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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.02%3A_Thermodynamics_of_Solutions/8.2.2C%3A_8.2.2C%3A_Solutions_of_Liquid_Solutes_in_Liquid_Solvents |
Whereas all gases will mix to form solutions regardless of the proportions, liquids are much more fussy. Some liquids, such as ethyl alcohol and water, are miscible in all proportions. Others, like the proverbial oil and water, are not; each liquid has only a limited solubility in the other, and once either of these limits is exceeded, the mixture separates into two phases. The reason for this variability is apparent from the table. Mixing of two liquids can be exothermic, endothermic, or without thermal effect, depending on the particular substances. Whatever the case, the energy factors are not usually very large, but neither is the increase in randomness; the two factors are frequently sufficiently balanced to produce limited miscibility. The range of possibilities is shown here in terms of the mole fractions of two liquids A and B. If A and B are only slightly miscible, they separate into two layers according to their relative densities. Note that when one takes into account trace levels, no two liquids are totally immiscible. A useful general rule is that liquids are completely miscible when their intermolecular forces are very similar in nature; “like dissolves like”. Thus water is miscible with other liquids that can engage in hydrogen bonding, whereas a hydrocarbon liquid in which London or dispersion forces are the only significant intermolecular effect will only be completely miscible with similar kinds of liquids. Substances such as the alcohols, CH (CH ) OH, which are hydrogen-bonding (and thus hydrophilic) at one end and hydrophobic at the other, tend to be at least partially miscible with both kinds of solvents. If is large, the hydrocarbon properties dominate and the alcohol has only a limited solubility in water. Very small values of allow the –OH group to dominate, so miscibility in water increases and becomes unlimited in ethanol ( = 1) and methanol ( = 0), but miscibility with hydrocarbons decreases owing to the energy required to break alcohol-alcohol hydrogen bonds when the non polar liquid is added. These considerations have become quite important in the development of alternative automotive fuels based on mixing these alcohols with gasoline. At ordinary temperatures the increased entropy of the mixture is great enough that the unfavorable energy factor is entirely overcome, and the mixture is completely miscible. At low temperatures, the entropy factor becomes less predominant, and the fuel mixture may separate into two phases, presenting severe problems to the fuel filter and carburetor. | 2,576 | 2,312 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/02%3A_Gas_Laws/2.03%3A_Avogadro's_Hypothesis |
Avogadro’s hypothesis is another classical gas law. It can be stated: At the same temperature and pressure, equal volumes of different gases contain the same number of molecules. When the mass, in grams, of an ideal gas sample is equal to the (traditionally called the ) of the gas, the number of molecules in the sample is equal to , \(\overline{N}\)\({}^{1}\). Avogadro’s number is the number of molecules in a . In the modern definition, one mole is the number of atoms of \(C^{12}\) in exactly 12 g of \(C^{12}\). That is, the number of atoms of \(C^{12}\) in exactly 12 g of \(C^{12}\) is Avogadro’s number. The currently accepted value is \(\mathrm{6.02214199\times }{\mathrm{10}}^{\mathrm{23}}\) molecules per mole. We can find the gram atomic mass of any other element by finding the mass of that element that combines with exactly 12 g of \(C^{12}\) in a compound whose molecular formula is known. The validity of Avogadro’s hypothesis follows immediately either from the fact that the Boyle’s law constant, \(\alpha (T)\), is the same for any gas or from the fact that the Charles’ law constants, \(\beta (P)\) and \(\gamma \left(P\right)\), are the same for any gas. However, this entails a significant circularity; these experiments can show that \(\alpha (T)\), \(\beta (P)\), and \(\gamma \left(P\right)\) are the same for any gas only if we know how to find the number of moles of each gas that we use. To do so, we must know the molar mass of each gas. Avogadro’s hypothesis is crucially important in the history of chemistry: Avogadro’s hypothesis made it possible to determine relative molar masses. This made it possible to determine molecular formulas for gaseous substances and to create the atomic mass scale. | 1,748 | 2,313 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/05%3A_Dioxygen_Reactions/5.04%3A_Cytochrome_c_Oxidase |
Most of the O consumed by aerobic organisms is used to produce energy in a process referred to as "oxidative phosphorylation," a series of reactions in which electron transport is coupled to the synthesis of ATP and in which the driving force for the reaction is provided by the four-electron oxidizing power of O (Reaction 5.1). (This subject is described in any standard text on biochemistry and will not be discussed in detail here.) The next to the last step in the electrontransport chain produces reduced cytochrome c, a water-soluble electron-transfer protein. Cytochrome c then transfers electrons to cytochrome c oxidase, where they are ultimately transferred to O . (Electron-transfer reactions are discussed in Chapter 6.) Cytochrome c oxidase is the terminal member of the respiratory chain in all animals and plants, aerobic yeasts, and some bacteria. This enzyme is always found associated with a membrane: the inner mitochondrial membrane in higher organisms or the cell membrane in bacteria. It is a large, complex, multisubunit enzyme whose characterization has been complicated by its size, by the fact that it is membrane-bound, and by the diversity of the four redox metal sites, i.e., two copper ions and two heme iron units, each of which is found in a different type of environment within the protein. Because of the complexity of this system and the absence of detailed structural information, spectroscopic studies of this enzyme and comparisons of spectral properties with O -binding proteins and with model iron-porphyrin and copper complexes have been invaluable in its characterization. Iron-porphyrin complexes of imidazole are a logical starting point in the search for appropriate spectroscopic models for heme centers in metalloproteins, since the histidyl imidazole side chain is the most common axial ligand bound to iron in such enzymes. Iron-porphyrin complexes with two axial imidazole ligands are known for both the ferrous and ferric oxidation states. \(\tag{5.32}\) Monoimidazole complexes of iron porphyrins are also known for both the ferrous and the ferric oxidation states. The design of these model complexes has been more challenging than for six-coordinate complexes because of the high affinity of the five-coordinate complexes for a sixth ligand. In the ferrous complex, five coordination has been achieved by use of 2-methylimidazole ligands, as described in Chapter 4. The ferrous porphyrin binds a single 2-methylimidazole ligand, and, because the Fe center is raised out of the plane of the porphyrin ring, the 2-methyl substituent suffers minimal steric interactions with the porphyrin. However, the affinity of the five-coordinate complex for another 2-methylimidazole ligand is substantially lower, because the Fe must drop down into the plane of the porphyrin to form the six-coordinate complexes, in which case the 2-methyl substitutents on both axial ligands suffer severe steric interactions with the porphyrin. Using this approach, five-coordinate monoimidazole complexes can be prepared. They are coordinatively unsaturated, and will bind a second axial ligand, such as O and CO. They have been extensively studied as models for O -binding heme proteins such as hemoglobin and myoglobin. Monoimidazole ferrous porphyrins thus designed are high-spin d6 with four unpaired electrons. They are even-spin systems and EPR spectra have not been observed. Five-coordinate monoimidazole ferric-porphyrin complexes have also been prepared in solution by starting with a ferric porphyrin complex with a very poorly coordinating anion, e.g., Fe P(SbF ). Addition of one equivalent of imidazole results in formation of the five-coordinate monoimidazole complex (Reaction 5.33). \[Fe^{III}(TPP)(SbF_{6}) + ImH \rightarrow [Fe^{III}(TPP)(ImH)]^{+} + SbF_{6}^{-} \tag{5.33}\] When imidazole is added to ferric-porphyrin complexes of other anionic ligands, e.g., CI , several equivalents of imidazole are required to displace the more strongly bound anionic ligand; consequently, only six-coordinate complexes are observed (Reaction 5.34). \[FE^{III}(TPP)Cl + 2 ImH \rightarrow [Fe^{III}(TPP)(ImH)_{2}]^{+} + Cl^{-} \tag{5.34}\] Monoimidazole ferric porphyrins are coordinatively unsaturated, readily bind a second axial ligand, and thus are appropriate models for methemoglobin or metmyoglobin. The five-coordinate complexes are high-spin d , but usually become low-spin upon binding another axial ligand to become six-coordinate. The oxidized form of cytochrome c oxidase contains two Cu and two Fe heme centers. It can be fully reduced to give a form of the enzyme containing two Cu and two Fe heme centers. The heme found in cytochrome c oxidase is different from that found in other heme proteins. It is heme a, closely related to heme b, which is found in hemoglobin, myoglobin, and cytochrome P-450, but has one of the vinyl groups replaced by a farnesyl substituent and one of the methyl groups replaced by a formyl substituent (see 5.35). Each of the four metal centers has a different coordination environment appropriate to its function. Cytochrome a and Cu appear solely to carry out an electron-transfer function without interacting directly with dioxygen. Cytochrome a and Cu appear to be part of a binuclear center that acts as the site for dioxygen binding and reduction. A schematic describing the probable nature of these four metal sites within cytochrome oxidase is given in Figure 5.3 and a description of the evidence supporting the formulation of each center then follows. Cytochrome a in both oxidation states has spectral characteristics that are entirely consistent with a low-spin ferric heme center with two axial imidazole ligands. In its oxidized form, it gives an EPR spectrum with g values similar to those obtained with model ferric-porphyrin complexes with two axial imidazole ligands (see above, Section IV.B.1). Moreover, addition of cyanide anion to the oxidized enzyme or CO to the reduced enzyme does not perturb this center, indicating that cyanide does not bind to the heme, again consistent with a six-coordinate heme. The absence of ligand binding is characteristic of six-coordinate heme sites found in electron-transfer proteins and suggests strongly that cytochrome a functions as an electron-transfer center within cytochrome c oxidase. Cu is also believed to act as an electron-transfer site. It has quite remarkable EPR spectroscopic characteristics, with g values at g = 2.18, 2.03, and 1.99, and no hyperfine splitting, resembling more an organic free radical than a typical Cu center (see Figures 5.2 and 5.4A). ENDOR studies of yeast cytochrome c oxidase containing H-cysteine or N-histidine (from yeast grown with the isotopically substituted amino acids) showed shifts relative to the unsubstituted enzyme, indicating that both of these ligands are bound to Cu . But the linear electric-field effect of Cu did not give the patterns characteristic of Cu -histidine complexes, indicating that the unpaired electron is not on the copper ion. The current hypothesis about this center is that copper is bonded in a highly covalent fashion to one, or more likely two, sulfur ligands, and that the unpaired electron density is principally on sulfur, i.e., [Cu - SR\(\leftrightarrow\)Cu - • SR]. Copper-thiolate model complexes with spectroscopic properties similar to Cu have never been synthesized, presumably because such complexes are unstable with respect to disulfide bond formation, i.e, 2 RS• \(\rightarrow\) RS-SR. In the enzyme, RS• radicals are presumably constrained in such a way that they cannot couple to form disulfide bonds. The other heme center, cytochrome a , does bind ligands such as cyanide to the Fe form and carbon monoxide to the Fe form, indicating that it is either five-coordinate or that it has a readily displaceable ligand. Reaction with CO, for example, produces spectral changes characteristic of a five-coordinate ferrous heme binding CO to give the six-coordinate carbonmonoxy product analogous to MbCO. The cytochrome a site is therefore an excellent candidate for O binding within cytochrome oxidase. The EPR spectrum of fully oxidized cytochrome c oxidase might be expected to give signals corresponding to two Cu centers and two ferric heme centers. In fact, all that is observed in the EPR spectrum of the oxidized enzyme is the typical low-spin six-coordinate ferric heme spectrum due to cytochrome G and the EPR signal attributed to Cu (see Figure 5.4A). The fact that signals attributable to cytochrome a and Cu are not observed in the EPR spectrum led to the suggestion that these two metal centers are antiferromagnetically coupled. The measured magnetic susceptibility for the isolated enzyme was found to be consistent with this hypothesis, suggesting that these two metal centers consist of an S = \(\frac{1}{2}\) Cu antiferromagnetically coupled through a bridging ligand to a high-spin S = \(\frac{5}{2}\) Fe to give an S = 2 binuclear unit. EXAFS measurements indicating a copper-iron separation of 3-4 Å as well as the strength of the magnetic coupling suggest that the metal ions are linked by a single-atom ligand bridge, but there is no general agreement as to the identity of this bridge. The cytochrome a -Cu coupling can be disrupted by reduction of the individual metal centers. In this fashion, a g = 6 ESR signal can be seen for cytochrome a or g = 2.053, 2.109, and 2.278 signals for Cu . Nitric-oxide binding to Cu also decouples the metals, allowing the g = 6 signal to be seen (see Figure 5.4B). Mössbauer spectroscopy also indicates that cytochrome a is high-spin in the oxidized as well as the reduced state. ENDOR studies suggest that Cu has three nitrogens from imidazoles bound to it with water or hydroxide as a fourth ligand. Studies using N-labeled histidine in yeast have demonstrated that histidine is a ligand to cytochrome a . All of these features have been incorporated into Figure 5.3. Before we consider the reactions of cytochrome c oxidase with dioxygen, it is instructive to review the reactions of dioxygen with iron porphyrins and copper complexes. Dioxygen reacts with ferrous-porphyrin complexes to make mononuclear dioxygen complexes (Reaction 5.36; see preceding chapter for discussion of this important reaction). Such dioxygen complexes react rapidly with another ferrous porphyrin, unless sterically prevented from doing so, to form binuclear peroxo-bridged complexes (Reaction 5.37). These peroxo complexes are stable at low temperature, but, when the temperature is raised, the O—O bond cleaves and two equivalents of an iron(IV) oxo complex are formed (Reaction 5.38). Subsequent reactions between the peroxo-bridged complex and the Fe oxo complex produce the\(\mu\)-oxo dimer (see Reactions 5.39-5.40). \[3Fe^{II}(P) + 3O_{2} \rightarrow 3Fe(P)(O_{2}) \tag{5.36}\] \[3Fe(P)(O_{2}) + 3Fe^{II}(P) \rightarrow 3(P)Fe^{III}—O—O—Fe^{III}(P) \tag{5.37}\] \[(P)Fe^{III}—O—O—Fe^{III}(P) \rightarrow 2Fe^{IV}(P)(O) \tag{5.38}\] \[2Fe^{IV}(P)(O) + 2(P)Fe^{III}—O—O—Fe^{III}(P) \rightarrow 2(P)Fe^{III}—O—Fe^{III}(P) + 2Fe(P)(O_{2}) \tag{5.39}\] \[2Fe(P)(O_{2}) \rightarrow 2 Fe^{II}(P) + 2O_{2} \tag{5.40}\] \[4Fe^{II}(P) + O_{2} \rightarrow 2(P)Fe^{III}—O—Fe^{III}(P) \tag{5.41}\] The reaction sequence (5.36) to (5.40) thus describes a four-electron reduction of O in which the final products, two oxide, O , ligands act as bridging ligands in binuclear ferric-porphyrin complexes (Reaction 5.41). Copper(l) complexes similarly react with dioxygen to form peroxo-bridged binuclear complexes. Such complexes do not readily undergo O—O bond cleavage, apparently because the copper(III) oxidation state is not as readily attainable as the Fe(IV) oxidation state in an iron-porphyrin complex. Nevertheless, stable peroxo complexes of copper(II) have been difficult to obtain, because, as soon as it is formed, the peroxo complex either is protonated to give free hydrogen peroxide or is itself reduced by more copper(l) (Reactions 5.42 to 5.46). \[2Cu^{I} + O_{2} \rightarrow Cu^{II}—O—O—Cu^{II} \tag{5.42}\] \[Cu^{II}—O—O—Cu^{II} + 2H^{+} \rightarrow 2 Cu^{II} + H_{2}O_{2} \tag{5.43}\] \[2Cu^{I} + H_{2}O_{2} + 2H^{+} \rightarrow 2 Cu^{II} + 2 H_{2}O \tag{5.44}\] or $$Cu^{II}—O—O—Cu^{II} + 2 Cu^{I} + 4 H^{+} \rightarrow 4 Cu^{II} + 2 H_{2}O \tag{5.45}\] \[4 Cu^{I} + O_{2} + 4 H^{+} \rightarrow 4 Cu^{II} + 2 H_{2}O \tag{5.46}\] Recently, however, examples of the long-sought stable binuclear copper(II) peroxo complex have been successfully synthesized and characterized, and interestingly enough, two entirely different structural types have been identified, i.e., \(\mu\)-1,2 and \(\mu\)-\(\eta^{2}\):\(\eta^{2}\) dioxygen complexes (see 5.47) \(\tag{5.47}\) A single turnover in the reaction of cytochrome c oxidase involves (1) reduction of the four metal centers by four equivalents of reduced cytochrome c, (2) binding of dioxygen to the partially or fully reduced enzyme, (3) transfer of four electrons to dioxygen, coupled with (4) protonation by four equivalents of protons to produce two equivalents of water, all without the leakage of any substantial amount of potentially harmful partially reduced dioxygen byproducts such as superoxide or hydrogen peroxide. At low temperatures, the reaction can be slowed down, so that the individual steps in the dioxygen reduction can be observed. Such experiments are carried out using the fully reduced enzyme to which CO has been bound. Binding of CO to the Fe heme center in reduced cytochrome c oxidase inhibits the enzyme and makes it unreactive to dioxygen. The CO-inhibited derivative can then be mixed with dioxygen and the mixture cooled. Photolysis of metal-CO complexes almost always leads to dissociation of CO, and CO-inhibited cytochrome c oxidase is no exception. Photolytic dissociation of CO frees the Fe heme, thereby initiating the reaction with dioxygen, which can then be followed spectroscopically. Dioxygen reacts very rapidly with the fully reduced enzyme to produce a species that appears to be the dioxygen adduct of cytochrome a (Reaction 5.48). Such a species is presumed to be similar to other mononuclear oxyheme derivatives. The dioxygen ligand in this species is then rapidly reduced to peroxide by the nearby Cu , forming what is believed to be a binuclear \(\mu\)-peroxo species (Reaction 5.49). These steps represent a two-electron reduction of dioxygen to the peroxide level, and are entirely analogous to the model reactions discussed above (Reactions 5.36 to 5.46), except that the binuclear intermediates contain one copper and one heme iron. The \(\mu\)-peroxo Fe - (O ) - Cu species is then reduced by a third electron, resulting in cleavage of the O—O bond (Reaction 5.50). One of the oxygen atoms remains with iron in the form of a ferryl complex, i. e., an Fe oxo, and the other is protonated and bound to copper in the form of a Cu aquo complex. Reduction by another electron leads to hydroxo complexes of both the Fe heme and the Cu centers (Reaction 5.51). Protonation then causes dissociation of two water molecules from the oxidized cytochrome a -Cu center (Reaction 5.52). \[(cyt\; a_{3})\overbrace{Fe^{II} \quad Cu_{B}^{I}} + O_{2} \rightarrow (cyt\; a_{3})\overbrace{Fe^{III}(O_{2}^{-}) \quad Cu_{B}^{I}} \tag{5.48}\] \[(cyt\; a_{3})\overbrace{Fe^{III}(O_{2}^{-}) \quad Cu_{B}^{I}} \rightarrow (cyt\; a_{3})\overbrace{Fe^{III}-(O_{2}^{2-}) \quad Cu_{B}^{II}} \tag{5.49}\] \[(cyt\; a_{3})\overbrace{Fe^{III}-(O_{2}^{2-})-Cu_{B}^{II}} + e^{-} + 2H^{+} \rightarrow (cyt\; a_{3})\overbrace{Fe^{IV}=O \quad H_{2}O-Cu_{B}^{II}} \tag{5.50}\] \[(cyt\; a_{3})\overbrace{Fe^{IV}=O \quad H_{2}O-Cu_{B}^{II}} +e^{-} \rightarrow (cyt\; a_{3})\overbrace{Fe^{III}-(OH^{-}) \quad (HO^{-})-Cu_{B}^{II}} \tag{5.51}\] \[(cyt\; a_{3})\overbrace{Fe^{III}-(OH^{-}) \quad (HO^{-})-Cu_{B}^{II}} + 2H^{+} \rightarrow (cyt\; a_{3})\overbrace{Fe^{III} \quad Cu_{B}^{II}} + 2 H_{2}O \tag{5.52}\] Several important questions remain to be resolved in cytochrome c oxidase research. One is the nature of the ligand bridge that links cytochrome a and Cu in the oxidized enzyme. Several hypotheses have been advanced (imidazolate, thiolate sulfur, and various oxygen ligands), but then discarded or disputed, and there is consequently no general agreement concerning its identity. However, EXAFS measurements of metal-metal separation and the strength of the magnetic coupling between the two metal centers provide evidence that a single atom bridges the two metals. Another issue, which is of great importance, is to find out how the energy released in the reduction of dioxygen is coupled to the synthesis of ATP. It is known that this occurs by coupling the electron-transfer steps to a proton-pumping process, but the molecular mechanism is unknown. Future research should provide some interesting insights into the mechanism of this still mysterious process. | 16,986 | 2,314 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/03%3A_Diffusion/13%3A_Friction_and_the_Langevin_Equation/13.01%3A_Langevin_Equation |
Consider the forces acting on a particle as we pull it through a fluid. We pull the particle with an external force \(f_{ext}\), which is opposed by a drag force from the fluid, \(f_d\). The drag or damping acts as resistance to motion of the particle, which results from trying to move the fluid out of the way. \[f_d = -\zeta v \qquad \qquad \quad \zeta \mathrm{(kg/s)} \] A drag force requires movement, so it is proportional to the velocity of the particle \(v=dx/dt=x\) and the friction coefficient \(\zeta\) is the proportionality constant that describes the magnitude of the damping. Newton’s second law relates the acceleration of this particle is to the sum of these forces: \[ ma = f_d+f_{ext} \nonumber \] Now microscopically, we also recognize that there are time-dependent random forces that the molecules of the fluid exert on a molecule (\(f_r\)). So that the specific molecular details of solute–solvent collisions can be averaged over, it is useful to think about a nanoscale solute in water (e.g., biological macromolecules) with dimensions large enough that its position is simultaneously influenced by many solvent molecules, but is also light enough that the constant interactions with the solvent leave an unbalanced force acting on the solute at any moment in time: \[\overline{f}_r(t)=-\sum_i \overline{f}_i(t) \nonumber\] Then Newton’s second law is \[ ma = f_d + f_{ext} +f_r(t) \nonumber \] The drag force is present regardless of whether an external force is present, so in the absence of external forces (\(f_{ext}=0\)) the equation of motion governing the spontaneous fluctuations of this solute is determined from the forces due to drag and the random fluctuations: \[ma = f_d + f_r(t) \label{13.1.1}\] \[m\ddot{x} + \zeta \dot{x} - f_r(t) = 0 \label{13.1.2}\] This equation of motion is the . An equation of motion such as this that includes a time-dependent random force is known as “ ”. Inserting a random process into a deterministic equation means that we need to use a statistical approach to solve this equation. We will be looking to describe the average and root-mean-squared position of the particle. First, what can we say about the random force? Although there may be momentary imbalances, on average the perturbations from the solvent on a larger particle will average to zero at equilibrium: \[ \langle f_r(t) \rangle = 0 \] Equation \ref{13.1.1} seems to imply that the drag force and the random force are independent, but in fact they originate in the same molecular forces. If the molecule of interest is a protein that experiences the fluctuations of many rapidly moving solvent molecules, then the averaged forces due to random fluctuations and the drag forces are related. The is the general relationship that relates the friction to the correlation function for the random force. In the Markovian limit this is \[ \langle f_r(t) f_r(t') \rangle = 2\zeta k_BT \delta (t-t')\] or \[ \zeta = \dfrac{\langle f_r^2 \rangle }{2k_BT} \nonumber \] Markovian indicates that no correlation exists between the random force for |t‒t′| > 0. More generally, we can recover the friction coefficient from the integral over the correlation function for the random force \[ \zeta = \dfrac{1}{2k_BT} \int^{+\infty}_{-\infty} dt \langle f_R(0)f_R(t) \rangle \nonumber \] To describe the time evolution of the position of our protein molecule, we would like to obtain an expression for mean-square displacement ⟨x (t)⟩. The position of the molecule can be described by integrating over its time-dependent velocity: \(x(t)=\int^t_0 dt' \dot{x}(t')\), so we can express the mean-square displacement in terms of the velocity autocorrelation function \[\langle x^2(t) \rangle = \int^t_0 dt' \int^t_0 dt'' \langle \dot{x}(t') \dot{x}(t'') \rangle \] Our approach to obtaining ⟨x (t)⟩ starts by multiplying eq. (13.1.2) by x and then ensemble averaging. \[ m \langle x \dfrac{d}{dt} \dot{x} \rangle + \zeta \langle x \dot{x} \rangle - \langle xf_r(t) \rangle =0 \] From eq. (13.1.3), the last term is zero, and from the chain rule we know \[ \dfrac{d}{dt} (x\dot{x}) = x\dfrac{d}{dt} \dot{x} + \dfrac{dx}{dt}\dot{x} \] Therefore, we can write eq. (13.1.6) as \[ m \left( \dfrac{d}{dt} \langle x\dot{x} \rangle - \langle \dot{x} \dot{x} \rangle \right) + \zeta \langle x \dot{x} \rangle = 0 \] Further, the equipartition theorem states that for each translational degree of freedom the kinetic energy is partitioned as \[ \dfrac{1}{2} m \langle \dot{x}^2 \rangle = \dfrac{k_BT}{2} \] So, \[m \dfrac{d}{dt} \langle x\dot{x} \rangle + \zeta \langle x \dot{x} \rangle = k_BT \] Here we are describing motion in 1D, but when fluctuations and displacement are included for 3D motion, then we switch x → r and k T→3k T. Integrating eq. (13.1.10) twice with respect to time, and using the initial condition x(0) = 0, we obtain \[ \langle x^2 \rangle = \dfrac{2k_BT}{\zeta } \left\{ t +\dfrac{m}{\zeta } \left[ \exp \left( -\dfrac{\zeta }{m}t \right) -1 \right] \right\} \] in 3D: \[\langle r^2 \rangle = \dfrac{6k_BT}{\zeta } \left\{ t +\dfrac{m}{\zeta } \left[ \exp \left( -\dfrac{\zeta }{m}t \right) -1 \right] \right\} \nonumber \] To investigate eq. (13.1.11), let’s consider two limiting cases. We see that m/ζhas units of time, and so we define the relaxation time \[ \tau_C = m/ \zeta \] and investigate time scale short and long compared to τ : 1) For \(t \ll \tau_C \), we can expand the exponential in eq. (11) and retain the first three terms, which leads to \[ \langle x^2 \rangle \approx \dfrac{k_BT}{m}t^2 = \langle v^2 \rangle t^2 \qquad \qquad \quad \text{(short time: inertial)}\] 2) For \(t \gg \tau_C\), eq. (11) is dominated by the leading term: \[ \langle x^2 \rangle = \dfrac{2k_BT}{\zeta} t \qquad \qquad \quad \text{(long time: diffusive)} \] In the diffusive limit the behavior of the molecule is governed entirely by the fluid, and its mass does not matter. The diffusive limit in a stochastic equation of motion is equivalent to setting m→0. We see that τ is a time-scale separating motion in the inertial and diffusive limits. It is a correlation time for the randomization of the velocity of the particle due to the random fluctuations of the environment. For very little friction or short time, the particle moves with traditional deterministic motion x = v t, where root-mean-square displacement x = ⟨x ⟩ and v comes from the average translational kinetic energy of the particle. For high-friction or long times, we see diffusive behavior with x ~t . Furthermore, by comparing eq. (13.1.14) to our earlier continuum result, ⟨x ⟩ = 2Dt, we see that the diffusion constant can be related to the friction coefficient by \[ D= \dfrac{k_BT}{\zeta} \quad \text{(in 1D)} \] This is the Einstein formula. For 3D problems, we replace k T with 3k T in the expressions above and find \( D_{3D}=3k_BT /\zeta\). \[ \tau_C = \dfrac{m}{\zeta} = \dfrac{mD}{k_BT} \quad \text{(in 1D)} \nonumber \] How long does it take to approach the diffusive regime? Very fast. Consider a 100 kDa protein with R = 3 nm in water at T = 300 K, we find a characteristic correlation time for randomizing velocities of \(\tau_C=3\mathrm{x}10^{-12}\) s, which corresponds to a distance of about 10 nm before the onset of diffusive behavior. We can find other relationships. Noting the relationship of ⟨x ⟩ to the velocity autocorrelation function in eq. (13.1.5), we find that the particle velocity is described by \[ \langle v_x(0)v_x(t) \rangle = \langle v_x^2\rangle e^{-\zeta t /m}= \langle v^2_x \rangle e^{-t/\tau_C} \qquad \qquad \qquad v_x = \dot{x} \nonumber \] which can be integrated over time to obtain the diffusion constant. \[ \int^{\infty}_0 \langle v_x(0)v_x(t) \rangle dt = \dfrac{k_BT}{\zeta} = D \] This expression is the Green–Kubo relationship. This is a practical way of analyzing molecular trajectories in simulations or using particle-tracking experiments to quantify diffusion constants or friction coefficients. | 7,987 | 2,315 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/01%3A_General_Techniques/1.04%3A_Heating_and_Cooling_Methods/1.4C%3A_Adjustable_Platforms |
It is quite useful for an apparatus to rest on a platform that can be adjusted up or down. For example, when heating a round bottomed flask in a distillation or reflux, the heat source should be held in such a way that it can be easily lowered and removed from the flask. This is an important safety measure, as it allows for adjustment if the system heats too rapidly (as is evidenced by bumping and foaming), or if anything unexpected occurs (smoking or charring). Removal of the heat source is also of course necessary at the end of a process, and it is best if the heat can be removed while leaving the apparatus intact to cool. Adjustable platforms come in many forms. A lab jack (Figure 1.43a) is the easiest to manipulate, and can be adjusted up or down by turning the knob. Unfortunately, lab jacks are expensive so are likely to be used in research settings but not in teaching labs. A simple platform can be made from anything stackable, such as wood blocks or KimWipe boxes (Figure 1.43b), although at some height they can be easily tipped. A more secure platform can be created by placing a wire mesh atop a ring clamp (Figure 1.43c). Adjustable platforms should be used underneath any flask that is clamped in an apparatus well above the benchtop and contains chemicals, especially if they are to be heated or are extremely reactive. If a clamp were to fail for some reason, a platform is a fail-safe and prevents hot or reactive chemicals from falling onto a heat source or splashing on the benchtop where they may become a hazard. Figure 1.43d shows withdrawal of an air-sensitive reagent by syringe, and the reagent bottle is secured by a clamp and supported with a ring clamp/wire mesh platform. If the reagent were to slip from the grip of the clamp, the platform ensures that the hazardous material will not fall. | 1,844 | 2,316 |
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The Langevin equation for the motion of a Brownian particle can be modified to account for an additional external force, in addition to the drag force and random force. From Newton’s Second Law: \[ m \ddot{x} = f_d + f_r(t)+ f_{ext}(t) \nonumber \] where the added force is obtained from the gradient of the potential it experiences: \[ f_{ext} = -\dfrac{\partial U}{\partial x} \] With the fluctuation-dissipation relation \( \langle f_r(t)f_r(t')\rangle = 2\zeta k_BT \delta (t-t')\), the Langevin equation becomes \[ m\ddot{x} + (\partial U/\partial x)+\zeta \dot{x} - \sqrt{2\zeta k_BT } R(t)=0 \] Here \(R(t)\) refers to a Gaussian distributed sequence of random numbers with \(⟨R(t)⟩ = 0\) and \(⟨R(t) R(t′)⟩ = δ(t ‒ t′)\). Brownian dynamics simulations are performed using this equation of motion in the diffusion-dominated, or strong friction limit \( |m\ddot{x}|\ll |\zeta \dot{x}|\). Then, we can neglect inertial motion, and set the acceleration of the particle to zero to obtain an expression for the velocity of the particle \[\dot{x} (t) = \dfrac{\dfrac{\partial U}{\partial x}}{\zeta} -\sqrt{2k_BT/\zeta} R(t) \nonumber \] We then integrate this equation of motion in the presence of random perturbations to determine the dynamics \(x(t)\). | 1,267 | 2,317 |
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Visible light, gamma rays, x-rays, ultraviolet (black) light, infrared radiation, microwaves, and radio waves are all related. Many of their properties can be explained by a wave theory: periodically varying electric and magnetic fields (electromagnet waves). Figure \(\Page {1}\) indicates the relationship of these fluctuating electric and magnetic fields. It also illustrates the maximum A and the λ. The intensity of a wave is associated with the square of its amplitude. Electromagnetic waves travel through a vacuum at the speed of light, c = 2.9979 × 10 m s . The entire wave shown in Figure 21.1 may be thought of as moving from left to right. Thus at position , where the electric field had maximum amplitude at the instant the figure was drawn there is a progressive decrease in amplitude with time. The amplitude reaches its smallest (most negative) value when the wave has moved a distance equal to that separating points ′ and . Eventually the amplitude increases to its maximum value again, corresponding to movement of a distance ″ to , or one wavelength λ. A moving wave can be characterized by the at which points of maximum amplitude pass a fixed position. The speed of the wave (distance traveled per unit time) must be the product of the wavelength (distance between maxima) and the frequency (number of maxima passing per unit time): \[ c= \lambda \upsilon \label{1} \] Since the speed of electromagnetic radiation in a vacuum is always the same, radiation may be characterized by specifying either λ or . The other quantity can always be calculated from Equation \(\ref{1}\). Specify the frequency, wavelength, and speed in a vacuum of each of the types of electromagnetic radiation listed below: In each case we make use of the relationship = λ = 2.998 × 10 m s . The results obtained in Example \(\Page {1}\) indicate that the frequency and wavelength of electromagnetic radiation can vary over a wide range. The experiments which did most to convince scientists that light could be described by a wave model are concerned with interference. In 1802 Thomas Young (1773 to 1829), an English physicist, allowed light of a single wavelength to pass through a pair of parallel slits very close to each other and then onto a screen. Young observed the interference pattern of alternating dark and bright strips, shown in Figure \(\Page {2}\). Instead of two strips of light, three appeared on the screen, the most prominent being in the center. The appearance of these bright and dark strips on the screen is easy to explain if light is regarded as a wave. The bright areas are the result of , while the dark ones result from . Constructive interference occurs when the crests of two waves reach the same point at the same time. The amplitudes of the two waves add together, giving a resultant larger than either. In the case of destructive interference a maximum in one wave and a minimum in the other reach the same point at the same time. Thus one cancels the effect of the other, and the resultant wave is smaller. This is illustrated in Figure 3. When destructive interference occurs between two waves which have the amplitude, the resultant wave has zero amplitude (and zero intensity). Hence the dark strips observed in the double-slit experiment. Although the behavior of light and other forms of electromagnetic radiation can usually be interpreted in terms of wave motion, this is not always so. When radiation is absorbed or emitted by matter, it is usually more convenient to regard it as a stream of particles called photons. Thus electromagnetic radiation has the same kind of we encountered in the case of . However, photons have some properties which are very different from those of electrons and other particles. Although photons have mass and energy, and although we can count them, they can travel . We cannot slow down a photon or stop it without changing it into something else. The wave-particle duality of photons and electromagnetic radiation is enshrined in an equation first proposed by the German physicist Max Planck (1858 to 1947). The energy of a photon and the frequency of the electromagnetic radiation associated with it are related in the following way: \[E=h \upsilon \label{2} \] where is a universal constant of nature called with the value 6.6262 × 10 J s. The application of Equation \(\ref{2}\) is best shown by an example. Calculate the energy of photons associated with each kind of electromagnetic radiation mentioned in Example 1. Compare each result with the mean bond enthalpy for a C—C single bond (348 kJ mol ). In each case use the formula = . If wavelengths are given instead of frequencies, the formula = /λ may be obtained by combining Equations \(\ref{1}\) and \(\ref{2}\). Since the quoted refers to 1 mol C—C bonds, we must divide by the Avogadro constant to obtain a quantity which is appropriate to compare with the energy of a single quantum of radiation: Enthalpy to dissociate one C—C bond is \[\dfrac{348kJmol^{-1}}{6.022\times10^{23}mol^{-1}} = 5.78 \times 10^{–19}\; \,J = 0.578\, aJ \nonumber \] Clearly the energies of visible photons are comparable with the energies of chemical bonds. Infrared and radio waves have far less energy per photon. Gamma-ray photons have enough energy to break open about 100 000 chemical bonds. As a consequence chemical changes often occur when gamma rays or other high-energy photons are absorbed by matter. Such changes are usually detrimental to living systems, and materials such as lead are used to shield humans from sources of high-energy radiation. The entire spectrum of electromagnetic radiation may be characterized in terms of wavelength, frequency, or energy per photon, as shown in Figure \(\Page {3}\). The example calculations we have done so far and the figure both indicate the broad range covered by λ, , and . Electromagnetic radiation which can be detected by the human retina is but a small slice out of the total available spectrum. | 6,033 | 2,318 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Substitution_Reactions/Carboxyl_Substitution/Comparative_Energies%3A_The_Ski_Hill |
In most cases, the reactivity of carboxyloids involves converting one carboxyloid into another. This is done by replacing one heteroatom substituent with another. For example, the chloride in an acid chloride may be replaced by an alcohol or alkoxide ion to make an ester. The fact that one carboxyloid can be converted into another suggests that there would be an equilibrium between them. The ratio of two carboxyloids at equilibrium would be determined by their relative stability, as well as the stability of other associated species in solution. We can map out the stability of carboxyloids on a potential energy surface, as shown below. The higher energy, less stable, more reactive carboxyloids are shown at the top of the potential energy curve. The lower energy, more stable, less reactive carboxyloids are found lower down on the potential energy curve. A potential energy curve showing relative reactivity of carboxloids. The heteroatom attached to the carbonyl in a carboxyloid is always an electronegative atom with a lone pair. Either of those two features might be useful in understanding the reactivity trend illustrated above. For example, an electronegative atom would make the carbonyl carbon more positive. That carbon is already very positive because of the double bond to oxygen. Adding an additional electronegative atom should make it even more so. The amount of positive charge on the carbonyl carbon would be even greater if the atom attached to it were exceptionally electronegative. On the other hand, a nearby lone pair might counteract the electron-attracting power of the carbonyl carbon. In a sense, we might think about that lone pair as competing with donation from a potential nucleophile. The ability of an atom to π-donate, then, might have an influence on how strongly the carbonyl will attract nucleophiles. Of course, there is some trade-off involved in π-donation. Usually the atom that donates must take on a positive charge, since it is lending a pair of its own electrons to another atom. Factors that influence how easily this may happen could be important in determining carboxyloid reactivity. Based on electronegativity of the atom attached to the carbonyl carbon, we might expect a specific trend in carboxyloid reactivity. Explain how this factor would affect electrophilicity at the carbonyl carbon and predict the corresponding trend in reactivity. Compare this trend with the information in Figure CX3.1. Lone pair donation from the atom attached to the carbonyl carbon could also influence carboxyloid reactivity. Explain how this factor would affect electrophilicity at the carbonyl carbon and predict the corresponding trend in reactivity. Compare this trend with the information in Figure CX3.1. Using the information in Figure CX3.1, explain why peptides (containing a number of amide bonds, R(C=O)N) are such a common structural feature in biology. The potential energy curve in Figure CX3.1 is a useful index for the interconversion of carboxyloids. In general, it is easy to go downhill on the curve, but more difficult to go uphill. That means that compounds lower down on the ski hill can be made easily from compounds farther up the ski hill. In general, pi donation from the heteroatom attached to the carbonyl is a primary factor that determines carboxyloid reactivity. The more able the heteroatom is to donate its pi electrons, the less electrophilic is the carbonyl. Nitrogen is very good at donating its lone pair. It is about the same size as the carbon atom it needs to donate to, and it only a little more electronegative than the carbon. Oxygen (in esters and carboxylic acids) is next in line, since oxygen is more electronegative than nitrogen. Chlorine and sulfur are a little too large to donate very well to a carbon atom. The size and energy mismatch between these atoms leads to poor pi bonding, and poor pi donation. , | 3,914 | 2,319 |
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The results of a scientific experiment must be communicated to be of value. This affords an opportunity for other scientists to check them. It also allows the scientific community, and sometimes the general public, to share new knowledge. Communication, however, is not always as straightforward as it might seem. Ambiguous terminology can often turn a seemingly clear statement into a morass of misunderstanding. As an example, consider someone tells you that the forecast is a high of 25°. Do you put on a winter jacket, or summer wear? If you are thinking winter, then you interpreted the temperature as 25°F. However if 25°C was meant, which is equal to 77°F, a winter jacket would be far too warm. Or consider filling up a car with gasoline. If you are in the US, you will be dealing in dollars per gallon, whereas, if you were in continental Europe, you will be dealing in Euros per liter. Given that the exchange rate from USD to Euros fluctuates and that there are roughly 3.79 liters in 1 gallon, it is difficult to simply compare numbers between gas prices in the USA, and say, France, if you don't know what units you are using. As a final example, consider the speed 24 meters/sec. Do you interpret this as close to highway speed limits in the US? It is the same speed as 53.7 miles per hour, likely a more familiar set of units for speed for people in the US. While some of these examples may seem tired and overdone, they still showcase that even if you know unit conversions well, the many systems for measuring quantities can make things complex and confusing. Even rocket scientists get it wrong sometimes, as was seen in 1999 when a orbiter sent to Mars was unable propel itself correctly into orbit because one team had used metric system measurements, whereas another had used English system measurements. If in the midst of this hodgepodge, you asked, “Would it not be easier to have a single unit for mass, a single unit for volume, and express all masses or volumes in these units,” you would not be the first person to have such an idea. The main difficulty is that it is hard to get everyone to agree on a single consistent set of units. Some units are especially convenient for some tasks. For example, a yard was originally defined as the distance from a man’s nose to the end of his thumb when his arm was held horizontally to one side. This made it easy to measure cloth or ribbon by holding one end to the nose and stretching an arm’s length with the other hand. Now that yardsticks, meter sticks, and other devices are readily available, the original utility of the yard is gone, but we still measure ribbon and cloth in that same unit. Many people would probably be distressed if a change were made. Scientists are not all that different from other people—they too have favorite units which are especially suited to certain areas of research. Nevertheless, scientists have constantly pressed for improvement and uniformity in systems of measurement. The first such action occurred nearly 200 years ago when, in the aftermath of the French Revolution, the spread over most of continental Europe and was adopted by scientists everywhere. The United States nearly followed suit, but in 1799 Thomas Jefferson was unsuccessful in persuading Congress that a system based on powers of 10 was far more convenient and would eventually become the standard of the world. The metric system has undergone continuous evolution and improvement since its original adoption by France. Beginning in 1899, a series of international conferences have been held for the purpose of redefining and regularizing the system of units. In 1960 the Eleventh Conference on Weights and Measures proposed major changes in the metric system and suggested a new name — the — for the revised metric system. (The abbreviation , from the French , is commonly used.) Scientific bodies such as the U.S. National Bureau of Standards and the International Union of Pure and Applied Chemistry have endorsed the SI. At the heart of the SI are the seven units, listed here. All other units are derived from these seven so-called . For example, units for area and volume may be derived by squaring or cubing the unit for length. Some of the base units are probably familiar to you, while others, such as the mole, candela, and kelvin, may be less so. Rather than defining each of them now, we shall wait until later chapters when the less familiar units, as well as the quantities they are used to measure, can be described in detail. The candela, which measures the intensity of light, is not used often by chemists, and so we shall pay no further attention to it. | 4,671 | 2,320 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Electrophilic_Addition_to_Alkenes/EA8._Solutions_for_Selected_Problems |
Two products are formed and they are enantiomers. They are diastereomers. One chiral center has the same configuration in both compounds but the others are opposite. The second bromine could occupy any of the secondary positions if there were a true carbocation. That doesn't happen; the second bromine occupies only the position next to the other bromine. The nucleophile in the second step changes under different conditions. Crowding is more severe in the structure on the left than in the structure on the right. The structure on the right, representing an approach to the transition state of the reaction, is more favourable than the other one. , | 664 | 2,321 |
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Compounds that are capable of accepting electrons, such as O or F , are called because they can oxidize other compounds. In the process of accepting electrons, an oxidant is reduced. Compounds that are capable of donating electrons, such as sodium metal or cyclohexane (C H ), are called because they can cause the reduction of another compound. In the process of donating electrons, a reductant is oxidized. These relationships are summarized in : oxidant + reductant → oxidation−reduction \[ \underset {gains\, e \, (is \, reduced)}{O_2 (g)} + \underset {loses \, e \, (is \, oxidized)}{4Na (s)} \rightarrow 2 \underset {redox \, reaction}{Na_2O (s)} \tag{3.30}\] Some oxidants have a greater ability than others to remove electrons from other compounds. Oxidants can range from very powerful, capable of oxidizing most compounds with which they come in contact, to rather weak. Both F and Cl are powerful oxidants: for example, F will oxidize H O in a vigorous, potentially explosive reaction. In contrast, S is a rather weak oxidant, and O falls somewhere in between. Conversely, reductants vary in their tendency to donate electrons to other compounds. Reductants can also range from very powerful, capable of giving up electrons to almost anything, to weak. The alkali metals are powerful reductants, so they must be kept away from atmospheric oxygen to avoid a potentially hazardous redox reaction. A , first introduced in , is an oxidation–reduction reaction in which the oxidant is O . One example of a combustion reaction is the burning of a candle, shown in . Consider, for example, the combustion of cyclohexane, a typical hydrocarbon, in excess oxygen. The balanced chemical equation for the reaction, with the oxidation state shown for each atom, is as follows: \[ \underset {-2}{C_6} \overset {+1}{H_{12}} + 9 \overset {0}{O_2} \rightarrow 6 \overset {+4}{C} \underset {-2}{O_2} + 6 \overset {+1}{H_2} \underset {-2}{O} \tag{3.31}\] If we compare the oxidation state of each element in the products and the reactants, we see that hydrogen is the only element whose oxidation state does not change; it remains +1. Carbon, however, has an oxidation state of −2 in cyclohexane and +4 in CO ; that is, each carbon atom changes its oxidation state by six electrons during the reaction. Oxygen has an oxidation state of 0 in the reactants, but it gains electrons to have an oxidation state of −2 in CO and H O. Because carbon has been oxidized, cyclohexane is the reductant; because oxygen has been reduced, it is the oxidant. All combustion reactions are therefore oxidation–reduction reactions. We described the defining characteristics of oxidation–reduction, or redox, reactions in . Most of the reactions we considered there were relatively simple, and balancing them was straightforward. When oxidation–reduction reactions occur in aqueous solution, however, the equations are more complex and can be more difficult to balance by inspection. Because a balanced chemical equation is the most important prerequisite for solving any stoichiometry problem, we need a method for balancing oxidation–reduction reactions in aqueous solution that is generally applicable. One such method uses , and a second is referred to as the method. We show you how to balance redox equations using oxidation states in this section; the half-reaction method is discussed in the second semester Species in high oxidation states act as oxidants, whereas species in low oxidation states act as reductants. When an aqueous solution of a compound that contains an element in a high oxidation state is mixed with an aqueous solution of a compound that contains an element in a low oxidation state, an oxidation–reduction reaction is likely to occur. | 3,777 | 2,323 |
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Your ability to interpret the numerical value of a quantity in terms of what it means in a practical sense is an essential part of developing a working understanding of Chemistry. This is particularly the case for equilibrium constants, whose values span the entire range of the positive numbers. Although there is no explicit rule, for most practical purposes you can say that equilibrium constants within the range of roughly 0.01 to 100 indicate that a chemically significant amount of all components of the reaction system will be present in an equilibrium mixture and that the reaction will be or “ ”. As an equilibrium constant approaches the limits of zero or infinity, the reaction can be increasingly characterized as a one-way process; we say it is “ ” or “ ”. The latter term must of course not be taken literally; the Le Chatelier principle still applies (especially insofar as temperature is concerned), but addition or removal of reactants or products will have less effect. Although it is by no means a general rule, it frequently happens that reactions having very large equilibrium constants are kinetically hindered, often to the extent that the reaction essentially does not take place. The examples in the following table are intended to show that numbers (values of ), no matter how dull they may look, do have practical consequences! The equilibrium expression for the synthesis of ammonia \[\ce{ 3 H2(g) + N2(g) -> 2 NH3(g)} \label{15.4.1}\] can be expressed as \[ K_p =\dfrac{P^2_{NH_3}}{P_{N_2}P^3_{H_2}} \label{15.4.2}\] or \[ K_c = \dfrac{[NH_3]^2}{[N_2] [H_2]^3} \label{15.4.3}\] so \(K_p\) for this process would appear to have units of atm , and \(K_c\) would be expressed in mol L . And yet these quantities are often represented as being dimensionless. Which is correct? The answer is that both forms are acceptable. There are some situations (which you will encounter later) in which ’s must be considered dimensionless, but in simply quoting the value of an equilibrium constant it is permissible to include the units, and this may even be useful in order to remove any doubt about the units of the individual terms in equilibrium expressions containing both pressure and concentration terms. In carrying out your own calculations, however, there is rarely any real need to show the units. ke \(\ce{CaF(s)}\), t is \([\ce{CaF2}]/[\ce{CaF2}]\) wh The magnitude of the equilibrium constant, \(K\), indicates the extent to which a reaction will proceed: Knowing the value of the equilibrium constant, \(K\), will allow us to determine: (1) he direction a reaction will proceed to achieve equilibrium and (2) the ratios of the concentrations of reactants and products when equilibrium is reached ( ) ) | 2,758 | 2,324 |
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A convenient way for chemists to report the results of a TLC plate in lab notebooks is through a " ",\(^2\) or , which quantitates a compound's movement (Equation \ref{2}). \[R_f = \dfrac{\text{distance traveled by the compound}}{\text{distance traveled by the solvent front}} \label{2}\] To measure how far a compound traveled, the distance is measured from the compound's original location (the baseline marked with pencil) to the compound's location after elution (the approximate middle of the spot, Figure 2.14a). Due to the approximate nature of this measurement, ruler values should be recorded only to the nearest millimeter. To measure how far the solvent traveled, the distance is measured from the baseline to the solvent front. The solvent front (Figure 2.14b) is essential to this \(R_f\) calculation. When removing a TLC plate from its chamber, the solvent front needs to be marked with pencil, as the solvent will often evaporate rapidly. The \(R_f\) value is a ratio, and it represents the relative distance the spot traveled compared to the distance it could have traveled if it moved with the solvent front. An \(R_f\) of 0.55 means the spot moved \(55\%\) as far as the solvent front, or a little more than halfway. Since an \(R_f\) is essentially a percentage, it is not particularly important to let a TLC run to any particular height on the TLC plate. In Figure 2.15, a sample of acetophenone was eluted to different heights, and the \(R_f\) was calculated in each case to be similar, although not . Slight variations in \(R_f\) arise from error associated with ruler measurements, but also different quantities of adsorbed water on the TLC plates that alter the properties of the adsorbent. \(R_f\) values should always be regarded as approximate. Although in theory a TLC can be run to any height, it's customary to let the solvent run approximately \(0.5 \: \text{cm}\) from the top of the plate to minimize error in the \(R_f\) calculations, and to achieve the best separation of mixtures. A TLC plate should not be allowed to run completely to the top of the plate as it may affect the results. However, if using a saturated, sealed TLC chamber, the \(R_f\) can still be calculated. \(^2\)Sometimes the \(R_f\) is called the retardation factor, as it is a measurement of how the movement of the spots is slowed, or retarded. | 2,367 | 2,325 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.04%3A_Solution_Formation_and_Equilibrium |
When a solute dissolves, its individual atoms, molecules, or ions interact with the solvent, become solvated, and are able to diffuse independently throughout the solution (part (a) in Figure \(\Page {1}\)). This is not, however, a unidirectional process. If the molecule or ion happens to collide with the surface of a particle of the undissolved solute, it may adhere to the particle in a process called crystallization. Dissolution and crystallization continue as long as excess solid is present, resulting in a dynamic equilibrium analogous to the equilibrium that maintains the vapor pressure of a liquid. We can represent these opposing processes as follows: \[ \text{solute} + \text{solvent} \ce{<=>[\ce{crystallization},\ce{dissolution}]} \text{solution} \label{13.4.1} \] Although the terms precipitation and crystallization are both used to describe the separation of solid solute from a solution, crystallization refers to the formation of a solid with a well-defined crystalline structure, whereas precipitation refers to the formation of any solid phase, often one with very small particles. Figure \(\Page {2}\) shows plots of the solubilities of several organic and inorganic compounds in water as a function of temperature. Although the solubility of a solid generally increases with increasing temperature, there is no simple relationship between the structure of a substance and the temperature dependence of its solubility. Many compounds (such as glucose and \(CH_3CO_2Na\)) exhibit a dramatic increase in solubility with increasing temperature. Others (such as \(NaCl\) and \(K_2SO_4\)) exhibit little variation, and still others (such as \(Li_2SO_4\)) become less soluble with increasing temperature. Notice in particular the curves for \(NH_4NO_3\) and \(CaCl_2\). The dissolution of ammonium nitrate in water is endothermic (\(ΔH_{soln} = +25.7\; kJ/mol\)), whereas the dissolution of calcium chloride is exothermic (\(ΔH_{soln} = −68.2 \;kJ/mol\)), yet Figure \(\Page {2}\) shows that the solubility of both compounds increases sharply with increasing temperature. In fact, the magnitudes of the changes in both enthalpy and entropy for dissolution are temperature dependent. Because the solubility of a compound is ultimately determined by relatively small differences between large numbers, there is generally no good way to predict how the solubility will vary with temperature. The variation of solubility with temperature has been measured for a wide range of compounds, and the results are published in many standard reference books. Chemists are often able to use this information to separate the components of a mixture by fractional crystallization the separation of compounds on the basis of their solubilities in a given solvent (Figure \(\Page {3}\)). For example, if we have a mixture of 150 g of sodium acetate (\(CH_3CO_2Na\)) and 50 g of \(KBr\), we can separate the two compounds by dissolving the mixture in 100 g of water at 80°C and then cooling the solution slowly to 0°C. According to the temperature curves in Figure \(\Page {2}\), both compounds dissolve in water at 80°C, and all 50 g of \(KBr\) remains in solution at 0°C. Only about 36 g of \(CH_3CO_2Na\) are soluble in 100 g of water at 0°C, however, so approximately 114 g (150 g − 36 g) of \(CH_3CO_2Na\) crystallizes out on cooling. The crystals can then be separated by filtration. Thus fractional crystallization allows us to recover about 75% of the original \(CH_3CO_2Na\) in essentially pure form in only one step. Fractional crystallization is a common technique for purifying compounds as diverse as those shown in Figure \(\Page {2}\) and from antibiotics to enzymes to drugs (Figure \(\Page {4}\)). For the technique to work properly, the compound of interest must be more soluble at high temperature than at low temperature, so that lowering the temperature causes it to crystallize out of solution. In addition, the impurities must be more soluble than the compound of interest (as was \(KBr\) in this example) and preferably present in relatively small amounts. When a solution contains the maximum amount of solute that can dissolve under a given set of conditions, it is a saturated solution. Otherwise, it is unsaturated. Supersaturated solutions, which contain more dissolved solute than allowed under particular conditions, are not stable; the addition of a seed crystal, a small particle of solute, will usually cause the excess solute to crystallize. A system in which crystallization and dissolution occur at the same rate is in dynamic equilibrium. The solubility of most substances depends strongly on the temperature. The solubility of most solid or liquid solutes increases with increasing temperature. The components of a mixture can often be separated using fractional crystallization, which separates compounds according to their solubilities. The solubility of a gas decreases with increasing temperature. | 4,959 | 2,326 |
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A Grignard reagent has a formula RMgX where X is a halogen, and R is an alkyl or aryl (based on a benzene ring) group. For the purposes of this page, we shall take R to be an alkyl group (e.g., CH CH MgBr). Everything must be perfectly dry because Grignard reagents react with water. Grignard These are reactions of the carbon-oxygen double bond, and so aldehydes and ketones react in exactly the same way - all that changes are the groups that happen to be attached to the carbon-oxygen double bond. It is much easier to understand what is going on by looking closely at the general case (using "R" groups rather than specific groups) - and then slotting in the various real groups as and when you need to. The "R" groups can be either hydrogen or alkyl in any combination. In the first stage, the Grignard reagent adds across the carbon-oxygen double bond: Dilute acid is then added to this to hydrolyse it. An alcohol is formed. One of the key uses of Grignard reagents is the ability to make complicated alcohols easily. What sort of alcohol you get depends on the carbonyl compound you started with - in other words, what R and R' are. Methanal is the simplest possible aldehyde with Assuming that you are starting with CH CH MgBr and using the general equation above, the alcohol you get always has the form: Since both R groups are hydrogen atoms, the final product will be: A primary alcohol is formed. A primary alcohol has only one alkyl group attached to the carbon atom with the -OH group on it. You could obviously get a different primary alcohol if you started from a different Grignard reagent. The next biggest aldehyde is ethanal with one of the R groups is hydrogen and the other CH . Again, think about how that relates to the general case. The alcohol formed is: So this time the final product has one CH group and one hydrogen attached: A secondary alcohol has two alkyl groups (the same or different) attached to the carbon with the -OH group on it. You could change the nature of the final secondary alcohol by either: Ketones have two alkyl groups attached to the carbon-oxygen double bond. The simplest one is propanone. This time when you replace the R groups in the general formula for the alcohol produced you get a tertiary alcohol. Jim Clark ( ) | 2,291 | 2,327 |
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Take some aluminum foil. Cut it in half. Now there are two smaller pieces of aluminum foil. Cut one of the pieces in half again. Cut one of those smaller pieces in half again. Continue cutting, making smaller and smaller pieces of aluminum foil. It should be obvious that the pieces are still aluminum foil; they are just becoming smaller and smaller. But how far can this exercise be taken, at least in theory? Can one continue cutting the aluminum foil into halves forever, making smaller and smaller pieces? Or is there some limit, some absolute smallest piece of aluminum foil? Thought experiments like this—and the conclusions based on them—were debated as far back as the fifth century BC. John Dalton (1766-1844) is the scientist credited for proposing the atomic theory. The theory explains several concepts that are relevant in the observable world: the composition of a pure gold necklace, what makes the pure gold necklace different than a pure silver necklace, and what occurs when pure gold is mixed with pure copper. This article explains the theories that Dalton used as a basis for his theory: "Nothing comes from nothing" is an important idea in ancient Greek philosophy that argues that what exists has always , since no new matter can come into existence where there was none before. Antoine Lavoisier (1743-1794) restated this principle for chemistry with the law of conservation of mass, which "means that the atoms of an object cannot be created or destroyed, but can be moved around and be changed into different particles." This law says that when a chemical reaction rearranges atoms into a new product, the mass of the reactants (chemicals before the chemical reaction) is the same as the mass of the products (the new chemicals made). More simply, whatever you do, you will still have the same amount of stuff (however, certain nuclear reactions like fusion and fission can convert a small part of the mass into energy. The law of conservation of mass states that the total mass present before a chemical reaction is the same as the total mass present after the chemical reaction; in other words, . The law of conservation of mass was formulated by Lavoisier as a result of his combustion experiment, in which he observed that the mass of his original substance—a glass vessel, tin, and air—was equal to the mass of the produced substance—the glass vessel, “tin calx”, and the remaining air. Historically, this was a difficult concept for scientists to grasp. If this law was true, then how could a large piece of wood be reduced to a small pile of ashes? The wood clearly has a greater mass than the ashes. From this observation scientists concluded that mass had been lost. However, Figure \(\Page {1}\) shows that the burning of word does follow the law of conservation of mass. Scientists did not account for the gases that play a critical role in this reaction. Joseph Proust (1754-1826) formulated the (also called the or ). This law states that if a compound is broken down into its constituent elements, the masses of the constituents will always have the same proportions, regardless of the quantity or source of the original substance. Joseph Proust based this law primarily on his experiments with basic copper carbonate. The illustration below depicts this law; 31 grams of H O and 8 grams of H O are made up of the same percent of hydrogen and oxygen. Oxygen makes up 88.8% of the mass of any sample of pure water, while hydrogen makes up the remaining 11.2% of the mass. You can get water by melting ice or snow, by condensing steam, from river, sea, pond, etc. It can be from different places: USA, UK, Australia, or anywhere. It can be made by chemical reactions like burning hydrogen in oxygen. However, if the water is , it will consist of 88.8 % oxygen by mass and 11.2 % hydrogen by mass, irrespective of its source or method of preparation. Many combinations of elements can react to form more than one compound. In such cases, this law states that the weights of one element that combine with a fixed weight of another of these elements are integer multiples of one another. It's easy to say this, but please make sure that you understand how it works. Nitrogen forms a very large number of oxides, five of which are shown here. The law of multiple proportions states that if two elements form more than one compound between them, the masses of one element combined with a fixed mass of the second element form in ratios of small integers. Consider two separate compounds are formed by only carbon and oxygen. The first compound contains 42.9% carbon and 57.1% oxygen (by mass) and the second compound contains 27.3% carbon and 72.7% oxygen (again by mass). Is this consistant with the law of multiple proportions? The states that the masses of one element which combine with a fixed mass of the second element are in a ratio of numbers. Hence, the masses of oxygen in the two compounds that combine with a fixed mass of carbon should be in a whole-number ratio. Thus for every 1 g of the first compound there are 0.57 g of oxygen and 0.429 g of carbon. The mass of oxygen per gram carbon is: \[ \dfrac{0.571\;g\; oxygen}{0.429 \;g \;carbon} = 1.33\; \dfrac{g oxygen}{g carbon}\] Similarly, for 1 g of the second compound, there are 0.727 g oxygen and 0.273 g of carbon. The ration of mass of oxygen per gram of carbon is \[ \dfrac{0.727\;g\; oxygen}{0.273 \;g \;carbon} = 2.66\; \dfrac{g oxygen}{g carbon}\] Dividing the mass of oxygen per g of carbon of the second compound: \[\dfrac{2.66}{1.33} = 2\] Hence the masses of oxygen combine with carbon in a 2:1 ratio which s consistent with the Law of Multiple Proportions since they are whole numbers. The modern atomic theory, proposed about 1803 by the English chemist John Dalton (Figure \(\Page {4}\)), is a fundamental concept that states that all elements are composed of atoms. Previously, an atom was defined as the smallest part of an element that maintains the identity of that element. Individual atoms are extremely small; even the largest atom has an approximate diameter of only 5.4 × 10 m. With that size, it takes over 18 million of these atoms, lined up side by side, to equal the width of the human pinkie (about 1 cm). Dalton’s ideas are called the atomic theory because the concept of atoms is very old. The Greek philosophers Leucippus and Democritus originally introduced atomic concepts in the fifth century BC. (The word comes from the Greek word , which means “indivisible” or “uncuttable.”) Dalton had something that the ancient Greek philosophers didn’t have, however; he had experimental evidence, such as the formulas of simple chemicals and the behavior of gases. In the 150 years or so before Dalton, natural philosophy had been maturing into modern science, and the scientific method was being used to study nature. When Dalton announced a modern atomic theory, he was proposing a fundamental theory to describe many previous observations of the natural world; he was not just participating in a philosophical discussion. Dalton's Theory was a powerful development as it explained the three laws of chemical combination (above) and recognized a workable distinction between the fundamental particle of an element (atom) and that of a compound (molecule). Six postulates are involved in Dalton's Atomic Theory: In light of the current state of knowledge in the field of Chemistry, Dalton’s theory had a few drawbacks. According to Dalton’s postulates, Despite these drawbacks, the importance of Dalton’s theory should not be underestimated. He displayed exceptional insight into the nature of matter. and his ideas provided a framework that was later modified and expanded by other. Consequentiually, John Dalton is often considered to be the father of modern atomic theory. Fundamental Experiments in Chemistry: This article explains the theories that Dalton used as a basis for his theory: | 7,957 | 2,328 |
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The hydrogen-oxygen bond of a hydroxyl group gives a characteristic absorption band in the but, as we may expect, this absorption is considerably influenced by hydrogen bonding. For example, in the vapor state (in which there is essentially no hydrogen bonding), ethanol gives an infrared spectrum with a fairly sharp absorption band at \(3700 \: \text{cm}^{-1}\), owing to a free or unassociated hydroxyl group (Figure 15-2a). In contrast, this band is barely visible at \(3640 \: \text{cm}^{-1}\) in the spectrum of a \(10\%\) solution of ethanol in carbon tetrachloride (Figure 15-2b). However, there is a relatively broad band around \(3350 \: \text{cm}^{-1}\), which is characteristic of hydrogen-bonded hydroxyl groups. The shift in frequency of about \(300 \: \text{cm}^{-1}\) arises because hydrogen bonding weakens the \(\ce{O-H}\) bond; its absorption frequency then will be lower. The association band is broad because the hydroxyl groups are associated in aggregates of various sizes and shapes. This produces a variety of different kinds of hydrogen bonds and therefore a spectrum of closely spaced \(\ce{O-H}\) absorption frequencies. In very dilute solutions of alcohols in nonpolar solvents, hydrogen bonding is minimized. However, as the concentration is increased, more and more of the molecules become associated and the intensity of the infrared absorption band due to associated hydroxyl groups increases at the expense of the free-hydroxyl band. Furthermore, the frequency of the association band is a measure of the strength of the hydrogen bond. The lower the frequency relative to the position of the free hydroxyl group, the stronger is the hydrogen bond. As we shall see in Chapter 18 the hydroxyl group in carboxylic acids \(\left( \ce{RCO_2H} \right)\) forms stronger hydrogen bonds than alcohols and accordingly absorbs at lower frequencies (lower by about \(400 \: \text{cm}^{-1}\), see Table 9-2). The infrared spectra of certain 1,2-diols (glycols) are interesting in that they show absorption due to hydrogen bonding. These usually are fairly sharp bands in the region \(3450 \: \text{cm}^{-1}\) to \(3570 \: \text{cm}^{-1}\), and, in contrast to bands due to intermolecular hydrogen bonding, they do not change in intensity with concentration. A typical example is afforded by -1,2-cyclopentanediol: Besides the \(\ce{O-H}\) stretching vibrations of alcohols, there is a bending \(\ce{O-H}\) vibration normally observed in the region \(1410\)-\(1260 \: \text{cm}^{-1}\). There also is a strong \(\ce{C-O}\) stretching vibration between \(1210 \: \text{cm}^{-1}\) and \(1050 \: \text{cm}^{-1}\). Both these bands are sensitive to structure as indicated below: The influence of hydrogen bonding on the of alcohols has been discussed previously ( ). You may recall that the chemical shift of the \(\ce{OH}\) proton is variable and depends on the extent of association through hydrogen bonding; generally, the stronger the association, the lower the field strength required to induce resonance. Alcohols also undergo intermolecular \(\ce{OH}\) proton exchange, and the rate of this exchange can influence the line-shape of the \(\ce{OH}\) resonance, the chemical shift, and the incidence of spin-spin splitting, as discussed in more detail in and . Concerning the protons on carbon bearing the hydroxyl group, that is, , they are deshielded by the electron-attracting oxygen atom and accordingly have chemical shifts some \(2.5\)-\(3.0 \: \text{ppm}\) to fields than alkyl protons. Perhaps you are curious as to why absorptions are observed in the infrared spectrum of alcohols that correspond to free and hydrogen-bonded hydroxyl groups, whereas only \(\ce{OH}\) resonance is observed in their proton nmr spectra. The explanation is that the lifetime of any molecule in either the free or the associated state is long enough to be detected by infrared absorption but much too short to be detected by nmr. Consequently, in the nmr one sees only the average \(\ce{OH}\) resonance of the nonhydrogen-bonded and hydrogen-bonded species present. The situation here is very much like that observed for conformational equilibration ( ). The longest-wavelength maxima of methanol and methoxymethane (dimethyl ether) are noted in Table 9-3. In each case the absorption maximum, which probably involves an \(n \rightarrow \sigma^*\) transition, occurs about \(184 \: \text{nm}\), well below the cut-off of the commonly available spectrometers. The of alcohols may not always show strong molecular ions. The reason is that the \(\ce{M^+}\) ions readily fragment by \(\alpha\) cleavage. The fragment ions are relatively stable and are the gaseous counterparts of protonated aldehydes and ketones: Ethers also fragment by \(\alpha\) cleavage: and (1977) | 4,812 | 2,329 |
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Like the alkali metals, the alkaline earth metals are so reactive that they are never found in elemental form in nature. Because they form +2 ions that have very negative reduction potentials, large amounts of energy are needed to isolate them from their ores. Four of the six group 2 elements—magnesium (Mg), calcium (Ca), strontium (Sr), and barium (Ba)—were first isolated in the early 19th century by Sir Humphry Davy, using a technique similar to the one he used to obtain the first alkali metals. In contrast to the alkali metals, however, compounds of the alkaline earth metals had been recognized as unique for many centuries. In fact, the name alkali comes from the Arabic al-qili, meaning “ashes,” which were known to neutralize acids. Medieval alchemists found that a portion of the ashes would melt on heating, and these substances were later identified as the carbonates of sodium and potassium (\(M_2CO_3\)). The ashes that did not melt (but did dissolve in acid), originally called alkaline earths, were subsequently identified as the alkaline earth oxides (MO). In 1808, Davy was able to obtain pure samples of Mg, Ca, Sr, and Ba by electrolysis of their chlorides or oxides. Beryllium (Be), the lightest alkaline earth metal, was first obtained in 1828 by Friedrich Wöhler in Germany and simultaneously by Antoine Bussy in France. The method used by both men was reduction of the chloride by the potent “new” reductant, potassium: Radium was discovered in 1898 by Pierre and Marie Curie, who processed tons of residue from uranium mines to obtain about 120 mg of almost pure \(RaCl_2\). Marie Curie was awarded the Nobel Prize in Chemistry in 1911 for its discovery. Because of its low abundance and high radioactivity however, radium has few uses. The alkaline earth metals are produced for industrial use by electrolytic reduction of their molten chlorides, as indicated in this equation for calcium: \[CaCl_{2\;(l)} \rightarrow Ca_{(l)} + Cl_{2\;(g)} \label{Eq2}\] The group 2 metal chlorides are obtained from a variety of sources. For example, \(BeCl_2\) is produced by reacting \(HCl\) with beryllia (\(BeO\)), which is obtained from the semiprecious stone beryl \([Be_3Al_2(SiO_3)_6]\). Chemical reductants can also be used to obtain the group 2 elements. For example, magnesium is produced on a large scale by heating a form of limestone called dolomite (CaCO ·MgCO ) with an inexpensive iron/silicon alloy at 1150°C. Initially \(CO_2\) is released, leaving behind a mixture of \(CaO\) and MgO; Mg is then reduced: \[2CaO·MgO_{(s)} + Fe/Si_{(s)} \rightarrow 2Mg(l) + Ca_2SiO_{4\;(s)} + Fe(s) \label{Eq3}\] An early source of magnesium was an ore called magnesite (\(MgCO_3\)) from the district of northern Greece called Magnesia. Strontium was obtained from strontianite (\(SrCO_3\)) found in a lead mine in the town of Strontian in Scotland. The alkaline earth metals are somewhat easier to isolate from their ores, as compared to the alkali metals, because their carbonate and some sulfate and hydroxide salts are insoluble. Several important properties of the alkaline earth metals are summarized in Table \(\Page {1}\). Although many of these properties are similar to those of the alkali metals (Table \(\Page {1}\)), certain key differences are attributable to the differences in the valence electron configurations of the two groups (ns for the alkaline earth metals versus ns for the alkali metals). As with the alkali metals, the atomic and ionic radii of the alkaline earth metals increase smoothly from Be to Ba, and the ionization energies decrease. As we would expect, the first ionization energy of an alkaline earth metal, with an ns valence electron configuration, is always significantly greater than that of the alkali metal immediately preceding it. The group 2 elements do exhibit some anomalies, however. For example, the density of Ca is less than that of Be and Mg, the two lightest members of the group, and Mg has the lowest melting and boiling points. In contrast to the alkali metals, the heaviest alkaline earth metal (Ba) is the strongest reductant, and the lightest (Be) is the weakest. The standard electrode potentials of Ca and Sr are not very different from that of Ba, indicating that the opposing trends in ionization energies and hydration energies are of roughly equal importance. One major difference between the group 1 and group 2 elements is their electron affinities. With their half-filled ns orbitals, the alkali metals have a significant affinity for an additional electron. In contrast, the alkaline earth metals generally have little or no tendency to accept an additional electron because their ns valence orbitals are already full; an added electron would have to occupy one of the vacant np orbitals, which are much higher in energy. With their low first and second ionization energies, the group 2 elements almost exclusively form ionic compounds that contain M ions. As expected, however, the lightest element (Be), with its higher ionization energy and small size, forms compounds that are largely covalent. Some compounds of Mg also have significant covalent character. Hence organometallic compounds like those discussed for Li in group 1 are also important for Be and Mg in group 2. The group 2 elements almost exclusively form ionic compounds containing M ions. All alkaline earth metals react vigorously with the halogens (group 17) to form the corresponding halides (MX ). Except for the beryllium halides, these compounds are all primarily ionic in nature, containing the M cation and two X anions. The beryllium halides, with properties more typical of covalent compounds, have a polymeric halide-bridged structure in the solid state, as shown for BeCl . These compounds are volatile, producing vapors that contain the linear X–Be–X molecules predicted by the valence-shell electron-pair repulsion (VSEPR) model. As expected for compounds with only four valence electrons around the central atom, the beryllium halides are potent Lewis acids. They react readily with Lewis bases, such as ethers, to form tetrahedral adducts in which the central beryllium is surrounded by an octet of electrons: \[ BeCl_{2(s)} + 2(CH_3CH_2)_2O_{(l)} \rightarrow BeCl_2[O(CH_2CH_3)_2]_{2(soln)} \label{Eq4}\] Because of their higher ionization energy and small size, both Be and Mg form organometallic compounds. The reactions of the alkaline earth metals with oxygen are less complex than those of the alkali metals. All group 2 elements except barium react directly with oxygen to form the simple oxide MO. Barium forms barium peroxide (BaO ) because the larger O ion is better able to separate the large Ba ions in the crystal lattice. In practice, only BeO is prepared by direct reaction with oxygen, and this reaction requires finely divided Be and high temperatures because Be is relatively inert. The other alkaline earth oxides are usually prepared by the thermal decomposition of carbonate salts: \[\mathrm{MCO_3(s)}\xrightarrow\Delta\mathrm{MO(s)}+\mathrm{CO_2(g)} \label{Eq5}\] The reactions of the alkaline earth metals with the heavier chalcogens (Y) are similar to those of the alkali metals. When the reactants are present in a 1:1 ratio, the binary chalcogenides (MY) are formed; at lower M:Y ratios, salts containing polychalcogenide ions (Y ) are formed. In the reverse of Equation \(\ref{Eq5}\), the oxides of Ca, Sr, and Ba react with CO to regenerate the carbonate. Except for BeO, which has significant covalent character and is therefore amphoteric, all the alkaline earth oxides are basic. Thus they react with water to form the hydroxides—M(OH) : \[MO_{(s)} + H_2O_{(l)} \rightarrow M^{2+}_{(aq)} + 2OH^−_{(aq)} \label{Eq6}\] and they dissolve in aqueous acid. Hydroxides of the lighter alkaline earth metals are insoluble in water, but their solubility increases as the atomic number of the metal increases. Because BeO and MgO are much more inert than the other group 2 oxides, they are used as refractory materials in applications involving high temperatures and mechanical stress. For example, MgO (melting point = 2825°C) is used to coat the heating elements in electric ranges. The carbonates of the alkaline earth metals also react with aqueous acid to give CO and H O: \[MCO_{3(s)} + 2H^+_{(aq)} \rightarrow M^{2+}_{(aq)} + CO_{2(g)} + H_2O_{(l)} \label{Eq7}\] The reaction in Equation \(\ref{Eq7}\) is the basis of antacids that contain MCO , which is used to neutralize excess stomach acid. The trend in the reactivities of the alkaline earth metals with nitrogen is the opposite of that observed for the alkali metals. Only the lightest element (Be) does not react readily with N to form the nitride (M N ), although finely divided Be will react at high temperatures. The higher lattice energy due to the highly charged M and N ions is apparently sufficient to overcome the chemical inertness of the N molecule, with its N≡N bond. Similarly, all the alkaline earth metals react with the heavier group 15 elements to form binary compounds such as phosphides and arsenides with the general formula M Z . Higher lattice energies cause the alkaline earth metals to be more reactive than the alkali metals toward group 15 elements. When heated, all alkaline earth metals, except for beryllium, react directly with carbon to form ionic carbides with the general formula MC . The most important alkaline earth carbide is calcium carbide (CaC ), which reacts readily with water to produce acetylene. For many years, this reaction was the primary source of acetylene for welding and lamps on miners’ helmets. In contrast, beryllium reacts with elemental carbon to form Be C, which formally contains the C ion (although the compound is covalent). Consistent with this formulation, reaction of Be C with water or aqueous acid produces methane: \[Be_2C_{(s)} + 4H_2O_{(l)} \rightarrow 2Be(OH)_{2(s)} + CH_{4(g)} \label{Eq8}\] Beryllium does not react with hydrogen except at high temperatures (1500°C), although BeH can be prepared at lower temperatures by an indirect route. All the heavier alkaline earth metals (Mg through Ba) react directly with hydrogen to produce the binary hydrides (MH ). The hydrides of the heavier alkaline earth metals are ionic, but both BeH and MgH have polymeric structures that reflect significant covalent character. All alkaline earth hydrides are good reducing agents that react rapidly with water or aqueous acid to produce hydrogen gas: \[CaH_{2(s)} + 2H_2O_{(l)} \rightarrow Ca(OH)_{2(s)} + 2H_{2(g)} \label{Eq9} \] Like the alkali metals, the heavier alkaline earth metals are sufficiently electropositive to dissolve in liquid ammonia. In this case, however, two solvated electrons are formed per metal atom, and no equilibriums involving metal dimers or metal anions are known. Also, like the alkali metals, the alkaline earth metals form a wide variety of simple ionic salts with oxoanions, such as carbonate, sulfate, and nitrate. The nitrate salts tend to be soluble, but the carbonates and sulfates of the heavier alkaline earth metals are quite insoluble because of the higher lattice energy due to the doubly charged cation and anion. The solubility of the carbonates and the sulfates decreases rapidly down the group because hydration energies decrease with increasing cation size. The solubility of alkaline earth carbonate and sulfates decrease down the group because the hydration energies decrease. Because of their higher positive charge (+2) and smaller ionic radii, the alkaline earth metals have a much greater tendency to form complexes with Lewis bases than do the alkali metals. This tendency is most important for the lightest cation (Be ) and decreases rapidly with the increasing radius of the metal ion. The alkaline earth metals have a substantially greater tendency to form complexes with Lewis bases than do the alkali metals. The chemistry of Be is dominated by its behavior as a Lewis acid, forming complexes with Lewis bases that produce an octet of electrons around beryllium. For example, Be salts dissolve in water to form acidic solutions that contain the tetrahedral [Be(H O) ] ion. Because of its high charge-to-radius ratio, the Be ion polarizes coordinated water molecules, thereby increasing their acidity: \[ [Be(H_2O)_4]^{2+}_{(aq)} \rightarrow [Be(H_2O)_3(OH)]^+_{(aq)} + H^+_{(aq)} \label{Eq10}\] Similarly, in the presence of a strong base, beryllium and its salts form the tetrahedral hydroxo complex: [Be(OH) ] . Hence beryllium oxide is amphoteric. Beryllium also forms a very stable tetrahedral fluoride complex: [BeF ] . Recall that beryllium halides behave like Lewis acids by forming adducts with Lewis bases (Equation \(\ref{Eq4}\)). The heavier alkaline earth metals also form complexes, but usually with a coordination number of 6 or higher. Complex formation is most important for the smaller cations (Mg and Ca ). Thus aqueous solutions of Mg contain the octahedral [Mg(H O) ] ion. Like the alkali metals, the alkaline earth metals form complexes with neutral cyclic ligands like the crown ethers and cryptands discussed in Section 21.3. Like the alkali metals, the lightest alkaline earth metals (Be and Mg) form the most covalent-like bonds with carbon, and they form the most stable organometallic compounds. Organometallic compounds of magnesium with the formula RMgX, where R is an alkyl or aryl group and X is a halogen, are universally called , after Victor Grignard (1871–1935), the French chemist who discovered them. Grignard reagents can be used to synthesize various organic compounds, such as alcohols, aldehydes, ketones, carboxylic acids, esters, thiols, and amines. Elemental magnesium is the only alkaline earth metal that is produced on a large scale (about 5 × 10 tn per year). Its low density (1.74 g/cm compared with 7.87 g/cm for iron and 2.70 g/cm for aluminum) makes it an important component of the lightweight metal alloys used in aircraft frames and aircraft and automobile engine parts (Figure \(\Page {1}\)). Most commercial aluminum actually contains about 5% magnesium to improve its corrosion resistance and mechanical properties. Elemental magnesium also serves as an inexpensive and powerful reductant for the production of a number of metals, including titanium, zirconium, uranium, and even beryllium, as shown in the following equation: \[TiCl_{4\;(l)} + 2Mg(s) \rightarrow Ti_{(s)} + 2MgCl_{2\;(s)} \label{11}\] The only other alkaline earth that is widely used as the metal is beryllium, which is extremely toxic. Ingestion of beryllium or exposure to beryllium-containing dust causes a syndrome called berylliosis, characterized by severe inflammation of the respiratory tract or other tissues. A small percentage of beryllium dramatically increases the strength of copper or nickel alloys, which are used in nonmagnetic, nonsparking tools (such as wrenches and screwdrivers), camera springs, and electrical contacts. The low atomic number of beryllium gives it a very low tendency to absorb x-rays and makes it uniquely suited for applications involving radioactivity. Both elemental Be and BeO, which is a high-temperature ceramic, are used in nuclear reactors, and the windows on all x-ray tubes and sources are made of beryllium foil. Millions of tons of calcium compounds are used every year. As discussed in earlier chapters, CaCl is used as “road salt” to lower the freezing point of water on roads in cold temperatures. In addition, CaCO is a major component of cement and an ingredient in many commercial antacids. “Quicklime” (CaO), produced by heating CaCO (Equation \(\ref{Eq5}\)), is used in the steel industry to remove oxide impurities, make many kinds of glass, and neutralize acidic soil. Other applications of group 2 compounds described in earlier chapters include the medical use of BaSO in “barium milkshakes” for identifying digestive problems by x-rays and the use of various alkaline earth compounds to produce the brilliant colors seen in fireworks. For each application, choose the most appropriate substance based on the properties and reactivities of the alkaline earth metals and their compounds. Explain your choice in each case. Use any tables you need in making your decision, such as K values (Table 17.1), lattice energies (Table 8.1), and band-gap energies. application and selected alkaline earth metals most appropriate substance for each application Based on the discussion in this section and any relevant information elsewhere in this book, determine which substance is most appropriate for the indicated use. Which of the indicated alkaline earth metals or their compounds is most appropriate for each application? Predict the products of each reaction and then balance each chemical equation. reactants products and balanced chemical equation Follow the procedure given in Example 3 to predict the products of each reaction and then balance each chemical equation. The balanced chemical equation is \[CaO_{(s)} + 2HCl_{(g)} → CaCl_{2(aq)} + H_2O_{(l)}\] We conclude that no reaction occurs. The balanced chemical equation is \(\mathrm{CaH_2(s)}+\mathrm{TiO_2(s)}\xrightarrow\Delta\mathrm{Ti(s)}+\mathrm{CaO(s)}+\mathrm{H_2O(l)}\) We could also write the products as Ti(s) + Ca(OH) (s). Predict the products of each reaction and then balance each chemical equation. Group 2 elements almost exclusively form ionic compounds containing the M ion, they are more reactive toward group 15 elements, and they have a greater tendency to form complexes with Lewis bases than do the alkali metals. Pure samples of most of the alkaline earth metals can be obtained by electrolysis of the chlorides or oxides. Beryllium was first obtained by the reduction of its chloride; radium chloride, which is radioactive, was obtained through a series of reactions and separations. In contrast to the alkali metals, the alkaline earth metals generally have little or no affinity for an added electron. All alkaline earth metals react with the halogens to produce the corresponding halides, with oxygen to form the oxide (except for barium, which forms the peroxide), and with the heavier chalcogens to form chalcogenides or polychalcogenide ions. All oxides except BeO react with CO to form carbonates, which in turn react with acid to produce CO and H O. Except for Be, all the alkaline earth metals react with N to form nitrides, and all react with carbon and hydrogen to form carbides and hydrides. Alkaline earth metals dissolve in liquid ammonia to give solutions that contain two solvated electrons per metal atom. The alkaline earth metals have a greater tendency than the alkali metals to form complexes with crown ethers, cryptands, and other Lewis bases. The most important alkaline earth organometallic compounds are Grignard reagents (RMgX), which are used to synthesize organic compounds. | 18,916 | 2,330 |
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A chemical equilibrium involving dissociation can be represented by the following reaction. \(\ce{AB \rightleftharpoons A + B}, \hspace{20px} K = \ce{\dfrac{[A,B]}{[AB]}}\) The concept equilibrium has been discussed in Mass Action Law and further discussed in Weak Acids and Bases, , , and , and . Further, if is the initial concentration of \(\ce{AB}\) before any dissociation takes place, and is the fraction of dissociated molecules, the concentration of \(\ce{AB}\) is (1- ) when the system has reached equilibrium. The concentration of \(\ce{A}\) and \(\ce{B}\) will each then be . Note that is also the total concentration. For convenience, we can write the concentration below the formula as: \[\begin{array}{cccccl}
\ce{AB &\rightleftharpoons &A &+ &B&}\\
(1-f)C &&fC &&fC &\leftarrow \ce{Concentration}
\end{array}\] and the equilibrium constant, \(K = \ce{\dfrac{[A,B]}{[AB]}}\) can be written as (using the concentration below the formula): \[K = \dfrac{(f\cancel{C})(fC)}{(1-f)\cancel{C}} = \dfrac{f^2 C}{1 - f} \label{2}\] How does \(f\) vary as a function of \(C\)? Common sense tells us that \(f\) has a value between 0 and 1 (0 < < 1). For dilute solutions, \(f \approx 1\), and for concentrated solutions, \(f \approx 0\). In solving Equation \(\ref{2}\) for , we obtain: \[f = \dfrac{ - K + \sqrt{K^2 + 4 K C}}{2 C}\] \(f\) is the fraction of molecules that have dissociated and is also called the . When converted to percentage, the term is used. Even with the given formulation, it is still difficult to see how \(f\) varies as \(C\) changes. Table \(\Page {1}\) illustrate the variation with a table below for a moderate value of \(K = 1.0 \times 10^{-5}\). There is little change in \(f\) when \(C\) decreases from \(1.0 \times 10^{-7}\) to \(1.0 \times 10^{-6}\), but the changes are rather somewhat regular for every 10 fold decrease in concentration. Please plot \(f\) against a log scale of \(C\) to see the shape of the variation as your activity. Normally, we will not encounter solution as dilute as = 1.0e-7, and we will never encounter solution as concentrated as 100 M either. \(\begin{array}{cccccl}
\ce{HB &\rightleftharpoons &H+ &+ &B- &}\\
0.05 &&0.05 &&0.05\: \ce M &\leftarrow \textrm{Concentrations at equilibrium}
\end{array}\) = ? | 2,331 | 2,331 |
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This page explains how to name some common complex metal ions. Although the names of complex ions can look long and worrying, the formulae are simply being coded in much the same way that organic names are coded. Once you have sorted out that code, the names are entirely descriptive. The table shows some common ligands and the code for them in the name of a complex ion. The old names sometimes differ by a letter or so, but never enough for it to be confusing. Take care with the code for ammonia as a ligand - it has 2 "m"s in its name. If you miss one of these out so that you are left with "amine" or "amino", you are refering to the NH group in an organic compound. This is probably the only point of confusion with these names. The normal prefixes apply if there is more than one ligand. For a complex ion containing only one type of ligand, there is no problem. For example: [Cu(H O) ] is called the hexaaquacopper(II) ion. (Don't worry about the copper(II) bit for the moment.) The fact that there are two "a"s next to each other in the name is OK. With more than one type of ligand in an ion, the ligands are named in alphabetical order - ignoring the prefixes. For example: [Cu(NH ) (H O) ] is called the tetraamminediaquacopper(II) ion. The "ammine" is named before the "aqua" because "am" comes before "aq" in the alphabet. The "tetra" and "di" are ignored. You might have thought that this was fairly obvious, but it isn't necessarily. It depends on whether the complex ion ends up as positively or negatively charged. A positively charged complex ion is called a cationic complex. A cation is a positively charged ion. The metal in this is named exactly as you would expect, with the addition of its oxidation state. Going back to a previous example, [Cu(H O) ] is called the hexaaquacopper Copper's oxidation is +2 because the original uncomplexed ion was Cu - NOT because the complex carries 2+ charges. The oxidation state is frequently left out if a metal only ever has one oxidation state. For example, in its compounds aluminium always has an oxidation state of +3. [Al(H O) ] is usually just called the hexaaquaaluminium ion rather than the hexaaquaaluminium(III) ion. A negatively charged complex ion is called an anionic complex. An anion is a negatively charged ion. In this case the name of the metal is modified to show that it has ended up in a negative ion. This is shown by the ending . With many metals, the basic name of the metal is changed as well - sometimes drastically! Common examples include: So, for example, suppose you bond 4 chloride ions around a Cu ion to give [CuCl ] . The name shows the 4 (tetra) chlorines (chloro) around a copper in an overall negative ion (cuprate). The copper has on oxidation state of +2. This is the tetrachlorocuprate(II) ion. Similarly, [Al(H O) (OH) ] is called the diaquatetrahydroxoaluminate ion. Take the name to pieces so that you can see exactly what refers to what. Don't forget that the two different ligands are named in alphabetical order - aqua before hydroxo - ignoring the prefixes, di and tetra. The oxidation state of the aluminum could be shown, but is not absolutely necessary because aluminum only has the one oxidation state in its compounds. The full name is the diaquatetrahydroxoaluminate(III) ion. Jim Clark ( ) | 3,329 | 2,333 |
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: A. Jan Hendriks and Nico van Straalen Aafke Schipper, John D. Stark and Thomas G. Preuss You should be able to intrinsic rate of increase, carrying capacity, exponential growth, Ecological risk assessment of toxicants usually focuses on the risks run by individuals, by comparing exposures with no-effect levels. However, in many cases it is not the protection of individual plants or animals that is of interest but the protection of a viable population of a species in an ecological context. Risk assessment generally does not take into account the quantitative dynamics of populations and communities. Yet, understanding and predicting effects of chemicals at levels beyond that of individuals is urgently needed for several reasons. First, we need to know whether quality standards are sufficiently but not overly protective at the population level, when extrapolated from toxicity tests. Second, responses of isolated, homogenous cohorts in the laboratory may be different from those of interacting, heterogeneous populations in the field. Third, to set the right priorities in management, we need to know the relative and cumulative effect of chemicals in relation to other environmental pressures. Ecological population models for algae, macrophytes, aquatic invertebrates, insects, birds and mammals have been widely used to address the risk of potentially toxic chemicals, however, until recently, these models were only rarely used in the regulatory risk assessment process due to a lack of connection between model output and risk assessment needs (Schmolke et al., 2010). Here, we will sketch the basic principles of population dynamics for environmental toxicology applications. Ecological textbooks usually start their chapter on population ecology by introducing exponential and logistic growth. Consider a population of size N. If resources are unlimited, and the (b) and (d) are constant in a population closed to migration, the number of individuals added to the population per time unit (dN/dt) can be written as: or where r is called the . This differential equation can be solved with boundary condition N(0) = N to yield Since toxicants will affect either reproduction or survival, or both, they will also affect the exponential growth rate (Figure 1a). This suggests that r can be considered a measure of population performance under toxic stress. But rather than from observed population trajectories, r is usually estimated from life-history data. We know from basic demographic theory that any organism with "time-invariant" vital rates (that is, fertility and survival may depend on age, but not on time), will be growing exponentially at rate r. The intrinsic rate of increase can be derived from age-specific survival and fertility rates using the so-called Euler-Lotka equation, which reads: in which x is age, x maximal age, l(x) survivorship from age zero to age x and m(x) the number of offspring produced per time unit at age x. Unfortunately this equation does not allow for a simple derivation of r; r must be obtained by iteration and the correct value is the one that, when combined with the l(x) and m(x) data, makes the integral equal to 1. Due to this complication approximate approaches are often applied. For example, in many cases a reasonably good estimate for r can be obtained from the age at first reproduction α, survival to first reproduction, S, and reproductive output, m, according to the following formula: This is due to the fact that for many animals in the environment, especially those with high reproductive output and low juvenile survivorship, age at first reproduction is the dominant variable determining population growth (Forbes and Calow, 1999). The classical demographic modelling approach, including the Euler-Lotka equation, considers time as a continuous variable and solves the equations by calculus. However, there is an equivalent formalism based on discrete time, in which population events are assumed to take place only at equidistant moments. The vital rates are then summarized in a so-called Leslie matrix, a table of survival and fertility scores for each age class, organized in such a way that when multiplied by the age distribution at any moment, the age distribution at the following time point is obtained. This type of modelling lends itself more easily to computer simulation. The outcome is much the same: if the Leslie matrix is time-invariant the population will grow each time step by a factor λ, which is related to r as ln λ = r (λ = 1 corresponds to r = 0). Mathematically speaking λ is the dominant eigenvalue of the Leslie matrix. The advantage of the discrete-time version is that λ can be more easily decomposed into its component parts, that is, the life-history traits that are affected by toxicants (Caswell, 1996). Hendriks et al. (2005) postulated that r should show a near-linear decrease with the concentration of a chemical, scaled to the LC (Figure 3). This relationship was confirmed in a meta-analysis of 200 laboratory experiments, mostly concerning invertebrate species (Figure 3). Anecdotal underpinning for large vertebrates comes from field cases where pollution limits population development. As exponentially growing populations are obviously rare, models that include some form of density-dependence are more realistic. One common approach is to assume that the birth rate b decreases with density due to increasing scarcity of resources. The simplest assumption is a linear decrease with N, expressed as follows: The question is, can the parameters of the logistic growth equation be used to measure population performance like in the case of exponential growth? Practical application is limited because the carrying capacity is difficult to measure under natural and contaminated conditions. Many field populations of arthropods, for example, fluctuate widely due to predator-prey dynamics, and hardly ever reach their carrying capacity within a growing season. An experimental study on the springtail (Noël et al., 2006) showed that zinc in the diet did not affect the carrying capacity of contained laboratory populations, although there were several interactions below K that were influenced by zinc, including hormesis (growth stimulation by low doses of a toxicant), and Allee effects (loss of growth potential at low density due to lower encounter rate). Density-dependence is expected to act as buffering mechanism at the population level because toxicity-induced population decline diminishes competition, however, the effects very much depend on the details of population regulation. This was demonstrated in a model for peregrine falcon exposed to DDE and PBDEs (Schipper et al., 2013). While the equilibrium size of the population declined by toxic exposure, the probability of individual birds finding a suitable territory increased. However, at the same time the number of non-breeding birds shifting to the breeding stage became limiting and this resulted in a strong decrease in the equilibrium number of breeders. To enhance the potential for application of population models in risk assessment, more ecological details of the species under consideration must be included, e.g. effects of dispersal, abiotic factors, predators and parasites, dispersal, landscape structure and many more. A further step is to track the physiology and ecology of each individual in the population. This is done in the dynamic energy budget modelling approach (DEB) developed by (Kooijman et al., 2009). By including such details, a model will become more realistic, and more precise predictions can be made on the effects of toxic exposures. These types of models are generally called "mechanistic effect models' (MEMs). They allow a causal link between the protection goal, a scenario of exposure to toxicants and the adverse population effects generated by model output (Hommen et al., 2015). The European Food Safety Authority (EFSA) in 2014 issued an opinion paper containing detailed guidelines on the development of such models and how to adjust them to be useful in the risk assessment of plant protection products. Caswell, H. (1996). Demography meets ecotoxicology: untangling the population level effects of toxic substances. In: Newman, M.C., Jagoe, C.H. (Eds.). . Lewis Publishers Boca Raton, pp. 255-292. Barata, C., Baird, D.G., Amata, F., Soares, A.M.V.M. (2000). Comparing population response to contaminants between laboratory and field: an approach using ephippial egg banks. Functional Ecology 14, 513-523. EFSA (2014). Scientific Opinion on good modeling practice in the context of mechanistic effect models for risk assessment of plant protection products. EFSA Panel on Plant Protection and their Residues (PPR). EFSA Journal 12, 3589. Forbes, V.E., Calow, P. (1999). Is the per capita rate of increase a good measure of population-level effects in ecotoxicology. Environmental Toxicology and Chemistry 18, 1544-1556. Hendriks, A.J., Maas, J.L., Heugens, E.H.W., Van Straalen, N.M. (2005). Meta-analysis of intrinsic rates of increase and carrying capacity of populations affected by toxic and other stressors. Environmental Toxicology and Chemistry 24, 2267-2277 Hommen, U., Forbes, V., Grimm, V., Preuss, T.G., Thorbek, P., Ducrot, V. (2015). How to use mechanistic effect models in environmental risk assessment of pesticides: case studies and recommendations from the SETAC workshop Modelink. Integrated Environmental Assessment and Management 12, 21-31. Kooijman, S.A.L.M., Baas, J., Bontje, D., Broerse, M., Van Gestel, C.A.M., Jager, T. (2009). Ecotoxicological Applications of Dynamic Energy Budget theory. In: Devillers 2, Springer, Dordrecht, pp. 237-260. Noël, H.L., Hopkin, S.P., Hutchinson, T.H., Williams, T.D., Sibly, R.M. (2006). Towards a population ecology of stressed environments: the effects of zinc on the springtail . Journal of Applied Ecology 43, 325-332. Schipper, A.M., Hendriks, H.W.M., Kaufmann, M.J., Hendriks, A.J., Huijbregts, M.A.J. (2013). Modelling interactions of toxicants and density dependence in wildlife populations. Journal of Applied Ecolog 50, 1469-1478. Schmolke, A., Thorbek, P., Chapman, P., Grimm, V. (2010) Ecological models and pesticide risk assessment: current modelling practice. Environmental Toxicology and Chemistry 29, 1006-1012. Stark, J.D., Banks, J.E. (2003) Population effects of pesticides and other toxicants on arthropods. Annual Review of Entomology 48, 505-519. Suhett, A.L. et al. (2015) An overview of the contribution of studies with cladocerans to environmental stress research. Acta Limnologica Brasiliensia 27, 145-159. | 10,676 | 2,334 |
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This page explains why complex ions of the type [M(H O) ] are acidic. The pH's of solutions containing hexaaqua ions vary a lot from one metal to another (assuming you are comparing solutions of equal concentrations). However, the underlying explanation is the same for all of them. We'll take the hexaaquairon(III) ion, [Fe(H O) ] as typical. The structure of the ion is: Each of the six water molecules are attached to the central iron(III) ion via a co-ordinate bond using one of the lone pairs on the oxygen. We'll choose one of these water molecules at random (it doesn't make any difference which one!), and look at the bonding in a bit more detail - showing all the bonds around the oxygen. Imagine for the moment that the 3+ charge is located entirely on the iron. When the lone pairs on the oxygens form co-ordinate bonds with the iron, there is obviously a movement of electrons towards the iron. That has an effect on the electrons in the O-H bonds. These electrons, in turn, get pulled towards the oxygen even more than usual. That leaves the hydrogen nuclei more exposed than normal. The overall effect is that each of the hydrogen atoms is more positive than it is in ordinary water molecules. The 3+ charge is no longer located entirely on the iron but spread out over the whole ion - much of it on the hydrogen atoms. The hydrogen atoms attached to the water ligands are sufficiently positive that they can be pulled off in a reaction involving water molecules in the solution. The first stage of this process is: \[ \ce{ Fe(H2O)6^{3+} + H2O <=> Fe(H2O)5(OH)^{2+} + H3O^{+}} \label{eq1}\] The complex ion is acting as an (Arrhenius) acid by donating a hydrogen ion to water molecules in the solution. The water is, of course, acting as a (Arrhenius) base by accepting the hydrogen ion. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify Equation \ref{eq1}: \[ \ce{[Fe(H_2O)_6]^{3+} (aq) \rightleftharpoons [Fe(H2O)_5(OH)]^{2+} (aq) + H^{+} (aq)} \label{eq2}\] However, if you write it like this, remember that the hydrogen ion is not just falling off the complex ion. It is being pulled off by a water molecule in the solution. Whenever you write "\(H^+(aq)\)" what you really mean is a hydroxonium ion, H O . The hexaaquairon(III) ion is quite strongly acidic giving solutions with pH's around 1.5, depending on concentration. You can get further loss of hydrogen ions as well, from a second and a third water molecule since the complex ion is a polyprotic acid. Losing a second hydrogen ion: \[ \ce{ [Fe(H2O)5(OH)]^{2+} (aq) \rightleftharpoons [Fe(H_2O)_4(OH)_2]^{+} (aq) + H^+ (aq)} \label{eq3}\] and a third one: \[ \ce{ [ Fe(H_2O)_4(OH)_2]^{+} (aq) \rightleftharpoons [ Fe(H_2O)_3(OH)_3] (s) + H^+ (aq)} \label{eq4}\] This time you end up with a neutral complex and because it has no charge, it does not dissolve to any extent in water, and a precipitate is formed. What do you actually get in solution if you dissolve an iron(III) salt in water? In fact you get a mixture of all the complexes that you have seen in the equations above. These reactions are all equilibria, so everything will be present. The proportions depend on how concentrated the solution is. The color of the solution is very variable and depends in part on the concentration of the solution. Dilute solutions containing iron(III) ions can be pale yellow. More concentrated ones are much more orange, and may even produce some orange precipitate. None of these colors represents the true color of the [Fe(H O) ] ion - which is a very pale lilac color and is only really easy to see in solids containing the ion. Looking at the equilibrium showing the loss of the first hydrogen ion: \[ \underbrace{\ce{ Fe(H2O)6^{3+}}}_{\text{pale lilac}} + \ce{H2O} \ce{<=>} \underbrace{\ce{Fe(H2O)5(OH)^{2+}}}_{\text{orange}} + \ce{H3O^{+}}\] The color of the new complex ion on the right-hand side is so strong that it completely masks the color of the hexaaqua ion. In concentrated solutions, the equilibrium position will be even further to the right-hand side (Le Chatelier's Principle), and so the color darkens. You will also get significant loss of other hydrogen ions (Equation \ref{eq3} and \ref{eq4}) leading to some formation of the neutral complex and thus some precipitate. The position of this equilibrium can be shifted by adding extra hydrogen ions from a concentrated acid (e.g., by adding concentrated nitric acid to a solution of iron(III) nitrate). The new hydrogen ions push the position of the equilibrium to the left so that you can see the color of the hexaaqua ion. This is slightly easier to follow if you write the simplified version of the equilibrium (Equation \ref{eq2}). Solutions containing 3+ hexaaqua ions tend to have pH's in the range from 1 to 3. Solutions containing 2+ ions have higher pH's - typically around 5 - 6, although they can go down to about 3. Remember that the reason that these ions are acidic is because of the pull of the electrons towards the positive central ion. An ion with 3+ charges on it is going to pull the electrons more strongly than one with only 2+ charges. In 3+ ions, the electrons in the O-H bonds will be pulled further away from the hydrogens than in 2+ ions. That means that the hydrogen atoms in the ligand water molecules will have a greater positive charge in a 3+ ion, and so will be more attracted to water molecules in the solution. If they are more attracted, they will be more readily lost - and so the 3+ ions are more acidic. If you have ions of the same charge, it seems reasonable that the smaller the volume this charge is packed into, the greater the distorting effect on the electrons in the O-H bonds. Ions with the same charge but in a smaller volume (a higher charge density) would be expected to be more acidic. You would therefore expect to find that the smaller the radius of the metal ion, the stronger the acid. Unfortunately, it's not that simple! Unfortunately, almost every data source that you refer to quotes different values for ionic radii. This is because the radius of a metal ion varies depending on what negative ion it is associated with. We are interested in what happens when the metal ion is bonded to water molecules, so haven't got simple ions at all! Whatever values we take are unlikely to represent the real situation. In the graphs which follow I have taken values for ionic radii from two common sources so that you can see what a lot of difference it makes to the argument. If it is true that the smaller the ionic radius, the stronger the acidity of the hexaaqua ion, you would expect some sort of regular increase in pK (showing weaker acids) as ionic radius increases. The following graphs plot pK against ionic radii for the 2+ ions of the elements in the first transition series from vanadium to copper. The first graph plots pK against ionic radii taken from Chemistry Data Book by Stark and Wallace. You can see that there is a trend for several of the ions, but it is completely broken by vanadium and chromium. The second graph uses ionic radii taken from the Nuffield Advanced Science Book of Data. There probably is a relationship between ionic radius and acid strength, but that it is nothing like as simple and straightforward as most books at this level pretend. The problem is that there are other more important effects operating as well (quite apart from differences in charge) that can completely swamp the effect of the changes in ionic radius. You have to look in far more detail at the bonding in the hexaaqua ions and the product ions. Jim Clark ( ) | 7,662 | 2,335 |
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Substitution of the hydroxyl hydrogens of alcohols by hydrocarbon groups gives compounds known as ethers. These compounds may be classified further as open-chain, cyclic, saturated, unsaturated, aromatic, and so on. For the naming of ethers, see and . The most generally useful methods of preparing ethers already have been discussed (Sections 8-7C, 8-7E, , and ). These and some additional special procedures are summarized in Table 15-4. In general, ethers are low on the scale of chemical reactivity because the carbon-oxygen bond is not cleaved readily. For this reason ethers frequently are employed as inert solvents in organic synthesis. Particularly important in this connection are diethyl ether, diisopropyl ether, tetrahydrofuran, and 1,4-dioxane. The mono- and dialkyl ethers of 1,2-ethanediol, 3-oxa-1,5-pentanediol, and related substances are useful high-boiling solvents. Unfortunately, their trade names are not very rational. Abbreviated names are in common use, such as “polyglymes,” “Cellosolves,” and “Carbitols.” For reference, are monoalkyl ethers of 1,2-ethanediol; are monoalkyl ethers of 3-oxa-1,5-pentanediol; are dimethyl ethers of 3-oxa-1,5-pentanediol or 3,6-dioxa-1,8-octanediol and are called and , respectively. The spectroscopic properties of ethers are unexceptional. Like alcohols, they have no electronic absorption beyond \(185 \: \text{nm}\); the important infrared bands are the \(\ce{C-O}\) stretching vibrations in the region \(1000\)-\(1230 \: \text{cm}^{-1}\); their proton nmr spectra show deshielding of the hydrogens by the ether oxygen \(\left( \delta_{HC_{\alpha}OC} \sim 3.4 \: \text{ppm} \right)\). The mass spectra of ethers and alcohols are very similar and give abundant ions of the type (\(\ce{R} = \ce{H}\) or alkyl) by \(\alpha\)-cleavage (see ). Unlike alcohols, ethers are not acidic and usually do not react with bases. However, exceptionally strong basic reagents, particularly certain alkali-metal alkyls, will react destructively with many ethers: Ethers, like alcohols, are weakly basic and are converted to highly reactive salts by strong acids (e.g., \(\ce{H_2SO_4}\), \(\ce{HClO_4}\), and \(\ce{HBr}\)) and to relatively stable coordination complexes with Lewis acids (e.g., \(\ce{BF_3}\) and \(\ce{RMgX}\)): Dimethyl ether is converted to trimethyloxonium fluoroborate by the combination of boron trifluoride and methyl fluoride: Both trimethyl- and triethyloxonium salts are fairly stable and can be isolated as crystalline solids. They are prepared more conveniently from the appropriate boron trifluoride etherate and chloromethyloxacyclopropane (epichlorohydrin). Trialkyloxonium ions are much more susceptible to nucleophilic displacement reactions than are neutral ether molecules. The reason is that \(\ce{ROR}\) is a better leaving group than \(\ce{RO}^\ominus\). In fact, trimethyloxonium salts are among the most effective methylating reagents known: Ethers can be cleaved under strongly acidic conditions by intermediate formation of dialkyloxonium salts. Hydrobromic and hydroiodic acids are especially useful for ether cleavage because both are strong acids and their anions are good nucleophiles. Tertiary alkyl ethers are very easily cleaved by acid reagents: Ethers are susceptible to attack by halogen atoms and radicals, and for this reason they are not good solvents for radical reactions. In fact, ethers are potentially hazardous chemicals, because in the presence of atmospheric oxygen radical-chain formation of peroxides occurs, and peroxides are unstable, explosion-prone compounds. This process is called and occurs not only with ethers but with many aldehydes and hydrocarbons. The reaction may be generalized in terms of the following steps involving initiation (1), propagation (2 and 3), and termination (4). The initiation and termination steps can occur in a variety of ways but it is the chain-carrying steps, 2 and 3, that effect the overall destruction of the compound. Commonly used ethers such as diethyl ether, diisopropyl ether, tetrahydrofuran, and 1,4-dioxane often become seriously contaminated with peroxides formed by autoxidation on prolonged storage and exposure to air and light. Purification of ethers frequently is necessary before use, and caution always should be exercised in the last stages of distilling them, because the distillation residues may contain dangerously high concentrations of explosive peroxides. and (1977) | 4,483 | 2,336 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/08%3A_Review_of_Chemical_Kinetics |
A reaction’s equilibrium position defines the extent to which the reaction can occur. For example, we expect a reaction with a large equilibrium constant, such as the dissociation of HCl in water \[\ce{HCl}(aq) + \ce{H2O}(l) ⇋ \ce{H3O+}(aq) + \ce{Cl-}(aq) \nonumber \] to proceed nearly to completion. A large equilibrium constant, however, does not guarantee that a reaction will reach its equilibrium position. Many reactions with large equilibrium constants, such as the reduction of MnO by H O \[\ce{4MnO4-}(aq) + \ce{2H2O}(l) ⇋ \ce{4MnO2}(s) + \ce{3O2}(g) + \ce{4OH-}(aq) \nonumber \] do not occur to an appreciable extent. The study of the rate at which a chemical reaction approaches its equilibrium position is called kinetics. A study of a reaction’s kinetics begins with the measurement of its reaction rate. Consider, for example, the general reaction shown below, involving the aqueous solutes A, B, C, and D, with stoichiometries of , , , and . \[a\ce{A} + b\ce{B} ⇋ c\ce{C} + d\ce{D}\tag{A17.1} \] The rate, or velocity, at which this reaction approaches its equilibrium position is determined by following the change in concentration of one reactant or one product as a function of time. For example, if we monitor the concentration of reactant A, we express the rate as \[R = -\dfrac{d[\ce A]}{dt}\tag{A17.2} \] where is the measured rate expressed as a change in concentration of A as a function of time. Because a reactant’s concentration decreases with time, we include a negative sign so that the rate has a positive value. We also can determine the rate by following the change in concentration of a product as a function of time, which we express as \[R´ = −\dfrac{d[\ce C]}{dt}\tag{A17.3} \] Rates determined by monitoring different species do not necessarily have the same value. The rate in equation A17.2 and the rate ´ in equation A17.3 have the same value only if the stoichiometric coefficients of A and C in reaction A17.1 are identical. In general, the relationship between the rates and ´ is \[R = \dfrac{a}{c} × R´ \nonumber \] A rate law describes how a reaction’s rate is affected by the concentration of each species in the reaction mixture. The rate law for reaction A17.1 takes the general form of \[R = k\mathrm{[A]^α[B]^β[C]^γ[D]^δ[E]^ε...}\tag{A17.4} \] where is the rate constant, and α, β, γ, δ, and ε are the reaction orders of the reaction for each species present in the reaction. There are several important points about the rate law in equation A17.4. First, a reaction’s rate may depend on the concentrations of both reactants and products, as well as the concentration of a species that does not appear in the reaction’s overall stoichiometry. Species E in equation A17.4, for example, may be a catalyst that does not appear in the reaction’s overall stoichiometry, but which increases the reaction’s rate. Second, the reaction order for a given species is not necessarily the same as its stoichiometry in the chemical reaction. Reaction orders may be positive, negative, or zero, and may take integer or non-integer values. Finally, the reaction’s overall reaction order is the sum of the individual reaction orders for each species. Thus, the overall reaction order for equation A17.4 is α + β + γ + δ + ε. In this section we review the application of kinetics to several simple chemical reactions, focusing on how we can use the integrated form of the rate law to determine reaction orders. In addition, we consider how we can determine the rate law for a more complex system. The simplest case we can treat is a first-order reaction in which the reaction’s rate depends on the concentration of only one species. The best example of a first-order reaction is an irreversible thermal decomposition of a single reactant, which we represent as \[\mathrm{A → Products}\tag{A17.5} \] with a rate law of \[R = -\dfrac{d[\ce A]}{dt} = k[\ce A]\tag{A17.6} \] The simplest way to demonstrate that a reaction is first-order in A, is to double the concentration of A and note the effect on the reaction’s rate. If the observed rate doubles, then the reaction must be first-order in A. Alternatively, we can derive a relationship between the concentration of A and time by rearranging equation A17.6 and integrating. \[\dfrac{d[\ce A]}{[\ce A]} = -k dt \nonumber \] \[\int_{[\ce A]_0}^{[\ce A]_t} \dfrac{d[\ce A]}{[\ce A]} = -k \int_{0}^{t}dt\tag{A17.7} \] Evaluating the integrals in equation A17.7 and rearranging \[\ln\dfrac{[\ce A]_t}{[\ce A]_0}= -kt\tag{A17.8} \] \[\ln[\ce A]_t = -kt+ \ln[\ce A]_0\tag{A17.9} \] shows that for a first-order reaction, a plot of ln[A] versus time is linear with a slope of – and a -intercept of ln[A] . Equation A17.8 and equation A17.9 are known as integrated forms of the rate law. Reaction A17.5 is not the only possible form of a first-order reaction. For example, the reaction \[\mathrm{A + B → Products}\tag{A17.10} \] will follow first-order kinetics if the reaction is first-order in A and if the concentration of B does not affect the reaction’s rate. This may happen if the reaction’s mechanism involves at least two steps. Imagine that in the first step, A slowly converts to an intermediate species, C, which rapidly reacts with the remaining reactant, B, in one or more steps, to form the products. \[\mathrm{A → B \hspace{20px}(slow)} \nonumber \] \[\mathrm{B + C → Products \hspace{20px} (fast)} \nonumber \] Because a reaction’s rate depends only on those species in the slowest step—usually called the rate-determining step—and any preceding steps, species B will not appear in the rate law. The simplest reaction demonstrating second-order behavior is \[\mathrm{2A → Products} \nonumber \] for which the rate law is \[R = -\dfrac{d[\ce A]}{dt}= k[\ce A]^2 \nonumber \] Proceeding as we did earlier for a first-order reaction, we can easily derive the integrated form of the rate law. \[\dfrac{d[\ce A]}{[\ce A]^2}= -k\, dt \nonumber \] \[\int_{[\ce A]_0}^{[\ce A]_t} \dfrac{d[\ce A]}{[\ce A]^2} = -k \int_{0}^{t}dt \nonumber \] \[\dfrac{1}{[\ce A]_t} = kt + \dfrac{1}{[\ce A]_0} \nonumber \] For a second-order reaction, therefore, a plot of [A] versus is linear with a slope of and a -intercept of [A] . Alternatively, we can show that a reaction is second-order in A by observing the effect on the rate when we change the concentration of A. In this case, doubling the concentration of A produces a four-fold increase in the reaction’s rate. time (min) % p-methyoxyphenylacetylene 67 85.9 161 70.0 241 57.6 381 40.7 479 32.4 545 27.7 604 24 Determine whether this reaction is first-order or second-order in -methoxyphenylacetylene. To determine the reaction’s order we plot ln(%pmethoxyphenylacetylene) versus time for a first-order reaction, and (%p-methoxyphenylacetylene) versus time for a second-order reaction (see Figure A17.1). Because a straight-line for the first-order plot fits the data nicely, we conclude that the reaction is first-order in -methoxyphenylacetylene. Note that when we plot the data using the equation for a second-order reaction, the data show curvature that does not fit the straight-line model. Unfortunately, most reactions of importance in analytical chemistry do not follow the simple first-order or second-order rate laws discussed above. We are more likely to encounter the second-order rate law given in equation A17.11 than that in equation A17.10. \[R = k\mathrm{[A,B]}\tag{A17.11} \] Demonstrating that a reaction obeys the rate law in equation A17.11 is complicated by the lack of a simple integrated form of the rate law. Often we can simplify the kinetics by carrying out the analysis under conditions where the concentrations of all species but one are so large that their concentrations remain effectively constant during the reaction. For example, if the concentration of B is selected such that [B] >> [A], then equation A17.11 simplifies to \[R = k´[\ce A] \nonumber \] where the rate constant ´ is equal to [B]. Under these conditions, the reaction appears to follow first-order kinetics in A and, for this reason we identify the reaction as pseudo-first-order in A. We can verify the reaction order for A using either the integrated rate law or by observing the effect on the reaction’s rate of changing the concentration of A. To find the reaction order for B, we repeat the process under conditions where [A] >> [B]. A variation on the use of pseudo-ordered reactions is the initial rate method. In this approach we run a series of experiments in which we change one at a time the concentration of those species expected to affect the reaction’s rate and measure the resulting initial rate. Comparing the reaction’s initial rate for two experiments in which only the concentration of one species is different allows us to determine the reaction order for that species. The application of this method is outlined in the following example. The following data was collected during a kinetic study of the iodation of acetone by measuring the concentration of unreacted I in solution. experiment number [C H O] (M) [H O ] (M) [I ] (M) Rate (M s ) 1 1.33 0.0404 6.65×10 1.78×10 2 1.33 0.0809 6.65×10 3.89×10 3 1.33 0.162 6.65×10 8.11×10 4 1.33 0.323 6.65×10 1.66×10 5 0.167 0.323 6.65×10 1.64×10 6 0.333 0.323 6.65×10 3.76×10 7 0.667 0.323 6.65×10 7.55×10 8 0.333 0.323 3.32×10 3.57×10 The order of the rate law with respect to the three reactants is determined by comparing the rates of two experiments in which there is a change in concentration for only one of the reactants. For example, in experiment 2 the [H O ] and the rate are approximately twice as large as in experiment 1, indicating that the reaction is first-order in [H O ]. Working in the same manner, experiments 6 and 7 show that the reaction is also first order with respect to [C H O], and experiments 6 and 8 show that the rate of the reaction is independent of the [I ]. Thus, the rate law is \[R = k\ce{[C3H6O,H3O+]} \nonumber \] To determine the value of the rate constant, we substitute the rate, the [C H O], and the [H O ] for each experiment into the rate law and solve for . Using the data from experiment 1, for example, gives a rate constant of 3.31×10 M sec . The average rate constant for the eight experiments is 3.49×10 M sec . | 10,323 | 2,337 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Diffraction_Scattering_Techniques/X-ray_Crystallography |
X-ray Crystallography is a scientific method used to determine the arrangement of atoms of a crystalline solid in three dimensional space. This technique takes advantage of the interatomic spacing of most crystalline solids by employing them as a diffraction gradient for x-ray light, which has wavelengths on the order of 1 angstrom (10 cm). In 1895, Wilhelm Rontgen discovered x- rays. The nature of x- rays, whether they were particles or electromagnetic radiation, was a topic of debate until 1912. If the wave idea was correct, researchers knew that the wavelength of this light would need to be on the order of 1 Angstrom (A) (10 cm). Diffraction and measurement of such small wavelengths would require a gradient with spacing on the same order of magnitude as the light. In 1912, Max von Laue, at the University of Munich in Germany, postulated that atoms in a crystal lattice had a regular, periodic structure with interatomic distances on the order of 1 A. Without having any evidence to support his claim on the periodic arrangements of atoms in a lattice, he further postulated that the crystalline structure can be used to diffract x-rays, much like a gradient in an infrared spectrometer can diffract infrared light. His postulate was based on the following assumptions: the atomic lattice of a crystal is periodic, x- rays are electromagnetic radiation, and the interatomic distance of a crystal are on the same order of magnitude as x- ray light. Laue's predictions were confirmed when two researchers: Friedrich and Knipping, successfully photographed the diffraction pattern associated with the x-ray radiation of crystalline \(CuSO_4 \cdot 5H_2O\). The science of x-ray crystallography was born. The arrangement of the atoms needs to be in an ordered, periodic structure in order for them to diffract the x-ray beams. A series of mathematical calculations is then used to produce a diffraction pattern that is characteristic to the particular arrangement of atoms in that crystal. X-ray crystallography remains to this day the primary tool used by researchers in characterizing the structure and bonding of organometallic compounds. Diffraction is a phenomena that occurs when light encounters an obstacle. The waves of light can either bend around the obstacle, or in the case of a slit, can travel through the slits. The resulting diffraction pattern will show areas of constructive interference, where two waves interact in phase, and destructive interference, where two waves interact out of phase. Calculation of the phase difference can be explained by examining Figure 1 below. In the figure below, two parallel waves, BD and AH are striking a gradient at an angle \(θ_o\). The incident wave BD travels farther than AH by a distance of CD before reaching the gradient. The scattered wave (depicted below the gradient) HF, travels father than the scattered wave DE by a distance of HG. So the total path difference between path AHGF and BCDE is CD - HG. To observe a wave of high intensity (one created through constructive interference), the difference CD - HG must equal to an integer number of wavelengths to be observed at the angle psi, \(CD - HG = n\lambda\), where \(\lambda\) is the wavelength of the light. Applying some basic trigonometric properties, the following two equations can be shown about the lines: \[CD = x \cos(θ o)\] and \[HG = x \cos (θ) \] where \(x\) is the distance between the points where the diffraction repeats. Combining the two equations, Diffraction of an x-ray beam, occurs when the light interacts with the electron cloud surrounding the atoms of the crystalline solid. Due to the periodic crystalline structure of a solid, it is possible to describe it as a series of planes with an equal interplaner distance. As an x-ray's beam hits the surface of the crystal at an angle ?, some of the light will be diffracted at that same angle away from the solid (Figure 2). The remainder of the light will travel into the crystal and some of that light will interact with the second plane of atoms. Some of the light will be diffracted at an angle \(theta\), and the remainder will travel deeper into the solid. This process will repeat for the many planes in the crystal. The x-ray beams travel different pathlengths before hitting the various planes of the crystal, so after diffraction, the beams will interact constructively only if the path length difference is equal to an integer number of wavelengths (just like in the normal diffraction case above). In the figure below, the difference in path lengths of the beam striking the first plane and the beam striking the second plane is equal to BG + GF. So, the two diffracted beams will constructively interfere (be in phase) only if \(BG + GF = n \lambda\). Basic trigonometry will tell us that the two segments are equal to one another with the interplaner distance times the sine of the angle \(\theta\). So we get: \[ BG = BC = d \sin \theta \label{1}\] Thus, \[ 2d \sin \theta = n \lambda \label{2}\] This equation is known as Bragg's Law, named after W. H. Bragg and his son, W. L. Bragg; who discovered this geometric relationship in 1912. {C}{C}Bragg's Law relates the distance between two planes in a crystal and the angle of reflection to the x-ray wavelength. The x-rays that are diffracted off the crystal have to be in-phase in order to signal. Only certain angles that satisfy the following condition will register: \[ \sin \theta = \dfrac{n \lambda}{2d} \label{3} \] For historical reasons, the resulting diffraction spectrum is represented as intensity vs. \(2θ\). The main components of an x-ray instrument are similar to those of many optical spectroscopic instruments. These include a source, a device to select and restrict the wavelengths used for measurement, a holder for the sample, a detector, and a signal converter and readout. However, for x-ray diffraction; only a source, sample holder, and signal converter/readout are required. x-ray tubes provides a means for generating x-ray radiation in most analytical instruments. An evacuated tube houses a tungsten filament which acts as a cathode opposite to a much larger, water cooled anode made of copper with a metal plate on it. The metal plate can be made of any of the following metals: chromium, tungsten, copper, rhodium, silver, cobalt, and iron. A high voltage is passed through the filament and high energy electrons are produced. The machine needs some way of controlling the intensity and wavelength of the resulting light. The intensity of the light can be controlled by adjusting the amount of current passing through the filament; essentially acting as a temperature control. The wavelength of the light is controlled by setting the proper accelerating voltage of the electrons. The voltage placed across the system will determine the energy of the electrons traveling towards the anode. X-rays are produced when the electrons hit the target metal. Because the energy of light is inversely proportional to wavelength (\(E=hc=h(1/\lambda\)), controlling the energy, controls the wavelength of the x-ray beam. Monochromators and filters are used to produce monochromatic x-ray light. This narrow wavelength range is essential for diffraction calculations. For instance, a zirconium filter can be used to cut out unwanted wavelengths from a molybdenum metal target (see figure 4). The molybdenum target will produce x-rays with two wavelengths. A zirconium filter can be used to absorb the unwanted emission with wavelength K while allowing the desired wavelength, K to pass through. The sample holder for an x-ray diffraction unit is simply a needle that holds the crystal in place while the x-ray diffractometer takes readings. In x-ray diffraction, the detector is a transducer that counts the number of photons that collide into it. This photon counter gives a digital readout in number of photons per unit time. Below is a figure of a typical x-ray diffraction unit with all of the parts labeled. In mathematics, a Fourier transform is an operation that converts one real function into another. In the case of FTIR, a Fourier transform is applied to a function in the time domain to convert it into the frequency domain. One way of thinking about this is to draw the example of music by writing it down on a sheet of paper. Each note is in a so-called "sheet" domain. These same notes can also be expressed by playing them. The process of playing the notes can be thought of as converting the notes from the "sheet" domain into the "sound" domain. Each note played represents exactly what is on the paper just in a different way. This is precisely what the Fourier transform process is doing to the collected data of an x-ray diffraction. This is done in order to determine the electron density around the crystalline atoms in real space. The following equations can be used to determine the electrons' position: \[p(x,y,z) = \sum_h \sum_k \sum_l F(hkl) e ^{-2\pi i (hx+ky+lz)} \label{1A}\] \[ \int _0^1 \int _0^1 \int _0^1 p(x,y,z) e ^{2\pi i (hx+ky+lz)} dx\;dy\;dz \label{2B}\] \[F(q) = | F(q) | e^{i \phi(q)} \label{3C}\] where \(p(xyz)\) is the electron density function, and \(F(hkl)\) is the electron density function in real space. Equation 1 represents the Fourier expansion of the electron density function. To solve for \(F(hkl)\), the equation 1 needs to be evaluated over all values of h, k, and l, resulting in Equation 2. The resulting function \(F(hkl)\) is generally expressed as a complex number (as seen in equation 3 above) with \(| F(q)|\) representing the magnitude of the function and \(\phi\) representing the phase. In order to run an x-ray diffraction experiment, one must first obtain a crystal. In organometallic chemistry, a reaction might work but when no crystals form, it is impossible to characterize the products. Crystals are grown by slowly cooling a supersaturated solution. Such a solution can be made by heating a solution to decrease the amount of solvent present and to increase the solubility of the desired compound in the solvent. Once made, the solution must be cooled gradually. Rapid temperature change will cause the compound to crash out of solution, trapping solvent and impurities within the newly formed matrix. Cooling continues as a seed crystal forms. This crystal is a point where solute can deposit out of the solution and into the solid phase. Solutions are generally placed into a freezer (-78 ºC) in order to ensure all of the compound has crystallized. One way to ensure gradual cooling in a -78 ºC freezer is to place the container housing the compound into a beaker of ethanol. The ethanol will act as a temperature buffer, ensuring a slow decrease in the temperature gradient between the flask and the freezer. Once crystals are grown, it is imperative that they remain cold as any addition of energy will cause a disruption of the crystal lattice, which will yield bad diffraction data. The result of an organometallic chromium compound crystallization can be seen below. Due to the air-sensitivity of most organometallic compounds, crystals must be transported in a highly viscous organic compound called paratone oil (Figure \(\Page {7}\)). Crystals are abstracted from their respective Schlenks by dabbing the end of a spatula with the paratone oil and then sticking the crystal onto the oil. Although there might be some exposure of the compounds to air and water, crystals can withstand more exposure than solution (of the preserved protein) before degrading. On top of serving to protect the crystal, the paratone oil also serves as the glue to bind the crystal to the needle. To describe the periodic, three dimensional nature of crystals, the are employed: \[ a(\cos \theta_o – \cos \theta) = h\lambda \label{eq1}\] \[b(\cos \theta_o – \cos \theta) = k\lambda \label{eq2}\] \[c(\cos \theta_o – \cos \theta) = l\lambda \label{eq3}\] where \(a\), \(b\), and \(c\) are the three axes of the unit cell, \(θ_o\), \(o\), \(?o\) are the angles of incident radiation, and ?, ?, ? are the angles of the diffracted radiation. A diffraction signal (constructive interference) will arise when \(h\), \(k\), and \(l\) are integer values. The rotating crystal method employs these equations. X-ray radiation is shown onto a crystal as it rotates around one of its unit cell axis. The beam strikes the crystal at a 90 degree angle. Using equation 1 above, we see that if \(\theta_o\) is 90 degrees, then \(\cos \theta_o = 0\). For the equation to hold true, we can set h=0, granted that \(\theta= 90^o\). The above three equations will be satisfied at various points as the crystal rotates. This gives rise to a diffraction pattern (shown in the image below as multiple h values). The cylindrical film is then unwrapped and developed. The following equation can be used to determine the length axis around which the crystal was rotated: \[ a = \dfrac{ch \lambda}{\sin \tan^{-1} (y/r}\] where \(a\) is the length of the axis, y is the distance from \(h=0\) to the \(h\) of interest, \(r\) is the radius of the firm, and ? is the wavelength of the x-ray radiation used. The first length can be determined with ease, but the other two require far more work, including remounting the crystal so that it rotates around that particular axis. The crystals that form are frozen in liquid nitrogen and taken to the synchrotron which is a highly powered tunable x-ray source. They are mounted on a goniometer and hit with a beam of x-rays. Data is collected as the crystal is rotated through a series of angles. The angle depends on the symmetry of the crystal. that are used for x-ray Crystallography studies. They are involved in many pathways in biology, often catalyzing reactions by increasing the reaction rate. Most scientists use x-ray Crystallography to solve the structures of protein and to determine functions of residues, interactions with substrates, and interactions with other proteins or nucleic acids. Proteins can be co - crystallized with these substrates, or they may be soaked into the crystal after crystallization. Proteins will solidify into crystals under certain conditions. These conditions are usually made up of salts, buffers, and precipitating agents. This is often the hardest step in x-ray crystallography. Hundreds of conditions varying the salts, pH, buffer, and precipitating agents are combined with the protein in order to crystallize the protein under the right conditions. This is done using 96 well plates; each well containing a different condition and crystals; which form over the course of days, weeks, or even months. The pictures below are crystals of APS Kinase D63N from taken at the Chemistry building at UC Davis after crystals formed over a period of a week. | 14,783 | 2,338 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/06%3A_Bonding_in_Organic_Molecules/6.02%3A_Hydrogenlike_Atomic_Orbitals |
With the modern concept of a hydrogen atom we do not visualize the orbital electron traversing a simple planetary orbit. Rather, we speak of an atomic orbital, in which there is only a of finding the electron in a particular volume a given distance and direction from the nucleus. The boundaries of such an orbital are not distinct because there always remains a finite, even if small, probability of finding the electron relatively far from the nucleus. There are several discrete atomic orbitals available to the electron of a hydrogen atom. These orbitals differ in energy, size, and shape, and mathematical descriptions for each are possible. Following is a qualitative description of the nature of some of the hydrogen atomic orbitals. The most stable or of a hydrogen atom is designated \(1s\).\(^1\) In the \(1s\) state the electron is, on the average, closest to the nucleus (i.e., it is the state with the smallest atomic orbital). . This means that the probability of finding the electron at a given distance \(r\) from the nucleus is independent of the direction from the nucleus. We shall represent the \(1s\) orbital as a sphere centered on the nucleus with a radius such that the probability of finding the electron within the boundary surface is high (0.80 to 0.95); see Figure 6-1. This may seem arbitrary, but an orbital representation that would have a probability of 1 for finding the electron within the boundary surface would have an infinite radius. The reason is that there is a finite, even if small, probability of finding the electron at given distance from the nucleus. The boundary surfaces we choose turn out to have sizes consistent with the distances between the nuclei of bonded atoms. The \(2s\) orbital is very much like the \(1s\) orbital except that it is larger and therefore more diffuse, and it has a higher energy. For principal quantum number 2, there are also three orbitals of equal energies called \(2p\) orbitals, which have different geometry than the \(s\) orbitals. These are shown in Figure 6-2, in which we see that the respective axes passing through the tangent spheres of the three \(p\) orbitals lie at right angles to one another. The \(p\) orbitals are spherically symmetrical. The \(3s\) and \(3p\) states are similar to the \(2s\) and \(2p\) states but are of higher energy. The \(3d\), \(4d\), \(4f\), \(\cdots\), orbitals have still higher energies and quite different geometries; they are not important for bonding in most organic substances, at least for carbon compounds with hydrogen and elements in the first main row (\(Li\)-\(Ne\)) of the periodic table. The sequence of orbital energies is shown in Figure 6-3. The famous states that no more than two elections can occupy a given orbital and then only if they differ with respect to a property of electrons called . An electron can have only one of two possible orientations of electron spin, as may be symbolized by \(\uparrow\) and \(\downarrow\). Two electrons with "paired" spins often are represented as \(\uparrow \downarrow\). Such a pair of electrons can occupy a orbital. The symbols \(\uparrow \uparrow\) (or \(\downarrow \downarrow\)) represent two electrons, which may go into a single orbital. If we assume that all atomic nuclei have orbitals like those of the hydrogen atom, we can see how atoms more complex than hydrogen can be built up by adding electrons to the orbitals in accord with the Pauli exclusion principle. The lowest-energy states will be those in which the electrons are added to the lowest-energy orbitals. For example, the electronic configuration of the lowest-energy state of a carbon atom is shown in Figure 6-4, which also shows the relative energies of the \(1s\) through \(4p\) atomic orbitals. The orbitals of lowest energy are populated with the proper number of electrons to balance the nuclear charge of \(+6\) for carbon and to preserve the Pauli condition of no more than two paired electrons per orbital. However, the two highest-energy electrons are put into \(2p\) orbitals with unpaired spins in accordance with . The rationale of Hund's rule depends on the fact that electrons come closer together. Now, suppose there are two electrons that can go into two different orbitals of the same energy (so-called ). Hund's rule tells us that the repulsion energy between these electrons will be less if they have unpaired spins (\(\uparrow \uparrow\)). Why is this so? Because if they have unpaired spins they cannot be in the orbital at the time. Therefore they will not be able to approach each other as closely as they would if they could be in the same orbital at the same time. For this reason the is expected to be more stable than the configuration if both orbitals have the energy. States such as the one shown in Figure 6-4 for carbon are built up through the following steps. Helium has two paired electrons in the \(1s\) orbital; its configuration can be written as \(\left( 1s \right)^2\), the superscript outside the parentheses denoting two paired electrons in the \(1s\) orbital. For lithium, we expect \(Li \: \left( 1s \right)^2 \left( 2s \right)^1\) to be the ground state, in which the \(1s\) electrons must be paired according to the exclusion principle. Continuing in this way, we can derive the electronic configurations for the elements in the first three rows of the periodic table, as shown in Table 6-1. These configurations conform to the \(^1\)The index number refers to the principal quantum number and corresponds to the "\(K\) shell" designation often used for the electron of the normal hydrogen atom. The principal quantum number 2 corresponds to the \(L\) shell, 2 to the \(M\) shell, and so on. The notation \(s\) (also \(p\), \(d\), \(f\) to come later) has been carried over from the early days of atomic spectroscopy and was derived from descriptions of spectroscopic lines as "sharp", "principal", "diffuse", and "fundamental," which once were used to identify transitions from particular atomic states. and (1977) | 6,064 | 2,339 |
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In this Module, the quantitative determination of a reaction rate is demonstrated. Reaction rates can be determined over particular time intervals or at a given point in time. A rate law describes the relationship between reactant rates and reactant concentrations. Reaction rates are usually expressed as the concentration of reactant consumed or the concentration of product formed per unit time. The units are thus moles per liter per unit time, written as M/s, M/min, or M/h. To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses, perhaps plot the concentration as a function of time on a graph, and then calculate the change in the concentration per unit time. The progress of a simple reaction (A → B) is shown in the beakers are snapshots of the composition of the solution at 10 s intervals. The number of molecules of reactant (A) and product (B) are plotted as a function of time in the graph. Each point in the graph corresponds to one beaker in . The reaction rate is the change in the concentration of either the reactant or the product over a period of time. The concentration of A decreases with time, while the concentration of B increases with time. \[\textrm{rate}=\dfrac{\Delta [\textrm B]}{\Delta t}=-\dfrac{\Delta [\textrm A]}{\Delta t} \label{Eq1}\] Square brackets indicate molar concentrations, and the capital Greek delta (Δ) means “change in.” Because chemists follow the convention of expressing all reaction rates as positive numbers, however, a negative sign is inserted in front of Δ[A]/Δt to convert that expression to a positive number. The reaction rate calculated for the reaction A → B using Equation \(\ref{Eq1}\) is different for each interval (this is not true for every reaction, as shown below). A greater change occurs in [A] and [B] during the first 10 s interval, for example, than during the last, meaning that the reaction rate is greatest at first. Reaction rates generally decrease with time as reactant concentrations decrease. We can use Equation \(\ref{Eq1}\) to determine the reaction rate of hydrolysis of aspirin, probably the most commonly used drug in the world (more than 25,000,000 kg are produced annually worldwide). Aspirin (acetylsalicylic acid) reacts with water (such as water in body fluids) to give salicylic acid and acetic acid, as shown in Figure \(\Page {2}\). Because salicylic acid is the actual substance that relieves pain and reduces fever and inflammation, a great deal of research has focused on understanding this reaction and the factors that affect its rate. Data for the hydrolysis of a sample of aspirin are in Table \(\Page {1}\) and are shown in the graph in . These data were obtained by removing samples of the reaction mixture at the indicated times and analyzing them for the concentrations of the reactant (aspirin) and one of the products (salicylic acid). The for a given time interval can be calculated from the concentrations of either the reactant or one of the products at the beginning of the interval (time = t ) and at the end of the interval (t ). Using salicylic acid, the reaction rate for the interval between t = 0 h and t = 2.0 h (recall that change is always calculated as final minus initial) is calculated as follows: The reaction rate can also be calculated from the concentrations of aspirin at the beginning and the end of the same interval, remembering to insert a negative sign, because its concentration decreases: If the reaction rate is calculated during the last interval given in Table \(\Page {1}\)(the interval between 200 h and 300 h after the start of the reaction), the reaction rate is significantly slower than it was during the first interval (t = 0–2.0 h): In the preceding example, the stoichiometric coefficients in the balanced chemical equation are the same for all reactants and products; that is, the reactants and products all have the coefficient 1. Consider a reaction in which the coefficients are not all the same, the fermentation of sucrose to ethanol and carbon dioxide: The coefficients indicate that the reaction produces four molecules of ethanol and four molecules of carbon dioxide for every one molecule of sucrose consumed. As before, the reaction rate can be found from the change in the concentration of any reactant or product. In this particular case, however, a chemist would probably use the concentration of either sucrose or ethanol because gases are usually measured as volumes and the volume of CO gas formed depends on the total volume of the solution being studied and the solubility of the gas in the solution, not just the concentration of sucrose. The coefficients in the balanced chemical equation tell us that the reaction rate at which ethanol is formed is always four times faster than the reaction rate at which sucrose is consumed: The concentration of the reactant—in this case sucrose— with time, so the value of Δ[sucrose] is negative. Consequently, a minus sign is inserted in front of Δ[sucrose] in so the rate of change of the sucrose concentration is expressed as a positive value. Conversely, the ethanol concentration with time, so its rate of change is automatically expressed as a positive value. Often the reaction rate is expressed in terms of the reactant or product with the smallest coefficient in the balanced chemical equation. The smallest coefficient in the sucrose fermentation reaction ( ) corresponds to sucrose, so the reaction rate is generally defined as follows: Consider the thermal decomposition of gaseous N O to NO and O via the following equation: Write expressions for the reaction rate in terms of the rates of change in the concentrations of the reactant and each product with time. balanced chemical equation reaction rate expressions Because O has the smallest coefficient in the balanced chemical equation for the reaction, define the reaction rate as the rate of change in the concentration of O and write that expression. The balanced chemical equation shows that 2 mol of N O must decompose for each 1 mol of O produced and that 4 mol of NO are produced for every 1 mol of O produced. The molar ratios of O to N O and to NO are thus 1:2 and 1:4, respectively. This means that the rate of change of [N O ] and [NO ] must be divided by its stoichiometric coefficient to obtain equivalent expressions for the reaction rate. For example, because NO is produced at four times the rate of O , the rate of production of NO is divided by 4. The reaction rate expressions are as follows: The key step in the industrial production of sulfuric acid is the reaction of SO with O to produce SO . \[2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}\] Write expressions for the reaction rate in terms of the rate of change of the concentration of each species. Using the reaction shown in Example \(\Page {1}\), calculate the reaction rate from the following data taken at 56°C: \[2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}\] balanced chemical equation and concentrations at specific times reaction rate Calculate the reaction rate in the interval between t = 240 s and t = 600 s. From Example \(\Page {1}\), the reaction rate can be evaluated using any of three expressions: Subtracting the initial concentration from the final concentration of N O and inserting the corresponding time interval into the rate expression for N O , Substituting actual values into the expression, Similarly, NO can be used to calculate the reaction rate: Allowing for experimental error, this is the same rate obtained using the data for N O . The data for O can also be used: Again, this is the same value obtained from the N O and NO data. Thus, the reaction rate does not depend on which reactant or product is used to measure it. Using the data in the following table, calculate the reaction rate of SO (g) with O (g) to give SO (g). \[2SO_{2(g)} + O_{2(g)} \rightarrow 2SO_{3(g)}\] 9.0 × 10 M/s The of a reaction is the reaction rate at any given point in time. As the period of time used to calculate an average rate of a reaction becomes shorter and shorter, the average rate approaches the instantaneous rate. Comparing this to calculus, the instantaneous rate of a reaction at a given time corresponds to the slope of a line tangent to the concentration-versus-time curve at that point—that is, the derivative of concentration with respect to time. \[ rate = \lim_{\Delta t \rightarrow 0} \dfrac{-\Delta [R]}{\Delta t} = - \dfrac{d[R]}{dt}\] The distinction between the instantaneous and average rates of a reaction is similar to the distinction between the actual speed of a car at any given time on a trip and the average speed of the car for the entire trip. Although the car may travel for an extended period at 65 mph on an interstate highway during a long trip, there may be times when it travels only 25 mph in construction zones or 0 mph if you stop for meals or gas. The average speed on the trip may be only only 50 mph, whereas the instantaneous speed on the interstate at a given moment may be 65 mph. Whether the car can be stopped in time to avoid an accident depends on its instantaneous speed, not its average speed. There are important differences between the speed of a car during a trip and the speed of a chemical reaction, however. The speed of a car may vary unpredictably over the length of a trip, and the initial part of a trip is often one of the slowest. In a chemical reaction, the initial interval typically has the fastest rate (though this is not always the case), and the reaction rate generally changes smoothly over time. Chemical kinetics generally focuses on one particular instantaneous rate, which is the initial reaction rate, t = 0. Initial rates are determined by measuring the reaction rate at various times and then extrapolating a plot of rate versus time to t = 0. | 9,959 | 2,341 |
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A common paradigm in solving for pHs in weak acids and bases is that the equilibria of solutions containing one weak acid or one weak base. In most cases, the amount of \(\ce{H+}\) from the autoionization of water is negligible. For very dilute solutions, the amount of \(\ce{H+}\) ions from the autoionization of water must also be taken into account. Thus, a strategy is given here to deal with these systems. When two or more acids are present in a solution, the concentration of \(\ce{H+}\) (or pH) of the solution depends on the concentrations of the acids and their acidic constants . The hydrogen ion is produced by the ionization of all acids, but the ionizations of the acids are governed by their equilibrium constants, 's. Similarly, the concentration of \(\ce{OH-}\) ions in a solution containing two or more weak bases depends on the concentrations and values of the bases. For simplicity, we consider two acids in this module, but the strategies used to discuss equilibria of two acids apply equally well to that of two bases. If the pH is between 6 and 8, the contribution due to autoionization of water to \(\ce{[H+]}\) should also be considered. When autoionization of water is considered, the method is called the or the . This method is illustrated below. When the contribution of pH due to self-ionization of water cannot be neglected, there are two coupled equilibria to consider: \[\ce{HA \rightleftharpoons H+ + A-} \] and \[\ce{H2O \rightleftharpoons H+ + OH-}\] Thus, \[\begin{align}
\ce{[H+]} &= ({\color{Red} x+y})\\
\ce{[A- ]} &= x\\
\ce{[OH- ]} &= y
\end{align}\] and the two equilibrium constants are \[K_{\large\textrm{a}} = \dfrac{ ({\color{Red} x + y})\, x}{C - x} \label{1}\] and \[K_{\large\textrm{w}} = ({\color{Red} x + y})\, y \label{2}\] Although you may use the method of successive approximation, the formula to calculate the pH can be derived directly from Equations \(\ref{1}\) and \(\ref{2}\). Solving for \(\color{ref} x\) from Equation \(\ref{2}\) gives \[x = \dfrac{K_{\large\textrm{w}}}{y} - y\] and substituting this expression into \(\ref{1}\) results in \[K_{\large\textrm{a}} = \dfrac{({\color{Red} x+y}) \left(\dfrac{K_{\large\textrm{w}}}{y} - y\right)}{C - \dfrac{K_{\large\textrm{w}}}{y} + y}\] Rearrange this equation to give: \[\begin{align}
\ce{[H+]} &= ({\color{Red} x+y})\\
&= \dfrac{C - \dfrac{K_{\large\textrm{w}}}{y} + y}{\dfrac{K_{\large\textrm{w}}}{y} - y}\, K_{\large\textrm{a}}
\end{align}\] Note that \[\dfrac{K_{\large\textrm w}}{y} = \ce{[H+]}\] so \[y = \dfrac{K_{\large\textrm w}}{\ce{[H+]}}.\] Thus, we get: \[\ce{[H+]} = \dfrac{ C - \ce{[H+]} + \dfrac{K_{\large\textrm{w}}}{\ce{[H+]}}}{\ce{[H+]} - \dfrac{K_{\large\textrm{w}}}{\ce{[H+]}}}\, K_{\large\textrm{a}} \label{Exact}\] As written, Equation \(\ref{Exact}\) is complicated, but can be put into a polynomial form \[\ce{[H+]^3} + K_{\large\textrm{a}} \ce{[H+]^2} - \left( K_{\large\textrm{w}} + C K_{\large\textrm{a}} \right) \ce{[H+]} - K_{\large\textrm{w}} K_{\large\textrm{a}} =0 \label{Exact2}\] Solving for the exact hydronium concentration requires solving a third-order polynomial. While this is , it is an awkward equation to handle. Instead, we often consider two approximations to Equation \(\ref{Exact2}\) that can made under limiting conditions. If \([H^+] > 1 \times 10^{-6}\), then \[\dfrac{K_w}{[H^+]} < 1 \times 10^{-8}.\] This is small indeed compared to \([H^+]\) and \(C\) in Equation \(\ref{Exact}\). Thus, \[[H^+] \approx \dfrac{C - [H^+]}{[H^+]} K_{\large\textrm{a}}\] \[[H^+]^2 + K_{\large\textrm{a}} [H^+] - C K_{\large\textrm{a}} \approx 0 \label{Quad}\] Equation \(\ref{Quad}\) is a quadratic equation with two solutions. However, only one will be positive and real: \[[H^+] \approx \dfrac{-K_{\large\textrm{a}} + \sqrt{K_{\large\textrm{a}}^2 + 4 C K_{\large\textrm{a}}}}{2}\] If \([H^+] \ll C\), then \[C - [H^+] \approx C\] Equation \(\ref{Exact}\) can be simplified \(\begin{align}
[H^+] &\approx \dfrac{C}{[H^+]}\: K_{\large\textrm{a}}\\
[H^+] &\approx \sqrt{C K_{\large\textrm{a}}}
\end{align}\) The treatment presented in deriving Equation \(\ref{Exact}\) is more general, and may be applied to problems involving two or more weak acids in one solution. Calculate the \(\ce{[H+]}\), \(\ce{[Ac- ]}\), and \(\ce{[Cc- ]}\) when the solution contains 0.200 M \(\ce{HAc}\) (\(K_a = 1.8 \times 10^{-5}\)), and 0.100 M \(\ce{HCc}\) (the acidity constant \(K_c = 1.4 \times 10^{-3}\)). (\(\ce{HAc}\) is acetic acid whereas \(\ce{HCc}\) is chloroacetic acid).
Assume x and y to be the concentrations of \(\ce{Ac-}\) and \(\ce{Cc-}\), respectively, and write the concentrations below the equations: \(\begin{array}{ccccc}
\ce{HAc &\rightleftharpoons &H+ &+ &Ac-}\\
0.200-x &&x &&x\\
\\
\ce{HCc &\rightleftharpoons &H+ &+ &Cc-}\\
0.100-y &&y &&y
\end{array}\) \[\ce{[H+]} = (x + y) \nonumber\] Thus, you have \[\dfrac{(x + y)\, x}{0.200 - x} = 1.8 \times 10^{-5} \label{Ex1.1}\] \[\dfrac{(x + y)\, y}{0.100 - y} = 1.4\times 10^{-3} \label{Ex1.2}\] Solving for x and y from Equations \(\ref{Ex1.1}\) and \(\ref{Ex1.2}\) may seem difficult, but you can often make some assumptions to simplify the solution procedure. Since \(\ce{HAc}\) is a weaker acid than is \(\ce{HCc}\), you expect x << y. Further, y << 0.100. Therefore, \(x + y \approx y\) and 0.100 - y => 0.100. Equation \(\ref{Ex1.2}\) becomes: \[\dfrac{ ( y)\, y}{0.100} = 1.4 \times 10^{-3} \label{Ex1.2a}\] which leads to \[\begin{align*} y &= (1.4 \times 10^{-3} \times 0.100)^{1/2}\\ &= 0.012 \end{align*}\] Substituting \(y\) in Equation \(\ref{Ex1.1}\) results in \[\dfrac{(x + 0.012)\, x}{0.200 - x} = 1.8 \times 10^{-5} \label{1'}\] This equation is easily solved, but you may further assume that \(0.200 - x \approx 0.200\), since \(x << 0.200\). Thus, \[\begin{align*}
x &= \dfrac{-0.012 + (1.44\times 10^{-4} + 1.44\times 10^{-5})^{1/2}}{2}\\
&= 2.9\times 10^{-4}\:\: \longleftarrow \textrm{Small indeed compared to 0.200}
\end{align*}\] You had a value of 0.012 for y by neglecting the value of x in Equation \(\ref{Ex1.2}\). You can now recalculate the value for y by substituting values for x and y in Equation \(\ref{Ex1.2}\). \[\dfrac{(2.9\times 10^{-4} + y)\, y}{0.100 - 0.012} = 1.4\times 10^{-3} \label{2"}\] Solving for y in the above equation gives \[y = 0.011 \nonumber\] You have improved the y value from 0.012 to 0.011. Substituting the new value for y in a successive approximation to recalculate the value for x improves its value from \(2.9 \times 10^{-4}\) to a new value of \(3.2 \times 10^{-4}\). Use your calculator to obtain these values. Further refinement does not lead to any significant changes for x or y. You should write down these calculations on your note pad, since reading alone does not lead to thorough understanding. A weak acid \(\ce{HA}\) has a \(K_a\) value of \(4.0 \times 10^{-11}\). What are the pH and the equilibrium concentration of \(\ce{A-}\) in a solution of 0.0010 M \(\ce{HA}\)?
For the solution of this problem, two methods are given here. If you like the x and y representation, you may use method (a). The two equilibrium equations are: \[\begin{array}{ccccc}
\ce{HA &\rightleftharpoons &H+ &+ &A-};\\
0.0010-x &&x &&x\\
\\
\ce{H2O &\rightleftharpoons &H+ &+ &OH-}\\
&&y &&y
\end{array}\] \[\ce{[H+]} = (x+y)\] \[\begin{align}
\dfrac{(x+y)\, x}{0.0010-x} &= 4.0\times 10^{-}11 \label{3}\\
\\
(x+y)\, y &= 1\times 10^{-}14 \label{4}
\end{align}\] Assume y << x, and x << 0.0010, then you have \[\begin{align}
\dfrac{(x )\, x}{0.0010} &= 4.0\times 10^{-11} \label{3'} \\
x &= (0.0010 \times 4.0e^{-11})^{1/2}\\
&= 2.0\times 10^{-7}
\end{align}\] Substituting \(2.0 \times ^{-7}\) for x in 4 and solving the quadratic equation for y gives, \[(2.0\times 10^{-}7+y)\, y = 1\times 10^{-14} \nonumber\] \[y = 4.1\times 10^{-8} \nonumber\] Substituting \(4.1 \times 10^{-8}\) in Equation \(\ref{3}\), but still approximating 0.0010-x by 0.0010: \[\dfrac{(x+4.1\times 10^{-8})\, x}{0.0010} = 4.0\times 10^{-11} \label{3''}\] Solving this quadratic equation for a positive root results in \[x = 1.8 \times 10^{-7} \;\text{M} \longleftarrow \textrm{Recall }x = \ce{[A- ]} \nonumber\] \[\begin{align*}
\ce{[H+]} &= x + y\\
&= (1.8 + 0.41)\,1\times 10^{-7}\\
&= 2.2\times 10^{-7}\\
\ce{pH} &= 6.65
\end{align*}\] The next method uses the formula derived earlier. Using the formula from the exact treatment, and using \(2 \times 10^{-7}\) for all the \(\ce{[H+]}\) values on the right hand side, you obtain a new value of \(\ce{[H+]}\) on the left hand side, \[\begin{align*}
\ce{[H+]} &= \dfrac{C - \ce{[H+]} + \dfrac{K_{\large\textrm w}}{\ce{[H+]}}}{\ce{[H+]} - \dfrac{K_{\large\textrm w}}{\ce{[H+]}}} K_{\large\textrm a}\\
&= 2.24\times 10^{-7}\\
\ce{pH} &= 6.65
\end{align*}\] The new \(\ce{[H+]}\) enables you to recalculate \(\ce{[A- ]}\) from the formula: \[\begin{align*}
(2.24\times 10^{-7}) \ce{[A- ]} &= C K_{\large\textrm a} \nonumber\\
\ce{[A- ]} &= \dfrac{(0.0010) (4.0\times 10^{-11})}{2.24\times 10^{-7}} \nonumber\\
&= 1.8\times 10^{-7} \nonumber
\end{align*}\] You may have attempted to use the approximation method: \[\begin{align*}
x &= (C K_{\large\textrm a})^{1/2} \nonumber\\
&= 2.0\times 10^{-7}\: \mathrm{M\: A^-,\: or\: H^+;\: pH = 6.70} \nonumber
\end{align*}\] and obtained a pH of 6.70, which is greater than 6.65 by less than 1%. However, when an approximation is made, you have no confidence in the calculated pH of 6.70. Water is both an acid and a base due to the autoionization, \[\ce{H2O \rightleftharpoons H+ + OH-} \nonumber\] However, the amount of \(\ce{H+}\) ions from water may be very small compared to the amount from an acid if the concentration of the acid is high. When calculating \(\ce{[H+]}\) in an acidic solution, approximation method or using the quadratic formula has been discussed in the modules on weak acids. | 9,903 | 2,342 |
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The kinetic theory of gases can be used to model the frequency of collisions between hard-sphere molecules, which is proportional to the reaction rate. Most systems undergoing a chemical reaction, however, are much more complex. The reaction rates may be dependent on specific interactions between reactant molecules, the phase(s) in which the reaction takes place, etc. The field of chemical kinetics is thus by-and-large based on empirical observations. From experimental observations, scientists have established that reaction rates almost always have a power-law dependence on the concentrations of one or more of the reactants. In the following sections, we will discuss different power laws that are commonly observed in chemical reactions. Consider a closed container initially filled with chemical species \(A\). At \(t = 0\), a stimulus, such as a change in temperature, the addition of a catalyst, or irradiation, causes an irreversible chemical reaction to occur in which \(A\) transforms into product \(B\): \[a \text{A} \longrightarrow b \text{B}\] The rate that the reaction proceeds, \(r\), can be described as the change in the concentrations of the chemical species with respect to time: \[r = -\dfrac{1}{a} \dfrac{d \left[ \text{A} \right]}{dt} = \dfrac{1}{b} \dfrac{d \left[ \text{B} \right]}{dt} \label{19.1}\] where \(\left[ \text{X} \right]\) denotes the molar concentration of chemical species \(\text{X}\) with units of \(\frac{\text{mol}}{\text{L}^3}\). Let us first examine a reaction \(\text{A} \longrightarrow \text{B}\) in which the reaction rate, \(r\), is constant with time: \[r = -\dfrac{d \left[ \text{A} \right]}{dt} = k \label{19.2}\] where \(k\) is a constant, also known as the rate constant with units of \(\dfrac{\text{mol}}{\text{m}^3 \text{s}}\). Such reactions are called zeroth order reactions because the reaction rate depends on the concentrations of species \(\text{A}\) and \(\text{B}\) to the \(0^{th}\) power. Integrating \(\left[ \text{A} \right]\) with respect to \(T\), we find that \[\left[ \text{A} \right] = -kt + c_1 \label{19.3}\] At \(t = 0\), \(\left[ \text{A} \right] \left( 0 \right) = \left[ \text{A} \right]_0\). Plugging these values into the equation, we find that \(c_1 = \left[ \text{A} \right]_0\). The final form of the equation is: \[\left[ \text{A} \right] = \left[ \text{A} \right]_0 - kt \label{19.4}\] A plot of the concentration of species \(\text{A}\) with time for a \(0^{th}\) order reaction is shown in Figure \(\Page {1}\), where the slope of the line is \(-k\) and the \(y\)-intercept is \(\left[ \text{A} \right]_0\). Such reactions in which the reaction rates are independent of the concentrations of products and reactants are rare in nature. An example of a system displaying \(0^{th}\) order kinetics would be one in which a reaction is mediated by a catalyst present in small amounts. Experimentally, it is observed than when a chemical reaction is of the form \[\sum_i \nu_i \text{A}_i = 0 \label{19.5}\] the reaction rate can be expressed as \[ r = k \prod_\text{reactants} \left[ \text{A}_i \right]^{\nu_i} \label{19.6}\] where it is assumed that the stoichiometric coefficients \(\nu_i\) of the reactants are all positive. Thus, or a reaction \(\text{A} \longrightarrow \text{B}\), the reaction rate depends on \(\left[ \text{A} \right]\) raised to the first power: \[r = \dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right] \label{19.7}\] For first order reactions, \(k\) has the units of \(\dfrac{1}{\text{s}}\). Integrating and applying the condition that at \(t = 0 \: \text{s}\), \(\left[ \text{A} \right] = \left[ \text{A} \right]_0\), we arrive at the following equation: \[\left[ \text{A} \right] = \left[ \text{A} \right] e^{-kt} \label{19.8}\] Figure \(\Page {2}\) displays the concentration profiles for species \(\text{A}\) and \(\text{B}\) for a first order reaction. To determine the value of \(k\) from experimental data, it is convenient to take the natural log of Equation \(\ref{19.8}\): \[\text{ln} \left( \left[ \text{A} \right] \right) = \text{ln} \left( \left[ \text{A} \right]_0 \right) - kt \label{19.9}\] For a first order irreversible reaction, a plot of \(\text{ln} \left( \left[ \text{A} \right] \right)\) vs. \(t\) is straight line with a slope of \(-k\) and a \(y\)-intercept of \(\text{ln} \left( \left[ \text{A} \right]_0 \right)\). Another type of reaction depends on the square of the concentration of species \(\text{A}\) - these are known as second order reactions. For a second order reaction in which \(2 \text{A} \longrightarrow \text{B}\), we can write the reaction rate to be \[r = -\dfrac{1}{2} \dfrac{d \left[ \text{A} \right]}{dt} = k \left[ \text{A} \right]^2 \label{19.10}\] For second order reactions, \(k\) has the units of \(\dfrac{\text{m}^3}{\text{mol} \cdot \text{s}}\). Integrating and applying the condition that at \(t = 0 \: \text{s}\), \(\left[ \text{A} \right] = \left[ \text{A} \right]_0\), we arrive at the following equation for the concentration of \(\text{A}\) over time: \[\left[ \text{A} \right] = \dfrac{1}{2kt + \dfrac{1}{\left[ \text{A} \right]_0}} \label{19.11}\] Figure \(\Page {3}\) shows concentration profiles of \(\text{A}\) and \(\text{B}\) for a second order reaction. To determine \(k\) from experimental data for second-order reactions, it is convenient to invert Equation \(\ref{19.11}\): \[\dfrac{1}{\left[ \text{A} \right]} = \dfrac{1}{\left[ \text{A} \right]_0} + 2kt \label{19.12}\] A plot of \(1/\left[ \text{A} \right]\) vs. \(t\) will give rise to a straight line with slope \(k\) and intercept \(1/\left[ \text{A} \right]_0\). Second order reaction rates can also apply to reactions in which two species react with each other to form a product: \[\text{A} + \text{B} \overset{k}{\longrightarrow} \text{C}\] In this scenario, the reaction rate will depend on the concentrations of both \(\text{A}\) and \(\text{B}\) to the first order: \[r = -\dfrac{d \left[ \text{A} \right]}{dt} = -\dfrac{d \left[ \text{B} \right]}{dt} = k \left[ \text{A} \right] \left[ \text{B} \right] \label{19.13}\] To integrate the above equation, we need to write it in terms of one variable. Since the concentrations of \(\text{A}\) and \(\text{B}\) are related to each other via the chemical reaction equation, we can write: \[\left[ \text{B} \right] = \left[ \text{B} \right]_0 - \left( \left[ \text{A} \right]_0 - \left[ \text{A} \right] \right) = \left[ \text{A} \right] + \left[ \text{B} \right]_0 - \left[ \text{A} \right]_0 \label{19.14}\] \[\dfrac{d \left[ \text{A} \right]}{dt} = -k \left[ \text{A} \right] \left( \left[ \text{A} \right] + \left[ \text{B} \right]_0 - \left[ \text{A} \right]_0 \right) \label{19.15}\] We can then use partial fractions to integrate: \[kdt = \dfrac{d \left[ \text{A} \right]}{\left[ \text{A} \right] \left( \left[ \text{A} \right] + \left[ \text{B} \right]_0 - \left[ \text{A} \right]_0 \right)} = \dfrac{1}{\left[ \text{B} \right]_0 - \left[ \text{A} \right]_0} \left( \dfrac{d \left[ \text{A} \right]}{\left[ \text{A} \right]} - \dfrac{d \left[ \text{A} \right]}{\left[ \text{B} \right]_0 - \left[ \text{A} \right]_0 + \left[ \text{A} \right]} \right) \label{19.16}\] \[kt = \dfrac{1}{\left[ \text{A} \right]_0 - \left[ \text{B} \right]_0} \text{ln} \dfrac{\left[ \text{A} \right] \left[ \text{B} \right]_0}{\left[ \text{B} \right] \left[ \text{A} \right]_0} \label{19.17}\] Figure \(\Page {4}\) displays the concentration profiles of species \(\text{A}\), \(\text{B}\), and \(\text{C}\) for a second order reaction in which the initial concentrations of \(\text{A}\) and \(\text{B}\) are not equal. ( ) | 7,633 | 2,343 |
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The chemical reactivity of benzene contrasts with that of the alkenes in that substitution reactions occur in preference to addition reactions, as illustrated in the following diagram (some comparable reactions of cyclohexene are shown in the green box). Many other substitution reactions of benzene have been observed, the five most useful are listed below (chlorination and bromination are the most common halogenation reactions). Since the reagents and conditions employed in these reactions are electrophilic, these reactions are commonly referred to as Electrophilic Aromatic Substitution. The catalysts and co-reagents serve to generate the strong electrophilic species needed to effect the initial step of the substitution. The specific electrophile believed to function in each type of reaction is listed in the right hand column. : Acylation:
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As a simple example of how the macroscopic properties of a substance can be explained on a microscopic level, consider the liquid mercury. Macroscopically, mercury at ordinary temperatures is a silvery which can be poured much like water—rather unusual for a metal. Mercury is also the heaviest known liquid. Its density is 13.6 g cm , as compared with only 1.0 g cm for water. When cooled below –38.9°C mercury solidifies and behaves very much like more familiar metals such as copper and iron. Mercury frozen around the end of a wooden stick can be used to hammer nails, as long as it is kept sufficiently cold. Solid mercury has a density of 14.1 g cm slightly greater than that of the liquid. When mercury is heated, it remains a liquid until quite a high temperature, finally boiling at 356.6°C to give an invisible vapor. Even at low concentrations gaseous mercury is extremely toxic if breathed into the lungs. It has been responsible for many cases of human poisoning. In other respects mercury vapor behaves much like any other . It is easily compressible. Even when quite modest pressures are applied, the volume decreases noticeably. Mercury vapor is also much less dense than the liquid or the solid. At 400°C and ordinary pressures, its density is 3.6 × 10 g cm about one four-thousandth that of solid or liquid mercury. A modern chemist would interpret these macroscopic properties in terms of a microscopic model involving atoms of mercury. As shown in the following figure, the atoms may be thought of as small, hard spheres. Like billiard balls they can move around and bounce off one another. In solid mercury the centers of adjacent atoms are separated by only 300 pm (300 × 10 m or 3.00Å). Although each atom can move around a little, the others surround it so closely that it cannot escape its allotted position. Hence the solid is rigid. Very few atoms move out of position even when it strikes a nail. As temperature increases, the atoms vibrate more violently, and eventually the solid melts. In liquid mercury, the regular, geometrically rigid structure is gone and the atoms are free to move about, but they are still rather close together and difficult to separate. This ability of the atoms to move past each other accounts for the fact that liquid mercury can flow and take the shape of its container. Note that the structure of the liquid is not as compact as that of the solid; a few gaps are present. These gaps explain why liquid mercury is less dense than the solid. In gaseous mercury, also called mercury vapor, the atoms are very much farther apart than in the liquid and they move around quite freely and rapidly. Since there are very few atoms per unit volume, the density is considerably lower than for the liquid and solid. By moving rapidly in all directions, the atoms of mercury (or any other gas for that matter) are able to fill any container in which they are placed. When the atoms hit a wall of the container, they bounce off. This constant bombardment by atoms on the sub-microscopic level accounts for the pressure exerted by the gas on the macroscopic level. The gas can be easily compressed because there is plenty of open space between the atoms. Reducing the volume merely reduces that empty space. The liquid and the solid are not nearly so easy to compress because there is little or no empty space between the atoms. You may have noticed that although our sub-microscopic model can explain many of the properties of solid, liquid, and gaseous mercury, it cannot explain all of them. Mercury’s silvery color and why the vapor is poisonous remain a mystery, for example. There are two approaches to such a situation. We might discard the idea of atoms in favor of a different theory that can explain more macroscopic properties. On the other hand it may be reasonable to extend the atomic theory so that it can account for more facts. The second approach has been followed by chemists. In the current section on as well as we shall discuss in more detail those facts that require only a simple atomic theory for their interpretation. Many of the subsequent sections will describe extensions of the atomic theory that allow interpretations of far more observations. | 4,248 | 2,346 |
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The localized uses a process called hybridization, in which atomic orbitals that are similar in energy, but not equivalent are combined mathematically to produce sets of equivalent orbitals that are properly oriented to form bonds. These new combinations are called hybrid atomic orbitals because they are produced by combining ( ) two or more atomic orbitals from the same atom. In BeH , we can generate two equivalent orbitals by combining the 2 orbital of beryllium and any one of the three degenerate 2 orbitals. By taking the sum and the difference of Be 2 and 2 atomic orbitals, for example, we produce two new orbitals with major and minor lobes oriented along the -axes, as shown in Figure \(\Page {0}\). This gives us Equation 11.3.1, where the value \(\frac{1}{\sqrt{2}}\) is needed mathematically to indicate that the 2 and 2 orbitals contribute equally to each hybrid orbital. \[sp = \dfrac{1}{\sqrt{2}} (2s + 2p_z) \label{11.3.1a}\] and \[sp = \dfrac{1}{\sqrt{2}} (2s - 2p_z) \label{11.3.1b}\] The nucleus resides just inside the minor lobe of each orbital. In this case, the new orbitals are called because they are formed from one and one orbital. The two new orbitals are equivalent in energy, and their energy is between the energy values associated with pure and orbitals, as illustrated in this diagram: Because both promotion and hybridization require an input of energy, the formation of a set of singly occupied hybrid atomic orbitals is energetically uphill. The overall process of forming a compound with hybrid orbitals will be energetically favorable if the amount of energy released by the formation of covalent bonds is greater than the amount of energy used to form the hybrid orbitals (Figure \(\Page {3}\)). As we will see, some compounds are highly unstable or do not exist because the amount of energy required to form hybrid orbitals is greater than the amount of energy that would be released by the formation of additional bonds. The concept of hybridization also explains why boron, with a 2 2 valence electron configuration, forms three bonds with fluorine to produce BF , as predicted by the Lewis and VSEPR approaches. With only a single unpaired electron in its ground state, boron should form only a single covalent bond. By the promotion of one of its 2 electrons to an unoccupied 2 orbital, however, followed by the hybridization of the three singly occupied orbitals (the 2 and two 2 orbitals), boron acquires a set of three equivalent hybrid orbitals with one electron each, as shown here: Looking at the 2 2 valence electron configuration of carbon, we might expect carbon to use its two unpaired 2 electrons to form compounds with only two covalent bonds. We know, however, that carbon typically forms compounds with four covalent bonds. We can explain this apparent discrepancy by the hybridization of the 2 orbital and the three 2 orbitals on carbon to give a set of four degenerate (“s-p-three” or “s-p-cubed”) hybrid orbitals, each with a single electron: The large lobes of the hybridized orbitals are oriented toward the vertices of a tetrahedron, with 109.5° angles between them (Figure \(\Page {5}\)). Like all the hybridized orbitals discussed earlier, the hybrid atomic orbitals are predicted to be equal in energy. In addition to explaining why some elements form more bonds than would be expected based on their valence electron configurations, and why the bonds formed are equal in energy, valence bond theory explains why these compounds are so stable: the amount of energy released increases with the number of bonds formed. In the case of carbon, for example, much more energy is released in the formation of four bonds than two, so compounds of carbon with four bonds tend to be more stable than those with only two. Carbon does form compounds with only two covalent bonds (such as CH or CF ), but these species are highly reactive, unstable intermediates that form in only certain chemical reactions. Valence bond theory explains the number of bonds formed in a compound and the relative bond strengths. The bonding in molecules such as NH or H O, which have lone pairs on the central atom, can also be described in terms of hybrid atomic orbitals. In NH , for example, N, with a 2 2 valence electron configuration, can hybridize its 2 and 2 orbitals to produce four hybrid orbitals. Placing five valence electrons in the four hybrid orbitals, we obtain three that are singly occupied and one with a pair of electrons: The three singly occupied lobes can form bonds with three H atoms, while the fourth orbital accommodates the lone pair of electrons. Similarly, H O has an hybridized oxygen atom that uses two singly occupied lobes to bond to two H atoms, and two to accommodate the two lone pairs predicted by the VSEPR model. Such descriptions explain the approximately tetrahedral distribution of electron pairs on the central atom in NH and H O. Unfortunately, however, recent experimental evidence indicates that in CH and NH , the hybridized orbitals are entirely equivalent in energy, making this bonding model an active area of research. Valence Bond Method & sp3 Hybridization: Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen. two chemical compounds number of electron pairs and molecular geometry, hybridization, and bonding Use the VSEPR model to predict the number of electron pairs and molecular geometry in each compound and then describe the hybridization and bonding of all atoms except hydrogen. The number of hybrid orbitals used by the central atom is the same as the number of electron pairs around the central atom. Hybridization is not restricted to the and atomic orbitals. The bonding in compounds with central atoms in the period 3 and below can also be described using hybrid atomic orbitals. In these cases, the central atom can use its valence ( − 1) orbitals as well as its and orbitals to form hybrid atomic orbitals, which allows it to accommodate five or more bonded atoms (as in PF and SF ). Using the orbital, all three orbitals, and one ( − 1) orbital gives a set of five hybrid orbitals that point toward the vertices of a trigonal bipyramid (part (a) in Figure \(\Page {6}\)). In this case, the five hybrid orbitals are all equivalent: three form a triangular array oriented at 120° angles, and the other two are oriented at 90° to the first three and at 180° to each other. Similarly, the combination of the orbital, all three orbitals, and orbitals gives a set of six equivalent hybrid orbitals oriented toward the vertices of an octahedron (part (b) in Figure \(\Page {6}\)). In the VSEPR model, PF and SF are predicted to be trigonal bipyramidal and octahedral, respectively, which agrees with a valence bond description in which or hybrid orbitals are used for bonding. What is the hybridization of the central atom in each species? Describe the bonding in each species. three chemical species hybridization of the central atom To accommodate five electron pairs, the sulfur atom must be hybridized. Filling these orbitals with 10 electrons gives four hybrid orbitals forming S–F bonds and one with a lone pair of electrons. What is the hybridization of the central atom in each species? Describe the bonding. Hybridization using orbitals allows chemists to explain the structures and properties of many molecules and ions. Like most such models, however, it is not universally accepted. Nonetheless, it does explain a fundamental difference between the chemistry of the elements in the period 2 (C, N, and O) and those in period 3 and below (such as Si, P, and S). Period 2 elements do not form compounds in which the central atom is covalently bonded to five or more atoms, although such compounds are common for the heavier elements. Thus whereas carbon and silicon both form tetrafluorides (CF and SiF ), only SiF reacts with F to give a stable hexafluoro dianion, SiF . Because there are no 2 atomic orbitals, the formation of octahedral CF would require hybrid orbitals created from 2 , 2 , and 3 atomic orbitals. The 3 orbitals of carbon are so high in energy that the amount of energy needed to form a set of hybrid orbitals cannot be equaled by the energy released in the formation of two additional C–F bonds. These additional bonds are expected to be weak because the carbon atom (and other atoms in period 2) is so small that it cannot accommodate five or six F atoms at normal C–F bond lengths due to repulsions between electrons on adjacent fluorine atoms. Perhaps not surprisingly, then, species such as CF have never been prepared. What is the hybridization of the oxygen atom in OF ? Is OF likely to exist? chemical compound hybridization and stability The VSEPR model predicts that OF will have five electron pairs, resulting in a trigonal bipyramidal geometry with four bonding pairs and one lone pair. To accommodate five electron pairs, the O atom would have to be hybridized. The only orbital available for forming a set of hybrid orbitals is a 3 orbital, which is higher in energy than the 2 and 2 valence orbitals of oxygen. As a result, the OF molecule is unlikely to exist. In fact, it has not been detected. What is the hybridization of the boron atom in \(BF_6^{3−}\)? Is this ion likely to exist? hybridization; no Expanded Octet Hybridization: The model (called ) assumes that covalent bonds are formed when atomic orbitals overlap and that the strength of a covalent bond is proportional to the amount of overlap. It also assumes that atoms use combinations of atomic orbitals ( ) to maximize the overlap with adjacent atoms. The formation of can be viewed as occurring via of an electron from a filled subshell to an empty or ( − 1) valence orbital, followed by , the combination of the orbitals to give a new set of (usually) equivalent orbitals that are oriented properly to form bonds. The combination of an and an orbital gives rise to two equivalent oriented at 180°, whereas the combination of an and two or three orbitals produces three equivalent or four equivalent , respectively. The bonding in molecules with more than an octet of electrons around a central atom can be explained by invoking the participation of one or two ( − 1) orbitals to give sets of five or six orbitals, capable of forming five or six bonds, respectively. The spatial orientation of the hybrid atomic orbitals is consistent with the geometries predicted using the VSEPR model. | 10,709 | 2,347 |
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In , we saw that a combination of high pressure and low temperature allows gases to be liquefied. As we increase the temperature of a gas, liquefaction becomes more and more difficult because higher and higher pressures are required to overcome the increased kinetic energy of the molecules. In fact, for every substance, there is some temperature above which the gas can no longer be liquefied, regardless of pressure. This temperature is the critical temperature ( ), the highest temperature at which a substance can exist as a liquid. Above the critical temperature, the molecules have too much kinetic energy for the intermolecular attractive forces to hold them together in a separate liquid phase. Instead, the substance forms a single phase that completely occupies the volume of the container. Substances with strong intermolecular forces tend to form a liquid phase over a very large temperature range and therefore have high critical temperatures. Conversely, substances with weak intermolecular interactions have relatively low critical temperatures. Each substance also has a critical pressure ( ), the minimum pressure needed to liquefy it at the critical temperature. The combination of critical temperature and critical pressure is called the critical point. The critical temperatures and pressures of several common substances are listed in . High-boiling-point, nonvolatile liquids have high critical temperatures and vice versa. To understand what happens at the critical point, consider the effects of temperature and pressure on the densities of liquids and gases, respectively. As the temperature of a liquid increases, its density decreases. As the pressure of a gas increases, its density increases. At the critical point, the liquid and gas phases have exactly the same density, and only a single phase exists. This single phase is called a supercritical fluid , which exhibits many of the properties of a gas but has a density more typical of a liquid. For example, the density of water at its critical point ( = 374°C, = 217.7 atm) is 0.32 g/mL, about one-third that of liquid water at room temperature but much greater than that of water vapor under most conditions. The transition between a liquid/gas mixture and a supercritical phase is demonstrated for a sample of chlorine in . At the critical temperature, the meniscus separating the liquid and gas phases disappears. Below the critical temperature the meniscus between the liquid and gas phases is apparent. At the critical temperature, the meniscus disappears because the density of the vapor is equal to the density of the liquid. Above , a dense homogeneous fluid fills the tube. In the last few years, supercritical fluids have evolved from laboratory curiosities to substances with important commercial applications. For example, carbon dioxide has a low critical temperature (31°C), a comparatively low critical pressure (73 atm), and low toxicity, making it easy to contain and relatively safe to manipulate. Because many substances are quite soluble in supercritical CO , commercial processes that use it as a solvent are now well established in the oil industry, the food industry, and others. Supercritical CO is pumped into oil wells that are no longer producing much oil to dissolve the residual oil in the underground reservoirs. The less-viscous solution is then pumped to the surface, where the oil can be recovered by evaporation (and recycling) of the CO . In the food, flavor, and fragrance industry, supercritical CO is used to extract components from natural substances for use in perfumes, remove objectionable organic acids from hops prior to making beer, and selectively extract caffeine from whole coffee beans without removing important flavor components. The latter process was patented in 1974, and now virtually all decaffeinated coffee is produced this way. The earlier method used volatile organic solvents such as methylene chloride (dichloromethane [CH Cl ], boiling point = 40°C), which is difficult to remove completely from the beans and is known to cause cancer in laboratory animals at high doses. Arrange methanol, -butane, -pentane, and N O in order of increasing critical temperatures. compounds order of increasing critical temperatures Identify the intermolecular forces in each molecule and then assess the strengths of those forces. Arrange the compounds in order of increasing critical temperatures. The critical temperature depends on the strength of the intermolecular interactions that hold a substance together as a liquid. In N O, a slightly polar substance, weak dipole–dipole interactions and London dispersion forces are important. Butane (C H ) and pentane (C H ) are larger, nonpolar molecules that exhibit only London dispersion forces. Methanol, in contrast, should have substantial intermolecular hydrogen bonding interactions. Because hydrogen bonds are stronger than the other intermolecular forces, methanol will have the highest . London forces are more important for pentane than for butane because of its larger size, so -pentane will have a higher than -butane. The only remaining question is whether N O is polar enough to have stronger intermolecular interactions than pentane or butane. Because the electronegativities of O and N are quite similar, the answer is probably no, so N O should have the lowest . We therefore predict the order of increasing critical temperatures as N O < -butane < -pentane < methanol. The actual values are N O (36.9°C) < -butane (152.0°C) < -pentane (196.9°C) < methanol (239.9°C). This is the same order as their normal boiling points—N O (−88.7°C) < -butane (−0.2°C) < -pentane (36.0°C) < methanol (65°C)—because both critical temperature and boiling point depend on the relative strengths of the intermolecular interactions. Exercise Arrange ethanol, methanethiol (CH SH), ethane, and -hexanol in order of increasing critical temperatures. ethane (32.3°C) < methanethiol (196.9°C) < ethanol (240.9°C) < -hexanol (336.9°C) Heating a salt to its melting point produces a molten salt . If we heated a sample of solid NaCl to its melting point of 801°C, for example, it would melt to give a stable liquid that conducts electricity. The characteristics of molten salts other than electrical conductivity are their high heat capacity, ability to attain very high temperatures (over 700°C) as a liquid, and utility as solvents because of their relatively low toxicity. Molten salts have many uses in industry and the laboratory. For example, in solar power towers in the desert of California, mirrors collect and focus sunlight to melt a mixture of sodium nitrite and sodium nitrate. The heat stored in the molten salt is used to produce steam that drives a steam turbine and a generator, thereby producing electricity from the sun for southern California. Due to their low toxicity and high thermal efficiency, molten salts have also been used in nuclear reactors to enable operation at temperatures greater than 750°C. One prototype reactor tested in the 1950s used a fuel and a coolant consisting of molten fluoride salts, including NaF, ZrF , and UF . Molten salts are also useful in catalytic processes such as coal gasification, in which carbon and water react at high temperatures to form CO and H . Molten salts are good electrical conductors, have a high heat capacity, can maintain a high temperature as a liquid, and are relatively nontoxic. Although molten salts have proven highly useful, more recently chemists have been studying the characteristics of ionic liquids , ionic substances that are at room temperature and pressure. These substances consist of small, symmetrical anions, such as PF and BF , combined with larger, asymmetrical organic cations that prevent the formation of a highly organized structure, resulting in a low melting point. By varying the cation and the anion, chemists can tailor the liquid to specific needs, such as using a solvent in a given reaction or extracting specific molecules from a solution. For example, an ionic liquid consisting of a bulky cation and anions that bind metal contaminants such as mercury and cadmium ions can remove those toxic metals from the environment. A similar approach has been applied to removing uranium and americium from water contaminated by nuclear waste. Ionic liquids consist of small, symmetrical anions combined with larger asymmetrical cations, which produce a highly polar substance that is a liquid at room temperature and pressure. The initial interest in ionic liquids centered on their use as a low-temperature alternative to molten salts in batteries for missiles, nuclear warheads, and space probes. Further research revealed that ionic liquids had other useful properties—for example, some could dissolve the black rubber of discarded tires, allowing it to be recovered for recycling. Others could be used to produce commercially important organic compounds with high molecular mass, such as Styrofoam and Plexiglas, at rates 10 times faster than traditional methods. A substance cannot form a liquid above its , regardless of the applied pressure. Above the critical temperature, the molecules have enough kinetic energy to overcome the intermolecular attractive forces. The minimum pressure needed to liquefy a substance at its critical temperature is its . The combination of the critical temperature and critical pressure of a substance is its . Above the critical temperature and pressure, a substance exists as a dense fluid called a , which resembles a gas in that it completely fills its container but has a density comparable to that of a liquid. A is a salt heated to its melting point, giving a stable liquid that conducts electricity. are ionic substances that are liquids at room temperature. Their disorganized structure results in a low melting point. Describe the changes that take place when a liquid is heated above its critical temperature. How does this affect the physical properties? What is meant by the term ? What is the effect of increasing the pressure on a gas to above its critical pressure? Would it make any difference if the temperature of the gas was greater than its critical temperature? Do you expect the physical properties of a supercritical fluid to be more like those of the gas or the liquid phase? Explain. Can an ideal gas form a supercritical fluid? Why or why not? What are the limitations in using supercritical fluids to extract organic materials? What are the advantages? Describe the differences between a molten salt and an ionic liquid. Under what circumstances would an ionic liquid be preferred over a molten salt? Thumbnail from | 10,712 | 2,348 |
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The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend upon the total concentration of solute species, regardless of their identities. These include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure. This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module. Several units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity (M) is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species: \[M=\dfrac{\text{mol solute}}{\text{L solution}} \label{11.5.1} \] Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different at different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are (introduced in the previous chapter on gases) and . The mole fraction, \(\chi\), of a component is the ratio of its molar amount to the total number of moles of all solution components: \[\chi_\ce{A}=\dfrac{\text{mol A}}{\text{total mol of all components}} \label{11.5.2} \] Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms: \[m=\dfrac{\text{mol solute}}{\text{kg solvent}} \label{11.5.3} \] Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module. The antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, C H (OH) , in a solution prepared from \(\mathrm{2.22 \times 10^3 \;g}\) of ethylene glycol and \(\mathrm{2.00 \times 10^3\; g}\) of water (approximately 2 L of glycol and 2 L of water)? (a) The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the unit definition. \(\mathrm{mol\:H_2O=2000\:g×\dfrac{1\:mol\:H_2O}{18.02\:g\:H_2O}=111\:mol\:H_2O}\) \(\chi_\mathrm{ethylene\:glycol}=\mathrm{\dfrac{35.8\:mol\:C_2H_4(OH)_2}{(35.8+111)\:mol\: total}=0.245}\) Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles). (b) To find molality, we need to know the moles of the solute and the mass of the solvent (in kg). First, use the given mass of ethylene glycol and its molar mass to find the moles of solute: \[\mathrm{2220\:g\:C_2H_4(OH)_2\left(\dfrac{mol\:C_2H_2(OH)_2}{62.07\:g}\right)=35.8\:mol\:C_2H_4(OH)_2} \nonumber \] Then, convert the mass of the water from grams to kilograms: \[\mathrm{2000\: g\:H_2O\left(\dfrac{1\:kg}{1000\:g}\right)=2\: kg\:H_2O} \nonumber \] Finally, calculate molarity per its definition: \[\begin{align*}
\ce{molality}&=\mathrm{\dfrac{mol\: solute}{kg\: solvent}}\\
\ce{molality}&=\mathrm{\dfrac{35.8\:mol\:C_2H_4(OH)_2}{2\:kg\:H_2O}}\\
\ce{molality}&=17.9\:m
\end{align*} \nonumber \] What are the mole fraction and molality of a solution that contains 0.850 g of ammonia, NH , dissolved in 125 g of water? 7.14 × 10 ; 0.399 Calculate the mole fraction of solute and solvent in a 3.0 solution of sodium chloride. Converting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as: \[\mathrm{\dfrac{3.0\;mol\; NaCl}{1.0\; kg\; H_2O}} \nonumber \] The numerator for this solution’s mole fraction is, therefore, 3.0 mol NaCl. The denominator may be computed by deriving the molar amount of water corresponding to 1.0 kg \[\mathrm{1.0\:kg\:H_2O\left(\dfrac{1000\:g}{1\:kg}\right)\left(\dfrac{mol\:H_2O}{18.02\:g}\right)=55\:mol\:H_2O} \nonumber \] and then substituting these molar amounts into the definition for mole fraction. \[\begin{align*}
X_\mathrm{H_2O}&=\mathrm{\dfrac{mol\:H_2O}{mol\: NaCl + mol\:H_2O}}\\
X_\mathrm{H_2O}&=\mathrm{\dfrac{55\:mol\:H_2O}{3.0\:mol\: NaCl+55\:mol\:H_2O}}\\
X_\mathrm{H_2O}&=0.95\\
X_\mathrm{NaCl}&=\mathrm{\dfrac{mol\: NaCl}{mol\: NaCl+mol\:H_2O}}\\
X_\mathrm{NaCl}&=\mathrm{\dfrac{3.0\:mol\:NaCl}{3.0\:mol\: NaCl+55\:mol\:H_2O}}\\
X_\mathrm{NaCl}&=0.052
\end{align*} \nonumber \] The mole fraction of iodine, \(\ce{I_2}\), dissolved in dichloromethane, \(\ce{CH_2Cl_2}\), is 0.115. What is the molal concentration, , of iodine in this solution? 1.50 As described in the chapter on liquids and solids, the equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates: \[ \text{liquid} \rightleftharpoons \text{gas} \label{11.5.4} \] Dissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid’s vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid's vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporization-condensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (Figure \(\Page {1}\)). While this kinetic interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of , a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid's vapor pressure, it is adequate to note that the greater entropy of a solution in comparison to its separate solvent and solute serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly higher boiling point as described in the next section of this module. The relationship between the vapor pressures of solution components and the concentrations of those components is described by : . \[P_\ce{A}=X_\ce{A}P^\circ_\ce{A} \label{11.5.5} \] where is the partial pressure exerted by component A in the solution, \(P^\circ_\ce{A}\) is the vapor pressure of pure A, and is the mole fraction of A in the solution. (Mole fraction is a concentration unit introduced in the chapter on gases.) Recalling that the total pressure of a gaseous mixture is equal to the sum of partial pressures for all its components (Dalton’s law of partial pressures), the total vapor pressure exerted by a solution containing components is \[P_\ce{solution}=\sum_{i}P_i=\sum_{i}X_iP^\circ_i \label{11.5.6} \] A nonvolatile substance is one whose vapor pressure is negligible ( ° ≈ 0), and so the vapor pressure above a solution containing only nonvolatile solutes is due only to the solvent: \[P_\ce{solution}=X_\ce{solvent}P^\circ_\ce{solvent} \label{11.5.7} \] Compute the vapor pressure of an ideal solution containing 92.1 g of glycerin, C H (OH) , and 184.4 g of ethanol, C H OH, at 40 °C. The vapor pressure of pure ethanol is 0.178 atm at 40 °C. Glycerin is essentially nonvolatile at this temperature. Since the solvent is the only volatile component of this solution, its vapor pressure may be computed per Raoult’s law as: \(P_\ce{solution}=X_\ce{solvent}P^\circ_\ce{solvent}\) First, calculate the molar amounts of each solution component using the provided mass data. \(\mathrm{92.1\cancel{g\:C_3H_5(OH)_3}×\dfrac{1\:mol\:C_3H_5(OH)_3}{92.094\cancel{g\:C_3H_5(OH)_3}}=1.00\:mol\:C_3H_5(OH)_3}\) Next, calculate the mole fraction of the solvent (ethanol) and use Raoult’s law to compute the solution’s vapor pressure. \(X_\mathrm{C_2H_5OH}=\mathrm{\dfrac{4.000\:mol}{(1.00\:mol+4.000\:mol)}=0.800}\) \(P_\ce{solv}=X_\ce{solv}P^\circ_\ce{solv}=\mathrm{0.800×0.178\:atm=0.142\:atm}\) A solution contains 5.00 g of urea, CO(NH ) (a nonvolatile solute) and 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution? 23.4 torr As described in the chapter on liquids and solids, the of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution’s boiling point will subsequently be increased. Compared to pure solvent, a solution, therefore, will require a higher temperature to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, \(ΔT_b\), is called boiling point elevation and is directly proportional to the molal concentration of solute species: \[ΔT_b=K_bm \label{11.5.8} \] where Boiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of for several solvents are listed in Table \(\Page {1}\). The extent to which the vapor pressure of a solvent is lowered and the boiling point is elevated depends on the total number of solute particles present in a given amount of solvent, not on the mass or size or chemical identities of the particles. A 1 aqueous solution of sucrose (342 g/mol) and a 1 aqueous solution of ethylene glycol (62 g/mol) will exhibit the same boiling point because each solution has one mole of solute particles (molecules) per kilogram of solvent. What is the boiling point of a 0.33 solution of a nonvolatile solute in benzene? Use the equation relating boiling point elevation to solute molality to solve this problem in two steps. What is the boiling point of the antifreeze described in Example \(\Page {4}\)? 109.2 °C Find the boiling point of a solution of 92.1 g of iodine, \(\ce{I2}\), in 800.0 g of chloroform, \(\ce{CHCl3}\), assuming that the iodine is nonvolatile and that the solution is ideal. We can solve this problem using four steps. Result: 0.363 mol Result: 0.454 Result: 1.65 °C Result: 62.91 °C Check each result as a self-assessment. What is the boiling point of a solution of 1.0 g of glycerin, \(\ce{C3H5(OH)3}\), in 47.8 g of water? Assume an ideal solution. 100.12 °C Distillation is a technique for separating the components of mixtures that is widely applied in both in the laboratory and in industrial settings. It is used to refine petroleum, to isolate fermentation products, and to purify water. This separation technique involves the controlled heating of a sample mixture to selectively vaporize, condense, and collect one or more components of interest. A typical apparatus for laboratory-scale distillations is shown in Figure \(\Page {2}\). Oil refineries use large-scale to separate the components of crude oil. The crude oil is heated to high temperatures at the base of a tall , vaporizing many of the components that rise within the column. As vaporized components reach adequately cool zones during their ascent, they condense and are collected. The collected liquids are simpler mixtures of hydrocarbons and other petroleum compounds that are of appropriate composition for various applications (e.g., diesel fuel, kerosene, gasoline), as depicted in Figure \(\Page {3}\). Solutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in “de-icing” schemes that use salt (Figure \(\Page {4}\)), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an “antifreeze” in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans). The decrease in freezing point of a dilute solution compared to that of the pure solvent, Δ , is called the and is directly proportional to the molal concentration of the solute \[ΔT_\ce{f}=K_\ce{f}m \label{11.5.9} \] where Just as for boiling point elevation constants, these are characteristic properties whose values depend on the chemical identity of the solvent. Values of for several solvents are listed in Table \(\Page {1}\). What is the freezing point of the 0.33 solution of a nonvolatile nonelectrolyte solute in benzene described in Example \(\Page {4}\)? Use the equation relating freezing point depression to solute molality to solve this problem in two steps. What is the freezing point of a 1.85 solution of a nonvolatile nonelectrolyte solute in nitrobenzene? −9.3 °C Sodium chloride and its group 2 analogs calcium and magnesium chloride are often used to de-ice roadways and sidewalks, due to the fact that a solution of any one of these salts will have a freezing point lower than 0 °C, the freezing point of pure water. The group 2 metal salts are frequently mixed with the cheaper and more readily available sodium chloride (“rock salt”) for use on roads, since they tend to be somewhat less corrosive than the NaCl, and they provide a larger depression of the freezing point, since they dissociate to yield three particles per formula unit, rather than two particles like the sodium chloride. Because these ionic compounds tend to hasten the corrosion of metal, they would not be a wise choice to use in antifreeze for the radiator in your car or to de-ice a plane prior to takeoff. For these applications, covalent compounds, such as ethylene or propylene glycol, are often used. The glycols used in radiator fluid not only lower the freezing point of the liquid, but they elevate the boiling point, making the fluid useful in both winter and summer. Heated glycols are often sprayed onto the surface of airplanes prior to takeoff in inclement weather in the winter to remove ice that has already formed and prevent the formation of more ice, which would be particularly dangerous if formed on the control surfaces of the aircraft ( \(\Page {1}\)). The colligative effects on vapor pressure, boiling point, and freezing point described in the previous section are conveniently summarized by comparing the phase diagrams for a pure liquid and a solution derived from that liquid. Phase diagrams for water and an aqueous solution are shown in Figure \(\Page {5}\). The liquid-vapor curve for the solution is located the corresponding curve for the solvent, depicting the vapor pressure , Δ , that results from the dissolution of nonvolatile solute. Consequently, at any given pressure, the solution’s boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, Δ , associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, Δ , that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the molecules that undergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed solvent only, and so transitions between these phases are not subject to colligative effects. A number of natural and synthetic materials exhibit , meaning that only molecules or ions of a certain size, shape, polarity, charge, and so forth, are capable of passing through (permeating) the material. Biological cell membranes provide elegant examples of selective permeation in nature, while dialysis tubing used to remove metabolic wastes from blood is a more simplistic technological example. Regardless of how they may be fabricated, these materials are generally referred to as . Consider the apparatus illustrated in Figure \(\Page {6}\), in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as . When osmosis is carried out in an apparatus like that shown in Figure \(\Page {6}\), the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the e (\(\Pi\)) of the solution. The osmotic pressure of a dilute solution is related to its solute molarity, , and absolute temperature, , according to the equation \[Π=MRT \label{11.5.10} \] where \(R\) is the universal gas constant. What is the osmotic pressure (atm) of a 0.30 solution of glucose in water that is used for intravenous infusion at body temperature, 37 °C? We can find the osmotic pressure, \( , using Equation \ref{11.5.10}, where is on the Kelvin scale (310 K) and the value of is expressed in appropriate units (0.08206 L atm/mol K). \[\begin{align*}
Π&=MRT\\
&=\mathrm{0.03\:mol/L×0.08206\: L\: atm/mol\: K×310\: K}\\
&=\mathrm{7.6\:atm}
\end{align*} \nonumber \] What is the osmotic pressure (atm) a solution with a volume of 0.750 L that contains 5.0 g of methanol, CH OH, in water at 37 °C? 5.3 atm If a solution is placed in an apparatus like the one shown in Figure \(\Page {7}\), applying pressure greater than the osmotic pressure of the solution reverses the osmosis and pushes solvent molecules from the solution into the pure solvent. This technique of reverse osmosis is used for large-scale desalination of seawater and on smaller scales to produce high-purity tap water for drinking. Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be with blood serum. If a less concentrated solution, a solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called . When a more concentrated solution, a solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called (Figure 11.5.8). Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the concentration of solute present. Consequently, we can use a measurement of one of these properties to determine the molar mass of the solute from the measurements. A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. What is the molar mass of this compound? We can solve this problem using the following steps. \(ΔT_\ce{f}=K \(m=\dfrac{ΔT_\ce{f}}{K_\ce{f}}=\dfrac{3.2\:°\ce C}{5.12\:°\ce C m^{−1}}=0.63\:m\) \(\ dfrac mol A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. What is the molar mass of this compound? 1.8 × 10 g/mol A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. What is the molar mass of hemoglobin? Here is one set of steps that can be used to solve the problem: \[\Pi=MRT \nonumber \] \(M=\dfrac{Π}{RT}=\mathrm{\dfrac{7.8×10^{−3}\:atm}{(0.08206\:L\: atm/mol\: K)(295\:K)}=3.2×10^{−4}\:M}\) \(\mathrm{molar\: mass=\dfrac{10.0\:g}{1.6×10^{−4}\:mol}=6.2×10^4\:g/mol}\) What is the molar mass of a protein if a solution of 0.02 g of the protein in 25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C? 2.7 × 10 g/mol As noted previously in this module, the colligative properties of a solution depend only on the number, not on the kind, of solute species dissolved. For example, 1 mole of any nonelectrolyte dissolved in 1 kilogram of solvent produces the same lowering of the freezing point as does 1 mole of any other nonelectrolyte. However, 1 mole of sodium chloride (an electrolyte) forms of ions when dissolved in solution. Each individual ion produces the same effect on the freezing point as a single molecule does. The concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Assume that each of the ions in the NaCl solution has the same effect on the freezing point of water as a nonelectrolyte molecule, and determine the freezing temperature the solution (which is approximately equal to the freezing temperature of seawater). We can solve this problem using the following series of steps. Check each result as a self-assessment. Assume that each of the ions in calcium chloride, CaCl , has the same effect on the freezing point of water as a nonelectrolyte molecule. Calculate the freezing point of a solution of 0.724 g of CaCl in 175 g of water. −0.208 °C Assuming complete dissociation, a 1.0 aqueous solution of NaCl contains 2.0 mole of ions (1.0 mol Na and 1.0 mol Cl ) per each kilogram of water, and its freezing point depression is expected to be \[ΔT_\ce{f}=\mathrm{2.0\:mol\: ions/kg\: water×1.86\:°C\: kg\: water/mol\: ion=3.7\:°C.} \label{11.5.11} \] When this solution is actually prepared and its freezing point depression measured, however, a value of 3.4 °C is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution. To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van’t Hoff is used. The is defined as the ratio of solute particles in solution to the number of formula units dissolved: \[i=\dfrac{\textrm{moles of particles in solution}}{\textrm{moles of formula units dissolved}} \label{11.5.12} \] Values for measured van’t Hoff factors for several solutes, along with predicted values assuming complete dissociation, are shown in Table \(\Page {2}\). In 1923, the chemists Peter and Erich proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that although interionic attraction in an aqueous solution is very greatly reduced by solvation of the ions and the insulating action of the polar solvent, it is not completely nullified. The residual attractions prevent the ions from behaving as totally independent particles (Figure \(\Page {9}\)). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the , or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van’t Hoff factors for the electrolytes in Table \(\Page {2}\) are for 0.05 solutions, at which concentration the value of for NaCl is 1.9, as opposed to an ideal value of 2. Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. The total concentration of solute particles in a solution also determines its osmotic pressure. This is the pressure that must be applied to the solution to prevent diffusion of molecules of pure solvent through a semipermeable membrane into the solution. Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted. | 26,720 | 2,349 |
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Previously, you learned that metal ions in aqueous solution are hydrated—that is, surrounded by a shell of usually four or six water molecules. A hydrated ion is one kind of a complex ion (or, simply, complex), a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons, such as the [Al(H O) ] ion. A complex ion forms from a metal ion and a ligand because of a Lewis acid–base interaction. The positively charged metal ion acts as a Lewis acid, and the ligand, with one or more lone pairs of electrons, acts as a Lewis base. Small, highly charged metal ions, such as Cu or Ru , have the greatest tendency to act as Lewis acids, and consequently, they have the greatest tendency to form complex ions. As an example of the formation of complex ions, consider the addition of ammonia to an aqueous solution of the hydrated Cu ion {[Cu(H O) ] }. Because it is a stronger base than H O, ammonia replaces the water molecules in the hydrated ion to form the [Cu(NH ) (H O) ] ion. Formation of the [Cu(NH ) (H O) ] complex is accompanied by a dramatic color change, as shown in . The solution changes from the light blue of [Cu(H O) ] to the blue-violet characteristic of the [Cu(NH ) (H O) ] ion. The replacement of water molecules from \(\ce{[Cu(H2O)6]^{2+}}\) by ammonia occurs in sequential steps. Omitting the water molecules bound to \(\ce{Cu^{2+}}\) for simplicity, we can write the equilibrium reactions as follows: \[\ce{Cu^{2+}(aq) + NH3(aq) <=> [Cu(NH3)]^{2+}(aq)} \tag{step 1}\] with \[K_1 = \dfrac{\ce{[Cu(NH3)]^{2+}(aq)}}{[\ce{Cu^{2+}(aq)},\ce{NH3(aq)}]} \nonumber\] \[\ce{[Cu(NH_3)]^{2+}(aq) + NH3(aq) <=> [Cu(NH3)2]^{2+}(aq)} \tag{step 2}\] with \[K_2= \dfrac{\ce{[Cu(NH3)2]^{2+}(aq)}}{[\ce{[Cu(NH3)]^{2+}(aq)},\ce{NH3(aq)}]} \nonumber\] \[\ce{ [Cu(NH3)2]^{2+}(aq) + NH3(aq) <=> [Cu(NH3)3]^{2+}(aq)} \tag{step 3}\] with \[K_3= \dfrac{\ce{[Cu(NH3)3]^{2+}(aq)}}{[\ce{[Cu(NH3)2]^{2+}(aq)},\ce{NH3(aq)}]} \nonumber\] \[\ce{[Cu(NH3)3]^{2+}(aq) + NH3(aq) <=> [Cu(NH3)4]^{2+}(aq)} \tag{step 4}\] with \[K_4= \dfrac{\ce{[Cu(NH3)4]^{2+}(aq)}}{[\ce{[Cu(NH3)3]^{2+}(aq)},\ce{NH3(aq)}]} \nonumber\] The sum of the stepwise reactions is the overall equation for the formation of the complex ion: The hydrated Cu ion contains six H O ligands, but the complex ion that is produced contains only four \(NH_3\) ligands, not six. \[\ce{Cu^{2+}(aq) + 4NH3(aq) <=> [Cu(NH3)4]^{2+}(aq)} \label{2}\] The equilibrium constant for the formation of the complex ion from the hydrated ion is called the (\(K_f\)). The equilibrium constant expression for has the same general form as any other equilibrium constant expression. In this case, the expression is as follows: \[\begin{align*} K_\textrm f &=\dfrac{\left[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}},\mathrm{NH_3}]^4} \\[4pt] &=2.1\times10^{13} \\[4pt] &=K_1 \times K_2 \times K_3 \times K_4 \label{3} \end{align*}\] The formation constant (\(K_f\)) has the same general form as any other equilibrium constant expression. Water, a pure liquid, does not appear explicitly in the equilibrium constant expression, and the hydrated Cu (aq) ion is represented as Cu for simplicity. As for any equilibrium, the larger the value of the equilibrium constant (in this case, ), the more stable the product. With = 2.1 × 10 , the [Cu(NH ) (H O) ] complex ion is very stable. The formation constants for some common complex ions are listed in . If 12.5 g of \(\ce{Cu(NO3)2•6H2O}\) is added to 500 mL of 1.00 M aqueous ammonia, what is the equilibrium concentration of \(\ce{Cu^{2+}(aq)}\)? mass of Cu salt and volume and concentration of ammonia solution equilibrium concentration of Cu (aq) Adding an ionic compound that contains \(\ce{Cu^{2+}(aq)}\) to an aqueous ammonia solution will result in the formation of [Cu(NH ) ] (aq), as shown in Equation \ref{2} We assume that the volume change caused by adding solid copper(II) nitrate to aqueous ammonia is negligible. The initial concentration of \(\ce{Cu^{2+}(aq)}\) from the amount of added copper nitrate prior to any reaction is as follows: \[12.5\mathrm{\;\cancel{g}\;Cu(NO_3)_2}\cdot\mathrm{6H_2O}\left(\dfrac{\textrm{1 mol}}{\textrm{295.65} \cancel{g}} \right )\left(\dfrac{1}{\textrm{500}\; \cancel{mL}} \right )\left(\dfrac{\textrm{1000}\; \cancel{mL}}{\textrm{1 L}} \right )=\textrm{0.0846 M} \nonumber\] Because the stoichiometry of the reaction is four NH to one Cu , the amount of NH required to react completely with the Cu is 4(0.0846) = 0.338 M. The concentration of ammonia after complete reaction is 1.00 M − 0.338 M = 0.66 M. These results are summarized in the first two lines of the following table. Because the equilibrium constant for the reaction is large (2.1 × 10 ), the equilibrium will lie far to the right. Thus we will assume that the formation of [Cu(NH ) ] in the first step is complete and allow some of it to dissociate into Cu and NH until equilibrium has been reached. If we define as the amount of Cu produced by the dissociation reaction, then the stoichiometry of the reaction tells us that the change in the concentration of [Cu(NH ) ] is − , and the change in the concentration of ammonia is +4 , as indicated in the table. The final concentrations of all species (in the bottom row of the table) are the sums of the concentrations after complete reaction and the changes in concentrations. \[\ce{Cu^{2+} + 4NH3 <=> [Cu(NH3)4]^{2+}} \nonumber\] Substituting the final concentrations into the expression for the formation constant (\(K_f\)) and assuming that \(x \ll 0.0846\), which allows us to remove \(x\) from the sum and difference, \[\begin{align*} K_\textrm f &=\dfrac{\left[[\mathrm{Cu(NH_3)_4}]^{2+}\right]}{[\mathrm{Cu^{2+}},\mathrm{NH_3}]^4} \\[4pt] &=\dfrac{0.0846-x}{x(0.66+4x)^4} \\[4pt] &\approx \dfrac{0.0846}{x(0.66)^4}=2.1\times10^{13} \\[4pt] x &=2.1\times10^{-14}\end{align*}\] The value of indicates that our assumption was justified. The equilibrium concentration of Cu (aq) in a 1.00 M ammonia solution is therefore 2.1 × 10 M. The ferrocyanide ion (\(\ce{[Fe(CN)6]^{4−}}\)) is very stable, with a \(K_f\) of \(1 × 10^{35}\). Calculate the concentration of cyanide ion in equilibrium with a 0.65 M solution of \(\ce{K4[Fe(CN)6]}\). 2 × 10 M What happens to the solubility of a sparingly soluble salt if a ligand that forms a stable complex ion is added to the solution? One such example occurs in conventional black-and-white photography. Recall that black-and-white photographic film contains light-sensitive microcrystals of AgBr, or mixtures of AgBr and other silver halides. AgBr is a sparingly soluble salt, with a of 5.35 × 10 at 25°C. When the shutter of the camera opens, the light from the object being photographed strikes some of the crystals on the film and initiates a photochemical reaction that converts AgBr to black Ag metal. Well-formed, stable negative images appear in tones of gray, corresponding to the number of grains of AgBr converted, with the areas exposed to the most light being darkest. To fix the image and prevent more AgBr crystals from being converted to Ag metal during processing of the film, the unreacted AgBr on the film is removed using a complexation reaction to dissolve the sparingly soluble salt. The reaction for the dissolution of silver bromide is as follows: \[\ce{AgBr(s) <=> Ag^+(aq) + Br^{−}(aq)} \label{4a}\] with \[K_{sp} = 5.35 \times 10^{−13} \text{ at 25°C} \label{4b}\] The equilibrium lies far to the left, and the equilibrium concentrations of Ag and Br ions are very low (7.31 × 10 M). As a result, removing unreacted AgBr from even a single roll of film using pure water would require tens of thousands of liters of water and a great deal of time. Le Chatelier’s principle tells us, however, that we can drive the reaction to the right by removing one of the products, which will cause more AgBr to dissolve. Bromide ion is difficult to remove chemically, but silver ion forms a variety of stable two-coordinate complexes with neutral ligands, such as ammonia, or with anionic ligands, such as cyanide or thiosulfate (S O ). In photographic processing, excess AgBr is dissolved using a concentrated solution of sodium thiosulfate. The reaction of Ag with thiosulfate is as follows: \[Ag^+_{(aq)} + 2S_2O^{2−}_{3(aq)} \rightleftharpoons [Ag(S_2O_3)_2]^{3−}_{(aq)} \label{5a}\] with \[K_f = 2.9 \times 10^{13} \label{5b}\] The magnitude of the equilibrium constant indicates that almost all Ag ions in solution will be immediately complexed by thiosulfate to form [Ag(S O ) ] . We can see the effect of thiosulfate on the solubility of \(\ce{AgBr}\) by writing the appropriate reactions and adding them together: \[\begin{align*}\mathrm{AgBr(s)} &\rightleftharpoons\mathrm{Ag^+(aq)}+\mathrm{Br^-(aq)} \quad K_{\textrm{sp}} =5.35\times10^{-13}
\\[4pt] \mathrm{Ag^+(aq)}+\mathrm{2S_2O_3^{2-}(aq)} &\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)} \quad K_\textrm f =2.9\times10^{13}
\\[4pt] \mathrm{AgBr(s)}+\mathrm{2S_2O_3^{2-}(aq)} &\rightleftharpoons\mathrm{[Ag(S_2O_3)_2]^{3-}(aq)}+\mathrm{Br^-(aq)} \quad K =K_{\textrm{sp}}K_{\textrm f}=15\end{align*} \] Comparing \(K\) with \(K_{sp}\) shows that the formation of the complex ion increases the solubility of \(\ce{AgBr}\) by approximately 3 × 10 . The dramatic increase in solubility combined with the low cost and the low toxicity explains why sodium thiosulfate is almost universally used for developing black-and-white film. If desired, the silver can be recovered from the thiosulfate solution using any of several methods and recycled. If a complex ion has a large , the formation of a complex ion can dramatically increase the solubility of sparingly soluble salts. Due to the common ion effect, we might expect a salt such as AgCl to be much less soluble in a concentrated solution of KCl than in water. Such an assumption would be incorrect, however, because it ignores the fact that silver ion tends to form a two-coordinate complex with chloride ions (AgCl ). Calculate the solubility of AgCl in each situation: At 25°C, = 1.77 × 10 for AgCl and = 1.1 × 10 for AgCl . of AgCl, of AgCl , and KCl concentration solubility of AgCl in water and in KCl solution with and without the formation of complex ions That is, AgCl dissolves in 1.0 M KCl to produce a 1.9 × 10 M solution of the AgCl complex ion. Thus we predict that AgCl has approximately the same solubility in a 1.0 M KCl solution as it does in pure water, which is 10 times greater than that predicted based on the common ion effect. (In fact, the measured solubility of AgCl in 1.0 M KCl is almost a factor of 10 greater than that in pure water, largely due to the formation of other chloride-containing complexes.) Calculate the solubility of mercury(II) iodide (\(\ce{HgI2}\)) in each situation: = 2.9 × 10 for HgI and = 6.8 × 10 for [HgI ] . 1.9 × 10 M 1.4 M Complexing agents, molecules or ions that increase the solubility of metal salts by forming soluble metal complexes, are common components of laundry detergents. Long-chain carboxylic acids, the major components of soaps, form insoluble salts with Ca and Mg , which are present in high concentrations in “hard” water. The precipitation of these salts produces a bathtub ring and gives a gray tinge to clothing. Adding a complexing agent such as pyrophosphate (O POPO , or P O ) or triphosphate (P O ) to detergents prevents the magnesium and calcium salts from precipitating because the equilibrium constant for complex-ion formation is large: \[\ce{Ca^{2+}(aq) + O_3POPO^{4−}4(aq) <=> [Ca(O3POPO3)]^{2−}(aq)} \nonumber\] with \[K_f = 4\times 10^4 \nonumber\] However, phosphates can cause environmental damage by promoting , the growth of excessive amounts of algae in a body of water, which can eventually lead to large decreases in levels of dissolved oxygen that kill fish and other aquatic organisms. Consequently, many states in the United States have banned the use of phosphate-containing detergents, and France has banned their use beginning in 2007. “Phosphate-free” detergents contain different kinds of complexing agents, such as derivatives of acetic acid or other carboxylic acids. The development of phosphate substitutes is an area of intense research. Commercial water softeners also use a complexing agent to treat hard water by passing the water over ion-exchange resins, which are complex sodium salts. When water flows over the resin, sodium ion is dissolved, and insoluble salts precipitate onto the resin surface. Water treated in this way has a saltier taste due to the presence of Na , but it contains fewer dissolved minerals. Another application of complexing agents is found in medicine. Unlike x-rays, magnetic resonance imaging (MRI) can give relatively good images of soft tissues such as internal organs. MRI is based on the magnetic properties of the H nucleus of hydrogen atoms in water, which is a major component of soft tissues. Because the properties of water do not depend very much on whether it is inside a cell or in the blood, it is hard to get detailed images of these tissues that have good contrast. To solve this problem, scientists have developed a class of metal complexes known as “MRI contrast agents.” Injecting an MRI contrast agent into a patient selectively affects the magnetic properties of water in cells of normal tissues, in tumors, or in blood vessels and allows doctors to “see” each of these separately ( ). One of the most important metal ions for this application is Gd , which with seven unpaired electrons is highly paramagnetic. Because Gd (aq) is quite toxic, it must be administered as a very stable complex that does not dissociate in the body and can be excreted intact by the kidneys. The complexing agents used for gadolinium are ligands such as DTPA (diethylene triamine pentaacetic acid), whose fully protonated form is shown in Figure \(\Page {2}\). The formation of complex ions can substantially increase the solubility of sparingly soluble salts if the complex ion has a large . A complex ion is a species formed between a central metal ion and one or more surrounding ligands, molecules or ions that contain at least one lone pair of electrons. Small, highly charged metal ions have the greatest tendency to act as Lewis acids and form complex ions. The equilibrium constant for the formation of the complex ion is the formation constant ( ). The formation of a complex ion by adding a complexing agent increases the solubility of a compound. | 14,601 | 2,350 |
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The human eye perceives a mixture of all the colors, in the proportions present in sunlight, as white light. Complementary colors, those located across from each other on a color wheel, are also used in color vision. The eye perceives a mixture of two complementary colors, in the proper proportions, as white light. Likewise, when a color is missing from white light, the eye sees its complement. For example, when red photons are absorbed from white light, the eyes see the color green. When violet photons are removed from white light, the eyes see lemon yellow (Figure \(\Page {1}\)). The blue color of the \(\ce{[Cu(NH3)4]^{2+}}\) ion results because this ion absorbs orange and red light, leaving the complementary colors of blue and green (Figure \(\Page {2}\)). If white light (ordinary sunlight, for example) passes through \(\ce{[Cu(NH3)4]SO4}\) solution, some wavelengths in the light are absorbed by the solution. The \(\ce{[Cu(NH3)4]^{2+}}\) ions in solution absorb light in the red region of the spectrum. The light which passes through the solution and out the other side will have all the colors in it except for the red. We see this mixture of wavelengths as pale blue (cyan). The diagram gives an impression of what happens if you pass white light through a \(\ce{[Cu(NH_3)_4]SO_4}\) solution. Working out what color you will see is not easy if you try to do it by imagining "mixing up" the remaining colors. You would not have thought that all the other colors apart from some red would look cyan, for example. Sometimes what you actually see is quite unexpected. Mixing different wavelengths of light doesn't give you the same result as mixing paints or other pigments. You can, however, sometimes get some estimate of the color you would see using the idea of complementary colors. Recall that the color we observe when we look at an object or a compound is due to light that is transmitted or reflected, not light that is absorbed, and that reflected or transmitted light is complementary in color to the light that is absorbed. Thus a green compound absorbs light in the red portion of the visible spectrum and vice versa, as indicated by the complementary color wheel. The striking colors exhibited by transition-metal complexes are caused by excitation of an electron from a lower-energy d orbital to a higher-energy d orbital, which is called a (Figure \(\Page {4}\)). For a photon to effect such a transition, its energy must be equal to the difference in energy between the two d orbitals, which depends on the magnitude of Δ . The octahedral complex [Ti(H O) ] has a single electron. To excite this electron from the ground state orbital to the orbital, this complex absorbs light from 450 to 600 nm. The maximum absorbance corresponds to Δ and occurs at 499 nm. Calculate the value of Δ in Joules and predict what color the solution will appear. We can convert wavelength to frequency: \[\begin{align*} \nu &=\dfrac{c}{λ} \\[4pt] &= \dfrac{3.00 \times 10^8 \, m/s} { (499\, \cancel{nm}) \times \left( \dfrac{1 \,m}{10^9\, \cancel{ nm}} \right)} \\[4pt] &=6.01 \times 10^{14} s^{-1} =6.01 \times 10^{14} \,Hz \end{align*}\] And then using Planck's equation that related the frequency of light to energy \[E=h \nu \nonumber\] so \[\begin{align*} E &= (6.63 \times 10^{−34} \textrm{J⋅s} ) \times (6.01 \times 10^{14} \,Hz) \\[4pt] &=3.99 \times 10^{−19} \,J \end{align*}\] Because the complex absorbs 600 nm (orange) through 450 (blue), the indigo, violet, and red wavelengths will be transmitted, and the complex will appear purple. Note: This is the energy for one transition (i.e., in one complex). If you want to calculate the energy in J/mol, then you have to multiply this by Avogadro's number (\(N_A\)). A complex that appears green, absorbs photons of what wavelengths? Small changes in the relative energies of the orbitals that electrons are transitioning between can lead to drastic shifts in the color of light absorbed. Therefore, the colors of coordination compounds depend on many factors. As shown in Figure \(\Page {4}\), different aqueous metal ions can have different colors. In addition, different oxidation states of one metal can produce different colors, as shown for the vanadium complexes in the link below. The specific ligands coordinated to the metal center also influence the color of coordination complexes. Because the energy of a photon of light is inversely proportional to its wavelength, the color of a complex depends on the magnitude of Δ , which depends on the structure of the complex. For example, the complex [Cr(NH ) ] has strong-field ligands and a relatively large Δ . Consequently, it absorbs relatively high-energy photons, corresponding to blue-violet light, which gives it a yellow color. A related complex with weak-field ligands, the [Cr(H O) ] ion, absorbs lower-energy photons corresponding to the yellow-green portion of the visible spectrum, giving it a deep violet color. For example, the iron(II) complex [Fe(H O) ]SO appears blue-green because the high-spin complex absorbs photons in the red wavelengths (Figure \(\Page {5}\)). In contrast, the low-spin iron(II) complex K [Fe(CN) ] appears pale yellow because it absorbs higher-energy violet photons. In general, strong-field ligands cause a large split in the energies of orbitals of the central metal atom (large Δ ). Transition metal coordination compounds with these ligands are yellow, orange, or red because they absorb higher-energy violet or blue light. On the other hand, coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light. Strong-field ligands cause a large split in the energies of orbitals of the central metal atom and transition metal coordination compounds with these ligands are typically yellow, orange, or red because they absorb higher-energy violet or blue light. Coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light. Identify the color (either blue, green, yellow, or orange) for the following complex ions formed with \(Co^{3+}\): Each of these complex ions has the same metal with the same oxidation state, so the ligand field is the relevant factors. Each of the complex ions also has an octahedral ligand field, so we only need to compare the strength of the ligands in determining \(\Delta_o\), which is determined by the spectrochemical series. The ligands for each complex ions are: (a) \(CN^-\), (b) \(NH_3\), (c) \(F^-\) and (d) \(H_2O\), which are ranked in increasing \(\Delta _o\) magnitude: \[ F^- < H_2O < NH_3 < CN^- \label{eq0}\] The relationship between the \(\Delta_o\) and the energy of the photons are absorbed in the d-d transition of \(\ce{Co^{3+}}\) is given by equating Planck's equation to the crystal field splitting parameter: \[E= h \nu= \dfrac{hc}{\lambda} = \Delta_o \label{eq1}\] Now, we need to get a relative correlation between observed color (to the eye) and the wavelength of the light that is absorbed. From the complementary color wheel in Figure \(\Page {1}\) we get the following relationships (arranged from highest energy absorbed to lowest): Of the four possible colors given in the problem (blue, green, yellow, and orange), the corresponding colors that are absorbed are (600 nm, 650 nm, 450 nm, and 430, respectively). From Equation \(\ref{eq1}\), the smaller \(\lambda\) of absorbed light corresponds to the higher energy photons, so we would correlate the four wavelengths of absorbing photons in terms of increasing energy to observed color: \[ \underbrace{650 \,nm }_{green} < \underbrace{600 \,nm}_{blue} < \underbrace{450 \, nm}_{yellow} < \underbrace{430 \, nm}_{orange} \label{eq3}\] Now, just correlate the ranking in Equation \(\ref{eq3}\) to the ranking in Equation \(\ref{eq0}\): \[ \underbrace{F^- }_{green} < \underbrace{ H_2O}_{blue} < \underbrace{NH_3}_{yellow} < \underbrace{CN^-}_{orange} \nonumber\] and more specifically in terms of the original question \[ \underbrace{[CoF_6]_3^{-4} }_{green} < \underbrace{ [Co(H_2O)_6]^{3+}}_{blue} < \underbrace{[Co(NH_3)_6]_3^{3+}}_{yellow} < \underbrace{[Co(CN)_6]_3^{-2}}_{orange} \nonumber\] If a complex ion \([M(NH_3)_6]^{2+}\) is red in solution, which of the following ligands, after a ligand exchange reaction to substitute for the ammine ligands, may shift the solution to be orange? More than one answer is possible. \(Cl^-\), \(CN^-\), \(CO\), \(F^-\) \(H_2O\), \(I^-\), \(en\), \(NO_2^-\), \(OH^-\), \(SCN^-\) A coordination compound of the Cu ion has a configuration, and all the orbitals are filled. To excite an electron to a higher level, such as the 4 orbital, photons of very high energy are necessary. This energy corresponds to very short wavelengths in the ultraviolet region of the spectrum. No visible light is absorbed, so the eye sees no change, and the compound appears white or colorless. A solution containing [Cu(CN) ] , for example, is colorless. On the other hand, octahedral Cu complexes have a vacancy in the orbitals, and electrons can be excited to this level. The wavelength (energy) of the light absorbed corresponds to the visible part of the spectrum, and Cu complexes are almost always colored—blue, blue-green violet, or yellow (Figure \(\Page {6}\)). Although CFT successfully describes many properties of coordination complexes, molecular orbital explanations (beyond the introductory scope provided here) are required to understand fully the behavior of coordination complexes. A ruby is a pink to blood-red colored gemstone that consist of trace amounts of chromium in the mineral corundum \(Al_2O_3\). In contrast, emeralds are colored green by trace amounts of chromium within a Be Al Si O matrix. We can now understand why emeralds and rubies have such different colors, even though both contain Cr in an octahedral environment provided by six oxide ions. Although the chemical identity of the six ligands is the same in both cases, the Cr–O distances are different because the compositions of the host lattices are different (Al O in rubies and Be Al Si O in emeralds). In ruby, the Cr–O distances are relatively short because of the constraints of the host lattice, which increases the d orbital–ligand interactions and makes Δ relatively large. Consequently, rubies absorb green light and the transmitted or reflected light is red, which gives the gem its characteristic color. In emerald, the Cr–O distances are longer due to relatively large [Si O ] silicate rings; this results in decreased d orbital–ligand interactions and a smaller Δ . Consequently, emeralds absorb light of a longer wavelength (red), which gives the gem its characteristic green color. It is clear that the environment of the transition-metal ion, which is determined by the host lattice, dramatically affects the spectroscopic properties of a metal ion. The absorbance spectrum of a ruby is shown Figure \(\Page {7; left}\). The number and positions of the peaks in the spectrum is determined by the electronic structure of the compound, which in this case depends upon the identity of the metal and the identities, number, and geometry of the surrounding ions. Crystal field theory may be used to predict the electronic structure and thus the absorbance spectrum. If white light is shown on the gem, the absorbance spectrum indicates which wavelengths of light are removed. In this case, there are strong bands centered at 414 and 561 nm. These wavelengths correspond with blue and yellow-green light, respectively. For the most part, these colors are not present in the light reaching ones eyes. An alternative way to express this concept is to recognize that the spectrum of light reaching the eye is the product of the spectrum of the incident light (white light) and the . For this ruby, the transmittance spectrum has a peak at 481 nm and a broad plateau past 620 nm. (Note that there is significant attenuation of the light across the entire spectrum.) Thus only light with wavelengths near 481 nm (cyan) and greater than 620 nm (red) reach the eye. A similar analysis of the spectrum of an emerald is possible. The absorbance spectrum (Figure \(\Page {8; left}\)) shows strong bands at 438 and 606 nm, which remove blue and orange light, respectively. The light that is not absorbed is shown by the transmittance spectrum, which indicates the dominant band of light reaching the eye is centered at 512 nm (green light) with smaller contributions from the far blue and red portions of the spectrum. This combination of wavelengths produces a deep green color. When atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the orbitals often allows photons in the visible range to be absorbed. ). ( ) | 13,161 | 2,351 |
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Aryl halides have a halogen directly bonded to a carbon of an aromatic ring. Examples are bromobenzene, fluorobenzene, and 2,4-dichloromethylbenzene: Some of the methods by which alkyl halides are prepared do not work for aryl halides because it is difficult to form \(\ce{C}\)-halogen bonds at aromatic ring carbons by nucleophilic displacement reactions. The most common ways of forming \(\ce{C}_\text{aryl}\)-halogen bonds are by substitution of \(\ce{C}_\text{aryl} \ce{-H}\) by electrophilic halogenating agents (e.g., \(\ce{Br_2}\) or \(\ce{Cl_2}\)), and by replacement of \(\ce{C-NH_2}\) by \(\ce{C}-\)halogen. These reactions are listed in Table 14-5 and will be discussed in more detail in Chapters 22 and 23. The carbon-halogen bonds of aryl halides are like those of alkenyl halides in being much stronger than those of alkyl halides (see Table 4-6). (Table 14-6). However, this low reactivity can be changed dramatically by changes in the reaction conditions and the structure of the aryl halide. In fact, nucleophilic displacement becomes quite rapid (a) when the aryl halide is activated by substitution with strongly electron-attracting groups such as \(\ce{NO_2}\), and (b) when very strongly basic nucleophilic reagents are used. Although the simple aryl halides are inert to the usual nucleophilic reagents, ortho para . For example, the displacement of chloride ion from 1-chloro-2,4-dinitrobenzene by dimethylamine occurs readily in ethanol solution at room temperature. Under the same conditions chlorobenzene completely fails to react; thus the activating influence of the two nitro groups amounts to a factor of at least \(10^8\): A related reaction is that of 2,4-dinitrofluorobenzene with the amino groups of peptides and proteins, and this reaction provides a means for analysis of the -terminal amino acids in polypeptide chains. (See .) In general, the reactions of activated aryl halides closely resemble the \(S_\text{N}2\)-displacement reactions of aliphatic halides. The same nucleophilic reagents are effective (e.g., \(\ce{CH_3O}^\ominus\), \(\ce{HO}^\ominus\), and \(\ce{RNH_2}\)); the reactions are second order overall (first order in halide and first order in nucleophile); and for a given halide the more nucleophilic the attacking reagent, the faster the reaction. However, there must be more than a subtle difference in mechanism because an aryl halide is unable to pass through the same type of transition state as an alkyl halide in \(S_\text{N}2\) displacements. The generally accepted mechanism of nucleophilic aromatic substitution of aryl halides carrying activating groups involves two steps that are closely analogous to those briefly described in for alkenyl and alkynyl halides. The first step involves attack of the nucleophile \(\ce{Y}^\ominus\) at the carbon bearing the halogen substituent to form an intermediate carbanion \(4\) (Equation 14-3). The aromatic system is destroyed on forming the anion, and the carbon at the reaction site changes from planar (\(sp^2\) bonds) to tetrahedral (\(sp^3\) bonds). In the second step, loss of an anion, \(\ce{X}^ominus\) or \(\ce{Y}^\ominus\), regenerates an aromatic system, and, if \(\ce{X}^\ominus\) is lost, the overall reaction is nucleophilic displacement of \(\ce{X}\) by \(\ce{Y}\) (Equation 14-4). In the case of a neutral nucleophilic reagent, \(\ce{Y}\) or \(\ce{HY}\), the reaction sequence would be the same except for the necessary adjustments in the charge of the intermediate: Why is this reaction pathway generally unfavorable for the simple aryl halides? The answer is that the intermediate \(4\), which we can express as a hybrid of the valence-bond structures \(4a\)-\(4c\), is too high in energy to be formed at any practical rate. Not only has \(4\) lost the aromatic stabilization of the benzene ring, but its formation results in transfer of negative charge to the ring carbons, which themselves are not very electronegative: However, when s ortho-para . As an example, consider the displacement of bromine by \(\ce{OCH_3}\) in the reaction of 4-bromonitrobenzene and methoxide ion: The anionic intermediate formed by of methoxide ion to the aryl halide can be described by the valence-bond structures \(5a\)-\(5d\). Of these structures \(5d\) is especially important because in it the charge is transferred from the ring carbons to the oxygen of the nitro substituent: . No formulas can be written analogous to \(5c\) and \(5d\) in which the negative charges are both on atoms next to positive nitrogen, \(\overset{\ominus}{\ce{C}} \overset{\oplus}{\ce{-N}-} \overset{\ominus}{\ce{O}}\) and \(\overset{\ominus}{\ce{O}} \overset{\oplus}{\ce{-N}-} \overset{\ominus}{\ce{O}}\), In a few instances, stable compounds resembling the postulated reaction intermediate have been isolated. One classic example is the complex \(7\) (isolated by J. Meisenheimer), which is the product of the reaction of either the methyl aryl ether 6\(6\) with potassium ethoxide, or the ethyl aryl ether \(8\) and potassium methoxide: The reactivities of aryl halides, such as the halobenzenes, are exceedingly low toward nucleophilic reagents that normally effect displacements with alkyl halides and activated aryl halides. Substitutions do occur under forcing conditions of either high temperatures or very strong bases. For example, chlorobenzene reacts with sodium hydroxide solution at temperatures around \(340^\text{o}\) and this reaction was once an important commercial process for the production of benzenol (phenol): In addition, aryl chlorides, bromides, and iodides can be converted to areneamines \(\ce{ArNH_2}\) by the conjugate bases of amines. In fact, the reaction of potassium amide with bromobenzene is extremely rapid, even at temperatures as low as \(-33^\text{o}\) with liquid ammonia as solvent: However, displacement reactions of this type differ from the previously discussed displacements of activated aryl halides in that rearrangement often occurs. That is, . For example, the hydrolysis of 4-chloromethylbenzene at \(340^\text{o}\) gives an equimolar mixture of 3- and 4-methylbenzenols: Even more striking is the exclusive formation of 3-methoxybenzenamine in the amination of 2-chloromethoxybenzene. Notice that this result is a violation of the principle of least structural change ( ): The mechanism of this type of reaction has been studied extensively, and much evidence has accumulated in support of a stepwise process, which proceeds first by base-catalyzed of hydrogen halide \(\left( \ce{HX} \right)\) from the aryl halide - as illustrated below for the amination of bromobenzene: The product of the elimination reaction is a highly reactive intermediate \(9\) called , or , which differs from benzene in having two less hydrogen and an extra bond between two ortho carbons. Benzyne reacts rapidly with any available nucleophile, in this case the solvent, ammonia, to give an addition product: The rearrangements in these reactions result from the attack of the nucleophile at one or the other of the carbons of the extra bond in the intermediate. With benzyne the symmetry is such that no rearrangement would be detected. With substituted benzynes isomeric products may result. Thus 4-methylbenzyne, \(10\), from the reaction of hydroxide ion with 4-chloro-1-methylbenzene gives both 3- and 4-methylbenzenols: In the foregoing benzyne reactions the base that produces the benzyne in the elimination step is derived from the nucleophile that adds in the addition step. This need not always be so, depending on the reaction conditions. In fact, the synthetic utility of aryne reactions depends in large part of the success with which the aryne can be generated by one reagent but captured by another. One such method will be discussed in and involves organometallic compounds derived from aryl halides. Another method is to generate the aryne by thermal decomposition of a 1,2-disubstituted arene compound such as \(11\), in which both substituents are leaving groups - one leaving with an electron pair, the other leaving without: When \(11\) decomposes in the presence of an added nucleophile, the benzyne intermediate is trapped by the nucleophile as it is formed. Or, if a conjugated diene is present, benzyne will react with it by a [4 + 2] cycloaddition. In the absence of other compounds with which it can react, benzyne will undergo [2 + 2] cycloaddition to itself: As with most organic halides, aryl halides most often are synthetic intermediates for the production of other useful substances. For example, chlorobenzene is the starting aryl halide for the synthesis of DDT; it also is a source of benzenol (phenol, ) which, in turn, has many uses ( ). Several aromatic chloro compounds are used extensively as insecticides, herbicides, fungicides, and bactericides. They also have acquired much notoriety because in some instances their indiscriminate usage has led to serious problems. For example, hexachlorophene is an external bactericide that until recently was used in cosmetic preparations such as soaps, deodorants, and so on. Its use has been discontinued because of compelling evidence that it can be absorbed through the skin in amounts that are dangerous, if not lethal, for infants and small children. Other pesticides, notably DDT and the herbicides 2,4-D and 2,4,5-T have been partially banned for different reasons. and (1977) | 9,424 | 2,352 |
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Make sure you thoroughly understand the following essential ideas: The stronger intermolecular forces in solids require more input of energy to disperse the molecular units into a liquid solution, but there is also a considerable increase in entropy that can more than compensate if the intermolecular forces are not too strong, and if the solvent has no strong hydrogen bonds that must be broken in order to introduce the solute into the liquid. For example, at 25° C and 1 atm pressure, 20 g of iodine crystals will dissolve in 100 ml of ethyl alcohol, but the same quantity of water will dissolve only 0.30 g of iodine. As the molecular weight of the solid increases, the intermolecular forces holding the solid together also increase, and solubilities tend to fall off; thus the solid linear hydrocarbons CH (CH ) CH ( > 20) show diminishing solubilities in hydrocarbon liquids. Since the Coulombic forces that bind ions and highly polar molecules into solids are quite strong, we might expect these solids to be insoluble in just about any solvent. Ionic solids are insoluble in most non-aqueous solvents, but the high solubility of some (including NaCl) in water suggests the need for some further explanation. The key factor here turns out to be the interaction of the ions with the solvent. The electrically-charged ions exert a strong coulombic attraction on the end of the water molecule that has the opposite partial charge. As a consequence, ions in solution are always ; that is, they are quite tightly bound to water molecules through ion-dipole interaction. The number of water molecules contained in the varies with the radius and charge of the ion.
Figure \(\Page {1}\): Hydration shells around some ions in a sodium chloride solution. The average time an ion spends in a shell is about 2-4 nanoseconds. But this is about two orders of magnitude longer than the lifetime of an individual \(H_2O–H_2O\) hydrogen bond. The dissolution of an ionic solid \(M\)X in water can be thought of as a sequence of two (hypothetical) steps: \[MX(s) \rightarrow M^+(g) + X^–(g) \] \[M^+(g) + X^–(g) + H_2O(l) \rightarrow M^+(aq) + X–(aq)\] The enthalpy difference of the first step is the lattice energy and is always positive; the enthalpy difference of the second step is the hydration energy and is always negative) Single-ion hydration energies (Table \(\Page {1}\)) cannot be observed directly, but are obtained from the differences in hydration energies of salts having the given ion in common. When you encounter tables such as the above in which numeric values are related to different elements, you should always stop and see if you can make sense of any obvious trends. In this case, the things to look for are the size and charge of the ions as they would affect the electrostatic interaction between two ions or between an ion and a [polar] water molecule. Lattice energies are not measured directly, but are estimates based on electrostatic calculations which are reliable only for simple salts. Enthalpies of solution are observable either directly or (for sparingly soluble salts,) indirectly. Hydration energies are not measurable; they are estimated as the sum the other two quantities. It follows that any uncertainty in the lattice energies is reflected in those of the hydration energies. For this reason, tabulated values of the latter will vary depending on the source. When calcium chloride, CaCl , is dissolved in water, will the temperature immediately after mixing rise or fall? Estimate the heat of solution of CaCl . (2258 – 2324) kJ mol = Since the process is exothermic, this heat will be released to warm the solution. As often happens for a quantity that is the sum of two large terms having opposite signs, the overall dissolution process can come out as either endothermic or exothermic, and examples of both kinds are common. Two common examples illustrate the contrast between exothermic and endothermic heats of solution of ionic solids: Hydration shells around some ions in a sodium chloride solution. The average time an ion spends in a shell is about 2-4 nanoseconds. But this is about two orders of magnitude longer than the lifetime of an individual \(H_2O\)–\(H_2O\) hydrogen bond. The balance between the lattice energy and hydration energy is a major factor in determining the solubility of an ionic crystal in water, but there is another factor to consider as well. We generally assume that there is a rather large increase in the entropy when a solid is dispersed into the liquid phase. However, in the case of ionic solids, each ion ends up surrounded by a shell of oriented water molecules. These water molecules, being constrained within the hydration shell, are unable to participate in the spreading of thermal energy throughout the solution, and reduce the entropy. In some cases this effect predominates so that dissolution of the salt leads to a net decrease in entropy. Recall that any process in which the the entropy diminishes becomes less probable as the temperature increases; this explains why the solubilities of some salts decrease with temperature. | 5,140 | 2,353 |
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As the density is increased the interactions between gas molecules become non-negligible. Deviations from the ideal gas law have been described in a large number of equations of state. The expresses the deviation from ideality in terms of a power series in the density: \[ \dfrac{P}{kT} = \rho + B_2(T)\rho^2 + B_3(T)\rho^3 + ... \nonumber \] The j virial coefficient can be calculated in terms of the interaction of j molecules in a volume \(V\). The second and third virial coefficients give most of the deviation from ideal (\(P/rkT\)) up to 100 atm. The is usually written as \(B\) or as \(B_2\). The second virial coefficient represents the initial departure from ideal-gas behavior. The second virial coefficient, in three dimensions, is given by: \[B_{2}(T)= - \dfrac{1}{2} \int \left( \exp\left(-\dfrac{\Phi_{12}({\mathbf r})}{k_BT}\right) -1 \right) 4 \pi r^2 dr \nonumber \] where \(\Phi_{12}({\mathbf r})\) is the intermolecular pair potential, \(T\) is the temperature, and \(k_B\) is the Boltzmann constant. Notice that the expression within the parenthesis of the integral is the . In practice, the integral is often to integrate analytically for anything other than, say, the hard sphere model, thus one numerically evaluates: \[B_{2}(T)= - \dfrac{1}{2} \int \left( \left\langle \exp\left(-\dfrac{\Phi_{12}({\mathbf r})}{k_BT}\right)\right\rangle -1 \right) 4 \pi r^2 dr \nonumber \] calculating: \[ \left\langle \exp\left(-\dfrac{\Phi_{12}({\mathbf r})}{k_BT}\right)\right\rangle \nonumber \] for each \(r\) using the numerical integration scheme proposed by Harold Conroy . The configuration integrals for \(Z_1\), \(Z_2\), and \(Z_3\) are: The series method allows the calculation of a number of virial coefficients. Recall that the second and third virial coefficient can account for the properties of gases up to hundreds of atmospheres. We will discuss the calculation of the second virial coefficient for a monatomic gas to illustrate the procedure. To calculate \(B_2(T)\) we need \(U_2\). For monatomic particles it is reasonable to assume that the potential depends only on the separation of the two particles so \(U_2 = u(r_{12})\), where \(r_{12} = |r_2 - r_1|\). Using a change of variables we can write this integral \(r_{12} = r_2 - r_1\) and after integration over \(r_1\) we can transform variables from \(dr_{12}\) to \(4\pi r^2dr\). The result is: \[ B_2(T) = -2 \pi \int [ e^{-\beta u(r)} - 1 ] r^2 \; dr \nonumber \] This expression can be used to obtain parameters from experiment. The second virial coefficient is tabulated for a number of gases. For a hard-sphere potential there is an infinite repulsive wall at a particle radius \(\sigma\). There is no attractive part. \[\begin{align*} B_2(T) &= -2 \pi \int_0^{\sigma} [ - 1 ] r^2 \; dr \\[4pt] &= \dfrac{2 \pi \sigma^3}{3} \end{align*} \] The Lennard-Jones potential cannot be calculated analytically, but the integral can be computed numerically. The second virial coefficient was a useful starting point for obtaining Lennard-Jones parameters that were used in simulations. The Isihara-Hadwiger formula was discovered simultaneously and independently by Isihara and the Swiss mathematician Hadwiger in 1950. The second virial coefficient for any hard convex body is given by the exact relation: \[B_2=RS+V \nonumber \] or: \[\dfrac{B_2}{V}=1+3 \alpha \nonumber \] where: \[\alpha = \dfrac{RS}{3V} \nonumber \] where \(V\) is the volume, \(S\), the surface area, and \(R\) the mean radius of curvature. For the hard sphere model one has: \[B_{2}(T)= - \dfrac{1}{2} \int_0^\sigma \left(\langle 0\rangle -1 \right) 4 \pi r^2 dr \nonumber \] leading to \[B_{2}= \dfrac{2\pi\sigma^3}{3} \nonumber \] Note that \(B_{2}\) for the hard sphere is independent of temperature. For the Van der Waals equation of state one has: \[B_{2}(T)= b -\dfrac{a}{RT} \nonumber \] The second virial coefficient can be computed from the expression: \[B_{2}= \dfrac{1}{2} \iint v_{\mathrm {excluded}} (\Omega,\Omega') f(\Omega) f(\Omega')~ {\mathrm d}\Omega {\mathrm d}\Omega' \nonumber \] where \(v_{\mathrm {excluded}}\) is the excluded volume. | 4,142 | 2,355 |
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In this chapter, we have described various galvanic cells in which a spontaneous chemical reaction is used to generate electrical energy. In an electrolytic cell, however, the opposite process, called , occurs: an external voltage is applied to drive a nonspontaneous reaction. In this section, we look at how electrolytic cells are constructed and explore some of their many commercial applications. If we construct an electrochemical cell in which one electrode is copper metal immersed in a 1 M Cu solution and the other electrode is cadmium metal immersed in a \(\,1\; M\, Cd^{2+}\) solution and then close the circuit, the potential difference between the two compartments will be 0.74 V. The cadmium electrode will begin to dissolve (Cd is oxidized to Cd ) and is the anode, while metallic copper will be deposited on the copper electrode (Cu is reduced to Cu), which is the cathode (Figure \(\Page {1a}\)). The overall reaction is as follows: \[ \ce{Cd (s) + Cu^{2+} (aq) \rightarrow Cd^{2+} (aq) + Cu (s)} \nonumber \] with \(E°_{cell} = 0.74\; V\) This reaction is thermodynamically spontaneous as written (\(ΔG^o < 0\)): \[ \begin{align*} \Delta G^\circ &=-nFE^\circ_\textrm{cell} \\[4pt] &=-(\textrm{2 mol e}^-)[\mathrm{96,485\;J/(V\cdot mol)}](\mathrm{0.74\;V}) \\[4pt] &=-\textrm{140 kJ (per mole Cd)} \end{align*} \nonumber \] In this direction, the system is acting as a galvanic cell. In an electrolytic cell, an external voltage is applied to drive a reaction. The reverse reaction, the reduction of Cd by Cu, is thermodynamically nonspontaneous and will occur only with an input of 140 kJ. We can force the reaction to proceed in the reverse direction by applying an electrical potential greater than 0.74 V from an external power supply. The applied voltage forces electrons through the circuit in the reverse direction, converting a galvanic cell to an electrolytic cell. Thus the copper electrode is now the anode (Cu is oxidized), and the cadmium electrode is now the cathode (Cd is reduced) (Figure \(\Page {1b}\)). The signs of the cathode and the anode have switched to reflect the flow of electrons in the circuit. The half-reactions that occur at the cathode and the anode are as follows: \[\ce{Cd^{2+}(aq) + 2e^{−} \rightarrow Cd(s)}\label{20.9.3} \] with \(E^°_{cathode} = −0.40 \, V\) \[\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2e^{−}} \label{20.9.4} \] with \(E^°_{anode} = 0.34 \, V \) \[\ce{Cd^{2+}(aq) + Cu(s) \rightarrow Cd(s) + Cu^{2+}(aq) } \label{20.9.5} \] with \(E^°_{cell} = −0.74 \: V\) Because \(E^°_{cell} < 0\), the overall reaction—the reduction of \(Cd^{2+}\) by \(Cu\)—clearly occur spontaneously and proceeds only when sufficient electrical energy is applied. The differences between galvanic and electrolytic cells are summarized in Table \(\Page {1}\). At sufficiently high temperatures, ionic solids melt to form liquids that conduct electricity extremely well due to the high concentrations of ions. If two inert electrodes are inserted into molten \(\ce{NaCl}\), for example, and an electrical potential is applied, \(\ce{Cl^{-}}\) is oxidized at the anode, and \(\ce{Na^{+}}\) is reduced at the cathode. The overall reaction is as follows: \[\ce{ 2NaCl (l) \rightarrow 2Na(l) + Cl2(g)} \label{20.9.6} \] This is the reverse of the formation of \(\ce{NaCl}\) from its elements. The product of the reduction reaction is liquid sodium because the melting point of sodium metal is 97.8°C, well below that of \(\ce{NaCl}\) (801°C). Approximately 20,000 tons of sodium metal are produced commercially in the United States each year by the electrolysis of molten \(\ce{NaCl}\) in a Downs cell (Figure \(\Page {2}\)). In this specialized cell, \(\ce{CaCl2}\) (melting point = 772°C) is first added to the \(\ce{NaCl}\) to lower the melting point of the mixture to about 600°C, thereby lowering operating costs. Similarly, in the Hall–Heroult process used to produce aluminum commercially, a molten mixture of about 5% aluminum oxide (Al O ; melting point = 2054°C) and 95% cryolite (Na AlF ; melting point = 1012°C) is electrolyzed at about 1000°C, producing molten aluminum at the cathode and CO gas at the carbon anode. The overall reaction is as follows: \[\ce{2Al2O3(l) + 3C(s) -> 4Al(l) + 3CO2(g)} \label{20.9.7} \] Oxide ions react with oxidized carbon at the anode, producing CO (g). There are two important points to make about these two commercial processes and about the electrolysis of molten salts in general. In the Hall–Heroult process, C is oxidized instead of O or F because oxygen and fluorine are more electronegative than carbon, which means that C is a weaker oxidant than either O or F . Similarly, in the Downs cell, we might expect electrolysis of a NaCl/CaCl mixture to produce calcium rather than sodium because Na is slightly less electronegative than Ca (χ = 0.93 versus 1.00, respectively), making Na easier to oxidize and, conversely, Na more difficult to reduce. In fact, the reduction of Na to Na is the observed reaction. In cases where the electronegativities of two species are similar, other factors, such as the formation of complex ions, become important and may determine the outcome. If a molten mixture of MgCl and KBr is electrolyzed, what products will form at the cathode and the anode, respectively? identity of salts electrolysis products The possible reduction products are Mg and K, and the possible oxidation products are Cl and Br . Because Mg is more electronegative than K (χ = 1.31 versus 0.82), it is likely that Mg will be reduced rather than K. Because Cl is more electronegative than Br (3.16 versus 2.96), Cl is a stronger oxidant than Br . Electrolysis will therefore produce Br at the anode and Mg at the cathode. Predict the products if a molten mixture of AlBr and LiF is electrolyzed. Br and Al Electrolysis can also be used to drive the thermodynamically nonspontaneous decomposition of water into its constituent elements: H and O . However, because pure water is a very poor electrical conductor, a small amount of an ionic solute (such as H SO or Na SO ) must first be added to increase its electrical conductivity. Inserting inert electrodes into the solution and applying a voltage between them will result in the rapid evolution of bubbles of H and O (Figure \(\Page {3}\)). The reactions that occur are as follows: For a system that contains an electrolyte such as Na SO , which has a negligible effect on the ionization equilibrium of liquid water, the pH of the solution will be 7.00 and [H ] = [OH ] = 1.0 × 10 . Assuming that \(P_\mathrm{O_2}\) = \(P_\mathrm{H_2}\) = 1 atm, we can use the standard potentials to calculate E for the overall reaction: \[\begin{align}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log(P_\mathrm{O_2}P^2_\mathrm{H_2}) \\ &=-\textrm{1.23 V}-\left(\dfrac{\textrm{0.0591 V}}{4}\right)\log(1)=-\textrm{1.23 V}\end{align} \label{20.9.11} \] Thus E is −1.23 V, which is the value of E° if the reaction is carried out in the presence of 1 M H rather than at pH 7.0. In practice, a voltage about 0.4–0.6 V greater than the calculated value is needed to electrolyze water. This added voltage, called an , represents the additional driving force required to overcome barriers such as the large activation energy for the formation of a gas at a metal surface. Overvoltages are needed in all electrolytic processes, which explain why, for example, approximately 14 V must be applied to recharge the 12 V battery in your car. In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. The p-block metals and most of the transition metals are in this category, but metals in high oxidation states, which form oxoanions, cannot be reduced to the metal by simple electrolysis. Active metals, such as aluminum and those of groups 1 and 2, react so readily with water that they can be prepared only by the electrolysis of molten salts. Similarly, any nonmetallic element that does not readily oxidize water to O can be prepared by the electrolytic oxidation of an aqueous solution that contains an appropriate anion. In practice, among the nonmetals, only F cannot be prepared using this method. Oxoanions of nonmetals in their highest oxidation states, such as NO , SO , PO , are usually difficult to reduce electrochemically and usually behave like spectator ions that remain in solution during electrolysis. In general, any metal that does not react readily with water to produce hydrogen can be produced by the electrolytic reduction of an aqueous solution that contains the metal cation. In a process called , a layer of a second metal is deposited on the metal electrode that acts as the cathode during electrolysis. Electroplating is used to enhance the appearance of metal objects and protect them from corrosion. Examples of electroplating include the chromium layer found on many bathroom fixtures or (in earlier days) on the bumpers and hubcaps of cars, as well as the thin layer of precious metal that coats silver-plated dinnerware or jewelry. In all cases, the basic concept is the same. A schematic view of an apparatus for electroplating silverware and a photograph of a commercial electroplating cell are shown in Figure \(\Page {4}\). The half-reactions in electroplating a fork, for example, with silver are as follows: The overall reaction is the transfer of silver metal from one electrode (a silver bar acting as the anode) to another (a fork acting as the cathode). Because \(E^o_{cell} = 0\, V\), it takes only a small applied voltage to drive the electroplating process. In practice, various other substances may be added to the plating solution to control its electrical conductivity and regulate the concentration of free metal ions, thus ensuring a smooth, even coating. If we know the stoichiometry of an electrolysis reaction, the amount of current passed, and the length of time, we can calculate the amount of material consumed or produced in a reaction. Conversely, we can use stoichiometry to determine the combination of current and time needed to produce a given amount of material. The quantity of material that is oxidized or reduced at an electrode during an electrochemical reaction is determined by the stoichiometry of the reaction and the amount of charge that is transferred. For example, in the reaction \[\ce{Ag^{+}(aq) + e^{−} → Ag(s)} \nonumber \] 1 mol of electrons reduces 1 mol of \(\ce{Ag^{+}}\) to \(\ce{Ag}\) metal. In contrast, in the reaction \[\ce{Cu^{2+}(aq) + 2e^{−} → Cu(s)} \nonumber \] 1 mol of electrons reduces only 0.5 mol of \(\ce{Cu^{2+}}\) to \(\ce{Cu}\) metal. Recall that the charge on 1 mol of electrons is 1 faraday (1 F), which is equal to 96,485 C. We can therefore calculate the number of moles of electrons transferred when a known current is passed through a cell for a given period of time. The total charge (\(q\) in coulombs) transferred is the product of the current (\(I\) in amperes) and the time (\(t\), in seconds): \[ q = I \times t \label{20.9.14} \] The stoichiometry of the reaction and the total charge transferred enable us to calculate the amount of product formed during an electrolysis reaction or the amount of metal deposited in an electroplating process. For example, if a current of 0.60 A passes through an aqueous solution of \(\ce{CuSO4}\) for 6.0 min, the total number of coulombs of charge that passes through the cell is as follows: \[\begin{align*} q &= \textrm{(0.60 A)(6.0 min)(60 s/min)} \\[4pt] &=\mathrm{220\;A\cdot s} \\[4pt] &=\textrm{220 C} \end{align*} \nonumber \] The number of moles of electrons transferred to \(\ce{Cu^{2+}}\) is therefore \[\begin{align*} \textrm{moles e}^- &=\dfrac{\textrm{220 C}}{\textrm{96,485 C/mol}} \\[4pt] &=2.3\times10^{-3}\textrm{ mol e}^- \end{align*} \nonumber \] Because two electrons are required to reduce a single Cu ion, the total number of moles of Cu produced is half the number of moles of electrons transferred, or 1.2 × 10 mol. This corresponds to 76 mg of Cu. In commercial electrorefining processes, much higher currents (greater than or equal to 50,000 A) are used, corresponding to approximately 0.5 F/s, and reaction times are on the order of 3–4 weeks. A silver-plated spoon typically contains about 2.00 g of Ag. If 12.0 h are required to achieve the desired thickness of the Ag coating, what is the average current per spoon that must flow during the electroplating process, assuming an efficiency of 100%? mass of metal, time, and efficiency current required We must first determine the number of moles of Ag corresponding to 2.00 g of Ag: \(\textrm{moles Ag}=\dfrac{\textrm{2.00 g}}{\textrm{107.868 g/mol}}=1.85\times10^{-2}\textrm{ mol Ag}\) The reduction reaction is Ag (aq) + e → Ag(s), so 1 mol of electrons produces 1 mol of silver. Using the definition of the faraday, The current in amperes needed to deliver this amount of charge in 12.0 h is therefore \[\begin{align*}\textrm{amperes} &=\dfrac{1.78\times10^3\textrm{ C}}{(\textrm{12.0 h})(\textrm{60 min/h})(\textrm{60 s/min})}\\
& =4.12\times10^{-2}\textrm{ C/s}=4.12\times10^{-2}\textrm{ A}\end{align*} \nonumber \] Because the electroplating process is usually much less than 100% efficient (typical values are closer to 30%), the actual current necessary is greater than 0.1 A. A typical aluminum soft-drink can weighs about 29 g. How much time is needed to produce this amount of Al(s) in the Hall–Heroult process, using a current of 15 A to reduce a molten Al O /Na AlF mixture? 5.8 h Electroplating: In electrolysis, an external voltage is applied to drive a reaction. The quantity of material oxidized or reduced can be calculated from the stoichiometry of the reaction and the amount of charge transferred. Relationship of charge, current and time: \[ q = I \times t \nonumber \] In electrolysis, an external voltage is applied to drive a nonspontaneous reaction. Electrolysis can also be used to produce H and O from water. In practice, an additional voltage, called an overvoltage, must be applied to overcome factors such as a large activation energy and a junction potential. Electroplating is the process by which a second metal is deposited on a metal surface, thereby enhancing an object’s appearance or providing protection from corrosion. The amount of material consumed or produced in a reaction can be calculated from the stoichiometry of an electrolysis reaction, the amount of current passed, and the duration of the electrolytic reaction. | 14,641 | 2,356 |
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Buffer solutions are extremely important in biology and medicine because most biological reactions and enzymes need very specific pH ranges in order to work properly. Human blood contains a buffer of carbonic acid (\(\ce{H2CO3}\)) and bicarbonate anion (\(\ce{HCO3^{-}}\)) in order to maintain blood pH between 7.35 and 7.45, as a value higher than 7.8 or lower than 6.8 can lead to death. In this buffer, hydronium and bicarbonate anion are in equilibrium with carbonic acid. Furthermore, the carbonic acid in the first equilibrium can decompose into \(\ce{CO2}\) gas and water, resulting in a second equilibrium system between carbonic acid and water. Because \(\ce{O2}\) is an important component of the blood buffer, its regulation in the body, as well as that of \(\ce{O2}\), is extremely important. The effect of this can be important when the human body is subjected to strenuous conditions. In the body, there exists another equilibrium between hydronium and oxygen which involves the binding ability of hemoglobin. An increase in hydronium causes this equilibrium to shift towards the oxygen side, thus releasing oxygen from hemoglobin molecules into the surrounding tissues/cells. This system continues during exercise, providing continuous oxygen to working tissues. In summation, the blood buffer is: \[\ce{H_3O^+ + HCO_3^- <=> H_2CO_3 + H_2O} \nonumber \] With the following simultaneous equilibrium: \[\ce{H_2CO_3 <=> H_2O + CO_2} \nonumber \] be especially useful when culturing bacteria, as their metabolic waste can affect the pH of their medium, consequently killing the sample. For example, a buffer of cacodylic acid (\(\ce{C2H7AsO2}\)) and its conjugate base is used to make samples which will undergo electron microscopy. Another buffer, tricine (\(\ce{C6H13NO5}\)), is used | 1,815 | 2,357 |
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Chemists study the structures, physical properties, and chemical properties of material substances. These consist of , which is anything that occupies space and has mass. Gold and iridium are matter, as are peanuts, people, and postage stamps. Smoke, smog, and laughing gas are matter. Energy, light, and sound, however, are not matter; ideas and emotions are also not matter. The of an object is the quantity of matter it contains. Do not confuse an object’s mass with its , which is a force caused by the gravitational attraction that operates on the object. Mass is a fundamental property of an object that does not depend on its location.In physical terms, the mass of an object is directly proportional to the force required to change its speed or direction. A more detailed discussion of the differences between weight and mass and the units used to measure them is included in Essential Skills 1 ( ). Weight, on the other hand, depends on the location of an object. An astronaut whose mass is 95 kg weighs about 210 lb on Earth but only about 35 lb on the moon because the gravitational force he or she experiences on the moon is approximately one-sixth the force experienced on Earth. For practical purposes, weight and mass are often used interchangeably in laboratories. Because the force of gravity is considered to be the same everywhere on Earth’s surface, 2.2 lb (a weight) equals 1.0 kg (a mass), regardless of the location of the laboratory on Earth. Under normal conditions, there are three distinct states of matter: solids, liquids, and gases. are relatively rigid and have fixed shapes and volumes. A rock, for example, is a solid. In contrast, have fixed volumes but flow to assume the shape of their containers, such as a beverage in a can. , such as air in an automobile tire, have neither fixed shapes nor fixed volumes and expand to completely fill their containers. Whereas the volume of gases strongly depends on their temperature and (the amount of force exerted on a given area), the volumes of liquids and solids are virtually independent of temperature and pressure. Matter can often change from one physical state to another in a process called a . For example, liquid water can be heated to form a gas called steam, or steam can be cooled to form liquid water. However, such changes of state do not affect the chemical composition of the substance. A pure chemical substance is any matter that has a fixed chemical composition and characteristic properties. Oxygen, for example, is a pure chemical substance that is a colorless, odorless gas at 25°C. Very few samples of matter consist of pure substances; instead, most are mixtures, which are combinations of two or more pure substances in variable proportions in which the individual substances retain their identity. Air, tap water, milk, blue cheese, bread, and dirt are all mixtures. If all portions of a material are in the same state, have no visible boundaries, and are uniform throughout, then the material is . Examples of homogeneous mixtures are the air we breathe and the tap water we drink. Homogeneous mixtures are also called solutions. Thus air is a solution of nitrogen, oxygen, water vapor, carbon dioxide, and several other gases; tap water is a solution of small amounts of several substances in water. The specific compositions of both of these solutions are not fixed, however, but depend on both source and location; for example, the composition of tap water in Boise, Idaho, is not the same as the composition of tap water in Buffalo, New York. Although most solutions we encounter are liquid, solutions can also be solid. The gray substance still used by some dentists to fill tooth cavities is a complex solid solution that contains 50% mercury and 50% of a powder that contains mostly silver, tin, and copper, with small amounts of zinc and mercury. Solid solutions of two or more metals are commonly called alloys. If the composition of a material is not completely uniform, then it is (e.g., chocolate chip cookie dough, blue cheese, and dirt). Mixtures that appear to be homogeneous are often found to be heterogeneous after microscopic examination. Milk, for example, appears to be homogeneous, but when examined under a microscope, it clearly consists of tiny globules of fat and protein dispersed in water. The components of heterogeneous mixtures can usually be separated by simple means. Solid-liquid mixtures such as sand in water or tea leaves in tea are readily separated by filtration, which consists of passing the mixture through a barrier, such as a strainer, with holes or pores that are smaller than the solid particles. In principle, mixtures of two or more solids, such as sugar and salt, can be separated by microscopic inspection and sorting. More complex operations are usually necessary, though, such as when separating gold nuggets from river gravel by panning. First solid material is filtered from river water; then the solids are separated by inspection. If gold is embedded in rock, it may have to be isolated using chemical methods. Homogeneous mixtures (solutions) can be separated into their component substances by physical processes that rely on differences in some physical property, such as differences in their boiling points. Two of these separation methods are distillation and crystallization. makes use of differences in volatility, a measure of how easily a substance is converted to a gas at a given temperature. A simple distillation apparatus for separating a mixture of substances, at least one of which is a liquid. The most volatile component boils first and is condensed back to a liquid in the water-cooled condenser, from which it flows into the receiving flask. If a solution of salt and water is distilled, for example, the more volatile component, pure water, collects in the receiving flask, while the salt remains in the distillation flask. Mixtures of two or more liquids with different boiling points can be separated with a more complex distillation apparatus. One example is the refining of crude petroleum into a range of useful products: aviation fuel, gasoline, kerosene, diesel fuel, and lubricating oil (in the approximate order of decreasing volatility). Another example is the distillation of alcoholic spirits such as brandy or whiskey. (This relatively simple procedure caused more than a few headaches for federal authorities in the 1920s during the era of Prohibition, when illegal stills proliferated in remote regions of the United States!) separates mixtures based on differences in solubility, a measure of how much solid substance remains dissolved in a given amount of a specified liquid. Most substances are more soluble at higher temperatures, so a mixture of two or more substances can be dissolved at an elevated temperature and then allowed to cool slowly. Alternatively, the liquid, called the solvent, may be allowed to evaporate. In either case, the least soluble of the dissolved substances, the one that is least likely to remain in solution, usually forms crystals first, and these crystals can be removed from the remaining solution by filtration. Most mixtures can be separated into pure substances, which may be either elements or compounds. An , such as gray, metallic sodium, is a substance that cannot be broken down into simpler ones by chemical changes; a , such as white, crystalline sodium chloride, contains two or more elements and has chemical and physical properties that are usually different from those of the elements of which it is composed. With only a few exceptions, a particular compound has the same elemental composition (the same elements in the same proportions) regardless of its source or history. The chemical composition of a substance is altered in a process called a . The conversion of two or more elements, such as sodium and chlorine, to a chemical compound, sodium chloride, is an example of a chemical change, often called a chemical reaction. Currently, about 118 elements are known, but millions of chemical compounds have been prepared from these 118 elements. The known elements are listed in . Different Definitions of Matter: In general, a reverse chemical process breaks down compounds into their elements. For example, water (a compound) can be decomposed into hydrogen and oxygen (both elements) by a process called electrolysis. In electrolysis, electricity provides the energy needed to separate a compound into its constituent elements ( ). A similar technique is used on a vast scale to obtain pure aluminum, an element, from its ores, which are mixtures of compounds. Because a great deal of energy is required for electrolysis, the cost of electricity is by far the greatest expense incurred in manufacturing pure aluminum. Thus recycling aluminum is both cost-effective and ecologically sound. The overall organization of matter and the methods used to separate mixtures are summarized in . Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution). : a chemical substance : its classification Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution). solution element heterogeneous mixture compound Different Definitions of Changes: Matter can be classified according to physical and chemical properties. Matter is anything that occupies space and has mass. The three states of matter are solid, liquid, and gas. A physical change involves the conversion of a substance from one state of matter to another, without changing its chemical composition. Most matter consists of mixtures of pure substances, which can be homogeneous (uniform in composition) or heterogeneous (different regions possess different compositions and properties). Pure substances can be either chemical compounds or elements. Compounds can be broken down into elements by chemical reactions, but elements cannot be separated into simpler substances by chemical means. The properties of substances can be classified as either physical or chemical. Scientists can observe physical properties without changing the composition of the substance, whereas chemical properties describe the tendency of a substance to undergo chemical changes (chemical reactions) that change its chemical composition. Physical properties can be intensive or extensive. Intensive properties are the same for all samples; do not depend on sample size; and include, for example, color, physical state, and melting and boiling points. Extensive properties depend on the amount of material and include mass and volume. The ratio of two extensive properties, mass and volume, is an important intensive property called density. ( ) | 10,780 | 2,358 |
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The open-chain hydrocarbons with one double bond have the general formula \(C_nH_{2n}\) and are called . The carbon-carbon double bond often is called an "olefinic linkage" and the alkenes designated as (oil-formers). These terms arose because the gaseous lower-molecular-weight alkenes yield "oily" products on treatment with chlorine or bromine. The term "unsaturated" hydrocarbon also is used - again because these substances normally react with bromine and chlorine and are hence "unsaturated" with reference to reagents of this type (also see ). According to the IUPAC system for naming alkenes, the longest continuous chain containing the double bond is given the name of the corresponding alkane with the ending changed to . This chain then is numbered so that the position of the carbon of the double bond is indicated by the lowest possible number: A few very common alkenes also are called "alkylenes" by appending the suffix to the name of the hydrocarbon radical with the same carbon skeleton. Examples are shown below with their alkylene names in parentheses. We shall continue to use the IUPAC names whenever possible. The hydrocarbon groups derived from alkenes have the suffix , as in alkenyl, and numbering of the group starts with the carbon atom attached to the main chain: A few alkenyl groups have trivial names that commonly are used in place of systematic names. These are vinyl, allyl, and isopropenyl. And again we shall avoid using these names, except parenthetically: Also, hydrogen atoms that are bonded directly to the unsaturated carbon atoms of a double bond often are called , although the term is more accurate and therefore preferable. are named by the system used for the open-chain alkenes, except that and continued around the ring the double bond so as to keep the index numbers as small as possible: When a hydrocarbon group is double-bonded to a single carbon of a cycloalkane ring, the suffix , as in alkylidene, is used: Many compounds contain two or more double bonds and are known as alkadienes, alkatrienes, alkatetraenes, and so on, the suffix denoting the number of double bonds. The location of each double bond is specified by appropriate numbers, as illustrated below: A further classification is used for the relationships of the double bonds to each other. Thus 1,2-alkadienes and similar substances are said to have double bonds: 1,3-Alkadienes and other compounds with alternating double and single bonds are said to have double bonds: Compounds with double bonds that are neither cumulated nor conjugated are classified as having double-bond systems: and (1977) | 2,662 | 2,360 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Chemiluminescence/4%3A_Instrumentation/4.05%3A_Chemiluminescence_Sensors |
Chemiluminescence has the advantage of lower background emission than fluorescence, avoiding noise caused by light scattering. However, because chemiluminescence reagents are irreversibly consumed, chemiluminescence sensors have shorter lifetimes than fluorescence sensors and their signals have a tendency to drift downwards due to consumption, migration and breakdown of reagents. Reagent immobilization onto suitable substrates plays an important role in the development of chemiluminescence sensors. Selectivity and sensitivity as well as lifetime of chemiluminescence sensors depends on the choice of reagent and substrate and on the method of immobilization . Chemiluminescence reagents are typically aqueous solutions of ions and so can be immobilized by convenient procedures onto ion exchange resins, giving high surface coverage, and released quantitatively by appropriate eluents. Analytes can also react directly with immobilized reagents. These properties have been widely used to prepare chemiluminescence sensors containing immobilized luminol or other reagents, which typically would be packed into a flow cell positioned in front of the window of a photomultiplier. “Bleeding” columns of anion/cation exchange resins with co-immobilized luminol and metal ions such as Co , Cu or [Fe(CN) ] can detect and measure analytes such as hydrogen peroxide, though this arrangement causes unnecessary dilution of samples and reagents, which impairs detectivity. Immobilized -(2, 2 -bipyridyl)ruthenium(II) can be regenerated from -(2, 2 -bipyridyl)ruthenium(III) and can be used for at least six months. Immobilization of enzymes can be used to produce highly active and selective chemiluminescence sensors from which enzyme is not consumed, though their operational stability is limited. Encapsulation of reagents in sol-gel silica involves little or no structural alteration and is very suitable for chemiluminescence sensors because of its optical transparency and chemical stability. For example, encapsulated horseradish peroxidase displays high activity and long life, as does sol-gel immobilized haemoglobin. Chemiluminescence sensors constructed from plant and animal tissues have advantages of cost, activity, stability and lifetime; examples are soyabean tissue in sensors for urea and spinach tissue in sensors for glycolic acid. Molecular imprinted polymers have been found to be very useful materials for fabrication of chemiluminescence sensors, both as molecular recognition agents and as chemiluminescence reaction media. Analytes that can be successfully detected in this way include 1,10-phenanthroline and dansylated amino-acids. Metal oxide particles can sometimes be entrapped onto membranes or in columns, including chemiluminescence flow cells. This affords a simple fabrication method producing long-lived sensors. Manganese dioxide has been immobilized in this way on sponge rubber for the assay of the drug, analgin, using manganese(IV) chemiluminescence. Chemiluminescence has been detected from surface reactions on nanoparticles, opening up the possibility of chemiluminescence nanosensors of good stability and durability. Coumarin C343, a fluorescent dye, has been conjugated to silica nanoparticles entrapped in sol-gel silica to produce nanosensors capable of enhancing the weak chemiluminescence associated with lipid peroxidation . | 3,395 | 2,361 |
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The importance of the issues raised in this chapter is evident if we examine environmental monitoring programs. The purpose of a monitoring program is to determine the present status of an environmental system, and to assess long term trends in the system’s health. These are broad and poorly defined goals. In many cases, an environmental monitoring program begins before the essential questions are known. This is not surprising since it is difficult to formulate questions in the absence of results. Without careful planning, however, a poor experimental design may result in data that has little value. These concerns are illustrated by the Chesapeake Bay Monitoring Program. This research program, designed to study nutrients and toxic pollutants in the Chesapeake Bay, was initiated in 1984 as a cooperative venture between the federal government, the state governments of Maryland, Virginia, and Pennsylvania, and the District of Columbia. A 1989 review of the program highlights the problems common to many monitoring programs [D’Elia, C. F.; Sanders, J. G.; Capone, D. G. , , 768–774]. At the beginning of the Chesapeake Bay monitoring program, little attention was given to selecting analytical methods, in large part because the eventual use of the data was not yet specified. The analytical methods initially chosen were standard methods already approved by the Environmental Protection Agency (EPA). In many cases these methods were not useful because they were designed to detect pollutants at their legally mandated maximum allowed concentrations. In unpolluted waters, however, the concentrations of these contaminants often are well below the detection limit of the EPA methods. For example, the detection limit for the EPA approved standard method for phosphate was 7.5 ppb. Since the actual phosphate concentrations in Chesapeake Bay were below the EPA method’s detection limit, it provided no useful information. On the other hand, the detection limit for a non-approved variant of the EPA method, a method routinely used by chemical oceanographers, was 0.06 ppb, a more realistic detection limit for their samples. In other cases, such as the elemental analysis for particulate forms of carbon, nitrogen and phosphorous, EPA approved procedures provided poorer reproducibility than nonapproved methods. | 2,337 | 2,363 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Chemiluminescence/4%3A_Instrumentation/4.06%3A_Chemiluminescence_Imaging |
Chemiluminescence imaging combines the sensitive detection of chemiluminescence with the ability to locate and quantify the light emission, but above all it massively provides parallel determinations of the analyte. A digital image is made up of thousands of pixels, each generated by an independent sensor, detecting and measuring the light that falls on it. This enables simultaneous measurement of multiple samples or analytes for high throughput screening. ‘ Chemiluminescence imaging microscopy detects labelled probes more simply and more accurately than does fluorescence. It could become an important tool for rapid, early diagnosis of a wide range of diseases. Whole animal in vivo chemiluminescence imaging makes possible real-time monitoring of pathological and biological phenomena and we may anticipate important advances of great impact in drug discovery, biotechnology and medicine . The last twenty years have witnessed a steady improvement in our ability to form images from analytical signals. Imaging makes use of the high sensitivity and specificity, low background and wide dynamic range of chemiluminescence to quantitate and localize analytes down to the level at which this can be achieved by emission of single photons. Early systems consisted of a low-light vacuum tube device connected to an optical microscope. CCD (charge coupled device) and CMOS (complementary metal oxide semiconductor) image sensors make use of different technologies, both invented in the late 1960s and 1970s, for capturing images digitally. Both types of imager convert light into electrical charge and process it into electronic signals, as in digital cameras. Each has characteristic strengths and weaknesses. CCD technology is in most respects the equal of CMOS. Costs are similar. It is central to chemiluminescence imaging that when the spatial distribution of the analyte is critical, a luminograph must be produced to adequately express the data. This can be done by a CCD, which must have a high light-collection efficiency. A CCD converts optical brightness into electrical amplitude signals. CCDs are arrays of semiconductor gates formed on an integrated circuit (IC) or chip. The gates individually collect, temporarily store and transfer charge, which represents a picture element or pixel of an image. When light falls on a CCD sensor, a small electrical charge is generated photoelectrically for each pixel; each charge is converted to voltage and output as an analogue signal, which can be converted to digital information by additional circuitry. All of the pixel can be devoted to light capture. A CCD camera, such as a digital camera includes a CCD imager IC located in the focal plane of the optical system and control circuits mounted on a printed assembly. Image data captured is stored in a storage medium such as a compact flash memory or an IC memory card and can be displayed on a monitor such as a liquid crystal display (LCD). CCDs have traditionally provided the highest image quality (as measured by quantum efficiency and noise). An intensified CCD (ICCD) camera is optically connected to an image intensifier. In an image intensifier, the photons which are coming from the light source fall onto the photocathode, thereby generating photoelectrons. The photoelectrons are accelerated towards a micro-channel plate (MCP) by a voltage applied between photocathode and MCP. The electrons are multiplied by the MCP and accelerated towards a phosphor screen, which converts them back to photons that are guided to the CCD by an optical fibre or lens. ICCD cameras permit high frame rates and real-time visualization, the limitation being the increased noise produced by amplification. Non-intensified slow scan CCD cameras, cooled to reduce thermal noise, permit integration of the signal over a relatively long time and are suitable for steady-state signals. CMOS image sensors have emerged as an alternative to CCD sensors. They consist of an integrated circuit (IC) containing an array of pixel sensors, each pixel containing a photodetector, an amplifier and additional circuitry to convert the charge (which represents light intensity) to a voltage. Amplification, noise-correction, and digitization often also occur on-chip. These other functions increase the design complexity and reduce the area available for light capture. Unlike CCD sensors, each pixel is doing its own conversion, so uniformity is lower, which affects image quality. But the chip requires less off-chip circuitry. The phrase "metal oxide semiconductor" implies transistor structure having a metal gate electrode on top of an oxide insulator, which in turn is on top of a semiconductor. CMOS sensors complement every n-type transistor with a p-type, connecting the pairs of gates and drains. Because ideally no current flows except when the inputs to the gates are being switched, this greatly reduces power consumption and avoids overheating, which is a major concern in all ICs. CMOS can potentially be implemented with fewer components, uses less power and provides faster readout than CCDs. CMOS imagers also offer greater integration (more functions on the chip) and smaller size. It is possible to use a CMOS sensor chip as a microscale contact imager and quantitative photometer for chemiluminescence assays. The applicability has been investigated for chemiluminescence detection of ATP by its reaction with a proprietary reagent in 1 mm diameter wells fabricated on a glass cover-slip placed directly onto the imaging sensor . Ambient light was excluded. For each well, chemiluminescence intensity was averaged over a 1 x 100 pixel region of interest and integrated over a 200 ms exposure. It correlated well with the ATP concentrations over a range of 0.1-1 mmol dm . The detectivity (<1 nmol amounts of ATP) is not as fine as can be obtained with a much more expensive CCD camera, but is susceptible to improvement. CMOS chips are suitable for droplet microfluidics or lab-on-a chip devices when the cost of the assay system is a factor that must be optimized, such as in “point-of-need” assays or diagnostics. Cameras suitable for luminescence imaging should be able to form an image at a brightness of 10 lux (1 lux = 1/621 W m of 550 nm light). The most sensitive detect single photons with an efficiency of about 20% and have an average noise level of 2 x 10 lux = 8 photons s cm . Available ultra-low light imaging systems include the imaging photon detector (IPD), used in conjunction with a microscope with high numerical aperture (NA) objective lens. A high NA lens collects more light than a low NA lens. The collected light is focussed onto the IPD. Photons are recorded and stored as a list of time and space coordinates created by the IPD processor. Images can be reconstituted as an array of dots over any desired time interval. This allows for continuous recording over any interval; 1 h at 100 photons per second can be stored in 1 Mb of memory (storage as images requires 0.3 Mb ). Chemiluminescence images are based on small numbers of photons, especially when exposure time is brief; whereas a ten minute exposure at 100 photons per second can build an image of 60000 photons, a one second exposure provides only 100 photons, so that the image comprises only 100 dots spread across the area of the image A high resolution (up to 1392 x 1040) single photon counting camera system is suitable for extremely low photon emission applications such as some chemiluminescence applications. The system includes a control unit with data acquisition and image processing software. Frame rates up to 100 Hz can be obtained. The conceptualization involved in the design of photon counting cameras is well illustrated by DELTA camera; initially designed for astronomy it has advantages for a wide range of high-resolution problems. It is a high sensitivity array detector, which yields the space and time coordinates of photon events at sustained count rates superior to one million per second . It has a flat field, very high resolution (for the prototype: 512 x 591 pixels in space and 2.6 μs in time) and high throughput. Each photon produces an intensified phosphor image which has the same position in a two-dimensional field as the photon. This image is focussed onto three one-dimensional CCDs, which record its position as three coordinates on axes mutually oriented on a plane at angles of 120°. Software converts these to (x, y) coordinates on orthogonal axes and a clock signal adds the time (t) of the event to produce (x, y, t) coordinates, which are listed and stored; artefacts due to excessive tolerance or to simultaneous photon events are also removed from the data A high resolution CCD camera (up to about 6.0 Mpixel), cooled to about -70ºC, gives the best quality images, better accuracy, longer exposure times (up to 24 hours), minimal dark noise and enhanced stability. There should be a light-tight dark chamber and a height adjustable sample platform with a numeric counter for exact positioning in specific, repeatable positions. The alternative is an advanced motorised robotic camera which is driven up and down allowing placement very close to the sample, outperforming a standard zoom lens, having a wider field of view, easier and faster operation, and better sensitivity. These benefits are especially useful for faint samples. The image acquisition and analysis software provides comprehensive tools for simple image capture and analysis of gels, plates and membranes as well as colony counting. Images can be enhanced, user preferences defined, reports generated and data exported. Some systems include an overhead white light so that chemiluminescence can be overlaid onto a reflected light image (the so-called “live image”) and combine facilities for fluorescence, chemiluminescence and colorimetric applications. Not all chemiluminescent reactions are suitable for imaging; the main requirement, especially for imaging microscopy, is micrometre scale localization . Excited species of short half-life are suitable and conditions (especially reactant concentrations) can be optimized to minimize the diffusion of excited products. Glow-type kinetics, arising from the attainment of a steady state, facilitates measurement procedures. Enzyme labels are widely used. Imaging is very suitable for high-throughput screening. Examples of assays with very high sample throughput include the determination of bioavailable mercury which has been determined in urine using E. coli expression of luciferase under the control of a mercury-inducible promoter. Throughput is more than 5000 samples per hour and the limit of detection is 10-13 mol dm . Acetylcholinesterase inhibitors can be assayed using acetylcholinesterase, choline oxidase and horseradish peroxidase (HRP). Kinetic analysis of luminol chemiluminescence is carried out with a throughput of 180-360 samples per hour. Imaging with flat field correction lenses can be used to read microtitre plates (up to 4 of 384 wells) faster than by a luminometer, but the latter is, however, more sensitive and has the ability to measure fast, flash-type reactions. Miniarrays of antibody or gene probes can be spotted onto 96-well microtitre plates and assayed with an enzyme-labelled detection reagent and a chemiluminescence substrate. The whole plate is imaged with a CCD camera to measure the light emission from each well. This principle has been applied to a sandwich-type enzyme-linked immunosorbent assay (ELISA) for cytokines and a hybidization-based mRNA assay with up to 16 spots in each well; a 16 x 96 array contains 1536 dots, making possible high-throughput multianalyte assays, using standard plates and their associated sample handling devices. Oligonucleotide probes specific for human papilloma virus (HPV) genotypes have been used for a multianalyte chemiluminescence imaging assay for the simultaneous determination of up to 7 HPV DNAs. Amplification by the polymerase chain reaction(PCR) in the presence of a digoxigenin-labelled nucleotide (dUTP) is followed by an ELISA using a novel polystyrene microtitre plate having an array of 24 main wells (containing digoxigenin-labelled PCR product) each divided into 7 subwells (containing the immobilized probes). The digoxigenin label was subsequently detected by peroxidase-labelled antibody and a chemiluminescent substrate. Imaging was performed using an ultrasensitive CCD camera . Results were comparable with conventional colorimetric PCR-ELISA. Thousands of simultaneous determinations can be made by high-resolution imaging of chemiluminescence at array densities of up to hundreds of spots per square centimetre. Array-based gene expression analysis is a good example. Protein analysis based on antigen-antibody or ligand-receptor interactions is increasingly used in clinical and research work and in drug discovery. As well as their use for protein expression profiling, there are high-throughput protein microarrays that detect up to 35 cytokines. Specific antibodies are spotted onto membranes, which are incubated with the samples and captured analytes are detected by enhanced chemiluminescence with HRP-labelled antibodies and an HRP-substrate. A protein chip for parallel ELISAs of tumour marker allows the discovery of patterns that can increase the sensitivity and specificity of the diagnosis . Immobilized on the chip were 12 monoclonal antibodies against different tumour markers captured by incubating the chip with serum samples. An HRP-conjugated second antibody was used for detection by chemiluminescence imaging. The chip has been successfully applied both for cancer diagnosis and for screening asymptomatic populations at high risk. Small scale analytical devices use extremely small sample volumes and so need very sensitive detection techniques; chemiluminescence imaging has the high resolution and high sensitivity necessary. Assays include the ELISA determinations of the herbicide 2,4-D in multiple samples using gold-coated surfaces or glass capillaries and of up to ten antibiotics in milk in five minutes: the sample is incubated with mixed monoclonal antibodies followed by detection with an HRP-labelled second antibody and a suitable chemiluminescent substrate. Multiple hybridizations can be performed in a three-dimensional chip incorporating an array of vertical glass channels. Specific gene probes are immobilized on the inner walls of the channels. This strengthens the signal by providing a larger area for probe-immobilization than is available in a two-dimensional microarray. The sample flows through the channels and the analyte is detected by an enzyme-labelled antibody followed by a chemiluminescent substrate. Lateral diffusion of the emitting species is prevented by the walls of the microchannel; this improves resolution of the image. Chemiluminescence imaging of miniaturized analytical devices is also useful for multiplexing (simultaneous quantitation of different analytes or on different samples) by integrating the chemiluminescence over a different target area for each analyte or sample. Chemiluminescence imaging detection with CCD cameras can be used for reactions that take place on gels and membranes. This allows intensity measurement over a wide dynamic range and software exists to compute the total emission from particular zones of the image for analytical purposes. Images can be stored on discs or printed out. Electrophoresis is the movement, and hence the separation, of charged molecules in an electrical field; electrophoresis on polyacrylamide gel (PAGE) is particularly good for separation of molecules at low concentrations. Separated molecules can be transferred onto a nitrocellulose membrane by electroblotting - electrophoresis in a direction at right angles to the gel surface; this is also called western blotting and it is used to detect specific proteins in a sample of tissue homogenate or extract. By similar procedures, Southern blotting identifies particular sequences of DNA within a complex mixture and northern blotting locates RNA. In western blotting, electroblotting is followed by immunostaining, in which particular proteins are identified by labelled antibodies. DNA and RNA can be similarly identified by hybridization with labelled probes. Dot blot is an immunological technique to detect with antibodies specific proteins in mixtures or in samples such as tissue lysates. It is based on western blotting but there is no separation of the protein on SDS-PAGE. One such assay is the detection of B19 parvovirus. After spotting samples onto a membrane, hybridization with digoxigenin-conjugated DNA probes and treatment with HRP- or AP-labelled anti-digoxigenin antibodies, chemiluminescence imaging gives a limit of detection ten times better than using colorimetry. Cytochromes can be separated by PAGE with sodium dodecyl sulfate (SDS-PAGE) and transferred to a nitrocellulose membrane; cytochrome c (which contains a catalytic haeme group) is detected by peroxidase-luminol chemiluminescence . CCD imaging results in detection 50 times more sensitive than the 3,3',5,5'-tetramethylbenzidine staining method. A sample of less than 1 ml of a bacterial culture is needed. A similar assay, based on luminol/hydrogen peroxide chemiluminescence with ammonium persulfate enhancement, detects haptoglobin phenotyping after PAGE on a 15 μL sample . Other iron-containing proteins, such as catalase and ferritin, can also be detected. The proposed detection is very fast, compared to traditional staining methods (minutes versus hours). MIPs have artificial recognition sites with shapes, sizes and functionalities complementary to the analyte, which is thus selected in preference to other closely related structures. They are cheaper and more robust than antibodies, enzymes and biological receptors and can serve when these biomolecules are not available. The recognition sites are fabricated around a suitable template, preferably the analyte itself, which is extracted after polymerization. Generally, when a template molecule and a functional monomer are mixed in an organic solvent a complex is formed between the template and the monomer through polar interactions. Polymerization with a cross-linker fixes the positions of the polar groups. Removal of the template with a suitable solvent leaves specific recognition sites. The functional monomers are chosen to promote hydrogen bonding with the template to obtain good selectivity and reversibility. Optimum binding occurs when the MIP is exposed to the same conditions as those used in polymerization, because it depends on the shape of the imprinted cavity and on the spatial positioning of the coordinated functional groups. Both of these depend on the conditions and are affected by swelling of the polymer, which can be exploited to achieve fast and controllable release of adsorbed molecules prior to detection. Molecular imprinting of polymers has been linked with chemiluminescence imaging detection to achieve chiral recognition of dansyl derivatives of phenylalanine (Phe) . The MIP microspheres were synthesized using precipitation polymerization (which produces uniform microspheres) with dansyl-L-Phe as template and the microspheres were immobilized on microtitre plates (96 wells) using poly(vinyl alcohol) (PVA) as glue. The analyte was selectively adsorbed onto the MIP microspheres. After washing, the bound fraction was quantified using peroxyoxalate chemiluminescence (POCL) analysis, a general method for all fluorescent and fluorescence-labelled analytes, which has a greater quantum yield than most other chemiluminescence systems. In the presence of dansyl-Phe, bis(2,4,6-trichlorophenyl)oxalate reacted with hydrogen peroxide (H O ) with chemiluminescence emission. The signal was detected and quantified with a highly sensitive cooled CCD. The intensity of the image of each well of the plate was determined using software to sum the intensities of all the pixels making up the spot. Chemiluminescence intensity increases with the proportion of the L-enantiomer in the sample. Chiral composition can thus be determined by comparison of the intensity for the mixture and for pure D- and L- enantiomers at the same concentration. The results show that MIP-based chemiluminescence imaging is useful for quick chiral recognition and, because the method can perform many independent measurements simultaneously in 30 min, high-throughput screening is possible. A simple, sensitive and specific method has been developed for high throughput detection of the vasodilator dipyridamole . The proposed method is based on a chemiluminescence imaging assay with MIP recognition providing selectivity. Molecularly imprinted microspheres were prepared using precipitation polymerization with methacrylic acid (MAA) as functional monomer, trimethylolpropane trimethacrylate (TRIM) as the crosslinker and dipyridamole as the template. Non-imprinted polymer (NIP) was prepared without template to use as a control. The microspheres were coated in 96-well microtitre plates using 0.1% PVA as glue. After incubation with the sample, the amount of polymer-bound dipyridamole was determined by POCL. The emitted light was measured with a cooled high-resolution CCD camera. The intensity of the image of each well was determined as in subsection c(i). Under the optimum conditions, there is a linear relationship between relative chemiluminescence intensity and concentration of dipyridamole ranging from 0.02 to 10 μg ml . The detection limit is 0.006 μg ml . The method was validated by measuring dipyridamole concentrations in spiked urine samples. High tolerance for a number of normal constituents of urine was demonstrated to be much greater in the presence of MIP rather than NIP. MIP-based chemiluminescence imaging exhibits high selectivity and sensitivity to dipyridamole, combined with high sample throughput and economy (50 μl/well) . Target molecules of chemiluminescence imaging include antigens, DNA sequences, enzymes and metabolites. Chemical processes in cells, tissues or whole animals may also be targeted. Methods used include imaging microscopy, immunohistochemistry (IHC), in situ hybridization (ISH); other chemical or enzymatic reactions may also be used. The chemiluminescence image is overlaid onto the visible light image and processed by background subtraction, contrast enhancement, pseudocolour and quantitation over defined areas; absolute quantitation needs reproducible conditions, a calibration system and appropriate sample properties. Chemiluminescence imaging microscopy uses ordinary microscopes with optimized light collection. Light loss is minimized by having a simple lens coupling system; coverslips are dispensed with. The microscope, or at least the sample, is contained in a dark box to exclude ambient light and has a motorized micrometric stage to permit automatic adjustment. The sample is incubated with the chemiluminescence reagent until a steady-state emission is obtained. The objective lens has the highest numerical aperture (NA) compatible with acceptable focal aberration and depth of field. Dry, rather than oil-immersion, objectives are used and give adequate magnification and spatial resolution for the localization of analytes in single cells or tissue sections ; the detection limit for HRP is about 500 molecules/μm . Chemiluminescence microscopy has become a standard tool in biomedical research. Photon detectors have been attached to microscopes and allow imaging of chemiluminescent probes and reporter genes in cells and tissues. Photon counting techniques allow days of continuous imaging without creating oversized files. Fluorescence imaging, however, gives better spatial resolution than chemiluminescence imaging and makes multiple determinations easier. Calcium can be determined in cytosol and in organelles by using the photoprotein aequorin, an intracellular calcium indicator extracted from the jellyfish . Natural aequorin consists of a polypeptide, apo-aequorin, covalently bound to a hydrophobic prosthetic group, coelenterazine. The principle of imaging free cytosolic calcium with aequorins is the conformational change of aequorin molecules on calcium binding, causing coelenterazine to be oxidized to coelenteramide with production of carbon dioxide and emission of blue light (466 nm). Aequorin cannot penetrate the plasma membrane of the cell. Microinjection is the method of choice for determining cytosolic calcium in large cells. For small cells, cloning and transfection of the cDNA of apo-aequorin makes microinjection unnecessary, greatly simplifying calcium recording. Genetically expressed apo-aequorin contains no coelenterazine and so does not emit light. It is reconstituted as aequorin by soaking the specimens with coelenterazine. Apo-aequorin can be targeted to specific organelles by incorporating signal translocation sequences in the polypeptide chain. Aequorin is sensitive and specific, though single cells, containing a low concentration, give feeble chemiluminescence. Intensity is proportional to cell volume and therefore to the cube of the diameter. Small cells present problems because the amount of aequorin is low. In a cell of 10 μm diameter, the resting calcium concentration leads to emission of less than one photon per hour – so fluorescence must be used instead. But elevated calcium concentrations or large numbers of cells can be imaged by chemiluminescence using photon-counting cameras. Natural aequorin accurately measures Ca concentrations in the range 0.5 to 10 μmol dm , which is suitable for transient changes but for higher concentrations, a mutant form has been constructed which, by raising its dissociation constant and thus lowering its affinity for calcium, extends the working range up to 100 to 1000 μmol dm . It has the advantage over calcium-specific fluorescent probes of permitting real-time measurements over a long period; this is possible as there is no disturbance of the intracellular environment (including Ca buffering capacity) because of the low aequorin concentration (about 5 nmol dm ) but it has poorer resolution and it is used up rapidly by high calcium concentrations. Aequorin chemiluminescence, however, has an excellent signal to noise ratio and extremely low background noise. Chemiluminescence calcium imaging using aequorin is the method of choice for exploratory studies, since it is extremely sensitive, can detect a broad range of calcium concentrations. The kinetic order with respect to calcium concentration of the chemiluminescence reaction is 2.1 or higher, which gives inherent contrast enhancement. Unlike fluorescence, it does not require the analyst to make preliminary predictions or assumptions which exclude calcium signals outside the expected range. But it cannot match the high spatial resolution of fluorescence methods. In addition, chemiluminescence microscopy uses a large depth of field and optical sections are not yet possible. In , the chemiluminescent calcium-binding protein, aequorin, is associated with GFP. Calcium-sensitive bioluminescent reporter genes have been constructed that fuse GFP and aequorin to increase the quantum yield of calcium-induced bioluminescence . Co-expression of GFP with free aequorin does not have the same effect. The constructs were varied by including different lengths of peptide spacer between the GFP and the aequorin; much more light was emitted in all cases and the constructs were much more stable in cytosol and more sensitive to calcium than recombinant apo-aequorin alone. Resonance (non-radiative) energy transfer to the GFP chromophore from the excited oxidation product of coelenterazine depends on their relative positions. The peptide spacer is therefore flexible and of variable length. The green:blue ratio (500 nm:450 nm) of the light emitted by different constructs was measured 48 h after the introduction of the reporter genes into the cells (transfection). The green:blue ratio was increased by the covalent attachment of GFP to aequorin and further increased when the linker was added; as the linker was made longer, the wavelength of maximum emission increased and the bandwidth of the spectrum decreased. The efficiency of intramolecular energy transfer is enhanced to a level comparable to that achieved by resonance energy transfer due to the more favourable configuration made possible by the linker. Using GFP-aequorin fusions it is possible to detect physiological calcium signals in single cells. Transfection of the cells is followed by aequorin reconstitution with coelenterazine. The result is calcium-induced photon emission detectable with a cooled, ICCD camera, using an integration time of only one second. Cytoplasmic aequorin had previously detected Ca activities only by the use of a photomultiplier, which is more sensitive but lacks any spatial resolution, or by using targeted fluorescence probes, which give a quicker response. The use of the transgenes in which aequorin reports Ca activity while GFP enhances bioluminescence could lead to real time imaging of calcium oscillations in integrated neural circuits in whole animals as well as in specific subcellular compartments. Aequorin and GFP-enhancement probes along with synthetic fluorescent dyes can be targeted to the endoplasmic reticulum (ER) , a membrane network within the cytoplasm of cells involved in the synthesis, modification, and transport of cellular materials; this has enabled the role of ER to be clarified. If the sample’s enzyme activity has been preserved and if the sample is in such a condition that the substrate has access to the active site, enzyme activity can be localized by chemiluminescence imaging. The best spatial resolution is obtained by applying a chemiluminescent substrate directly to an enzyme, e.g., alkaline phosphatise can be detected by dioxetane phosphate. Enzyme reactions coupled with chemiluminescence can be used for metabolite mapping but give lower resolution. Metabolites can be determined in shock frozen tissue biopsies at femtomole levels and with micrometre resolution. The tissue is frozen as soon as possible to stop enzyme activity and fix the metabolites. The specimen is then placed on a temperature-controlled microscope stage and the chemiluminescence reagent is added. Emission intensity, recorded as soon as the temperature rises sufficiently, is converted into metabolite concentrations. Measurement of the spatial distribution of metabolites, such as ATP, glucose, and lactate, in rapidly frozen tissue is based on enzymatic reactions that link the metabolites to luciferase with subsequent light emission. Using an array, cryosections are brought into contact with the enzymes in a reproducible way inducing emission of light from the section in proportion to the metabolite concentration, with high spatial resolution. There is a close correlation between the distribution of ATP and cell viability; there are also distribution differences between tumours and normal tissue. ATP, glucose, glycogen and lactate have been determined at microscopic levels and at high spatial resolution in arterial wall cryosections using luciferase-based chemiluminescence imaging , which is a powerful tool to measure energy metabolites. It has been used to quantify local metabolite concentrations in artery rings. Distributions of energy metabolites are heterogeneous under hypoxic conditions. Diffusion distances for oxygen and nutrients can be long and might make vessels prone to develop local deficiencies in energy metabolism that could contribute to atherogenesis. Luciferases are enzymes that emit light in the presence of oxygen and a luciferin (ADD LINK). They have been used for real-time, low-light imaging of gene expression; coding sequences have been detected by luciferase-labelled gene probes . These labels include bacterial lux and eukaryotic luciferase luc and ruc genes. Different luciferases differ in the stability/variability of the emitted signal. Luciferases have served as reporters in a number of promoter search and targeted gene expression experiments. Photon-counting CCD imaging of luciferase has been used, for example, to show promoter activity in single pancreatic islet β-cells and regulation of human immunodeficiency virus (HIV) and cytomegalovirus. Luciferase imaging has also been used to trace bacterial and viral infection and to visualize the proliferation of tumour cells in animal models. Infected cells are readily detectable at an incidence of one in a million cells. Single bacterial cells, whether transformed or naturally luminescent, have also been imaged and variation in expression over time, due to fluctuations in metabolic activity, has been demonstrated. Low-light CCD imaging is in itself a non-invasive technique that is useful for observing (see subsection b(iv). Documentation of gels and membranes ADD LINK) intracellular gene expression and small-scale assays such as hybridization (ISH) as well as for immunoassays, gels and blots [ADD LINK], DNA probes and imaging (see section D6h, ADD LINK). Slow-scan liquid nitrogen cooled CCD cameras are preferable for high resolution imaging with long exposures, but photon-counting CCD cameras are better for shorter exposure times. Flashing at frequencies greater than 1 Hz can be detected by ICCD cameras. Bioluminescence imaging has been applied in experimental biomedical research, e.g., development of necrosis, and in other areas of biology . It has also been used particularly on tumour biopsies in clinical oncology. In combination with immunohistochemistry,autoradiography or hybridization it can be particularly powerful. It has been shown for squamous cell carcinomas that accumulation of lactate in the primary lesions is associated with a high risk of metastasis. In this way, metabolic mapping indicates the degree of malignancy and the prognosis of tumours; it has stimulated a number of fundamental investigations. There are numerous other examples of methods to determine metabolites in living cells and tissues, including real-time imaging of metabolite production. Endogenous acetylcholinesterase (ACE) activity has been detected in rat coronal brain slices using coupled reactions with choline oxidase and horseradish peroxidase . The reagent is optimized to minimize diffusion of emitting species, giving sharp localization and a very low background. This imaging assay is more predictive than systems; and can be used to determine pathophysiological changes in ACE distribution or the effect of ACE inhibitors, which could be useful for screening candidate drugs. Nitric oxide (NO) released from cell cultures and living tissue has been visualized by a reaction with luminol and hydrogen peroxide to yield photons which were counted using a microscope coupled to a photon counting camera, giving new insight into release time course and diffusion profile . The method allowed integration times in the order of minutes to improve signal-to-noise ratio. However, the high sensitivity of this method also makes it possible to generate an image in seconds, allowing the production of real time moving pictures. This method has demonstrated potential for real time imaging of NO formation, with high temporal and spatial resolution. There was little earlier knowledge of this phenomenon due to the short half-life of NO. ISH and IHC are techniques that localize analytes in a wide range of suitable specimens such as cellular smears, or frozen or paraffin-embedded sections. Chemiluminescence detection does not require any special specimen preparation, but an accurately controlled section thickness of from 3 to 5 μm is necessary for reproducibility. Incorporating chemiluminescence detection (CL-IHC and CL-ISH) increases sensitivity compared with colorimetry or fluorometry. This adds reliable and accurate quantitative evaluation of spatial distribution to the specificity of the probe. The “theoretical” limit of detection of the enzyme label by chemiluminescence is 10 to 10 mol; as a detector for ISH,chemiluminescence is almost as sensitive as S autoradiography giving a nontoxic alternative to the use of radioactivity . Localization within cells of nucleic acid sequences, e.g., the sites of genes in chromosomes, can be achieved by hybridization to complementary nucleic acid probes. The two general types of hybridization involve nuclear DNA and cellular RNA respectively; they are conceptually similar but differ in practical detail. The technique is usually performed on specimens prepared for light microscopy. It is claimed that little or no microscopic training is necessary to evaluate the chemiluminescence images. Fig. D6.1 – Flow chart of operations for the performance of in situ hybridization. (All incubations are at room temperature). ISH involves nucleic acid probe hybridization with DNA or RNA, endogenous to the specimen or exogenous (viral/bacterial). The procedure is summarized in Fig. D6.1. Sensitivity is increased by indirect labelling, in which the label binds with a biospecific chemiluminescence reagent, e.g., biotin binds to streptavidin; fluorescein or digoxigenin bind to their respective antibodies, the chemiluminescence reagent having a covalently-bound signalling group (usually AP or HRP). ISH of HPV can be imaged using a digoxigenin-labelled gene probe followed by an HRP-labelled anti-digoxigenin antibody and a chemiluminescence reagent . To localize the virus, the chemiluminescence image (with intensities represented by pseudocolours) is overlayed onto a transmitted light image. ISH has also been performed on three human carcinoma cell lines and 40 biopsy specimens of human cervical neoplastic and preneoplastic lesions by using biotin-labelled complementary DNA probes of HPV, detected by HRP-labelled secondary antibodies; the chemiluminescence was detected by an ICCD camera . After only 10 min of photon accumulation, on cell line smears as well as on serial tissue sections, chemiluminescence gave comparable results to those obtained by a 3-week exposure for S-autoradiography. CL-ISH is quantitative because chemiluminescence is proportional to the enzyme activity of the label, and to the number of gene copies per cell. In a separate study of , chemiluminescence was proportional to the number of cells infected (following virus replication). An early ISH assay of DNA in infected human fibroblasts used dioxigenin-labelled probes and AP-labelled anti-digoxigenin antibody. Employing a low-light imaging luminograph 400 amol of AP were detected using 1,2-dioxetanes . Chemiluminescence was intense and stable, making possible quantitation within single cells, with a spatial resolution of 1 μm and very low background. Multiplexed CL-ISH assays have been developed in which probes with different enzyme labels detect different targets. One example of such techniques localizes the DNA of and in the same specimen using the following protocol. The faster HRP/luminol is added to the specimen and chemiluminescence is imaged, the specimen is given a short wash, then AP/dioxetane is added, and a second chemiluminescence image is recorded. A longer wash is needed if AP is added first. Human B19 is responsible for wide range of diseases. CL-ISH gives high resolution, providing precise localization and quantitative detection of the viral nucleic acids in single cells in cultures at different times after infection, giving an objective evaluation of the infection process with higher sensitivity than colorimetric ISH detection assessed by panels of observers. The improved sensitivity of CL-ISH detects more positive cells per sample, making possible earlier diagnosis. A peptide nucleic acid (PNA) has been developed which has improved specificity and faster, stronger binding than other DNA probes. The assay is based on the use of a biotin-labelled PNA probe which is detected by a streptavidin-linked alkaline phosphatase (AP), using the well-known biotin-streptavidin affinity: PNA–biotin + streptavidin–AP → PNA–biotin–streptavidin–AP adamantyl 1,2-dioxetane phosphate + AP → excited fragmentation products → light The chemiluminescence signal which arises was quantified and imaged with an ultrasensitive nitrogen-cooled CCD camera connected to an epifluorescence microscope with high-transmission optics and modified for acquisition of chemiluminescence. A threshold signal (representing non-specific binding of the probe and endogenous alkaline phosphatase activity) was established using mock-infected cells as negative controls. Following a B19 virus infectious cycle the percentage of infected cells, which reached its maximum at 24 h after infection, could be accurately monitored. The advantages of chemiluminescence detection (high detectability and wide linear range) allow the quantitative analysis of viral nucleic acids in infected single cells, showing a continuous increase with time after infection. Such investigations could be powerful tools for the assessment and diagnosis of viral infections and for measuring the virus load of infected cells . IHC involves the use of antibodies that bind to endogenous, viral or bacterial antigens (a protein usually) with subsequent detection by enzyme-conjugated antibodies. CL-IHC detects epithelium in thyroid tissue by HRP-labelled antibodies and luminol/H O , with adequate resolution and greater sensitivity than colorimetry or fluorescence. CL-IHC can also with advantage be applied to Interleukin 8 (IL-8) localization in gastric biopsy specimens infected by , an organism asssociate with gastric ulcers. It shows with greater sensitivity than other detection systems the variability of the IL-8 concentration in the mucosa and the foci of high concentration in the epithelial cells. Fig. D6.2 – Flow chart of operations for the performance of immunohistochemistry. (All incubations are at room temperature). Cervical cancers (cervical intraepithelial neoplasms, CIN) are classified into low- (CIN1) or high-grade (CIN2 or CIN3) in order to predict the risk of progression of early lesions and enable decisions to be made concerning surgical intervention. Judgments based on histology are imprecise in that different observers assign different grades to the same biopsy specimen. One way of overcoming this difficulty is to redefine the diagnostic criteria in terms of analytical chemistry. An immunohistochemical assay (see Fig. D6.2) with chemiluminescence detection (CL-IHC) has been used to quantitatively evaluate the overexpression of the protein p16INK4A and its localization in the epithelium of samples from cervical cancers and from non-cancerous cervical lesions. Fig. D6.3 shows that chemiluminescence (and hence p16INK4A protein content) generally increases from left to right. High-grade lesions give generally more intense chemiluminescence signals in the epithelium than low-grade and show a different distribution of p16INK4A protein. From the intensity of the chemiluminescence signal and the percentage of the epithelium involved in the overexpression of p16INK4A an expression score was obtained which discriminated well among different lesions. A cut-off value was determined to distinguish between low and high grades. The differences between the average scores of different CIN grades were statistically significant . The determination of p16 overexpression by CL-IHC used AP as the enzyme label and then, after washing in buffer, HPV DNA was determined by CL-ISH (see Fig. D6.1 and section f(i) ADD LINKS) using HRP as label to avoid interference between the two assays . To circumvent the non-equivalence of consecutive tissue sections, the two assays were carried out on the same sample. The assays cannot be carried out in the reverse order as the high-temperature step in ISH denatures the p16 protein to be determined by IHC. The high detectability of chemiluminescence gives improved discrimination between lesions as non-cancerous, CIN1 or high-grade CIN. This could become an objective and accurate diagnostic test. High-risk (HR) mucosal human papilloma virus (HPV) is strongly associated with cancer. It has been found in primary melanoma and in pigmented skin blemishes (birthmarks, moles) but has rarely been reported in normal skin, which is instead commonly infected with other relatively harmless strains of HPV. HPV DNA in skin cancer has been detected by polymerase chain reaction (PCR). In order to understand the relationship between HPVs and primary melanoma, it is necessary to know whether the presence of HPV is localized in cancer cells rather than in normal skin cells present in the tumour biopsy, what proportion of the cells harbours the virus or whether it can be due to contamination of the tumour surface by viruses from healthy skin. Because PCR methods measure only total DNA they are not suitable to ascertain this. To localize HR-HPV a rapid, specific and very sensitive method has been developed that combines an enzyme-amplified fluorescence in situ hybridization (FL-ISH, see Fig. D6.1 ADD LINK) for the detection of HPV nucleic acids (types 16 and 18, which are the types most likely to cause cancer) with a chemiluminescence immunohistochemistry (CL-IHC, see Fig. D6.2 ADD LINK) method for the detection sequentially in the same section of the tumoural melanocytic marker HMB-45. It is necessary to use the same section because the melanoma cells are distributed heterogeneously in the specimens. HMB-45 determination is an indicator of melanoma cell differentiation and is widely used in diagnostic pathology. Digital images of FL-ISH and CL-IHC were separately recorded, assigned different pseudocolours (see Fig. D6.5) and merged using specific software for image analysis. The results demonstrated a sharp colocalization (to an extent of about 70% of the total luminescent area of the specimen) of HPV nucleic acids and the melanoma marker in the same biopsy sections. In smaller areas, HPV was detected without HMB-45 (9.5% of total) or HMB-45 without HPV (20.5%). This demonstrates that viral nucleic acids were specifically present in melanoma cells and supports a possible active role of HPV in malignant melanoma . Fluorescence detection remains of great value for hybridization and in immunohistochemistry, particularly because of its greater precision of spatial localization compared with chemiluminescence. Chemiluminescence can, however, be harnessed as a means of exciting fluorescent probes and labels alternative to photo-excitation. Two examples of the principle are considered in this section. Using imaging chip-based devices, detection of aqueous peroxyoxalate chemiluminescence (POCL) from oxamide I in aqueous environments has been reported for fluorescence-labelled analytes and proved to be at least as sensitive as that using direct fluorescence detection requiring a light source for excitation. Using a CCD camera to record the chemiluminescence intensity from a 1000-fold range of analyte concentrations, POCL detection sensitivity of fluorescence-labelled immunoglobulins on a nitrocellulose membrane was investigated. Aqueous POCL of enterotoxin B (SEB) and its antibody were also used to demonstrate immuno- and affinity-detection using a CCD camera. SEB was detected by an immune sandwich assay in which SEB was captured by sheep polyclonal antibody spotted onto a nitrocellulose membrane and subsequently captured mouse monoclonal antibody, which was detected by fluorescence-labelled anti-mouse antibody. Affinity detection of biotin-labelled anti-SEB antibody used fluorescence-labelled streptavidin. Simultaneous detection by POCL of bovine serum albumin labelled with two different fluorescent labels has been demonstrated, using contact imaging with a CMOS colour imaging chip (ADD LINK). The proteins were spotted onto a membrane disc fixed to a cover slip which was placed on the sensing surface of the chip. They were visible on 8 s exposure as red and green spots respectively; a mixture of the labelled samples emitted yellow light. This procedure might be applicable to reading microarrays. A self-illuminating fluorescence reporter, comprising a dye conjugated to AP, has been demonstrated “in principle” for imaging detection using a CCD camera or a CMOS colour chip, making possible the imaging of fluorescent signals without the need for an external light source or sophisticated optics . It is based on bioluminescence resonance energy transfer (BRET), an example of which, already cited, is the use of GFP to enhance the light emission from the photoprotein aequorin (see section d(iii) ADD LINK). The efficiency of BRET is increased by minimizing the distance between the bioluminescent energy donor and the fluorescent acceptor and is also found to depend on the ratio of AP to fluorophore in the conjugate, on the fluorescent dye used and on the chemiluminescent substrate. Chemiluminescence detection is low-cost, suitable for low concentrations and portable but diffusion of the luminescent products leads to poor spatial resolution. BRET is a potential solution to this problem, but it has not yet been applied to a real analytical problem. In the demonstration, antibody, immobilized on the CMOS surface, captured a biotin-labelled target molecule that was then bound to the streptavidin-labelled AP-dye conjugate. The AP was used to generate light and the captured array images were viewed on a computer monitor. Images were also obtained by using a CCD camera. The chemiluminescent substrate for AP emitted at 450 nm; the energy from this emission was transferred to the fluorescent dye. This resulted in a second light emission with a longer wavelength (580 nm), which was localized at the position of target molecules, avoiding the problem of diffusion of the chemiluminescent product. In this way, image spatial resolution was greatly improved compared with conventional chemiluminescence detection. The shorter wavelength first emission that escaped absorption by the dye was removed by a high pass filter. The use of cameras remote from the site of light emission makes it possible to image events occurring in the interior of organs and organisms. This technique can be applied to the study of a wide range of phenomena such as tumour growth, metastasis and drug efficacy, assessed by injecting and imaging recombinant light-emitting tumour cells , which can be used as probes for tumour location. Another application of molecular imaging techniques is the non-invasive monitoring of transplanted embryonic cardiomyoblasts expressing firefly luciferase (Fluc) reporter gene . The movement in the rat gut of bioluminesent (expressing luciferase and the enzyme necessary for substrate synthesis) has also been imaged . Luciferase enzymes to label cells, pathogens, and genes are internal indicators that can be detected externally. Transgenic organisms have been produced in which the gene for the luciferase of functions stably in tobacco ( ), tomato ( ) and potato ( ). Strong light emission was imaged with a low-light video camera after only a few seconds immersion of leaves, slices and seedlings (see Fig. D6.6) in 3 μmol dm 2-benzyl luciferin solution; at this concentration, the substrate was nontoxic and no other abnormalities were apparent . Luciferase imaging enables complex gene activation effects to be modelled and observed in live animals . Bioluminescent reporters for given biological processes have been used widely in cell biology; in whole animal models, including light-producing transgenic animals as models of disease, they are useful in drug discovery and development. imaging of intact organs also furthers the understanding of biological processes. The application of this technology to living animal models of infectious disease has provided insights into disease processes, therapeutic efficacy and new mechanisms by which pathogens may avoid host defences . Progress of infections and efficacy of treatment are assessed by bioluminescent markers, e.g., light-emitting pathogens. Rapid, accessible high throughput screening is effective for pharmacokinetics, toxicology and target validation. Study of spatio-temporal patterns helps to characterize the site and time of action of drugs. There is also a possible clinical use in gene therapy and assessing gene vaccine delivery and efficacy. There are several advantages in the use of imaging. In continuous monitoring each animal serves as its own control – introducing less variability than comparing groups of animals each analyzed at a different time; in addition, fewer animals are used in the experiments. Multiplex assays are also possible, using two or more reporters in the same animal. Chemiluminescence imaging is validated by the correlation of bioluminescence with culture cell counts (e.g., of ). Investigations of this kind increase the number of data and can guide tissue sampling for subsequent biochemistry or histology. There are also several drawbacks. Red and infrared light (590-800 nm) penetrates tissue well, but blue/green light (400-590nm), usually the bulk of the bioluminescence emission, is strongly attenuated. Luciferase, however, is the most widely-used reporter and has a broad emission including red light. The spatial resolution (3-5 mm) is worse than in magnetic resonance imaging or computed tomography. Some pathological processes result spontaneously in light production. This is due to a weak spontaneous photon emission associated with oxidative phenomena, such as oxygen free radical (OFR) formation (ADD LINK), in whole organs removed from living animals. OFR have been imaged in rat livers that have been subjected to ischaemia (occlusion of blood supply) and reperfusion (restoration of blood supply), showing distribution in space and time of superoxide radicals on the liver surface and the effects on them of antioxidants, which remove OFR and can be screened by this model. The roles of aging, ethanol consumption and fat deposition on OFR formation in the liver have also been assessed. This system can be used to monitor the storage of organs for transplantation and to test agents and procedures for preserving them. | 54,339 | 2,364 |
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Like nonconjugated dienes, conjugated dienes are subject to attack by electrophiles. In fact, conjugated electrophiles experience relatively greater kinetic reactivity when reacted with electrophiles than nonconjugated dienes do. Upon electrophilic addition, the conjugated diene forms a mixture of two products—the kinetic product and the thermodynamic product—whose ratio is determined by the conditions of reaction. A reaction yielding more thermodynamic product is under thermodynamic control, and likewise, a reaction that yields more kinetic product is under kinetic control. The reaction of one equivalent of hydrogen bromide with 1,3-butadiene gives different products at under different conditions This is a classic example of the concept of versus control of a reaction. Take a look at this energy profile diagram in Figure 1. In this scenario, the starting material \(\ce{A}\) can react to form either \(\ce{B}\) (to the left) or \(\ce{C}\) (to the right). The formation of the product \(\ce{C}\) involves evolution over barrier (assuming a two step process) with , which means that it will form faster (ignoring ): \[E_{\mathrm{a},\ce{AB}} > E_{\mathrm{a},\ce{AC}} \qquad \Longrightarrow \qquad k_\ce{AB} < k_\ce{AC}\] If we keep the temperature sufficiently , the molecules of \(\ce{C}\), which are inevitably formed faster, will probably not have enough energy to overcome the reverse activation barrier (i.e., \(\ce{C} \rightarrow \ce{A}\)) to regenerate \(\ce{A}\). The forward reactions \(\ce{A->B}\) and \(\ce{A->C}\) are, under such conditions, effectively irreversible. Since the formation of \(\ce{C}\) is faster, it will predominate, and the major product formed will be \(\ce{C}\). This is known as and \(\ce{C}\) is the . At elevated temperatures, \(\ce{C}\) is still going to be the product that is formed . However, it also means that all the reactions will be reversible. This means that molecules of \(\ce{C}\) can revert back to \(\ce{A}\). Since the system is no longer limited by temperature, the system will minimize its Gibbs free energy, which is the thermodynamic criterion for chemical equilibrium. This means that, as the most thermodynamically stable molecule, \(\ce{B}\) will be predominantly formed. The reaction is said to be under and \(\ce{B}\) is the . A simple definition is that the is the product that is formed , and the is the product that is more . This is precisely what is happening here. The kinetic product is 3-bromobut-1-ene, and the thermodynamic product is 1-bromobut-2-ene (specifically, the isomer). Note that not every reaction has an energy profile diagram like Figure 1, and not every reaction has different thermodynamic and kinetic products! If the transition states leading to the formation of \(\ce{C}\) (e.g., ) were to be higher in energy than that leading to \(\ce{B}\) (e.g., ), then \(\ce{B}\) would simultaneously be both the thermodynamic and kinetic product. There are plenty of reactions in which the more stable product ( ) is also formed faster ( ). The first step is the protonation of one of the \(\ce{C=C}\) double bonds. In butadiene ( ), both double bonds are the same, so it does not matter which one you protonate. The protonation occurs regioselectively to give the more stable carbocation: The more stable cation is not only secondary, but also , and therefore enjoys stabilization via resonance (or conjugation). This is depicted in the and above. This allylic carbocation, more properly denoted as the resonance hybrid , has two carbons which have significant positive charge, and the bromide ion (here denoted as \(\ce{X-}\)) can attack either carbon. Attacking the central carbon, adjacent to the site of protonation, leads to the kinetic product (called the 1,2-adduct); attacking the terminal carbon, distant from the site of protonation, leads to the thermodynamic product (called the 1,4-adduct). There are some people who write that results from attack of \(\ce{X-}\) on resonance form , and from attack of \(\ce{X-}\) on resonance form . Resonance forms do not separately exist, and they are not distinct species that rapidly interconvert. As such, one cannot speak of undergoing a reaction. Now, why is the thermodynamic product, and why is the kinetic product for this reaction? It is perhaps simple enough to see why is more stable than . It has an internal, disubstituted double bond, and we know that as a general rule of thumb, the thermodynamic stability of an alkene increases with increasing substitution. So, compared to the terminal, monosubstituted alkene , is more stable. Both the isomer as well as the isomer can be formed via attack of the nucleophile at the terminal carbon, and both are disubstituted alkenes. However, the isomer is more stable than the isomer , because there is less steric repulsion between the two substituents on the double bond. As such, is the thermodynamic product. Several explanations may be proposed to explain the nature of the kinetic product. The worst possible argument argues that the resonance form , being an allylic carbocation, is more stable than resonance form , which is an allylic carbocation. Therefore, resonance form exists in greater relative proportion (i.e., more molecules will look like than ), and the nucleophile preferentially reacts with this specific carbocation, leading to the formation of . However, this is incorrect, since individual resonance forms do not exist. Moreover, such an argument suggests that we are looking for the more stable (I or I in Figure 1). In fact, we should be looking for the more stable ( in Figure 1). The carbocation is an , and not a . The most common argument is since resonance form is more stable than , is that it contributes more towards the . As such, the positive charge on the internal carbon is greater than the positive charge on the terminal carbon. The nucleophile, being negatively charged, is more strongly attracted to the more positively charged or more electrophilic carbon, and therefore attack there occurs faster (the transition state being stabilized by greater electrostatic interactions). That's actually a very sensible explanation; with only the data that has been presented so far, we would not be able to disprove it, and it was indeed the accepted answer for quite a while. In 1979, Nordlander DCl 3-pentadiene Resonance forms and are both allylic and secondary. There is a very minor difference in their stabilities arising from the different hyperconjugative ability of \(\ce{C-D}\) \(\ce{C-H}\) bonds, but in any case, it is not very large. Therefore, if we adopt the explanation in the previous section, one would expect there not to be any major pathway, and both 1,2- and 1,4-addition products ( and ) would theoretically be formed roughly equally. Instead, it was found that the 1,2-addition product was favored over the 1,4-addition product. For example, at \(-78\ ^\circ\mathrm{C}\) in the absence of solvent, there was a roughly \(75:25\) ratio of 1,2- to 1,4-addition products. Clearly, there is a factor that favors 1,2-addition that does not depend on the electrophilicity of the carbon being attacked! The authors attributed this effect to an mechanism. This means that, after the double bond is protonated (deuterated in this case), the chloride counterion remains in close proximity to the carbocation generated. Immediately following dissociation of \(\ce{DCl}\), the chloride ion is going to be much closer to \(\ce{C-2}\) than it is to \(\ce{C-4}\), and therefore attack at \(\ce{C-2}\) is much faster. In fact, normal electrophilic addition of \(\ce{HX}\) to conjugated alkenes in polar solvents can also proceed via similar ion pair mechanisms. This is reflected by the greater proportion of addition products to such substrates. The mechanism that favors 1,2-addition clearly does not depend on the electrophilicity of the carbon being attacked. | 8,024 | 2,365 |
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The formation of carbocations is sometimes accompanied by a structural rearrangement. Such rearrangements take place by a shift of a neighboring alkyl group or hydrogen, and are favored when the rearranged carbocation is more stable than the initial cation. The addition of HCl to 3,3-dimethyl-1-butene, for example, leads to an unexpected product, 2-chloro-2,3-dimethylbutane, in somewhat greater yield than 3-chloro-2,2-dimethylbutane, the expected product. This surprising result may be explained by a carbocation rearrangement of the initially formed 2º-carbocation to a 3º-carbocation by a 1,2-shift of a methyl group. Another factor that may induce rearrangement of carbocation intermediates is strain. The addition of HCl to α-pinene, the major hydrocarbon component of turpentine, gives the rearranged product, bornyl chloride, in high yield. As shown in the following equation, this rearrangement converts a 3º-carbocation to a 2º-carbocation, a transformation that is normally unfavorable. However, the rearrangement also expands a strained four-membered ring to a much less-strained five-membered ring, and this relief of strain provides a driving force for the rearrangement. The atom numbers (colored red) for the pinene structure are retained throughout the rearrangement to help orient the viewer. The green numbers in the final product represent the proper numbering of this . The propensity for structural rearrangement shown by certain molecular constitutions, as illustrated above, serves as a useful probe for the intermediacy of carbocations in a reaction. We shall use this test later. | 1,622 | 2,366 |
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Micelles are formed by amphiphiles that want to bury hydrophobic chains and expose charged head groups to water. Since a cavity must be formed for the micelle, the resulting surface tension of the cavity (the hydrophobic effect) results in the system trying to minimize its surface area, and thereby the number of molecules in the micelle. At the same time, the electrostatic repulsion between headgroups results in driving force to increase the surface area per headgroup. These competing effects result in an optimal micelle size. We start by defining the chemical potential Δμ , which is the free energy per mole of amphiphilic molecule A to assemble a micelle with n molecules. Instead of using n, we will try to express the size of the micelle in terms of its surface area a and assume that it is spherical. Then, the free energy for forming a cavity for the micelle grows as γa, where γ is the surface tension. The surface area is expressed as an average surface area spanned by the charged headgroup of a monomer unit: \[ a_e = a/n \] The repulsion term is hard to predict and depends on many variables. There are the electrostatic repulsions between head groups, but there is also the entropic penalty for forming the micelle that depends on size. As an approximation, we anticipate that the free energy should be inversely proportional to surface area. Then the free energy for forming a micelle with n molecules is \[ \begin{aligned} \Delta G_n &= \gamma a + \dfrac{x}{a} \\ &= \gamma n a_e + \dfrac{x}{na_e} \end{aligned} \] where x is a constant. Solving for Δμ=∂ΔG/∂n, differentiating it with respect to a , and setting to zero, we find the optimal micelle size, a , is \[ a_0 = \sqrt{\dfrac{x}{\gamma n^2}} \] Solving for x and substituting in eq. (4), we obtain the chemical potential as: \[ \Delta \mu = \dfrac{\gamma}{a_e}(a^2_e + a^2_0 ) = 2\gamma a_0 +\dfrac{\gamma}{a_e}(a_e-a_0)^2 \] It has a parabolic shape with a minimum at a . Next, we can obtain the probability distribution for the micelle size as a function of head group surface area and aggregation number \[ P_n = \exp (-n\Delta \mu /k_BT) \] \[ P_n(a_e) ~\exp \left( - \dfrac{n\gamma (a_e - a_0)^2}{a_ek_BT} \right) \] The relative populations of micelles are distributed in a Gaussian distribution about a . The distribution of sizes has a standard deviation (or polydispersity) given by \[ \sigma = \sqrt{\dfrac{na_ek_BT}{2\gamma}} \] From a = 4πr = na , we predict that the breadth of the micelle size distribution will scale linearly in the micelle radius, and as the square root of temperature and molecule number. ___________________________________________ K. Dill and S. Bromberg, Molecular Driving Forces: Statistical Thermodynamics in Biology, Chemistry, Physics, and Nanoscience. (Taylor & Francis Group, New York, 2010); J. N. Israelachvili, Intermolecular and Surface Forces, 3rd ed. (Academic Press, Burlington, MA, 2011), Ch. 20. | 2,939 | 2,367 |
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In the mid 1920's the German physicist Werner Heisenberg showed that if we try to locate an electron within a region \(Δx\); e.g. by scattering light from it, some momentum is transferred to the electron, and it is not possible to determine exactly how much momentum is transferred, even in principle. Heisenberg showed that consequently there is a relationship between the uncertainty in position \(Δx\) and the uncertainty in momentum \(Δp\). \[\Delta p \Delta x \ge \frac {\hbar}{2} \label {5-22}\] You can see from Equation \(\ref{5-22}\) that as \(Δp\) approaches 0, \(Δx\) must approach ∞, which is the case of the free particle discussed . This uncertainty principle, which also is discussed in , is a consequence of the wave property of matter. A wave has some finite extent in space and generally is not localized at a point. Consequently there usually is significant uncertainty in the position of a quantum particle in space. Activity 1 at the end of this chapter illustrates that a reduction in the spatial extent of a wavefunction to reduce the uncertainty in the position of a particle increases the uncertainty in the momentum of the particle. This illustration is based on the ideas described in the next section. Compare the minimum uncertainty in the positions of a baseball (mass = 140 gm) and an electron, each with a speed of 91.3 miles per hour, which is characteristic of a reasonable fastball, if the standard deviation in the measurement of the speed is 0.1 mile per hour. Also compare the wavelengths associated with these two particles. Identify the insights that you gain from these comparisons. | 1,643 | 2,368 |
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The compression factor (\(Z\)) for O at 200 K is measured to have the following values: Using numerical integration, calculate the fugacity constant for O at 200 K from these data. The normal boiling point of ethanol is 78.4 C. Its enthalpy of vaporization is 38.6 kJ/mol. Estimate the vapor pressure of ethanol at 24.4 C. When 20.0 grams of an unknown nonelectrolyte compound are dissolved in 500.0 grams of benzene, the freezing point of the resulting solution is 3.77 °C. The freezing point of pure benzene is 5.444 °C and the cryoscopic constant (\(K_f\)) for benzene is 5.12 °C/m. What is the molar mass of the unknown compound? Consider a mixture of two volatile liquids, A and B. The vapor pressure of pure liquid A is 324.3 Torr and that of pure liquid B is 502.3 Torr. What is the total vapor pressure over a mixture of the two liquids for which x = 0.675? Consider the following expression for osmotic pressure \[\pi V = \chi_BRT\] where \(\pi\) is the osmotic pressure, \(V\) is the molar volume of the solvent, \(\chi_B\) is the mole fraction of the solute, \(R\) is the gas law constant, and \(T\) is the temperature (in Kelvin). The molar volume of a particular solvent is 0.0180 L/mol. 0.200 g of a solute (B) is dissolved in 1.00 mol of the solvent. The osmotic pressure of the solvent is then measured to be 0.640 atm at 298 K. Calculate the molar mass of the solute. At 300 K, the vapor pressure of HCl(g) over a solution of \(HCl\) in \(GeCl_4\) are summarized in the following table. Calculate the Henry’s Law constant for HCl based on these data. Consider the mixing of 1.00 mol of hexane (C H ) with 1.00 mole of benzene (C H ). Calculate \(\ H\), \(\ S\), and \(\ G\) of mixing, of the mixing occurs ideally at 298 K. | 1,760 | 2,369 |
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A phase diagram for two immiscible solids and the liquid phase (which is miscible in all proportions) is shown in Figure \(\Page {1}\). The point labeled “e ” is the , meaning the composition for which the mixture of the two solids has the lowest melting point. The four main regions can be described as below: The unlabeled regions on the sides of the diagram indicate regions where one solid is so miscible in the other, that only a single phase solid forms. This is different than the “two-phase solid” region where there are two distinct phases, meaning there are regions (crystals perhaps) that are distinctly A or B, even though they are intermixed within on another. Region I contains two phases: a solid phase that is mostly compound A, and a liquid phase which contains both A and B. A sample in region II (such as the temperature/composition combination depicted by point b) will consist of two phases: 1 is a liquid mixture of A and B with a composition given by that at point a, and the other is a single phase solid that is mostly pure compound B, but with traces of A entrained within it. As always, the lever rule applies in determining the relative amounts of material in the two phases. In the case where the widths of the small regions on either side of the phase diagram are negligibly small, a simplified diagram with a form similar to that shown in Figure \(\Page {2}\) can be used. In this case, it is assumed that the solids never form a single phase! The tin-lead system exhibits such behavor. Another important case is that for which the two compounds A and B can react to form a third chemical compound C. If the compound C is stable in the liquid phase (does not decompose upon melting), the phase diagram will look like Figure \(\Page {3}\). In this diagram, the vertical boundary at \(\chi_B = 0.33\) is indicative of the compound \(C\) formed by \(A\) and \(B\). From the mole fraction of \(B\), it is evident that the formula of compound \(C\) is \(A_2B\). The reaction that forms compound C is \[2 A + B \rightarrow C \nonumber \] Thus, at overall compositions where \(\chi_B < 0.33\), there is excess compound A (B is the limiting reagent) and for \(\chi_B \) there is an excess of compound \(B\) (\(A\) is now the limiting reagent.) With this in mind, the makeup of the sample in each region can be summarized as Zinc and Magnesium are an example of two compounds that demonstrate this kind of behavior, with the third compound having the formula \(Zn_2Mg\) (Ghosh, Mezbahul-Islam, & Medraj, 2011). Oftentimes, the stable compound formed by two solids is only stable in the solid phase. In other words, it will decompose upon melting. As a result, the phase diagram will take a lightly different form, as is shown in Figure \(\Page {4}\). In this diagram, the formula of the stable compound is \(AB_3\) (consistent with \(\chi_B < 0.75\)). But you will notice that the boundary separating the two two-phase solid regions does not extend all of the way to the single phase liquid portion of the diagram. This is because the compound will decompose upon melting. The process of decomposition upon melting is also called . The makeup of each region can be summarized as There are many examples of pairs of compounds that show this kind of behavior. One combination is sodium and potassium, which form a compound (\(Na_2K\)) that is unstable in the liquid phase and so it melts incongruently (Rossen & Bleiswijk, 1912). | 3,463 | 2,370 |
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All the functional groups described in this chapter are found in the organic molecules that are constantly synthesized and destroyed by every living organism on Earth. A detailed understanding of the reactions that occur in living organisms is the goal of biochemistry, which deals with a wide variety of organic structures and reactions. The most abundant substances found in living systems belong to four major classes: proteins, carbohydrates, lipids, and nucleic acids. Here we briefly describe the structure and some functions of these biological molecules. In , we described proteins as biologically active polymers formed from amino acids linked together by amide bonds. In addition to an amine group and a carboxylic acid group, each amino acid contains a characteristic R group ( ). In the simplest amino acid, glycine, the R group is hydrogen (–H), but in other naturally occurring amino acids, the R group may be an alkyl group or a substituted alkyl group, a carboxylic group, or an aryl group. The nature of the R group determines the particular chemical properties of each amino acid. In , all the amino acids found in proteins except glycine are chiral compounds, which suggests that their interactions with other chiral compounds are selective. Some proteins, called enzymes, catalyze biological reactions, whereas many others have structural, contractile, or signaling functions. Because we have described proteins previously, we will not discuss them further. An amino acid is chiral except when R is an H atom. are the most abundant of the organic compounds found in nature. They constitute a substantial portion of the food we consume and provide us with the energy needed to support life. Table sugar, milk, honey, and fruits all contain low-molecular-mass carbohydrates that are easily assimilated by the human body. In contrast, the walls of plant cells and wood contain high-molecular-mass carbohydrates that we cannot digest. Once thought to be hydrates of carbon with the general formula C (H O) , carbohydrates are actually polyhydroxy aldehydes or polyhydroxy ketones (i.e., aldehydes or ketones with several –OH groups attached to the parent hydrocarbon). The simplest carbohydrates consist of unbranched chains of three to eight carbon atoms: one carbon atom is part of a carbonyl group, and some or all of the others are bonded to hydroxyl groups. The structure of a carbohydrate can be drawn either as a hydrocarbon chain, using a , or as a ring, using a ( ). The Haworth projection is named after the British chemist Sir Walter Norman Haworth, who was awarded a Nobel Prize in Chemistry in 1937 for his discovery that sugars exist mainly in their cyclic forms, as well as for his collaboration on the synthesis of vitamin C. The cyclic form is the product of nucleophilic attack by the oxygen of a hydroxyl group on the electrophilic carbon of the carbonyl group within the same molecule, producing a stable ring structure composed of five or six carbons that minimizes bond strain ( ). The substituents on the right side of the carbon chain in a Fischer projection are in the “down” position in the corresponding Haworth projection. Attack by the hydroxyl group on either side of the carbonyl group leads to the formation of two cyclic forms, called : an α form, with the –OH in the “down” position, and a β form, with the –OH in the “up” position. At age 14, Walter Norman Haworth left school to join his father to learn linoleum design and manufacturing, but he became interested in chemistry through his use of dyes. Private tutoring enabled him to pass the entrance exam of the University of Manchester, where he received his doctorate in 1911. During World War I, Haworth organized the laboratories at St. Andrews for the production of chemicals and drugs, returning to the investigation of carbohydrates after the war. Carbohydrates are classified according to the number of single , or sugar, units they contain (from the Latin , meaning “sugar”). The simplest are ; a consists of two linked monosaccharide units; a has three linked monosaccharide units; and so forth. Glucose is a monosaccharide, and sucrose (common table sugar) is a disaccharide. The hydrolysis of sucrose produces glucose and another monosaccharide, fructose, in a reaction catalyzed by an enzyme or by acid: hydrolyze to produce more than 10 monosaccharide units. The common monosaccharides contain several chiral carbons and exist in several isomeric forms. One isomer of glucose, for example, is galactose, which differs from glucose in the position of the –OH bond at carbon-4: Because carbons-2, -3, -4, and -5 of glucose are chiral, changing the position of the –OH on carbon-4 does not produce an enantiomer of glucose but a different compound, galactose, with distinct physical and chemical properties. Galactose is a hydrolysis product of lactose, a disaccharide found in milk. People who suffer from lack the enzyme needed to convert galactose to glucose, which is then metabolized to CO and H O, releasing energy. Galactose accumulates in their blood and tissues, leading to mental retardation, cataracts, and cirrhosis of the liver. Because carbohydrates have a carbonyl functional group and several hydroxyl groups, they can undergo a variety of biochemically important reactions. The carbonyl group, for example, can be oxidized to form a carboxylic acid or reduced to form an alcohol. The hydroxyl groups can undergo substitution reactions, resulting in derivatives of the original compound. One such derivative is Sucralose, an artificial sweetener that is six times sweeter than sucrose; it is made by replacing two of the hydroxyl groups on sucrose with chlorine. Carbohydrates can also eliminate hydroxyl groups, producing alkenes. Because carbohydrates have a carbonyl functional group and several hydroxyl groups, they can undergo a variety of reactions. Two familiar polysaccharides are and , which both hydrolyze to produce thousands of glucose units. They differ only in the connection between glucose units and the amount of branching in the molecule ( ). Starches can be coiled or branched and are hydrolyzed by the enzymes in our saliva and pancreatic juices. Animal starch, called , is stored in the liver and muscles. It consists of branched glucose units linked by bonds that produce a coiled structure. The glucose units in cellulose, in contrast, are linked to give long, unbranched chains. The chains in cellulose stack in parallel rows held together by hydrogen bonds between hydroxyl groups. This arrangement produces a rigid structure that is insoluble in water. Cellulose is the primary structural material of plants and one of the most abundant organic substances on Earth. Because our enzymes are not able to hydrolyze the bonds between the glucose units in cellulose, we are unable to digest it. A recently marketed product containing a high percentage of cellulose was sold as a dietetic substance for rapid weight loss, but those who consumed it experienced severe intestinal discomfort because the cellulose could not be digested. The product was quickly removed from the market. The Fischer projection of xylose, found in many varieties of apples, is shown. Draw the ring form (Haworth projection) of xylose. Fischer projection of a sugar cyclic structure Identify the nucleophile and the electrophile. Indicate the point of attack, remembering that cyclic structures are most stable when they contain at least five atoms in the ring to prevent bond strain from bond angles that are too small. Draw the cyclic form of the structure. The carbonyl carbon (C1) is a good electrophile, and each oxygen is a good nucleophile. Nucleophilic attack occurs from the –OH group on C4, producing a stable five-membered ring. Because of rotation about the bond between C1 and C2, ring formation gives both a and b anomers, with the following structures (H atoms have been omitted for clarity): Exercise Draw the cyclic form(s) of galactose, whose Fischer projection is shown in the previous discussion. Lipids (from the Greek , meaning “fat” or “lard”) are characterized by their insolubility in water. They form a family of compounds that includes fats, waxes, some vitamins, and steroids. , the simplest lipids, have a long hydrocarbon chain that ends with a carboxylic acid functional group. In saturated fatty acids, the hydrocarbon chains contain only C–C bonds, so they can stack in a regular array (part (a) in ). In contrast, unsaturated fatty acids have a single double bond in the hydrocarbon chain (monounsaturated) or more than one double bond (polyunsaturated). These double bonds give fatty acid chains a kinked structure, which prevents the molecules from packing tightly (part (b) in ). As a result of reduced van der Waals interactions, the melting point of an unsaturated fatty acid is lower than that of a saturated fatty acid of comparable molecule mass, thus producing an oil rather than a solid. (For more information on van der Waals interactions, see .) Fish oils and vegetable oils, for example, have a higher concentration of unsaturated fatty acids than does butter. The double bonds of unsaturated fatty acids can by hydrogenated in an addition reaction that produces a saturated fatty acid:
\( \underset{unsaturated\;fatty\;acid}{-CH=CH-} + H_{2}\rightarrow \underset{saturated\;fatty\;acid}{-CH_{2}CH_{2}-} \tag{23.6.1} \) They can also be oxidized to produce an aldehyde or carboxylic acid. (For more information on hydrogenation, see .) Unsaturated fatty acids are the starting compounds for the biosynthesis of . These hormone-like substances are involved in regulating blood pressure, tissue inflammation, and contracting and relaxing smooth muscles. Drugs such as aspirin and ibuprofen inhibit the production of prostaglandins, thereby reducing inflammation. are esters produced by the nucleophilic attack of an alcohol on the carbonyl carbon of a long-chain carboxylic acid ( ). For example, the wax used in shoe polish and wax paper, which is derived from beeswax, is formed from a straight-chain alcohol with 15 carbon atoms and a fatty acid with 31 carbon atoms. are a particularly important type of ester in living systems; they are used by the body to store fats and oils. These compounds are formed from one molecule of glycerol (1,2,3-trihydroxypropane) and three fatty acid molecules. During warmer months of the year, animals that hibernate consume large quantities of plants, seeds, and nuts that have a high fat and oil content. They convert the fatty acids to triacylglycerols and store them. Hydrolysis of stored triacylglycerols during hibernation (the reverse of 23.5.4) releases alcohols and carboxylic acids that the animal uses to generate energy for maintaining cellular activity, respiration, and heart rate. Derivatives of triacylglycerols with a phosphate group are major components of all cell membranes. are lipids whose structure is composed of three cyclohexane rings and one cyclopentane ring fused together. The presence of various substituents, including double bonds, on the basic steroid ring structure produces a large family of steroid compounds with different biological activities. For example, , a steroid found in cellular membranes, contains a double bond in one ring and four substituents: a hydroxyl group, two methyl groups, and a hydrocarbon chain. Cholesterol is the starting point for the biosynthesis of steroid hormones, including testosterone, the primary male sex hormone, and progesterone, which helps maintain pregnancy. These cholesterol derivatives lack the long hydrocarbon side chain, and most contain one or more ketone groups. Cholesterol is synthesized in the human body in a multistep pathway that begins with a derivative of acetic acid. We also consume cholesterol in our diets: eggs, meats, fish, and diary products all contain cholesterol, but vegetables and other plant-derived foods do not contain cholesterol. Excess cholesterol in the human body can cause gallstones, which are composed of nearly 100% cholesterol, or lipid deposits called plaque in arteries. A buildup of plaque can block a coronary artery and result in a heart attack ( ). Plaque, a lipid deposit, forms from excess cholesterol in the body. This artery is nearly blocked by a thick deposit of plaque, which greatly increases the risk of a heart attack due to reduced blood flow to the heart. Nucleic acids are the basic structural components of DNA (deoxyribonucleic acid) and RNA (ribonucleic acid), the biochemical substances found in the nuclei of all cells that transmit the information needed to direct cellular growth and reproduction. Their structures are derived from cyclic nitrogen-containing compounds called and , which can engage in hydrogen bonding through the lone electron pair on nitrogen (in pyrimidine and purine) or through the hydrogen of the amine (in purine): The same cyclic structures are found in substances such as caffeine, a purine that is a stimulant, and the antifungal agent flucytosine, a pyrimidine. (For more information on the structure of caffeine, see .) When a pyrimidine or a purine is linked to a sugar by a bond called a , a is formed. Adding a phosphoric acid group to the sugar then produces a (part (a) in ). The linkage of nucleotides forms a polymeric chain that consists of alternating sugar and phosphate groups, which is the backbone of DNA and RNA (part (b) in ). While the function of DNA is to preserve genetic information, RNA translates the genetic information in DNA and carries that information to cellular sites where proteins are synthesized. Many antibiotics function by interfering with the synthesis of proteins in one or more kinds of bacteria. Chloramphenicol, for example, is used against infections of the eye or outer ear canal; it inhibits the formation of peptide bonds between amino acids in a protein chain. Puromycin, which is used against herpes simplex type I, interrupts extension of a peptide chain, causing the release of an incomplete protein and the subsequent death of the virus. Mutations in the DNA of an organism may lead to the synthesis of defective proteins. ( ), for example, is a condition caused by a defective enzyme. Left untreated, it produces severe brain damage and mental retardation. is caused by a defective enzyme that is unable to produce melanin, the pigment responsible for the color of skin and hair. , the most common inherited disease in the United States, blocks pancreatic function and causes thick mucus secretions that make breathing difficult. An area of intense research in combating cancer involves the synthesis of drugs that stop uncontrolled cell growth by interfering with DNA replication. The four major classes of organic compounds found in biology are proteins, carbohydrates, lipids, and nucleic acids. Their structures and reactivity are determined by the functional groups present. are biologically active polymers formed from amino acids linked together by amide bonds. All the amino acids in proteins are chiral compounds except glycine. The most common organic compounds found in nature are the , polyhydroxy aldehydes or polyhydroxy ketones in unbranched chains of three to eight carbons. They are classified according to the number of sugar, or saccharide, units, and they can be drawn as a chain in a Fischer projection or in a cyclic form called a Haworth projection. The two cyclic forms in a Haworth projection are called anomers. Many sugars contain at least one chiral center. With their carbonyl and hydroxyl functional groups, carbohydrates can undergo a variety of biochemically relevant reactions. Starch and cellulose differ only in the connectivity between glucose units. Starches can be branched or unbranched, but cellulose, the structural material of plants, is unbranched, and cannot be digested by humans. are insoluble in water. The simplest lipids, fatty acids, have a long hydrocarbon chain ending in a carboxylic acid functional group. Their physical properties depend on the number of double bonds in the chain. Prostaglandins, hormone-like substances, are formed from unsaturated fatty acids, and waxes are long-chain esters of saturated fatty acids. Triacylglycerols, which the body uses to store fats and oils, consist of glycerol esterified to three fatty acid molecules. Steroids, which include cholesterol and the steroid hormones, are characterized by three cyclohexane rings and one cyclopentane ring fused together. The basic structural units of DNA and RNA are the , whose structures are derived from nitrogen-containing cyclic compounds called pyrimidines and purines. These structures are linked to a sugar through a glycosidic bond, forming a nucleoside. Adding a phosphoric acid group produces a nucleotide. Nucleotides link to form a polymeric chain that is the backbone of DNA and RNA. What are the strengths and limitations of using a Haworth projection? of using a Fischer projection? Nutritionists will often state that a leafy salad contains no calories. Do you agree? Would you expect margarine, a polyunsaturated fat, to have a higher or lower melting point than butter, a saturated fat? Propose a method for synthesizing the dipeptide alanylglycine (Ala-Gly), starting with the individual amino acids ( ). Are all the naturally occurring amino acids chiral compounds? Do you expect proteins to contain both enantiomers of alanine and other amino acids? Explain your answer. The structures of cholesterol and testosterone were shown in this section. Identify the functional groups in each. The structures of glucose and purine were shown in this section. Identify the functional groups in each. Use a condensation reaction: | 17,789 | 2,371 |
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An electrolyte solution is a solution that generally contains ions, atoms or molecules that have lost or gained electrons, and is electrically conductive. For this reason they are often called ionic solutions, however there are some cases where the electrolytes are not ions. For this discussion we will only consider solutions of ions. A basic principle of electrostatics is that opposite charges attract and like charges repel. It also takes a great deal of force to overcome this electrostatic attraction. The general form of Coulomb's law describes the force of attraction between charges: \[F=k\frac{q_1mq_2}{r^2}\] However, we must make some changes to this physics formula to be able to use it for a solution of oppositely charged ions. In Coulomb's Law, the constant \[k=\frac{1}{4\pi\varepsilon_{0}}\], where \(\varepsilon_{0}\) is the permittivity of free space, such as in a vacuum. However, since we are looking at a solution, we must consider the effect that the medium (the solvent in this case) has on the electrostatic force, which is represented by the dielectric constant \(\varepsilon\): \[F=\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}\varepsilon r^{2}}\] Polar substances such as water have a relatively high dielectric constant. Ions are not stable on their own, and thus no ions can ever be studied separately. Particularly in biology, all ions in a certain cell or tissue have a counterion that balances this charge. Therefore, we cannot measure the enthalpy or entropy of a single ion as we can atoms of a pure element. So we define a reference point. The \(\Delta_{f}\overline{H}^{\circ} \) of a hydrogen ion \(H^+\) is equal to zero, as are the other thermodynamic quantities. \[\Delta_{f}\overline{H}^{\circ}[H^{+}(aq)]=0\] \[\Delta_{f}\overline{G}^{\circ}[H^{+}(aq)]=0\] \[\overline{S}^{\circ}[H^{+}(aq)]=0\] When studying the formation of ionic solutions, the most useful quantity to describe is chemical potential \(\mu\), defined as the partial molar Gibbs energy of the ith component in a substance: \[\mu_{i}=\overline{G}_{i}=\left(\frac{\partial G}{\partial n_{i}}\right)_{T,P,n_{j}}=\mu_{i}^{\circ}+RT\ln x_{i}\] where \(x_{i}\) can be any unit of concentration of the component: mole fraction, molality, or for gases, the partial pressure divided by the pressure of pure component. To express the chemical potential of an electrolyte in solution in terms of molality, let us use the example of a dissolved salt such as magnesium chloride, \(MgCl_{2}\). \[MgCl_{2}\rightleftharpoons Mg^{2+}+2Cl^{-} \label{1}\] We can now write a more general equation for a dissociated salt: \[M_{\nu+}X_{\nu-}\rightleftharpoons\nu_{+}M^{z+}+\nu_{-}X^{z-} \label{2} \] where \(\nu_{\pm}\) represents the stoichiometric coefficient of the cation or anion and \(z_\pm\) represents the charge, and M and X are the metal and halide, respectively. The total chemical potential for these anion-cation pair would be the sum of their individual potentials multiplied by their stoichiometric coefficients: \[\mu=\nu_{+}\mu_{+}+\nu_{-}\mu_{-} \label{3} \] The chemical potentials of the individual ions are: \[\mu_{+} = \mu_+^{\circ}+RT\ln m_+ \label{4} \] \[\mu_{-} = \mu_-^{\circ}+RT\ln m_- \label{5} \] And the molalities of the individual ions are related to the original molality of the salt m by their stoichiometric coefficients \[m_{+}=\nu_{+}m\] Substituting Equations \(\ref{4}\) and \(\ref{5}\) into Equation \(\ref{3}\), \[ \mu=\left( \nu_+\mu_+^{\circ}+\nu_- mu_-^{\circ}\right)+RT\ln\left(m_+^{\nu+}m_-^{\nu-}\right) \label{6} \] since the total number of moles \(\nu=\nu_{+}+\nu_{-}\), we can define the mean ionic molality as the geometric average of the molarity of the two ions: \[ m_{\pm}=(m_+^{\nu+}m_-^{\nu-})^{\frac{1}{\nu}}\] then Equation \(\ref{6}\) becomes \[\mu=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+\nu RT\ln m_{\pm} \label{7} \] We have derived this equation for a ideal solution, but ions in solution exert electrostatic forces on one another to deviate from ideal behavior, so instead of molarities we must use the activity a to represent how the ion is behaving in solution. Therefore the mean ionic activity is defined as \[a_{\pm}=(a_{+}^{\nu+}+a_{-}^{\nu-})^{\frac{1}{\nu}}\] where \[a_{\pm}=\gamma m_{\pm} \label{mean}\] and \(\gamma_{\pm}\) is the , which is dependent on the substance.
Substituting the mean ionic activity of \Equation \(\ref{mean}\) into Equation \(\ref{7}\), \[\mu=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+\nu RT\ln a_{\pm}=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+RT\ln a_{\pm}^{\nu}=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+RT \ln a \label{11}\] when \(a=a_{\pm}^{\nu}\). Equation \(\ref{11}\) then represents the chemical potential of a nonideal electrolyte solutions. To calculate the mean ionic activity coefficient requires the use of the Debye-Hückel limiting law, part of the . Let us now write out the chemical potential in terms of molality of the salt in our first example, \(MgCl_{2}\). First from Equation \(\ref{1}\), the stoichiometric coefficients of the ions are: \[\nu_{+} = 1,\nu_{-} = 2,\nu\; = 3 \nonumber\] The mean ionic molality is \[\begin{align*} m_{\pm} &= (m_{+}^{1}m_{-}^{2})^{\frac{1}{3}} \\[4pt] &= (\nu_{+}m\times\nu_{-}m)^{\frac{1}{3}} \\[4pt] &=m(1^{1}2^{2})^{\frac{1}{3}} \\[4pt] &=1.6\, m \end{align*}\] The expression for the chemical potential of \[MgCl_{2}\] is \[\mu_{MgCl_{2}}=\mu_{MgCl_{2}}^{\circ}+3RT\ln 1.6\m m \nonumber\] | 5,484 | 2,372 |
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Crystalline solids have regular ordered arrays of components held together by uniform intermolecular forces, whereas the components of amorphous solids are not arranged in regular arrays. The learning objective of this module is to know the characteristic properties of crystalline and amorphous solids. With few exceptions, the particles that compose a solid material, whether ionic, molecular, covalent, or metallic, are held in place by strong attractive forces between them. When we discuss solids, therefore, we consider the positions of the atoms, molecules, or ions, which are essentially fixed in space, rather than their motions (which are more important in liquids and gases). The constituents of a solid can be arranged in two general ways: they can form a regular repeating three-dimensional structure called a crystal lattice, thus producing a crystalline solid, or they can aggregate with no particular order, in which case they form an amorphous solid (from the Greek ámorphos, meaning “shapeless”). (Right) Crystalline solids, or crystals, have distinctive internal structures that in turn lead to distinctive flat surfaces, or faces. The faces intersect at angles that are characteristic of the substance. When exposed to x-rays, each structure also produces a distinctive pattern that can be used to identify the material. The characteristic angles do not depend on the size of the crystal; they reflect the regular repeating arrangement of the component atoms, molecules, or ions in space. When an ionic crystal is cleaved (Figure 12.1), for example, repulsive interactions cause it to break along fixed planes to produce new faces that intersect at the same angles as those in the original crystal. In a covalent solid such as a cut diamond, the angles at which the faces meet are also not arbitrary but are determined by the arrangement of the carbon atoms in the crystal. Crystals tend to have relatively sharp, well-defined melting points because all the component atoms, molecules, or ions are the same distance from the same number and type of neighbors; that is, the regularity of the crystalline lattice creates local environments that are the same. Thus the intermolecular forces holding the solid together are uniform, and the same amount of thermal energy is needed to break every interaction simultaneously. Amorphous solids have two characteristic properties. When cleaved or broken, they produce fragments with irregular, often curved surfaces; and they have poorly defined patterns when exposed to x-rays because their components are not arranged in a regular array. An amorphous, translucent solid is called a glass. Almost any substance can solidify in amorphous form if the liquid phase is cooled rapidly enough. Some solids, however, are intrinsically amorphous, because either their components cannot fit together well enough to form a stable crystalline lattice or they contain impurities that disrupt the lattice. For example, although the chemical composition and the basic structural units of a quartz crystal and quartz glass are the same—both are SiO and both consist of linked SiO tetrahedra—the arrangements of the atoms in space are not. Crystalline quartz contains a highly ordered arrangement of silicon and oxygen atoms, but in quartz glass the atoms are arranged almost randomly. When molten SiO is cooled rapidly (4 K/min), it forms quartz glass, whereas the large, perfect quartz crystals sold in mineral shops have had cooling times of thousands of years. In contrast, aluminum crystallizes much more rapidly. Amorphous aluminum forms only when the liquid is cooled at the extraordinary rate of 4 × 10 K/s, which prevents the atoms from arranging themselves into a regular array. In an amorphous solid, the local environment, including both the distances to neighboring units and the numbers of neighbors, varies throughout the material. Different amounts of thermal energy are needed to overcome these different interactions. Consequently, amorphous solids tend to soften slowly over a wide temperature range rather than having a well-defined melting point like a crystalline solid. If an amorphous solid is maintained at a temperature just below its melting point for long periods of time, the component molecules, atoms, or ions can gradually rearrange into a more highly ordered crystalline form. Solids are characterized by an extended three-dimensional arrangement of atoms, ions, or molecules in which the components are generally locked into their positions. The components can be arranged in a regular repeating three-dimensional array (a crystal lattice), which results in a crystalline solid, or more or less randomly to produce an amorphous solid. Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points. In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures. 1. Compare the solid and liquid states in terms of a. rigidity of structure. b. long-range order. c. short-range order. 2. How do amorphous solids differ from crystalline solids in each characteristic? Which of the two types of solid is most similar to a liquid? a. rigidity of structure b. long-range order c. short-range order 3. Why is the arrangement of the constituent atoms or molecules more important in determining the properties of a solid than a liquid or a gas? 4. Why are the structures of solids usually described in terms of the positions of the constituent atoms rather than their motion? 5. What physical characteristics distinguish a crystalline solid from an amorphous solid? Describe at least two ways to determine experimentally whether a material is crystalline or amorphous. 6. Explain why each characteristic would or would not favor the formation of an amorphous solid. a. slow cooling of pure molten material b. impurities in the liquid from which the solid is formed c. weak intermolecular attractive forces 7. A student obtained a solid product in a laboratory synthesis. To verify the identity of the solid, she measured its melting point and found that the material melted over a 12°C range. After it had cooled, she measured the melting point of the same sample again and found that this time the solid had a sharp melting point at the temperature that is characteristic of the desired product. Why were the two melting points different? What was responsible for the change in the melting point? 3. The arrangement of the atoms or molecules is more important in determining the properties of a solid because of the greater persistent long-range order of solids. Gases and liquids cannot readily be described by the spatial arrangement of their components because rapid molecular motion and rearrangement defines many of the properties of liquids and gases. 7. The initial solid contained the desired compound in an amorphous state, as indicated by the wide temperature range over which melting occurred. Slow cooling of the liquid caused it to crystallize, as evidenced by the sharp second melting point observed at the expected temperature. | 7,175 | 2,373 |
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Enolate ions are just another example of anionic nucelophiles. The reason they get a page to themselves is that they are especially important, especially in biological chemistry. They are also important in the synthesis of organic compounds, such as in the pharmaceutical industry. An enolate ion is the anion that forms when a proton is removed next to a carbonyl. The carbon next to the carbonyl is called the α-position (alpha position). The alpha position is acidic both because of the amount of positive charge on a proton in that position and because of the stability of the anion that results if that proton is removed. Show why an enolate ion, such as the one formed from 2-propanone, above, is particularly stable. Show a mechanism, with curved arrows, for the formation of the enolate ion from 2-propanone, above. In the example above, 2-propanone is deprotonated at the α position to form the corresponding enolate ion. Note that sodium hydroxide is not a strong enough base to convert all of the 2-propanone to its enolate. The resulting enolate is basic enough to pull a proton from a water molecule, so an equilibrium results. Negative charges are fairly stable on oxygen atoms. That allows this particular reaction to shift back to the left again. To make the reaction go all the way to the right, we would need a less stable anion on the left. That would make that anion more basic. Can you think of atoms that would be less stable as anions than oxygen? The most commonly used very strong bases in synthetic chemistry involve anions of carbon, nitrogen or hydrogen. Some examples of compounds used as very strong bases are sodium hydride (NaH), sodium amide (NaNH ), lithium diisopropylamide (LiN[CH(CH ) ] ), and butyllithium (CH CH CH CH Li). However, it is sometimes really useful to have an equilibrium between a carbonyl compound and its enolate. That situation allows both a ketone (the 2-propanone, left) and its enolate (right) to be present at the same time. That means there is both a nucleophile and an electrophile (the ketone and the enolate). They will be able to react together. The reaction of an enolate nucleophile with another carbonyl compound is called an aldol reaction. A simple example of this reaction is shown here. This example involves the reaction of 2-propanone with its enolate. Provide a mechanism, with curved arrows, for the aldol reaction of 2-propanone, above. The biosynthesis of sugars, such as fructose, involves coupling smaller sugars together. If one sugar is converted into a nucleophile, it can donate electrons to the carbonyl on the other sugar, forming a new C-C bond. The carbonyl on the second sugar becomes a hydroxyl group in the new, larger sugar. In the cell, sugars are typically in a phosphorylated form when they react in this way. Phosphorylation is often an important step in activating molecules for biochemical reactions. Show the mechanism for the formation of the phosphorylated fructose shown above. Sometimes, aldol reactions are followed by a subsequent reaction, called an elimination reaction. That reaction formally produces a molecule of water. Early studies of this reaction would result in droplets of condensation on the glassware in which the reaction occured; hence, it is sometimes called a condensation reaction. Provide a mechanism for the aldol condensation shown above. It can be hard to predict the outcome of an aldol reaction because of the fact that there are two possible products from an aldol reaction (one with a new hydroxyl and one with a new double bond). A chemist might try to make one product in the laboratory, and end up with the other. This process can be difficult to control. However, in general, the elimination reaction is encouraged by heating the reaction. The reaction sometimes occurs without elimination if the reaction is kept cool. However, there are also other factors that may come into play. Predict the products of the following aldol reactions. The following compound would give multiple products through different aldol condensation reactions. Show the products. Sometimes, two different compounds may react together in an aldol reaction. The following compounds would give multiple products through different aldol condensation reactions. Show the products. Only some of the following compounds may undergo aldol reactions. Select which ones may not undergo the reaction, and explain what factor prevents them from reacting. Fill in the products of the following aldol condensations. Aldol reactions do not just occur with enolate anions, however. Enols are the neutral form of enolates, protonated on the oxygen instead of the alpha carbon. Enols are also good nucleophiles. In an enol nucleophile, the pi bond acts as the electron source, rather than the lone pair. However, the pi bond gets a boost from the lone pair on the oxygen. Enols are always present in equilibrium with aldehydes and ketones. An enol is a simple tautomer of a carbonyl compound. To get from one to the other, a proton is simply transferred from one position to the other. Enamines are also good nucleophiles for aldol-type reactions. Show the subsequent protonation step in the reactions involving the enol and the enamine above. An enamine reaction is usually followed by hydrolysis of the C=N bond in the iminium ion. Show the mechanism for conversion of the iminium ion to the carbonyl. , | 5,413 | 2,375 |
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Based on the nature of the forces that hold the component atoms, molecules, or ions together, solids may be formally classified as , , ( ), or . The variation in the relative strengths of these four types of interactions correlates nicely with their wide variation in properties. You learned in 4 that an ionic solid consists of positively and negatively charged ions held together by electrostatic forces. (For more information about ionic solids, see .) The strength of the attractive forces depends on the charge and size of the ions that compose the lattice and determines many of the physical properties of the crystal. The , the energy required to separate 1 mol of a crystalline ionic solid into its component ions in the gas phase, is directly proportional to the product of the ionic charges and inversely proportional to the sum of the radii of the ions. For example, NaF and CaO both crystallize in the face-centered cubic (fcc) sodium chloride structure, and the sizes of their component ions are about the same: Na (102 pm) versus Ca (100 pm), and F (133 pm) versus O (140 pm). Because of the higher charge on the ions in CaO, however, the lattice energy of CaO is almost four times greater than that of NaF (3401 kJ/mol versus 923 kJ/mol). The forces that hold Ca and O together in CaO are much stronger than those that hold Na and F together in NaF, so the heat of fusion of CaO is almost twice that of NaF (59 kJ/mol versus 33.4 kJ/mol), and the melting point of CaO is 2927°C versus 996°C for NaF. In both cases, however, the values are large; that is, simple ionic compounds have high melting points and are relatively hard (and brittle) solids. Molecular solids consist of atoms or molecules held to each other by dipole–dipole interactions, London dispersion forces, or hydrogen bonds, or any combination of these, which were discussed in . The arrangement of the molecules in solid benzene is as follows: In solid benzene, the molecules are not arranged with their planes parallel to one another but at 90° angles. Because the intermolecular interactions in a molecular solid are relatively weak compared with ionic and covalent bonds, molecular solids tend to be soft, low melting, and easily vaporized (Δ and Δ are low). For similar substances, the strength of the London dispersion forces increases smoothly with increasing molecular mass. For example, the melting points of benzene (C H ), naphthalene (C H ), and anthracene (C H ), with one, two, and three fused aromatic rings, are 5.5°C, 80.2°C, and 215°C, respectively. The enthalpies of fusion also increase smoothly within the series: benzene (9.95 kJ/mol) < naphthalene (19.1 kJ/mol) < anthracene (28.8 kJ/mol). If the molecules have shapes that cannot pack together efficiently in the crystal, however, then the melting points and the enthalpies of fusion tend to be unexpectedly low because the molecules are unable to arrange themselves to optimize intermolecular interactions. Thus toluene (C H CH ) and -xylene [ -C H (CH ) ] have melting points of −95°C and −48°C, respectively, which are significantly lower than the melting point of the lighter but more symmetrical analog, benzene. Self-healing rubber is an example of a molecular solid with the potential for significant commercial applications. The material can stretch, but when snapped into pieces it can bond back together again through reestablishment of its hydrogen-bonding network without showing any sign of weakness. Among other applications, it is being studied for its use in adhesives and bicycle tires that will self-heal. The methyl groups attached to the phenyl ring in toluene and -xylene prevent the rings from packing together as in solid benzene. Covalent solids are formed by networks or chains of atoms or molecules held together by covalent bonds. A perfect single crystal of a covalent solid is therefore a single giant molecule. For example, the structure of diamond, shown in part (a) in , consists of hybridized carbon atoms, each bonded to four other carbon atoms in a tetrahedral array to create a giant network. The carbon atoms form six-membered rings. The unit cell of diamond can be described as an fcc array of carbon atoms with four additional carbon atoms inserted into four of the tetrahedral holes. It thus has the zinc blende structure described in , except that in zinc blende the atoms that compose the fcc array are sulfur and the atoms in the tetrahedral holes are zinc. Elemental silicon has the same structure, as does silicon carbide (SiC), which has alternating C and Si atoms. The structure of crystalline quartz (SiO ), shown in , can be viewed as being derived from the structure of silicon by inserting an oxygen atom between each pair of silicon atoms. All compounds with the diamond and related structures are hard, high-melting-point solids that are not easily deformed. Instead, they tend to shatter when subjected to large stresses, and they usually do not conduct electricity very well. In fact, diamond (melting point = 3500°C at 63.5 atm) is one of the hardest substances known, and silicon carbide (melting point = 2986°C) is used commercially as an abrasive in sandpaper and grinding wheels. It is difficult to deform or melt these and related compounds because strong covalent (C–C or Si–Si) or polar covalent (Si–C or Si–O) bonds must be broken, which requires a large input of energy. Other covalent solids have very different structures. For example, graphite, the other common allotrope of carbon, has the structure shown in part (b) in . It contains planar networks of six-membered rings of hybridized carbon atoms in which each carbon is bonded to three others. This leaves a single electron in an unhybridized 2 orbital that can be used to form C=C double bonds, resulting in a ring with alternating double and single bonds. Because of its resonance structures, the bonding in graphite is best viewed as consisting of a network of C–C single bonds with one-third of a π bond holding the carbons together, similar to the bonding in benzene. To completely describe the bonding in graphite, we need a molecular orbital approach similar to the one used for benzene in . In fact, the C–C distance in graphite (141.5 pm) is slightly longer than the distance in benzene (139.5 pm), consistent with a net carbon–carbon bond order of 1.33. In graphite, the two-dimensional planes of carbon atoms are stacked to form a three-dimensional solid; only London dispersion forces hold the layers together. As a result, graphite exhibits properties typical of both covalent and molecular solids. Due to strong covalent bonding within the layers, graphite has a very high melting point, as expected for a covalent solid (it actually sublimes at about 3915°C). It is also very soft; the layers can easily slide past one another because of the weak interlayer interactions. Consequently, graphite is used as a lubricant and as the “lead” in pencils; the friction between graphite and a piece of paper is sufficient to leave a thin layer of carbon on the paper. Graphite is unusual among covalent solids in that its electrical conductivity is very high parallel to the planes of carbon atoms because of delocalized C–C π bonding. Finally, graphite is black because it contains an immense number of alternating double bonds, which results in a very small energy difference between the individual molecular orbitals. Thus light of virtually all wavelengths is absorbed. Diamond, on the other hand, is colorless when pure because it has no delocalized electrons. compares the strengths of the intermolecular and intramolecular interactions for three covalent solids, showing the comparative weakness of the interlayer interactions. Metals are characterized by their ability to reflect light, called luster , their high electrical and thermal conductivity, their high heat capacity, and their malleability and ductility. Every lattice point in a pure metallic element is occupied by an atom of the same metal. The packing efficiency in metallic crystals tends to be high, so the resulting metallic solids are dense, with each atom having as many as 12 nearest neighbors. Bonding in metallic solids is quite different from the bonding in the other kinds of solids we have discussed. Because all the atoms are the same, there can be no ionic bonding, yet metals always contain too few electrons or valence orbitals to form covalent bonds with each of their neighbors. Instead, the valence electrons are delocalized throughout the crystal, providing a strong cohesive force that holds the metal atoms together. Valence electrons in a metallic solid are delocalized, providing a strong cohesive force that holds the atoms together. The strength of metallic bonds varies dramatically. For example, cesium melts at 28.4°C, and mercury is a liquid at room temperature, whereas tungsten melts at 3680°C. Metallic bonds tend to be weakest for elements that have nearly empty (as in Cs) or nearly full (Hg) valence subshells, and strongest for elements with approximately half-filled valence shells (as in W). As a result, the melting points of the metals increase to a maximum around group 6 and then decrease again from left to right across the block. Other properties related to the strength of metallic bonds, such as enthalpies of fusion, boiling points, and hardness, have similar periodic trends. A somewhat oversimplified way to describe the bonding in a metallic crystal is to depict the crystal as consisting of positively charged nuclei in an electron sea ( ). In this model, the valence electrons are not tightly bound to any one atom but are distributed uniformly throughout the structure. Very little energy is needed to remove electrons from a solid metal because they are not bound to a single nucleus. When an electrical potential is applied, the electrons can migrate through the solid toward the positive electrode, thus producing high electrical conductivity. The ease with which metals can be deformed under pressure is attributed to the ability of the metal ions to change positions within the electron sea without breaking any specific bonds. The transfer of energy through the solid by successive collisions between the metal ions also explains the high thermal conductivity of metals. This model does not, however, explain many of the other properties of metals, such as their metallic luster and the observed trends in bond strength as reflected in melting points or enthalpies of fusion. A more complete description of metallic bonding is presented in . An alloy is a mixture of metals with metallic properties that differ from those of its constituent elements. Brass (Cu and Zn in a 2:1 ratio) and bronze (Cu and Sn in a 4:1 ratio) are examples of substitutional alloys , which are metallic solids with large numbers of substitutional impurities. In contrast, small numbers of interstitial impurities, such as carbon in the iron lattice of steel, give an interstitial alloy . Because scientists can combine two or more metals in varying proportions to tailor the properties of a material for particular applications, most of the metallic substances we encounter are actually alloys. Examples include the low-melting-point alloys used in solder (Pb and Sn in a 2:1 ratio) and in fuses and fire sprinklers (Bi, Pb, Sn, and Cd in a 4:2:1:1 ratio). The compositions of most alloys can vary over wide ranges. In contrast, intermetallic compounds consist of certain metals that combine in only specific proportions. Their compositions are largely determined by the relative sizes of their component atoms and the ratio of the total number of valence electrons to the number of atoms present (the ). The structures and physical properties of intermetallic compounds are frequently quite different from those of their constituent elements, but they may be similar to elements with a similar valence electron density. For example, Cr Pt is an intermetallic compound used to coat razor blades advertised as “platinum coated”; it is very hard and dramatically lengthens the useful life of the razor blade. With similar valence electron densities, Cu and PdZn have been found to be virtually identical in their catalytic properties. Some general properties of the four major classes of solids are summarized in . The general order of increasing strength of interactions in a solid is molecular solids < ionic solids ≈ metallic solids < covalent solids. Classify Ge, RbI, C (CH ) , and Zn as ionic, molecular, covalent, or metallic solids and arrange them in order of increasing melting points. compounds classification and order of melting points Locate the component element(s) in the periodic table. Based on their positions, predict whether each solid is ionic, molecular, covalent, or metallic. Arrange the solids in order of increasing melting points based on your classification, beginning with molecular solids. Germanium lies in the block just under Si, along the diagonal line of semimetallic elements, which suggests that elemental Ge is likely to have the same structure as Si (the diamond structure). Thus Ge is probably a covalent solid. RbI contains a metal from group 1 and a nonmetal from group 17, so it is an ionic solid containing Rb and I ions. The compound C (CH ) is a hydrocarbon (hexamethylbenzene), which consists of isolated molecules that stack to form a molecular solid with no covalent bonds between them. Zn is a -block element, so it is a metallic solid. Arranging these substances in order of increasing melting points is straightforward, with one exception. We expect C (CH ) to have the lowest melting point and Ge to have the highest melting point, with RbI somewhere in between. The melting points of metals, however, are difficult to predict based on the models presented thus far. Because Zn has a filled valence shell, it should not have a particularly high melting point, so a reasonable guess is C (CH ) < Zn ~ RbI < Ge. The actual melting points are C (CH ) , 166°C; Zn, 419°C; RbI, 642°C; and Ge, 938°C. This agrees with our prediction. Exercise Classify C , BaBr , GaAs, and AgZn as ionic, covalent, molecular, or metallic solids and then arrange them in order of increasing melting points. C (molecular) < AgZn (metallic) ~ BaBr (ionic) < GaAs (covalent). The actual melting points are C , about 300°C; AgZn, about 700°C; BaBr , 856°C; and GaAs, 1238°C. The major types of solids are ionic, molecular, covalent, and metallic. consist of positively and negatively charged ions held together by electrostatic forces; the strength of the bonding is reflected in the lattice energy. Ionic solids tend to have high melting points and are rather hard. are held together by relatively weak forces, such as dipole–dipole interactions, hydrogen bonds, and London dispersion forces. As a result, they tend to be rather soft and have low melting points, which depend on their molecular structure. consist of two- or three-dimensional networks of atoms held together by covalent bonds; they tend to be very hard and have high melting points. have unusual properties: in addition to having high thermal and electrical conductivity and being malleable and ductile, they exhibit luster, a shiny surface that reflects light. An is a mixture of metals that has bulk metallic properties different from those of its constituent elements. Alloys can be formed by substituting one metal atom for another of similar size in the lattice ( ), by inserting smaller atoms into holes in the metal lattice ( ), or by a combination of both. Although the elemental composition of most alloys can vary over wide ranges, certain metals combine in only fixed proportions to form with unique properties. Four vials labeled A–D contain sucrose, zinc, quartz, and sodium chloride, although not necessarily in that order. The following table summarizes the results of the series of analyses you have performed on the contents: Match each vial with its contents. Do ionic solids generally have higher or lower melting points than molecular solids? Why? Do ionic solids generally have higher or lower melting points than covalent solids? Explain your reasoning. The strength of London dispersion forces in molecular solids tends to increase with molecular mass, causing a smooth increase in melting points. Some molecular solids, however, have significantly lower melting points than predicted by their molecular masses. Why? Suppose you want to synthesize a solid that is both heat resistant and a good electrical conductor. What specific types of bonding and molecular interactions would you want in your starting materials? Explain the differences between an interstitial alloy and a substitutional alloy. Given an alloy in which the identity of one metallic element is known, how could you determine whether it is a substitutional alloy or an interstitial alloy? How are intermetallic compounds different from interstitial alloys or substitutional alloys? In a substitutional alloy, the impurity atoms are similar in size and chemical properties to the atoms of the host lattice; consequently, they simply replace some of the metal atoms in the normal lattice and do not greatly perturb the structure and physical properties. In an interstitial alloy, the impurity atoms are generally much smaller, have very different chemical properties, and occupy holes between the larger metal atoms. Because interstitial impurities form covalent bonds to the metal atoms in the host lattice, they tend to have a large effect on the mechanical properties of the metal, making it harder, less ductile, and more brittle. Comparing the mechanical properties of an alloy with those of the parent metal could be used to decide whether the alloy were a substitutional or interstitial alloy. Will the melting point of lanthanum(III) oxide be higher or lower than that of ferrous bromide? The relevant ionic radii are as follows: La , 104 pm; O , 132 pm; Fe , 83 pm; and Br , 196 pm. Explain your reasoning. Draw a graph showing the relationship between the electrical conductivity of metallic silver and temperature. Which has the higher melting point? Explain your reasoning in each case. Draw a graph showing the relationship between the electrical conductivity of a typical semiconductor and temperature. | 18,370 | 2,376 |
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A student goes to the hospital and has blood work done and the results state that he has a b The molar mass of bromide is \(79.90\frac{g}{mol}\). \(1L=10dL\), \(1g=1000mg\)
\[\left(\dfrac{\textrm{3.17} \cancel{mg}\;}{\textrm{1} \cancel{dL}\;}\right)\left(\dfrac{\textrm{1} \cancel{g}\;}{\textrm{1000} \cancel{mg}\;}\right)\left(\dfrac{10\cancel{dL}}{\textrm{1} \textrm{L}\;}\right )\left(\dfrac{\textrm{1} \textrm{mol}\;}{\textrm{79.90} \cancel{g}\;}\right)=3.97\times10^{-4}\ \frac{mol}{L}\nonumber \] Assume there is 1L of solution. Modality (m) is calculated by:
\[\textrm{3.17 x 10}^{-2} \cancel{g}\;\left(\dfrac{\textrm{1} \textrm{ mol}\;}{\textrm{79.90} \cancel{g}\;}\right)\left(\dfrac{1\textrm{ }}{\textrm{1.06} \textrm{ kg}\;}\right )=3.74\times10^{-4}\frac{mol}{kg}\nonumber \] If we know that there is \(3.17\times 10^{-2}g\) of bromide in \(1L\) of blood, so all that needs to be done is divide \(3g\) by \(3.17\times 10^{-2}\ \frac{g}{L}\):
Volume of Blood = 94.64 L Given a 3.1416 M aqueous solution of sucrose, \(\ce{C12H22O11}\) with a density of \(1.5986\ \mathrm{\frac{g}{mL}}\). Calculate the molality of this solution, remember that the molar mass of sucrose is \(342.297 \mathrm{\frac{g}{mol}}\). Before we calculate the molality of this solution, don't forget that molarity is just mol/L so we will multiply the molarity of the solution by 1 L to isolate the moles. \[1.00\; \cancel{L}\; \text{Water} = \dfrac{3.1416\: mol\ \text{Sucrose}}{1.00\; \cancel{L}\; \text{Water}}= 3.1416\; mol\ \text{Sucrose} \nonumber \] Once we have the moles of Sucrose we will multiply it by the molar mass of sucrose in order to find the number of sucrose grams present in the solution: \[3.1416\; \cancel{mol}\; \text{Sucrose} =\dfrac{342.297\: g\ \text{Sucrose}}{1.00\; \cancel{mol}\; \text{Sucrose}}= 1075.4\; g\ \text{Sucrose}\nonumber \] We next will multiply the density by 1000 mL so as get the number of grams per liter, this way we can find the total grams of the solution and can subtract the sucrose grams from the solution grams to find the grams of water. \[1000\; \cancel{mL}\; =\dfrac{1.5986\, g \: \text{SucroseSoln}}{1.00\; \cancel{mL}\; \text{SucroseSoln}}= 1598.6\; g/L\ \text{SucroseSoln} \nonumber \] \[\text{Water grams} =[{1598.6g\; \text{Sucrose Soln.}}]-[{1075.4g\; \text{Sucrose} }]={523.2g}\; \text{Water} \; ={0.5232\; kg\; \text{Water} \;}\nonumber \] Lastly we divide the moles of Sucrose by the kg of water in order to get molality. \[\dfrac{3.1416\: mol\ \text{Sucrose}}{0.5232\; kg\; \text{Water}}= 6.005\; molal\nonumber \] A solution contains 75% of ethanol (\(\ce{C2H6O}\)) by mass and the rest is water. a) For easy calculation, assume that there is 1 L of the solution. Using: \[\text{mol} = \text{molarity} \times \text{volume}\] The mol of 15M ethanol in 1L is 15 mol. The molar mass of ethanol is \(46.068\frac{g}{mol}\). Using the equation below: \[\text{ethanol mol} = \dfrac{\text{mass ethanol}}{\text{ethanol molar mass}}\] \[\text{ethanol mass in the solution} = \text{ethanol mol} \times \text{molar mass} = (15mol)(46.068\dfrac{g}{mol})=691.02g\] Ethanol is 75% of the solution by mass, therefore: \[\text{mass of solution} \times 0.75= \text{mass of ethanol} \] \[\text{mass of solution} \times 0.75= 691.02g\] \[\text{mass of solution} = \dfrac{691.02g}{0.75}=921.36g\] The equation to calculate density is: \[\text{Density} = \dfrac{\text{mass}}{\text{volume}}\] \[\text{Density} = \dfrac{921.36g}{1L} = 0.921 \dfrac{g}{ml}\] b)To prepare a 3.5L of 2M ethanol; \[\text{ethanol mol needed} = \text{molarity} \times \text{volume} = (2 \, \text{M}) (3.5 \, \text{L})=7 \, \text{mol}\] \[V_{\text{ethanol needed}}=\dfrac{\text{mol needed}}{\text{molarity}}=\dfrac{7 \, \text{mol}}{15 \, \text{mol}} \times 1 \, \text{L} = 0.467 \, \text{L}\] 0.467L of the solution is needed, which is 467mL. with \(\ce{KOH}\), The balanced equation of \(\ce{H2SO4}\) with KOH is: \[ \ce{H2SO4 (aq) + 2 KOH (aq) -> K2SO4 (aq) + 2 H2O (l)} \nonumber \] Calculate the amount of moles of \(\ce{H2SO4}\) being neutralized: \[ n_{\ce{H2SO4}} = 36.2 \; \text{g} \; \ce{H2SO4} \times \dfrac{1 \; \text{mol} \; \ce{H2SO4}}{98.079 \; \text{g} \; \ce{H2SO4}} = 0.369 \; \text{mol} \; \ce{H2SO4} \nonumber \] Using stoichiometry to find the amount of moles of \(\ce{KOH}\) needed to neutralize the amount of \(\ce{H2SO4}\) present: \[ n_{\ce{KOH}} = 0.369 \; \text{mol} \; \ce{H2SO4} \times \dfrac{2 \; \text{mol} \; \ce{KOH}}{1 \; \text{mol} \; \ce{H2SO4}} = 0.738 \; \text{mol} \; \ce{KOH} \nonumber \] Calculate the volume of the o neutralize the amount of \(\ce{H2SO4}\) present: \[ V_{\ce{KOH}} = 0.738 \; \text{mol} \; \ce{KOH} \times \dfrac{1 \; \text{L}}{5.49 \; \text{mol} \; \ce{KOH}} = 0.134 \; \text{L} = 134 \; \text{mL} \nonumber \] Therefore, approximately 134 mL of the 5.49 M \(\ce{KOH}\) solution is needed to neutralize 36.2 g \(\ce{H2SO4}\). Phosphoric acid is made industrially as a by-product of the extraction of calcium phosphate \[ \ce{Ca_3(PO_4)_2 + 6HNO_3 + 12H_2O -> 2H_3PO_4 + 3Ca(NO_3)_2 + 12H_2O} \nonumber\] What volume of 8.5 M phosphoric acid is generated by the reaction of 4.5 metric tons (4500 kg) of calcium phosphate? \[4500\,\text{kg} \ \ce{Ca3(PO4)2} \times \left(\dfrac{1000\,\text{g}}{1\ \text{kg}} \right) \times \left( \dfrac{1\ \text{mol}}{310 \ \text{g}} \right) \times \left( \dfrac{2 \, \ce{H3PO4}}{1 \, \ce{Ca3(PO4)2}} \right) \times \left( \dfrac{1\ \text{L}}{8.5\ \text{moles}} \right) =3415.56 \, \text{L} \ \ce{H3PO4} \nonumber \] Write a balanced equation for the acid-base reaction that leads to the production of each of the following salts. Hydrogen selenide is classified as a binary acid that has similar properties as hydrogen sulfide. Write down the balanced equation of hydrogen selenide reacting with the base potassium hydroxide and name the salt product of the reaction. When the acid \(\ce{H2S}\) reacts with the base \(\ce{KOH}\), the reaction goes as follows: \[\ce{H2S + 2 KOH -> K2S + 2 H2O} \nonumber\] Forming potassium sulfide. Similarly, since \(\ce{H2Se}\) has similar properties as \(\ce{H2S}\), a similar reaction occurs: \[\ce{H2Se + 2 KOH -> K2Se + 2 H2O} \nonumber\] Potassium selenide is the salt product of the reaction. A student working with an unknown solution of hydrochloric acid is assigned the task of finding its concentration. The student places a 200.0 mL sample of the solution in a 350.0 mL Erlenmeyer flask and titrates the solution with a 0.1234 M solution of sodium hydroxide, where a phenolphthalein indicator is used to find the endpoint. The endpoint is found to be 31.416 mL, what is the concentration of the original hydrochloric acid solution sample? Remember that in titration of a strong acid and base, when the endpoint is reached the number of moles used is equal, so the number of moles of NaOH at the endpoint are equal to that of HCl. Use this information to find moles of NaOH from the concentration and volume used to titrate. \[ \text{Concentration of X} = \dfrac{\text{moles of X}}{\text{liters of X}}\nonumber \] Use dimensional analysis to convert concentration of NaOH to moles using the known amount of volume titrated, in liters: \[0.031416\; \cancel{\text{liter}}\; \ce{NaOH} =\dfrac{0.1234\: \text{mol}\, \ce{NaOH}}{1.00\; \cancel{\text{liter}}\; \ce{NaOH}}= 0.003877\; \text{mol} \nonumber \] Now because we know that the moles of NaOH are equal to the moles of HCl we simply plug in our calculated moles and known original sample volume, in liters, to find the concentration of HCl: \[ \text{Concentration of HCl} = \dfrac{0.003877\: \text{mol}\ \ce{HCl}}{0.200\: \text{liters} \ \ce{HCl}}= 0.019385\; \text{M} \; \ce{HCl} \nonumber \] The concentration of the original hydrochloric acid solution sample was, therefore, 0.019385 M. The vapor pressure of salicylic acid \(\ce{C7H6O3}\) at 200 0.1598 atm. A 19.0 g sample of ethanol \(\ce{C2H6O}\) is dissolved in 60.0 g of salicylic acid. Calculate the vapor pressure of salicylic acid above the resulting solution. First find the molar masses of salicylic acid and ethanol. Then use them to compute the chemical amounts of the two in solution. \[n_{\ce{C2H6O}}=19.0\;g\; \ce{C2H6O} \times\dfrac{1\;mol\;\ce{C2H6O}}{46.068\;g\;\ce{C2H6O}}=0.4124\;mol\;\ce{C2H6O}\nonumber \] \[n_{\ce{C7H6O3}}=60.0\;g\;\ce{C7H6O3}\times\dfrac{1\;mol\;\ce{C7H6O3}}{138.121\;g\;\ce{C7H6O3}}=0.4344\;mol\;\ce{C7H6O3}\nonumber \] The mole fraction of ethanol in the solution equals \[X_{\ce{C2H6O}}=\dfrac{0.4124}{0.4344+0.4124}=0.4870\nonumber \] The change in vapor pressure of the salicylic acid due to the presence of of the ethanol is The final vapor pressure of the salicylic acid equals its original vapor pressure P When 53 mol of an unknown compound was added to 100 g of water, the normal boiling point of water increased to 101.1 C. What is the boiling point elevation constant of water? The compound does not dissociate in solution. The relevant relationship here is the colligative property associated with boiling point elevation \[\Delta T_{b}=iK_{b}m\] \[\Delta T_{b}=101.1 ^{o}\,C-100 ^{o}\,C=1.1^{o}\,C\] Since the compound was told to not dissociation, we know that the van't Hoff factors is 1: \[i=1\] \[m=\dfrac{\text{mol solute}}{\text{kg solvent}}=\dfrac{53\,mol}{0.100\,kg}=0.53\frac{mol}{g} \] \[1.1^{o}C=(1)\times K_{b}\times (0.53\frac{mol}{g})\] \[K_{b}=\dfrac{1.1^{o}C}{0.53\dfrac{mol}{kg}}=2.075\dfrac{^{o}C\ kg}{mol}\] When 5.82 g of uniformly dissociating unknown salt, YX, is dissolved in 100.0 g of water, the boiling point of the water is raised by 0.20°C. When 7.09 g of uniformly dissociating unknown salt, ZX, is dissolved in 200.0 g of water, the boiling point of the water is raised by 0.30°C. Assuming both have a van't hoff factor of 2, identify the elements that X, Y, and Z represent. For both salts, one can utilize the boiling point elevation expression (including the van't hoff factor): \(\Delta{T}= i\times K_b\times m\) and by substituting moles solute over mass solvent into the molality, m, one can derive that: \(n_{\text{salt}}= \dfrac{\Delta{T}\times m_{\text{water}}}{i\times K_b}\) From this we get the moles of each salt present, and can therefore get their molar mass by dividing their mass by moles present. Next, subtracting the heavier salt by the lighter salt will the molar mass of X, \(\sim 35.5 \mathrm{\frac{g}{mol}}\). Subtracting this from the molar masses of each salt will then Y and Z respectively. Camphor is a useful agent for determining solute’s molar mass. Camphor’s melting point and \(K_f\) are given at \(451.55 \, \mathrm{K}\) and \(37.7 \mathrm{\frac{K\ kg}{mol}}\). Find the freezing point of the solution of 0.72 g of glucose is dissolved in the 30.0 g of camphor. Since we are evaluating glucose as the solute, \(i = 1\). \[\Delta T = -iK_{f}m\] First find number of moles of glucose in the solution using molar mass. \[\text{Moles}_{\text{glucose}} = \frac{(0.72g)}{(180.1559\frac{g}{mol})}= 0.0040mol\] Molality then can be determined. \[m= \frac{(0.0040mol)}{(0.030kg)}= 0.13\frac{mol}{kg}\] Freezing then can be easily found as such: \[\Delta T = -iK_{f}m=-(1)(37.7\frac{^oC\ kg}{mol})(0.13\frac{mol}{kg})=-4.9^oC\] The freezing is \(451.55K - 4.9K = 446.65K = 173.5^oC\). At what temperature does the first ice crystals begin to form in a 27% salt (by mass) aqueous solution of NaCl? As the crystallization of water carries on, the remaining solution becomes more concentrated, so what happens to the freezing point of the solution? Given that the \((K_f)\) of water is \(1.86\frac{^oC\ kg}{mol}\). \[27 \%\: \ce{NaCl (aq)} = \dfrac{27 \; g \; \ce{NaCl}}{73 \; g \; \ce{H2O}}\] \[ molality = \dfrac{27 \; g \; \ce{NaCl}}{73 \; g \; \ce{H2O}} \times \dfrac{mol \; \ce{NaCl}}{58.44 \; g} \times \dfrac{1000 \; g}{1 \; kg} = 6.33 \; m \] \[ \Delta T = -i\times m\times K_{p} = -2 \times 6.33 \dfrac{mol}{kg} \times 1.86 \dfrac {^{o}C \cdot kg}{mol} = -23.5^{o}C \] The first ice crystal begins to appear at \(-23.5^oC\), as the solution becomes more concentrated, its freezing point decreases further because freezing point depression is a colligative property, so t When homemade ice cream is being made, the temperature ranging downward from -3°C are needed. Ice cubes from a freezer have a temperature of about -12°C, which is cold enough; however, when the ice cream mixture is mixed with the ice cubes, the liquid balances out to 0°C, which is too warm. To obtain a liquid that is cold enough, salt NaCl is dissolved in water and ice is added to the saltwater. The salt lowers the freezing point of the water enough so that it can freeze the liquid inside the ice cream maker. The instruction for the ice cream maker say to add one part salt to 11 parts water (by mass). What is the freezing point of this solution in °C? Assume the NaCl dissociates fully into ions, and that the solution is ideal. \[\Delta T_{f} = -i k_{f} m\] \[\dfrac{1.00\ g\ \ce{NaCl}}{53.5\frac{g}{mol}}=0.01869\ mol\ \ce{NaCl}\] \[m=\dfrac{0.01869\ mol}{11.00 g(\dfrac{1\ kg}{1000\ g})}=1.699 m\] \[x - (-12°C) = - (2)\times (1.86\frac{°C}{m})\times (1.699 m)\] \[x = -18.32 °C\] A sample of a purified unknown compound is dissolved in toluene, diluting the solvent to a volume of 1.05 mL. The resulting solution has an osmotic pressure of 0.025 atm at 275 K. If the solute has a molar mass of 46.06 g/mol, how many grams of the unknown compound were added? \(\Pi\) = \(\dfrac{nRT}{V}\) and we know \(T= 275K\), \(R=0.0821\dfrac{L\ atm}{mol\ K}\) and \(\Pi\)=0.025 atm Thus V= 1.05 mL Since we need V in liters because the units of R are \(\dfrac{L\ atm}{mol\ K}\) , we use unit conversions: \(1.05\ mL\times \dfrac{1 L}{1000 mL} = 0.00105\ L\) Plug in the values we know to find n \(0.025= \dfrac{n\times 0.0821\times 275}{0.00105}\) \(\dfrac{0.025\times 0.00105}{0.0821\times 275} = n\) \(n=1.163\times 10^{-6}\) Now that we know number of moles we just have to multiply the number of moles by the molar mass to find grams of substance. \((1.163\times 10^{-6}) \times 46.06844\frac{g}{mol}=5.356\times 10^{-5}g\) An amount of hydrogen gas (\(\ce{H_2}\)) with a partial pressure of 4.6 atm is dissolved in water and this solution is sealed. The Henry's Law constant K for this solution at \(25^oC\) is \(7.8\times 10^{-4} \mathrm{\frac{M}{atm}}\). a) According to , \[C = k \times P_{\text{gas}}\nonumber \] In this equation, C = concentration of gas (M), k = Henry's Law constant \((\frac{M}{atm})\), and \(P_{\text{gas}}\) = partial pressure of gas. So, given k and \(P_{\text{gas}}\): \[C = 7.8 \times 10^{-4} \dfrac{M}{atm} \times 4.6 \; atm\nonumber \] \[C = 0.003588 \dfrac{mol \; \ce{H_2}}{L_{\text{water}}}\nonumber \] b) The partial pressure of \(\ce{H_2}\) in the Earth's atmosphere is less than 1 atm. When the solution is unsealed, the \(\ce{H_2}\) molecules will no longer be under enough pressure to dissolve in water, so they will quickly bubble out of the solution and escape as gas into the surroundings. produce propanol: \[ \ce{C3H6 +H2O -> CH3(CH2)2OH} \nonumber\] What suggestions can you give to the company regarding the conditions of pressure and temperature that will have a maximum yield of propanol at equilibrium without calculating the actual data? \[\Delta H^{o}_{f(\text{propene})}=20.4\dfrac{kJ}{mol}\] \[\Delta H^{o}_{f(\text{water})}=-241.8\dfrac{kJ}{mol}\] \[\Delta H^{o}_{f(\ce{CH3(CH2)2OH})}=-104.6\dfrac{kJ}{mol}\] \[\Delta H^o = \Delta H^o_{f(\ce{CH3(CH2)2OH})} - \Delta H^o_{f(\text{propene})} - \Delta H^{o}_{f(\text{water})}\] \[=-104.6\dfrac{kJ}{mol} - (-241.8\dfrac{kJ}{mol}) - 20.4\dfrac{kJ}{mol}\\=116.8\dfrac{kJ}{mol}>0\] It is an endothermic reaction. As a result, a higher temperature is required to maximize the yield of product. Also, according to the ideal gas law, because the moles of gas are proportional to the volume, and the volume is inverse proportional to the pressure; a higher pressure is required to maximize the yield of product. \[\left (\dfrac{X_A\: mol}{X_A + X_B\: mol} \right )\nonumber \] \[(10.0\; \cancel{g\; \ce{C6H6}}) \left(\dfrac{1\: mol}{78.108\: \cancel{ g\; \ce{C6H6}}} \right) = 0.128\; mol \nonumber \] \[(30.0\; \cancel{g \; \ce{C3H8}} ) \left (\dfrac{1\: mol}{44.094\: \cancel{g\; \ce{C3H8}}} \right ) = 0.680\; mol \nonumber \] \[ \left (\dfrac{0.128\: mol}{0.128 + 0.680\: mol} \right )= 0.158 \nonumber \] Here we have chosen to use units of kPa: \(m_{\ce{Cl}}=2g\ \ce{CaCl2} (\frac{2(35.5g\ \ce{Cl})}{110.98g\ \ce{CaCl2}}) + 2g\ \ce{MgCl2} (\frac{2(35.5g\ \ce{Cl})}{95.12g\ \ce{MgCl2}})\) \(m_{\ce{Cl}}=2.77g\ \ce{Cl}\) \(\text{mass of chlorine per liter: } \frac{2.77g\ \ce{Cl}}{150 \times 10^{-3} L}=18.5 \frac{g}{L}\) Chromium ion, \(\ce{Cr^{2+}}\), is a good reducing agent, often being itself turned into \(\ce{Cr^{3+}}\). Suppose that 10 mL of 0.1 M \(\ce{Cr(OH)2}\) was needed to reduce completely an 0.05 grams of unknown substance X. (a) Number of moles of \(\ce{Cr(OH)2}\) used = \(M\times V\) \(= 0.1\times 10\ \mathrm{mL} = 1\ \mathrm{mmol} = 10^{-3}\ \mathrm{moles}\) Since only one mole of X was used for every mole of Chromium Hydroxide, the number of moles reacted are equal. \(\frac{\text{Mass}}{\text{Molar mass}}\) = number of moles, hence for X, \(\frac{0.05\ \mathrm{g}}{(0.01 \mathrm{L} \times 0.1 \mathrm{M})}= 50 \mathrm{\frac{g}{mol}}\) (b) \(\frac{0.05\ \mathrm{g}}{(0.01 \mathrm{L} \times 0.1 \mathrm{M} \times 5)}= 10 \mathrm{\frac{g}{mol}}\) A 0.200 g sample of chloride with unknown purity was dissolved in 200 mL of water was then titrated with \(\ce{AgNO3}\) solution with a molarity of 0.1 M. After the titration it was determined that 28 mL of \(\ce{AgNO3}\) was required to fully titrate the sample of chloride. What's the mass percent of chloride in the original dry sample? First write the precipitation reaction: \[\ce{Ag+ (aq) + Cl- (aq) -> AgCl (s)} \nonumber \] Next we need to determine the amount of moles of \(\ce{Ag}\) used during the titration: \[\textrm{mol Ag}^{+}=\textrm{28} \cancel{mL}\;\left(\dfrac{\textrm{1} \cancel{L}\;}{\textrm{1000} \cancel{mL}\;}\right)\left(\dfrac{0.1\textrm{ mol}}{\textrm{1} \cancel{L}\;}\right )=2.80\times10^{-3}\textrm{ mol Ag}^{+}\nonumber \] Now that we know the amount of moles of Ag reacted because Ag and Cl combine at a one to one ration we can find the mass of Cl used: \[\textrm{mass Cl}^{-}=2.80\times10^{-3}\cancel{mol}\;\left(\dfrac{35.45\textrm{ g}}{\textrm{1} \cancel{mol}\;}\right )=9.92\times10^{-2}\textrm{ g Cl}^{-}\nonumber \] Now that we know the mass of Cl in the unknown we can calculate the mass percent: \[100\textrm{%}\;\left(\dfrac{0.0992\textrm{ g}}{\textrm{0.2} \textrm{g}\;}\right )=49.63\textrm{%}\nonumber \] At 20°C, the vapor pressure of pure water is 17.5 mmHg. When 110 g of a substance \(\ce{X}_y\) is dissolved in 500 g of water, the vapor pressure of the solution is 15.0 mmHg. Given that this substance has a molecular mass of 12.011 g/mol, what is the molecular formula of this substance \(\ce{X}_y\). (i.e. what is the number value of subscript y). For this question, we can use \[P_{\text{solution}} = \chi_{\text{solvent}} \times P°_{\text{solvent}}\nonumber \] First, we plug in the known values of the pressure of the solution and the pressure of the pure solvent, water, to find the mole fraction. \(15.0 \; \mathrm{mmHg} = \chi_{\text{solvent}} \times 17.5 \; \mathrm{mmHg}\) \(\chi_{\text{solvent}} = 0.8571\) Next, we can calculate the the number of moles of the solvent from its mass and molecular weight. \(\text{Moles of solvent} = 500 \; \mathrm{g} \times \dfrac {1}{18.015 \; \mathrm{g/mol}} = 27.75 \; \mathrm{moles}\) Next, we can calculate the moles of the solute from the mole fraction of the solvent \(0.8571 = \dfrac{27.75 \; \mathrm{moles}}{27.75 \; \mathrm{moles} \; + \text{ moles of } X_y}\) \(\text{moles of } X_y = 4.63 \; \mathrm{moles}\) From here, we can calculate the molecular formula of the substance by dividing the amount of grams per mole used in the solution by the molecular mass of the substance. \(\dfrac {110 \; \mathrm{g}}{4.63 \; \mathrm{moles}} = 23.76 \; \mathrm{g/mol}\) \((23.76 \; \mathrm{g/mol}) \times (\dfrac {1}{12.011 \; \mathrm{g/mol}}) = 1.98\) \(1.98 \approx 2\) Thus, the molecular formula of our substance is \(\ce{X_2}\). \[-0.0218=-1.86 K \cdot kg \cdot mol^{-1} m \nonumber \] \[m=0.015\dfrac{kg}{mol}\nonumber \] \[ \begin{align*} \pi&=cRT \\[4pt] &=(0.15M)(0.08206\frac{L\ atm}{mol\ K})(303.15K) \\[4pt] &=0.374atm \end{align*} \] Jim loves the smell of grass. He loves it so much that he frequently sniffs the fumes from 1-Hexanol, which is a chemical compound that is known to smell like grass trimmings. Because of his love, he decides to make his garage into a “gas den”, and aimed to have 60% of the air in it as 1-Hexanol gas. He achieves this by filling a pool with a mixture of 1-Hexanol and Water. Assume that the air in the “gas den” consists of only the fumes from this mixture. If the pool has 10000L of water in it, how many litres of 1-Hexanol does he need? Use Henry’s Law to determine this amount. 1-Hexanol: Water: This problem asks for the amount, in litres, of 1-Hexanol that is required to make Jim’s dreams come true. The information given is the density of 1-Hexanol and its molar mass, but since they don’t have very much to do with pressure, they can be set aside for now. The question explicitly mentions that Henry’s Law should be used, so start be identifying what Henry’s Law actually does: \[C = kP \nonumber \] Where C is concentration, k is Henry’s Law constant, and P is partial pressure. The question gives out the fact that Jim wants the air in his garage to be 60% 1-Hexanol vapor. Remember the total pressure of air is 1 atm ≈ 1 bar (ignore the error here). Therefore, the partial pressure of 1-hexanol is \(1 \, \text{bar} \times 60\% = 0.6 \, \text{bar}\). The Henry’s Law constant is also provided, and by plugging those two in, the concentration of 1-Hexanol in the desired “solution” (more accurately, giant pool of chemicals) is found. This process is shown below: \[C= kP = (64\dfrac{molal}{bar})\times 0.6 \, \text{bar} = 38.4\nonumber \] This concentration, from the units of k, is in mo which is \(\dfrac{mol_{\text{solute}}}{kg_{\text{solvent}}}\). Now all that is left is to convert molality to litres. By utilizing the molar mass of 1-Hexanol: \[38.4\dfrac{mol_{\text{solute}}}{kg_{\text{solvent}}}\times{107.1748\dfrac{g}{mol}}\approx{4115.51\dfrac{g_{\text{solute}}}{kg_{\text{solvent}}}}\nonumber \] We know that the solvent is water, and that water has a density of \(1000\dfrac{kg}{m^3}\). Also, the conversion rate of \(m^{3}\) to \(L\) is 1:1000. Therefore, the equation can be solved as follows. \[4115.51 \dfrac{g_{\text{solute}}}{kg_{\text{solvent}}} \times 1000 \dfrac{kg_{\ce{H2O}}}{m^3} \times \dfrac{m^3}{1000 \, L_{\ce{H2O}}} \approx 4115.51 \dfrac{g_{\text{solute}}}{L_{\text{solvent}}}\nonumber \] We now have the value for how many grams of solute (1-Hexanol) there should be per litre of solvent. simply divide this value by the density to obtain litres of 1-Hexanol per litre of water. Multiply this number by the amount of water in the pool to get the desired amount of solute that is needed for the solution! (note that the density was converted to \(\dfrac{g}{L}\), \[4115.51 \dfrac{g_{\text{solute}}}{L_{\text{solvent}}} \div \left( 814000\dfrac{g_{\text{1-hexanol}}}{L} \right) \times 10000L_{\ce{H2O}}=50.56 \, \text{L} \nonumber \] Therefore, Jim needs 50.56L of 1-Hexanol to put in his pool. Please note that, while various units do appear in the conversions, \(\ce{H2O}\) can be considered as solute and 1-Hexanol as the solvent, so the units can cancel. : Utilize henry's law to find concentration (molality). From that, convert molality to \(g_{\text{solute}}\) to \(L_{\text{solute}}\). A 3.5 gram sample is decomposed into a compound containing nitrogen, hydrogen, and oxygen, together creating 4.00 g of \(\ce{NO2}\) and 7.00 g of \(\ce{H2O}\). The molar mass of the sample is \(78.5 \mathrm{\frac{g}{mol}}\). Find the molecular formula of this sample. First, the moles of each Nitrogen, Hydrogen, and Oxygen in solution are calculated: \(mol_N = 4g\ \ce{NO2} \times (\frac{1\ mol\ \ce{NO2}}{46.005\frac{g}{mol}})\times (\frac{1\ mol\ N}{1\ mol\ \ce{NO2}}) = 0.0869\ mol\) \(mass_N = (14.00\frac{g}{mol})(0.0869mol) = 1.2172g\ \ce{N}\) \(mol_H = 7g\ \ce{H_2O} \times (\frac{1\ mol\ H_2O}{18.02\frac{g}{mol}})\times (\frac{2\ mol\ H}{1\ mol\ \ce{H_2O}}) = 0.7769\ mol\) \(mass_H = (1.01\frac{g}{mol})(0.7769 mol) = 0.7847g\ \ce{H}\) From these values, moles of \(O\) can be calculated to be: \(3.5\ gram\ sample = grams_N + grams_H + grams_O\) \(3.5 = 1.2172 g + 0.7847 g + grams_O\) \(grams_O = 1.4981g\) \(mol_O = \frac{1.4981g}{15.9994\frac{g}{mol}} = 0.0936\ mol\) The mole values for each element create a ratio of \(\mathrm{H}_{0.7769}\,\mathrm{N}_{0.0869}\,\mathrm{O}_{0.0936}\), creating an empirical formula \(\ce{H9NO}\) From here, the molecular formula can be established using the molar mass: \(\text{Empirical formula mass} = 9(1.01\frac{g}{mol\ H}) + (14.00\frac{g}{mol\ N}) + (15.9994\frac{g}{mol\ O})\) \(\text{Empirical formula mass} = 39.0894\frac{g}{mol}\) \(\text{Sample Mass} = 78.5\frac{g}{mol}\) \(\frac{78.5\frac{g}{mol}}{39.084\frac{g}{mol}} = 2.00\) \(2(\ce{H9NO}) = \ce{H18N2O2} = \text{Molecular Formula}\) The generic equation for this reaction is \[\ce{C_{x} H_{y} O_{z} + O_{2} -> H_{2}O + CO_{2}} \nonumber \] The molar masses of carbon, hydrogen, and oxygen are 12.01 g/mol, 1.008 g/mol, and 16.00 g/mol respectively. Therefore the molar masses of water and carbon dioxide are \(18.016\frac{g}{mol}\) and \(44.01\frac{g}{mol}\). These values can be used to determine the mass percent of hydrogen in water \[\dfrac {2 (1.008)\; \frac{g}{mol}} {18.016\; \frac{g}{mol}} \times 100 =11.2\% \nonumber \] And the mass percent of carbon in carbon dioxide \[\dfrac{12.01\; \frac{g}{mol}} {44.01\; \frac{g}{mol}} \times 100 =27.3\% \nonumber \] The mass percents can be used to determine the amount of hydrogen and carbon are present. \[3.00\; g \times 0.112 = 0.336\; g\; \text {of Hydrogen} \nonumber \] \[7.33\; g \times 0.273 = 2.001\; g\; \text {of Carbon} \nonumber \] By subtracting the masses of Carbon and Hydrogen from the original mass of the substance, the mass of oxygen present in the substance can be found \[5\; g - (0.336\; g + 2.001\;g) = 2.663\; g\; \text {of Oxygen} \nonumber \] By dividing the masses present by the molar mass, the amount of moles of each element can be determined \[ \dfrac{0.112\; g}{1.008\; \frac{g}{mol}} = 0.333\; mol\; \text {of Hydrogen} \nonumber \] \[ \dfrac{2.001\; g}{12.01\; \frac{g}{mol}} = 0.167\; mol\; \text {of Carbon} \nonumber \] \[\dfrac{2.663\; g}{16.00\; \frac{g}{mol}} = 0.166\; mol\; \text {of Oxygen} \nonumber \] By looking at the amount of moles of each element, it can be noticed that there are equal amounts of carbon and oxygen and twice as much hydrogen as carbon or oxygen. Therefore the ratio of \(\ce{C}:\ce{H}:\ce{O}\) in the substance is \(1:2:1\). However, the molar mass of the substance remains unknown, but it can be found using the information about the freezing point depression. From the formula for freezing point depression \[\Delta T=-i \times k_{f} \times m\nonumber \] \[-0.0605°C = -1 \times 0.512°C\; m^{-1} \times m \nonumber \] \[m=\dfrac{0.0605°C}{0.512\frac{°C}{m}}=0.118\; \frac{mol}{kg}\nonumber \] By multiply by the mass of solvent (water), we can determine the amount of moles that was dissolved in the water \[0.118\; \frac{mol}{kg} \times 0.5\; kg = 0.059\; mol\nonumber \] The last step in finding the molar mass of the substance is to divide the mass dissolved by the moles dissolved \[\dfrac{10.64\;g}{0.059\; mol} = 180.3\; \frac{g}{mol}\nonumber \] Now we can find the formula of the substance. The first step is to find the number of moles of the substance combusted \[\dfrac {5\;g} {180.3\; \frac{g}{mol}} = 0.0277\; mol\nonumber \] We can compare this to the number of carbon moles combusted to determine the number of carbon atoms per molecule of the substance \[\dfrac {0.167\; g \text{ of Carbon}\;} {0.0277\; mol} = 6 \text{ Carbon atoms} \nonumber \] When we apply the number of carbon atoms to the ratio of elements found earlier, the chemical formula comes to be \(\ce{C_{6}H_{12}O_{6}}\) which is glucose. Answer: Glucose | 28,414 | 2,377 |
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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 | 2,378 |
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Boiling solutions always have the potential to "bump", where bubbles vigorously erupt from superheated areas of the solution: areas where the temperature is above the boiling point of the solvent, but gas bubbles have not yet formed due to lack of a nucleation site. Bumping can splash hot material out of a flask: onto your hand or onto a hotplate surface where it might start a fire. Bumping is hazardous, not to mention frightening when a bubble unexpectedly erupts. Several methods can be used to prevent bumping and ensure smooth boiling. Boiling stones (or boiling chips) are small pieces of black porous rock (often silicon carbide) that are added to a solvent or solution. They contain trapped air that bubbles out as a liquid is heated, and have high surface area that can act as nucleation sites for formation of solvent bubbles. They should be added to a cool liquid, not one that is near its boiling point, or a vigorous eruption of bubbles may ensue. When a liquid is brought to a boil using boiling stones, the bubbles tend to originate primarily from the stones (Figure 1.39b). Boiling stones cannot be reused, as after one use, their crevices fill with solvent and they can no longer create bubbles. Boiling stones should not be used when heating concentrated solutions of sulfuric or phosphoric acid, as they may degrade and contaminate the solution. For example, Figure 1.40 shows a Fischer esterification reaction that uses concentrated sulfuric acid. When a stir bar is used for bump prevention, the solution remains colorless (Figure 1.40a). When the same reaction is conducted using a boiling stone, the solution darkens during heating (Figure 1.40b) and eventually turns the entire solution a deep purple-brown color (Figure 1.40c). Besides contaminating the solution, the dark color makes manipulation of the material with a separatory funnel difficult: two layers are present in Figure 1.40d, although it is very difficult to see. "Boiling sticks" (wood splints) are also used to encourage smooth boiling. They are plunged directly into a solvent or solution, and act much the same as boiling stones: they too are highly porous and contain nucleation sites. When a liquid is brought to a boil using a boiling stick, the bubbles tend to originate primarily from the surface of the stick (Figure 1.41b). When choosing between boiling stones and boiling sticks, the main advantage of boiling stones is that they are small and so will fit in any flask. They also absorb very little compound, unlike boiling sticks. The main advantage of boiling sticks is that they can be easily removed from a solution. This is useful in crystallization, as the stick can be easily removed before crystals form (Figure 1.41 c+d). Stir bars (stirring bars, or spin vanes in microscale work, Figure 1.42a) are Teflon-coated magnets that can be made to spin with a magnetic stir plate (Figure 1.42b). Stirring is often used with heating as stirring encourages homogeneity, allowing for liquids to more quickly heat or cool, and disrupts superheated areas. In the context of chemical reactions, stirring also increases the rate of mixing (especially for heterogeneous mixtures) and thus increases reaction rates. A stir bar can be removed from a flask with a magnet, called a " " (Figure 1.42d). If a solution is to be subsequently poured through a funnel and into a separatory funnel, they are also easily removed by the funnel. | 3,442 | 2,379 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/05%3A_The_Second_Law/5.05%3A_Comparing_the_System_and_the_Surroundings |
It is oftentimes important (for reasons that will be discussed in the next section) to calculate both the entropy change of the system as well as that of the surroundings. Depending on the size of the surroundings, they can provide or absorb as much heat as is needed for a process without changing temperature. As such, it is oftentimes a very good approximation to consider the changes to the surroundings as happening isothermally, even though it may not be the case for the system (which is generally smaller.) Consider 18.02 g (1.00 mol) of ice melting at 273 K in a room that is 298 K. Calculate S for the ice, the surrounding room, and of the universe. ( H = 6.01 kJ/mol) For the process under constant pressure: \(q_{ice} = -q_{room}\): \[ q = n \Delta H_{fus} = (1.00 \, mol) (6010 \, J/mol) = 6010 \,J \nonumber \] For the ice: \[ \Delta S_{ice} = \dfrac{q_{ice}}{T_{ice}} = \dfrac{6010\,J}{273\,K} = 22.0\, J/K \nonumber \] For the room: \[ \Delta S_{room} = \dfrac{q_{room}}{T_{room}} = \dfrac{-6010\,J}{298\,K} = -20.2\, J/K \nonumber \] For the universe: \[\begin{align*} \Delta S_{univ} &=\Delta S_{ice} + \Delta S_{room} \\[4pt] &= 22.0 J/K - 20.2\,J/K = 1.8\,J/K \end{align*} \] : \(\Delta S_{univ}\) is positive, which is characteristic of a spontaneous change! A 10.0 g piece of metal (C = 0.250 J/g °C) initially at 95 °C is placed in 25.0 g of water initially at 15 °C in an insulated container. Calculate the final temperature of the metal and water once the system has reached thermal equilibrium. Also, calculate the entropy change for the metal, the water, and the entire system. Heat will be transferred from the hot metal to the cold water. Since it has nowhere else to go, the final temperature can be calculated from the expression \[q_w = -q_m \nonumber \] where \(q_w\) is the heat absorbed by the water, and \(q_m\) is the heat lost by the metal. And since \[q = mC\Delta T \nonumber \] it follows that \[ (25\,g)(4.184\,J/g\, °C)) (T_f - 15 \,°C) = -(10.0\,g)(0.250 \, J/g\, °C))(T_f-95\,J/g\, °C) \nonumber \] A bit of algebra determines the final temperature to be: \[T_f = 16.9 \, °C. \nonumber \] To get the entropy changes, use the expression: \[\Delta S= m C_p \ln \left( \dfrac{T_f}{T_i} \right) \nonumber \] So, for the water: \[ \begin{align*} \Delta S_{water} &= (25\,g)(4.184\,J/g\, °C)) \ln \left( \dfrac{289.9\,K}{288\,K} \right) \\[4pt] &= 0.689\, J/K \end{align*} \] And for the metal: \[ \begin{align*} \Delta S_{metal} &= (10.0\,g)(0.250\,J/g\, °C)) \ln \left( \dfrac{289.9\,K}{368\,K} \right) \\[4pt] &= - 0.596 \, J/K \end{align*} \] For the system: \[ \begin{align*}\Delta S_{sys} &= \Delta S_{water} + \Delta S_{metal} \\[4pt] &= 0.689\, J/K + - 0.596 \, J/K = 0.093 \, J/K \end{align*} \] The total entropy change is positive, suggesting that this will be a spontaneous process. This should make some sense since one expects heat to flow from the hot metal to the cool water rather than the other way around. Also, note that the sign of the entropy change is positive for the part of the system that is absorbing the heat, and negative for the part losing the heat. In summary, \(\Delta S\) can be calculated for a number of pathways fairly conveniently. And \[\Delta S_{univ} = \Delta S_{sys} + \Delta S_{surr}. \nonumber \] This calculation is important as \(\Delta S_{univ}\) provides the criterion for spontaneity for which we were searching from the outset. This also suggests a new way to state the second law: The entropy of the universe increases in any spontaneous change. If we think of “the direction of spontaneous” to be the natural direction of chance, we can see that entropy and the second law are tied inexorably with the natural direction of the flow of time. Basically, we can expect the entropy of the universe to continue to increase as time flows into the future. We can overcome this natural tendency to greater entropy by doing work on a system. This is why it requires such great effort, for example, to straighten a messy desk, but little effort for the desk to get messy over time. The Second Law can be summed up in a very simple mathematical expression called the . \[ \Delta S_{universe} \ge 0 \nonumber \] which must be true for any spontaneous process. It is not the most convenient criterion for spontaneity, but it will do for now. In the next chapter, we will derive a criterion which is more useful to us as chemists, who would rather focus on the system itself rather than both the system and its surroundings. Another statement of the Clausius theorem is \[\oint \dfrac{dq}{T} \ge 0 \nonumber \] with the only condition of the left hand side equaling zero is if the system transfers all heat reversibly. | 4,717 | 2,380 |
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 | 2,381 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Oxidative_Addition_and_Reductive_Elimination/OA5._Coupling_Reactions |
Oxidative addition and reductive elimination are key steps in industrial catalysis. For example, both steps are featured in palladium-catalyzed cross-coupling reactions, the subject of the 2010 Nobel Prize in Chemistry. The prize was awarded to Richard Heck of the University of Delaware, Ei-Ichi Negishi of Purdue University and Akira Suzuki of Hokkaido University. With these reactions, workers in a variety of fields can make molecules that otherwise would be quite difficult to make. These molecules in turn may be important pharmaceuticals or useful compounds for electronic displays in computers and other devices, to name just a couple. The involves catalytic addition of alkylzinc nucleophiles to vinyl halides. The catalytic cycle believed to operate for this reaction involves the crucial oxidative addition of the vinyl halide to the metal. The alkylzinc probably delivers the alkyl nucleophile to the metal via more conventional nucleophilic substitution. Once both pieces are both on the metal, they can reductively eliminate together. Explain why the vinyl halide pictured in the example above would not react directly with the alkylzinc reagent. Label each step of the catalytic cycle for the Negishi reaction with the appropriate term (oxidative addition, etc). Draw the mechanism for the Negishi reaction using curved arrow notation. There are many other examples of coupling reactions in organic synthesis. The is somewhat similar to the Negishi reaction. The involves activation of a vinylic or aryl C-H bond. By analogy with the Negishi reaction, propose a catalytic cycle for the Suzuki reaction. Propose a catalytic cycle for the Heck reaction. Read more about these Nobel Prize-winning reactions at the Nobel Prize website. , | 1,767 | 2,382 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Principles_of_Modern_Chemistry_(Oxtoby_et_al.)/Unit_3%3A_The_States_of_Matter/11%3A_Solutions/11.7%3A_Colloidal_Suspensions |
Suspensions and colloids are two common types of mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures. A suspension is a heterogeneous mixture of particles with diameters of about 1 µm (1000 nm) that are distributed throughout a second phase. Common suspensions include paint, blood, and hot chocolate, which are solid particles in a liquid, and aerosol sprays, which are liquid particles in a gas. If the suspension is allowed to stand, the two phases will separate, which is why paints must be thoroughly stirred or shaken before use. A colloid is also a heterogeneous mixture, but the particles of a colloid are typically smaller than those of a suspension, generally in the range of 2 to about 500 nm in diameter. Colloids include fog and clouds (liquid particles in a gas), milk (solid particles in a liquid), and butter (solid particles in a solid). Other colloids are used industrially as catalysts. Unlike in a suspension, the particles in a colloid do not separate into two phases on standing. The only combination of substances that cannot produce a suspension or a colloid is a mixture of two gases because their particles are so small that they always form true solutions. The properties of suspensions, colloids, and solutions are summarized in Table \(\Page {1}\). Colloids were first characterized in about 1860 by Thomas Graham, who also gave us Graham’s law of diffusion and effusion. Although some substances, such as starch, gelatin, and glue, appear to dissolve in water to produce solutions, Graham found that they diffuse very slowly or not at all compared with solutions of substances such as salt and sugar. Graham coined the word colloid (from the Greek kólla, meaning “glue”) to describe these substances, as well as the words sol and gel to describe certain types of colloids in which all of the solvent has been absorbed by the solid particles, thus preventing the mixture from flowing readily, as we see in Jell-O. Two other important types of colloids are aerosols, which are dispersions of solid or liquid particles in a gas, and emulsions, which are dispersions of one liquid in another liquid with which it is immiscible. Colloids share many properties with solutions. For example, the particles in both are invisible without a powerful microscope, do not settle on standing, and pass through most filters. However, the particles in a colloid scatter a beam of visible light, a phenomenon known as the Tyndall effect,The effect is named after its discoverer, John Tyndall, an English physicist (1820–1893). whereas the particles of a solution do not. The Tyndall effect is responsible for the way the beams from automobile headlights are clearly visible from the side on a foggy night but cannot be seen from the side on a clear night. It is also responsible for the colored rays of light seen in many sunsets, where the sun’s light is scattered by water droplets and dust particles high in the atmosphere. An example of the Tyndall effect is shown in Figure \(\Page {1}\). Although colloids and suspensions can have particles similar in size, the two differ in stability: the particles of a colloid remain dispersed indefinitely unless the temperature or chemical composition of the dispersing medium is changed. The chemical explanation for the stability of colloids depends on whether the colloidal particles are hydrophilic or hydrophobic. Most proteins, including those responsible for the properties of gelatin and glue, are hydrophilic because their exterior surface is largely covered with polar or charged groups. Starch, a long-branched polymer of glucose molecules, is also hydrophilic. A hydrophilic colloid particle interacts strongly with water, resulting in a shell of tightly bound water molecules that prevents the particles from aggregating when they collide. Heating such a colloid can cause aggregation because the particles collide with greater energy and disrupt the protective shell of solvent. Moreover, heat causes protein structures to unfold, exposing previously buried hydrophobic groups that can now interact with other hydrophobic groups and cause the particles to aggregate and precipitate from solution. When an egg is boiled, for example, the egg white, which is primarily a colloidal suspension of a protein called albumin, unfolds and exposes its hydrophobic groups, which aggregate and cause the albumin to precipitate as a white solid. In some cases, a stable colloid can be transformed to an aggregated suspension by a minor chemical modification. Consider, for example, the behavior of hemoglobin, a major component of red blood cells. Hemoglobin molecules normally form a colloidal suspension inside red blood cells, which typically have a “donut” shape and are easily deformed, allowing them to squeeze through the capillaries to deliver oxygen to tissues. In a common inherited disease called sickle-cell anemia, one of the amino acids in hemoglobin that has a hydrophilic carboxylic acid side chain (glutamate) is replaced by another amino acid that has a hydrophobic side chain (valine). Under some conditions, the abnormal hemoglobin molecules can aggregate to form long, rigid fibers that cause the red blood cells to deform, adopting a characteristic sickle shape that prevents them from passing through the capillaries (Figure \(\Page {2}\)). The reduction in blood flow results in severe cramps, swollen joints, and liver damage. Until recently, many patients with sickle-cell anemia died before the age of 30 from infection, blood clots, or heart or kidney failure, although individuals with the sickle-cell genetic trait are more resistant to malaria than are those with “normal” hemoglobin. Aggregation and precipitation can also result when the outer, charged layer of a particle is neutralized by ions with the opposite charge. In inland waterways, clay particles, which have a charged surface, form a colloidal suspension. High salt concentrations in seawater neutralize the charge on the particles, causing them to precipitate and form land at the mouths of large rivers, as seen in the satellite view in Figure \(\Page {3}\). Charge neutralization is also an important strategy for precipitating solid particles from gaseous colloids such as smoke, and it is widely used to reduce particulate emissions from power plants that burn fossil fuels. Emulsions are colloids formed by the dispersion of a hydrophobic liquid in water, thereby bringing two mutually insoluble liquids, such as oil and water, in close contact. Various agents have been developed to stabilize emulsions, the most successful being molecules that combine a relatively long hydrophobic “tail” with a hydrophilic “head”: Examples of such emulsifying agents include soaps, which are salts of long-chain carboxylic acids, such as sodium stearate \(\ce{[CH_3(CH_2)_{16}CO_2−Na^{+}]}\), and detergents, such as sodium dodecyl sulfate \(\ce{[CH_3(CH_2)_{11}OSO_3−Na^{+}]}\), whose structures are as follows: When you wash your laundry, the hydrophobic tails of soaps and detergents interact with hydrophobic particles of dirt or grease through dispersion forces, dissolving in the interior of the hydrophobic particle. The hydrophilic group is then exposed at the surface of the particle, which enables it to interact with water through ion–dipole forces and hydrogen bonding. This causes the particles of dirt or grease to disperse in the wash water and allows them to be removed by rinsing. Similar agents are used in the food industry to stabilize emulsions such as mayonnaise. A related mechanism allows us to absorb and digest the fats in buttered popcorn and French fries. To solubilize the fats so that they can be absorbed, the gall bladder secretes a fluid called bile into the small intestine. Bile contains a variety of bile salts, detergent-like molecules that emulsify the fats. Detergents and soaps are surprisingly soluble in water in spite of their hydrophobic tails. The reason for their solubility is that they do not, in fact, form simple solutions. Instead, above a certain concentration they spontaneously form micelles, which are spherical or cylindrical aggregates that minimize contact between the hydrophobic tails and water. In a micelle, only the hydrophilic heads are in direct contact with water, and the hydrophobic tails are in the interior of the aggregate (Figure \(\Page {4a}\)). A large class of biological molecules called phospholipids consists of detergent-like molecules with a hydrophilic head and two hydrophobic tails, as can be seen in the molecule of phosphatidylcholine. The additional tail results in a cylindrical shape that prevents phospholipids from forming a spherical micelle. Consequently, phospholipids form bilayers, extended sheets consisting of a double layer of molecules. As shown in Figure \(\Page {4b}\), the hydrophobic tails are in the center of the bilayer, where they are not in contact with water, and the hydrophilic heads are on the two surfaces, in contact with the surrounding aqueous solution. A cell membrane is essentially a mixture of phospholipids that form a phospholipid bilayer. One definition of a cell is a collection of molecules surrounded by a phospholipid bilayer that is capable of reproducing itself. The simplest cells are bacteria, which consist of only a single compartment surrounded by a single membrane. Animal and plant cells are much more complex, however, and contain many different kinds of compartments, each surrounded by a membrane and able to carry out specialized tasks. A suspension is a heterogeneous mixture of particles of one substance distributed throughout a second phase; the dispersed particles separate from the dispersing phase on standing. In contrast, the particles in a colloid are smaller and do not separate on standing. A colloid can be classified as a sol, a dispersion of solid particles in a liquid or solid; a gel, a semisolid sol in which all of the liquid phase has been absorbed by the solid particles; an aerosol, a dispersion of solid or liquid particles in a gas; or an emulsion, a dispersion of one liquid phase in another. A colloid can be distinguished from a true solution by its ability to scatter a beam of light, known as the Tyndall effect. Hydrophilic colloids contain an outer shell of groups that interact favorably with water, whereas hydrophobic colloids have an outer surface with little affinity for water. Emulsions are prepared by dispersing a hydrophobic liquid in water. In the absence of a dispersed hydrophobic liquid phase, solutions of detergents in water form organized spherical aggregates called micelles. Phospholipids are a class of detergent-like molecules that have two hydrophobic tails attached to a hydrophilic head. A bilayer is a two-dimensional sheet consisting of a double layer of phospholipid molecules arranged tail to tail with a hydrophobic interior and a hydrophilic exterior. Cells are collections of molecules that are surrounded by a phospholipid bilayer called a cell membrane and are able to reproduce themselves. | 11,122 | 2,383 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Electrophilic_Aromatic_Substitution/AR4._Activation_and_Deactivation |
Because the benzene acts as a nucleophile in electrophilic aromatic substitution, substituents that make the benzene more electron-rich can accelerate the reaction. Substituents that make the benzene moor electron-poor can retard the reaction. In the mid-twentieth century, physical organic chemists including Christopher Ingold conducted a number of kinetic studies on electrophilic aromatic substitution reactions. In table 1, you can see that some substituents confer a rate of reaction that is much higher than that of benzene (R = H). Phenol, C H OH, undergoes nitration a thousand times faster than benzene does. Nitrobenzene, C H NO , undergoes the reaction millions of times more slowly. These observations are consistent with the role of the aromatic as a nucleophile in this reaction. Substituents that draw electron density away from the aromatic ring slow the reaction down. These groups are called deactivating groups in this reaction. Substituents that readily donate electron desnity to the ring, or that effectively stabilize the cationic intermediate, promote the reaction. These groups are called activating groups in this reaction. The roles of these groups are related to their electronic interactions with the electrons in the ring. Some groups might be π-donors, providing additional electron density to the benzene ring via conjugation. Other groups may be π-acceptors, drawing electron density away from the ring via conjugation. Still others may be σ-acceptors, drawing electron density away from the ring via a simple inductive effect which arises from the electronegativity of a substituent. In some cases, there may be multiple effects, and the overall influence of the substituents is determined by the balance of the effects. One effect may be stronger in one case than the other, so it wins out in one case and loses in another. Explain why a fluorine atom would slow down an electrophilic substitution on an adjacent benzene ring. Show, with structures, how the OH group in phenol makes the benzene ring more nucleophilic. Show, with structures, how the CO Et group makes the benzene ring less nucleophilic. Show, with structures, how a methyl group stabilizes the cationic intermediate during a nitration reaction. In general, deactivating groups fall into two classes. Π-acceptors, such as carbonyls, if placed directly adjacent to the aromatic ring, slow down the reaction. Highly electronegative atoms, typically halogens, attached directly to the aromatic ring also slow down the reaction. Activating groups also fall into two categories. Π-donors, typically oxygen or nitrogen atoms, accelerate the reaction. This observation is true even though these atoms are also highly electronegative. Alkyl groups attached to the aromatic ring also accelerate the reaction. Note that halogens are also π-donors, but they are less effective in this regard than nitrogen or oxygen. That's because nitrogen and oxygen are similar in size to carbon, and they form effective π-overlap with the adjacent carbon on the benzene ring. In the halogens, electronegativity wins by default, because their π-donating effects are not good enough to make them activators. Conversely, nitrogen and oxygen are both very electronegative, but their exceptional π-donating ability makes them activators rather than deactivators. Thus, in many cases, there is a subtle balance between activating and directing effects. In some cases, the activating effect is more pronounced, and that is what is observed. In other cases, it is the deactivating effect that wins out. Alkyl groups behave almost as sigma donors, although that may be a misleading way to think about them. Instead, their mild activating effect arises from hyperconjugation, in which a pair of C-H bonding electrons can weakly interact with a cationic site, providing a little extra stability to the cation. One of the groups in the table, CH Cl, does not quite fit the general rules. It is very slightly deactivating. Explain why this group acts in this way. Predict whether each of the following groups would be activating or deactivating towards electrophilic aromatic substitution. a) NH b) CN c) OCH d) SMe e) C(O)CH Explain the trend in activating effects among the different halogens. , | 4,278 | 2,384 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.07%3A_Essential_Skills_7 |
Previous Essential Skills sections introduced many of the mathematical operations you need to solve chemical problems. We now introduce the quadratic formula, a mathematical relationship involving sums of powers in a single variable that you will need to apply to solve some of the problems in this chapter. Mathematical expressions that involve a sum of powers in one or more variables (e.g., ) multiplied by coefficients (such as ) are called . Polynomials of a single variable have the general form \[a_nx^n + ... + a_2x^2 + a_1x + a_0 \tag{15.7.1} \] The highest power to which the variable in a polynomial is raised is called its . Thus the polynomial shown here is of the order. For example, if were 3, the polynomial would be third order. A is a second-order polynomial equation in a single variable : \[ax^2 + bx + c = 0 \tag{15.7.2} \] According to the fundamental theorem of algebra, a second-order polynomial equation has two solutions—called —that can be found using a method called . In this method, we solve for by first adding − to both sides of the quadratic equation and then divide both sides by : \[x^2+\dfrac{bx}{a} =−\dfrac{c}{a} \tag{15.7.3} \] We can convert the left side of this equation to a perfect square by adding /4 , which is equal to ( /2 ) : Left side: \[x^2+\dfrac{b}{a}x+\dfrac{b^2}{4a^2}=(x+\dfrac{b}{2a})^2 \tag{15.7.4} \] Having added a value to the left side, we must now add that same value, b ⁄ 4a , to the right side: \[(x+\dfrac{b}{2a})^2=−\dfrac{c}{a}+\dfrac{b^2}{4a^2} \tag{15.7.5} \] The common denominator on the right side is 4 . Rearranging the right side, we obtain the following: \[(x+\dfrac{b}{2a})^2=\dfrac{b^2−4ac}{4a^2} \tag{15.7.6} \] Taking the square root of both sides and solving for , This equation, known as the , has two roots: Thus we can obtain the solutions to a quadratic equation by substituting the values of the coefficients ( , , ) into the quadratic formula. When you apply the quadratic formula to obtain solutions to a quadratic equation, it is important to remember that one of the two solutions may not make sense or neither may make sense. There may be times, for example, when a negative solution is not reasonable or when both solutions require that a square root be taken of a negative number. In such cases, we simply discard any solution that is unreasonable and only report a solution that is reasonable. Skill Builder ES1 gives you practice using the quadratic formula. Use the quadratic formula to solve for in each equation. Report your answers to three significant figures. | 2,599 | 2,385 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/21%3A_Chemistry_of_The_Main-Group_Elements_I/21.1%3A_Periodic_Trends_and_Charge_Density |
As we begin our summary of periodic trends, recall that the single most important unifying principle in understanding the chemistry of the elements is the systematic increase in atomic number, accompanied by the orderly filling of atomic orbitals by electrons, which leads to periodicity in such properties as atomic and ionic size, ionization energy, electronegativity, and electron affinity. The same factors also lead to periodicity in valence electron configurations, which for each group results in similarities in oxidation states and the formation of compounds with common stoichiometries. The most important periodic trends in atomic properties are summarized in Figure \(\Page {1}\). Recall that these trends are based on periodic variations in a single fundamental property, the (\(Z_{eff}\)), which increases from left to right and from top to bottom in the periodic table. The diagonal line in Figure \(\Page {1}\) separates the metals (to the left of the line) from the nonmetals (to the right of the line). Because metals have relatively low electronegativities, they tend to lose electrons in chemical reactions to elements that have relatively high electronegativities, forming compounds in which they have positive oxidation states. Conversely, nonmetals have high electronegativities, and they therefore tend to gain electrons in chemical reactions to form compounds in which they have negative oxidation states. The semimetals lie along the diagonal line dividing metals and nonmetals. It is not surprising that they tend to exhibit properties and reactivities intermediate between those of metals and nonmetals. Because the elements of groups 13, 14, and 15 span the diagonal line separating metals and nonmetals, their chemistry is more complex than predicted based solely on their valence electron configurations. The chemistry of the second-period element of each group (n = 2: Li, Be, B, C, N, O, and F) differs in many important respects from that of the heavier members, or congeners, of the group. Consequently, the elements of the third period (n = 3: Na, Mg, Al, Si, P, S, and Cl) are generally more representative of the group to which they belong. The anomalous chemistry of second-period elements results from three important characteristics: small radii, energetically unavailable d orbitals, and a tendency to form pi (π) bonds with other atoms. In contrast to the chemistry of the second-period elements, the chemistry of the third-period elements is more representative of the chemistry of the respective group. Due to their small radii, second-period elements have electron affinities that are less negative than would be predicted from general periodic trends. When an electron is added to such a small atom, increased electron–electron repulsions tend to destabilize the anion. Moreover, the small sizes of these elements prevent them from forming compounds in which they have more than four nearest neighbors. Thus BF forms only the four-coordinate, tetrahedral BF ion, whereas under the same conditions AlF forms the six-coordinate, octahedral AlF ion. Because of the smaller atomic size, simple binary ionic compounds of second-period elements also have more covalent character than the corresponding compounds formed from their heavier congeners. The very small cations derived from second-period elements have a high charge-to-radius ratio and can therefore polarize the filled valence shell of an anion. As such, the bonding in such compounds has a significant covalent component, giving the compounds properties that can differ significantly from those expected for simple ionic compounds. As an example, LiCl, which is partially covalent in character, is much more soluble than NaCl in solvents with a relatively low dielectric constant, such as ethanol (ε = 25.3 versus 80.1 for H O). Because d orbitals are never occupied for principal quantum numbers less than 3, the valence electrons of second-period elements occupy 2s and 2p orbitals only. The energy of the 3d orbitals far exceeds the energy of the 2s and 2p orbitals, so using them in bonding is energetically prohibitive. Consequently, electron configurations with more than four electron pairs around a central, second-period element are simply not observed. You may recall that the role of d orbitals in bonding in main group compounds with coordination numbers of 5 or higher remains somewhat controversial. In fact, theoretical descriptions of the bonding in molecules such as SF have been published without mentioning the participation of d orbitals on sulfur. Arguments based on d-orbital availability and on the small size of the central atom, however, predict that coordination numbers greater than 4 are unusual for the elements of the second period, which is in agreement with experimental results. One of the most dramatic differences between the lightest main group elements and their heavier congeners is the tendency of the second-period elements to form species that contain multiple bonds. For example, N is just above P in group 15: N contains an N≡N bond, but each phosphorus atom in tetrahedral P forms three P–P bonds. This difference in behavior reflects the fact that within the same group of the periodic table, the relative energies of the π bond and the sigma (σ) bond differ. A C=C bond, for example, is approximately 80% stronger than a C–C bond. In contrast, an Si=Si bond, with less p-orbital overlap between the valence orbitals of the bonded atoms because of the larger atomic size, is only about 40% stronger than an Si–Si bond. Consequently, compounds that contain both multiple and single C to C bonds are common for carbon, but compounds that contain only sigma Si–Si bonds are more energetically favorable for silicon and the other third-period elements. The refers to the empirical observation that the heavier elements of groups 13–17 often have oxidation states that are lower by 2 than the maximum predicted for their group. For example, although an oxidation state of +3 is common for group 13 elements, the heaviest element in group 13, thallium (Tl), is more likely to form compounds in which it has a +1 oxidation state. There appear to be two major reasons for the inert-pair effect: increasing ionization energies and decreasing bond strengths. In moving down a group in the p-block, increasing ionization energies and decreasing bond strengths result in an inert-pair effect. The ionization energies increase because filled (n − 1)d or (n − 2)f subshells are relatively poor at shielding electrons in ns orbitals. Thus the two electrons in the ns subshell experience an unusually high effective nuclear charge, so they are strongly attracted to the nucleus, reducing their participation in bonding. It is therefore substantially more difficult than expected to remove these ns electrons, as shown in Table \(\Page {1}\) by the difference between the first ionization energies of thallium and aluminum. Because Tl is less likely than Al to lose its two ns electrons, its most common oxidation state is +1 rather than +3. Source of data: John A. Dean, Lange’s Handbook of Chemistry, 15th ed. (New York: McGraw-Hill, 1999). Going down a group, the atoms generally became larger, and the overlap between the valence orbitals of the bonded atoms decreases. Consequently, bond strengths tend to decrease down a column. As shown by the M–Cl bond energies listed in Table \(\Page {1}\), the strength of the bond between a group 13 atom and a chlorine atom decreases by more than 30% from B to Tl. Similar decreases are observed for the atoms of groups 14 and 15. The net effect of these two factors—increasing ionization energies and decreasing bond strengths—is that in going down a group in the p-block, the additional energy released by forming two additional bonds eventually is not great enough to compensate for the additional energy required to remove the two ns electrons. Based on the positions of the group 13 elements in the periodic table and the general trends outlined in this section, positions of elements in the periodic table classification, oxidation-state stability, and chemical reactivity From the position of the diagonal line in the periodic table separating metals and nonmetals, classify the group 13 elements. Then use the trends discussed in this section to compare their relative stabilities and chemical reactivities. Based on the positions of the group 14 elements C, Si, Ge, Sn, and Pb in the periodic table and the general trends outlined in this section, The most important unifying principle in describing the chemistry of the elements is that the systematic increase in atomic number and the orderly filling of atomic orbitals lead to periodic trends in atomic properties. The most fundamental property leading to periodic variations is the effective nuclear charge (Z ). Because of the position of the diagonal line separating metals and nonmetals in the periodic table, the chemistry of groups 13, 14, and 15 is relatively complex. The second-period elements (n = 2) in each group exhibit unique chemistry compared with their heavier congeners because of their smaller radii, energetically unavailable d orbitals, and greater ability to form π bonds with other atoms. Increasing ionization energies and decreasing bond strengths lead to the inert-pair effect, which causes the heaviest elements of groups 13–17 to have a stable oxidation state that is lower by 2 than the maximum predicted for their respective groups. | 9,535 | 2,386 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/19%3A_The_First_Law_of_Thermodynamics/19.08%3A_Pressure-Volume_Work |
An important point is that pressure-volume work \(-PdV\) is only one kind of work. It is the important one for gases but for most other systems we are interested in other kinds of work (e.g. electrical work in a battery). A good way to measure \(ΔU\)'s is to make sure there are no work terms at all. If so: \[ΔU_{no work} = q +w = q+0 = q \nonumber \] However, this means that the \(-PdV\) volume work term should also be zero and this implies we must keep volumes the same. That can actually be hard. Therefore we define at new state function ENTHALPY \[H ≡ U + PV \nonumber \] ( The \(≡\) symbol is used to show that this equality is actually a definition.) If we differentiate we get: \[dH = dU + d(PV) = dU + PdV + VdP \nonumber \] We know that under reversible conditions we have \[dU = δw +δq = -PdV + δq \nonumber \] (+ other work terms that we assume zero) Thus, \[dH = -PdV +δq + PdV + VdP \nonumber \] \[dH = δq + VdP \nonumber \] That means that as long as there is no other work and we keep the pressure constant: \[ΔH = q_P \nonumber \] instead of \[ΔU= q_V \nonumber \] Working at constant \(P\) is a lot easier to do than at constant \(V\). This means that the enthalpy is a much easier state function to deal with than the energy U. For example when we melt ice volumes change whether we like or not, but at long as the weather does not change too much pressure is constant. So if we measure how much heat we need to add to melt a mole of ice we get the molar heat of fusion: Such enthalpies are measured and tabulated. In this case the volume change is actually quite small, as it usually is for condensed matter. Only if we are dealing with gases is the difference between enthalpy and energy really important So, \(U ~ H\) for condensed matter, but \(U\) and \(H\) differ for gases. A good example of this is the difference between the heat capacity at constant V and at constant P. For most materials there is not much of a difference, but for an ideal gas we have \[C_p = C-V + nR \nonumber \] Needless to say that the heat capacity is a path function: it depends on what you keep constant. | 2,130 | 2,388 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.25%3A_Boiling-Point_Elevation_and_Freezing-Point_Depression |
We often encounter solutions in which the solute has such a low vapor pressure as to be negligible. In such cases the vapor above the solution consists only of solvent molecules and the vapor pressure is always lower than that of the pure solvent. Consider, for example, the solution obtained by dissolving 0.020 mol sucrose (C H O ) in 0.980 mol H O at 100°C. The sucrose will contribute nothing to the vapor pressure, while we can expect the water vapor, by Raoult’s law, to contribute \[P_{\text{H}_2 \text{O}}=P^*_{\text{H}_2\text{O}} * x_{ \text{H}_2 \text{O}} = 760 \text{mmHg} * 0.980 = 744.8 \text{mmHg} \nonumber \] We would thus expect the vapor pressure to be 744.8 mmHg, in reasonable agreement with the observed value of 743.3 mmHg. A direct consequence of the lowering of the vapor pressure by a nonvolatile solute is of the solution relative to that of the solvent. We can see why this is so by again using the sucrose solution as an example. At 100°C this solution has a vapor pressure which is lower than atmospheric pressure, and therefore it will not boil. In order to increase the vapor pressure from 743.3 to 760 mmHg so that boiling will occur, we need to raise the temperature. Experimentally we find that the temperature must be raised to 100.56°C. We say that the Δ is 0.56 K. The image below demonstrates a solute dissolved in water. As you can see in the image, the presence of the solute interferes with the ability of the water to "escape" into the gaseous phase. This interference can be physical (as seen in the oversimplified image above) or it can be due to more complex factors, like intermolecular forces. Therefore, the solution has a lower vapor pressure for any given temperature. Thus, the mixture will require a higher temperature overall than a pure solution in order to boil, which is evident as a higher boiling boiling point. A second result of the lowering of the vapor pressure is a depression of the freezing point of the solution. Any aqueous solution of a nonvolatile solute, for example, will have a vapor pressure at 0°C which is less than the vapor pressure of ice (0.006 atm or 4.6 mmHg) at this temperature. Accordingly, ice and the aqueous solution will not be in equilibrium. If the temperature is lowered, though, the vapor pressure of the ice decreases more rapidly than that of the solution and a temperature is soon reached when both ice and the solution have the same vapor pressure. Since both phases are now in equilibrium, this lower temperature is also the freezing point of the solution. In the case of the sucrose solution of mole fraction 0.02 described above, we find experimentally that the freezing point is –2.02°C. We say that the Δ is 2.02 K. The image below demonstrates how this freezing point depression works on a very simple level. The salt (purple) in the water prevents the rigid, ordered arrangement of a solid from being achieved. The particles of solute block the crystallization of water, at least up until a certain point, causing a depression in the freezing point of the water. The depression of the freezing point of water by a solute explains why the sea does not freeze at 0°C. Because of its high salt content the sea has a freezing point of –2.2°C. If the sea froze at 0°C, larger stretches of ocean would turn into ice and the climate of the earth would be very different. We can now also understand why we add ethylene glycol, CH OHCH OH, to water in a car radiator in winter. Without any additive the water would freeze at 0°C and the resulting increase in volume would crack the radiator. Since ethylene glycol is very soluble in water, it can form a solution with a freezing point low enough to prevent freezing even on the coldest winter day. Both the freezing-point depression and the boiling-point elevation of a solution were once important methods for determining the molar mass of a newly prepared compound. Nowadays a is usually used for this purpose, often on an assembly-line basis. Many chemists send samples of newly prepared compounds to a laboratory specializing in these determinations in much the same way as a medical doctor will send a sample of your blood to a laboratory for analysis. The reason we can use the boiling-point elevation and the freezing-point depression to determine the molar mass of the solute is that both properties are . The actual relationship, which we will not derive, is \[x_{A}=\frac{\Delta H_{m}}{RT^{\text{2}}}\text{ }\Delta T \nonumber \] where is the mole fraction of the solute and Δ is the boiling-point elevation or freezing-point depression. indicates either the boiling point or freezing point of the pure solvent, and Δ is the molar enthalpy of vaporization or fusion, whichever is appropriate. This relationship tells us that we can measure the mole fraction of the solute in a solution merely by finding its boiling point or freezing point of the solution. A solution of sucrose in water boils at 100.56°C and freezes at –2.02°C. Calculate the mole fraction of the solution from each temperature. For boiling we have, from the Table of Molar Enthalpies of Fusion and Vaporization, \[\triangle H_m = 40.7 \dfrac{\text{kJ}}{\text{mol}} \nonumber \] and \(T=373.15 \text{K}\) As in previous examples the units of should be compatible with the other units appearing in the equation. In this case since Δ is given in units of kJ mol , = 8.314 J K mol is most appropriate. Since ΔT = 0.56 K, we have \[x_{\text{sucrose}}=\frac{\Delta H_{m}}{RT^{\text{2}}}\Delta T=\frac{\text{40}\text{.7 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}\text{ }\times \text{ 0}\text{.56 K}}{\text{8}\text{.314 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{ (373}\text{.15)}^{\text{2}}}=\text{0}\text{.0197} \nonumber \] \[\triangle H_m = 6.01 \dfrac{kJ}{mol} \nonumber \] \(T=273.15 \text{K}\) \(\triangle T = 2.02 \text{K}\) so that \[x_{\text{sucrose}}=\dfrac{\text{6}\text{.01 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}\text{ }\times \text{ 2}\text{.02 K}}{\text{8}\text{.314 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{ }\times \text{ (273}\text{.15 K)}^{\text{2}}}=\text{0}\text{.0196} \nonumber \] The two values are in reasonable agreement. Once we know the mole fraction of the solute, its molar mass is easily calculated from the mass composition of the solution. 33.07 g sucrose is dissolved in 85.27 g H O. The resulting solution freezes at –2.02°C. Calculate the molar mass of sucrose. Since the freezing point of the solution is the same as in part of Example 1, the mole fraction must be the same. Thus \[x_{\text{sucrose}}=\text{0}\text{.0196}=\dfrac{n_{\text{sucrose}}}{n_{\text{sucrose}}\text{ + }n_{\text{H}_{\text{2}}\text{O}}} \nonumber \] Furthermore \[\text{ }n_{\text{H}_{\text{2}}\text{O}}=\dfrac{\text{85}\text{.27 g}}{\text{18}\text{.02 g mol}^{-\text{1}}}=\text{4}\text{.732 mol} \nonumber \] Thus \[\text{0}\text{.0196}=\dfrac{n_{\text{sucrose}}}{n_{\text{sucrose}}\text{ + 4}\text{.732 mol}} \nonumber \] From which \[\text{ }M_{\text{sucrose}}=\dfrac{\text{33}\text{.07 g}}{\text{0}\text{.0946 mol}}=\text{350 g mol}^{-\text{1}} \nonumber \] | 7,233 | 2,389 |
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As we have talked about using Lewis structures to depict the bonding in organic compounds, we have been very vague in our language about the actual nature of the chemical bonds themselves. We know that a covalent bond involves the ‘sharing’ of a pair of electrons between two atoms - but how does this happen, and how does it lead to the formation of a bond holding the two atoms together? In this chapter, the is introduced to describe bonding in organic molecules. In this model, bonds are considered to form from the overlapping of two atomic orbitals on different atoms, each orbital containing a single electron. In looking at simple inorganic molecules such as H or HF, our present understanding of s and p atomic orbitals will suffice. In order to explain the bonding in organic molecules, however, we will need to introduce the concept of . Valence bond theory is adequate for describing many aspects of organic structure. In some cases, however, chemists need to use a different model, called , to talk about covalent bonds in which electrons are shared not just between two atoms, but between several, or even over an entire molecule. The simplest case to consider is the hydrogen molecule, H . When we say that the two electrons from each of the hydrogen atoms are shared to form a covalent bond between the two atoms, what we mean in valence bond theory terms is that the two spherical 1 orbitals overlap, allowing the two electrons to form a pair within the two overlapping orbitals. These two electrons are now attracted to the positive charge of of the hydrogen nuclei, with the result that they serve as a sort of ‘chemical glue’ holding the two nuclei together. How far apart are the two nuclei? That is a very important issue to consider. If they are too far apart, their respective 1 orbitals cannot overlap, and thus no covalent bond can form - they are still just two separate hydrogen atoms. As they move closer and closer together, orbital overlap begins to occur, and a bond begins to form. This lowers the potential energy of the system, as new, positive-negative electrostatic interactions become possible between the nucleus of one atom and the electron of the second. But something else is happening at the same time: as the atoms get closer, the positive-positive interaction between the two nuclei also begins to increase. At first this repulsion is more than offset by the attraction between nuclei and electrons, but at a certain point, as the nuclei get even closer, the repulsive forces begin to overcome the attractive forces, and the potential energy of the system rises quickly. When the two nuclei are ‘too close’, we have a very unstable, high-energy situation. There is a defined optimal distance between the nuclei in which the potential energy is at a minimum, meaning that the combined attractive and repulsive forces add up to the greatest overall attractive force. This optimal internuclear distance is the . For the H molecule, this distance is 74 x 10 meters, or 0.74 Å (Å means angstrom, or 10 meters). Likewise, the difference in potential energy between the lowest state (at the optimal internuclear distance) and the state where the two atoms are completely separated is called the . For the hydrogen molecule, this energy is equal to about 104 kcal/mol. Every covalent bond in a given molecule has a characteristic length and strength. In general, carbon-carbon single bonds are about 1.5 Å long (Å means angstrom, or 10 meters) while carbon-carbon double bonds are about 1.3 Å, carbon-oxygen double bonds are about 1.2 Å, and carbon-hydrogen bonds are in the range of 1.0 – 1.1 Å. Most covalent bonds in organic molecules range in strength from just under 100 kcal/mole (for a carbon-hydrogen bond in ethane, for example) up to nearly 200 kcal/mole. You can refer to tables in reference books such as the CRC Handbook of Chemistry and Physics for extensive lists of bond lengths, strengths, and many other data for specific organic compounds. It is not accurate, however, to picture covalent bonds as rigid sticks of unchanging length - rather, it is better to picture them as which have a defined length when relaxed, but which can be compressed, extended, and bent. This ‘springy’ picture of covalent bonds will become very important, when we study the analytical technique known as infrared (IR) spectroscopy. | 4,404 | 2,390 |
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\[-F_x = ma_x = m\dfrac{\Delta v_x}{\Delta t} \label{2.1}\] \[-F_x = -2m\dfrac{v_x}{\Delta t} \label{2.2}\] Thus, the force that the particle exerts on the wall is \[F_x = \dfrac{mv_x^2}{a} \label{2.4}\] The mechanical definition of pressure is \[P = \dfrac{\langle F \rangle}{A} \label{2.5}\] \[P = \dfrac{N \langle F_x \rangle}{a^2} \label{2.6}\] because we have \(N\) particles hitting the wall. Hence, \[P = \dfrac{N m \langle v_x^2 \rangle}{a^3} \label{2.7}\] from our study of the Maxwell-Boltzmann distribution, we found that \[\langle v_x ^2 \rangle = \dfrac{k_B T}{m} \label{2.8}\] Hence, since \(a^3 = V\), \[P = \dfrac{N k_B T}{V} = \dfrac{n R T}{V} \label{2.9}\] \[\dfrac{P V}{n R T} = \dfrac{P \bar{V}}{R T} = \dfrac{P}{\rho R T} = 1 \label{2.10}\] \[Z = \dfrac{P V}{n R T} = \dfrac{P \bar{V}}{R T} = \dfrac{P}{\rho R T} \label{2.11}\] ( ) | 876 | 2,391 |
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Work in general is defined as a product of a force \(\textbf{F}\) and a path element \(\textbf{ds}\). Both are vectors and work is computed by integrating over their inner product: \[w = \int \textbf{F} \cdot \textbf{ds} \nonumber \] Moving an object against the force of friction as done in the above dissipation experiment is but one example of work: \[w_{friction} = \int \textbf{F}_\text{friction} \cdot \textbf{ds} \nonumber \] We could also think of work. In that case we would be moving a charge (e.g. the negative charge of an electron) against an electrical (vector) field \(\textbf{E}\). The work would be: \[w_{electical} = \int e \textbf{E} \cdot \textbf{ds} \nonumber \] Other examples are the stretching of a rubber band against the elastic force or moving a magnet in a magnetic field etc, etc. In the case of a cylinder with a piston, the pressure of gas molecules on the inside of the cylinder, \(P\), and the gas molecules external to the piston, \(P_\text{ext}\) both exert a force against each other. Pressure, (\P\), is the force, \(F\), being exerted by the particles per area, \(A\): \[P=\frac{F}{A} \nonumber \] We can assume that all the forces generated by the pressure of the particles operate parallel to the direction of motion of the piston. That is, the force moves the piston up or down as the movement of the piston is constrained to one direction. The piston moves as the molecules of the gas rapidly equilibrate to the applied pressure such that the internal and external pressures are the same. The result of this motion is work: \[w_{volume} = \int \left( \dfrac{F}{A} \right) (A\,ds) = \int P\,dV \label{Volume work}\] This particular form of work is called and will play an important role in the development of our theory. Notice however that volume work is only of work. It is important to create a sign convention at this point: positive heat, positive work is always energy you put in into the system. If the system decides to remove energy by giving off heat or work, that gets a minus sign. In other words: . To comply with this convention we need to rewrite volume work (Equation \(\ref{Volume work}\)) as \[w_{PV} = - \int \left( \dfrac{F}{A} \right) (A\,ds) = - \int P\,dV \nonumber \] Hence, to decrease the volume of the gas (\(\Delta V\) is negative), we must put in (positive) work. Thermodynamics would not have come very far without cylinders to hold gases, in particular steam. The following figure shows when the external pressure, \(P_\text{ext}\), is greater than and less than the internal pressure, \(P\), of the piston. If the pressure, \(P_\text{ext}\), being exerted on the system is constant, then the integral becomes: \[w = -P_\text{ext}\int_{V_{initial}}^{V_{final}}dV = -P_\text{ext}\Delta V \label{irreversible PV work} \] Since the system pressure (inside the piston) is not the same as the pressure exerted on the system, the system is not in a state of equilibrium and cannot be shown directly on and \(PV\) diagram. This type of process is called an irreversible process. For a system that undergoes irreversible work at constant external pressure, we can show the amount of work being done on a \(PV\) diagram despite not being able to show the process itself. Note that the external pressure, \(P_\text{ext}\), exerted on the system is constant. If the external pressure changes during the compression, we must over the whole range: \[ w = - \int_{V_{initial}}^{V_{final}}P_\text{ext}(V)\, dV \nonumber \] | 3,508 | 2,392 |
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In we examined several methods for separating an analyte from potential interferents. For example, in a liquid–liquid extraction the analyte and interferent initially are present in a single liquid phase. We add a second, immiscible liquid phase and thoroughly mix them by shaking. During this process the analyte and interferents partition between the two phases to different extents, effecting their separation. After allowing the phases to separate, we draw off the phase enriched in analyte. Despite the power of liquid–liquid extractions, there are significant limitations. Suppose we have a sample that contains an analyte in a matrix that is incompatible with our analytical method. To determine the analyte’s concentration we first separate it from the matrix using a simple liquid–liquid extraction. If we have several analytes, we may need to complete a separate extraction for each analyte. For a complex mixture of analytes this quickly becomes a tedious process. This is one limitation to a liquid–liquid extraction. A more significant limitation is that the extent of a separation depends on the distribution ratio of each species in the sample. If the analyte’s distribution ratio is similar to that of another species, then their separation becomes impossible. For example, let’s assume that an analyte, , and an interferent, , have distribution ratios of, respectively, 5 and 0.5. If we use a liquid–liquid extraction with equal volumes of sample and extractant, then it is easy to show that a single extraction removes approximately 83% of the analyte and 33% of the interferent. Although we can remove 99% of the analyte with three extractions, we also remove 70% of the interferent. In fact, there is no practical combination of number of extractions or volumes of sample and extractant that produce an acceptable separation. From we know that the distribution ratio, , for a solute, , is \[D=\frac{[S]_{\mathrm{ext}}}{[S]_{\mathrm{samp}}} \nonumber\] where [ ] is its equilibrium concentration in the extracting phase and [ ] is its equilibrium concentration in the sample. We can use the distribution ratio to calculate the fraction of that remains in the sample, , after an extraction \[q_{\text {samp}}=\frac{V_{\text {samp }}}{D V_{\text {ext }}+V_{\text {samp }}} \nonumber\] where is the volume of sample and is the volume of the extracting phase. For example, if = 10, = 20, and = 5, the fraction of remaining in the sample after the extraction is \[q_{\text { sanp }}=\frac{20}{10 \times 5+20}=0.29 \nonumber\] or 29%. The remaining 71% of the analyte is in the extracting phase. The problem with a liquid–liquid extraction is that the separation occurs in one direction only: from the sample to the extracting phase. Let’s take a closer look at the liquid–liquid extraction of an analyte and an interferent with distribution ratios of, respectively, 5 and 0.5. Figure 12.1.1
shows that a single extraction using equal volumes of sample and extractant transfers 83% of the analyte and 33% of the interferent to the extracting phase. If the original concentrations of and are identical, then their concentration ratio in the extracting phase after one extraction is \[\frac{[A]}{[I]}=\frac{0.83}{0.33}=2.5 \nonumber\] A single extraction, therefore, enriches the analyte by a factor of \(2.5 \times\). After completing a second extraction (Figure 12.1.1
) and combining the two extracting phases, the separation of the analyte and the interferent, surprisingly, is less efficient. \[\frac{[A]}{[I]}=\frac{0.97}{0.55}=1.8 \nonumber\] Figure 12.1.1
makes it clear why the second extraction results in a poorer overall separation: the second extraction actually favors the interferent! We can improve the separation by first extracting the solutes from the sample into the extracting phase and then extracting them back into a fresh portion of solvent that matches the sample’s matrix (Figure 12.1.2
). Because the analyte has the larger distribution ratio, more of it moves into the extractant during the first extraction and less of it moves back to the sample phase during the second extraction. In this case the concentration ratio in the extracting phase after two extractions is significantly greater. \[\frac{[A]}{[I]}=\frac{0.69}{0.11}=6.3 \nonumber\] Not shown in Figure 12.2 is that we can add a fresh portion of the extracting phase to the sample that remains after the first extraction (the bottom row of the first stage in Figure 12.2, beginning the process anew. As we increase the number of extractions, the analyte and the interferent each spread out in space over a series of stages. Because the interferent’s distribution ratio is smaller than the analyte’s, the interferent lags behind the analyte. With a sufficient number of extractions—that is, a sufficient number of stages—a complete separation of the analyte and interferent is possible. This process of extracting the solutes back and forth between fresh portions of the two phases, which we call a , was developed by Craig in the 1940s [Craig, L. C. , , 519–534]. The same phenomenon forms the basis of modern chromatography. See for a more detailed consideration of the mathematics behind a countercurrent extraction. In we pass a sample-free phase, which we call the , over a second sample-free that remains fixed in space (Figure 12.1.3
). We inject or place the sample into the mobile phase. As the sample moves with the mobile phase, its components partition between the mobile phase and the stationary phase. A component whose distribution ratio favors the stationary phase requires more time to pass through the system. Given sufficient time and sufficient stationary and mobile phase, we can separate solutes even if they have similar distribution ratios. There are many ways in which we can identify a chromatographic separation: by describing the physical state of the mobile phase and the stationary phase; by describing how we bring the stationary phase and the mobile phase into contact with each other; or by describing the chemical or physical interactions between the solute and the stationary phase. Let’s briefly consider how we might use each of these classifications. We can trace the history of chromatography to the turn of the century when the Russian botanist Mikhail Tswett used a column packed with calcium carbonate and a mobile phase of petroleum ether to separate colored pigments from plant extracts. As the sample moved through the column, the plant’s pigments separated into individual colored bands. After effecting the separation, the calcium carbonate was removed from the column, sectioned, and the pigments recovered. Tswett named the technique chromatography, combining the Greek words for “color” and “to write.” There was little interest in Tswett’s technique until Martin and Synge’s pioneering development of a theory of chromatography (see Martin, A. J. P.; Synge, R. L. M. “A New Form of Chromatogram Employing Two Liquid Phases,” , , 1358–1366). Martin and Synge were awarded the 1952 Nobel Prize in Chemistry for this work. The mobile phase is a liquid or a gas, and the stationary phase is a solid or a liquid film coated on a solid substrate. We often name chromatographic techniques by listing the type of mobile phase followed by the type of stationary phase. In gas–liquid chromatography, for example, the mobile phase is a gas and the stationary phase is a liquid film coated on a solid substrate. If a technique’s name includes only one phase, as in gas chromatography, it is the mobile phase. There are two common methods for bringing the mobile phase and the stationary phase into contact. In we pack the stationary phase into a narrow column and pass the mobile phase through the column using gravity or by applying pressure. The stationary phase is a solid particle or a thin liquid film coated on either a solid particulate packing material or on the column’s walls. In the stationary phase is coated on a flat surface—typically, a glass, metal, or plastic plate. One end of the plate is placed in a reservoir that contains the mobile phase, which moves through the stationary phase by capillary action. In paper chromatography, for example, paper is the stationary phase. The interaction between the solute and the stationary phase provides a third method for describing a separation (Figure 12.1.4
). In , solutes separate based on their ability to adsorb to a solid stationary phase. In , the stationary phase is a thin liquid film on a solid support. Separation occurs because there is a difference in the equilibrium partitioning of solutes between the stationary phase and the mobile phase. A stationary phase that consists of a solid support with covalently attached anionic (e.g., \(-\text{SO}_3^-\) ) or cationic (e.g., \(-\text{N(CH}_3)_3^+\)) functional groups is the basis for ion-exchange chromatography in which ionic solutes are attracted to the stationary phase by electrostatic forces. In size-exclusion chromatography the stationary phase is a porous particle or gel, with separation based on the size of the solutes. Larger solutes are unable to penetrate as deeply into the porous stationary phase and pass more quickly through the column. There are other interactions that can serve as the basis of a separation. In affinity chromatography the interaction between an antigen and an antibody, between an enzyme and a substrate, or between a receptor and a ligand forms the basis of a separation. See this chapter’s for some suggested readings. In chromatography, a separation occurs because there is a difference in the equilibrium partitioning of solutes between the mobile phase and the stationary phase. Equilibrium partitioning, however, is not the only basis for effecting a separation. In an electrophoretic separation, for example, charged solutes migrate under the influence of an applied potential. A separation occurs because of differences in the charges and the sizes of the solutes (Figure 12.1.5
). | 10,048 | 2,393 |
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The proteins in all living species, from bacteria to humans, are constructed from the same set of 20 amino acids, so called because each contains an amino group attached to a carboxylic acid. The amino acids in proteins are α-amino acids, which means the amino group is attached to the α-carbon of the carboxylic acid. Humans can synthesize only about half of the needed amino acids; the remainder must be obtained from the diet and are known as essential amino acids. However, two additional amino acids have been found in limited quantities in proteins: Selenocysteine was discovered in 1986, while pyrrolysine was discovered in 2002. The amino acids are colorless, nonvolatile, crystalline solids, melting and decomposing at temperatures above 200°C. These melting temperatures are more like those of inorganic salts than those of amines or organic acids and indicate that the structures of the amino acids in the solid state and in neutral solution are best represented as having both a negatively charged group and a positively charged group. Such a species is known as a zwitterion. In addition to the amino and carboxyl groups, amino acids have a side chain or R group attached to the α-carbon. Each amino acid has unique characteristics arising from the size, shape, solubility, and ionization properties of its R group. As a result, the side chains of amino acids exert a profound effect on the structure and biological activity of proteins. Although amino acids can be classified in various ways, one common approach is to classify them according to whether the functional group on the side chain at neutral pH is nonpolar, polar but uncharged, negatively charged, or positively charged. The structures and names of the 20 amino acids, their one- and three-letter abbreviations, and some of their distinctive features are given in Table \(\Page {1}\). The first amino acid to be isolated was asparagine in 1806. It was obtained from protein found in asparagus juice (hence the name). Glycine, the major amino acid found in gelatin, was named for its sweet taste (Greek , meaning “sweet”). In some cases an amino acid found in a protein is actually a derivative of one of the common 20 amino acids (one such derivative is hydroxyproline). The modification occurs the amino acid has been assembled into a protein. Notice in Table \(\Page {1}\) that glycine is the only amino acid whose α-carbon is . Therefore, with the exception of glycine, the amino acids could theoretically exist in either the D- or the L-enantiomeric form and rotate plane-polarized light. As with sugars, chemists used L-glyceraldehyde as the reference compound for the assignment of absolute configuration to amino acids. Its structure closely resembles an amino acid structure except that in the latter, an amino group takes the place of the group on the chiral carbon of the L-glyceraldehyde and a carboxylic acid replaces the aldehyde. Modern stereochemistry assignments using the Cahn-Ingold-Prelog priority rules used ubiquitously in chemistry show that all of the naturally occurring chiral amino acids are S except Cys which is R. We learned that all naturally occurring sugars belong to the D series. It is interesting, therefore, that nearly all known plant and animal proteins are composed entirely of L-amino acids. However, certain bacteria contain D-amino acids in their cell walls, and several antibiotics (e.g., actinomycin D and the gramicidins) contain varying amounts of D-leucine, D-phenylalanine, and D-valine. Amino acids can be classified based on the characteristics of their distinctive side chains as nonpolar, polar but uncharged, negatively charged, or positively charged. The amino acids found in proteins are L-amino acids. Each of the thousands of naturally occurring proteins has its own characteristic amino acid composition and sequence that result in a unique three-dimensional shape. Since the 1950s, scientists have determined the amino acid sequences and three-dimensional conformation of numerous proteins and thus obtained important clues on how each protein performs its specific function in the body. Proteins are compounds of high molar mass consisting largely or entirely of chains of amino acids. Because of their great complexity, protein molecules cannot be classified on the basis of specific structural similarities, as carbohydrates and lipids are categorized. The two major structural classifications of proteins are based on far more general qualities: whether the protein is (1) fiberlike and insoluble or (2) globular and soluble. Some proteins, such as those that compose hair, skin, muscles, and connective tissue, are fiberlike. These fibrous proteins are insoluble in water and usually serve structural, connective, and protective functions. Examples of fibrous proteins are keratins, collagens, myosins, and elastins. Hair and the outer layer of skin are composed of keratin. Connective tissues contain collagen. Myosins are muscle proteins and are capable of contraction and extension. Elastins are found in ligaments and the elastic tissue of artery walls. Globular proteins, the other major class, are soluble in aqueous media. In these proteins, the chains are folded so that the molecule as a whole is roughly spherical. Familiar examples include egg albumin from egg whites and serum albumin in blood. Serum albumin plays a major role in transporting fatty acids and maintaining a proper balance of osmotic pressures in the body. Hemoglobin and myoglobin, which are important for binding oxygen, are also globular proteins. The structure of proteins is generally described as having four organizational levels. The first of these is the primary structure, which is the number and sequence of amino acids in a protein’s polypeptide chain or chains, beginning with the free amino group and maintained by the peptide bonds connecting each amino acid to the next. The primary structure of insulin, composed of 51 amino acids, is shown in Figure \(\Page {1}\). A protein molecule is not a random tangle of polypeptide chains. Instead, the chains are arranged in unique but specific conformations. The term secondary structure refers to the fixed arrangement of the polypeptide backbone. On the basis of X ray studies, Linus Pauling and Robert Corey postulated that certain proteins or portions of proteins twist into a spiral or a helix. This helix is stabilized by hydrogen bonding between the carbonyl oxygen atom of one amino acid and the amide hydrogen atom four amino acids up the chain (located on the next turn of the helix) and is known as a right-handed α-helix. X ray data indicate that this helix makes one turn for every 3.6 amino acids, and the side chains of these amino acids project outward from the coiled backbone (Figure \(\Page {2}\)). The α-keratins, found in hair and wool, are exclusively α-helical in conformation. Some proteins, such as gamma globulin, chymotrypsin, and cytochrome c, have little or no helical structure. Others, such as hemoglobin and myoglobin, are helical in certain regions but not in others. Another common type of secondary structure, called the , is a sheetlike arrangement in which two or more extended polypeptide chains (or separate regions on the same chain) are aligned side by side. The aligned segments can run either parallel or antiparallel—that is, the N-terminals can face in the same direction on adjacent chains or in different directions—and are connected by hydrogen bonding (Figure \(\Page {3}\)). The β-pleated sheet is particularly important in structural proteins, such as silk fibroin. It is also seen in portions of many enzymes, such as carboxypeptidase A and lysozyme. Tertiary structure refers to the unique three-dimensional shape of the protein as a whole, which results from the folding and bending of the protein backbone. The tertiary structure is intimately tied to the proper biochemical functioning of the protein. Figure \(\Page {4}\) shows a depiction of the three-dimensional structure of insulin. Four major types of attractive interactions determine the shape and stability of the tertiary structure of proteins. You studied several of them . When a protein contains more than one polypeptide chain, each chain is called a . The arrangement of multiple subunits represents a fourth level of structure, the quaternary structure of a protein. Hemoglobin, with four polypeptide chains or subunits, is the most frequently cited example of a protein having quaternary structure (Figure \(\Page {6}\)). The quaternary structure of a protein is produced and stabilized by the same kinds of interactions that produce and maintain the tertiary structure. A schematic representation of the four levels of protein structure is in Figure \(\Page {7}\). Image from the ( ) of PDB 1I3D (R.D. Kidd, H.M. Baker, A.J. Mathews, T. Brittain, E.N. Baker (2001) Oligomerization and ligand binding in a homotetrameric hemoglobin: two high-resolution crystal structures of hemoglobin Bart's (gamma(4)), a marker for alpha-thalassemia. Protein Sci. 1739–1749). The consists of the specific amino acid sequence. The resulting peptide chain can twist into an α-helix, which is one type of . This helical segment is incorporated into the of the folded polypeptide chain. The single polypeptide chain is a subunit that constitutes the of a protein, such as hemoglobin that has four polypeptide chains. The highly organized structures of proteins are truly masterworks of chemical architecture. But highly organized structures tend to have a certain delicacy, and this is true of proteins. Denaturation is the term used for any change in the three-dimensional structure of a protein that renders it incapable of performing its assigned function. A denatured protein cannot do its job. (Sometimes denaturation is equated with the precipitation or coagulation of a protein; our definition is a bit broader.) A wide variety of reagents and conditions, such as heat, organic compounds, pH changes, and heavy metal ions can cause protein denaturation (Figure \(\Page {1}\)). Anyone who has fried an egg has observed denaturation. The clear egg white turns opaque as the albumin denatures and coagulates. No one has yet reversed that process. However, given the proper circumstances and enough time, a protein that has unfolded under sufficiently gentle conditions can refold and may again exhibit biological activity (Figure \(\Page {8}\)). Such evidence suggests that, at least for these proteins, the primary structure determines the secondary and tertiary structure. A given sequence of amino acids seems to adopt its particular three-dimensional arrangement naturally if conditions are right. The primary structures of proteins are quite sturdy. In general, fairly vigorous conditions are needed to hydrolyze peptide bonds. At the secondary through quaternary levels, however, proteins are quite vulnerable to attack, though they vary in their vulnerability to denaturation. The delicately folded globular proteins are much easier to denature than are the tough, fibrous proteins of hair and skin. Proteins can be divided into two categories: fibrous, which tend to be insoluble in water, and globular, which are more soluble in water. A protein may have up to four levels of structure. The primary structure consists of the specific amino acid sequence. The resulting peptide chain can form an α-helix or β-pleated sheet (or local structures not as easily categorized), which is known as secondary structure. These segments of secondary structure are incorporated into the tertiary structure of the folded polypeptide chain. The quaternary structure describes the arrangements of subunits in a protein that contains more than one subunit. Four major types of attractive interactions determine the shape and stability of the folded protein: ionic bonding, hydrogen bonding, disulfide linkages, and dispersion forces. A wide variety of reagents and conditions can cause a protein to unfold or denature. | 12,046 | 2,394 |
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Two common analytical problems are matrix components that interfere with an analyte’s analysis and an analyte with a concentration that is too small to analyze accurately. As we have learned in this chapter, we can use a separation to solve the first problem. Interestingly, we often can use a separation to solve the second problem as well. For a separation in which we recover the analyte in a new phase, it may be possible to increase the analyte’s concentration if we can extract the analyte from a larger volume into a smaller volume. This step in an analytical procedure is known as a . An example from the analysis of water samples illustrates how we can simultaneously accomplish a separation and a preconcentration. In the gas chromatographic analysis for organophosphorous pesticides in environmental waters, the analytes in a 1000-mL sample are separated from their aqueous matrix by a solid-phase extraction that uses 15 mL of ethyl acetate [Aguilar, C.; Borrul, F.; Marcé, R. M. , , 1048–1054]. After the extraction, the analytes in the ethyl acetate have a concentration that is 67 times greater than that in the original sample (assuming the extraction is 100% efficient). | 1,203 | 2,397 |
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A macroscale column is much too big for very small quantities of material \(\left( < 20 \: \text{mg} \right)\). Instead, a column can be constructed using a disposable Pasteur pipette. In order to get good separation, it is ideal if the desired component has an and is separated from other components by at least 0.2 \(R_f\) units.\(^{11}\) If the spots to be separated are very close (< 0.2 \(\Delta R_f\)), it's best if the middle of the spots has an \(R_f\) of 0.35. An \(R_f\) near 0.35 is ideal because it is slow enough that stationary-mobile phase equilibration can occur, but fast enough to minimize band widening from diffusion. The column pictured in this section shows purification of a drop of dilute purple food dye (made from 1 drop red dye, 1 drop blue dye and 15 drops water). The dye is separated as best as possible into its three components: blue, red and pink (as seen in the TLC plate of Figure 2.68a). A 2.5" column of silica gel is used and eluted with a solution made from a 1:3:1 volume ratio of \(6 \: \text{M} \: \ce{NH_4OH}\):1-pentanol:ethanol. Wedge a bit of cotton into the bottom of a pipette. Use a scooping method to fill silica or alumina to 2-2.5 inches high. Remember to break the seal before letting go of the pipette bulb, or suction will ruin the column. Refill the column with eluent as necessary. Adjust the eluent level to the sand layer, and then delicately add the sample. Use pressure to push the eluent down onto the silica/alumina layer. Rinse with one portion of eluent and push the solvent onto the column. Fill the pipette with eluent and apply pressure to elute the column. Collect liquid into test tubes. (refill whenever the eluent level nears the sand layer). Keep fractions organized in a test tube rack in the order they are eluted. Use TLC to determine the purity of the fractions, and combine appropriate fractions. Remove the solvent with the rotary evaporator. \(^{11}\)W.C. Still, M. Kahn, A. Mitra, , Vol. 43, No. 14, . | 1,999 | 2,398 |
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High Performance Liquid Chromotagraphy (HPLC) is an analytical technique used for the separation of compounds soluble in a particular solvent. Liquid chromatography was initially discovered as an analytical technique in the early twentieth century and was first used as a method of separating colored compounds. This is where the name means color, means writing, was derived. A Russian botanist named Mikhail S. Tswett used a rudimentary form of chromatographic separation to purify mixtures of plant pigments into the pure constituents. He separated the pigments based on their interaction with a , which is essential to any chromatographic separation. The stationary phase he used was powdered chalk and aluminia, the mobile phase in his separation was the solvent. After the solid stationary phase was packed into a glass column (essentially a long, hollow, glass tube) he poured the mixture of plant pigments and solvent in the top of the column. He then poured additional solvent into the column until the samples were eluted at the bottom of the column. The result of this process most crucial to his investigation was that the plant pigments separated into bands of pure components as they passed through the stationary phase. Modern high performance liquid chromatography or HPLC has its roots in this separation, the first form of liquid chromatography. The chromatographic process has been significantly improved over the last hundred years, yielding greater separation efficiency, versatility and speed. All chromatographic separations, including HPLC operate under the same basic principle; every compound interacts with other chemical species in a characteristic manner. Chromatography separates a sample into its constituent parts because of the difference in the relative affinities of different molecules for the mobile phase and the stationary phase used in the separation. All chemical reactions have a characteristic equilibrium constant. For the reaction \[ A_{aq} + B_s \rightleftharpoons AB_s \label{1}\] There is a chemical equilibrium constant that dictates what percentage of compound will be in solution and what percentage will be bound to the stationary compound . During a chromatographic separation, there is similar relationship between compound and the solvent, or , . This will yield an overall equilibrium equation which dictates the quantity of that will be associated with the stationary phase and the quantity of that will be associated with the mobile phase. \[ A_{mobile} \rightleftharpoons A_{stationary} \label{2}\] The equilibrium between the mobile phase and stationary phase is given by the constant . \[ K_c = \dfrac{(a_A )_S}{(a_A )_M} \approx \dfrac{c_S}{c_M} \label{3}\] Where , the , is the ratio of the activity of compound in the stationary phase and activity of compound in the mobile phase. In most separations, which contain low concentrations of the species to be separated, the activity of in each is approximately equal to the concentration of in that state. The distribution constant indicates the amount of time that compound spends adsorbed to the stationary phase as the opposed to the amount of time spends solvated by the mobile phase. This relationship determines the amount of time it will take for compound to travel the length of the column. The more time spends adsorbed to the stationary phase, the more time compound will take to travel the length of the column. The amount of time between the injection of a sample and its elution from the column is known as the ; it is given the symbol . The amount of time required for a sample that does not interact with the stationary phase, or has a equal to zero, to travel the length of the column is known as the , . No compound can be eluted in less than the void time. Since is a factor that is wholly dependent on a particular column and solvent flow rate, a quantitative measure of the affinity of a compound for a particular set of mobile and stationary phases that does not depend on the column geometry is useful. The , , can be derived from and is independent of the column size and the solvent flow rate. \[ k_C = \dfrac{K_C V_S }{V_M } \label{4}\] The retention factor is calculated by multiplying the distribution constant by the volume of stationary phase in the column and dividing by the volume of mobile phase in the column. In order to separate two compounds, their respective retention factors must be different, otherwise both compounds would be eluted simultaneously; the factor is the ratio of the retention factors. \[ \alpha = \dfrac{k_B }{k_A} \label{5}\] Where is the compound that is retained more strongly by the column and is the compound with the faster elution time. As a compound passes through the column it slowly diffuses away from the initial injection band, which is the area of greatest concentration. The initial, narrow, band that contained all of the sample becomes broader the longer the analyte remains in the column. This increases the time required for complete elution of a particular compound and is generally undesirable. It must be minimized so that overly broad elution bands do not overlap with one another. We will see how this is measured quantitatively when we discuss momentarily. The overriding purpose of a chromatographic separation is just that, to separate two or more compounds contained in solution. In analytical chemistry, a quantitative metric of every experimental parameter is desired, and so separation efficiency is measured in plates. The concept of plates as a separation metric arose from the original method of , where compounds were separated based on their volatilities through many simultaneous , each simple distillation occurred on one of many distillation plates. In chromatography, no actual plates are used, but the concept of a , as a distinct region where a single equilibrium is maintained, remains. In a particular liquid chromatographic separation, the number of theoretical plates and the are related simply by the length of the column \[ N = \dfrac{L}{H} \label{6}\] Where is the number of theoretical plates, is the length of the column, and is the height equivalent to a theoretical plate. The plate height is given by the (standard deviation squared) of an elution peak divided by the length of the column. \[ H = \dfrac{\sigma ^2}{L} \label{7}\] The standard deviation of an elution peak can be approximated by assuming that a Gaussian elution peak is roughly triangular, in that case the plate height can be given by the width of the elution peak squared times the length of the column over the retention time of the that peak squared times 16. \[ H = \dfrac{LW^2 }{16t_R^2} \label{8}\] Using the relationship between plate height and number of plates, the number of plates can also be found in terms of retention time and peak width. \[ N = 16 \left( \dfrac{t_R}{W} \right)^2\label{9} \] In order to optimize separation efficiency, it is necessary in maximize the number of theoretical plates, which requires reducing the plate height. The plate height is related to the of the mobile phase, so for a fixed set of mobile phase, stationary phase, and analytes; separation efficiency can be maximized by optimizing flow rate as dictated by the . \[ H = A + \dfrac{B}{v} + Cv \label{10}\] The three constants in the van Deemter equation are factors that describe possible causes of band broadening in a particular separation. \(A\) is a constant which represents the different possible paths that can be taken by the analyte through the stationary phase, it decreases if the packing of the column is kept as small as possible. \(B\) is a constant that describes the longitudinal diffusion that occurs in the system. \(C\) is a constant that describes the rate of adsorption and desorption of the analyte to the stationary phase. \(A\), \(B\) and \(C\) are constant for any given system (with constant analyte, stationary phase, and mobile phase), so flow rate must be optimized accordingly. If the flow rate is too low, the (\(\dfrac{B}{v}\)) will increase significantly, which will increase plate height. At low flow rates, the analyte spends more time at rest in the column and therefore longitudinal diffusion in a more significant problem. If the flow rate is too high, the (\(Cv\)) will increase and reduce column efficiency. At high flow rates the adsorption of the analyte to the stationary phase results in some of the sample lagging behind, which also leads to band broadening. The of a elution is a quantitative measure of how well two elution peaks can be differentiated in a chromatographic separation. It is defined as the difference in retention times between the two peaks, divided by the combined widths of the elution peaks. \[ R_S = \dfrac{2\left[ {\left( {t_R } \right)_B - \left( {t_R } \right)_A } \right]}{W_B + W_A} \label{11}\] Where is the species with the longer retention time, and and are the retention time and elution peak width respectively. If the resolution is greater than one, the peaks can usually be differentiated successfully.
While all of these basic principles hold true for all chromatographic separations, HPLC was developed as method to solve some of the shortcomings of standard . Classic liquid chromatography has several severe limitations as a separation method. When the solvent is driven by gravity, the separation is very slow, and if the solvent is driven by vacuum, in a standard packed column, the plate height increases and the effect of the vacuum is negated. The limiting factor in liquid chromatography was originally the size of the column packing, once columns could be packed with particles as small as 3 µm, faster separations could be performed in smaller, narrower, columns. High pressure was required to force the mobile phase and sample through these new columns, and previously unneeded apparatus was required to maintain reproducibility of results in this new instruments. The use of high pressures in a narrow column allowed for a more effective separation to be achieved in much less time than was required for previous forms of liquid chromatography. Specialized apparatus is required for an HPLC separation because of the high pressures and low tolerances under which the separation occurs. If the results are to be reproducible, then the conditions of the separation must also be reproducible. Thus HPLC equipment must be of high quality; it is therefore expensive. The , or solvent, in HPLC is usually a mixture of polar and non-polar liquid components whose respective concentrations are varied depending on the composition of the sample. As the solvent is passed through a very narrow bore column, any contaminants could at worst plug the column, or at the very least add variability to the retention times during repeated different trials. Therefore HPLC solvent must be kept free of dissolved gases, which could come out of solution mid-separation, and particulates. In the , the components of the sample separate based on their differing interactions with the column packing. If a species interacts more strongly with the stationary phase in the column, it will spend more time adsorbed to the column's adsorbent and will therefore have a greater retention time. Columns can be packed with solids such as silica or alumina; these columns are called . If stationary phase in the column is a liquid, the column is deemed a . Bonded columns contain a liquid stationary phase bonded to a sold support, which is again usually silica or alumina. The value of the constant described in the equation is proportional, in HPLC, to the diameter of the particles that constitute the column's packing material. The drives the solvent and sample through the column. To reduce variation in the elution, the pump must maintain a constant, pulse free, flow rate; this is achieved with . The presence of two pistons allows the flow rate to be controlled by one piston as the other recharges. A can be used for even greater control of flow rate; however, the syringe pump is unable to produce as much pressure as a piston pump, so it cannot be used in all HPLC applications. The , located at the end of the column, must register the presence of various components of the sample, but must not detect the solvent. For that reason there is no universal detector that works for all separations. A common HPLC detector is a , as most medium to large molecules absorb UV radiation. Detectors that measure fluorescence and refractive index are also used for special applications. A relatively new development is the combination of an HPLC separation with an . This allows the pure components of the sample to be identified and quantified by nuclear magnetic resonance after having been separated by HPLC, in one integrated process. If the stationary phase is more polar than the mobile phase, the separation is deemed . If the stationary phase is less polar than the mobile phase, the separation is . In reverse phase HPLC the retention time of a compound increases with decreasing polarity of the particular species. The key to an effective and efficient separation is to determine the appropriate ratio between polar and non-polar components in the mobile phase. The goal is for all the compounds to elute in as short a time as possible, while still allowing for the resolution of individual peaks. Typical columns for normal phase separation are packed with alumina or silica. Alkyl, aliphatic or phenyl bonded phases are typically used for reverse phase separation. If the composition of the mobile phase remains constant throughout the HPLC separation, the separation is deemed an . Often the only way to elute all of the compounds in the sample in a reasonable amount of time, while still maintaining peak resolution, is to change the ratio of polar to non-polar compounds in the mobile phase during the sample run. Known as , this is the technique of choice when a sample contains components of a wide range of polarities. For a , the solvent starts out relatively polar and slowly becomes more non-polar. The gradient elution offers the most complete separation of the peaks, without taking an inordinate amount of time. A sample containing compounds of a wide range of polarities can be separated by a gradient elution in a shorter time period without a loss of resolution in the earlier peaks or excessive broadening of later peaks. However, gradient elution requires more complex and expensive equipment and it is more difficult to maintain a constant flow rate while there are constant changes in mobile phase composition. Gradient elution, especially at high speeds, brings out the limitations of lower quality experimental apparatus, making the results obtained less reproducible in equipment already prone to variation. If the flow rate or mobile phase composition fluctuates, the results will not be reproducible. HPLC can be used in both qualitative and quantitative applications, that is for both compound identification and quantification. Normal phase HPLC is only rarely used now, almost all HPLC separation can be performed in reverse phase. Reverse phase HPLC (RPLC) is ineffective in for only a few separation types; it cannot separate inorganic ions (they can be separated by ion exchange chromatography). It cannot separate polysaccharides (they are too hydrophilic for any solid phase adsorption to occur), nor polynucleotides (they adsorb irreversibly to the reverse phase packing). Lastly, incredibly hydrophobic compounds cannot be separated effectively by RPLC (there is little selectivity). Aside from these few exceptions, RPLC is used for the separation of almost all other compound varieties. RPLC can be used to effectively separate similar simple and aromatic hydrocarbons, even those that differ only by a single methylene group. RPLC effectively separates simple amines, sugars, lipids, and even pharmaceutically active compounds. RPLC is also used in the separation of amino acids, peptides, and proteins. Finally RPLC is used to separate molecules of biological origin. The determination of caffeine content in coffee products is routinely done by RPLC in commercial applications in order to guarantee purity and quality of ground coffee. HPLC is a useful addition to an analytical arsenal, especially for the separation of a sample before further analysis. | 16,503 | 2,399 |
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This page looks at the reaction of aldehydes and ketones with 2,4-dinitrophenylhydrazine (Brady's reagent) as a test for the carbon-oxygen double bond. It also looks briefly at some other similar reactions which are all known as addition-elimination (or condensation) reactions. 2,4-dinitrophenylhydrazine is often abbreviated to 2,4-DNP or 2,4-DNPH. A solution of 2,4-dinitrophenylhydrazine in a mixture of methanol and sulfuric acid is known as Brady's reagent. Although the name sounds complicated, and the structure of 2,4-dinitrophenylhydrazine looks quite complicated, it is actually very easy to work out. Start with the formula of hydrazine. That's almost all you need to remember! Hydrazine is: In phenylhydrazine, one of the hydrogens is replaced by a phenyl group, C H . This is based on a benzene ring. In 2,4-dinitrophenylhydrazine, there are two nitro groups, NO , attached to the phenyl group in the 2- and 4- positions. The corner with the nitrogen attached is counted as the number 1 position, and you just number clockwise around the ring. Details vary slightly depending on the nature of the aldehyde or ketone, and the solvent that the 2,4-dinitrophenylhydrazine is dissolved in. Assuming you are using Brady's reagent (a solution of the 2,4-dinitrophenylhydrazine in methanol and sulphuric acid). Add either a few drops of the aldehyde or ketone, or possibly a solution of the aldehyde or ketone in methanol, to the Brady's reagent. A bright orange or yellow precipitate shows the presence of the carbon-oxygen double bond in an aldehyde or ketone. This is the simplest test for an aldehyde or ketone. The overall reaction is given by the equation: R and R' can be any combination of hydrogen or hydrocarbon groups (such as alkyl groups). If at least one of them is a hydrogen, then the original compound is an aldehyde. If both are hydrocarbon groups, then it is a ketone. Look carefully at what has happened. The product is known as a "2,4-dinitrophenylhydrazone". Notice that all that has changed is the ending from "-ine" to "-one". The product from the reaction with ethanal would be called ethanal 2,4-dinitrophenylhydrazone; from propanone, you would get propanone 2,4-dinitrophenylhydrazone - and so on. The reaction is known as a condensation reaction. A condensation reaction is one in which two molecules join together with the loss of a small molecule in the process. In this case, that small molecule is water. In terms of mechanisms, this is a nucleophilic addition-elimination reaction. The 2,4-dinitrophenylhydrazine first adds across the carbon-oxygen double bond (the addition stage) to give an intermediate compound which then loses a molecule of water (the elimination stage). The reaction has two uses in testing for aldehydes and ketones. The precipitate is filtered and washed with, for example, methanol and then recrystallised from a suitable solvent which will vary depending on the nature of the aldehyde or ketone. For example, you can recrystallise the products from the small aldehydes and ketones from a mixture of ethanol and water. The crystals are dissolved in the minimum quantity of hot solvent. When the solution cools, the crystals are re-precipitated and can be filtered, washed with a small amount of solvent and dried. They should then be pure. If you then find the melting point of the crystals, you can compare it with tables of the melting points of 2,4-dinitrophenylhydrazones of all the common aldehydes and ketones to find out which one you are likely to have got. If you go back and look at the equations, nothing in the 2,4-dinitrophenylhydrazine changes during the reaction apart from the -NH group. You can get a similar reaction if the -NH group is attached to other things. In each case, the reaction would look like this: In what follows, all that changes is the nature of the "X". with hydrazine The product is a "hydrazone". If you started from propanone, it would be propanone hydrazone. with phenylhydrazine The product is a "phenylhydrazone". with hydroxylamine The product is an "oxime" - for example, ethanal oxime. Jim Clark ( ) | 4,125 | 2,400 |
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Electrochemiluminescence is chemiluminescence arising as a result of electrochemical reactions. It includes electrochemical initiation of ordinary chemiluminescent reactions, electrochemical modification of an analyte enabling it to take part in a chemiluminescent reaction, or electron transfer reactions between radicals or ions generated at electrodes. Prominent in the work done on electrochemiluminescence are reactions involving polyaromatic hydrocarbons or transition metal complexes, especially those of ruthenium, palladium, osmium and platinum. Applications have made use of the sensitivity, selectivity and wide working range of analytical chemiluminescence, but electrochemiluminescence offers additional advantages without adding much to the inexpensive instrumentation . Electrodes can be designed to achieve maximum detection of the light emitted and electrochemical measurements can be made simultaneously with the light output. Generation of chemiluminescence reagents at electrodes gives control over the course of light producing reactions, which can effectively be switched on and off by alteration of the applied potential; this is particularly useful when using unstable reagents or intermediates. Other possible benefits include generation of reagents from inactive precursors and regeneration of reagents, which permits the use of lower concentrations or immobilization of the reagents on the electrode. Analytes can also be regenerated, so that each analyte molecule can produce many photons, increasing sensitivity, or they can be modified to make them detectable by the chemiluminescence reaction in use. Electrochemiluminescence can be coupled with high performance liquid chromatography or with capillary electrophoresis. The usefulness of -(2, 2 -bipyridyl)ruthenium(II) (discussed in chapter B9 ADD LINK) in electrochemiluminescence rests on its activity with very high efficiency at easily accessible potentials and ambient temperature in aqueous buffer solutions in the presence of dissolved oxygen and other impurities. The reaction sequence that leads to electrochemiluminescence is shown in equations D7.1 to D7.4: (D7.1) Oxidation: [Ru(bipy) ] ─ e → [Ru(bipy) ] (D7.2) Reduction by analyte: [Ru(bipy) ] + e → [Ru(bipy) ] (D7.3) Electron transfer: [Ru(bipy) ] + [Ru(bipy) ] → [Ru(bipy) ] + [Ru(bipy) ] (D7.4) Chemiluminescence: [R(bipy) ] → [Ru(bipy) ] + light The oxidation occurs electrochemically at the anode, whereas the reduction is brought about chemically by the analyte in the free solution. Electron transfer and subsequent chemiluminescence also occur in the free solution close to the anode, where the [Ru(bipy) ] is concentrated. Other analytes, e.g. alkylamines, are oxidized at the anode to form a highly reducing radical intermediate that reacts with [Ru(bipy) ] to form [Ru(bipy) ] *, which emits light. Oxalates, on the other hand, are oxidized by [Ru(bipy) ] to radicals that then reduce more [Ru(bipy) ] to give [Ru(bipy) ] * and chemiluminescence. Instrumentation for electrochemiluminescence differs from that for other chemiluminescence only in having a flow cell provided with working, counter and reference electrodes, regulated by a potentiostat, which is in turn controlled by the computer that receives input from the photomultiplier or other transducer that receives the light signals. Figure D7.1 shows the usual flow injection manifold used for measuring electrochemiluminescence. The flow cell is in a light-tight box to exclude ambient light. A more portable alternative is a probe containing a set of electrodes and a fibre optic bundle to carry emitted light to a photomultiplier. Ambient light is excluded by means of baffles in the channels that admit the test solution. Because it can be electrochemically regenerated, it is useful to immobilize [Ru(bipy) ] in a cation exchange resin deposited on the electrode to form a sensor that does not need a continual reagent supply. | 3,982 | 2,401 |
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Cycloalkanes are very important in components of food, pharmaceutical drugs, and much more. However, to use cycloalkanes in such applications, we must know the effects, functions, properties, and structures of cycloalkanes. Cycloalkanes are alkanes that are in the form of a ring; hence, the prefix cyclo-. Stable cycloalkanes cannot be formed with carbon chains of just any length. Recall that in alkanes, carbon adopts the sp tetrahedral geometry in which the angles between bonds are 109.5°. For some cycloalkanes to form, the angle between bonds must deviate from this ideal angle, an effect known as . Additionally, some hydrogen atoms may come into closer proximity with each other than is desirable (become eclipsed), an effect called . These destabilizing effects, angle strain and torsional strain are known together as . The smaller cycloalkanes, cyclopropane and cyclobutane, have particularly high ring strains because their bond angles deviate substantially from 109.5° and their hydrogens eclipse each other. Cyclopentane is a more stable molecule with a small amount of ring strain, while is able to adopt the perfect geometry of a cycloalkane in which all angles are the ideal 109.5° and no hydrogens are eclipsed; it has no ring strain at all. Cycloalkanes larger than cyclohexane have ring strain and are not commonly encountered in organic chemistry. There are many forms of cycloalkanes, such as cyclopropane, cyclobutane, cyclopentane, cyclohexane, among others. The process of naming cycloalkanes is the same as naming alkanes but the addition of the prefix cyclo- is required. Cyclobutane is in a form of a square, which is highly unfavorable and unstable (this will be explained soon). There are different drawings for cyclobutane, but they are equivalent to each other. Cyclobutane can reduce the ring string by puckering the square cyclobutane. Cyclopentane takes the shape of a pentagon and cyclohexane is in the shape of a hexagon. There are two ways to draw cyclohexane because it can be in a hexagon shape or in a different conformational form called the conformation and the conformation. Although there are multiple ways to draw cyclohexane, the most stable and major conformer is the chair because is has a lower activation barrier from the energy diagram. The transition state structure is called a . This energy diagram shows that the chair conformation is lower in energy; therefore, it is more stable. The chair conformation is more stable because it does not have any steric hindrance or steric repulsion between the hydrogen bonds. By drawing cyclohexane in a chair conformation, we can see how the H's are positioned. There are two positions for the H's in the chair conformation, which are in an axial or an equitorial formation. This is how a chair conformation looks, but you're probably wondering which H's are in the equitorial and axial form. Here are more pictures to help. These are hydrogens in the axial form. These hydrogens are in an equitorial form. Of these two positions of the H's, the equitorial form will be the most stable because the hydrogen atoms, or perhaps the other substituents, will not be touching each other. This is the best time to build a chair conformation in an equitorial and an axial form to demonstrate the stability of the equitorial form. Cycloalkanes tend to give off a very high and non-favorable energy, and the spatial orientation of the atoms is called the ring strain. When atoms are close together, their proximity is highly unfavorable and causes steric hindrance. The reason we do not want ring strain and steric hindrance is because heat will be released due to an increase in energy; therefore, a lot of that energy is stored in the bonds and molecules, causing the ring to be unstable and reactive. Another reason we try to avoid ring strain is because it will affect the structures and the conformational function of the smaller cycloalkanes. One way to determine the presence of ring strain is by its heat of combustion. By comparing the heat of combustion with the value measured for the straight chain molecule, we can determine the stability of the ring. There are two types of strain, which are eclipsing/torsional strain and bond angle strain. Bond angle strain causes a ring to have a poor overlap between the atoms, resulting in weak and reactive C-C bonds. An eclipsed spatial arrangement of the atoms on the cycloalkanes results in high energy. With so many cycloalkanes, which ones have the highest ring strain and are very unlikely to stay in its current form? The figures below show cyclopropane, cyclobutane, and cyclopentane, respectively. Cyclopropane is one of the cycloalkanes that has an incredibly high and unfavorable energy, followed by cyclobutane as the next strained cycloalkane. Any ring that is small (with three to four carbons) has a significant amount of ring strain; cyclopropane and cyclobutane are in the category of small rings. A ring with five to seven carbons is considered to have minimal to zero strain, and typical examples are cyclopentane, cyclohexane, and cycloheptane. However, a ring with eight to twelve carbons is considered to have a moderate strain, and if a ring has beyond twelve carbons, it has minimal strain. There are different types of ring strain: Most of the time, cyclohexane adopts the fully staggered, ideal angle . In the chair conformation, if any carbon-carbon bond were examined, it would be found to exist with its substituents in the staggered conformation and all bonds would be found to possess an angle of 109.5°. In the chair conformation, hydrogen atoms are labeled according to their location. Those hydrogens which exist above or below the plane of the molecule (shown with red bonds above) are called . Those hydrogens which exist in the plane of the molecule (shown with blue bonds above) are called . Although the chair conformation is the most stable conformation that cyclohexane can adopt, there is enough thermal energy for it to also pass through less favorable conformations before returning to a different chair conformation. When it does so, the axial and equatorial substituents change places. The passage of cyclohexane from one chair conformation to another, during which the axial substituents switch places with the equatorial substituents, is called a . Methylcyclohexane is cyclohexane in which one hydrogen atom is replaced with a methyl group substituent. Methylcyclohexane can adopt two basic chair conformations: one in which the methyl group is axial, and one in which it is equatorial. Methylcyclohexane strongly prefers the equatorial conformation. In the axial conformation, the methyl group comes in close proximity to the axial hydrogens, an energetically unfavorable effect known as a (Figure 3). Thus, the equatorial conformation is preferred for the methyl group. In most cases, if the cyclohexane ring contains a substituent, the substituent will prefer the equatorial conformation. 3. | 7,017 | 2,402 |
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Convert the temperatures indicated to complete the following table Make a graph representing the potential energy of a harmonic oscillator as a function of displacement from equilibrium. On the same graph, include a function describing the kinetic energy as a function of displacement from equilibrium as well as the total energy of the system. Calculate the work required to move a 3.2 kg mass 10.0 m against a resistive force of 9.80 N. Calculate the work needed for a 22.4 L sample of gas to expand to 44.8 L against a constant external pressure of 0.500 atm. If the internal and external pressure of an expanding gas are equal at all points along the entire expansion pathway, the expansion is called “reversible.” Calculate the work of a reversible expansion for 1.00 mol of an ideal gas expanding from 22.4 L at 273 K to a final volume of 44.8 L. | 870 | 2,403 |
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The relationship for chemical potential \[\mu = \mu^o + RT \ln \left( \dfrac{p}{p^o} \right) \nonumber \] was derived assuming behavior. But for real gases that deviate widely from ideal behavior, the expression has only limited applicability. In order to use the simple expression on real gases, a “fudge” factor is introduced called . Using fugacity instead of pressure, the chemical potential expression becomes \[\mu = \mu^o + RT \ln \left( \dfrac{f}{f^o} \right) \nonumber \] where \(f\) is the fugacity. Fugacity is related to pressure, but contains all of the deviations from ideality within it. To see how it is related to pressure, consider that a change in chemical potential for a single component system can be expressed as \[d\mu - Vdp - SdT \nonumber \] and so \[ \left(\dfrac{\partial \mu}{\partial p} \right)_T = V \label{eq3} \] Differentiating the expression for chemical potential above with respect to pressure at constant volume results in \[ \left(\dfrac{\partial \mu}{\partial p} \right)_T = \left \{ \dfrac{\partial}{\partial p} \left[ \mu^o + RT \ln \left( \dfrac{f}{f^o} \right) \right] \right \} \nonumber \] which simplifies to \[ \left(\dfrac{\partial \mu}{\partial p} \right)_T = RT \left[ \dfrac{\partial \ln (f)}{\partial p} \right]_T = V \nonumber \] Multiplying both sides by \(p/RT\) gives \[\left[ \dfrac{\partial \ln (f)}{\partial p} \right]_T = \dfrac{pV}{RT} =Z \nonumber \] where \(Z\) is the compression factor as discussed previously. Now, we can use the expression above to obtain the \(\gamma\), as defined by \[ f= \gamma p \nonumber \] Taking the natural logarithm of both sides yields \[ \ln f= \ln \gamma + \ln p \nonumber \] or \[ \ln \gamma = \ln f - \ln p \nonumber \] Using some calculus and substitutions from above, \[ \int \left(\dfrac{\partial \ln \gamma}{\partial p} \right)_T dp = \int \left(\dfrac{\partial \ln f}{\partial p} - \dfrac{\partial \ln p }{\partial p} \right)_T dp \nonumber \] \[= \int \left(\dfrac{Z}{\partial p} - \dfrac{1}{\partial p} \right)_T dp \nonumber \] Finally, integrating from \(0\) to \(p\) yields \[\ln \gamma = \int_0^{p} \left( \dfrac{ Z-1}{p}\right)_T dp \nonumber \] If the gas behaves ideally, \(\gamma = 1\). In general, this will be the limiting value as \(p \rightarrow 0\) since all gases behave ideal as the pressure approaches 0. The advantage to using the fugacity in this manner is that it allows one to use the expression \[ \mu = \mu^o + RT \ln \left( \dfrac{f}{f^o}\right) \nonumber \] to calculate the chemical potential, insuring that Equation \ref{eq3} holds even for gases that deviate from ideal behavior! | 2,637 | 2,404 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Environmental_Toxicology_(van_Gestel_et_al.)/05%3A_Population_Community_and_Ecosystem_Ecotoxicology/5.06%3A_Metapopulations |
: Nico van den Brink : Michiel Kraak, Heikki Setälä, You should be able to Populations can be defined as a group of organisms from the same species which live in a specific geographical area. These organisms interact and breed with each other. At a higher level, one can define meta-populations which can be described as a set of spatially separated populations which interact to a certain extent. The populations may function separately, but organisms can migrate between the populations. Generally the individual populations occur in more or less favourable habitat patches which may be separated by less favourable areas. However, in between populations, good habitats may also occur, where populations have not yet established , or the local populations may have gone extinct. Exchange between populations within a meta-population depends on i) the distances between the individual populations, ii) the quality of the habitat between the populations, e.g. the availability of so-called stepping stones, areas where organisms may survive for a while but which are too small or of too low habitat quality to support a local population and iii) the dispersal potential of the species. Due to the interactions between the different populations within a meta-population, chemicals may affect species at levels higher than the (local) population, also at non-contaminated sites. An important effect of chemicals at meta-population scale is that local populations may act as a source or sink for other populations within the meta-population. When a chemical affects the survival of organisms in a local population, the local population densities decline. This may increase the immigration of organisms from neighbouring populations within the meta-population. Decrease of local densities would decrease emigration, resulting in a net-influx of organisms into the contaminated site. This is the case when organisms do not sense the contaminants, or that the contaminants do not alter the habitat quality for the organisms. In case the immigration rate at the delivering/source population to replace the populations is too high to replace the leaving organisms, population densities in neighbouring populations may decline, even at the non-contaminated sites. Consequently, local populations at contaminated sites may act as a for other populations within the meta-population, so chemicals may have a much broader impact than just local. On the contrary, when the local population is relatively small, or the chemical stress is not chronic, meta-population dynamics may also mitigate local chemical stress. Population level impacts of chemicals may be minimised by influx of organisms of neighbouring populations, potentially the population densities prior to the chemical stress. Such recovery depends on the extent and duration of the chemical impact on the local populations and the capacity of the other populations to replenish the loss of the organisms in the affected population. Meta-population dynamics may thus alter the extent to which contaminants may affect populations _ through migration between populations. However, chemicals may affect the total carrying capacity of the meta-population as a whole. This can be illustrated by the modelling approach developed by Levins in the late 1960s (Levins 1969). A first assumption in this model is that not all patches that can potentially carry a local population are actually occupied, so let F be the fraction of occupied patches (1-F being the fraction not occupied). Populations have an average change of extinction being (day when calculating on a daily base), while non-occupied patches have a change of of being populated (day ) from the populated patches. The daily change in numbers of occupied patches is therefore: In this formula *F*(1-F) equals the number of non-occupied patches that are being occupied from the occupied patches, while * F equals the fraction of patches that go extinct during the day. This can be recalculated to a carrying capacity (CC) of while the growth rate (GR) of the meta-population can be calculated by In case chemicals increase extinction risk ( ), or decrease the chance on establishment in a new patch ( ) this will affect the CC (which will decrease because / will increase) as well as the GR (will decrease, may even go below 0). However, this model uses average coefficients, which may not be directly applicable to individual contaminated sites within a meta-population. More (complex) recent approaches include the possibility to use local-population specific parameters and even more, such model include stochasticity, increasing their environmental relevance. Besides affecting populations directly in their habitats, chemicals may also affect the areas between habitat patches. This may affect the potential of organisms to migrate between patches. This may decrease the chances of organisms to repopulate non-occupied patches, i.e. decrease , and as such both CC and GR. Hence, in a meta-population setting chemicals even in a non-preferred habitat may affect long term meta-population dynamics. Levins, R. (1969). Some demographic and genetic consequences of environmental heterogeneity for biological control. Bulletin of the Entomological Society of America 15, 237-240 What are drivers of recovery of local populations that are affected by a chemical stressor in a meta-population setting? In what type of meta-population would a local population be less affected, with smaller number of local populations which are relatively large, or a setup with lot of small local populations? | 5,627 | 2,405 |
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The idea that a molecule could be held together by a shared pair of electrons was first suggested by Lewis in 1916. Although Lewis never won the Nobel prize for this or his many other theories, the shared pair of electrons is nevertheless one of the most significant contributions to chemistry of all time. Wave mechanics was still 10 years in the future, and so Lewis was unable to give any mathematical description of exactly how electron sharing was possible. Instead of the detailed picture presented in the previous section, Lewis indicated the formation of a hydrogen molecule from two hydrogen atoms with the aid of his electron-dot diagrams as follows: Lewis also suggested that the tendency to acquire a noble-gas structure is not confined to ionic compounds but occurs among covalent compounds as well. In the hydrogen molecule, for example, each hydrogen atom acquires some control over electrons, thus achieving something resembling the helium structure. Similarly the formation of a fluorine molecule from its atoms can be represented by Again a pair of electrons is shared, enabling each atom to attain a neon structure with eight electrons (i.e., an octet) in its valence shell. Similar diagrams can be used to describe the other halogen molecules: In each case a shared pair of electrons contributes to a noble-gas electron configuration on atoms. Since only the valence electrons are shown in these diagrams, the attainment of a noble-gas structure is easily recognized as the attainment of a full complement of eight electron dots (an ) around each symbol. In other words as well as ionic compounds obey the octet rule. The octet rule is very useful, though by no means infallible, for predicting the formulas of many covalent compounds, and it enables us to explain the usual valence exhibited by many of the representative elements. According to Lewis’ theory, hydrogen and the halogens each exhibit a valence of 1 because the atoms of hydrogen and the halogens each contain one less electron than a noble-gas atom. In order to attain a noble-gas structure, therefore, they need only to participate in the sharing of pair of electrons. If we identify a shared pair of electrons with a chemical bond, these elements can only form bond. A similar argument can be extended to oxygen and the group VI elements to explain their valence of 2. Here electrons are needed to complete a noble-gas configuration. By sharing two pairs of electrons, i.e., by forming two bonds, an octet is attained: Nitrogen and the group V elements likewise require three electrons to complete their octets, and so can participate in three shared pairs: Finally, since carbon and the group IV elements have four vacancies in their valence shells, they are able to form four bonds:
Draw Lewis structures and predict the formulas of compounds containing (a) P and Cl; (b) Se and H. Draw Lewis diagrams for each atom. Since the P atom can share three electrons and the Cl atom only one, three Cl atoms will be required, and the formula is Since Se is in periodic group VI, it lacks two electrons of a noble-gas configuration and thus has a valence of 2. The formula is In drawing Lewis structures, the bonding pairs of electrons are often indicated by a connecting the atoms they hold together. Electrons which are not involved in bonding are usually referred to as or . Lone pairs are often omitted from Lewis diagrams, or they may also be indicated by lines. Here are some of the alternative ways in which H , F , and PCl can be written. | 3,586 | 2,406 |
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This page gives you the facts and a simple, uncluttered mechanism for the elimination reaction between a simple halogenoalkane like 2-bromopropane and hydroxide ions (from, for example, sodium hydroxide) to give an alkene like propene. 2-bromopropane is heated 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. Everything else present (including anything formed in the alternative substitution reaction) will be trapped in the flask. In elimination reactions, the hydroxide ion acts as a base - removing a hydrogen as a hydrogen ion from the carbon atom next door to the one holding the bromine. The resulting re-arrangement of the electrons expels the bromine as a bromide ion and produces propene. Jim Clark ( ) | 972 | 2,407 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/05%3A_Dioxygen_Reactions/5.02%3A_Chemistry_of_Dioxygen |
The reduction potential for the four-electron reduction of dioxygen (Reaction 5.1) is a measure of the great oxidizing power of the dioxygen molecule. However, the reaction involves the transfer of four electrons, a process that rarely, if ever, occurs in one concerted step, as shown in Reaction (5.2). \[O_{2} \xrightarrow{e^{-}} O_{2}^{-} \xrightarrow{e^{-}, 2H^{+}} H_{2}O_{2} \xrightarrow{e^{-}, H^{+}} H_{2}O + OH \xrightarrow{e^{-}, H^{+}} 2 H_{2}O \tag{5.2}\] \[dioxygen \xrightarrow{e^{-}} superoxide \xrightarrow{e^{-}, 2H^{+}} hydrogen\; peroxide \xrightarrow{e^{-}, H^{+}} water + hyrdoxyl\; radical \xrightarrow{e^{-}, H^{+}} water\] Since most reducing agents can transfer at most one or two electrons at a time to an oxidizing agent, the thermodynamics of the one- and two-electron reductions of dioxygen must be considered in order to understand the overall mechanism. In aqueous solution, the most common pathway for dioxygen reduction in the absence of any catalyst is one-electron reduction to give superoxide. But this is the least favorable of the reaction steps that make up the full four-electron reduction (see Table 5.1) and requires a moderately strong reducing agent. Thus if only one-electron pathways are available for dioxygen reduction, the low reduction potential for one-electron reduction of O to O presents a barrier that protects vulnerable species from the full oxidizing power of dioxygen that comes from the subsequent steps. If superoxide is formed (Reaction 5.3), however, it disproportionates quite rapidly in aqueous solution (except at very high pH) to give hydrogen peroxide and dioxygen (Reaction 5.4). The stoichiometry of the overall reaction is therefore that of a net two-electron reduction (Reaction 5.5). It is thus impossible under normal conditions to distinguish one-electron and two-electron reaction pathways for the reduction of dioxygen in aqueous solution on the basis of stoichiometry alone. \[2O_{2} + 2e^{-} \rightarrow 2 O_{2}^{-} \tag{5.3}\] \[2O_{2}^{-} + 2 H^{+} \rightarrow H_{2}O_{2} + O_{2} \tag{5.4}\] \[O_{2} + 2 e^{-} + 2 H^{+} \rightarrow H_{2}O_{2} \tag{5.5}\] The thermodynamics of dioxygen reactions with organic substrates is also of importance in understanding dioxygen reactivity. The types of reactions that are of particular interest to us here are hydroxylation of aliphatic and aromatic C—H bonds and epoxidation of olefins, since these typical reactions of oxygenase enzymes are ones that investigators are trying to mimic using synthetic reagents. Some of the simpler examples of such reactions (plus the reaction of H for comparison) are given in the reactions in Table 5.2. It is apparent that all these reactions of dioxygen with various organic substrates in Table 5.2 are thermodynamically favorable. However, reactions of dioxygen with organic substrates in the absence of a catalyst are generally very slow, unless the substrate is a particularly good reducing agent. To understand the sluggishness of dioxygen reactions with organic substrates, we must consider the kinetic barriers to these reactions. The principal kinetic barrier to direct reaction of dioxygen with an organic substrate arises from the fact that the ground state of the dioxygen molecule is triplet, i.e., contains two unpaired electrons. Typical organic molecules that are representative of biological substrates have singlet ground states, i.e., contain no unpaired electrons, and the products resulting from their oxygenation also have singlet ground states. Reactions between molecules occur in shorter times than the time required for conversions from triplet to singlet spin. Therefore the number of unpaired electrons must remain the same before and after each elementary step of a chemical reaction. For these reasons, we know that it is impossible for Reaction (5.6) to go in one fast, concerted step. \[\frac{1}{2} \;^{3}O_{2} + \;^{1}X \rightarrow \;^{1}XO \tag{5.6}\] \[\qquad \uparrow \uparrow \qquad \downarrow \uparrow \quad \qquad \downarrow \uparrow\] The arrows represent electron spins: \(\downarrow \uparrow\) represents a singlet molecule with all electron spins paired; \(\uparrow \uparrow\) represents a triplet molecule with two unpaired electrons; and \(\uparrow\) (which we will see in Reaction 5.13) represents a doublet molecule, also referred to as a free radical, with one unpaired electron. The pathways that do not violate the spin restriction are all costly in energy, resulting in high activation barriers. For example, the reaction of ground-state triplet dioxygen, i.e., O , with a singlet substrate to give the excited triplet state of the oxygenated product (Reaction 5.7) is spin-allowed, and one could imagine a mechanism in which this process is followed by a slow spin conversion to a singlet product (Reaction 5.8). \[\frac{1}{2} \;^{3}O_{2} + \;^{1}X \rightarrow \;^{3}XO \tag{5.7}\] \[\qquad \uparrow \uparrow \qquad \downarrow \uparrow \qquad \quad \uparrow \uparrow\] \[ \;^{3}XO \xrightarrow{slow} \;^{1}XO \tag{5.38}\] \[ \; \uparrow \uparrow \qquad \qquad \downarrow \uparrow\] But such a reaction pathway would give a high activation barrier, because the excited triplet states of even unsaturated molecules are typically 40-70 kcal/mol less stable than the ground state, and those of saturated hydrocarbons are much higher. Likewise, a pathway in which O is excited to a singlet state that then reacts with the substrate would be spin-allowed (Reactions 5.9 and 5.10). The high reactivity of singlet dioxygen, generated by photochemical or chemical means, is well-documented. However, such a pathway for a reaction of dioxygen, which is initially in its ground triplet state, would also require a high activation energy, since the lowest-energy singlet excited state of dioxygen is 22.5 kcal/mol higher in energy than ground-state triplet dioxygen. \[\;^{3}O_{2} + 22.5\; kcal/mol \rightarrow \;^{1}O_{2} \tag{5.9}\] \[ \uparrow \uparrow \qquad \qquad \qquad \qquad \qquad \downarrow \uparrow\] \[\frac{1}{2} \;^{1}O_{2} + \;^{1}X \rightarrow \;^{1}XO \tag{5.10}\] \[\quad \downarrow \uparrow \qquad \downarrow \uparrow \quad \qquad \downarrow \uparrow\] Moreover, the products of typical reactions of singlet-state dioxygen with organic substrates (Reactions 5.11 and 5.12, for example) are quite different in character from the reactions of dioxygen with organic substrates catalyzed by oxygenase enzymes (see Section V): \(\tag{5.11}\) \(\tag{5.12}\) One pathway for a direct reaction of triplet ground-state dioxygen with a singlet ground-state organic substrate that can occur readily without a catalyst begins with the one-electron oxidation of the substrate by dioxygen. The products of such a reaction would be two doublets, i.e., superoxide and the oneelectron oxidized substrate, each having one unpaired electron (Reaction 5.13). These free radicals can diffuse apart and then recombine with their spins paired (Reaction 5. 14). \[\;^{3}O_{2} + \;^{1}X \rightarrow \;^{2}O_{2}^{-} + \;^{2}X^{+} \tag{5.13}\] \[\uparrow \uparrow \qquad \downarrow \uparrow \quad \quad \uparrow \qquad \uparrow\] \[\;^{2}O_{2}^{-} + \;^{2}X^{+} \rightarrow \;^{2}O_{2}^{-} + 2X^{+} \rightarrow \;^{1}XO_{2} \tag{5.14}\] \[\uparrow \qquad \uparrow \qquad \qquad \uparrow \qquad \downarrow \qquad \qquad \downarrow \uparrow\] Such a mechanism has been shown to occur for the reaction of dioxygen with reduced flavins shown in Reaction (5.15). \(\tag{5.15}\) However, this pathway requires that the substrate be able to reduce dioxygen to superoxide, a reaction that requires an unusually strong reducing agent (such as a reduced flavin), since dioxygen is not a particularly strong one-electron oxidizing agent (see Table 5.1 and discussion above). Typical organic substrates in enzymatic and nonenzymatic oxygenation reactions usually are not sufficiently strong reducing agents to reduce dioxygen to superoxide; so this pathway is not commonly observed. The result of these kinetic barriers to dioxygen reactions with most organic molecules is that uncatalyzed reactions of this type are usually quite slow. An exception to this rule is an oxidation pathway known as free-radical autoxidation. The term free-radical autoxidation describes a reaction pathway in which dioxygen reacts with an organic substrate to give an oxygenated product in a free-radical chain process that requires an initiator in order to get the chain reaction started. (A free-radical initiator is a compound that yields free radicals readily upon thermal or photochemical decomposition.) The mechanism of free radical autoxidation is as shown in Reactions (5.16) to (5.21). Initiation: $$X_{2} \rightarrow 2X \cdotp \tag{5.16}\] \[X \cdotp + RH \rightarrow XH + R \cdotp \tag{5.17}\] Propagation: $$R \cdotp + O_{2} \rightarrow ROO \cdotp \tag{5.18}\] \[\downarrow \qquad \uparrow \uparrow \qquad \quad \uparrow\] \[ROO \cdotp + RH \rightarrow ROOH + R \cdotp \tag{5.19}\] Termination: $$R \cdotp + ROO \cdotp \rightarrow ROOR \tag{5.20}\] \[2 ROO \cdotp \rightarrow ROOOOR \rightarrow O_{2} + ROOR \tag{5.21}\] (plus other oxidized products, such as ROOH, ROH, RC(O)R, RC(O)H). This reaction pathway results in oxygenation of a variety of organic substrates, and is not impeded by the spin restriction, because triplet ground-state dioxygen can react with the free radical R• to give a free-radical product ROO•, in a spin-allowed process (Reaction 5.18). It is a chain reaction, since R• is regenerated in Reaction (5.19), and it frequently occurs with long chain lengths prior to the termination steps, resulting in a very efficient pathway for oxygenation of some organic substrates, such as, for example, the oxidation of cumene to give phenol and acetone (Reaction 5.22). \(\tag{5.22}\) When free-radical autoxidation is used for synthetic purposes, initiators are intentionally added. Common initiators are peroxides and other compounds capable of fragmenting readily into free radicals. Free-radical autoxidation reactions are also frequently observed when no initiator has been intentionally added, because organic substrates frequently contain peroxidic impurities that may act as initiators. Investigators have sometimes been deceived into assuming that a metal-complex catalyzed reaction of dioxygen with an organic substrate occurred by a nonradical mechanism. In such instances, the reactions later proved, upon further study, to be free-radical autoxidations, the role of the metal complex having been to generate the initiating free radicals. Although often useful for synthesis of oxygenated derivatives of relatively simple hydrocarbons, free-radical autoxidation lacks selectivity and therefore, with more complex substrates, tends to give multiple products. In considering possible mechanisms for biological oxidation reactions used for biosynthesis or energy production, free-radical autoxidation is not an attractive possibility, because such a mechanism requires diffusion of highly reactive free radicals. Such radicals, produced in the cell, will react indiscriminately with vulnerable sites on enzymes, substrates, and other cell components, causing serious damage. In fact, free-radical autoxidation is believed to cause certain deleterious reactions of dioxygen in biological systems, for example the oxidation of lipids in membranes. It is also the process that causes fats and oils to become rancid (Reaction 5.23). \(\tag{5.23}\) We see then the reasons that uncatalyzed reactions of dioxygen are usually either slow or unselective. The functions of the metalloenzymes for which dioxygen is a substrate are, therefore, to overcome the kinetic barriers imposed by spin restrictions or unfavorable one-electron reduction pathways, and, for the oxygenase enzymes, to direct the reactions and make them highly specific. It is instructive to consider (1) how these metalloenzymes function to lower the kinetic barriers to dioxygen reactivity, and (2) how the oxygenase enzymes redirect the reactions along different pathways so that very different products are obtained. The first example given below is cytochrome c oxidase. This enzyme catalyzes the four-electron reduction of dioxygen. It overcomes the kinetic barriers to dioxygen reduction by binding dioxygen to two paramagnetic metal ions at the dioxygen binding site, thus overcoming the spin restriction, and by reducing dioxygen in a two-electron step to peroxide, thus bypassing the unfavorable one-electron reduction to form free superoxide. The reaction occurs in a very controlled fashion, so that the energy released by dioxygen reduction can be used to produce ATP. A second example is provided by the catechol dioxygenases, which appear to represent substrate rather than dioxygen activation, and in which dioxygen seems to react with the substrate while it is complexed to the paramagnetic iron center. Another example given below is the monooxygenase enzyme cytochrome PASO, which catalyzes the reaction of dioxygen with organic substrates. It binds dioxygen at the paramagnetic metal ion at its active site, thus overcoming the spin restriction, and then carries out what can be formally described as a multielectron reduction of dioxygen to give a highly reactive high-valent metal-oxo species that has reactivity like that of the hydroxyl radical. Unlike a free hydroxyl radical, however, which would be highly reactive but nonselective, the reaction that occurs at the active site of cytochrome P-450 can be highly selective and stereospecific, because the highly reactive metal-oxo moiety is generated close to a substrate that is bound to the enzyme in such a way that it directs the reactive oxygen atom to the correct position. Thus, metalloenzymes have evolved to bind dioxygen and to increase while controlling its reactivity. | 13,842 | 2,409 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Electrophilic_Aromatic_Substitution/AR3._Formation_of_Electrophiles |
The mechanism of electrophilic aromatic substitution follows two elementary steps. First, donation of a pair of π electrons to the electrophile results in a loss of aromaticity and formation of a cation. Second, removal of a proton from that cation restores aromaticity. How does the electrophile form in the first place? The details of that part of the reaction vary from case to case. With the catalysed bromine reaction, the Lewis acid activates the halogen to render it more electrophilic. The activation may even go so far as to form a bromine cation, as suggested earlier. Otherwise, the positive charge on the bromine atom that ligates the Lewis acid can be nullified, indirectly, when the arene donates to the terminal bromine atom. The appearance of a bromide ion to deprotonate the cation simply results fom the equilibrium of the Lewis acid-base complex. Show the mechanism for chlorination of benzene in the presence of ferric chloride. The reactions of alkyl and acyl halides also involve Lewis acid catalysts; frequently, aluminum chloride (AlCl ) is employed. These two reactions are called - reactions after the French and American co-discoverers of the reaction. Typically, Friedel-Crafts reactions are believed to occur through initial formation of cationic electrophiles, which then react with aromatics in the same way as halogen electrophiles. Because Friedel-Crafts alkylations occur via alkyl cations, the reactions of primary alkyl halides are generally pretty slow. Also, in some cases, multiple products may result via rearrangements. These observations provide additional evidence for the cationic nature of the intermediate. Show the mechanism for the Friedel-Crafts alkylation of benzene with 2-chloropropane in the presence of aluminum chloride. Why is the Friedel-Crafts reaction of 1-chloropropane so much slower than the reaction of 2-chloropropane? Explain using a mechanism and intermediates. Show why Friedel-Crafts alkylation of benzene with 2-chloropentane results in the formation of two different products. Show the mechanism for the Friedel-Crafts acylation of benzene with ethanoyl chloride (acetyl chloride) in the presence of aluminum chloride. Show why the Friedel-Crafts acylation of benzene with pentanoyl chloride results in only one product, with no rearrangement. Nitration and sulfonation reactions differ from the other substitutions that we have seen because they do not utilize Lewis acid catalysis. These reactions depend on equilibria that occur in strongly acidic media. When nitric acid is dissolved in sulfuric acid, there is spectroscopic evidence than NO forms, providing an electrophile. Similarly, when sulfuric acid is concentrated by boiling off residual water, sulfur trioxide results. The latter probably forms via SO H , the electrophile in sulfonation. Provide a mechanism for the formation of NO from nitric and sulfuric acid. Provide a mechanism for the formation of SO H from sulfuric acid. , | 2,982 | 2,410 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/05%3A_Solutions/5.E%3A_Solutions_(Exercises) |
Calculate the molality for a 0.005M NH SO . What is the difference between the molarity and molality of the solution? Calculate the molality for a 0.05M ammonium sulfite [NH SO ] solution. The molarity of a solution is the ratio between the number of moles of solute and the solution. The molality of a solution is the ratio between the number of moles of solute and the mass of solvent: After taking tylenol for stomach pains, a man has 0.300 mg of acetaminophen in his 5.0 L blood. After 4 hours, there is 0.010 mg left in his blood. Calculate the number of moles of acetaminophen per ml of blood and the total number of moles and grams before and after. (Molar Mass of Acetaminophen: 151.2 g/mol) Before: \(n=.300\times 10^{-3}g \times \dfrac{1mol}{151.2g}=1.98 \times 10^{-6}mol\) \(M=\dfrac{1.98\times10^{-6}mol}{5.0L}(\dfrac{1000ml}{1L})=3.96\times 10^{-4}\dfrac{mol}{ml}\) After: \(n=.010\times 10^{-3}g \times \dfrac{1mol}{151.2g}=6.61 \times 10^{-8}mol\) \(M=\dfrac{6.61\times10^{-8}mol}{5.0L}(\dfrac{1000ml}{1L})=1.32\times 10^{-5}\dfrac{mol}{ml}\) Charlie goes on a submarine ride with his family. On their submarine ride, they have dinner which Charlie washes down with a Coke. Charlie notes that upon the removal of the cap from the Coke bottle, it does not make much of a noise that he has grown accustomed to hearing when opening a soft drink bottle. Soon after their dinner, Charlie and his family make their way back up to the surface to disembark from their submarine journey. As they were ascending, Charlie started to burp. What has just occurred? When on a submarine, you submerge into the water and are below sea level. When below sea level, the total and partial pressures of CO increases in comparison to when you are at or above sea level. Since Charlie's family has brought the Coke down from land (at sea level), the solubility of the CO is increased and the concentration of the CO in solution is raised. When Charlie returns to sea level, the conditions go back to normal which causes both the pressure and the solubility to decrease and the CO inside of Charlie comes out and causes him to burp in order to release the CO gas. Calculate the Henry's Law constant for O in blood at a partial pressure of 0.25. There's about 2.5 kg of blood in the individual and 15 g of \(O_2\) present. \[P_{O_2}=k\cdot m \rightarrow k=\dfrac{P_{0_2}}{m}\] \(k=\dfrac{.25}{.188\dfrac{mol}{kg}}\) \(M=(15.0g)(\dfrac{1mol}{32.0g})\) Imagine the ideal gas \(F_2\) (at 25°C) has a solubility of \(123 \times 10^{-4}\; mol/L\) at a partial pressure of 0.5 atm, and \(321 \times 10^{-4}\) at 5 atm. Use Henry’s law to determine the number of moles of \(F_2\) at the given pressures. Write the equation showing the relationship between the molality and the change of temperature. Show possible reasons that leading to the derivation of this equation. \[ \Delta T = K_b \times m_2\] Molality is independent of temperature. There are two liquids X and Y. The boiling point of liquid X is 125 C while the boiling point of liquid Y is 250 C. If liquids X and Y were mixed to form an ideal solution, would the boiling point of the solution be less than, greater than, or equal to that of (a) liquid X? (b) liquid Y? Why don't you keep your honey in the fridge? Explain using the terms hypertonic, and hypotonic. Honey is a hypertonic solution and bacteria prefer a hypotonic environment. The freezing point depression measurement of Compound X in Reagent A yields a molar mass of 144 g; the same measurement in Reagent Y gives a value of 72 g. What is the reason for this discrepancy? (Hint: Consider solvent-solute interactions). What would be the volume of 20 moles of sugar (C12H22O11) dissolved in 0.5 g/cm^-3 of water at 10K? Given the freezing point depression constant for water is 1.86 K/mol kg, what would be the molality of the sugar? You have a boiling pot with 4 L of water. You add 1000 grams of NaCl. Water boils at 100 C and freezes at 0 C. To what temperature is the boiling point of water elevated to with the addition of NaCl? To what temperature is the freezing point depressed to? Blood plasma, an enriched with platelets, contains several growth factors that help healing of tissues and bones. There are a lot of medical treatments which inject the comparable concentration of a certain solution into human blood. Explain why ? The injected solution should be comparable to that of blood plasma. There is a certain amount of platelets which keep the state the blood plasma in equilibrium. If we inject too much, it significantly affects to human blood by changing the concentration of platelets. Immerse a human cell in hypertonic or hypotonic solution can cause harmful consequences. Explain what would happen in each situation and the case when we got injected with a high concentrated solutions in our vein? The Davis water tower is 250 ft tall (about 76 meters). What is the osmotic pressure required to push water from the ground to the top of the tower? \[\pi = h\rho g\] \[\pi = (76m)(1000kg \ m^{-3})(9.81 m \ s^{-2})\] \[\pi = (7.456\ast 10^{5}N \ m^{-2})(\dfrac{1 atm}{1.013 \ast 10^{5}N\ m^{-2}})\] \[\pi = 7.36 atm\] A solution of liquid A is mixed with liquid B and exerts ideal behavior. At \(103^oC\) the total vapor pressure of a solution is 413mmHg containing 1.4 moles of A and 2.0 moles of B. After you add another mole of A to the solution, the vapor pressure increases to 486mmHg. At \(103^oC\) calculate the vapor pressure of pure A and B. Step1: (Hint: Calculate the mole fraction of A and B) \[x_A=(\dfrac{molesA}{(molesA+molesB)})=(\dfrac{1.4molesA}{(1.4molesA+2.0molesB)})=.42\] \[x_B=(\dfrac{molesB}{(molesA+molesB)})=(\dfrac{2.0molesB}{(1.4molesA+2.0molesB)})=.58\] Step 2: (Hint: Set up total pressure equation) \[P=x_AP^*_A+x_BP^*_B\] \[413=(.42)P^*_A+(.58)P^*_B\] Step 3: (Hint: Calculate the mole fraction of A and B with the addition of 1 mole of A) \[x_A=(\dfrac{molesA}{(molesA+molesB)})=(\dfrac{2.4molesA}{(2.4molesA+2.0molesB)})=.55\] \[x_B=(\dfrac{molesB}{(molesA+molesB)})=(\dfrac{2.0molesB}{(2.4molesA+2.0molesB)})=.45\] Step 4: (Hint: Set up the second total pressure equation) \[486=(.55)P^*_A+(.45)P^*_B\] Step 5: (Hint: Solve for \(P^*_A\) and \(P^*_B\) by substitution of the two total pressure equations) \[413=(.42)P^*_A+(.58)P^*_B\] \[486=(.55)P^*_A+(.45)P^*_B\] Rearrange equation 1 \[P^*_A=983.33-1.38095P^*_B\] Substitute into any equation and solve for \(P^*_B\) \[486=(.55)(983.33-1.38095P^*_B)+(.45)P^*_B\] \[P^*_B=177.149mmHg\] Substitute \(P^*_B\) into any equation to get \(P^*_A\) \[P^*_A=983.33-1.38095P^*_B\] \[=738.696mmHg\] An ideal solution is composed of equal amount of component A with molar mass of 28.05g/mol and component B with molar mass of 80.91g/mol. At 35°C, component A has a vapor pressure of 0.2 atm and component B has a vapor pressure of 0.65 atm. (a) remember mole fraction = (mole of solute) / (total mole of solution) ; \[n_A=\dfrac{weight_A}{28.05g/mol}\\
n_B=\dfrac{weight_B}{80.91g/mol}\\x_A=\dfrac{n_A}{n_A+n_B}=\dfrac{\dfrac{weight_A}{28.05g/mol}}{\dfrac{weight_A}{28.05g/mol}+\dfrac{weight_B}{80.91g/mol}}=\dfrac{1/28.05}{(1/28.05)+(1/80.91)}=0.7426\\
x_B=\dfrac{n_B}{n_A+n_B}=\dfrac{\dfrac{weight_B}{80.91g/mol}}{\dfrac{weight_A}{28.05g/mol}+\dfrac{weight_B}{80.91g/mol}}=\dfrac{1/80.91}{(1/28.05)+(1/80.91)}=0.2574\] (b)\(P_A=x_A*P_A^*=0.7426*0.2atm=0.1485atm\\
P_B=x_B*P_B^*=0.2574*0.62atm=0.1596atm\) (c)\(x_A=\dfrac{P_A}{P_A+P_B}=\dfrac{0.1485atm}{0.1485atm+0.1596atm}=0.4820\\
x_B=\dfrac{P_B}{P_A+P_B}=\dfrac{0.1596atm}{0.1485atm+0.1596atm}=0.5180\\
P_A=x_A*P_A^*=(0.4820)(0.2atm)=0.0964atm\\
P_B=x_B*P_B^*=(0.5180)(0.62atm)=0.3212atm\) Give a brief explanation for the lower vapor pressure of a solvent in presence of a solute from the entropy point of view. (Hint: Look at colligative property of Describe the process of vapor-pressure lowering from the colligative properties of solutions. Suppose you have an ideal solution to which you can apply Raoult's law: (Note: This solution has both a solvent and a solute which is nonvolatile). Next, since we know that , we can ultimately rearrange the equation to be: In the equation above, is directly proportional to the mole fraction of the solute. Sugar is added to water. Would there be more water molecules in the vapor above the sugar solution or in the vapor above the pure water? We would expect more water molecules in the vapor above the pure water. 4.0 moles of a compound was added to 500 g of HCl. What was the freezing point of the pure solvent if the freezing point of the solution is 188.6 K. (K =\(.55\dfrac{K\cdot kg}{mol})\) \(T_f= 188.6K\\ m= \dfrac{4mol}{.500kg}= 8\dfrac{mol}{kg}\) \[\bigtriangleup T= -(K_f\cdot m) \rightarrow T_i=T_f+[(k_f)(m)]= 188.6K+(.55\dfrac{K\cdot kg}{mol})(8\dfrac{mol}{kg}= 193K\] After dissolving a substance weighing 0.246 grams in 100.0 g of water, the freezing point of the solution went down 1K below that of the pure solvent. What would be the molar mass of this substance? The following information is needed. \[ \triangle T=K_fm_2 \\ K_{fH_2O} = 1.86 \] Now solve for the molar mass of the solute. \[ 1 \ K = 1.86 m_2 \\ m_2 = 0.538 \\ m_2 = mol \div kg \ of \ solvent \\ mol_2 = 0.538 \ mol \ kg^{-1} \cdot 0.1 \ kg \ of \ water \\ mol_2 = 0.0538 \ moles \ of \ substance \\ Molar \ Mass = 0.246 \ grams \ of \ substance \ \div \ 0.0538 \ moles \ of \ substance \\ \underline{Molar \ Mass = 4.572 \ g/mol} \] for more information on freezing point depression. 5.525 grams of toluene is mixed with 20.0 grams of acetone. Calculate the freezing point of acetone. The freezing point of the solution is found to be -102.6 C. The K of acetone is 2.40 C kg/mol, and the molar mass of toluene is 92.14 g/mol We can relate the molality of the solution to the freezing point depression: \[m=\dfrac{\Delta T}{K_{f}}\] Rearranged: \[\Delta T=K_{f}m\] \[T_{acetone}-T_{solution}=K_{f}m\] \[T_{acetone}=T_{solution}+K_{f}\dfrac{mols \,of \,toluene}{kg\,of\, solvent}\] \[=-102.6^{\circ}C+(2.40\dfrac{^{\circ}C}{kg\cdot mol})\dfrac{(5.252g\times \dfrac{1 mol}{92.14g})}{20.0\times 10^{-3}kg}\] \[=-95.8^{\circ}C\] Two aqueous urea solutions have osmotic pressures of 1.4 atm and 3.0 atm, respectively, at a certain temperature. What is the osmotic pressure of a solution prepared by mixing equal volumes of these two solutions at the same temperature? A research scientist used 50 mL of sterilized water to conduct experiments and obtained odd results. She suspected that it had been contaminated with another compound and determined that the boiling point of the solution had increased to 105 C. How many moles of the compound (assume it dissolves as a molecule) had been added to the water? Time-released pills are designed to have longer effect at a constant rate. The active ingredient is concealed by insoluble substances so that the active ingredient has to be dissolved by getting through the holes of insoluble layers covering outside. Explain how the insoluble layers regulate the absorption of the active ingredient. Acetic Acid can make hydrogen bonds with water molecules and also with benzene. A solution of 4.0 g of acetic acid in 95g \(C_6H_6\) has a freezing point of \(3.5^oC\). What is the molar mass of the solute? \[\Delta T=K_fm\] =molality\) \[3.5^oC=(5.12^oC\,kg\,mol^{-1})(m)\] \[m=.68359mol\,kg^{-1}\] \[m=\dfrac{moles\,solvent}{kg\,solvent}\] \[moles\,solvent=.71757moles\] \[mol\,acetic\,acid=\dfrac{g\,acetic\,acid}{molar\,mass,acetic\,acid}\] \[7.1957\,moles=\dfrac{4.0g\,acetic\,acid}{.555887g\,mol^{-1}}\] Determine on whether each of the following statements is true or false and briefly explain your answers. (a) Colligative properties depends on both the number of solute molecules and the size of the solute molecules. (b) The addition of a nonvolatile solute to a solvent only changes the chemical potential of liquid \((\mu _1(l))\). (c) The mole fraction of a component is never the same as the activity of the same component. Provide an example of a nonideal solution explain why it is nonideal.(Hint: An example of a non-ideal solution is a mixture of diethyl ether and water. Diethyl ether is a non-polar solvent used for dissolving organic molecules whereas water is polar. Intermolecular forces between these two components are weak; hence they will form two layers upon mixing; making the mixture a non-ideal solution. An unknown substance has a molal boiling-point elevation constant, K of 1.23 K/m. Find the molar mass of this substance. The enthalpy of vaporization for this substance is 38.56 kJ/mol at 78.37 C. For any substance: \[K_{b}=\dfrac{RT_{0}^{2}\mathfrak{M}_{1}}{\Delta _{vap}\bar{H}}\] Rearranging this equation, we get: \[\mathfrak{M}_{1}=\dfrac{K_{b}\Delta _{vap}\bar{H}}{RT_{0}^{2}}\] \[=\dfrac{(1.23\dfrac{K}{mol})(38.56\times 10^{3}\dfrac{J}{mol})}{(8.3145\dfrac{J}{K\cdot mol})(78.37+276.15)^{2}K}\] \[=0.04616 \dfrac{kg}{mol}=46.16\dfrac{g}{mol}\] Find K (the molal boiling-point elevation constant) for acetone solution at 56ºC, given that: \[ \triangle _{vap}\overline{H}^{\circ}_{Acetone} = 31.3 \ kJ \ mol^{-1}\] You need to use the following equation: \[ K_b = \dfrac{RT_{b}^{2}M_{molar}}{\triangle_{vap}\overline{H}}\] Plug in values and solve for the K . \[ \begin{align}
&K_b = \dfrac{(\dfrac{8.3145}{1000} \ kJ \ mol^{-1} \ K^{-1})(298K)^{2}(\dfrac{58.08}{1000}\ kg \ mol^{-1})}{31.3 \ kJ \ mol^{-1}} \\
& \underline{K_b = 1.37 K \ mol^{-1} \ kg}
\end{align} \] for more information on boiling point elevation. Describe the difference between a hypertonic, and hypotonic. Why does a cell shrink in a hypertonic solution, and expand in a hypotonic solution. The following data give the pressures for water-methanol solution at 39.9 C. Find the activity coefficients of both components based on: A solution is made up of some amount of sucrose (C H O ) and 1.03 kg water. Calculate the amount of sucrose (in grams) in solution if the solution freezes at -2 C and the activity coefficient of sucrose is 0.789. \[\Delta T = K_{b}m\] \[m = \dfrac{\Delta T}{K_{b}} = \dfrac{2K}{1.86\ K\ mol^{-1}\ kg}\] \[m = 1.075 mol kg^{-1}\] a is effective molar concentration so a = m in: \[a = \gamma x\] x is the actual amount of sucrose in solution, so \[x = \dfrac{a}{\gamma} = \dfrac{1.075}{0.789}\] \[x = 1.36 mol \ kg^{-1}\] \[mass\ sucrose = (\dfrac{1.36 mol}{kg\ H_{2}O})(1.03 kg\ H_{2}O)(\dfrac{342.3 g}{mol})\] \[mass\ sucrose = 479 g\] You know that the mean activity coefficient is 0.42. From this information find what the ionic strength of the compound LiNO is. I=0.548 A 0.15 m Ca(NO ) solution has a mean ionic activity coefficient of 0.17 at 25°C. Calculate the mean molality, the mean ionic activity, and the activity of the compound. In 0.03m aqueous NaNO solution at temperature 300K., the size of the ionic atmosphere is 1/k, also known as Debye radius, is 27.8x10 m. Using the Debye radius formula to calculate the ionic strength I. For a 0.0030m aqueous solution of \(CaCl_2\) at 298K, calculate the ionic strength. (a) Calculate the activity coefficients of \(Ca^{2+}\) and \(Cl^-\) ions in the solution and (b) calculate the mean ionic activity coefficients of these ions. (Hint: Use the Debye-Huckel limiting law) \[I=\dfrac{1}{2}\Sigma _im_iz^2_i\] \[=\dfrac{1}{2}[(0.0030m)(2)^{2}+(0.0060m)(-1)^{2}]\] \[=-.009m\] (a) \[log\gamma_i=-0.509z^2_i\sqrt{I}\] For \(Ca^{2+}\), \[log\gamma_+=-0.5092^2\sqrt{0.009}\] \[log\gamma_+=-0.193\] \[\gamma_+=.640985\] For \(Cl^{-}\), \[log\gamma_-=-0.509(-1)^2\sqrt{.009}\] \[log\gamma_-=-0.048288\] \[\gamma_-=.89477\] (b) \[\gamma _{\pm}=(\gamma _+^{v_{-}}\gamma _{-}^{v_{-}})^{\dfrac{1}{v}}=[(.640985)^{1}(.89477)^2]^{\dfrac{1}{3}}=.8006\] For an ideal mixture, what mole fraction of each gas maximizes the entropy of mixing: \[\Delta _{mix}S=-nR(x_{1}lnx_{1}+x_{2}lnx_{2})\] If we take the partial derivative of the : \[\Delta _{mix}S=-nR(x_{1}lnx_{1}+x_{2}lnx_{2})\] \[\left [\dfrac{\delta \Delta _{mix}S}{\delta x_{1}} \right ]_{_{x_2}, n, R} =\left [-\dfrac{\delta }{\delta x_{1}}nR(x_{1}lnx_{1}+x_{2}lnx_{2}) \right ]_{x_{2}, n, R}\] If we rewrite the function in terms of x \[=\left [-\dfrac{\delta }{\delta x_{1}}nR(x_{1}lnx_{1}+(1-x_{1})ln(1-x_{1}) \right ]_{x_{2}, n, R}\] and derive using the chain rule: \[=-\left (x_{1}\dfrac{1}{x_{1}}+lnx_{1}+(1-x_{1})\left (-\dfrac{1}{1-x_{1}} \right )+ln(1-x_{1})(-1) \right )\] \[=1-lnx_{1}+1+ln(1-x_{1})\] \[=-lnx_{1}+ln(1-x_{1})\] \[=-ln\dfrac{x_{1}}{1-x_{1}}\] To find the local maximum, we set this to 0 and solve for x \[0=-ln\dfrac{x_{1}}{1-x_{1}}\] \[e^{0}=\dfrac{x_{1}}{1-x_{1}}\] \[1=\dfrac{x_{1}}{1-x_{1}}\] \[1-x_{1}=x_{1}\] \[x_{1}=0.5\] To check that this is a maximum and not a minimum, we plug in values to the left and right of the given x value (ex. 0.4 and 0.6). We then find that at x <0.5 the derivative, or slope, is positive and at x 0.5m the slope is negative. This is further confirmed by the following diagram, where is maximized at mole fraction 0.5. The osmotic pressure arbitrary non-ideal diluted solution is measured at 2 different concentrations at 300K. At concentration of 5.20 g/L, osmotic pressure is 10.20 x 10 atm and at concentration of 8.90 g/L the osmotic pressure is 2.30 x 10 atm. Use this information to estimate the molar mass of the solute. (Hint: For dilute solutions, only the second virial coefficient is concerned. Look at equation for and make a graph of slope of the graph of \[slope = \dfrac{(2.584 \times 10^{-6} - 2.038 \times 10^{-6}) \dfrac{atm \cdot L}{g}}{(8.90 - 5.20) \dfrac{g}{L}} = 1.476 \times 10^{-7} \dfrac{atm \cdot L^2}{g^2}\] \[ \dfrac{\pi}{c} =\dfrac{RT}{M}(1+Bc) = \dfrac{RT}{M} + (\dfrac{RTB}{M})c \] \[slope = \dfrac{RBT}{M} = 1.476 \times 10^{-7} \dfrac{atm \cdot L^{2}}{g^{2}}\] \[2.584 \times 10^{-6} = \dfrac{RT}{M} + ( 1.476 \times 10^{-7})(8.90) \] \[ \Rightarrow \dfrac{RT}{M} = 1.271 \times 10^{-6} \dfrac{atm \cdot L}{g} \] \[M = RT (1.271 \times 10^{-6} \dfrac{atm \cdot L}{g}) = (0.08206 \dfrac{L \cdot}{K \cdot mol})(300K)(1.271 \times 10^{-6} \dfrac{atm \cdot L}{g}) = {\color{red}3.13 \times 10^{-5} \dfrac{g}{mol}}\] Assume that you have made an ideal solution of an unknown compound made of two components. Why does the mole fraction 0.5 achieve the maximum \(Δ_{mix}S\)? Use the equation. \[ \Delta_{mix} S = -nR ( x_1 \ln x_1+x_2 \ln x_2 ) \] Find the partial derivative since we want to find out the max value for this equation for x . We find x since it will show us x . \[\begin{align} [\dfrac{\partial }{\partial x_1} \Delta_{mix} S]_{x_2,n,R} &= [\dfrac{\partial }{\partial x_1}-nR \left( x_1 \ln x_1+x_2 \ln x_2 \right) ]_{x_2,n,R} \\ since & \ x_1+x_2 =1 \\ &= - [\dfrac{\partial }{\partial x_1}nR \left( x_1 \ln x_1+(1-x_1) \ln (1-x_1) \right) ]_{x_2,n,R} \\ derive & \\ &= -\left [(x_1 \cdot \dfrac{1}{x_1})+(\ln x_1)+ \left((1-x_1) \cdot \dfrac{-1}{1-x_1} \right ) + \left( (-1) \cdot \ln (1-x_1) \right ) \right] \\ &= -1- \ln x_1+1+ \ln (1-x_1) \\ &= \ln (1-x_1) - \ln x_1 \\ &= \ln \dfrac{1-x_1}{x_1} \end{align}\] Now that we have the partial derivative, we set it to 0 and solve to attempt to find the maximum value. \[\begin{align} & 0 = \ln \dfrac{1-x_1}{x_1} \\ &e^{0} = e^{\ln \dfrac{1-x_1}{x_1}} \\ &1 =\dfrac{1-x_1}{x_1} \\ &x_1 = 1-x_1 \\ &2x_1 = 1 \\ &\underline{x_1 = 0.5} \end{align}\] Now we know that x = 0.5 is either a min or max. If we plug in a number smaller then 0.5 and bigger than 0.5 into the partial derivative we found earlier, we can conclude that the is to the left of 0.5 and to the right of the equation. This is characteristic of maximum values. Thus x = 0.5 and x = 0.5 attain the maximum values of Δ Δ | 19,748 | 2,411 |
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Boiling commences when the vapor pressure of a liquid or solution equals the external or applied pressure (often atmospheric pressure). Thus, (see the graph for cinnamyl alcohol in Figure 5.47a). This behavior occurs because a lower vapor pressure is necessary for boiling, which can be achieved at a lower temperature. For example, water produces a vapor pressure of \(760 \: \text{mm} \: \ce{Hg}\) at \(100^\text{o} \text{C}\) (see Figure 5.47b), and so water boils at sea level at \(100^\text{o} \text{C}\). However, if water was subjected to a reduced pressure of only \(100 \: \text{mm} \: \ce{Hg}\), it would boil at roughly \(50^\text{o} \text{C}\) as this is the temperature that it produces a vapor pressure of \(100 \: \text{mm} \: \ce{Hg}\) (see Figure 5.47b). This trend follow the . The dependence of boiling point on applied pressure can be exploited in the distillation of very high boiling compounds (normal b.p. \(> 150^\text{o} \text{C}\)), which may decompose if heated to their normal boiling point. A vacuum distillation is performed by applying a vacuum source to the vacuum adapter of either a simple or fractional distillation (Figure 5.48). When the pressure is lowered inside the apparatus, solutions boil at a lower temperature. \(^{12}\)Data from , 12\(^\text{th}\) edition, Merck Research Laboratories, . \(^{13}\)Data from J. A. Dean, , 15\(^\text{th}\) ed., McGraw-Hill, , Sect. 5.28. | 1,434 | 2,413 |
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\[ 2 \times \text { atomic mass of carbon} = 2 \, atoms \left ( {12.011 \, amu \over atoms } \right ) = 24.022 \,amu \] \[ 6 \times \text { atomic mass of hydrogen} = 2 \, atoms \left ( {1.0079 \, amu \over atoms } \right ) = 6.0474 \,amu \] \[ 1 \times \text { atomic mass of oxygen} = 1 \, atoms \left ( {15.9994 \, amu \over atoms } \right ) = 15.994 \,amu \] Adding together the masses gives the molecular mass: \[ 24.022 \,amu + 6.0474 \,amu + 15.9994 \,amu = 46.069 \,amu\] Alternatively, we could have used unit conversions to reach the result in one step: \[ \left [ 2 \, atoms C \left ( {12.011 \, amu \over 1 \, atom C} \right ) \right ] + \left [ 6 \, atoms H \left ( {1.0079 \, amu \over 1 \, atom H} \right ) \right ] + \left [ 1 \, atoms C \left ( {15.9994 \, amu \over 1 \, atom 0} \right ) \right ] = 46.069 \, amu \] The same calculation can also be done in a tabular format, which is especially helpful for more complex molecules: \[ 2 C \, \, \, (2\, atoms) (12.011 \, amu/atom ) = 24.022 \, amu \] \[ 6 H \, \, \, (6\, atoms) (1.0079 \, amu/atom ) = 6.0474 \, amu \] \[ + 1O \, \, \, (1\, atoms) (15.9994 \, amu/atom ) = 15.9994 \, amu \] \[ C_2H_6O \, \, \, \, \, \text {molecular mass of ethanol} = 46.069 \, amu \] Calculate the molecular mass of trichlorofluoromethane, also known as Freon-11, whose condensed structural formula is \(CCl_3F\). Until recently, it was used as a refrigerant. The structure of a molecule of Freon-11 is as follows: : 137.368 \,amu Unlike molecules, which form covalent bonds, ionic compounds do not have a readily identifiable molecular unit. Therefore, for ionic compounds, the formula mass (also called the empirical formula mass) of the compound is used instead of the molecular mass. The formula mass is the sum of the atomic masses of all the elements in the empirical formula, each multiplied by its subscript (written or implied). It is directly analogous to the molecular mass of a covalent compound. The units are atomic mass units. Atomic mass, molecular mass, and formula mass all have the same units: atomic mass units. Determining the Molar Mass of a Molecule: Calculate the formula mass of Ca (PO ) , commonly called calcium phosphate. This compound is the principal source of calcium found in bovine milk. : ionic compound : formula mass : : The empirical formula—Ca (PO ) —indicates that the simplest electrically neutral unit of calcium phosphate contains three Ca ions and two PO ions. The formula mass of this molecular unit is calculated by adding together the atomic masses of three calcium atoms, two phosphorus atoms, and eight oxygen atoms. Taking atomic masses from the periodic table, we obtain \[ 3 \times \text {atomic mass of calcium} = 3 \, atoms \left ( {40.078 \, amu \over atom } \right ) = 120.234 \, amu \] \[ 2 \times \text {atomic mass of phosphorus} = 2 \, atoms \left ( {30.973761 \, amu \over atom } \right ) = 61.947522 \, amu \] \[ 8 \times \text {atomic mass of oxygen} = 8 \, atoms \left ( {15.9994 \, amu \over atom } \right ) = 127.9952 \, amu \] Adding together the masses gives the formula mass of Ca (PO ) : \[120.234 \,amu + 61.947522 \, amu + 127.9952 \, amu = 310.177 \, amu \] We could also find the formula mass of Ca3(PO4)2 in one step by using unit conversions or a tabular format: \[ \left [ 3 \, atoms Ca \left ({40.078 \, amu \over 1 \, atom Ca } \right ) \right ] + \left [ 2 \, atoms P \left ({30.973761 \, amu \over 1 \, atom P } \right ) \right ] + \left [ 8 \, atoms O \left ({15.9994 \, amu \over 1 \, atom O } \right ) \right ] \] \[= 310.177 \,amu \] \[ 2P \, \, \, \, (2\, atoms)(30.973761 \, amu/atom) = 61.947522 \, amu \] \[ + 8O \, \, \, \, (8\, atoms)(15.9994 \, amu/atom) = 127.9952 \, amu \] \[ Ca_3P_2O_8 \, \, \, \, \text {formula mass of Ca}_3(PO_4)_2 = 310.177 \, amu \] \[(moles)(molar mass) \rightarrow mass \label{2.7.1}\] or, more specifically, \[ moles \left ( {grams \over mole } \right ) = grams \] Conversely, to convert the mass of a substance to moles: \[ \left ( { grams \over grams/mole} \right ) = grams \left ( {mole \over grams } \right ) = moles \label{2.7.2B}\] \[2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}\] the production of two moles of water would require the consumption of 2 moles of \(H_2\) and one mole of \(O_2\). Therefore, when considering particular reaction would be considered to be . These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of \(H_2O\) would be produced from 1.57 moles of \(O_2\)? \[ (1.57\; mol\; O_2) \left( \dfrac{2\; mol H_2O}{1\;mol\;O_2} \right) = 3.14\; mol\; H_2O\] The ratio \( \left( \dfrac{2\; mol\; H_2O}{1\;mol\;O_2} \right)\) is the stoichiometric relationship between \(H_2O\) and \(O_2\) from the balanced equation for this reaction. For the combustion of butane (\(C_4H_{10}\)) the balanced equation is: \[2C_4H_{10(l)} + 13O_{2(g)} \rightarrow 8CO_{2(g)} + 10H_2O_{(l)}\] Calculate the mass of \(CO_2\) that is produced in burning 1.00 gram of \(C_4H_{10}\). Thus, the overall sequence of steps to solve this problem is: First of all we need to calculate how many moles of butane we have in a 1.00 gram sample: \[ (1.00\; g\; C_4H_{10}) \left(\dfrac{1\; mol\; C_4H_{10}}{58.0\;g\; C_4H_{10}}\right) = 1.72 \times 10^{-2} \; mol\; C_4H_{10}\] Now, the stoichiometric relationship between \(C_4H_{10}\) and \(CO_2\) is: \[\left( \dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}}\right)\] Therefore: \[ \left(\dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}} \right) \times 1.72 \times 10^{-2} \; mol\; C_4H_{10} = 6.88 \times 10^{-2} \; mol\; CO_2\] The question called for the determination of the mass of \(CO_2\) produced, thus we have to convert moles of \(CO_2\) into grams (by using the of \(CO_2\)): \[ 6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2\] Be sure to pay attention to the units when converting between mass and moles. is a flowchart for converting between mass; the number of moles; and the number of atoms, molecules, or formula units. The use of these conversions is illustrated in Example \(\Page {3}\) and Example \(\Page {4}\). Conversions Between Grams, Mol, & Atoms: For 35.00 g of ethylene glycol (HOCH CH OH), which is used in inks for ballpoint pens, calculate the number of : mass and molecular formula : number of moles and number of molecules : : a. The molecular mass of ethylene glycol can be calculated from its molecular formula using the method illustrated in Example \(\Page {2}\).7.1: \[ 2C (2 \,atoms )(12.011 \, amu/atom) = 24.022 \, amu \] \[ 6H (6 \,atoms )(1.0079 \, amu/atom) = 6.0474 \, amu \] \[ 2O (2 \,atoms )(15.9994 \, amu/atom) = 31.9988 \, amu \] The molar mass of ethylene glycol is 62.068 g/mol. The number of moles of ethylene glycol present in 35.00 g can be calculated by dividing the mass (in grams) by the molar mass (in grams per mole): \[ { \text {mass of ethylene glycol (g)} \over \text {molar mass (g/mol)} } = \text {moles ethylene glycol (mol) }\] So \[ 35.00 \, g \text {ethylene glycol} \left ( {1 \, mole \text {ethylene glycol} \over 62.068 \, g \text {ethylene glycol } } \right ) = 0.5639 \,mol \text {ethylene glycol} \] It is always a good idea to estimate the answer before you do the actual calculation. In this case, the mass given (35.00 g) is less than the molar mass, so the answer should be less than 1 mol. The calculated answer (0.5639 mol) is indeed less than 1 mol, so we have probably not made a major error in the calculations. b. To calculate the number of molecules in the sample, we multiply the number of moles by Avogadro’s number: \[ \text {molecules of ethylene glycol} = 0.5639 \, mol \left ( {6.022 \times 10^{23} \, molecules \over 1 \, mol } \right ) \] \[ = 3.396 \times 10^{23} \, molecules \] Because we are dealing with slightly more than 0.5 mol of ethylene glycol, we expect the number of molecules present to be slightly more than one-half of Avogadro’s number, or slightly more than 3 × 10 molecules, which is indeed the case. \[ 2S (2 \, atoms)(32.065 \, amu/atom ) = 64.130 \, amu \] \[+ 2Cl (2 \, atoms )(35.453 \, amu/atom ) = 70.906 \, amu \] \[ S_2 Cl_2 \text {molecular mass of } S_2Cl_2 = 135.036 \, amu \] The molar mass of S Cl is 135.036 g/mol. The mass of 1.75 mol of S Cl is calculated as follows: \[moles S_2Cl_2 \left [\text {molar mass}\left ({ g \over mol} \right )\right ] \rightarrow mass of S_2Cl_2 \, (g) \] \[ 1.75 \, mol S_2Cl_2\left ({135.036 \, g S_2Cl_2 \over 1 \, mol S_2Cl_2 } \right ) = 236 \, g S_2Cl_2 \] b. The formula mass of Ca(ClO) is obtained as follows: \[1Ca (1 \, atom)(40.078 \, amu/atom) = 40.078 \, amu \] \[2Cl (2 \, atoms)(35.453 \, amu/atom) = 70.906 \, amu \] \[+ 2O (2 \, atoms)(15.9994 \, amu/atom) = 31.9988 \, amu \] \[ Ca (ClO)_2 \text { formula mass of } Ca (ClO)_2 = 142.983 \, amu\] The molar mass of Ca(ClO) 142.983 g/mol. The mass of 1.75 mol of Ca(ClO) is calculated as follows: \[ 1.75 \, mol Ca(ClO)_2 \left [ {142.983 \, g Ca(ClO)_2 \over 1 \, mol Ca(ClO)_2 } \right ] = 250 \, g Ca(ClO)_2 \] Because 1.75 mol is less than 2 mol, the final quantity in grams in both cases should be less than twice the molar mass, which it is. Calculate the mass of 0.0122 mol of each compound. : The coefficients in a balanced chemical equation can be interpreted both as the relative numbers of molecules involved in the reaction and as the relative number of moles. For example, in the equation: \[2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}\] the production of two moles of water would require the consumption of 2 moles of \(H_2\) and one mole of \(O_2\). Therefore, when considering particular reaction would be considered to be . These stoichiometric relationships, derived from balanced equations, can be used to determine expected amounts of products given amounts of reactants. For example, how many moles of \(H_2O\) would be produced from 1.57 moles of \(O_2\)? \[ (1.57\; mol\; O_2) \left( \dfrac{2\; mol H_2O}{1\;mol\;O_2} \right) = 3.14\; mol\; H_2O\] The ratio \( \left( \dfrac{2\; mol\l H_2O}{1\;mol\;O_2} \right)\) is the stoichiometric relationship between \(H_2O\) and \(O_2\) from the balanced equation for this reaction. For the combustion of butane (\(C_4H_{10}\)) the balanced equation is: \[2C_4H_{10(l)} + 13O_{2(g)} \rightarrow 8CO_{2(g)} + 10H_2O_{(l)}\] Calculate the mass of \(CO_2\) that is produced in burning 1.00 gram of \(C_4H_{10}\). First of all we need to calculate how many moles of butane we have in a 1.00 gram sample: \[ (1.00\; g\; C_4H_{10}) \left(\dfrac{1\; mol\; C_4H_{10}}{58.0\;g\; C_4H_{10}}\right) = 1.72 \times 10^{-2} \; mol\; C_4H_{10}\] Now, the stoichiometric relationship between \(C_4H_{10}\) and \(CO_2\) is: \[\left( \dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}}\right)\] Therefore: \[ \left(\dfrac{8\; mol\; CO_2}{2\; mol\; C_4H_{10}} \right) \times 1.72 \times 10^{-2} \; mol\; C_4H_{10} = 6.88 \times 10^{-2} \; mol\; CO_2\] The question called for the determination of the mass of \(CO_2\) produced, thus we have to convert moles of \(CO_2\) into grams (by using the of \(CO_2\)): \[ 6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2\] Thus, the overall sequence of steps to solve this problem were: In a similar way we could determine the mass of water produced, or oxygen consumed, etc. The molecular mass and the formula mass of a compound are obtained by adding together the atomic masses of the atoms present in the molecular formula or empirical formula, respectively; the units of both are atomic mass units (amu). The mole is a unit used to measure the number of atoms, molecules, or (in the case of ionic compounds) formula units in a given mass of a substance. The mole is defined as the amount of substance that contains the number of carbon atoms in exactly 12 g of carbon-12, Avogadro’s number (6.022 × 10 ) of atoms of carbon-12. The molar mass of a substance is defined as the mass of 1 mol of that substance, expressed in grams per mole, and is equal to the mass of 6.022 × 10 atoms, molecules, or formula units of that substance. | 12,245 | 2,414 |
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If benzene is forced to react by increasing the temperature and/or by addition of a catalyst, It undergoes rather than the addition reactions that are typical of alkenes. This further confirms the previous indication that the six-carbon benzene core is unusually stable to chemical modification. The conceptual contradiction presented by a high degree of unsaturation (low H:C ratio) and high chemical stability for benzene and related compounds remained an unsolved puzzle for many years. Eventually, the presently accepted structure of a regular-hexagonal, planar ring of carbons was adopted, and the exceptional thermodynamic and chemical stability of this system was attributed to of a conjugated cyclic triene. Here, two structurally and energetically equivalent electronic structures for a stable compound are written, but no single structure provides an accurate or even an adequaterepresentation of the true molecule. The six-membered ring in benzene is a perfect hexagon (all carbon-carbon bonds have an identical length of 1.40 Å). The cyclohexatriene contributors would be expected to show alternating bond lengths, the double bonds being shorter (1.34 Å) than the single bonds (1.54 Å). An alternative representation for benzene (circle within a hexagon) emphasizes the pi-electron delocalization in this molecule, and has the advantage of being a single diagram. In cases such as these, the electron delocalization described by resonance enhances the stability of the molecules, and compounds composed of such molecules often show exceptional stability and related properties. Evidence for the enhanced thermodynamic stability of benzene was obtained from measurements of the heat released when double bonds in a six-carbon ring are hydrogenated (hydrogen is added catalytically) to give cyclohexane as a common product. In the following diagram cyclohexane represents a low-energy reference point. Addition of hydrogen to cyclohexene produces cyclohexane and releases heat amounting to 28.6 kcal per mole. If we take this value to represent the energy cost of introducing one double bond into a six-carbon ring, we would expect a cyclohexadiene to release 57.2 kcal per mole on complete hydrogenation, and 1,3,5-cyclohexatriene to release 85.8 kcal per mole. These would reflect the relative thermodynamic stability of the compounds. In practice, 1,3-cyclohexadiene is slightly more stable than expected, by about 2 kcal, presumably due to conjugation of the double bonds. . This sort of stability enhancement is now accepted as a characteristic of all aromatic compounds. A of benzene provides a more satisfying and more general treatment of "aromaticity". We know that benzene has a planar hexagonal structure in which all the carbon atoms are sp hybridized, and all the carbon-carbon bonds are equal in length. As shown below, the remaining cyclic array of six p-orbitals ( one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π ) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or , that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases. | 3,763 | 2,417 |
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Below are two videos of a process, one going forward in time, and the other rewinding back through time. It is clear that the first video of nitrogen triiodide exploding at the touch of a feather is going forward in time. It is also as immediately obvious that the second video, of nitrogen triiodide "unexploding" is going backward. We know instinctively from everyday experience that such a process would be unnatural, and simply would not occur. Exploding is not the only process for which we have this basic intuition on. Below is another video going backward in time to show a processes we know is unnatural and would not occur spontaneously in real life: In the above video, a dye is "unmixed." The dye goes from being fully dispersed in solution to returning back to a concentrated drop separate from the rest of the solution. The mixing of the dye is spontaneous, we expect it to happen, so we know that the video must be in reverse, showing a process we intuitively know is unnatural. These two examples show processes we know from basic world experience go in one direction, and do not go in the opposite (reverse) direction. There must be an underlying reason for one direction being spontaneous while the other is not. Understanding this reason will allow one to determine which direction a process will go when it is less obvious than the explosion in the first example. Following the chemist's view of the world, we can understand what is occurring macroscopically by . | 1,496 | 2,418 |
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Alkenes can coordinate to transition metals to form alkene complexes. In some cases, coordination of the alkene to a metal leaves it susceptible to reaction with a nucleophile such as water. The classic case of nucleophilic donation to a coordinated alkene occurs with mercury (II) salts such as mercuric chloride, HgCl , or mercuric acetate, Hg(OAc) . The reaction, or rather the sequence of reactions, is called oxymercuration - demercuration or oxymercuration - reduction. Compare the product of the reaction above to that obtained from treatment of 1-pentene with aqueous sulfuric acid. We will break the two different reactions in this sequence apart and focus only on the first one: oxymercuration. This reaction qualifies as an electrophilic addition because, as in the previous cases, it begins with donation of a π-bonding pair to an electrophile. In this case, we will consider the electrophile to be aqueous Hg ion. That electrophilic addition (from the alkene's perspective) results in the formation of an alkene complex. In reality, the mercury ion is also coordinated by several water molecules, but we will ignore them for simplicity. You may know that alkene complexes are not observed with d transition metals. Although π-to-metal donation is the key event in the formation of such complexes, the alkene is just a little more sticky if the metal has d electrons. These electrons are able to "back-donate" into the alkene portion of the complex, adding extra stability to the interaction. This situation is something like formation of a cyclic bromonium ion. Note that the overall transfer of electrons is still from alkene to metal. That imbalance isn't apparent in a Lewis structure sense, in which case you can draw the structure so that there appears to be an equal trade. In a computational chemistry approach, in which we rely on basic principles of quantum mechanics and let computers churn out high-level calculations, we would still predict a little bit of positive charge on the alkene. In the structure below, we have over-emphasized that charge, just to see what happens next in the reaction. When we draw it that way, it looks a lot more like simple addition of electrophile, such as H , to alkene. We know it's more subtle than that. We'll get back to the real mechanism after a small detour. Of course, the next step is donation of a lone pair from a nucleophile to the almost-cationic carbon. That looks easy. After that, deprotonation would result in the formation of a hydroxy group. Suppose deprotonation is carried out by the acetate ion in solution. Draw a mechanism for this step. How do we know the reaction doesn't happen through this simple cation? Partly we know that because we know about alkene complexes. There are thousands of examples, structurally characterized by NMR spectroscopy and X-ray crystallography. In addition, we know it isn't a simple cation because nothing like the following scenario plays out during oxymercuration. There are no hydride shifts. The cation stays put. The hydroxy group forms right where the alkene used to be. That means there is not a full carbocation like the one shown above. If there isn't a real carbocation, though, why does the nucleophile end up at one particular end of the alkene? The hydroxy does end up at the position that would form the more stable cation. ( In other words, this reaction results in what is called "Markovnikov addition".) There are a couple of reasons that could play a role. Foremost, the alkene isn't bound symmetrically. One end is held a little closer to the mercury than the other. Mostly that's because of sterics. Any other ligands on the mercury (such as those water molecules) push that more crowded end away a little bit. That slight asymmetry allows a little more charge to build on the more substitutted end of the alkene, which is therefore more electrophilic. The final part of the reaction sequence is displacement of mercury from the hydroxyalkylmercury complex, effected through addition of sodium borohydride. The details of the reaction are usually dismissed in textbooks because they have little to do with electrophilic addition, the topic we are focusing on. However, the result is that the mercury is replaced by a hydrogen atom. The metal is converted to silvery, liquid, elemental mercury. Suppose the demercuration reaction takes place via addition of hydride nucleophile to mercury, followed by reductive elimination. Draw this mechanism. When mercuration takes place in a ethanol instead of water, an ether product results rather than an alcohol. Work through the mechanism and show the result of mercuration-demercuration in ethanol. Show the products of the following reactions. , | 4,737 | 2,419 |
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The first important distinction we will make is among the terms analysis, determination, and measurement. An analysis provides chemical or physical information about a sample. The component in the sample of interest to us is called the , and the remainder of the sample is the matrix. In an analysis we determine the identity, the concentration, or the properties of an analyte. To make this determination we measure one or more of the analyte’s chemical or physical properties. An example will help clarify the difference between an , a and a . In 1974 the federal government enacted the Safe Drinking Water Act to ensure the safety of the nation’s public drinking water supplies. To comply with this act, municipalities monitor their drinking water supply for potentially harmful substances, such as fecal coliform bacteria. Municipal water departments collect and analyze samples from their water supply. To determine the concentration of fecal coliform bacteria an analyst passes a portion of water through a membrane filter, places the filter in a dish that contains a nutrient broth, and incubates the sample for 22–24 hrs at 44.5 C ± 0.2 C. At the end of the incubation period the analyst counts the number of bacterial colonies in the dish and reports the result as the number of colonies per 100 mL (Figure 3.1.1
). Thus, a municipal water department analyzes samples of water to determine the concentration of fecal coliform bacteria by measuring the number of bacterial colonies that form during a carefully defined incubation period A fecal coliform count provides a general measure of the presence of pathogenic organisms in a water supply. For drinking water, the current maximum contaminant level (MCL) for total coliforms, including fecal coliforms is less than 1 colony/100 mL. Municipal water departments must regularly test the water supply and must take action if more than 5% of the samples in any month test positive for coliform bacteria. | 1,980 | 2,420 |
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In proposing his theory that octets can be completed by two atoms sharing electron pairs, Lewis provided scientists with the first description of covalent bonding. In this section, we expand on this and describe some of the properties of covalent bonds. The general properties of Ionic substances are: However, the vast majority of chemical substances are in nature. G.N. Lewis reasoned that an atom might attain a noble gas electron configuration by electrons A chemical bond formed by sharing a pair of electrons is called a covalent bond We begin our discussion of the relationship between structure and bonding in covalent compounds by describing the interaction between two identical neutral atoms—for example, the H molecule, which contains a purely covalent bond. Each hydrogen atom in H contains one electron and one proton, with the electron attracted to the proton by electrostatic forces. As the two hydrogen atoms are brought together, additional interactions must be considered (Figure \(\Page {1}\)): A plot of the potential energy of the system as a function of the internuclear distance (Figure \(\Page {2}\)) shows that the system becomes more stable (the energy of the system decreases) as two hydrogen atoms move toward each other from = ∞, until the energy reaches a minimum at = (the observed internuclear distance in H is 74 pm). Thus at intermediate distances, proton–electron attractive interactions dominate, but as the distance becomes very short, electron–electron and proton–proton repulsive interactions cause the energy of the system to increase rapidly. Notice the similarity between Figure \(\Page {2}\) and Figure 8.1, which described a system containing two oppositely charged . The shapes of the energy versus distance curves in the two figures are similar because they both result from attractive and repulsive forces between charged entities. At long distances, both attractive and repulsive interactions are small. As the distance between the atoms decreases, the attractive electron–proton interactions dominate, and the energy of the system decreases. At the observed bond distance, the repulsive electron–electron and proton–proton interactions just balance the attractive interactions, preventing a further decrease in the internuclear distance. At very short internuclear distances, the repulsive interactions dominate, making the system less stable than the isolated atoms. Lewis structures (also known as Lewis dot diagrams, electron dot diagrams, Lewis dot formulas, Lewis dot structures, and electron dot structures) are diagrams that show the bonding between atoms of a molecule and the lone pairs of electrons that may exist in the molecule. Lewis structures show each atom and its position in the structure of the molecule using its chemical symbol. Lines are drawn between atoms that are bonded to one another (pairs of dots can be used instead of lines). Excess electrons that form lone pairs are represented as pairs of dots, and are placed next to the atoms. The diatomic hydrogen molecule (H ) is the simplest model of a covalent bond, and is represented in Lewis structures as: The of electrons provides hydrogen atom with two electrons in its valence shell (the 1 ) orbital. In a sense, each hydrogen atoms has the electron configuration of the noble gas helium (the octet rule). When two chlorine atoms covalently bond to form \(Cl_2\), the following sharing of electrons occurs: Each chlorine atom shared the bonding pair of electrons and achieves the electron configuration of the noble gas argon. In Lewis structures the bonding pair of electrons is usually displayed as a line, and the unshared electrons as dots: The shared electrons are not located in a fixed position between the nuclei. In the case of the \(H_2\) compound, the electron density is concentrated between the two nuclei: The two atoms are bound into the \(H_2\) molecule mainly due to the attraction of the positively charged nuclei for the negatively charged electron cloud located between them. Examples of hydride compounds of the above elements (covalent bonds with hydrogen: The sharing of a pair of electrons represents a single covalent bond, usually just referred to as a . However, in many molecules atoms attain complete octets by sharing more than one pair of electrons between them: Because each nitrogen contains 5 valence electrons, they need to share 3 pairs to each achieve a valence octet. N is fairly inert, due to the strong triple bond between the two nitrogen atoms and the N - N bond distance in N is 1.10 Å (fairly short). From a study of various Nitrogen containing compounds bond distance as a function of bond type can be summarized as follows: For the nonmetals (and the 's' block metals) the number of valence electrons is equal to the group number: F 7 1 O 6 2 N 5 3 C 4 4 Lewis Structure of Molecules: Thus, the Lewis bonds successfully describe the covalent interactions between various nonmetal elements. When we draw Lewis structures, we place one, two, or three pairs of electrons between adjacent atoms. In the Lewis bonding model, the number of electron pairs that hold two atoms together is called the bond order. For a single bond, such as the C–C bond in H C–CH , the bond order is one. For a double bond (such as H C=CH ), the bond order is two. For a triple bond, such as HC≡CH, the bond order is three. When analogous bonds in similar compounds are compared, bond length decreases as bond order increases. The bond length data in Table \(\Page {1}\), for example, show that the C–C distance in H C–CH (153.5 pm) is longer than the distance in H C=CH (133.9 pm), which in turn is longer than that in HC≡CH (120.3 pm). Additionally, as noted in Section 8.5, molecules or ions whose bonding must be described using resonance structures usually have bond distances that are intermediate between those of single and double bonds, as we demonstrated with the C–C distances in benzene. The relationship between bond length and bond order is not linear, however. A double bond is not half as long as a single bond, and the length of a C=C bond is not the average of the lengths of C≡C and C–C bonds. Nevertheless, as bond orders increase, bond lengths generally decrease. Sources: Data from (2004); (2005); h . As a general rule, the distance between bonded atoms decreases as the number of shared electron pairs increases Bond Lengths: As shown in Table \(\Page {1}\), triple bonds between like atoms are shorter than double bonds, and because more energy is required to completely break all three bonds than to completely break two, a triple bond is also stronger than a double bond. Similarly, double bonds between like atoms are stronger and shorter than single bonds. Bonds of the same order between atoms show a wide range of bond energies, however. Data from J. E. Huheey, E. A. Keiter, and R. L. Keiter, , 4th ed. (1993). Table \(\Page {2}\) lists the average values for some commonly encountered bonds. Although the values shown vary widely, we can observe four trends: Bonds between hydrogen and atoms in a given column in the periodic table are weaker down the column; bonds between like atoms usually become weaker down a column. Bond strengths increase as bond order , while bond distances . Bond energy is defined as the energy required to break a particular bond in a molecule in the gas phase. Its value depends on not only the identity of the bonded atoms but also their environment. Thus the bond energy of a C–H single bond is the same in all organic compounds. For example, the energy required to break a C–H bond in methane varies by as much as 25% depending on how many other bonds in the molecule have already been broken (Table \(\Page {3}\)); that is, the C–H bond energy depends on its molecular environment. Except for diatomic molecules, the bond energies listed in Table \(\Page {2}\) are values for all bonds of a given type in a range of molecules. Even so, they are not likely to differ from the actual value of a given bond by more than about 10%. Data from (2004). We can estimate the enthalpy change for a chemical reaction by adding together the average energies of the bonds broken in the reactants and the average energies of the bonds formed in the products and then calculating the difference between the two. If the bonds formed in the products are stronger than those broken in the reactants, then energy will be released in the reaction (Δ < 0): \[ ΔH_{rxn} \approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \label{8.3.1}\] The ≈ sign is used because we are adding together bond energies; hence this approach does not give exact values for Δ . Let’s consider the reaction of 1 mol of -heptane (C H ) with oxygen gas to give carbon dioxide and water. This is one reaction that occurs during the combustion of gasoline: \[CH_3(CH_2)_5CH_{3(l)} + 11 O_{2(g)} \rightarrow 7 CO_{2(g)} + 8 H_2O_{(g)} \label{8.3.2}\] In this reaction, 6 C–C bonds, 16 C–H bonds, and 11 O=O bonds are broken per mole of -heptane, while 14 C=O bonds (two for each CO ) and 16 O–H bonds (two for each H O) are formed. The energy changes can be tabulated as follows: The bonds in the products are stronger than the bonds in the reactants by about 4444 kJ/mol. This means that \(ΔH_{rxn}\) is approximately −4444 kJ/mol, and the reaction is highly exothermic (which is not too surprising for a combustion reaction). If we compare this approximation with the value obtained from measured \(ΔH_f^o\) values (\(ΔH_{rxn} = −481\;7 kJ/mol\)), we find a discrepancy of only about 8%, less than the 10% typically encountered. Chemists find this method useful for calculating approximate enthalpies of reaction for molecules whose actual \(ΔH^ο_f\) values are unknown. These approximations can be important for predicting whether a reaction is or —and to what degree. The compound RDX (Research Development Explosive) is a more powerful explosive than dynamite and is used by the military. When detonated, it produces gaseous products and heat according to the following reaction. Use the approximate bond energies in Table \(\Page {2}\) to estimate the \(ΔH_{rxn}\) per mole of RDX. chemical reaction, structure of reactant, and Table \(\Page {2}\). \(ΔH_{rxn}\) per mole We must add together the energies of the bonds in the reactants and compare that quantity with the sum of the energies of the bonds in the products. A nitro group (–NO ) can be viewed as having one N–O single bond and one N=O double bond, as follows: In fact, however, both N–O distances are usually the same because of the presence of two equivalent resonance structures. We can organize our data by constructing a table: From Equation 8.3.1, we have \[ ΔH_{rxn} \approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}}\] \[= 7962 \; kJ/mol − 10,374 \; kJ/mol\] \[=−2412 \;kJ/mol] Thus this reaction is also highly exothermic. The molecule HCFC-142b, a hydrochlorofluorocarbon used in place of chlorofluorocarbons (CFCs) such as the Freons, can be prepared by adding HCl to 1,1-difluoroethylene: Use tabulated bond energies to calculate \(ΔH_{rxn}\). −54 kJ/mol Bond Energies: \[ ΔH_{rxn} \approx \sum{\text{(bond energies of bonds broken)}}−\sum{\text{(bond energies of bonds formed)}} \label{8.3.1}\] is the number of electron pairs that hold two atoms together. Single bonds have a bond order of one, and multiple bonds with bond orders of two (a double bond) and three (a triple bond) are quite common. In closely related compounds with bonds between the same kinds of atoms, the bond with the highest bond order is both the shortest and the strongest. In bonds with the same bond order between different atoms, trends are observed that, with few exceptions, result in the strongest single bonds being formed between the smallest atoms. Tabulated values of average bond energies can be used to calculate the enthalpy change of many chemical reactions. If the bonds in the products are stronger than those in the reactants, the reaction is exothermic and vice versa. ( ) | 12,184 | 2,421 |
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Amino acids are the products of stage 1 of protein catabolism. Usually, amino acids are used to synthesize proteins and other substances, but when they are in excess they are degraded to produce energy. Also when there as shortage of carbohydrates and fats, amino acids are used as an alternate source of energy. Degradation of amino acids has two sets of catabolic pathways: one that deals with the processing of \(\ce{N}\) and the other that deals with the processing of the remaining \(\ce{C}\) skeleton of the amino acids. Processing of \(\ce{N}\) usually happens in the liver and has three major stages: i) transamination, ii) oxidative deamination, and iii) urea cycle, as illustrated with some examples in Figure \(\Page {1}\). is the process of exchange of an ammonium (\(\ce{-NH3^{+}}\)) group of amino acids with ketone (\(\ce{C=O}\)) group of an \(\alpha\)-ketoacid. Usually \(\alpha\)-amino acids exchange there \(\ce{-NH3^{+}}\) group with ketone (\(\ce{C=O}\)) group of an \(\alpha\)-ketoglutarate. For example, alanine exchanges its \(\ce{-NH3^{+}}\) group with a \(\ce{C=O}\) of an \(\alpha\)-ketoglutarate producing a new \(\alpha\)-keto acid (pyruvate in this example) and a new \(\alpha\)-amino acids (L-glutamate in this example), as shown in reaction 1 in Figure \(\Page {1}\). Other \(\alpha\)-keto acids can also receive \(\ce{-NH3^{+}}\) of amino acids also. For example, L-glutamate can transfer its \(\ce{-NH3^{+}}\) to oxaloacetate that regenerates \(\alpha\)-ketoglutarate and produces aspartate, as shown in reaction 3 in Figure \(\Page {1}\). reaction replaces \(\ce{-NH3^{+}}\) with a \(\ce{C=O}\) producing an \(\alpha\)-keto acid and an ammonium (\(\ce{NH4^{+}}\)) ion, as shown in reaction 2 in Figure \(\Page {1}\). This is an oxidation reaction that is coupled with the reduction of \(\ce{NAD^{+}}\) to \(\ce{NADH}\). It happens in mitochondria where \(\ce{NADH}\) enters oxidative phosphorylation pathway to produce \(\ce{ATP's}\). The other product, i.e., \(\ce{NH4^{+}}\), is a toxic substance that needs to be disposed of. Some of \(\ce{NH4^{+}}\) may be used in anabolic reactions, e.g., reactions 4 and 5 in Figure \(\Page {1}\), but most of it enters the urea cycle. Ammonium ion \(\ce{NH4^{+}}\) is a toxic substance that is converted to less toxic urea in the urea cycle by the following overall reaction. \(\ce{2NH4^{+} + CO2 -> \underbrace{H2N-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-NH2}_{Urea} + 2H^{+} + H2O}\). Before entry into the urea cycle, \(\ce{NH4^{+}}\) reacts with two \(\ce{ATP's}\) and bicarbonate (\(\ce{CO3H^{-}}\)) producing carbamoyl phosphate, two \(\ce{ATP's}\) and phosphate (\(\ce{P_i}\), as shown below. This reaction is catalyzed by an enzyme carbamoyl phosphate synthetase I. Blood contains \(\ce{CO3H^{-}}\) which is the product of dissolution of \(\ce{CO2}\). The product of the above reaction i.e., Carbamoyl phosphate goes through a set of four reactions called the illustrated in Figure \(\Page {2}\). Since \(\ce{CO3H^{-}}\) is produced by the dissolution of \(\ce{CO2}\) in water, the \(\ce{NH4^{+}}\) + \(\ce{CO3H^{-}}\) are equivalent to \(\ce{NH3}\) + \(\ce{CO2}\) + \(\ce{H2O}\). So, the overall equation of the urea cycle becomes: \(\ce{NH3 + CO2 + H2O + aspartate + 3ATP + 3H2O -> urea + fumarate + 2ADP + 2P_{i} + H2O}\) Since \(\ce{NH3}\) is removed to convert aspartate into fumarate along with \(\ce{PP_{i} + H2O -> 2P_{i}}\), substituting these for aspartate and fumarate simplifies the above equation into: \(\ce{2NH3 + CO2 + 3ATP + 3H2O -> urea + 2ADP + 4P_{i} + H2O}\) One \(\ce{NADH}\) is produced during oxidative deamination of L-glutamate (reaction 2 in Figure \(\Page {1}\)). Another \(\ce{NADH}\) is produced when fumarate from reaction 3 of the urea cycle is processed in the citric acid cycle, i.e., fumarate is converted into malate by enzyme fumarase and then malate is oxidized to oxaloacetate at the expense of reduction of \(\ce{NAD^{+}}\) to \(\ce{NADH}\) by enzyme malate dehydrogenase. Adding these reactions to the above reaction of the urea cycle results in the following overall reaction. \(\ce{CO2 + glutamate + aspartate + 3ATP + 2NAD^{+} + 3H2O -> urea + \alpha{-ketoglutarate} + 2ADP + 2P_{i} + AMP + 2NADH}\) Three \(\ce{ATP}\) are consumed but two \(\ce{NADH}\) enter oxidative phosphorylation and yield ~5\(\ce{ATP}\), i.e., there is net production of two \(\ce{ATP}\) in the process of \(\ce{N}\) of an amino acid. Urea that is released into the blood is later filtered out by the kidneys and excreted with urine. An adult excretes about 25 to 30 g of urea in urine per day. If urea is not eliminated properly it builds up to toxic levels and needs medical treatment, like dialysis. Reducing protein intake also lowers urea output. The urea cycle and citric acid cycle are two separate cycles but are linked with each other. Aspartate that provides one \(\ce{N}\) in the urea cycle is produced by the transamination of oxaloacetate which is an intermediate in the citric acid cycle. Fumarate is a product of the urea cycle that is an intermediate of the citric acid cycle and returns to it. \(\alpha\)-Ketoacids are left as \(\ce{C}\) skeleton of amino acid after transamination or oxidative deamination. For example, pyruvate is left after transamination of alanine and \(\alpha\)-ketoglutarate is left after oxidative deamination of glutamate, as shown in reactions 1 and 2, respectively, in Figure \(\Page {1}\). \(\alpha\)-Ketoacids are either used as precursors for the synthesis of other compounds are they enter the citric acid cycle to produce \(\ce{CO2}\), \(\ce{H2O}\), and energy. \(\alpha\)-Ketoglutarate and oxaloacetate are \(\alpha\)-ketoacids that are intermediates in the citric acid cycle and directly enter the cycle. Other \(\alpha\)-ketoacids go through a series of reactions to convert into one of the intermediates in the citric acid cycle or they convert to pyruvate or acetyl-(\ce{CoA}\) that enter the citric acid cycle, as illustrated in Figure \(\Page {3}\). Details of the conversation of \(\alpha\)-ketoacids into intermediates of the citric acid cycle or pyruvate or acetyl-(\ce{CoA}\) are not described here. Amino acids that can degrade to pyruvate or oxaloacetate are called because these products have the ability to form glucose through the glucogenesis pathway. For example, alanine yields pyruvate, and aspartate yield oxaloacetate, as shown in reactions 1 and 3 in Figure \(\Page {1}\). Amino acids that degrade to acetyl-(\ce{CoA}\) or acetoacetic acid which can not form glucose but can be converted into ketone bodies, are called . Lysine and leucine are ketogenic amino acids. Several amino acids can catabolize to produce both glugogenic and ketogenic intermediates and they fall into both classes, as shown in Figure \(\Page {3}\). | 6,849 | 2,422 |
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According to , the product is a constant at any given temperature, so a plot of as a function of the pressure of an ideal gas yields a horizontal straight line. This implies that any increase in the pressure of the gas is exactly counteracted by a decrease in the volume as the molecules are crowded closer together. But we know that the molecules themselves are finite objects having volumes of their own, and this must place a lower limit on the volume into which they can be squeezed. So we must reformulate the ideal gas equation of state as a relation that is true only in the limiting case of zero pressure: \[\lim_{P \rightarrow 0} PV=nRT\] So what happens when a real gas is subjected to a very high pressure? The outcome varies with both the molar mass of the gas and its temperature, but in general we can see the the effects of both repulsive and attractive intermolecular forces: The universal attractive force described above is known as the , or force. There may also be additional (and usually stronger) attractive forces related to charge imbalance in the molecule or to hydrogen bonding. These various attractive forces are often referred to collectively as . A plot of as a function of pressure is a very sensitive indicator of deviations from ideal behavior, since such a plot is just a horizontal line for an ideal gas. The two illustrations below show how these plots vary with the nature of the gas, and with temperature. , which generally increase with molecular weight, cause the product to decrease as higher pressures bring the molecules closer together and thus within the range of these attractive forces; the effect is to cause the volume to decrease more rapidly than it otherwise would. The repulsive forces always eventually win out. But as the molecules begin to intrude on each others' territory, the stronger repulsive forces cause the curve to bend upward. The makes a big difference! At higher temperatures, increased thermal motions overcome the effects of intermolecular attractions which normally dominate at lower pressures. So all gases behave more ideally at higher temperatures. For any gas, there is a special temperature (the ) at which attractive and repulsive forces exactly balance each other at zero pressure. As you can see in this plot for methane, this some of this balance does remain as the pressure is increased. How might we modify the ideal gas equation of state to take into account the effects of intermolecular interactions? The first and most well known answer to this question was offered by the Dutch scientist J.D. van der Waals (1837-1923) in 1873. The ideal gas model assumes that the gas molecules are merely points that occupy no volume; the " " term in the equation is the volume of the container and is independent of the nature of the gas. van der Waals recognized that the molecules themselves take up space that subtracts from the volume of the container, so that the “volume of the gas” in the ideal gas equation should be replaced by the term ( ), in which relates to the , typically of the order of 20-100 cm mol . The excluded volume surrounding any molecule defines the closest possible approach of any two molecules during collision. Note that the excluded volume is greater then the volume of the molecule, its radius being half again as great as that of a spherical molecule. The other effect that van der Waals needed to correct for are the intermolecular attractive forces. These are ignored in the ideal gas model, but in real gases they exert a small cohesive force between the molecules, thus helping to hold the gas together and reducing the pressure it exerts on the walls of the container. Because this pressure depends on both the frequency the intensity of collisions with the walls, the reduction in pressure is proportional to the of the number of molecules per volume of space, and thus for a fixed number of molecules such as one mole, the reduction in pressure is inversely proportional to the square of the volume of the gas. The smaller the volume, the closer are the molecules and the greater will be the effect. The van der Waals equation replaces the \(P\) term in the ideal gas equation with \(P + (a / V^2)\) in which the magnitude of the constant a increases with the strength of the intermolecular attractive forces. The complete van der Waals equation of state can be written as Although most students are not required to memorize this equation, you are expected to understand it and to explain the significance of the terms it contains. You should also understand that the van der Waals constants \(a\) and \(b\) must be determined empirically for every gas. This can be done by plotting the behavior of the gas and adjusting the values of and until the van der Waals equation results in an identical plot. The constant \(a\) is related in a simple way to the molecular radius; thus the determination of \(a\) constitutes an indirect measurement of an important microscopic quantity. The van der Waals equation is only one of many equations of state for real gases. More elaborate equations are required to describe the behavior of gases over wider pressure ranges. These generally take account of higher-order nonlinear attractive forces, and require the use of more empirical constants. Although we will make no use of them in this course, they are widely employed in chemical engineering work in which the behavior of gases at high pressures must be accurately predicted. ) | 5,536 | 2,423 |
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TLC is an excellent analytical tool for separating mixtures in a sample. In this section are discussed the details of the separation, and expand upon the general discussion of . In all forms of chromatography, samples equilibrate between stationary and mobile phases. In almost all applications of TLC, the stationary phase is a silica or alumina adsorbent and the mobile phase is an organic solvent or solvent mixture (the " ") that rises up the plate (equation 3). \[\ce{X}_\text{(silica/alumina)} \rightleftharpoons \ce{X}_\text{(solvent)} \label{3}\] Silica gel (shown in Figure 2.16) is composed of a network of silicon-oxygen bonds, with \(\ce{O-H}\) bonds on its surface, as well as a layer of water molecules. Silica gel \(\left( \ce{SiO_2} \cdot x \ce{H_2O} \right)\) is used in this discussion, but is structurally analogous to alumina \(\left( \ce{Al_2O_3} \cdot x \ce{H_2O} \right)\). This very polar stationary phase is paired with a relatively nonpolar mobile phase (an organic solvent or solution), in what is referred to as "normal phase" TLC. Although this is the most common form of TLC (and what will be focused on in this section), "reverse phase" TLC (with a nonpolar stationary phase and a polar mobile phase) is sometimes used. Figure 2.16 shows how acetophenone would cling to the surface of silica gel through (IMF's). In this case, acetophenone can hydrogen bond (the IMF indicated in Figure 2.16a) to the silica surface through its oxygen atom. As eluent flows over the sample (Figure 2.16b), an equilibrium is established between the sample being adsorbed on the stationary phase and dissolved in the mobile phase. When in the mobile phase, the compound moves up the plate with the flow of liquid (Figure 2.16c) to later readsorb on the stationary phase further up the plate. The resulting \(R_f\) of the compound is dependent on the amount of time spent in the stationary and mobile phases. The equilibrium distribution between the two phases depends on several factors: The degree of attraction by a compound to the stationary and mobile phases lead to the same conclusion: (summarized in Figure 2.17). To demonstrate the effect of structural features on \(R_f\), an eluted TLC plate of benzyl alcohol, benzaldehyde, and ethylbenzene is shown in Figure 2.18. The relative order of \(R_f\) reflects the polarity trend in the series. Benzyl alcohol and benzaldehyde have polar functional groups so thus had lower \(R_f\) values than ethylbenzene, which is completely nonpolar. Both compounds are able to hydrogen bond to the polar stationary phase (Figure 2.19a+b), so are more strongly attracted to the stationary phase than ethylbenzene, which interacts only through weak London dispersion forces (Figure 2.19c). As the least "polar" of the series, ethylbenzene is also the best dissolved by the weakly polar eluent. For these reasons, ethylbenzene spent the least time in the stationary phase and the most time in the mobile phase, which is why it traveled the furthest up the plate and had the highest \(R_f\) of the series. Both benzaldehyde and benzyl alcohol are capable of hydrogen bonding with the stationary phase, but benzyl alcohol had the lower \(R_f\) because it can form hydrogen bonds (through both the oxygen and hydrogen atoms of the \(\ce{OH}\) group, Figure 2.19a). This caused benzyl alcohol to be more strongly adhered to the silica/alumina than benzaldehyde, causing it to spend more time in the stationary phase. To demonstrate a different structural effect on \(R_f\), an eluted TLC plate of acetophenone and benzophenone is shown in Figure 2.20. Both compounds are similar in that they can hydrogen bond to the stationary phase through their oxygen atom. However, the larger size of benzophenone causes it to have a slightly higher \(R_f\) than acetophenone. This result can be explained in multiple ways: The ability of chromatography to separate components in a mixture depends on equilibration of a compound between the stationary and mobile phases. Since the mobile phase is an important factor, it is possible to change the \(R_f\) of a compound by changing the polarity of the mobile phase. This general trend is demonstrated in Figure 2.21b+c, where the TLC of three UV-active compounds (lanes 2-4) was run using two different mixed solvents. The first plate was run using a 6:1 hexane:ethyl acetate mixture, which means the solvent was created by using 6 volumes of hexane for every 1 volume of ethyl acetate. This mixed solvent is mostly nonpolar due to the high percentage of hexane, but is more polar than straight hexane, due to the presence of some ethyl acetate (which has polar bonds, Figure 2.21a). The second plate was run using a 3:2 hexane:ethyl acetate mixture, which is more polar than the 6:1 mixture because there is a higher percentage of ethyl acetate present. Note that in Figure 2.21c all spots maintained their relative order but traveled to a greater height on the plate and increased their \(R_f\) values (Table 2.2) in the more polar eluent. An increase in solvent polarity increases \(R_f\) values for two reasons: | 5,118 | 2,424 |
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Although you have been introduced to some of the interactions that hold molecules together in a liquid, we have not yet discussed the consequences of those interactions for the bulk properties of liquids. We now turn our attention to three unique properties of liquids that intimately depend on the nature of intermolecular interactions: If liquids tend to adopt the shapes of their containers, then why do small amounts of water on a freshly waxed car form raised droplets instead of a thin, continuous film? The answer lies in a property called , which depends on intermolecular forces. Surface tension is the energy required to increase the surface area of a liquid by a unit amount and varies greatly from liquid to liquid based on the nature of the intermolecular forces, e.g., water with hydrogen bonds has a surface tension of 7.29 x 10 J/m (at 20°C), while mercury with metallic bonds has as surface tension that is 15 times higher: 4.86 x 10 J/m (at 20°C). Figure \(\Page {1}\) presents a microscopic view of a liquid droplet. A typical molecule in the of the droplet is surrounded by other molecules that exert attractive forces from all directions. Consequently, there is no force on the molecule that would cause it to move in a particular direction. In contrast, a molecule on the experiences a net attraction toward the drop because there are no molecules on the outside to balance the forces exerted by adjacent molecules in the interior. Because a sphere has the smallest possible surface area for a given volume, intermolecular attractive interactions between water molecules cause the droplet to adopt a spherical shape. This maximizes the number of attractive interactions and minimizes the number of water molecules at the surface. Hence raindrops are almost spherical, and drops of water on a waxed (nonpolar) surface, which does not interact strongly with water, form round beads. A dirty car is covered with a mixture of substances, some of which are polar. Attractive interactions between the polar substances and water cause the water to spread out into a thin film instead of forming beads. The same phenomenon holds molecules together at the surface of a bulk sample of water, almost as if they formed a skin. When filling a glass with water, the glass can be overfilled so that the level of the liquid actually extends the rim. Similarly, a sewing needle or a paper clip can be placed on the surface of a glass of water where it “floats,” even though steel is much denser than water. Many insects take advantage of this property to walk on the surface of puddles or ponds without sinking. This is even observable in the zero gravity conditions of space as shown in Figure \(\Page {2}\) (and more so in the video link) where water wrung from a wet towel continues to float along the towel's surface! Such phenomena are manifestations of surface tension, which is defined as the energy required to increase the surface area of a liquid by a specific amount. Surface tension is therefore measured as energy per unit area, such as joules per square meter (J/m ) or dyne per centimeter (dyn/cm), where 1 dyn = 1 × 10 N. The values of the surface tension of some representative liquids are listed in Table \(\Page {1}\). Note the correlation between the surface tension of a liquid and the strength of the intermolecular forces: the stronger the intermolecular forces, the higher the surface tension. For example, water, with its strong intermolecular hydrogen bonding, has one of the highest surface tension values of any liquid, whereas low-boiling-point organic molecules, which have relatively weak intermolecular forces, have much lower surface tensions. Mercury is an apparent anomaly, but its very high surface tension is due to the presence of strong metallic bonding. Adding soaps and detergents that disrupt the intermolecular attractions between adjacent water molecules can reduce the surface tension of water. Because they affect the surface properties of a liquid, soaps and detergents are called surface-active agents, or surfactants. In the 1960s, Navy researchers developed a method of fighting fires aboard aircraft carriers using “foams,” which are aqueous solutions of fluorinated surfactants. The surfactants reduce the surface tension of water below that of fuel, so the fluorinated solution is able to spread across the burning surface and extinguish the fire. Such foams are now used universally to fight large-scale fires of organic liquids. Intermolecular forces also cause a phenomenon called capillary action, which is the tendency of a polar liquid to rise against gravity into a small-diameter tube (a ), as shown in Figure \(\Page {3}\). When a glass capillary is is placed in liquid water, water rises up into the capillary. The height to which the water rises depends on the diameter of the tube and the temperature of the water but on the angle at which the tube enters the water. The smaller the diameter, the higher the liquid rises. Capillary action is the net result of two opposing sets of forces: cohesive forces, which are the intermolecular forces that hold a liquid together, and adhesive forces, which are the attractive forces between a liquid and the substance that composes the capillary. Water has both strong adhesion to glass, which contains polar SiOH groups, and strong intermolecular cohesion. When a glass capillary is put into water, the surface tension due to cohesive forces constricts the surface area of water within the tube, while adhesion between the water and the glass creates an upward force that maximizes the amount of glass surface in contact with the water. If the adhesive forces are stronger than the cohesive forces, as is the case for water, then the liquid in the capillary rises to the level where the downward force of gravity exactly balances this upward force. If, however, the cohesive forces are stronger than the adhesive forces, as is the case for mercury and glass, the liquid pulls itself down into the capillary below the surface of the bulk liquid to minimize contact with the glass (Figure \(\Page {4}\)). The upper surface of a liquid in a tube is called the meniscus, and the shape of the meniscus depends on the relative strengths of the cohesive and adhesive forces. In liquids such as water, the meniscus is concave; in liquids such as mercury, however, which have very strong cohesive forces and weak adhesion to glass, the meniscus is convex (Figure \(\Page {4}\)). Polar substances are drawn up a glass capillary and generally have a concave meniscus. Fluids and nutrients are transported up the stems of plants or the trunks of trees by capillary action. Plants contain tiny rigid tubes composed of cellulose, to which water has strong adhesion. Because of the strong adhesive forces, nutrients can be transported from the roots to the tops of trees that are more than 50 m tall. Cotton towels are also made of cellulose; they absorb water because the tiny tubes act like capillaries and “wick” the water away from your skin. The moisture is absorbed by the entire fabric, not just the layer in contact with your body. (η) is the resistance of a liquid to flow. Some liquids, such as gasoline, ethanol, and water, flow very readily and hence have a . Others, such as motor oil, molasses, and maple syrup, flow very slowly and have a . The two most common methods for evaluating the viscosity of a liquid are (1) to measure the time it takes for a quantity of liquid to flow through a narrow vertical tube and (2) to measure the time it takes steel balls to fall through a given volume of the liquid. The higher the viscosity, the slower the liquid flows through the tube and the steel balls fall. Viscosity is expressed in units of the poise (mPa•s); the higher the number, the higher the viscosity. The viscosities of some representative liquids are listed in Table 11.3.1 and show a correlation between viscosity and intermolecular forces. Because a liquid can flow only if the molecules can move past one another with minimal resistance, strong intermolecular attractive forces make it more difficult for molecules to move with respect to one another. The addition of a second hydroxyl group to ethanol, for example, which produces ethylene glycol (HOCH CH OH), increases the viscosity 15-fold. This effect is due to the increased number of hydrogen bonds that can form between hydroxyl groups in adjacent molecules, resulting in dramatically stronger intermolecular attractive forces. There is also a correlation between viscosity and molecular shape. Liquids consisting of long, flexible molecules tend to have higher viscosities than those composed of more spherical or shorter-chain molecules. The longer the molecules, the easier it is for them to become “tangled” with one another, making it more difficult for them to move past one another. London dispersion forces also increase with chain length. Due to a combination of these two effects, long-chain hydrocarbons (such as motor oils) are highly viscous. Viscosity increases as intermolecular interactions or molecular size increases. Discussing Surface Tension and Viscosity. Link: Motor oils and other lubricants demonstrate the practical importance of controlling viscosity. The oil in an automobile engine must effectively lubricate under a wide range of conditions, from subzero starting temperatures to the 200°C that oil can reach in an engine in the heat of the Mojave Desert in August. Viscosity decreases rapidly with increasing temperatures because the kinetic energy of the molecules increases, and higher kinetic energy enables the molecules to overcome the attractive forces that prevent the liquid from flowing. As a result, an oil that is thin enough to be a good lubricant in a cold engine will become too “thin” (have too low a viscosity) to be effective at high temperatures. The viscosity of motor oils is described by an (Society of Automotive Engineers) rating ranging from SAE 5 to SAE 50 for engine oils: the lower the number, the lower the viscosity (Figure \(\Page {5}\)). So-called can cause major problems. If they are viscous enough to work at high operating temperatures (SAE 50, for example), then at low temperatures, they can be so viscous that a car is difficult to start or an engine is not properly lubricated. Consequently, most modern oils are , with designations such as SAE 20W/50 (a grade used in high-performance sports cars), in which case the oil has the viscosity of an SAE 20 oil at subzero temperatures (hence the W for winter) and the viscosity of an SAE 50 oil at high temperatures. These properties are achieved by a careful blend of additives that modulate the intermolecular interactions in the oil, thereby controlling the temperature dependence of the viscosity. Many of the commercially available oil additives “for improved engine performance” are highly viscous materials that increase the viscosity and effective SAE rating of the oil, but overusing these additives can cause the same problems experienced with highly viscous single-grade oils. Based on the nature and strength of the intermolecular cohesive forces and the probable nature of the liquid–glass adhesive forces, predict what will happen when a glass capillary is put into a beaker of SAE 20 motor oil. Will the oil be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? (Hint: the surface of glass is lined with Si–OH groups.) substance and composition of the glass surface behavior of oil and the shape of meniscus Motor oil is a nonpolar liquid consisting largely of hydrocarbon chains. The cohesive forces responsible for its high boiling point are almost solely London dispersion forces between the hydrocarbon chains. Such a liquid cannot form strong interactions with the polar Si–OH groups of glass, so the surface of the oil inside the capillary will be lower than the level of the liquid in the beaker. The oil will have a convex meniscus similar to that of mercury. Predict what will happen when a glass capillary is put into a beaker of ethylene glycol. Will the ethylene glycol be pulled up into the tube by capillary action or pushed down below the surface of the liquid in the beaker? What will be the shape of the meniscus (convex or concave)? Capillary action will pull the ethylene glycol up into the capillary. The meniscus will be concave. Surface tension, capillary action, and viscosity are unique properties of liquids that depend on the nature of intermolecular interactions. is the energy required to increase the surface area of a liquid by a given amount. The stronger the intermolecular interactions, the greater the surface tension. are molecules, such as soaps and detergents, that reduce the surface tension of polar liquids like water. is the phenomenon in which liquids rise up into a narrow tube called a capillary. It results when , the intermolecular forces in the liquid, are weaker than , the attraction between a liquid and the surface of the capillary. The shape of the , the upper surface of a liquid in a tube, also reflects the balance between adhesive and cohesive forces. The of a liquid is its resistance to flow. Liquids that have strong intermolecular forces tend to have high viscosities. | 13,415 | 2,425 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.03%3A_Alkanes |
We begin our study of organic chemistry with the hydrocarbons, the simplest organic compounds, which are composed of carbon and hydrogen atoms only. As we noted, there are several different kinds of hydrocarbons. They are distinguished by the types of bonding between carbon atoms and the properties that result from that bonding. Hydrocarbons with only carbon-to-carbon single bonds (C–C) and existing as a continuous chain of carbon atoms also bonded to hydrogen atoms are called alkanes (or saturated hydrocarbons). , in this case, means that each carbon atom is bonded to four other atoms (hydrogen or carbon)—the most possible; there are no double or triple bonds in the molecules. The word has the same meaning for hydrocarbons as it does for the dietary fats and oils: the molecule has no carbon-to-carbon double bonds (C=C). We previously introduced the three simplest alkanes—methane (CH ), ethane (C H ), and propane (C H ) and they are shown again in Figure \(\Page {1}\). The flat representations shown do not accurately portray bond angles or molecular geometry. Methane has a tetrahedral shape that chemists often portray with wedges indicating bonds coming out toward you and dashed lines indicating bonds that go back away from you. An ordinary solid line indicates a bond in the plane of the page. Recall that the VSEPR theory correctly predicts a tetrahedral shape for the methane molecule (Figure \(\Page {2}\)). Methane (CH ), ethane (C H ), and propane (C H ) are the beginning of a series of compounds in which any two members in a sequence differ by one carbon atom and two hydrogen atoms—namely, a CH unit. The first 10 members of this series are given in Table \(\Page {1}\). Consider the series in Figure \(\Page {3}\). The sequence starts with C H , and a CH unit is added in each step moving up the series. Any family of compounds in which adjacent members differ from each other by a definite factor (here a CH group) is called a homologous series. The members of such a series, called , have properties that vary in a regular and predictable manner. The principle of gives organization to organic chemistry in much the same way that the periodic table gives organization to inorganic chemistry. Instead of a bewildering array of individual carbon compounds, we can study a few members of a homologous series and from them deduce some of the properties of other compounds in the series. The principle of homology allows us to write a general formula for alkanes: C H . Using this formula, we can write a molecular formula for any alkane with a given number of carbon atoms. For example, an alkane with eight carbon atoms has the molecular formula C H = C H . | 2,717 | 2,427 |
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