\), \(\g_i\) also approaches 1. We can use this limit to evaluate \(C_i\): \begin{equation} C_i = \lim_{x_i \ra 1}\frac{\g_i x_i}{\fug_i} = \frac{1}{\fug_i^*} \tag{9.6.4} \end{equation} Here \(\fug_i^*\) is the fugacity of \(i\) in a gas phase equilibrated with pure liquid \(i\) at the temperature and pressure of the mixture. Then substitution of this value of \(C_i\) (which is independent of \(x_i\)) in Eq. 9.6.3 gives us an expression for \(\g_i\) at any liquid composition: \begin{equation} \g_i=\frac{\fug_i}{x_i\fug_i^*} \tag{9.6.5} \end{equation} We can follow the same procedure for a solvent or solute of a liquid solution. We replace the left side of Eq. 9.6.1 with an expression from among Eqs. 9.5.15–9.5.18, then derive an expression analogous to Eq. 9.6.3 for the activity coefficient with a composition-independent factor, and finally apply the limiting conditions that cause the activity coefficient to approach unity (Eqs. 9.5.21–9.5.24) and allow us to evaluate the factor. When we take the limits that cause the solute activity coefficients to approach unity, the ratios \(\fug\B/x\B\), \(\fug\B/c\B\), and \(\fug\B/m\B\) become Henry’s law constants (Eqs. 9.4.19–9.4.21). The resulting expressions for activity coefficients as functions of fugacity are listed in Table 9.4. Figure 9.11(a) shows the function \((\phi_m - 1)/m\B\) for aqueous sucrose solutions over a wide range of molality. The dependence of the solute activity coefficient on molality, generated from Eq. 9.6.20, is shown in Fig. 9.11(b). Figure 9.11(c) is a plot of the effective sucrose molality \(\g\mbB m\B\) as a function of composition. Note how the activity coefficient becomes greater than unity beyond the ideal-dilute region, and how in consequence the effective molality \(\g\mbB m\B\) becomes considerably greater than the actual molality \(m\B\). Section 9.6.1 described the evaluation of the activity coefficient of a constituent of a liquid mixture from its fugacity in a gas phase equilibrated with the mixture. Section 9.6.3 mentioned the use of solvent fugacities in gas phases equilibrated with pure solvent and with a solution, in order to evaluate the osmotic coefficient of the solution. Various experimental methods are available for measuring a partial pressure in a gas phase equilibrated with a liquid mixture. A correction for gas nonideality, such as that given by Eq. 9.3.16, can be used to convert the partial pressure to fugacity. If the solute of a solution is nonvolatile, we may pump out the air above the solution and use a manometer to measure the pressure, which is the partial pressure of the solvent. Dynamic methods involve passing a stream of inert gas through a liquid mixture and analyzing the gas mixture to evaluate the partial pressures of volatile components. For instance, we could pass dry air successively through an aqueous solution and a desiccant and measure the weight gained by the desiccant. The is one of the most useful methods for determining the fugacity of H\(_2\)O in a gas phase equilibrated with an aqueous solution. This is a comparative method using a binary solution of the solute of interest, B, and a nonvolatile reference solute of known properties. Some commonly used reference solutes for which data are available are sucrose, NaCl, and CaCl\(_2\). In this method, solute B can be either a nonelectrolyte or electrolyte. Dishes, each containing water and an accurately weighed sample of one of the solutes, are placed in wells drilled in a block made of metal for good thermal equilibration. The assembly is placed in a gas-tight chamber, the air is evacuated, and the apparatus is gently rocked in a thermostat for a period of up to several days, or even weeks. During this period, H\(_2\)O is transferred among the dishes through the vapor space until the chemical potential of the water becomes the same in each solution. The solutions are then said to be . Finally, the dishes are removed from the apparatus and weighed to establish the molality of each solution. The H\(_2\)O fugacity is known as a function of the molality of the reference solute, and is the same as the H\(_2\)O fugacity in equilibrium with the solution of solute B at its measured molality. The isopiestic vapor pressure method can also be used for nonaqueous solutions. | 13,288 | 2,593 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Introduction_to_Alkyl_Halides/Nucleophiles |
Nucleophilic functional groups are those which have electron-rich atoms able to donate a pair of electrons to form a new covalent bond. In both laboratory and biological organic chemistry, the most relevant nucleophilic atoms are oxygen, nitrogen, and sulfur, and the most common nucleophilic functional groups are water, alcohols, phenols, amines, thiols, and occasionally carboxylates. More specifically in laboratory reactions, halide and azide (N ) anions are commonly seen acting as nucleophiles. Of course, carbons can also be nucleophiles - otherwise how could new carbon-carbon bonds be formed in the synthesis of large organic molecules like DNA or fatty acids? Enolate ions (section 7.5) are the most common carbon nucleophiles in biochemical reactions, while the cyanide ion (CN ) is just one example of a carbon nucleophile commonly used in the laboratory. Reactions with carbon nucleophiles will be dealt with in chapters 13 and 14, however - in this chapter and the next, we will concentrate on non-carbon nucleophiles. When thinking about nucleophiles, the first thing to recognize is that, for the most part, the same quality of 'electron-richness' that makes a something nucleophilic also makes it basic: . It should not be surprising, then, that most of the trends in basicity that we have already discussed also apply to nucleophilicity. Now, lets discuss some of the major factors that affect nucleophilicity. The protonation state of a nucleophilic atom has a very large effect on its nucleophilicity. This is an idea that makes intuitive sense: a hydroxide ion is much more nucleophilic (and basic) than a water molecule, because the negatively charged oxygen on the hydroxide ion carries greater electron density than the oxygen atom of a neutral water molecule. In practical terms, this means that a hydroxide nucleophile will react in an S 2 reaction with methyl bromide much faster ( about 10,000 times faster) than a water nucleophile. There are predictable periodic trends in nucleophilicity. Moving horizontally across the second row of the table, the trend in nucleophilicity parallels the trend in basicity: The reasoning behind the horizontal nucleophilicity trend is the same as the reasoning behind the basicity trend: more electronegative elements hold their electrons more tightly, and are less able to donate them to form a new bond. This horizontal trends also tells us that amines are more nucleophilic than alcohols, although both groups commonly act as nucleophiles in both laboratory and biochemical reactions. Recall that the basicity of atoms decreases as we move vertically down a column on the periodic table: thiolate ions are less basic than alkoxide ions, for example, and bromide ion is less basic than chloride ion, which in turn is less basic than fluoride ion. Recall also that this trend can be explained by considering the increasing size of the 'electron cloud' around the larger ions: the electron density inherent in the negative charge is spread around a larger area, which tends to increase stability (and thus reduce basicity). The vertical periodic trend for nucleophilicity is somewhat more complicated that that for basicity: depending on the solvent that the reaction is taking place in, the nucleophilicity trend can go in either direction. Let's take the simple example of the SN2 reaction below: . . .where Nu is one of the halide ions: fluoride, chloride, bromide, or iodide, and the leaving group I* is a radioactive isotope of iodine (which allows us to distinguish the leaving group from the nucleophile in that case where both are iodide). If this reaction is occurring in a (that is, a solvent that has a hydrogen bonded to an oxygen or nitrogen - water, methanol and ethanol are the most important examples), then the reaction will go fastest when iodide is the nucleophile, and slowest when fluoride is the nucleophile, reflecting the relative strength of the nucleophile. This of course, is opposite that of the vertical periodic trend for basicity, where iodide is the basic. What is going on here? Shouldn't the stronger base, with its more reactive unbonded valence electrons, also be the stronger nucleophile? As mentioned above, it all has to do with the solvent. Remember, we are talking now about the reaction running in a solvent like ethanol. Protic solvent molecules form very strong ion-dipole interactions with the negatively-charged nucleophile, essentially creating a 'solvent cage' around the nucleophile: In order for the nucleophile to attack the electrophile, it must break free, at least in part, from its solvent cage. The lone pair electrons on the larger, less basic iodide ion interact less tightly with the protons on the protic solvent molecules - thus the iodide nucleophile is better able to break free from its solvent cage compared the smaller, more basic fluoride ion, whose lone pair electrons are bound more tightly to the protons of the cage. The picture changes if we switch to a , such as acetone, in which there is a molecular dipole but . Now, fluoride is the best nucleophile, and iodide the weakest. The reason for the reversal is that, with an aprotic solvent, the ion-dipole interactions between solvent and nucleophile are much weaker: the positive end of the solvent's dipole is hidden in the interior of the molecule, and thus it is shielded from the negative charge of the nucleophile. A weaker solvent-nucleophile interaction means a weaker solvent cage for the nucleophile to break through, so the solvent effect is much less important, and the more basic fluoride ion is also the better nucleophile. Why not use a completely nonpolar solvent, such as hexane, for this reaction, so that the solvent cage is eliminated completely? The answer to this is simple - the nucleophile needs to be in solution in order to react at an appreciable rate with the electrophile, and a solvent such as hexane will not solvate an a charged (or highly polar) nucleophile at all. That is why chemists use polar aprotic solvents for nucleophilic substitution reactions in the laboratory: they are polar enough to solvate the nucleophile, but not so polar as to lock it away in an impenetrable solvent cage. In addition to acetone, three other commonly used polar aprotic solvents are acetonitrile, dimethylformamide (DMF), and dimethyl sulfoxide (DMSO). In biological chemistry, where the solvent is protic (water), the most important implication of the periodic trends in nucleophilicity is that thiols are more powerful nucleophiles than alcohols. The thiol group in a , for example, is a powerful nucleophile and often acts as a nucleophile in enzymatic reactions, and of course negatively-charged thiolates (RS ) are even more nucleophilic. This is not to say that the hydroxyl groups on serine, threonine, and tyrosine do not also act as nucleophiles - they do. Resonance effects also come into play when comparing the inherent nucleophilicity of different molecules. The reasoning involved is the same as that which we used to understand resonance effects on basicity. If the electron lone pair on a heteroatom is delocalized by resonance, it is inherently less reactive - meaning less nucleophilic, and also less basic. An alkoxide ion, for example, is more nucleophilic and more basic than a carboxylate group, even though in both cases the nucleophilic atom is a negatively charged oxygen. In the alkoxide, the negative charge is localized on a single oxygen, while in the carboxylate the charge is delocalized over two oxygen atoms by resonance. The nitrogen atom on an amide is less nucleophilic than the nitrogen of an amine, due to the resonance stabilization of the nitrogen lone pair provided by the amide carbonyl group. Steric hindrance is an important consideration when evaluating nucleophility. For example, -butanol is less potent as a nucleophile than methanol. This is because the comparatively bulky methyl groups on the tertiary alcohol effectively block the route of attack by the nucleophilic oxygen, slowing the reaction down considerably (imagine trying to walk through a narrow doorway while carrying three large suitcases!). It is not surprising that it is more common to observe serines acting as nucleophiles in enzymatic reactions compared to threonines - the former is a primary alcohol, while the latter is a secondary alcohol. | 8,401 | 2,594 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/21%3A_Chemistry_of_The_Main-Group_Elements_I/21.5%3A_Group_14%3A_The_Carbon_Family |
The elements of group 14 show a greater range of chemical behavior than any other family in the periodic table. Three of the five elements—carbon, tin, and lead—have been known since ancient times. For example, some of the oldest known writings are Egyptian hieroglyphics written on papyrus with ink made from lampblack, a finely divided carbon soot produced by the incomplete combustion of hydrocarbons (Figure \(\Page {1}\)). Activated carbon is an even more finely divided form of carbon that is produced from the thermal decomposition of organic materials, such as sawdust. Because it adsorbs many organic and sulfur-containing compounds, activated carbon is used to decolorize foods, such as sugar, and to purify gases and wastewater. Tin and lead oxides and sulfides are easily reduced to the metal by heating with charcoal, a discovery that must have occurred by accident when prehistoric humans used rocks containing their ores for a cooking fire. However, because tin and copper ores are often found together in nature, their alloy—bronze—was probably discovered before either element, a discovery that led to the Bronze Age. The heaviest element in group 14, lead, is such a soft and malleable metal that the ancient Romans used thin lead foils as writing tablets, as well as lead cookware and lead pipes for plumbing. (Recall that the atomic symbols for tin and lead come from their Latin names: Sn for stannum and Pb for plumbum.) Although the first glasses were prepared from silica (silicon oxide, SiO ) around 1500 BC, elemental silicon was not prepared until 1824 because of its high affinity for oxygen. Jöns Jakob Berzelius was finally able to obtain amorphous silicon by reducing Na SiF with molten potassium. The crystalline element, which has a shiny blue-gray luster, was not isolated until 30 yr later. The last member of the group 14 elements to be discovered was germanium, which was found in 1886 in a newly discovered silver-colored ore by the German chemist Clemens Winkler, who named the element in honor of his native country. The natural abundance of the group 14 elements varies tremendously. Elemental carbon, for example, ranks only 17th on the list of constituents of Earth’s crust. Pure graphite is obtained by reacting coke, an amorphous form of carbon used as a reductant in the production of steel, with silica to give silicon carbide (SiC). This is then thermally decomposed at very high temperatures (2700°C) to give graphite: \[\mathrm{SiO_2(s)}+\mathrm{3C(s)}\xrightarrow{\Delta}\mathrm{SiC(s)}+\mathrm{2CO(g)} \label{\(\Page {1}\)}\] One allotrope of carbon, diamond, is metastable under normal conditions, with a ΔG° of 2.9 kJ/mol versus graphite. At pressures greater than 50,000 atm, however, the diamond structure is favored and is the most stable form of carbon. Because the structure of diamond is more compact than that of graphite, its density is significantly higher (3.51 g/cm versus 2.2 g/cm ). Because of its high thermal conductivity, diamond powder is used to transfer heat in electronic devices. The most common sources of diamonds on Earth are ancient volcanic pipes that contain a rock called kimberlite, a lava that solidified rapidly from deep inside the Earth. Most kimberlite formations, however, are much newer than the diamonds they contain. In fact, the relative amounts of different carbon isotopes in diamond show that diamond is a chemical and geological “fossil” older than our solar system, which means that diamonds on Earth predate the existence of our sun. Thus diamonds were most likely created deep inside Earth from primordial grains of graphite present when Earth was formed (part (a) in Figure \(\Page {2}\)). Gem-quality diamonds can now be produced synthetically and have chemical, optical, and physical characteristics identical to those of the highest-grade natural diamonds. Although oxygen is the most abundant element on Earth, the next most abundant is silicon, the next member of group 14. Pure silicon is obtained by reacting impure silicon with Cl to give SiCl , followed by the fractional distillation of the impure SiCl and reduction with H : Several million tons of silicon are annually produced with this method. Amorphous silicon containing residual amounts of hydrogen is used in photovoltaic devices that convert light to electricity, and silicon-based solar cells are used to power pocket calculators, boats, and highway signs, where access to electricity by conventional methods is difficult or expensive. Ultrapure silicon and germanium form the basis of the modern electronics industry (part (b) in Figure \(\Page {2}\)). In contrast to silicon, the concentrations of germanium and tin in Earth’s crust are only 1–2 ppm. The concentration of lead, which is the end product of the nuclear decay of many radionuclides, is 13 ppm, making lead by far the most abundant of the heavy group 14 elements. No concentrated ores of germanium are known; like indium, germanium is generally recovered from flue dusts obtained by processing the ores of metals such as zinc. Because germanium is essentially transparent to infrared radiation, it is used in optical devices. Tin and lead are soft metals that are too weak for structural applications, but because tin is flexible, corrosion resistant, and nontoxic, it is used as a coating in food packaging. A “tin can,” for example, is actually a steel can whose interior is coated with a thin layer (1–2 µm) of metallic tin. Tin is also used in superconducting magnets and low-melting-point alloys such as solder and pewter. Pure lead is obtained by heating galena (PbS) in air and reducing the oxide (PbO) to the metal with carbon, followed by electrolytic deposition to increase the purity: \[\mathrm{PbS(s)}+\frac{3}{2}\mathrm{O_2(g)}\xrightarrow{\Delta}\mathrm{PbO(s)}+\mathrm{SO_2(g)} \label{\(\Page {4}\)}\] or By far the single largest use of lead is in lead storage batteries. The group 14 elements all have ns np valence electron configurations. All form compounds in which they formally lose either the two np and the two ns valence electrons or just the two np valence electrons, giving a +4 or +2 oxidation state, respectively. Because covalent bonds decrease in strength with increasing atomic size and the ionization energies for the heavier elements of the group are higher than expected due to a higher Z , the relative stability of the +2 oxidation state increases smoothly from carbon to lead. The relative stability of the +2 oxidation state increases, and the tendency to form catenated compounds decreases, from carbon to lead in group 14. Recall that many carbon compounds contain multiple bonds formed by π overlap of singly occupied 2p orbitals on adjacent atoms. Compounds of silicon, germanium, tin, and lead with the same stoichiometry as those of carbon, however, tend to have different structures and properties. For example, CO is a gas that contains discrete O=C=O molecules, whereas the most common form of SiO is the high-melting solid known as quartz, the major component of sand. Instead of discrete SiO molecules, quartz contains a three-dimensional network of silicon atoms that is similar to the structure of diamond but with an oxygen atom inserted between each pair of silicon atoms. Thus each silicon atom is linked to four other silicon atoms by bridging oxygen atoms. The tendency to catenate—to form chains of like atoms—decreases rapidly as we go down group 14 because bond energies for both the E–E and E–H bonds decrease with increasing atomic number (where E is any group 14 element). Consequently, inserting a CH group into a linear hydrocarbon such as n-hexane is exergonic (ΔG° = −45 kJ/mol), whereas inserting an SiH group into the silicon analogue of n-hexane (Si H ) actually costs energy (ΔG° ≈ +25 kJ/mol). As a result of this trend, the thermal stability of catenated compounds decreases rapidly from carbon to lead. In Table \(\Page {1}\) "Selected Properties of the Group 14 Elements" we see, once again, that there is a large difference between the lightest element (C) and the others in size, ionization energy, and electronegativity. As in group 13, the second and third elements (Si and Ge) are similar, and there is a reversal in the trends for some properties, such as ionization energy, between the fourth and fifth elements (Sn and Pb). As for group 13, these effects can be explained by the presence of filled (n − 1)d and (n − 2)f subshells, whose electrons are relatively poor at screening the outermost electrons from the higher nuclear charge. The group 14 elements follow the same pattern as the group 13 elements in their periodic properties. Carbon is the building block of all organic compounds, including biomolecules, fuels, pharmaceuticals, and plastics, whereas inorganic compounds of carbon include metal carbonates, which are found in substances as diverse as fertilizers and antacid tablets, halides, oxides, carbides, and carboranes. Like boron in group 13, the chemistry of carbon differs sufficiently from that of its heavier congeners to merit a separate discussion. The structures of the allotropes of carbon—diamond, graphite, fullerenes, and nanotubes—are distinct, but they all contain simple electron-pair bonds (Figure 7.18). Although it was originally believed that fullerenes were a new form of carbon that could be prepared only in the laboratory, fullerenes have been found in certain types of meteorites. Another possible allotrope of carbon has also been detected in impact fragments of a carbon-rich meteorite; it appears to consist of long chains of carbon atoms linked by alternating single and triple bonds, (–C≡C–C≡C–) . Carbon nanotubes (“buckytubes”) are being studied as potential building blocks for ultramicroscale detectors and molecular computers and as tethers for space stations. They are currently used in electronic devices, such as the electrically conducting tips of miniature electron guns for flat-panel displays in portable computers. Although all the carbon tetrahalides (CX ) are known, they are generally not obtained by the direct reaction of carbon with the elemental halogens (X ) but by indirect methods such as the following reaction, where X is Cl or Br: \[CH_{4(g)} + 4X_{2(g)} \rightarrow CX_{4(l,s)} + 4HX_{(g)} \label{\(\Page {7}\)}\] The carbon tetrahalides all have the tetrahedral geometry predicted by the valence-shell electron-pair repulsion (VSEPR) model, as shown for CCl and CI . Their stability decreases rapidly as the halogen increases in size because of poor orbital overlap and increased crowding. Because the C–F bond is about 25% stronger than a C–H bond, fluorocarbons are thermally and chemically more stable than the corresponding hydrocarbons, while having a similar hydrophobic character. A polymer of tetrafluoroethylene (F C=CF ), analogous to polyethylene, is the nonstick Teflon lining found on many cooking pans, and similar compounds are used to make fabrics stain resistant (such as Scotch-Gard) or waterproof but breathable (such as Gore-Tex). The stability of the carbon tetrahalides decreases with increasing size of the halogen due to increasingly poor orbital overlap and crowding. Carbon reacts with oxygen to form either CO or CO , depending on the stoichiometry. Carbon monoxide is a colorless, odorless, and poisonous gas that reacts with the iron in hemoglobin to form an Fe–CO unit, which prevents hemoglobin from binding, transporting, and releasing oxygen in the blood (see Figure 23.26). In the laboratory, carbon monoxide can be prepared on a small scale by dehydrating formic acid with concentrated sulfuric acid: Carbon monoxide also reacts with the halogens to form the oxohalides (COX ). Probably the best known of these is phosgene (Cl C=O), which is highly poisonous and was used as a chemical weapon during World War I: Despite its toxicity, phosgene is an important industrial chemical that is prepared on a large scale, primarily in the manufacture of polyurethanes. Carbon dioxide can be prepared on a small scale by reacting almost any metal carbonate or bicarbonate salt with a strong acid. As is typical of a nonmetal oxide, CO reacts with water to form acidic solutions containing carbonic acid (H CO ). In contrast to its reactions with oxygen, reacting carbon with sulfur at high temperatures produces only carbon disulfide (CS ): The selenium analog CSe is also known. Both have the linear structure predicted by the VSEPR model, and both are vile smelling (and in the case of CSe , highly toxic), volatile liquids. The sulfur and selenium analogues of carbon monoxide, CS and CSe, are unstable because the C≡Y bonds (Y is S or Se) are much weaker than the C≡O bond due to poorer π orbital overlap. \(\pi\) bonds between carbon and the heavier chalcogenides are weak due to poor orbital overlap. Binary compounds of carbon with less electronegative elements are called carbides. The chemical and physical properties of carbides depend strongly on the identity of the second element, resulting in three general classes: ionic carbides, interstitial carbides, and covalent carbides. The reaction of carbon at high temperatures with electropositive metals such as those of groups 1 and 2 and aluminum produces ionic carbides, which contain discrete metal cations and carbon anions. The identity of the anions depends on the size of the second element. For example, smaller elements such as beryllium and aluminum give methides such as Be C and Al C , which formally contain the C ion derived from methane (CH ) by losing all four H atoms as protons. In contrast, larger metals such as sodium and calcium give carbides with stoichiometries of Na C and CaC . Because these carbides contain the C ion, which is derived from acetylene (HC≡CH) by losing both H atoms as protons, they are more properly called acetylides. Reacting ionic carbides with dilute aqueous acid results in protonation of the anions to give the parent hydrocarbons: CH or C H . For many years, miners’ lamps used the reaction of calcium carbide with water to produce a steady supply of acetylene, which was ignited to provide a portable lantern. The reaction of carbon with most transition metals at high temperatures produces interstitial carbides. Due to the less electropositive nature of the transition metals, these carbides contain covalent metal–carbon interactions, which result in different properties: most interstitial carbides are good conductors of electricity, have high melting points, and are among the hardest substances known. Interstitial carbides exhibit a variety of nominal compositions, and they are often nonstoichiometric compounds whose carbon content can vary over a wide range. Among the most important are tungsten carbide (WC), which is used industrially in high-speed cutting tools, and cementite (Fe C), which is a major component of steel. Elements with an electronegativity similar to that of carbon form covalent carbides, such as silicon carbide (SiC; Equation \(\ref{Eq1}\)) and boron carbide (B C). These substances are extremely hard, have high melting points, and are chemically inert. For example, silicon carbide is highly resistant to chemical attack at temperatures as high as 1600°C. Because it also maintains its strength at high temperatures, silicon carbide is used in heating elements for electric furnaces and in variable-temperature resistors. Carbides formed from group 1 and 2 elements are ionic. Transition metals form interstitial carbides with covalent metal–carbon interactions, and covalent carbides are chemically inert. For each reaction, explain why the given product forms. balanced chemical equations why the given products form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form. Predict the products of the reactions and write a balanced chemical equation for each reaction. Although silicon, germanium, tin, and lead in their +4 oxidation states often form binary compounds with the same stoichiometry as carbon, the structures and properties of these compounds are usually significantly different from those of the carbon analogues. Silicon and germanium are both semiconductors with structures analogous to diamond. Tin has two common allotropes: white (β) tin has a metallic lattice and metallic properties, whereas gray (α) tin has a diamond-like structure and is a semiconductor. The metallic β form is stable above 13.2°C, and the nonmetallic α form is stable below 13.2°C. Lead is the only group 14 element that is metallic in both structure and properties under all conditions. Based on its position in the periodic table, we expect silicon to be amphoteric. In fact, it dissolves in strong aqueous base to produce hydrogen gas and solutions of silicates, but the only aqueous acid that it reacts with is hydrofluoric acid, presumably due to the formation of the stable SiF ion. Germanium is more metallic in its behavior than silicon. For example, it dissolves in hot oxidizing acids, such as HNO and H SO , but in the absence of an oxidant, it does not dissolve in aqueous base. Although tin has an even more metallic character than germanium, lead is the only element in the group that behaves purely as a metal. Acids do not readily attack it because the solid acquires a thin protective outer layer of a Pb salt, such as PbSO . All group 14 dichlorides are known, and their stability increases dramatically as the atomic number of the central atom increases. Thus CCl is dichlorocarbene, a highly reactive, short-lived intermediate that can be made in solution but cannot be isolated in pure form using standard techniques; SiCl can be isolated at very low temperatures, but it decomposes rapidly above −150°C, and GeCl is relatively stable at temperatures below 20°C. In contrast, SnCl is a polymeric solid that is indefinitely stable at room temperature, whereas PbCl is an insoluble crystalline solid with a structure similar to that of SnCl . The stability of the group 14 dichlorides increases dramatically from carbon to lead. Although the first four elements of group 14 form tetrahalides (MX ) with all the halogens, only fluorine is able to oxidize lead to the +4 oxidation state, giving PbF . The tetrahalides of silicon and germanium react rapidly with water to give amphoteric oxides (where M is Si or Ge): \[MX_{4(s,l)} + 2H_2O_{(l)} \rightarrow MO_{2(s)} + 4HX_{(aq)} \label{\(\Page {1}\)1}\] In contrast, the tetrahalides of tin and lead react with water to give hydrated metal ions. Because of the stability of its +2 oxidation state, lead reacts with oxygen or sulfur to form PbO or PbS, respectively, whereas heating the other group 14 elements with excess O or S gives the corresponding dioxides or disulfides, respectively. The dioxides of the group 14 elements become increasingly basic as we go down the group. The dioxides of the group 14 elements become increasingly basic down the group. Because the Si–O bond is even stronger than the C–O bond (~452 kJ/mol versus ~358 kJ/mol), silicon has a strong affinity for oxygen. The relative strengths of the C–O and Si–O bonds contradict the generalization that bond strengths decrease as the bonded atoms become larger. This is because we have thus far assumed that a formal single bond between two atoms can always be described in terms of a single pair of shared electrons. In the case of Si–O bonds, however, the presence of relatively low-energy, empty d orbitals on Si and nonbonding electron pairs in the p or sp hybrid orbitals of O results in a partial π bond (Figure \(\Page {3}\)). Due to its partial π double bond character, the Si–O bond is significantly stronger and shorter than would otherwise be expected. A similar interaction with oxygen is also an important feature of the chemistry of the elements that follow silicon in the third period (P, S, and Cl). Because the Si–O bond is unusually strong, silicon–oxygen compounds dominate the chemistry of silicon. Because silicon–oxygen bonds are unusually strong, silicon–oxygen compounds dominate the chemistry of silicon. Compounds with anions that contain only silicon and oxygen are called silicates, whose basic building block is the SiO unit: The number of oxygen atoms shared between silicon atoms and the way in which the units are linked vary considerably in different silicates. Converting one of the oxygen atoms from terminal to bridging generates chains of silicates, while converting two oxygen atoms from terminal to bridging generates double chains. In contrast, converting three or four oxygens to bridging generates a variety of complex layered and three-dimensional structures, respectively. In a large and important class of materials called aluminosilicates, some of the Si atoms are replaced by Al atoms to give aluminosilicates such as zeolites, whose three-dimensional framework structures have large cavities connected by smaller tunnels (Figure \(\Page {4}\)). Because the cations in zeolites are readily exchanged, zeolites are used in laundry detergents as water-softening agents: the more loosely bound Na ions inside the zeolite cavities are displaced by the more highly charged Mg and Ca ions present in hard water, which bind more tightly. Zeolites are also used as catalysts and for water purification. Silicon and germanium react with nitrogen at high temperature to form nitrides (M N ): \[ 3Si_{(l)} + 2N_{2(g)} \rightarrow Si_3N_{4(s)} \label{\(\Page {1}\)2}\] Silicon nitride has properties that make it suitable for high-temperature engineering applications: it is strong, very hard, and chemically inert, and it retains these properties to temperatures of about 1000°C. Because of the diagonal relationship between boron and silicon, metal silicides and metal borides exhibit many similarities. Although metal silicides have structures that are as complex as those of the metal borides and carbides, few silicides are structurally similar to the corresponding borides due to the significantly larger size of Si (atomic radius 111 pm versus 87 pm for B). Silicides of active metals, such as Mg Si, are ionic compounds that contain the Si ion. They react with aqueous acid to form silicon hydrides such as SiH : \[Mg_2Si_{(s)} + 4H^+_{(aq)} \rightarrow 2Mg^{2+}_{(aq)} + SiH_{4(g)} \label{\(\Page {1}\)3}\] Unlike carbon, catenated silicon hydrides become thermodynamically less stable as the chain lengthens. Thus straight-chain and branched silanes (analogous to alkanes) are known up to only n = 10; the germanium analogues (germanes) are known up to n = 9. In contrast, the only known hydride of tin is SnH , and it slowly decomposes to elemental Sn and H at room temperature. The simplest lead hydride (PbH ) is so unstable that chemists are not even certain it exists. Because E=E and E≡E bonds become weaker with increasing atomic number (where E is any group 14 element), simple silicon, germanium, and tin analogues of alkenes, alkynes, and aromatic hydrocarbons are either unstable (Si=Si and Ge=Ge) or unknown. Silicon-based life-forms are therefore likely to be found only in science fiction. The stability of group 14 hydrides decreases down the group, and E=E and E≡E bonds become weaker. The only important organic derivatives of lead are compounds such as tetraethyllead [(CH CH ) Pb]. Because the Pb–C bond is weak, these compounds decompose at relatively low temperatures to produce alkyl radicals (R·), which can be used to control the rate of combustion reactions. For 60 yr, hundreds of thousands of tons of lead were burned annually in automobile engines, producing a mist of lead oxide particles along the highways that constituted a potentially serious public health problem. (Example 6 in Section 22.3 examines this problem.) The use of catalytic converters reduced the amount of carbon monoxide, nitrogen oxides, and hydrocarbons released into the atmosphere through automobile exhausts, but it did nothing to decrease lead emissions. Because lead poisons catalytic converters, however, its use as a gasoline additive has been banned in most of the world. Compounds that contain Si–C and Si–O bonds are stable and important. High-molecular-mass polymers called silicones contain an (Si–O–) backbone with organic groups attached to Si (Figure \(\Page {5}\)). The properties of silicones are determined by the chain length, the type of organic group, and the extent of cross-linking between the chains. Without cross-linking, silicones are waxes or oils, but cross-linking can produce flexible materials used in sealants, gaskets, car polishes, lubricants, and even elastic materials, such as the plastic substance known as Silly Putty. For each reaction, explain why the given products form. balanced chemical equations why the given products form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the observed reaction products form. Predict the products of the reactions and write a balanced chemical equation for each reaction. The group 14 elements show the greatest diversity in chemical behavior of any group; covalent bond strengths decease with increasing atomic size, and ionization energies are greater than expected, increasing from C to Pb. The group 14 elements show the greatest range of chemical behavior of any group in the periodic table. Because the covalent bond strength decreases with increasing atomic size and greater-than-expected ionization energies due to an increase in Z , the stability of the +2 oxidation state increases from carbon to lead. The tendency to form multiple bonds and to catenate decreases as the atomic number increases. The stability of the carbon tetrahalides decreases as the halogen increases in size because of poor orbital overlap and steric crowding. Carbon forms three kinds of carbides with less electronegative elements: ionic carbides, which contain metal cations and C (methide) or C (acetylide) anions; interstitial carbides, which are characterized by covalent metal–carbon interactions and are among the hardest substances known; and covalent carbides, which have three-dimensional covalent network structures that make them extremely hard, high melting, and chemically inert. Consistent with periodic trends, metallic behavior increases down the group. Silicon has a tremendous affinity for oxygen because of partial Si–O π bonding. Dioxides of the group 14 elements become increasingly basic down the group and their metallic character increases. Silicates contain anions that consist of only silicon and oxygen. Aluminosilicates are formed by replacing some of the Si atoms in silicates by Al atoms; aluminosilicates with three-dimensional framework structures are called zeolites. Nitrides formed by reacting silicon or germanium with nitrogen are strong, hard, and chemically inert. The hydrides become thermodynamically less stable down the group. Moreover, as atomic size increases, multiple bonds between or to the group 14 elements become weaker. Silicones, which contain an Si–O backbone and Si–C bonds, are high-molecular-mass polymers whose properties depend on their compositions. | 27,255 | 2,595 |
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Solubility equilibria defined by a single, simple expression are probably the exception rather than the rule. Such equilibria are often in competition with other reactions with such species as H or OH , complexing agents, oxidation-reduction, formation of other sparingly soluble species or, in the case of carbonates and sulfites, of gaseous products. The exact treatments of these systems can be extremely complicated, involving the solution of large sets of simultaneous equations. For most practical purposes it is sufficient to recognize the general trends, and to carry out approximate calculations. The solubility product of an ionic compound describes the concentrations of ions in equilibrium with a solid, but what happens if some of the cations become associated with anions rather than being completely surrounded by solvent? Then predictions of the total solubility of the compound based on the assumption that the solute exists solely as discrete ions would differ substantially from the actual solubility, as would predictions of ionic concentrations. In general, four situations explain why the solubility of a compound may be other than expected: ion pair formation, the incomplete dissociation of molecular solutes, the formation of complex ions, and changes in pH. The Diverse Ion Effect typically has the opposite effect on solubility than the common ion effect. How does the solubility of \(AgCl\) change in absence and presence of \(NaNO_3\), given that the molar solubility product of \(AgCl\) is \(1.76 \times 10^{−10}\). The Uncommon Ion Effect is a "real" effect and requires the definition of \(K_{sp}\) to address. The exact definition of the solubility constant is: \[K_{sp}=a\{Ag^+\} \cdot a\{Cl^−\}=[Ag+] \gamma(Ag^+) [Cl^−] \gamma (Cl^−) \nonumber\] Where \(a\{Ag^+\}\) and \(a\{Cl^−\}\) are the activities of these ions in the presence of the \(NaNO_3\). The \(\gamma\) values are the are the corresponding activity coefficients for the ions. Hence, in the presence of the \(NaNO_3\) soluble salt, the molar solubility of the slightly soluble \(AgCl\) salt . An opposite effect is expected if there were a common ions between the two salts. An consists of a cation and an anion that are in intimate contact in solution, rather than separated by solvent ( ). The ions in an ion pair are held together by the same attractive electrostatic force in ionic solids. As a result, the ions in an ion pair migrate as a single unit, whose net charge is the sum of the charges on the ions. In many ways, we can view an ion pair as a species intermediate between the ionic solid (in which each ion participates in many cation–anion interactions that hold the ions in a rigid array) and the completely dissociated ions in solution (where each is fully surrounded by water molecules and free to migrate independently). As illustrated for calcium sulfate in the following equation, a second equilibrium must be included to describe the solubility of salts that form ion pairs: \[\mathrm{CaSO_4(s)}\rightleftharpoons\mathrm{Ca^{2+}}\cdot\underset{\textrm{ion pair}}{\mathrm{SO_4^{2-}(aq)}}\rightleftharpoons \mathrm{Ca^{2+}(aq)}+\mathrm{SO_4^{2-}(aq)} \label{17.5.3.1}\] The ion pair is represented by the symbols of the individual ions separated by a dot, which indicates that they are associated in solution. The formation of an ion pair is a dynamic process, just like any other equilibrium, so a particular ion pair may exist only briefly before dissociating into the free ions, each of which may later associate briefly with other ions. Ion-pair formation can have a major effect on the measured solubility of a salt. For example, the measured for calcium sulfate is 4.93 × 10 at 25°C. The solubility of CaSO should be 7.02 × 10 M if the only equilibrium involved were as follows: \[CaSO_{4(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO^{2−}_{4(aq)} \label{17.5.2}\] In fact, the experimentally measured solubility of calcium sulfate at 25°C is 1.6 × 10 M, almost twice the value predicted from its . The reason for the discrepancy is that the concentration of ion pairs in a saturated CaSO solution is almost as high as the concentration of the hydrated ions. Recall that the magnitude of attractive electrostatic interactions is greatest for small, highly charged ions. Hence ion pair formation is most important for salts that contain M and M ions, such as Ca and La , and is relatively unimportant for salts that contain monopositive cations, except for the smallest, Li . We therefore expect a saturated solution of CaSO to contain a high concentration of ion pairs and its solubility to be greater than predicted from its . A molecular solute may also be more soluble than predicted by the measured concentrations of ions in solution due to incomplete dissociation. This is particularly common with weak organic acids. Although strong acids (HA) dissociate completely into their constituent ions (H and A ) in water, weak acids such as carboxylic acids do not ( = 1.5 × 10 ). However, the molecular (undissociated) form of a weak acid (HA) is often quite soluble in water; for example, acetic acid (CH CO H) is completely miscible with water. Many carboxylic acids, however, have only limited solubility in water, such as benzoic acid (C H CO H), with = 6.25 × 10 . Just as with calcium sulfate, we need to include an additional equilibrium to describe the solubility of benzoic acid: \[ C_6H_5CO_2H_{(s)} \rightleftharpoons C_6H_5CO_2H_{(aq)} \rightleftharpoons C_6H_5CO^−_{2(aq)} + H^+_{(aq)} \label{17.5.3}\] In a case like this, measuring only the concentration of the ions grossly underestimates the total concentration of the organic acid in solution. In the case of benzoic acid, for example, the pH of a saturated solution at 25°C is 2.85, corresponding to [H ] = [C H CO ] = 1.4 × 10 M. The total concentration of benzoic acid in the solution, however, is 2.8 × 10 M. Thus approximately 95% of the benzoic acid in solution is in the form of hydrated neutral molecules—\(C_6H_5CO_2H_{(aq)}\)—and only about 5% is present as the dissociated ions (Figure \(\Page {3}\)). Incomplete dissociation of a molecular solute that is miscible with water can increase the solubility of the solute. Although ion pairs, such as Ca •SO , and undissociated electrolytes, such as C H CO H, are both electrically neutral, there is a major difference in the forces responsible for their formation. Simple electrostatic attractive forces between the cation and the anion hold the ion pair together, whereas a polar covalent O−H bond holds together the undissociated electrolyte. \[CdI_{2(s)} \rightleftharpoons Cd^{2+} + 2 I^–\] \[Cd^{2+}_{(aq)} + 2 I^–_{(aq)} → CdI_{2(aq)}\] \(CdI^–_{(aq)}\) As a consequence, the concentration of "free" Cd in an aqueous cadmium iodide solution is only about 2% of the value you would calculate by taking as the solubility product. The principal component of such as solution is actually [covalently-bound] CdI . It turns out that The solubility constant, as with all equilibrium constants, are properly defined in terms of (activities) and the use of true concentrations to approximate activities in equilibrium constants can often fail. As with all approximations, it is important to understand the conditions that it will fail. three examples are discussed. The salt effect refers to the fact that the presence of a salt which has no ion in common with the solute, has an effect on the ionic strength of the solution and hence on activity coefficients, so that the equilibrium constant, expressed as a concentration quotient, changes. ) | 7,639 | 2,596 |
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The Woodward-Hoffman rules (R. B. Woodward was one of the U.S.'s most successful and renowned organic chemists; Roald Hoffmann is a theoretical chemist. Hoffmann shared the 1981 Nobel Prize with Kenichi Fukui) allow us to follow the symmetry of the occupied molecular orbitals along a suggested reaction path connecting reactants to products and to then suggest whether a symmetry-imposed additional energy barrier (i.e., above any thermodynamic energy requirement) would be expected. For example, let us consider the disrotatory closing of 1,3-butadiene to produce cyclobutene. The four \(p\) orbitals of butadiene, denoted \(\pi_1\) through \(\pi_4\) in the figure shown below, evolve into the \(\sigma\) and \(\sigma^{*}\) and \(\pi\) and \(\pi^{*}\) orbitals of cyclobutene as the closure reaction proceeds. Along the disrotatory reaction path, a plane of symmetry is preserved (i.e., remains a valid symmetry operation) so it can be used to label the symmetries of the four reactant and four product orbitals. Labeling them as odd (o) or even (e) under this plane and then connecting the reactant orbitals to the product orbitals according to their symmetries, produces the orbital correlation diagram shown below. The four \(\pi\) electrons of 1,3-butadiene occupy the \(\pi_1\) and \(\pi_2\) orbitals in the ground electronic state of this molecule. Because these two orbitals do not correlate directly to the two orbitals that are occupied (\(\sigma\) and \(\pi\)) in the ground state of cyclobutene, the Woodward-Hoffmann rules suggest that this reaction, along a disrotatory path, will encounter a symmetry-imposed energy barrier. In contrast, the excited electronic state in which one electron is promoted from \(\pi_2\) to \(\pi_3\) (as, for example, in a photochemical excitation experiment) does correlate directly to the lowest energy excited state of the cyclobutene product molecule, which has one electron in its orbital and one in its * orbital. Therefore, these same rules suggest that photo-exciting butadiene in the particular way in which the excitation promotes one electron from \(\pi_2\) to \(\pi_3\) will produce a chemical reaction to generate cyclobutene. | 2,199 | 2,597 |
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Einstein’s photons of light were individual packets of energy having many of the characteristics of particles. Recall that the collision of an electron (a particle) with a sufficiently energetic photon can eject a from the surface of a metal. Any excess energy is transferred to the electron and is converted to the kinetic energy of the ejected electron. Einstein’s hypothesis that energy is concentrated in localized bundles, however, was in sharp contrast to the classical notion that energy is spread out uniformly in a wave. We now describe Einstein’s theory of the relationship between energy and mass, a theory that others built on to develop our current model of the atom. Einstein initially assumed that photons had zero mass, which made them a peculiar sort of particle indeed. In 1905, however, he published his special theory of relativity, which related energy and mass according to the famous equation: \[ E=h u=h\dfrac{c}{\lambda }=mc^{2} \label{6.4.1} \] According to this theory, a photon of wavelength \(λ\) and frequency \( u\) has a nonzero mass, which is given as follows: \[ m=\dfrac{E}{c^{2}}=\dfrac{h u }{c^{2}}=\dfrac{h}{\lambda c} \label{6.4.2} \] That is, light, which had always been regarded as a wave, also has properties typical of particles, a condition known as wave–particle duality (a principle that matter and energy have properties typical of both waves and particles). Depending on conditions, light could be viewed as either a wave or a particle. In 1922, the American physicist Arthur Compton (1892–1962) reported the results of experiments involving the collision of x-rays and electrons that supported the particle nature of light. At about the same time, a young French physics student, Louis de Broglie (1892–1972), began to wonder whether the converse was true: Could particles exhibit the properties of waves? In his PhD dissertation submitted to the Sorbonne in 1924, de Broglie proposed that a particle such as an electron could be described by a wave whose wavelength is given by \[\lambda =\dfrac{h}{mv} \label{6.4.3} \] where This revolutionary idea was quickly confirmed by American physicists Clinton Davisson (1881–1958) and Lester Germer (1896–1971), who showed that beams of electrons, regarded as particles, were diffracted by a sodium chloride crystal in the same manner as x-rays, which were regarded as waves. It was proven experimentally that electrons do exhibit the properties of waves. For his work, de Broglie received the Nobel Prize in Physics in 1929. If particles exhibit the properties of waves, why had no one observed them before? The answer lies in the numerator of de Broglie’s equation, which is an extremely small number. As you will calculate in Example \(\Page {1}\), Planck’s constant (6.63 × 10 J•s) is so small that the wavelength of a particle with a large mass is too short (less than the diameter of an atomic nucleus) to be noticeable. Calculate the wavelength of a baseball, which has a mass of 149 g and a speed of 100 mi/h. mass and speed of object wavelength The wavelength of a particle is given by \(λ = h/mv\). We know that = 0.149 kg, so all we need to find is the speed of the baseball: \( v=\left ( \dfrac{100\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right ) \) Recall that the joule is a derived unit, whose units are (kg•m )/s . Thus the wavelength of the baseball is \[ \lambda =\dfrac{6.626\times 10^{-34}\; J\cdot s}{\left ( 0.149\; kg \right )\left ( 44.69\; m\cdot s \right )}= \dfrac{6.626\times 10^{-34}\; \cancel{kg}\cdot m{^\cancel{2}\cdot \cancel{s}{\cancel{^{-2}}\cdot \cancel{s}}}}{\left ( 0.149\; \cancel{kg} \right )\left ( 44.69\; \cancel{m}\cdot \cancel{s^{-1}} \right )}=9.95\times 10^{-35}\; m \nonumber \] (You should verify that the units cancel to give the wavelength in meters.) Given that the diameter of the nucleus of an atom is approximately 10 m, the wavelength of the baseball is almost unimaginably small. Calculate the wavelength of a neutron that is moving at 3.00 × 10 m/s. 1.32 Å, or 132 pm As you calculated in Example \(\Page {1}\), objects such as a baseball or a neutron have such short wavelengths that they are best regarded primarily as particles. In contrast, objects with very small masses (such as photons) have large wavelengths and can be viewed primarily as waves. Objects with intermediate masses, however, such as electrons, exhibit the properties of both particles waves. Although we still usually think of electrons as particles, the wave nature of electrons is employed in an , which has revealed most of what we know about the microscopic structure of living organisms and materials. Because the wavelength of an electron beam is much shorter than the wavelength of a beam of visible light, this instrument can resolve smaller details than a light microscope can (Figure \(\Page {1}\)). A wave is a disturbance that travels in space. The magnitude of the wave at any point in space and time varies sinusoidally. While the absolute value of the magnitude of one wave at any point is not very important, the displacement of two waves, called the phase difference, is vitally important because it determines whether the waves reinforce or interfere with each other. Figure \(\Page {2A}\) shows an arbitrary phase difference between two wave and Figure \(\Page {2B}\) shows what happens when the two waves are 180 degrees out of phase. The green line is their sum. Figure \(\Page {2C}\) shows what happens when the two lines are in phase, exactly superimposed on each other. Again, the green line is the sum of the intensities. A pattern of constructive and destructive interference is obtained when two (or more) diffracting waves interact with each other. This principle of diffraction and interference was used to prove the wave properties of electrons and is the basis for how electron microscopes work. For a mathematical analysis of phase aspects in sinusoids, check the math Libretexts library. De Broglie also investigated why only certain orbits were allowed in Bohr’s model of the hydrogen atom. He hypothesized that the electron behaves like a (a wave that does not travel in space). An example of a standing wave is the motion of a string of a violin or guitar. When the string is plucked, it vibrates at certain fixed frequencies because it is fastened at both ends (Figure \(\Page {3}\)). If the length of the string is \(L\), then the lowest-energy vibration (the fundamental) has wavelength \[ \begin{align} \dfrac{\lambda }{2} & =L \nonumber \\ \lambda &= 2L \nonumber \end{align} \label{6.4.4} \] Higher-energy vibrations are called (the vibration of a standing wave that is higher in energy than the fundamental vibration) and are produced when the string is plucked more strongly; they have wavelengths given by \[ \lambda=\dfrac{2L}{n} \label{6.4.5} \] where n has any integral value. When plucked, all other frequencies die out immediately. Only the resonant frequencies survive and are heard. Thus, we can think of the resonant frequencies of the string as being quantized. Notice in Figure \(\Page {3}\) that all overtones have one or more nodes, points where the string does not move. The amplitude of the wave at a node is zero. Quantized vibrations and overtones containing nodes are not restricted to one-dimensional systems, such as strings. A two-dimensional surface, such as a drumhead, also has quantized vibrations. Similarly, when the ends of a string are joined to form a circle, the only allowed vibrations are those with wavelength \[2πr = nλ \label{6.4.6} \] where \(r\) is the radius of the circle. De Broglie argued that Bohr’s allowed orbits could be understood if the electron behaved like a (Figure \(\Page {4}\)). The standing wave could exist only if the circumference of the circle was an integral multiple of the wavelength such that the propagated waves were all in phase, thereby increasing the net amplitudes and causing . Otherwise, the propagated waves would be out of phase, resulting in a net decrease in amplitude and causing The nonresonant waves interfere with themselves! De Broglie’s idea explained Bohr’s allowed orbits and energy levels nicely: in the lowest energy level, corresponding to \(n = 1\) in Equation \(\ref{6.4.6}\), one complete wavelength would close the circle. Higher energy levels would have successively higher values of with a corresponding number of nodes. Like all analogies, although the standing wave model helps us understand much about why Bohr's theory worked, it also, if pushed too far, can mislead. As you will see, some of de Broglie’s ideas are retained in the modern theory of the electronic structure of the atom: the wave behavior of the electron and the presence of nodes that increase in number as the energy level increases. Unfortunately, his (and Bohr's) explanation also contains one major feature that we now know to be incorrect: in the currently accepted model, the electron in a given orbit is always at the same distance from the nucleus. Because a wave is a disturbance that travels in space, it has no fixed position. One might therefore expect that it would also be hard to specify the exact position of a that exhibits wavelike behavior. A characteristic of light is that is can be bent or spread out by passing through a narrow slit. You can literally see this by half closing your eyes and looking through your eye lashes. This reduces the brightness of what you are seeing and somewhat fuzzes out the image, but the light bends around your lashes to provide a complete image rather than a bunch of bars across the image. This is called diffraction. This behavior of waves is captured in (1870 or so) for electromagnetic waves and was and is well understood. An "uncertainty principle" for light is, if you will, merely a conclusion about the nature of electromagnetic waves and nothing new. De Broglie's idea of wave-particle duality means that particles such as electrons which exhibit wavelike characteristics will also undergo diffraction from slits whose size is on the order of the electron wavelength. This situation was described mathematically by the German physicist Werner Heisenberg (1901–1976; Nobel Prize in Physics, 1932), who related the position of a particle to its momentum. Referring to the electron, Heisenberg stated that “at every moment the electron has only an inaccurate position and an inaccurate velocity, and between these two inaccuracies there is this uncertainty relation.” Mathematically, the states that the uncertainty in the position of a particle (Δ ) multiplied by the uncertainty in its momentum [Δ( )] is greater than or equal to Planck’s constant divided by 4π: \[ \left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\ge \dfrac{h}{4\pi } \label{6.4.7} \] Because Planck’s constant is a very small number, the Heisenberg uncertainty principle is important only for particles such as electrons that have very low masses. These are the same particles predicted by de Broglie’s equation to have measurable wavelengths. If the precise position \(x\) of a particle is known absolutely (Δ = 0), then the uncertainty in its momentum must be infinite: \[ \left ( \Delta \left [ mv \right ] \right )= \dfrac{h}{4\pi \left ( \Delta x \right ) }=\dfrac{h}{4\pi \left ( 0 \right ) }=\infty \label{6.4.8} \] Because the mass of the electron at rest (\(m\)) is both constant and accurately known, the uncertainty in \(Δ(mv)\) must be due to the \(Δv\) term, which would have to be infinitely large for \(Δ(mv)\) to equal infinity. That is, according to Equation \(\ref{6.4.8}\), the more accurately we know the exact position of the electron (as \(Δx → 0\)), the less accurately we know the speed and the kinetic energy of the electron (1/2 ) because \(Δ(mv) → ∞\). Conversely, the more accurately we know the precise momentum (and the energy) of the electron [as \(Δ(mv) → 0\)], then \(Δx → ∞\) and we have no idea where the electron is. Bohr’s model of the hydrogen atom violated the Heisenberg uncertainty principle by trying to specify both the position (an orbit of a particular radius) and the energy (a quantity related to the momentum) of the electron. Moreover, given its mass and wavelike nature, the electron in the hydrogen atom could not possibly orbit the nucleus in a well-defined circular path as predicted by Bohr. You will see, however, that the of the electron in the hydrogen atom is exactly the one predicted by Bohr’s model. Calculate the minimum uncertainty in the position of the pitched baseball from Example \(\ref{6.4.1}\) that has a mass of exactly 149 g and a speed of 100 ± 1 mi/h. mass and speed of object minimum uncertainty in its position The Heisenberg uncertainty principle (Equation \ref{6.4.7}) tells us that \[(Δx)(Δ(mv)) = h/4π \nonumber \]. Rearranging the inequality gives \( \Delta x \ge \left( {\dfrac{h}{4\pi }} \right)\left( {\dfrac{1}{\Delta (mv)}} \right)\) We know that = 6.626 × 10 J•s and = 0.149 kg. Because there is no uncertainty in the mass of the baseball, Δ( ) = Δ and Δ = ±1 mi/h. We have \[ \Delta u =\left ( \dfrac{1\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1\; \cancel{min}}{60\; s} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )=0.4469\; m/s \nonumber \] \[ \Delta x \ge \left ( \dfrac{6.626\times 10^{-34}\; J\cdot s}{4\left ( 3.1416 \right )} \right ) \left ( \dfrac{1}{\left ( 0.149\; kg \right )\left ( 0.4469\; m\cdot s^{-1} \right )} \right ) \nonumber \] Inserting the definition of a joule (1 J = 1 kg•m /s ) gives \[ \Delta x \ge \left ( \dfrac{6.626\times 10^{-34}\; \cancel{kg} \cdot m^{\cancel{2}} \cdot s}{4\left ( 3.1416 \right )\left ( \cancel{s^{2}} \right )} \right ) \left ( \dfrac{1\; \cancel{s}}{\left ( 0.149\; \cancel{kg} \right )\left ( 0.4469\; \cancel{m} \right )} \right ) \nonumber \] \[ \Delta x \ge 7.92 \pm \times 10^{-34}\; m \nonumber \] This is equal to \(3.12 \times 10^{−32}\) inches. We can safely say that if a batter misjudges the speed of a fastball by 1 mi/h (about 1%), he will not be able to blame Heisenberg’s uncertainty principle for striking out. Calculate the minimum uncertainty in the position of an electron traveling at one-third the speed of light, if the uncertainty in its speed is ±0.1%. Assume its mass to be equal to its mass at rest. 6 × 10 m, or 0.6 nm (about the diameter of a benzene molecule) An electron possesses both particle and wave properties. The modern model for the electronic structure of the atom is based on recognizing that an electron possesses particle and wave properties, the so-called . Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle. \[\lambda =\dfrac{h}{mv} \nonumber \] The electron in Bohr’s circular orbits could thus be described as a , one that does not move through space. Standing waves are familiar from music: the lowest-energy standing wave is the vibration, and higher-energy vibrations are and have successively more , points where the amplitude of the wave is always zero. Werner Heisenberg’s states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior. \[ \left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\geqslant \dfrac{h}{4\pi } \nonumber \] ( ) | 15,646 | 2,598 |
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Although some people fall into the trap of visualizing atoms and ions as small, hard spheres similar to miniature table-tennis balls or marbles, the quantum mechanical model tells us that their shapes and boundaries are much less definite than those images suggest. As a result, atoms and ions cannot be said to have exact sizes. In this section, we discuss how atomic and ion “sizes” are defined and obtained. Recall that the probability of finding an electron in the various available orbitals falls off slowly as the distance from the nucleus increases. This point is illustrated in Figure \(\Page {1}\) which shows a plot of total electron density for all occupied orbitals for three noble gases as a function of their distance from the nucleus. Electron density diminishes gradually with increasing distance, which makes it impossible to draw a sharp line marking the boundary of an atom. Figure \(\Page {1}\) also shows that there are distinct peaks in the total electron density at particular distances and that these peaks occur at different distances from the nucleus for each element. Each peak in a given plot corresponds to the electron density in a given principal shell. Because helium has only one filled shell ( = 1), it shows only a single peak. In contrast, neon, with filled = 1 and 2 principal shells, has two peaks. Argon, with filled = 1, 2, and 3 principal shells, has three peaks. The peak for the filled = 1 shell occurs at successively shorter distances for neon ( = 10) and argon ( = 18) because, with a greater number of protons, their nuclei are more positively charged than that of helium. Because the 1 shell is closest to the nucleus, its electrons are very poorly shielded by electrons in filled shells with larger values of . Consequently, the two electrons in the = 1 shell experience nearly the full nuclear charge, resulting in a strong electrostatic interaction between the electrons and the nucleus. The energy of the = 1 shell also decreases tremendously (the filled 1 orbital becomes more stable) as the nuclear charge increases. For similar reasons, the filled = 2 shell in argon is located closer to the nucleus and has a lower energy than the = 2 shell in neon. Figure \(\Page {1}\) illustrates the difficulty of measuring the dimensions of an individual atom. Because distances between the nuclei in pairs of covalently bonded atoms can be measured quite precisely, however, chemists use these distances as a basis for describing the approximate sizes of atoms. For example, the internuclear distance in the diatomic Cl molecule is known to be 198 pm. We assign half of this distance to each chlorine atom, giving chlorine a covalent atomic radius ( ) of 99 pm or 0.99 Å (part (a) in Figure \(\Page {2}\)). In a similar approach, we can use the lengths of carbon–carbon single bonds in organic compounds, which are remarkably uniform at 154 pm, to assign a value of 77 pm as the covalent atomic radius for carbon. If these values do indeed reflect the actual sizes of the atoms, then we should be able to predict the lengths of covalent bonds formed between different elements by adding them. For example, we would predict a carbon–chlorine distance of 77 pm + 99 pm = 176 pm for a C–Cl bond, which is very close to the average value observed in many organochlorine compounds. Covalent atomic radii can be determined for most of the nonmetals, but how do chemists obtain atomic radii for elements that do not form covalent bonds? For these elements, a variety of other methods have been developed. With a metal, for example, the metallic atomic radius (r ) is defined as half the distance between the nuclei of two adjacent metal atoms (part (b) in Figure \(\Page {2}\)). For elements such as the noble gases, most of which form no stable compounds, we can use what is called the van der Waals atomic radius (r ), which is half the internuclear distance between two nonbonded atoms in the solid (part (c) in Figure \(\Page {2}\)). This is somewhat difficult for helium which does not form a solid at any temperature. An atom such as chlorine has both a covalent radius (the distance between the two atoms in a \(Cl_2\) molecule) and a van der Waals radius (the distance between two Cl atoms in different molecules in, for example, \(Cl_{2(s)}\) at low temperatures). These radii are generally not the same (part (d) in Figure \(\Page {2}\)). Because it is impossible to measure the sizes of both metallic and nonmetallic elements using any one method, chemists have developed a self-consistent way of calculating atomic radii using the quantum mechanical functions. Although the radii values obtained by such calculations are not identical to any of the experimentally measured sets of values, they do provide a way to compare the intrinsic sizes of all the elements and clearly show that atomic size varies in a periodic fashion (Figure \(\Page {3}\)). In the periodic table, atomic radii decrease from left to right across a row and increase from top to bottom down a column. Because of these two trends, the largest atoms are found in the lower left corner of the periodic table, and the smallest are found in the upper right corner (Figure \(\Page {4}\)). Atomic radii decrease from left to right across a row and increase from top to bottom down a column. Trends in atomic size result from differences in the experienced by electrons in the outermost orbitals of the elements. For all elements except H, the effective nuclear charge is always than the actual nuclear charge because of shielding effects. The greater the effective nuclear charge, the more strongly the outermost electrons are attracted to the nucleus and the smaller the atomic radius. The atoms in the second row of the periodic table (Li through Ne) illustrate the effect of electron shielding. All have a filled 1 inner shell, but as we go from left to right across the row, the nuclear charge increases from +3 to +10. Although electrons are being added to the 2 and 2 orbitals, . Thus the single 2 electron in lithium experiences an effective nuclear charge of approximately +1 because the electrons in the filled 1 shell effectively neutralize two of the three positive charges in the nucleus. (More detailed calculations give a value of = +1.26 for Li.) In contrast, the two 2 electrons in beryllium do not shield each other very well, although the filled 1 shell effectively neutralizes two of the four positive charges in the nucleus. This means that the effective nuclear charge experienced by the 2 electrons in beryllium is between +1 and +2 (the calculated value is +1.66). Consequently, beryllium is significantly smaller than lithium. Similarly, as we proceed across the row, the increasing nuclear charge is not effectively neutralized by the electrons being added to the 2 and 2 orbitals. The result is a steady increase in the effective nuclear charge and a steady decrease in atomic size. The increase in atomic size going down a column is also due to electron shielding, but the situation is more complex because the principal quantum number is not constant. As we saw in Chapter 2, the size of the orbitals increases as increases, . In group 1, for example, the size of the atoms increases substantially going down the column. It may at first seem reasonable to attribute this effect to the successive addition of electrons to orbitals with increasing values of . However, it is important to remember that the radius of an orbital depends dramatically on the nuclear charge. As we go down the column of the group 1 elements, the principal quantum number increases from 2 to 6, but the nuclear charge increases from +3 to +55! As a consequence the radii of the in Cesium are much smaller than those in lithium and the electrons in those orbitals experience a much larger force of attraction to the nucleus. That force depends on the effective nuclear charge experienced by the the inner electrons. If the outermost electrons in cesium experienced the full nuclear charge of +55, a cesium atom would be very small indeed. In fact, the effective nuclear charge felt by the outermost electrons in cesium is much less than expected (6 rather than 55). This means that cesium, with a 6 valence electron configuration, is much larger than lithium, with a 2 valence electron configuration. The effective nuclear charge changes relatively little for electrons in the outermost, or valence shell, from lithium to cesium because . Even though cesium has a nuclear charge of +55, it has 54 electrons in its filled 1 2 2 3 3 4 3 4 5 4 5 shells, abbreviated as [Xe]5 4 5 , which effectively neutralize most of the 55 positive charges in the nucleus. The same dynamic is responsible for the steady increase in size observed as we go down the other columns of the periodic table. Irregularities can usually be explained by variations in effective nuclear charge. Electrons in the same principal shell are not very effective at shielding one another from the nuclear charge, whereas electrons in filled inner shells are highly effective at shielding electrons in outer shells from the nuclear charge. On the basis of their positions in the periodic table, arrange these elements in order of increasing atomic radius: aluminum, carbon, and silicon. three elements arrange in order of increasing atomic radius These elements are not all in the same column or row, so we must use pairwise comparisons. Carbon and silicon are both in group 14 with carbon lying above, so carbon is smaller than silicon (C < Si). Aluminum and silicon are both in the third row with aluminum lying to the left, so silicon is smaller than aluminum (Si < Al) because its effective nuclear charge is greater. Combining the two inequalities gives the overall order: C < Si < Al. On the basis of their positions in the periodic table, arrange these elements in order of increasing size: oxygen, phosphorus, potassium, and sulfur. O < S < P < K Atomic Radius: An ion is formed when either one or more electrons are removed from a neutral atom (cations) to form a positive ion or when additional electrons attach themselves to neutral atoms (anions) to form a negative one. The designations cation or anion come from the early experiments with electricity which found that positively charged particles were attracted to the negative pole of a battery, the cathode, while negatively charged ones were attracted to the positive pole, the anode. Ionic compounds consist of regular repeating arrays of alternating positively charged cations and negatively charges anions. Although it is not possible to measure an ionic radius directly for the same reason it is not possible to directly measure an atom’s radius, it possible to measure the distance between the nuclei of a cation and an adjacent anion in an ionic compound to determine the ionic radius (the radius of a cation or anion) of one or both. As illustrated in Figure \(\Page {6}\) , the internuclear distance corresponds to the of the radii of the cation and anion. A variety of methods have been developed to divide the experimentally measured distance proportionally between the smaller cation and larger anion. These methods produce sets of ionic radii that are internally consistent from one ionic compound to another, although each method gives slightly different values. For example, the radius of the Na ion is essentially the same in NaCl and Na S, as long as the same method is used to measure it. Thus despite minor differences due to methodology, certain trends can be observed. A comparison of ionic radii with atomic radii (Figure \(\Page {7}\)) . When one or more electrons is removed from a neutral atom, two things happen: (1) repulsions between electrons in the same principal shell decrease because fewer electrons are present, and (2) the effective nuclear charge felt by the remaining electrons increases because there are fewer electrons to shield one another from the nucleus. Consequently, the size of the region of space occupied by electrons decreases (compare Li at 167 pm with Li at 76 pm). If different numbers of electrons can be removed to produce ions with different charges, the ion with the greatest positive charge is the smallest (compare Fe at 78 pm with Fe at 64.5 pm). Conversely, adding one or more electrons to a neutral atom causes electron–electron repulsions to increase and the effective nuclear charge to decrease, so the size of the probability region increases (compare F at 42 pm with F at 133 pm). Cations are smaller than the neutral atom, and anions are larger. Because most elements form either a cation or an anion but not both, there are few opportunities to compare the sizes of a cation and an anion derived from the same neutral atom. A few compounds of sodium, however, contain the Na ion, allowing comparison of its size with that of the far more familiar Na ion, which is found in compounds. The radius of sodium in each of its three known oxidation states is given in Table \(\Page {1}\). All three species have a nuclear charge of +11, but they contain 10 (Na ), 11 (Na ), and 12 (Na ) electrons. The Na ion is significantly smaller than the neutral Na atom because the 3 electron has been removed to give a closed shell with = 2. The Na ion is larger than the parent Na atom because the additional electron produces a 3 valence electron configuration, while the nuclear charge remains the same. Ionic radii follow the same vertical trend as atomic radii; that is, for ions with the same charge, the ionic radius increases going down a column. The reason is the same as for atomic radii: shielding by filled inner shells produces little change in the effective nuclear charge felt by the outermost electrons. Again, principal shells with larger values of lie at successively greater distances from the nucleus. Because elements in different columns tend to form ions with different charges, it is not possible to compare ions of the same charge across a row of the periodic table. Instead, elements that are next to each other tend to form ions with the same number of electrons but with different overall charges because of their different atomic numbers. Such a set of species is known as an isoelectronic series. For example, the isoelectronic series of species with the neon closed-shell configuration (1 2 2 ) is shown in Table 7.3 The sizes of the ions in this series decrease smoothly from N to Al . All six of the ions contain 10 electrons in the 1 , 2 , and 2 orbitals, but the nuclear charge varies from +7 (N) to +13 (Al). As the positive charge of the nucleus increases while the number of electrons remains the same, there is a greater electrostatic attraction between the electrons and the nucleus, which causes a decrease in radius. Consequently, the ion with the greatest nuclear charge (Al ) is the smallest, and the ion with the smallest nuclear charge (N ) is the largest. The neon atom in this isoelectronic series is not listed in Table \(\Page {3}\), because neon forms no covalent or ionic compounds and hence its radius is difficult to measure. Table Radius of Ions with the Neon Closed-Shell Electron Configuration. R. D. Shannon, “Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides,” Acta Crystallographica 32, no. 5 (1976): 751–767. Based on their positions in the periodic table, arrange these ions in order of increasing radius: Cl , K , S , and Se . four ions order by increasing radius We see that S and Cl are at the right of the third row, while K and Se are at the far left and right ends of the fourth row, respectively. K , Cl , and S form an isoelectronic series with the [Ar] closed-shell electron configuration; that is, all three ions contain 18 electrons but have different nuclear charges. Because K has the greatest nuclear charge ( = 19), its radius is smallest, and S with = 16 has the largest radius. Because selenium is directly below sulfur, we expect the Se ion to be even larger than S . The order must therefore be K < Cl < S < Se . Based on their positions in the periodic table, arrange these ions in order of increasing size: Br , Ca , Rb , and Sr . Ca < Sr < Rb < Br A variety of methods have been established to measure the size of a single atom or ion. The is half the internuclear distance in a molecule with two identical atoms bonded to each other, whereas the is defined as half the distance between the nuclei of two adjacent atoms in a metallic element. The of an element is half the internuclear distance between two nonbonded atoms in a solid. Atomic radii decrease from left to right across a row because of the increase in effective nuclear charge due to poor electron screening by other electrons in the same principal shell. Moreover, atomic radii increase from top to bottom down a column because the effective nuclear charge remains relatively constant as the principal quantum number increases. The of cations and anions are always smaller or larger, respectively, than the parent atom due to changes in electron–electron repulsions, and the trends in ionic radius parallel those in atomic size. A comparison of the dimensions of atoms or ions that have the same number of electrons but different nuclear charges, called an , shows a clear correlation between increasing nuclear charge and decreasing size. ( ) | 17,496 | 2,599 |
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We now turn from an overview of periodic trends to a discussion of the s-block elements, first by focusing on hydrogen, whose chemistry is sufficiently distinct and important to be discussed in a category of its own. Most versions of the periodic table place hydrogen in the upper left corner immediately above lithium, implying that hydrogen, with a 1s electron configuration, is a member of group 1. In fact, the chemistry of hydrogen does not greatly resemble that of the metals of Group 1. Indeed, some versions of the periodic table place hydrogen above fluorine in because the addition of a single electron to a hydrogen atom completes its valence shell. Although hydrogen has an ns electron configuration, its chemistry does not resemble that of the . Hydrogen, the most abundant element in the universe, is the ultimate source of all other elements by the process of nuclear fusion. Table \(\Page {1}\) "The Isotopes of Hydrogen" compares the three isotopes of hydrogen, all of which contain one proton and one electron per atom. The most common isotope is ( H or H), followed by ( H or D), which has an additional neutron. The rarest isotope of hydrogen is ( H or T), which is produced in the upper atmosphere by a nuclear reaction when cosmic rays strike nitrogen and other atoms; it is then washed into the oceans by rainfall. Tritium is radioactive, decaying to He with a half-life of only 12.32 years. Consequently, the atmosphere and oceans contain only a very low, steady-state level of tritium. The term hydrogen and the symbol H normally refer to the naturally occurring mixture of the three isotopes. The different masses of the three isotopes of hydrogen cause them to have different physical properties. Thus H , D , and T differ in their melting points, boiling points, densities, and heats of fusion and vaporization. In 1931, Harold Urey and coworkers discovered deuterium by slowly evaporating several liters of liquid hydrogen until a volume of about 1 mL remained. When that remaining liquid was vaporized and its emission spectrum examined, they observed new absorption lines in addition to those previously identified as originating from hydrogen. The natural abundance of tritium, in contrast, is so low that it could not be detected by similar experiments; it was first prepared in 1934 by a nuclear reaction. Urey won the Nobel Prize in Chemistry in 1934 for his discovery of deuterium ( H). Urey was born and educated in rural Indiana. After earning a BS in zoology from the University of Montana in 1917, Urey changed career directions. He earned his PhD in chemistry at Berkeley with G. N. Lewis and subsequently worked with Niels Bohr in Copenhagen. During World War II, Urey was the director of war research for the Atom Bomb Project at Columbia University. In later years, his research focused on the evolution of life. In 1953, he and his graduate student, Stanley Miller, showed that organic compounds, including amino acids, could be formed by passing an electric discharge through a mixture of compounds thought to be present in the atmosphere of primitive Earth. Because the normal boiling point of D O is 101.4°C (compared to 100.0°C for H O), evaporation or fractional distillation can be used to increase the concentration of deuterium in a sample of water by the selective removal of the more volatile H O. Thus bodies of water that have no outlet, such as the Great Salt Lake and the Dead Sea, which maintain their level solely by evaporation, have significantly higher concentrations of deuterated water than does lake or seawater with at least one outlet. A more efficient way to obtain water highly enriched in deuterium is by prolonged electrolysis of an aqueous solution. Because a deuteron (D ) has twice the mass of a proton (H ), it diffuses more slowly toward the electrode surface. Consequently, the gas evolved at the cathode is enriched in H, the species that diffuses more rapidly, favoring the formation of H over D or HD. Meanwhile, the solution becomes enriched in deuterium. Deuterium-rich water is called heavy water because the density of D O (1.1044 g/cm at 25°C) is greater than that of H O (0.99978 g/cm ). Heavy water was an important constituent of early nuclear reactors. Because deuterons diffuse so much more slowly, D O will not support life and is actually toxic if administered to mammals in large amounts. The rate-limiting step in many important reactions catalyzed by enzymes involves proton transfer. The transfer of D is so slow compared with that of H because bonds to D break more slowly than those to H, so the delicate balance of reactions in the cell is disrupted. Nonetheless, deuterium and tritium are important research tools for biochemists. By incorporating these isotopes into specific positions in selected molecules, where they act as labels, or tracers, biochemists can follow the path of a molecule through an organism or a cell. Tracers can also be used to provide information about the mechanism of enzymatic reactions. The 1s electron configuration of hydrogen indicates a single valence electron. Because the 1s orbital has a maximum capacity of two electrons, hydrogen can form compounds with other elements in three ways (Figure \(\Page {1}\)): Hydrogen can also act as a bridge between two atoms. One familiar example is the , an electrostatic interaction between a hydrogen bonded to an electronegative atom and an atom that has one or more lone pairs of electrons (Figure \(\Page {2}\)). An example of this kind of interaction is the hydrogen bonding network found in water (Figure \(\Page {2}\)). Hydrogen can also form a , in which a hydride bridges two electropositive atoms. Compounds that contain hydrogen bonded to boron and similar elements often have this type of bonding. The B–H–B units found in boron hydrides cannot be described in terms of localized electron-pair bonds. Because the H atom in the middle of such a unit can accommodate a maximum of only two electrons in its 1s orbital, the B–H–B unit can be described as containing a hydride that interacts simultaneously with empty sp orbitals on two boron atoms (Figure \(\Page {3}\)). In these bonds, only two bonding electrons are used to hold three atoms together, making them electron-deficient bonds. You encountered a similar phenomenon in the discussion of π bonding in ozone and the nitrite ion. Recall that in both these cases, we used the presence of two electrons in a π molecular orbital extending over three atoms to explain the fact that the two O–O bond distances in ozone and the two N–O bond distances in nitrite are the same, which otherwise can be explained only by the use of resonance structures. Hydrogen can lose its electron to form H , accept an electron to form H , share its electron, hydrogen bond, or form a three-center bond. The first known preparation of elemental hydrogen was in 1671, when Robert Boyle dissolved iron in dilute acid and obtained a colorless, odorless, gaseous product. Hydrogen was finally identified as an element in 1766, when Henry Cavendish showed that water was the sole product of the reaction of the gas with oxygen. The explosive properties of mixtures of hydrogen with air were not discovered until early in the 18th century; they partially caused the spectacular explosion of the hydrogen-filled dirigible Hindenburg in 1937 (Figure \(\Page {4}\)). Due to its extremely low molecular mass, hydrogen gas is difficult to condense to a liquid (boiling point = 20.3 K), and solid hydrogen has one of the lowest melting points known (13.8 K). The most common way to produce small amounts of highly pure hydrogen gas in the laboratory was discovered by Boyle: reacting an active metal (M), such as iron, magnesium, or zinc, with dilute acid: \[M_{(s)} + 2H^+_{(aq)} \rightarrow H_{2(g)} + M^{2+}_{(aq)} \label{21.1}\] Hydrogen gas can also be generated by reacting metals such as aluminum or zinc with a strong base: \[\mathrm{Al(s)}+\mathrm{OH^-(aq)}+\mathrm{3H_2O(l)}\rightarrow\frac{3}{2}\mathrm{H_2(g)}+\mathrm{[Al(OH)_4]^-(aq)} \label{21.2}\] Solid commercial drain cleaners such as Drano use this reaction to generate gas bubbles that help break up clogs in a drainpipe. Hydrogen gas is also produced by reacting ionic hydrides with water. Because ionic hydrides are expensive, however, this reaction is generally used for only specialized purposes, such as producing HD gas by reacting a hydride with D O: \[MH_{(s)} + D_2O(l) \rightarrow HD_{(g)} + M^+(aq) + OD^−_{(aq)} \label{21.3}\] On an industrial scale, H is produced from methane by means of catalytic steam reforming, a method used to convert hydrocarbons to a mixture of CO and H known as synthesis gas, or syngas. The process is carried out at elevated temperatures (800°C) in the presence of a nickel catalyst: \[\mathrm{CH_4(g)}+\mathrm{H_2O(g)}\xrightarrow{\mathrm{Ni}}\mathrm{CO(g)}+\mathrm{3H_2(g)} \label{21.4}\] Most of the elements in the periodic table form binary compounds with hydrogen, which are collectively referred to as hydrides. Binary hydrides in turn can be classified in one of three ways, each with its own characteristic properties. Covalent hydrides contain hydrogen bonded to another atom via a covalent bond or a polar covalent bond. Covalent hydrides are usually molecular substances that are relatively volatile and have low melting points. Ionic hydrides contain the hydride ion as the anion with cations derived from electropositive metals. Like most ionic compounds, they are typically nonvolatile solids that contain three-dimensional lattices of cations and anions. Unlike most ionic compounds, however, they often decompose to H (g) and the parent metal after heating. Metallic hydrides are formed by hydrogen and less electropositive metals such as the transition metals. The properties of metallic hydrides are usually similar to those of the parent metal. Consequently, metallic hydrides are best viewed as metals that contain many hydrogen atoms present as interstitial impurities. Covalent hydrides are relatively volatile and have low melting points; ionic hydrides are generally nonvolatile solids in a lattice framework. Hydrogen can lose an electron to form a proton, gain an electron to form a hydride ion, or form a covalent bond or polar covalent electron-pair bond. The three isotopes of hydrogen—protium ( H or H), deuterium ( H or D), and tritium ( H or T)—have different physical properties. Deuterium and tritium can be used as tracers, substances that enable biochemists to follow the path of a molecule through an organism or a cell. Hydrogen can form compounds that contain a proton (H ), a hydride ion (H ), an electron-pair bond to H, a hydrogen bond, or a three-center bond (or electron-deficient bond), in which two electrons are shared between three atoms. Hydrogen gas can be generated by reacting an active metal with dilute acid, reacting Al or Zn with a strong base, or industrially by catalytic steam reforming, which produces synthesis gas, or syngas. | 11,052 | 2,600 |
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Transition metal complexes which undergo rapid substitution of one ligand for another are , whereas complexes in which substitution proceed slowly or not at all are . For an inert complex, it is a large activation energy which prevents ligand substitution. Inert complexes are therefore compounds. As a classic example, the substitution reaction: \[[Co(NH_3)_6]^{3+}_{(aq)} + 6H_3O^+_{(aq)} \rightarrow [Co(H_2O)_6]^{3+}_{(aq)} + 6NH^+_{4(aq)} \label{Eq1}\] has an equilibrium constant of \(10^{64}\). The large equilibrium constant suggests that the complex ion \([Co(NH_3)_6]^{3+}\) is . This reaction is highly thermodynamically favored, yet the inert \([Co(NH_3)_6]^{3+}\) complex ion lasts for weeks in acidic solutions because the rate of the reaction is very slow. Thus, the large activation energy acts as an efficient barrier for ligand substitution rendering the \([Co(NH_3)_6]^{3+}\) ion . Although diamond is thermodynamically less stable than graphite, diamonds last a long time (~100 million years) since the conversion of diamond into graphite occurs extremely slowly. Here is an example of a complex (notice difference in oxidation state for \(Co\) in Equation \(\ref{Eq1}\): \[[Co(NH_3)_6]^{2+}_{(aq)} + 6H_3O^+_{(aq)} \rightarrow [Co(H_2O)_6]^{2+}_{(aq)} + 6NH^+_{4(aq)}\] This reaction is virtually complete in a few seconds. The \([Co(NH_3)_6]^{2+}\) complex is thermodynamically unstable and also labile. Notice that the only difference between the two ammine cobalt complexes shown in the examples above is the oxidation number of the cobalt atom. The inert complex has Co(III) while the labile one has Co(II). Both ammine complexes are octahedral and in the case of Co(III), a d species, the t levels are filled. Co(II), on the other hand, has partially filled e orbitals. It is straightforward to demonstrate lability since changes will occur and ligand substitution in complexes is normally accompanied by a color change. Inertness, however, is somewhat dull since nothing happens. In this experiment the inertness of ligand substitution by chromium (III) ions is compared with other reactions that do proceed at reasonably fast rates. Henry Taube (Nobel Prize, 1983) tried to understand lability by comparing the factors that govern bond strengths in ionic complexes to observations about the rates of reaction of coordination complexes. He saw some things that were unsurprising. Taube observed that many M ions (M = metal) are more labile than many M ions, in general. That is not too surprising, since metal ions function Lewis acids and ligands function Lewis bases in forming coordination complexes. In other words, metals with higher charges ought to be stronger Lewis acids, and so they should bind ligands more tightly. However, there were exceptions to that general rule. For example, Taube also observed that Mo(V) compounds are more labile than Mo(III) compounds. That means there is more going on here than just charge effects. Another factor that governs ionic bond strengths is the size of the ion. Typically, ions with smaller atomic radii form stronger bonds than ions with larger radii. Taube observed that Al , V , Fe and Ga ions are all about the same size. All these ions exchange ligands at about the same rate. However, other factors are in play that are outside the scope of this Module. In which compound from each pair would you expect the strongest ionic bonds? Why? | 3,452 | 2,601 |
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Textmap \[\displaystyle\textit{mass% of component} = \frac{\textit{mass of component in soln}}{\textit{total mass of soln}}\times 100 \nonumber \] \[\displaystyle\textit{ppm of component} = \frac{\textit{mass of component in soln}}{\textit{total mass of soln}}\times 10^6 \nonumber \] \[\displaystyle\textit{mole fraction of component} = \frac{\textit{moles of component}}{\textit{total moles of all components}} \nonumber \] Colligative properties are physical properties that depend on quantity Colloidal dispersions (colloids) are intermediate types of dispersions or suspensions | 602 | 2,602 |
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An analysis requires a sample and how we acquire that sample is critical. The samples we collect must accurately represent their target population, and our sampling plan must provide a sufficient number of samples of appropriate size so that uncertainty in sampling does not limit the precision of our analysis. A complete sampling plan requires several considerations, including the type of sample to collect (random, judgmental, systematic, systematic–judgmental, stratified, or convenience); whether to collect grab samples, composite samples, or in situ samples; whether the population is homogeneous or heterogeneous; the appropriate size for each sample; and the number of samples to collect. Removing a sample from its population may induce a change in its composition due to a chemical or physical process. For this reason, we collect samples in inert containers and we often preserve them at the time of collection. When an analytical method’s selectivity is insufficient, we may need to separate the analyte from potential interferents. Such separations take advantage of physical properties—such as size, mass or density—or chemical properties. Important examples of chemical separations include masking, distillation, and extractions. centrifugation convenience sampling distillation extraction efficiency grab sample homogeneous laboratory sample Nyquist theorem purge-and-trap recrystallization secondary equilibrium reaction size exclusion chromatography sublimation systematic–judgmental sampling composite sample density gradient centrifugation distribution ratio filtrate gross sample
in situ sampling masking
partition coefficient random sampling retentate selectivity coefficient Soxhlet extractor subsamples systematic sampling coning and quartering dialysis extraction
filtration heterogeneous judgmental sampling masking agents preconcentration recovery sampling plan separation factor stratified sampling supercritical fluid target population | 2,017 | 2,603 |
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Anabolic reactions are those that lead to the synthesis of biomolecules. In contrast to the catabolic reactions just discussed (glycolysis, TCA cycle and electron transport/oxidative phosphorylation) which lead to the oxidative degradation of carbohydrates and fatty acids and energy release, anabolic reactions lead to the synthesis of more complex biomolecules including biopolymers (glycogen, proteins, nucleic acids) and complex lipids. Many biosynthetic reactions, including those for fatty acid synthesis, are reductive and hence require reducing agents. Reductive biosynthesis and complex polymer formation require energy input, usually in the form of ATP whose exergonic cleavage is coupled to endergonic biosynthesis. Cells have evolved interesting mechanism so as not to have oxidative degradation reactions (which release energy) proceed at the same time and in the same cell as reductive biosynthesis (which requires energy input). Consider this scenario. You dive into a liver cell and find palmitic acid, a 16C fatty acid. From where did it come? Was it just synthesized by the liver cell or did it just enter the cell from a distant location such as adipocytes (fat cells). Should it be oxidized, which should happen if there is a demand for energy production by the cell, or should the liver cell export it, perhaps to adipocytes, which might happen if there is an excess of energy storage molecules? Cells have devised many ways to distinguish these opposing needs. One is by using a slightly different pool of redox reagents for anabolic and catabolic reactions. Oxidative degradation reactions typically use the redox pair NAD /NADH (or FAD/FADH ) while reductive biosynthesis often uses phosphorylated variants of NAD , NADP /NADPH. In addition, cells often carry out competing reactions in different cellular compartments. Fatty acid oxidation of our example molecule (palmitic acid) occurs in the mitochondrial matrix, while reductive fatty acid synthesis occurs in the cytoplasm of the cell. Fatty acids entering the cell destined for oxidative degradation are transported into the mitochondria by the carnitine transport system. This transport system is inhibited under conditions when fatty acid synthesis is favored. We will discuss the regulation of metabolic pathways in a subsequent section. One of the main methods, as we will see, is to activate or inhibit key enzymes in the pathways under a given set of cellular conditions. The key enzyme in fatty acid synthesis, acetyl-CoA carboxylase, is inhibited when cellular conditions require fatty acid oxidation. The following examples give short descriptions of anabolic pathways. Compare them to the catabolic pathways from the previous section. Now its time to see how the various pathways fit together to form an integrated set of pathways. , | 2,853 | 2,604 |
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Chromatography and electrophoresis are powerful analytical techniques that both separate a sample into its components and provide a means for determining each component’s concentration. Chromatographic separations utilize the selective partitioning of the sample’s components between a stationary phase that is immobilized within a column and a mobile phase that passes through the column. The effectiveness of a chromatographic separation is described by the resolution between two chromatographic bands and is a function of each component’s retention factor, the column’s efficiency, and the column’s selectivity. A solute’s retention factor is a measure of its partitioning into the stationary phase, with larger retention factors corresponding to more strongly retained solutes. The column’s selectivity for two solutes is the ratio of their retention factors, providing a relative measure of the column’s ability to retain the two solutes. Column efficiency accounts for those factors that cause a solute’s chromatographic band to increase in width during the separation. Column efficiency is defined in terms of the number of theoretical plates and the height of a theoretical plate, the later of which is a function of a number of parameters, most notably the mobile phase’s flow rate. Chromatographic separations are optimized by increasing the number of theoretical plates, by increasing the column’s selectivity, or by increasing the solute retention factor. In gas chromatography the mobile phase is an inert gas and the stationary phase is a nonpolar or polar organic liquid that either is coated on a particulate material and packed into a wide-bore column, or coated on the walls of a narrow-bore capillary column. Gas chromatography is useful for the analysis of volatile components. In high-performance liquid chromatography the mobile phase is either a nonpolar solvent (normal phase) or a polar solvent (reversed-phase). A stationary phase of opposite polarity, which is bonded to a particulate material, is packed into a wide-bore column. HPLC is applied to a wider range of samples than GC; however, the separation efficiency for HPLC is not as good as that for capillary GC. Together, GC and HPLC account for the largest number of chromatographic separations. Other separation techniques, however, find special- ized applications: of particular importance are ion-exchange chromatography for separating anions and cations; size-exclusion chromatography for separating large molecules; and supercritical fluid chromatography for the analysis of samples that are not easily analyzed by GC or HPLC. In capillary zone electrophoresis a sample’s components are separated based on their ability to move through a conductive medium under the influence of an applied electric field. Positively charged solutes elute first, with smaller, more highly charged cations eluting before larger cations of lower charge. Neutral species elute without undergoing further separation. Finally, anions elute last, with smaller, more negatively charged anions being the last to elute. By adding a surfactant, neutral species can be separated by micellar electrokinetic capillary chromatography. Electrophoretic separations also can take advantage of the ability of polymeric gels to separate solutes by size (capillary gel electrophoresis), and the ability of solutes to partition into a stationary phase (capillary electrochromatography). In comparison to GC and HPLC, capillary electrophoresis provides faster and more efficient separations. adjusted retention time baseline width capillary column capillary gel electrophoresis chromatography cryogenic focusing electroosmotic flow velocity electrophoresis exclusion limit
gas chromatography general elution problem headspace sampling inclusion limit isocratic elution Kovat’s retention index loop injector mass transfer mobile phase nonretained solutes open tubular column peak capacity porous-layer open tubular column retention factor selectivity factor split injection stationary phase tailing thermal conductivity detector wall-coated open-tubular column adsorption chromatography bleed capillary electrochromatography capillary zone electrophoresis column chromatography electrokinetic injection electron capture detector electrophoretic mobility flame ionization detector gas–liquid chromatography guard column high-performance liquid chromatography ion-exchange chromatography isothermal liquid–solid adsorption chromatography mass spectrometer micelle monolithic column normal-phase chromatography packed columns planar chromatography purge-and-trap retention time single-column ion chromatography splitless injection supercritical fluid chromatography temperature programming van Deemter equation zeta potential band broadening bonded stationary phase capillary electrophoresis chromatogram counter-current extraction electroosmotic flow electropherogram electrophoretic velocity fronting
gas–solid chromatography gradient elution hydrodynamic injection ion suppressor column Joule heating longitudinal diffusion mass spectrum micellar electrokinetic capillary chromatography multiple paths on-column injection partition chromatography polarity index resolution reversed-phase chromatography solid-phase microextraction stacking support-coated open tubular column theoretical plate void time | 5,390 | 2,605 |
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There are several different ways to quantitatively describe the concentration of a solution. For example, molarity is a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example \(\Page {1}\) reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known. Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL. : mass of substance and mass and density of solution : molarity and mole fraction : : The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass. \[ \begin{align*} \text{moles } \ce{CH_3CO_2H} &=\dfrac{3.78\; \cancel{\ce{g}}\; \ce{CH_3CO_2H}}{60.05\; \cancel{\ce{g}}/\ce{mol}} \\[4pt] &=0.0629 \; \ce{mol} \end{align*} \nonumber \] The volume of the solution equals its mass divided by its density. \[ \begin{align*} \text{volume} &=\dfrac{\text{mass}}{\text{density}} \\[4pt] &=\dfrac{100.0\; \cancel{\ce{g}}\; \text{solution}}{1.00\; \cancel{\ce{g}}/\ce{mL}}=100\; mL\nonumber \end{align*} \nonumber \] Then calculate the molarity directly. \[ \begin{align*} \text{molarity of } \ce{CH_3CO_2H} &=\dfrac{\text{moles } \ce{CH3CO2H} }{\text{liter solution}} \\[4pt] &=\dfrac{0.0629\; mol\; \ce{CH_3CO_2H}}{(100\; \cancel{\ce{mL}})(1\; L/1000\; \cancel{\ce{mL}})}=0.629\; M \; \ce{CH_3CO_2H} \end{align*} \nonumber \] This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than \(\ce{ 1/2}\) mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL. To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have \[moles\; \ce{H_2O}=\dfrac{96.2\; \cancel{\ce{g}}\; \ce{H_2O}}{18.02\; \cancel{\ce{g}}/mol}=5.34\; mol\; \ce{H_2O}\nonumber \] The mole fraction \(\chi\) of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present: \[ \begin{align*} \chi_{\ce{CH3CO2H}} &=\dfrac{moles\; \ce{CH_3CO_2H}}{moles \; \ce{CH_3CO_2H} + moles\; \ce{H_2O}} \\[4pt] &=\dfrac{0.0629\; mol}{0.0629 \;mol + 5.34\; mol} \\[4pt] &=0.0116=1.16 \times 10^{−2} \end{align*} \nonumber \] This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01. A solution of \(\ce{HCl}\) gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of \(\ce{HCl}\) per 100.0 g of solution, and its density is 1.10 g/mL. 6.10 M HCl \(\chi_{HCl} = 0.111\) The concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent: \[ \text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \label{Eq1} \] Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of \(H_2O\) under these conditions is very close to 1.0 L, and a 0.50 M solution of \(KBr\) in water, for example, has approximately the same concentration as a 0.50 m solution. Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb): \[ \begin{align} \text{mass percentage}&=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \label{Eq2} \\[4pt] \text{parts per million (ppm)} &=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \label{Eq3} \\[4pt] \text{parts per billion (ppb)}&=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \label{Eq4} \end{align} \] In the health sciences, the concentration of a solution is often expressed as parts per thousand (ppt), indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution. The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of \(\ce{H_2SO_4}\) per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL). Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL. : volume of sample, solute concentration, and density of solution : molarity of solute and mass of solute in 250 mL : : a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore \[\begin{align*} \text{molarity}&=\dfrac{\text{moles}}{\text{liter solution}} \\[4pt] &=\dfrac{(12.7\; \cancel{mg}) \left(\frac{1\; \cancel{g}}{1000\; \cancel{mg}}\right)\left(\frac{1\; mol}{78.114\; \cancel{g}}\right)}{1.00\; L} \\[4pt] &=1.63 \times 10^{-4} M\end{align*} \nonumber \] b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is \[\begin{align*} \text{mass of benzene} &=\dfrac{(12.7\; mg\; \text{benzene})(250\; \cancel{mL})}{1000\; \cancel{mL}} \\[4pt] &=3.18\; mg \\[4pt] &=3.18 \times 10^{-3}\; g\; \text{benzene} \end{align*} \nonumber \] The maximum allowable concentration of lead in drinking water is 9.0 ppb. 4.3 × 10 M 2 × 10 g How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature. Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as discussed previously. Mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies. Table \(\Page {1}\) summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example \(\Page {3}\). Vodka is essentially a solution of ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka. : volume percent and density : mass percentage, mole fraction, molarity, and molality : : The key to this problem is to use the density of pure ethanol to determine the mass of ethanol (\(CH_3CH_2OH\)), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units. A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density: \[mass\; of\; EtOH=(40.0\; \cancel{mL})\left(\dfrac{0.789\; g}{\cancel{mL}}\right)=31.6\; g\; EtOH\nonumber \] If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution. B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage: \[ \begin{align*} \%EtOH &=\left(\dfrac{mass\; of\; EtOH}{mass\; of\; solution}\right)(100) \\[4pt] &=\left(\dfrac{31.6\; \cancel{g}\; EtOH}{31.6\; \cancel{g} \;EtOH +60.0\; \cancel{g} \; H_2O} \right)(100) \\[4pt]&= 34.5\%\end{align*} \nonumber \] C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL: \[ \begin{align*} moles\; \ce{CH_3CH_2OH}&=(31.6\; \cancel{g\; \ce{CH_3CH_2OH}}) \left(\dfrac{1\; mol}{46.07\; \cancel{g\; \ce{CH_3CH_2OH}}}\right) \\[4pt] &=0.686 \;mol\; \ce{CH_3CH_2OH} \end{align*} \nonumber \] Similarly, the number of moles of water is \[ moles \;\ce{H_2O}=(60.0\; \cancel{g \; \ce{H_2O}}) \left(\dfrac{1 \;mol\; \ce{H_2O}}{18.02\; \cancel{g\; \ce{H_2O}}}\right)=3.33\; mol\; \ce{H_2O}\nonumber \] The mole fraction of ethanol is thus \[ \chi_{\ce{CH_3CH_2OH}}=\dfrac{0.686\; \cancel{mol}}{0.686\; \cancel{mol} + 3.33\;\cancel{ mol}}=0.171\nonumber \] D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward: \[ m_{\ce{CH_3CH_2OH}}=\left(\dfrac{0.686\; mol\; EtOH}{60.0\; \cancel{g}\; H_2O } \right) \left(\dfrac{1000\; \cancel{g}}{kg}\right)=\dfrac{11.4\; mol\; EtOH}{kg\; H_2O}=11.4 \;m\nonumber \] A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene. mass percentage toluene = 24.8% \(\chi_{toluene} = 0.219\) 2.35 M toluene 3.59 m toluene A Discussing Different Measures of Concentration. Link: A Discussing how to Convert Measures of Concentration. Link: Different units are used to express the concentrations of a solution depending on the application. The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb), micrograms of solute per kilogram of solution; and molality (m), the number of moles of solute per kilogram of solvent. | 13,796 | 2,607 |
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Photochemical smog is a type of air pollution due to the reaction of solar radiation with airborne pollutant mixtures of nitrogen oxides (NOx) and volatile organic compounds (hydrocarbons). Smog is a byproduct of modern industrialization. Due to industry and the number of motor vehicles, this is more of a problem in large cities that have a warm, sunny and dry climate. The different reactions involved in the formation of photochemical smog are given below. People begin driving in the morning, nitrogen is burned or oxidized \[N_2 + O_2 \rightarrow 2NO \nonumber \] After a few hours, NO combines with O , in another oxidation reaction \[2NO + O_2 \rightarrow 2NO_2 \nonumber \] Nitrogen dioxide absorbs light energy, resulting in a reduction reaction \[NO_2 \rightarrow NO + O \nonumber \] In sunlight, atomic oxygen combines with oxygen gas to form ozone \[O + O_2 \rightarrow O_3 \nonumber \] Reaction is temperature and sunlight dependent \[O_3 + NO \rightleftharpoons NO_2 + O_2 \nonumber \] NO and NO can also react with the hydrocarbons instead of ozone to form other volatile compounds known as PAN (peroxyacetyl nitrate) as shown in Figure . The accumulation of ozone and volatile organic compounds along with the energy from the sun forms the brown, photochemical smog seen on hot, sunny days. Panoramic view of Santiago covered by a layer of smog on May 10, 2006. The Metropolitan Region of Santiago facing the driest autumn last 28 years due to lack of rainfall, which coupled with poor air circulation, causes an increase in smog. Because ozone is highly reactive, it has the ability to oxidize and destroy lung tissue. Short term exposures to elevated levels of ozone (above .75 ppm) have been linked to a host of respiratory irritations including coughing, wheezing, substernal soreness, pharyngitis, and dyspnea. Prolonged exposure to the molecule has been proven to cause a permanent reduction in lung function, as well as elevate the risk of developing asthma. Sulfur dioxide is a common component of London smog. Epidemiological studies have linked short term sulfur dioxide exposure to respiratory irritations including coughing, wheezing, and pharyngitis. Plants are harmed by exposure to nitrogen oxides, ozone, and peroxyacetyl nitrate (PAN, see above), all oxidants present in a smoggy atmosphere. PAN is the most harmful of these constituents, damaging younger plant leaves, especially. Ozone exposure causes formation of yellow spots on leaves, a condition called chlorotic stippling. Some plant species, including sword-leaf lettuce, black nightshade, quickweed, and double-fortune tomato, are extremely susceptible to damage by oxidant species in smog and are used as bioindicators of the presence of smog. Costs of crop and orchard damage by smog run into millions of dollars per year in areas prone to this kind of air pollution, such as southern California. Materials that are adversely affected by smog are generally those that are attacked by oxidants. The best example of such a material is rubber, especially natural rubber, which is attacked by ozone. Indeed, the hardening and cracking of natural rubber has been used as a test for atmospheric ozone. Visibility-reducing atmospheric aerosol particles are the most common manifestation of the harm done to atmospheric quality by smog. The smog-forming process occurs by the oxidation of organic materials in the atmosphere, and carbon-containing organic materials are the most common constituents of the aerosol particles in an atmosphere afflicted by smog. Conifer trees(pine and cypress) and citrus trees are major contributors to the organic hydrocarbons that are precursors to organic particle formation in smog. Every new vehicle sold in the United States must include a catalytic converter to reduce photochemical emissions. Catalytic converters force CO and incompletely combusted hydrocarbons to react with a metal catalyst, typically platinum, to produce CO and H O. Additionally, catalytic converters reduce nitrogen oxides from exhaust gases into O and N , eliminating the cycle of ozone formation. Many scientists have suggested that pumping gas at night could reduce photochemical ozone formation by limiting the amount of exposure VOCs have with sunlight. Smog is basically a chemical problem, which would indicate that it should be amenable to chemical solutions. Indeed, the practice of green chemistry and the application of the principles of industrial ecology can help to reduce smog. This is due in large part to the fact that a basic premise of green chemistry is to avoid the generation and release of chemical species with the potential to harm the environment. The best way to prevent smog formation is to avoid the release of nitrogen oxides and organic vapors that enable smog to form. At an even more fundamental level, measures can be taken to avoid the use of technologies likely to release such substances, for example, by using alternatives to polluting automobiles for transportation. The evolution of automotive pollution control devices to reduce smog provides an example of how green chemistry can be used to reduce pollution. The first measures taken to reduce hydrocarbon and nitrogen oxide emissions from automobiles were very much command-and-control and “end-of-pipe” measures. These primitive measures implemented in the early 1970s did reduce emissions, but with a steep penalty in fuel consumption and in driving performance of vehicles. However, over the last three decades, the internal combustion automobile engine has evolved into a highly sophisticated computer-controlled machine that generally performs well, emits few air pollutants, and is highly efficient. (And it would be much more efficient if those drivers who feel that they must drive “sport utility” behemoths would switch to vehicles of a more sensible size.) This change has required an integrated approach involving reformulation of gasoline. The first major change was elimination from gasoline of tetraethyllead, an organometallic compound that poisoned automotive exhaust catalysts (and certainly was not good for people). Gasoline was also reformulated to eliminate excessively volatile hydrocarbons and unsaturated hydrocarbons (those with double bonds between carbon atoms) that are especially reactive in forming photochemical smog. An even more drastic approach to eliminating smog-forming emissions is the use of electric automobiles that do not burn gasoline. These vehicles certainly do not pollute as they are being driven, but they suffer from the probably unsolvable problem of a very limited range between charges and the need for relatively heavy batteries. However, hybrid automobiles using a small gasoline or diesel engine that provides electricity to drive electric motors propelling the automobile and to recharge relatively smaller batteries can largely remedy the emission and fuel economy problems with automobiles. The internal combustion engine on these vehicles runs only as it is needed to provide power and, in so doing, can run at a relatively uniform speed that provides maximum economy with minimum emissions. Another approach that is being used on vehicles as large as buses that have convenient and frequent access to refueling stations is the use of fuel cells that can generate electricity directly from the catalytic combination of elemental hydrogen and oxygen, producing only harmless water as a product . There are also catalytic process that can generate hydrogen from liquid fuels, such as methanol, so that vehicles carrying such a fuel can be powered by electricity generated in fuel cells. Green chemistry can be applied to devices and processes other than automobiles to reduce smog-forming emissions. This is especially true in the area of organic solvents used for parts cleaning and other industrial operations, vapors of which are often released to the atmosphere. The substitution of water with proper additives or even the use of supercritical carbon dioxide fluid can eliminate such emissions. Ed Vitz (Kutztown University), (University of ( ) | 8,136 | 2,608 |
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At the beginning of , we noted that there are several different types of solute/stationary phase interactions in liquid chromatography, but limited our discussion to liquid–liquid chromatography. In this section we turn our attention to liquid chromatography techniques in which partitioning occurs by liquid–solid adsorption, ion-exchange, and size exclusion. In (LSC) the column packing also serves as the stationary phase. In Tswett’s original work the stationary phase was finely divided CaCO , but modern columns employ porous 3–10 μm particles of silica or alumina. Because the stationary phase is polar, the mobile phase usually is a nonpolar or a moderately polar solvent. Typical mobile phases include hexane, isooctane, and methylene chloride. The usual order of elution—from shorter to longer retention times—is olefins < aromatic hydrocarbons < ethers < esters, aldehydes, ketones < alcohols, amines < amide < carboxylic acids Nonpolar stationary phases, such as charcoal-based absorbents, also are used. For most samples, liquid–solid chromatography does not offer any special advantages over liquid–liquid chromatography. One exception is the analysis of isomers, where LSC excels. In (IEC) the stationary phase is a cross-linked polymer resin, usually divinylbenzene cross-linked polystyrene, with covalently attached ionic functional groups (see Figure 12.6.1
and Table 12.6.1
). The counterions to these fixed charges are mobile and are displaced by ions that compete more favorably for the exchange sites. Ion-exchange resins are divided into four categories: strong acid cation exchangers; weak acid cation exchangers; strong base anion exchangers; and weak base anion exchangers. . Structures of styrene, divinylbenzene, and a styrene–divinylbenzene co-polymer modified for use as an ion-exchange resin are shown on the left. The ion-exchange sites, indicated by R and shown in , are mostly in the position and are not necessarily bound to all styrene units. The cross-linking is shown in . The photo on the right shows an example of the polymer beads. These beads are approximately 0.30–0.85 mm in diameter. Resins for use in ion-exchange chromatography typically are 5–11 μm in diameter. \(-\text{SO}_3^-\) \(-\text{CH}_2\text{CH}_2\text{SO}_3^-\) \(-\text{COO}^-\) \(-\text{CH}_2\text{COO}^-\) \(-\text{CH}_2\text{N(CH}_3)_3^+\) \(-\text{CH}_2\text{CH}_2\text{N(CH}_2\text{CH}_3)_3^+\) \(-\text{NH}_4^+\) \(-\text{CH}_2\text{CH}_2\text{NH(CH}_2\text{CH}_3)_3^+\) Strong acid cation exchangers include a sulfonic acid functional group that retains it anionic form—and thus its capacity for ion-exchange—in strongly acidic solutions. The functional groups for a weak acid cation exchanger, on the other hand, are fully protonated at pH levels less then 4 and lose their exchange capacity. The strong base anion exchangers include a quaternary amine, which retains a positive charge even in strongly basic solutions. Weak base anion exchangers remain protonated only at pH levels that are moderately basic. Under more basic conditions a weak base anion exchanger loses a proton and its exchange capacity. The ion-exchange reaction of a monovalent cation, M , exchange site is \[-\mathrm{SO}_{3}^{-} \mathrm{H}^{+}(s)+\mathrm{M}^{+}(a q)\rightleftharpoons-\mathrm{SO}_{3}^{-} \mathrm{M}^{+}(s)+\mathrm{H}^{+}(a q) \nonumber\] The equilibrium constant for this ion-exchange reaction, which we call the selectivity coefficient, , is \[K=\frac{\left\{-\mathrm{SO}_{3}^{-} \mathrm{M}^{+}\right\}\left[\mathrm{H}^{+}\right]}{\left\{-\mathrm{SO}_{3}^{-} \mathrm{H}^{+}\right\}\left[\mathrm{M}^{+}\right]} \label{12.1}\] where we use curly brackets, { }, to indicate a surface concentration instead of a solution concentration. We don’t usually think about a solid’s concentration. There is a good reason for this. In most cases, a solid’s concentration is a constant. If you break a piece of chalk into two parts, for example, the mass and the volume of each piece retains the same proportional relationship as in the original piece of chalk. When we consider an ion binding to a reactive site on the solid’s surface, however, the fraction of sites that are bound, and thus the concentration of bound sites, can take on any value between 0 and some maximum value that is proportional to the density of reactive sites. Rearranging Equation \ref{12.1} shows us that the distribution ratio, , for the exchange reaction \[D=\frac{\text { amount of } \mathrm{M}^{+} \text { in the stationary phase }}{\text { amount of } \mathrm{M}^{+} \text { in the mobile phase }} \nonumber\] \[D=\frac{\left\{-\mathrm{SO}_{3}^{-} \mathrm{M}^{+}\right\}}{\left[\mathrm{M}^{+}\right]}=K \times \frac{\left\{-\mathrm{SO}_{3}^{-} \mathrm{H}^{+}\right\}}{\left[\mathrm{H}^{+}\right]} \label{12.2}\] is a function of the concentration of H and, therefore, the pH of the mobile phase. An ion-exchange resin’s selectivity is somewhat dependent on whether it includes strong or weak exchange sites and on the extent of cross-linking. The latter is particularly important as it controls the resin’s permeability, and, therefore, the accessibility of exchange sites. An approximate order of selectivity for a typical strong acid cation exchange resin, in order of decreasing , is Al > Ba > Pb > Ca > Ni > Cd > Cu > Co > Zn > Mg > Ag > K > \(\text{NH}_4^+\) > Na > H > Li Note that highly charged cations bind more strongly than cations of lower charge, and that for cations of similar charge, those with a smaller hydrated radius (see in Chapter 6), or that are more polarizable, bind more strongly. For a strong base anion exchanger the general elution order is \(\text{SO}_4^{2-}\) > I > \(\text{HSO}_4^-\) > \(\text{NO}_3^-\) > Br > \(\text{NO}_2^-\) > Cl > \(\text{HCO}_3^-\) > CH3COO > OH > F Anions of higher charge and of smaller hydrated radius bind more strongly than anions with a lower charge and a larger hydrated radius. The mobile phase in IEC usually is an aqueous buffer, the pH and ionic composition of which determines a solute’s retention time. Gradient elutions are possible in which the mobile phase’s ionic strength or pH is changed with time. For example, an IEC separation of cations might use a dilute solution of HCl as the mobile phase. Increasing the concentration of HCl speeds the elution rate for more strongly retained cations because the higher concentration of H allows it to compete more successfully for the ion-exchange sites. From Equation \ref{12.2}, a cation’s distribution ratio, , becomes smaller when the concentration of H in the mobile phase increases. An ion-exchange resin is incorporated into an HPLC column either as 5–11 μm porous polymer beads or by coating the resin on porous silica particles. Columns typically are 250 mm in length with internal diameters ranging from 2–5 mm. Measuring the conductivity of the mobile phase as it elutes from the column serves as a universal detector for cationic and anionic analytes. Because the mobile phase contains a high concentration of ions—a mobile phase of dilute HCl, for example, contains significant concentrations of H and Cl —we need a method for detecting the analytes in the presence of a significant background conductivity. To minimize the mobile phase’s contribution to conductivity, an is placed between the analytical column and the detector. This column selectively removes mobile phase ions without removing solute ions. For example, in cation-exchange chromatography using a dilute solution of HCl as the mobile phase, the suppressor column contains a strong base anion-exchange resin. The exchange reaction \[\mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)+\mathrm{Resin}^{+} \mathrm{OH}^{-}(s)\rightleftharpoons\operatorname{Resin}^{+} \mathrm{Cl}^{-}(s)+\mathrm{H}_{2} \mathrm{O}(l ) \nonumber\] replaces the mobile phase ions H and Cl with H O. A similar process is used in anion-exchange chromatography where the suppressor column contains a cation-exchange resin. If the mobile phase is a solution of Na CO , the exchange reaction \[2 \mathrm{Na}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q)+2 \operatorname{Resin}^{-} \mathrm{H}^{+}(s)\rightleftharpoons2 \operatorname{Resin}^{-} \mathrm{Na}^{+}(s)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \nonumber\] replaces a strong electrolyte, Na CO , with a weak electrolyte, H CO . Ion-suppression is necessary when the mobile phase contains a high concentration of ions. , in which an ion-suppressor column is not needed, is possible if the concentration of ions in the mobile phase is small. Typically the stationary phase is a resin with a low capacity for ion-exchange and the mobile phase is a very dilute solution of methane sulfonic acid for cationic analytes, or potassium benzoate or potassium hydrogen phthalate for anionic analytes. Because the background conductivity is sufficiently small, it is possible to monitor a change in conductivity as the analytes elute from the column. A UV/Vis absorbance detector can be used if the analytes absorb ultraviolet or visible radiation. Alternatively, we can detect indirectly analytes that do not absorb in the UV/Vis if the mobile phase contains a UV/Vis absorbing species. In this case, when a solute band passes through the detector, a decrease in absorbance is measured at the detector. Ion-exchange chromatography is an important technique for the analysis of anions and cations in water. For example, an ion-exchange chromatographic analysis for the anions F , Cl , Br , \(\text{NO}_2^-\), \(\text{NO}_3^-\), \(\text{PO}_4^{3-}\), and \(\text{SO}_4^{2-}\) takes approximately 15 minutes (Figure 12.6.2
). A complete analysis of the same set of anions by a combination of potentiometry and spectrophotometry requires 1–2 days. Ion-exchange chromatography also is used for the analysis of proteins, amino acids, sugars, nucleotides, pharmaceuticals, consumer products, and clinical samples. We have considered two classes of micron-sized stationary phases in this section: silica particles and cross-linked polymer resin beads. Both materials are porous, with pore sizes ranging from approximately 5–400 nm for silica particles, and from 5 nm to 100 μm for divinylbenzene cross-linked polystyrene resins. In —which also is known by the terms molecular-exclusion or gel permeation chromatography—the separation of solutes depends upon their ability to enter into the pores of the stationary phase. Smaller solutes spend proportionally more time within the pores and take longer to elute from the column. A stationary phase’s size selectivity extends over a finite range. All solutes significantly smaller than the pores move through the column’s entire volume and elute simultaneously, with a retention volume, , of \[V_{r}=V_{i}+V_{o} \label{12.3}\] where is the volume of mobile phase occupying the stationary phase’s pore space and is volume of mobile phase in the remainder of the column. The largest solute for which Equation \ref{12.3} holds is the column’s inclusion limit, or permeation limit. Those solutes too large to enter the pores elute simultaneously with an retention volume of \[V_{r} = V_{o} \label{12.4}\] Equation \ref{12.4} defines the column’s . For a solute whose size is between the inclusion limit and the exclusion limit, the amount of time it spends in the stationary phase’s pores is proportional to its size. The retention volume for these solutes is \[V_{r}=DV_{i}+V_{o} \label{12.5}\] where is the solute’s distribution ratio, which ranges from 0 at the exclusion limit to 1 at the inclusion limit. Equation \ref{12.5} assumes that size-exclusion is the only interaction between the solute and the stationary phase that affects the separation. For this reason, stationary phases using silica particles are deactivated as described earlier, and polymer resins are synthesized without exchange sites. Size-exclusion chromatography provides a rapid means for separating larger molecules, including polymers and biomolecules. A stationary phase for proteins that consists of particles with 30 nm pores has an inclusion limit of 7500 g/mol and an exclusion limit of \(1.2 \times 10^6\) g/mol. Mixtures of proteins that span a wider range of molecular weights are separated by joining together in series several columns with different inclusion and exclusion limits. Another important application of size-exclusion chromatography is the estimation of a solute’s molecular weight (MW). Calibration curves are prepared using a series of standards of known molecular weight and measuring each standard’s retention volume. As shown in Figure 12.6.3
, a plot of log(MW) versus is roughly linear between the exclusion limit and the inclusion limit. Because a solute’s retention volume is influenced by both its size and its shape, a reasonably accurate estimation of molecular weight is possible only if the standards are chosen carefully to minimize the effect of shape. Size-exclusion chromatography is carried out using conventional HPLC instrumentation, replacing the HPLC column with an appropriate size-exclusion column. A UV/Vis detector is the most common means for obtaining the chromatogram. Although there are many analytical applications of gas chromatography and liquid chromatography, they can not separate and analyze all types of samples. Capillary column GC separates complex mixtures with excellent resolution and short analysis times. Its application is limited, however, to volatile analytes or to analytes made volatile by a suitable derivatization reaction. Liquid chromatography separates a wider range of solutes than GC, but the most common detectors—UV, fluorescence, and electrochemical— have poorer detection limits and smaller linear ranges than GC detectors, and are not as universal in their selectivity. For some samples, (SFC) provides a useful alternative to gas chromatography and liquid chromatography. The mobile phase in supercritical fluid chromatography is a gas held at a temperature and pressure that exceeds its critical point (Figure 12.6.4
). Under these conditions the mobile phase is neither a gas nor a liquid. Instead, the mobile phase is a supercritical fluid. Some properties of a supercritical fluid, as shown in Table 12.6.2
, are similar to a gas; other properties, however, are similar to a liquid. The viscosity of a supercritical fluid, for example, is similar to a gas, which means we can move a supercritical fluid through a capillary column or a packed column without the high pressures needed in HPLC. Analysis time and resolution, although not as good as in GC, usually are better than in conventional HPLC. The density of a supercritical fluid, on the other hand, is much closer to that of a liquid, which explains why supercritical fluids are good solvents. In terms of its separation power, a mobile phase in SFC behaves more like the liquid mobile phase in HPLC than the gaseous mobile phase in GC. The most common mobile phase for supercritical fluid chromatography is CO . Its low critical temperature of 31.1 C and its low critical pressure of 72.9 atm are relatively easy to achieve and maintain. Although supercritical CO is a good solvent for nonpolar organics, it is less useful for polar solutes. The addition of an organic modifier, such as methanol, improves the mobile phase’s elution strength. Other common mobile phases and their critical temperatures and pressures are listed in Table 12.6.3
. The instrumentation for supercritical fluid chromatography essentially is the same as that for a standard HPLC. The only important additions are a heated oven for the column and a pressure restrictor downstream from the column to maintain the critical pressure. Gradient elutions are accomplished by changing the applied pressure over time. The resulting change in the mobile phase’s density affects its solvent strength. Detection is accomplished using standard GC detectors or HPLC detectors. Supercritical fluid chromatography has many applications in the analysis of polymers, fossil fuels, waxes, drugs, and food products. | 16,065 | 2,609 |
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Lewis formulas are misleading in the sense that atoms and electrons are shown as being static. By being essentially two-dimensional representations they also fail to give an accurate idea of the three-dimensional features of the molecule, such as actual bond angles and topography of the molecular frame. Furthermore, a given compound can have several valid Lewis formulas. For example CH CNO can be represented by at least three different but valid Lewis structures called , shown below. However, a stable compound such as the above does not exist in multiple states represented by structures I, or II, or III. The compound exists in a single state called a of all three structures. That is, it contains of all three resonance forms, much like a person might have physical features inherited from each parent to varying degrees. In the resonance forms shown above the atoms remain in one place. The basic bonding pattern, or , is the same in all structures, but some electrons have changed locations. This means that there are certain rules for electron mobility that enable us to “push” electrons around to arrive from one resonance structure to another. These rules will be examined in detail in a later paper. By convention, we use double-headed arrows to indicate that several resonance structures contribute to the same hybrid. Continuing with the example we’ve been using, the resonance structures for CH CNO should be written in this way if we want to emphasize that they represent the same hybrid. Do not confuse double-headed arrows with double arrows. A double arrow indicates that two or more species are in equilibrium with each other and therefore have a separate existence. Double-headed arrows indicate resonance structures that do not exist by themselves. They simply represent features that the actual molecule, the hybrid, possesses to one extent or another. When writing resonance structures keep in mind that THEY ALL MUST BE VALID LEWIS FORMULAS. The factors that make up valid Lewis formulas are as follows. 1. Observe the rules of covalent bonding, including common patterns as discussed previously. Make sure to show all single, double, and triple bonds. 2. Account for the total number of valence electrons being shared (from all the elements), including bonding and nonbonding electrons. Make sure to show these nonbonding electrons. 3. Account for the net charge of the molecule or species, showing formal charges where they belong. 4. Observe the octet rule as much as possible, but also understand that there are instances where some atoms may not fulfill this rule. 5. Avoid having unpaired electrons (single electrons with no partners) unless the total number of valence electrons for all elements is an odd number. This is not a very frequent occurrence, but the following example shows a species that could exist as a reaction intermediate in some high energy environments. The total number of valence electrons being shared for all atoms is 4 from carbon and 3 from the three hydrogens, for a total of 7. Because it is an odd number, it is impossible to have all these electrons paired. Therefore the presence of a single electron cannot be avoided. Notice that there is no formal charge on carbon, since it has no surplus or deficit of valence electrons. From the examples given so far it can be seen that some resonance forms are structurally equivalent and others are not. . If the Lewis structures are not equivalent, then the potential energy associated with them is most likely different. This means that equivalent resonance structures are also equivalent in stability and nonequivalent structures have different stabilities. This in turn means that equivalent structures contribute equally to the hybrid and nonequivalent structures do not contribute equally to the hybrid. In the example below, the two structures are equivalent. Therefore they make equal contributions to the hybrid. It is very difficult to accurately represent the hybrid with drawings because it is a composite of all the resonance contributors. Some representations such as the first one shown below are sometimes given. In this case the broken line represents the electrons and the negative charge which are spread over three atoms (O-C-O). Another commonly used representation of the hybrid is given on the right, showing that each oxygen atom shares a -1/2 charge. However, none of them accurately conveys the true picture. For convenience, the unshared electrons are sometimes omitted from some representations, just like it’s done with hydrogen atoms. This must be kept in mind when examining the different structures. There is a third resonance form that can be drawn for the acetate ion hybrid. The structure shown below is structurally different from the ones shown above. This means that it is of different energy and therefore does not contribute to the hybrid to the same extent as the others. In this case, it happens to be less stable than the other two and therefore does not make a significant contribution to the hybrid. What factors dictate the relative stabilities of different resonance structures? is an important one. In the example just given there are too many charges present in the structure (even though the net charge is still the same). This results in a structure with higher potential energy than others with fewer or no formal charges. This structure is less stable and its contribution to the hybrid is probably minor. Another important factor that increases potential energy (lowers stability) is the presence of atoms with an . In the example below structure I has a carbon atom with a positive charge and therefore an incomplete octet. Based on this criterion, this structure is less stable than structure II and makes a less significant contribution to the hybrid. Finally, another factor that comes into play when determining the relative energies of resonance structures is the More electronegative atoms are more comfortable with negative charges. Less electronegative atoms are more comfortable with positive charges. In the example below, structure I is less stable than II. The negative charge is on carbon, which is less electronegative than oxygen. Therefore structure II makes a larger contribution to the hybrid. Some texts refer to the different contributors as “important” or “unimportant.” This can be misleading, because “importance” is context-dependent. In the example above, structure I is less important in terms of its contribution to the hybrid, but in terms of reactivity, it is very important. It will be seen later that this structure provides a better indication of how this species reacts with electrophiles in certain types of reactions. Based on the charge separation criterion, structure II below might be labeled “unimportant,” or “less important” because it is less stable than I. But if we are trying to assess the polarity of this molecule, structure II becomes very important because it reveals that the carbon atom has positive character and the oxygen has negative character. The hybrid representation on the right then portrays the molecule as a polar molecule, which structure I alone does not give much indication of. The following additional examples further illustrate the relative importance of the factors that contribute to structural energy and stability. | 7,373 | 2,610 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/02%3A_Introduction_to_Organic_Structure_and_Bonding_II |
While organic and biological chemistry is a very diverse field of study, one fundamental question that interests all organic chemists is how the structure of an organic molecule determines its physical properties. We will look more closely at the nature of single and double covalent bonds, using the concepts of 'hybrid orbitals' and 'resonance' to attempt to explain how orbital overlap results in characteristic geometries and rotational behavior for single and double bonds, as well as bonds that have characteristics of somewhere in between single and double. Then we will move on to a review of the noncovalent interactions between molecules - Van der Waals, ion-ion, dipole-dipole and ion-dipole interactions, and hydrogen bonds - and how they are manifested in the observable physical properties of all organic substances. | 842 | 2,611 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/15%3A_Alcohols_and_Ethers/15.10%3A_Protection_of_Hydroxyl_Groups |
By now it should be apparent that hydroxyl groups are very reactive to many reagents. This is both an advantage and a disadvantage in synthesis. To avoid interference by hydroxyl groups, it often is necessary to protect (or mask) them by conversion to less reactive functions. The general principles of how functional groups are protected were outlined and illustrated in . In the case of alcohols the hydroxyl group may be protected by formation of an ether, an ester, or an acetal. A good protecting group is one that does everything you want it to do you want it to. It must be easily put into place, stable to the reagents from which protection is required, easily removed when desired. For this reason simple ethers such as methyl or ethyl ethers usually are not suitable protecting groups because they cannot be removed except under rather drastic conditions ( ). More suitable ethers are phenylmethyl and trimethylsilyl ethers: Both of these ethers are prepared easily by nucleophilic displacements (Equations 15-7 and 15-8) and can be converted back to the parent alcohol under mild conditions, by catalytic hydrogenation for phenylmethyl ethers (Equation 15-9), or by mild acid hydrolysis for trimethylsilyl ethers (Equation 15-10): Esters are formed from the alcohol and acyl halide, anhydride, or acid ( ). The alcohol can be regenerated easily by either acid or base hydrolysis of the ester: We have seen that alcohols can be converted to acetals under acidic conditions ( ). The acetal function is a very suitable protecting group for alcohols under basic conditions, but is not useful under acidic conditions because acetals are not stable to acids: An excellent reagent to form acetals is the unsaturated cyclic ether, \(16\). This ether adds alcohols in the presence of an acid catalyst to give the acetal \(17\): The 3-oxacyclohexene (dihydropyran) protecting group can be removed readily by treating the acetal, \(17\), with aqueous acid: An example of the use of \(16\) to protect an \(\ce{OH}\) function is given in . and (1977) | 2,075 | 2,613 |
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is the study of matter and the changes that material substances undergo. Of all the scientific disciplines, it is perhaps the most extensively connected to other fields of study. Geologists who want to locate new mineral or oil deposits use chemical techniques to analyze and identify rock samples. Oceanographers use chemistry to track ocean currents, determine the flux of nutrients into the sea, and measure the rate of exchange of nutrients between ocean layers. Engineers consider the relationships between the structures and the properties of substances when they specify materials for various uses. Physicists take advantage of the properties of substances to detect new subatomic particles. Astronomers use chemical signatures to determine the age and distance of stars and thus answer questions about how stars form and how old the universe is. The entire subject of environmental science depends on chemistry to explain the origin and impacts of phenomena such as air pollution, ozone layer depletion, and global warming. The disciplines that focus on living organisms and their interactions with the physical world rely heavily on , the application of chemistry to the study of biological processes. A living cell contains a large collection of complex molecules that carry out thousands of chemical reactions, including those that are necessary for the cell to reproduce. Biological phenomena such as vision, taste, smell, and movement result from numerous chemical reactions. Fields such as medicine, pharmacology, nutrition, and toxicology focus specifically on how the chemical substances that enter our bodies interact with the chemical components of the body to maintain our health and well-being. For example, in the specialized area of sports medicine, a knowledge of chemistry is needed to understand why muscles get sore after exercise as well as how prolonged exercise produces the euphoric feeling known as “runner’s high.” Examples of the practical applications of chemistry are everywhere ( ). Engineers need to understand the chemical properties of the substances when designing biologically compatible implants for joint replacements or designing roads, bridges, buildings, and nuclear reactors that do not collapse because of weakened structural materials such as steel and cement. Archaeology and paleontology rely on chemical techniques to date bones and artifacts and identify their origins. Although law is not normally considered a field related to chemistry, forensic scientists use chemical methods to analyze blood, fibers, and other evidence as they investigate crimes. In particular, matching—comparing biological samples of genetic material to see whether they could have come from the same person—has been used to solve many high-profile criminal cases as well as clear innocent people who have been wrongly accused or convicted. Forensics is a rapidly growing area of applied chemistry. In addition, the proliferation of chemical and biochemical innovations in industry is producing rapid growth in the area of patent law. Ultimately, the dispersal of information in all the fields in which chemistry plays a part requires experts who are able to explain complex chemical issues to the public through television, print journalism, the Internet, and popular books. By this point, it shouldn’t surprise you to learn that chemistry was essential in explaining a pivotal event in the history of Earth: the disappearance of the dinosaurs. Although dinosaurs ruled Earth for more than 150 million years, fossil evidence suggests that they became extinct rather abruptly approximately 66 million years ago. Proposed explanations for their extinction have ranged from an epidemic caused by some deadly microbe or virus to more gradual phenomena such as massive climate changes. In 1978 Luis Alvarez (a Nobel Prize–winning physicist), the geologist Walter Alvarez (Luis’s son), and their coworkers discovered a thin layer of sedimentary rock formed 66 million years ago that contained unusually high concentrations of iridium, a rather rare metal (part (a) in ). This layer was deposited at about the time dinosaurs disappeared from the fossil record. Although iridium is very rare in most rocks, accounting for only 0.0000001% of Earth’s crust, it is much more abundant in comets and asteroids. Because corresponding samples of rocks at sites in Italy and Denmark contained high iridium concentrations, the Alvarezes suggested that the impact of a large asteroid with Earth led to the extinction of the dinosaurs. When chemists analyzed additional samples of 66-million-year-old sediments from sites around the world, all were found to contain high levels of iridium. In addition, small grains of quartz in most of the iridium-containing layers exhibit microscopic cracks characteristic of high-intensity shock waves (part (b) in ). These grains apparently originated from terrestrial rocks at the impact site, which were pulverized on impact and blasted into the upper atmosphere before they settled out all over the world. Scientists calculate that a collision of Earth with a stony asteroid about 10 kilometers (6 miles) in diameter, traveling at 25 kilometers per second (about 56,000 miles per hour), would almost instantaneously release energy equivalent to the explosion of about 100 million megatons of (trinitrotoluene). This is more energy than that stored in the entire nuclear arsenal of the world. The energy released by such an impact would set fire to vast areas of forest, and the smoke from the fires and the dust created by the impact would block the sunlight for months or years, eventually killing virtually all green plants and most organisms that depend on them. This could explain why about 70% of all species—not just dinosaurs—disappeared at the same time. Scientists also calculate that this impact would form a crater at least 125 kilometers (78 miles) in diameter. Recently, satellite images from a Space Shuttle mission confirmed that a huge asteroid or comet crashed into Earth’s surface across the Yucatan’s northern tip in the Gulf of Mexico 65 million years ago, leaving a partially submerged crater 180 kilometers (112 miles) in diameter ( ). Thus simple chemical measurements of the abundance of one element in rocks led to a new and dramatic explanation for the extinction of the dinosaurs. Though still controversial, this explanation is supported by additional evidence, much of it chemical. This is only one example of how chemistry has been applied to an important scientific problem. Other chemical applications and explanations that we will discuss in this text include how astronomers determine the distance of galaxies and how fish can survive in subfreezing water under polar ice sheets. We will also consider ways in which chemistry affects our daily lives: the addition of iodine to table salt; the development of more effective drugs to treat diseases such as cancer, (acquired immunodeficiency syndrome), and arthritis; the retooling of industry to use nonchlorine-containing refrigerants, propellants, and other chemicals to preserve Earth’s ozone layer; the use of modern materials in engineering; current efforts to control the problems of acid rain and global warming; and the awareness that our bodies require small amounts of some chemical substances that are toxic when ingested in larger doses. By the time you finish this text, you will be able to discuss these kinds of topics knowledgeably, either as a beginning scientist who intends to spend your career studying such problems or as an informed observer who is able to participate in public debates that will certainly arise as society grapples with scientific issues. An understanding of chemistry is essential for understanding much of the natural world and is central to many other disciplines. Chemistry is the study of matter and the changes material substances undergo. It is essential for understanding much of the natural world and central to many other scientific disciplines, including astronomy, geology, paleontology, biology, and medicine. | 8,125 | 2,614 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/10%3A_Enzyme_Kinetics/10.E%3A_Exercises |
One day in class about enzyme kinetics, Jack comes over to you and asks, "I know enzymes are biological catalysts, but I do not understand how it works. Can you explain how enzymes make reactions go faster? And is it only faster in one direction?" Because the activation energy is the energy hill between reactants and products, enzymes decreasing the size of the hill also decreases the amount of energy needed for reactions to go in either direction. A smaller energy hill allows reactants and products to overcome the barrier quicker, resulting a faster reaction rate. If your student colleagues argues that a catalysts affects only the rate of only one direction of a reaction. Explain why he is correct or not. False. Catalysts affect rate by providing an alternative mechanism which has a lower transition state energy. It's impossible to lower transition state energy for only one direction of a reaction. (It'd be like making a hill shorter from the north, but keeping it the same height from the south.) The dissociation constant k = k / k =6x10 s / 4x10 M s = 0.15 M The Michaelis constant k = k +k / k 6x10 s + 2.0x10 s / 4x10 M s =0.155M The two constant are not equal. Therefore, the binding does not follow the equilibrium scheme. Is it appropriate to use the rapid equilibrium scheme to model the kinetics of a catalyzed reaction with the following rate constants? \[ k_1 = 7 \times 10^7\ M^{-1}\ s^{-1} \] \[ k_{-1} = 8 \times 10^5\ s^{-1} \] \[ k_2 = 5 \times 10^4\ s^{-1} \] Assuming rapid equilibrium is appropriate when \(k_{-1} \gg k_2\), so that K = K . In this case, the values of k and k are not dramatically different, so we can calculate K and K and compare. 1. Calculate \(K_M\): \[ K_M = \dfrac{k_{-1} + k_2}{k_1} \] \[ K_M = \dfrac{8 \times 10^5\ s^{-1} + 5 \times 10^4\ s^{-1}}{7 \times 10^7\ M^{-1}\ s^{-1}} \] \[ K_M = 0.01\ M \] 2. Calculate \(K_S\): \[ K_S = \dfrac{k_{-1}}{k_1} \] \[ K_M = \dfrac{8 \times 10^5\ s^{-1}}{7 \times 10^7\ M^{-1}\ s^{-1}} \] \[ K_M = 0.01\ M \] Since \(K_M = K_S\), it is appropriate to assume rapid equilibrium. Given k = 7 x 10^6 M-1s-1, k =6 x 10^4 s , k = 2 x 10^3 s , determine if the enzyme substance binding follow the equilibrium or steady state scheme? \[ \begin{align} K_s &= \dfrac{k_{-1}}{k_1} \\ &= \dfrac{6 \times 10^4\, s^{-1}}{7 \times 10^6 M^{-1}s^{-1}} \\ &=8.8 \times 10^{-3} M \end{align}\] \[\begin{align} K_M &= k_{-1}+k_{2} \\ &= \dfrac{6 \times 10^4 s^{-1} +2 \times 10^3s^{-1}}{ 7 \times 10^6 M ^{-1} s^{-1}} \\ &= 8.9 \times 10^{-3} M \end{align}\] The substrate N-acetylglycine ethyl ester can be catalyzed by the enzyme . This enzyme has a turnover rate of 30,000 s . Determine how long it will take carbonic anhydrase to cleave the substrate. We already know the turnover number (\(k_{cat}\)). The amount of time necessary to cleave the substrate is the reciprocal of the turnover rate. \[t= \dfrac{1}{k} = \dfrac{1}{30,000 \;s^{-1}} = 3.33 \times 10^{-5}\] \[(5 \;minutes)(7.1 \times 10^{-6}) = 3.5 \times 10^{-5}\] RuBisCO is an enzyme in the Calvin cycle that fixes atmospheric carbon and has a turnover rate of 3.3 s . How long does it take RuBisCO to fix one molecule of carbon dioxide? The turnover number is the number of molecules of substrate per unit time (when the enzyme is fully saturated). So simply take the reciprocal to find the time per molecule of substrate. \[ \dfrac{1}{3.3\ s^{-1}} = 0.30\ s \] Note: RuBisCO is a notoriously slow enzyme. The catalyze of acetylcholine has a rate 50000 s . Calculate the time for the enzyme to cleave one Ach molecule. p-nitrophenyl acetate(PNPA) is catalyzed by chymotrypsin to yield p-nitrophenolate ion and acetate ion. The turnover rate of that enzyme is 40,000 s . How long will it take for the enzyme to produce 1 mole of Nitrophenyl acetate? It takes 1 second to convert all 40,000 molecule substrate into the product, so: \(t=\dfrac{40,000}{9.5x10^5}\) = 3.8 sec. What is plotted on the x and y axes on a Lineweaver-Burk plot? Show how to derive the equation for the plot from the equation \[ v_0 = \dfrac{V_{max}[S]}{K_M + [S]} \] and explain how V and K can be found from the graph's intercepts. The x-axis is 1/ν , and the x-axis is 1/V . 1. Take the reciprocal of both sides of the equation. \[ \dfrac{1}{v_0} = \dfrac{K_M + [S]}{V_{max} [S]} \] \[ \dfrac{1}{v_0} = \dfrac{K_M}{V_{max}} \dfrac{1}{[S]} + \dfrac{1}{V_{max}} \] 2. Set 1/[S] = 0 to find the y-intercept, and show that it relates to V . \[ Yint = \dfrac{1}{V_{max}} \] \[ V_{max} = \dfrac{1}{Yint} \] 3. Set 1/ν = 0 to find the x-intercept, and show that it relates to K . \[ 0 = \dfrac{K_M}{V_{max}} Xint + \dfrac{1}{V_{max}} \] \[ \dfrac{K_M}{V_{max}} Xint = - \dfrac{1}{V_{max}} \] \[ Xint = -\dfrac{1}{K_M} \] \[ K_M = -\dfrac{1}{Xint} \] Derive the Michaelis-Menten equation by assuming rapid equilibrium. \[ E + S \overset{K_{1}}{\rightleftharpoons} ES \overset{K_{2}}{\rightarrow} E + P\] \(\dfrac{[E,S]}{dt} = K_{-1}[ES]\) \(\dfrac{[ES]}{dt} = K_{1}[E,S]\) \(\dfrac{[E,S]}{dt}=\dfrac{[ES]}{dt}\) \(K_{-1}[ES]=K_{1}[E,S]\) \(K_{M}\dfrac{k_{-1}}{K_{1}}=\dfrac{[E,S]}{[ES]}\) \(K_{M}=\dfrac{[E]_{0}-[ES,S]}{[ES]}\) \(K_{M}=\dfrac{[E]_{0}[S]-[ES,S]}{[ES]}\) \(K_{M}\times[ES]=[E]_{0}[S]-[ES,S]\) \(K_{M}\times[ES]+[ES,S]=[E]_{0}[S]\) \((K_{M}+[S])[ES]=[E_{0},S]\) \([ES]=\dfrac{[E_{0},S]}{K_{M}+[S]}\) \(v_{0}=\dfrac{K_{2}[E_{0},S]}{K_{M}+[S]}\) \(v_{0} = \dfrac{V_{max}[S]}{K_{M}+[S]}\) We know that the Michaelis Menten derivation for the following reaction: \[\ce{E + S \rightleftharpoons ES -> E + P}\] However, what if the reaction took place in a different scenario whereby: \[\ce{E + S \rightleftharpoons ES1 -> ES2 -> E + P}\] What would be the corresponding Michaelis-Menten Ebe quation now? This is the an outline for determining an expression for the rate of substrate conversion in the given case: Prove that \(K_s\) equals the concentration S when the initial rate is half its maximum value. An enzyme that has a \(K_m\) value of \(4.6 \times 10^{-5}\; M\) is studied at an initial substrate concentration of 0.041 M. After a minute, it is found that 7.3 uM of product has been produced. Calculate the value of Vmax and the amount of product formed after 4.5 minutes. An solution initially contains a catalytic amount of an enzyme with K = 1.5 mM, 0.25 M of substrate, and no product. After 45 seconds, the solution contains 25 µM of product. Find V and the concentration of product after 2.0 minutes. 1. Find the initial velocity. \[ v_0 = \dfrac{25\ \mu M}{0.75\ min} \] \[ v_0 = 33.3\ \mu M/min \] 2. Use v , [S], and K to solve for V . \[ v_0 = \dfrac{V_{max} [S]}{K_M + [S]} \] \[ V_{max} = v_0 \left ( \dfrac{K_M}{[S]} + 1 \right ) \] \[ V_{max} = 33.3\ \mu M/min \left ( \dfrac{1.5\ mM}{0.25\ M} \times \dfrac{M}{1000\ mM} + 1 \right ) \] \[ V_{max} = 33.3\ \mu M/min \] Notice that ν = V . Since [S] >>K , the reaction will continue with a velocity of V for the remainder of the two minutes. 3. Predict [P] after 2.0 minutes at the rate V . \[ [P] = V_{max} \times t \] \[ [P] = 33.3\ \mu M/min \times 2.0\ min \] \[ [P] = 66.6\ \mu M \] A particular enzyme at a research facility is being studied by a group of graduate students. This enzyme has a K value of 5.0 X 10 M. The students study this enzyme with an initial substrate concentration of 0.055 M. At one minute, 7 µM of product was made. What is the amount of product produced after 5 minutes. What is the V 7.0 X 10 M = V (0.055 M) / (5.0 X 10 M + .055 M) V = 7.1 X 10 M/min At 5 minutes the amount of product formed is: Calculate the value if an enzyme has , value and Given the values, [S]/10^-4 M 3.0 4.6 10.5 16.5 vo/10^-6 M * min^-1. 2.64 3.5 6.2 7.8 construct a Lineweaver-Burk plot, and assuming Michaelis-Menten kinetics, calculate the values of Vmax, Km, and k2 using the constructed plot. The data below represents the data recorded after the hydrolysis of a substrate by an enzyme. Calculate V , K and K using a Lineweaver-Burk plot. Assume Michaelis. Given [E} is 5.0 X 10 M Figure: Lineweaver-Burk Plot The linear equation for the graph above is: y=176x+4.23X10 To solve for V use: Intercept= 1/V V = 1/(.4.23X10 ) = 2.36 X 10 To solve for K use Slope= K /V (176)( 2.36 X 10 ) = 4.2 X 10 To solve for K use K = V / [E} = (2.36 X 10 )(5.0 X 10 M)= 4.72 min Using the table below, calculate the K , V , and slope. slope=\(\dfrac{5.1\times10^{-3}-2.9\times10^{-3}}{3.3\times10^{2}-0.3\times10^{2}}=6.73\times10^{-6}\) \(6.73\times10^{-6}\times0.3\times10^{2}+\dfrac{1}{V_{max}}=2.9\times10^{-3}\) \(\dfrac{1}{V_{max}}=0.00270\) \(V_{max}=370.62Ms^{-1}\) \(6.73\times10^{-6}\times\dfrac{-1}{K_{M}}+0.00270=0\) \(\dfrac{-1}{K_{M}}=-0.00943\) \(K_{M}=106.044\) Given the value = 0.00032 and . Find the ratio between and Divided both side by 1: Divided both side by 1 again: From this graph determine the K and V ? From his graph we can see that the value K is 2. Then we look to see where K is half. At that point, we see that K /2 is 1 and the x-value for that coordinate is 1. This means V Neutral sphingomyelinase 2 converts sphingomyelin into ceramide and phosphocholine. Assume its V is 35 µM min . When you provide 3 x 10 M of sphingomyelin, you observe an initial velocity of 6.0 µM min . Calculate the K for this reaction, rounding to 3 significant figures. \[\dfrac{1}{V_{o}} = \dfrac{K_{M}}{V_{max}[S]}+\dfrac{1}{V_{max}}\] \[\dfrac{1}{6.0} = \dfrac{K_{M}}{(35)(30)}+\dfrac{1}{35}\] \[\dfrac{1}{6}-\dfrac{1}{35} = \dfrac{K_{M}}{1050}\] We are given a second order equation: r=k[A,B]. The concentration of A is 0.05g and the concentration of B is 2.5g. We decide this difference is great enough to treat this as a pseudo first reaction? What concentration is held constant and why? Write the new equation. We hold B constant because the concentration is so much larger, so it should be close to constant for the reaction. The new equation would be r=k'[a] Name and briefly describe two types of reactions that do not follow Michaelis-Menten kinetics. Irreversible inhibition - the inhibitor binds covalently and irreversibly to the enzyme Allosteric interactions - the binding of effectors at allosteric sites (away from the active site) influence substrate binding. For the reaction mechanism below, how does the concentration of \(C\) affect the concentration of \(B\)? \[A\leftrightharpoons B\rightarrow C\] Because the second state is (i.e., since arrow), it does not matter if you have a large or little concentration of \(C\), it would not affect \(B\) and hence the kinetics of the reaction. However, the concentration of \(A\0 would clearly affect the concentration of \(B\). | 10,684 | 2,615 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Mossbauer_Spectroscopy |
Mössbauer is a versatile technique used to study nuclear structure with the absorption and re-emission of gamma rays, part of the . The technique uses a combination of the Mössbauer effect and Doppler shifts to probe the hyperfine transitions between the excited and ground states of the nucleus. Mössbauer spectroscopy requires the use of solids or crystals which have a probability to absorb the photon in a recoilless manner, many isotopes exhibit Mössbauer characteristics but the most commonly studied isotope is Fe. Rudolf L. Mössbauer became a physics student at Technical University in Munich at the age of 20. After passing his intermediate exams Mössbauer began working on his thesis and doctorate work in 1955, while working as an assistant lecturer at Institute for Mathematics. In 1958 at the age of 28 Mössbauer graduated, and also showed experimental evidence for recoilless resonant absorption in the nucleus, later to be called the Mössbauer Effect. In 1961 Mössbauer was awarded the Nobel Prize in physics and, under the urging of Richard Feynman, accepted the position of Professor of Physics at the California Institute of Technology. The recoil energy associated with absorption or emission of a can be described by the conservation of momentum. In it we find that the recoil energy depends inversely on the mass of the system. For a gas the mass of the single nucleus is small compared to a solid. The solid or crystal absorbs the energy as phonons, quantized vibration states of the solid, but there is a probability that no phonons are created and the whole lattice acts as the mass, resulting in a recoilless emission of the gamma ray. The new radiation is at the proper energy to excite the next ground state nucleus. The probability of recoilless events increases with decreasing transition energy. \[P_R = P_{\gamma} \nonumber \] \[P^2_\gamma = P^2_{\gamma} \nonumber \] \[2 M E_R = \dfrac{E^2_{\gamma}}{c^2} \nonumber \] \[E_R = \dfrac{E^2_\gamma}{2M{c^2}} \nonumber \] The Doppler shift describes the change in frequency due to a moving source and a moving observer. \(f\) is the frequency measured at the observer, \(v\) is the velocity of the wave so for our case this is the speed of light \(c\), \(v_r\) is the velocity of the observer, \(v_s\) is the velocity of the source which is positive when heading away from the observer, and \(f_0\) is the initial frequency. \[f = {\left (\dfrac{v+v_r}{v+v_s}\right)} f_0 \nonumber \] \[f = {\left (\dfrac{c}{c+v_s}\right)} f_0 \nonumber \] In the case where the source is moving toward a stationary observer the perceived frequency is higher. For the opposite situation where the source travels away from the observer frequencies recorded at the observer will be of lower compared to the initial wave. \[E = \dfrac{hc}{\lambda} = hv \nonumber \] The energy differences between hyperfine states are minimal (fractions of an eV) and the energy variation is achieved by the moving the source toward and away from the sample in an oscillating manner, commonly at a velocity of a few mm/s. The transmittance is then plotted against the velocity of the source and a peak is seen at the energy corresponding to the resonance energy. In the above spectrum the emission and absorption are both estimated by the Lorentzian distribution. By far the most common isotopes studied using Mössbauer spectroscopy is Fe, but many other isotopes have also displayed a Mössbauer spectrum. Two criteria for functionality are Both conditions are met by Fe and it is thus used extensively in Mössbauer spectroscopy. In the figure to the right the red colored boxes of the periodic table of elements indicate all elements that have isotopes visible using the Mössbauer technique. Mössbauer spectroscopy allows the researcher to probe structural elements of the nucleus in several ways, termed isomer shift, quadrupole interactions, and magnetic splitting. These are each explained by the following sections as individual graphs, but in practice Mössbauer spectrum are likely to contain a combination of all effects. An isomeric shift occurs when non identical atoms play the role of source and absorber, thus the radius of the source, \(R_s\), is different that of the absorber, \(R_a\), and the same holds that the electron density of each species is different. The Coulombic interactions affects the ground and excited state differently leading to a energy difference that is not the same for the two species. This is best illustrated with the equation: \[R_A \neq R_S \nonumber \] \[\rho_S \neq \rho_S \nonumber \] \[E_A \neq E_S \nonumber \] \[\delta = E_A-E_S = \dfrac{2}{3}nZ{e^2}{(\rho_A - \rho_S)}(R^2_{es} - R^2_{gs}) \nonumber \] Where delta represents the change in energy necessary to excite the absorber, which is seen as a shift from the Doppler speed 0 to V . The isomer shift depends directly on the s-electrons and can be influenced by the shielding p, d, f electrons. From the measured delta shift there is information about the valance state of the absorbing atom The energy level diagram for \(\delta\) shift shows the change in source velocity due to different sources used. The shift may be either positive or negative. The Hamiltonian for quadrupole interaction using \({}^{57}Fe\) nuclear excited state is given by \[H_Q = \dfrac{eQV_{ZZ}}{12}[3I^2_Z-I(I+1) + \eta(I^2_X-I^2_y)] \nonumber \] where the nuclear excited states are split into two degenerate doublets in the absence of magnetic interactions. For the asymmetry parameter \(\eta = 0\) doublets are labeled with magnetic quantum numbers \(m_{es} = \pm 3/2\) and \(m_{es} = \pm 1/2\), where the \(m_{es} = \pm 3/2\) doublet has the higher energy. The energy difference between the doublets is thus \[\Delta{EQ} = \dfrac{eQV_{zz}}{2}\sqrt{1+\dfrac{\eta^2}{3}} \nonumber \] The energy diagram and corresponding spectrum can be seen as Magnetic splitting of seen in Mössbauer spectroscopy can be seen because the nuclear spin moment undergoes dipolar interactions with the magnetic field \[E(m_I) = -g_n{\beta_n}{B_{eff}}m_I \tag{14} \] where \(g_n\) is the nuclear g-factor and \(\beta_n\) is the nuclear magneton. In the absence of quadrupole interactions the Hamiltonian splits into equally spaced energy levels of The allowed gamma stimulated transitions of nuclear excitation follows the magnetic dipole transition selection rule: \[ \Delta I = 1 \nonumber \] and \[\Delta m_I = 0, \pm 1 \nonumber \] where \(m_I\) is the magnetic quantum number and the direction of \(\beta\) defines the nuclear quantization axis. If we assume \(g\) and \(A\) are (direction independent) where \(g_x = g_y = g_z\) and \(B\) is actually a combination of the applied and internal magnetic fields: \[H = g\beta{S}\centerdot{B}+AS\centerdot{I} - g_n\beta_nB\centerdot{I} \nonumber \] The electronic Zeeman term is far larger then the nuclear Zeeman term, meaning the electronic term dominates the equation so \(S\) is approximated by \( \langle S \rangle\) and \[ \langle S_z\rangle = m_s = \pm \dfrac{1}{2} \nonumber \] and \[\langle S_x \rangle = \langle S_y \rangle \approx 0 \nonumber \] \[H_n = A \langle S \rangle \centerdot{I} - g_n\beta_nB\centerdot{I} \nonumber \] Pulling out a \(-g_n\beta_n\) followed by \(I\) leaves \[H_n = -g_n\beta_n \left( -\dfrac{A \langle S \rangle}{g_n\beta_n} + B\right){I} \nonumber \] Substituting the internal magnetic field with \[B_{int} = -\dfrac{A \langle S \rangle }{g_n\beta_n} \nonumber \] results in a combined magnetic field term involving both the applied magnetic field and the internal magnetic field \[H_n = -g_n\beta_n(B_{int} + B)\centerdot{I} \nonumber \] which is simplified by using the effective magnetic field \(B_{eff}\) \[H_n = -g_n\beta_nB_{eff}\centerdot{I} \nonumber \] | 7,787 | 2,616 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Exemplars/Batteries%3A_Electricity_though_chemical_reactions |
Batteries consist of one or more electrochemical cells that store chemical energy for later conversion to electrical energy. Batteries are used in many day-to-day devices such as cellular phones, laptop computers, clocks, and cars. Batteries are composed of at least one electrochemical cell which is used for the storage and generation of electricity. Though a variety of electrochemical cells exist, batteries generally consist of at least one voltaic cell. Voltaic cells are also sometimes referred to as galvanic cells. Chemical reactions and the generation of electrical energy is spontaneous within a voltaic cell, as opposed to the reactions electrolytic cells and . It was while conducting experiments on electricity in 1749 that Benjamin Franklin first coined the term "battery" to describe linked capacitors. However his battery was not the first battery, just the first ever referred to as such. Rather it is believed that the Baghdad Batteries, discovered in 1936 and over 2,000 years old, were some of the first ever batteries, though their exact purpose is still debated. Luigi Galvani (for whom the galvanic cell is named) first described "animal electricity" in 1780 when he created an electrical current through a frog. Though he was not aware of it at the time, this was a form of a battery. His contemporary Alessandro Volta (for whom the voltaic cell and voltaic pile are named) was convinced that the "animal electricity" was not coming from the frog, but something else entirely. In 1800, his produced the first real battery: the voltaic pile. In 1836, John Frederic Daniell created the Daniell cell when researching ways to overcome some of the problems associated with Volta's voltaic pile. This discovery was followed by developments of the Grove cell by William Robert Grove in 1844; the first rechargeable battery, made of a lead-acid cell in 1859 by Gaston Plante; the gravity cell by Callaud in the 1860s; and the Leclanche cell by Georges Leclanche in 1866. Until this point, all batteries were wet cells. Then in 1887 Carl Gassner created the first dry cell battery, made of a zinc-carbon cell. The nickel-cadmium battery was introduced in 1899 by Waldmar Jungner along with the nickel-iron battery. However Jungner failed to patent the nickel-iron battery and in 1903, Thomas Edison patented a slightly modified design for himself. A major breakthrough came in 1955 when Lewis Urry, an employee of what is now know as Energizer, introduced the common alkaline battery. The 1970s led to the nickel hydrogen battery and the 1980s to the nickel metal-hydride battery. Lithium batteries were first created as early as 1912, however the most successful type, the lithium ion polymer battery used in most portable electronics today, was not released until 1996. Voltaic cells are composed of two half-cell reactions (oxidation-reduction) linked together via a semipermeable membrane (generally a salt bath) and a wire (Figure 1). Each side of the cell contains a metal that acts as an electrode. One of the electrodes is termed the cathode, and the other is termed the anode. The side of the cell containing the cathode is reduced, meaning it gains electrons and acts as the oxidizing agent for the anode. The side of the cell containing the anode is where oxidation occurs, meaning it loses electrons and acts as the reducing agent for the cathode. The two electrodes are each submerged in an electrolyte, a compound that consists of ions. This electrolyte acts as a concentration gradient for both sides of the half reaction, facilitating the process of the electron transfer through the wire. This movement of electrons is what produces energy and is used to power the battery. The cell is separated into two compartments because the chemical reaction is spontaneous. If the reaction was to occur without this separation, energy in the form of heat would be released and the battery would not be effective. Primary batteries are non-rechargeable and disposable. The electrochemical reactions in these batteries are non-reversible. The materials in the electrodes are completely utilized and therefore cannot regenerate electricity. Primary batteries are often used when long periods of storage are required, as they have a much lower discharge rate than secondary batteries. Use of primary batteries is exemplified by smoke detectors, flashlights, and most remote controls. Secondary batteries are rechargeable. These batteries undergo electrochemical reactions that can be readily reversed. The chemical reactions that occur in secondary batteries are reversible because the components that react are not completely used up. Rechargeable batteries need an external electrical source to recharge them after they have expended their energy. Use of secondary batteries is exemplified by car batteries and portable electronic devices. Wet cell batteries contain a liquid electrolyte. They can be either primary or secondary batteries. Due to the liquid nature of wet cells, insulator sheets are used to separate the anode and the cathode. Types of wet cells include Daniell cells, Leclanche cells (originally used in dry cells), Bunsen cells, Weston cells, Chromic acid cells, and Grove cells. The lead-acid cells in automobile batteries are wet cells. In dry cell batteries, no free liquid is present. Instead the electrolyte is a paste, just moist enough to allow current flow. This allows the dry cell battery to be operated in any position without worrying about spilling its contents. This is why dry cell batteries are commonly used in products which are frequently moved around and inverted, such as portable electronic devices. Dry cell batteries can be either primary or secondary batteries. The most common dry cell battery is the Leclanche cell. The capacity of a battery depends directly on the quantity of electrode and electrolyte material inside the cell. Primary batteries can lose around 8% to 20% of their charge over the course of a year without any use. This is caused by side chemical reactions that do not produce current. The rate of side reactions can be slowed by lowering temperature. Warmer temperatures can also lower the performance of the battery, by speeding up the side chemical reactions. Primary batteries become polarized with use. This is when hydrogen accumulates at the cathode, reducing the battery's effectiveness. Depolarizers can be used to remove this build up of hydrogen. Secondary batteries self-discharge even more rapidly. They usually lose about 10% of their charge each month. Rechargeable batteries gradually lose capacity after every recharge cycle due to deterioration. This is caused by active materials falling off the electrodes or electrolytes moving away from the electrodes. can be used to approximate relationships between current, capacity, and discharge time. This is represented by the equation \[t = \dfrac{Q_p}{K^k}\] where is the current, is a constant of about 1.3, is the time the battery can sustain the current, and is the capacity when discharged at a rate of 1 amp. There is a significant correlation between a cell's current and voltage. Current, as the name implies, is the flow of electrical charge. Voltage is how much current can potentially flow through the system. Figure 4 illustrates the difference between current and voltage. Standard reduction potential, E , is a measurement of voltage. Standard reduction potential can be calculated with the knowledge that it is the difference in energy potentials between the cathode and the anode E = E − E . For standard conditions, the electrode potentials for the half cells can be determined by using a table of standard electrode potentials. For nonstandard conditions, determining the electrode potential for the cathode and the anode is not as simple as looking at a table. Instead, the Nernst equation must be used in to determine E for each half cell. The Nernst equation is represented by , where is the universal gas constant (8.314 J K mol ), is the temperature in Kelvin, is the number of moles of electrons transferred in the half reaction, is the Faraday constant (9.648 x 10 C mol ), and is the reaction quotient. Batteries vary both in size and voltage due to the chemical properties and contents within the cell. However, batteries of different sizes may have the same voltage. The reason for this phenomenon is that the standard cell potential does not depend on the size of a battery but rather on its internal content. Therefore, batteries of different sizes can have the same voltage (Figure 5). Additionally, there are ways in which batteries can amplify their voltages and current. When batteries are lined up in a series of rows it increases their voltage, and when batteries are lined up in a series of columns it can increases their current. Figure 5: Four batteries of different sizes all of 1.5 voltage Batteries can explode through misuse or malfunction. By attempting to overcharge a rechargeable battery or charging it at an excessive rate, gases can build up in the battery and potentially cause a rupture. A short circuit can also lead to an explosion. A battery placed in a fire can also lead to an explosion as steam builds up inside the battery. Leakage is also a concern, because chemicals inside batteries can be dangerous and damaging. Leakage emitted from the batteries can ruin the device they are housed in, and is dangerous to handle. There are numerous environmental concerns with the widespread use of batteries. The production of batteries consumes many resources and involves the handling of many dangerous chemicals. Used batteries are often improperly disposed of and contribute to electronic waste. The materials inside batteries can potentially be toxic pollutants, making improper disposal especially dangerous. Through electronic recycling programs, toxic metals such as lead and mercury are kept from entering and harming the environment. Consumption of batteries is harmful and can lead to death. Any liquid or moist object that has enough ions to be electrically conductive can be used to make a battery. It is even possible to generate small amounts of electricity by inserting electrodes of different metals into potatoes, lemons, bananas, or carbonated cola. A voltaic pile can be created using two coins and a paper dipped in salt water. Stacking multiple coins in a series can results in an increase in current. 1. Yes/No 2. T/F 3. Determine the standard electrode potential of a voltaic cell within a Leclanche (Dry) cell with half cell voltages of .875V at the graphite cathode and .253V at the zinc anode. 4. Determine the standard electrode potential with given half cell voltages of .987V at the cathode and .632V at the anode. 5. Explain why rechargeable batteries might be advantageous over disposable batteries. 1. Yes/No 2. T/F 3. E =E (cathode)-E (anode) E l= 0.875V - 0.253V = 0.622v 4. E =E (cathode)-E (anode) E = 0.987V -0.632V = 0.355V 5. Even though disposable batteries are cheaper initially and easier to make, the longer lifespan of rechargeable batteries is often more efficient and useful. Rechargeable batteries mean less waste, as less batteries need to be made and less are disposed of in land-fills or through recycling programs. Rechargeable batteries are also more convenient as changing batteries is no longer required. This is especially beneficial in portable electronic devices. Also, because the components in a secondary cell are reusable, rechargeable batteries will generally cost less than disposable batteries in the long run. | 11,634 | 2,617 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Simple_Addition_Reactions |
This page looks at the addition of hydrogen cyanide and sodium hydrogensulphite (sodium bisulphite) to aldehydes and ketones. Hydrogen cyanide adds across the carbon-oxygen double bond in aldehydes and ketones to produce compounds known as hydroxynitriles. These used to be known as . For example, with ethanal (an aldehyde) you get 2-hydroxypropanenitrile: With propanone (a ketone) you get 2-hydroxy-2-methylpropanenitrile: The reaction isn't normally done using hydrogen cyanide itself, because this is an extremely poisonous gas. Instead, the aldehyde or ketone is mixed with a solution of sodium or potassium cyanide in water to which a little sulphuric acid has been added. The pH of the solution is adjusted to about 4 - 5, because this gives the fastest reaction. The reaction happens at room temperature.The solution will contain hydrogen cyanide (from the reaction between the sodium or potassium cyanide and the sulphuric acid), but still contains some free cyanide ions. This is important for the mechanism. The product molecules contain two functional groups: For example, starting from a hydroxynitrile made from an aldehyde, you can quite easily produce relatively complicated molecules like 2-amino acids - the which are used to construct proteins. Sodium hydrogensulphite used to be known as sodium bisulphite, and you might well still come across it in organic textbooks under this name - or using the bisulfite spelling. This reaction only works well for aldehydes. In the case of ketones, one of the hydrocarbon groups attached to the carbonyl group needs to be a methyl group. Bulky groups attached to the carbonyl group get in the way of the reaction happening. The aldehyde or ketone is shaken with a saturated solution of sodium hydrogensulphite in water. Where the product is formed, it separates as white crystals. In the case of ethanal, the equation is: and with propanone, the equation is: These compounds are rarely named systematically, and are usually known as "hydrogensulphite (or bisulphite) addition compounds". The reaction is usually used during the purification of aldehydes (and any ketones that it works for). The addition compound can be split easily to regenerate the aldehyde or ketone by treating it with either dilute acid or dilute alkali. If you have an impure aldehyde, for example, you could shake it with a saturated solution of sodium hydrogensulphite to produce the crystals. These crystals could easily be filtered and washed to remove any other impurities. Addition of dilute acid, for example, would then regenerate the original aldehyde. It would, of course, still need to be separated from the excess acid and assorted inorganic products of the reaction. | 2,726 | 2,618 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Amino_Acids/Properties_of_Amino_Acids/Stereochemistry_of_Amino_Acids |
With the exception of glycine, all the 19 other common amino acids have a uniquely different functional group on the central tetrahedral alpha carbon (i.e. \(C_{\alpha}\)). The \(C_{\alpha}\) is termed "chiral" to indicate there are four different constituents and that the Ca is asymmetric. Since the \(C_{\alpha}\) is asymmetric there exists two possible, non-superimposable, mirror images of the amino acids: How are these two uniquely different structures in the figure ago distinguished? Based off the stereochemistyr at the \(C_{\alpha}\). Glyceraldehyde contains a chiral carbon, and therefore, there are two enantiomers of this molecule. One is labeled the "L" form, and the other the "D" form. This is the frame of reference used to describe amino acid enantiomers as being either the "L" or "D" form Even though the two enantiomers would seem to be essentially equivalent to each other, all common amino acids are found in the "L" enantiomer in living systems. When looking down the H-C, a bond towards the \(C_{\alpha}\) there is a mnemonic to identify the L-enantiomer of amino acids (note: in this view the three functional groups are pointing away from you, and not towards you; the H atom is omitted for clarity - but it would be in front of the C) Starting with the carbonyl functional group, and going clockwise around the \(C_{\alpha}\) of the L-enantiomer, the three functional groups spell out the word CORN. If you follow the above instructions, it will spell out CONR (a silly, meaningless word) for the D-enantiomer Enantiomeric molecules have an optical property known as optical activity - the ability to rotate the plane of plane polarized light. Clockwise rotation is known as "dextrorotatory" behavior and counterclockwise rotation is known as "levorotatory" behavior. All common amino acids are the L-enantiomer (i.e. their \(C_{\alpha}\) chiral center is the L-enantiomer), based on the structural comparison with L-glyceraldehyde. However, not all L-amino acids are Levorotatory, some are actually Dextrorotatory with regard to their optical activity. To (attempt) to avoid confusion, the optical activities are given as (+) for dextrorotatory, and (-) for levorotatory Multiple chiral centers A relative ranking of the "priority" of various functional groups is given as: \[\ce{SH > OH > NH2 > COOH > CHO > CH2OH > CH3 > H}\] This refers to the ability of amino acids to absorb or emit electromagnetic energy at different wavelengths (i.e. energies) The 20 common amino acids differ from one another in several important ways. Here are just two: Amino Acid Mass (Da) Isoelectric Point Amino Acid Mass (Da) Isoelectric Point Aspartic Acid 114.11 2.98 Isoleucine 113.16 6.038 Glutamic Acid 129.12 3.08 Glycine 57.05 6.064 Cysteine 103.15 5.02 Alanine 71.09 6.107 Tyrosine 163.18 5.63 Proline 97.12 6.3 Serine 87.08 5.68 Histidine 137.14 7.64 Methionine 131.19 5.74 Lysine 128.17 9.47 Tryptophan 186.12 5.88 Arginine 156.19 10.76 Phenylalanine 147.18 5.91 Threonine 101.11 - Valine 99.14 6.002 Asparagine 115.09 - Leucine 113.16 6.036 Glutamine 128.14 - We can use these differences in physical properties to fractionate complex mixtures of amino acids into individual amino acids | 3,228 | 2,620 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/01%3A_Structure_of_Solid_Surfaces/1.09%3A_Other_Single_Crystal_Surfaces |
Although some of the more common metallic surface structures have been discussed in previous sections (1.2-1.4), there are many other types of single crystal surface which may be prepared and studied. These include These will not be covered in any depth, but a few illustrative examples are given below to give you a flavor of the additional complexity involved when considering such surfaces. High index surfaces are those for which one or more of the are relatively large numbers. The most commonly studied surfaces of this type are which are cut at a relatively small angle to one of the low index surfaces. The ideal surfaces can then be considered to consist of terraces which have an atomic arrangement identical with the corresponding low index surface, separated by monatomic steps (steps which are a single atom high). As seen above, the ideal fcc(775) surface has a regular array of such steps and these steps are both straight and parallel to one another. What is the coordination number of a step atom on this surface ? The coordination number of the atoms at the steps is 7 Rationale: Each step atom has four nearest neighbors in the surface layer of terrace atoms which terminates at the step, and another three in the layer immediately below ; a total of 7. This contrasts with the CN of surface atoms on the terraces which is 9. By contrast a surface for which all the Miller indices differ must not only exhibit steps but must also contain kinks in the steps. An example of such a surface is the fcc(10.8.7) surface - the ideal geometry of which is shown below. What is the lowest coordination number exhibited by any of the atoms on this surface? The lowest coordination number is 6 which is that exhibited by atoms at the kinks in the steps. Rationale: The lowest coordination number is exhibited by atoms "on the outside" of the kinks in the steps. Such atoms have only three nearest neighbors in the surface layer of terrace atoms which terminates at the step, and another three in the layer immediately below ; a total of 6. This contrasts with the surface atoms on the terraces which have a coordination number of 9 and the normal step atoms which have a coordination number of 7. Real vicinal surfaces do not, of course, exhibit the completely regular array of steps and kinks illustrated for the ideal surface structures, but they do exhibit this type of step and terrace morphology. The special adsorption sites available adjacent to the steps are widely recognized to have markedly different characteristics to those available on the terraces and may thus have an important role in certain types of catalytic reaction. The ideal surface structures of the low index planes of compound materials can be easily deduced from the bulk structures in exactly the same way as can be done for the basic metal structures. For example, the NaCl(100) surface that would be expected from the bulk structure is shown below: In addition to the relaxation and reconstruction exhibited by elemental surfaces, the surfaces of compounds may also show deviations from the bulk stoichiometry due to surface localised reactions (e.g. surface reduction) and/or surface segregation of one or more components of the material. | 3,242 | 2,621 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Fundamentals/Dexter_Energy_Transfer |
Dexter energy transfer is sometimes called short-range, collisional or exchange energy transfer which is a non-radiative process with electron exchange. Dexter Energy transfer although similar to , differs greatly in length scale and underlying mechanism. The energy transfer could take place via the interaction between an excited chemical group, say D*, and a ground-state chemical group, say A, without emitting a photon when transferring energy. Förster provided a model saying that the energy released from an excited donor could simultaneously excite the ground-state acceptor based on the Coulombic interaction between these two chemical groups. Independently, David L. Dexter provided another mechanism that an excited donor group and an acceptor group might indeed exchange electrons to accomplish the non-radiative process. The Dexter exchange energy transfer is generally associated with quenching. The term “quenching” means any physical process or molecular state that decreases the molecular fluorescence. Previously, the Dexter energy transfer was treated a fundamental phenomena in photochemistry field. Recently, it is greatly applied to novel emitting materials, such as white organic light-emitting diodes and energy up-conversion systems. (Blue light emission and white light emission. Dexter energy transfer is a process that two molecules (intermolecular) or two parts of a molecule (intramolecular) bilaterally exchange their electrons. Unlike the sixth-power dependence of Förster energy transfer, the reaction rate constant of Dexter energy transfer exponentially decays as the distance between these two parties increases. On account of the exponential relationship to the distance, the exchange mechanism typically occurs within 10 Angstroms. Hence, the exchange mechanism is also called the short-range energy transfer. The exchange mechanism is based on the Wigner spin conservation rule; thus, the spin-allow process could be: Singlet-singlet energy transfer: \[\ce{^{1}D^{*} + ^1A -> ^{1}D + ^{1}A^{*}}\] It could be understood by: a singlet group produces another singlet group. Triplet-triplet energy transfer: \[\ce{^{3}D^{*} + ^{1}A → ^{1}D + ^{3}A^{*}}\] It could be understood by: a triplet group produces another triplet group. The singlet-singlet energy transfer can happen when undergoing the Coulombic interaction. However, the Coulombic interaction will not involve the triplet-triplet energy transfer because that violates the . As the figure 1 implies, the Dexter energy transfer is a process that the donor and the acceptor exchange their electron. In other words, the exchanged electrons should occupy the orbital of the other party. Hence, besides the overlap of emission spectra of \(D\) and absorption spectra of \(A\), the exchange energy transfer needs the overlap of wavefunctions. In the popular words, it needs the overlap of the electron cloud. The overlap of wavefunctions also implies that the excited donor and ground-state acceptor should be close enough so the exchange could happen. If \(D\) and \(A\) are different molecules, collisions can greatly make them bump into each other. The short distance that makes energy transfer happen is almost comparable to the collisional diameter. This is the reason why the exchange energy transfer is always referred to the term “collision.” The rate constant of exchange energy transfer is given by \[ k_{dexter}= KJ \; \text{exp} \left (\dfrac {-2R_{DA}}{L} \right ) \] while the rate constant of exchange energy transfer is given by \[ k_{dexter} \propto \dfrac {D_{D}^{2} D_{A}^{2}}{R_{DA}^{6}} \] \(J\) is the normalized spectral overlap integral. The term “normalized” here means making the absorption spectra and the emission spectra on the same scale and have the same highest level (For real examples, please refer to ) \(K\) is an experimental factor, \(R_{DA}\) is the distance between \(D\) and \(A\), and \(L\) is the sum of van der Waals radius; \(D_D\) is the transition dipole of donor, \(D_A\) is the transition dipole of acceptor. Comparing rate constants of different energy transfer models, we can find that the rate constant of exchange energy transfer decays steeply because of its intrinsic exponential relationship, which is the reason that the exchange energy transfer is also called the short-range energy transfer and the Förster mechanism is called the long-range energy transfer. According to the Figure \(\Page {1}\), by exchanging electrons, a triplet excited donor could also excite the acceptor to its triplet state. This process is called “sensitization” process; hence, the triplet excited-state donor is the sensitizer. Generally speaking, the time scale for fluorescing a photon is about several nanosecond. The intersystem crossing happens in this time scale, too. However, roughly speaking, fluorescence generally happens faster that the intersystem crossing, which means an electron would fluoresce before it proceeds to occupy a triplet state through thermal relaxation. The (TTA) is an important case of exchange energy transfer. Two triplet chemical groups, D* and A*, react to produce two singlet states. Generally, the energy gap between S and T is bigger than the energy gap between T and S . Hence, if two triplet excited-state molecules encounter, the process might have enough energy to excite one of them to the higher singlet states, which can make the fluorescence happen in this system. After the annihilation process, the energy level that the electron occupies will be twice of the lowest triplet energy gap. The Dexter energy transfer is a fundamental phenomenon in photochemistry. The difference between Förster and Dexter mechanism include: The figure above is the energy levels that stands for an upconversion fluorescence system. Surprisingly, the Dexter energy transfer and the triplet-triplet annihilation mechanisms provide a fantastic way to produce high-energy light simply by using medium-energy light. If choosing the donor and the acceptor properly, one can produce three kinds of wavelengths of fluorescence, say red, green, and blue. After mixing these light, this system might produce white light if the ingredient light is well mixed. Today, one can even simply predict th photophysical properties by using commercial quantum chemistry software, before starting to do the experiment. Below are some questions behind this figure. (These question are adapted from Chemical Communication, 2009, 27, 4064 and Physical Review B, 2008, 78, 195112) Question 1: What does the worm-like arrow "A" stand for? Question 2: What does the worm-like arrow "B" stand for? Question 3: What does the process "C" stand for? Question 4: What does the process "D" stand for? Question 5: The original compound absorbs the light with frequency v1 and ultimately emits the light with frequency v2. Which frequency is bigger? So this research group used low-energy light to produce high-energy light or use inverse idea? Does it make sense? Answer 1: Internal conversion or vibrational relaxation. Answer 2: Intersystem crossing. Answer 3: Dexter exchange energy transfer. Answer 4: Triplet-triplet . | 7,174 | 2,622 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/09%3A_Chemical_Equilibria/9.01%3A_Prelude_to_Chemical_Equilibria |
The small is great, the great is small; all is in equilibrium in necessity... As was discussed in Chapter 6, the natural tendency of chemical systems is to seek a state of minimum Gibbs function. Once the minimum is achieved, movement in any chemical direction will not be spontaneous. It is at this point that the system achieves a state of equilibrium. From the diagram above, it should be clear that the direction of spontaneous change is determined by minimizing \[\left(\frac{\partial G}{\partial \xi}\right)_{p,T}. \nonumber \] If the slope of the curve is negative, the reaction will favor a shift toward products. And if it is positive, the reaction will favor a shift toward reactants. This is a non-trivial point, as it underscores the importance of the composition of the reaction mixture in the determination of the direction of the reaction. | 867 | 2,623 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/10%3A_Electrochemistry/10.04%3A_Entropy_of_Electrochemical_Cells |
The Gibbs function is related to entropy through its temperature dependence \[ \left( \dfrac{\partial \Delta G}{\partial T} \right)_p = - \Delta S \nonumber \] A similar relationship can be derived for the temperature variance of \(E^o\). \[ nF \left( \dfrac{\partial E^o}{\partial T} \right)_p = \Delta S \label{eq2} \] Consider the following data for the Daniel cell (Buckbeei, Surdzial, & Metz, 1969) which is defined by the following reaction \[Zn(s) + Cu^{2+}(aq) \rightleftharpoons Zn^{2+}(aq) + Cu(s) \nonumber \] From a fit of the data to a quadratic function, the temperature dependence of \[\left( \dfrac{\partial E^o}{\partial T} \right)_p \nonumber \] is easily established. The quadratic fit to the data results in \[\left( \dfrac{\partial E^o}{\partial T} \right)_p = 3.8576 \times 10^{-6} \dfrac{V}{°C^2}(T) - 6.3810 \times 10^{-4} \dfrac{V}{°C} \nonumber \] So, at 25 °C, \[\left( \dfrac{\partial E^o}{\partial T} \right)_p = -54166 \times 10^{-4} V/K \nonumber \] noting that \(K\) can be substituted for \(°C\) since in difference they have the same magnitude. So the entropy change is calculated (Equation \ref{eq2}) is \[ \Delta S = nF \left( \dfrac{\partial E^o}{\partial T} \right)_p = (2\,mol)(95484\,C/mol) (-5.4166 \times 10^{-4} V/K) \nonumber \] Because \[ 1\,C \times 1\,V = 1\,J \nonumber \] The standard entropy change for the Daniel cell reaction at 25 °C is It is the negative entropy change that leads to an increase in standard cell potential at lower temperatures. For a reaction such as \[Pb(s) + 2 H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) \nonumber \] which has a large increase in entropy (due to the production of a gas-phase product), the standard cell potential decreases with decreasing temperature. As this is the reaction used in most car batteries, it explains why it can be difficult to start ones car on a very cold winter morning. The topic of temperature dependence of several standard cell potentials is reported and discussed by Bratsch (Bratsch, 1989). | 2,019 | 2,624 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Surface_Science_(Nix)/01%3A_Structure_of_Solid_Surfaces/1.07%3A_Relaxation_and_Reconstruction |
The phenomena of relaxation and reconstruction involve rearrangements of surface ( and near surface ) atoms, this process being driven by the energetics of the system i.e. the desire to reduce the surface free energy (see ). As with all processes, there may be kinetic limitations which prevent or hinder these rearrangements at low temperatures. Both processes may occur with clean surfaces in ultrahigh vacuum, but it must be remembered that adsorption of species onto the surface may enhance, alter or even reverse the process ! Relaxation is a small and subtle rearrangement of the surface layers which may nevertheless be significant energetically, and seems to be commonplace for metal surfaces. It involves adjustments in the layer spacings perpendicular to the surface, there is change either in the periodicity parallel to the surface or to the symmetry of the surface. \(d_{1-2} < d_{bulk}\). The right picture shows the relaxed surface: the first layer of atoms is typically drawn in slightly towards the second layer (i.e. d < d ). We can consider what might be the driving force for this process at the atomic level. If we use a localized model for the bonding in the solid then it is clear that an atom in the bulk is acted upon by a balanced, symmetrical set of forces. On the other hand, an atom at the unrelaxed surface suffers from an imbalance of forces and the surface layer of atoms may therefore be pulled in towards the second layer. (Whether this is a reasonable model for bonding in a metal is open to question !) The magnitude of the contraction in the first layer spacing is generally small ( < 10 % )- compensating adjustments to other layer spacings may extend several layers into the solid. The reconstruction of surfaces is a much more readily observable effect, involving larger (yet still atomic scale) displacements of the surface atoms. It occurs with many of the less stable metal surfaces (e.g. it is frequently observed on fcc(110) surfaces), but is much more prevalent on semiconductor surfaces. Unlike relaxation, the phenomenon of reconstruction involves a change in the periodicity of the surface structure - the diagram below shows a surface, viewed from the side, which corresponds to an unreconstructed termination of the bulk structure. This may be contrasted with the following picture which shows a schematic of a reconstructed surface - this particular example is similar to the "missing row model" proposed for the structure of a number of reconstructed (110) fcc metal surfaces. Since reconstruction involves a change in the periodicity of the surface and in some cases also a change in surface symmetry, it is readily detected using surface diffraction techniques (e.g. LEED & RHEED ). The overall driving force for reconstruction is once again the minimization of the surface free energy - at the atomic level, however, it is not always clear why the reconstruction should reduce the surface free energy. For some metallic surfaces, it may be that the change in periodicity of the surface induces a splitting in surface-localized bands of energy levels and that this can lead to a lowering of the total electronic energy when the band is initially only partly full. In the case of many semiconductors, the simple reconstructions can often be explained in terms of a "surface healing" process in which the co-ordinative unsaturation of the surface atoms is reduced by bond formation between adjacent atoms. For example, the formation of a Si(100) surface requires that the bonds between the Si atoms that form the new surface layer and those that were in the layer immediately above in the solid are broken - this leaves two "dangling bonds" per surface Si atom. A relatively small co-ordinated movement of the atoms in the topmost layer can reduce this unsatisfied co-ordination - pairs of Si atoms come together to form surface "Si dimers", leaving only one dangling bond per Si atom. This process leads to a change in the surface periodicity: the period of the surface structure is doubled in one direction giving rise to the so-called (2x1) reconstruction observed on all clean Si(100) surfaces [ ]. More examples:
Si(111)-(7x7) From the web-pages of the Omicron NanoTechnology GmbH (showing data courtesy of Prof. Hongiun Gao´s group, Institute of Physics, CAS, Beijing, China). In this section, attention has been concentrated on the reconstruction of clean surfaces. It is, however, worth noting that reconstruction of the substrate surface is frequently induced by the adsorption of molecular or atomic species onto the surface - this phenomenon is known as (see Section 2.5 for some examples). The minimization of surface energy means that even single crystal surfaces will not exhibit the ideal geometry of atoms to be expected by truncating the bulk structure of the solid parallel to a particular plane. The differences between the real structure of the clean surface and the ideal structure may be imperceptibly small (e.g. a very slight ) or much more marked and involving a change in the surface periodicity in one or more of the main symmetry directions ( ). | 5,171 | 2,625 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Group_Theory/Group_Theory%3A_Theory |
Symmetry can help resolve many chemistry problems and usually the first step is to determine the symmetry. If we know how to determine the symmetry of small molecules, we can determine symmetry of other targets which we are interested in. Therefore, this module will introduce basic concepts of group theory and after reading this module, you will know how to determine the symmetries of small molecules. Symmetry is very important in chemistry researches and group theory is the tool that is used to determine symmetry. Usually, it is not only the symmetry of molecule but also the symmetries of some local atoms, molecular orbitals, rotations and vibrations of bonds, etc. that are important. For example, if the symmetries of molecular orbital wave functions are known, we can find out information about the binding. Also, by the selection rules that are associated with symmetries, we can explain whether the is forbidden or not and also we can predict and interpret the bands we can observe in or spectrum. Symmetry operations and symmetry elements are two basic and important concepts in group theory. When we perform an operation to a molecule, if we cannot tell any difference before and after we do the operation, we call this operation a This means that the molecule seems unchanged before and after a symmetry operation. As Cotton defines it in his book, when we do a symmetry operation to a molecule, every points of the molecule will be in an equivalent position. For different molecules, there are different kinds of symmetry operations we can perform. To finish a symmetry operation, we may rotate a molecule on a line as an axis, reflect it on a mirror plane, or invert it through a point located in the center. These lines, planes, or points are called symmetry elements. There may be more then one symmetry operations associated with a particular symmetry The molecule does not move and all atoms of the molecule stay at the same place when we apply an identity operation, E, on it. All molecules have the identity operation. Identity operation can also be a combination of different operations when the molecule returns to its original position after these operations are performed. This will be demonstrated later. | 2,260 | 2,626 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/04%3A_Putting_the_First_Law_to_Work/4.04%3A_The_Joule_Experiment |
Going back to the expression for changes in internal energy that stems from assuming that \(U\) is a function of \(V\) and \(T\) (or \(U(V, T)\) for short) \[ dU = \left( \dfrac{\partial U}{\partial V} \right)_TdV+ \left( \dfrac{\partial U}{\partial T} \right)_V dT \nonumber \] one quickly recognizes one of the terms as the constant volume heat capacity, \(C_V\). And so the expression can be re-written \[ dU = \left( \dfrac{\partial U}{\partial V} \right)_T dV + C_V dT \nonumber \] But what about the first term? The partial derivative is a coefficient called the “internal pressure”, and given the symbol \(\pi_T\). \[ \pi_T = \left( \dfrac{\partial U}{\partial V} \right)_T \nonumber \] James Prescott Joule (1818-1889) recognized that \(\pi_T\) should have units of pressure (Energy/volume = pressure) and designed an experiment to measure it. He immersed two copper spheres, A and B, connected by a stopcock. Sphere A is filled with a sample of gas while sphere B was evacuated. The idea was that when the stopcock was opened, the gas in sphere A would expand (\(\Delta V > 0\)) against the vacuum in sphere B (doing no work since \(p_{ext} = 0\). The change in the internal energy could be expressed \[ dU = \pi_T dV + C_V dT \nonumber \] But also, from the first law of thermodynamics \[ dU = dq + dw \nonumber \] Equating the two \[ \pi_T dV + C_V dT = dq + dw \nonumber \] and since \(dw = 0\) \[ \pi_T dV + C_V dT = dq \nonumber \] Joule concluded that \(dq = 0\) (and \(dT = 0\) as well) since he did not observe a temperature change in the water bath which could only have been caused by the metal spheres either absorbing or emitting heat. And because \(dV > 0\) for the gas that underwent the expansion into an open space, \(\pi_T\) must also be zero! In truth, the gas did undergo a temperature change, but it was too small to be detected within his experimental precision. Later, we (once we develop the ) will show that \[ \left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial p}{\partial T} \right)_V -p \label{eq3} \] For an ideal gas \(p = RT/V\), so it is easy to show that \[\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{R}{V} \label{eq4} \] so combining Equations \ref{eq3} and \ref{eq4} together to get \[ \left( \dfrac{\partial U}{\partial V} \right)_T = \dfrac{RT}{V} - p \label{eq5} \] And since also becuase \(p = RT/V\), then Equation \ref{eq5} simplifies to \[ \left( \dfrac{\partial U}{\partial V} \right)_T = p -p = 0 \nonumber \] So while Joule’s observation was consistent with limiting ideal behavior, his result was really an artifact of his experimental uncertainty masking what actually happened. For a van der Waals gas, \[ p = \dfrac{RT}{V-b} - \dfrac{a}{V^2} \label{eqV1} \] so \[\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{R}{V-b} \label{eqV2} \] and \[ \left( \dfrac{\partial U}{\partial V} \right)_T = T\dfrac{R}{V-b} - p \label{eqV3} \] Substitution of the expression for \(p\) (Equation \ref{eqV1}) into this Equation \ref{eqV3} \[ \left( \dfrac{\partial U}{\partial V} \right)_T = \dfrac{a}{V^2} \nonumber \] In general, it can be shown that \[\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{\alpha}{\kappa_T} \nonumber \] And so the internal pressure can be expressed entirely in terms of measurable properties \[ \left( \dfrac{\partial U}{\partial V} \right)_T = T \dfrac{\alpha}{\kappa_T}-p \nonumber \] and need not apply to only gases (real or ideal)! | 3,485 | 2,627 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Bruice)/15%3A_Aromaticity_(Reactions_of_Benzene)/15.15%3A_The_Friedel-Crafts_Alkylation_of_Benzene |
This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and chloromethane in the presence of an aluminium chloride catalyst. Any other chloroalkane would work similarly. Alkylation means substituting an alkyl group into something - in this case into a benzene ring. A hydrogen on the ring is replaced by a group like methyl or ethyl and so on. Benzene is treated with a chloroalkane (for example, chloromethane or chloroethane) in the presence of aluminum chloride as a catalyst. On this page, we will look at substituting a methyl group, but any other alkyl group could be used in the same way. Substituting a methyl group gives methylbenzene - once known as toluene. \[C_6H_6 + CH_3Cl \rightarrow C_6H_5CH_3 + HCl\] or better: The aluminium chloride isn't written into these equations because it is acting as a . If you wanted to include it, you could write AlCl over the top of the arrow. The electrophile is CH . It is formed by reaction between the chloromethane and the aluminum chloride catalyst. \[ CH_3Cl + AlCl_3 \rightarrow ^+CH_3 + AlCl_4^-\] Stage one Stage two The hydrogen is removed by the \(AlCl_4^-\) ion wh ich was formed at the same time as the \(CH_3^+\) ele ctrophile . The aluminum chloride catalyst is re-generated in this second stage. Jim Clark ( ) | 1,364 | 2,628 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/05%3A_Dioxygen_Reactions/5.03%3A_Copper-zinc_Superoxide_Dismutase |
Two families of metalloproteins are excellent catalysts for the disproportionation of superoxide (Reaction 5.95). \[2O_{2}^{-} + 2 H^{+} \xrightarrow{SOD} O_{2} + H_{2}O_{2} \tag{5.95}\] These are (1) the copper-zinc superoxide dismutases, CuZnSOD, found in almost all eukaryotic cells and a very few prokaryotes, and (2) the manganese and iron superoxide dismutases, MnSOD and FeSOD, the former found in the mitochondria of eukaryotic cells, and both found in many prokaryotes. Recent studies of bacterial and yeast mutants that were engineered to contain no superoxide dismutases demonstrated that the cells were unusually sensitive to dioxygen and that the sensitivity to dioxygen was relieved when an SOD gene was reintroduced into the cells. These results indicate that the superoxide dismutase enzymes playa critical role in dioxygen metabolism, but they do not define the chemical agent responsible for dioxygen toxicity (see Section III). Several transition-metal complexes have been observed to catalyze superoxide disproportionation; in fact, aqueous copper ion, Cu , is an excellent SOD catalyst, comparable in activity to CuZnSOD itself! Free aqueous Cu would not itself be suitable for use as an SOD , however, because it is too toxic (see Section III) and because it binds too strongly to a large variety of cellular components and thus would not be present as the free ion. (Most forms of complexed cupric ion show much less superoxide dismutase activity than the free ion.) Aside from aqueous copper ion, few other complexes are as effective as the SOD enzymes. Two mechanisms (Reactions 5.96 to 5.99) have been proposed for catalysis of superoxide disproportionation by metal complexes and metalloenzymes. Mechanism I: $$M^{n+} + O_{2}^{-} \rightarrow M^{(n-1)+} + O_{2} \tag{5.96}\] \[M^{(n-1)+} + O_{2}^{-} \rightarrow M^{n+}(O_{2}^{2-}) \xrightarrow{2H^{+}} M^{n+} + H_{2}O_{2} \tag{5.97}\] Mechanism II: $$M^{n+} + O_{2}^{-} \rightarrow M^{n+} (O_{2}^{-}) \tag{5.98}\] \[M^{n+} (O_{2}^{-}) + O_{2}^{-} \rightarrow M^{n+}(O_{2}^{2-}) \xrightarrow{2 H^{+}} M^{n} + H_{2}O_{2} \tag{5.99}\] \[+O_{2}\] In Mechanism I, which is favored for the SOD enzymes and most redox-active metal complexes with SOD activity, superoxide reduces the metal ion in the first step, and then the reduced metal ion is reoxidized by another superoxide, presumably via a metal-peroxo complex intermediate. In Mechanism II, which is proposed for nonredox metal complexes but may be operating in other situations as well, the metal ion is never reduced, but instead forms a superoxo complex, which is reduced to a peroxo complex by a second superoxide ion. In both mechanisms, the peroxo ligands are protonated and dissociate to give hydrogen peroxide. Analogues for each of the separate steps of Reactions (5.96) to (5.99) have been observed in reactions of superoxide with transition-metal complexes, thereby establishing the feasibility of both mechanisms. For example, superoxide was shown to reduce Cu phen) to give Cu (phen) (phen = 1,10-phenanthroline), a reaction analogous to Reaction (5.96). On the other hand, superoxide reacts with Cu (tet b) to form a superoxo complex (a reaction analogous to Reaction 5.98), presumably because Cu (tet b) is not easily reduced to the cuprous state, because the ligand cannot adjust to the tetrahedral geometry that Cu prefers. \(\tag{5.100}\) Reaction of superoxide with a reduced metal-ion complex to give oxidation of the complex and release of hydrogen peroxide (analogous to Reaction 5.97) has been observed in the reaction of Fe EDTA with superoxide. Reduction of a Co superoxo complex by free superoxide to give a peroxo complex (analogous to Reaction 5.99) has also been observed. If a metal complex can be reduced by superoxide and if its reduced form can be oxidized by superoxide, both at rates competitive with superoxide disproportionation, the complex can probably act as an SOD by Mechanism I. Mechanism II has been proposed to account for the apparent catalysis of superoxide disproportionation by Lewis acidic nonredox-active metal ions under certain conditions. However, this mechanism should probably be considered possible for redox metal ions and the SOD enzymes as well. It is difficult to distinguish the two mechanisms for redox-active metal ions and the SOD enzymes unless the reduced form of the catalyst is observed directly as an intermediate in the reaction. So far it has not been possible to observe this intermediate in the SOD enzymes or the metal complexes. The x-ray crystal structure of the oxidized form of CuZnSOD from bovine erythrocytes shows a protein consisting of two identical subunits held together almost entirely by hydrophobic interactions. Each subunit consists of a flattened cylindrical barrel of \(\beta\)-pleated sheet from which three external loops of irregular structure extend (Figure 5.15). The metal-binding region of the protein binds Cu and Zn in close proximity to each other, bridged by the imidazolate ring of a histidyI side chain. Figure 5.16 represents the metal-binding region. The Cu ion is coordinated to four histidyl imidazoles and a water in a highly distorted square-pyramidal geometry with water at the apical position. The Zn ion is coordinated to three histidyl imidazoles (including the one shared with copper) and an aspartyl carboxylate group, forming a distorted tetrahedral geometry around the metal ion. One of the most unusual aspects of the structure of this enzyme is the occurrence of the bridging imidazolate ligand, which holds the copper and zinc ions 6 Å apart. Such a configuration is not unusual for imidazole complexes of metal ions, which sometimes form long polymeric imidazolate-bridged structures. \(\tag{5.101}\) However, no other imidazolate-bridged bi- or polymetallic metalloprotein has yet been identified. The role of the zinc ion in CuZnSOD appears to be primarily structural. There is no evidence that water, anions, or other potential ligands can bind to the zinc, so it is highly unlikely that superoxide could interact with that site. Moreover, removal of zinc under conditions where the copper ion remains bound to the copper site does not significantly diminish the SOD activity of the enzyme. However, such removal does result in a diminished thermal stability, i.e., the zinc-depleted protein denatures at a lower temperature than the native protein, supporting the hypothesis that the role of the zinc is primarily structural in nature. The copper site is clearly the site of primary interaction of superoxide with the protein. The x-ray structure shows that the copper ion lies at the bottom of a narrow channel that is large enough to admit only water, small anions, and similarly small ligands (Figure 5.17). In the lining of the channel is the positively charged side chain of an arginine residue, 5 Å away from the copper ion and situated in such a position that it could interact with superoxide and other anions when they bind to copper. Near the mouth of the channel, at the surface of the protein, are two positively charged lysine residues, which are believed to play a role in attracting anions and guiding them into the channel. Chemical modification of these lysine or arginine residues substantially diminishes the SOD activity, supporting their role in the mechanism of reaction with superoxide. The x-ray structural results described above apply only to the oxidized form of the protein, i.e., the form containing Cu . The reduced form of the enzyme containing Cu is also stable and fully active as an SOD. If, as is likely, the mechanism of CuZnSOD-catalyzed superoxide disproportionation is Mechanism I (Reactions 5.96-5.97), the structure of the reduced form is of critical importance in understanding the enzymatic mechanism. Unfortunately, that structure is not yet available. The mechanism of superoxide disproportionation catalyzed by CuZnSOD is generally believed to go by Mechanism I (Reactions 5.96-5.97), i.e., reduction of Cu to Cu by superoxide with the release of dioxygen, followed by reoxidation of Cu to Cu by a second superoxide with the release of HO or H O . The protonation of peroxide dianion, O , prior to its release from the enzyme is required, because peroxide dianion is highly basic and thus too unstable to be released in its unprotonated form. The source of the proton that protonates peroxide in the enzymatic mechanism is the subject of some interest. Reduction of the oxidized protein has been shown to be accompanied by the uptake of one proton per subunit. That proton is believed to protonate the bridging imidazolate in association with the breaking of the bridge upon reduction of the copper. Derivatives with Co substituted for Zn at the native zinc site have been used to follow the process of reduction of the oxidized Cu form to the reduced Cu form. The Co in the zinc site does not change oxidation state, but acts instead as a spectroscopic probe of changes occurring at the native zinc-binding site. Upon reduction (Reaction 5.102), the visible absorption band due to Co shifts in a manner consistent with a change occurring in the ligand environment of Co . The resulting spectrum of the derivative containing Cu in the copper site and Co in the zinc site is very similar to the spectrum of the derivative in which the copper site is empty and the zinc site contains Co . This result suggests strongly that the imidazolate bridge is cleaved and protonated and that the resulting imidazole ligand is retained in the coordination sphere of Co (Reaction 5.102). \(\tag{5.102}\) The same proton is thus an attractive possibility for protonation of peroxide as it is formed in the enzymatic mechanism (Reactions 5.103 and 5.104). \(\tag{5.103}\) \(\tag{5.104}\) Attractive as this picture appears, there are several uncertainties about it. For example, the turnover of the enzyme may be too fast for protonation and deprotonation of the bridging histidine to occur. Moreover, the mechanism proposed would require the presence of a metal ion at the zinc site to hold the imidazole in place and to regulate the pK of the proton being transferred. The observation that removal of zinc gives a derivative with almost full SOD activity is thus surprising and may also cast some doubt on this mechanism. Other criticisms of this mechanism have been recently summarized. Studies of CuZnSOD derivatives prepared by site-directed mutagenesis are also providing interesting results concerning the SOD mechanism. For example, it has been shown that mutagenized derivatives of human CuZnSOD with major differences in copper-site geometry relative to the wild-type enzyme may nonetheless remain fully active. Studies of these and similar derivatives should provide considerable insight into the mechanism of reaction of CuZnSOD with superoxide. Studies of the interaction of CuZnSOD and its metal-substituted derivatives with anions have been useful in predicting the behavior of the protein in its reactions with its substrate, the superoxide anion, O . Cyanide, azide, cyanate, and thiocyanate bind to the copper ion, causing dissociation of a histidyl ligand and the water ligand from the copper. Phosphate also binds to the enzyme at a position close to the Cu center, but it apparently does not bind directly to it as a ligand. Chemical modification of Arg-141 with phenylglyoxal blocks the interaction of phosphate with the enzyme, suggesting that this positively charged residue is the site of interaction with phosphate. Electrostatic calculations of the charges on the CuZnSOD protein suggest that superoxide and other anions entering into the vicinity of the protein will be drawn toward and into the channel leading down to the copper site by the distribution of positive charges on the surface of the protein, the positively charged lysines at the mouth of the active-site cavity, and the positively charged arginine and copper ion within the active-site region. Some of the anions studied, e.g., CN , F , N ,and phosphate, have been shown to inhibit the SOD activity of the enzyme. The source of the inhibition is generally assumed to be competition with superoxide for binding to the copper, but it may sometimes result from a shift in the redox potential of copper, which is known to occur sometimes when an anion binds to copper. In the example described above, studies of a metal-substituted derivative helped in the evaluation of mechanistic possibilities for the enzymatic reaction. In addition, studies of such derivatives have provided useful information about the environment of the metal-ion binding sites. For example, metal-ion-substituted derivatives of CuZnSOD have been prepared with Cu , Cu , Zn , Ag , Ni , or Co bound to the native copper site, and with Zn , Cu , Cu , Co , Hg , Cd , Ni , or Ag bound to the native zinc site. The SOD activities of these derivatives are interesting; only those derivatives with copper in the copper site have a high degree of SOD activity, whereas the nature of the metal ion in the zinc site or even its absence has little or no effect. Derivatives of CuZnSOD are known with Cu ion bound either to the native copper site or to the native zinc site. The electronic absorption spectra of these derivatives indicate that the ligand environments of the two sites are very different. Copper(II) is a d transition-metal ion, and its d-d transitions are usually found in the visible and near-IR regions of the spectrum. Copper(II) complexes with coordinated nitrogen ligands are generally found to have an absorption band between 500 and 700 nm, with an extinction coefficient below 100 M cm . Bands in the absorption spectra of complexes with geometries that are distorted away from square planar tend to be red-shifted because of a smaller d-d splitting, and to have higher extinction coefficients because of the loss of centrosymmetry. Thus the optical spectrum of CuZnSOD with an absorption band with a maximum at 680 nm (14,700 cm ; see Figure 5.18A) and an extinction coefficient of 155 M cm per Cu is consistent with the crystal structural results that indicate that copper(II) is bound to four imidazole nitrogens and a water molecule in a distorted square-pyramidal geometry. Metal-substituted derivatives with Cu at the native copper site but with Co , Cd , Hg , or Ni substituted for Zn at the native zinc site all have a band at 680 nm, suggesting that the substitution of another metal ion for zinc perturbs the copper site very little, despite the proximity of the two metal sites. The absorption spectra of native CuZnSOD and these CuMSOD derivatives also have a shoulder at 417 nm (24,000 cm ; see Figure 5.18A), which is at lower energy than normal imidazole-to-Cu charge-transfer transitions, and has been assigned to an imidazolate-to-Cu charge transfer, indicating that the imidazolate bridge between Cu and the metal ion in the native zinc site is present, as observed in the crystal structure of CuZnSOD. Derivatives with the zinc site empty, which therefore cannot have an imidazolate bridge, are lacking this 417 nm shoulder. Small but significant changes in the absorption spectrum are seen when the metal ion is removed from the zinc site, e.g., in copper-only SOD (Figure 5.18B). The visible absorption band shifts to 700 nm (14,300 cm ), presumably due to a change in ligand field strength upon protonation of the bridging imidazolate. In addition, the shoulder at 417 nm has disappeared, again due to the absence of the imidazolate ligand. The spectroscopic properties due to copper in the native zinc site are best observed in the derivative Ag CuSOD, which has Ag in the copper site and Cu in the zinc site (see Figure 5.18C), since the d Ag ion is spectroscopically silent. In this derivative, the d-d transition is markedly red-shifted from the visible region of the spectrum into the near-IR, indicating that the ligand environment of Cu in that site is either tetrahedral or five coordinate. The EPR properties of Cu in this derivative are particularly interesting (as discussed below). The derivative with Cu bound at both sites, CuCuSOD, has a visible-near IR spectrum that is nearly a superposition of the spectra of CuZnSOD and Ag CuSOD (see Figure 5.19), indicating that the geometry of Cu in each of these sites is little affected by the nature of the metal ion in the other site. EPR spectroscopy has also proven to be particularly valuable in characterizing the metal environments in CuZnSOD and derivatives. The EPR spectrum of native CuZnSOD is shown in Figure 5.20A. The g resonance is split by the hyperfine coupling between the unpaired electron on Cu and the I = \(\frac{3}{2}\) nuclear spin of copper. The A value, 130 G, is intermediate between the larger A typical of square-planar Cu complexes with four nitrogen donor ligands and the lower A observed in blue copper proteins (see Chapter 6). The large linewidth seen in the g region indicates that the copper ion is in a rhombic (i.e., distorted) environment. Thus, the EPR spectrum is entirely consistent with the distorted square-pyramidal geometry observed in the x-ray structure. Removal of zinc from the native protein to give copper-only SOD results in a perturbed EPR spectrum, with a narrower g resonance and a larger A value (142 G) more nearly typical of Cu in an axial N environment (Figure 5.20B). Apparently the removal of zinc relaxes some constraints imposed on the geometry of the active-site ligands, allowing the copper to adopt to a geometry closer to its preferred tetragonal arrangement. The EPR spectrum due to Cu in the native Zn site in the Ag CuSOD derivative indicates that Cu is in a very different environment than when it is in the native copper site (Figure 5.20C). The spectrum is strongly rhombic, with a low value of A (97 G), supporting the conclusion based on the visible spectrum that copper is bound in a tetrahedral or five-coordinate environment. This type of site is unusual either for copper coordination complexes or for copper proteins in general, but does resemble the Cu EPR signal seen when either laccase or cytochrome c oxidase is partially reduced (see Figure 5.21). Partial reduction disrupts the magnetic coupling between these Cu centers that makes them EPR-silent in the fully oxidized protein. The EPR spectrum of CuCuSOD is very different from that of any of the other copper-containing derivatives (Figure 5.22) because the unpaired spins on the two copper centers interact and magnetically couple across the imidazolate bridge, resulting in a triplet EPR spectrum. This spectrum is virtually identical with that of model imidazolate-bridged binuclear copper complexes. Electronic absorption and EPR studies of derivatives of CuZnSOD containing Cu have provided useful information concerning the nature of the metal binding sites of those derivatives. H NMR spectra of those derivatives are generally not useful, however, because the relatively slowly relaxing paramagnetic Cu center causes the nearby proton resonances to be extremely broad. This difficulty has been overcome in two derivatives, CuCoSOD and CuNiSOD, in which the fast-relaxing paramagnetic Co and Ni centers at the zinc site interact across the imidazolate bridge and increase the relaxation rate of the Cu center, such that well-resolved paramagnetically shifted H NMR spectra of the region of the proteins near the two paramagnetic metal centers in the protein can be obtained and the resonances assigned. The use of H NMR to study CuCoSOD derivatives of CuZnSOD in combination with electronic absorption and EPR spectroscopies has enabled investigators to compare active-site structures of a variety of wild-type and mutant CuZnSOD proteins in order to find out if large changes in active-site structure have resulted from replacement of nearby amino-acid residues. | 19,934 | 2,629 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO2._General_Reactivity_Patterns |
In the following pictures, a number of anions are added to a simple carbonyl compound, a ketone (2-propanone, or acetone). In each case, addition of the nucleophile is followed by addition of a proton source. Note that, overall, the reaction involves addition of the nucleophile to the carbonyl carbon and addition of the proton to the carbonyl oxygen. Figure CO2.1. Addition of some anionic (and "semi-anionic") nucleophiles to a ketone. Look at the way the reaction is presented in each case. The organic (carbon-based) starting material is presented on the left hand side of the reaction arrow. The reagent added to this starting material is often shown over the arrow. This reagent transforms the starting material into something else. That something else, the product, is shown to the right of the arrow. Very often, the solvent for the reaction is shown underneath the arrow. The solvent is the liquid that is used to dissolve the starting material and reagents. This is done for a number of reasons. First, reactions generally happen much more quickly in solution than they do without a solvent. When dissolved, the reactants can move around more easily and bump into each other, as if they are swimming. Also, most useful reactions generate heat, and the solvent acts as a heat sink, carrying the excess heat away. (People who have not thought about the importance of solvent sometimes accidentally start fires as a result.) However, there are exceptions, and not all reactions need solvent. These reactions shown above do need solvent, but the solvent is not shown for other reasons. These reagents must be added in a particular order: first the nucleophile and then the base. The nucleophile and base cannot be allowed to mix before the nucleophile has a chance to react with the carbonyl. The pattern of reactivity is very different with another class of nucleophile. These could be called neutral nucleophiles (as opposed to anionic ones). Take another look at the general pattern of reactivity for the anionic nucleophiles and the neutral nucleophiles. In the case of the anionic nucleophiles, the pattern is relatively easy to discover. The product has incorporated the nucleophile into its structure (or at least the anionic part of the nucleophile, which you will soon learn about). The nucleophile has attached at the carbonyl carbon. The carbonyl oxygen has become part of a hydroxyl group. These are very common patterns in the addition of nucleophiles to carbonyls. In the case of the neutral nucleophiles, there are some similarities and some differences. The nucleophile is still incorporated into the product structure. It has added at the carbonyl position in the electrophile. However, the fate of the carbonyl oxygen is a little bit different with neutral nucleophiles. Generally, this atom is lost as a water molecule in these cases. If you look closely, you will be able to tell where the two protons come from in each case in order to form the water molecule. It's not really the HCl, which is only added in very tiny amounts and acts catalytically. The protons come from other positions in the nucleophile, and sometimes from the electrophile, too. This chapter will help you to develop skills so that you can recognize where nucleophilic additions have taken place in reactions. You will also be able to predict what products may result from a nucleophilic addition. Much of the chapter will focus on mechanisms of reaction. A reaction mechanism is, at the very least, the series of elementary steps needed to accomplish an overall reaction, and all of the intermediate structures that would be formed on the way from the reactants to the products. An elementary reaction is typically a bond-forming or a bond-breaking step. In a bond-forming step, a pair of electrons are donated from one atom to another. In a bond-breaking step, a pair of electrons that were shared between two atoms are drawn to one end of the bond or the other, so that the bond breaks and the electrons end up on one atom only. Very often, curved arrows are used to show the path that electrons take in these elementary steps. These arrows are always drawn from the source of the electrons to the place to which the electrons are attracted. These arrows help to illustrate bond-making and bond-breaking steps and also serve a book-keeping function, helping us to keep track of electrons over the course of the reaction. Often, only one arrow is required in showing an elementary step, but not always. Sometimes, a bond-making step can happen at the same time as a bond-breaking step. This usually happens when an atom isn't large enough to accommodate the electrons from the new bond and sill keep the electrons from an old bond. In this case, two pairs of electrons move in the same elementary step, so two curved arrows are shown. Very rareley, more than two curved arrows are needed to show the events in one elementary step. Sometimes other information is displayed in a reaction mechanism. Computational chemists will often leave out the curved arrow notation but will instead indicate the relative energy differences between all the intermediate structures along the reaction pathway. These energies may be experimentally determined (i.e. they may be based on the measurement of real reactions) or they may be calculated using an appropriate level of quantum theory. The energies may be displayed numerically, possibly in a table, or they may be illustrated using a picture, such as a reaction profile. , | 5,527 | 2,630 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Elimination_Reactions/Elimination_Reactions/E._The_Dehydration_of_Ethanol |
This page looks at the mechanism for the acid catalysed dehydration of a simple primary alcohol like ethanol to give an alkene like ethene. Ethanol can be dehydrated to give ethene by heating it with an excess of concentrated sulphuric acid at about 170°C. Concentrated phosphoric(V) acid, H PO , can be used instead. The acids aren't written into the equation because they serve as catalysts. If you like, you could write, for example, "conc H SO " over the top of the arrow. You will find two versions of the mechanism for the dehydration of primary alcohols on the web and in various textbooks. One of these is exactly the same as the mechanism for the reaction involving propan-2-ol and other secondary or tertiary alcohols (known technically as an E1 mechanism), but the other is different (known as an E2 mechanism). The more reliable sources give the E2 mechanism for the dehydration of primary alcohols including ethanol. I am going to treat this as the "correct" version. If you have read the page on the dehydration of propan-2-ol, you will know that it involves the formation of a carbocation (a carbonium ion). If ethanol used the same mechanism, you would get a primary carbocation formed, CH CH , but this is much less stable than a secondary or tertiary carbocation. That would lead to a very high activation energy for the reaction. The alternative mechanism avoids the formation of the carbocation, and so avoids the high activation energy. We are going to discuss the mechanism using sulphuric acid. In the first stage, one of the lone pairs of electrons on the oxygen picks up a hydrogen ion from the sulphuric acid. The alcohol is said to be protonated. That is exactly the same as happens with propan-2-ol and the other secondary and tertiary alcohols. In the mechanism we have already looked at with propan-2-ol, the next thing to happen was loss of water to form a carbocation, followed by removal of a hydrogen ion from the carbocation and the formation of a double bond. In this case, instead of happening in two separate steps, this all happens at the same time in one smooth operation. By doing that, you avoid the formation of an unstable primary carbocation. I am not giving a simplified version of this mechanism just in terms of hydrogen ions. If you don't show something removing the hydrogen ion from the protonated alcohol, you are really missing an important feature of the reaction. This is the mechanism that looks justs like the one involving propan-2-ol, and involves the formation of an unstable primary carbocation. This is the version that you would have found on Chemguide up to December 2012, and was originally included because it was the one wanted by one of the UK Exam Boards at the time I wrote the page in 2000. It is possible that your examiners may still want it. As far as I am aware, the only syllabus which I track which still asks this topic is IB, and one of their mark schemes gave the version below. If you need to know about this, you need to check what your examiners are currently expecting. | 3,067 | 2,631 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.02%3A_Glassware_and_Equipment/1.2F%3A_Drying_Glassware |
If dry glassware is not needed right away, it should be rinsed with distilled water and allowed to dry overnight (in a locker). If dry glassware is promptly needed, glassware can be rinsed with acetone and the residual acetone allowed to evaporate. Rinsing with acetone works well because water is miscible with acetone, so much of the water is removed in the rinse waste. Evaporation of small amounts of residual acetone can be expedited by placing the rinsed glassware in a warm oven for a short amount of time or by using suction from a tube connected to the water aspirator. Residual acetone should be evaporated inside a hot oven (>\(100^\text{o} \text{C}\)) as acetone may polymerize and/or ignite under these conditions. It should also be evaporated using the house compressed air lines, as this is likely to contaminate the glassware with dirt, oil, and moisture from the air compressor. Glassware that appears "dry" actually contains a thin film of water condensation on its surface. When using reagents that react with water (sometimes violently!), this water layer needs to be removed. To evaporate the water film, glassware can be placed in a \(110^\text{o} \text{C}\) oven overnight, or at the least for several hours. The water film can also be manually evaporated using a burner or heat gun, a process called " ". Both methods result in extremely hot glassware that must be handled carefully with tongs or thick gloves. To flame dry glassware, first remove any vinyl sleeves on an extension clamp (Figure 1.11a), as these can melt or catch on fire. Clamp the flask to be dried, including a stir bar if using (Figure 1.11b). Apply the burner or heat gun to the glass, and initially fog will be seen as water vaporizing from one part of the glassware condenses elsewhere (Figure 1.11c). Continue waving the heat source all over the glassware for several minutes until the fog is completely removed and glassware is scorching hot (Figure 1.11d). If the glass is only moderately hot, water will condense from the air before you are able to fully exclude it. Glassware will be hot after flame drying. Regardless of the manner in which glassware is heated (oven or flame drying), allow the glassware to cool in a water-free environment (in a desiccator, under a stream of inert gas, or with a drying tube, Figure 1.12) before obtaining a mass or adding reagents. A drying tube is used when moderately but not meticulously dry conditions are desired in an apparatus. If meticulously dry conditions are necessary, glassware should be oven or flame dried, then the air displaced with a dry, inert gas. Drying tubes are pieces of glassware that can be filled with a drying agent (often anhydrous \(\ce{CaCl_2}\) or \(\ce{CaSO_4}\) in the pellet form) and connected to an apparatus either through a thermometer adapter (Figures 1.13 b+c) or rubber tubing (Figure 1.13d). Air passing through the tube is removed of water when it comes in contact with the drying agent. Since it is important that air can flow through the drying tube, especially so the apparatus is not a closed system, the drying agent should be fresh as used drying agents can sometimes harden into a plug that restricts air flow. Drying tubes can also be filled with basic solids such as \(\ce{Na_2CO_3}\) to neutralize acidic gases. | 3,323 | 2,633 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_A_Molecular_Approach_(Tro)/05%3A_Gases/5.06%3A_Mixtures_of_Gases_and_Partial_Pressures |
In our use of the ideal gas law thus far, we have focused entirely on the properties of pure gases with only a single chemical species. But what happens when two or more gases are mixed? In this section, we describe how to determine the contribution of each gas present to the total pressure of the mixture. The ideal gas law that all gases behave identically and that their behavior is independent of attractive and repulsive forces. If volume and temperature are held constant, the ideal gas equation can be rearranged to show that the pressure of a sample of gas is directly proportional to the number of moles of gas present: \[P=n \left(\dfrac{RT}{V}\right) = n \times \rm const. \label{10.6.1} \] Nothing in the equation depends on the of the gas—only the amount. With this assumption, let’s suppose we have a mixture of two ideal gases that are present in equal amounts. What is the total pressure of the mixture? Because the pressure depends on only the total number of particles of gas present, the total pressure of the mixture will simply be twice the pressure of either component. More generally, the total pressure exerted by a mixture of gases at a given temperature and volume is the sum of the pressures exerted by each gas alone. Furthermore, if we know the volume, the temperature, and the number of moles of each gas in a mixture, then we can calculate the pressure exerted by each gas individually, which is its partial pressure, the pressure the gas would exert if it were the only one present (at the same temperature and volume). To summarize, . This law was first discovered by John Dalton, the father of the atomic theory of matter. It is now known as . We can write it mathematically as where \(P_{tot}\) is the total pressure and the other terms are the partial pressures of the individual gases (up to \(n\) component gases). For a mixture of two ideal gases, \(A\) and \(B\), we can write an expression for the total pressure: More generally, for a mixture of \(n\) component gases, the total pressure is given by Equation \(\ref{10.6.2b}\) restates Equation \(\ref{10.6.3}\) in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of dozens or even hundreds of gaseous species. For Equation \(\ref{10.6.2b}\) to be valid, the identity of the particles present cannot have an effect. Thus an ideal gas must be one whose properties are not affected by either the size of the particles or their intermolecular interactions because both will vary from one gas to another. The calculation of total and partial pressures for mixtures of gases is illustrated in Example \(\Page {1}\). Deep-sea divers must use special gas mixtures in their tanks, rather than compressed air, to avoid serious problems, most notably a condition called “the bends.” At depths of about 350 ft, divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of \(O_2\) and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature? masses of components, total volume, and temperature partial pressures and total pressure The number of moles of \(\ce{He}\) is \[n_{\rm He}=\rm\dfrac{326.4\;g}{4.003\;g/mol}=81.54\;mol \nonumber \] The number of moles of \(\ce{O_2}\) is \[n_{\rm O_2}=\rm \dfrac{51.2\;g}{32.00\;g/mol}=1.60\;mol \nonumber \] We can now use the ideal gas law to calculate the partial pressure of each: \[P_{\rm He}=\dfrac{n_{\rm He}RT}{V}=\rm\dfrac{81.54\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=196.2\;atm \nonumber \] \[P_{\rm O_2}=\dfrac{n_{\rm O_2} RT}{V}=\rm\dfrac{1.60\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=3.85\;atm \nonumber \] The total pressure is the sum of the two partial pressures: \[P_{\rm tot}=P_{\rm He}+P_{\rm O_2}=\rm(196.2+3.85)\;atm=200.1\;atm \nonumber \] A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder. \(P_{CH_4}=137 \; atm\); \(P_{C_2H_6}=13.4\; atm\); \(P_{tot}=151\; atm\) The composition of a gas mixture can be described by the mole fractions of the gases present. The mole fraction (\(\chi\)) of any component of a mixture is the ratio of the number of moles of that component to the total number of moles of all the species present in the mixture (\(n_{tot}\)): \[\chi_A=\dfrac{\text{moles of A}}{\text{total moles}}= \dfrac{n_A}{n_{tot}} =\dfrac{n_A}{n_A+n_B+\cdots}\label{10.6.5} \] The mole fraction is a dimensionless quantity between 0 and 1. If \(\chi_A = 1.0\), then the sample is pure \(A\), not a mixture. If \(\chi_A = 0\), then no \(A\) is present in the mixture. The sum of the mole fractions of all the components present must equal 1. To see how mole fractions can help us understand the properties of gas mixtures, let’s evaluate the ratio of the pressure of a gas \(A\) to the total pressure of a gas mixture that contains \(A\). We can use the ideal gas law to describe the pressures of both gas \(A\) and the mixture: \(P_A = n_ART/V\) and \(P_{tot} = n_tRT/V\). The ratio of the two is thus \[\dfrac{P_A}{P_{tot}}=\dfrac{n_ART/V}{n_{tot}RT/V} = \dfrac{n_A}{n_{tot}}=\chi_A \label{10.6.6} \] Rearranging this equation gives \[P_A = \chi_AP_{tot} \label{10.6.7} \] That is, the partial pressure of any gas in a mixture is the total pressure multiplied by the mole fraction of that gas. This conclusion is a direct result of the ideal gas law, which assumes that all gas particles behave ideally. Consequently, the pressure of a gas in a mixture depends on only the percentage of particles in the mixture that are of that type, not their specific physical or chemical properties. By volume, Earth’s atmosphere is about 78% \(N_2\), 21% \(O_2\), and 0.9% \(Ar\), with trace amounts of gases such as \(CO_2\), \(H_2O\), and others. This means that 78% of the particles present in the atmosphere are \(N_2\); hence the mole fraction of \(N_2\) is 78%/100% = 0.78. Similarly, the mole fractions of \(O_2\) and \(Ar\) are 0.21 and 0.009, respectively. Using Equation \ref{10.6.7}, we therefore know that the partial pressure of N is 0.78 atm (assuming an atmospheric pressure of exactly 760 mmHg) and, similarly, the partial pressures of \(O_2\) and \(Ar\) are 0.21 and 0.009 atm, respectively. We have just calculated the partial pressures of the major gases in the air we inhale. Experiments that measure the composition of the air we yield different results, however. The following table gives the measured pressures of the major gases in both inhaled and exhaled air. Calculate the mole fractions of the gases in exhaled air. pressures of gases in inhaled and exhaled air mole fractions of gases in exhaled air Calculate the mole fraction of each gas using Equation \(\ref{10.6.7}\). The mole fraction of any gas \(A\) is given by \[\chi_A=\dfrac{P_A}{P_{tot}} \nonumber \] where \(P_A\) is the partial pressure of \(A\) and \(P_{tot}\) is the total pressure. For example, the mole fraction of \(\ce{CO_2}\) is given as: \[\chi_{\rm CO_2}=\rm\dfrac{48\;mmHg}{767\;mmHg}=0.063 \nonumber \] The following table gives the values of \(\chi_A\) for the gases in the exhaled air. is an inhospitable place, with a surface temperature of 560°C and a surface pressure of 90 atm. The atmosphere consists of about 96% CO and 3% N , with trace amounts of other gases, including water, sulfur dioxide, and sulfuric acid. Calculate the partial pressures of CO and N . \[P_{\rm CO_2}=\rm86\; atm \nonumber \] \[P_{\rm N_2}=\rm2.7\;atm \nonumber \] The partial pressure of each gas in a mixture is proportional to its mole fraction. The pressure exerted by each gas in a gas mixture (its ) is independent of the pressure exerted by all other gases present. Consequently, the total pressure exerted by a mixture of gases is the sum of the partial pressures of the components ( ). The amount of gas present in a mixture may be described by its partial pressure or its mole fraction. The of any component of a mixture is the ratio of the number of moles of that substance to the total number of moles of all substances present. In a mixture of gases, the partial pressure of each gas is the product of the total pressure and the mole fraction of that gas. | 8,605 | 2,634 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/08%3A_Advanced_Theories_of_Covalent_Bonding/8.E%3A_Advanced_Theories_of_Covalent_Bonding_(Exercises) |
1. : Both types of bonds result from overlap of atomic orbitals on adjacent atoms and contain a maximum of two electrons. : σ bonds are stronger and result from end-to-end overlap and all single bonds are σ bonds; π bonds between the same two atoms are weaker because they result from side-by-side overlap, and multiple bonds contain one or more π bonds (in addition to a σ bond). 2 When H and Cl are separate (the x axis) the energy is at a particular value. As they approach, it decreases to a minimum at 127 pm (the bond distance), and then it increases sharply as you get closer. \[−184.7 kJ/mol=(ΔH∘BDE(H–H)+ΔH∘BDE(Cl–Cl))−(2ΔH∘BDE(H–Cl))\] \[H–H is 436 kJ/mol and Cl–Cl is 243\] \[–184.7 kJ/mol = (436 + 243) – 2x = 679 – 2x\] \[2x = 863.7 kJ/mol\] \[x = 432\; kJ/mol\] This is very close to the value from part (a). 3. The specific average bond distance is the distance with the lowest energy. At distances less than the bond distance, the positive charges on the two nuclei repel each other, and the overall energy increases. 4. The single bond present in each molecule results from overlap of the relevant orbitals: F 2 orbitals in F , the H 1 and F 2 orbitals in HF, and the Cl 3 orbital and Br 4 orbital in ClBr. 5. Bonding: One σ bond and one π bond. The orbitals are filled and do not overlap. The orbitals overlap along the axis to form a σ bond and side by side to form the π bond. 6. \(\ce{H–C≡N}\) has two σ (H–C and C–N) and two π (making the CN triple bond). 7. No, two of the orbitals (one on each N) will be oriented end-to-end and will form a σ bond. 8. (a) 2 σ 2 π; (b) 1 σ 2 π; Why is the concept of hybridization required in valence bond theory? Hybridization is introduced to explain the geometry of bonding orbitals in valance bond theory. Give the shape that describes each hybrid orbital set: (a) (b) (c) (d) Explain why a carbon atom cannot form five bonds using hybrid orbitals. There are no orbitals in the valence shell of carbon. What is the hybridization of the central atom in each of the following? (a) BeH (b) SF (c) \(\ce{PO4^3-}\) (d) PCl A molecule with the formula AB could have one of four different shapes. Give the shape and the hybridization of the central A atom for each. trigonal planar, ; trigonal pyramidal (one lone pair on A) ; T-shaped (two lone pairs on A , or (three lone pairs on A) Methionine, CH SCH CH CH(NH )CO H, is an amino acid found in proteins. Draw a Lewis structure of this compound. What is the hybridization type of each carbon, oxygen, the nitrogen, and the sulfur? Sulfuric acid is manufactured by a series of reactions represented by the following equations: \(\ce{S8}(s)+\ce{8O2}(g)⟶\ce{8SO2}(g)\) \(\ce{2SO2}(g)+\ce{O2}(g)⟶\ce{2SO3}(g)\) \(\ce{SO3}(g)+\ce{H2O}(l)⟶\ce{H2SO4}(l)\) Draw a Lewis structure, predict the molecular geometry by VSEPR, and determine the hybridization of sulfur for the following: (a) circular S molecule (b) SO molecule (c) SO molecule (d) H SO molecule (the hydrogen atoms are bonded to oxygen atoms) (a) Each S has a bent (109°) geometry, (b) Bent (120°), (c) Trigonal planar, (d) Tetrahedral, Two important industrial chemicals, ethene, C H , and propene, C H , are produced by the steam (or thermal) cracking process: For each of the four carbon compounds, do the following: (a) Draw a Lewis structure. (b) Predict the geometry about the carbon atom. (c) Determine the hybridization of each type of carbon atom. For many years after they were discovered, it was believed that the noble gases could not form compounds. Now we know that belief to be incorrect. A mixture of xenon and fluorine gases, confined in a quartz bulb and placed on a windowsill, is found to slowly produce a white solid. Analysis of the compound indicates that it contains 77.55% Xe and 22.45% F by mass. (a) What is the formula of the compound? (b) Write a Lewis structure for the compound. (c) Predict the shape of the molecules of the compound. (d) What hybridization is consistent with the shape you predicted? (a) XeF (b) (c) linear (d) Consider nitrous acid, HNO (HONO). (a) Write a Lewis structure. (b) What are the electron pair and molecular geometries of the internal oxygen and nitrogen atoms in the HNO molecule? (c) What is the hybridization on the internal oxygen and nitrogen atoms in HNO ? Strike-anywhere matches contain a layer of KClO and a layer of P S . The heat produced by the friction of striking the match causes these two compounds to react vigorously, which sets fire to the wooden stem of the match. KClO contains the \(\ce{ClO3-}\) ion. P S is an unusual molecule with the skeletal structure. (a) Write Lewis structures for P S and the \(\ce{ClO3-}\) ion. (b) Describe the geometry about the P atoms, the S atom, and the Cl atom in these species. (c) Assign a hybridization to the P atoms, the S atom, and the Cl atom in these species. (d) Determine the oxidation states and formal charge of the atoms in P S and the \(\ce{ClO3-}\) ion. (a) (b) P atoms, trigonal pyramidal; S atoms, bent, with two lone pairs; Cl atoms, trigonal pyramidal; (c) Hybridization about P, S, and Cl is, in all cases, ; (d) Oxidation states P +1, S \(−1\dfrac{1}{3}\), Cl +5, O –2. Formal charges: P 0; S 0; Cl +2: O –1 Identify the hybridization of each carbon atom in the following molecule. (The arrangement of atoms is given; you need to determine how many bonds connect each pair of atoms.) Write Lewis structures for NF and PF . On the basis of hybrid orbitals, explain the fact that NF , PF , and PF are stable molecules, but NF does not exist. Phosphorus and nitrogen can form hybrids to form three bonds and hold one lone pair in PF and NF , respectively. However, nitrogen has no valence orbitals, so it cannot form a set of hybrid orbitals to bind five fluorine atoms in NF . Phosphorus has orbitals and can bind five fluorine atoms with hybrid orbitals in PF . In addition to NF , two other fluoro derivatives of nitrogen are known: N F and N F . What shapes do you predict for these two molecules? What is the hybridization for the nitrogen in each molecule? The bond energy of a C–C single bond averages 347 kJ mol ; that of a C≡C triple bond averages 839 kJ mol . Explain why the triple bond is not three times as strong as a single bond. A triple bond consists of one σ bond and two π bonds. A σ bond is stronger than a π bond due to greater overlap. For the carbonate ion, \(\ce{CO3^2-}\), draw all of the resonance structures. Identify which orbitals overlap to create each bond. A useful solvent that will dissolve salts as well as organic compounds is the compound acetonitrile, H CCN. It is present in paint strippers. (a) Write the Lewis structure for acetonitrile, and indicate the direction of the dipole moment in the molecule. (b) Identify the hybrid orbitals used by the carbon atoms in the molecule to form σ bonds. (c) Describe the atomic orbitals that form the π bonds in the molecule. Note that it is not necessary to hybridize the nitrogen atom. (a) (b) The terminal carbon atom uses hybrid orbitals, while the central carbon atom is hybridized. (c) Each of the two π bonds is formed by overlap of a 2 orbital on carbon and a nitrogen 2 orbital. For the molecule allene, \(\mathrm{H_2C=C=CH_2}\), give the hybridization of each carbon atom. Will the hydrogen atoms be in the same plane or perpendicular planes? Identify the hybridization of the central atom in each of the following molecules and ions that contain multiple bonds: (a) ClNO (N is the central atom) (b) CS (c) Cl CO (C is the central atom) (d) Cl SO (S is the central atom) (e) SO F (S is the central atom) (f) XeO F (Xe is the central atom) (g) \(\ce{ClOF2+}\) (Cl is the central atom) (a) ; (b) ; (c) ; (d) ; (e) ; (f) ; (g) Describe the molecular geometry and hybridization of the N, P, or S atoms in each of the following compounds. (a) H PO , phosphoric acid, used in cola soft drinks (b) NH NO , ammonium nitrate, a fertilizer and explosive (c) S Cl , disulfur dichloride, used in vulcanizing rubber (d) K [O POPO ], potassium pyrophosphate, an ingredient in some toothpastes For each of the following molecules, indicate the hybridization requested and whether or not the electrons will be delocalized: (a) ozone (O ) central O hybridization (b) carbon dioxide (CO ) central C hybridization (c) nitrogen dioxide (NO ) central N hybridization (d) phosphate ion (\(\ce{PO4^3-}\)) central P hybridization (a) , delocalized; (b) , localized; (c) , delocalized; (d) , delocalized For each of the following structures, determine the hybridization requested and whether the electrons will be delocalized: (a) Hybridization of each carbon (b) Hybridization of sulfur (c) All atoms Draw the orbital diagram for carbon in CO showing how many carbon atom electrons are in each orbital. Each of the four electrons is in a separate orbital and overlaps with an electron on an oxygen atom. Sketch the distribution of electron density in the bonding and antibonding molecular orbitals formed from two orbitals and from two orbitals. How are the following similar, and how do they differ? (a) σ molecular orbitals and π molecular orbitals (b) for an atomic orbital and for a molecular orbital (c) bonding orbitals and antibonding orbitals (a) Similarities: Both are bonding orbitals that can contain a maximum of two electrons. Differences: σ orbitals are end-to-end combinations of atomic orbitals, whereas π orbitals are formed by side-by-side overlap of orbitals. (b) Similarities: Both are quantum-mechanical constructs that represent the probability of finding the electron about the atom or the molecule. Differences: for an atomic orbital describes the behavior of only one electron at a time based on the atom. For a molecule, represents a mathematical combination of atomic orbitals. (c) Similarities: Both are orbitals that can contain two electrons. Differences: Bonding orbitals result in holding two or more atoms together. Antibonding orbitals have the effect of destabilizing any bonding that has occurred. If molecular orbitals are created by combining five atomic orbitals from atom A and five atomic orbitals from atom B combine, how many molecular orbitals will result? Can a molecule with an odd number of electrons ever be diamagnetic? Explain why or why not. An odd number of electrons can never be paired, regardless of the arrangement of the molecular orbitals. It will always be paramagnetic. Can a molecule with an even number of electrons ever be paramagnetic? Explain why or why not. Why are bonding molecular orbitals lower in energy than the parent atomic orbitals? Bonding orbitals have electron density in close proximity to more than one nucleus. The interaction between the bonding positively charged nuclei and negatively charged electrons stabilizes the system. Calculate the bond order for an ion with this configuration: Explain why an electron in the bonding molecular orbital in the H molecule has a lower energy than an electron in the 1 atomic orbital of either of the separated hydrogen atoms. The pairing of the two bonding electrons lowers the energy of the system relative to the energy of the nonbonded electrons. Predict the valence electron molecular orbital configurations for the following, and state whether they will be stable or unstable ions. (a) \(\ce{Na2^2+}\) (b) \(\ce{Mg2^2+}\) (c) \(\ce{Al2^2+}\) (d) \(\ce{Si2^2+}\) (e) \(\ce{P2^2+}\) (f) \(\ce{S2^2+}\) (g) \(\ce{F2^2+}\) (h) \(\ce{Ar2^2+}\) Determine the bond order of each member of the following groups, and determine which member of each group is predicted by the molecular orbital model to have the strongest bond. (a) H , \(\ce{H2+}\), \(\ce{H2-}\) (b) O , \(\ce{O2^2+}\), \(\ce{O2^2-}\) (c) Li , \(\ce{Be2+}\), Be (d) F , \(\ce{F2+}\), \(\ce{F2-}\) (e) N , \(\ce{N2+}\), \(\ce{N2-}\) (a) H bond order = 1, \(\ce{H2+}\) bond order = 0.5, \(\ce{H2-}\) bond order = 0.5, strongest bond is H ; (b) O bond order = 2, \(\ce{O2^2+}\) bond order = 3; \(\ce{O2^2-}\) bond order = 1, strongest bond is \(\ce{O2^2+}\); (c) Li bond order = 1, \(\ce{Be2+}\) bond order = 0.5, Be bond order = 0, strongest bond is \(\ce{Li2}\);(d) F bond order = 1, \(\ce{F2+}\) bond order = 1.5, \(\ce{F2-}\) bond order = 0.5, strongest bond is \(\ce{F2+}\); (e) N bond order = 3, \(\ce{N2+}\) bond order = 2.5, \(\ce{N2-}\) bond order = 2.5, strongest bond is N For the first ionization energy for an N molecule, what molecular orbital is the electron removed from? Compare the atomic and molecular orbital diagrams to identify the member of each of the following pairs that has the highest first ionization energy (the most tightly bound electron) in the gas phase: (a) H and H (b) N and N (c) O and O (d) C and C (e) B and B (a) H ; (b) N ; (c) O; (d) C ; (e) B Which of the period 2 homonuclear diatomic molecules are predicted to be paramagnetic? A friend tells you that the 2 orbital for fluorine starts off at a much lower energy than the 2 orbital for lithium, so the resulting σ molecular orbital in F is more stable than in Li . Do you agree? Yes, fluorine is a smaller atom than Li, so atoms in the 2 orbital are closer to the nucleus and more stable. True or false: Boron contains 2 2 valence electrons, so only one orbital is needed to form molecular orbitals. What charge would be needed on F to generate an ion with a bond order of 2? 2+ Predict whether the MO diagram for S would show s-p mixing or not. Explain why \(\ce{N2^2+}\) is diamagnetic, while \(\ce{O2^4+}\), which has the same number of valence electrons, is paramagnetic. N has s-p mixing, so the π orbitals are the last filled in \(\ce{N2^2+}\). O does not have s-p mixing, so the σ orbital fills before the π orbitals. Using the MO diagrams, predict the bond order for the stronger bond in each pair: (a) B or \(\ce{B2+}\) (b) F or \(\ce{F2+}\) (c) O or \(\ce{O2^2+}\) (d) \(\ce{C2+}\) or \(\ce{C2-}\) | 13,973 | 2,636 |
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Potassium permanganate, KMnO , is a powerful oxidizing agent, and has many uses in organic chemistry. Of all the oxidizing agents discussed in organic chemistry textbooks, potassium permanganate, KMnO , is probably the most common, and also the most applicable. As will be shown below, KMnO can be utilized to oxidize a wide range of organic molecules. The products that are obtained can vary depending on the conditions, but because KMnO is such a strong oxidizing agent, the final products are often carboxylic acids. Mn(VII) is reduced under acidic conditions to Mn(IV) or Mn(II) according to the s shown below, with the indicated cell potentials \[MnO_4^- + 4H^+ + 3e^- \rightarrow MnO_2 + 2H_2O\;\;\;\;E^o = 1.68\,V\] \[MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O\;\;\;\;E^o = 1.5\, V\] \[MnO_4^- + 2H_2O + 3e^- \rightarrow Mn^{2+} + 4OH^-\;\;\;\;E^o = 0.6\, V\] KMnO is able to oxidize carbon atoms if they contain sufficiently weak bonds, including KMnO also oxidizes phenol to para-benzoquinone. Examples of carbons that are not oxidized Exhaustive oxidation of organic molecules by KMnO will proceed until the formation of carboxylic acids. Therefore, alcohols will be oxidized to carbonyls (aldehydes and ketones), and aldehydes (and some ketones, as in (3) above) will be oxidized to carboxylic acids. Using the principles above, we expect KMno to react with alkenes, alkynes, alcohols, aldehydes and aromatic side chains. Examples are provided below. It is easiest to start at the top. Aldehydes RCHO are readily oxidized to carboxylic acids. Unless great efforts are taken to maintain a neutral pH, KMnO oxidations tend to occur under basic conditions. In fact, the most effective conditions for aldehyde oxidation by KMnO involves -butanol as solvent with a NaH PO buffer. The reactions above are deliberately not balanced equations. would involve using the methods learned in general chemistry, requiring half reactions for all processes. Primary alcohols such as octan-1-ol can be oxidized efficiently by KMnO , in the presence of basic copper salts. However, the product is predominantly octanoic acid, with only a small amount of aldehyde, resulting from overoxidation. Although overoxidation is less of a problem with secondary alcohols, KMnO is still not considered generally well-suited for conversions of alcohols to aldehydes or ketones. Under mild conditions, potassium permanganate can effect conversion of alkenes to glycols. It is, however, capable of further oxidizing the glycol with cleavage of the carbon-carbon bond, so careful control of the reaction conditions is necessary. A cyclic manganese diester is an intermediate in these oxidations, which results in glycols formed by addition. With addition of heat and/or more concentrated KMnO , the glycol can be further oxidized, cleaving the C-C bond. More substituted olefins will terminate at the ketone Oxidative cleavage of the diol can be carried out more mildly by using IO as the oxidant. The cleavage of alkenes to ketones/carboxylic acids can be used to determine the position of double bonds in organic molecules. Instead of bis-hydroxylation that occurs with alkenes, permanganate oxidation of alkynes initially leads to the formation of diones. Under harsher conditions, the dione is cleaved to form two carboxylic acids. Treatment of an alkylbenzene with potassium permanganate results in oxidation to give the benzoic acid. The position directly adjacent to an aromatic group is called the “benzylic” position. The reaction only works if there is at least one hydrogen attached to the carbon. However, if there is at least one hydrogen, the oxidation proceeds all the way to the carboxylic acid. Notes: Note that in example 2 the extra carbons are cleaved to give the same product as in example 1. And in example 3, two benzoic acids are formed. Finally, when no hydrogens are present on the benzylic carbon, no reaction occurs (example 4). The oxidation of alkyl side-chains to form benzoic acids was historically used in qualitative analysis to determine the positions of alkyl groups in substituted aromatic systems. Alkyl-substituted rings can be coverted to poly-acids, which can be distinguished on the basis of their pKas | 4,267 | 2,637 |
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The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. In contrast to molecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, molecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances . The properties of liquids are intermediate between those of gases and solids, but are more similar to solids. Intermolecular forces determine bulk properties, such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid. Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures. In this section, we explicitly consider three kinds of intermolecular interactions. The first two are often described collectively as van der Waals forces. Polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a ). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in Figure \(\Page {1a}\). These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (Figure \(\Page {1c}\)). Hence dipole–dipole interactions, such as those in Figure \(\Page {1b}\), are , whereas those in Figure \(\Page {1d}\) are . Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in Figure \(\Page {2}\). On average, however, the attractive interactions dominate. Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions. Recall that the attractive energy between two ions is proportional to 1/ , where is the distance between the ions. Doubling the distance ( → 2 ) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/ , so doubling the distance between the dipoles decreases the strength of the interaction by 2 , or 8-fold. Thus a substance such as \(\ce{HCl}\), which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure. Conversely, \(\ce{NaCl}\), which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in Table \(\Page {1}\). The attractive energy between two ions is proportional to 1/r, whereas the attractive energy between two dipoles is proportional to 1/r6. Discussing Dipole Intermolecular Forces. Arrange ethyl methyl ether (CH OCH CH ), 2-methylpropane [isobutane, (CH ) CHCH ], and acetone (CH COCH ) in order of increasing boiling points. Their structures are as follows: compounds. order of increasing boiling points. Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points. The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds. The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point. Ethyl methyl ether has a structure similar to H O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point. Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point. Thus we predict the following order of boiling points: This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D. Arrange carbon tetrafluoride (CF ), ethyl methyl sulfide (CH SC H ), dimethyl sulfoxide [(CH ) S=O], and 2-methylbutane [isopentane, (CH ) CHCH CH ] in order of decreasing boiling points. dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C) Thus far, we have considered only interactions between polar molecules. Other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature; why others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both (Table \(\Page {2}\)). What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments, which produce attractive forces called London dispersion forces between otherwise nonpolar substances. Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in Figure \(\Page {3}\), the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole, in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/ . Doubling the distance therefore decreases the attractive energy by 2 , or 64-fold. Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H molecules in part (b) in Figure \(\Page {3}\), tends to become more pronounced as atomic and molecular masses increase (Table \(\Page {2}\)). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1 electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability. Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more than lighter ones. For similar substances, London dispersion forces get stronger with increasing molecular size. The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles. Thus, London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in Figure \(\Page {4}\)). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in Figure \(\Page {4}\) shows 2,2-dimethylpropane (neopentane) and -pentane, both of which have the empirical formula C H . Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas -pentane has an extended conformation that enables it to come into close contact with other -pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of -pentane (36.1°C). All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate. Discussing London/Dispersion Intermolecular Forces. Arrange -butane, propane, 2-methylpropane [isobutene, (CH ) CHCH ], and -pentane in order of increasing boiling points. compounds order of increasing boiling points Determine the intermolecular forces in the compounds, and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and -pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and -butane has the more extended shape. Consequently, we expect intermolecular interactions for -butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < -butane (−0.5°C) < -pentane (36.1°C). Arrange GeH , SiCl , SiH , CH , and GeCl in order of decreasing boiling points. GeCl (87°C) > SiCl (57.6°C) > GeH (−88.5°C) > SiH (−111.8°C) > CH (−161°C) Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent, Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in Figure \(\Page {5}\). Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H Te and H Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C. Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds, as shown for ice in Figure \(\Page {6}\). A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the ) and the atom that has the lone pair of electrons (the ). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two O⋅⋅⋅H hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water, rather than sinks. Each water molecule accepts two hydrogen bonds from two other water molecules and donates two hydrogen atoms to form hydrogen bonds with two more water molecules, producing an open, cagelike structure. The structure of liquid water is very similar, but in the liquid, the hydrogen bonds are continually broken and formed because of rapid molecular motion. Hydrogen bond formation requires both a hydrogen bond donor and a hydrogen bond acceptor. Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” and why unprotected pipes in houses break if they are allowed to freeze. Discussing Hydrogen Bonding Intermolecular Forces. Considering CH , C H , Xe, and (CH ) N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures. compounds formation of hydrogen bonds and structure Of the species listed, xenon (Xe), ethane (C H ), and trimethylamine [(CH ) N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors. The one compound that can act as a hydrogen bond donor, methanol (CH OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows: Considering CH CO H, (CH ) N, NH , and CH F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures. CH CO H and NH ; Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as can form only two hydrogen bonds at a time as can, on average, pure liquid NH . Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms hydrogen bonds at a time. Arrange C (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N O in order of increasing boiling points. compounds. order of increasing boiling points. Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N O have very similar molar masses (40 and 44 g/mol, respectively), but N O is polar while Ar is not. Consequently, N O should have a higher boiling point. A C molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses: He (−269°C) < Ar (−185.7°C) < N O (−88.5°C) < C (>280°C) < NaCl (1465°C). Arrange 2,4-dimethylheptane, Ne, CS , Cl , and KBr in order of decreasing boiling points. KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS (46.6°C) > Cl (−34.6°C) > Ne (−246°C) Identify the most significant intermolecular force in each substance. Identify the most significant intermolecular force in each substance. hydrogen bonding dipole-dipole interactions Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid, or the liquid and gas phases, are due to changes in intermolecular interactions, but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as ), and hydrogen bonds. arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/ , where is the distance between dipoles. are due to the formation of in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an in adjacent molecules; their energy falls off as 1/ . Larger atoms tend to be more than smaller ones, because their outer electrons are less tightly bound and are therefore more easily perturbed. are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the ) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the ). Because of strong O⋅⋅⋅H hydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cagelike structure that is less dense than liquid water. | 22,288 | 2,638 |
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The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles. The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates: Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of ideal gases. Postulates 1 and 4 state that gas molecules are in constant motion and collide frequently with the walls of their containers. The collision of molecules with their container walls results in a (impulse) from molecules to the walls (Figure 10.7.1). The to the wall perpendicular to \(x\) axis as a molecule with an initial velocity \(u_x\) in \(x\) direction hits is expressed as: \[\Delta p_x=2mu_x \tag{10.7.1} \] The , a number of collisions of the molecules to the wall per unit area and per second, increases with the molecular speed and the number of molecules per unit volume. \[f\propto (u_x) \times \Big(\dfrac{N}{V}\Big) \tag{10.7.2}\] The pressure the gas exerts on the wall is expressed as the product of impulse and the collision frequency. \[P\propto (2mu_x)\times(u_x)\times\Big(\dfrac{N}{V}\Big)\propto \Big(\dfrac{N}{V}\Big)mu_x^2 \tag{10.7.3}\] At any instant, however, the molecules in a gas sample are traveling at different speed. Therefore, we must replace \(u_x^2\) in the expression above with the average value of \(u_x^2\), which is denoted by \(\overline{u_x^2}\). The overbar designates the average value over all molecules. The exact expression for pressure is given as : \[P=\dfrac{N}{V}m\overline{u_x^2} \tag{10.7.4}\] Finally, we must consider that there is nothing special about \(x\) direction. We should expect that \(\overline{u_x^2}= \overline{u_y^2}=\overline{u_z^2}=\dfrac{1}{3}\overline{u^2}\). Here the quantity \(\overline{u^2}\) is called the defined as the average value of square-speed (\(u^2\)) over all molecules. Since \(u^2=u_x^2+u_y^2+u_z^2\) for each molecule, \(\overline{u^2}=\overline{u_x^2}+\overline{u_y^2}+\overline{u_z^2}\). By substituting \(\dfrac{1}{3}\overline{u^2}\) for \(\overline{u_x^2}\) in the expression above, we can get the final expression for the pressure: \[P=\dfrac{1}{3}\dfrac{N}{V}m\overline{u^2} \tag{10.7.5}\] Anything that increases the frequency with which the molecules strike the walls or increases the momentum of the gas molecules (i.e., how hard they hit the walls) increases the pressure; anything that decreases that frequency or the momentum of the molecules decreases the pressure. Because volumes and intermolecular interactions are negligible, postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules. This is the essence of the ideal gas law, which treats all gases as collections of particles that are identical in all respects except mass. Postulate 2 also explains why it is relatively easy to compress a gas; you simply decrease the distance between the gas molecules. Postulate 5 provides a molecular explanation for the temperature of a gas. It refers to the kinetic energy of the molecules of a gas which can be represented as \((\overline{e_K})\), \((T)\)
\[\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \tag{10.7.6}\] where \(N_A\) is the Avogadro's constant. The total translational kinetic energy of 1 mole of molecules can be obtained by multiplying the equation by \(N_A\): \[N_A\overline{e_K}=\dfrac{1}{2}M\overline{u^2}=\dfrac{3}{2}RT \tag{10.7.7}\] where \(M\) is the molar mass of the gas molecules and is related to the molecular mass by \(M=N_Am\). By rearranging the equation, we can get the relationship between the root-mean square speed (\(u_{\rm rms}\)) and the temperature. The rms speed (\(u_{\rm rms}\)) is the square root of the sum of the squared speeds divided by the number of particles: \[u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \tag{10.7.8}\] where \(N\) is the number of particles and \(u_i\) is the speed of particle \(i\). The relationship between \(u_{\rm rms}\) and the temperature is given by: \[u_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \tag{10.7.9}\] In this equation, \(u_{\rm rms}\) has units of meters per second; consequently, the units of molar mass \(M\) are kilograms per mole, temperature \(T\) is expressed in kelvins, and the ideal gas constant \(R\) has the value 8.3145 J/(K·mol). The equation shows that \(u_{\rm rms}\) of a gas is proportional to the square root of its Kelvin temperature and inversely proportional to the square root of its molar mass. The root mean-square speed of a gas increase with increasing temperature. At a given temperature, heavier gas molecules have slower speeds than do lighter ones. The rms speed and the average speed do not differ greatly (typically by less than 10%). The distinction is important, however, because the rms speed is the speed of a gas particle that has average kinetic energy. Particles of different gases at the same temperature have the same average kinetic energy, the same average speed. In contrast, the most probable speed ( ) is the speed at which the greatest number of particles is moving. If the average kinetic energy of the particles of a gas increases linearly with increasing temperature, then tells us that the rms speed must also increase with temperature because the mass of the particles is constant. At higher temperatures, therefore, the molecules of a gas move more rapidly than at lower temperatures, and increases. At a given temperature, all gaseous particles have the same average kinetic energy but not the same average speed. The speeds of eight particles were found to be 1.0, 4.0, 4.0, 6.0, 6.0, 6.0, 8.0, and 10.0 m/s. Calculate their average speed (\(v_{\rm av}\)) root mean square speed ( ), and most probable speed ( ). particle speeds average speed (\(v_{\rm av}\)) root mean square speed ( ), and most probable speed ( ) Use to calculate the average speed and to calculate the rms speed. Find the most probable speed by determining the speed at which the greatest number of particles is moving. The average speed is the sum of the speeds divided by the number of particles: \[v_{\rm av}=\rm\dfrac{(1.0+4.0+4.0+6.0+6.0+6.0+8.0+10.0)\;m/s}{8}=5.6\;m/s \notag \] The rms speed is the square root of the sum of the squared speeds divided by the number of particles: \[v_{\rm rms}=\rm\sqrt{\dfrac{(1.0^2+4.0^2+4.0^2+6.0^2+6.0^2+6.0^2+8.0^2+10.0^2)\;m^2/s^2}{8}}=6.2\;m/s \notag \] The most probable speed is the speed at which the greatest number of particles is moving. Of the eight particles, three have speeds of 6.0 m/s, two have speeds of 4.0 m/s, and the other three particles have different speeds. Hence = 6.0 m/s. The of the particles, which is related to the average kinetic energy, is greater than their average speed. Exercise Ten particles were found to have speeds of 0.1, 1.0, 2.0, 3.0, 3.0, 3.0, 4.0, 4.0, 5.0, and 6.0 m/s. Calculate their average speed (\(v_{\rm av}\)) root mean square speed ( ), and most probable speed ( ). \[ \bar{v} = 3.1 \; m/s;\; v_{rms}=3.5 \; m/s; v_{p}=3.0 m/s \notag \] At any given time, what fraction of the molecules in a particular sample has a given speed? Some of the molecules will be moving more slowly than average, and some will be moving faster than average, but how many in each situation? Answers to questions such as these can have a substantial effect on the amount of product formed during a chemical reaction, as you will learn in . This problem was solved mathematically by Maxwell in 1866; he used statistical analysis to obtain an equation that describes the distribution of molecular speeds at a given temperature. Typical curves showing the distributions of speeds of molecules at several temperatures are displayed in . Increasing the temperature has two effects. First, the peak of the curve moves to the right because the most probable speed increases. Second, the curve becomes broader because of the increased spread of the speeds. Thus increased temperature increases the of the most probable speed but decreases the relative number of molecules that have that speed. Although the mathematics behind curves such as those in were first worked out by Maxwell, the curves are almost universally referred to as Boltzmann distributions , after one of the other major figures responsible for the kinetic molecular theory of gases. We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. At constant temperature, the kinetic energy of the molecules of a gas and hence the rms speed remain unchanged. If a given gas sample is allowed to occupy a larger volume, then the speed of the molecules does not change, but the density of the gas (number of particles per unit volume) decreases, and the average distance between the molecules increases. Hence the molecules must, on average, travel farther between collisions. They therefore collide with one another and with the walls of their containers less often, leading to a decrease in pressure. Conversely, increasing the pressure forces the molecules closer together and increases the density, until the collective impact of the collisions of the molecules with the container walls just balances the applied pressure. Raising the temperature of a gas increases the average kinetic energy and therefore the rms speed (and the average speed) of the gas molecules. Hence as the temperature increases, the molecules collide with the walls of their containers more frequently and with greater force. This increases the pressure, the volume increases to reduce the pressure, as we have just seen. Thus an increase in temperature must be offset by an increase in volume for the net impact (pressure) of the gas molecules on the container walls to remain unchanged. Postulate 3 of the kinetic molecular theory of gases states that gas molecules exert no attractive or repulsive forces on one another. If the gaseous molecules do not interact, then the presence of one gas in a gas mixture will have no effect on the pressure exerted by another, and Dalton’s law of partial pressures holds. The temperature of a 4.75 L container of N gas is increased from 0°C to 117°C. What is the qualitative effect of this change on the temperatures and volume effect of increase in temperature Use the relationships among pressure, volume, and temperature to predict the qualitative effect of an increase in the temperature of the gas. Exercise A sample of helium gas is confined in a cylinder with a gas-tight sliding piston. The initial volume is 1.34 L, and the temperature is 22°C. The piston is moved to allow the gas to expand to 2.12 L at constant temperature. What is the qualitative effect of this change on the a. no change; b. no change; c. no change; d. no change; e. decreases; f. decreases; g. decreases As you have learned, the molecules of a gas are stationary but in constant motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment. shows white fumes of solid ammonium chloride (NH Cl) forming when containers of aqueous ammonia and HCl are placed near each other, even with no draft to stir the air. This phenomenon suggests that NH and HCl molecules (as well as the more complex organic molecules responsible for the aromas of pizza and perfumes) move without assistance. < Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. As we shall see in , , and , diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. We can describe the phenomenon shown in by saying that the molecules of HCl and NH are able to diffuse away from their containers, and that NH Cl is formed where the two gases come into contact. Similarly, we say that a perfume or an aroma diffuses throughout a room or a house. The related process, effusion , is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space. The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship, \[ rate\propto \dfrac{1}{\sqrt{M}} \notag \] is referred to as Graham’s law , after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses. If is the effusion rate and is the molar mass, then \[ \dfrac{r_{1}}{r_{2}}=\dfrac{\sqrt{M_{2}}}{\sqrt{M_{1}}} \tag{10.7.10} \] Although diffusion and effusion are different phenomena, the rate of diffusion is closely approximated using that is, if < , then gas #1 will diffuse more rapidly than gas #2. This point is illustrated by the experiment shown in , which is a more quantitative version of the case shown in . The reaction is the same [NH (aq) + HCl(aq) → NH Cl(g)], but in this experiment, two cotton balls containing aqueous ammonia and HCl are placed along a meter stick in a draft-free environment, and the position at which the initial NH Cl fumes appear is noted. The white cloud forms much nearer the HCl-containing ball than the NH -containing ball. Because ammonia ( = 17.0 g/mol) diffuses much faster than HCl ( = 36.5 g/mol), the NH Cl fumes form closer to HCl because the HCl molecules travel a shorter distance. The ratio of the distances traveled by NH and HCl in is about 1.7, in reasonable agreement with the ratio of 1.47 predicted by their molar masses [(36.5/17.0) = 1.47]. Heavy molecules effuse through a porous material more slowly than light molecules, as illustrated schematically in for ethylene oxide and helium. Helium ( = 4.00 g/mol) effuses much more rapidly than ethylene oxide ( = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon \[ \sqrt{\dfrac{29}{4.00}}=2.7 \notag \] times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Mylar balloons can retain their helium for days. At a given temperature, heavier molecules move more slowly than lighter molecules. During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of U. Naturally occurring uranium is only 0.720% U, whereas most of the rest (99.275%) is U, which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound UF (boiling point = 56°C). isotopic content of naturally occurring uranium and atomic masses of U and U ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure UF Calculate the molar masses of UF and UF , and then use Graham’s law to determine the ratio of the effusion rates. Use this value to determine the isotopic content of UF after a single effusion step. Divide the final purity by the initial purity to obtain a value for the number of separation steps needed to achieve the desired purity. Use a logarithmic expression to compute the number of separation steps required. The molar mass of UF is The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law using : Thus passing UF containing a mixture of the two isotopes through a single porous barrier gives an enrichment of 1.0043, so after one step the isotopic content is (0.720%)(1.0043) = 0.723% UF . In this case, 0.990 = (0.00720)(1.0043) , which can be rearranged to give Taking the logarithm of both sides gives Thus at least a thousand effusion steps are necessary to obtain highly enriched U. shows a small part of a system that is used to prepare enriched uranium on a large scale. Exercise Helium consists of two isotopes: He (natural abundance = 0.000134%) and He (natural abundance = 99.999866%). Their atomic masses are 3.01603 and 4.00260, respectively. Helium-3 has unique physical properties and is used in the study of ultralow temperatures. It is separated from the more abundant He by a process of gaseous effusion. a. ratio of effusion rates = 1.15200; one step gives 0.000154% He; b. 96 steps Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses: \[ \overline{KE}=\dfrac{1}{2}M_{1}v_{rms_{1}}^{2}=\dfrac{1}{2}M_{2}v_{rms_{2}}^{2} \tag{10.7.11}\] Multiplying both sides by 2 and rearranging give \[ \dfrac{v_{rms_{2}}^{2}}{v_{rms_{1}}^{2}}=\dfrac{M_{1}}{M_{2}} \tag{10.7.12}\] Taking the square root of both sides gives \[ \dfrac{v_{rms_{2}}}{v_{rms_{1}}}=\sqrt{\dfrac{M_{1}}{M_{2}}} \tag{10.7.13} \] Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy. Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in for some common gases. Because all gases have the same average kinetic energy, according to the Boltzmann distribution, molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds. The postulates of the kinetic molecular theory of gases lead to the following equation, which directly relates molar mass, temperature, and rms speed: \[v_{\rm rms}=\sqrt{\dfrac{3RT}{M}} \tag{10.7.14}\] In this equation, has units of meters per second; consequently, the units of molar mass are kilograms per mole, temperature is expressed in kelvins, and the ideal gas constant has the value 8.3145 J/(K·mol). The lightest gases have a wider distribution of speeds and the highest average speeds. Molecules with lower masses have a wider distribution of speeds and a higher average speed. Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 10 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in The average distance traveled by a molecule between collisions is the mean free path . The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 10 m (about 6 million miles). The denser the gas, the shorter the mean free path. Calculate the rms speed of a sample of -2-butene (C H ) at 20°C. compound and temperature rms speed Calculate the molar mass of -2-butene. Be certain that all quantities are expressed in the appropriate units and then use to calculate the rms speed of the gas. To use , we need to calculate the molar mass of -2-butene and make sure that each quantity is expressed in the appropriate units. Butene is C H , so its molar mass is 56.11 g/mol. Thus \[ v_{rms}=\sqrt{\dfrac{3RT}{M}} \tag{10.7.14}\] Exercise Calculate the rms speed of a sample of radon gas at 23°C \[ M = 56.11 \;g/mol = 56.11 \times 10^{-3} \; kg/mol \notag \] \[ T= 20+273 = 293 \;K \notag \] \[ R= 8.3145 \; J/\left ( K\cdot mol \right ) = 8.3145 \; \left ( kg\cdot m^{2} \right )/ \left ( s^{2\cdot }K\cdot mol \right ) \notag \] \[ v_{rms}=\sqrt{\dfrac{3RT}{M}}=\sqrt{\dfrac{3 \cdot 8.3145 \; \left (\cancel{kg}\cdot m^{2} \right )/\left ( s^{2}\cdot \cancel{K \cdot \cancel{mol}} \right )\cdot 293 \; K}{56.11\times 10^{-3} \; \cancel{kg}/\cancel{mol}}} = 3.61\times 10^{2} \; m/s \notag \]. 1.82 × 10 m/s (about 410 mi/h) The kinetic molecular theory of gases demonstrates how a successful theory can explain previously observed empirical relationships (laws) in an intuitively satisfying way. Unfortunately, the actual gases that we encounter are not “ideal,” although their behavior usually approximates that of an ideal gas. In we explore how the behavior of real gases differs from that of ideal gases. The behavior of ideal gases is explained by the . Molecular motion, which leads to collisions between molecules and the container walls, explains pressure, and the large intermolecular distances in gases explain their high compressibility. Although all gases have the same average kinetic energy at a given temperature, they do not all possess the same . The actual values of speed and kinetic energy are not the same for all particles of a gas but are given by a , in which some molecules have higher or lower speeds (and kinetic energies) than average. is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass ( ), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The of a molecule is the average distance it travels between collisions. : \[\overline{e_K}=\dfrac{1}{2}m\overline{u^2}=\dfrac{3}{2}\dfrac{R}{N_A}T \] : \[u_{\rm rms}=\sqrt{\overline{u^2}}=\sqrt{\dfrac{u_1^2+u_2^2+\cdots u_N^2}{N}} \tag{10.7.8}\] .10 \[ \dfrac{r_{1}}{r_{2}}=\dfrac{\sqrt{M_{2}}}{\sqrt{M_{1}}} \tag{10.7.10} \] : \[ v_{rms}=\sqrt{\dfrac{3RT}{M}} \] Which of the following processes represents effusion, and which represents diffusion? Which postulate of the kinetic molecular theory of gases most readily explains the observation that a helium-filled balloon is round? Why is it relatively easy to compress a gas? How does the compressibility of a gas compare with that of a liquid? A solid? Why? Which of the postulates of the kinetic molecular theory of gases most readily explains these observations? What happens to the average kinetic energy of a gas if the rms speed of its particles increases by a factor of 2? How is the rms speed different from the average speed? Which gas—radon or helium—has a higher average kinetic energy at 100°C? Which has a higher average speed? Why? Which postulate of the kinetic molecular theory of gases most readily supports your answer? What is the relationship between the average speed of a gas particle and the temperature of the gas? What happens to the distribution of molecular speeds if the temperature of a gas is increased? Decreased? Qualitatively explain the relationship between the number of collisions of gas particles with the walls of a container and the pressure of a gas. How does increasing the temperature affect the number of collisions? What happens to the average kinetic energy of a gas at constant temperature if the What happens to the density of a gas at constant temperature if the Use the kinetic molecular theory of gases to describe how a decrease in volume produces an increase in pressure at constant temperature. Similarly, explain how a decrease in temperature leads to a decrease in volume at constant pressure. Graham’s law is valid only if the two gases are at the same temperature. Why? If we lived in a helium atmosphere rather than in air, would we detect odors more or less rapidly than we do now? Explain your reasoning. Would we detect odors more or less rapidly at sea level or at high altitude? Why? At a given temperature, what is the ratio of the rms speed of the atoms of Ar gas to the rms speed of molecules of H gas? At a given temperature, what is the ratio of the rms speed of molecules of CO gas to the rms speed of molecules of H S gas? What is the ratio of the rms speeds of argon and oxygen at any temperature? Which diffuses more rapidly? What is the ratio of the rms speeds of Kr and NO at any temperature? Which diffuses more rapidly? Deuterium (D) and tritium (T) are heavy isotopes of hydrogen. Tritium has an atomic mass of 3.016 amu and has a natural abundance of 0.000138%. The effusion of hydrogen gas (containing a mixture of H , HD, and HT molecules) through a porous membrane can be used to obtain samples of hydrogen that are enriched in tritium. How many membrane passes are necessary to give a sample of hydrogen gas in which 1% of the hydrogen molecules are HT? Samples of HBr gas and NH gas are placed at opposite ends of a 1 m tube. If the two gases are allowed to diffuse through the tube toward one another, at what distance from each end of the tube will the gases meet and form solid NH Br? At any temperature, the rms speed of hydrogen is 4.45 times that of argon. Tumbnail from | 28,041 | 2,639 |
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Formation of an ion pair by transfer of an electron from an Li atom to an H atom results in an overall of the two nuclei and four electrons involved. How and why this occurs is best seen if we break ion-pair formation into three simpler steps and consider the energy change involved in each. The three steps are The energy required in step 1 to remove an electron completely from an isolated atom is called the . The ionization energy of lithium is 520 kJ mol . In other words, 520 kJ of energy is needed to remove a mole of 2 electrons from a mole of isolated lithium atoms in order to form a mole of isolated lithium ions. Alternatively we can say that 520 kJ is needed to a mole of lithium atoms. See Figure \(\Page {1}\) for a visual representation of the energy "cost" of ionization. While energy is needed to accomplish step 1, we find that energy is in step 2 when a hydrogen atom accepts an electron and becomes a hydride ion. The reason for this can be seen in the dot density diagram. The second electron acquired by the hydrogen atom can pair up with the electron already in the 1 orbital without contradicting the Pauli exclusion principle. As a result, the new electron can move in close enough to the hydrogen nucleus to be held fairly firmly, lowering its energy significantly. Although the paired electrons repel each other somewhat, this is not enough to offset the attraction of the nucleus for both. Since the energy of the electron is , the law of conservation of energy requires that the same quantity of energy must be when a hydrogen atom is transformed into a hydride ion. The energy released when an electron is acquired by an atom is called the . The electron affinity of hydrogen is 73 kJ mol indicating that 73 kJ of energy is released when 1 mol of isolated hydrogen atoms each accepts an electron and is converted into a hydride ion. Since 520 kJ mol is to remove an electron from a lithium atom, while 73 kJ mol is when the electron is donated to a hydrogen atom, it follows that transfer of an electron from a lithium to a hydrogen atom requires (520 – 73) kJ mol = 447 kJ mol . This net energy change can best be seen in Figure \(\Page {1}\). At room temperature processes which require such a large quantity of energy are extremely unlikely. Indeed the transfer of the electron would be impossible if it were not for step 3, the close approach of the two ions. When oppositely charged particles move closer to each other, their potential energy decreases and they release energy. The energy released when lithium ions and hydride ions come together to 160 pm under the influence of their mutual attraction is 690 kJ mol , more than enough to offset the 447 kJ mol needed to transfer the electron. Thus there is a net release of (690 – 447) kJ mol = 243 kJ mol from the overall process. The transfer of the electron from lithium to hydrogen and the formation of an Li H ion pair results in an overall lowering of energy, as seen in Figure \(\Page {1}\) below. In the above figure, the energy change in each step and the overall change are illustrated diagrammatically. As in the case of atomic structure, where electrons occupy orbitals having the lowest allowable energy, a collection of atoms tends to rearrange its constituent electrons so as to minimize its total energy. Formation of a lithium hydride ion pair is energetically “downhill” and therefore favored. A more complete picture of the energies involved is provided by the "Born-Haber Cycle" diagram of LiF below. It emphasizes that the EA and IP apply to single atoms in the gas phase, and that other energies may be more important in determining the exothermicity of formation of ionic compounds. The greatest energy released is not the EA, but the lattice energy. Other energy costs involve formation of single atoms in the gas phase by vaporization of Li (s) (ΔH ) to give Li (g), and the dissociation energy of F (ΔH ) to give 2F. The sum of all the energies is the energy of formation of the ionic compound, according to Hess' Law. Basically, the Born Haber Cycle shows all the details of the formation of LiF , while the diagram we looked at earlier takes a big picture view. | 4,219 | 2,640 |
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This page looks at esterification - the reaction between alcohols and carboxylic acids to make esters. Esters have a hydrocarbon group of some sort replacing the hydrogen in the -COOH group of a carboxylic acid. We shall just be looking at cases where it is replaced by an alkyl group, but it could equally well be an aryl group (one based on a benzene ring). The most commonly discussed ester is ethyl ethanoate. In this case, the hydrogen in the -COOH group has been replaced by an ethyl group. The formula for ethyl ethanoate is: Notice that the ester is named the opposite way around from the way the formula is written. The "ethanoate" bit comes from ethanoic acid. The "ethyl" bit comes from the ethyl group on the end. In each case, be sure that you can see how the names and formulae relate to each other. Remember that the acid is named by counting up the total number of carbon atoms in the chain - including the one in the -COOH group. So, for example, CH CH COOH is propanoic acid, and CH CH COO is the propanoate group. 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 containing a benzene ring). 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: 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 sulfuric 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 sulfuric acid in the mixture by fractional distillation. Jim Clark ( ) | 2,550 | 2,641 |
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The described previously can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. The specific three dimensional arrangement of atoms in molecules is referred to as . We also define molecular geometry as the positions of the atomic nuclei in a molecule. There are various instrumental techniques such as and other experimental techniques which can be used to tell us where the atoms are located in a molecule. Using advanced techniques, very complicated structures for proteins, enzymes, , and have been determined. Molecular geometry is associated with the chemistry of vision, smell, taste, drug reactions, and enzyme controlled reactions to name a few. The Lewis structure of carbon tetrachloride provides information about connectivities, provides information about valence orbitals, and provides information about bond character. However, the Lewis structure provides no information about the shape of the molecule, which is defined by t . For carbon tetrachloride, each C-Cl bond length is 1.78Å and each Cl-C-Cl bond angle is 109.5°. Hence, carbon tetrachloride is in structure: Molecular geometry is associated with the specific orientation of bonding atoms. A careful analysis of electron distributions in orbitals will usually result in correct molecular geometry determinations. In addition, the simple writing of Lewis diagrams can also provide important clues for the determination of molecular geometry. Molecular shapes, or geometries, are critical to molecular recognition and function. Table \(\Page {1}\) shows some examples of geometries where a central atom \(A\) is bonded to two or more \(X\) atoms. As indicated in several of the geometries below, non-bonding electrons \(E\) can strongly influence the molecular geometry of the molecule; this is discussed in more details in . These structures can generally be predicted, when A is a nonmetal, using the "valence-shell electron-pair repulsion model (VSEPR) discussed in the next section. ( ) | 2,175 | 2,642 |
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Assuming the form of the Maxwell distribution allowing for motion in three directions to be \[ f(v) = N v^2 \text{exp} \left( \dfrac{m v^2}{2 k_BT} \right) \label{MBFullN}\] derive the correct expression for \(N\) such that the distribution is normalized. Hint: a table of definite integrals indicates \[ \int_0^{\infty} x^{2n} e^{-ax^2} dx = \dfrac{1}{4} \dfrac{\sqrt{\pi}}{a^{3/2}}\] Dry ice (solid CO ) has a density of 1.6 g/cm . Assuming spherical molecules, estimate the collisional cross section for \(CO_2\). How does it compare to the value listed in the text? Calculate the pressure exerted by 1.00 mol of Ar, N , and CO as an ideal gas, a van der Waals gas, and a Redlich-Kwong gas, at 25 °C and 24.4 L. The compression factor \(Z\) for CO at 0 °C and 100 atm is 0.2007. Calculate the volume of a 2.50 mole sample of CO at 0 °C and 100 atm. What is the maximum pressure that will afford a N molecule a mean-free-path of at least 1.00 m at 25 °C? In a Knudsen cell, the effusion orifice is measured to be 0.50 mm . If a sample of naphthalene is allowed to effuse for 1.0 hr at a temperature of 40.3 °C, the cell loses 0.0236 g. From this data, calculate the vapor pressure of naphthalene at this temperature. The vapor pressure of scandium was determined using a Knudsen cell [Kirkorian, , , 1586 (1963)]. The data from the experiment are given below. From this data, find the vapor pressure of scandium at 1555.4 K. A thermalized sample of gas is one that has a distribution of molecular speeds given by the Maxwell-Boltzmann distribution. Considering a sample of N at 25 C what fraction of the molecules have a speed less than Assume that a person has a body surface area of 2.0 m . Calculate the number of collisions per second with the total surface area of this person at 25 °C and 1.00 atm. (For convenience, assume air is 100% N ) Two identical balloons are inflated to a volume of 1.00 L with a particular gas. After 12 hours, the volume of one balloon has decreased by 0.200 L. In the same time, the volume of the other balloon has decreased by 0.0603 L. If the lighter of the two gases was helium, what is the molar mass of the heavier gas? Assuming it is a van der Waals gas, calculate the critical temperature, pressure and volume for \(CO_2\). Find an expression in terms of van der Waals coefficients for the Boyle temperature. ( : use the viral expansion of the van der Waals equation to find an expression for the second viral coefficient!) Consider a gas that follows the equation of state \[p =\dfrac{RT}{V_m - b}\] Using a virial expansion, find an expression for the second virial coefficient. Consider a gas that obeys the equation of state \[ p =\dfrac{nRT}{V_m - b}\] where a and b are non-zero constants. Does this gas exhibit critical behavior? If so, find expressions for \(p_c\), \(V_c\), and \(T_c\) in terms of the constants \(a\), \(b\), and \(R\). Consider a gas that obeys the equation of state \[ p = \dfrac{nRT}{V- nB}-\dfrac{an}{V}\] | 3,007 | 2,643 |
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The alkali metals are so reactive that they are never found in nature in elemental form. Although some of their ores are abundant, isolating them from their ores is somewhat difficult. For these reasons, the group 1 elements were unknown until the early 19th century, when Sir Humphry Davy first prepared sodium (Na) and potassium (K) by passing an electric current through molten alkalis. (The ashes produced by the combustion of wood are largely composed of potassium and sodium carbonate.) Lithium (Li) was discovered 10 years later when the Swedish chemist Johan Arfwedson was studying the composition of a new Brazilian mineral. Cesium (Cs) and rubidium (Rb) were not discovered until the 1860s, when Robert Bunsen conducted a systematic search for new elements. Known to chemistry students as the inventor of the Bunsen burner, Bunsen’s spectroscopic studies of ores showed sky blue and deep red emission lines that he attributed to two new elements, Cs and Rb, respectively. Francium (Fr) is found in only trace amounts in nature, so our knowledge of its chemistry is limited. All the isotopes of Fr have very short half-lives, in contrast to the other elements in group 1. Davy was born in Penzance, Cornwall, England. He was a bit of a wild man in the laboratory, often smelling and tasting the products of his experiments, which almost certainly shortened his life. He discovered the physiological effects that cause nitrous oxide to be called “laughing gas” (and became addicted to it!), and he almost lost his eyesight in an explosion of nitrogen trichloride (NCl ), which he was the first to prepare. Davy was one of the first to recognize the utility of Alessandro Volta’s “electric piles” (batteries). By connecting several “piles” in series and inserting electrodes into molten salts of the alkali metals and alkaline earth metals, he was able to isolate six previously unknown elements as pure metals: sodium, potassium, calcium, strontium, barium, and magnesium. He also discovered boron and was the first to prepare phosphine (PH ) and hydrogen telluride (H Te), both of which are highly toxic. Bunsen was born and educated in Göttingen, Germany. His early work dealt with organic arsenic compounds, whose highly toxic nature and explosive tendencies almost killed him and did cost him an eye. He designed the Bunsen burner, a reliable gas burner, and used it and emission spectra to discover cesium (named for its blue line) and rubidium (named for its red line). Because the alkali metals are among the most potent reductants known, obtaining them in pure form requires a considerable input of energy. Pure lithium and sodium for example, are typically prepared by the electrolytic reduction of molten chlorides: \[\mathrm{LiCl(l)}\rightarrow\mathrm{Li(l)}+\frac{1}{2}\mathrm{Cl_2(g)} \label{21.15}\] In practice, CaCl is mixed with LiCl to lower the melting point of the lithium salt. The electrolysis is carried out in an argon atmosphere rather than the nitrogen atmosphere typically used for substances that are highly reactive with O and water because Li reacts with nitrogen gas to form lithium nitride (Li N). Metallic sodium is produced by the electrolysis of a molten mixture of NaCl and CaCl . In contrast, potassium is produced commercially from the reduction of KCl by Na, followed by the fractional distillation of K(g). Although rubidium and cesium can also be produced by electrolysis, they are usually obtained by reacting their hydroxide salts with a reductant such as Mg: \[2RbOH_{(s)} + Mg_{(s)} \rightarrow 2Rb_{(l)} + Mg(OH)_{2(s)} \label{21.6}\] Massive deposits of essentially pure NaCl and KCl are found in nature and are the major sources of sodium and potassium. The other alkali metals are found in low concentrations in a wide variety of minerals, but ores that contain high concentrations of these elements are relatively rare. No concentrated sources of rubidium are known, for example, even though it is the 16th most abundant element on Earth. Rubidium is obtained commercially by isolating the 2%–4% of Rb present as an impurity in micas, minerals that are composed of sheets of complex hydrated potassium–aluminum silicates. Alkali metals are recovered from silicate ores in a multistep process that takes advantage of the pH-dependent solubility of selected salts of each metal ion. The steps in this process are leaching, which uses sulfuric acid to dissolve the desired alkali metal ion and Al from the ore; basic precipitation to remove Al from the mixture as Al(OH) ; selective precipitation of the insoluble alkali metal carbonate; dissolution of the salt again in hydrochloric acid; and isolation of the metal by evaporation and electrolysis. Figure \(\Page {1}\) illustrates the isolation of liquid lithium from a lithium silicate ore by this process. Various properties of the group 1 elements are summarized in Table \(\Page {1}\). In keeping with overall periodic trends, the atomic and ionic radii increase smoothly from Li to Cs, and the first ionization energies decrease as the atoms become larger. As a result of their low first ionization energies, the alkali metals have an overwhelming tendency to form ionic compounds where they have a +1 charge. All the alkali metals have relatively high electron affinities because the addition of an electron produces an anion (M−) with an ns electron configuration. The densities of the elements generally increase from Li to Cs, reflecting another common trend: because the atomic masses of the elements increase more rapidly than the atomic volumes as you go down a group, the densest elements are near the bottom of the periodic table. An unusual trend in the group 1 elements is the smooth decrease in the melting and boiling points from Li to Cs. As a result, Cs (melting point = 28.5°C) is one of only three metals (the others are Ga and Hg) that are liquids at body temperature (37°C). The standard reduction potentials (E°) of the alkali metals do not follow the trend based on ionization energies. Unexpectedly, lithium is the strongest reductant, and sodium is the weakest (Table \(\Page {1}\)). Because Li is much smaller than the other alkali metal cations, its hydration energy is the highest. The high hydration energy of Li more than compensates for its higher ionization energy, making lithium metal the strongest reductant in aqueous solution. This apparent anomaly is an example of how the physical or the chemical behaviors of the elements in a group are often determined by the subtle interplay of opposing periodic trends. All alkali metals are electropositive elements with an ns valence electron configuration, forming the monocation (M ) by losing the single valence electron. Because removing a second electron would require breaking into the (n − 1) closed shell, which is energetically prohibitive, the chemistry of the alkali metals is largely that of ionic compounds that contain M ions. However, as we discuss later, the lighter group 1 elements also form a series of organometallic compounds that contain polar covalent M–C bonds. All the alkali metals react vigorously with the halogens (group 17) to form the corresponding ionic halides, where \(X\) is a halogen: \[2M_{(s)} + X_{2(s, l, g)} \rightarrow 2M^+X^−_{(s)} \label{21.7}\] Similarly, the alkali metals react with the heavier chalcogens (sulfur, selenium, and tellurium in group 16) to produce metal chalcogenides, where Y is S, Se, or Te: \[2M_{(s)} + Y_{(s)} \rightarrow M_2Y_{(s)} \label{21.8}\] When excess chalcogen is used, however, a variety of products can be obtained that contain chains of chalcogen atoms, such as the sodium polysulfides (Na S , where n = 2–6). For example, Na S contains the S ion, which is V shaped with an S–S–S angle of about 103°. The one-electron oxidation product of the trisulfide ion (S ) is responsible for the intense blue color of the gemstones lapis lazuli and blue ultramarine (Figure \(\Page {2}\)). Reacting the alkali metals with oxygen, the lightest element in group 16, is more complex, and the stoichiometry of the product depends on both the metal:oxygen ratio and the size of the metal atom. For instance, when alkali metals burn in air, the observed products are Li O (white), Na O (pale yellow), KO (orange), RbO (brown), and CsO (orange). Only Li O has the stoichiometry expected for a substance that contains two M cations and one O ion. In contrast, Na O contains the O (peroxide) anion plus two Na cations. The other three salts, with stoichiometry MO , contain the M cation and the O (superoxide) ion. Because O is the smallest of the three oxygen anions, it forms a stable ionic lattice with the smallest alkali metal cation (Li ). In contrast, the larger alkali metals—potassium, rubidium, and cesium—react with oxygen in air to give the metal superoxides. Because the Na cation is intermediate in size, sodium reacts with oxygen to form a compound with an intermediate stoichiometry: sodium peroxide. Under specific reaction conditions, however, it is possible to prepare the oxide, peroxide, and superoxide salts of all five alkali metals, except for lithium superoxide (LiO ). The chemistry of the alkali metals is largely that of ionic compounds containing the M ions. The alkali metal peroxides and superoxides are potent oxidants that react, often vigorously, with a wide variety of reducing agents, such as charcoal or aluminum metal. For example, Na O is used industrially for bleaching paper, wood pulp, and fabrics such as linen and cotton. In submarines, Na O and KO are used to purify and regenerate the air by removing the CO produced by respiration and replacing it with O . Both compounds react with CO in a redox reaction in which O or O is simultaneously oxidized and reduced, producing the metal carbonate and O : \[2Na_2O_{2(s)} + 2CO_{2(g)} \rightarrow 2Na_2CO_{3(s)} + O_{2(g)} \label{21.9}\] \[4KO_{2(s)} + 2CO_{2(g)} \rightarrow 2K_2CO_{3(s)} + 3O_{2(g)} \label{21.10}\] The presence of water vapor, the other product of respiration, makes KO even more effective at removing CO because potassium bicarbonate, rather than potassium carbonate, is formed: \[4KO_{2(s)} + 4CO_{2(g)} + 2H_2O_{(g)} \rightarrow 4KHCO_{3(s)} + 3O_{2(g)} \label{21.11}\] Notice that 4 mol of CO are removed in this reaction, rather than 2 mol in Equation 21.10. Lithium, the lightest alkali metal, is the only one that reacts with atmospheric nitrogen, forming lithium nitride (Li N). Lattice energies again explain why the larger alkali metals such as potassium do not form nitrides: packing three large K cations around a single relatively small anion is energetically unfavorable. In contrast, all the alkali metals react with the larger group 15 elements phosphorus and arsenic to form metal phosphides and arsenides (where Z is P or As): \[12M_{(s)} + Z_{4(s)} \rightarrow 4M_3Z_{(s)} \label{21.12}\] Because of lattice energies, only lithium forms a stable oxide and nitride. The alkali metals react with all group 14 elements, but the compositions and properties of the products vary significantly. For example, reaction with the heavier group 14 elements gives materials that contain polyatomic anions and three-dimensional cage structures, such as K Si whose structure is shown here. In contrast, lithium and sodium are oxidized by carbon to produce a compound with the stoichiometry M C (where M is Li or Na): \[ 2M_{(s)} + 2C_{(s)} \rightarrow M_2C_{2(s)} \label{21.13}\] The same compounds can be obtained by reacting the metal with acetylene (C H ). In this reaction, the metal is again oxidized, and hydrogen is reduced: \[ 2M_{(s)} + C_2H_{2(g)} \rightarrow M_2C_{2(s)} + H_{2(g)} \label{21.14}\] The acetylide ion (C ), formally derived from acetylene by the loss of both hydrogens as protons, is a very strong base. Reacting acetylide salts with water produces acetylene and MOH(aq). The heavier alkali metals (K, Rb, and Cs) also react with carbon in the form of graphite. Instead of disrupting the hexagonal sheets of carbon atoms, however, the metals insert themselves between the sheets of carbon atoms to give new substances called (part (a) in Figure \(\Page {3}\)). The stoichiometries of these compounds include MC and MC , which are black/gray; MC and MC , which are blue; and MC , which is bronze (part (b) in Figure \(\Page {3}\)). The remarkably high electrical conductivity of these compounds (about 200 times greater than graphite) is attributed to a net transfer of the valence electron of the alkali metal to the graphite layers to produce, for example, K C . All the alkali metals react directly with gaseous hydrogen at elevated temperatures to produce ionic hydrides (M H ): \[2M_{(s)} + H_{2(g)} \rightarrow 2MH_{(s)} \label{21.15a}\] All are also capable of reducing water to produce hydrogen gas: \[\mathrm{M(s)}+\mathrm{H_2O(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{MOH(aq)} \label{21.16}\] Alkali metal cations are found in a wide variety of ionic compounds. In general, any alkali metal salt can be prepared by reacting the alkali metal hydroxide with an acid and then evaporating the water: \[2MOH_{(aq)} + H_2SO_{4(aq)} \rightarrow M_2SO_{4(aq)} + 2H_2O_{(l)} \label{21.17}\] \[MOH_{(aq)} + HNO_{3(aq)} \rightarrow MNO_{3(aq)} + H_2O_{(l)} \label{21.18}\] Hydroxides of alkali metals also can react with organic compounds that contain an acidic hydrogen to produce a salt. An example is the preparation of sodium acetate (CH CO Na) by reacting sodium hydroxide and acetic acid: \[CH_3CO_2H_{(aq)} + NaOH_{(s)} \rightarrow CH_3CO_2Na_{(aq)} + H_2O_{(l)} \label{21.19}\] Soap is a mixture of the sodium and potassium salts of naturally occurring carboxylic acids, such as palmitic acid [CH (CH ) CO H] and stearic acid [CH (CH ) CO H]. Lithium salts, such as lithium stearate [CH (CH ) CO Li], are used as additives in motor oils and greases. Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to react with simple Lewis bases to form metal complexes. Complex formation is most significant for the smallest cation (Li ) and decreases with increasing radius. In aqueous solution, for example, Li forms the tetrahedral [Li(H O) ] complex. In contrast, the larger alkali metal cations form octahedral [M(H O) ] complexes. Complex formation is primarily due to the electrostatic interaction of the metal cation with polar water molecules. Because of their high affinity for water, anhydrous salts that contain Li and Na ions (such as Na SO ) are often used as drying agents. These compounds absorb trace amounts of water from nonaqueous solutions to form hydrated salts, which are then easily removed from the solution by filtration. Because of their low positive charge (+1) and relatively large ionic radii, alkali metal cations have only a weak tendency to form complexes with simple Lewis bases. Electrostatic interactions also allow alkali metal ions to form complexes with certain cyclic polyethers and related compounds, such as crown ethers and cryptands. As discussed in Chapter 13, are cyclic polyethers that contain four or more oxygen atoms separated by two or three carbon atoms. All crown ethers have a central cavity that can accommodate a metal ion coordinated to the ring of oxygen atoms, and crown ethers with rings of different sizes prefer to bind metal ions that fit into the cavity. For example, 14-crown-4, with the smallest cavity that can accommodate a metal ion, has the highest affinity for Li , whereas 18-crown-6 forms the strongest complexes with K . are more nearly spherical analogues of crown ethers and are even more powerful and selective complexing agents. Cryptands consist of three chains containing oxygen that are connected by two nitrogen atoms (part (b) in Figure 13.7). They can completely surround (encapsulate) a metal ion of the appropriate size, coordinating to the metal by a lone pair of electrons on each O atom and the two N atoms. Like crown ethers, cryptands with different cavity sizes are highly selective for metal ions of particular sizes. Crown ethers and cryptands are often used to dissolve simple inorganic salts such as KMnO in nonpolar organic solvents. A remarkable feature of the alkali metals is their ability to dissolve reversibly in liquid ammonia. Just as in their reactions with water, reacting alkali metals with liquid ammonia eventually produces hydrogen gas and the metal salt of the conjugate base of the solvent—in this case, the amide ion (NH ) rather than hydroxide: \[\mathrm{M(s)}+\mathrm{NH_3(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{M^+(am)}+\mathrm{NH_2^-(am)} \label{21.20}\] where the (am) designation refers to an ammonia solution, analogous to (aq) used to indicate aqueous solutions. Without a catalyst, the reaction in Equation 21.20 tends to be rather slow. In many cases, the alkali metal amide salt (MNH ) is not very soluble in liquid ammonia and precipitates, but when dissolved, very concentrated solutions of the alkali metal are produced. One mole of Cs metal, for example, will dissolve in as little as 53 mL (40 g) of liquid ammonia. The pure metal is easily recovered when the ammonia evaporates. Solutions of alkali metals in liquid ammonia are intensely colored and good conductors of electricity due to the presence of solvated electrons (e , NH ), which are not attached to single atoms. A solvated electron is loosely associated with a cavity in the ammonia solvent that is stabilized by hydrogen bonds. Alkali metal–liquid ammonia solutions of about 3 M or less are deep blue (Figure \(\Page {5}\)) and conduct electricity about 10 times better than an aqueous NaCl solution because of the high mobility of the solvated electrons. As the concentration of the metal increases above 3 M, the color changes to metallic bronze or gold, and the conductivity increases to a value comparable with that of the pure liquid metals. In addition to solvated electrons, solutions of alkali metals in liquid ammonia contain the metal cation (M ), the neutral metal atom (M), metal dimers (M ), and the metal anion (M ). The anion is formed by adding an electron to the singly occupied ns valence orbital of the metal atom. Even in the absence of a catalyst, these solutions are not very stable and eventually decompose to the thermodynamically favored products: M NH and hydrogen gas (Equation 21.20). Nonetheless, the solvated electron is a potent reductant that is often used in synthetic chemistry. Compounds that contain a metal covalently bonded to a carbon atom of an organic species are called . The properties and reactivities of organometallic compounds differ greatly from those of either the metallic or organic components. Because of its small size, lithium, for example, forms an extensive series of covalent organolithium compounds, such as methyllithium (LiCH ), which are by far the most stable and best-known group 1 organometallic compounds. These volatile, low-melting-point solids or liquids can be sublimed or distilled at relatively low temperatures and are soluble in nonpolar solvents. Like organic compounds, the molten solids do not conduct electricity to any significant degree. Organolithium compounds have a tendency to form oligomers with the formula (RLi) , where R represents the organic component. For example, in both the solid state and solution, methyllithium exists as a tetramer with the structure shown in Figure \(\Page {6}\), where each triangular face of the Li tetrahedron is bridged by the carbon atom of a methyl group. Effectively, the carbon atom of each CH group is using a single pair of electrons in an sp hybrid lobe to bridge three lithium atoms, making this an example of two-electron, four-center bonding. Clearly, such a structure, in which each carbon atom is apparently bonded to six other atoms, cannot be explained using any of the electron-pair bonding schemes. Molecular orbital theory can explain the bonding in methyllithium, but the description is beyond the scope of this text. Because sodium remains liquid over a wide temperature range (97.8–883°C), it is used as a coolant in specialized high-temperature applications, such as nuclear reactors and the exhaust valves in high-performance sports car engines. Cesium, because of its low ionization energy, is used in photosensors in automatic doors, toilets, burglar alarms, and other electronic devices. In these devices, cesium is ionized by a beam of visible light, thereby producing a small electric current; blocking the light interrupts the electric current and triggers a response. Compounds of sodium and potassium are produced on a huge scale in industry. Each year, the top 50 industrial compounds include NaOH, used in a wide variety of industrial processes; Na CO , used in the manufacture of glass; K O, used in porcelain glazes; and Na SiO , used in detergents. Several other alkali metal compounds are also important. For example, Li CO is one of the most effective treatments available for manic depression or bipolar disorder. It appears to modulate or dampen the effect on the brain of changes in the level of neurotransmitters, which are biochemical substances responsible for transmitting nerve impulses between neurons. Consequently, patients who take “lithium” do not exhibit the extreme mood swings that characterize this disorder. For each application, choose the more appropriate substance based on the properties and reactivities of the alkali metals and their compounds. Explain your choice in each case. application and selected alkali metals appropriate metal for each application Use the properties and reactivities discussed in this section to determine which alkali metal is most suitable for the indicated application. Indicate which of the alternative alkali metals or their compounds given is more appropriate for each application. Predict the products of each reaction and then balance each chemical equation. reactants products and balanced chemical equation Determine whether one of the reactants is an oxidant or a reductant or a strong acid or a strong base. If so, a redox reaction or an acid–base reaction is likely to occur. Identify the products of the reaction. If a reaction is predicted to occur, balance the chemical equation. The balanced chemical equation is 2Na(s) + O (g) → Na O (s). The balanced chemical equation is Li O(s) + H O(l) → 2LiOH(aq). The balanced chemical equation is as follows: \(\mathrm{K(s)}+\mathrm{CH_3OH(l)}\rightarrow\frac{1}{2}\mathrm{H_2(g)}+\mathrm{CH_3OK(soln)}\). Two moles of lithium are required to balance the equation: 2Li(s) + CH Cl(l) → LiCl(s) + CH Li(soln). We conclude that the two substances will not react with each other. Predict the products of each reaction and balance each chemical equation. The alkali metals are potent reductants whose chemistry is largely that of ionic compounds containing the M ion. Alkali metals have only a weak tendency to form complexes with simple Lewis bases. The first alkali metals to be isolated (Na and K) were obtained by passing an electric current through molten potassium and sodium carbonates. The alkali metals are among the most potent reductants known; most can be isolated by electrolysis of their molten salts or, in the case of rubidium and cesium, by reacting their hydroxide salts with a reductant. They can also be recovered from their silicate ores using a multistep process. Lithium, the strongest reductant, and sodium, the weakest, are examples of the physical and chemical effects of opposing periodic trends. The alkali metals react with halogens (group 17) to form ionic halides; the heavier chalcogens (group 16) to produce metal chalcogenides; and oxygen to form compounds, whose stoichiometry depends on the size of the metal atom. The peroxides and superoxides are potent oxidants. The only alkali metal to react with atmospheric nitrogen is lithium. Heavier alkali metals react with graphite to form graphite intercalation compounds, substances in which metal atoms are inserted between the sheets of carbon atoms. With heavier group 14 elements, alkali metals react to give polyatomic anions with three-dimensional cage structures. All alkali metals react with hydrogen at high temperatures to produce the corresponding hydrides, and all reduce water to produce hydrogen gas. Alkali metal salts are prepared by reacting a metal hydroxide with an acid, followed by evaporation of the water. Both Li and Na salts are used as drying agents, compounds that are used to absorb water. Complexing agents such as crown ethers and cryptands can accommodate alkali metal ions of the appropriate size. Alkali metals can also react with liquid ammonia to form solutions that slowly decompose to give hydrogen gas and the metal salt of the amide ion (NH ). These solutions, which contain unstable solvated electrons loosely associated with a cavity in the solvent, are intensely colored, good conductors of electricity, and excellent reductants. Alkali metals can react with organic compounds that contain an acidic proton to produce salts. They can also form organometallic compounds, which have properties that differ from those of their metallic and organic components. | 25,295 | 2,644 |
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There are two mechanistic models for how an alkyl halide can undergo nucleophilic substitution. In the first picture, the reaction takes place in a single step, and bond-forming and bond-breaking occur simultaneously. (In all figures in this section, 'X' indicates a halogen substituent). This is called an ' mechanism. In the term S 2, S stands for 'substitution', the subscript N stands for 'nucleophilic', and the number 2 refers to the fact that this is a : the overall rate depends on a step in which two separate molecules (the nucleophile and the electrophile) collide. A potential energy diagram for this reaction shows the transition state (TS) as the highest point on the pathway from reactants to products. If you look carefully at the progress of the S 2 reaction, you will realize something very important about the outcome. The nucleophile, being an electron-rich species, must attack the electrophilic carbon from the relative to the location of the leaving group. Approach from the front side simply doesn't work: the leaving group - which is also an electron-rich group - blocks the way. The result of this backside attack is that the stereochemical configuration at the central carbon as the reaction proceeds. In a sense, the molecule is turned inside out. At the transition state, the electrophilic carbon and the three 'R' substituents all lie on the same plane. What this means is that S 2 reactions whether enzyme catalyzed or not, are inherently stereoselective: when the substitution takes place at a stereocenter, we can confidently predict the stereochemical configuration of the product. Below is an animation illustrating the principles we have just learned, showing the S 2 reaction between hydroxide ion and methyl iodide. Notice how backside attack by the hydroxide nucleophile results in inversion at the tetrahedral carbon electrophile. Predict the structure of the product in this S 2 reaction. Be sure to specify stereochemistry. The rate of an SN2 reaction is significantly influenced by the solvent in which the reaction takes place. The use of (those, such as water or alcohols, with hydrogen-bond donating capability) decreases the power of the nucleophile, because of strong hydrogen-bond interactions between solvent protons and the reactive lone pairs on the nucleophile. A less powerful nucleophile in turn means a slower SN2 reaction. SN2 reactions are faster in those that lack hydrogen-bond donating capability. Below are several polar aprotic solvents that are commonly used in the laboratory: These aprotic solvents are polar but, because they do not form hydrogen bonds with the anionic nucleophile, there is a relatively weak interaction between the aprotic solvent and the nucleophile. 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 model for a nucleophilic substitution reaction is called the ' , or ' mechanism: in this picture, the C-X bond breaks , before the nucleophile approaches: This results in the formation of a carbocation: because the central carbon has only three bonds, it bears a formal charge of +1. Recall that a carbocation should be pictured as hybridized, with trigonal planar geometry. Perpendicular to the plane formed by the three hybrid orbitals is an empty, unhybridized orbital. In the second step of this two-step reaction, the nucleophile attacks the empty, 'electron hungry' orbital of the carbocation to form a new bond and return the carbon to tetrahedral geometry. We saw that S 2 reactions result specifically in inversion of stereochemistry at the electrophilic carbon center. What about the stereochemical outcome of S 1 reactions? In the model S 1 reaction shown above, the leaving group dissociates completely from the vicinity of the reaction before the nucleophile begins its attack. Because the leaving group is no longer in the picture, the nucleophile is free to attack from either side of the planar, -hybridized carbocation electrophile. This means that about half the time the product has the same stereochemical configuration as the starting material (retention of configuration), and about half the time the stereochemistry has been inverted. In other words, has occurred at the carbon center. As an example, the tertiary alkyl bromide below would be expected to form a racemic mix of and alcohols after an S 1 reaction with water as the incoming nucleophile. The S 1 reaction we see an example of a reaction intermediate, a very important concept in the study of organic reaction mechanisms that was introduced earlier in the module on organic reactivity Recall that many important organic reactions do not occur in a single step; rather, they are the sum of two or more discreet bond-forming / bond-breaking steps, and involve transient intermediate species that go on to react very quickly. In the S 1 reaction, the carbocation species is a reaction intermediate. A potential energy diagram for an S 1 reaction shows that the carbocation intermediate can be visualized as a kind of valley in the path of the reaction, higher in energy than both the reactant and product but lower in energy than the two transition states. Recall that the first step of the reaction above, in which two charged species are formed from a neutral molecule, is much the slower of the two steps, and is therefore rate-determining. This is illustrated by the energy diagram, where the activation energy for the first step is higher than that for the second step. Also recall that an S 1 reaction has kinetics, because the rate determining step involves one molecule splitting apart, not two molecules colliding. Consider two nucleophilic substitutions that occur uncatalyzed in solution. Assume that reaction A is S 2, and reaction B is S 1. Predict, in each case, what would happen to the rate of the reaction if the concentration of the nucleophile were doubled, while all other conditions remained constant. Draw a complete curved-arrow mechanism for the methanolysis reaction of allyl bromide shown above. One more important point must be made before continuing: nucleophilic substitutions as a rule occur at sp -hybridized carbons, and where the leaving group is attached to an sp -hybridized carbon:: Bonds on sp -hybridized carbons are inherently shorter and stronger than bonds on sp -hybridized carbons, meaning that it is harder to break the C-X bond in these substrates. S 2 reactions of this type are unlikely also because the (hypothetical) electrophilic carbon is protected from nucleophilic attack by electron density in the p bond. S 1 reactions are highly unlikely, because the resulting carbocation intermediate, which would be sp-hybridized, would be very unstable (we’ll discuss the relative stability of carbocation intermediates in a later section of this module). Before we look at some real-life nucleophilic substitution reactions in the next chapter, we will spend some time in the remainder of this module focusing more closely on the three principal partners in the nucleophilic substitution reaction: the nucleophile, the electrophile, and the leaving group. | 7,412 | 2,645 |
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Group 13 is the first group to span the dividing line between metals and nonmetals, so its chemistry is more diverse than that of groups 1 and 2, which include only metallic elements. Except for the lightest element (boron), the group 13 elements are all relatively electropositive; that is, they tend to lose electrons in chemical reactions rather than gain them. Although group 13 includes aluminum, the most abundant metal on Earth, none of these elements was known until the early 19th century because they are never found in nature in their free state. Elemental boron and aluminum, which were first prepared by reducing B O and AlCl , respectively, with potassium, could not be prepared until potassium had been isolated and shown to be a potent reductant. Indium (In) and thallium (Tl) were discovered in the 1860s by using spectroscopic techniques, long before methods were available for isolating them. Indium, named for its indigo (deep blue-violet) emission line, was first observed in the spectrum of zinc ores, while thallium (from the Greek thallos, meaning “a young, green shoot of a plant”) was named for its brilliant green emission line. Gallium (Ga; Mendeleev’s eka-aluminum) was discovered in 1875 by the French chemist Paul Émile Lecoq de Boisbaudran during a systematic search for Mendeleev’s “missing” element in group 13. Group 13 elements are never found in nature in their free state. As reductants, the group 13 elements are less powerful than the alkali metals and alkaline earth metals. Nevertheless, their compounds with oxygen are thermodynamically stable, and large amounts of energy are needed to isolate even the two most accessible elements—boron and aluminum—from their oxide ores. Although boron is relatively rare (it is about 10,000 times less abundant than aluminum), concentrated deposits of borax [Na B O (OH) ·8H O] are found in ancient lake beds (Figure \(\Page {1}\)) and were used in ancient times for making glass and glazing pottery. Boron is produced on a large scale by reacting borax with acid to produce boric acid [B(OH) ], which is then dehydrated to the oxide (B O ). Reduction of the oxide with magnesium or sodium gives amorphous boron that is only about 95% pure: \[\mathrm{Na_2B_4O_5(OH)_4\cdot8H_2O(s)}\xrightarrow{\textrm{acid}}\mathrm{B(OH)_3(s)}\xrightarrow{\Delta}\mathrm{B_2O_3(s)} \label{Eq1}\] \[\mathrm{B_2O_3(s)}+\mathrm{3Mg(s)}\xrightarrow{\Delta}\mathrm{2B(s)}+\mathrm{3MgO(s)} \label{Eq2}\] Pure, crystalline boron, however, is extremely difficult to obtain because of its high melting point (2300°C) and the highly corrosive nature of liquid boron. It is usually prepared by reducing pure BCl with hydrogen gas at high temperatures or by the thermal decomposition of boron hydrides such as diborane (B H ): \[\mathrm{BCl_3(g)}+\frac{3}{2}\mathrm{H_2(g)}\rightarrow\mathrm{B(s)}+\mathrm{3HCl(g)} \label{Eq3}\] \[B_2H_{6(g)} \rightarrow 2B_{(s)} + 3H_{2(g)} \label{Eq4}\] The reaction shown in Equation \(\ref{Eq3}\) is used to prepare boron fibers, which are stiff and light. Hence they are used as structural reinforcing materials in objects as diverse as the US space shuttle and the frames of lightweight bicycles that are used in races such as the Tour de France. Boron is also an important component of many ceramics and heat-resistant borosilicate glasses, such as Pyrex, which is used for ovenware and laboratory glassware. In contrast to boron, deposits of aluminum ores such as bauxite, a hydrated form of Al O , are abundant. With an electrical conductivity about twice that of copper on a weight for weight basis, aluminum is used in more than 90% of the overhead electric power lines in the United States. However, because aluminum–oxygen compounds are stable, obtaining aluminum metal from bauxite is an expensive process. Aluminum is extracted from oxide ores by treatment with a strong base, which produces the soluble hydroxide complex [Al(OH) ] . Neutralization of the resulting solution with gaseous CO results in the precipitation of Al(OH) : \[2[Al(OH)_4]^−_{(aq)} + CO_{2(g)} \rightarrow 2Al(OH)_{3(s)} + CO^{2−}_{3(aq)} + H_2O_{(l)} \label{Eq5}\] Thermal dehydration of Al(OH) produces Al O , and metallic aluminum is obtained by the electrolytic reduction of Al O using the . Of the group 13 elements, only aluminum is used on a large scale: for example, each Boeing 777 airplane is about 50% aluminum by mass. The other members of group 13 are rather rare: gallium is approximately 5000 times less abundant than aluminum, and indium and thallium are even scarcer. Consequently, these metals are usually obtained as by-products in the processing of other metals. The extremely low melting point of gallium (29.6°C), however, makes it easy to separate from aluminum. Due to its low melting point and high boiling point, gallium is used as a liquid in thermometers that have a temperature range of almost 2200°C. Indium and thallium, the heavier group 13 elements, are found as trace impurities in sulfide ores of zinc and lead. Indium is used as a crushable seal for high-vacuum cryogenic devices, and its alloys are used as low-melting solders in electronic circuit boards. Thallium, on the other hand, is so toxic that the metal and its compounds have few uses. Both indium and thallium oxides are released in flue dust when sulfide ores are converted to metal oxides and SO . Until relatively recently, these and other toxic elements were allowed to disperse in the air, creating large “dead zones” downwind of a smelter. The flue dusts are now trapped and serve as a relatively rich source of elements such as In and Tl (as well as Ge, Cd, Te, and As). Table \(\Page {1}\) summarizes some important properties of the group 13 elements. Notice the large differences between boron and aluminum in size, ionization energy, electronegativity, and standard reduction potential, which is consistent with the observation that boron behaves chemically like a nonmetal and aluminum like a metal. All group 13 elements have ns np valence electron configurations, and all tend to lose their three valence electrons to form compounds in the +3 oxidation state. The heavier elements in the group can also form compounds in the +1 oxidation state formed by the formal loss of the single np valence electron. Because the group 13 elements generally contain only six valence electrons in their neutral compounds, these compounds are all moderately strong Lewis acids. Neutral compounds of the group 13 elements are electron deficient, so they are generally moderately strong Lewis acids. In contrast to groups 1 and 2, the group 13 elements show no consistent trends in ionization energies, electron affinities, and reduction potentials, whereas electronegativities actually increase from aluminum to thallium. Some of these anomalies, especially for the series Ga, In, Tl, can be explained by the increase in the effective nuclear charge (Z ) that results from poor shielding of the nuclear charge by the filled (n − 1)d and (n − 2)f subshells. Consequently, although the actual nuclear charge increases by 32 as we go from indium to thallium, screening by the filled 5d and 4f subshells is so poor that Z increases significantly from indium to thallium. Thus the first ionization energy of thallium is actually greater than that of indium. Anomalies in periodic trends among Ga, In, and Tl can be explained by the increase in the effective nuclear charge due to poor shielding. Elemental boron is a semimetal that is remarkably unreactive; in contrast, the other group 13 elements all exhibit metallic properties and reactivity. We therefore consider the reactions and compounds of boron separately from those of other elements in the group. All group 13 elements have fewer valence electrons than valence orbitals, which generally results in delocalized, metallic bonding. With its high ionization energy, low electron affinity, low electronegativity, and small size, however, boron does not form a metallic lattice with delocalized valence electrons. Instead, boron forms unique and intricate structures that contain multicenter bonds, in which a pair of electrons holds together three or more atoms. Elemental boron forms multicenter bonds, whereas the other group 13 elements exhibit metallic bonding. The basic building block of elemental boron is not the individual boron atom, as would be the case in a metal, but rather the B icosahedron. Because these icosahedra do not pack together very well, the structure of solid boron contains voids, resulting in its low density (Figure \(\Page {3}\)). Elemental boron can be induced to react with many nonmetallic elements to give binary compounds that have a variety of applications. For example, plates of boron carbide (B C) can stop a 30-caliber, armor-piercing bullet, yet they weigh 10%–30% less than conventional armor. Other important compounds of boron with nonmetals include boron nitride (BN), which is produced by heating boron with excess nitrogen (Equation \(\ref{Eq22.6}\)); boron oxide (B O ), which is formed when boron is heated with excess oxygen (Equation \(\ref{Eq22.7}\)); and the boron trihalides (BX ), which are formed by heating boron with excess halogen (Equation \(\ref{Eq22.8}\)). \[\mathrm{2B(s)}+\mathrm{N_2(g)}\xrightarrow{\Delta}\mathrm{2BN(s)} \label{Eq22.6}\] \[\mathrm{4B(s)} + \mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2B_2O_3(s)}\label{Eq22.7}\] \[\mathrm{2B(s)} +\mathrm{3X_2(g)}\xrightarrow{\Delta}\mathrm{2BX_3(g)}\label{Eq22.8}\] Boron nitride is similar in many ways to elemental carbon. With eight electrons, the B–N unit is isoelectronic with the C–C unit, and B and N have the same average size and electronegativity as C. The most stable form of BN is similar to graphite, containing six-membered B N rings arranged in layers. At high temperature and pressure, hexagonal BN converts to a cubic structure similar to diamond, which is one of the hardest substances known. Boron oxide (B O ) contains layers of trigonal planar BO groups (analogous to BX ) in which the oxygen atoms bridge two boron atoms. It dissolves many metal and nonmetal oxides, including SiO , to give a wide range of commercially important borosilicate glasses. A small amount of CoO gives the deep blue color characteristic of “cobalt blue” glass. At high temperatures, boron also reacts with virtually all metals to give metal borides that contain regular three-dimensional networks, or clusters, of boron atoms. The structures of two metal borides—ScB and CaB —are shown in Figure \(\Page {4}\). Because metal-rich borides such as ZrB and TiB are hard and corrosion resistant even at high temperatures, they are used in applications such as turbine blades and rocket nozzles. Boron hydrides were not discovered until the early 20th century, when the German chemist Alfred Stock undertook a systematic investigation of the binary compounds of boron and hydrogen, although binary hydrides of carbon, nitrogen, oxygen, and fluorine have been known since the 18th century. Between 1912 and 1936, Stock oversaw the preparation of a series of boron–hydrogen compounds with unprecedented structures that could not be explained with simple bonding theories. All these compounds contain multicenter bonds. The simplest example is diborane (B H ), which contains two bridging hydrogen atoms (part (a) in Figure \(\Page {5}\). An extraordinary variety of polyhedral boron–hydrogen clusters is now known; one example is the B H ion, which has a polyhedral structure similar to the icosahedral B unit of elemental boron, with a single hydrogen atom bonded to each boron atom. A related class of polyhedral clusters, the carboranes, contain both CH and BH units; an example is shown here. Replacing the hydrogen atoms bonded to carbon with organic groups produces substances with novel properties, some of which are currently being investigated for their use as liquid crystals and in cancer chemotherapy. The enthalpy of combustion of diborane (B H ) is −2165 kJ/mol, one of the highest values known: \[B_2H_{6(g)} + 3O_{2(g)} \rightarrow B_2O_{3(s)} + 3H_2O(l)\;\;\; ΔH_{comb} = −2165\; kJ/mol \label{Eq 22.9}\] Consequently, the US military explored using boron hydrides as rocket fuels in the 1950s and 1960s. This effort was eventually abandoned because boron hydrides are unstable, costly, and toxic, and, most important, B O proved to be highly abrasive to rocket nozzles. Reactions carried out during this investigation, however, showed that boron hydrides exhibit unusual reactivity. Because boron and hydrogen have almost identical electronegativities, the reactions of boron hydrides are dictated by minor differences in the distribution of electron density in a given compound. In general, two distinct types of reaction are observed: electron-rich species such as the BH ion are reductants, whereas electron-deficient species such as B H act as oxidants. For each reaction, explain why the given products form. balanced chemical equations why the given products form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form. Predict the products of the reactions and write a balanced chemical equation for each reaction. All four of the heavier group 13 elements (Al, Ga, In, and Tl) react readily with the halogens to form compounds with a 1:3 stoichiometry: \[ 2M_{(s)} + 3X_{2(s,l,g)} \rightarrow 2MX_{3(s)} \text{ or } M_2X_6 \label{Eq10}\] The reaction of Tl with iodine is an exception: although the product has the stoichiometry TlI , it is not thallium(III) iodide, but rather a thallium(I) compound, the Tl salt of the triiodide ion (I ). This compound forms because iodine is not a powerful enough oxidant to oxidize thallium to the +3 oxidation state. Of the halides, only the fluorides exhibit behavior typical of an ionic compound: they have high melting points (>950°C) and low solubility in nonpolar solvents. In contrast, the trichorides, tribromides, and triiodides of aluminum, gallium, and indium, as well as TlCl and TlBr , are more covalent in character and form halogen-bridged dimers (part (b) in Figure \(\Page {4}\)). Although the structure of these dimers is similar to that of diborane (B H ), the bonding can be described in terms of electron-pair bonds rather than the delocalized electron-deficient bonding found in diborane. Bridging halides are poor electron-pair donors, so the group 13 trihalides are potent Lewis acids that react readily with Lewis bases, such as amines, to form a Lewis acid–base adduct: \[Al_2Cl_{6(soln)} + 2(CH_3)_3N_{(soln)} \rightarrow 2(CH_3)_3N:AlCl_{3(soln)} \label{Eq11}\] In water, the halides of the group 13 metals hydrolyze to produce the metal hydroxide (\[M(OH)_3\)): \[MX_{3(s)} + 3H_2O_{(l)} \rightarrow M(OH)_{3(s)} + 3HX_{(aq)} \label{Eq12}\] In a related reaction, Al (SO ) is used to clarify drinking water by the precipitation of hydrated Al(OH) , which traps particulates. The halides of the heavier metals (In and Tl) are less reactive with water because of their lower charge-to-radius ratio. Instead of forming hydroxides, they dissolve to form the hydrated metal complex ions: [M(H O) ] . Of the group 13 halides, only the fluorides behave as typical ionic compounds. Like boron (Equation \(\ref{Eq22.7}\)), all the heavier group 13 elements react with excess oxygen at elevated temperatures to give the trivalent oxide (M O ), although Tl O is unstable: \[\mathrm{4M(s)}+\mathrm{3O_2(g)}\xrightarrow{\Delta}\mathrm{2M_2O_3(s)} \label{Eq13}\] Aluminum oxide (Al O ), also known as alumina, is a hard, high-melting-point, chemically inert insulator used as a ceramic and as an abrasive in sandpaper and toothpaste. Replacing a small number of Al ions in crystalline alumina with Cr ions forms the gemstone ruby, whereas replacing Al with a mixture of Fe , Fe , and Ti produces blue sapphires. The gallium oxide compound MgGa O gives the brilliant green light familiar to anyone who has ever operated a xerographic copy machine. All the oxides dissolve in dilute acid, but Al O and Ga O are amphoteric, which is consistent with their location along the diagonal line of the periodic table, also dissolving in concentrated aqueous base to form solutions that contain M(OH) ions. Group 13 trihalides are potent Lewis acids that react with Lewis bases to form a Lewis acid–base adduct. Aluminum, gallium, and indium also react with the other group 16 elements (chalcogens) to form chalcogenides with the stoichiometry M Y . However, because Tl(III) is too strong an oxidant to form a stable compound with electron-rich anions such as S , Se , and Te , thallium forms only the thallium(I) chalcogenides with the stoichiometry Tl Y. Only aluminum, like boron, reacts directly with N (at very high temperatures) to give AlN, which is used in transistors and microwave devices as a nontoxic heat sink because of its thermal stability; GaN and InN can be prepared using other methods. All the metals, again except Tl, also react with the heavier group 15 elements (pnicogens) to form the so-called III–V compounds, such as GaAs. These are semiconductors, whose electronic properties, such as their band gaps, differ from those that can be achieved using either pure or doped group 14 elements. For example, nitrogen- and phosphorus-doped gallium arsenide (GaAs P N ) is used in the displays of calculators and digital watches. All group 13 oxides dissolve in dilute acid, but Al O and Ga O are amphoteric. Unlike boron, the heavier group 13 elements do not react directly with hydrogen. Only the aluminum and gallium hydrides are known, but they must be prepared indirectly; AlH is an insoluble, polymeric solid that is rapidly decomposed by water, whereas GaH is unstable at room temperature. Boron has a relatively limited tendency to form complexes, but aluminum, gallium, indium, and, to some extent, thallium form many complexes. Some of the simplest are the hydrated metal ions [M(H O) ], which are relatively strong Brønsted–Lowry acids that can lose a proton to form the M(H O) (OH) ion: \[[M(H_2O)_6]^{3+}_{(aq)} \rightarrow M(H_2O)_5(OH)^{2+}_{(aq)} + H^+_{(aq)} \label{Eq14}\] Group 13 metal ions also form stable complexes with species that contain two or more negatively charged groups, such as the oxalate ion. The stability of such complexes increases as the number of coordinating groups provided by the ligand increases. For each reaction, explain why the given products form. balanced chemical equations why the given products form Classify the type of reaction. Using periodic trends in atomic properties, thermodynamics, and kinetics, explain why the reaction products form. Predict the products of the reactions and write a balanced chemical equation for each reaction. Compounds of the group 13 elements with oxygen are thermodynamically stable. Many of the anomalous properties of the group 13 elements can be explained by the increase in Z moving down the group. Isolation of the group 13 elements requires a large amount of energy because compounds of the group 13 elements with oxygen are thermodynamically stable. Boron behaves chemically like a nonmetal, whereas its heavier congeners exhibit metallic behavior. Many of the inconsistencies observed in the properties of the group 13 elements can be explained by the increase in Z that arises from poor shielding of the nuclear charge by the filled (n − 1)d and (n − 2)f subshells. Instead of forming a metallic lattice with delocalized valence electrons, boron forms unique aggregates that contain multicenter bonds, including metal borides, in which boron is bonded to other boron atoms to form three-dimensional networks or clusters with regular geometric structures. All neutral compounds of the group 13 elements are electron deficient and behave like Lewis acids. The trivalent halides of the heavier elements form halogen-bridged dimers that contain electron-pair bonds, rather than the delocalized electron-deficient bonds characteristic of diborane. Their oxides dissolve in dilute acid, although the oxides of aluminum and gallium are amphoteric. None of the group 13 elements reacts directly with hydrogen, and the stability of the hydrides prepared by other routes decreases as we go down the group. In contrast to boron, the heavier group 13 elements form a large number of complexes in the +3 oxidation state. | 20,479 | 2,646 |
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When reactants are not present in stoichiometric quantities, the limiting reactant determines the maximum amount of product that can be formed from the reactants. The amount of product calculated in this way is the theoretical yield, the amount obtained if the reaction occurred perfectly and the purification method were 100% efficient. In reality, less product is always obtained than is theoretically possible because of mechanical losses (such as spilling), separation procedures that are not 100% efficient, competing reactions that form undesired products, and reactions that simply do not run to completion, resulting in a mixture of products and reactants; this last possibility is a common occurrence. Therefore, the actual yield, the measured mass of products obtained from a reaction, is almost always less than the theoretical yield (often much less). The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, multiplied by 100 to give a percentage: \[ percent \, yield = {actual \, yield \, (g) \over theoretical \, yield \, (g) } \times 100 \tag{4.5.1} \] The method used to calculate the percent yield of a reaction is illustrated in Example \(\Page {1}\). Procaine is a key component of Novocain, an injectable local anesthetic used in dental work and minor surgery. Procaine can be prepared in the presence of H2SO4 (indicated above the arrow) by the reaction \[ \underset {p-amino benzoic\,acid}{C_7H_7NO_2} + \underset {2-diethylaminoethanol}{C_6H_{15}NO}\,\,\underrightarrow {H_2SO_4} \,\, \underset {procaine}{C_{13}H_{20}N_2O_2} + H_2O\] If this reaction was carried out using 10.0 g of p-aminobenzoic acid and 10.0 g of 2-diethylaminoethanol, and 15.7 g of procaine were isolated, what is the percent yield? : masses of reactants and product : percent yield : : From the formulas given for the reactants and the products, we see that the chemical equation is balanced as written. According to the equation, 1 mol of each reactant combines to give 1 mol of product plus 1 mol of water. To determine which reactant is limiting, we need to know their molar masses, which are calculated from their structural formulas: p-aminobenzoic acid (C H NO ), 137.14 g/mol; 2-diethylaminoethanol (C H NO), 117.19 g/mol. Thus the reaction used the following numbers of moles of reactants: \[ moles \, p-aminobenzoic \, acid = 10.0 \, g \, \times \, {1 \, mol \over 137.14 \, g } = 0.0729 \, mol \, p-aminbenzoic\, acid \] The actual yield was only 15.7 g of procaine, so the percent yield is \[ percent \, yield = {15.7 \, g \over 17.2 \, g } \times 100 = 91.3 \% \] (If the product were pure and dry, this yield would indicate very good lab technique!) Lead was one of the earliest metals to be isolated in pure form. It occurs as concentrated deposits of a distinctive ore called galena (PbS), which is easily converted to lead oxide (PbO) in 100% yield by roasting in air via the following reaction: \[ 2PbS_{(s)} + 3O_2 \rightarrow 2PbO_{ (s)} + 2SO_{2 (g)} \] The resulting PbO is then converted to the pure metal by reaction with charcoal. Because lead has such a low melting point (327°C), it runs out of the ore-charcoal mixture as a liquid that is easily collected. The reaction for the conversion of lead oxide to pure lead is as follows: \[ PbO_{(s)} + C_{(s)} \rightarrow Pb_{(l)} + CO_{(g)} \] If 93.3 kg of PbO is heated with excess charcoal and 77.3 kg of pure lead is obtained, what is the percent yield? : 89.2% Percent yield can range from 0% to 100%. In the laboratory, a student will occasionally obtain a yield that appears to be greater than 100%. This usually happens when the product is impure or is wet with a solvent such as water. If this is not the case, then the student must have made an error in weighing either the reactants or the products. The law of conservation of mass applies even to undergraduate chemistry laboratory experiments. A 100% yield means that everything worked perfectly, and the chemist obtained all the product that could have been produced. Anyone who has tried to do something as simple as fill a salt shaker or add oil to a car’s engine without spilling knows the unlikelihood of a 100% yield. At the other extreme, a yield of 0% means that no product was obtained. A percent yield of 80%–90% is usually considered good to excellent; a yield of 50% is only fair. In part because of the problems and costs of waste disposal, industrial production facilities face considerable pressures to optimize the yields of products and make them as close to 100% as possible. Most of time it takes more than one steps to get desired product. Two more complex reactions are the consecutive and simultaneous reactions. Reactions that are carried out one after another in sequence to yield a final product are called consecutive reactions (e.g., occurring sequenceally): \[ A \rightarrow Y \rightarrow Z \tag{4.5.2}\] Any substance that is produced in one step and is consumed in another step of a multistep process is called an (e.g., \(Y\) in Equation 4.5.2). The is the chemical equation that expresses all the reactions occurring in a single overall equation. \[2TiO2_{(\text{impure solid})} + 3C_{(s)} +4Cl_{2(g)} \rightarrow 2TiCl_{4(g)} + CO_{2(g)} + 2CO_{(g)}\] \[2\times [ TiCl_{4(g)} + O_{2(g)} \rightarrow TiO_{2(s)} +2Cl_{2(g)}] \] Overall: \[3C_{(s)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2CO_{(g)}\] \(TiCl_4\) is an intermediate in this reaction How many gram of \(NaBr\) is produced if 55.85g of \(Fe\) is consumed the the following series of reactions? \[ 3Fe + 3Br_2 \rightarrow 3FeBr_2 \] \[ 3FeBr_2 + Br_2 \rightarrow Fe_3Br_8\] \[ Fe_3Br_8 + 4Na_2CO_3 \rightarrow 8NaBr + 4CO_2 + Fe_3O_4\] Overall reaction: \[3Fe + 4Br_2 + 4Na_2CO_3 \rightarrow 8NaBr + 4CO_2 + Fe_3O_4\] In simultaneous reactions, two or more substances react independently of one another in separate reactions occurring simultaneously \[ A \rightarrow Y \tag{4.5.3a}\] and \[ B \rightarrow B \tag{4.5.3b}\] Equations 4.5.3 involve two reactions evolving in parallel, however sometimes there exist a coupling between the two (or more) reactions with shared reactant, intermediates or products. is competition involved, like in the scheme: \[ A + B \rightarrow Y \tag{4.5.4a}\] \[ A + C \rightarrow Z \tag{4.5.3b}\] In the above simultaneous reaction, both \(B\) and \(C\) compete with each another for reacting \(A\). This is superficially similar to a . Magnalium is an aluminum alloy with magnesium that exhibits greater strength, greater corrosion resistance, and lower density than pure aluminum. How many grams of \(H_{2(g)}\) are generated if a 0.710 g piece of magnalium (with a 70% Al and 30.0% Mg by mass composition) with excess \(HCl_{(aq)}\)? The stoichiometry of a reaction describes the relative amounts of reactants and products in a balanced chemical equation. A stoichiometric quantity of a reactant is the amount necessary to react completely with the other reactant(s). If a quantity of a reactant remains unconsumed after complete reaction has occurred, it is in excess. The reactant that is consumed first and limits the amount of product(s) that can be obtained is the limiting reactant. To identify the limiting reactant, calculate the number of moles of each reactant present and compare this ratio to the mole ratio of the reactants in the balanced chemical equation. The maximum amount of product(s) that can be obtained in a reaction from a given amount of reactant(s) is the theoretical yield of the reaction. The actual yield is the amount of product(s) actually obtained in the reaction; it cannot exceed the theoretical yield. The percent yield of a reaction is the ratio of the actual yield to the theoretical yield, expressed as a percentage. ( ) | 7,778 | 2,647 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/Appendix_I%3A_Index_of_enzymatic_reactions_by_pathway |
While is organized, like most sophomore-level organic chemistry texts, around a structural and mechanistic framework, students of biochemistry may often want to clarify the mechanism of an enzymatic reaction which they encounter when studying the central metabolic pathways. An excellent resource for this purpose is John McMurry's (Roberts and Company, 2005), but also helpful will be this index of enzymatic reactions organized by pathway, with links to sections/problems in this text where the reaction mechanism is addressed. Sections/problems listed with an asterisk (*) do not discuss the exact reaction indicated, but do discuss a closely related reaction. are provided whenever possible, with links to the corresponding page in the database of enzymes. ing on the 'reaction flask' icon on a BRENDA page brings up the reaction diagram. : content below redirects to an older edition of the text, which differs from the current version in some content and organization.
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Atomic_Theory/Isotopes/Isotopes_II |
Although all atoms of an element have the same number of protons, individual atoms may have different numbers of neutrons. These differing atoms are called isotopes. All atoms of chlorine (Cl) have 17 protons, but there are chlorine isotopes with 15 to 23 neutrons. Only two chlorine isotopes exist in significant amounts in nature: those with 18 neutrons (75.53% of all chlorine atoms found in nature), and those with 20 neutrons (24.47%). To write the symbol for an isotope, place the atomic number as a subscript and the mass number (protons plus neutrons) as a superscript to the left of the atomic symbol. The symbols for the two naturally occurring isotopes of chlorine are written as follows: \( ^{35}_{17}{\rm Cl}\) and \( ^{37}_{17}{\rm Cl}\). The subscript is somewhat unnecessary, because all atoms of chlorine have 17 protons; isotope symbols are usually written without the subscript, as in Cl and Cl. In discussing these isotopes, the terms chlorine-35 and chlorine-37 are used to differentiate between them. In general, for an atom to be stable, it must have more neutrons than protons. Nuclei with too many of either kind of fundamental particle are unstable, and break down radioactively. How many protons, neutrons, and electrons are there in an atom of uranium-238? Write the symbol for this isotope. The atomic number of uranium (see periodic table) is 92, and the mass number of the isotope is given as 238. Therefore, it has 92 protons, 92 electrons, and 238 — 92 : 146 neutrons. Its symbol is \(\ce{^{238}_{92}U} \) (or U). The total mass of an atom is called its atomic weight, and is the Calculate the mass that is lost when an atom of carbon-12 is formed from protons, electrons, and neutrons. Because the atomic number of every carbon atom is 6, carbon-12 has 6 protons and therefore 6 electrons. To find the number of neutrons, we subtract the number of protons from the mass number: 12 – 6 = 6 neutrons. The data in Table 1-1 can be used to calculate the total mass of these particles: Protons: 6 x 1.00728 amu = 6.04368 u
Neutrons: 6 x 1.00867 amu = 6.05202 u
Total particle mass = 12.09900 u However, by the definition of the scale of atomic mass units, the mass of one carbon-12 atom is exactly 12 amu. Therefore, 0.0990 u of mass has disappeared in the process of building the atom from its particles. Each isotope of an element is characterized by an atomic number (the number of protons), a mass number (the total number of protons and neutrons), and an atomic weight (mass of atom in atomic mass units). Because mass losses upon formation of an atom are small, the mass number is usually the same as the atomic weight rounded to the nearest integer (for example, the atomic weight of chlorine-37 is 36.966, which is rounded to 37). If there are several isotopes of an element in nature, then the experimentally observed atomic weight (the natural atomic weight) is the weighted average of the isotope weights. The average is weighted according to the percent abundance of the isotopes. Chlorine occurs in nature as 75.53% chlorine-35 (34.97 u) and 24.47% chlorine-37 (36.97 u), so the weighted average of the isotope weights is (07553 x 34.97 u) + (0.2447 x 36.97 u) = 35.46 u. The atomic weights found in periodic tables are all weighted averages of the isotopes occurring in nature, and these are the figures used for the remainder of this article, except when discussing one isotope specifically. In general, all isotopes of an element behave the same way chemically. Their behaviors differ with regard to mass-sensitive properties such as diffusion rates. Magnesium (Mg) has three significant natural isotopes: 78.70% of all magnesium atoms have an atomic weight of 23.985 u, 10.13% have an atomic weight of 24.986 u, and 11.17% have an atomic weight of 25.983 u. How many protons and neutrons are present in each of these three isotopes? How are the symbols for each isotope written? Finally, what is the weighted average of the atomic weights? There are 12 protons in all magnesium isotopes. The isotope whose atomic weight is 23.985 u has a mass number of 24 (protons and neutrons), so 24 - 12 protons gives 12 neutrons. The symbol for this isotope is Mg. Similarly, the isotope whose atomic weight is 24.986 amu has a mass number of 25, 13 neutrons, and Mg as a symbol. The third isotope (25.983 amu) has a mass number of 26, 14 neutrons, and Mg as a symbol. The average atomic weight is calculated as follows: (0.7870 x 23.985) + (0.1013 x 24.986) + (0.1117 x 25.983) = 24.31 u Boron has two naturally occurring isotopes, B and B. In nature, 80.22% of its atoms are B, with atomic weight 11.009 u. From the natural atomic weight, calculate the atomic weight of the B isotope. If 80.22% of all boron atoms are B, then 100.00 — 80.22, or 19.78%, are the unknown isotope. In the periodic table the atomic weight of boron is found to be 10.81 u. We can use to represent the unknown atomic weight in our calculation: \[ (0.8022 \times 11.009) + (0.1978 \times W) = 10.81 {\rm u} \quad {\rm (natural~atomic~weight)} \] \[ W=\dfrac{10.81-8.831}{0.1978}=10.01 {\rm u} \] | 5,143 | 2,649 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Amino_Acids/Properties_of_Amino_Acids/1._Backgrounds_of_Amino_Acids |
This page explains what amino acids are, concentrating on the 2-amino acids that are biologically important. It looks in some detail at their simple physical properties such as solubility and melting points. Amino acids are exactly what they say they are! They are compounds containing an amino group, -NH , and a carboxylic acid group, -COOH. The biologically important amino acids have the amino group attached to the carbon atom next door to the -COOH group. They are known as 2-amino acids. They are also known (slightly confusingly) as alpha-amino acids. These are the ones we will concentrated on. The two simplest of these amino acids are 2-aminoethanoic acid and 2-aminopropanoic acid. Because of the biological importance of molecules like these, they are normally known by their traditional biochemical names. 2-aminoethanoic acid, for example, is usually called glycine, and 2-aminopropanoic acid is usually known as alanine. The general formula for a 2-amino acid is: . . . where "R" can be quite a complicated group containing other active groups like -OH, -SH, other amine or carboxylic acid groups, and so on. It is definitely NOT necessarily a simple hydrocarbon group! The amino acids are crystalline solids with surprisingly high melting points. It is difficult to pin the melting points down exactly because the amino acids tend to decompose before they melt. Decomposition and melting tend to be in the 200 - 300°C range. For the size of the molecules, this is very high. Something unusual must be happening. If you look again at the general structure of an amino acid, you will see that it has both a basic amine group and an acidic carboxylic acid group. There is an internal transfer of a hydrogen ion from the -COOH group to the -NH group to leave an ion with both a negative charge and a positive charge. This is called a zwitterion. A zwitterion is a compound with no overall electrical charge, but which contains separate parts which are positively and negatively charged. This is the form that amino acids exist in even in the solid state. Instead of the weaker hydrogen bonds and other intermolecular forces that you might have expected, you actually have much stronger ionic attractions between one ion and its neighbors. These ionic attractions take more energy to break and so the amino acids have high melting points for the size of the molecules. Amino acids are generally soluble in water and insoluble in non-polar organic solvents such as hydrocarbons. This again reflects the presence of the zwitterions. In water, the ionic attractions between the ions in the solid amino acid are replaced by strong attractions between polar water molecules and the zwitterions. This is much the same as any other ionic substance dissolving in water. The extent of the solubility in water varies depending on the size and nature of the "R" group. The lack of solubility in non-polar organic solvents such as hydrocarbons is because of the lack of attraction between the solvent molecules and the zwitterions. Without strong attractions between solvent and amino acid, there won't be enough energy released to pull the ionic lattice apart. If you look yet again at the general formula for an amino acid, you will see that (apart from glycine, 2-aminoethanoic acid) the carbon at the centre of the structure has four different groups attached. In glycine, the "R" group is another hydrogen atom. This is equally true if you draw the structure of the zwitterion instead of this simpler structure. Because of these four different groups attached to the same carbon atom, amino acids (apart from glycine) are chiral. The lack of a plane of symmetry means that there will be two stereoisomers of an amino acid (apart from glycine) - one the non-superimposable mirror image of the other. For a general 2-amino acid, the isomers are: All the naturally occurring amino acids have the right-hand structure in this diagram. This is known as the "L-" configuration. The other one is known as the "D-" configuration. If you read around the other groups in a clockwise direction, you get the word CORN. You can't tell by looking at a structure whether that isomer will rotate the plane of polarisation of plane polarised light clockwise or anticlockwise. All the naturally occurring amino acids have the same L- configuration, but they include examples which rotate the plane clockwise (+) and those which do the opposite (-). For example: It is quite common for natural systems to only work with one of the optical isomers (enantiomers) of an optically active substance like the amino acids. It isn't too difficult to see why that might be. Because the molecules have different spatial arrangements of their various groups, only one of them is likely to fit properly into the active sites on the enzymes they work with. | 4,842 | 2,650 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_v2.0_(Soderberg)/01%3A_Introduction_to_Organic_Structure_and_Bonding_I |
In this chapter, you will be introduced to some of the most fundamental principles of organic chemistry. With the concepts we learn about, we can begin to understand how carbon and a very small number of other elements in the periodic table can combine in predictable ways to produce a virtually limitless chemical repertoire. As you read through, you will recognize that the chapter contains a lot of review of topics you have probably learned already in an introductory chemistry course, but there will likely also be a few concepts that are new to you, as well as some topics which are already familiar to you but covered at a greater depth and with more of an emphasis on biologically relevant organic compounds.
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https://chem.libretexts.org/Bookshelves/Organic_Chemistry/OCLUE%3A_Organic_Chemistry_Life_the_Universe_and_Everything_(Copper_and_Klymkowsky)/04%3A_Nucleophilic_Substitution_Part_II |
In Chapter \(1\), we learned about one of the most fundamental reactions in organic chemistry: nucleophilic substitution. Before we move on, it is important to make sure that you have a good understanding of what the terms nucleophile, electrophile, and leaving group mean and that you are able to predict the products for a range of substrate molecules (electrophiles) with different leaving groups and nucleophiles. In this section, we move forward and look at nucleophilic substitution reactions in more detail by examining the evidence that leads us to understand how the mechanisms of nucleophilic substitutions were determined.
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/05%3A_Solutions/5.08%3A_Ionic_Activity |
The properties of electrolyte solutions can significantly deviate from the laws used to derive chemical potential of solutions. In nonelectrolyte solutions, the intermolecular forces are mostly comprised of weak Van der Waals interactions, which have a \(r^{-7}\) dependence, and for practical purposes this can be considered ideal. In ionic solutions, however, there are significant electrostatic interactions between solute-solvent as well as solute-solute molecules. These electrostatic forces are governed by Coulomb's law, which has a \(r^{-2}\) dependence. Consequently, the behavior of an electrolyte solution deviates considerably from that an ideal solution. Indeed, this is why we utilize the activity of the individual components and not the concentration to calculate deviations from ideal behavior. In 1923, Peter Debye and Erich Hückel developed a theory that would allow us to calculate the mean ionic activity coefficient of the solution, \(\gamma_{\pm} \), and could explain how the behavior of ions in solution contribute to this constant. The Debye-Hückel theory is based on three assumptions of how ions act in solution: Debye and Hückel developed the following equation to calculate the mean ionic activity coefficient \(\gamma_{\pm}\): \[ \log\gamma_{\pm}=-\dfrac{1.824\times10^{6}}{(\varepsilon T)^{3/2}}\mid z_{+}z_{-}\mid\sqrt{I} \label{1}\] where The Equation \(\ref{1}\) is known as the . The ionic strength is calculated by the following relation: \[I=\dfrac{1}{2}\sum_{i}m_{i}z_{i}^{2} \label{2}\] where \(m_{i}\) and \(z_{i}\) are the molality and the charge of the ith ion in the electrolyte. Since most of the electrolyte solutions we study are aqueous \((\varepsilon=78.54)\) and have a temperature of 298 K, the Limiting Law in Equation \ref{1} reduces to \[\log \gamma_{\pm}=-0.509\mid z_{+}z_{-}\mid\sqrt{I} \label{3}\] Calculate ionic strength, mean ionic activity coefficient \(\gamma_{\pm}\), and the mean ionic molality \(m_{\pm}\) for a 0.02 aqueous solution of zinc chloride, \(\ce{ZnCl2}\). Zinc chloride will dissolve as \[\ce{ZnCl2 -> Zn^{2+}(aq) + 2Cl^{-}(aq)} \nonumber \] The concentrations of the zinc and chloride ions will then be 0.02 and 0.04 molal, respectively. First calculate the mean ionic molality. The mean ionic molality is defined as the average molality of the two ions (see ): \[m_{\pm}=(m_{+}^{\nu+}m_{-}^{\nu-})^{\frac{1}{\nu}} \nonumber \] where \(\nu\) is the stoichiometric coefficient of the ions, and the total of the coefficients in the exponent. In our case, the mean ionic molality is \[\begin{align*} m_{\pm} &=(m_{Zn}^{\nu(Zn)} m_{Cl}^{\nu(Cl)})^{\frac{1}{\nu (Zn) + \nu (Cl)}} \\[4pt] &=[(0.02)^{1}(0.04)^{2}]^{\frac{1}{3}} \\[4pt] &=\left[(0.02)(0.0016)\right]^{\frac{1}{3}} \\[4pt] &=3.17 \times 10^{-2} \end{align*} \] To calculate the mean ionic activity coefficient, we first need the ionic strength of the solution from Equation \(\ref{2}\): \[\begin{align*} I &=\dfrac{1}{2}[(0.02)(+2)^{2}+(0.04)(-1)^{2}] \\[4pt] &=\dfrac{1}{2}(0.08+0.04)=0.06 \end{align*} \] Now we can use Equation \(\ref{3}\) to calculate the activity coefficient: \[\begin{align*} \log\gamma_{\pm} &=-0.509\mid(+2)(-1)\mid\sqrt{0.06} \\[4pt] &=(-0.509)(2)(0.245) \\[4pt] &=-0.250 \\[4pt] \gamma_{\pm}&=10^{-0.25} \\[4pt] &=0.1627 \end{align*} \] Determine the mean ionic activity coefficient and mean activity of a 0.004 molal of \(\ce{Ba(HCO3)2}\)? \[γ_±=10^{−0.10188}≈0.7909 \nonumber\] \[a_±=0.7909 \times 0.2785\, m= 0.2203 \,m \nonumber\] The kinetic salt effect is the effect of salts preset in solution on the rate of a reaction. In biological systems, salts influence how well proteins and DNA function. Salts are formed by ionic bonds, between a metal and an electromagnetic atom(s). Some examples of salts include NaCl, KCl, and Na SO . Salt molecules are able to disassociate, forming cations and anions. An increase in the charge (- or +) of a transition state or an activated complex results in an increase in solvation (creating more order in the system), and causes a decrease in the change of entropy (ΔS). In contrast, a decrease in the charge of the transition state causes an increase in ΔS. \[I^- + C^+ \rightleftharpoons E^o \rightarrow Product \label{2A}\] The kinetic salt effect describes the way salts stabilize reactants. For example, in the above reaction, each reactant has a charge. The negatively-charged reactant is stabilized by the positive charges from the salt, and the positively-charged reactant is stabilized by the negative charges from the salt. As a result, the rate at which the reactants come together decreases, thus decreasing the rate at which E forms. Because a charged intermediate is also stabilized in the solution, the half life of the intermediate at equilibrium increases, shifting the reaction toward product formation. Because the rate of the product formation is higher due to increased amounts of the intermediate present on the solution, first order kinetics is used to derive the rate constant equation: \[\log K_{TS} = \log K_{TS^o} + 2Z_AZ_B\sqrt{I} \label{3A}\] where The relationship between Z Z , I, and the rate of the reaction is presented in tabular form below: Note: I=0 at very dilute salt concentrations or if the salt is inert. | 5,276 | 2,655 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Path_Functions/Heat |
\[q_{System}+q_{Surroundings} = 0\] her, heat, as energy, will transfer from the to the . As One can also look at phase diagrams to see how heat and pressure affect the changes in phases. A phase diagram is a graphical representation of the conditions of temperature and pressure at which solids, liquids, and gases exist, either as single phases, or states of matter or as two or more phases in equilibrium with one another. The following picture is the example of a phase diagram: One can also look at heating curves to see the change in temperature as more heat is added. The following is an example of a heating curve. Notice that as the phase changes occur, there is no change in temperature. 1 kCal = 1000 cal) is the unit of heat that we usually encounter in daily life, such as on the back of a cereal box. 1 cal = 4.184 Joules 1 BTU = 252 calories = 1055 Joules specific heat capacity We can now use the following formula to solve for the amount of heat needed. isn't necessary information. This is when you would use the second equation. Answers are in the attached files section. | 1,113 | 2,656 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Potentials/Factors_that_Influence_Reduction_Potential |
In general, the ions of very late transition metals -- those towards the right-hand end of the transition metal block, such as copper, silver and gold -- have high reduction potentials. In other words, their ions are easily reduced. Alkali metal ions -- on the very left edge of the periodic table, such as potassium or cesium -- have very negative reduction potentials. These ions are very difficult to reduce. These trends are not surprising, because alkali metals are generally at the lower end of the electronegativity scale (Table A2) and are typically found as cations, not as neutral atoms. Late transition metals are electronegative in this case, and so we would expect their ions to attract electrons more easily than alkali metal ions. The nice thing about redox is you can always look at it from either direction. Oxidation is simply the opposite of reduction. How easily does an alkali metal lose an electron? If the standard reduction potential of lithium is very negative, then the oxidation potential of lithium ion is very positive. If it is uphill to transfer an electron from hydrogen to lithium cation, it must be downhill to transfer an electron from a lithium atom to a proton. After all, hydrogen is more electronegative than any of the alkalis. Of course, since a late transition metal is generally more electronegative than an alkali metal, copper or silver or gold ought to be more difficult to oxidize than sodium or potassium. The large trends in redox chemistry are not surprising, then. It's simply a matter of the electron moving to a lower energy level on another atom. If we look a little more closely, though, there are plenty of exceptions to the general trend. For example, in the coinage triad (Group 11), gold has the most positive reduction potential, followed by silver, then copper. That's exactly the of expectations; copper, at the top of the column, should be the most electronegative and have the most positive reduction potential, not the least. What's going on in those cases? Well, there's more going on than just moving an electron. Remember, in the measurement of a reduction potential, we are generally working with a metal electrode in an aqueous solution of ions. What else is going on in this reaction? Well, the atom that gets reduced starts out as an ion in water, but an ion in water does not sit around on its own. It's a Lewis acid, an electrophile. Water is a nucleophile, a potential ligand. So the ion in solution is actually a coordination complex. It swims around for a while, then bumps into the cathode, where it picks up the electron. But the resulting ion does not stay in solution; it gets deposited at the electrode, along with others of its kind. It becomes part of a metal solid. So there are three different things happening here: If we could get some physical data on each of those events, we might be able to explain why these reduction potentials are contrary to expectations. The kinds of data we have available for these individual steps may actually fit the opposite reaction better. We can estimate the energy involved in the removal of a metal atom from the solid, the loss of an electron from the metal, and the binding of water to the resulting ion. These data come from measurement of the heat of vaporization of the metal, the ionization energy of the metal, and the enthalpy of hydration of the metal. The trouble is, these data all involve the gas phase. If they really applied to this situation, it would be as if metal atoms sprayed out into the air above the electrode, shot their electrons back, grabbed some water molecules that drifted by, and then dropped down into the solution. Of course that does not happen; we do not see a little, sparkly, metallic mist appear when we connect the circuit, or little lightning bolts from the cloud of metal atoms to the electrode, and we do not see a splash or a fizz or little tendrils of steam as the resulting ions drop into the water. That does not matter. The data we have here are still very useful.That's because what we are looking at -- the energy difference between two states -- is a state function. That means it does not matter how we get from one state to the other; the overall difference will always be the same. So if the reaction did happen via the gas phase, the energy change would be exactly as it is when it happens directly at the electrode - solution interface. We can do a sort of thought experiment using the data we know, and even though those steps do not really happen the way they do in the experiments that gave rise to the data, they will eventually lead to the right place. This sort of imaginary path to mimic a reaction we want to know more about employs an idea called "Hess's Law". It is frequently used to gain insight into reactions throughout chemistry. Taking all of these data together, we can get a better picture of the overall energy changes that would occur during a reduction or, more directly, an oxidation. The first thing to note is that copper has a higher ionization energy than silver. As expected, Cu is harder to form than Ag , because copper is more electronegative than silver. However, Au appears to be the hardest to form of all three. It is as if gold were the most electronegative of these three elements, but it's at the bottom of this column. Gold really is more electronegative than copper or silver (Table A2). There are a few deviations from expectation in periodic trends, but this one is probably attributable to a phenomenon called "the lanthanide contraction." Notice the covalent radii of gold and silver in the table above. Normally, we expect atoms to get bigger row by row, as additional layers of electrons are filled in. Not so for the third row of transition metals. To see the probable reason for that, you have to look at the whole periodic table, and remember for the first time ever that the lanthanides and actinides -- the two orphaned rows at the bottom -- actually fit in the middle of the periodic table. The lanthanides, in particular (lanthanum, La, to ytterbium, Yb), go in between lutetium (Lu) and Hafnium (Hf). As a result, the third row of transition metals contain many more protons in their nuclei, compared to the second row transition metals of the same column. Silver has ten more protons in its nucleus than rubidium, the first atom in the same row as silver, but gold has twenty four more than cesium. The third row "contracts" because of these additional protons. So the exceptionally positive reduction potential of Au (and, by relation, the exceptionally negative oxidation potential of gold metal) may be a result of the lanthanide contraction. What about copper versus silver? Copper still has a higher electronegativity than silver, but copper metal is more easily oxidized. It's not that copper is more easily pulled away from the metallic bonds holding it in the solid state; copper's heat of vaporization is a little higher than silver's. That leaves hydration. In fact, copper ion does have a higher enthalpy of hydration than silver; more energy is released when water binds to copper than when water binds to silver. The difference between these two appears to be all about the solvation of the copper ion, which is more stable with respect to the metal than is silver ion. Why would that be? Well, copper is smaller than silver. A simple look at Coulomb's law reminds us that the closer the electrons of the donor ligand are to the cation, the more tightly bound they will be. Looking at it a slightly different way, copper is smaller and "harder" than silver, and forms a stronger bond with water, which is a "hard" ligand. Taking a look at a Hess's Law cycle for a redox reaction is a useful approach to get some additional insight into the reaction. It lets us use data to assess the influence of various aspects of the reaction that we can't evaluate directly from the reduction potential, because in the redox reaction all of these factors are conflated into one number. , | 8,042 | 2,657 |
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Although we are studying only about 20 amino acids, there are about six more found in the body. Many others are also known from a variety of sources. Amino acids are the building blocks used to make proteins and peptides. The different amino acids have interesting properties because they have a variety of structural parts which result in different polarities and solubilities. Each amino acid has at least one amine and one acid functional group as the name implies. See graphic on the left. The different properties result from variations in the structures of different R groups. The R group is often referred to as the amino acid "side chain". Amino acids have special common names, however, a three letter abbreviation for the name is used most of the time. Consult the amino acid table on the next page for structure, names, and abbreviations. Amino acid physical properties indicate a "salt-like" behavior. Amino acids are crystalline solids with relatively high melting points, and most are quite soluble in water and insoluble in non-polar solvents. In solution, the amino acid molecule appears to have a charge which changes with pH. An intramolecular neutralization reaction leads to a salt-like ion called a . The accepted practice is to show the amino acids in the zwitterion form: | 1,316 | 2,658 |
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The choice of what fractionating column to use for which application depends in part on availability and the task at hand. Several columns are shown in Figure 5.39: a) Vigreux column with glass indentations, b) Steel wool column made simply be loosely inserting steel wool into the cavity of a fractionating column (similar to a West condenser, but wider), c) Glass beads filled in the cavity of a fractionating column. These columns have different surface areas and numbers of theoretical plates, and thus differ in their ability to separate close-boiling components. They also differ somewhat in the quantity of compound that will be sequestered through wetting the column. A Vigreux column has the least surface area, making it the least capable of separating close-boiling components. And yet with the lowest surface area, it can have the highest recovery, making it the optimal choice if a separation is not particularly difficult. Glass beads have a high surface area, so are a good choice for separation of close-boiling components. And yet, beads will suffer the greatest loss of material. Steel wool columns have intermediate surface areas and their effectiveness can depend on how tightly the wool is packed in the column. Steel wool columns also cannot be used with corrosive vapors like those containing acid. To demonstrate the effectiveness of different fractionating columns, a mixture containing \(75 \: \text{mol}\%\) ethylbenzene (normal b.p. \(136^\text{o} \text{C}\)) and \(25 \: \text{mol}\%\) -cymene (normal b.p. \(177^\text{o} \text{C}\)) was distilled using several methods. The first two \(\text{mL}\) of distillate was collected in each process (Figure 5.40). Gas chromatograph analysis of the initial mixture and distillates allowed for quantitation of the mixtures, although a calibration curve was necessary to translate the reported percentages into meaningful values. As the two components had only a \(41^\text{o} \text{C}\) difference in boiling points, a fractional distillation was more effective than a simple distillation, although in some cases only slightly (see Table 5.7). The distillate was never \(> 98\%\) pure with any method as there was a large quantity of the minor component in the distilling pot. {{template.HideTOC()}] | 2,283 | 2,659 |
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The metal hydride reductions and organometallic additions to aldehydes and ketones, described above, both decrease the carbonyl carbon's oxidation state, and may be classified as reductions. As noted, they proceed by attack of a strong nucleophilic species at the electrophilic carbon. Other useful reductions of carbonyl compounds, either to alcohols or to hydrocarbons, may take place by different mechanisms. For example, hydrogenation (Pt, Pd, Ni or Ru catalysts), reaction with diborane, and reduction by lithium, sodium or potassium in hydroxylic or amine solvents have all been reported to convert carbonyl compounds into alcohols. However, the complex metal hydrides are generally preferred for such transformations because they give cleaner products in high yield. Aldehydes and ketones may also be reduced by hydride transfer from alkoxide salts. The reductive conversion of a carbonyl group to a methylene group requires complete removal of the oxygen, and is called . In the shorthand equation shown here the symbol refers to unspecified reduction conditions which effect the desired change. Three very different methods of accomplishing this transformation will be described here. R C=O + R CH + H O Reaction of an aldehyde or ketone with excess hydrazine generates a , which on heating with base gives the corresponding hydrocarbon. A high-boiling hydroxylic solvent, such as diethylene glycol, is commonly used to achieve the temperatures needed. The following diagram shows how this reduction may be used to convert cyclopentanone to cyclopentane. A second example, in which an aldehyde is similarly reduced to a methyl group, also illustrates again the use of an acetal protective group. The mechanism of this useful transformation involves tautomerization of the initially formed hydrazone to an azo isomer, and will be displayed on pressing the "Show Mechanism" button. The strongly basic conditions used in this reaction preclude its application to base sensitive compounds. This alternative reduction involves heating a carbonyl compound with finely divided, amalgamated zinc in a hydroxylic solvent (often an aqueous mixture) containing a mineral acid such as HCl. The mercury alloyed with the zinc does not participate in the reaction, it serves only to provide a clean active metal surface. The first example below shows a common application of this reduction, the conversion of a Friedel-Crafts acylation product to an alkyl side-chain. The second example illustrates the lability of functional substituents alpha to the carbonyl group. Substituents such as hydroxyl, alkoxyl & halogens are reduced first, the resulting unsubstituted aldehyde or ketone is then reduced to the parent hydrocarbon. In contrast to the previous two procedures, this method of carbonyl deoxygenation requires two separate steps. It does, however, avoid treatment with strong base or acid. The first step is to convert the aldehyde or ketone into a thioacetal: These derivatives may be isolated and purified before continuing the reduction. The second step involves refluxing an acetone solution of the thioacetal over a reactive nickel catalyst, called Raney Nickel. All carbon-sulfur bonds undergo hydrogenolysis (the C–S bonds are broken by addition of hydrogen). In the following example, 1,2-ethanedithiol is used for preparing the thioacetal intermediate, because of the high yield this reactant usually affords. The bicyclic compound shown here has two carbonyl groups, one of which is sterically hindered (circled in orange). Consequently, a mono-thioacetal is easily prepared from the less-hindered ketone, and this is reduced without changing the remaining carbonyl function. The carbon atom of a carbonyl group has a relatively high oxidation state. This is reflected in the fact that most of the reactions described thus far either cause no change in the oxidation state (e.g. acetal and imine formation) or effect a reduction ( organometallic additions and deoxygenations). The most common and characteristic oxidation reaction is the conversion of aldehydes to carboxylic acids. In the shorthand equation shown here the symbol refers to unspecified oxidation conditions which effect the desired change. Several different methods of accomplishing this transformation will be described here. RCH=O + RC(OH)=O In discussing the oxidations of 1º and 2º-alcohols, we noted that Jones' reagent (aqueous chromic acid) converts aldehydes to carboxylic acids, presumably via the hydrate. Other reagents, among them aqueous potassium permanganate and dilute bromine, effect the same transformation. Even the oxygen in air will slowly oxidize aldehydes to acids or peracids, most likely by a radical mechanism. Useful tests for aldehydes, , & , take advantage of this ease of oxidation by using Ag and Cu as oxidizing agents (oxidants). RCH=O + 2 [ OH ] RC(OH)=O + 2 (metallic mirror) + H O When silver cation is the oxidant, as in the above equation, it is reduced to metallic silver in the course of the reaction, and this deposits as a beautiful mirror on the inner surface of the reaction vessel. The Fehling and Benedict tests use cupric cation as the oxidant. This deep blue reagent is reduced to cuprous oxide, which precipitates as a red to yellow solid. All these cation oxidations must be conducted under alkaline conditions. To avoid precipitation of the insoluble metal hydroxides, the cations must be stabilized as complexed ions. Silver is used as its ammonia complex, Ag(NH ) , and cupric ions are used as citrate or tartrate complexes. Saturated ketones are generally inert to oxidation conditions that convert aldehydes to carboxylic acids. Nevertheless, under vigorous acid-catalyzed oxidations with nitric or chromic acids ketones may undergo carbon-carbon bond cleavage at the carbonyl group. The reason for the vulnerability of the alpha-carbon bond will become apparent in the following section. | 5,961 | 2,660 |
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Phenols are enols, and enols normally are unstable with respect to the corresponding carbonyl compounds ( ). Thus
The situation is different for phenols because of the inclusion of the carbon-carbon double bond into the aromatic ring and the associated aromatic stabilization. Phenol (benzenol) exists exclusively in the enol form: The of some representative phenols are summarized in Table 26-1. In general, phenols are somewhat more polar than the corresponding saturated alcohols. The magnitudes of the differences are well illustrated by comparison of the physical properties of benzenol and cyclohexanol, shown in Table 26-2. The determining factor appears to be the greater acidity of the phenolic hydroxyl group, which means that, in the undissociated form, the \(\ce{O-H}\) bond is more strongly polarized as \(\overset{\delta \ominus}{\ce{O}}-\overset{\delta \oplus}{\ce{H}}\) than for alcohols. Phenols therefore form stronger hydrogen bonds than alcohols, thereby resulting in higher boiling points, higher water solubility, and increased ability to act as solvents for reasonably polar organic molecules. The wavelengths of the ultraviolet absorption maxima of the arenols shown in Table 26-1 indicate a considerable effect of substituents on these absorptions, which correspond to the \(200\)-\(\text{nm}\) and \(255\)-\(\text{nm}\) absorptions of benzene ( ). Substances such as 2-hydroxybenzaldehyde, 2-hydroxybenzoic acid, and 2-nitrobenzenol form rather than intermolecular hydrogen bonds. This effectively reduces intermolecular attraction, thereby reducing boiling points and increasing solubility in nonpolar solvents as compared to the meta and para isomers, which only form molecular hydrogen bonds: Benzenol and the 2-, 3-, and 4-methylbenzenols (cresols) can be isolated from coal tar ( ). Benzenol itself is used commercially in such large quantities that alternate methods of preparation are necessary and most of these start with benzene or alkylbenzenes. Direct oxidation of benzene is not satisfactory because benzenol is oxidized more readily than is benzene. At one time, benzenol was made industrially by sulfonating or chlorinating benzene and then introducing the hydroxyl group by nucleophilic substitution with strong alkali: Current commercial syntheses of benzenol involve oxidation of methylbenzene or isopropylbenzene ( ). Oxidation of isopropylbenzene is economically feasible for the production of benzenol because 2-propanone (acetone) also is a valuable product: A common laboratory procedure converts an aromatic amine to a phenol by way of the arenediazonium salt, \(\ce{ArNH_2} \rightarrow \ce{ArN_2^+} \rightarrow \ce{ArOH}\) ( ). The reactions of the hydroxyl groups of phenols, wherein the \(\ce{O-H}\) bonds are broken, are similar to those of alcohols. Thus phenols are weak acids (\(K_a = 10^{-10}\) to \(10^{-8}\); Table 26-1), intermediate in strength between carboxylic acids and alcohols. Enols are stronger acids than alcohols because of the increase in electron delocalization in enolate anions as compared to the neutral enols (see ). The stabilization energy of benzenol (Table 21-1) is \(48 \: \text{kcal mol}^{-1}\), \(5 \: \text{kcal}\) greater than that of benzene. We can ascribe this increase to delocalization of an unshared electron pair from oxygen: When the \(\ce{OH}\) proton is removed by a base the resulting anion has even greater stabilization, because the corresponding valence-bond structures do not involve charge separation: We can be confident that substituent groups that stabilize the anion will increase the acidity. Thus 4-nitrobenzenol is about 500 times stronger as an acid than benzenol, because of the increased delocalization of charge to the nitro group: It is possible to prepare esters of phenols with carboxylic acid anhydrides or acid halides, and phenyl ethers by reaction of benzenolate anion with halides, sulfate esters, sulfonates, or other alkyl derivatives that react well by the \(S_\text{N}2\) mechanism: Phenols (like carboxylic acids; and Table 18-6) are converted to methoxy derivatives with diazomethane: Almost all phenols and enols (such as those of 1,3-diketones) give colors with ferric chloride in dilute water or alcohol solutions. Benzenol itself produces a violet color with ferric chloride, and the methylbenzenols give a blue color. The products apparently are ferric arenolate salts, which absorb visible light to give an excited state having electrons delocalized over both the iron atom and the unsaturated system. In general, it is very difficult to break the aromatic \(\ce{C-O}\) bond of arenols. Thus concentrated halogen acids do convert simple arenols to aryl halides, and alkoxyarenes are cleaved with hydrogen bromide or hydrogen iodide in the manner \(\ce{ArO-R}\) rather than \(\ce{Ar-OR}\): (Diaryl ethers, such as diphenyl ether, do not react with hydrogen iodide even at \(200^\text{o}\).) There is no easy way to convert arenols to aryl halides, except where activation is provided by 2- or 4-nitro groups. Thus 2,4-dinitrobenzenol is converted to 1-chloro-2,4-dinitrobenzene with phosphorus pentachloride: An exception to the generalization that \(\ce{C-O}\) bonds to aromatic systems are difficult both to make and to break is provided by reversible conversion of benzenediols and 1- or 2-naphthalenols to the corresponding amines, usually at elevated temperatures with sodium hydrogen sulfite or an acidic catalyst. The sodium hydrogen sulfite-induced reaction is called the : These reactions do not work well with simple benzenols because the key step is formation of the keto isomer of the arenol - a process that is unfavorable for simple benzenols. \(\tag{26-1}\) The role of the hydrogen sulfite is participation in a reversible 1,4-addition to the unsaturated ketone to hold it in the ketone form that then is converted to the imine by \(\ce{NH_3}\) ( ) and hence to the arenamine: The electron-rich \(\pi\)-orbital systems of arenols and especially of arenolate ions make these compounds very susceptible to electrophilic substitution. Arenols typically react rapidly with bromine in aqueous solution to substituted the positions ortho or para to the hydroxyl group. Benzenol itself gives 2,4,6-tribromobenzenol in high yield: Several important reactions of arenols involve aromatic substitution of arenolate ions with electrophiles. In a sense, these reactions are alkylation and acylation reactions as discussed for arenes ( and ). In another sense, they are alkylation and acylation reactions of and therefore could give rise to products by \(\ce{C}\)- and \(\ce{O}\)-alkylation, or \(\ce{C}\)- and \(\ce{O}\)-acylation (Section 17-4). Thus: In most cases, \(\ce{O}\)-alkylation predominates. However, with 2-propenyl halides either reaction can be made essentially the exclusive reaction by proper choice of solvent. With sodium benzenolate the more polar solvents, such as 2-propanone, lead to 2-propenyloxybenzene, whereas in nonpolar solvents, such as benzene, 2-(2-propenyl)benzenol is the favored product: It should be noted that formation of 2-(2-propenyl)benzenol in nonpolar solvents is the result of \(\ce{O}\)-propenylation by rearrangement, even though the \(\ce{C}\)-propenylation product is thermodynamically more stable. Rearrangement in fact does occur, but at much higher temperatures (above \(200^\text{o}\)) than required to propenylate sodium benzenolate: Such rearrangements are quite general for aryl allyl ethers and are called . They are examples of the pericyclic reactions discussed in . The produces \(\ce{O}\)- and \(\ce{C}\)-carboxylation through the reaction of carbon dioxide with sodium benzenolate at \(125^\text{o}\): Sodium benzenolate absorbs carbon dioxide at room temperature to form sodium phenyl carbonate (\(\ce{O}\)-carboxylation) and, when this is heated to \(125^\text{o}\) under a pressure of several atmospheres of carbon dioxide, it rearranges to sodium 2-hydroxybenzoate (sodium salicylate). However, there is no evidence that this reaction is other than a dissociation-recombination process, in which the important step involves electrophilic attack by carbon dioxide on the aromatic ring of the benzenolate ion (\(\ce{C}\)-carboxylation): With the sodium benzenolate at temperatures of \(125^\text{o}\) to \(150^\text{o}\), ortho substitution occurs; at higher temperatures (\(250^\text{o}\) to \(300^\text{o}\)), particularly with the potassium salt, the para isomer is favored. The Kolbe-Schmitt reaction is related to enzymatic carboxylations as of \(D\)-ribulose-1,5-diphosphate with carbon dioxide, a key step in photosynthesis ( ). The overall result is \(\ce{C-C}\) bond formation by addition of \(\ce{CO_2}\) to an enolate salt or its enamine equivalent. In the somewhat related , sodium benzenolate with trichloromethane in alkaline solution forms the sodium salt of 2-hydroxybenzenecarbaldehyde (salicylaldehyde). The electrophile in this case probably is dichlorocarbene ( ): Many phenols undergo aldol-like addition reactions with carbonyl compounds in the presence of acids or bases. Thus benzenol reacts with methanal under mild alkaline conditions to form (4-hydroxyphenyl)methanol: The use of this type of reaction in the formation of polymers will be discussed in Chapter 29. Arenols usually will undergo diazo coupling reactions with aryldiazonium salts at pH values high enough to convert some of the arenol to the more powerfully nucleophilic arenolate anions: However, if the pH is too high, coupling is inhibited because the diazonium salt is transformed into \(\ce{ArN=N-O}^\ominus\), which is the nonelectrophilic conjugate base of a diazotic acid (Table 23-4). Arenols can be reduced successfully with hydrogen over nickel catalysts to the corresponding cyclohexanols. A variety of alkyl-substituted cyclohexanols can be prepared in this way: Benzenol can be oxidized to 1,4-benzenedione ( -benzoquinone) by chromic acid. The reaction may proceed by way of phenyl hydrogen-chromate ( ) as follows: Oxidation reactions of arenols with other oxidants are complex. Oxidative attack seems to involve, as the first step, removal of the hydroxyl hydrogen to yield a phenoxy radical: The subsequent course depends upon the substituents on the aromatic ring. With 2,4,6-tri- -butylbenzenol, the radical is reasonably stable in benzene solution and its presence is indicated by both its dark-blue color and the fact that it adds to 1,3-butadiene: Apparently, dimerization of the above phenoxy radical through either oxygen or the ring is inhibited by the bulky -butyl groups. With fewer or smaller substituents, the phenoxy radicals may form dimerization or disproportionation products. Examples of these reactions follow. Several important aromatic compounds have more than one arene hydroxyl group. These most often are derivatives of the following dihydric and trihydric arenols, all of which have commonly used (but poorly descriptive) names: All are exceptionally reactive towards electrophilic reagents, particularly in alkaline solution, and all are readily oxidized. The 1,2- and 1,4-benzenediols, but not 1,3-benzenediol, are oxidized to quinones: The preparation of these substances can be achieved by standard methods for synthesizing arenols, but most of them actually are made on a commercial scale by rather special procedures, some of which are summarized as follows: The gallic acid used in the preparation of 1,2,3-benzenetriol can be obtained by microbial degradation of , which are complex combinations of glucose and gallic acid obtained from oak bark and gallnuts. A few other representatives of the many types of naturally occurring derivatives of polyhydric arenols are and (1977) | 11,768 | 2,661 |
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Optical isomerism is a form of stereoisomerism. This page explains what stereoisomers are and how you recognize the possibility of optical isomers in a molecule. Isomers are molecules that have the same molecular formula, but have a different arrangement of the atoms in space. That excludes any different arrangements which are simply due to the molecule rotating as a whole, or rotating about particular bonds. Where the atoms making up the various isomers are joined up in a different order, this is known as structural isomerism. Structural isomerism is a form of stereoisomerism, which involve the atoms of the complex bonded in the same order, but in different spatial arrangements. Optical isomerism is one form of stereoisomerism; geometric isomers are a second type. Optical isomers are named like this because of their effect on plane polarized light. Simple substances which show optical isomerism exist as two isomers known as enantiomers. The examples of organic optical isomers contain a carbon atom joined to different groups. These two models each have the same groups joined to the central carbon atom, but still manage to be different: Obviously as they are drawn, the orange and blue groups are not aligned the same way. Could you get them to align by rotating one of the molecules? The next diagram shows what happens if you rotate molecule B. They still are not the same - and there is no way that you can rotate them so that they look exactly the same. These are isomers of each other. They are described as being non-superimposable in the sense that (if you imagine molecule B being turned into a ghostly version of itself) you couldn't slide one molecule exactly over the other one. Something would always be pointing in the wrong direction. What happens if two of the groups attached to the central carbon atom are the same? The next diagram shows this possibility. The two models are aligned exactly as before, but the orange group has been replaced by another pink one. Rotating molecule this time shows that it is exactly the same as molecule A. You only get optical isomers if all four groups attached to the central carbon are different. The essential difference between the two examples we've looked at lies in the symmetry of the molecules. If there are two groups the same attached to the central carbon atom, the molecule has a plane of symmetry. If you imagine slicing through the molecule, the left-hand side is an exact reflection of the right-hand side. Where there are four groups attached, there is no symmetry anywhere in the molecule A molecule which has no plane of symmetry is described as chiral. The carbon atom with the four different groups attached which causes this lack of symmetry is described as a chiral center or as an asymmetric carbon atom. The molecule on the left above (with a plane of symmetry) is described as achiral. Only chiral molecules have optical isomers. One of the enantiomers is simply a non-superimposable mirror image of the other one. In other words, if one isomer looked in a mirror, what it would see is the other one. The two isomers (the original one and its mirror image) have a different spatial arrangement, and so cannot be superimposed on each other. If an achiral molecule (one with a plane of symmetry) looked in a mirror, you would always find that by rotating the image in space, you could make the two look identical. It would be possible to superimpose the original molecule and its mirror image. The asymmetric carbon atom in a compound (the one with four different groups attached) is often shown by a star. It's extremely important to draw the isomers correctly. Draw one of them using standard bond notation to show the 3-dimensional arrangement around the asymmetric carbon atom. Then draw the mirror to show the examiner that you know what you are doing, and then the mirror image. Notice that you don't literally draw the mirror images of all the letters and numbers! It is, however, quite useful to reverse large groups - look, for example, at the ethyl group at the top of the diagram. It doesn't matter in the least in what order you draw the four groups around the central carbon. As long as your mirror image is drawn accurately, you will automatically have drawn the two isomers. So which of these two isomers is (+)butan-2-ol and which is (-)butan-2-ol? There is no simple way of telling that. For A'level purposes, you can just ignore that problem - all you need to be able to do is to draw the two isomers correctly. Once again the chiral center is shown by a star. The two enantiomers are: It is important this time to draw the COOH group backwards in the mirror image. If you don't there is a good chance of you joining it on to the central carbon wrongly. If you draw it like this in an exam, you will not get the mark for that isomer even if you have drawn everything else perfectly. This is typical of naturally-occurring amino acids. Structurally, it is just like the last example, except that the -OH group is replaced by -NH The two enantiomers are: Only one of these isomers occurs naturally: the (+) form. You cannot tell just by looking at the structures which this is. It has, however, been possible to work out which of these structures is which. Naturally occurring alanine is the right-hand structure, and the way the groups are arranged around the central carbon atom is known as an configuration. Notice the use of the capital L. The other configuration is known as . So you may well find alanine described as L-(+)alanine. That means that it has this particular structure and rotates the plane of polarization clockwise. Even if you know that a different compound has an arrangement of groups similar to alanine, you still cannot say which way it will rotate the plane of polarization. The other amino acids, for example, have the same arrangement of groups as alanine does (all that changes is the CH group), but some are (+) forms and others are (-) forms. It's quite common for natural systems to only work with one of the enantiomers of an optically active substance. It is not too difficult to see why that might be. Because the molecules have different spatial arrangements of their various groups, only one of them is likely to fit properly into the active sites on the enzymes they work with. In the lab, it is quite common to produce equal amounts of both forms of a compound when it is synthesized. This happens just by chance, and you tend to get racemic mixtures. A skeletal formula is the most stripped-down formula possible. Look at the structural formula and skeletal formula for butan-2-ol. Notice that in the skeletal formula all of the carbon atoms have been left out, as well as all of the hydrogen atoms attached to carbons. In a skeletal diagram of this sort: We have already discussed the butan-2-ol case further up the page, and you know that it has optical isomers. The second carbon atom (the one with the -OH attached) has four different groups around it, and so is a chiral center. Is this obvious from the skeletal formula? Well, it is, provided you remember that each carbon atom has to have 4 bonds going away from it. Since the second carbon here only seems to have 3, there must also be a hydrogen attached to that carbon. So it has a hydrogen, an -OH group, and two different hydrocarbon groups (methyl and ethyl). Four different groups around a carbon atom means that it is a chiral center. The diagrams show an uncluttered skeletal formula, and a repeat of it with two of the carbons labeled. Look first at the carbon atom labeled 2. Is this a chiral center? No, it is not. Two bonds (one vertical and one to the left) are both attached to methyl groups. In addition, of course, there is a hydrogen atom and the more complicated hydrocarbon group to the right. It doesn't have 4 different groups attached, and so is not a chiral center. What about the number 3 carbon atom? This has a methyl group below it, an ethyl group to the right, and a more complicated hydrocarbon group to the left. Plus, of course, a hydrogen atom to make up the 4 bonds that have to be formed by the carbon. That means that it is attached to 4 different things, and so is a chiral center. We will start with a fairly simple ring compound: When you are looking at rings like this, as far as optical isomerism is concerned, you don't need to look at any carbon in a double bond. You also don't need to look at any junction which only has two bonds going away from it. In that case, there must be 2 hydrogens attached, and so there cannot possibly be 4 different groups attached. In this case, that means that you only need to look at the carbon with the -OH group attached. It has an -OH group, a hydrogen (to make up the total number of bonds to four), and links to two carbon atoms. How does the fact that these carbon atoms are part of a ring affect things? You just need to trace back around the ring from both sides of the carbon you are looking at. Is the arrangement in both directions exactly the same? In this case, it is not. Going in one direction, you come immediately to a carbon with a double bond. In the other direction, you meet two singly bonded carbon atoms, and then one with a double bond. That means that you haven't got two identical hydrocarbon groups attached to the carbon you are interested in, and so it has 4 different groups in total around it. It is asymmetric - a chiral center. What about this near-relative of the last molecule? In this case, everything is as before, except that if you trace around the ring clockwise and counter-clockwise from the carbon at the bottom of the ring, there is an identical pattern in both directions. You can think of the bottom carbon being attached to a hydrogen, an -OH group, and two identical hydrocarbon groups. It therefore is a chiral center. The other thing which is very noticeable about this molecule is that there is a plane of symmetry through the carbon atom we are interested in. If you chopped it in half through this carbon, one side of the molecule would be an exact reflection of the other. In the first ring molecule above, that is not the case. If you can see a plane of symmetry through the carbon atom it will not be a chiral center. If there is not a plane of symmetry, it will be a chiral center. The skeletal diagram shows the structure of cholesterol. Some of the carbon atoms have been numbered for discussion purposes below. These are part of the normal system for numbering the carbon atoms in cholesterol. Before you read on, look carefully at each of the numbered carbon atoms, and decide which of them are chiral centers. The other carbon atoms in the structure cannot be chiral centers, because they are either parts of double bonds, or are joined to either two or three hydrogen atoms. So . . . how many chiral centers did you find? In fact, there are 8 chiral centers out of the total of 9 carbons marked. If you didn't find all eight, go back and have another look before you read any further. It might help to sketch the structure on a piece of paper and draw in any missing hydrogens attached to the numbered carbons, and write in the methyl groups at the end of the branches as well. This is done for you below, but it would be a lot better if you did it yourself and then checked your sketch afterwards . Starting with the easy one - it is obvious that carbon 9 has two methyl groups attached. It doesn't have 4 different groups, and so cannot be chiral. If you take a general look at the rest, it is fairly clear that none of them has a plane of symmetry through the numbered carbons. Therefore they are all likely to be chiral centers. But it's worth checking to see what is attached to each of them. This all looks difficult at first glance, but it is not. You do, however, have to take a great deal of care in working through it - it is amazingly easy to miss one out. Jim Clark ( ) | 11,952 | 2,662 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Equilibria/Complex_Equilibrium |
The lone pairs of electrons in water molecules make water an excellent ligand. Metal ions in any body of water in the environment usually are coordinated to 6 water molecules, e.g. Ag(H O) , Cu(H O) , and Cr(H O) , etc. In the literature, these ions are represented by Ag (aq), Cu (aq), and Cr (aq) and very often notations (aq) is also omitted. When a stronger coordinating ligand such as \(NH_3\) is present, its molecules compete with water for the coordination site. \[Ag(H_2O)_6^+ + NH_3 \rightleftharpoons Ag(NH_3)(H_2O)_5^+\] \[Ag(NH_3)(H_2O)_5^+ + NH_3 \rightleftharpoons Ag(NH_3)_2(H_2O)_4^+\] And these are simply represented by the following with the equilibrium constant as indicated. \[Ag^+ + NH_3\rightleftharpoons Ag(NH_3)^+ \;\;\;K_1 = 2.0 \times 10^3\] \[Ag(NH_3)^+ + NH_3 \rightleftharpoons Ag(NH_3)_2^+ \;\;\;K_2 = 8.0 \times 10^3\] Where \(K_1\) and \(K_2\) are called as opposed to the , \(\beta_1\) and \(\beta_2\). Note, however, that \[K_1 = \dfrac{[Ag(NH_3)^+]}{[Ag^+] [NH_3]}\] \[\beta_1 = K_1\] \[K_2 = \dfrac{[Ag(NH_3)^{+2}]}{[Ag(NH_3)^+,NH_3]}\] \[\beta_2 = \dfrac{[Ag(NH_3)^{+2}]}{[Ag^+,NH_3]^2} = K_1 K_2\] A mole fraction (\(X\)) distribution diagram for various complexes of silver can be used to show their variations. The total concentration of silver-ion-containing species \(C_0\) is, \[C_0 = [Ag^+] + [Ag(NH_3)^+] + [Ag(NH_3)_2^+]\] \[ = [Ag^+] \left(1 + \beta_1 [NH_3] + \beta_2 [NH_3]^2\right)\] And the mole fraction, \(X\), of a species \(A\), \(X(A)\) is easily calculated using the following equations. \[X(Ag^+) = \dfrac{[Ag^+]}{C_0}\] \[ = \dfrac{1}{1 + \beta_1 [NH_3] + \beta_2 [NH_3]^2}\] \[X(Ag(NH_3)+) = \dfrac{[Ag(NH_3)^+]}{C_0}\] = \beta_1 [NH ] / (1 + \beta_1 [NH ] + \beta_2 [NH ] )
= \beta_1 [NH ] (Ag )
(Ag(NH ) ) = [Ag(NH ) ] /
= \beta_2 [NH ] / (1 + \beta_1 [NH ] + \beta_2 [NH ] )
= \beta_2 [NH ] (Ag ) Evaluate X(Ag+), X(Ag(NH3)+), and X(Ag(NH3)2+) for a solution containing 0.20 M Ag+ ions mixed with an equal volume of 2.00 M NH3 solution. Assume the volume doubles when the solutions are mixed. We have, \[\beta_1 = K_1 = 2.0 \times 10^3\] \[\beta_2 = K_1 K_2 = 1.6 \times 10^7\] \[[NH_3] = 1.00\; M\] \[C_0 = [Ag^+] = 0.10\; M\] \[X(Ag^+) = \dfrac{1}{1 + \beta_1 [NH_3] + \beta_2 [NH_3]^2}\] \[= 1 / 1.6 \times 10^7 = 6.25 \times 10^{-8}\] \[X(Ag(NH_3)^+) = \beta_1 [NH_3] X(Ag^+)\] \[ = 1.25 \times 10^{-4}\] \[X(Ag(NH_3)_2^+) = \beta_2 [NH_3]_2 X(Ag^+)\] \[ = 1\]
More precisely [NH ] = 0.80 M, but the results are not changed. Since [NH ] is >> [Ag ], the complex with the highest number of ligands is the dominating species. What are the concentrations of NH when These are cross points for the mole fraction distribution lines.
Early on, we have derived these relationships. \[X(Ag(NH_3)^+) = \beta_1 [NH_3] X(Ag^+)\] \[X(Ag(NH_3)_2^+) = \beta_2 [NH_3]^2 X(Ag^+)\] When the mole fractions are equal, their concentrations are also equal. Thus, when the mole fraction of Ag is the same as the mole fraction of Ag(NH ) , \[X(Ag^+) = X(Ag(NH_3)^+)\] We have \[\beta_1[NH_3] = 1\]
\[[NH_3] = \dfrac{1}{\beta_1}\] By the same arguments, the cross point of (Ag(NH ) ) line and (Ag(NH ) ) line occur when [NH3] = 1 / (\beta_2)1/2 Note that [NH ] refers to the concentration of the free ligand, not the total concentration of NH . For a chemical engineering application, much more details must be revealed in order to understand the complexity of coordination equilibrium. Here is a case for a simulation model either implemented using spread sheet or programming. In the case of \(Cu^{2+}\) ions, the equilibria are more complicated, because the of ligands can be as high as 6. Consider the following data: \[Cu^{2+} + NH_3 \rightarrow Cu(NH_3)^{2+} \] with \(K_1 = 1.1 \times 10^4 \label{1}\) \[Cu(NH_3)^{2+} + NH_3 \rightarrow Cu(NH_3)_2^{2+} \] with \(K_2 = 2.7 \times 10^3 \label{2}\) \[Cu(NH_3)_2^{2+} + NH_3 \rightarrow Cu(NH_3)_3^{2+} \] with \(K_3 = 6.3 \times 10^2 \label{3}\) \[Cu(NH_3)_3^{2+} + NH_3 \rightarrow Cu(NH_3)_4^{2+}\] with \(K_4 = 30 \label{4}\) At high concentration of \(NH_3\) more of \(Cu(NH_3)_4^{2+}\) complexes are formed. The higher the number of ammonia is coordinated to the cupric ion, the deeper is the color blue. What we have done for the silver ion complexes can be applied to the cupric complexes, and the calculation is a little more complicated. However, the theory and method are the same, and you may applied the discussion on silver ion to that of the cupric ion cases. | 4,552 | 2,664 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/Accelerator_Mass_Spectroscopy |
Accelerator Mass Spectroscopy (AMS) is a highly sensitive technique that is useful in isotopic analysis of specific elements in small samples (1mg or less of sample containing 10 atoms or less of the isotope of interest). AMS requires a particle accelerator, originally used in nuclear physics research, which limits its widespread use due to high costs and technical complexity. Fortunately, UC Davis researchers have access to the Lawrence Livermore National Laboratory Center for Accelerator Mass Spectrometry (CAMS LLNL), one of over 180 AMS research facilities in the world. AMS is distinct from conventional Mass Spectrometry (MS) because it accelerates ions to extremely high energies (millions of electron volts) compared to the thousands of electron volts in MS (1keV=1.6×10 J). This allows AMS to resolve ambiguities that arise in MS due to atomic and molecular ions of the same mass. AMS is most widely used for isotope studies of C, which has applications in a variety of fields such as radiocarbon dating, climate studies, and biomedical analysis. Some of the most fascinating applications of AMS range from exposure dating of surface rocks, C labeled drug tracer studies, and even radiocarbon dating of artifacts such as the Shroud of Turin and the Dead Sea Scrolls. In conventional atomic mass spectrometry, samples are atomized and ionized, separated by their mass-to-charge ratio, then measured and/or counted by a detector. Rare isotopes such as C present a challenge to conventional MS due to their low natural abundance and high background levels. Researchers were challenged by isobaric interference (interference from equal mass isotopes of different elements exemplified by N in C analysis), isotopic interference (interference from equal mass to charge isotopes of different elements), and molecular interference (interference from equal mass to charge molecules, such as CH , CD, or CH in C analysis). Most AMS systems employ an electrostatic tandem accelerator that has a direct improvement in background rejection, resulting in a 10 time increase in the sensitivity of isotope ratio measurements. As the natural abundance of C in modern carbon is 10 (isotopic ratio of C: C), a sensitivity of 10 is a prerequisite for C analysis. Figure 1, above, starts with a negative ion sputter source, which commonly consists of a stream of Cesium ions (Cs ) with energies of 2-3 keV focused on the surface of a solid sample in order to transfer enough energy to the target material to produce free atoms and ions of the sample material. This process, called sputtering, separates neutral, as well as positive and negative ions from the sample surface. The sample is held at a negative potential, and negatively charged ions are accelerated away from the sample, resulting in a beam of negative ions (Figure 2, below). Cs is particularly useful in C studies because it does not form a negative ion from N, thereby eliminating isobar interference. It is important to have a beam of negative ions entering the accelerator because the negative ions are attacted to the high -voltage terminal which results in their net acceleration. The low energy (~5-10 keV) diverging beam that leaves the ion source is accelerated, focused and transported to the accelerator by the injector system. CAMS LLNL employs a low-energy mass spectrometer that selects for the desired atomic mass that separates ions by their mass to charge ratio ( C, C, and C ions pass through separately). Most AMS systems use sequential injection, a process that switches between stable and rare isotopes via the application of varying voltages to the electrically insulated vacuum chamber of the analyzer magnet. In sequential injection, typical injection repetition rates are 10 sec to minimize variations in the electrical load. This process allows the development of more versatile systems, allowing for analysis of a wide range of isotopes. The alternative to sequential injection is simultaneous injection, a process adopted in accelerators dedicated to C analysis. A recombinator is used following sequential injection, which is a sequence of magnetic analyzers and quadrupole lenses that focus the stable and rare isotopes so they recombine and enter the accelerator together. The traditional accelerator was first developed in the early 1930s for nuclear physics research. In 1939, UC Berkeley scientists Luis Alvarez and Robert Cornog were the first to use AMS in the detection of He in nature using the 88-inch Berkeley cylclotron. Now, over 70 years later, cyclotrons have been replaced by an accelerator type with greater energy stability: the tandem electrostatic accelerator. An electrostatic accelerator works by accellerating particles though a magnetic field generated by high voltages using a mechanic transport system that continuously transports charges from ground to the insulated high-voltage terminal. All tandem accelerators with a maximum terminal voltage above 5 MV use such a mechanical system. The negative ions that enter the accelerator are attracted to the high-voltage terminal, which is what accellerates theCAMS LLNL employs a tandem Van de Graaff accelerator, in which a second acceleration of millions of volts is applied. In all tandem accelerators, atoms are stripped at the high-voltage terminal using either a thin Carbon foil or Argon gas. Stripping is the process in which two or more electrons are removed. The Van de Graaff accelerator removes at least four electrons. It is preferrable to remove at least three electrons because by this process that molecular isobars of C (such as CH , CD, or CH ) are destroyed due to the high instability of their positively charged forms, and atomic C ions such as C , C , and C are separated due to their different mass to charge ratios. The negative ions are changed to positively charged ions and are thus accelerated back to the ground potential in the high-energy part of the accelerator. Transmission through a foil changes with time due to radiation damage and foil thickening, thus gas strippers are used in all modern analyzers due to their increased transmission stability. Magnetic lenses focus the high energy particles leaving the accelerator into a magnetic dipole, (the high energy analyzing magnet). Stable isotopes can be collected at off-axis beam stops where secondary focusing lenses and additional analyzing equipment remove unwanted ions and molecular fragments to eliminate background. At CAMS LLNL, a magnetic quadrupole lens focuses the desired isotope and charge state to a high-energy mass spectrometer which passes C and C into Faraday cups and further focuses and stabilizes C in a quadrupole/electrostatic cylindrical analyzer that leads to a gas ionization detector. The magnetic quadrupole and electrostatic selectors coupled together ensure high selectivity and sensitivity, respectively. Other detectors commonly found in AMS systems include surface barrier, time-of-flight, gas filled magnets, and x-ray detectors. Rare isotopes analyzed by AMS are always measured as a ratio of a stable, more abundant (but not too abundant) isotope. For example, the ratio in C studies is generally shown as C/ C. Less abundant isotopes are preferable in AMS because the decreased flux of ions reduces background and wear on the instrument, which is of particular concern due to the quick deterioration of particle detectors (performance deteriorates at rates higher than a few thousand particles per second ). Common radioisotope elements measured with AMS and their applications are shown in Table 1 , below. Because C analysis is by far the most popular application of AMS, the methods discussed below are all techniques used involving C. Radiocarbon dating is an analytical method based on the rate of decay of C, a radioactive carbon isotope formed in the atmosphere by the reaction between neutrons from cosmic rays and N (neutron + N = C + proton). Resultant C atoms are taken up by plants in the form of CO , then transferred to animals though the food chain. When animals and plants die, they cease to uptake C, and a steady decay of C continues in their tissues over time. C atoms decay via electron emission (β radiation) to form N, a process which has a half life of 5,730 years. Radiocarbon levels in the atmosphere change according to complex patterns which are affected by a variety of fluctuations ranging from the sun’s solar activity and the earth’s magnetic field, to ocean ventilation rate and climate. C analysis of tree rings, corals, lake sediments, ice cores, and other sources has led to a detailed record of C variations through time, allowing researchers to establish an official radiocarbon calibration curve (also referred to as a radiocarbon clock) dating back 26,000 calendar years. In the 1960s, nuclear weapons testing released large amounts of neutrons into the atmosphere, nearly doubling C activity. Samples taken after this time period can be radiocarbon dated using a C bomb curve like the peak shown below in Figure 3, can retrieve very precise dates (within 1 year at the steepest part of the curve). C analysis provides valuable information in the radiocarbon dating of the world’s most priceless artifacts. One such example of the monumental impact of C AMS is the radiocarbon dating of the Dead Sea Scrolls to dates from 300 BC to AD 61 by labs in Zurich and Arizona. AMS has also contributed greatly to environmental and atmospheric studies by providing information regarding particle composition and origin. In the biochemical field, synthesized C labeled compounds can be administered as a tracer dose for in-vivo human metabolic and drug studies which require AMS analysis of graphitized biological samples. AMS is a highly sensitive method for isotopic analysis that has numerous key applications that are only growing with advances in technology. High costs and technical complexities that arise with the use of a particle accelerator are the only limits to the widespread use of AMS. Recent times have seen the emergence of commercially available compact accelerators that use as low as 200 kV for radiocarbon dating and biomedical applications, and as particle accelerators become more commonplace, modifications to the instrument have also broadened the number of isotopes the instrument can measure. | 10,425 | 2,666 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Basics_of_Electrochemistry/Electrochemistry/Nernst_Equation |
Electrochemistry deals with cell potential as well as energy of chemical reactions. The energy of a chemical system drives the charges to move, and the driving force gives rise to the cell potential of a system called galvanic cell. The energy aspect is also related to the chemical equilibrium. All these relationships are tied together in the concept of the . Walther H. Nernst (1864-1941) received the Nobel prize in 1920 " ". His contribution to chemical thermodynamics led to the well known equation correlating chemical energy and the electric potential of a galvanic cell or battery. Energy takes many forms: mechanical work (potential and kinetic energy), heat, radiation (photons), chemical energy, nuclear energy (mass), and electric energy. A summary is given regarding the evaluation of electric energy, as this is related to electrochemistry. Energy drives all changes including chemical reactions. In a redox reaction, the energy released in a reaction due to movement of charged particles gives rise to a . The maximum potential difference is called the (EMF), , and the maximum electric work is the product of charge in Coulomb (C), and the potential in Volt (= J / C) or EMF. \(W \mathrm{\color{Periwinkle} J} = q\: \textrm D E\: \mathrm{\color{Periwinkle} C\: J/C\: (units)}\) Note that the EMF is determined by the nature of the reactants and electrolytes, not by the size of the cell or amounts of material in it. The amount of reactants is proportional to the charge and available energy of the galvanic cell. The is the negative value of maximum electric work, \(\begin{align}
\ce \Delta G &= - W\\
&= - q \:\ce \Delta E
\end{align}\) A redox reaction equation represents definite amounts of reactants in the formation of also definite amounts of products. The number ( ) of electrons in such a reaction equation is related to the amount of charge transferred when the reaction is completed. Since each mole of electron has a charge of 96485 C (known as the Faraday's constant, ), \(q = n F\) and, \(\ce \Delta G = - n F \:\ce \Delta E\) At standard conditions, \(\ce \Delta G^\circ = - n F \:\ce \Delta E^\circ\) The general Nernst equation correlates the Gibb's Free Energy and the EMF of a chemical system known as the galvanic cell. For the reaction \(\ce{a\, A + b\, B \rightleftharpoons c\, C + d\, D}\) and \(Q = \mathrm{\dfrac{[C]^c [D]^d}{[A]^a [B]^b}}\) It has been shown that \(\textrm D G = \textrm D G^\circ + R T \ln Q\) and \(\textrm D G = - n F\:\textrm D E\) Therefore \(- n F \:\textrm D E = - n F \: \textrm D E^\circ + R T \ln Q\) where and are the gas constant (8.314 J mol K ), temperature (in K), reaction quotient, and Faraday constant (96485 C) respectively. Thus, we have \(\textrm D E = \textrm D E^\circ - \dfrac{R T}{n F} \ln \mathrm{\dfrac{[C]^c [D]^d}{[A]^a [B]^b}}\) This is known as the . The equation allows us to calculate the cell potential of any galvanic cell for any concentrations. Some examples are given in the next section to illustrate its application. It is interesting to note the relationship between equilibrium and the Gibb's free energy at this point. When a system is at , = 0, and = . Therefore, we have, \(\textrm D E^\circ = \dfrac{R T}{n F} \ln \mathrm{\dfrac{[C]^c [D]^d}{[A]^a [B]^b}},\: \textrm{(for equilibrium concentrations)}\) Thus, the equilibrium constant and ° are related. At any specific temperature, the Nernst equation derived above can be reduced into a simple form. For example, at the standard condition of 298 K (25°), the Nernst equation becomes \(\textrm D E = \textrm D E^\circ - \dfrac{0.0592\: \textrm V}{n} \log \mathrm{\dfrac{[C]^c [D]^d}{[A]^a [B]^b}}\) Please note that log is the logarithm function based 10, and ln, the natural logarithm function. For the cell \(\ce{Zn \,|\, Zn^2+ \,||\, H+ \,|\, H2 \,|\, Pt}\) we have a net chemical reaction of \(\ce{Zn_{\large{(s)}} + 2 H+ \rightarrow Zn^2+ + H_{2\large{(g)}}}\) and the standard cell potential ° = 0.763. If the concentrations of the ions are not 1.0 M, and the \(\ce{H2}\) pressure is not 1.0 atm, then the cell potential may be calculated using the Nernst equation: \(\textrm D E = \textrm D E^\circ - \dfrac{0.0592\: \ce V}{n} \log \ce{\dfrac{P(H2) [Zn^2+]}{[H+]^2}}\) with = 2 in this case, because the reaction involves 2 electrons. The numerical value is 0.0592 only when T = 298 K. This constant is temperature dependent. Note that the reactivity of the solid \(\ce{Zn}\) is taken as 1. If the \(\ce{H2}\) pressure is 1 atm, the term \(\ce{P(H2)}\) may also be omitted. The expression for the argument of the log function follows the same rules as those for the expression of equilibrium constants and reaction quotients. Indeed, the argument for the log function is the expression for the equilibrium constant , or reaction quotient . When a cell is at equilibrium, = 0.00 and the expression becomes an equilibrium constant , which bears the following relationship: \(\log K = \dfrac{n \textrm D E^\circ}{0.0592}\) where ° is the difference of standard potentials of the half cells involved. A battery containing any voltage is not at equilibrium. The Nernst equation also indicates that you can build a battery simply by using the same material for both cells, but by using different concentrations. Cells of this type are called . \(\mathrm{Zn_{\large{(s)}} \,|\, Zn^{2+}\: (0.024\: M) \,||\, Zn^{2+}\: (2.4\: M) \,|\, Zn_{\large{(s)}}}\) \(\mathrm{Zn^{2+}\: (2.4\: M) + 2 e^- \rightarrow Zn \hspace{30px} Reduction}\\
\mathrm{\underline{Zn \rightarrow Zn^{2+}\: (0.024\: M) + 2 e^- \hspace{15px} Oxidation \hspace{35px}}}\\
\mathrm{Zn^{2+}\: (2.4\: M) \rightarrow Zn^{2+}\: (0.024\: M)},\:\:\: \textrm D E^\circ = 0.00 \leftarrow \textrm{Net reaction}\) Using the Nernst equation: \(\begin{align}
\textrm D E &= 0.00 - \dfrac{0.0592}{2} \log \dfrac{0.024}{2.4}\\
&= (-0.296)(-2.0)\\
&= \textrm{0.0592 V}
\end{align}\) Understandably, the \(\ce{Zn^2+}\) ions try to move from the concentrated half cell to a dilute solution. That driving force gives rise to 0.0592 V. From here, you can also calculate the energy of dilution. If you write the equation in the reverse direction, \(\mathrm{Zn^{2+}\: (0.024\: M) \rightarrow Zn^{2+}\: (2.4\: M)}\), its voltage will be -0.0592 V. At equilibrium concentrations in the two half cells will have to be equal, in which case the voltage will be zero.
Assume that you have the cell \(\ce{Mg \,|\, Mg^2+ \,||\, Ag+ \,|\, Ag}\) and the reaction is: \(\ce{Mg + 2 Ag+ \rightarrow Mg^2+ + 2 Ag}\) Using the Nernst equation \(\textrm D E = \textrm D E^\circ - \dfrac{0.0592}{2} \log \ce{\dfrac{[Mg^2+]}{[Ag+]^2}}\) If you multiply the equation of reaction by 2, you will have \(\ce{2 Mg + 4 Ag+ \rightarrow 2 Mg^2+ + 4 Ag}\) Note that there are 4 electrons involved in this equation, and = 4 in the Nernst equation: \(\textrm D E = \textrm D E^\circ - \dfrac{0.0592}{4} \log \ce{\dfrac{[Mg^2+]^2}{[Ag+]^4}}\) which can be simplified as \(\textrm D E = \textrm D E^\circ - \dfrac{0.0592}{2} \log \ce{\dfrac{[Mg^2+]}{[Ag+]^2}}\) Thus, the cell potential is not affected. \(\ce{Fe + Zn^2+ \rightarrow Zn + Fe^2+}\) \(\ce{Zn^2+}\) \(\ce{Fe^2+}\)
The equilibrium constant may be calculated using \(\begin{align}
K &= 10^{\large{(n \textrm D E^\circ)/0.0592}}\\
&= 10^{-11.93}\\
&= 1.2\times10^{-12}\\
&= \mathrm{[Fe^{2+}]/[Zn^{2+}]}
\end{align}\) Since \(\mathrm{[Zn^{2+}] = 1\: M}\), it is evident that
\(\ce{[Fe^2+]} = \textrm{1.2E-12 M}\). \(\ce{AgCl \rightarrow Ag+ + Cl-}\)
There are \(\ce{Ag+}\) and \(\ce{AgCl}\) involved in the reaction, and from the table of standard reduction potentials, you will find: \(\ce{AgCl + e^- \rightarrow Ag + Cl-},\hspace{15px} E^\circ = \mathrm{0.2223\: V} \tag{1}\) Since this equation does not contain the species \(\ce{Ag+}\), you need, \(\ce{Ag+ + e^- \rightarrow Ag}, \hspace{15px} E^\circ = \mathrm{0.799\: V} \tag{2}\) Subtracting (2) from (1) leads to, \(\ce{AgCl \rightarrow Ag+ + Cl-} \hspace{15px} \textrm D E^\circ = - 0.577\) Let be the solubility product, and employ the Nernst equation, \(\begin{align}
\log K_{\ce{sp}} &= \dfrac{-0.577}{0.0592} = -9.75\\
K_{\ce{sp}} &= 10^{-9.75} = 1.8\times10^{-10}
\end{align}\) This is the value that you have been using in past tutorials. Now, you know that is not always measured from its solubility. \(\ce{Pb \,|\, PbSO4 \,|\, H2SO4 \,|\, PbSO4,\: PbO2 \,|\, Pb}\) \(\ce{Zn_{\large{(s)}} \,|\, Zn^2+ \,||\, Cu^2+ \,|\, Cu_{\large{(s)}}}\). \(\ce{Zn_{\large{(s)}} \,|\, Zn^2+ \,||\, Cu^2+ \,|\, Cu_{\large{(s)}}}\). \(\ce{Zn_{\large{(s)}} \,|\, Zn^2+ \,||\, Cu^2+ \,|\, Cu_{\large{(s)}}} \hspace{15px} \textrm D E^\circ = \mathrm{1.100\: V}\) \(\ce{Pb + PbO2 + 2 HSO4- + 2 H+ \rightarrow 2 PbSO4 + 2 H2O}\) \(\textrm D E = \textrm D E^\circ - \left(\dfrac{0.0592}{2}\right)\log\ce{\dfrac{1}{[HSO4- ]^2[H+]^2}}\).
A likely wrong result is 1.041 V.
The term that modifies is \(-\left(\dfrac{0.059}{n}\right)\log\ce{\dfrac{[Zn^2+]}{[Cu^2+]}}\) (n = 2 in this case).
Understandably, if the concentration of \(\ce{Zn^2+}\) is low, there is more tendency for the reaction, \(\ce{Zn \rightarrow Zn^2+ + 2 e^-}\). \(0 = 1.100 - 0.0296 \log \left(\ce{\dfrac{[Zn^2+]}{[Cu^2+]}}\right)\) The Nernst equation is useful for the determination of equilibrium constants. Understanding is the key. Take time to understand it; there is no point in rushing. | 9,446 | 2,667 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Medicinal_Chemistry/Cholinergic_Drugs_I_-_Nicotinic_and_Muscarinic_Receptors |
Neurotransmitters released from nerve terminals bind to specific receptors, which are specialized macromolecules embedded in the cell membrane. The binding action initiates a series of specific biochemical reactions in the target cell that produce a physiological response. These effects can be modified by various drugs that act as agonists or antagonists. The autonomic system consists of two major divisions the Sympathetic Nervous System and the Parasympathetic Nervous System. These often function in antagonistic ways. A signal is transmitted from the spinal cord to peripheral areas through two successive neurons. The first neuron (preganglionic), which originates in the spinal cord, will synapse with the second neuron (postganglionic) in a ganglion. Parasympathetic ganglia tend to lie close to or within the organs or tissues that their neurons innervate, whereas sympathetic ganglia lie at a more distant site from their target organs. Both systems have associated sensory fibers that send feedback information into the central nervous system regarding the functional condition of target tissues. The significant difference between the two systems is that their postganglionic fibers secrete different neurotransmitters. Those of the parasympathetic system secrete acetylcholine (ACh), hence the name cholinergic, whereas the postganglionic fibers secrete norepinephrine (NE), hence the name adrenergic. The preganglionic fibers of both systems secrete ACh; therefore, both preganglionic fibers are cholinergic. Motor neurons which are not part of the autonomic nervous system also release acetylcholine (see Figure 1). Acetylcholine acts on more than one type of receptor. Henry Dale, a British physiologist working in London in 1914, found that two foreign substances, nicotine and muscarine, could each mimic some, but not all, of the parasympathetic effects of acetylcholine. It was found that Nicotine stimulates receptors on skeletal muscle and sympathetic and parasympathetic postganglionic neurons, however, muscarine stimulates receptor sites located only at the junction between postganglionic parasympathetic neurons and the target organ. Dale therefore classified the many actions of acetylcholine into nicotinic effects and muscarinic effects. It has subsequently become clear that there are two distinct types of acetylcholine receptors affected by either muscarine or nicotine. To restate this again, nicotinic receptors cause sympathetic postganglionic neurons and parasympathetic postganglionic neurons to fire and release their chemicals and skeletal muscle to contract. Muscarinic receptors are associated mainly with parasympathetic functions and stimulates receptors located in peripheral tissues (e.g., glands, smooth muscle). Acetylcholine activates all of these sites. Advanced biochemical techniques have now shown a more fundamental difference in the two types of cholinergic receptors. The nicotinic receptor is a channel protein that, upon binding by acetylcholine, opens to allow diffusion of cations. The muscarinic receptor, on the other hand, is a membrane protein upon stimulation by neurotransmitter, it causes the opening of ion channels indirectly, through a second messenger. For this reason, the action of a muscarinic synapse is relatively slow. Muscarinic receptors predominate at higher levels of the central nervous system, while nicotinic receptors, which are much faster acting, are more prevalent at neurons of the spinal cord and at neuromuscular junctions in skeletal muscle. A cholinergic drug is any of various drugs that inhibit, enhance, or mimic the action of the neurotransmitter acetylcholine within the body. Acetylcholine stimulation of the parasympathetic nervous system helps contract smooth muscles, dilate blood vessels, increase secretions, and slow the heart rate. Some cholinergic drugs, such as muscarine, pilocarpine, and arecoline, mimic the activity of acetylcholine in stimulating the parasympathetic nervous system. These drugs, however, have few therapeutic uses. Other cholinergic drugs, such as atropine and scopolamine, inhibit the action of acetylcholine and thus suppress all the actions of the parasympathetic nervous system. These drugs help dry up such bodily secretions as saliva and mucus and relax smooth-muscle walls. They are used therapeutically to relieve spasms of the smooth-muscle walls of the intestines, to relieve bronchial spasms, to diminish salivation and bronchial secretions during anesthesia, and to dilate the pupil during ophthalmological procedures. Nicotine is an organic compound that is the principal alkaloid of tobacco. Nicotine occurs throughout the tobacco plant and especially in the leaves. The compound constitutes about 5 percent of the plant by weight. Both the tobacco plant ( ) and the compound are named for Jean Nicot, a French ambassador to Portugal, who sent tobacco seeds to Paris in 1550. Crude nicotine was known by 1571, and the compound was obtained in purified form in 1828; the correct molecular formula was established in 1843, and the first laboratory synthesis was reported in 1904. Nicotine is one of the few liquid alkaloids. In its pure state it is a colorless, volatile base (pKa -8.5) with an oily consistency, but when exposed to light or air, it acquires a brown color and gives off a strong odor of tobacco. The complex and often unpredictable changes that occur in the body after administration of nicotine are due not only to its actions on a variety of neuroeffector and chemosensitive sites but also to the fact that the alkaloid has both stimulant and depressant phases of action. The ultimate response of any one system represents the summation of the several different and opposing effects of nicotine. For example, the drug can increase the heart rate by excitation of sympathetic cardiac ganglia, and it can slow down the heart rate by stimulation of parasympathetic cardiac ganglia. In addition, the effects of the drug on the chemoreceptors of the carotid and aortic bodies and on medullary centers influence heart rate, as do also the cardiovascular compensatory reflexes resulting from changes in blood pressure caused by nicotine. Finally, nicotine causes a discharge of epinephrine from the adrenal medulla, and this hormone accelerates cardiac rate and raises blood pressure. Nicotine is unique in its biphasic effects. In the medulla, small doses of nicotine evoke the discharge of catacholamines, and in larger doses prevent their release in response to splanic nerve stimulation. Its biphasic effect causes a stimulant effect when inhaled in short puffs, but when smoked in deep drags it can have a tranquilizing effect. This is why smoking can feel invigorating at some times and can seem to block stressful stimuli at others. Nicotine markedly stimulates the central nervous system (CNS). Appropriate doses produce tremors in both man and laboratory animals; with somewhat larger dose, the tremor is followed by convulsions. The excitation of respiration is a prominent action of nicotine; although large doses act directly on the medulla oblongata, smaller doses augment respiration reflexly by excitation of the chemoreceptors of the carotid and aortic bodies. Stimulation of the CNA is followed by depression, and death usually results from failure of respiration due to both central analysis and peripheral blockade of muscles of respiration. Nicotine also causes vomiting by central and peripheral actions. The central component of the vomiting response is due to stimulation of the chemoreceptor trigger zone is in the medulla.oblongata. In addition, nicotine activates vagal and spinal afferent nerves that from the sensory input of the reflex pathways involved in the act of vomiting. Although acetylcholine causes vasodilation and a decrease in heart rate, when administered intravenously to the dog, nicotine characteristically produces an increase in heart rate and blood pressure. This is because in general, the cardiovascular responses to nicotine are due to stimulation of the sympathetic ganglia and the adrenal medulla, together with the discharge of catacholamines from sympathetic nerve endings. Nicotine is commercially obtained from tobacco scraps and is used as an insecticide and as a veterinary vermifuge (wormer). Nitric acid or other oxidizing agents convert nicotine to nicotinic acid, or niacin, which is used as a food supplement. Muscarine, and alkaloid obtained from the poisonous mushroom , produces the effects predictable from stimulation of postgangiolinc parasympathetic fibers. The symptoms usually occur within 15-30 minutes of ingestion or injection, and are focused on the involuntary nervous system. The muscarinic alkaloids stimulate the smooth muscle and therby increase motility; large doses cause spasm and severe diarrhea. The bronchial musculature is also stimulated, causing asmatic-like attacks. Excessive salivation, sweating, tears, lactation (in pregnant women), plus severe vomiting also occur. The most prominent cardiovascular effects are the a marked fall in the blood pressure and a slowing or temporarily cessation of the heart. Victims normally recover within 24 hours, but severe cases may result in death due to respiratory failure. All effects of muscarine-like drugs are prevented by the alkaloid atropine. Furthermore, neither atropine-like nor muscarine-like drugs show effects at the neuromuscular junction. Although muscarine and muscarine like alkaloids are of great value as pharmacological tools, present clinical use is largely restricted. Since evidence is beginning to accumulate that there are distinct subtypes of muscarinic receptors, there has been a renewed interest in synthetic analogs that may enhance the tissue selectivity of muscarinic agonists. | 9,822 | 2,668 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/10%3A_Electrolyte_Solutions/10.07%3A_Chapter_10_Problems |
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\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) An underlined problem number or problem-part letter indicates that the numerical answer appears in Appendix I. The mean ionic activity coefficient of NaCl in a 0.100 molal aqueous solution at \(298.15\K\) has been evaluated with measurements of equilibrium cell potentials, with the result \(\ln\g_{\pm}=-0.2505\) (R. A. Robinson and R. H. Stokes, , 2nd edition, Butterworths, London, 1959, Table 9.3). Use this value in Eq. 10.6.9, together with the values of osmotic coefficients in Table 10.1, to evaluate \(\g_{\pm}\) at each of the molalities shown in the table; then plot \(\g_{\pm}\) as a function of \(m\B\). | 7,953 | 2,669 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/08%3A_Phase_Equilibrium/8.03%3A_Criterion_for_Phase_Equilibrium |
The thermodynamic criterion for phase equilibrium is simple. It is based upon the chemical potentials of the components in a system. For simplicity, consider a system with only one component. For the overall system to be in equilibrium, the chemical potential of the compound in each phase present must be the same. Otherwise, there will be some mass migration from one phase to another, decreasing the total chemical potential of the phase from which material is being removed, and increasing the total chemical potential of the phase into which the material is being deposited. So for each pair of phases present (\(\alpha\) and \(\beta\)) the following must be true: \[\mu_\alpha = \mu_\beta \nonumber \] The describes the number of compositional and phase variables that can be varied freely for a system at equilibrium. For each phase present in a system, the mole fraction of all but one component can be varied independently. However, the relationship \[\sum_i \chi_i =1 \nonumber \] places a constraint on the last mole fraction. As such, there are \(C – 1\) compositional degrees of freedom for each phase present, where \(C\) is the number of components in the mixture. Similarly, all but one of the chemical potentials of each phase present must be equal, leaving only one that can be varied independently, leading to \(P – 1\) thermodynamic constraints placed on each component. Finally, there are two state variables that can be varied (such as pressure and temperature), adding two additional degrees of freedom to the system. The net number of degrees of freedom is determined by adding all of the degrees of freedom and subtracting the number of thermodynamic constraints. \[\begin{align} F &= 2+ P(C-1) - C(P-1) \nonumber \\[4pt] &= 2 + PC - P -PC +C \nonumber \\[4pt] &= 2+C-P \label{Phase} \end{align} \] Equation \ref{Phase} is the Gibbs phase rule. Show that the maximum number of phases that can co-exist at equilibrium for a single component system is \(P = 3\). The maximum number of components will occur when the number of degrees of freedom is zero. \[ \begin{align*} 0 &= 2+1 -P \\[4pt] P&=3 \end{align*} \] Note: This shows that there can never be a “quadruple point” for a single component system! Because a system at its triple point has no degrees of freedom, the triple point makes a very convenient physical condition at which to define a temperature. For example, the International Practical Temperature Scale of 1990 (IPT-90) uses the triple points of hydrogen, neon, oxygen, argon, mercury, and water to define several low temperatures. (The calibration of a platinum resistance thermometer at the triple point of argon, for example, is described by Strouse (Strouse, 2008)). The advantage to using a triple point is that the compound sets both the temperature and pressure, rather than forcing the researcher to set a pressure and measure the temperature of a phase change, introducing an extra parameter than can introduce uncertainty into the measurement. | 3,015 | 2,671 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/09%3A_Chemical_Equilibria/9.10%3A_Solubility_of_Ionic_Compounds |
The solubility of ionic compounds in water can also be described using the concepts of equilibrium. If you consider the dissociation of a generic salt MX \[MX(s) \rightleftharpoons M^+(aq) + X^-(aq) \nonumber \] The equilibrium expression is \[K_{sp} = [M^+,X^-] \nonumber \] \(K_{sp}\) is the and is the equilibrium constant that describes the solubility of an electrolyte. And again, the pure solid MX is not included in the expression since it has unit activity throughout the establishment of equilibrium. What is the maximum solubility of CuS at 25 °C? (\(K_{sp} = 1 \times 10^{-36}\, M^2\)) Yup – time for an ICE table. So the equilibrium expression is \[ 1 \times 10^{-36} M^2 = x^2 \nonumber \] \[ x = \sqrt{ 1 \times 10^{-36} \,M^2 } = 1 \times 10^{-18}\, M \nonumber \] What is the maximum solubility of \(\ce{CuS}\) at 25 °C in 0.100 M \(\ce{NaS}\) with (\(K_{sp} = 1 \times 10^{-36}\, M^2\))? In this problem we need to consider the existence of S (aq) from the complete dissociation of the strong electrolyte NaS. An ICE table will help, as usual. Given the miniscule magnitude of the solubility product, x will be negligibly small compared to 0.100 MS the equilibrium expression is \[ 1 \times 10^{-36} M^2 = x(0.100\,M) \nonumber \] \[ 1 \times 10^{-35} \,M \nonumber \] The huge reduction in solubility is due to the common ion effect. The existence of sulfide in the solution due to sodium sulfide greatly reduces the solutions capacity to support additional sulfide due to the dissociation of \(\ce{CuS}\). | 1,540 | 2,672 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.02%3A_Covalent_Compounds_and_Intermolecular_Forces |
The ionic compounds are almost all solids with melting temperatures above 600°C. By contrast, most substances which contain simple molecules are either gases or liquids at room temperature. They can only be persuaded to solidify at rather low temperatures. The reason for this contrasting behavior is easily explained on the microscopic level. Oppositely charged ions attract each other very strongly and usually require energies of 400 kJ mol or more in order to be separated. On the other hand, molecules are electrically neutral and scarcely attract each other at all. The energy needed to separate two simple molecules is usually less than a hundredth of that needed to separate ions. For example, only 1.23 kJ mol is needed to separate two molecules of methane, CH . At room temperature virtually all molecules are moving around with energies in excess of this, so that methane is ordinarily a gas. Only if we cool the gas to quite a low temperature can we slow down the molecules to a point where they find it difficult to acquire an energy of 1.23 kJ mol . At such a temperature, the molecules will be difficult to separate and the substance will become a liquid or a solid. Experimentally, we find that methane condenses to a liquid at –162°C, and this liquid freezes at –182°C. | 1,300 | 2,673 |
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