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https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.04%3A_Colligative_Properties-_Boiling_Point_Elevation_and_Freezing_Point_Depression |
Make sure you thoroughly understand the following essential ideas: The colligative properties really depend on the escaping tendency of solvent molecules from the liquid phase. You will recall that the vapor pressure is a direct measure of escaping tendency, so we can use these terms more or less interchangeably. If addition of a nonvolatile solute lowers the vapor pressure of the solution via Raoult's law , then it follows that the temperature must be raised to restore the vapor pressure to the value corresponding to the pure solvent. In particular, the temperature at which the vapor pressure is 1 atm will be higher than the normal boiling point by an amount known as the boiling point elevation . The exact relation between the boiling point of the solution and the mole fraction of the solvent is rather complicated, but for dilute solutions the elevation of the boiling point is directly proportional to the concentration of the solute: Bear in mind that the proportionality constant is a property of the because this is the only component that contributes to the vapor pressure in the model we are considering in this section. Sucrose (C O H , 342 g mol ), like many sugars, is highly soluble in water; almost 2000 g will dissolve in 1 L of water, giving rise to what amounts to pancake syrup. Estimate the boiling point of such a sugar solution. moles of sucrose: \[ \dfrac{2000\, g}{342\, g\, mol^{–1}} = 5.8\; mol\] mass of water: assume 1000 g (we must know the density of the solution to find its exact value) The molality of the solution is (5.8 mol) ÷ (1.0 kg) = 5.8 m. Using the value of from the table, the boiling point will be raised by (0.514 K mol kg) × (5.8 mol kg ) = 3.0 K, so the boiling point will be 103° C. The freezing point of a substance is the temperature at which the solid and liquid forms can coexist indefinitely — that is, they are in equilibrium. Under these conditions molecules pass between the two phases at equal rates because their escaping tendencies from the two phases are identical. Suppose that a liquid solvent and its solid (water and ice, for example) are in equilibrium ( below), and we add a non-volatile solute (such as salt, sugar, or automotive antifreeze liquid) to the water. This will have the effect of reducing the mole fraction of H O molecules in the liquid phase, and thus reduce the tendency of these molecules to escape from it, not only into the vapor phase (as we saw above), but also into the solid (ice) phase. This will have no effect on the rate at which H O molecules escape from the ice into the water phase, so the system will no longer be in equilibrium and the ice will begin to melt . If we wish to keep the solid from melting, the escaping tendency of molecules from the solid must be reduced. This can be accomplished by reducing the ; this lowers the escaping tendency of molecules from both phases, but it affects those in the solid more than those in the liquid, so we eventually reach the new, lower freezing point where the two quantities are again in exact balance and both phases can coexist . If you prefer to think in terms of vapor pressures, you can use the same argument if you bear in mind that the vapor pressures of the solid and liquid must be the same at the freezing point. Dilution of the liquid (the solvent) by the nonvolatile solute reduces the vapor pressure of the solvent according to Raoult’s law, thus reducing the temperature at which the vapor pressures of the liquid and frozen forms of the solution will be equal. As with boiling point elevation, in dilute solutions there is a simple linear relation between the freezing point depression and the molality of the solute: \[ \Delta T_f = K_f \dfrac{\text{moles of solute}}{\text{kg of solvent}}\] Note that values are all ! The use of salt to de-ice roads is a common application of this principle. The solution formed when some of the salt dissolves in the moist ice reduces the freezing point of the ice. If the freezing point falls below the ambient temperature, the ice melts. In very cold weather, the ambient temperature may be below that of the salt solution, and the salt will have no effect. The effectiveness of a de-icing salt depends on the number of particles it releases on dissociation and on its solubility in water: Automotive radiator antifreezes are mostly based on ethylene glycol, (CH OH) . Owing to the strong hydrogen-bonding properties of this double alcohol, this substance is miscible with water in all proportions, and contributes only a very small vapor pressure of its own. Besides lowering the freezing point, antifreeze also raises the boiling point, increasing the operating range of the cooling system. The pure glycol freezes at –12.9°C and boils at 197°C, allowing water-glycol mixtures to be tailored to a wide range of conditions. Estimate the freezing point of an antifreeze mixture is made up by combining one volume of ethylene glycol (MW = 62, density 1.11 g cm ) with two volumes of water. Assume that we use 1 L of glycol and 2 L of water (the actual volumes do not matter as long as their ratios are as given.) The mass of the glycol will be 1.10 kg and that of the water will be 2.0 kg, so the total mass of the solution is 3.11 kg. We then have: Any ionic species formed by dissociation will also contribute to the freezing point depression. This can serve as a useful means of determining the fraction of a solute that is dissociated. An aqueous solution of nitrous acid (HNO , MW = 47) freezes at –0.198 .C. If the solution was prepared by adding 0.100 mole of the acid to 1000 g of water, what percentage of the HNO is dissociated in the solution? The nominal molality of the solution is (.001 mol) ÷ (1.00 kg) = 0.001 mol kg . But the molality according to the observed Δ value is given by Δ ÷ = (–.198 K) ÷(–1.86 K kg mol ) = 0.106 mol kg ; this is the total number of moles of species present after the dissociation reaction HNO → H + NO has occurred. If we let = [H ] = [NO ], then by stoichiometry, [HNO ] = 0.100 - x and .106 - = 2x and = .0355. The fraction of HNO that is dissociated is .0355 ÷ 0.100 = .355, corresponding to 35.5% dissociation of the acid. A simple can provide more insight into these phenomena. You may already be familiar with the phase map for water below. The normal boiling point of the pure solvent is indicated by point where the vapor pressure curve intersects the 1-atm line — that is, where the escaping tendency of solvent molecules from the liquid is equivalent to 1 atmosphere pressure. Addition of a non-volatile solute reduces the vapor pressures to the values given by the blue line. This shifts the boiling point to the right , corresponding to the increase in temperature Δ required to raise the escaping tendency of the H O molecules back up to 1 atm. To understand freezing point depression, notice that the vapor pressure line intersects the curved black vapor pressure line of the solid (ice), which corresponds to a new triple point at which all three phases (ice, water vapor, and liquid water) are in equilibrium and thus exhibit equal escaping tendencies. This point is by definition the origin of the freezing (solid-liquid) line, which intersects the 1-atm line at a reduced freezing point Δ , indicated by . Note that the above analysis assumes that the solute is soluble only in the liquid solvent, but not in its solid form. This is generally more or less true. For example, when arctic ice forms from seawater, the salts get mostly "squeezed" out. This has the interesting effect of making the water that remains more saline, and hence more dense, causing it to sink to the bottom part of the ocean where it gets taken up by the south-flowing deep current. Those readers who have some knowledge of thermodynamics will recognize that what we have been referring to as "escaping" tendency is really a manifestation of the . This schematic plot shows how the 's for the solid, liquid, and gas phases of a typical substance vary with the temperature. The rule is that the phase with the most negative free energy rules. The phase that is most stable (and which therefore is the only one that exists) is always the one having the most negative free energy (indicated here by the thicker portions of the plotted lines.) The melting and boiling points correspond to the respective temperatures where the solid and liquid and liquid and vapor have identical free energies. As we saw above, adding a solute to the liquid dilutes it, making its free energy more negative, with the result that the freezing and boiling points are shifted to the left and right, respectively. The relationships shown in these plots depend on the differing slopes of the lines representing the free energies of the phases as the temperature changes. These slopes are proportional to the entropy of each phase. Because gases have the highest entropies, the slope of the "gaseous solvent" line is much greater than that of the others. Note that this plot is not to scale. | 9,067 | 3,040 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_OpenStax/06%3A_Electronic_Structure_and_Periodic_Properties_(Exercises) |
The light produced by a red neon sign is due to the emission of light by excited neon atoms. Qualitatively describe the spectrum produced by passing light from a neon lamp through a prism. The spectrum consists of colored lines, at least one of which (probably the brightest) is red. An FM radio station found at 103.1 on the FM dial broadcasts at a frequency of 1.031 × 10 s (103.1 MHz). What is the wavelength of these radio waves in meters? \(λ=\dfrac{c}{ν}\) \(λ = \dfrac{2.998 \times10^{8}\: \dfrac{m}{s}}{1.031 \times10^{8}\: \dfrac{1}{s}} = 2.908\:m\) FM-95, an FM radio station, broadcasts at a frequency of 9.51 × 10 s (95.1 MHz). What is the wavelength of these radio waves in meters? \(λ=\dfrac{c}{ν}\) \(λ = \dfrac{2.998 \times10^{8}\: \dfrac{m}{s}}{9.51 \times10^{7}\: \dfrac{1}{s}} = 3.15 \:m\) A bright violet line occurs at 435.8 nm in the emission spectrum of mercury vapor. What amount of energy, in joules, must be released by an electron in a mercury atom to produce a photon of this light? \(E= \dfrac{hc}{λ}\) \(E= \dfrac{(2.998 \times 10^{8} \dfrac{m}{s})\: (6.6262 \times 10^{-34} Js)}{4.358 \times 10^{-7} m}\) \(=4.56\times 10^{-19}J\) Light with a wavelength of 614.5 nm looks orange. What is the energy, in joules, per photon of this orange light? What is the energy in eV (1 eV = 1.602 × 10 J)? 1.) First convert 614.5 nm into meters \(6.145 nm\) = \(6.145 \times10^{-7} m\) 2.) Then calculate the amount of energy this wavelength of light contains using the equations: \(E = hν\) and \(ν=\dfrac{c}{λ}\) which can be manipulated algebraically into: \(E=\dfrac{hc}{λ}\) \[= \dfrac{(6.6262\times 10^{−34}\;Js) \ (2.998\times 10^{8}\; \dfrac {m}{s}) } {6.145 \times 10^{−7}m \ } =3.233 \times 10^{-19} J\] 3.) Then convert Joules into eV: \[=(3.233 \times 10^{−19}J) \times\dfrac{1eV}{1.602\times 10^{-19}J} \]\[=2.018 eV\] Heated lithium atoms emit photons of light with an energy of 2.961 × 10 J. Calculate the frequency and wavelength of one of these photons. What is the total energy in 1 mole of these photons? What is the color of the emitted light? 1.) \(E = hν\) and \(ν=\dfrac{E}{h}\) Frequency: \(ν=\dfrac{ 2.961\times 10^{-19}J}{6.6262\times 10^{-34}Js}\ = 4.469\times 10^{14} Hz\) 2.) \(λ=\dfrac{c}{ν}\) c = Speed of light → \(2.998\times 10^{8}\; \dfrac {m}{s}\) Wavelength: \(λ=\dfrac{ 2.998\times 10^{8}\dfrac {m}{s}}{4.469\times 10^{14}\dfrac {1}{s}}\ = 6.709\times 10^{-7} m\) = Total energy: \(E= \dfrac{2.961 \times 10^{-19}J}{1\: photon}\times\dfrac{6.022\times 10^{23}\: photons}{1\: mole} = 1.783\times 10^{5} J \) A photon of light produced by a surgical laser has an energy of 3.027 × 10 J. Calculate the frequency and wavelength of the photon. What is the total energy in 1 mole of photons? What is the color of the emitted light? = 4.568 × 10 s; λ = 656.3 nm; Energy mol = 1.823 × 10 J mol ; red When rubidium ions are heated to a high temperature, two lines are observed in its line spectrum at wavelengths (a) 7.9 × 10 m and (b) 4.2 × 10 m. What are the frequencies of the two lines? What color do we see when we heat a rubidium compound? The emission spectrum of cesium contains two lines whose frequencies are (a) 3.45 × 10 Hz and (b) 6.53 × 10 Hz. What are the wavelengths and energies per photon of the two lines? What color are the lines? (a) = 8.69 × 10 m; = 2.29 × 10 J; (b) = 4.59 × 10 m; = 4.33 × 10 J; The color of (a) is red; (b) is blue. Photons of infrared radiation are responsible for much of the warmth we feel when holding our hands before a fire. These photons will also warm other objects. How many infrared photons with a wavelength of 1.5 × 10 m must be absorbed by the water to warm a cup of water (175 g) from 25.0 °C to 40 ° 1.) First we must use the equation: q=mCΔT to calculate the amount of Energy in Joules (J) to warm 175g of H O a total of 15 Celsius \[q = 175g\times \dfrac {4.184 J}{g°C}\times 15° C\] q= 11,000 J 2.) Now we need to calculate the amount of energy in Joules (J) that one Photon with a wavelength of 1.5x10 m contains. We will use the equations: \(E = hν\) and \(ν=\dfrac{c}{λ}\) which can be manipulated algebraically into: \(E=\dfrac{hc}{λ}\) \[= \dfrac{(6.6262\times 10^{−34}\;Js) \ (2.998\times 10^{8}\; \dfrac {m}{s}) } {1.5 \times 10^{−6}m \ } =1.3 \times 10^{-19} J\] This is the energy in one photon, so now we have to see how many times 1.3x10 J fits into the 11,000 J from our first calculation: \[= \dfrac{11,000\;J} {1.3 \times 10^{−19}\; \dfrac {J}{photon} } =8.3\times 10^{22} photons\] \[= 8.3\times 10^{22} photons\] One of the radiographic devices used in a dentist's office emits an X-ray of wavelength 2.090 × 10 m. What is the energy, in joules, and frequency of this X-ray? = 9.502 × 10 J; = 1.434 × 10 s The eyes of certain reptiles pass a single visual signal to the brain when the visual receptors are struck by photons of a wavelength of 850 nm. If a total energy of 3.15 × 10 J is required to trip the signal, what is the minimum number of photons that must strike the receptor? 1.) First we need to convert nanometers to meters 850nm = 8.5 x 10 m 2. Then calculate Energy in Joules a photon of this wavelength (λ) produces: \[(E=\dfrac{hc}{λ}) = \dfrac{(6.6262\times 10^{−34}\;Js) \ (2.998\times 10^{8}\; \dfrac {m}{s}) } {8.5 \times 10^{−7}m \ } = 2.3 \times 10^{-19} \dfrac {J}{photon}\] 3.) Then we need to find out how many of these photons it will take to trip the visual signal to the brain: \[ \begin{align} &= \dfrac{3.15\times 10^{-14}\;\cancel{J}} {2.3 \times 10^{−19}\; \dfrac {\cancel{J}}{photon} } \\[5pt] &=1.3\times 10^{5}\, photons \end{align}\] RGB color television and computer displays use cathode ray tubes that produce colors by mixing red, green, and blue light. If we look at the screen with a magnifying glass, we can see individual dots turn on and off as the colors change. Using a spectrum of visible light, determine the approximate wavelength of each of these colors. What is the frequency and energy of a photon of each of these colors? Somewhat different numbers are also possible. Answer the following questions about a Blu-ray laser: What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 10 s ejects a photon with 7.74 × 10 J kinetic energy? Will the photoelectric effect be observed if sodium is exposed to orange light? 5.49 × 10 s ; no Why is the electron in a Bohr hydrogen atom bound less tightly when it has a quantum number of 3 than when it has a quantum number of 1? What does it mean to say that the energy of the electrons in an atom is quantized? Quantized energy means that the electrons can possess only certain discrete energy values; values between those quantized values are not permitted. Using the Bohr model, determine the energy, in joules, necessary to ionize a ground-state hydrogen atom. Show your calculations. The electron volt (eV) is a convenient unit of energy for expressing atomic-scale energies. It is the amount of energy that an electron gains when subjected to a potential of 1 volt; \(1 \;eV = 1.602 \times 10^{-19}\; J\). Using the Bohr model, determine the energy, in electron volts, of the photon produced when an electron in a hydrogen atom moves from the orbit with \(n = 5\) to the orbit with \(n = 2\). Show your calculations. \[E=E_2−E_5=2.179 \times 10^{−18} \left (\dfrac{1}{n_2^2}−\dfrac{1}{n^2_5}\right) \; J\] \[ = 2.179 \times 10^{-18} \left (\dfrac{1}{2^2}−\dfrac{1}{5^2}\right)=4.576 \times 10^{−19}\; J\] \[= \dfrac{4.576\times 10^{−19}\;J} {1.602 \times 10^{−19}\;J \; eV^{−1} } =2.856\; eV\] Using the Bohr model, determine the lowest possible energy for the electron in the \(He^+\) ion. Using the Bohr model, determine the energy of an electron with \(n = 6\) in a hydrogen atom. −8.716 × 10−18 J Using the Bohr model, determine the energy of an electron with \(n = 8\) in a hydrogen atom. How far from the nucleus in angstroms (1 angstrom = \(1 \times 10^{–10}\; \ce m\)) is the electron in a hydrogen atom if it has an energy of \(-8.72 \times 10^{-20}\; \ce J\)? \(−3.405 \times 10^{−20} J\) What is the radius, in angstroms, of the orbital of an electron with n = 8 in a hydrogen atom? Using the Bohr model, determine the energy in joules of the photon produced when an electron in a \(\ce{He^{+}}\) ion moves from the orbit with n = 5 to the orbit with n = 2. Using the Bohr model, determine the energy in joules of the photon produced when an electron in a Li2+ ion moves from the orbit with n = 2 to the orbit with n = 1. 1.471 × 10−17 J Consider a large number of hydrogen atoms with electrons randomly distributed in the n = 1, 2, 3, and 4 orbits How are the Bohr model and the Rutherford model of the atom similar? How are they different? The spectra of hydrogen and of calcium are shown in What causes the lines in these spectra? Why are the colors of the lines different? Suggest a reason for the observation that the spectrum of calcium is more complicated than the spectrum of hydrogen Both involve a relatively heavy nucleus with electrons moving around it, although strictly speaking, the Bohr model works only for one-electron atoms or ions. According to classical mechanics, the Rutherford model predicts a miniature “solar system” with electrons moving about the nucleus in circular or elliptical orbits that are confined to planes. If the requirements of classical electromagnetic theory that electrons in such orbits would emit electromagnetic radiation are ignored, such atoms would be stable, having constant energy and angular momentum, but would not emit any visible light (contrary to observation). If classical electromagnetic theory is applied, then the Rutherford atom would emit electromagnetic radiation of continually increasing frequency (contrary to the observed discrete spectra), thereby losing energy until the atom collapsed in an absurdly short time (contrary to the observed long-term stability of atoms). The Bohr model retains the classical mechanics view of circular orbits confined to planes having constant energy and angular momentum, but restricts these to quantized values dependent on a single quantum number, n. The orbiting electron in Bohr’s model is assumed not to emit any electromagnetic radiation while moving about the nucleus in its stationary orbits, but the atom can emit or absorb electromagnetic radiation when the electron changes from one orbit to another. Because of the quantized orbits, such “quantum jumps” will produce discrete spectra, in agreement with observations. How are the Bohr model and the quantum mechanical model of the hydrogen atom similar? How are they different? Both models have a central positively charged nucleus with electrons moving about the nucleus in accordance with the Coulomb electrostatic potential. The Bohr model that the electrons move in circular orbits that have quantized energies, angular momentum, and radii that are specified by a single quantum number, = 1, 2, 3, …, but this quantization is an ad hoc assumption made by Bohr to incorporate quantization into an essentially classical mechanics description of the atom. Bohr also assumed that electrons orbiting the nucleus normally do not emit or absorb electromagnetic radiation, but do so when the electron switches to a different orbit. In the quantum mechanical model, the electrons do not move in precise orbits (such orbits violate the Heisenberg uncertainty principle) and, instead, a probabilistic interpretation of the electron’s position at any given instant is used, with a mathematical function called a wavefunction that can be used to determine the electron’s spatial probability distribution. These wavefunctions, or orbitals, are three-dimensional stationary waves that can be specified by three quantum numbers that arise naturally from their underlying mathematics (no ad hoc assumptions required): the principal quantum number, (the same one used by Bohr), which specifies shells such that orbitals having the same all have the same energy and approximately the same spatial extent; the angular momentum quantum number , which is a measure of the orbital’s angular momentum and corresponds to the orbitals’ general shapes, as well as specifying subshells such that orbitals having the same (and ) all have the same energy; and the orientation quantum number , which is a measure of the component of the angular momentum and corresponds to the orientations of the orbitals. The Bohr model gives the same expression for the energy as the quantum mechanical expression and, hence, both properly account for hydrogen’s discrete spectrum (an example of getting the right answers for the wrong reasons, something that many chemistry students can sympathize with), but gives the wrong expression for the angular momentum (Bohr orbits necessarily all have non-zero angular momentum, but some quantum orbitals [ orbitals] can have zero angular momentum). What are the allowed values for each of the four quantum numbers: , , , and ? Describe the properties of an electron associated with each of the following four quantum numbers: , , , and . determines the general range for the value of energy and the probable distances that the electron can be from the nucleus. determines the shape of the orbital. determines the orientation of the orbitals of the same value with respect to one another. determines the spin of an electron. Answer the following questions: Identify the subshell in which electrons with the following quantum numbers are found: (a) 2 ; (b) 4 ; (c) 6 Which of the subshells described in Question 5 contain degenerate orbitals? How many degenerate orbitals are in each? Identify the subshell in which electrons with the following quantum numbers are found: (a) 3 (b) 1 (c) 4 Which of the subshells described in Question 7 contain degenerate orbitals? How many degenerate orbitals are in each? Sketch the boundary surface of a \(d_{x^2−y^2}\) and a orbital. Be sure to show and label the axes. Sketch the and orbitals. Be sure to show and label the coordinates. Consider the orbitals shown here in outline. (a) x. 2, y. 2, z. 2; (b) x. 1, y. 3, z. 0; (c) x. 4 0 0 \(\dfrac{1}{2}\), y. 2 1 0 \(\dfrac{1}{2}\), z. 3 2 0 \(\dfrac{1}{2}\); (d) x. 1, y. 2, z. 3; (e) x. = 0, = 0, y. = 1, –1, 0, or 1, z. = 2, = –2, –1, 0, +1, +2 State the Heisenberg uncertainty principle. Describe briefly what the principle implies. How many electrons could be held in the second shell of an atom if the spin quantum number could have three values instead of just two? (Hint: Consider the Pauli exclusion principle.) 12 Which of the following equations describe particle-like behavior? Which describe wavelike behavior? Do any involve both types of behavior? Describe the reasons for your choices. Write a set of quantum numbers for each of the electrons with an of 4 in a Se atom. Read the labels of several commercial products and identify monatomic ions of at least four transition elements contained in the products. Write the complete electron configurations of these cations. Read the labels of several commercial products and identify monatomic ions of at least six main group elements contained in the products. Write the complete electron configurations of these cations and anions. For example, Na : 1 2 2 ; Ca : 1 2 2 ; Sn : 1 2 2 3 3 3 4 4 4 5 ; F : 1 2 2 ; O : 1 2 2 ; Cl : 1 2 2 3 3 . Using complete subshell notation (not abbreviations, 1 2 2 , and so forth), predict the electron configuration of each of the following atoms: a.) 1s 2s 2p b.) 1s 2s 2p 3s 3p c.) 1s 2s 2p 3s 3p 3d 4s d.) 1s 2s 2p 3s 3p 3d 4s 4p 4d 5s 5p e.) 1s 2s 2p 3s 3p 3d 4s 4p 4d 4f 5s 5p 6s Using complete subshell notation (1 2 2 , and so forth), predict the electron configuration of each of the following atoms: Is 1 2 2 the symbol for a macroscopic property or a microscopic property of an element? Explain your answer. What additional information do we need to answer the question “Which ion has the electron configuration 1 2 2 3 3 ”? The charge on the ion. Draw the orbital diagram for the valence shell of each of the following atoms: Use an orbital diagram to describe the electron configuration of the valence shell of each of the following atoms: (a) Using complete subshell notation (1 2 2 , and so forth), predict the electron configurations of the following ions. Which atom has the electron configuration: 1 2 2 3 3 4 3 4 5 4 ? Zr Which atom has the electron configuration: 1 2 2 3 3 3 4 ? Co ; Cobalt Which ion with a +1 charge has the electron configuration 1 2 2 3 3 3 4 4 ? Which ion with a –2 charge has this configuration? Rb , Se Which of the following atoms contains only three valence electrons: Li, B, N, F, Ne? B ; Boron Which of the following has two unpaired electrons? Although both (b) and (c) are correct, (e) encompasses both and is the best answer. Which atom would be expected to have a half-filled 6 subshell? Bi ; Bismuth Which atom would be expected to have a half-filled 4 subshell? In one area of Australia, the cattle did not thrive despite the presence of suitable forage. An investigation showed the cause to be the absence of sufficient cobalt in the soil. Cobalt forms cations in two oxidation states, Co and Co . Write the electron structure of the two cations. Thallium was used as a poison in the Agatha Christie mystery story “The Pale Horse.” Thallium has two possible cationic forms, +1 and +3. The +1 compounds are the more stable. Write the electron structure of the +1 cation of thallium. 1 2 2 3 3 3 4 4 4 5 5 6 4 5 Write the electron configurations for the following atoms or ions: Cobalt–60 and iodine–131 are radioactive isotopes commonly used in nuclear medicine. How many protons, neutrons, and electrons are in atoms of these isotopes? Write the complete electron configuration for each isotope. Co has 27 protons, 27 electrons, and 33 neutrons: 1 2 2 3 3 4 3 . I has 53 protons, 53 electrons, and 78 neutrons: 1 2 2 3 3 3 4 4 4 5 5 . Write a set of quantum numbers for each of the electrons with an of 3 in a Sc atom. Based on their positions in the periodic table, predict which has the smallest atomic radius: Mg, Sr, Si, Cl, I. Cl Based on their positions in the periodic table, predict which has the largest atomic radius: Li, Rb, N, F, I. Based on their positions in the periodic table, predict which has the largest first ionization energy: Mg, Ba, B, O, Te. O Based on their positions in the periodic table, predict which has the smallest first ionization energy: Li, Cs, N, F, I. Based on their positions in the periodic table, rank the following atoms in order of increasing first ionization energy: F, Li, N, Rb Rb < Li < N < F Based on their positions in the periodic table, rank the following atoms or compounds in order of increasing first ionization energy: Mg, O, S, Si Atoms of which group in the periodic table have a valence shell electron configuration of ? 15 (5A) Atoms of which group in the periodic table have a valence shell electron configuration of ? Based on their positions in the periodic table, list the following atoms in order of increasing radius: Mg, Ca, Rb, Cs. Mg < Ca < Rb < Cs Based on their positions in the periodic table, list the following atoms in order of increasing radius: Sr, Ca, Si, Cl. Based on their positions in the periodic table, list the following ions in order of increasing radius: K , Ca , Al , Si . Si < Al < Ca < K List the following ions in order of increasing radius: Li , Mg , Br , Te . Which atom and/or ion is (are) isoelectronic with Br : Se , Se, As , Kr, Ga , Cl ? Se, As Which of the following atoms and ions is (are) isoelectronic with S : Si , Cl , Ar, As , Si, Al ? Compare both the numbers of protons and electrons present in each to rank the following ions in order of increasing radius: As , Br , K , Mg . Mg < K < Br < As Of the five elements Al, Cl, I, Na, Rb, which has the most exothermic reaction? (E represents an atom.) What name is given to the energy for the reaction? Hint: note the process depicted does correspond to electron affinity \(\ce{E+}(g)+\ce{e-}⟶\ce{E}(g)\) Of the five elements Sn, Si, Sb, O, Te, which has the most endothermic reaction? (E represents an atom.) What name is given to the energy for the reaction? \[\ce{E}(g)⟶\ce{E+}(g)+\ce{e-}\] O, IE The ionic radii of the ions S , Cl , and K are 184, 181, 138 pm respectively. Explain why these ions have different sizes even though they contain the same number of electrons. Which main group atom would be expected to have the lowest second ionization energy? Ra Explain why Al is a member of group 13 rather than group 3? | 20,768 | 3,042 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/05%3A_Dioxygen_Reactions/5.01%3A_Catalase_and_Peroxidase |
Catalase and peroxidase are heme enzymes that catalyze reactions of hydrogen peroxide. In catalase, the enzymatic reaction is the disproportionation of hydrogen peroxide (Reaction 5.82) and the function of the enzyme appears to be prevention of any buildup of that potentially dangerous oxidant (see the discussion of dioxygen toxicity in Section III). \[2H_{2}O_{2} \xrightarrow{catalase} 2H_{2}O + O_{2} \tag{5.82}\] Peroxidase reacts by mechanisms similar to catalase, but the reaction catalyzed is the oxidation of a wide variety of organic and inorganic substrates by hydrogen peroxide (Reaction 5.83). \[H_{2}O_{2} + AH_{2} \xrightarrow{peroxidase} 2H_{2}O + A \tag{5.83}\] (The catalase reaction can be seen to be a special case of Reaction 5.83 in which the substrate, AH , is hydrogen peroxide.) Some examples of peroxidases that have been characterized are horseradish peroxidase, cytochrome c peroxidase, glutathione peroxidase, and myeloperoxidase. X-ray crystal structures have been determined for beef-liver catalase and for horseradish peroxidase in the resting, high-spin ferric state. In both, there is a single heme b group at the active site. In catalase, the axial ligands are a phenolate from a tyrosyl residue, bound to the heme on the side away from the active-site cavity, and water, bound to heme within the cavity and presumably replaced by hydrogen peroxide in the catalytic reaction. In horseradish peroxidase, the axial ligand is an imidazole from a histidyl residue. Also within the active-site cavity are histidine and aspartate or asparagine side chains that appear to be ideally situated to interact with hydrogen peroxide when it is bound to the iron. These residues are believed to play an important part in the mechanism by facilitating O—O bond cleavage (see Section VI.B below). Three other forms of catalase and peroxidase can be generated, which are referred to as compounds I, II, and III. Compound I is generated by reaction of the ferric state of the enzymes with hydrogen peroxide. Compound I is green and has spectral characteristics very similar to the Fe (P )(O) complex prepared at low temperatures by reaction of ferric porphyrins with single-oxygenatom donors (see Section V.C.1.a). Titrations with reducing agents indicate that it is oxidized by two equivalents above the ferric form. It has been proposed (see 5.84) that the anionic nature of the tyrosinate axial ligand in catalase may serve to stabilize the highly oxidized iron center in compound I of that enzyme, and furthermore that the histidyl imidazole ligand in peroxidase may deprotonate, forming imidazolate, or may be strongly hydrogen bonded, thus serving a similar stabilizing function for compound I in that enzyme. \(\tag{5.84}\) Reduction of compound I by one electron produces compound II, which has the characteristics of a normal ferryl-porphyrin complex, analogous to , i.e., (L)Fe (P)(O). Reaction of compound II with hydrogen peroxide produces compound III, which can also be prepared by reaction of the ferrous enzyme with dioxygen. It is an oxy form, analogous to oxymyoglobin, and does not appear to have a physiological function. The reactions producing these three forms and their proposed formulations are summarized in Reactions (5.85) to (5.88). \[Fe^{III}(P)^{+} + H_{2}O_{2} \rightarrow Fe^{IV}(P^{-})(O)^{+} + H_{2}O \tag{5.85}\] \[ferric\; form \quad \qquad \qquad Compound\; I \qquad \qquad \] \[Fe^{IV}(P^{\cdotp -})(O)^{+} + e^{-} \rightarrow Fe^{IV}(P)(O) \tag{5.86}\] \[Compound\; I \qquad \qquad Compound\; II \quad \] \[Fe^{IV}(P)(O) + H_{2}O_{2} \rightarrow Fe(P)O_{2} + H_{2}O \tag{5.87}\] \[Compound\; II \qquad \qquad Compound\; III \qquad \qquad \] \[Fe{II}(P) + O_{2} \rightarrow Fe(P)O_{2} \tag{5.88}\] \[ferrous\; form \qquad Compound\; III\] The accepted mechanisms for catalase and peroxidase are described in Reactions (5.89) to (5.94). \[Fe^{III}(P)^{+} + H_{2}O_{2} \rightarrow Fe^{III}(P)(H_{2}O_{2})^{+} \rightarrow Fe^{IV}(P^{\cdotp -})(O)^{+} + H_{2}O \tag{5.89}\] \[\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad Compound\; I\] catalase: \[ Fe^{IV}(P^{\cdotp -})(O)^{+} + H_{2}O_{2} \rightarrow Fe^{III}(P)^{+} + H_{2}O + O_{2} \tag{5.90}\] \[Compound\; I \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \] peroxidase: \[ Fe^{IV}(P^{\cdotp -})(O)^{+} + AH_{2} \rightarrow Fe^{IV}(P)(O) +HA^{\cdotp} + H^{+} \tag{5.91}\] \[Compound\; I \qquad \qquad \qquad \qquad Compound\; II\] \[Fe^{IV}(P)(O) + AH_{2} \rightarrow Fe^{III}(P)^{+} +HA^{\cdotp} + OH^{-} \tag{5.92}\] \[Compound\; II \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \] \[2HA^{\cdotp} \rightarrow A + AH_{2} \tag{5.93}\] or \[2HA^{\cdotp} \rightarrow HA - AH \tag{5.94}\] In the catalase reaction, it has been established by use of H O that the dioxygen formed is derived from hydrogen peroxide, i.e., that O—O bond cleavage does not occur in Reaction (5.90), which is therefore a two-electron reduction of compound I by hydrogen peroxide, with the oxo ligand of the former being released as water. For the peroxidase reaction under physiological conditions, it is believed that the oxidation proceeds in one-electron steps (Reactions 5.91 and 5.92), with the final formation of product occurring by disproportionation (Reaction 5.93) or coupling (Reaction 5.94) of the one-electron oxidized intermediate. The proposal that these three enzymes all go through a similar high-valent oxo intermediate, i.e., or compound I, raises two interesting questions. The first of these is why the same high-valent metal-oxo intermediate gives two very different types of reactions, i.e., oxygen-atom transfer with cytochrome P-450 and electron transfer with catalase and peroxidase. The answer is that, although the high-valent metal-oxo heme cores of these intermediates are in fact very similar, the substrate-binding cavities seem to differ substantially in how much access the substrate has to the iron center. With cytochrome P-450, the substrate is jammed right up against the location where the oxo ligand must reside in the high-valent oxo intermediate. But the same location in the peroxidase enzymes is blocked by the protein structure so that substrates can interact only with the heme edge. Thus oxidation of the substrate by electron transfer is possible for catalase and peroxidase, but the substrate is too far away from the oxo ligand for oxygen-atom transfer. The second question is about how the the high-valent oxo intermediate forms in both enzymes. For catalase and peroxidase, the evidence indicates that hydrogen peroxide binds to the ferric center and then undergoes heterolysis at the O—O bond. Heterolytic cleavage requires a significant separation of positive and negative charge in the transition state. In catalase and peroxidase, analysis of the crystal structure indicates strongly that amino-acid side chains are situated to aid in the cleavage by stabilizing a charge-separated transition state (Figure 5.14). In cytochrome P-450, as mentioned in Section V.C.1, no such groups are found in the hydrophobic substrate-binding cavity. It is possible that the cysteinyl axial ligand in cytochrome P-450 plays an important role in O—O bond cleavage, and that the interactions found in catalase and peroxidase that appear to facilitate such cleavage are therefore not necessary. | 7,402 | 3,043 |
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It was initially believed that the noble gases could not form compounds due to their full valence shell of electrons that rendered them very chemically stable and unreactive. All noble gases have full s and p outer electron shells (except helium, which has no p sublevel), and so do not form chemical compounds easily. Because of their high ionization energy and almost zero electron affinity, they were not expected to be reactive. The heavier noble gases have more electron shells than the lighter ones. Hence, the outermost electrons experience a shielding effect from the inner electrons that makes them more easily ionized, since they are less strongly attracted to the positively charged nucleus. This results in an ionization energy low enough to form stable compounds with the most electronegative elements, fluorine and oxygen, and even with less electronegative elements such as nitrogen and carbon under certain circumstances. These compounds are listed in order of decreasing order of the atomic weight of the noble gas, which generally reflects the priority of their discovery, and the breadth of available information for these compounds. | 1,170 | 3,044 |
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Essential Skills 3 in , introduced the common, or base-10, logarithms and showed how to use the properties of exponents to perform logarithmic calculations. In this section, we describe natural logarithms, their relationship to common logarithms, and how to do calculations with them using the same properties of exponents. Many natural phenomena exhibit an exponential rate of increase or decrease. Population growth is an example of an exponential rate of increase, whereas a runner’s performance may show an exponential decline if initial improvements are substantially greater than those that occur at later stages of training. Exponential changes are represented logarithmically by , where is an irrational number whose value is approximately 2.7183. The , abbreviated as ln, is the power to which must be raised to obtain a particular number. The natural logarithm of is 1 (ln = 1). Some important relationships between base-10 logarithms and natural logarithms are as follows: According to these relationships, ln 10 = 2.303 and log 10 = 1. Because multiplying by 1 does not change an equality, Substituting any value for 10 gives Other important relationships are as follows: Entering a value , such as 3.86, into your calculator and pressing the “ln” key gives the value of ln , which is 1.35 for = 3.86. Conversely, entering the value 1.35 and pressing “ ” key gives an answer of 3.86. Hence Calculate the natural logarithm of each number and express each as a power of the base . What number is each value the natural logarithm of? Like common logarithms, natural logarithms use the properties of exponents. We can compare the properties of common and natural logarithms: \[ \dfrac{10^{a}}{10^{b}}=10^{a-b} \notag \] \[ \dfrac{e^{a}}{e^{b}}=e^{a-b} \notag \] \[ log \left (\dfrac{1}{a} \right )=-log\left ( a \right ) \notag \] \[ ln \left (\dfrac{1}{x} \right )=-ln\left ( x \right ) \notag \] The number of significant figures in a number is the same as the number of digits after the decimal point in its logarithm. For example, the natural logarithm of 18.45 is 2.9151, which means that is equal to 18.45. Calculate the natural logarithm of each number. The answers obtained using the two methods may differ slightly due to rounding errors. Calculate the natural logarithm of each number. | 2,338 | 3,045 |
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To study the flow of energy during a chemical reaction, we need to distinguish between a system, the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its surroundings, the rest of the universe, including the container in which the reaction is carried out (Figure \(\Page {1}\)). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa. Three kinds of systems are important in chemistry. An can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, , which must be true if . The state of a system is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A state function is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are . For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (Figure \(\Page {2}\)). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose. The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel (Figure \(\Page {3}\)). The balanced chemical equation for the reaction is as follows: \[\ce{ 2Al(s) + Fe_2O_3(s) -> 2Fe(s) + Al_2O_3(s)} \label{5.2.1} \] We can also write this chemical equation as \[\ce{2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s)} + \text{heat} \label{5.2.2} \] to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called . In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings. A process in which heat ( ) is transferred a system its surroundings is described as exothermic. By convention, \(q < 0\) for an exothermic reaction. When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation: \[ \ce{heat + H_2O(s) \rightarrow H_2O(l)} \label{5.2.3} \] When heat is transferred a system its surroundings, the process is endothermic. By convention, \(q > 0\) for an endothermic reaction. Technically, it is poor form to have a \(heat\) term in the chemical reaction like in Equations \(\ref{5.2.2}\) and \(\ref{5.2.3}\) since is it not a true species in the reaction. However, this is a convenient approach to represent exothermic and endothermic behavior and is commonly used by chemists. The relationship between the energy change of a system and that of its surroundings is given by the , which states that the energy of the universe is constant. We can express this law mathematically as follows: \[U_{univ}=ΔU_{sys}+ΔU_{surr}=0 \label{5.2.4a} \] \[\Delta{U_{sys}}=−ΔU_{surr} \label{5.2.4b} \] where the subscripts univ, sys, and surr refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings. The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy. An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the \(CO_2/H_2O\) product mixture is less than that of the isooctane/ \(O_2\) reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred (q) and the work done (w): \[ΔU_{sys} = q + w \label{5.2.5} \] Although \(q\) and \(w\) are not state functions on their own, their sum (\(ΔU_{sys}\)) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed. Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings. Although \(q\) and \(w\) are not state functions, their sum (\(ΔU_{sys}\)) is independent of the path taken and therefore is a state function. Thus, because of the first law, we can determine \(ΔU\) for any process if we can measure both \(q\) and \(w\). Heat, \(q\), may be calculated by measuring a change in temperature of the surroundings. Work, \(w\), may come in different forms, but it too can be measured. One important form of work for chemistry is done by an expanding gas. At a constant external pressure (for example, atmospheric pressure) \[w = −PΔV \label{5.2.6} \] The negative sign associated with \(PV\) work done indicates that the system loses energy when the volume increases. That is, an expanding gas does work on its surroundings, while a gas that is compressed has work done on it by the surroundings. A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (ΔU) of the gas in joules? : initial volume, final volume, external pressure, and quantity of energy transferred as heat Asked for: total change in internal energy A From Equation \(\ref{5.2.5}\), we know that ΔU = q + w. We are given the magnitude of q (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, q is negative by convention. B Because the gas is being compressed, we know that work is being done on the system, so \(w\) must be positive. From Equation \(\ref{5.2.5}\), \[w=-P_{\textrm{ext}}\Delta V=-8.00\textrm{ atm}(\textrm{0.0500 L} - \textrm{0.400 L})\left(\dfrac{\textrm{101.3 J}}{\mathrm{L\cdot atm}} \right)=284\textrm{ J} \nonumber \] Thus \[\begin{align*} ΔU &= q + w \\[4pt] &= −140 \,J + 284\, J \\[4pt] &= 144\, J\end{align*} \nonumber \] In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs. A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (ΔU) of the gas in joules? −216 J By convention (to chemists), both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa. In chemistry, the small part of the universe that we are studying is the , and the rest of the universe is the . can exchange both matter and energy with their surroundings, can exchange energy but not matter with their surroundings, and can exchange neither matter nor energy with their surroundings. A is a property of a system that depends on only its present , not its history. A reaction or process in which heat is transferred from a system to its surroundings is . A reaction or process in which heat is transferred to a system from its surroundings is . The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. The heat flow is equal to the change in the internal energy of the system plus the work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔU. | 11,805 | 3,046 |
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The state exhibited by a given sample of matter depends on the identity, temperature, and pressure of the sample. A phase diagram is a graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system. A typical phase diagram consists of discrete regions that represent the different phases exhibited by a substance (Figure \(\Page {1}\)). Each region corresponds to the range of combinations of temperature and pressure over which that phase is stable. The combination of high pressure and low temperature (upper left of Figure \(\Page {1}\)) corresponds to the solid phase, whereas the gas phase is favored at high temperature and low pressure (lower right). The combination of high temperature and high pressure (upper right) corresponds to a supercritical fluid. The solid phase is favored at low temperature and high pressure; the gas phase is favored at high temperature and low pressure. The lines in a phase diagram correspond to the combinations of temperature and pressure at which two phases can coexist in equilibrium. In Figure \(\Page {1}\), the line that connects points A and D separates the solid and liquid phases and shows how the melting point of a solid varies with pressure. The solid and liquid phases are in equilibrium all along this line; crossing the line horizontally corresponds to melting or freezing. The line that connects points A and B is the vapor pressure curve of the liquid, which we discussed in . It ends at the critical point, beyond which the substance exists as a supercritical fluid. The line that connects points A and C is the vapor pressure curve of the phase. Along this line, the solid is in equilibrium with the vapor phase through sublimation and deposition. Finally, point A, where the solid/liquid, liquid/gas, and solid/gas lines intersect, is the triple point; it is the combination of temperature and pressure at which all three phases (solid, liquid, and gas) are in equilibrium and can therefore exist simultaneously. Because no more than three phases can ever coexist, a phase diagram can never have more than three lines intersecting at a single point. Remember that a phase diagram, such as the one in Figure \(\Page {1}\), is for a single pure substance in a closed system, not for a liquid in an open beaker in contact with air at 1 atm pressure. In practice, however, the conclusions reached about the behavior of a substance in a closed system can usually be extrapolated to an open system without a great deal of error. Figure \(\Page {2}\) shows the phase diagram of water and illustrates that the triple point of water occurs at 0.01°C and 0.00604 atm (4.59 mmHg). Far more reproducible than the melting point of ice, which depends on the amount of dissolved air and the atmospheric pressure, the triple point (273.16 K) is used to define the absolute (Kelvin) temperature scale. The triple point also represents the lowest pressure at which a liquid phase can exist in equilibrium with the solid or vapor. At pressures less than 0.00604 atm, therefore, ice does not melt to a liquid as the temperature increases; the solid sublimes directly to water vapor. Sublimation of water at low temperature and pressure can be used to “freeze-dry” foods and beverages. The food or beverage is first cooled to subzero temperatures and placed in a container in which the pressure is maintained below 0.00604 atm. Then, as the temperature is increased, the water sublimes, leaving the dehydrated food (such as that used by backpackers or astronauts) or the powdered beverage (as with freeze-dried coffee). The phase diagram for water illustrated in Figure \(\Page {2b}\) shows the boundary between ice and water on an expanded scale. The melting curve of ice slopes up and slightly to the left rather than up and to the right as in Figure \(\Page {1}\); that is, the melting point of ice with increasing pressure; at 100 MPa (987 atm), ice melts at −9°C. Water behaves this way because it is one of the few known substances for which the crystalline solid is than the liquid (others include antimony and bismuth). Increasing the pressure of ice that is in equilibrium with water at 0°C and 1 atm tends to push some of the molecules closer together, thus decreasing the volume of the sample. The decrease in volume (and corresponding increase in density) is smaller for a solid or a liquid than for a gas, but it is sufficient to melt some of the ice. In Figure \(\Page {2b}\) point A is located at = 1 atm and = −1.0°C, within the solid (ice) region of the phase diagram. As the pressure increases to 150 atm while the temperature remains the same, the line from point A crosses the ice/water boundary to point B, which lies in the liquid water region. Consequently, applying a pressure of 150 atm will melt ice at −1.0°C. We have already indicated that the pressure dependence of the melting point of water is of vital importance. If the solid/liquid boundary in the phase diagram of water were to slant up and to the right rather than to the left, ice would be denser than water, ice cubes would sink, water pipes would not burst when they freeze, and antifreeze would be unnecessary in automobile engines. Until recently, many textbooks described ice skating as being possible because the pressure generated by the skater’s blade is high enough to melt the ice under the blade, thereby creating a lubricating layer of liquid water that enables the blade to slide across the ice. Although this explanation is intuitively satisfying, it is incorrect, as we can show by a simple calculation. Recall that pressure ( ) is the force ( ) applied per unit area ( ): \[P=\dfrac{F}{A} \nonumber \] To calculate the pressure an ice skater exerts on the ice, we need to calculate only the force exerted and the area of the skate blade. If we assume a 75.0 kg (165 lb) skater, then the force exerted by the skater on the ice due to gravity is \[ F = mg \nonumber \] where is the mass and is the acceleration due to Earth’s gravity (9.81 m/s ). Thus the force is \[F = (75.0\; kg)(9.81\; m/s^2) = 736\; (kg•m)/s^2 = 736 N \nonumber \] If we assume that the skate blades are 2.0 mm wide and 25 cm long, then the area of the bottom of each blade is \[ A = (2.0 \times 10^{−3}\; m)(25 \times 10^{−2}\; m) = 5.0 \times 10^{−4} m^2 \nonumber \] If the skater is gliding on one foot, the pressure exerted on the ice is \[ P= \dfrac{736\;N}{5.0 \times 10^{-4} \; m^2} = 1.5 \times 10^6 \; N/m^2 = 1.5 \times 10^6\; Pa =15 \; atm \nonumber \] The pressure is much lower than the pressure needed to decrease the melting point of ice by even 1°C, and experience indicates that it is possible to skate even when the temperature is well below freezing. Thus pressure-induced melting of the ice cannot explain the low friction that enables skaters (and hockey pucks) to glide. Recent research indicates that the surface of ice, where the ordered array of water molecules meets the air, consists of one or more layers of almost liquid water. These layers, together with melting induced by friction as a skater pushes forward, appear to account for both the ease with which a skater glides and the fact that skating becomes more difficult below about −7°C, when the number of lubricating surface water layers decreases. Referring to the phase diagram of water in Figure \(\Page {2}\): phase diagram, temperature, and pressure physical form and physical changes Referring to the phase diagram of water in Figure \(\Page {2}\), predict the physical form of a sample of water at −0.0050°C as the pressure is gradually increased from 1.0 mmHg to 218 atm. The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid. In contrast to the phase diagram of water, the phase diagram of CO (Figure \(\Page {3}\)) has a more typical melting curve, sloping up and to the right. The triple point is −56.6°C and 5.11 atm, which means that liquid CO cannot exist at pressures lower than 5.11 atm. At 1 atm, therefore, solid CO sublimes directly to the vapor while maintaining a temperature of −78.5°C, the normal sublimation temperature. Solid CO is generally known as dry ice because it is a cold solid with no liquid phase observed when it is warmed. Also notice the critical point at 30.98°C and 72.79 atm. Supercritical carbon dioxide is emerging as a natural refrigerant, making it a low carbon (and thus a more environmentally friendly) solution for domestic heat pumps. As the phase diagrams above demonstrate, 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 Figure \(\Page {1}\). High-boiling-point, nonvolatile liquids have high critical temperatures and vice versa. A Discussing Phase Diagrams. | 10,092 | 3,048 |
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The concept of valence implies that atoms of each element have a characteristic number of sites by which they can be connected to atoms of other elements. Carbon, as seen below, has 4 'connection' or valence sites. Here Carbon is connected to 4 Hydrogen atoms, forming CH , better known as methane. Note that the symbolic and 3D representations below both depict the same molecule, but the symbolic representation clearly shows the valence sites, while the 3D representation accurately shows how those valence sites are arranged in 3D space. The number of valence sites repeats periodically as atomic weight increases, and occasionally even this regular repetition is imperfect. Atoms of similar atomic weight often have quite different properties, while some which differ widely in relative mass behave almost the same. Dalton’s atomic theory considers atoms to be indestructible spheres whose most important property is mass. This is clearly inadequate to account for the macroscopic observations of the elements. In order to continue using the atomic theory, we must attribute some underlying structure to atoms. If both valence and atomic weight are determined by that structure, we should be able to account for the close but imperfect relationship between these two properties. The next section will describe some of the experiments which led to current theories about just what this atomic structure is like. | 1,427 | 3,049 |
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Oxides are chemical compounds with one or more atoms combined with another element (e.g. Li O). Oxides are binary compounds of oxygen with another element, e.g., CO , SO , CaO, CO, ZnO, BaO , H O, etc. These are termed as oxides because here, oxygen is in combination with only one element. Based on their acid-base characteristics oxides are classified as acidic, basic, or neutral: There are different properties which help distinguish between the three types of oxides. The term ("without water") refers to compounds that assimilate H O to form either an acid or a base upon the addition of water. Acidic oxides are the oxides of non-metals ( ) and these form acids with water: \[\ce{SO_2 + H_2O \rightarrow H_2SO_3} \label{1}\] \[\ce{ SO_3 + H_2O \rightarrow H_2SO_4} \label{2}\] \[ \ce{CO_2 + H_2O \rightarrow H_2CO_3} \label{3}\] Acidic oxides are known as acid anhydrides (e.g., sulfur dioxide is sulfurous anhydride and sulfur trioxide is sulfuric anhydride) and when combined with bases, they produce salts, e.g., \[\ce{ SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O} \label{4}\] Generally and elements form bases called base anhydrides or basic oxides e.g., \[ \ce{K_2O \; (s) + H_2O \; (l) \rightarrow 2KOH \; (aq) } \label{5}\] Basic oxides are the oxides of metals. If soluble in water, they react with water to produce hydroxides (alkalies) e.g., \[ \ce{ CaO + H_2O \rightarrow Ca(OH)_2} \label{6}\] \[ \ce{ MgO + H_2O \rightarrow Mg(OH)_2} \label{7}\] \[ \ce{ Na_2O + H_2O \rightarrow 2NaOH } \label{8}\] These metallic oxides are known as basic anhydrides. They react with acids to produce salts, e.g., \[ \ce{ MgO + 2HCl \rightarrow MgCl_2 + H_2O } \label{9}\] \[ \ce{ Na_2O + H_2SO_4 \rightarrow Na_2SO_4 + H_2O} \label{10}\] An amphoteric solution is a substance that can chemically react as either acid or base. For example, when HSO reacts with water it will make both hydroxide and ions: \[ HSO_4^- + H_2O \rightarrow SO_4^{2^-} + H_3O^+ \label{11}\] \[ HSO_4^- + H_2O \rightarrow H_2SO_4 + OH^- \label{12}\] Amphoteric oxides exhibit both basic as well as acidic properties. When they react with an acid, they produce salt and water, showing basic properties. While reacting with alkalies they form salt and water showing acidic properties. \[ZnO + 2HCl \rightarrow \underset{\large{zinc\:chloride}}{ZnCl_2}+H_2O\,(basic\: nature) \label{13}\] \[ZnO + 2NaOH \rightarrow \underset{\large{sodium\:zincate}}{Na_2ZnO_2}+H_2O\,(acidic\: nature) \label{14}\] \[Al_2O_3 + 3H_2SO_4 \rightarrow Al_2(SO_4)_3+3H_2O\,(basic\: nature) \label{15}\] \[Al_2O_3 + 2NaOH \rightarrow 2NaAlO_2+H_2O\,(acidic\: nature) \label{16}\] Neutral oxides show neither basic nor acidic properties and hence do not form salts when reacted with acids or bases, e.g., carbon monoxide (CO); nitrous oxide (N O); nitric oxide (NO), etc., are neutral oxides. : metals react rapidly with oxygen to produce several different ionic oxides, usually in the form of \( M_2O \). With the oyxgen exhibiting an of -2. \[ 4 Li + O_2 \rightarrow 2Li_2O \label{19} \] : Often and reacts with oxygen to produce the peroxide, \( M_2O_2 \). with the oxidation number of the oxygen equal to -1. \[ H_2 + O_2 \rightarrow H_2O_2 \label{20}\] : Often , Rubidium, and Cesium react with excess oxygen to produce the superoxide, \( MO_2 \). with the oxidation number of the oxygen equal to -1/2. \[ Cs + O_2 \rightarrow CsO_2 \label{21}\] A peroxide is a metallic oxide which gives hydrogen peroxide by the action of dilute acids. They contain more oxygen than the corresponding basic oxide, e.g., sodium, calcium and barium peroxides. \[BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 \label{22}\] \[Na_2O_2 + H_2SO_4 \rightarrow Na_2SO_4 + H_2O_2 \label{23}\] Dioxides like PbO and MnO also contain higher percentage of oxygen like peroxides and have similar molecular formulae. These oxides, however, do not give hydrogen peroxide by action with dilute acids. Dioxides on reaction with concentrated HCl yield Cl and on reacting with concentrated H SO yield O . \[PbO_2 + 4HCl \rightarrow PbCl_2 + Cl_2 + 2H_2O \label{24}\] \[2PbO_2 + 2H_2SO_4 \rightarrow 2PbSO_4 + 2H_2O + O_2 \label{25}\] Compound oxides are metallic oxides that behave as if they are made up of two oxides, one that has a lower oxidation and one with a higher oxidation of the same metal, e.g., \[\textrm{Red lead: } Pb_3O_4 = PbO_2 + 2PbO \label{26}\] \[\textrm{Ferro-ferric oxide: } Fe_3O_4 = Fe_2O_3 + FeO \label{27}\] On treatment with an acid, compound oxides give a mixture of salts. \[\underset{\text{Ferro-ferric oxide}}{Fe_3O_4} + 8HCl \rightarrow \underset{\text{ferric chloride}}{2FeCl_3} + \underset{\text{ferrous chloride}}{FeCl_2} + 4H_2O \label{28}\] Oxides can be generated via multiple reactions. Below are a few. Many metals and non-metals burn rapidly when heated in oxygen or air, producing their oxides, e.g., \[2Mg + O_2 \xrightarrow{Heat} 2MgO\] \[2Ca + O_2 \xrightarrow{Heat} 2CaO\] \[S + O_2 \xrightarrow{Heat} SO_2\] \[P_4 + 5O_2 \xrightarrow{Heat} 2P_2O_5\] At higher temperatures, oxygen also reacts with many compounds forming oxides, e.g., \[2PbS + 3O_2 \xrightarrow{\Delta} 2PbO + 2SO_2\] \[2ZnS + 3O_2 \xrightarrow{\Delta} 2ZnO + 2SO_2\] \[C_2H_5OH + 3O_2 \rightarrow 2CO_2 + 3H_2O\] \[CaCO_3 \xrightarrow{\Delta} CaO + CO_2\] \[2Cu(NO_3)_2 \xrightarrow{\Delta} 2CuO + 4NO_2 + O_2\] \[Cu(OH)_2 \xrightarrow{\Delta} CuO + H_2O\] \[2Cu + 8HNO_3 \xrightarrow{Heat} 2CuO + 8NO_2 + 4H_2O + O_2\] \[Sn + 4HNO_3 \xrightarrow{Heat} SnO_2 + 4NO_2 + 2H_2O\] \[C + 4HNO_3 \rightarrow CO_2 + 4NO_2 + 2H_2O\] The oxides of elements in a period become progressively more acidic as one goes from left to right in a period of the periodic table. For example, in third period, the behavior of oxides changes as follows: \(\underset{\large{Basic}}{\underbrace{Na_2O,\: MgO}}\hspace{20px}
\underset{\large{Amphoteric}}{\underbrace{Al_2O_3,\: SiO_2}}\hspace{20px}
\underset{\large{Acidic}}{\underbrace{P_4O_{10},\: SO_3,\:Cl_2O_7}}\hspace{20px}\) If we take a closer look at a specific period, we may better understand the acid-base properties of oxides. It may also help to examine the , but it is not necessary. Metal oxides on the left side of the periodic table produce basic solutions in water (e.g. Na O and MgO). Non-metal oxides on the right side of the periodic table produce acidic solutions (e.g. Cl O, SO , P O ). There is a trend within acid-base behavior: basic oxides are present on the left side of the period and acidic oxides are found on the right side. Aluminum oxide shows acid and basic properties of an oxide, it is amphoteric. Thus Al O entails the marking point at which a change over from a basic oxide to acidic oxide occurs. It is important to remember that the trend only applies for oxides in their highest oxidation states. The individual element must be in its highest possible oxidation state because the trend does not follow if all oxidation states are included. Notice how the amphoteric oxides (shown in blue) of each period signify the change from basic to acidic oxides, The figure above show oxides of the and elements.
Water as such is a neutral stable molecule. It is difficult to break the covalent O-H bonds easily. Hence, electrical energy through the electrolysis process is applied to separate dioxygen from water. When a small amount of acid is added to water ionization is initiated which helps in electrochemical reactions as follows. \[[H_2O\:(acidulated)\rightleftharpoons H^+\,(aq)+OH]^-\times4\] At cathode: \[[H^+\,(aq)+e^-\rightarrow\dfrac{1}{2}H_2(g)]\times4\] At anode: \[4OH^-\,(aq)\rightarrow O_2+2H_2O + 4e^-\] Net reaction: \[2H_2O \xrightarrow{\large{electrolysis}} 2H_2\,(g) + O_2\,(g)\] Oxygen can thus be obtained from acidified water by its electrolysis. | 7,818 | 3,051 |
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The coinage metals—copper, silver, and gold—occur naturally (like the gold nugget shown here); consequently, these were probably the first metals used by ancient humans. For example, decorative gold artifacts dating from the late Stone Age are known, and some gold Egyptian coins are more than 5000 yr old. Copper is almost as ancient, with objects dating to about 5000 BC. Bronze, an alloy of copper and tin that is harder than either of its constituent metals, was used before 3000 BC, giving rise to the Bronze Age. Deposits of silver are much less common than deposits of gold or copper, yet by 3000 BC, methods had been developed for recovering silver from its ores, which allowed silver coins to be widely used in ancient times. Deposits of gold and copper are widespread and numerous, and for many centuries it was relatively easy to obtain large amounts of the pure elements. For example, a single gold nugget discovered in Australia in 1869 weighed more than 150 lb. Because the demand for these elements has outstripped their availability, methods have been developed to recover them economically from even very low-grade ores (as low as 1% Cu content for copper) by operating on a vast scale, as shown in the photo of an open-pit copper mine. Copper is used primarily to manufacture electric wires, but large quantities are also used to produce bronze, brass, and alloys for coins. Much of the silver made today is obtained as a by-product of the manufacture of other metals, especially Cu, Pb, and Zn. In addition to its use in jewelry and silverware, silver is used in Ag/Zn and Ag/Cd button batteries. Gold is typically found either as tiny particles of the pure metal or as gold telluride (AuTe ). It is used as a currency reserve, in jewelry, in the electronics industry for corrosion-free contacts, and, in very thin layers, as a reflective window coating that minimizes heat transfer. Some properties of the coinage metals are listed in Table \(\Page {1}\). The electronegativity of gold (χ = 2.40) is close to that of the nonmetals sulfur and iodine, which suggests that the chemistry of gold should be somewhat unusual for a metal. The coinage metals have the highest electrical and thermal conductivities of all the metals, and they are also the most ductile and malleable. With an ns (n − 1)d valence electron configuration, the chemistry of these three elements is dominated by the +1 oxidation state due to losing the single ns electron. Higher oxidation states are also known, however: +2 is common for Cu and, to a lesser extent, Ag, and +3 for Au because of the relatively low values of the second and (for Au) third ionization energies. All three elements have significant electron affinities due to the half-filled ns orbital in the neutral atoms. As a result, gold reacts with powerful reductants like Cs and solutions of the alkali metals in liquid ammonia to produce the gold anion Au with a 6s 5d valence electron configuration. | 2,982 | 3,052 |
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We have seen that the strengths of acids and bases vary over many orders of magnitude. In this section, we explore some of the structural and electronic factors that control the acidity or basicity of a molecule. In general, the stronger the \(\ce{A–H}\) or \(\ce{B–H^+}\) bond, the less likely the bond is to break to form \(H^+\) ions and thus the less acidic the substance. This effect can be illustrated using the hydrogen halides: The trend in bond energies is due to a steady decrease in overlap between the 1s orbital of hydrogen and the valence orbital of the halogen atom as the size of the halogen increases. The larger the atom to which H is bonded, the weaker the bond. Thus the bond between H and a large atom in a given family, such as I or Te, is weaker than the bond between H and a smaller atom in the same family, such as F or O. As a result, acid strengths of binary hydrides increase as we go down a column of the periodic table. For example, the order of acidity for the binary hydrides of elements is as follows, with \(pK_a\) values in parentheses: \[H_2O (14.00 = pK_w) < H_2S (7.05) < H_2Se (3.89) < H_2Te (2.6) \label{1}\] Whether we write an acid–base reaction as \(AH \rightleftharpoons A^−+H^+\) or as \(BH^+ \rightleftharpoons B+H^+\), the conjugate base (\(A^−\) or \(B\)) contains one more lone pair of electrons than the parent acid (\(AH\) or \(BH^+\)). Any factor that stabilizes the lone pair on the conjugate base favors dissociation of \(H^+\) and makes the parent acid a stronger acid. Let’s see how this explains the relative acidity of the binary hydrides of the elements in the second row of the periodic table. The observed order of increasing acidity is the following, with pKa values in parentheses: \[CH_4 (~50) \ll NH_3 (~36) < H_2O (14.00) < HF (3.20) \label{2}\] Consider, for example, the compounds at both ends of this series: methane and hydrogen fluoride. The conjugate base of \(CH_4\) is \(CH_3^−\), and the conjugate base of \(HF\) is \(F^−\). Because fluorine is much more electronegative than carbon, fluorine can better stabilize the negative charge in the \(F^−\) ion than carbon can stabilize the negative charge in the CH3− ion. Consequently, \(\ce{HF}\) has a greater tendency to dissociate to form \(H^+\) and \(F^−\) than does methane to form \(H^+\) and \(CH_3^−\), making HF a much stronger acid than \(CH_4\). The same trend is predicted by analyzing the properties of the conjugate acids. For a series of compounds of the general formula \(HE\), as the electronegativity of E increases, the E–H bond becomes more polar, favoring dissociation to form \(E^−\) and \(H^+\). Due to both the increasing stability of the conjugate base and the increasing polarization of the E–H bond in the conjugate acid, acid strengths of binary hydrides increase as we go from left to right across a row of the periodic table. Acid strengths of binary hydrides increase as we go down a column or from left to right across a row of the periodic table. The stornger acid, the weaker the covalent bond to a hydrogen atom. So the strongest acid possible is the molecule with the weakest bond. That is the hydrohelium (1+) cation, \(\ce{HeH^{+}}\), which is a positively charged ion formed by the reaction of a proton with a helium atom in the gas phase. It was first produced in the laboratory in 1925 and is isoelectronic with molecular hydrogen (\ce{H2}}). It is the strongest known acid, with a proton affinity of 177.8 kJ/mol. \(\ce{HeH^{+}}\) cannot be prepared in a condensed phase, as it would protonate any anion, molecule or atom with which it were associated. However it is possible to estimate a aqueous acidity using : A free energy change of dissociation of −360 kJ/mol is equivalent to a p of −63. It has been suggested that \(\ce{HeH^{+}}\) should occur naturally in the interstellar medium, but it has not yet been detected. Atoms or groups of atoms in a molecule other than those to which H is bonded can induce a change in the distribution of electrons within the molecule. This is called an inductive effect, and, much like the coordination of water to a metal ion, it can have a major effect on the acidity or basicity of the molecule. For example, the hypohalous acids (general formula HOX, with X representing a halogen) all have a hydrogen atom bonded to an oxygen atom. In aqueous solution, they all produce the following equilibrium: \[ HOX_{(aq)} \rightleftharpoons H^+_{(aq)} + OX^−{(aq)} \label{3}\] The acidities of these acids vary by about three orders of magnitude, however, due to the difference in electronegativity of the halogen atoms: As the electronegativity of \(X\) increases, the distribution of electron density within the molecule changes: the electrons are drawn more strongly toward the halogen atom and, in turn, away from the H in the O–H bond, thus weakening the O–H bond and allowing dissociation of hydrogen as \(H^+\). The acidity of oxoacids, with the general formula \(HOXO_n\) (with \(n\) = 0−3), depends strongly on the number of terminal oxygen atoms attached to the central atom \(X\). As shown in Figure \(\Page {1}\), the \(K_a\) values of the oxoacids of chlorine increase by a factor of about \(10^4\) to \(10^6\) with each oxygen as successive oxygen atoms are added. The increase in acid strength with increasing number of terminal oxygen atoms is due to both an inductive effect and increased stabilization of the conjugate base. Any inductive effect that withdraws electron density from an O–H bond increases the acidity of the compound. Because oxygen is the second most electronegative element, adding terminal oxygen atoms causes electrons to be drawn away from the O–H bond, making it weaker and thereby increasing the strength of the acid. The colors in Figure \(\Page {1}\) show how the electrostatic potential, a measure of the strength of the interaction of a point charge at any place on the surface of the molecule, changes as the number of terminal oxygen atoms increases. In Figure \(\Page {1}\) and Figure \(\Page {2}\), blue corresponds to low electron densities, while red corresponds to high electron densities. The oxygen atom in the O–H unit becomes steadily less red from \(HClO\) to \(HClO_4\) (also written as \(HOClO_3\), while the H atom becomes steadily bluer, indicating that the electron density on the O–H unit decreases as the number of terminal oxygen atoms increases. The decrease in electron density in the O–H bond weakens it, making it easier to lose hydrogen as \(H^+\) ions, thereby increasing the strength of the acid. At least as important, however, is the effect of delocalization of the negative charge in the conjugate base. As shown in Figure \(\Page {2}\), the number of resonance structures that can be written for the oxoanions of chlorine increases as the number of terminal oxygen atoms increases, allowing the single negative charge to be delocalized over successively more oxygen atoms. Electron delocalization in the conjugate base increases acid strength. The electrostatic potential plots in Figure \(\Page {2}\) demonstrate that the electron density on the terminal oxygen atoms decreases steadily as their number increases. The oxygen atom in ClO− is red, indicating that it is electron rich, and the color of oxygen progressively changes to green in \(ClO_4^+\), indicating that the oxygen atoms are becoming steadily less electron rich through the series. For example, in the perchlorate ion (\(ClO_4^−\)), the single negative charge is delocalized over all four oxygen atoms, whereas in the hypochlorite ion (\(OCl^−\)), the negative charge is largely localized on a single oxygen atom (Figure \(\Page {2}\)). As a result, the perchlorate ion has no localized negative charge to which a proton can bind. Consequently, the perchlorate anion has a much lower affinity for a proton than does the hypochlorite ion, and perchloric acid is one of the strongest acids known. As the number of terminal oxygen atoms increases, the number of resonance structures that can be written for the oxoanions of chlorine also increases, and the single negative charge is delocalized over more oxygen atoms. As these electrostatic potential plots demonstrate, the electron density on the terminal oxygen atoms decreases steadily as their number increases. As the electron density on the oxygen atoms decreases, so does their affinity for a proton, making the anion less basic. As a result, the parent oxoacid is more acidic. Similar inductive effects are also responsible for the trend in the acidities of oxoacids that have the same number of oxygen atoms as we go across a row of the periodic table from left to right. For example, \(H_3PO_4\) is a weak acid, \(H_2SO_4\) is a strong acid, and \(HClO_4\) is one of the strongest acids known. The number of terminal oxygen atoms increases steadily across the row, consistent with the observed increase in acidity. In addition, the electronegativity of the central atom increases steadily from P to S to \(Cl\), which causes electrons to be drawn from oxygen to the central atom, weakening the \(\ce{O–H}\) bond and increasing the strength of the oxoacid. Careful inspection of the data in Table \(\Page {1}\) shows two apparent anomalies: carbonic acid and phosphorous acid. If carbonic acid \((H_2CO_3\)) were a discrete molecule with the structure \(\ce{(HO)_2C=O}\), it would have a single terminal oxygen atom and should be comparable in acid strength to phosphoric acid (\(H_3PO_4\)), for which pKa1 = 2.16. Instead, the tabulated value of \(pK_{a1}\) for carbonic acid is 6.35, making it about 10,000 times weaker than expected. As we shall see, however, \(H_2CO_3\) is only a minor component of the aqueous solutions of \(CO_2\) that are referred to as carbonic acid. Similarly, if phosphorous acid (\(H_3PO_3\)) actually had the structure \((HO)_3P\), it would have no terminal oxygen atoms attached to phosphorous. It would therefore be expected to be about as strong an acid as \(HOCl\) (pKa = 7.40). In fact, the \(pK_{a1}\) for phosphorous acid is 1.30, and the structure of phosphorous acid is \(\ce{(HO)_2P(=O)H}\) with one H atom directly bonded to P and one \(\ce{P=O}\) bond. Thus the pKa1 for phosphorous acid is similar to that of other oxoacids with one terminal oxygen atom, such as \(H_3PO_4\). Fortunately, phosphorous acid is the only common oxoacid in which a hydrogen atom is bonded to the central atom rather than oxygen. Inductive effects are also observed in organic molecules that contain electronegative substituents. The magnitude of the electron-withdrawing effect depends on both the nature and the number of halogen substituents, as shown by the pKa values for several acetic acid derivatives: \[pK_a CH_3CO_2H 4.76< CH_2ClCO_2H 2.87<CHCl_2CO_2H 1.35<CCl_3CO_2H 0.66<CF_3CO_2H 0.52 \nonumber\] As you might expect, fluorine, which is more electronegative than chlorine, causes a larger effect than chlorine, and the effect of three halogens is greater than the effect of two or one. Notice from these data that inductive effects can be quite large. For instance, replacing the \(\ce{–CH_3}\) group of acetic acid by a \(\ce{–CF_3}\) group results in about a 10,000-fold increase in acidity! Arrange the compounds of each series in order of increasing acid or base strength. The structures are shown here. : series of compounds : relative acid or base strengths : Use relative bond strengths, the stability of the conjugate base, and inductive effects to arrange the compounds in order of increasing tendency to ionize in aqueous solution. : Although both sulfuric acid and sulfurous acid have two –OH groups, the sulfur atom in sulfuric acid is bonded to two terminal oxygen atoms versus one in sulfurous acid. Because oxygen is highly electronegative, sulfuric acid is the stronger acid because the negative charge on the anion is stabilized by the additional oxygen atom. In comparing sulfuric acid and fluorosulfonic acid, we note that fluorine is more electronegative than oxygen. Thus replacing an –OH by –F will remove more electron density from the central S atom, which will, in turn, remove electron density from the S–OH bond and the O–H bond. Because its O–H bond is weaker, \(FSO_3H\) is a stronger acid than sulfuric acid. The predicted order of acid strengths given here is confirmed by the measured pKa values for these acids: \[pKa H_2SO_3 1.85<H_2SO_4^{−2} < FSO_3H−10 \nonumber\] The structures of both trifluoramine and hydroxylamine are similar to that of ammonia. In trifluoramine, all of the hydrogen atoms in NH3 are replaced by fluorine atoms, whereas in hydroxylamine, one hydrogen atom is replaced by OH. Replacing the three hydrogen atoms by fluorine will withdraw electron density from N, making the lone electron pair on N less available to bond to an \(H^+\) ion. Thus \(NF_3\) is predicted to be a much weaker base than \(NH_3\). Similarly, because oxygen is more electronegative than hydrogen, replacing one hydrogen atom in \(NH_3\) by \(OH\) will make the amine less basic. Because oxygen is less electronegative than fluorine and only one hydrogen atom is replaced, however, the effect will be smaller. The predicted order of increasing base strength shown here is confirmed by the measured \(pK_b\) values: \[pK_bNF_3—<<NH_2OH 8.06<NH_3 4.75 \nonumber\] Trifluoramine is such a weak base that it does not react with aqueous solutions of strong acids. Hence its base ionization constant has never been measured. Arrange the compounds of each series in order of \(HClO-3 > CH_3PO_3H_2 > H_3PO_4\) \(CF_3S^− < CH_3S^− < OH^−\) Inductive effects and charge delocalization significantly influence the acidity or basicity of a compound. The acid–base strength of a molecule depends strongly on its structure. The weaker the A–H or B–H+ bond, the more likely it is to dissociate to form an \(H^+\) ion. In addition, any factor that stabilizes the lone pair on the conjugate base favors the dissociation of \(H^+\), making the conjugate acid a stronger acid. Atoms or groups of atoms elsewhere in a molecule can also be important in determining acid or base strength through an inductive effect, which can weaken an \(\ce{O–H}\) bond and allow hydrogen to be more easily lost as \(H^+\) ions. | 14,283 | 3,053 |
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We often find that there is a limit to the quantity of solute which will dissolve in a given quantity of solvent. This is especially true when solids dissolve in liquids. For example, if 36 g KCl crystals is shaken with 100 g H O at 25°C only 35.5 g of the solid dissolves. If we raise the temperature somewhat, all the KCl will dissolve, but on cooling to 25°C again, the extra 0.5 g KC1 will precipitate, leaving exactly 35.5 g of the salt dissolved. We describe this phenomenon by saying that at 25°C the of KCl in H O is 35.5 g KC1 per 100 g H O. A solution of this composition is also described as a solution since it can accommodate no more KCl. Under some circumstances it is possible to prepare a solution which behaves anomalously and contains more solute than a saturated solution. Such a solution is said to be . A good example of supersaturation is provided by Na S O , sodium thiosulfate, whose solubility at 25°C is 50 g Na S O per 100 g H O. If 70 g Na S O crystals is dissolved in 100 g hot H O and the solution cooled to room temperature, the extra 20 g Na S O usually does not precipitate. The resulting solution is supersaturated; consequently it is also unstable. It can be “seeded” by adding a crystal of Na S O , whereupon the excess salt suddenly crystallizes and heat is given off. After the crystals have settled and the temperature has returned to 25°C, the solution above the crystals is a saturated solution—it contains 50 g Na S O . Another example of crystallizing salt out of a supersaturated solution can be seen in the following video. In this case, a supersaturated solution of sodium acetate is poured over a crystal of sodium acetate. These crystals provide the lattice structure "seed" which causes the sodium acetate ions in solution to crystallize out. The salt begins to crystallize, forming a large sodium acetate structure from the precipitation of the ions out of solution. When the sodium acetate crystallizes, the oppositely charged ions are brought closer together by the crystal structure. Since formation of a lowers potential energy by placing like charges close together, the system releases the excess energy in the crystallization process. Thus, the structure ends up being warm to the touch from this excess energy. | 2,290 | 3,054 |
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After selecting a method, the next step is to develop a procedure that accomplish our goals for the analysis. In developing a procedure we give attention to compensating for interferences, to selecting and calibrating equipment, to acquiring a representative sample, and to validating the method. A method’s accuracy depends on its selectivity for the analyte. Even the best method, however, may not be free from interferents that contribute to the measured signal. Potential interferents may be present in the sample itself or in the reagents used during the analysis. When the sample is free of interferents, the total signal, , is a sum of the signal due to the analyte, , and the signal due to interferents in the reagents, , \[S_{total} = S_A + S_{reag} = k_A n_A + S_{reag} \label{3.1}\] \[S_{total} = S_A + S_{reag} = k_A C_A + S_{reag} \label{3.2}\] Without an independent determination of we cannot solve Equation \ref{3.1} or \ref{3.2} for the moles or concentration of analyte. To determine the contribution of in Equations \ref{3.1} and \ref{3.2} we measure the signal for a , a solution that does not contain the sample. Consider, for example, a procedure in which we dissolve a 0.1-g sample in a portion of solvent, add several reagents, and dilute to 100 mL with additional solvent. To prepare the method blank we omit the sample and dilute the reagents to 100 mL using the solvent. Because the analyte is absent, for the method blank is equal to . Knowing the value for makes it is easy to correct for the reagent’s contribution to the total signal; thus \[(S_{total} - S_{reag}) = S_A = k_A n_A \nonumber\] \[(S_{total} - S_{reag}) = S_A = k_A C_A \nonumber\] By itself, a method blank cannot compensate for an interferent that is part of the sample’s matrix. If we happen to know the interferent’s identity and concentration, then we can be add it to the method blank; however, this is not a common circumstance and we must, instead, find a method for separating the analyte and interferent before continuing the analysis. A method blank also is known as a reagent blank. When the sample is a liquid, or is in solution, we use an equivalent volume of an inert solvent as a substitute for the sample. A simple definition of a quantitative analytical method is that it is a mechanism for converting a measurement, the signal, into the amount of analyte in a sample. Assuming we can correct for interferents, a quantitative analysis is nothing more than solving or for or for . To solve these equations we need the value of . For a total analysis method usually we know the value of because it is defined by the stoichiometry of the chemical reactions responsible for the signal. For a concentration method, however, the value of usually is a complex function of experimental conditions. A is the process of experimentally determining the value of by measuring the signal for one or more standard samples, each of which contains a known concentration of analyte. With a single standard we can calculate the value of using or . When using several standards with different concentrations of analyte, the result is best viewed visually by plotting versus the concentration of analyte in the standards. Such a plot is known as a , an example of which is shown in Figure 3.5.1
. Selecting an appropriate method and executing it properly helps us ensure that our analysis is accurate. If we analyze the wrong sample, however, then the accuracy of our work is of little consequence. A proper sampling strategy ensures that our samples are representative of the material from which they are taken. Biased or nonrepresentative sampling, and contaminating samples during or after their collection are two examples of sampling errors that can lead to a significant error in accuracy. It is important to realize that sampling errors are independent of errors in the analytical method. As a result, we cannot correct a sampling error in the laboratory by, for example, evaluating a reagent blank. provides a more detailed discussion of sampling, including strategies for obtaining representative samples. If we are to have confidence in our procedure we must demonstrate that it can provide acceptable results, a process we call . Perhaps the most important part of validating a procedure is establishing that its precision and accuracy are appropriate for the problem we are trying to solve. We also ensure that the written procedure has sufficient detail so that different analysts or laboratories will obtain comparable results. Ideally, validation uses a standard sample whose composition closely matches the samples we will analyze. In the absence of appropriate standards, we can evaluate accuracy by comparing results to those obtained using a method of known accuracy. You will find more details about validating analytical methods in . | 4,889 | 3,055 |
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Stoichiometry is a section of chemistry that involves using relationships between reactants and/or products in a chemical reaction to determine desired quantitative data. In Greek, means element and means measure, so stoichiometry literally translated means the measure of elements. In order to use stoichiometry to run calculations about chemical reactions, it is important to first understand the relationships that exist between products and reactants and why they exist, which require understanding how to balance reactions. In chemistry, chemical reactions are frequently written as an equation, using chemical symbols. The reactants are displayed on the left side of the equation and the products are shown on the right, with the separation of either a single or double arrow that signifies the direction of the reaction. The significance of single and double arrow is important when discussing solubility constants, but we will not go into detail about it in this module. To balance an equation, it is necessary that there are the same number of atoms on the left side of the equation as the right. One can do this by raising the coefficients. A chemical equation is like a recipe for a reaction so it displays all the ingredients or terms of a chemical reaction. It includes the elements, molecules, or ions in the reactants and in the products as well as their states, and the proportion for how much of each particle reacts or is formed relative to one another, through the stoichiometric coefficient. The following equation demonstrates the typical format of a chemical equation: \[\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber\] In the above equation, the elements present in the reaction are represented by their chemical symbols. Based on the , which states that matter is neither created nor destroyed in a chemical reaction, every chemical reaction has the same elements in its reactants and products, though the elements they are paired up with often change in a reaction. In this reaction, sodium (\(Na\)), hydrogen (\(H\)), and chloride (\(Cl\)) are the elements present in both reactants, so based on the law of conservation of mass, they are also present on the product side of the equations. Displaying each element is important when using the chemical equation to convert between elements. In a balanced reaction, both sides of the equation have the same number of elements. The stoichiometric coefficient is the number written in front of atoms, ion and molecules in a chemical reaction to balance the number of each element on both the reactant and product sides of the equation. Though the stoichiometric coefficients can be fractions, whole numbers are frequently used and often preferred. This stoichiometric coefficients are useful since they establish the mole ratio between reactants and products. In the balanced equation: \[\ce{2 Na(s) + 2HCl(aq) \rightarrow 2NaCl(aq) + H2(g)} \nonumber\] we can determine that 2 moles of \(HCl\) will react with 2 moles of \(Na_{(s)}\) to form 2 moles of \(NaCl_{(aq)}\) and 1 mole of \(H_{2(g)}\). If we know how many moles of \(Na\) reacted, we can use the ratio of 2 moles of \(NaCl\) to 2 moles of Na to determine how many moles of \(NaCl\) were produced or we can use the ratio of 1 mole of \(H_2\) to 2 moles of \(Na\) to convert to \(NaCl\). This is known as the coefficient factor. The balanced equation makes it possible to convert information about the change in one reactant or product to quantitative data about another reactant or product. Understanding this is essential to solving stoichiometric problems. Lead (IV) hydroxide and sulfuric acid react as shown below. Balance the reaction. \[\ce{Pb(OH)4 + H2SO4 \rightarrow Pb(SO4)2 +H2O} \nonumber\] Start by counting the number of atoms of each element. Pb 1 1 O 8 9 H 6 2 S 1 2 The reaction is not balanced; the reaction has 16 reactant atoms and only 14 product atoms and does not obey the conservation of mass principle. Stoichiometric coefficients must be added to make the equation balanced. In this example, there are only one sulfur atom present on the reactant side, so a coefficient of 2 should be added in front of \(H_2SO_4\) to have an equal number of sulfur on both sides of the equation. Since there are 12 oxygen on the reactant side and only 9 on the product side, a 4 coefficient should be added in front of \(H_2O\) where there is a deficiency of oxygen. Count the number of elements now present on either side of the equation. Since the numbers are the same, the equation is now balanced. \[\ce{ Pb(OH)4 + 2 H2SO4 \rightarrow Pb(SO4)2 + 4H2O} \nonumber\] Pb 1 1 O 12 12 H 8 8 S 2 2 Balancing reactions involves finding least common multiples between numbers of elements present on both sides of the equation. In general, when applying coefficients, add coefficients to the molecules or unpaired elements last. A balanced equation ultimately has to satisfy two conditions. In stoichiometry, balanced equations make it possible to compare different elements through the discussed earlier. This is the mole ratio between two factors in a chemical reaction found through the ratio of stoichiometric coefficients. Here is a real world example to show how stoichiometric factors are useful. There are 12 party invitations and 20 stamps. Each party invitation needs 2 stamps to be sent. How many party invitations can be sent? The equation for this can be written as \[\ce{I + 2S \rightarrow IS2}\nonumber\] where Based on this, we have the ratio of 2 stamps for 1 sent invite, based on the balanced equation. Invitations Stamps Party Invitations Sent In this example are all the reactants (stamps and invitations) used up? No, and this is normally the case with chemical reactions. There is often excess of one of the reactants. The limiting reagent, the one that runs out first, prevents the reaction from continuing and determines the maximum amount of product that can be formed. What is the in this example? Stamps, because there was only enough to send out invitations, whereas there were enough invitations for 12 complete party invitations. Aside from just looking at the problem, the problem can be solved using factors. 12 I x ( /1I) = 12 IS possible 20 S x ( /2S) = 10 IS possible When there is no limiting reagent because the ratio of all the reactants caused them to run out at the same time, it is known as . There are 6 basic types of reactions. Before applying stoichiometric factors to chemical equations, you need to understand molar mass. Molar mass is a useful chemical ratio between mass and moles. The atomic mass of each individual element as listed in the periodic table established this relationship for atoms or ions. For compounds or molecules, you have to take the sum of the atomic mass times the number of each atom in order to determine the molar mass What is the molar mass of H O? \[\text{Molar mass} = 2 \times (1.00794\; g/mol) + 1 \times (15.9994\; g/mol) = 18.01528\; g/mol\] Using molar mass and coefficient factors, it is possible to convert mass of reactants to mass of products or vice versa. Propane (\(\ce{C_3H_8}\)) burns in this reaction: \[\ce{C_3H_8 + 5O_2 \rightarrow 4H_2O + 3CO_2} \nonumber\] If 200 g of propane is burned, how many g of \(H_2O\) is produced? Steps to getting this answer: Since you cannot calculate from grams of reactant to grams of products you must convert from grams of \(C_3H_8\) to moles of \(C_3H_8\) then from moles of \(C_3H_8\) to moles of \(H_2O\). Then convert from moles of \(H_2O\) to grams of \(H_2O\). Almost every quantitative relationship can be converted into a ratio that can be useful in data analysis. Density (\(\rho\)) is calculated as mass/volume. This ratio can be useful in determining the volume of a solution, given the mass or useful in finding the mass given the volume. In the latter case, the inverse relationship would be used. Volume x (Mass/Volume) = Mass Mass x (Volume/Mass) = Volume Percents establish a relationship as well. A percent mass states how many grams of a mixture are of a certain element or molecule. The percent X% states that of every 100 grams of a mixture, X grams are of the stated element or compound. This is useful in determining mass of a desired substance in a molecule. A substance is 5% carbon by mass. If the total mass of the substance is 10 grams, what is the mass of carbon in the sample? How many moles of carbon are there? 10 g sample x (5 g carbon/100 g sample) = 0.5 g carbon 0.5g carbon x (1 mol carbon/12.011g carbon) = 0.0416 mol carbon Molarity (moles/L) establishes a relationship between moles and liters. Given volume and molarity, it is possible to calculate mole or use moles and molarity to calculate volume. This is useful in chemical equations and dilutions. How much 5 M stock solution is needed to prepare 100 mL of 2 M solution? 100 mL of dilute solution (1 L/1000 mL)(2 mol/1L solution)(1 L stock solution/5 mol solution)(1000 ml stock solution/1L stock solution) = 40 mL stock solution. These ratios of molarity, density, and mass percent are useful in complex examples ahead. An empirical formula can be determined through chemical stoichiometry by determining which elements are present in the molecule and in what ratio. The ratio of elements is determined by comparing the number of moles of each element present. 1.000 gram of an organic molecule burns completely in the presence of excess oxygen. It yields 0.0333 mol of CO and 0.599 g of H O. What is the empirical formula of the organic molecule? This is a combustion reaction. The problem requires that you know that organic molecules consist of some combination of carbon, hydrogen, and oxygen elements. With that in mind, write the chemical equation out, replacing unknown numbers with variables. Do not worry about coefficients here. \[ \ce{C_xH_yO_z(g) + O_2(g) \rightarrow CO_2(g) + H_2O(g)} \nonumber\] Since all the moles of C and H in CO and H O, respectively have to have came from the 1 gram sample of unknown, start by calculating how many moles of each element were present in the unknown sample. 0.0333mol CO ( C/ CO ) = 0.0333mol C in unknown 0.599g H O ( H O/ 18.01528g H O)( H/ H O) = 0.0665 H in unknown Calculate the final moles of oxygen by taking the sum of the moles of oxygen in CO and H O. This will give you the number of moles from both the unknown organic molecule and the O so you must subtract the moles of oxygen transferred from the O . 0.0333mol CO ( O/ CO ) = 0.0666 O 0.599g H O ( H O/18.01528 g H O)( O/ H O) = 0.0332 O Using the Law of Conservation, we know that the mass before a reaction must equal the mass after a reaction. With this we can use the difference of the final mass of products and initial mass of the unknown organic molecule to determine the mass of the O reactant. 0.333mol CO (44.0098g CO / CO ) = 1.466g CO 1.466g CO + 0.599g H O - 1.000g unknown organic = 1.065g O 1.065g O ( O / 31.9988g O )( O/ O ) = 0.0666mol O (0.0666mol O + 0.0332 O) - 0.0666mol O = 0.0332 O Construct a mole ratio for C, H, and O in the unknown and divide by the smallest number. (1/0.0332)(0.0333mol C : 0.0665mol H : 0.0332 O) => C: 2 H: 1 O From this ratio, the empirical formula is calculated to be CH O. To determine a molecular formula, first determine the empirical formula for the compound as shown in the section above and then determine the molecular mass experimentally. Next, divide the molecular mass by the molar mass of the empirical formula (calculated by finding the sum the total atomic masses of all the elements in the empirical formula). Multiply the subscripts of the molecular formula by this answer to get the molecular formula. In the example above, it was determined that the unknown molecule had an empirical formula of CH O. 1. Find the molar mass of the empircal formula CH O. 12.011g C + (1.008 g H) * (2 H) + 15.999g O = 30.026 g/mol CH O 2. Determine the molecular mass experimentally. For our compound, it is 120.056 g/mol. 3. Divide the experimentally determined molecular mass by the mass of the empirical formula. (120.056 g/mol) / (30.026 g/mol) = 3.9984 4. Since 3.9984 is very close to four, it is possible to safely round up and assume that there was a slight error in the experimentally determined molecular mass. If the answer is not close to a whole number, there was either an error in the calculation of the empirical formula or a large error in the determination of the molecular mass. 5. Multiply the ratio from step 4 by the subscripts of the empirical formula to get the molecular formula. CH O * 4 = ? C: 1 * 4 = 4 H: 2 * 4 = 8 O 1 * 4 = 4 CH O * 4 = C H O 6. Check your result by calculating the molar mass of the molecular formula and comparing it to the experimentally determined mass. molar mass of C H O = 120.104 g/mol experimentally determined mass = 120.056 g/mol % error = | theoretical - experimental | / theoretical * 100% % error = | 120.104 g/mol - 120.056 g/mol | / 120.104 g/mol * 100% % error = 0.040 % An amateur welder melts down two metals to make an alloy that is 45% copper by mass and 55% iron(II) by mass. The alloy's density is 3.15 g/L. One liter of alloy completely fills a mold of volume 1000 cm . He accidentally breaks off a 1.203 cm piece of the homogenous mixture and sweeps it outside where it reacts with acid rain over years. Assuming the acid reacts with all the iron(II) and not with copper, how many grams of H (g) are released into the atmosphere because of the amateur's carelessness? (Note that the situation is fiction.) : Write a balanced equation after determining the products and reactants. In this situation, since we assume copper does not react, the reactants are only H (aq) and Fe(s). The given product is H2(g) and based on knowledge of reactions, the other product must be Fe (aq). \[\ce{Fe(s) + 2H^{+}(aq) \rightarrow H2(g) + Fe^{2+}(aq)} \nonumber\] Write down all the given information Alloy density = (3.15g alloy/ 1L alloy) x grams of alloy = 45% copper = ( Cu(s)/ alloy) x grams of alloy = 55% iron(II) = ( Fe(s)/ alloy) 1 liter alloy = alloy alloy sample = 1.203cm alloy Answer the question of what is being asked. The question asks how much H2(g) was produced. You are expected to solve for the amount of product formed. Start with the compound you know the most about and use given ratios to convert it to the desired compound. Convert the given amount of alloy reactant to solve for the moles of Fe(s) reacted. 1.203cm alloy( alloy/ alloy)(3.15g alloy/ alloy)( Fe(s)/ alloy)( Fe(s)/55.8g Fe(s))=3.74 x 10 Fe(s) The balanced equation must now be used to convert moles of Fe(s) to moles of H (g). Remember that the balanced equation's state the factor or mole ratio of reactants and products. 3.74 x 10 Fe (s) ( H (g)/ Fe(s)) = 3.74 x 10 H (g) Check units The question asks for how many of H (g) were released so the moles of H (g) must still be converted to grams using the molar mass of H (g). Since there are two H in each H , its molar mass is twice that of a single H atom. molar mass = 2(1.00794g/ ) = 2.01588g/ 3.74 x 10 H (g) (2.01588g H (g)/ H (g)) = Stoichiometry and balanced equations make it possible to use one piece of information to calculate another. There are countless ways stoichiometry can be used in chemistry and everyday life. Try and see if you can use what you learned to solve the following problems. 1) Why are the following equations not considered balanced? 2) Hydrochloric acid reacts with a solid chunk of aluminum to produce hydrogen gas and aluminum ions. Write the balanced chemical equation for this reaction. 3) Given a 10.1M stock solution, how many mL must be added to water to produce 200 mL of 5M solution? 4) If 0.502g of methane gas react with 0.27g of oxygen to produce carbon dioxide and water, what is the limiting reagent and how many moles of water are produced? The unbalanced equation is provided below. \[\ce{CH4(g) + O2(g) \rightarrow CO2(g) + H2O(l)} \nonumber\] 5) A 0.777g sample of an organic compound is burned completely. It produces 1.42g CO and 0.388g H O. Knowing that all the carbon and hydrogen atoms in CO and H O came from the 0.777g sample, what is the empirical formula of the organic compound? | 16,301 | 3,056 |
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As we have seen carbonyl compounds undergo both acid and base catalyzed reactions involving nucleophilic attack at the carbonyl carbon (or at the beta carbon of conjugated carbonyls). This reaction, in its many guises, can produce an impressive range of products from the formation of an ester from an acid to the production of alcohols from carbonyls. While these reactions may seem superficially different, if you understand the underlying mechanism involved, it is possible to make plausible predictions for the outcome for literally thousands of reactions. By this time, you should have come to understand such processes. Now, it is time to reconsider carbonyl groups in light of the fact that there is a completely different set of reactions that involve the reactivity of the alpha carbon of the carbonyl groups. Carbonyl compounds typically exist in two tautomeric forms: the keto and enol forms. The keto form is usually the major tautomer and there is always some enol present as well. The structure of the enol form can provide clues about its different reactivity, which is distinct from that of the keto form. The enol form consists of an alcohol directly attached to a \(\mathrm{C}\) that is involved in a double bond. As we know, alkenes are electron-rich and tend to undergo electrophilic attack; the presence of an attached \(\mathrm{-OH}\) group makes such an electrophilic attack even more likely. Just like an \(\mathrm{-OH}\) group on an aromatic ring, the \(\mathrm{OH}\) can donate electrons through resonance with the \(\mathrm{-C=C-}\) and make the enol more reactive. Carbonyl groups can react, through the small percentage of the enol form present, to undergo electrophilic attack at the alpha carbon. We have already seen that aldehydes and ketones exist as keto-enol tautomers, but, in fact, carboxylic acids, esters, and other acid derivatives also have the potential to exist in the corresponding enol form. Another implication of the alcohol-nature of an enol is that we expect it to be acidic—and indeed it is. The conjugate base of the enol is called the enolate ion and it is resonance-stabilized so that the negative charge is delocalized on both the oxygen and on the alpha carbon. The \(\mathrm{pK}_{\mathrm{a}}\) of acetone is \(19\)—somewhat higher than a typical alcohol (\(\mathrm{pK}_{\mathrm{a}} \sim 15\)). In the enolate form, the majority of the charge sits on the more electronegative oxygen, but a significant proportion of the negative charge is associated with the alpha carbon: the enolate ion is a stabilized carbanion. The enolate anion is often written in its carbanion form because this is the form that produces most of the interesting chemistry. Treatment of a carbonyl compound with a base, such as an alkoxide, results in the reversible formation of a small amount of the enolate ion (although the equilibrium lies on the side of the unprotonated form). Similarly, many carbonyl compounds can be deprotonated to give the corresponding enolate anion. For example, esters can be treated with a base to give the corresponding enolate anion. Ethyl acetate has a \(\mathrm{pK}_{\mathrm{a}}\) of around \(25\) (less acidic than acetone: \(\mathrm{pK}_{\mathrm{a }} 19\)), but still well within reach of many of the strong bases. For example, sodium amide (\(\mathrm{NaNH}_{2}\)), the conjugate base of ammonia (\(\mathrm{pK}_{\mathrm{a }} 33\)), is strong enough to deprotonate the ester. In fact, we typically use what is known as a hindered base, such as lithium di-isopropylamide (\(\mathrm{LDA}\)), which is similar to sodium amide but the nitrogen has two bulky isopropyl groups attached to it. Since \(\mathrm{LDA}\) is such a strong base, treatment of most carbonyl compounds with \(\mathrm{LDA}\) produces essentially 100% of the corresponding enolate anion. However, there are exceptions. Any carbonyl compound that has a more acidic proton than the \(\mathrm{H}\) associated with the alpha carbon will not undergo this reaction. For example, treatment of carboxylic acids with \(\mathrm{LDA}\) will merely result in the loss of the acidic proton from the carboxylic acid OH group. Most carbonyl compounds have \(\mathrm{pK}_{\mathrm{a}}\)‘s between \(19\) and \(25\). Compounds that have carbonyl groups that are beta to each other (that is, separated by an intervening carbon), have significantly lower \(\mathrm{pK}_{\mathrm{a}}\)‘s (around \(9\)), because the resulting anion can be stabilized on both carbonyl oxygens. They can be easily deprotonated by bases such as sodium ethoxide or sodium hydroxide.
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Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode. Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties. When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, \(\ce{NaCl}\), and \(\ce{CaCl_2}\). Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both \(\ce{NaCl}\) and \(\ce{CaCl_2}\) are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of \(\ce{NaCl}\) contains 0.01 M \(\ce{Na^{+}}\) ions and 0.01 M \(\ce{Cl^{−}}\) ions, for a total particle concentration of 0.02 M. Similarly, the \(\ce{CaCl_2}\) solution contains 0.01 M \(\ce{Ca^{2+}}\) ions and 0.02 M \(\ce{Cl^{−}}\) ions, for a total particle concentration of 0.03 M. These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete. At (typically >1 M), especially with salts of small, highly charged ions (such as \(\ce{Mg^{2+}}\) or \(\ce{Al^{3+}}\)), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind. Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure \(\Page {1}\), which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent. Figure \(\Page {2}\) shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant. If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore \[P_A=\chi_AP^0_A \label{13.5.1} \] where \(P_A\) is the vapor pressure of component A of the solution (in this case the solvent), \(\chi_A\) is the mole fraction of \(A\) in solution, and \(P^0_A\) is the vapor pressure of pure \(A\). Equation \(\ref{13.5.1}\) is known as , after the French chemist who developed it. If the solution contains only a single nonvolatile solute (B), then \(\chi_A + \chi_B = 1\), and we can substitute \(\chi_A = 1 − \chi_B\) to obtain \[\begin{align} P_A &=(1−\chi_B)P^0_A \\[4pt] &=P^0_A−\chi_BP^0_A \label{13.5.2} \end{align} \] Rearranging and defining \(ΔP_A=P^0_A−P_A\), we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute: \[ \begin{align} P^0_A−P_A &=ΔP_A \\[4pt] &=\chi_BP^0_A \label{13.5.3} \end{align} \] We can solve vapor pressure problems in either of two ways: by using Equation \(\ref{13.5.1}\) to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation \(\ref{13.5.3}\) to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute. Ethylene glycol (\(\ce{HOCH_2CH_2OH}\)), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter. : identity of solute, percentage by mass, and vapor pressure of pure solvent : vapor pressure of solution : : A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present: \[moles \;EG=(302 \;\cancel{g}) \left( \dfrac{1\; mol}{62.07\; \cancel{g}} \right)=4.87\; mol\; EG \nonumber \] \[moles \; H_2O=(698 \;\cancel{g}) \left( \dfrac{1\; mol}{18.02\; \cancel{g}} \right)=38.7\; mol\; H_2O \nonumber \] The mole fraction of water is thus \[\chi_{H_2O}=\dfrac{38.7\; \cancel{mol} \; H_2O}{38.7\; \cancel{mol}\; H_2O +4.87 \cancel{mol}\; EG} =0.888 \nonumber \] From Raoult’s law (Equation \(\ref{13.5.1}\)), the vapor pressure of the solution is \[ \begin{align*} P_{H_2O} &=(\chi_{H2_O})(P^0_{H_2O}) \\[4pt] &=(0.888)(760\; mmHg) =675 \;mmHg \end{align*} \nonumber \] Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation \(\ref{13.5.3}\) to calculate the resulting decrease in vapor pressure: \[\chi_{EG}=\dfrac{4.87\; mol\; EG}{4.87\; mol\; EG+38.7\; mol\; H_2O}=0.112 \nonumber \] \[ΔP_{H2_O}=(\chi_{EG})(P^0_{H_2O})=(0.112)(760\; mmHg)=85.1\; mmHg \nonumber \] \[P_{H_2O}=P^0_{H_2O}−ΔP_{H_2O}=760\; mmHg−85.1\; mmHg=675\; mmHg \nonumber \] The same result is obtained using either method. Seawater is an approximately 3.0% aqueous solution of \(\ce{NaCl}\) by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of \(\ce{NaCl}\), remembering that 1 mol of \(\ce{NaCl}\) produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg. 0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling. Even when a solute is volatile, meaning that it has a measurable vapor pressure, we can still use Raoult’s law. In this case, we calculate the vapor pressure of each component separately. The total vapor pressure of the solution (\(P_{tot}\)) is the sum of the vapor pressures of the components: \[P_{tot}=P_A+P_B=\chi_AP^0_A+\chi_BP^0_B \label{13.5.4} \] Because \(\chi_B = 1 − \chi_A\) for a two-component system, \[P_{tot}=\chi_AP^0_A+(1−\chi_A)P^0_B \label{13.5.5} \] Thus we need to specify the mole fraction of only one of the components in a two-component system. Consider, for example, the vapor pressure of solutions of benzene and toluene of various compositions. At 20°C, the vapor pressures of pure benzene and toluene are 74.7 and 22.3 mmHg, respectively. The vapor pressure of benzene in a benzene–toluene solution is \[P_{C_6H_6}=\chi_{C_6H_6}P^0_{C_6H_6} \label{13.5.6} \] and the vapor pressure of toluene in the solution is \[P{C_6H_5CH_3}=\chi_{C_6H_5CH_3}P^0_{C_6H_5CH3} \label{13.5.7} \] Equations \(\ref{13.5.6}\) and \(\ref{13.5.7}\) are both in the form of the equation for a straight line: \(y = mx + b\), where \(b = 0\). Plots of the vapor pressures of both components versus the mole fractions are therefore straight lines that pass through the origin, as shown in Figure \(\Page {3}\). Furthermore, a plot of the total vapor pressure of the solution versus the mole fraction is a straight line that represents the sum of the vapor pressures of the pure components. Thus the vapor pressure of the solution is always greater than the vapor pressure of either component. A solution of two volatile components that behaves like the solution in Figure \(\Page {3}\), which is defined as a solution that obeys Raoult’s law. Like an ideal gas, an ideal solution is a hypothetical system whose properties can be described in terms of a simple model. Mixtures of benzene and toluene approximate an ideal solution because the intermolecular forces in the two pure liquids are almost identical in both kind and magnitude. Consequently, the change in enthalpy on solution formation is essentially zero (\(ΔH_{soln} ≈ 0\)), which is one of the defining characteristics of an ideal solution. Ideal solutions and ideal gases are both simple models that ignore intermolecular interactions. Most real solutions, however, do not obey Raoult’s law precisely, just as most real gases do not obey the ideal gas law exactly. Real solutions generally deviate from Raoult’s law because the intermolecular interactions between the two components A and B differ. We can distinguish between two general kinds of behavior, depending on whether the intermolecular interactions between molecules A and B are stronger or weaker than the A–A and B–B interactions in the pure components. If the A–B interactions are stronger than the A–A and B–B interactions, each component of the solution exhibits a lower vapor pressure than expected for an ideal solution, as does the solution as a whole. The favorable A–B interactions effectively stabilize the solution compared with the vapor. This kind of behavior is called a negative deviation from Raoult’s law. Systems stabilized by hydrogen bonding between two molecules, such as acetone and ethanol, exhibit from Raoult’s law. Conversely, if the A–B interactions are weaker than the A–A and B–B interactions yet the entropy increase is enough to allow the solution to form, both A and B have an increased tendency to escape from the solution into the vapor phase. The result is a higher vapor pressure than expected for an ideal solution, producing a positive deviation from Raoult’s law. In a solution of \(\ce{CCl_4}\) and methanol, for example, the nonpolar \(\ce{CCl_4}\) molecules interrupt the extensive hydrogen bonding network in methanol, and the lighter methanol molecules have weaker London dispersion forces than the heavier \(\ce{CCl_4}\) molecules. Consequently, solutions of \(\ce{CCl_4}\) and methanol exhibit from Raoult’s law. For each system, compare the intermolecular interactions in the pure liquids and in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation). : identity of pure liquids : predicted deviation from Raoult’s law (Equation \ref{13.5.1}) : Identify whether each liquid is polar or nonpolar, and then predict the type of intermolecular interactions that occur in solution. : For each system, compare the intermolecular interactions in the pure liquids with those in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation): approximately equal positive deviation (vapor pressure greater than predicted) negative deviation (vapor pressure less than predicted) A Discussing Roult's Law. Link: A Discussing How to find the Vapor Pressure of a Solution. Link: Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure \(\Page {4}\)). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to P = 1 atm at a higher temperature than does the curve for pure water. Figure \(\Page {4}\): Phase Diagrams of Pure Water and an Aqueous Solution of a Nonvolatile Solute. The vaporization curve for the solution lies below the curve for pure water at all temperatures, which results in an increase in the boiling point and a decrease in the freezing point of the solution. The boiling point of a solution with a nonvolatile solute is greater than the boiling point of the pure solvent. The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure \(\Page {5}\)). We can define the boiling point elevation (\(ΔT_b\)) as the difference between the boiling points of the solution and the pure solvent: \[ΔT_b=T_b−T^0_b \label{13.5.8} \] where \(T_b\) is the boiling point of the solution and \(T^0_b\) is the boiling point of the pure solvent. We can express the relationship between \(ΔT_b\) and concentration as follows \[ΔT_b = mK_b \label{13.5.9} \] where m is the concentration of the solute expressed in molality, and \(K_b\) is the of the solvent, which has units of °C/m. Table \(\Page {1}\) lists characteristic \(K_b\) values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration. The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros. According to Table \(\Page {1}\), the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous \(\ce{NaCl}\) solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved ions. Hence a 1.00 m \(\ce{NaCl}\) solution will have a boiling point of about 101.02°C. In Example \(\Page {1}\), we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution. : composition of solution : boiling point : Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation \(\ref{13.5.9}\) to calculate the increase in boiling point. : From Example \(\Page {1}\), we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus \[\text{molality of ethylene glycol}= \left(\dfrac{4.87 \;mol}{698 \; \cancel{g} \;H_2O} \right) \left(\dfrac{1000\; \cancel{g}}{1 \;kg} \right)=6.98\, m \nonumber \] From Equation \(\ref{13.5.9}\), the increase in boiling point is therefore \[ΔT_b=m K_b=(6.98 \cancel{m})(0.51°C/\cancel{m})=3.6°C \nonumber \] The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation \(\ref{13.5.9}\) may not be valid. Assume that a tablespoon (5.00 g) of \(\ce{NaCl}\) is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil? 100.04°C, or 100°C to three significant figures. (Recall that 1 mol of \(\ce{NaCl}\) produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.) A Discussing Boiling Point Elevation and Freezing Point Depression. Link: The phase diagram in Figure \(\Page {4}\) shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to \(P = 1\,atm\) at a lower temperature than the curve for pure water. We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice. By analogy to our treatment of boiling point elevation,the freezing point depression (\(ΔT_f\)) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution: \[ ΔT_f=T^0_f−T_f \label{13.5.10} \] where \(T^0_f\) is the freezing point of the pure solvent and \(T_f\) is the freezing point of the solution. The order of the terms is reversed compared with Equation \(\ref{13.5.8}\) to express the freezing point depression as a positive number. The relationship between \(ΔT_f\) and the solute concentration is given by an equation analogous to Equation \(\ref{13.5.9}\): \[ΔT_f = mK_f \label{13.5.11} \] where \(m\) is the molality of the solution and \(K_f\) is the molal freezing point depression constant for the solvent (in units of °C/m). Like \(K_b\), each solvent has a characteristic value of \(K_f\) (see Table \(\Page {1}\)). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous \(\ce{NaCl}\) solution has twice as large a freezing point depression as a glucose solution of the same molality. People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing. The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles. In colder regions of the United States, \(\ce{NaCl}\) or \(\ce{CaCl_2}\) is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in the Figure below to estimate the concentrations of two saturated solutions at 0°C, one of \(\ce{NaCl}\) and one of \(\ce{CaCl_2}\), and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice. : solubilities of two compounds : concentrations and freezing points : : A From Figure above, we can estimate the solubilities of \(\ce{NaCl}\) and \(\ce{CaCl_2}\) to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are \[m_{NaCl}=\left(\dfrac{36 \; \cancel{g \;NaCl}}{100 \;\cancel{g} \;H_2O}\right)\left(\dfrac{1\; mol\; NaCl}{58.44\; \cancel{ g\; NaCl}}\right)\left(\dfrac{1000\; \cancel{g}}{1\; kg}\right)=6.2\; m \nonumber \] \[m_{CaCl_2}=\left(\dfrac{60\; \cancel{g\; CaCl_2}}{100\;\cancel{g}\; H_2O}\right)\left(\dfrac{1\; mol\; CaCl_2}{110.98\; \cancel{g\; CaCl_2}}\right)\left(\dfrac{1000 \;\cancel{g}}{1 kg}\right)=5.4\; m \nonumber \] The lower formula mass of \(\ce{NaCl}\) more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than \(CaCl_2\). B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of \(\ce{NaCl}\) and \(CaCl_2\), respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for \(\ce{NaCl}\) and 3 × 5.4 m = 16 m for \(CaCl_2\). The resulting freezing point depressions can be calculated using Equation \(\ref{13.5.11}\): \[\ce{NaCl}: ΔT_f=mK_f=(12\; \cancel{m})(1.86°C/\cancel{m})=22°C \nonumber \] \[\ce{CaCl2}: ΔT_f=mK_f=(16\;\cancel{m})(1.86°C/\cancel{m})=30°C \nonumber \] Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that \(CaCl_2\) is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, \(CaCl_2\) is the salt usually sold for home use, and it is also often used on highways. Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Examples \(\Page {5}\) and \(\Page {5}\). −13.0°C Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m \(KCl\), 0.1 m glucose, 0.1 m SrCl2, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m HCl. : molalities of six solutions relative freezing points : : Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. \(\ce{KCl}\), \(\ce{SrCl_2}\), and \(\ce{HCl}\) are , producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule). The molalities of the solutions in terms of the total particles of solute are: \(\ce{KCl}\) and \(\ce{HCl}\), 0.2 m; \(SrCl_2\), 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > \(\ce{HCl}\) = \(\ce{KCl}\) > \(\ce{SrCl_2}\). Arrange these aqueous solutions in order of increasing freezing points: 0.2 m \(\ce{NaCl}\), 0.3 m acetic acid, 0.1 m \(\ce{CaCl_2}\), and 0.2 m sucrose. 0.2 m \(\ce{NaCl}\) (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m \(\ce{CaCl_2}\) < 0.2 m sucrose (highest freezing point) Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing \(K_b\) and \(K_f\) values in Table \(\Page {1}\), we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of \(K_f\) (37.8°C/m), d-(+)-camphor (Table \(\Page {1}\)) is often used to determine the molar mass of organic compounds by this method. A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of \(CS_2\) to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved \(\ce{S_n}\) molecules (i.e., what is the value of \(n\)?). : masses of solute and solvent and freezing point : molar mass and number of \(\ce{S}\) atoms per molecule : : A The first step is to calculate the freezing point depression using Equation \(\ref{13.5.10}\): \[ΔT_f=T^0_f−T_f=−112.1°C−(−113.5°C)=1.4°C \nonumber \] Then Equation \(\ref{13.5.11}\) gives B The total number of moles of solute present in the solution is \[\text{moles solute}=\left(\dfrac{0.37 mol}{\cancel{kg}}\right) (75.0\; g) \left(\dfrac{1 kg}{1000\; g}\right)=0.028 \;mol \nonumber \] C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus \[\text{molar mass}=\dfrac{7.08\; g}{0.028\; mol}=260\; g/mol \nonumber \] The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of \(\ce{S_8}\). One of the byproducts formed during the synthesis of \(C_{60}\) is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of \(CCl_4\) has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance? 847 g/mol; \(\ce{C_{70}}\) A Discussing how to find the Molecular Weight of an Unknown using Colligative Properties. Link: Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration. Osmosis can be demonstrated using a U-tube like the one shown in Figure \(\Page {6}\), which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure (\(\Pi\)) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution. Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation: \[\Pi=\dfrac{nRT}{V}=MRT \label{13.5.12} \] where \(M\) is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), \(R\) is the ideal gas constant, and \(T\) is the absolute temperature. As shown in Example \(\Page {7}\), osmotic pressures tend to be quite high, even for rather dilute solutions. When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% \(\ce{NaCl}\) by mass; the solution density is 1.02 g/mL at 25°C. : concentration, density, and temperature of \(\ce{NaCl}\) solution; internal osmotic pressure of cell : osmotic pressure of \(\ce{NaCl}\) solution and concentration of glycerol needed : : A The solution contains 4.0 g of \(\ce{NaCl}\) per 100 g of solution. Using the formula mass of \(\ce{NaCl}\) (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity: \[ \begin{align*} M_{NaCl} &=\dfrac{moles\; NaCl}{\text{liter solution}} \\[4pt] &=\left(\dfrac{4.0 \; \cancel{g} \;NaCl}{58.44\; \cancel{g}/mol\; NaCl}\right)\left(\dfrac{1}{100\; \cancel{g \;solution}}\right)\left(\dfrac{1.02\; \cancel{g\; solution}}{1.00\; \cancel{mL}\; solution}\right)\left(\dfrac{1000\; \cancel{mL}}{1\; L}\right) \\[4pt] &= 0.70\; M\; \ce{NaCl} \end{align*} \nonumber \] Because 1 mol of \(\ce{NaCl}\) produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M. B Now we can use Equation \ref{13.5.12} to calculate the osmotic pressure of the solution: \[ \begin{align*} \Pi &=MRT \\[4pt] &=(1.4 \;mol/L)\left[ 0.0821\; (L⋅atm)/(K⋅mol) \right ] (298\; K)\\[4pt] &=34 \;atm\end{align*} \nonumber \] C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation \ref{13.5.12} for the molarity corresponding to this osmotic pressure: \[ \begin{align*} M&=\dfrac{\Pi}{RT}\\[4pt] &=\dfrac{27\; \cancel{atm}}{[0.0821(L⋅\cancel{atm})/(\cancel{K}⋅mol)] (298 \;\cancel{K})}\\[4pt] &=1.1 \;M \;\text{glycerol} \end{align*} \nonumber \] In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the \(\ce{NaCl}\) solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol. Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them. 24 atm Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins. The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure \(\Page {7}\), which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells. In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves. Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure \(\Page {8}\), applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on Navy lifeboats. A Discussing Osmotic Pressure. Link: Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of \(\ce{NaCl}\) in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of \(\ce{NaCl}\) and \(\ce{KCl}\) are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution. The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van’t Hoff factor (\(i\)) and is defined as follows: \[i=\dfrac{\text{apparent number of particles in solution}}{\text{ number of moles of solute dissolved}} \label{13.5.13} \] Named for Jacobus Hendricus van’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions. As the solute concentration increases, the van’t Hoff factor decreases. The van’t Hoff factor is therefore . The lower the van’t Hoff factor, the greater the deviation. As the data in Table \(\Page {2}\) show, the van’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution. Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure \(\Page {9}\)). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as \(Mg^{2+}\), \(Al^{3+}\), \(SO_4^{2−}\), and \(PO_4^{3−}\) have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations. A 0.0500 M aqueous solution of \(FeCl_3\) has an osmotic pressure of 4.15 atm at 25°C. Calculate the van’t Hoff factor \(i\) for the solution. : solute concentration, osmotic pressure, and temperature : van’t Hoff factor : : A If \(\ce{FeCl_3}\) dissociated completely in aqueous solution, it would produce four ions per formula unit [Fe3+(aq) plus 3Cl−(aq)] for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be \[\begin{align*} \Pi &=MRT \\[4pt] &=(0.200 \;mol/L) \left[0.0821\;(L⋅atm)/(K⋅mol) \right] (298\; K)=4.89\; atm \end{align*} \nonumber \] B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 particles per mole of \(\ce{FeCl_3}\) dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm: \[4.15\; atm=M \left( \dfrac{0.0821 \;(L⋅atm)}{(K⋅mol)}\right) (298 \;K) \nonumber \] or after rearranging \[M = 0.170 mol \nonumber \] The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 particles per mole of \(\ce{FeCl_3}\) dissolved. From Equation \(\ref{13.5.13}\), the van’t Hoff factor for the solution is \[i=\dfrac{\text{3.40 particles observed}}{\text{1 formula unit}\; \ce{FeCl_3}}=3.40 \nonumber \] Calculate the van’t Hoff factor for a 0.050 m aqueous solution of \(MgCl_2\) that has a measured freezing point of −0.25°C. 2.7 (versus an ideal value of 3). A Discussing the Colligative Properties in Solutions. Link: The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as . Solutions that obey Raoult’s law are called ideal solutions. Most real solutions exhibit positive or negative deviations from Raoult’s law. The boiling point elevation (\(ΔT_b\)) and freezing point depression (\(ΔT_f\)) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt. | 43,430 | 3,058 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Atomic_Theory/Atomic_Mass |
. The mass of an atom or a molecule is referred to as the The atomic mass is used to find the average mass of elements and molecules and to solve stoichiometry problems. In chemistry, there are many different concepts of mass. It is often assumed that is the mass of an atom indicated in . However, the book published by the IUPAC clearly states: The name "atomic mass" is used for historical reasons, and originates from the fact that chemistry was the first science to investigate the same physical objects on macroscopic and microscopic levels. In addition, the situation is rendered more complicated by the isotopic distribution. On the macroscopic level, most mass measurements of pure substances refer to a mixture of isotopes. This means that from a physical stand point, these mixtures are not pure. For example, the macroscopic mass of oxygen (O ) does not correspond to the microscopic mass of O . The former usually implies a certain isotopic distribution, whereas the latter usually refers to the most common isotope ( O ). Note that the former is now often referred to as the "molecular weight" or "atomic weight". These concepts are further explained below. Isotopes are atoms with the same atomic number, but different mass numbers. A different mass size is due to the difference in the number of neutrons that an atom contains. Although mass numbers are whole numbers, the actual masses of individual atoms are never whole numbers (except for carbon-12, by definition). This explains how lithium can have an atomic mass of 6.941 Da. The atomic masses on the periodic table take these isotopes into account, weighing them based on their abundance in nature; more weight is given to the isotopes that occur most frequently in nature. Average mass of the element E is defined as: \[ m(E) = \sum_{n=1} m(I_n) \times p(I_n) \] where ∑ represents a n-times summation over all isotopes \(I_n\) of element E, and p(I) represents the relative abundance of the isotope I. Find the average atomic mass of boron using the Table 1 below: The average mass of Boron is: \[ m(B) = (10.013\ Da)(0.199) + (11.009\ Da)(0.801) = 1.99\ Da + 8.82\ Da = 10.81\ Da \] Traditionally it was common practice in chemistry to avoid using any units when indicating atomic masses (e.g. masses on microscopic scale). Even today, it is common to hear a chemist say, " C has exactly mass 12". However, because mass is not a dimensionless quantity, it is clear that a mass indication needs a unit. Chemists have tried to rationalize the omission of a unit; the result is the concept of relative mass, which strictly speaking is not even a mass but a ratio of two masses. Rather than using a unit, these chemists claim to indicate the ratio of the mass they want to indicate and the atomic mass constant m which is defined analogous to the unit they want to avoid. Hence the relative atomic mass of the mass m is defined as: \[A_r = \dfrac{m}{m_u} \] The quantity is now dimensionless. As this unit is confusing and against the standards of modern metrology, the use of relative mass is discouraged. Until recently, the concept of mass was not clearly distinguished from the concept of weight. In colloquial language this is still the case. Many people indicate their "weight" when they actually mean their mass. Mass is a fundamental property of objects, whereas weight is a force. Weight is the force F exerted on a mass m by a gravitational field. The exact definition of the weight is controversial. The weight of a person is different on ground than on a plane. Strictly speaking, . When discussing atoms and molecules, the mass of a molecule is often referred to as the "molecular weight". There is no univerally-accepted definition of this term; however, mosts chemists agree that it means an average mass, and many consider it dimensionless. This would make "molecular weight" a synonym to "average relative mass". Because the proton and the neutron have similar mass, and the electron has a very small mass compared to the former, most molecules have a mass that is close to an integer value when measured in daltons. Therefore it is quite common to only indicate the of molecules. Integer mass is only meaningful when using dalton (or u) units. Many mass spectrometers can determine the mass of a molecule with accuracy exceeding that of the integer mass. This measurement is therefore called the of the molecule. Isotopes (and hence molecules) have atomic masses that are not integer masses due to a mass defect caused by binding energy in the nucleus. The atomic mass is usually measured in the units (u), or dalton (Da). Both units are derived from the carbon-12 isotope, as 12 u is the exact atomic mass of that isotope. So 1 u is 1/12 of the mass of a carbon-12 isotope: 1 u = 1 Da = m( C)/12 The first scientists to measure atomic mass were John Dalton (between 1803 and 1805) and Jons Jacoband Berzelius (between 1808 and 1826). Early atomic mass theory was proposed by the English chemist William Prout in a series of published papers in 1815 and 1816. Known was Prout's Law, Prout suggested that the known elements had atomic weights that were whole number multiples of the atomic mass of hydrogen. Berzelius demonstrated that this is not always the case by showing that chlorine (Cl) has a mass of 35.45, which is not a whole number multiple of hydrogen's mass. Some chemists use the atomic mass unit (amu). The amu was defined differently by physicists and by chemists: Chemists used oxygen in the naturally occurring isotopic distribution as the reference. Because the isotopic distribution in nature can change, this definition is a moving target. Therefore, both communities agreed to the compromise of using m( C)/12 as the new unit, naming it the "unified atomic mass unit" (u). Hence, the amu is no longer in use; those who still use it do so with the definition of the u in mind. For this reason, the dalton (Da) is increasingly recommended as the accurate mass unit. Neither u nor Da are SI units, but both are recognized by the SI. The molar mass is the mass of one of substance, whether the substance is an element or a compound. A mole of substance is equal to Avogadro's number (6.023×10 ) of that substance. The molar mass has units of g/mol or kg/mol. When using the unit g/mol, the numerical value of the molar mass of a molecule is the same as its average mass in daltons: This allows for a smooth transition from the microscopic world, where mass is measured in daltons, to the macroscopic world, where mass is measured in kilograms. What is the molar mass of phenol, C H OH? Average mass m = 6 × 12.011 Da + 6 × 1.008 Da + 1 × 15.999 Da = 94.113 Da Molar mass = 94.113 g/mol = 0.094113 kg/mo Masses of atoms and molecules are measured by mass spectrometry. Mass spectrometry is a technique that measures the mass-to-charge ratio (m/q) of ions. It requires that all molecules and atoms to be measured be ionized. The ions are then separated in a mass analyzer according to their mass-to-charge ratio. The charge of the measured ion can then be determined, because it is a multiple of the elementary charge. The the ion's mass can be deduced. The average masses indicated in the periodic table are then calculated using the isotopic abundances, as explained above. The masses of all isotopes have been measured with very high accuracy. Therefore, it is much simpler and more accurate to calculate the mass of a molecule of interest as a sum of its isotopes than measuring it with a commercial mass spectrometer. Note that the same is not true on the nucleon scale. The mass of an isotope cannot be calculated accurately as the sum of its particles (given in the table below); this would ignore the mass defect caused by the binding energy of the nucleons, which is significant. As shown in Table 2, the mass of an electron is relatively small; it contributes less than 1/1000 to the overall mass of the atom. The atomic mass found on the (below the element's name) is the average atomic mass. For example, for Lithium: The indicates the atomic mass of lithium. As shown in Table 2 above and mathematically explained below, the masses of a protons and neutrons are about 1u. This, however, does not explain why lithium has an atomic mass of 6.941 Da where 6 Da is expected. This is true for all elements on the periodic table. The atomic mass for lithium is actually the average atomic mass of its isotopes. This is discussed further in the next section. One particularly useful way of writing an isotope is as follows: Note: One particularly important relationship is illustrated by the fact that an atomic mass unit is equal to 1.66 × 10 g. This is the reciprocal of Avogadro's constant, and it is no coincidence: \[\dfrac{\rm Atomic~Mass~(g)}{1 {\rm g}} \times \dfrac{1 {\rm mol}}{6.022 \times 10^{23}} = \dfrac{\rm Mass~(g)}{1 {\rm atom}}\] \[1\ u = \frac{M_u\ (molar\ mass\ unit)}{N_A\ (Avogadro's\ Number)} = 1\ \frac{g}{mol\ N_A} \] N known as Avogadro's number (Avogadro's constant) is equal to 6.023×10 atoms. Atomic mass is particularly important when dealing with . a) Molecular mass of Ra(HCO ) = 226 + 2(1.01 u + 12.01 u + (16.00 u)(3)) = 348 u or g/mol b) Number of neutrons: Be, N, Na, Cl Atomic Mass: Be, N, Na, Cl Note: It is the same increasing order for both number of neutrons and atomic mass because more neutrons means more mass. c) Atomic mass of Zenium: (59 u)(0.62) + (61 u)(0.27) + (67 u)(0.11) = 37 u + 16 u + 7.4 u = 60.4 u or g/mol d) Mn e) (3.71 moles F )(19 × 2 g/mol F ) = (3.71 mol F )(38 g/mol F ) = 140 g F f) (4.3 × 10 molecules POCI )(1 mol/6.022 × 10 molecules POCI3)(30.97 + 16.00 + 35.45 x 3 g/mol POCI ) = (4.3 × 10 molecules POCI )(mol/6.022 × 10 molecules POCI3)(153.32 g/mol POCI ) = 11 g POCI g) (23 g Na CO )(1 mol/22.99 × 2 + 12.01 + 16.00 × 3 g Na CO ) = (23 g Na CO )(1 mol/105.99 g Na CO ) = (0.22 mol Na CO ) | 9,930 | 3,059 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/12%3A_Chemistry_of_Earth/12.04%3A_Earth's_Dwindling_Resources |
( ), commonly known as or in the United States and in Britain, is a waste type consisting of everyday items that are discarded by the public. "Garbage" can also refer specifically to food waste, as in a garbage disposal; the two are sometimes collected separately. In the European Union, the semantic definition is 'mixed municipal waste,' given waste code 20 03 01 in the European Waste Catalog. Although the waste may originate from a number of sources that has nothing to do with a municipality, the traditional role of municipalities in collecting and managing these kinds of waste have produced the particular etymology 'municipal.' The composition of municipal solid waste varies greatly from municipality to municipality, and it changes significantly with time. In municipalities which have a well developed waste recycling system, the waste stream mainly consists of intractable wastes such as plastic film and non-recyclable packaging materials. At the start of the 20th century, the majority of domestic waste (53%) in the UK consisted of coal ash from open fires. In developed areas without significant recycling activity it predominantly includes food wastes, market wastes, yard wastes, plastic containers and product packaging materials, and other miscellaneous solid wastes from residential, commercial, institutional, and industrial sources. Most definitions of municipal solid waste do not include industrial wastes, agricultural wastes, medical waste, radioactive waste or sewage sludge. Waste collection is performed by the municipality within a given area. The term relates to waste left from household sources containing materials that have not been separated out or sent for processing. Waste can be classified in several ways but the following list represents a typical classification: For example, typical municipal solid waste in China is composed of 55.9% food residue, 8.5% paper, 11.2% plastics, 3.2% textiles, 2.9% wood waste, 0.8% rubber, and 18.4% non-combustibles. The breakdown of municipal waste by material is shown below (Figure \(\Page {1}\)). The most effective way to reduce waste is to not create it in the first place. Making a new product requires a lot of materials and energy - raw materials must be extracted from the earth, and the product must be fabricated then transported to wherever it will be sold. As a result, reduction and reuse are the most effective ways you can save natural resources, protect the environment and save money. Recycling is the process of collecting and processing materials that would otherwise be thrown away as trash and turning them into new products. Recycling can benefit your community and the environment. EPA released significant findings on the economic benefits of the recycling industry with an update to the national Recycling Economic Information (REI) Study in 2016. This study analyzes the numbers of jobs, wages and tax revenues attributed to recycling. The study found that in a single year, recycling and reuse activities in the United States accounted for: This equates to 1.57 jobs, $76,000 in wages, and $14,101 in tax revenues for every 1,000 tons of materials recycled. ( ) US EPA (Environmental Protection Agency) | 3,246 | 3,060 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/15%3A_Alcohols_and_Ethers/15.08%3A_Polyhydric_Alcohols |
The simplest example of an alcohol with more than one hydroxyl group is methanediol or methylene glycol, \(\ce{HOCH_2OH}\). The term “glycol” indicates a , which is a substance with two alcoholic hydroxyl groups. Methylene glycol is reasonably stable in water solution, but attempts to isolate it lead only to its dehydration product, methanal (formaldehyde): This behavior is rather typical of -diols ( geminal, that is, with both hydroxyl groups on the carbon atom). The few -diols of this kind that can be isolated are those that carry strongly electron-attracting substituents such as the following: Polyhydric alcohols in which the hydroxyl groups are situated on different carbons are relatively stable, and, as we might expect for substances with multiple polar groups, they have high boiling points and considerable water solubility, but low solubility in nonpolar solvents: 1,2-Diols are prepared from alkenes by oxidation with reagents such as , potassium permanganate, or hydrogen peroxide ( ). However, ethylene glycol is made on a commercial scale from oxacyclopropane, which in turn is made by air oxidation of ethene at high temperatures over a silver oxide catalyst ( ). Ethylene glycol has important commercial uses. It is an excellent permanent antifreeze for automotive cooling systems because it is miscible with water in all proportions and a \(50\%\) solution freezes at \(-34^\text{o}\) \(\left( -29^\text{o} \text{F} \right)\). It also is used as a solvent and as an intermediate in the production of polymers (polyesters) and other products (Chapter 29). The trihydric alcohol, 1,2,3-propanetriol (glycerol), is a nontoxic, water-soluble, viscous, hygroscopic liquid that is used widely as a humectant (moistening agent). It is an important component of many food, cosmetic, and pharmaceutical preparations. At one time, glycerol was obtained on a commercial scale only as a by-product of soap manufacture through hydrolysis of fats, which are glyceryl esters of long-chain alkanoic acids. The major present source is by synthesis from propene ( ). The trinitrate ester of glycerol (nitroglycerin) is an important but shock-sensitive explosive: Dynamite is a much safer and more controllable explosive, and is made by absorbing nitroglycerin in porous material such as sawdust or diatomaceous earth. Dynamite has largely been replaced by cheaper explosives containing ammonium nitrate as the principal ingredient. Glycerol, as a constituent of fats and lipids, plays an important role in animal metabolism. and (1977) | 2,568 | 3,061 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Significant_Digits/Uncertainties_in_Measurements |
All measurements have a degree of uncertainty regardless of precision and accuracy. This is caused by two factors, the limitation of the measuring instrument (systematic error) and the skill of the experimenter making the measurements (random error). The graduated buret in Figure 1 contains a certain amount of water (with yellow dye) to be measured. The amount of water is somewhere between 19 ml and 20 ml according to the marked lines. By checking to see where the bottom of the meniscus lies, referencing the ten smaller lines, the amount of water lies between 19.8 ml and 20 ml. The next step is to estimate the uncertainty between 19.8 ml and 20 ml. Making an approximate guess, the level is less than 20 ml, but greater than 19.8 ml. We then report that the measured amount is approximately 19.9 ml. The graduated cylinder itself may be distorted such that the graduation marks contain inaccuracies providing readings slightly different from the actual volume of liquid present. The diagram below illustrates the distinction between and errors. When we use tools meant for measurement, we assume that they are correct and accurate, however measuring tools are not always right. In fact, they have errors that naturally occur called . Systematic errors tend to be consistent in magnitude and/or direction. If the magnitude and direction of the error is known, accuracy can be improved by additive or proportional corrections. involves adding or subtracting a constant adjustment factor to each measurement; involves multiplying the measurement(s) by a constant. s: Sometimes called human error, random error is determined by the experimenter's skill or ability to perform the experiment and read scientific measurements. These errors are random since the results yielded may be too high or low. Often random error determines the precision of the experiment or limits the precision. For example, if we were to time a revolution of a steadily rotating turnable, the random error would be the reaction time. Our reaction time would vary due to a delay in starting (an underestimate of the actual result) or a delay in stopping (an overestimate of the actual result). Unlike systematic errors, random errors vary in magnitude and direction. It is possible to calculate the average of a set of measured positions, however, and that average is likely to be more accurate than most of the measurements. | 2,430 | 3,062 |
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Laboratory courses can be magical.They can be enlightening experiences that open your eye to a big picture. A laboratory experience should work in tandem with a lecture course, and fully realize concepts, techniques and reactions that you have heard of. Unfortunately, they can also be discouraging experiences. One of the key factors that dictates which experience you will have, is preparation. A rewarding aspect of a well-prepared experiment is that it can firmly cement the information that you have obtained through studying in a way that is far superior to simply reading about it. Your knowledge evolves beyond routine memorizing to real understanding, because you have seen the reaction and principles with your own eyes. The synergy between a lecture course and a lab component should not be underestimated. In this chapter I outline a systematic way of preparing for any organic chemistry experiment to ensure that you succeed in the laboratory and that you leave with an optimal experience. Seeing, after all, is believing. I discuss how to make and use the flow diagrams, how to obtain relevant safety information about the chemicals you are handling, and how to use your note-book to prepare efficiently.
| 1,233 | 3,063 |
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If the reaction follows a second order rate law, the some methodology can be employed. The rate can be written as \[ -\dfrac{d[A]}{dt} = k [A]^2 \label{eq1A} \] The separation of concentration and time terms (this time keeping the negative sign on the left for convenience) yields \[ -\dfrac{d[A]}{[A]^2} = k \,dt \nonumber \] The integration then becomes \[ - \int_{[A]_o}^{[A]} \dfrac{d[A]}{[A]^2} = \int_{t=0}^{t}k \,dt \label{eq1} \] And noting that \[ - \dfrac{dx}{x^2} = d \left(\dfrac{1}{x} \right) \nonumber \] the result of integration Equation \ref{eq1} is \[ \dfrac{1}{[A]} -\dfrac{1}{[A]_o} = kt \nonumber \] or \[ \dfrac{1}{[A]} = \dfrac{1}{[A]_o} + kt \nonumber \] And so a plot of \(1/[A]\) as a function of time should produce a linear plot, the slope of which is \(k\), and the intercept of which is \(1/[A]_0\). Other 2 order rate laws are a little bit trickier to integrate, as the integration depends on the actual stoichiometry of the reaction being investigated. For example, for a reaction of the type \[A + B \rightarrow P \nonumber \] That has rate laws given by \[ -\dfrac{d[A]}{dt} = k [A,B] \nonumber \] and \[ -\dfrac{d[B]}{dt} = k [A,B] \nonumber \] the integration will depend on the decrease of [A] and [B] (which will be related by the stoichiometry) which can be expressed in terms the concentration of the product [P]. \[[A] = [A]_o – [P] \label{eqr1} \] and \[[B] = [B]_o – [P]\label{eqr2} \] The concentration dependence on \(A\) and \(B\) can then be eliminated if the rate law is expressed in terms of the production of the product. \[ \dfrac{d[P]}{dt} = k [A,B] \label{rate2} \] Substituting the relationships for \([A]\) and \([B]\) (Equations \ref{eqr1} and \ref{eqr2}) into the rate law expression (Equation \ref{rate2}) yields \[ \dfrac{d[P]}{dt} = k ( [A]_o – [P]) ([B] = [B]_o – [P]) \label{rate3} \] Separation of concentration and time variables results in \[\dfrac{d[P]}{( [A]_o – [P]) ([B] = [B]_o – [P])} = k\,dt \nonumber \] Noting that at time \(t = 0\), \([P] = 0\), the integrated form of the rate law can be generated by solving the integral \[\int_{[A]_o}^{[A]} \dfrac{d[P]}{( [A]_o – [P]) ([B]_o – [P])} = \int_{t=0}^{t} k\,dt \nonumber \] Consulting a table of integrals reveals that for \(a \neq b\) , \[ \int \dfrac{dx}{(a-x)(b-x)} = \dfrac{1}{b-a} \ln \left(\dfrac{b-x}{a-x} \right) \nonumber \] Applying the definite integral (as long as \([A]_0 \neq [B]_0\)) results in \[ \left. \dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B]_0-[P]}{[A]_0-[P]} \right) \right |_0^{[A]} = \left. k\, t \right|_0^t \nonumber \] \[ \dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B]_0-[P]}{[A]_0-[P]} \right) -\dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B]_0}{[A]_0} \right) =k\, t \label{finalint} \] Substituting Equations \ref{eqr1} and \ref{eqr2} into Equation \ref{finalint} and simplifying (combining the natural logarithm terms) yields \[\dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B,A]_o}{[A,B]_o} \right) = kt \nonumber \] For this rate law, a plot of \(\ln([B]/[A])\) as a function of time will produce a straight line, the slope of which is \[ m = ([B]_0 – [A]_0)k. \nonumber \] In the limit at \([A]_0 = [B]_0\), then \([A] = [B]\) at all times, due to the stoichiometry of the reaction. As such, the rate law becomes \[ \text{rate} = k [A]^2 \nonumber \] and integrate direct like in Equation \ref{eq1A} and the integrated rate law is (as before) \[ \dfrac{1}{[A]} = \dfrac{1}{[A]_o} + kt \nonumber \] Consider the following kinetic data. Use a graph to demonstrate that the data are consistent with second order kinetics. Also, if the data are second order, determine the value of the rate constant for the reaction. The plot looks as follows: From this plot, it can be seen that the rate constant is 0.2658 M s . The concentration at time \(t = 0\) can also be inferred from the intercept. This integral form can be generated by using the method of partial fractions. See (House, 2007) for a full derivation. | 3,986 | 3,065 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Vitamins_Cofactors_and_Coenzymes/Vitamin_B%3A_Cobalamin/Cobalamin_1 |
Cobalamin or Vitamin B a water soluble vitamin that has known functions for improving brain and nerve cells, and the production of adequate blood cells.1 It is commonly found in meats, poultry, dairy products, eggs and seafood. However, organisms such as bacteria and algae are also known to produce the active form of vitamin B through fermentation. The structure of cobalamin is unique with the central atom; cobalt, that has potential for metalloenzyme active sites. In particular, coenzyme B or adenosylcobalamin (AdoCbl) is an essential for several enzymes such as methylmaonly-CoA mutase, diol dehydratase, and ethanolamine ammonia lyase. Cobalamin is an important biologically active, though small, enzyme involved in several configurational changes on its active site. As the name suggests, Cobalamin incorporates several structural elements surrounding a cobalt atom as the metalloenzyme active site. The vitamin-B12 configuration of Cobalamin that is of importance to biological life, primarily in the function of mechanisms in the liver, is that of Cyanocobalamin, one of the rare cases where a cyanide group is presently ingested by living organisms where otherwise would be toxic. Four main forms of Cobalamin exist and are interchanged in the body as the B12 performs various functions. The Cyanocobalamin used as the B12 vitamin supplement however is not naturally occurring, and must be generated then ingested in order to configurationally lose its cyanide group for a methyl group. Various functions in cell metabolism are involved with the transfer of the removable ligand group on the Cobalamin active site, most importantly is the transfer of methyl groups thus acting as a catalyst between configurational changes in many enzymes. Figure 1. Adenosylcobalamin Cobalamin has a formula C63H88CoN14o14P and molecular mass 1355.37 g/mol. This is a fairly large molecule, thus, analyzing the structure confirmation can determine the point group. The structure contains sigma vertical planes and has no sigma horizontal plane. The central atom, Cobalt, has an R attached that makes the molecule unique in several enzymatic catalyses. The point group of Cobalamin is assigned as C4v. Figure 2. Minimal mechanism of diol dehydratase reaction. The mechanism of diol dehydrates is known to catalyzes the conversion of 1,2-diols to the corresponding aldehydes. Figure 2 shows a mechanism for this enzymatic dehydration that involves the hydrogen atom abstraction from C1 (1,2-diol) and the migration of an OH group from C2 to C1 of 1,2-propanediol. The adenosyl (AdoCH2) radical that is generated by the hemolytic cleavage of the Co-C covalent bond in AdoCbl plays an essential role in this OH group migration and thus effectively promotes this chemically difficult reaction.2 Therefore, the OH group on C2 migrates to C1 leading to a formation of a 1,1-diol radical, which leads to the formation of the 1-1-diol and the regeneration of AdoCH2 radical. The crystal structure of diol dehydratase with cyanocobalamin and adeninlypentylcobalamin have shown that both the OH groups of substrate coordinate directly to K+ ion at the active site, which implies the participation of K+ ion in the OH group migration.3 Kamachi and colleagues performed density functional theory to reveal the catalytic roles of K+ ion in the diol dehydratase reaction. As a result, the course of a reaction the substrate and the radical intermediates are always bound to K+ ion until the release of product aldehyde from the active site and that OH group proceed with the aid of K+. Therefore, the role of K+ ion have suggest that it is the most important role in the reaction to fix the substrate and the intermediates in a proper position in order to ensure the hydrogen abstraction and recombination. Figure 2 shows the optimized structure of the enzyme in the QM region. K+ ion is corresponding to the five oxygen atoms originated from the side chain of Gln141, Glu170, Glu221, Gln296, and the carbonyl group of Ser352.2 The sixth and the seventh coordination positions are occupied by O1 and O2 of the substrates (S)-1,2-propanediol (PDO); the S-enantiomer is preferred in the binding by enzyme.2 The ribose moiety of 5’-deoxyadenosyl radical and the side chain of His143 are also involved in the QM region. The interaction of the migrating OH group with the imidazolum ion of His143 has been considered to be essential for the stabilization of the transition state for the OH migration. The Ribosyl rotation for the radical transfer from AdoCbl to substrate can essentially promote the Co-C cleavage upon binding to apodiol dehydratase, where adeninylpropylcobalamin (AdePeCbl) and other longer chain homologues cannot.2 The presence of the adenine-binding site in dio dehydratase was recently determined by the crystal structure analysis of the diol hydratase-AdePeCbl complex.8 The crystal structure shows that the adenine moiety of this analogue is trapped by hydrogen-bonding network with a water molecule and surrounding amion acid residues, Ser224, Ser229, Ser301, and Gly261.2 For this reason, the adenine-binding pocket fixes the adenine ring to allow tight binding of adenylpentyl group to the Co atom at a distance of 1.89 A, which is the main reason for the catalytic inactivity of the analogue. [NEED TO RELOAD THIS IMAGE PROPERLY] Figure 4. Active site structure of diol dehydratase. Figure 4, part A shows an X-ray structure of the diol dehydratase-AdePeCbl complex. Part B shows Diol dehydratase-AdoCbl model complex produced by replacing the pentyl moiety of A with ribose. Part C shows the optimized structure of the diol dehydratase- AdoCbl complex model after the rotation of the ribose moiety.4 However, there is still argument whether the K+ ion in the active site remains with the substrate and radical intermediates through the reaction. Further studies could contribute to the understanding of hydrogen bonds to the active site residues, hydrogen abstraction and the steroselective hydrogen recombination.
Where R = -OH, Hydroxycobalamin, -CN, Cyanocobalamin, -Me, Methylcobalamin, -Ado, 5-deoxyadensosine. The Cyanocobalamin is generated in-situ by bacteria in the gastrointestinal systems of many mammals, or by the carbonization of Hydroxycobalamin created by other types of bacteria when exposed to a charcoal environment. While the B12 ingested may be the Cyanocobalamin, the cyanide group is removed in the body when absorbed and is decomposed in the process of removal. Once in the body, the B12 structure is used in the transport of methyl groups in DNA construction as well as 5-deoxyadensosine in mitochondrial energy production in cells. However important this B12 interaction is in the human body, humans don't naturally utilize nor produce any of this B12, rather the functional use of Folic Acid in the body is merely replaced by Cobalamin. As a result, many health effects of both deficiency of B12 or folic acid can be rectified by the addition of the other if need be. Deficiency of either B12 or folic acid can result in several neurological disorders and lack of motivation or onset of depression, which can possibly be related to the available energy production in cells being altered or slowed down due to this deficiency. However, too much Cobalamin in the human blood stream can potentially lead to several serious diseases, many effects of which are still being researched and less understood due to its homogenous behavior alongside other compounds already present in the human body, such as folic acid. Several of these diseases are both the cause of overutilization of Cobalamin, and the resulting effect of which, including several types of leukemia, resulting in the high levels of Cobalamin being stored in tissues. This large amount of stored Cobalamin being involved in the high presence of haptocorrin, from the corrin portion of the cobalamin structure, leading to several, some life threatening, liver diseases. Even though
The active site of the cobalt metal in Cobalamin possesses an octahedral configuration, forming primary sigma bonds with the transfer ligand, R, and the amine ring in the Corrin ring structure, while secondary pi-type bonding occurs in the three planar imidazole-type rings of the Corrin structure, and with the imidazole-type rings below axis, as shown above. This structural configuration allows a unique electron configuration along with nearby pi-system stabilization making the cobalt atom active site a preferred and semi-stable target for transferring the hydroxyl, methyl, cyano, and 5-deoxyadensosine in the various functions of its enzyme catalysis action. If only the three planar and one sub-axial ligand bonds from the nitrogens in the imidazole-like ring structures are taken as identical, and the sigma bonding amine group and 'R' active site group are taken as separate, the cobalt and its immediate ligand environment can be seen to possibly possess a Cs type symmetry with one mirror plane running through the R and amine ligands. Though the above Cs symmetry applies if only the simple immediate ligand environment is considered, because of the complexity in the structure of the actual surrounding Corrin ring and Nucleotide loop in the plane and below the active site of the imidazole-type structures can't in reality be taken as identical once the structure is extended beyond a couple bond lengths away. Though this may be the case, the immediate electron density of the supposed identical imidazole-type ring groups connected to the cobalt metal may be considered near identical enough to promote the rotational configuration of the 'R' group attached to be related to the amine ligand link instead. For the R groups being -OH, -CN, or Me this has no effect whatsoever on the bonding rotation as these ligands are considered symmetric along the attachment site, however the 5-deoxyadensosine link used extensively in the energy production in cells will be effected by this symmetry of the active site, and could potentially be one reason this Cobalamin performs well in the transfer of these groups. Cobalamin is an important compound used in the human body which is used in various configurations for specific tasks in enzymatic catalysis in subgroup transfer between systems, though not naturally utilized, it is more of a replacement for Folic Acid in the human body for these same functions, making it an important vitamin in sufficient, but not over extensive, quantities. | 10,536 | 3,066 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.00%3A_Prelude_to_Solutions |
If you watch any of the medical dramas on television, you may have heard a doctor (actually an actor) call for an intravenous solution of “Ringer’s lactate” (or “lactated Ringer’s” or "Ri-Lac"). So what is Ringer’s lactate? Intravenous (IV) solutions are administered for two main reasons: Many people with acute or long-term medical conditions have received some type of an IV solution. One basic IV solution, called a , is simply a dilute solution of NaCl dissolved in water. Normal saline is 9.0 g of NaCl dissolved in each liter of solution. is a normal saline solution that also has small amounts of potassium and calcium ions mixed in. In addition, it contains about 2.5 g of lactate ions (C H O ) per liter of solution. The liver metabolizes lactate ions into bicarbonate (HCO ) ions, which help maintain the acid-base balance of blood. Many medical problems, such as heart attacks and shock, affect the acid-base balance of blood, and the presence of lactate in the IV solution eases problems caused by this imbalance. Physicians can select from a range of premade IV solutions, in accordance with a patient’s particular needs. Ringer’s lactate is commonly used when a patient’s blood volume must be increased quickly. Another frequently used IV solution, called D5W, is a 5% solution of dextrose (a form of sugar) in water. | 1,347 | 3,067 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Structure_and_Bonding_in_Ethene-The_Pi_Bond |
Ethene is the formal IUPAC name for H C=CH , but it also goes by a common name: Ethylene. The name Ethylene is used because it is like an ethyl group (\(CH_2CH_3\)) but there is a double bond between the two carbon atoms in it. Ethene has the formula \(C_2H_4\) and is the simplest alkene because it has the fewest carbons (two) necessary for a carbon-carbon double bond. Bonding in carbon is covalent, containing either sigma or \(\pi\) bonds. Carbon can make single, double, or triple bonds. The number of bonds it makes determines the structure. With four single bonds, carbon has a tetrahedral structure, while with one double bond it's structure is trigonal planar, and with a triple bond it has a linear structure. A solitary carbon atom has four electrons, two in the 2s orbital, and one in each of the 2\(p_x\) and 2\(p_y\) orbitals, leaving the \(2p_z\) orbital empty. A single carbon atom can make up to four bonds, but by looking at its electron configuration this would not be possible because there are only two electrons available to bond with. The other two are in a lone pair state, making them much less reactive to another electron that is by itself. Well it is, in order to make the four bonds, the carbon atom promotes one of the 2s electrons into the empty \(2p_z\) orbital, leaving the carbon with four unpaired electrons allowing it to now form four bonds. The electron is not promoted spontaneously. It becomes promoted when a photon of light with the correct wavelength hits the carbon atom. When this photon hits the carbon atom it gives the atom enough energy to promote one of the lone pair electrons to the \(2p_z\) orbital. All the bonds in Ethene are covalent, meaning that they are all formed by two adjacent atoms sharing their valence electrons. As opposed to ionic bonds which hold atoms together through the attraction of two ions of opposite charges. Sigma bonds are created when there is overlap of similar orbitals, orbitals that are aligned along the inter-nuclear axis. Common sigma bonds are \(s+s\), \(p_z+p_z\) and \(s+p_z\), \(z\) is the axis of the bond on the xyz-plane of the atom. \(\pi\) bonds are created when there is adequate overlap of similar, adjacent \(p\) orbitals, such as \(p_x\)+\(p_x\) and \(p_y\)+\(p_y\). Each p orbital has two lobes, one usually indicated by a + and the other indicated by a - (sometimes one may be shaded while the other is not). This + and - (shaded, not shaded) are only meant to indicate the opposite phase \(\phi\) the wave functions, they indicate any type of electrical charge. For a \(\pi\) bond to form both lobes of the \(p\) orbital must overlap, + with + and - with -. When a + lobe overlaps with a - lobe this creates an anti-bonding orbital interaction which is much higher in energy, and therefore not a desirable interaction. Usually there can be no \(\pi\) bonds between two atoms without having at least one sigma bond present first. But there are special cases such as dicarbon (\(C_2\)) where the central bond is a \(\pi\) bond not a sigma bond, but in cases like these the two atoms want to have as much orbital overlap as possible so the bond lengths between the atoms are smaller than what is normally expected. The \(\pi\) bond in ethene is weak compared to the sigma bond between the two carbons. This weakness makes the \(\pi\) bond and the overall molecule a site of comparatively high chemical reactivity to an array of different substances. This is due to the high electron density in the \(\pi\) bond, and because it is a weak bond with high electron density the \(\pi\) bond will easily break in order to form two separate sigma bonds. Sites such as these are referred to as functional groups or functionalities. These groups have characteristic properties and they control the reactivity of the molecule as a whole. How these functional groups and other reactants form various products are an important concept in organic chemistry. Ethene is made up of four 1s Hydrogen atoms and two 2s 2\(p_x\) 2\(p_y\) carbon atoms. These carbon atoms already have four electrons, but they each want to get four more so that they have a full eight in the valence shell. Having eight valence electrons around carbon gives the atom itself the same electron configuration as neon, a noble gas. Carbon wants to have the same configuration as Neon because when it has eight valence electrons carbon is at its most stable, lowest energy state, it has all of the electrons that it wants, so it is no longer reactive. Ethene is not a very complicated molecule. It contains two carbon atoms that are double bonded to each other, with each of these atoms also bonded to two Hydrogen atoms. This forms a total of three bonds to each carbon atom, giving them an \(sp^2\) hybridization. Since the carbon atom is forming three sigma bonds instead of the four that it can, it only needs to hybridize three of its outer orbitals, instead of four. It does this by using the \(2s\) electron and two of the \(2p\) electrons, leaving the other unchanged. This new orbital is called an \(sp^2\) hybrid because that's exactly what it is, it is made from one s orbital and two p orbitals. When atoms are an \(sp^2\) hybrid they have a trigonal planar structure. These structures are very similar to a 'peace' sign, there is a central atom with three atoms around it, all on one plane. Trigonal planar molecules have an ideal bond angle of 120° on each side. The H-C-H bond angle is 117°, which is very close to the ideal 120° of a carbon with \(sp^2\) hybridization. The other two angles (H-C=C) are both 121.5°. There is rigidity in the Ethene molecule due to the double-bonded carbons. In Ethane there are two carbons that share a single bond, this allows the two Methyl groups to rotate with respect to each other. These different conformations result in higher and lower energy forms of Ethane. In Ethene there is no free rotation about the carbon-carbon sigma bond. There is no rotation because there is also a \(\pi\) bond along with the sigma bond between the two carbons. A \(\pi\) bond is only formed when there is adequate overlap between both top and bottom p-orbitals. In order for there to be free rotation the p-orbitals would have to go through a phase where they are 90° from each other, which would break the \(\pi\) bond because there would be no overlap. Since the \(\pi\) bond is essential to the structure of Ethene it must not break, so there can be not free rotation about the carbon-carbon sigma bond. 1. H CCH 2. 3. 4. C-H: 1.076 angstroms, C-C: 1.54 angstroms, C=C: 1.330 angstroms 5. the carbons cannot freely rotate about the carbon-carbon double bond because in order to rotate the p-orbitals would have to pass through a 90° point where there would no longer be any overlap, so the \(\pi\) bond would have to break for there to be free rotation. | 6,878 | 3,069 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Molecular_Geometry/Trigonal_Pyramidal_Molecular_Geometry |
An example of trigonal pyramid molecular geometry that results from tetrahedral electron pair geometry is NH . The nitrogen has 5 valence electrons and thus needs 3 more electrons from 3 hydrogen atoms to complete its octet. This then leaves a lone electron pair that is not bonded to any other atom. The three hydrogen atoms and the lone electron pair are as far apart as possible at nearly 109 bond angle. This is geometry. The lone electron pairs exerts a little extra repulsion on the three bonding hydrogen atoms to create a slight compression to a 107 bond angle.The molecule is trigonal pyramid molecular geometry because the lone electron pair, although still exerting its influence, is invisible when looking at molecular geometry. The molecule is three dimensional as opposed to the case which was a flat trigonal planar molecular geometry because it did not have a lone electron pair. In this example, H O , the Lewis diagram shows O at the center with one lone electron pair and three hydrogen atoms attached. Compare this with ammonia, NH , which also has a lone pair. Compare it to the which has 2 hydrogen atoms and 2 lone electron pairs.. The third hydrogen bonds to the water molecule as a hydrogen ion (no electrons) bonding to the lone pair on the oxygen. This shows tetrahedral geometry for the electron pair geometry and and trigonal pyramid the molecular geometry. Hydronium ion is a more accurate method to depict the hydrogen ion associated with acid properties of some molecules in water solution. In this example, SO , the Lewis diagram shows sulfur at the center with one lone electron pair. The sulfur and and one oxygen are bonded through a double bond which counts as "one electron pair". Hence the molecule has four electron pairs and is tetrahedral. The Lewis diagram is as follows:
S = 6 e-
O = 6e- x 3 = 18e-
2- charge = 2e-
Total electrons = 26 Sulfur atoms and all oxygen atoms have an octet of electrons. Sulfite and bisulfite ions are used as a preservative in wines. It is also found as a component of acid rain, formed by the interaction of sulfur dioxide and water molecules. | 2,147 | 3,070 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Properties_of_Alkenes/Nuclear_Magnetic_Resonance_(NMR)_of_Alkenes |
In the previous section, we learned about the physical properties of alkenes (Physical Properties of Alkenes). The double bonded carbons in alkene molecules also have an effect of shifts shown in H and C nuclear magnetic resonance spectr. For background information of H NMR, you can refer H Nuclear Magnetic Resonance from the last chapter. In H NMR spectrum, hydrogen atoms bound to a carbon consisting of a double bond (these hydrogens are called ) are typically found in low field of the NMR spectrum, which is the left side, and the hydrogens are said to be . The cause for this is due to the movement of the electrons in the pi bond of the carbon-carbon double bond. Alkenyl hydrogens create an external magnetic field that is perpendicular to the double bond axis and causes the electrons in the pi bond to enter a circular motion (shown in red). The circular motion actually reinforces the external field at the edge of the double bond on both sides of the pi bond but creates a local field (shown in purple and green) that opposes the external field in the center of the double bond. Because of this pulling force within the pi bond across the double bond which reinforces the regions occupied by alkenyl hydrogens, the alkenyl hydrogens are strongly deshielded. Additionally, alkenyl hydrogens do not have to be all chemical-shift equivalent, and when they aren't, coupling will be observed which is the different peaks in an MNR spectrum. Here are a couple of terms to know: When alkynel hydrogen atoms are on a double bonded carbon, the hydrogens of a cis and trans isomer will yield a different shift on the NMR spectrum. Because the coupling constant is smaller in a cis isomer than in a trans isomer, the NMR spectrums of the two isomers are different conveying the to-- the right of the spectrum-- and Sometimes coupling will lead to very complicated patterns as a result of the J values that vary widely due to the relationship between the hydrogens involved. When this occurs, information can still be derived to determine the structure of a molecule by looking at the number of signals, the chemical shift of each one, integration, and splitting patterns similarly to identifying alkane NMR. For background information on C NMR, please refer to C Nuclear Magnetic Resonance from the previous chapter. Compared to alkane carbons with one bond, alkene carbons show a relatively low field shift on the C NMR spectrum and absorb about 100 ppm lower field. Also, in broad-band decoupled C NMR, sp carbons absorb as sharp single lines so with these two methods, it is easy to determine the presence of a double bond in C NMR spectrum. Here are the common C Chemical Shift Ranges: Note that the carbon-carbon double bonds are found in the range between 100-170 ppm. Carbon atoms on alkenes that are attatched to another carbon group are found more downfield than carbon alkenes attatched to hydrogens. Remember from previous sections that to solve an NMR spectrum with double bonds, we must know the Degrees of Unsaturation. From this, we get degrees of unsaturation= (9-7)/2=1 so there is one pi bond or ring in our molecule. Next we must look at the integration of the NMR spectrums. There are three ways to attach the discovery we made, but only one of them are the correct answer. Since the coupling of the two alkene hydrogens are small whereas vicinal hydrogens tend to be large, we conclude that the hydrogens are geminal and appear on the same carbon. This leaves the two other groups to be located on the other alkene carbon. | 3,583 | 3,071 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Chromatography/Chromatographic_Columns |
Chromatography is an analytical technique that separates components in a mixture. Chromatographic columns are part of the instrumentation that is used in chromatography. Five chromatographic methods that use columns are (GC), (LC), Ion exchange chromatography (IEC), size exclusion chromatography (SEC), and chiral chromatography. The basic principals of chromatography can be applied to all five methods. In the mobile phase is a gas. Gas chromatographic columns are usually between 1 and 100 meters long. The liquid stationary phase is bonded or adsorbed onto the surface of an open tubular (capillary) column, or onto a packed solid support inside the column. Matching the polarities of the analyte and stationary phase is not an exact science. The two should have similar polarities. The thickness of the stationary phase ranges between 0.1 and 8 µm. The thicker the layer the more volatile the analyte can be. (HPLC) is a type of that uses a liquid moblie phase. The same basic principals from gas chromatography are applied to liquid chromatography. There are three basic types of liquid chromatographic columns: liquid-liquid, liquid-solid, and ion-exchange. Liquid-liquid chromatographic columns have the liquid stationary phase bonded or absorbed to the surface of the column, or packed material. liquid-liquid chromatographic columns are not as popular because they have limited stability and they are inconvenient. Partitioning occurs between the two different liquids of the mobile and stationary phases. In liquid-solid chromatographic columns the stationary phase is a solid and the analyte absorbs onto the stationary phase which separates the components of the mixture. In ion-exchange chromatographic columns the stationary phase is an ion-exchange resin and partitioning occurs with ion exchanges that occur between the analyte and stationary phase. Usually HPLC has a guard column ahead of the analytical column to protect and extend the life of the analytical column. The guard column removes particulate matter, contaminants, and molecules that bind irreversibly to the column. The guard column has a stationary phase similar to the analytical column. The most common HPLC columns are made from stainless steel, but they can be also made out of thick glass, polymers such as polyetherethelketone, a combination of stainless steel and glass, or a combination of stainless steel and polymers. Typical HPLC analytical columns are between 3 and 25 cm long and have a diameter of 1 to 5 mm. The columns are usually straight unlike GC columns. Particles that pack the columns have a typical diameter between 3 to 5 µm. Liquid chromatographic columns will increase in efficiency when the diameter of the packed particles inside the column decreases. HPLC columns are usually packed with pellicular, or porous particles. Pellicular particles are made from polymer, or glass beads. Pellicular particles are surrounded by a thin uniform layer of silica, polystyrene-divinyl-benzene synthetic resin, alumina, or other type of ion-exchange resin. The diameter of the pellicular beads is between 30 and 40 µm. Porous particles are more commonly used and have diameters between 3 to 10 µm. Porous particles are made up silica, polystyrene-divinyl-benzene synthetic resin, alumina, or other type of ion-exchange resin. Silica is the most common type of porous particle packing material. Partition HPLC uses liquid bonded phase columns, where the liquid stationary phase is chemically bonded to the packing material. The packing material is usually hydrolyzed silica which reacts with the bond-phase coating. Common bond phase coatings are siloxanes. The relative structure of the siloxane is shown in Figure \(\Page {1}\). A polar stationary phase and a non-polar mobile phase are used for normal phase HPLC. In normal phase, the most common R groups attached to the siloxane are: diol, amino, cyano, inorganic oxides, and dimethylamino. Normal phase is also a form of liquid-solid chromatography. The most non-polar compounds will elute first when doing normal phase HPLC. Reverse phase HPLC uses a polar mobile phase and a non-polar stationary phase. Reverse phase HPLC is the most common liquid chromatography method used. The R groups usually attached to the siloxane for reverse phase HPLC are: C , C ,or any hydrocarbon. Reverse phase can also use water as the mobile phase, which is advantageous because water is cheap, nontoxic, and invisible in the UV region. The most polar compounds will elute first when performing reverse phase HPLC. Check the animation on the principle of reversed-phase chromatography to understand its principle. Ion exchange columns are used to separate ions and molecules that can be easily ionized. Separation of the ions depends on the ion's affinity for the stationary phase, which creates an ion exchange system. The electrostatic interactions between the analytes, moble phase, and the stationary phase, contribute to the separation of ions in the sample. Only positively or negatively charged complexes can interact with their respective cation or anion exchangers. Common packing materials for ion exchange columns are amines, sulfonic acid, diatomaceous earth, styrene-divinylbenzene, and cross-linked polystyrene resins. Some of the first ion exchangers used were inorganic and made from aluminosilicates (zeolites). Although aluminosilicates are not widely used as ion exchange resins used. separate molecules based upon their size, not molecular weight. A common packing material for these columns is molecular sieves. Zeolites are a common molecular sieve that is used. The molecular sieves have pores that small molecules can go into, but large molecules cannot. This allows the larger molecules to pass through the column faster than the smaller ones. Other packing materials for size exclusion chromatographic columns are polysaccharides and other polymers, and silica. The pore size for size exclusion separations varies between 4 and 200 nm. are used to separate enantiomers. Separation of chiral molecules is based upon steriochemistry. These columns have a stationary phase that selectively interacts with one enantiomer over the other. These types of columns are very useful for separating racemic mixtures. Some Stationary Phases Used to Separate Enantiomer are show in Table \(\Page {2}\). Peak or band broadening causes the column to be less efficient. The ideal situation would to have sharp peaks that are resolved. The longer a substance stays in the column it will cause the peaks to widen. Lengthening the column is a way to improve the separation of different species in the column. A column usually needs to remain at a constant temperature to remain efficient. Plate height and number of theoretical plates determines the efficiency of the column. Improving the efficiency would be to increase the number of plates and decrease the plate height. The number of plates can be determined from the equation: \[N=L/H\] where L is the length of the column and H is the height of each plate. N can also be determined from the equation: \[N=16\left( \dfrac{t_R}{W}\right)^2 \] or \[N=5.54\left(\dfrac{t_R}{W_{1/2}}\right)^2\] where \(t_R\) is the retention time, \(W\) is the width of the peak and \(W_{1/2}\) is half the width of the peak. Height equivalent to a theoretical plate (HETP) is determined from the equation: \[H=L/N\] or HETP can also be determined by the : \[H=A+\dfrac{B}{u}+Cu\] where H equals HETP, A is the term for eddy diffusion, B is the term for longitudinal diffusion, C is the coefficient for mass-transfer between the stationary and mobile phases, and u is the linear velocity. The equation for HETP is often used to describe the efficiency of the column. An efficient column would have a minimum HETP value. Gas chromatographic columns have plate heights that are at least one order of magnitude greater than liquid chromatographic column plates. However GC columns are longer, which causes them to be more efficient. LC columns have a maximum length of 25 cm whereas GC columns can be 100 meters long. | 8,123 | 3,072 |
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Define . Give several examples of solutions. What is the difference between a solvent and a solute? Can a solution have more than one solute in it? Can you give an example? Does a solution have to be a liquid? Give several examples to support your answer. Give at least two examples of solutions found in the human body. Which substances will probably be soluble in water, a very polar solvent? Which substances will probably be soluble in toluene (C H CH ), a nonpolar solvent? The solubility of alcohols in water varies with the length of carbon chain. For example, ethanol (CH CH OH) is soluble in water in any ratio, while only 0.0008 mL of heptanol (CH CH CH CH CH CH CH OH) will dissolve in 100 mL of water. Propose an explanation for this behavior. Dimethyl sulfoxide [(CH ) SO] is a polar liquid. Based on the information in Exercise 9, which do you think will be more soluble in it—ethanol or heptanol? a homogeneous mixture A solvent is the majority component of a solution; a solute is the minority component of a solution. A solution does not have to be liquid; air is a gaseous solution, while some alloys are solid solutions (answers will vary). 9. Small alcohol molecules have strong polar intermolecular interactions, so they dissolve in water. In large alcohol molecules, the nonpolar end overwhelms the polar end, so they do not dissolve very well in water. 10. Ethanol is a smaller molecule. It will be more soluble in water than heptanol. What are some of the units used to express concentration? Distinguish between the terms and . % m/m, % m/v, ppm, ppb, molarity, and Eq/L (answers will vary) Solubility is typically a limit to how much solute can dissolve in a given amount of solvent. Concentration is the quantitative amount of solute dissolved at any concentration in a solvent. Define . Do all solutes have the same solubility? Explain why the terms or are of limited usefulness in describing the concentration of solutions. If the solubility of sodium chloride (NaCl) is 30.6 g/100 mL of H O at a given temperature, how many grams of NaCl can be dissolved in 250.0 mL of H O? If the solubility of glucose (C H O ) is 120.3 g/100 mL of H O at a given temperature, how many grams of C H O can be dissolved in 75.0 mL of H O? How many grams of sodium bicarbonate (NaHCO ) can a 25.0°C saturated solution have if 150.0 mL of H O is used as the solvent? If 75.0 g of potassium bromide (KBr) are dissolved in 125 mL of H O, is the solution saturated, unsaturated, or supersaturated? Calculate the mass/mass percent of a saturated solution of NaCl. Use the data from Table \(\Page {1}\) "Solubilities of Various Solutes in Water at 25°C (Except as Noted)", assume that masses of the solute and the solvent are additive, and use the density of H O (1.00 g/mL) as a conversion factor. Calculate the mass/mass percent of a saturated solution of MgCO Use the data from Table \(\Page {1}\) "Solubilities of Various Solutes in Water at 25°C (Except as Noted)", assume that masses of the solute and the solvent are additive, and use the density of H O (1.00 g/mL) as a conversion factor. Only 0.203 mL of C H will dissolve in 100.000 mL of H O. Assuming that the volumes are additive, find the volume/volume percent of a saturated solution of benzene in water. Only 35 mL of aniline (C H NH ) will dissolve in 1,000 mL of H O. Assuming that the volumes are additive, find the volume/volume percent of a saturated solution of aniline in water. A solution of ethyl alcohol (C H OH) in water has a concentration of 20.56% v/v. What volume of C H OH is present in 255 mL of solution? What mass of KCl is present in 475 mL of a 1.09% m/v aqueous solution? The average human body contains 5,830 g of blood. What mass of arsenic is present in the body if the amount in blood is 0.55 ppm? The Occupational Safety and Health Administration has set a limit of 200 ppm as the maximum safe exposure level for carbon monoxide (CO). If an average breath has a mass of 1.286 g, what is the maximum mass of CO that can be inhaled at that maximum safe exposure level? Which concentration is greater—15 ppm or 1,500 ppb? Express the concentration 7,580 ppm in parts per billion. What is the molarity of 0.500 L of a potassium chromate solution containing 0.0650 mol of K CrO ? What is the molarity of 4.50 L of a solution containing 0.206 mol of urea [(NH ) CO]? What is the molarity of a 2.66 L aqueous solution containing 56.9 g of NaBr? If 3.08 g of Ca(OH) is dissolved in enough water to make 0.875 L of solution, what is the molarity of the Ca(OH) ? What mass of HCl is present in 825 mL of a 1.25 M solution? What mass of isopropyl alcohol (C H O) is dissolved in 2.050 L of a 4.45 M aqueous C H O solution? What volume of 0.345 M NaCl solution is needed to obtain 10.0 g of NaCl? How many milliliters of a 0.0015 M cocaine hydrochloride (C H ClNO ) solution is needed to obtain 0.010 g of the solute? Aqueous calcium chloride reacts with aqueous silver nitrate according to the following balanced chemical equation: How many moles of AgCl(s) are made if 0.557 L of 0.235 M CaCl react with excess AgNO ? How many grams of AgCl are made? Sodium bicarbonate (NaHCO ) is used to react with acid spills. The reaction with sulfuric acid (H SO ) is as follows: If 27.6 mL of a 6.25 M H SO solution were spilled, how many moles of NaHCO would be needed to react with the acid? How many grams of NaHCO is this? The fermentation of glucose to make ethanol and carbon dioxide has the following overall chemical equation: If 1.00 L of a 0.567 M solution of C H O were completely fermented, what would be the resulting concentration of the C H OH solution? How many moles of CO would be formed? How many grams is this? If each mole of CO had a volume of 24.5 L, what volume of CO is produced? Aqueous sodium bisulfite gives off sulfur dioxide gas when heated: If 567 mL of a 1.005 M NaHSO solution were heated until all the NaHSO had reacted, what would be the resulting concentration of the Na SO solution? How many moles of SO would be formed? How many grams of SO would be formed? If each mole of SO had a volume of 25.78 L, what volume of SO would be produced? What is the concentration of a 1.0 M solution of K (aq) ions in equivalents/liter? What is the concentration of a 1.0 M solution of SO (aq) ions in equivalents/liter? A solution having initial concentration of 0.445 M and initial volume of 45.0 mL is diluted to 100.0 mL. What is its final concentration? A 50.0 mL sample of saltwater that is 3.0% m/v is diluted to 950 mL. What is its final mass/volume percent? 0.345 mol; 29.0 g Explain how the solvation process describes the dissolution of a solute in a solvent. Each particle of the solute is surrounded by particles of the solvent, carrying the solute from its original phase. Describe what happens when an ionic solute like Na SO dissolves in a polar solvent. Describe what happens when a molecular solute like sucrose (C H O ) dissolves in a polar solvent. Classify each substance as an electrolyte or a nonelectrolyte. Each substance dissolves in H O to some extent. Classify each substance as an electrolyte or a nonelectrolyte. Each substance dissolves in H O to some extent. Will solutions of each solute conduct electricity when dissolved? Will solutions of each solute conduct electricity when dissolved? Each ion of the ionic solute is surrounded by particles of solvent, carrying the ion from its associated crystal. 5. 6. What are the colligative properties of solutions? Explain how the following properties of solutions differ from those of the pure solvent: vapor pressure, boiling point, freezing point, and osmotic pressure. Colligative properties are characteristics that a solution has that depend on the number, not the identity, of solute particles. In solutions, the vapor pressure is lower, the boiling point is higher, the freezing point is lower, and the osmotic pressure is higher. In each pair of aqueous systems, which will have the lower vapor pressure? In each pair of aqueous systems, which will have the lower vapor pressure? In each pair of aqueous systems, which will have the higher boiling point? In each pair of aqueous systems, which will have the higher boiling point? Estimate the boiling point of each aqueous solution. The boiling point of pure water is 100.0°C. Estimate the freezing point of each aqueous solution. The freezing point of pure water is 0.0°C. Explain why salt (NaCl) is spread on roads and sidewalks to inhibit ice formation in cold weather. Salt (NaCl) and calcium chloride (\(\ce{CaCl2}\)) are used widely in some areas to minimize the formation of ice on sidewalks and roads. One of these ionic compounds is better, mole for mole, at inhibiting ice formation. Which is that likely to be? Why? What is the osmolarity of each aqueous solution? What is the osmolarity of each aqueous solution? 11. A 1.0 M solution of an unknown soluble salt has an osmolarity of 3.0 osmol. What can you conclude about the salt? 12. A 1.5 M NaCl solution and a 0.75 M Al(NO ) solution exist on opposite sides of a semipermeable membrane. Determine the osmolarity of each solution and the direction of solvent flow, if any, across the membrane. 3. 6. 7. NaCl lowers the freezing point of water, so it needs to be colder for the water to freeze. 9. 10. 11. It must separate into three ions when it dissolves. 12. Both NaCl and Al(NO ) have 3.0 osmol. There will be no net difference in the solvent flow. 1. What is chemical equilibrium? 2. What does the equilibrium constant tell us? 1. The rate of the forward reaction equals the rate of the reverse reaction. 2. The ratio of products and reactants when the system is at equilibrium. 1. If the reaction H + I ⇌2HI is at equilibrium, do the concentrations of HI, H , and I have to be equal? 2. Do the concentrations at equilibrium depend upon how the equilibrium was reached? 3. What does it mean if the K is > 1? 4. What does it mean if the K is < 1? 5. Does the equilibrium state depend on the starting concentrations? 6. Write an expression for the equilibrium constant K equation. a. PCl ⇄ PCl + Cl b. \(2\,O_{3}\,(g) \rightleftharpoons 3\,O_{2}\,(g)\) 7. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: 3C H (g)⟶C H (g). Which value of would make this reaction most useful commercially? Explain your answer. a. ≈ 0.01 b. ≈ 1 c. ≈ 10. 8. Tell whether the reactants or the products are favored at equilibrium: a. 2NH (g) ⇌ N (g) + 3H (g) = 172 b. 2SO (g) ⇌ 2SO (g) + O (g) = 0.230 c. 2NO(g) + Cl (g) ⇌ 2NOCl(g) = 4.6×10 d. N (g) + O (g) ⇌ 2NO(g) = 0.050 1. No, the concentrations are constant but the concentrations do not have to be equal. 2. No. 3. More products than reactants are present at equilibrium. 4. More reactants than products present at equilibrium. 5. No. The equilibrium ratio does not depend on the initial concentrations. 6. a. \(K=\dfrac{[PCl_3,Cl_2]}{[PCl_5]}\) b. \(K=\dfrac{[O_2]^3}{[O_3]^2}\) 7. The answer is c. ≈ 10. Since \(K=\dfrac{[C_6H_6]^2}{[C_2H_2]^3}\) (K≈10), this means that C H predominates over C H . In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable. 8. a. products b. reactants c. products d. reactants 1. Define Le Chatelier’s principle. 2. List the three factors types of changes that can disturb the equilibrium of a system. 1. Le Chatelier’s principle states that a system at equilibrium is disturbed, it will respond in a way to minimize the disturbance. 2. temperature, change in amount of substance, change in pressure through change in volume 1. How will each change affect the reaction? PCl ( ) + heat ⇌PCl ( ) + Cl ( ) 2. How will each change affect the reaction? HNO ( ) ⇌H ( ) + NO ( ) 3. How will each change affect the reaction? CO ( ) + C( ) ⇌2CO( ) ΔH=172.5kJ 4. How will each change affect the reaction? H ( ) + I ( ) ⇌2HI( ) ΔH=−9.48kJ 1. 2. 3. 4. 1. What are some of the features of a semipermeable membrane? 2. What do the prefixes hyper, hypo, and iso mean? 1. A semipermeable membrane allows some substances to pass through but not others. 2. hyper – higher; hypo – lower; iso - same 1. Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water. 2. Two solutions with different concentrations and compositions are separated by a semipermeable membrane. The left-hand solution is a .50 M solution of MgSO , while the right-hand solution contains CaCl at a concentration of .40 M. Determine the direction of the flow of solvent, left or right. 3. Given the following situations, wherein two tanks of different solutions are separated by a semipermeable membrane, determine the direction of the flow of solvent (water). a. Solution A contains a 0.40 M concentration of CaCl , while Solution B contains a 0.45 M concentration of KI b. Solution A contains a 1.00 M concentration of NH Cl, while Solution B contains a 1.00 M concentration of CH O 4. Cells are placed in a solution and the cells then undergo hemolysis. What can be said about the relative concentrations of solute in the cell and the solution? 5. Describe the relative concentrations inside and outside a red blood cell when crenation occurs. 6. A saltwater fish is placed in a freshwater tank. What will happen to the fish? Describe the flow of water molecules to explain the outcome. 7. What makes up the "head" region of a phospholipid? Is it hydrophobic or hydrophilic? 8. What makes up the "tail" region of a phospholipid? Is it hydrophobic or hydrophilic? 1. Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water. 2. Water (solvent) flows from left to right. 3. a. Water flows from Solution B to Solution A. b. Water flows from Solution B to Solution A. 4. Cells contain fluid with higher concentration than solution outside the cell. 5. Cells contain fluid with a lower concentration than the solution outside the cell. 6. Water molecules will flow from the tank water into the fish because the fish has a higher concentration of salt. If the fish absorbs too much water, it will die. 7. The "head" region is a phosphate group and it is hydrophilic. 8. The "tail" is a hydrocarbon tail and it is hydrophobic. , Ph.D. (Department of Chemistry, University of Kentucky) Calcium nitrate reacts with sodium carbonate to precipitate solid calcium carbonate: \[Ca(NO_3)_{2(aq)} + Na_2CO_{3(aq)} \rightarrow CaCO_{3(s)} + NaNO_{3(aq)}\] The compound HCl reacts with sodium carbonate to generate carbon dioxide gas: \[HCl_{(aq)} + Na_2CO_{3(aq)} \rightarrow H_2O_{(ℓ)} + CO_{2(g)} + NaCl_{(aq)}\] Estimate the freezing point of concentrated aqueous HCl, which is usually sold as a 12 M solution. Assume complete ionization into H and Cl ions. Estimate the boiling point of concentrated aqueous H SO , which is usually sold as an 18 M solution. Assume complete ionization into H and HSO ions. Seawater can be approximated by a 3.0% m/m solution of NaCl in water. Determine the molarity and osmolarity of seawater. Assume a density of 1.0 g/mL. Human blood can be approximated by a 0.90% m/m solution of NaCl in water. Determine the molarity and osmolarity of blood. Assume a density of 1.0 g/mL. How much water must be added to 25.0 mL of a 1.00 M NaCl solution to make a resulting solution that has a concentration of 0.250 M? Sports drinks like Gatorade are advertised as capable of resupplying the body with electrolytes lost by vigorous exercise. Find a label from a sports drink container and identify the electrolytes it contains. You should be able to identify several simple ionic compounds in the ingredients list. Occasionally we hear a sensational news story about people stranded in a lifeboat on the ocean who had to drink their own urine to survive. While distasteful, this act was probably necessary for survival. Why not simply drink the ocean water? (Hint: See Exercise 5 and Exercise 6 above. What would happen if the two solutions in these exercises were on opposite sides of a semipermeable membrane, as we would find in our cell walls?) | 16,341 | 3,073 |
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Hydrogen and iodine gases react to form hydrogen iodide according to the following reaction: \[\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightleftharpoons 2 \ce{HI} \left( g \right) \nonumber \] \[\begin{align} &\text{Forward reaction:} \: \: \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right) \\ &\text{Reverse reaction:} \: \: 2 \ce{HI} \left( g \right) \rightarrow \ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \end{align} \nonumber \] Initially, only the forward reaction occurs because no \(\ce{HI}\) is present. As soon as some \(\ce{HI}\) has formed, it begins to decompose back into \(\ce{H_2}\) and \(\ce{I_2}\). Gradually, the rate of the forward reaction decreases while the rate of the reverse reaction increases. Eventually the rate of combination of \(\ce{H_2}\) and \(\ce{I_2}\) to produce \(\ce{HI}\) becomes equal to the rate of decomposition of \(\ce{HI}\) into \(\ce{H_2}\) and \(\ce{I_2}\). When the rates of the forward and reverse reactions have become equal to one another, the reaction has achieved a state of balance. is the state of a system in which the rate of the forward reaction is equal to the rate of the reverse reaction. Chemical equilibrium can be attained whether the reaction begins with all reactants and no products, all products and no reactants, or some of both. The figure below shows changes in concentration of \(\ce{H_2}\), \(\ce{I_2}\), and \(\ce{HI}\) for two different reactions. In the reaction depicted by the graph on the left (A), the reaction begins with only \(\ce{H_2}\) and \(\ce{I_2}\) present. There is no \(\ce{HI}\) initially. As the reaction proceeds towards equilibrium, the concentrations of the \(\ce{H_2}\) and \(\ce{I_2}\) gradually decrease, while the concentration of the \(\ce{HI}\) gradually increases. When the curve levels out and the concentrations all become constant, equilibrium has been reached. At equilibrium, concentrations of all substances are constant. In reaction B, the process begins with only \(\ce{HI}\) and no \(\ce{H_2}\) or \(\ce{I_2}\). In this case, the concentration of \(\ce{HI}\) gradually decreases while the concentrations of \(\ce{H_2}\) and \(\ce{I_2}\) gradually increase until equilibrium is again reached. Notice that in both cases, the relative position of equilibrium is the same, as shown by the relative concentrations of reactants and products. The concentration of \(\ce{HI}\) at equilibrium is significantly higher than the concentrations of \(\ce{H_2}\) and \(\ce{I_2}\). This is true whether the reaction began with all reactants or all products. The position of equilibrium is a property of the particular reversible reaction and does not depend upon how equilibrium was achieved. It may be tempting to think that once equilibrium has been reached, the reaction stops. Chemical equilibrium is a dynamic process. The forward and reverse reactions continue to occur even after equilibrium has been reached. However, because the rates of the reactions are the same, there is no change in the relative concentrations of reactants and products for a reaction that is at equilibrium. The conditions and properties of a system at equilibrium are summarized below. The description of equilibrium in this concept refers primarily to equilibrium between reactants and products in a chemical reaction. Other types of equilibrium include phase equilibrium and solution equilibrium. A phase equilibrium occurs when a substance is in equilibrium between two states. For example, a stoppered flask of water attains equilibrium when the rate of evaporation is equal to the rate of condensation. A solution equilibrium occurs when a solid substance is in a saturated solution. At this point, the rate of dissolution is equal to the rate of recrystallization. Although these are all different types of transformations, most of the rules regarding equilibrium apply to any situation in which a process occurs reversibly. Red blood cells transport oxygen to the tissues so they can function. In the absence of oxygen, cells cannot carry out their biochemical responsibilities. Oxygen moves to the cells attached to hemoglobin, a protein found in the red cells. In cases of carbon monoxide poisoning, \(\ce{CO}\) binds much more strongly to the hemoglobin, blocking oxygen attachment and lowering the amount of oxygen reaching the cells. Treatment involves the patient breathing pure oxygen to displace the carbon monoxide. The equilibrium reaction shown below illustrates the shift toward the right when excess oxygen is added to the system: \[\ce{Hb(CO)_4} \left( aq \right) + 4 \ce{O_2} \left( g \right) \rightleftharpoons \ce{Hb(O_2)_4} \left( aq \right) + 4 \ce{CO} \left( g \right) \nonumber \] Consider the hypothetical reversible reaction in which reactants \(\ce{A}\) and \(\ce{B}\) react to form products \(\ce{C}\) and \(\ce{D}\). This equilibrium can be shown below, where the lowercase letters represent the coefficients of each substance. \[a \ce{A} + b \ce{B} \rightleftharpoons c \ce{C} + d \ce{D} \nonumber \] As we have established, the rates of the forward and reverse reactions are the same at equilibrium, and so the concentrations of all of the substances are constant. Since that is the case, it stands to reason that a ratio of the concentration for any given reaction at equilibrium maintains a constant value. The is the ratio of the mathematical product of the products of a reaction to the mathematical product of the concentrations of the reactants of the reaction. Each concentration is raised to the power of its coefficient in the balanced chemical equation. For the general reaction above, the equilibrium constant expression is written as follows: \[K_\text{eq} = \frac{\left[ \ce{C} \right]^c \left[ \ce{D} \right]^d}{\left[ \ce{A} \right]^a \left[ \ce{B} \right]^b} \nonumber \] The concentrations of each substance, indicated by the square brackets around the formula, are measured in molarity units \(\left( \text{mol/L} \right)\). The value of the equilibrium constant for any reaction is only determined by experiment. As detailed in the above section, the position of equilibrium for a given reaction does not depend on the starting concentrations and so the value of the equilibrium constant is truly constant. It does, however, depend on the temperature of the reaction. This is because equilibrium is defined as a condition resulting from the rates of forward and reverse reactions being equal. If the temperature changes, the corresponding change in those reaction rates will alter the equilibrium constant. For any reaction in which a \(K_\text{eq}\) is given, the temperature should be specified. The equilibrium constant can vary over a wide range of values. Because products are in the numerator of the equilibrium constant expression and reactants are in the denominator, values of K greater than \(10^3\) indicate a strong tendency for reactants to form products. In this case, chemists say that equilibrium lies to the right as written, favoring the formation of products. An example is the reaction between \(H_2\) and \(Cl_2\) to produce \(HCl\), which has an equilibrium constant of \(1.6 \times 10^{33}\) at 300 K. Because \(H_2\) is a good reducing agent and \(Cl_2\) is a good oxidizing agent, the reaction proceeds essentially to completion. \(H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)}\) \(K\)=\(1.6 \times 10^{33}\) at 300K In contrast, values of \(K\) less than \(10^{-3}\) indicate that the ratio of products to reactants at equilibrium is very small. That is, reactants do not tend to form products readily, and the equilibrium lies to the left as written, favoring the formation of reactants. An example is the decomposition of water. No wonder you won’t find water a very good source of oxygen gas and hydrogen gas at ordinary temperatures! \(H_2O_{(g)} \rightleftharpoons H_{2(g)} + ½ O_{2(g)}\) \(K\)=\(8 \times 10^{-41}\) at 25°C Figure \(\Page {4}\) summarizes the relationship between the magnitude of K and the relative concentrations of reactants and products at equilibrium for a general reaction, written as reactants \(\rightleftharpoons\) products. A large value of the equilibrium constant \(K\) means that products predominate at equilibrium; a small value means that reactants predominate at equilibrium. Write the expression for the equilibrium constant K for the following reaction: N + 3 H ⇄ 2 NH The equilibrium constant equation is the ratio of the concentration of the products (NH ) to the concentration of the reactants (N and H ), each raised to the power of its coefficient in the balanced chemical equation. \(K=\dfrac{[NH_3]^2}{[N_2,H_2]^3}\) Write the expression of the equilibrium constant K for the following reaction:. 2 NOCl ⇄ 2 NO + Cl \(K=\dfrac{[NO]^2[Cl_2]}{[NOCl]^2}\) Predict which systems at equilibrium will (a) contain essentially only products, (b) contain essentially only reactants, and (c) contain appreciable amounts of both products and reactants. Use the value of the equilibrium constant to determine whether the equilibrium mixture will contain essentially only products, essentially only reactants, or significant amounts of both. Hydrogen and nitrogen react to form ammonia according to the following balanced chemical equation: \[3H_{2(g)}+N_{2(g)} \rightleftharpoons 2NH_{3(g)} \nonumber \] Values of the equilibrium constant at various temperatures were reported as a. 327°C, where \(K\) is smallest b. 25°C The reaction quotient, \(Q\), is used when questioning if we are at equilibrium. The calculation for \(Q\) is the same as for \(K\) but we can only use \(K\) when we know we are at equilibrium. Comparing \(Q\) and \(K\) allows the direction of the reaction to be predicted. , Ph.D. (Department of Chemistry, University of Kentucky) | 9,861 | 3,074 |
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Based on the thermodynamic criterion for equilibrium, it is possible to draw some conclusions about the state variables \(p\) and \(T\) and how they are related along phase boundaries. First, the chemical potentials of the two phases \(\alpha\) and \(\beta\) in with one another must be equal. \[\mu_{\alpha} = \mu_{\beta} \label{eq1} \] Also, any infinitesimal changes to the chemical potential of one phase must be offset by an infinitesimal change to the chemical potential of the other phase that is equal in magnitude. \[ \mu_{\alpha} + d\mu_{\alpha} = \mu_{\beta}+ d\mu_{\beta} \label{eq2} \] Taking the difference between these Equations \ref{eq1} and \ref{eq2} shows that \[ d\mu_{\alpha} = d\mu_{\beta} \nonumber \] And since \(d\mu\) can be expressed in terms of molar volume and molar entropy \[d\mu = Vdp - SdT \nonumber \] It is clear that there will be constraints placed on changes of temperature and pressure while maintaining equilibrium between the phases. \[V_{\alpha} dP - S_{\alpha} dT = V_{\beta} dP - S_{\beta} dT \nonumber \] Gathering pressure terms on one side and temperature terms on the other \[ (V_{\alpha} - V_{\beta} ) dP = (S_{\alpha} - S_{\beta}) dT \nonumber \] The differences \(V_{\alpha} - V_{\beta}\) and \(S_{\alpha} - S_{\beta}\) are the changes in molar volume and molar entropy for the phase changes respectively. So the expression can be rewritten \[\Delta V dp = \Delta S dT \nonumber \] or \[\dfrac{dp}{dT} = \dfrac{\Delta S}{\Delta V} \label{clap1} \] Equation \ref{clap1} is the . This expression makes it easy to see how the phase diagram for water is qualitatively different than that for most substances. Specifically, the negative slope of the solid-liquid boundary on a pressure-temperature phase diagram for water is very unusual, and arises due to the fact that for water, the molar volume of the liquid phase is smaller than that of the solid phase. Given that for a phase change \[\Delta S_{phase} = \dfrac{\Delta H_{phase}}{T} \nonumber \] the Clapeyron equation is sometimes written \[\dfrac{dp}{dT} = \dfrac{\Delta H}{T \Delta V} \label{clap2} \] Calculate the magnitude of the change in freezing point for water (\(\Delta H_{fus} = 6.009\, kJ/mol\)) and the density of ice is \(\rho_{ice} = 0.9167\, g/cm^3\) while that for liquid water is \(\rho_{liquid} = 0.9999\, g/cm^3\)) for an increase in pressure of \(1.00\, atm\) at \(273\, K\). The molar volume of ice is given by \[ \left( 0.9167 \, \dfrac{g}{cm^3} \right) \left(\dfrac{1\,mol}{18.016\, g} \right)\left(\dfrac{1000\,cm^3}{1\, L} \right) = 50.88 \, \dfrac{L}{mol} \nonumber \] The molar volume of liquid water at 0 C is given by \[ \left( 0.9999 \, \dfrac{g}{cm^3} \right) \left(\dfrac{1\,mol}{18.016\, g} \right)\left(\dfrac{1000\,cm^3}{1\, L} \right) = 55.50 \, \dfrac{L}{mol} \nonumber \] So \(\Delta V\) for the phase change of \(\text{solid} \rightarrow \text{liquid}\) (which corresponds to an endothermic change) is \[ 50.88 \, \dfrac{L}{mol} - 55.50 \, \dfrac{L}{mol} = -4.62 \, \dfrac{L}{mol} \nonumber \] To find the change in temperature, use the (Equation \ref{clap2}) and separating the variables \[dp = \dfrac{\Delta H_{fus}}{\Delta V} \dfrac{dt}{T} \nonumber \] Integration (with the assumption that \(\Delta H_{fus}/\Delta V\) does not change much over the temperature range) yields \[\int_{p1}^{p2} dp = \dfrac{\Delta H_{fus}}{\Delta V} \int_{T1}^{T2}\dfrac{dt}{T} \nonumber \] \[p_2-p_1 = \Delta p = \dfrac{\Delta H_{fus}}{\Delta V} \ln \left( \dfrac{T_2}{T_1} \right) \nonumber \] or \[ T_2 = T_1\, \text{exp} \left(\dfrac{\Delta V \Delta p}{\Delta H_{fus}} \right) \nonumber \] so \[T_2 = (273\,K) \, \text{exp} \left(\dfrac{(1\, atm)\left(-4.62 \, \dfrac{L}{mol} \right) }{6009 \dfrac{J}{mol} } \underbrace{\left( \dfrac{8.314\,J}{0.08206 \, atm\,L} \right)}_{\text{conversion factor}} \right) \nonumber \] \[ T_2 = 252.5\,K \nonumber \] \[\Delta T = T_2-T_1 = 252.5\,K - 273\,K = -20.5 \,K \nonumber \] So the melting point will decrease by 20.5 K. Note that the phase with the smaller molar volume is favored at the higher pressure (as expected from Le Chatelier's principle)! | 4,143 | 3,076 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/05%3A_The_Second_Law/5.08%3A_Adiabatic_Compressibility |
In Chapter 4, we learned about the isothermal compressibility, \(\kappa_T\), which is defined as \[ \kappa_T = - \dfrac{1}{V} \left(\dfrac{\partial V}{\partial p} \right)_T \nonumber \] \(\kappa_T\) is a very useful quantity, as it can be measured for many different substances and tabulated. Also, as we will see in the next chapter, it can be used to evaluate several different partial derivatives involving thermodynamic variables. In his seminal work, (Newton, 1723), Isaac Newton (1643 - 1727) (Doc) calculated the speed of sound through air, assuming that sound was carried by isothermal compression waves. His calculated value of 949 m/s was about 15% smaller than experimental determinations. He accounted for the difference by pointing to “non-ideal effects”. But it turns out that his error, albeit an understandable one (since sound waves do not appear to change bulk air temperatures) was that the compression waves are adiabatic, rather than isothermal. As such, there are small temperature oscillations that occur due to the adiabatic compression followed by expansion of the gas carrying the sound waves. The oversight was correct by Pierre-Simon Laplace (1749 – 1827) (O'Connor & Robertson, Pierre-Simon Laplace, 1999). LaPlace modeled the compression waves using the , \(\kappa_S\) defined by \[ \kappa_S =- \dfrac{1}{V} \left(\dfrac{\partial V}{\partial p} \right)_S \nonumber \] Since the entropy is defined by \[ dS = \dfrac{dq_{rev}}{T} \nonumber \] it follows that any adiabatic pathway (\(dq = 0\)) is also (\(dS = 0\)), or proceeds at constant entropy. Adiabatic pathways are also isentropic. A couple of interesting conclusions can be reached by following the derivation of an expression for the speed of sound where the sound waves are modeled as adiabatic compression waves. We can begin by expanding the description of \(\kappa_S\) by using Partial Derivative Transformation Type II. Applying this, the adiabatic compressibility can be expressed \[ \kappa_S =\dfrac{1}{V} \left(\dfrac{\partial V}{\partial S} \right)_p \left(\dfrac{\partial S}{\partial p} \right)_V \nonumber \] or by using transformation type I \[ \kappa_S =\dfrac{1}{V} \dfrac{ \left(\dfrac{\partial S}{\partial p }\right)_V}{ \left(\dfrac{\partial S}{\partial V} \right)_p} \nonumber \] Using a simple chain rule, the partial derivatives can be expanded to get something a little easier to evaluate: \[ \kappa_S =\dfrac{1}{V} \dfrac{ \left(\dfrac{\partial S}{\partial T }\right)_V \left(\dfrac{\partial T}{\partial p }\right)_V }{ \left(\dfrac{\partial S}{\partial T} \right)_p \left(\dfrac{\partial T}{\partial V }\right)_p} \label{eq10} \] The utility here is that \[\left(\dfrac{\partial S}{\partial T }\right)_V = \dfrac{C_V}{T} \label{Note1} \] \[\left(\dfrac{\partial S}{\partial T }\right)_p = \dfrac{C_p}{T} \label{Note2} \] This means that Equation \ref{eq10} simplifies to \[ \kappa_S = \dfrac{C_V}{C_p} \left( \dfrac{1}{V} \dfrac{ \left(\dfrac{\partial T}{\partial p }\right)_V }{ \left(\dfrac{\partial T}{\partial V }\right)_p} \right) \nonumber \] Simplifying what is in the parenthesis yields \[ \kappa_S = \dfrac{C_V}{C_p} \left( \dfrac{1}{V} \left(\dfrac{\partial T}{\partial p }\right)_V \left(\dfrac{\partial V}{\partial T }\right)_p \right) \nonumber \] \[ \kappa_S = \dfrac{C_V}{C_p} \left( - \dfrac{1}{V} \left(\dfrac{\partial V}{\partial p }\right)_T \right) \nonumber \] \[ \kappa_S = \dfrac{C_V}{C_p} \kappa_T \nonumber \] As will be shown in the next chapter, \(C_p\) is always bigger than \(C_V\), so \(\kappa_S\) is always smaller than \(\kappa_T\). But there is more! We can use this methodology to revisit how pressure affects volume along an adiabat. In order to do this, we would like to evaluate the partial derivative \[ \left(\dfrac{\partial V}{\partial p }\right)_S \nonumber \] This can be expanded in the same way as above \[ \left(\dfrac{\partial V}{\partial p }\right)_S = - \dfrac{ \left(\dfrac{\partial V}{\partial S}\right)_p }{ \left(\dfrac{\partial p}{\partial S }\right)_V } \nonumber \] And further expand \[ \left(\dfrac{\partial V}{\partial p }\right)_S = - \dfrac{ \left(\dfrac{\partial V}{\partial T}\right)_p \left(\dfrac{\partial T}{\partial S}\right)_p}{ \left(\dfrac{\partial p}{\partial T}\right)_V \left(\dfrac{\partial T}{\partial S}\right)_V} \label{eq20} \] And as before, noting that the relationships in Equations \ref{Note1} and \ref{Note2}, Equation \ref{eq20} can be simplified to \[ \left(\dfrac{\partial V}{\partial p }\right)_S= - \dfrac{C_V}{C_p} \left(\dfrac{\partial V}{\partial T}\right)_p \left(\dfrac{\partial T}{\partial p}\right)_V \nonumber \] \[ = \dfrac{C_V}{C_p} \left(\dfrac{\partial V}{\partial p}\right)_T \label{eq22} \] Or defining \(\gamma = C_p/C_V\), Equation \ref{eq22} can be easily rearranged to \[ \gamma \left(\dfrac{\partial V}{\partial p}\right)_S = \left(\dfrac{\partial V}{\partial p}\right)_T \nonumber \] The right-hand derivative is easy to evaluate if we assume a specific equation of state. For an ideal gas, \[ \left(\dfrac{\partial V}{\partial p}\right)_T = - \dfrac{nRT}{p^2} = - \dfrac{V}{p} \nonumber \] Substitution yields \[ \gamma \left(\dfrac{\partial V}{\partial p}\right)_S = - \dfrac{V}{p} \nonumber \] which is now looking like a form that can be integrated. Separation of variables yields \[ \gamma \dfrac{dV}{V} = \dfrac{dP}{p} \nonumber \] And integration (assuming that is independent of volume) yields \[ \gamma \int_{V_1}^{V_2} \dfrac{dV}{V} = \int_{p_1}^{p_2} \dfrac{dP}{p} \nonumber \] or \[ \gamma \ln \left( \dfrac{V_2}{V_1} \right) =\ln \left( \dfrac{p_2}{p_1} \right) \nonumber \] which is easily manipulated to show that \[p_1V_1^{\gamma} = p_2V_2^{\gamma} \nonumber \] or \[pV^{\gamma} = \text{constant} \nonumber \] which is what we previously determined for the behavior of an ideal gas along an adiabat. Finally, it should be noted that the correct expression for the speed of sound is given by \[v_{sound} = \sqrt{\dfrac{1}{\rho \, \kappa_S}} \nonumber \] where \(\rho\) is the density of the medium. For an ideal gas, this expression becomes \[v_{sound} = \sqrt{\dfrac{\gamma RT}{M}} \nonumber \] where \(M\) is the molar mass of the gas. Isaac Newton’s derivation, based on the idea that sound waves involved isothermal compressions, would produce a result which is missing the factor of \(\gamma\), accounting for the systematic deviation from experiment which he observed. | 6,438 | 3,077 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Electrophilic_Addition_Reactions_of_Alkynes |
Addition of strong acids to alkynes is quite similar to the addition of strong acids to alkenes. The regiochemistry follows , but the stereochemistry is often different. Addition to alkenes is usually not stereospecific, whereas alkynes usually undergo addition. Strong acids usually protonate alkenes on their less substituted carbon thus developing a positive charge on the more substituted carbon of the reacting double bond. This intermediate carbocation may rearrange or simply pick up a nucleophile to form the expected addition product. The situation with alkynes is different. Similar protonation of the \(\pi\) bond of an alkyne would produce a very unstable vinyl cation so this does not happen in most cases. A vinyl cation can only have one alkyl group adjacent to the vacant p-orbital to stabilize its positive charge so the stability of the vinyl cation is similar to the mono-substituted primary alkyl cation (See below). As a result, a strong acid can form a \(\pi\) complex with the alkyne \(\pi\) bond but proton transfer to the alkyne is extremely difficult. The strong acid, however, does polarize the \(\pi\) bond developing positive charge at both carbons of the \(\pi\) bond. The more substituted carbon holds most of the positive charge and becomes the site of attack by any nucleophile present in solution. The nucleophile bonds to the more substituted carbon allowing the acid to protonate the less substituted carbon of the \(\pi\) bond. In reactions, a nucleophile attacks an electron poor species. Now we will take a look at electrophilic addition reactions, particularly of alkynes. The reaction mechanisms, as you will notice, are quite similar to the electrophilic addition reactions of alkenes. The triple bonds of alkynes, because of its high electron density, are easily attacked by electrophiles, but less reactive than alkenes due to the compact C-C electron cloud.As with electrophilic addition to unsymmetrical alkenes, the Markovnikov rule is followed, adding the electrophile to the less substituted carbon. Here we will go through the following reactions listed below to learn the mechanisms behind these electrophilic additions of alkynes: (1) HX Addition to , (2) Halogenation of and (3) Hydration of Alkynes. Figure 1 The addition of an electrophile to either an alkene or an alkyne will undergo the same steps listed below. Make Note: The way to determine where the addition of the proton and the halide takes place is based on Markovnikov’s Rule, which states that the proton adds onto the carbon with the most hydrogens and the halogen prefers the most substituted carbon. Now let's take a look at our first reaction, the addition of Hydrogen Halide. Reactivity order of : HI > HB r> HCl > HF. Follows Markovnikov’s rule: As described in Figure 1, the \(\pi\) electrons will attack the hydrogen of the HBr and because this is a symmetric molecule it does not matter which carbon it adds to, but in an asymmetric molecule the hydrogen will covalently bond to the carbon with the most hydrogens. Once the hydrogen is covalently bonded to one of the carbons, you will get a carbocation intermediate (not shown, but will look the same as depicted in Figure 1) on the other carbon. Again, this is a symmetric molecule and if it were asymmetric, which carbon would have the positive charge? The final step is the addition of the Bromine, which is a good nucleophile because it has electrons to donate or share. Bromine, therefore attacks the carbocation intermediate placing it on the highly substituted carbon. As a result, you get 2-bromobutene from your 2-butyne reactant, as shown below. Now, what if you have excess HBr? Here, the electrophilic addition proceeds with the same steps used to achieve the product in Addition of a HX to an Internal Alkyne. The \(\pi\) electrons attacked the hydrogen, adding it to the carbon on the left (shown in blue). Why was hydrogen added to the carbon on left and the one on the right bonded to the Bromine? Now, you will have your carbocation intermediate, which is followed by the attack of the Bromine to the carbon on the right resulting in a haloalkane product. The \(\pi\) electrons are attacking the hydrogen, depicted by the electron pushing arrows and the Bromine gains a negative charge. The carbocation intermediate forms a positive charge on the left carbon after the hydrogen was added to the carbon with the most hydrogen substituents. The Bromine, which has a negative charge, attacks the positively charged carbocation forming the final product with the nucleophile on the more substituted carbon. First, you see the polarized Br being attacked by the \(\pi\) electrons. Once you form the C-Br bond, the other bromine is released as a bromine ion. The intermediate here is a bromonium ion, which is electrophilic and reacts with the bromine ion giving you the dibromo product. With the addition of water, alkynes can be hydrated to form enols that spontaneously tautomerize to ketones. Reaction is catalyzed by mercury ions. Follows Markovnikov’s Rule: Terminal alkynes give methyl ketones Now let's look at some Hydration Reactions. Just as described in Figure 7 the \(\pi\) electrons will attack a proton, forming a carbocation, which then gets attacked by the nucleophilic water molecules. After deprotination, we generate an enol, which then tautomerizes into the ketone form shown. As you can see here, the \(\pi\) electrons of the triple bond are attacking the proton, which forms a covalent bond on the carbon with the most hydrogen substituents. Once the hydrogen is bound you have a carbocation, which gets attacked by the water molecule. Now you have a positive charge on the oxygen which results in a base coming in and deprotinating the molecule. Once deprotonated, you have an enol, which then gets tautomerized. Tautomerism is shown here when the proton gets attacked by the double bond \(\pi\) electrons forming a covalent bond between the carbon and the hydrogen on the less substituted carbon. Electrons from the Oxygen end up moving to the carbon, forming a double bond with carbon and giving itself a positive charge, which then gets attacked by the base. The base deprotonates the oxygen resulting in the more stable final product at equilibrium, which is a ketone. What product would result from the reaction of 1-Pentyne with Hydrogen Bromide? Take your answer from 1a and add an excess of Hydrogen Bromide. What would be the product? What is the mechanism for this reaction? What product would result from the reaction of 1-Pentyne with Br ? Take your answer from 2a and add an excess of Br . What would be the product? What is the mechanism for this reaction? If 1-Pentyne were to react with mercury(I) sulfate, water, and sulfuric acid, what would be the product? What is the product when 3-methylbutyne reacts with HCl? Include in your answer:
a) the reaction mechanism
b) a clear explanation for why the hydrogen and chlorine bind to where they do
a) Draw the enol form of the Ketone below. b) Draw the Keto form from the enol below. | 7,101 | 3,078 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/05%3A_The_Second_Law/5.E%3A_The_Second_Law_(Exercises) |
What is the minimum amount of work needed to remove 10.0 J of energy from a freezer at -10.0 °C, depositing the energy into a room that is 22.4 °C? Consider the isothermal, reversible expansion of 1.00 mol of a monatomic ideal gas (C = 3/2 R) from 10.0 L to 25.0 L at 298 K. Calculate \(q\), \(w\), \(\Delta U\), \(\Delta H\), and \(\Delta S\) for the expansion. Consider the isobaric, reversible expansion of 1.00 mol of a monatomic ideal gas (C = 5/2 R) from 10.0 L to 25.0 L at 1.00 atm. Calculate \(q\), \(w\), \(\Delta U\), \(\Delta H\), and \(\Delta S\) for the expansion. Consider the isochoric, reversible temperature increase of 1.00 mol of a monatomic ideal gas (C = 3/2 R) °Ccupying 25.0 L from 298 K to 345 K. Calculate \(q\), \(w\), \(\Delta U\), \(\Delta H\), and \(\Delta S\) for this process. Consider the adiabatic expansion of 1.00 mol of a monatomic ideal gas (C = 3/2 R) from 10.0 L at 273 K to a final volume of 45.0 L. Calculate \(\Delta T\), \(q\), \(w\), \(\Delta U\), \(\Delta H\), and \(\Delta S\) for the expansion. 15.0 g of ice (\(\Delta H_{fus} = 6.009\, kJ/mol\)) at 0 °C sits in a room that is at 21 °C. The ice melts to form liquid at 0 °C. Calculate the entropy change for the ice, the room, and the universe. Which has the largest magnitude? 15.0 g of liquid water (C = 75.38 J mol °C ) at 0 °C sits in a room that is at 21 °C. The liquid warms from 0 °C to 21 °C. Calculate the entropy change for the liquid, the room, and the universe. Which has the largest magnitude? Calculate the entropy change for taking 12.0 g of H O from the solid phase (C = 36.9 J mol K ) at -12.0 °C to liquid (C = 75.2 J mol K ) at 13.0 °C. The enthalpy of fusion for water is \(\Delta H_{fus} = 6.009 \,kJ/mol\). Using , calculate the standard reaction entropies (\(\Delta S^o\)) for the following reactions at 298 K. 1.00 mole of an ideal gas is taken through a cyclic process involving three steps: 2.00 moles of a monatomic ideal gas (C = 3/2 R) initially exert a pressure of 1.00 atm at 300.0 K. The gas undergoes the following three steps, all of which are reversible: \[p = a+bT\] where \(a\) and \(b\) are constants. Sketch the cycle on a pressure-temperature plot, and calculate \(\ U\) and \(\ S\) for each of the legs. Are \(\Delta U\) and \(\Delta S\) zero for the sum of the three legs? A 10.0 g piece of iron (C = 0.443 J/g °C) initially at 97.6 °C is placed in 50.0 g of water (C = 4.184 J/g °C) initially at 22.3 °C in an insulated container. The system is then allowed to come to thermal equilibrium. Assuming no heat flow to or from the surroundings, calculate Considers a crystal of \(CHFClBr\) as having four energetically equivalent orientations for each molecule. What is the expected residual entropy at 0 K for 2.50 mol of the substance? A sample of a certain solid is measured to have a constant pressure heat capacity of 0.436 J mol K at 10.0 K. Assuming the Debeye extrapolation model \[ C_p(T) = aT^3\] holds at low temperatures, calculate the molar entropy of the substance at 12.0 K. | 3,056 | 3,079 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_Lab_Techniques_(Nichols)/05%3A_Distillation/5.05%3A_Steam_Distillation/5.5C%3A_Separation_Theory |
Although steam distillation appears almost identical to "regular" distillation, the principles behind the separation of components are quite different. In a "regular" distillation, separation is attempted on a mixture of components that dissolve in one another. The vapor produced from these mixtures can be described by a combination of and , as shown in Equation \ref{14} for a two-component mixture. \[\text{Miscible components:} \: \: \: \: \: P_\text{solution} =P_A + P_B = P_A^o \chi_A + P_B^o \chi_B \label{14}\] When the components in the distilling flask do dissolve in one another, such as when water and nonpolar organic compounds are present, the vapor produced from these mixtures is different. The components act independently from one another (which makes sense considering they do not mix), and the partial pressure from each component is no longer determined by its mole fraction. The partial pressure of each component is simply its vapor pressure, and the vapor composition for a two-component mixture is described by Equation \ref{15}. Although the components do not mix in the liquid phase, they do in the gas phase, which allows for co-distillation of "incompatible components". \[\text{Immiscible components:} \: \: \: \: \: P_\text{solution} = P_A^o + P_B^o \label{15}\] The implications of Equation \ref{15} are several. First, since mole fraction is not a factor, it is possible that a minor component in the distilling flask can be a major component in the distillate if it has an appreciable vapor pressure. In the steam distillation of volatile plant materials, this means that the distillate composition is independent of the quantity of water or steam used in the distilling flask. Secondly, since the vapor pressures of each component add, the . For example, at \(100^\text{o} \text{C}\) water has a vapor pressure of \(760 \: \text{torr}\) since it is at its normal boiling point. If another volatile, non-water-soluble component was present with the water in a distilling flask at \(100^\text{o} \text{C}\) (for example toluene, which has a boiling point of \(111^\text{o} \text{C}\)), it too would produce vapors that contributed to the total pressure. As boiling occurs when the pressure matches the atmospheric pressure (let's say \(760 \: \text{torr}\) for the sake of this calculation), boiling would occur below \(100^\text{o} \text{C}\). For example, a mixture of toluene and water boils at \(85^\text{o} \text{C}\), as shown in Equation (16). \[\text{Water/toluene mix at } 85^\text{o} \text{C}: \: \: \: \: \: \begin{align} P_\text{solution} &= P_\text{water}^o + P_\text{toluene}^o \\ &= \left( 434 \: \text{torr} \right) + \left( 326 \: \text{torr} \right) = 760 \: \text{torr} \end{align} \label{16}\] The distilling temperature in steam distillation is always below \(100^\text{o} \text{C}\) (the boiling point of water), although in many cases the distilling temperature is very near or just under \(100^\text{o} \text{C}\). This feature allows for plant essential oils (complex mixtures that often include components with very high boiling points, \(> 250^\text{o} \text{C}\)), to be extracted at lower temperatures than their normal boiling points, and thus without decomposition. | 3,249 | 3,080 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Substitution_Reactions_of_Benzene_Derivatives |
Substituted rings are divided into two groups based on the type of the substituent that the ring carries: Examples of activating groups in the relative order from the most activating group to the least activating: -NH , -NR > -OH, -OR> -NHCOR> -CH and other alkyl groups with R as alkyl groups (C H ) Examples of deactivating groups in the relative order from the most deactivating to the least deactivating: -NO , -CF > -COR, -CN, -CO R, -SO H > Halogens with R as alkyl groups (C H ) The order of reactivity among Halogens from the more reactive (least deactivating substituent) to the least reactive (most deactivating substituent) halogen is: F> Cl > Br > I The order of reactivity of the benzene rings toward the electrophilic substitution when it is substituted with a halogen groups, follows the order of electronegativity. The ring that is substituted with the most electronegative halogen is the most reactive ring ( less deactivating substituent ) and the ring that is substituted with the least electronegatvie halogen is the least reactive ring ( more deactivating substituent ), when we compare rings with halogen substituents. Also the size of the halogen effects the reactivity of the benzene ring that the halogen is attached to. As the size of the halogen increase, the reactivity of the ring decreases. The activating group directs the reaction to the ortho or para position, which means the electrophile substitute the hydrogen that is on carbon 2 or carbon 4. The deactivating group directs the reaction to the meta position, which means the electrophile substitute the hydrogen that is on carbon 3 with the exception of the halogens that is a deactivating group but directs the ortho or para substitution. Resonance effect is the conjugation between the ring and the substituent, which means the delocalizing of the \(\pi\) electrons between the ring and the substituent. Inductive effect is the withdraw of the sigma ( the single bond ) electrons away from the ring toward the substituent, due to the higher of the substituent compared to the carbon of the ring. When the substituents like -OH have an unshared pair of electrons, the resonance effect is stronger than the inductive effect which make these substituents stronger activators, since this resonance effect direct the electron toward the ring. In cases where the subtituents is esters or amides, they are less activating because they form resonance structure that pull the electron density away from the ring. By looking at the mechanism above, we can see how groups donating electron direct the ortho, para electrophilic substition. Since the electrons locatinn transfer between the ortho and para carbons, then the electrophile prefer attacking the carbon that has the free electron. Inductive effect of alkyl groups activates the direction of the ortho or para substitution, which is when s electrons gets pushed toward the ring. The deactivating groups deactivate the ring by the inductive effect in the presence of an electronegative atom that withdraws the electrons away from the ring. we can see from the mechanism above that when there is an electron withdraw from the ring, that leaves the carbons at the ortho, para positions with a positive charge which is unfavorable for the electrophile, so the electrophile attacks the carbon at the meta positions. Halogens are an exception of the deactivating group that directs the ortho or para substitution. The halogens deactivate the ring by inductive effect not by the resonance even though they have an unpaired pair of electrons. The unpaired pair of electrons gets donated to the ring, but the inductive effect pulls away the electrons from the ring by the electronegativity of the halogens. The reaction of a substituted ring with an activating group is faster than benzene. On the other hand, a substituted ring with a deactivated group is slower than benzene. Activating groups speed up the reaction because of the resonance effect. The presence of the unpaired electrons that can be donated to the ring, stabilize the carbocation in the transition state. Thus; stabilizing the intermediate step, speeds up the reaction; and this is due to the decrease of the activating energy. On the other hand, the deactivating groups, withdraw the electrons away from the carbocation formed in the intermediate step, thus; the activation energy is increased which slows down the reaction. 2. Which nitration product is going to form faster? nitration of aniline or nitration of nitrobenzene? 3. Predict the product of the following two sulfonation reactions: A. 4. Classify these two groups as activating or deactivating groups: A. alcohol B. ester 5. By which effect does trichloride effect a monosubstituted ring? 1. The first substitution is going to be ortho and/or para substitution since we have a halogen subtituent. The second substition is going to be ortho and/or para substitution also since we have an alkyl substituent. 2. The nitration of aniline is going to be faster than the of nitrobenzene, since the aniline is a ring with NH substituent and nitrobenzene is a ring with NO substiuent. As described above NH is an activating group which speeds up the reaction and NO is deactivating group that slows down the reaction. 3. A. the product is B. the product is 4. A. alcohol is an activating group. B. ester is a deactivating group. 5. Trichloride deactivate a monosubstitued ring by inductive effect. | 5,482 | 3,081 |
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A great many processes involve proton transfer, or acid-base types of reactions. As many biological systems depend on carefully controlled pH, these types of processes are extremely important. The pH is defined by \[ pH \equiv -\log a(\ce{H^{+}}) \approx -\log [\ce{H^{+}}] \label{eq1} \] where \(a\) is the activity of hydronium ions and \([\ce{H^{+}}]\) is the true concentration of hydronium ions (both in mol/L). The dissociation of a weak acid in water is governed by the equilibrium defined by \[HA(aq) \rightleftharpoons H^+(aq) + A^-(aq) \nonumber \] The equilibrium constant for such a reaction, \(K_a\), takes the form \[K_a = \dfrac{[H^+,A^-]}{[HA]} \label{eq3} \] As is the case for all thermodynamic equilibrium constants, the concentrations are replaced by activities and the equilibrium constant is unitless. However, if all species behave ideally (have unit activity coefficients) the units can be used as a very useful guide in solving problems. What is the pH of a 0.200 M HOAc (acetic acid) solution? (K = 1.8 x 10 M) An ICE table will come in very handy here! The equilibrium problem can then be set up as \[ K_a = \dfrac{[H^+,OAc^-]}{[HOAc]} \nonumber \] Substituting the values that are known \[ 1.8 \times 10^{-5} =\dfrac{ x^2}{0.200 \, M -x} \nonumber \] This produces a quadratic equation, and thus two values of \(x\) which satisfy the relationship. \[x_1 = -0.001906 \,M \nonumber \] \[x_2 = 0.001888 \,M \nonumber \] The negative root is not physically meaningful since the concentrations of \(H^+\) and \(OAc^-\) cannot be negative. Using the value of \(x_2\) as \([H^+]\), the pH is then calculated (via Equation \ref{eq1}) to be \[ pH \approx -\log_{10} (0.001888) = 2.72 \nonumber \] Water is a very important solvent as water molecules have large dipole moments which create favorable interactions with ionic compounds. Water also has a large dielectric constant which damps the electric field generated by ions in solutions, making the comparative interactions with water more favorable than with other ions in solution in many cases. But water also dissociates into ions through the reaction \[H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq) \label{eqA} \] The equilibrium constant governing this dissociation is highly temperature dependent. The data below are presented by Bandura and Lvov (Bandura & Lvov, 2006) From these data, a van’t Hoff plot can be constructed. There is some curvature to the line, suggesting some (albeit small) temperature dependence for \(\Delta H_{rxn}\) for Equation \ref{eqA}. However, from the fit of these data, a value of \(\Delta H_{rxn}\) can be determined to be 52.7 kJ/mol. Of particular note is that the dissociation is , so increases in temperature will lead to a greater degree of dissociation. What is the pH of neutral water at 37 °C (normal human body temperature)? Neutral water no excess of \([H^+]\) over \([OH^-]\) or vice versa. From the best-fit line in the van’t Hoff plot of Figure \(\Page {1}\), the value of \(K_w\) can be calculated: \[ \ln (K_w) = - \dfrac{6338 \,K}{310\,K} - 11.04 \nonumber \] \[ K_w= 2.12 \times 10^{-14}\,M^2 \nonumber \] Since \(K_w\) gives the product of \([H^+]\) and \([OH^-]\) (which must be equal in a neutral solution), \[H^+] = \sqrt{2.12 \times 10^{-14}\,M^2} = 1.456 \times 10^{-7}\,M \nonumber \] And the pH is given by (Equation \ref{eq1}): \[ pH = -\log_{10} (1.456 \times 10^{-7}) = 6.84 \nonumber \] : This is slightly less than a pH of 7.00, which is normally considered to be “neutral.” But a pH of 7.00 is only neutral at 25 °C! At higher temperatures, neutral pH is a lower value due to the endothermic nature of the auto-ionization water. While it has a nigher \([H^+]\) concentration, it also has a higher \([OH^-]\) and at the same level, so it is still technically neutral. Hydrolysis is defined as a reaction with water that splits a water molecule. The hydrolysis of a weak base defines the equilibrium constant K . \[A^- + H_2O \rightleftharpoons HA + OH^- \nonumber \] For this reaction, the equilibrium constant \(K_b\) is given by \[ K_b = \dfrac{[HA,OH^-]}{[A^-]} \nonumber \] The concentration (or activity) of the pure compound H O is not included in the equilibrium expression because, being a pure compound in its standard state, it has unit activity throughout the process of establishing equilibrium. Further, it should be noted that when K is combined with the expression for K for the weak acid HA (Equation \ref{eq3}), \[K_a K_b = \left( \dfrac{[H^+,A^-]}{[HA]} \right) \left( \dfrac{[HA,OH^-]}{[A^-]} \right) = [H^+,OH^-] = K_w \nonumber \] As a consequence, if one knows \(K_a\) for a weak acid, one also knows \(K_b\) for its conjugate base, since the product results in \(K_w\). What is the pH of a 0.150 M solution of KF? (For HF, pK = 3.17 at 25 °C) The problem involves the hydrolysis of the conjugate base of HF, F . The hydrolysis reaction is \[F^- + H_2O \rightleftharpoons HF + OH^- \nonumber \] An ICE table is in order here. So the expression for \(K_b\) is \[ K_b = \dfrac{K_w}{K_a} = \dfrac{1.0 \times 10^{-14} M^2}{10^{3.17} M} = \dfrac{x^2}{1.50 \,M-x} \nonumber \] In this case, the small value of \(K_b\) insures that the value of x will be negligibly small compared to 0.150 M. In this limit, the value of \(x\) (which is equal to [OH ]) \[ x = [OH^-] = 1.49 \times 10^{-6}\,M \nonumber \] So \([H^+]\) is given by \[ [H^+] =\dfrac{K_w}{[OH^-]} = \dfrac{10^{-14} M^2}{1.49 \times 10^{-6} M} = 6.71 \times 10^{-9} \,M \nonumber \] And the pH is given by (Equation \ref{eq1}): \[ pH = -\log_{10} (6.71 \times 10^{-9}) = 8.17 \nonumber \] : The pH of this salt solution is slightly basic. This is to be expected as KF can be thought of being formed in the reaction of a weak acid (HF) with a strong base (KOH). In the competition to control the pH, the strong base ends up winning the battle. | 5,881 | 3,082 |
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In addition to polymers, other materials, such as ceramics, high-strength alloys, and composites, play a major role in almost every aspect of our lives. Until relatively recently, steel was used for any application that required an especially strong and durable material, such as bridges, automobiles, airplanes, golf clubs, and tennis rackets. In the last 15 to 20 years, however, graphite or boron fiber golf clubs and tennis rackets have made wood and steel obsolete for these items. Likewise, a modern jet engine now is largely composed of Ti and Ni by weight rather than steel ( ). The percentage of iron in wings and fuselages is similarly low, which indicates the extent to which other materials have supplanted steel. The Chevrolet Corvette introduced in 1953 was considered unusual because its body was constructed of fiberglass, a composite material, rather than steel; by 1992, Jaguar fabricated an all-aluminum limited-edition vehicle. In fact, the current models of many automobiles have engines that are made mostly of aluminum rather than steel. In this section, we describe some of the chemistry behind three classes of contemporary materials: ceramics, superalloys, and composites. A ceramic is any nonmetallic, inorganic solid that is strong enough for use in structural applications. Traditional ceramics, which are based on metal silicates or aluminosilicates, are the materials used to make pottery, china, bricks, and concrete. Modern ceramics contain a much wider range of components and can be classified as either , which are based on metal oxides such as alumina (Al O ), zirconia (ZrO ), and beryllia (BeO), or , which are based on metal carbides such as silicon carbide (carborundum, SiC) and tungsten carbide (WC), or nitrides like silicon nitride (Si N ) and boron nitride (BN). All modern ceramics are hard, lightweight, and stable at very high temperatures. Unfortunately, however, they are also rather brittle, tending to crack or break under stresses that would cause metals to bend or dent. Thus a major challenge for materials scientists is to take advantage of the desirable properties of ceramics, such as their thermal and oxidative stability, chemical inertness, and toughness, while finding ways to decrease their brittleness to use them in new applications. Few metals can be used in jet engines, for example, because most lose mechanical strength and react with oxygen at the very high operating temperatures inside the engines (approximately 2000°C). In contrast, ceramic oxides such as Al O cannot react with oxygen regardless of the temperature because aluminum is already in its highest possible oxidation state (Al ). Even nonoxide ceramics such as silicon and boron nitrides and silicon carbide are essentially unreactive in air up to about 1500°C. Producing a high-strength ceramic for service use involves a process called sintering, which fuses the grains into a dense and strong material ( ). Ceramics are hard, lightweight, and able to withstand high temperatures, but they are also brittle. One of the most widely used raw materials for making ceramics is clay. Clay minerals consist of hydrated alumina (Al O ) and silica (SiO ) that have a broad range of impurities, including barium, calcium, sodium, potassium, and iron. Although the structures of clay minerals are complicated, they all contain layers of metal atoms linked by oxygen atoms. Water molecules fit between the layers to form a thin film of water. When hydrated, clays can be easily molded, but during high-temperature heat treatment, called , a dense and strong ceramic is produced. Because ceramics are so hard, they are easily contaminated by the material used to grind them. In fact, the ceramic often grinds the metal surface of the mill almost as fast as the mill grinds the ceramic! The sol-gel process was developed to address this problem. In this process, a water-soluble precursor species, usually a metal or semimetal alkoxide [M(OR) ] undergoes a hydrolysis reaction to form a cloudy aqueous dispersion called a . The sol contains particles of the metal or semimetal hydroxide [M(OH) ], which are typically 1–100 nm in diameter. As the reaction proceeds, molecules of water are eliminated from between the M(OH) units in a condensation reaction, and the particles fuse together, producing oxide bridges, M–O–M. Eventually, the particles become linked in a three-dimensional network that causes the solution to form a , similar to a gelatin dessert. Heating the gel to 200°C–500°C causes more water to be eliminated, thus forming small particles of metal oxide that can be amazingly uniform in size. This chemistry starts with highly pure SiCl and proceeds via the following reactions starting with the alkoxide formation
\[ SiCl_{4}\left ( s \right )+4CH_{3}CH_{2}OH\left ( l \right )+4NH_{3}\left ( g \right ){\rightarrow}SiO\left ( OCH_{2}CH_{3} \right )_{4}\left ( s \right )+4NH_{4}Sl\left ( s \right ) \label{12.8.1} \] and then the hydrolysis of the alkoxide
\[ SiO\left ( OCH_{2}CH_{3} \right )_{4}\left ( s \right )+4H_{2}O\left ( l \right ) {\rightarrow} \left ( HO \right )_{3}Si-OH\left ( s \right )+ 4CH_{3}CH_{2}OH\left ( aq \right ) \label{12.8.2} \] ending with the condensation \[ \left ( HO_{3} \right )Si-OH\left ( s \right )+nHO-Si\left ( OH \right )_{3}\left ( s \right )\rightarrow \left ( HO_{3} \right )Si\left ( -O-Si\left ( OH \right )_{3} \right )_{n}\left ( s \right )+nH_{2}O\left ( l \right ) \ \label{12.8.3} \] Nature uses the same process to create opal gemstones. Superalloys are high-strength alloys, often with a complex composition, that are used in systems requiring mechanical strength, high surface stability (minimal flaking or pitting), and resistance to high temperatures. The aerospace industry, for example, requires materials that have high strength-to-weight ratios to improve the fuel efficiency of advanced propulsion systems, and these systems must operate safely at temperatures greater than 1000°C. Superalloys are used in systems requiring mechanical strength, minimal flaking or pitting, and high-temperature resistance. Although most superalloys are based on nickel, cobalt, or iron, other metals are used as well. Pure nickel or cobalt is relatively easily oxidized, but adding small amounts of other metals (Al, Co, Cr, Mo, Nb, Ti, and W) results in an alloy that has superior properties. Consequently, most of the internal parts of modern gas turbine jet engines are now made of superalloys based on either nickel (used in blades and disks) or cobalt (used in vanes, combustion chamber liners, and afterburners). The cobalt-based superalloys are not as strong as the nickel-based ones, but they have excellent corrosion resistance at high temperatures. Other alloys, such as aluminum–lithium and alloys based on titanium, also have applications in the aerospace industry. Because aluminum–lithium alloys are lighter, stiffer, and more resistant to fatigue at high temperatures than aluminum itself, they are used in engine parts and in the metal “skins” that cover wings and bodies. Titanium’s high strength, corrosion resistance, and lightweight properties are equally desirable for applications where minimizing weight is important (as in airplanes). Unfortunately, however, metallic titanium reacts rapidly with air at high temperatures to form TiN and TiO . The welding of titanium or any similar processes must therefore be carried out in an argon or inert gas atmosphere, which adds significantly to the cost. Initially, titanium and its alloys were primarily used in military applications, but more recently, they have been used as components of the airframes of commercial planes, in ship structures, and in biological implants. Composite materials have at least two distinct components: the matrix (which constitutes the bulk of the material) and fibers or granules that are embedded within the matrix and limit the growth of cracks by pinning defects in the bulk material ( ). The resulting material is stronger, tougher, stiffer, and more resistant to corrosion than either component alone. Composites are thus the nanometer-scale equivalent of reinforced concrete, in which steel rods greatly increase the mechanical strength of the cement matrix, and are extensively used in the aircraft industry, among others. For example, the Boeing 777 is 9% composites by weight, whereas the newly developed Boeing 787 is 50% composites by weight. Not only does the use of composite materials reduce the weight of the aircraft, and therefore its fuel consumption, but it also allows new design concepts because composites can be molded. Moreover, by using composites in the Boeing 787 multiple functions can be integrated into a single system, such as acoustic damping, thermal regulation, and the electrical system. Three distinct types of composite material are generally recognized, distinguished by the nature of the matrix. These are polymer-matrix composites, metal-matrix composites, and ceramic-matrix composites. Composites are stronger, tougher, stiffer, and more resistant to corrosion than their components alone. Fiberglass is a polymer-matrix composite that consists of glass fibers embedded in a polymer, forming tapes that are then arranged in layers impregnated with epoxy. The result is a strong, stiff, lightweight material that is resistant to chemical degradation. It is not strong enough, however, to resist cracking or puncturing on impact. Stronger, stiffer polymer-matrix composites contain fibers of carbon (graphite), boron, or polyamides such as Kevlar. High-tech tennis rackets and golf clubs as well as the skins of modern military aircraft such as the “stealth” F-117A fighters and B-2 bombers are made from both carbon fiber–epoxy and boron fiber–epoxy composites. Compared with metals, these materials are 25%–50% lighter and thus reduce operating costs. Similarly, the space shuttle payload bay doors and panels are made of a carbon fiber–epoxy composite. The structure of the Boeing 787 has been described as essentially one giant macromolecule, where everything is fastened through cross-linked chemical bonds reinforced with carbon fiber. Metal-matrix composites consist of metals or metal alloys reinforced with fibers. They offer significant advantages for high-temperature applications but pose major manufacturing challenges. For example, obtaining a uniform distribution and alignment of the reinforcing fibers can be difficult, and because organic polymers cannot survive the high temperatures of molten metals, only fibers composed of boron, carbon, or ceramic (such as silicon carbide) can be used. Aluminum alloys reinforced with boron fibers are used in the aerospace industry, where their strength and lightweight properties make up for their relatively high cost. The skins of hypersonic aircraft and structural units in the space shuttle are made of metal-matrix composites. Ceramic-matrix composites contain ceramic fibers in a ceramic matrix material. A typical example is alumina reinforced with silicon carbide fibers. Combining the two very high-melting-point materials results in a composite that has excellent thermal stability, great strength, and corrosion resistance, while the SiC fibers reduce brittleness and cracking. Consequently, these materials are used in very high-temperature applications, such as the leading edge of wings of hypersonic airplanes and jet engine parts. They are also used in the protective ceramic tiles on the space shuttle, which contain short fibers of pure SiO mixed with fibers of an aluminum–boron–silicate ceramic. These tiles are excellent thermal insulators and extremely light (their density is only about 0.2 g/cm ). Although their surface reaches a temperature of about 1250°C during reentry into Earth’s atmosphere, the temperature of the underlying aluminum alloy skin stays below 200°C. An engineer is tasked with designing a jet ski hull. What material is most suited to this application? Why? design objective most suitable material Determine under what conditions the design will be used. Then decide what type of material is most appropriate. A jet ski hull must be lightweight to maximize speed and fuel efficiency. Because of its use in a marine environment, it must also be resistant to impact and corrosion. A ceramic material provides rigidity but is brittle and therefore tends to break or crack under stress, such as when it impacts waves at high speeds. Superalloys provide strength and stability, but a superalloy is probably too heavy for this application. Depending on the selection of metals, it might not be resistant to corrosion in a marine environment either. Composite materials, however, provide strength, stiffness, and corrosion resistance; they are also lightweight materials. This is not a high-temperature application, so we do not need a metal-matrix composite or a ceramic-matrix composite. The best choice of material is a polymer-matrix composite with Kevlar fibers to increase the strength of the composite on impact. In designing a new generation of space shuttle, National Aeronautics and Space Administration (NASA) engineers are considering thermal-protection devices to protect the skin of the craft. Among the materials being considered are titanium- or nickel-based alloys and silicon-carbide ceramic reinforced with carbon fibers. Why are these materials suitable for this application? Ti- or Ni-based alloys have a high strength-to-weight ratio, resist corrosion, and are safe at high temperatures. Reinforced ceramic is lightweight; has high thermal and oxidative stability; and is chemically inert, tough, and impact resistant. are nonmetallic, inorganic solids that are typically strong; they have high melting points but are brittle. The two major classes of modern ceramics are ceramic oxides and nonoxide ceramics, which are composed of nonmetal carbides or nitrides. The production of ceramics generally involves pressing a powder of the material into the desired shape and sintering at a temperature just below its melting point. The necessary fine powders of ceramic oxides with uniformly sized particles can be produced by the . are new metal phases based on cobalt, nickel, or iron that exhibit unusually high temperature stability and resistance to oxidation. consist of at least two phases: a matrix that constitutes the bulk of the material and fibers or granules that act as a reinforcement. have reinforcing fibers embedded in a polymer matrix. have a metal matrix and fibers of boron, graphite, or ceramic. use reinforcing fibers, usually also ceramic, to make the matrix phase less brittle. Can a compound based on titanium oxide qualify as a ceramic material? Explain your answer. What features make ceramic materials attractive for use under extreme conditions? What are some potential drawbacks of ceramics? How do composite materials differ from the other classes of materials discussed in this chapter? What advantages do composites have versus other materials? How does the matrix control the properties of a composite material? What is the role of an additive in determining the properties of a composite material? | 15,212 | 3,083 |
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Alkynes can be a useful functional group to synthesize due to some of their antibacterial, antiparasitic, and antifungal properties. One simple method for alkyne synthesis is by double elimination from a dihaloalkane. One case in which elimination can occur is when a haloalkane is put in contact with a nucleophile. The table below is used to determine which situations will result in elimination and the formation of a \(\pi\) bond. To synthesize alkynes from dihaloalkanes we use dehydrohalogenation. The majority of these reactions take place using alkoxide bases (other strong bases can also be used) with high temperatures. This combination results in the majority of the product being from the E2 mechanism. Recall that the E2 mechanism is a concerted reaction (occurs in 1 step). However, in this 1 step there are 3 different changes in the molecule. This is the reaction between 2-Bromo-2-methylpropane and sodium hyrdroxide. Figure: Step 1: the base (blue) will deprotonate the haloalkane; Step 2: the leaving group (red) will depart from the molecule; Step 3: the deprotonated carbon will rehybridize from \(sp^3\) to \(sp^2\) via the formation of a \(pi\) bond. This is a brief review of the E2 reaction. For further information on why the reaction proceeds as it does visit . Now, if we apply this concept using 2 halides on vicinal or geminal carbons, the E2 reaction will take place twice resulting in the formation of 2 Pi bonds and thus an Alkyne. This is a general picture of the reaction taking place without any of the mechanisms shown. or Lets look at the mechanism of a reaction between 2,3-Dibromopentane with sodium amide in liquid ammonia. Lastly, we will briefly look at how to prepare alkynes from alkenes. This is a simple process using first halogenation of the alkene bond to form the dihaloalkane, and next, using the double elimination process to protonate the alkane and from the 2 Pi bonds. This first process is gone over in much greater detail in the page on . In general, chlorine or bromine is used with an inert halogenated solvent like chloromethane to create a vicinal dihalide from an alkene. The vicinal dihalide formed is the reactant needed to produce the alkyene using double elimination, as covered previously on this page. Due to the strong base and high temperatures needed for this reaction to take place, the triple bond may change positions. An example of this is when reactants that should form a terminal alkyne, form a 2-Alkyne instead. The use of NaNH in liquid NH is used in order to prevent this from happening due to its lower reacting temperature. Even so, most chemists will prefer to use nucleophilic substitution instead of elimination when trying to form a terminal alkyne. Remember that hydrogen atoms on terminal alkynes make the alkyne acidic. One of the base molecules will pull off the terminal hydrogen instead of one of the halides like we want. a.) 1-Pentyne b.) 1-Pentyne Bromine or chlorine can be used with different inert solvents for the halogenation. This can be done using many different bases. Liquid ammonia is used as a solvent and needs to be followed by an aqueous work-up. | 3,181 | 3,084 |
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One of the most useful aspects of theory only becomes apparent when we consider . According to molecular-orbital theory, occupy orbitals which are . That is, the orbitals spread over the entire molecule. An electron is not confined only to the vicinity of one or two atomic nuclei as in the lone pairs and bond pairs of valence-bond theory. As seen in , there are some molecules, such as O and C H , which cannot be described adequately by a single . In such cases valence-bond theory resorts to resonance hybrids, such as for ozone. The same molecules can be handled by MO theory without the need for several contributing structures, because electrons can occupy orbitals belonging to the molecule as a whole. As an example of this we shall apply MO theory to the pi orbitals in ozone. The sigma bonds in this molecule may be attributed to overlap of hybrid orbitals on each of the three oxygen atoms. You will recall that hybrids are directed toward the corners of an equilateral triangle, in reasonable agreement with the 117° angle in ozone. For the sigma bonds and lone pairs, then, we have the Lewis diagram This involves only 14 of the 18 valence electrons of the three oxygens, and so 4 electrons are left for pi bonding. Furthermore, each oxygen has a orbital perpendicular to the plane of its hybrids (that is, perpendicular to the plane of the three oxygen atoms), which has not yet been used for bonding. These three remaining orbitals can overlap sideways to form molecular orbitals, which are included in the jmol shown here. Using the scroll bar, you may select which molecular orbital to display for the ozone molecule. In order to demonstrate delocalized electrons, we will focus on three of the molecular orbitals shown. The first of these orbitals to highlight is seen by selecting the orbital labeled in the scroll menu. This orbital shows a concentrated electron density between the central oxygen and each of the other two, and is therefore a bonding MO. The second to focus on is orbital , which looks very similar to the original atomic orbitals on the end oxygen atoms. An electron in this orbital neither strengthens nor weakens the attractions between the atoms, and so the MO is said to be nonbonding. Finally, there is an antibonding MO which has nodes between each pair of oxygen atoms and can be seen by selecting orbital . Since there are only four electrons, only the two lower-energy MO’s (the bonding and nonbonding) will be occupied. The two electrons in the bonding MO provide a bond order of 1, but this is spread over both O—O bonds, and so it contributes a bond order of one-half between each pair of oxygens. Similarly the two electrons in the nonbonding MO correspond to a lone pair, half on the left oxygen and half on the right. Including the sigma bonding framework, this gives 1½ bonds between each pair of oxygens and 2½ lone pairs on each end oxygen. This is exactly the average of the two resonance structures already given. The MO treatment can also be used to interpret the spectrum of ozone. Ozone in the earth’s stratosphere absorbs much solar ultraviolet radiation which would otherwise cause damage to the biosphere. This absorption is due to a band centered around 255 nm which corresponds to excitation of an electron to the unfilled antibonding pi orbital shown in Figure \(\Page {1}\) . Another important molecule to which MO theory can be applied usefully is benzene. As described benzene can be represented by the resonance hybrid This indicates that all C—C bonds are equivalent and intermediate between a single and a double bond. As in the case of ozone, we can treat the sigma bonds of benzene in valence-bond terms, dealing only with pi bonding by the molecular-orbital, method. The sigma bonding framework is This involves 24 electrons in bonds formed by overlap of hybrid orbitals on the carbon atoms with 1s orbitals on each hydrogen or with other hybrids on other carbons. This gives a planar framework and leaves six orbitals (one on each carbon) which are perpendicular to the molecular plane and can overlap sideways to form six pi molecular orbitals. Only the three lowest-energy MO’s are occupied, and their electron clouds are shown in Figure \(\Page {2}\). All these MO’s are bonding, while the three not shown are anti- bonding. When the six valence electrons not used for sigma bonding occupy the pi-bonding MO’s, they are evenly distributed around the ring of carbon atoms. This contributes a bond order of one-half between each pair of carbons, or a total bond order of 1½ when the sigma bond is included. Thus the C—C bond is intermediate between a single and double bond, in accord with the resonance hybrid. | 4,737 | 3,085 |
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We take advantage of changes between the gas, liquid, and solid states to cool a drink with ice cubes (solid to liquid), cool our bodies by perspiration (liquid to gas), and cool food inside a refrigerator (gas to liquid and vice versa). We use dry ice, which is solid CO , as a refrigerant (solid to gas), and we make artificial snow for skiing and snowboarding by transforming a liquid to a solid. In this section, we examine what happens when any of the three forms of matter is converted to either of the other two. These changes of state are often called phase changes . The six most common phase changes are shown in . Phase changes are accompanied by a change in the energy of a system. For example, converting a liquid, in which the molecules are close together, to a gas, in which the molecules are, on average, far apart, requires an input of energy (heat) to give the molecules enough kinetic energy to allow them to overcome the intermolecular attractive forces. The stronger the attractive forces, the more energy is needed to overcome them. Solids, which are highly ordered, have the strongest intermolecular interactions, whereas gases, which are very disordered, have the weakest. Thus any transition from a more ordered to a less ordered state (solid to liquid, liquid to gas, or solid to gas) requires an input of energy; it is . Conversely, any transition from a less ordered to a more ordered state (liquid to solid, gas to liquid, or gas to solid) releases energy; it is . The energy change associated with each common phase change is shown in . In , we defined the enthalpy changes associated with various chemical and physical processes. The melting points and molar , the energy required to convert from a solid to a liquid, a process known as fusion (or melting) , as well as the normal boiling points and enthalpies of vaporization (Δ ) of selected compounds are listed in . The substances with the highest melting points usually have the highest enthalpies of fusion; they tend to be ionic compounds that are held together by very strong electrostatic interactions. Substances with high boiling points are those with strong intermolecular interactions that must be overcome to convert a liquid to a gas, resulting in high enthalpies of vaporization. The enthalpy of vaporization of a given substance is much greater than its enthalpy of fusion because it takes more energy to completely separate molecules (conversion from a liquid to a gas) than to enable them only to move past one another freely (conversion from a solid to a liquid). Δ is positive for any transition from a more ordered to a less ordered state and negative for a transition from a less ordered to a more ordered state. The direct conversion of a solid to a gas, without an intervening liquid phase, is called sublimation . The amount of energy required to sublime 1 mol of a pure solid is the enthalpy of sublimation (Δ ) . Common substances that sublime at standard temperature and pressure (STP; 0°C, 1 atm) include CO (dry ice); iodine ( ); naphthalene, a substance used to protect woolen clothing against moths; and 1,4-dichlorobenzene. As shown in the enthalpy of sublimation of a substance is the sum of its enthalpies of fusion and vaporization provided all values are at the same ; this is an application of Hess’s law. (For more information about Hess’s law, see ).
\[ \Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap} \tag{11.5.1} \] When solid iodine is heated at ordinary atmospheric pressure, it sublimes. When the I vapor comes in contact with a cold surface, it deposits I crystals. Fusion, vaporization, and sublimation are endothermic processes; they occur only with the absorption of heat. Anyone who has ever stepped out of a swimming pool on a cool, breezy day has felt the heat loss that accompanies the evaporation of water from the skin. Our bodies use this same phenomenon to maintain a constant temperature: we perspire continuously, even when at rest, losing about 600 mL of water daily by evaporation from the skin. We also lose about 400 mL of water as water vapor in the air we exhale, which also contributes to cooling. Refrigerators and air-conditioners operate on a similar principle: heat is absorbed from the object or area to be cooled and used to vaporize a low-boiling-point liquid, such as ammonia or the chlorofluorocarbons (CFCs) and the hydrofluorocarbons (HCFCs) discussed in in connection with the ozone layer. The vapor is then transported to a different location and compressed, thus releasing and dissipating the heat. Likewise, ice cubes efficiently cool a drink not because of their low temperature but because heat is required to convert ice at 0°C to liquid water at 0°C, as demonstrated later in Example 8. The processes on the right side of —freezing, condensation, and deposition, which are the reverse of fusion, sublimation, and vaporization—are exothermic. Thus heat pumps that use refrigerants are essentially air-conditioners running in reverse. Heat from the environment is used to vaporize the refrigerant, which is then condensed to a liquid in coils within a house to provide heat. The energy changes that occur during phase changes can be quantified by using a heating or cooling curve. shows a heating curve , a plot of temperature versus heating time, for a 75 g sample of water. The sample is initially ice at 1 atm and −23°C; as heat is added, the temperature of the ice increases linearly with time. The slope of the line depends on both the mass of the ice and the specific heat ( ) of ice, which is the number of joules required to raise the temperature of 1 g of ice by 1°C. As the temperature of the ice increases, the water molecules in the ice crystal absorb more and more energy and vibrate more vigorously. At the melting point, they have enough kinetic energy to overcome attractive forces and move with respect to one another. As more heat is added, the temperature of the system does increase further but remains constant at 0°C until all the ice has melted. Once all the ice has been converted to liquid water, the temperature of the water again begins to increase. Now, however, the temperature increases more slowly than before because the specific heat capacity of water is than that of ice. When the temperature of the water reaches 100°C, the water begins to boil. Here, too, the temperature remains constant at 100°C until all the water has been converted to steam. At this point, the temperature again begins to rise, but at a rate than seen in the other phases because the heat capacity of steam is than that of ice or water. Thus . In this example, as long as even a tiny amount of ice is present, the temperature of the system remains at 0°C during the melting process, and as long as even a small amount of liquid water is present, the temperature of the system remains at 100°C during the boiling process. The rate at which heat is added does affect the temperature of the ice/water or water/steam mixture because the added heat is being used exclusively to overcome the attractive forces that hold the more condensed phase together. Many cooks think that food will cook faster if the heat is turned up higher so that the water boils more rapidly. Instead, the pot of water will boil to dryness sooner, but the temperature of the water does not depend on how vigorously it boils. The temperature of a sample does not change during a phase change. If heat is added at a constant rate, as in , then the length of the horizontal lines, which represents the time during which the temperature does not change, is directly proportional to the magnitude of the enthalpies associated with the phase changes. In 5.3, the horizontal line at 100°C is much longer than the line at 0°C because the enthalpy of vaporization of water is several times greater than the enthalpy of fusion. A superheated liquid is a sample of a liquid at the temperature and pressure at which it should be a gas. Superheated liquids are not stable; the liquid will eventually boil, sometimes violently. The phenomenon of superheating causes “bumping” when a liquid is heated in the laboratory. When a test tube containing water is heated over a Bunsen burner, for example, one portion of the liquid can easily become too hot. When the superheated liquid converts to a gas, it can push or “bump” the rest of the liquid out of the test tube. Placing a stirring rod or a small piece of ceramic (a “boiling chip”) in the test tube allows bubbles of vapor to form on the surface of the object so the liquid boils instead of becoming superheated. Superheating is the reason a liquid heated in a smooth cup in a microwave oven may not boil until the cup is moved, when the motion of the cup allows bubbles to form. The cooling curve , a plot of temperature versus cooling time, in plots temperature versus time as a 75 g sample of steam, initially at 1 atm and 200°C, is cooled. Although we might expect the cooling curve to be the mirror image of the heating curve in , the cooling curve is an identical mirror image. As heat is removed from the steam, the temperature falls until it reaches 100°C. At this temperature, the steam begins to condense to liquid water. No further temperature change occurs until all the steam is converted to the liquid; then the temperature again decreases as the water is cooled. We might expect to reach another plateau at 0°C, where the water is converted to ice; in reality, however, this does not always occur. Instead, the temperature often drops below the freezing point for some time, as shown by the little dip in the cooling curve below 0°C. This region corresponds to an unstable form of the liquid, a supercooled liquid . If the liquid is allowed to stand, if cooling is continued, or if a small crystal of the solid phase is added (a seed crystal ), the supercooled liquid will convert to a solid, sometimes quite suddenly. As the water freezes, the temperature increases slightly due to the heat evolved during the freezing process and then holds constant at the melting point as the rest of the water freezes. Subsequently, the temperature of the ice decreases again as more heat is removed from the system. Supercooling effects have a huge impact on Earth’s climate. For example, supercooling of water droplets in clouds can prevent the clouds from releasing precipitation over regions that are persistently arid as a result. Clouds consist of tiny droplets of water, which in principle should be dense enough to fall as rain. In fact, however, the droplets must aggregate to reach a certain size before they can fall to the ground. Usually a small particle (a ) is required for the droplets to aggregate; the nucleus can be a dust particle, an ice crystal, or a particle of silver iodide dispersed in a cloud during (a method of inducing rain). Unfortunately, the small droplets of water generally remain as a supercooled liquid down to about −10°C, rather than freezing into ice crystals that are more suitable nuclei for raindrop formation. One approach to producing rainfall from an existing cloud is to cool the water droplets so that they crystallize to provide nuclei around which raindrops can grow. This is best done by dispersing small granules of solid CO (dry ice) into the cloud from an airplane. Solid CO sublimes directly to the gas at pressures of 1 atm or lower, and the enthalpy of sublimation is substantial (25.3 kJ/mol). As the CO sublimes, it absorbs heat from the cloud, often with the desired results. If a 50.0 g ice cube at 0.0°C is added to 500 mL of tea at 20.0°C, what is the temperature of the tea when the ice cube has just melted? Assume that no heat is transferred to or from the surroundings. The density of water (and iced tea) is 1.00 g/mL over the range 0°C–20°C, the specific heats of liquid water and ice are 4.184 J/(g·°C) and 2.062 J/(g·°C), respectively, and the enthalpy of fusion of ice is 6.01 kJ/mol. mass, volume, initial temperature, density, specific heats, and Δ final temperature Substitute the values given into the general equation relating heat gained to heat lost ( ) to obtain the final temperature of the mixture. Recall from that when two substances or objects at different temperatures are brought into contact, heat will flow from the warmer one to the cooler. The amount of heat that flows is given by where is heat, is mass, is the specific heat, and Δ is the temperature change. Eventually, the temperatures of the two substances will become equal at a value somewhere between their initial temperatures. Calculating the temperature of iced tea after adding an ice cube is slightly more complicated. The general equation relating heat gained and heat lost is still valid, but in this case we also have to take into account the amount of heat required to melt the ice cube from ice at 0.0°C to liquid water at 0.0C: \[ q_{lost}=-q_{gained} \notag \] \[ m_{iced \; tea}C_{s}\left ( H_{2}O \right )\Delta T_{iced \; tea}=-\left [ m_{ice}C_{s}\left ( H_{2}O \right )\Delta T_{ice} \right ]+mol_{ice} \Delta H_{fus}\left ( H_{2}O \right ) \notag \] \[ 500 \; \cancel{g} \left [ 4.184 \; J/\left ( \cancel{g}\cdot ^{o}C \right ) \right ]\left ( T_{f}-20.0 \; ^{o}C \right ) = -50 \; \cancel{g} \left [ 4.184 \; J/\left ( \cancel{g}\cdot ^{o}C \right ) \right ]\left ( T_{f}-0.0 \; ^{o}C \right )-\dfrac{50.0 \; \cancel{g}}{18.0 \; \cancel{g}/\cancel{mol}}6.01\times 10^{3} \; J/\cancel{mol} \notag \] \[ \left ( 2090 \; J/^{o}C \right )\left ( T_{f}-20.0 \; ^{o}C \right )= -\left ( 209 \; J/^{o}C \right )T_{f} - 1.67\times 10^{4} J \notag \] \[ 2.53 \times 10^{4} J= \left ( 2310 \; J/^{o}C \right )T_{f} \notag \] \[ 11.0 \; ^{o}C=T_{f} \notag \] Exercise Suppose you are overtaken by a blizzard while ski touring and you take refuge in a tent. You are thirsty, but you forgot to bring liquid water. You have a choice of eating a few handfuls of snow (say 400 g) at −5.0°C immediately to quench your thirst or setting up your propane stove, melting the snow, and heating the water to body temperature before drinking it. You recall that the survival guide you leafed through at the hotel said something about not eating snow, but you can’t remember why—after all, it’s just frozen water. To understand the guide’s recommendation, calculate the amount of heat that your body will have to supply to bring 400 g of snow at −5.0°C to your body’s internal temperature of 37°C. Use the data in Example 8 200 kJ (4.1 kJ to bring the ice from −5.0°C to 0.0°C, 133.6 kJ to melt the ice at 0.0°C, and 61.9 kJ to bring the water from 0.0°C to 37°C), which is energy that would not have been expended had you first melted the snow. Changes of state are examples of , or . All phase changes are accompanied by changes in the energy of a system. Changes from a more-ordered state to a less-ordered state (such as a liquid to a gas) are . Changes from a less-ordered state to a more-ordered state (such as a liquid to a solid) are always . The conversion of a solid to a liquid is called . The energy required to melt 1 mol of a substance is its enthalpy of fusion (Δ ). The energy change required to vaporize 1 mol of a substance is the enthalpy of vaporization (Δ ). The direct conversion of a solid to a gas is . The amount of energy needed to sublime 1 mol of a substance is its and is the sum of the enthalpies of fusion and vaporization. Plots of the temperature of a substance versus heat added or versus heating time at a constant rate of heating are called . Heating curves relate temperature changes to phase transitions. A , a liquid at a temperature and pressure at which it should be a gas, is not stable. A is not exactly the reverse of the heating curve because many liquids do not freeze at the expected temperature. Instead, they form a , a metastable liquid phase that exists below the normal melting point. Supercooled liquids usually crystallize on standing, or adding a of the same or another substance can induce crystallization. In extremely cold climates, snow can disappear with no evidence of its melting. How can this happen? What change(s) in state are taking place? Would you expect this phenomenon to be more common at high or low altitudes? Explain your answer. Why do car manufacturers recommend that an automobile should not be left standing in subzero temperatures if its radiator contains only water? Car manufacturers also warn car owners that they should check the fluid level in a radiator only when the engine is cool. What is the basis for this warning? What is likely to happen if it is ignored? Use Hess’s law and a thermochemical cycle to show that, for any solid, the enthalpy of sublimation is equal to the sum of the enthalpy of fusion of the solid and the enthalpy of vaporization of the resulting liquid. Three distinct processes occur when an ice cube at −10°C is used to cool a glass of water at 20°C. What are they? Which causes the greatest temperature change in the water? When frost forms on a piece of glass, crystals of ice are deposited from water vapor in the air. How is this process related to sublimation? Describe the energy changes that take place as the water vapor is converted to frost. What phase changes are involved in each process? Which processes are exothermic, and which are endothermic? What phase changes are involved in each process? Which processes are exothermic, and which are endothermic? Why do substances with high enthalpies of fusion tend to have high melting points? Why is the enthalpy of vaporization of a compound invariably much larger than its enthalpy of fusion? What is the opposite of fusion, sublimation, and condensation? Describe the phase change in each pair of opposing processes and state whether each phase change is exothermic or endothermic. Draw a typical heating curve (temperature versus amount of heat added at a constant rate) for conversion of a solid to a liquid and then to a gas. What causes some regions of the plot to have a positive slope? What is happening in the regions of the plot where the curve is horizontal, meaning that the temperature does not change even though heat is being added? If you know the mass of a sample of a substance, how could you use a heating curve to calculate the specific heat of the substance, as well as the change in enthalpy associated with a phase change? Draw the heating curve for a liquid that has been superheated. How does this differ from a normal heating curve for a liquid? Draw the cooling curve for a liquid that has been supercooled. How does this differ from a normal cooling curve for a liquid? When snow disappears without melting, it must be subliming directly from the solid state to the vapor state. The rate at which this will occur depends solely on the partial pressure of water, not on the total pressure due to other gases. Consequently, altitude (and changes in atmospheric pressure) will not affect the rate of sublimation directly. The general equations and enthalpy changes for the changes of state involved in converting a solid to a gas are: The relationship between these enthalpy changes is shown schematically in the thermochemical cycle below: The formation of frost on a surface is an example of deposition, which is the reverse of sublimation. The change in enthalpy for deposition is equal in magnitude, but opposite in sign, to Δ , which is a positive number: Δ = Δ + Δ . The enthalpy of vaporization is larger than the enthalpy of fusion because vaporization requires the addition of enough energy to disrupt all intermolecular interactions and create a gas in which the molecules move essentially independently. In contrast, fusion requires much less energy, because the intermolecular interactions in a liquid and a solid are similar in magnitude in all condensed phases. Fusion requires only enough energy to overcome the intermolecular interactions that lock molecules in place in a lattice, thereby allowing them to move more freely. The portions of the curve with a positive slope correspond to heating a single phase, while the horizontal portions of the curve correspond to phase changes. During a phase change, the temperature of the system does not change, because the added heat is melting the solid at its melting point or evaporating the liquid at its boiling point. A superheated liquid exists temporarily as liquid with a temperature above the normal boiling point of the liquid. When a supercooled liquid boils, the temperature drops as the liquid is converted to vapor. Conversely, a supercooled liquid exists temporarily as a liquid with a temperature lower than the normal melting point of the solid. As shown below, when a supercooled liquid crystallizes, the temperature increases as the liquid is converted to a solid. The density of oxygen at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of oxygen. The density of propane at 1 atm and various temperatures is given in the following table. Plot the data and use your graph to predict the normal boiling point of propane. Draw the cooling curve for a sample of the vapor of a compound that has a melting point of 34°C and a boiling point of 77°C as it is cooled from 100°C to 0°C. Propionic acid has a melting point of −20.8°C and a boiling point of 141°C. Draw a heating curve showing the temperature versus time as heat is added at a constant rate to show the behavior of a sample of propionic acid as it is heated from −50°C to its boiling point. What happens above 141°C? A 0.542 g sample of I requires 96.1 J of energy to be converted to vapor. What is the enthalpy of sublimation of I ? A 2.0 L sample of gas at 210°C and 0.762 atm condenses to give 1.20 mL of liquid, and 476 J of heat is released during the process. What is the enthalpy of vaporization of the compound? One fuel used for jet engines and rockets is aluminum borohydride [Al(BH ) ], a liquid that readily reacts with water to produce hydrogen. The liquid has a boiling point of 44.5°C. How much energy is needed to vaporize 1.0 kg of aluminum borohydride at 20°C, given a Δ of 30 kJ/mol and a molar heat capacity ( ) of 194.6 J/(mol·K)? How much energy is released when freezing 100.0 g of dimethyl disulfide (C H S ) initially at 20°C? Use the following information: melting point = −84.7°C, Δ = 9.19 kJ/mol, = 118.1 J/(mol·K). Δ Δ How much heat is released in the conversion of 1.00 L of steam at 21.9 atm and 200°C to ice at −6.0°C and 1 atm? How much heat must be applied to convert a 1.00 g piece of ice at −10°C to steam at 120°C? How many grams of boiling water must be added to a glass with 25.0 g of ice at −3°C to obtain a liquid with a temperature of 45°C? How many grams of ice at −5.0°C must be added to 150.0 g of water at 22°C to give a final temperature of 15°C? The transition from a liquid to a gaseous phase is accompanied by a drastic decrease in density. According to the data in the table and the plot, the boiling point of liquid oxygen is between 90 and 100 K (actually 90.2 K). 45.0 kJ/mol 488 kJ 32.6 kJ 57 g 11.5.3 from the @ YouTube Thumbnail from | 23,234 | 3,086 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.13%3A_Balancing_Chemical_Equations |
We have now determined symbols and formulas for all the ingredients of chemical equations, but one important step remains. We must be sure that the law of conservation of mass is obeyed. . This reflects our belief in Dalton’s third postulate that atoms are neither created, destroyed, nor changed from one kind to another during a chemical process. When the law of conservation of mass is obeyed, the equation is said to be The same number of atoms (or moles of atoms) of a given type must appear on each side of the equation. As a simple example of how to balance an equation, let us take the reaction which occurs when a large excess of mercury combines with bromine. The video above shows the liquid bromine (a dark brown) being combined with the shiny silver (also liquid) Mercury. In this case the product is a white solid which does not melt but instead changes to a gas when heated above 345°C. It is insoluble in water and turns a salmon color in the presence of UV light. From these properties it can be identified as mercurous bromide, Hg Br . The equation for the reaction would look like this: \[ \text{Hg} + \text{Br}_{2} \rightarrow \text{Hg}_{2}\text{Br}_{2} \label{1} \] but it is not balanced because there are 2 mercury atoms (in Hg Br ) on the right side of the equation and only 1 on the left. An incorrect way of obtaining a balanced equation is to change this to \[ \text{Hg} + \text{Br}_{2} \rightarrow \cancel{\text{Hg}\text{Br}_{2}} \label{2} \] This equation is wrong because we had already determined from the properties of the product that the product was Hg Br . Equation \(\ref{2}\) is balanced, but it refers to a different reaction which produces a different product. The equation might also be incorrectly written as \[ \cancel{\text{Hg}_{2}} + \text{Br}_{2} \rightarrow \text{Hg}_{2}\text{Br}_{2}\label{3} \] The formula Hg suggests that molecules containing 2 mercury atoms each were involved, but our previous microscopic experience with this element indicates that such molecules do not occur. In balancing an equation you must remember that the in the formulas have been determined . Changing them indicates a change in the nature of the reactants or products. It is permissible, however, to change the amounts of reactants or products involved. For example, the equation in question is correctly balanced as follows: \[ \underline{2}\text{Hg} + \text{Br}_{2} \rightarrow \text{Hg}_{2}\text{Br}_{2} \label{4} \] image credits: By CCoil (talk) - Own work, CC BY- , The 2 written before the symbol Hg is called a . It indicates that on the microscopic level 2Hg atoms are required to react with the molecule. On a macroscopic scale the coefficient 2 means that 2 Hg atoms are required to react with 1 mol Br molecules. Notice as well that Br has a coefficient of 1, meaning that on a microscopic level, 1 molecule of Br reacts with every 2 atoms of Hg, and on a macroscopic level, 1 mole of Br is required for every 2 moles of Hg. Finally, on the microscopic level, 1 molecule of Hg Br can be thought of as a single unit in the lattice structure shown above. On a macroscopic level, 1 mole of Hg Br is 6.02 x 10 molecules of Mercury (I) Bromide arranged in the lattice structure shown above. To summarize: Once the formulas (subscripts) have been determined, . Nothing else may be changed. Balance the equation \( \ce{Hg2Br2 + Cl2 -> HgCl2 + Br2}\) Although Br and Cl are balanced, Hg is not. A coefficient of 2 with HgCl is needed: \( \text{Hg}_{2}\text{Br}_{2} + \text{Cl}_{2} \rightarrow \underline{2}\text{HgCl}_{2} + \text{Br}_{2}\) Now Cl is not balanced. We need 2 Cl molecules on the left: \( \text{Hg}_{2}\text{Br}_{2} + \underline{2}\text{Cl}_{2} \rightarrow \text{2HgCl}_{2} + \text{Br}_{2}\) We now have 2Hg atoms, 2Br atoms, and 4Cl atoms on each side, and so balancing is complete. Most chemists use several techniques for balancing equations. For example, it helps to know which element you should balance first. When each chemical symbol appears in a single formula on each side of the equation (as Example \(\Page {1}\) ), you can start wherever you want and the process will work. When a symbol appears in three or more formulas, however, that particular element will be more difficult to balance and should usually be left until last. When butane (C H ) is burned in oxygen gas (O ), the only products are carbon dioxide(CO ) and water. Write a balanced equation to describe this reaction. First write an unbalanced equation showing the correct formulas of all the reactants and products: \( \ce{C4H10 + O2 -> CO2 + H2O}\) We note that O atoms appear in three formulas, one on the left and two on the right. Therefore we balance C and H first. The formula C H determines how many C and H atoms must remain after the reaction, and so we write coefficients of 4 for CO and 5 for H O: \( \text{C}_{4}\text{H}_{10} + \text{O}_{2} \rightarrow \underline{4}\text{CO}_{2} + \underline{5}\text{H}_{2}\text{O}\) We now have a total of 13 O atoms on the right-hand side, and the equation can be balanced by using a coefficient of \(\frac{13}{2}\) in front of O : \( \text{C}_{4}\text{H}_{10} + \frac{13}{2}\text{O}_{2} \rightarrow \text{4CO}_{2} + \text{5H}_{2}\text{O}\) Usually it is preferable to remove fractional coefficients since they might be interpreted to mean a fraction of a molecule. (One-half of an O molecule would be an O atom, which has quite different chemical reactivity.) Therefore we multiply all coefficients on both sides of the equation by two to obtain the final result: \( \underline{2}\text{C}_{4}\text{H}_{10} + \underline{13}\text{O}_{2} \rightarrow \underline{8}\text{CO}_{2} + \underline{10}\text{H}_{2}\text{O}\) (Sometimes, when we are interested in moles rather than individual molecules, it may be useful to omit this last step. Obviously the idea of half a mole of O molecules, that is, 3.011 × 10 molecules, is much more tenable than the idea of half a molecule.) Another useful technique is illustrated in Example \(\Page {2}\). When an element (such as O ) appears by itself, it is usually best to choose its coefficient last. Furthermore, groups such as NO , SO , etc., often remain unchanged in a reaction and can be treated as if they consisted of a single atom. When such a group of atoms is enclosed in parentheses followed by a subscript, the subscript applies to all of them. That is, the formula involves Ca(NO ) involves 1Ca, 2N and 2 × 3 = 6 O atoms. Balance the equation \( \ce{NaMnO4 + H2O2 + H2SO4 -> MnSO4 + Na2SO4 + O2 + H2O}\) We note that oxygen atoms are found in every one of the seven formulas in the equation, making it especially hard to balance. However, Na appears only in two formulas: \( \underline{2}\text{NaMnO}_{4} + \text{H}_{2}\text{O}_{2} + \text{H}_{2}\text{SO}_{4} \rightarrow \text{MnSO}_{4} + \text{Na}_{2}\text{SO}_{4} + \text{O}_{2} + {H}_{2}\text{O}\) as does manganese, Mn: \( \text{2NaMnO}_{4} + \text{H}_{2}\text{O}_{2} + \text{H}_{2}\text{SO}_{4} \rightarrow \underline{2}\text{MnSO}_{4} + \text{Na}_{2}\text{SO}_{4} + \text{O}_{2} + \text{H}_{2}\text{O}\) We now note that the element S always appears with 4 O atoms, and so we balance the SO groups: \( \text{2NaMnO}_{4} + \text{H}_{2}\text{O}_{2} + \underline{3}\text{H}_{2}\text{SO}_{4} \rightarrow \text{2MnSO}_{4} + \text{Na}_{2}\text{SO}_{4} + \text{O}_{2} + \text{H}_{2}\text{O}\) Now we are in a position to balance hydrogen: \( \text{2NaMnO}_{4} + \text{H}_{2}\text{O}_{2} + \text{3H}_{2}\text{SO}_{4} \rightarrow \text{2MnSO}_{4} + \text{Na}_{2}\text{SO}_{4} + \text{O}_{2} + \underline{4}\text{H}_{2}\text{O}\) and finally oxygen. (We are aided by the fact that it appears as the element.) \( \text{2NaMnO}_{4} + \text{H}_{2}\text{O}_{2} + \text{3H}_{2}\text{SO}_{4} \rightarrow \text{2MnSO}_{4} + \text{Na}_{2}\text{SO}_{4} + \underline{3}\text{O}_{2} + \text{4H}_{2}\text{O}\) Notice that in this example we followed the rule of balancing first those elements whose symbols appeared in the smallest number of formulas: Na and Mn in two each, S (or SO ) and H in three each, and finally O. Even using this rule, however, equations in which one or more elements appear in four or more formulas are difficult to balance without some additional techniques which we will develop when we investigate . The balancing of chemical equations has an important environmental message for us. If atoms are conserved in a chemical reaction, then we cannot get rid of them. In other words . There are only two things we can do with atoms: Move them from place to place or from compound to compound. Thus when we "dispose" of something by burning it, dumping it, or washing it down the sink, we have not really gotten rid of it at all. The atoms which constituted it are still around someplace, and it is just as well to know where they are and what kind of molecule they are in. Discarded atoms in places where we do not want them and in undesirable molecules are known as pollution. | 9,003 | 3,089 |
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 | 3,090 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/13%3A_Solutions_and_their_Physical_Properties/13.07%3A_Osmotic_Pressure |
Osmotic pressure is a of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration. Osmosis can be demonstrated using a U-tube like the one shown in Figure \(\Page {1}\), which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure (\(\Pi\)) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution. The osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation: \[ \Pi=\dfrac{nRT}{V}=MRT \label{eq1}\] where As shown in Example \(\Page {1}\), osmotic pressures tend to be quite high, even for rather dilute solutions. When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% \(NaCl\) by mass; the solution density is 1.02 g/mL at 25°C. Calculate the osmotic pressure of a 4.0% aqueous \(NaCl\) solution at 25°C. If the normal osmotic pressure inside a yeast cell is 7.3 atm, corresponding to a total concentration of dissolved particles of 0.30 M, what concentration of glycerol must the cells synthesize to exactly balance the external osmotic pressure at 25°C? : concentration, density, and temperature of \(NaCl\) solution; internal osmotic pressure of cell : osmotic pressure of \(NaCl\) solution and concentration of glycerol needed : : A The solution contains 4.0 g of \(NaCl\) per 100 g of solution. Using the formula mass of \(NaCl\) (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity: \[\begin{align} M_{NaCl} &=\dfrac{moles\; NaCl}{\text{liter solution}} \nonumber \\[4pt] &=\left(\dfrac{4.0 \; \cancel{g} \;NaCl}{58.44\; \cancel{g}/mol\; NaCl}\right)\left(\dfrac{1}{100\; \cancel{g \;solution}}\right)\left(\dfrac{1.02\; \cancel{g\; solution}}{1.00\; \cancel{mL}\; solution}\right)\left(\dfrac{1000\; \cancel{mL}}{1\; L}\right) \nonumber \\[4pt] &= 0.70\; M\; NaCl \nonumber \end{align} \nonumber\] Because 1 mol of \(NaCl\) produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M. B Now we can use Equation \ref{eq1} to calculate the osmotic pressure of the solution: \[\begin{align*} \Pi &=MRT \\[4pt] &=(1.4 \;mol/L)\left[ 0.0821\; (L⋅atm)/(K⋅mol) \right ] (298\; K) \\[4pt] &= 34 \;atm \end{align*}\] C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation \ref{eq1} for the molarity corresponding to this osmotic pressure: \[\begin{align*} M &=\dfrac{\Pi}{RT} \\[4pt] &= \dfrac{27\; \cancel{atm}}{[0.0821(L⋅\cancel{atm})/(\cancel{K}⋅mol)] (298 \;\cancel{K})} \\[4pt] &= 1.1 \;M \;glycerol \end{align*}\] In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the \(NaCl\) solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol. Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them. : 24 atm Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins. The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure \(\Page {2}\), which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells. In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves. Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure \(\Page {3}\), applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats. Osmotic Pressure: When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. | 7,928 | 3,091 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.03%3A_Group_II-_Alkaline_Earths |
Group IIA also known as the , include beryllium, magnesium, calcium, strontium, barium, and radium. The last member of the group, Ra, is radioactive and will not be considered here. All alkaline earths are silvery-gray metals which are ductile and relatively soft. However, the following table shows that they are much denser than the group IA metals, and their melting points are significantly higher. They are also harder than the alkali metals. This may be attributed to the general valence electron configuration for the alkaline earths, which involves two electrons per metal atom in metallic bonding (instead of just one as in an alkali metal). First and second ionization energies for the alkaline earths (corresponding to removal of the first and second valence electrons) are relatively small, but the disruption of an octet by removal of a third electron is far more difficult. Like the alkali metals, the alkaline-earth atoms lose electrons easily, and so they are good reducing agents. Other trends among the data in the table are what we would expect. Ionization energies and electronegativities decrease from top to bottom of the group, and atomic and ionic radii increase. The radii of +2 alkaline-earth ions are much smaller than the +1 alkali-metal ions of the same period, because the greater nuclear charge holds the inner shells more tightly. This effect is sufficiently large that an alkaline earth below and to the right of a given alkali metal in the periodic table often has nearly the same ionic radius. Thus Na (95 pm), can fit into exactly the same type of crystal lattice as Ca (99 pm), and these two elements are often found in the same minerals. The same is true of K and Ba . Below is the table for alkali metals, to compare with the table of alkaline earth metals. Similarity of ionic radii also leads to related properties for Li and Mg. Since these two elements are adjacent along a diagonal line from the upper left to the lower right in the periodic table, their similarity is called a . Diagonal relationships are mainly evident in the second and third periods: Be is similar to Al, and B is like Si in many ways. Farther toward the right-hand side of the table such relationships are less pronounced. The most striking similarity between Li and Mg is their ability to form covalent bonds with elements of average electronegativity, such as C, while forming fairly ionic compounds with more electronegative elements, such as O or F. Two examples of covalent compounds are ethyllithium, CH CH Li, and diethylmagnesium, (CH CH ) Mg. Such compounds are likely in the case of Li and Mg but not the alkali or alkaline earths below them, because Li and Mg are small enough to be strongly polarizing and thus form bonds with considerable covalent character. The alkaline earth metals react directly with most nonmetallic elements. forming Except for beryllium, the alkaline earths react directly with hydrogen gas to form hydrides, MH ; M = Mg, Ca, Sr, Ba, or Ra. Beryllium hydride, BeH can also be prepared, but not directly from the elements. Alkaline-earth metals combine readily with oxygen from the air to form oxides, MO. This follows the general reaction: \[\text{2M}(s) + \text{O}_2(g) \rightarrow \text{2MO}_2(s) \nonumber \] M = Be, Mg, Ca, Sr, Ba, or Ra The following video shows the reaction of magnesium with oxygen: In the video, magnesium is burned in air, and emits a bright white flame. A white powder of MgO remains after the reaction described by the equation: \[\text{2Mg}(s) + \text{O}_2(g) \rightarrow \text{2MgO}_2(s) \nonumber \] It should also be noted that while MgO is the main product, nitrogen is also present in the air, and so some magnesium nitride is also produced according to the chemical equation: \[\text{Mg}(s) + \text{N}_2(g) \rightarrow \text{Mg}_3\text{N}_2(s) \nonumber \] These oxides will coat the surface of the metal and prevent other substances from contacting and reacting with it. A good example of the effect of such an oxide coating is the reaction of alkaline-earth metals with water. Beryllium and magnesium react much more slowly than the others because their oxides are insoluble and prevent water from contacting the metal. Alkaline-earth metals react directly with halogens to form salts: \[\text{M}(s) + \text{Cl}_2(g) \rightarrow \text{MCl}_2(s) \nonumber \] M = Be, Mg, Ca, Sr, Ba, or Ra Salt obtained by evaporating seawater (sea salt) contains a good deal of magnesium chloride and calcium chloride as well as sodium chloride. It also has small traces of iodide salts, accounting for the absence of simple goiter in communities which obtain their salt from the oceans. Simple goiter is an enlargement of the thyroid gland caused by iodine deficiency. Alkaline earths also form sulfides: MS. In all these compounds the alkaline-earth elements occur as dipositive ions, Mg , Ca , Sr , or Ba . Similar compounds of Be can be formed by roundabout means, but not by direct combination of the elements. Moreover, the Be compounds are more covalent than ionic. The Be ion has a very small radius (31 pm) and is therefore capable of distorting (polarizing) the electron cloud of an anion in its vicinity. Therefore all bonds involving Be have considerable covalent character, and the chemistry of Be is significantly different from that of the other members of group IIA. As in the case of the alkali metals, the most important and abundant alkaline earths, Mg and Ca, are in the third and fourth periods. Be is rare, although its strength and low density make it useful in certain special alloys. Sr and Ba occur naturally as the relatively insoluble sulfates SrSO (strontianite) and BaSO (barite), but these two elements are of minor commercial importance. The most common ores of Mg and Ca are dolomite, MgCO •CaCO , after which an entire mountain range in Italy is named, and limestone, CaCO , an important building material. Mg is also recovered from seawater on a wide scale. The oxides of the alkaline earths are commonly obtained by heating the carbonates. For example, lime, CaO, is obtained from limestone as follows: \[\text{CaCO}_3(s) \xrightarrow{\Delta } \text{CaO}(s) + \text{CO}_2(g) \nonumber \] Except for BeO, which is covalently bonded, alkaline-earth oxides contain O ions and are strongly basic. When treated with water (a process known as ), they are converted to hydroxides: \[\text{CaO}(s) + \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(s) \nonumber \] Ca(OH) (slaked lime) is an important strong base for industrial applications, because it is cheaper than NaOH. MgO has an extremely high melting point (2800°C) because of the close approach and large charges of its constituent Mg and O ions in the crystal lattice. As a solid it is a good electrical insulator, and so it is used to surround metal-resistance heating wires in electric ranges. MgO is also used to line high-temperature furnaces. When converted to the hydroxide, Mg finds a different use. Mg(OH) is quite insoluble in water, and so it does not produce a high enough concentration of hydroxide ions to be caustic. It is basic, however, and gram for gram can neutralize nearly twice the quantity of acid that NaOH can. Consequently a suspension of Mg(OH) in water (milk of magnesia) makes an excellent antacid, for those who can stand its taste. Because the carbonate ion behaves as a Brönstedt-Lowry base, carbonate salts dissolve in acidic solutions. In nature, water often becomes acidic because the acidic oxide CO is present in the atmosphere. When CO from the air dissolves in water, it can help dissolve limestone: \[\text{CO}_2(g) + \text{H}_2\text{O}(l) + \text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{HCO}_3^{-}(aq) \nonumber \] This reaction often occurs underground as rainwater saturated with CO seeps through a layer of limestone. Caves from which the limestone has been dissolved are often prevalent in areas where there are large deposits of CaCO . In addition, the groundwater and well water in such areas becomes hard. contains appreciable concentrations of Ca , Mg , and certain other metal ions. These form insoluble compounds with soap, causing curdy, scummy precipitates. Hard water can be softened by adding Na CO , washing soda, which precipitates CaCO , or by , a process in which the undesirable Ca and Mg ions are replaced in solution by Na ions, which do not precipitate soap. Most home water softeners work on the latter principle. | 8,498 | 3,093 |
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Because of its highly polar structure, liquid water can act as either an acid (by donating a proton to a base) or a base (by using a lone pair of electrons to accept a proton). For example, when a strong acid such as HCl dissolves in water, it dissociates into chloride ions (\(Cl^−\)) and protons (\(H^+\)). The proton, in turn, reacts with a water molecule to form the hydronium ion (\(H_3O^+\)): \[\underset{aicd}{\ce{HCl(aq)}} + \underset{base}{\ce{H2O(l)}} \rightarrow \underset{acid}{\ce{H3O^{+}(aq)}} + \underset{base}{\ce{Cl^{-}(aq)}} \label{16.3.1a}\] In this reaction, \(\ce{HCl}\) is the acid, and water acts as a base by accepting an \(\ce{H^{+}}\) ion. The reaction in Equation \(\ref{16.3.1a}\) is often written in a simpler form by removing \(\ce{H2O}\) from each side: \[ \ce {HCl (aq) \rightarrow H^{+} (aq) + Cl^{-} (aq)} \label{16.3.1b}\] In Equation \(\ref{16.3.1b}\), the hydronium ion is represented by \(\ce{H^{+}(aq)}\), although free \(\ce{H^{+}}\) ions do not exist in liquid water as this reaction demonstrates: \[ \ce{H^{+}(aq) + H2O(l) \rightarrow H3O^{+}(aq)}\] Water can also act as an acid, as shown in Equation \(\ref{16.3.2}\). In this equilibrium reaction, \(H_2O\) donates a proton to \(NH_3\), which acts as a base: \[\underset{aicd}{\ce{H2O(aq)}} + \underset{base}{\ce{NH3(aq)}} \rightleftharpoons \underset{acid}{\ce{NH^{+}4 (aq)}} + \underset{base}{\ce{OH^{-}(aq)}} \label{16.3.2}\] Water is thus termed , meaning that it can behave as either an acid or a base, depending on the nature of the other reactant. Notice that Equation \(\ref{16.3.2}\) is an equilibrium reaction as indicated by the double arrow and hence has an equilibrium constant associated with it. Because water is amphiprotic, one water molecule can react with another to form an \(OH^−\) ion and an \(H_3O^+\) ion in an autoionization process: \[\ce{2H2O(l) <=> H3O^{+}(aq) + OH^{−}(aq)} \label{16.3.3}\] In pure water, a very small fraction of water molecules donate protons to other water molecules to form hydronium ions and hydroxide ions: The equilibrium constant \(K\) for this reaction can be written as follows: \[K_{a}=\dfrac{a_{H_3O^+}·a_{OH^-}}{a_{H_2O}^2} \approx \frac{[H_{3}O^{+},HO^{-}]}{(1)^{2}}=[H_{3}O^{+},HO^{-}] \label{16.3.4}\] where \(a\) is the activity of a species. Because water is the solvent, and the solution is assumed to be dilute, the activity of the water is approximated by the activity of pure liquid water, which is defined as having a value of 1. The activity of each solute is approximated by the molarity of the solute. It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that K = K [H O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation K = K (\(\textit{a}_{H_2O}\)). Because \(\textit{a}_{H_2O}\) = 1 for a dilute solution, K = K (1), or K = K . In this reaction, one water molecule acts as an acid and one water molecule acts as a base. Thus, this reaction actually can be designated as the \(K_a\) of water and as the \(K_b\) of water. It is most common, however, to designate this reaction and the associated law of mass action as the \(K_w\) of water: \[K_{w}=[H_{3}O^{+},HO^{-}] \label{16.3.5}\] When pure liquid water is in equilibrium with hydronium and hydroxide ions at 25 °C, the concentrations of the hydronium ion and the hydroxide ion are equal: \[[H_3O^+] = [OH^−] = 1.003 \times 10^{−7}\; M \label{16.3.6}\] Thus the number of dissociated water molecules is very small indeed, approximately 2 ppb. Substituting the values for \([H_3O^+]\) and \([OH^−]\) at 25°C into this expression \[K_w=(1.003 \times 10^{−7})(1.003 \times 10^{−7})=1.006 \times 10^{−14} \label{16.3.7}\] The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 °C.Thus, to three significant figures, \(K_w = 1.01 \times 10^{−14}\; M\) . Like any other equilibrium constant, \(K_w\) varies with temperature, ranging from \(1.15 \times 10^{−15}\) at 0°C to \(4.99 \times 10^{−13}\) at 100°C. In pure water, the concentrations of the hydronium ion and the hydroxide ion are equal, and the solution is therefore neutral. If \([H_3O^+] > [OH^−]\), however, the solution is acidic, whereas if \([H_3O^+] < [OH^−]\), the solution is basic. For an aqueous solution, the \(H_3O^+\) concentration is a quantitative measure of acidity: the higher the \(H_3O^+\) concentration, the more acidic the solution. Conversely, the higher the \(OH^−\) concentration, the more basic the solution. In most situations that you will encounter, the \(H_3O^+\) and \(OH^−\) concentrations from the dissociation of water are so small (\(1.003 \times 10^{−7} M\)) that they can be ignored in calculating the \(H_3O^+\) or \(OH^−\) concentrations of solutions of acids and bases, but this is not always the case. What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 °C? The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, \(\ce{[H_3O^+]} = \ce{[OH^- ]}\). At 25 °C: \[\begin{align*} K_\ce{w} &=\ce{[H_3O^+,OH^- ]} \\[4pt]&=\ce{[H_3O^+]^2} \\[4pt] &=\ce{[OH^- ]^2} \\[4pt] &=1.0 \times 10^{−14} \end{align*}\] So: \[\ce{[H_3O^+]}=\ce{[OH^- ]}=\sqrt{1.0 \times 10^{−14}} =1.0 \times 10^{−7}\; M \nonumber\] The hydronium ion concentration and the hydroxide ion concentration are the same, and we find that both equal \(1.0 \times 10^{−7}\; M\). The ion product of water at 80 °C is \(2.4 \times 10^{−13}\). What are the concentrations of hydronium and hydroxide ions in pure water at 80 °C? \(\ce{[H_3O^+]} = \ce{[OH^- ]} = 4.9 \times 10^{−7}\; M\) It is important to realize that the autoionization equilibrium for water is established in all aqueous solutions. Adding an acid or base to water will not change the position of the equilibrium. Example \(\Page {2}\) demonstrates the quantitative aspects of this relation between hydronium and hydroxide ion concentrations. A solution of carbon dioxide in water has a hydronium ion concentration of \(2.0 \times 10^{−6}\; M\). What is the concentration of hydroxide ion at 25 °C? We know the value of the ion-product constant for water at 25 °C: \[\ce{2 H_2O}_{(l)} \rightleftharpoons \ce{H_3O^+}_{(aq)} + \ce{OH^-}_{(aq)} \nonumber\] and \[K_\ce{w}=\ce{[H3O+,OH^- ]}=1.0 \times 10^{−14} \nonumber\] Thus, we can calculate the missing equilibrium concentration. Rearrangement of the \(k_w\) expression yields that \([\ce{OH^- }]\) is directly proportional to the inverse of [H O ]: \[[\ce{OH^- }]=\dfrac{K_{\ce w}}{[\ce{H_3O^+}]}=\dfrac{1.0 \times 10^{−14}}{2.0 \times 10^{−6}}=5.0 \times 10^{−9} \nonumber\] The hydroxide ion concentration in water is reduced to \(5.0 \times 10^{−9}\: M\) as the hydrogen ion concentration increases to \(2.0 \times 10^{−6}\; M\). This is expected from Le Chatelier’s principle; the autoionization reaction shifts to the left to reduce the stress of the increased hydronium ion concentration and the \(\ce{[OH^- ]}\) is reduced relative to that in pure water. A check of these concentrations confirms that our arithmetic is correct: \[K_\ce{w}=\ce{[H_3O^+,OH^- ]}=(2.0 \times 10^{−6})(5.0 \times 10^{−9})=1.0 \times 10^{−14} \nonumber\] What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M at 25 °C? \[\ce{[H3O+]} = 1 \times 10^{−11} M \nonumber\] Self-Ionization of Water (Kw): The pH scale is a concise way of describing the \(H_3O^+\) concentration and hence the acidity or basicity of a solution. Recall that pH and the \(H^+\) (\(H_3O^+\)) concentration are related as follows: \[\begin{align} pH &=−\log_{10}[\ce{H^{+}}] \label{16.3.8} \\[4pt] [\ce{H^{+}}] &=10^{−pH} \label{16.3.9} \end{align}\] Because the scale is logarithmic, a pH difference of 1 between two solutions corresponds to a difference of a factor of 10 in their hydronium ion concentrations. Recall also that the pH of a neutral solution is 7.00 (\([H_3O^+] = 1.0 \times 10^{−7}\; M\)), whereas acidic solutions have pH < 7.00 (corresponding to \([\ce{H3O^{+}}] > 1.0 \times 10^{−7}\)) and basic solutions have pH > 7.00 (corresponding to \([\ce{H3O^{+}}] < 1.0 \times 10^{−7}\)). Similar notation systems are used to describe many other chemical quantities that contain a large negative exponent. For example, chemists use an analogous pOH scale to describe the hydroxide ion concentration of a solution. The pOH and \([\ce{OH^{−}}]\) are related as follows: \[\begin{align} pOH &=−\log_{10}[\ce{OH^{−}}] \label{16.3.10} \\[4pt] [\ce{OH^{−}}] &=10^{−pOH} \label{16.3.11} \end{align}\] The constant \(K_w\) can also be expressed using this notation, where \(pK_w = −\log\; K_w\). Because a neutral solution has \([OH^−] = 1.0 \times 10^{−7}\), the pOH of a neutral solution is 7.00. Consequently, the sum of the pH and the pOH for a neutral solution at 25°C is 7.00 + 7.00 = 14.00. We can show that the sum of pH and pOH is equal to 14.00 for any aqueous solution at 25°C by taking the negative logarithm of both sides of Equation \(\ref{16.3.6b}\): \[\begin{align} −\log K_w &= pK_w \\[4pt] &=−\log([H_3O^+,OH^−]) \\[4pt] &= (−\log[H_3O^+]) + (−\log[OH^−]) \\[4pt] &= pH + pOH \label{16.3.12} \end{align}\] Thus at any temperature, \(pH + pOH = pK_w\), so at 25°C, where \(K_w = 1.0 \times 10^{−14}\) and \(pH + pOH = 14.00\). More generally, the pH of any neutral solution is half of the \(pK_w\) at that temperature. The relationship among pH, pOH, and the acidity or basicity of a solution is summarized graphically in Figure \(\Page {1}\) over the common pH range of 0 to 14. Notice the inverse relationship between the pH and pOH scales. For any neutral solution, pH + pOH = 14.00 (at 25°C) with pH=pOH=7. The \(k_w\) for water at 100°C is \(4.99 \times 10^{−13}\). Calculate \(pK_w\) for water at this temperature and the pH and the pOH for a neutral aqueous solution at 100°C. Report pH and pOH values to two decimal places. : \(K_w\) : \(pK_w\), \(pH\), and \(pOH\) : : Because \(pK_w\) is the negative logarithm of Kw, we can write \[pK_w = −\log K_w = −\log(4.99 \times 10^{−13}) = 12.302 \nonumber\] The answer is reasonable: \(K_w\) is between \(10^{−13}\) and \(10^{−12}\), so \(pK_w\) must be between 12 and 13. Equation \ref{16.3.6b} shows that \(K_w = [H_3O^+,OH^−]\). Because \([H_3O^+] = [OH^−]\) in a neutral solution, we can let \(x = [H_3O^+] = [OH^−]\): \[\begin{align*} K_w &= [\ce{H3O^{+}},\ce{OH^{−}}] \\[4pt] &= (x)(x) \\[4pt] &=x^2 \\[4pt] x &=\sqrt{K_w} \\[4pt] &=\sqrt{4.99 \times 10^{−13}} \\[4pt] &=7.06 \times 10^{−7}\; M \end{align*}\] Because \(x\) is equal to both \([\ce{H3O^{+}}]\) and \([\ce{OH^{−}}]\), \[\begin{align*} pH = pOH &= −\log(7.06 \times 10^{−7}) \\[4pt] &= 6.15 \, \text{(to two decimal places)} \end{align*}\] We could obtain the same answer more easily (without using logarithms) by using the \(pK_w\). In this case, we know that \(pK_w = 12.302\), and from Equation \ref{16.3.12}, we know that \(pK_w = pH + pOH\). Because \(pH = pOH\) in a neutral solution, we can use Equation \ref{16.3.12} directly, setting \(pH = pOH = y\). Solving to two decimal places we obtain the following: \[pK_w = pH + pOH = y + y = 2y \nonumber\] \[y=\dfrac{pK_w}{2}=\dfrac{12.302}{2}=6.15=pH=pOH \nonumber\] Humans maintain an internal temperature of about 37°C. At this temperature, \(K_w = 3.55 \times 10^{−14}\). Calculate \(pK_w\) and the pH and the pOH of a neutral solution at 37°C. Report \(pH\) and \(pOH\) values to two decimal places. \(pK_w = 13.45\) and \(pH = pOH = 6.73\) Introduction to pH: Water is amphiprotic: it can act as an acid by donating a proton to a base to form the hydroxide ion, or as a base by accepting a proton from an acid to form the hydronium ion (\(H_3O^+\)). The autoionization of liquid water produces \(OH^−\) and \(H_3O^+\) ions. The equilibrium constant for this reaction is called the ion-product constant of liquid water (\(K_w\)) and is defined as \(K_w = [H_3O^+,OH^−]\). At 25°C, \(K_w\) is \(1.01 \times 10^{−14}\); hence \(pH + pOH = pK_w = 14.00\). The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion . For example, at 100 °C, the value for \(K_\ce{w}\) is approximately \(5.1 \times 10^{−13}\), roughly 100-times larger than the value at 25 °C. ). | 12,563 | 3,094 |
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Approximately one-third of the compounds produced industrially are organic compounds. All living organisms are composed of organic compounds, as are most foods, medicines, clothing fibers, and plastics. The detection of organic compounds is useful in many fields. In one recently developed application, scientists have devised a new method called “material degradomics” to monitor the degradation of old books and historical documents. As paper ages, it produces a familiar “old book smell” from the release of organic compounds in gaseous form. The composition of the gas depends on the original type of paper used, a book’s binding, and the applied media. By analyzing these organic gases and isolating the individual components, preservationists are better able to determine the condition of an object and those books and documents most in need of immediate protection. The simplest class of organic compounds is the hydrocarbons, which consist entirely of carbon and hydrogen. Petroleum and natural gas are complex, naturally occurring mixtures of many different hydrocarbons that furnish raw materials for the chemical industry. The four major classes of hydrocarbons are the following: the alkanes, which contain only carbon–hydrogen and carbon–carbon single bonds; the alkenes, which contain at least one carbon–carbon double bond; the alkynes, which contain at least one carbon–carbon triple bond; and the aromatic hydrocarbons, which usually contain rings of six carbon atoms that can be drawn with alternating single and double bonds. Alkanes are also called hydrocarbons, whereas hydrocarbons that contain multiple bonds (alkenes, alkynes, and aromatics) are . The simplest alkane is methane (CH ), a colorless, odorless gas that is the major component of natural gas. In larger alkanes whose carbon atoms are joined in an unbranched chain ( ), each carbon atom is bonded to at most two other carbon atoms. The structures of two simple alkanes are shown in , and the names and condensed structural formulas for the first 10 straight-chain alkanes are in . The names of all alkanes end in - , and their boiling points increase as the number of carbon atoms increases. Alkanes with four or more carbon atoms can have more than one arrangement of atoms. The carbon atoms can form a single unbranched chain, or the primary chain of carbon atoms can have one or more shorter chains that form branches. For example, butane (C H ) has two possible structures. butane (usually called -butane) is CH CH CH CH , in which the carbon atoms form a single unbranched chain. In contrast, the condensed structural formula for is (CH ) CHCH , in which the primary chain of three carbon atoms has a one-carbon chain branching at the central carbon. Three-dimensional representations of both structures are as follows: The systematic names for branched hydrocarbons use the lowest possible number to indicate the position of the branch along the longest straight carbon chain in the structure. Thus the systematic name for isobutane is 2-methylpropane, which indicates that a methyl group (a branch consisting of –CH ) is attached to the second carbon of a propane molecule. Similarly, states that one of the major components of gasoline is commonly called isooctane; its structure is as follows: The compound has a chain of five carbon atoms, so it is a derivative of pentane. There are two methyl group branches at one carbon atom and one methyl group at another. Using the lowest possible numbers for the branches gives 2,2,4-trimethylpentane for the systematic name of this compound. The simplest alkenes are , C H or CH =CH , and , C H or CH CH=CH (part (a) in ). The names of alkenes that have more than three carbon atoms use the same stems as the names of the alkanes ( ) but end in - instead of - . As with alkanes, more than one structure is possible for alkenes with four or more carbon atoms. For example, an alkene with four carbon atoms has three possible structures. One is CH =CHCH CH (1-butene), which has the double bond between the first and second carbon atoms in the chain. The other two structures have the double bond between the second and third carbon atoms and are forms of CH CH=CHCH (2-butene). All four carbon atoms in 2-butene lie in the same plane, so there are two possible structures (part (a) in ). If the two methyl groups are on the same side of the double bond, the compound is -2-butene (from the Latin , meaning “on the same side”). If the two methyl groups are on opposite sides of the double bond, the compound is -2-butene (from the Latin , meaning “across”). These are distinctly different molecules: -2-butene melts at −138.9°C, whereas -2-butene melts at −105.5°C. Just as a number indicates the positions of branches in an alkane, the number in the name of an alkene specifies the position of the carbon atom of the double bond. The name is based on the lowest possible number starting from of the carbon chain, so CH CH CH=CH is called 1-butene, 3-butene. Note that CH =CHCH CH and CH CH CH=CH are different ways of writing the (1-butene) in two different orientations. The positions of groups or multiple bonds are always indicated by the lowest number possible. The simplest alkyne is , C H or HC≡CH (part (b) in ). Because a mixture of acetylene and oxygen burns with a flame that is hot enough (>3000°C) to cut metals such as hardened steel, acetylene is widely used in cutting and welding torches. The names of other alkynes are similar to those of the corresponding alkanes but end in - . For example, HC≡CCH is , and CH C≡CCH is because the multiple bond begins on the second carbon atom. The number of bonds between carbon atoms in a hydrocarbon is indicated in the suffix: In a cyclic hydrocarbon, the ends of a hydrocarbon chain are connected to form a ring of covalently bonded carbon atoms. Cyclic hydrocarbons are named by attaching the prefix - to the name of the alkane, the alkene, or the alkyne. The simplest cyclic alkanes are (C H ) a flammable gas that is also a powerful anesthetic, and (C H ) (part (c) in ). The most common way to draw the structures of cyclic alkanes is to sketch a polygon with the same number of vertices as there are carbon atoms in the ring; each vertex represents a CH unit. The structures of the cycloalkanes that contain three to six carbon atoms are shown schematically in . Alkanes, alkenes, alkynes, and cyclic hydrocarbons are generally called aliphatic hydrocarbons. The name comes from the Greek , meaning “oil,” because the first examples were extracted from animal fats. In contrast, the first examples of , also called , were obtained by the distillation and degradation of highly scented (thus ) resins from tropical trees. The simplest aromatic hydrocarbon is (C H ), which was first obtained from a coal distillate. The word now refers to benzene and structurally similar compounds. As shown in part (a) in , it is possible to draw the structure of benzene in two different but equivalent ways, depending on which carbon atoms are connected by double bonds or single bonds. is similar to benzene, except that one hydrogen atom is replaced by a –CH group; it has the formula C H (part (b) in ). The chemical behavior of aromatic compounds differs from the behavior of aliphatic compounds. Benzene and toluene are found in gasoline, and benzene is the starting material for preparing substances as diverse as aspirin and nylon. illustrates two of the molecular structures possible for hydrocarbons that have six carbon atoms. As shown, compounds with the same molecular formula can have very different structures. Write the condensed structural formula for each hydrocarbon. : name of hydrocarbon : condensed structural formula : : a. The prefix hept- tells us that this hydrocarbon has seven carbon atoms, and n- indicates that the carbon atoms form a straight chain. The suffix -ane tells that it is an alkane, with no carbon–carbon double or triple bonds. The condensed structural formula is CH CH CH CH CH CH CH , which can also be written as \(CH_3(CH_2)_5CH_3\). b. The prefix pent- tells us that this hydrocarbon has five carbon atoms, and the suffix -ene indicates that it is an alkene, with a carbon–carbon double bond. The 2- tells us that the double bond begins on the second carbon of the five-carbon atom chain. The condensed structural formula of the compound is therefore CH CH=CHCH CH . c. The prefix but- tells us that the compound has a chain of four carbon atoms, and the suffix -yne indicates that it has a carbon–carbon triple bond. The 2- tells us that the triple bond begins on the second carbon of the four-carbon atom chain. So the condensed structural formula for the compound is CH C≡CCH . d. A The prefix cyclo- tells us that this hydrocarbon has a ring structure, and oct- indicates that it contains eight carbon atoms, which we can draw as The suffix -ene tells us that the compound contains a carbon–carbon double bond, but where in the ring do we place the double bond? B Because all eight carbon atoms are identical, it doesn’t matter. We can draw the structure of cyclooctene as Write the condensed structural formula for each hydrocarbon. : The general name for a group of atoms derived from an alkane is an . The name of an alkyl group is derived from the name of the alkane by adding the suffix - . Thus the –CH fragment is a group, the –CH CH fragment is an group, and so forth, where the dash represents a single bond to some other atom or group. Similarly, groups of atoms derived from aromatic hydrocarbons are , which sometimes have unexpected names. For example, the –C H fragment is derived from benzene, but it is called a group. In general formulas and structures, alkyl and aryl groups are often abbreviated as R. . Replacing one or more hydrogen atoms of a hydrocarbon with an –OH group gives an alcohol, represented as ROH. The simplest alcohol (CH OH) is called either (its systematic name) or (its common name). Methanol is the antifreeze in automobile windshield washer fluids, and it is also used as an efficient fuel for racing cars, most notably in the Indianapolis 500. Ethanol (or ethyl alcohol, CH CH OH) is familiar as the alcohol in fermented or distilled beverages, such as beer, wine, and whiskey; it is also used as a gasoline additive ( ). The simplest alcohol derived from an aromatic hydrocarbon is C H OH, (shortened from alcohol), a potent disinfectant used in some sore throat medications and mouthwashes. Ethanol, which is easy to obtain from fermentation processes, has successfully been used as an alternative fuel for several decades. Although it is a “green” fuel when derived from plants, it is an imperfect substitute for fossil fuels because it is less efficient than gasoline. Moreover, because ethanol absorbs water from the atmosphere, it can corrode an engine’s seals. Thus other types of processes are being developed that use bacteria to create more complex alcohols, such as octanol, that are more energy efficient and that have a lower tendency to absorb water. As scientists attempt to reduce mankind’s dependence on fossil fuels, the development of these so-called is a particularly active area of research. The simplest organic compounds are the , which contain carbon and hydrogen. contain only carbon–hydrogen and carbon–carbon single bonds, contain at least one carbon–carbon double bond, and contain one or more carbon–carbon triple bonds. Hydrocarbons can also be , with the ends of the chain connected to form a ring. Collectively, alkanes, alkenes, and alkynes are called . , or , are another important class of hydrocarbons that contain rings of carbon atoms related to the structure of benzene (C H ). A derivative of an alkane or an arene from which one hydrogen atom has been removed is called an or an , respectively. are another common class of organic compound, which contain an –OH group covalently bonded to either an alkyl group or an aryl group (often abbreviated ). | 12,115 | 3,095 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/13%3A_Air/13.08%3A_Stratospheric_Ozone-_Earth's_Vital_Shield |
The earth's stratospheric ozone layer plays a critical role in absorbing ultraviolet radiation emitted by the sun. In the last thirty years, it has been discovered that stratospheric ozone is depleting as a result of anthropogenic pollutants. There are a number of chemical reactions that can deplete stratospheric ozone; however, some of the most significant of these involves the catalytic destruction of ozone by halogen radicals such as chlorine and bromine. The atmosphere of the Earth is divided into five layers. In order of closest and thickest to farthest and thinnest the layers are listed as follows: troposphere, stratosphere, mesosphere, thermosphere and exosphere. The majority of the ozone in the atmosphere resides in the stratosphere, which extends from six miles above the Earth’s surface to 31 miles. Humans rely heavily on the absorption of ultraviolet B rays by the ozone layer because UV-B radiation causes skin cancer and can lead to genetic damage. The ozone layer has historically protected the Earth from the harmful UV rays, although in recent decades this protection has diminished due to stratospheric ozone depletion. Ozone depletion is largely a result of man-made substances. Humans have introduced gases and chemicals into the atmosphere that have rapidly depleted the ozone layer in the last century. This depletion makes humans more vulnerable to the UV-B rays which are known to cause skin cancer as well as other genetic deformities. The possibility of ozone depletion was first introduced by scientists in the late 1960's as dreams of super sonic transport began to become a reality. Scientists had long been aware that nitric oxide (NO) can catalytically react with ozone (\(O_3\)) to produce \(O_2\) molecules; however, \(NO\) molecules produced at ground level have a half life far too short to make it into the stratosphere. It was not until the advent of commercial super sonic jets (which fly in the stratosphere and at an altitude much higher then conventional jets) that the potential for \(NO\) to react with stratospheric ozone became a possibility. The threat of ozone depletion from commercial super sonic transport was so great that it is often cited as the main reason why the US federal government pulled support for its development in 1971. Fear of ozone depletion was abated until 1974 when Sherwood Rowland and Mario Molina discovered that chlorofluorocarbons could be photolyzed by high energy photons in the stratosphere. They discovered that this process could releasing chlorine radicals that would catalytically react with \(O_3\) and destroy the molecule. This process is called the Rowland-Molina theory of \(O_3\) depletion. The stratosphere is in a constant cycle with oxygen molecules and their interaction with ultraviolet rays. This process is considered a cycle because of its constant conversion between different molecules of oxygen. The ozone layer is created when ultraviolet rays react with oxygen molecules (O ) to create ozone (O ) and atomic oxygen (O). This process is called the . Step 1: An oxygen molecules is photolyzed by solar radiation, creating two oxygen radicals: \[ h\nu + O_2 \rightarrow 2O^. \nonumber \] Step 2: Oxygen radicals then react with molecular oxygen to produce ozone: \[O_2 + O^. \rightarrow O_3 \nonumber \] Step 3: Ozone then reacts with an additional oxygen radical to form molecular oxygen: \[O_3 + O^. \rightarrow 2O_2 \nonumber \] Step 4: Ozone can also be recycled into molecular oxygen by reacting with a photon: \[O_3 + h\nu \rightarrow O_2 + O^. \nonumber \] It is important to keep in mind that ozone is constantly being created and destroyed by the Chapman cycle and that these reactions are natural processes, which have been taking place for millions of years. Because of this, the thickness the ozone layer at any particular time can vary greatly. It is also important to know that O is constantly being introduced into the atmosphere through photosynthesis, so the ozone layer has the capability of regenerating itself. CFC molecules are made up of chlorine, fluorine and carbon atoms and are extremely stable. This extreme stability allows CFC's to slowly make their way into the stratosphere (most molecules decompose before they can cross into the stratosphere from the troposphere). This prolonged life in the atmosphere allows them to reach great altitudes where photons are more energetic. When the CFC's come into contact with these high energy photons, their individual components are freed from the whole. The following reaction displays how Cl atoms have an ozone destroying cycle: \[Cl + O_3 \rightarrow ClO + O_2 \tag{step 1} \] \[ClO + O^. \rightarrow Cl + O_2 \tag{step 2} \] \[O_3 + O^. \rightarrow 2O_2 \tag{Overall reaction} \] Chlorine is able to destroy so much of the ozone because it acts as a catalyst. Chlorine initiates the breakdown of ozone and combines with a freed oxygen to create two oxygen molecules. After each reaction, chlorine begins the destructive cycle again with another ozone molecule. One chlorine atom can thereby destroy thousands of ozone molecules. Because ozone molecules are being broken down they are unable to absorb any ultraviolet light so we experience more intense UV radiation at the earths surface. From 1985 to 1988, researchers studying atmospheric properties over the south pole continually noticed significantly reduced concentrations of ozone directly over the continent of Antarctica. For three years it was assumed that the ozone data was incorrect and was due to some type of instrument malfunction. In 1988, researchers finally realized their error and concluded that an enormous hole in the ozone layer had indeed developed over Antarctica. Examination of NASA satellite data later showed that the hole had begun to develop in the mid 1970's. The ozone hole over Antarctica is formed by a slew of unique atmospheric conditions over the continent that combine to create an ideal environment for ozone destruction. As spring comes to Antarctica, the PSC's melt in the stratosphere and release all of the halogenated compounds that were previously absorbed to the cloud. In the antarctic summer, high energy photons are able to photolyze the halogenated compounds, freeing halogen radicals that then catalytically destroy O . Because Antarctica is constantly surrounded by a polar vortex, radical halogens are not able to be diluted over the entire globe. The ozone hole develops as result of this process. Resent research suggests that the strength of the polar vortex from any given year is directly correlated to the size of the ozone hole. In years with a strong polar vortex, the ozone hole is seen to expand in diameter, whereas in years with a weaker polar vortex, the ozone hole is noted to shrink The following substances are listed as ozone depleting substances under Title VI of the United State Clean Air Act: ( ) | 6,933 | 3,096 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.05%3A_The_Equilibrium_Constant_in_Terms_of_Pressure |
Some equilibria involve physical instead of chemical processes. One example is the equilibrium between liquid and vapor in a closed container. we stated that the vapor pressure of a liquid was always the same at a given temperature, regardless of how much liquid was present. This can be seen to be a consequence of if we recognize that the pressure of a gas is related to its concentration through . Rearranging PV = RT we obtain \[P=\frac{n}{V}RT=cRT\label{1} \] since = amount of substance/volume = / . Thus if the vapor pressure is constant at a given temperature, the concentration must be constant also. Equation \(\ref{1}\) also allows us to relate the equilibrium constant to the vapor pressure. In the case of water, for example, the equilibrium reaction and are given by: \[\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{O}(g) \nonumber \] \[\text{K}_c = [\text{H}_2\text{O}(g)] \nonumber \] Substituting for the concentration of water vapor from Equation \(\ref{1}\), we obtain \[K_{c}=\frac{P_{\text{H}_{\text{2}}\text{O}}}{RT} \nonumber \] At 25°C for example, the vapor pressure of water is 17.5 mmHg (2.33 kPa), and so we can calculate \[K_{c}=\frac{\text{2}\text{.33 kPa}}{\text{(8}\text{.314 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{)(298}\text{.15 K)}} \nonumber \] \[=\text{9}\text{.40 }\times \text{ 10}^{-\text{4}}\text{ mol/L} \nonumber \] For some purposes it is actually more useful to express the equilibrium law for gases in terms of rather than in terms of concentrations. In the general case: \[a\text{A}(g) + b\text{B}(g) \rightleftharpoons c\text{C}(g) + d\text{D}(g) \nonumber \] The is defined by the relationship: \[K_{p}=\frac{p_{\text{C}}^{c}p_{\text{D}}^{d}}{p_{\text{A}}^{a}p_{\text{B}}^{b}} \nonumber \] where is the partial pressure of component A, of component B, and so on. Since = [A] × RT, = [B] × RT, and so on, we can also write as follows: \[\begin{align} K_{p} & =\frac{p_{\text{C}}^{c}p_{\text{D}}^{d}}{p_{\text{A}}^{a}p_{\text{B}}^{b}}=\frac{\text{( }\!\![\!\!\text{ C }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{c}\text{( }\!\![\!\!\text{ D }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{d}}{\text{( }\!\![\!\!\text{ A }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{a}\text{( }\!\![\!\!\text{ B }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{b}} \\ & =\frac{\text{ }\!\![\!\!\text{ C }\!\!]\!\!\text{ }^{c}\text{ }\!\![\!\!\text{ D }\!\!]\!\!\text{ }^{d}}{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }^{a}\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }^{b}}\text{ }\times \text{ }\frac{\text{(}RT\text{)}^{c}\text{(}RT\text{)}^{d}}{\text{(}RT\text{)}^{a}\text{(}RT\text{)}^{b}} \\ & =K_{c}\text{ }\times \text{ (}RT\text{)}^{\text{(}c\text{ + }d\text{ }-\text{ }a\text{ }-\text{ }b\text{ )}} \\ & =K_{c}\text{ }\times \text{ (}RT\text{)}^{\Delta n} \end{align} \nonumber \] Again \(Δn\) is the increase in the number of gaseous molecules represented in the equilibrium equation. If the number of gaseous molecules does not change, Δn = 0, = , and both equilibrium constants are dimensionless quantities. In what SI units will the equilibrium constant be measured for the following reactions? Also predict for which reactions = . We apply the rule that the units are given by (mol dm ) . | 3,296 | 3,097 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/09%3A_Gases/9.17%3A_Kinetic_Theory_of_Gases-_The_Distribution_of_Molecular_Speeds |
Molecules in the same sample of gas may have widely varying speeds. Much the same way that raindrops rolling down a window vary in speed, even though have the same composition and are exposed to the same conditions, particles within a gas vary in speed. In some cases, we shall find it useful to know just how popular certain speeds are—that is, we might want to know what fraction of the molecules in a sample have speeds between 0 and 400 m s , or how many are going 3600 to 4000 m s . Such information can be obtained from a graph showing the number of molecules within speed on one axis and number of molecules on the other axis. To imagine what such a graph would look like, let us study Figure 9.16.1.c in some detail. The figure depicts H (g) at 300 K. Since the tails on the molecules indicate their velocities, we could count how many are traveling between 0 and 400 m s , how many between 400 and 800 m s , and so on. If these results are plotted as a histogram (bar graph), Figure 2 is obtained. You can see from the histogram, for example, that 20 of the 100 H molecules in Figure \(\Page {1}\) are traveling between 1200 and 1600 m s . Of course any real sample of gas at 300 K and normal atmospheric pressure would contain far more than the 100 molecules in Figure 9.16.1. When the distribution of molecular speeds is measured for such a large number of molecules, the curve printed in red over the histogram is obtained. The curve is known as the of molecular speeds. The most important aspect of the Maxwell-Boltzmann curve is that it is lopsided in favor of higher speeds. The maximum of the first curve is at 1414 m s . This is the most probable speed for an H molecule at 300 K. It is less than the root-mean-square velocity of 1926 m s calculated in . More molecules move faster than the most probable speed than move slower. Moreover, there is a small but very important fraction of molecules with very large velocities. Three have speeds between 3600 and 4000 m s , and one is moving faster than 4000 m s . This is about 3 times the most probable speed and more than double the average rms velocity. Raising the temperature of a gas raises the average speed of its molecules, shifting the Maxwell-Boltzmann curve toward higher speeds, as shown in Figure \(\Page {2}\) b. Increasing the temperature from 300 to 373 K increases the most probable speed from 1414 to 1577 m s and the rms velocity from 1926 to 2157 m s , both increases of 11 percent. The number of really fast molecules goes up much more significantly, however. In the range 3600 to 4000 m s the number of molecules increases from 3 to 5 (67 percent), and in the range 4000 to 4400 m s , the number increases from 1 to 3 (300 percent). If we apply the Maxwell-Boltzmann distribution to 1 mole H ( ) (instead of the 100 molecules in Figure \(\Page {2}\) ), we find 0.75 million molecules moving faster than 10 000 m s at 300 K. At 373 K this number has increased to 465 million—more than 60 000 percent. There are many things in nature which depend on the average (rms) velocity of molecules. A mercury thermometer is one, and the pressure of a gas is another. Many other things, however, are influenced by the number of very fast molecules rather than by the average velocity. One example is the human finger—in a sample of H ( ) at 300 K it feels pleasantly warm, but at 373 K it will blister. This important difference in behavior is not caused by an 11-percent increase in the average velocity of the molecules. It is caused by a dramatic increase in the number of very energetic molecules, which occurs when the temperature is raised from 300 to 373 K. The rates of chemical reactions respond to the number of really fast molecules instead of the average molecular velocity, just as your finger would. Most chemical reactions can be speeded up tremendously by raising the temperature, and this is why chemists so often boil things in flasks to get reactions to occur. | 3,987 | 3,098 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.07%3A_A_Particulate_Model_for_Gases-_Kinetic_Molecular_Theory |
The Learning Objective of this Module is to understand the significance of the kinetic molecular theory of gases. The laws that describe the behavior of gases were well established long before anyone had developed a coherent model of the properties of gases. In this section, we introduce a theory that describes why gases behave the way they do. The theory we introduce can also be used to derive laws such as the ideal gas law from fundamental principles and the properties of individual particles. The kinetic molecular theory of gases explains the laws that describe the behavior of gases. Developed during the mid-19th century by several physicists, including the Austrian Ludwig Boltzmann (1844–1906), the German Rudolf Clausius (1822–1888), and the Englishman James Clerk Maxwell (1831–1879), who is also known for his contributions to electricity and magnetism, this theory is based on the properties of individual particles as defined for an ideal gas and the fundamental concepts of physics. Thus the kinetic molecular theory of gases provides a molecular explanation for observations that led to the development of the ideal gas law. The kinetic molecular theory of gases is based on the following five postulates: Figure 10.7.1 Visualizing molecular motion. Molecules of a gas are in constant motion and collide with one another and with the container wall. Although the molecules of real gases have nonzero volumes and exert both attractive and repulsive forces on one another, for the moment we will focus on how the kinetic molecular theory of gases relates to the properties of gases we have been discussing. In Section 10.8, we explain how this theory must be modified to account for the behavior of real gases. We now describe how the kinetic molecular theory of gases explains some of the important relationships we have discussed previously. \[P_{1}V_{1}=P_{2}V_{2}\] \[\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\] \[P_{total}=P_{a}+P_{b}+P_{c} +...\] | 1,996 | 3,099 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/01%3A_Matter-_Its_Properties_And_Measurement/1.7%3A_Significant_Figures |
The numerical values we deal with in science (and in many other aspects of life) represent measurements whose values are never known exactly. Our pocket-calculators or computers don't know this; they treat the numbers we punch into them as "pure" mathematical entities, with the result that the operations of arithmetic frequently yield answers that are physically ridiculous even though mathematically correct. The purpose of this unit is to help you understand why this happens, and to show you what to do about it. Consider the two statements shown below: Which of these would you be justified in dismissing immediately? Certainly not the second one, because it probably comes from a database which contains one record for each voter, so the number is found simply by counting the number of records. The first statement cannot possibly be correct. Even if a city’s population could be defined in a precise way (Permanent residents? Warm bodies?), how can we account for the minute-by minute changes that occur as people are born and die, or move in and move away? What is the difference between the two population numbers stated above? The first one expresses a quantity that cannot be known exactly — that is, it carries with it a degree of uncertainty. It is quite possible that the last census yielded precisely 157,872 records, and that this might be the “population of the city” for legal purposes, but it is surely not the “true” population. To better reflect this fact, one might list the population (in an atlas, for example) as or even . These two quantities have been rounded off to four and three significant figures, respectively, and the have the following meanings: Which of these two values we would report as “the population” will depend on the degree of confidence we have in the original census figure; if the census was completed last week, we might round to four significant digits, but if it was a year or so ago, rounding to three places might be a more prudent choice. In a case such as this, there is no really objective way of choosing between the two alternatives. This illustrates an important point: the concept of has less to do with mathematics than with our confidence in a measurement. This confidence can often be expressed numerically (for example, the height of a liquid in a measuring tube can be read to ±0.05 cm), but when it cannot, as in our population example, we must depend on our personal experience and judgment. So, what is a significant digit? According to the usual definition, it is all the numerals in a measured quantity (counting from the left) whose values are considered as known exactly, plus one more whose value could be one more or one less: Although , what we are getting rid of can be considered to be “numeric noise” that does not contribute to the quality of the measurement. The purpose in rounding off is to avoid expressing a value to a greater degree of precision than is consistent with the uncertainty in the measurement. If you know that a balance is accurate to within 0.1 mg, say, then the uncertainty in any measurement of mass carried out on this balance will be ±0.1 mg. Suppose, however, that you are simply told that an object has a length of 0.42 cm, with no indication of its precision. In this case, all you have to go on is the number of digits contained in the data. Thus the quantity “0.42 cm” is specified to 0.01 unit in 0 42, or one part in 42 . The implied relative uncertainty in this figure is 1/42, or about 2%. The precision of any numeric answer calculated from this value is therefore limited to about the same amount. It is important to understand that the number of significant digits in a value provides only a rough indication of its precision, and that information is lost when rounding off occurs. Suppose, for example, that we measure the weight of an object as 3.28 g on a balance believed to be accurate to within ±0.05 gram. The resulting value of 3.28±.05 gram tells us that the true weight of the object could be anywhere between 3.23 g and 3.33 g. The absolute uncertainty here is 0.1 g (±0.05 g), and the relative uncertainty is 1 part in 32.8, or about 3 percent. How many significant digits should there be in the reported measurement? Since only the left most “3” in “3.28” is certain, you would probably elect to round the value to 3.3 g. So far, so good. But what is someone else supposed to make of this figure when they see it in your report? The value “3.3 g” suggests an of 3.3±0.05 g, meaning that the true value is likely between 3.25 g and 3.35 g. This range is 0.02 g below that associated with the original measurement, and so rounding off has introduced a bias of this amount into the result. Since this is less than half of the ±0.05 g uncertainty in the weighing, it is not a very serious matter in itself. However, if several values that were rounded in this way are combined in a calculation, the rounding-off errors could become significant. The standard rules for rounding off are well known. Before we set them out, let us agree on what to call the various components of a numeric value. Students are sometimes told to increment the least significant digit by 1 if it is odd, and to leave it unchanged if it is even. One wonders if this reflects some idea that even numbers are somehow “better” than odd ones! (The ancient superstition is just the opposite, that only the odd numbers are "lucky".) In fact, you could do it equally the other way around, incrementing only the even numbers. If you are only rounding a single number, it doesn't really matter what you do. However, when you are rounding a series of numbers that will be used in a calculation, if you treated each first nonsignificant 5 in the same way, you would be over- or understating the value of the rounded number, thus accumulating round-off error. Since there are equal numbers of even and odd digits, incrementing only the one kind will keep this kind of error from building up. You could do just as well, of course, by flipping a coin! Suppose that an object is found to have a weight of 3.98 ± 0.05 g. This would place its true weight somewhere in the range of g to g. In judging how to round this number, you count the number of digits in “3.98” that are known exactly, and you find none! Since the “4” is the left most digit whose value is uncertain, this would imply that the result should be rounded to one significant figure and reported simply as 4 g. An alternative would be to bend the rule and round off to two significant digits, yielding 4.0 g. How can you decide what to do? In a case such as this, you should look at the implied uncertainties in the two values, and compare them with the uncertainty associated with the original measurement. Clearly, rounding off to two digits is the only reasonable course in this example. Observed values should be rounded off to the number of digits that most accurately conveys the uncertainty in the measurement. When carrying out calculations that involve multiple steps, you should avoid doing any rounding until you obtain the final result. Suppose you use your calculator to work out the area of a rectangle: Your calculator is of course correct as far as the pure numbers go, but you would be wrong to write down "1.57676 cm " as the answer. Two possible options for rounding off the calculator answer are shown at the right. It is clear that neither option is entirely satisfactory; rounding to 3 significant digits overstates the precision of the answer, whereas following the rule and rounding to the two digits in ".42" has the effect of throwing away some precision. In this case, it could be argued that rounding to three digits is justified because the implied relative uncertainty in the answer, 0.6%, is more consistent with those of the two factors. The "rules" for rounding off are generally useful, convenient guidelines, but they do not always yield the most desirable result. When in doubt, it is better to rely on relative implied uncertainties. In operations involving significant figures, the answer is reported in such a way that it reflects the reliability of the operation. An answer is no more precise that the least precise number used to get the answer. When adding or subtracting, we go by the number of (i.e., the number of digits on the right side of the decimal point) rather than by the number of significant digits. Identify the quantity having the smallest number of decimal places, and use this number to set the number of decimal places in the answer. The result must contain the same number of significant figures as in the value having the least number of significant figures. If a number is expressed in the form × 10 ("scientific notation") with the additional restriction that the coefficient is no less than 1 and less than 10, the number is in its form. Express the base-10 logarithm of a value using the same number of significant figures as is present in the of that value. Similarly, for antilogarithms (numbers expressed as powers of 10), use the same number of significant figures as are in that power. The following examples will illustrate the most common problems you are likely to encounter in rounding off the results of calculations. They deserve your careful study! A certain book has a thickness of 117 mm; find the height of a stack of 24 identical books: The last of the examples shown above represents the very common operation of converting one unit into another. There is a certain amount of ambiguity here; if we take "9 in" to mean a distance in the range 8.5 to 9.5 inches, then the implied uncertainty is ±0.5 in, which is 1 part in 18, or about ± 6%. The relative uncertainty in the answer must be the same, since all the values are multiplied by the same factor, 2.54 cm/in. In this case we are justified in writing the answer to two significant digits, yielding an uncertainty of about ±1 cm; if we had used the answer "20 cm" (one significant digit), its implied uncertainty would be ±5 cm, or ±25%. When the appropriate number of significant digits is in question, calculating the relative uncertainty can help you decide. | 10,189 | 3,100 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/01%3A_Introduction_to_Analytical_Chemistry/1.02%3A_The_Analytical_Perspective |
Having noted that each area of chemistry brings a unique perspective to the study of chemistry, let’s ask a second deceptively simple question: What is the analytical perspective? Many analytical chemists describe this perspective as an analytical approach to solving problems. For different viewpoints on the analytical approach see (a) Beilby, A. L. , , 237-238; (b) Lucchesi, C. A. , October, 112-119; (c) Atkinson, G. F. , , 201-202; (d) Pardue, H. L.; Woo, J. , , 409-412; (e) Guarnieri, M. , , 201-203, (f) Strobel, H. A. , October, 17-24. Although there likely are as many descriptions of the analytical approach as there are analytical chemists, it is convenient to define it as the five-step process shown in Figure 1.2.1
. Three general features of this approach deserve our attention. First, in steps 1 and 5 analytical chemists have the opportunity to collaborate with individuals outside the realm of analytical chemistry. In fact, many problems on which analytical chemists work originate in other fields. Second, the heart of the analytical approach is a feedback loop (steps 2, 3, and 4) in which the result of one step requires that we reevaluate the other steps. Finally, the solution to one problem often suggests a new problem. Analytical chemistry begins with a problem, examples of which include evaluating the amount of dust and soil ingested by children as an indicator of environmental exposure to particulate based pollutants, resolving contradictory evidence regarding the toxicity of perfluoro polymers during combustion, and developing rapid and sensitive detectors for chemical and biological weapons. At this point the analytical approach involves a collaboration between the analytical chemist and the individual or agency working on the problem. Together they determine what information is needed and clarify how the problem relates to broader research goals or policy issues, both essential to the design of an appropriate experimental procedure. These examples are taken from a series of articles, entitled the “Analytical Approach,” which for many years was a regular feature of the journal . To design the experimental procedure the analytical chemist considers criteria, such as the required accuracy, precision, sensitivity, and detection limit, the urgency with which results are needed, the cost of a single analysis, the number of samples to analyze, and the amount of sample available for analysis. Finding an appropriate balance between these criteria frequently is complicated by their interdependence. For example, improving precision may require a larger amount of sample than is available. Consideration also is given to how to collect, store, and prepare samples, and to whether chemical or physical interferences will affect the analysis. Finally a good experimental procedure may yield useless information if there is no method for validating the results. The most visible part of the analytical approach occurs in the laboratory. As part of the validation process, appropriate chemical and physical standards are used to calibrate equipment and to standardize reagents. The data collected during the experiment are then analyzed. Frequently the data first is reduced or transformed to a more readily analyzable form and then a statistical treatment of the data is used to evaluate accuracy and precision, and to validate the procedure. Results are compared to the original design criteria and the experimental design is reconsidered, additional trials are run, or a solution to the problem is proposed. When a solution is proposed, the results are subject to an external evaluation that may result in a new problem and the beginning of a new cycle. introduces you to the language of analytical chemistry. You will find terms such accuracy, precision, and sensitivity defined there. introduces the statistical analysis of data. Calibration and standardization methods, including a discussion of linear regression, are covered in . See for a discussion of how to collect, store, and prepare samples for analysis. See for a discussion about how to validate an analytical method. As noted earlier some scientists question whether the analytical approach is unique to analytical chemistry. Here, again, it helps to distinguish between a chemical analysis and analytical chemistry. For an analytically-oriented scientist, such as a physical organic chemist or a public health officer, the primary emphasis is how the analysis supports larger research goals that involve fundamental studies of chemical or physical processes, or that improve access to medical care. The essence of analytical chemistry, however, is in developing new tools for solving problems, and in defining the type and quality of information available to other scientists. As an exercise, let’s adapt our model of the analytical approach to the development of a simple, inexpensive, portable device for completing bioassays in the field. Before continuing, locate and read the article “Simple Telemedicine for Developing Regions: Camera Phones and Paper-Based Microfluidic Devices for Real-Time, Off-Site Diagnosis” by Andres W. Martinez, Scott T. Phillips, Emanuel Carriho, Samuel W. Thomas III, Hayat Sindi, and George M. Whitesides. You will find it on pages 3699-3707 in Volume 80 of the journal , which was published in 2008. As you read the article, pay particular attention to how it emulates the analytical approach and consider the following questions: Don’t let the technical details in the paper overwhelm you; if you skim over these you will find the paper both well-written and accessible. A medical diagnoses often relies on the results of a clinical analysis. When you visit a doctor, they may draw a sample of your blood and send it to the lab for analysis. In some cases the result of the analysis is available in 10-15 minutes. What is possible in a developed country, such as the United States, may not be feasible in a country with less access to expensive lab equipment and with fewer trained personnel available to run the tests and to interpret the results. The problem addressed in this paper, therefore, is the development of a reliable device for rapidly performing a clinical assay under less than ideal circumstances. In considering a solution to this problem, the authors identify seven important criteria for the analytical method: (1) it must be inexpensive; (2) it must operate without the need for much electricity, so that it can be used in remote locations; (3) it must be adaptable to many types of assays; (4) its must not require a highly skilled technician; (5) it must be quantitative; (6) it must be accurate; and (7) it must produce results rapidly. The authors describe how they developed a paper-based microfluidic device that allows anyone to run an analysis simply by dipping the device into a sample (synthetic urine, in this case). The sample moves by capillary action into test zones containing reagents that react with specific species (glucose and protein, for this prototype device). The reagents react to produce a color whose intensity is proportional to the species’ concentration. A digital photograph of the microfluidic device is taken using a cell phone camera and sent to an off-site physician who uses image editing software to analyze the photograph and to interpret the assay’s result. In developing this analytical method the authors considered several chemical or physical interferences. One concern was the possibility of non-specific interactions between the paper and the glucose or protein, which might lead to non-uniform image in the test zones. A careful analysis of the distribution of glucose and protein in the text zones showed that this was not a problem. A second concern was the possibility that particulate materials in the sample might interfere with the analyses. Paper is a natural filter for particulate materials and the authors found that samples containing dust, sawdust, and pollen do not interfere with the analysis for glucose. Pollen, however, is an interferent for the protein analysis, presumably because it, too, contains protein. To calibrate the device the authors analyzed a series of standard solutions that contained known concentrations of glucose and protein. Because an image’s intensity depends upon the available light, a standard sample is run with the test samples, which allows a single calibration curve to be used for samples collected under different lighting conditions. The test device contains two test zones for each analyte, which allows for duplicate analyses and provides one level of experimental validation. To further validate the device, the authors completed 12 analyses at each of three known concentrations of glucose and protein, obtaining acceptable accuracy and precision in all cases. Developing this analytical method required several cycles through steps 2, 3, and 4 of the analytical approach. Examples of this feedback loop include optimizing the shape of the test zones and evaluating the importance of sample size. Yes. The authors were successful in meeting their goals by developing and testing an inexpensive, portable, and easy-to-use device for running clinical samples in developing countries. This exercise provides you with an opportunity to think about the analytical approach in the context of a real analytical problem. Practice exercises such as this provide you with a variety of challenges ranging from simple review problems to more open-ended exercises. You will find answers to practice exercises at the end of each chapter. Use this to access the article’s abstract from the journal’s web site. If your institution has an on-line subscription you also will be able to download a PDF version of the article. | 9,808 | 3,102 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/09%3A_Diffusion |
Diffusion can be described as the random movement of particles through space, usually due to a concentration gradient. Diffusion is a spontaneous process and is a result of the random thermal motions between two particles. The diffusion coefficient (\(D\)) can be solved for with Fick’s laws of diffusion, which are broken up into two laws. Fick's first law of diffusion is given by the following equation: \[ J = -D \dfrac{dc}{dx} \label{1} \] where Equation \ref{1} indicates that if the flux and the change in the concentration over time are known, then the diffusion coefficient can be calculated. The negative sign indicates that the concentration gradient is negative. The first law can only be applied to systems in which the conditions remain the same— in other words, if the flux coming into the system equals the flux going out. Fick’s second law is more applicable to physical science and other systems that are changing. This second law is applied to systems in which the condition are not steady, or the solution in not equal throughout. \[ \dfrac{dc}{dt} = D \dfrac{d^2c}{dx^2} \label{2} \] Where \( \frac{dc}{dt} \) represents the rate of change of concentration in a certain area and \( \frac{d^2c}{dx^2} \) represents the changes that the change in concentration can take; this would not be a smooth curve. This term accounts for a varying concentration in the system. If the concentration does not vary, then \[ \dfrac{d^2c}{dx^2} = 0 \label{3} \] Diffusion can be thought of as a series of random steps that the particle takes as it moves from where it started. These steps can either lead the particle away from where it started, or lead back to where it started. This means that the overall path that a particle takes can look like line overlapping itself multiple times. A third way to calculate the diffusion concentration is through the : \[ D = \dfrac{\lambda^2}{2\tau} \label{4} \] where \(λ\) is the length that each step takes, and \(τ\) is the time that each step takes. In this particular model, each step is the same distance. The diffusion coefficient is useful because it can tell you something about the system. For example, different substances have different diffusion coefficients, so knowing this can give you an idea of the substance. Ions at room temperature usually have a diffusion coefficient of 0.6×10 to 2×10 m /s, and biological molecules fall in the range 10 to 10 m /s. The diffusion coefficient changes as the properties of the system change. For example, at higher temperatures, the diffusion coefficient is greater because the molecules have more thermal motion. The diffusion coefficient is also related to the viscosity of the solution. The greater the diffusion coefficient, the lower the viscosity. Because the rate of diffusion depends on the temperature of the system, the can be applied. Applying this equation gives: \[ D = [D]_{o}e^{-Ea/RT} \label{5} \] The dependence of the diffusion coefficient on the viscosity can be modeled by the Stokes-Einstein relation: \[ D = \dfrac{kT}{6\pi\eta{a}} \label{6} \] where a is the radius of the molecule and η is the coefficient of viscosity and is defined by: \[ \eta = \eta_o{e^{Ea/RT}} \label{7} \] This equation demonstrates the dependence of viscosity on temperature. | 3,303 | 3,103 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/10%3A_Gases/10.03%3A_The_Ideal_Gas_Law |
In a gas, molecules freely move, filling any volume that they occupy. The kinetic energy of the molecules greatly exceeds any potential energy of attraction or repulsion between molecules and the size of the molecules are miniscule compared to the average space between them. Contrast this with solids, where the molecules are held in place by the attractive forces and the comparatively small kinetic energy only results in the molecules jiggling in place when thermally excited. Again, in a solid, the molecules are packed so closely together that the volume of the solid is essentially that of the sum of the molecular volumes of all of the molecules in it. Liquids represent a situation where molecular attraction and kinetic energy are balanced and the spacing between molecules is higher than in solids but much less than in gases. At high temperatures and low pressures, where the kinetic energy and the spacing between molecules are both large, one may neglect both the miniscule attractive forces and molecular volume. Under such conditions the properties of the gas, the pressure, P, volume, V, number of moles, n, and the temperature, T are independent of the mixture of molecules in the gas. In that limit the behavior of the gas is called ideal, and governed by the ideal gas law No gases are truly ideal, but while no model is perfect, some are useful, and under most commonly encountered conditions, the ideal gas law is very useful indeed to describe the behavior of gases. It is also easy to predict that the conditions where the ideal gas law would have trouble describing the behavior of a gas would be low temperature (small molecular kinetic energy) and high pressure (little space between molecules). The behavior of gases under such conditions becomes increasingly like that of liquids and will be discussed in the next chapter. The ideal gas law
\( PV = nRT \tag{10.3.1} \) relates the pressure, volume, temperature and number of moles in a gas to each other. R is a constant called the gas constant. The ideal gas law is what is called an equation of state because it is a complete description of the gas's thermodynamic state. No other information is needed to calculate any other thermodynamic variable, and, since the equation relates four variables, a knowledge of any three of them is sufficient. Pressure, volume and number of moles, the latter sometime called extent, share an important property, they can never be negative. What would a negative volume be, or an absolute negative pressure or extent? The concepts do not even exist. This means that temperature in the ideal gas law is similarly limited. We can determine what zero temperature on the ideal gas scale is by holding the number of moles and the pressure constant and extrapolate the temperature measured in Celcius to what its value would be at zero volume °C, which is called absolute zero because no lower temperature is possible (unless, of course you can come up with a negative volumes, but you cannot). For convenience we set the degrees on the Kelvin scale to the size of the degree on the Celcius scale. In other words \( 1 \; K = 1 ^{o}C \tag{10.3.2} \) Before we can use the ideal gas law, however, we need to know the value of the gas constant . Its form depends on the units used for the other quantities in the expression. If is expressed in liters (L), in atmospheres (atm), in kelvins (K), and in moles (mol), then
\( R = 0.08206 \;L \cdot atm /\left (K\cdot ·mol \right ) \tag{10.3.3} \) Because the product has the units of energy, as described in and Essential Skills 4 ( ), can also have units of J/(K·mol) or cal/(K·mol):
\( R = 8.3145 \;J/\left (K \cdot mol \right ) = 1.9872 cal/\left (K \cdot mol \right )
\tag{10.3.4} \) Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and 1 atm pressure, referred to as standard temperature and pressure (STP) This can be confusing because as discussed in the previous chapter,the standard state for enthalpies of reaction and formation is 298K. Be sure to use the appropriate standard state for enthalpies (298 K) and gas problems (273.15 K) We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in :
\( V = \dfrac{nRT}{P}= \dfrac{\left (1.000 \; \cancel{mol} \left ( 0.082057 \; L\cdot \cancel{atm}/\cancel{K}\cdot \cancel{mol} \right )\right )\left ( 273.15 \; \cancel{K} \right )}{1.000\; \cancel{atm}}=22.31\; L \tag{10.3.5} \) Thus the volume of 1 mol of an ideal gas at 0°C and 1 atm pressure is 22.41 L, approximately equivalent to the volume of three basketballs. The quantity 22.41 L is called the standard molar volume of an ideal gas. The molar volumes of several real gases at STP are given in , which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at STP. The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables ( , , , and ). It also allows us to predict the of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters ( , , , and ) are specified for an Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample. The balloon that Charles used for hisflight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft ), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon? volume, temperature, and pressure amount of gas Solve the ideal gas law for the unknown quantity, in this case . Make sure that all quantities are given in units that are compatible with the units of the gas constant. If necessary, convert them to the appropriate units, insert them into the equation you have derived, and then calculate the number of moles of hydrogen gas needed. We are given values for , , and and asked to calculate . If we solve the ideal gas law ( .1) for , we obtain \( n = \dfrac{PV}{RT} \) and are given in units that are not compatible with the units of the gas constant [ = 0.082057 (L·atm)/(K·mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres: \[\rm745\; \cancel{mmHg} \times\dfrac{1\;atm}{760\; \cancel{mmHg}}=0.980\;atm\] \[T=273+30=303{\rm K}\] Substituting these values into the expression we derived for , we obtain \[n=\dfrac{PV}{RT}=\rm\dfrac{0.980\; \cancel{atm} \times31150\; \cancel{L} }{0.08206\dfrac{\cancel{atm} \cdot \cancel{L} }{\rm mol\cdot \cancel{K}}\times 303\; \cancel{K}}=1.23\times10^3\;mol\] Exercise Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO . What is the pressure of the gas at 25°C? 1.5 atm In Example 5, we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of in any of the specified conditions on any of the other parameters, as shown in Example 6. The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain
\[\dfrac{n}{V}=\dfrac{P}{RT}\tag{10.3.6}\] The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass ( , in grams) divided by its molar mass ( , in grams per mole):
\[n=\dfrac{m}{M}\tag{10.3.7}\] Substituting this expression for into .12 gives
\[\dfrac{m}{MV}=\dfrac{P}{RT}\tag{10.3.8}\] Because / is the density of a substance, we can replace / by and rearrange to give
\[ d=\dfrac{m}{V}=\dfrac{MP}{RT}\tag{10.3.9} \] The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL). Calculate the density of butane at 25°C and a pressure of 750 mmHg. compound, temperature, and pressure density Calculate the molar mass of butane and convert all quantities to appropriate units for the value of the gas constant. Substitute these values into to obtain the density. The molar mass of butane (C H ) is \notag Using 0.082057 (L·atm)/(K·mol) for means that we need to convert the temperature from degrees Celsius to kelvins ( = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres: \[P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm \notag \] Substituting these values into gives \[d=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L \notag \] Exercise Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics. radon, 9.23 g/L; N , 1.17 g/L A common use of is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and re-weighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example 10. The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound. pressure, temperature, mass, and volume molar mass and chemical formula Solve for the molar mass of the gas and then calculate the density of the gas from the information given. Convert all known quantities to the appropriate units for the gas constant being used. Substitute the known values into your equation and solve for the molar mass. Propose a reasonable empirical formula using the atomic masses of nitrogen and oxygen and the calculated molar mass of the gas. Solving for the molar mass gives \[M=\dfrac{mRT}{PV}=\dfrac{dRT}{P} \notag \] Density is the mass of the gas divided by its volume: \[d=\dfrac{m}{V}=\dfrac{0.289\rm g}{0.17\rm L}=1.84 \rm g/L \notag \] We must convert the other quantities to the appropriate units before inserting them into the equation: \[T=18+273=291 K \notag \]
\[P=727\rm mmHg\times\dfrac{1\rm atm}{760\rm mmHg}=0.957\rm atm \notag \] The molar mass of the unknown gas is thus \[d=\rm\dfrac{1.84\;g/L\times0.08206\dfrac{L\cdot atm}{K\cdot mol}\times291\;K}{0.957\;atm}=45.9 g/mol \notag \] The atomic masses of N and O are approximately 14 and 16, respectively, so we can construct a list showing the masses of possible combinations: \[M({\rm NO})=14 + 16=30 \rm\; g/mol \notag \]
\[M({\rm N_2O})=(2)(14)+16=44 \rm\;g/mol \notag \]
\[M({\rm NO_2})=14+(2)(16)=46 \rm\;g/mol \notag \] The most likely choice is NO which is in agreement with the data. The red-brown color of smog also results from the presence of NO gas. Exercise You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it. 44 g/mol; CO The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the , = . The proportionality constant, , is called the and has the value 0.08206 (L·atm)/(K·mol), 8.3145 J/(K·mol), or 1.9872 cal/(K·mol), depending on the units used. The ideal gas law describes the behavior of an , a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the . All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity ( , , , or ) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of , , , and ) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured. : \( = where \(R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol}=8.3145 \dfrac{\rm J}{\rm K\cdot mol}\) : \(d=\dfrac{MP}{RT}\) For an ideal gas, is volume directly proportional or inversely proportional to temperature? What is the volume of an ideal gas at absolute zero? What is meant by STP? If a gas is at STP, what further information is required to completely describe the state of the gas? Given the following initial and final values, what additional information is needed to solve the problem using the ideal gas law? Given the following information and using the ideal gas law, what equation would you use to solve the problem? Using the ideal gas law as a starting point, derive the relationship between the density of a gas and its molar mass. Which would you expect to be denser—nitrogen or oxygen? Why does radon gas accumulate in basements and mine shafts? Use the ideal gas law to derive an equation that relates the remaining variables for a sample of an ideal gas if the following are held constant. Tennis balls that are made for Denver, Colorado, feel soft and do not bounce well at lower altitudes. Use the ideal gas law to explain this observation. Will a tennis ball designed to be used at sea level be harder or softer and bounce better or worse at higher altitudes? Calculate the number of moles in each sample at STP. Calculate the number of moles in each sample at STP. Calculate the mass of each sample at STP. Calculate the mass of each sample at STP. Calculate the volume in liters of each sample at STP. Calculate the volume in liters of each sample at STP. Calculate the volume of each gas at STP. Calculate the volume of each gas at STP. One method for preparing hydrogen gas is to pass HCl gas over hot aluminum; the other product of the reaction is AlCl . If you wanted to use this reaction to fill a balloon with a volume of 28,500 L at sea level and a temperature of 78°F, what mass of aluminum would you need? What volume of HCl at STP would you need? An 3.50 g sample of acetylene is burned in excess oxygen according to the following reaction: At STP, what volume of CO (g) is produced? Calculate the density of ethylene (C H ) under each set of conditions. Determine the density of O under each set of conditions. At 140°C, the pressure of a diatomic gas in a 3.0 L flask is 635 kPa. The mass of the gas is 88.7 g. What is the most likely identity of the gas? What volume must a balloon have to hold 6.20 kg of H for an ascent from sea level to an elevation of 20,320 ft, where the temperature is −37°C and the pressure is 369 mmHg? What must be the volume of a balloon that can hold 313.0 g of helium gas and ascend from sea level to an elevation of 1.5 km, where the temperature is 10.0°C and the pressure is 635.4 mmHg? The average respiratory rate for adult humans is 20 breaths per minute. If each breath has a volume of 310 mL of air at 20°C and 0.997 atm, how many moles of air does a person inhale each day? If the density of air is 1.19 kg/m , what is the average molecular mass of air? 281 mmHg 2174 | 17,148 | 3,104 |
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Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 3,105 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./07.E%3A_Periodic_Table_and_Periodic_Trands_(Exercises) |
" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact ♦ Most plants, animals, and bacteria use oxygen as the terminal oxidant in their respiration process. In a few locations, however, whole ecosystems have developed in the absence of oxygen, containing creatures that can use sulfur compounds instead of oxygen. List the common oxidation states of sulfur, provide an example of any oxoanions containing sulfur in these oxidation states, and name the ions. Which ion would be the best oxidant? Titanium is currently used in the aircraft industry and is now used in ships, which operate in a highly corrosive environment. Interest in this metal is due to the fact that titanium is strong, light, and corrosion resistant. The densities of selected elements are given in the following table. Why can an element with an even lower density such as calcium not be used to produce an even lighter structural material? ♦ The compound Fe O was called lodestone in ancient times because it responds to Earth’s magnetic field and can be used to construct a primitive compass. Today Fe O is commonly called magnetite because it contains both Fe and Fe , and the unpaired electrons on these ions align to form tiny magnets. How many unpaired electrons does each ion have? Would you expect to observe magnetic behavior in compounds containing Zn ? Why or why not? Would you expect Fe or Zn to have the lower third ionization energy? Why? ♦ Understanding trends in periodic properties allows us to predict the properties of individual elements. For example, if we need to know whether francium is a liquid at room temperature (approximately 20°C), we could obtain this information by plotting the melting points of the other alkali metals versus atomic number. Based on the data in the following table, would you predict francium to be a solid, a liquid, or a gas at 20°C? Francium is found in minute traces in uranium ores. Is this consistent with your conclusion? Why or why not? Why would francium be found in these ores, but only in small quantities? Due to its 3 3 electron configuration, sulfur has three common oxidation states: +6, +4, and −2. Examples of each are: −2 oxidation state, the sulfide anion, S or hydrogen sulfide, H S; +4 oxidation state, the sulfite ion, SO ; +6 oxidation state, the sulfate ion, SO . The sulfate ion would be the best biological oxidant, because it can accept the greatest number of electrons. Iron(II) has four unpaired electrons, and iron(III) has five unpaired electrons. Compounds of Zn do not exhibit magnetic behavior, because the Zn ion has no unpaired electrons. The third ionization potential of zinc is larger than that of iron, because removing a third electron from zinc requires breaking into the closed 3 subshell. | 2,850 | 3,107 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/03%3A_Chemical_Compounds/3.2%3A_The_Mole_Concept_and_Chemical_Compounds |
\[ 2 \times \text { atomic mass of carbon} = 2 \, atoms \left ( {12.011 \, amu \over atoms } \right ) = 24.022 \,amu \] \[ 6 \times \text { atomic mass of hydrogen} = 2 \, atoms \left ( {1.0079 \, amu \over atoms } \right ) = 6.0474 \,amu \] \[ 1 \times \text { atomic mass of oxygen} = 1 \, atoms \left ( {15.9994 \, amu \over atoms } \right ) = 15.994 \,amu \] 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. 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: \[ \left [ 3 \, atoms Ca \left ({40.078 \, amu \over 1 \, atom Ca } \right ) \right ] + \left [ 2 \, atoms P \left ({30.973761 \, amu \over 1 \, atom P } \right ) \right ] + \left [ 8 \, atoms O \left ({15.9994 \, amu \over 1 \, atom O } \right ) \right ] \] \[= 310.177 \,amu \] \[ 2P \, \, \, \, (2\, atoms)(30.973761 \, amu/atom) = 61.947522 \, amu \] \[ + 8O \, \, \, \, (8\, atoms)(15.9994 \, amu/atom) = 127.9952 \, amu \] \[ Ca_3P_2O_8 \, \, \, \, \text {formula mass of Ca}_3(PO_4)_2 = 310.177 \, amu \] Molar Masses of Compounds: \[(moles)(molar mass) \rightarrow mass \label{3.2.1}\] or, more specifically, \[ moles \left ( {grams \over mole } \right ) = grams \] Conversely, to convert the mass of a substance to moles: \[ \left ( { grams \over grams/mole} \right ) = grams \left ( {mole \over grams } \right ) = moles \label{3.2.2B}\] \[2H_{2(g)} + O_{2(g)} \rightarrow 2H_2O_{(l)}\] 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 \(CO_2\)): \[ 6.88 \times 10^{-2} \; mol\; CO_2 \left( \dfrac{44.0\; g\; CO_2}{1\; mol\; CO_2} \right) = 3.03\;g \; CO_2\] Conversions Between Grams, Mol, & Atoms: 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 \(\Page {4}\). 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 \(\Page {3}\).2.1: \[ 2C (2 \,atoms )(12.011 \, amu/atom) = 24.022 \, amu \] \[ 6H (6 \,atoms )(1.0079 \, amu/atom) = 6.0474 \, amu \] \[ 2O (2 \,atoms )(15.9994 \, amu/atom) = 31.9988 \, amu \] 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. Finding Mols and Masses of Reactants and Products Using Stoichiometric Factors (Mol Ratios): 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. | 12,321 | 3,108 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Vitamins_Cofactors_and_Coenzymes/Flavin_Adenine_Dinucleotide_(FAD) |
The structure shown on the left is for FAD and is similar to in that it contains a vitamin-riboflavin, adenine, ribose, and phosphates. As shown it is the diphosphate, but is also used as the monophosphate (FMN). In the form of FMN it is involved in the first enzyme complex 1 of the . A FMN (flavin adenine mononucleotide) as an oxidizing agent is used to react with NADH for the second step in the electron transport chain. The simplified reaction is: NADH + H + FMN → FMNH + NAD Red.Ag. Ox.Ag. Note the fact that the two hydrogens and 2e are "passed along" from NADH to FFMN. Also note that NAD as a product is back to its original state as an oxidizing agent ready to begin the cycle again. The FMN has now been converted to the reducing agent and is the starting point for the third step. Ubiquinone: As its name suggests, is very widely distributed in nature. There are some differences in the length of the isoprene unit (in bracket on left) side chain in various species. All the natural forms of CoQ are insoluble in water, but soluble in membrane lipids where they function as a mobile electron carrier in the . The long hydrocarbon chain gives the non-polar property to the molecule. CoQ acts as a bridge between enzyme complex 1 and 3 or between complex 2 and 3. Electrons are transferred from NADH along with two hydrogens to the double bond oxygens in the benzene ring. These in turn convert to alcohol groups. The electrons are then passed along to the cytochromes in enzyme complex 3. Although not used in the electron transport chain, Coenzyme A is a major cofactor which is used to transfer a two carbon unit commonly referred to as the The structure has many common features with NAD and FAD in that it has the diphosphate, ribose, and adenine. In addition it has a vitamin called pantothenic acid, and finally terminated by a thiol group. The thiol (-SH) is the sulfur analog of an alcohol (-OH). The acetyl group (CH C=O) is attached to the sulfur of the CoA through a thiol ester type bond. Acetyl CoA is important in the breakdown of fatty acids and is a starting point in the citric acid cycle. | 2,147 | 3,109 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Amino_Acids/Reactions_of_Amino_Acids/Amino_Acid_Reactions |
Amino acids react with each other in a typical acid-base neutralization reaction to form a salt. The acid group becomes negative, and the amine nitrogen becomes positive because of the positive hydrogen ion. For example in the graphic on the left - top, glycine (gly) and alanine (ala) may just interact in the zwitterion form by an attraction of the positive (amine) of the alanine and negative (carboxyl acid) charges to form the salt. A more important interaction for protein tertiary structure is the interaction of the acid and base "side chains". If the amino acid has an extra acid or amine on the "side chain", these are used in the salt formation. For example in the left-bottom graphic, Aspartic acid (asp) has a side chain that forms a salt with the amine on the lysine (lys) side chain. The hydrogen ion (red) moves to the amine nitrogen resulting in the salt with the attraction of the positive and negative charges. The amino acid cysteine undergoes oxidation and reduction reactions involving the -SH (sulfhydryl group). The oxidation of two sulfhydryl groups results in the formation of a by the removal of two hydrogens. The of two cysteine amino acids is shown in the graphic on the left. An unspecified oxidizing agent (O) provides an oxygen which reacts with the hydrogen (red) on the -SH group to form water. The sulfurs (yellow) join to make the . This is an important bond to recognize in protein tertiary structure. The reduction of a disulfide bond is the opposite reaction which again leads to two separate cysteine molecules. Remember that reduction is the addition of hydrogen. | 1,628 | 3,110 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/09%3A_Mixtures/9.02%3A_Partial_Molar_Quantities |
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\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) The symbol \(X_i\), where \(X\) is an extensive property of a homogeneous mixture and the subscript \(i\) identifies a constituent species of the mixture, denotes the of species \(i\) defined by \begin{gather} \s{ X_i \defn \Pd{X}{n_i}{T,p,n_{j \ne i}} } \tag{9.2.1} \cond{(mixture)} \end{gather} This is the rate at which property \(X\) changes with the amount of species \(i\) added to the mixture as the temperature, the pressure, and the amounts of all other species are kept constant. A partial molar quantity is an state function. Its value depends on the temperature, pressure, and composition of the mixture. Keep in mind that as a practical matter, a macroscopic amount of a charged species (i.e., an ion) cannot be added by itself to a phase because of the huge electric charge that would result. Thus if species \(i\) is charged, \(X_i\) as defined by Eq. 9.2.1 is a theoretical concept whose value cannot be determined experimentally. An older notation for a partial molar quantity uses an overbar: \(\overline{X}_i\). The notation \(X_i'\) was suggested in the first edition of the IUPAC Green Book (Ian Mills et al, , Blackwell, Oxford, 1988, p. 44), but is not mentioned in later editions. In order to gain insight into the significance of a partial molar quantity as defined by Eq. 9.2.1, let us first apply the concept to the of an open single-phase system. Volume has the advantage for our example of being an extensive property that is easily visualized. Let the system be a binary mixture of water (substance A) and methanol (substance B), two liquids that mix in all proportions. The partial molar volume of the methanol, then, is the rate at which the system volume changes with the amount of methanol added to the mixture at constant temperature and pressure: \(V\B=\pd{V}{n\B}{T,p,n\A}\). At \(25\units{\(\degC\)}\) and \(1\br\), the molar volume of pure water is \(V\mA^* = 18.07\units{cm\(^3\) mol\(^{-1}\)}\) and that of pure methanol is \(V\mB^* = 40.75\units{cm\(^3\) mol\(^{-1}\)}\). If we mix \(100.0\units{cm\(^3\)}\) of water at \(25\units{\(\degC\)}\) with \(100.0\units{cm\(^3\)}\) of methanol at \(25\units{\(\degC\)}\), we find the volume of the resulting mixture at \(25\units{\(\degC\)}\) is not the sum of the separate volumes, \(200.0\units{cm\(^3\)}\), but rather the slightly smaller value \(193.1\units{cm\(^3\)}\). The difference is due to new intermolecular interactions in the mixture compared to the pure liquids. Let us calculate the mole fraction composition of this mixture: \begin{equation} n\A = \frac{V\A^*}{V\mA^*} = \frac{100.0\units{cm\(^3\)}}{18.07\units{cm\(^3\) mol\(^{-1}\)}} = 5.53\mol \tag{9.2.2} \end{equation} \begin{equation} n\B = \frac{V\B^*}{V\mB^*} = \frac{100.0\units{cm\(^3\)}}{40.75\units{cm\(^3\) mol\(^{-1}\)}} = 2.45\mol \tag{9.2.3} \end{equation} \begin{equation} x\B = \frac{n\B}{n\A+n\B} = \frac{2.45\mol}{5.53\mol + 2.45\mol} = 0.307 \tag{9.2.4} \end{equation} See Fig. 9.3(a) for an example. In this figure, the tangent to the curve drawn at the point on the curve at the composition of interest (the composition used as an illustration in Sec. 9.2.1) intercepts the vertical line where \(x\B\) equals \(0\) at \(V/n = V\A = 17.7\units{cm\(^3\) mol\(^{-1}\)}\), and intercepts the vertical line where \(x\B\) equals \(1\) at \(V/n = V\B = 38.8\units{cm\(^3\) mol\(^{-1}\)}\). To derive this property of a tangent line for the plot of \(V/n\) versus \(x\B\), we use Eq. 9.2.9 to write \begin{equation} \begin{split} (V/n) & = \frac{ V\A n\A + V\B n\B}{n} = V\A x\A + V\B x\B \cr & =V\A(1-x\B) + V\B x\B = (V\B - V\A)x\B + V\A \end{split} \tag{9.2.15} \end{equation} When we differentiate this expression for \(V/n\) with respect to \(x\B\), keeping in mind that \(V\A\) and \(V\B\) are functions of \(x\B\), we obtain \begin{equation} \begin{split} \frac{\dif (V/n)}{\dx\B} & = \frac{\dif[(V\B - V\A)x\B + V\A]}{\dx\B}\cr & = V\B - V\A + \left( \frac{\dif V\B}{\dx\B} - \frac{\dif V\A}{\dx\B} \right)x\B + \frac{\dif V\A}{\dx\B} \cr & = V\B - V\A + \left(\frac{\dif V\A}{\dx\B}\right)(1-x\B) + \left(\frac{\dif V\B}{\dx\B}\right)x\B \cr & = V\B - V\A + \left(\frac{\dif V\A}{\dx\B}\right)x\A + \left(\frac{\dif V\B}{\dx\B}\right)x\B \end{split} \tag{9.2.16} \end{equation} The differentials \(\dif V\A\) and \(\dif V\B\) are related to one another by the Gibbs–Duhem equation (Eq. 9.2.13): \(x\A\dif V\A + x\B\dif V\B = 0\). We divide both sides of this equation by \(\dx\B\) to obtain \begin{equation} \left(\frac{\dif V\A}{\dx\B}\right)x\A + \left(\frac{\dif V\B}{\dx\B}\right)x\B = 0 \tag{9.2.17} \end{equation} and substitute in Eq. 9.2.16 to obtain \begin{equation} \frac{\dif(V/n)}{\dx\B} = V\B - V\A \tag{9.2.18} \end{equation} Let the partial molar volumes of the constituents of a binary mixture of arbitrary composition \(x'\B\) be \(V'\A\) and \(V'\B\). Equation 9.2.15 shows that the value of \(V/n\) at the point on the curve of \(V/n\) versus \(x\B\) where the composition is \(x'\B\) is \((V'\B-V'\A)x'\B+V'\A\). Equation 9.2.18 shows that the tangent to the curve at this point has a slope of \(V'\B - V'\A\). The equation of the line that passes through this point and has this slope, and thus is the tangent to the curve at this point, is \(y=(V'\B - V'\A)x\B + V'\A\), where \(y\) is the vertical ordinate on the plot of \((V/n)\) versus \(x\B\). The line has intercepts \(y{=}V'\A\) at \(x\B{=}0\) and \(y{=}V'\B\) at \(x\B{=}1\). A variant of the method of intercepts is to plot the molar integral volume of mixing given by \begin{equation} \Del V\m\mix = \frac{\Del V\mix}{n} = \frac{V-n\A V\mA^*-n\B V\mB^*}{n} \tag{9.2.19} \end{equation} versus \(x\B\), as illustrated in Fig. 9.3(b). \(\Del V\mix\) is the integral volume of mixing—the volume change at constant \(T\) and \(p\) when solvent and solute are mixed to form a mixture of volume \(V\) and total amount \(n\) (see Sec. 11.1.1). The tangent to the curve at the composition of interest has intercepts \(V\A-V\mA^*\) at \(x\B{=}0\) and \(V\B-V\mB^*\) at \(x\B{=}1\). To see this, we write \begin{equation} \begin{split} \Del V\m\mix & = (V/n) - x\A V\mA^* - x\B V\mB^* \cr & = (V/n) - (1- x\B)V\mA^* - x\B V\mB^* \end{split} \tag{9.2.20} \end{equation} We make the substitution \((V/n)=(V\B-V\A)x\B+V\A\) from Eq. 9.2.15 and rearrange: \begin{equation} \Del V\m\mix = \left[\left( V\B-V\mB^* \right) - \left( V\A-V\mA^* \right) \right]x\B + \left( V\A-V\mA^* \right) \tag{9.2.21} \end{equation} Differentiation with respect to \(x\B\) yields \begin{equation} \begin{split} \frac{\dif \Del V\m\mix}{\dx\B} & = \left( V\B-V\mB^* \right) - \left( V\A-V\mA^* \right) + \left( \frac{\dif V\B}{\dx\B}-\frac{\dif V\A}{\dx\B} \right)x\B + \frac{\dif V\A}{\dx\B} \cr & = \left( V\B-V\mB^* \right) - \left( V\A-V\mA^* \right) + \left(\frac{\dif V\A}{\dx\B}\right)(1-x\B) + \left(\frac{\dif V\B}{\dx\B}\right)x\B \cr & = \left( V\B-V\mB^* \right) - \left( V\A-V\mA^* \right) + \left(\frac{\dif V\A}{\dx\B}\right)x\A + \left(\frac{\dif V\B}{\dx\B}\right)x\B \end{split} \tag{9.2.22} \end{equation} With a substitution from Eq. 9.2.17, this becomes \begin{equation} \frac{\dif \Del V\m\mix}{\dx\B} = \left( V\B-V\mB^* \right) - \left( V\A-V\mA^* \right) \tag{9.2.23} \end{equation} Equations 9.2.21 and 9.2.23 are analogous to Eqs. 9.2.15 and 9.2.18, with \(V/n\) replaced by \(\Del V\m\mix\), \(V\A\) by \((V\A-V\mA^*)\), and \(V\B\) by \((V\B-V\mB^*)\). Using the same reasoning as for a plot of \(V/n\) versus \(x\B\), we find the intercepts of the tangent to a point on the curve of \(\Del V\m\mix\) versus \(x\B\) are at \(V\A-V\mA^*\) and \(V\B-V\mB^*\). Figure 9.3 shows smoothed experimental data for water–methanol mixtures plotted in both kinds of graphs, and the resulting partial molar volumes as functions of composition. Note in Fig. 9.3(c) how the \(V\A\) curve mirrors the \(V\B\) curve as \(x\B\) varies, as predicted by the Gibbs–Duhem equation. The minimum in \(V\B\) at \(x\B {\approx} 0.09\) is mirrored by a maximum in \(V\A\) in agreement with Eq. 9.2.14; the maximum is much attenuated because \(n\B/n\A\) is much less than unity. Macroscopic measurements are unable to provide unambiguous information about molecular structure. Nevertheless, it is interesting to speculate on the implications of the minimum observed for the partial molar volume of methanol. One interpretation is that in a mostly aqueous environment, there is association of methanol molecules, perhaps involving the formation of dimers. The discussion above of partial molar volumes used the notation \(V\mA^*\) and \(V\mB^*\) for the molar volumes of pure A and B. The partial molar volume of a pure substance is the same as the molar volume, so we can simplify the notation by using \(V\A^*\) and \(V\B^*\) instead. Hereafter, this e-book will denote molar quantities of pure substances by such symbols as \(V\A^*\), \(H\B^*\), and \(S_i^*\). The relations derived above for the volume of a binary mixture may be generalized for any extensive property \(X\) of a mixture of any number of constituents. The partial molar quantity of species \(i\), defined by \begin{equation} X_i \defn \Pd{X}{n_i}{T,p,n_{j\ne i}} \tag{9.2.24} \end{equation} is an intensive property that depends on \(T\), \(p\), and the composition of the mixture. The additivity rule for property \(X\) is \begin{gather} \s{ X = \sum_i n_i X_i } \tag{9.2.25} \cond{(mixture)} \end{gather} and the Gibbs–Duhem equation applied to \(X\) can be written in the equivalent forms \begin{gather} \s{ \sum_i n_i \dif X_i = 0 } \tag{9.2.26} \cond{(constant \(T\) and \(p\))} \end{gather} and \begin{gather} \s{ \sum_i x_i \dif X_i = 0 } \tag{9.2.27} \cond{(constant \(T\) and \(p\))} \end{gather} These relations can be applied to a mixture in which each species \(i\) is a nonelectrolyte substance, an electrolyte substance that is dissociated into ions, or an individual ionic species. In Eq. 9.2.27, the mole fraction \(x_i\) must be based on the different species considered to be present in the mixture. For example, an aqueous solution of NaCl could be treated as a mixture of components A=H\(_2\)O and B=NaCl, with \(x\B\) equal to \(n\B/(n\A+n\B)\); or the constituents could be taken as H\(_2\)O, Na\(^+\), and Cl\(^-\), in which case the mole fraction of Na\(^+\) would be \(x_+=n_+/(n\A+n_++n_-)\). A general method to evaluate the partial molar quantities \(X\A\) and \(X\B\) in a binary mixture is based on the variant of the method of intercepts described in Sec. 9.2.3. The molar mixing quantity \(\Del X\mix/n\) is plotted versus \(x\B\), where \(\Del X\mix\) is \((X{-}n\A X\A^*{-}n\B X\B^*)\). On this plot, the tangent to the curve at the composition of interest has intercepts equal to \(X\A{-}X\A^*\) at \(x\B{=}0\) and \(X\B{-}X\B^*\) at \(x\B{=}1\). We can obtain experimental values of such partial molar quantities of an uncharged species as \(V_i\), \(C_{p,i}\), and \(S_i\). It is not possible, however, to evaluate the partial molar quantities \(U_i\), \(H_i\), \(A_i\), and \(G_i\) because these quantities involve the internal energy brought into the system by the species, and we cannot evaluate the absolute value of internal energy (Sec. 2.6.2). For example, while we can evaluate the difference \(H_i-H_i^*\) from calorimetric measurements of enthalpies of mixing, we cannot evaluate the partial molar enthalpy \(H_i\) itself. We can, however, include such quantities as \(H_i\) in useful theoretical relations. A partial molar quantity of a species is something else we cannot evaluate. It is possible, however, to obtain values relative to a reference ion. Consider an aqueous solution of a fully-dissociated electrolyte solute with the formula \(\tx{M}_{\nu_+}\tx{X}_{\nu_-}\), where \(\nu_+\) and \(\nu_-\) are the numbers of cations and anions per solute formula unit. The partial molar volume \(V\B\) of the solute, which can be determined experimentally, is related to the (unmeasurable) partial molar volumes \(V_+\) and \(V_-\) of the constituent ions by \begin{equation} V\B = \nu_+ V_+ + \nu_- V_- \tag{9.2.28} \end{equation} For aqueous solutions, the usual reference ion is H\(^+\), and the partial molar volume of this ion at infinite dilution is arbitrarily set equal to zero: \(V\subs{H\(^+\)}^{\infty}=0\). For example, given the value (at \(298.15\K\) and \(1\br\)) of the partial molar volume at infinite dilution of aqueous hydrogen chloride \begin{equation} V\subs{HCl}^{\infty}=17.82\units{cm\(^3\) mol\(^{-1}\)} \tag{9.2.29} \end{equation} we can find the so-called “conventional” partial molar volume of Cl\(^-\) ion: \begin{equation} V\subs{Cl\(^-\)}^{\infty}=V\subs{HCl}^{\infty}-V\subs{H\(^+\)}^{\infty}=17.82\units{cm\(^3\) mol\(^{-1}\)} \tag{9.2.30} \end{equation} Going one step further, the measured value \(V\subs{NaCl}^{\infty}=16.61\units{cm\(^3\) mol\(^{-1}\)}\) gives, for Na\(^+\) ion, the conventional value \begin{equation} V\subs{Na\(^+\)}^{\infty}=V\subs{NaCl}^{\infty}-V\subs{Cl\(^-\)}^{\infty} = (16.61-17.82)\units{cm\(^3\) mol\(^{-1}\)} = -1.21\units{cm\(^3\) mol\(^{-1}\)} \tag{9.2.31} \end{equation} A of a substance is the partial molar quantity divided by the molar mass, and has dimensions of volume divided by mass. For example, the partial specific volume \(v\B\) of solute B in a binary solution is given by \begin{equation} v\B=\frac{V\B}{M\B} = \bPd{V}{m(\tx{B})}{T,p,m(\tx{A})} \tag{9.2.32} \end{equation} where \(m(\tx{A})\) and \(m(\tx{B})\) are the masses of solvent and solute. Although this e-book makes little use of specific quantities and partial specific quantities, in some applications they have an advantage over molar quantities and partial molar quantities because they can be evaluated without knowledge of the molar mass. For instance, the value of a solute’s partial specific volume is used to determine its molar mass by the method of sedimentation equilibrium (Sec. 9.8.2). The general relations in Sec. 9.2.4 involving partial molar quantities may be turned into relations involving partial specific quantities by replacing amounts by masses, mole fractions by mass fractions, and partial molar quantities by partial specific quantities. Using volume as an example, we can write an additivity relation \(V=\sum_i m(i)v_i\), and Gibbs–Duhem relations \(\sum_i m(i)\dif v_i=0\) and \(\sum_i w_i\dif v_i=0\). For a binary mixture of A and B, we can plot the specific volume \(v\) versus the mass fraction \(w\B\); then the tangent to the curve at a given composition has intercepts equal to \(v\A\) at \(w\B{=}0\) and \(v\B\) at \(w\B{=}1\). A variant of this plot is \(\left(v-w\A v\A^*-w\B v\B^*\right)\) versus \(w\B\); the intercepts are then equal to \(v\A-v\A^*\) and \(v\B-v\B^*\). Just as the molar Gibbs energy of a pure substance is called the and given the special symbol \(\mu\), the partial molar Gibbs energy \(G_i\) of species \(i\) in a mixture is called the of species \(i\), defined by \begin{gather} \s{ \mu_i \defn \Pd{G}{n_i}{T,p,n_{j \ne i}} } \tag{9.2.33} \cond{(mixture)} \end{gather} If there are work coordinates for nonexpansion work, the partial derivative is taken at constant values of these coordinates. The chemical potential of a species in a phase plays a crucial role in equilibrium problems, because it is a measure of the escaping tendency of the species from the phase. Although we cannot determine the absolute value of \(\mu_i\) for a given state of the system, we are usually able to evaluate the difference between the value in this state and the value in a defined reference state. In an open single-phase system containing a mixture of \(s\) different nonreacting species, we may in principle independently vary \(T\), \(p\), and the amount of each species. This is a total of \(2 + s\) independent variables. The total differential of the Gibbs energy of this system is given by Eq. 5.5.9, often called the Gibbs fundamental equation: \begin{gather} \s{ \dif G = -S\dif T + V\difp + \sum_{i=1}^{s} \mu_i \dif n_i } \tag{9.2.34} \cond{(mixture)} \end{gather} Consider the special case of a mixture containing species, for example an aqueous solution of the electrolyte KCl. We could consider the constituents to be either the substances H\(_2\)O and KCl, or else H\(_2\)O and the species K\(^+\) and Cl\(^-\). Any mixture we can prepare in the laboratory must remain electrically neutral, or virtually so. Thus, while we are able to independently vary the amounts of H\(_2\)O and KCl, we cannot in practice independently vary the amounts of K\(^+\) and Cl\(^-\) in the mixture. The chemical potential of the K\(^+\) ion is defined as the rate at which the Gibbs energy changes with the amount of K\(^+\) added at constant \(T\) and \(p\) while the amount of Cl\(^-\) is kept constant. This is a hypothetical process in which the net charge of the mixture increases. The chemical potential of a ion is therefore a valid but purely theoretical concept. Let A stand for H\(_2\)O, B for KCl, \(+\) for K\(^+\), and \(-\) for Cl\(^-\). Then it is theoretically valid to write the total differential of \(G\) for the KCl solution either as \begin{equation} \dif G = -S\dif T + V\difp + \mu\A\dif n\A + \mu\B\dif n\B \tag{9.2.35} \end{equation} or as \begin{equation} \dif G = -S\dif T + V\difp + \mu\A\dif n\A + \mu_+\dif n_+ + \mu_-\dif n_- \tag{9.2.36} \end{equation} This section extends the derivation described in Sec. 8.1.2, which was for equilibrium conditions in a multiphase system containing a single substance, to a more general kind of system: one with two or more homogeneous phases containing mixtures of nonreacting species. The derivation assumes there are no internal partitions that could prevent transfer of species and energy between the phases, and that effects of gravity and other external force fields are negligible. The system consists of a reference phase, \(\pha'\), and other phases labeled by \(\pha{\ne}\pha'\). Species are labeled by subscript \(i\). Following the procedure of Sec. 8.1.1, we write for the total differential of the internal energy \begin{equation} \begin{split} \dif U & = \dif U\aphp + \sum_{\pha\ne\pha'}\dif U\aph \cr & = T\aphp\dif S\aphp - p\aphp\dif V\aphp + \sum_i\mu_i\aphp\dif n_i\aphp \cr & \quad + \sum_{\pha\ne\pha'}\left(T\aph\dif S\aph - p\aph\dif V\aph + \sum_i\mu_i\aph\dif n_i\aph\right) \end{split} \tag{9.2.37} \end{equation} The conditions of isolation are \begin{equation} \dif U = 0 \qquad \tx{(constant internal energy)} \tag{9.2.38} \end{equation} \begin{equation} \dif V\aphp + \sum_{\pha\ne\pha'}\dif V\aph = 0 \qquad \tx{(no expansion work)} \tag{9.2.39} \end{equation} \begin{equation} \begin{split} &\tx{For each species \(i\):} \cr &\dif n_i\aphp + \sum_{\pha\ne\pha'}\dif n_i\aph = 0 \qquad \tx{(closed system)} \end{split} \tag{9.2.40} \end{equation} We use these relations to substitute for \(\dif U\), \(\dif V\aphp\), and \(\dif n_i\aphp\) in Eq. 9.2.37. After making the further substitution \(\dif S\aphp = \dif S - \sum_{\pha\ne\pha'}\dif S\aph\) and solving for \(\dif S\), we obtain \begin{equation} \dif S = \sum_{\pha\ne\pha'}\frac{T\aphp-T\aph}{T\aphp}\dif S\aph - \sum_{\pha\ne\pha'}\frac{p\aphp-p\aph}{T\aphp}\dif V\aph + \sum_i\sum_{\pha\ne\pha'}\frac{\mu_i\aphp-\mu_i\aph}{T\aphp}\dif n_i\aph \tag{9.2.41} \end{equation} This equation is like Eq. 8.1.6 with provision for more than one species. In the equilibrium state of the isolated system, \(S\) has the maximum possible value, \(\dif S\) is equal to zero for an infinitesimal change of any of the independent variables, and the coefficient of each term on the right side of Eq. 9.2.41 is zero. We find that in this state each phase has the same temperature and the same pressure, and for each species the chemical potential is the same in each phase. Suppose the system contains a species \(i'\) that is effectively excluded from a particular phase, \(\pha''\). For instance, sucrose molecules dissolved in an aqueous phase are not accommodated in the crystal structure of an ice phase, and a nonpolar substance may be essentially insoluble in an aqueous phase. We can treat this kind of situation by setting \(\dif n^{\pha''}_{i'}\) equal to zero. Consequently there is no equilibrium condition involving the chemical potential of this species in phase \(\pha''\). To summarize these conclusions: In an equilibrium state of a multiphase, multicomponent system without internal partitions, the temperature and pressure are uniform throughout the system, and each species has a uniform chemical potential except in phases where it is excluded. This statement regarding the uniform chemical potential of a species applies to both a substance and an ion, as the following argument explains. The derivation in this section begins with Eq. 9.2.37, an expression for the total differential of \(U\). Because it is a total differential, the expression requires the amount \(n_i\) of each species \(i\) in each phase to be an independent variable. Suppose one of the phases is the aqueous solution of KCl used as an example at the end of the preceding section. In principle (but not in practice), the amounts of the species H\(_2\)O, K\(^+\), and Cl\(^-\) can be varied independently, so that it is valid to include these three species in the sums over \(i\) in Eq. 9.2.37. The derivation then leads to the conclusion that K\(^+\) has the same chemical potential in phases that are in transfer equilibrium with respect to K\(^+\), and likewise for Cl\(^-\). This kind of situation arises when we consider a Donnan membrane equilibrium (Sec. 12.7.3) in which transfer equilibrium of ions exists between solutions of electrolytes separated by a semipermeable membrane. Here we derive several useful relations involving partial molar quantities in a single-phase system that is a mixture. The independent variables are \(T\), \(p\), and the amount \(n_i\) of each constituent species \(i\). From Eqs. 9.2.26 and 9.2.27, the Gibbs–Duhem equation applied to the chemical potentials can be written in the equivalent forms \begin{gather} \s{ \sum_i n_i \dif\mu_i = 0 } \tag{9.2.42} \cond{(constant \(T\) and \(p\))} \end{gather} and \begin{gather} \s{ \sum_i x_i \dif\mu_i = 0 } \tag{9.2.43} \cond{(constant \(T\) and \(p\))} \end{gather} These equations show that the chemical potentials of different species cannot be varied independently at constant \(T\) and \(p\). A more general version of the Gibbs–Duhem equation, without the restriction of constant \(T\) and \(p\), is \begin{equation} S\dif T - V\difp + \sum_i n_i \dif\mu_i = 0 \tag{9.2.44} \end{equation} This version is derived by comparing the expression for \(\dif G\) given by Eq. 9.2.34 with the differential \(\dif G {=} \sum_i \mu_i\dif n_i {+} \sum_i n_i\dif\mu_i\) obtained from the additivity rule \(G {=} \sum_i \mu_i n_i\). The Gibbs energy is defined by \(G=H-TS\). Taking the partial derivatives of both sides of this equation with respect to \(n_i\) at constant \(T\), \(p\), and \(n_{j \ne i}\) gives us \begin{equation} \Pd{G}{n_i}{T,p,n_{j \ne i}} = \Pd{H}{n_i}{T,p,n_{j \ne i}} - T\Pd{S}{n_i}{T,p,n_{j \ne i}} \tag{9.2.45} \end{equation} We recognize each partial derivative as a partial molar quantity and rewrite the equation as \begin{equation} \mu_i = H_i -TS_i \tag{9.2.46} \end{equation} This is analogous to the relation \(\mu=G/n=H\m-TS\m\) for a pure substance. From the total differential of the Gibbs energy, \(\dif G = -S\dif T + V\difp + \sum_i \mu_i\dif n_i\) (Eq. 9.2.34), we obtain the following reciprocity relations: \begin{equation} \Pd{\mu_i}{T}{p,\allni} = -\Pd{S}{n_i}{T,p,n_{j \ne i}} \qquad \Pd{\mu_i}{p}{T,\allni} = \Pd{V}{n_i}{T,p,n_{j \ne i}} \tag{9.2.47} \end{equation} The symbol \(\allni\) stands for the set of amounts of all species, and subscript \(\allni\) on a partial derivative means the amount of species is constant—that is, the derivative is taken at constant composition of a closed system. Again we recognize partial derivatives as partial molar quantities and rewrite these relations as follows: \begin{equation} \Pd{\mu_i}{T}{p,\allni} = -S_i \tag{9.2.48} \end{equation} \begin{equation} \Pd{\mu_i}{p}{T,\allni} = V_i \tag{9.2.49} \end{equation} These equations are the equivalent for a mixture of the relations \(\pd{\mu}{T}{p} = -S\m\) and \(\pd{\mu}{p}{T} = V\m\) for a pure phase (Eqs. 7.8.3 and 7.8.4). Taking the partial derivatives of both sides of \(U=H-pV\) with respect to \(n_i\) at constant \(T\), \(p\), and \(n_{j \ne i}\) gives \begin{equation} U_i = H_i - pV_i \tag{9.2.50} \end{equation} Finally, we can obtain a formula for \(C_{p,i}\), the partial molar heat capacity at constant pressure of species \(i\), by writing the total differential of \(H\) in the form \begin{equation} \begin{split} \dif H & = \Pd{H}{T}{p,\allni}\dif T + \Pd{H}{p}{T,\allni}\difp + \sum_i \Pd{H}{n_i}{T,p,n_{j \ne i}}\dif n_i \cr & = C_p\dif T + \Pd{H}{p}{T,\allni}\difp + \sum_i H_i\dif n_i \end{split} \tag{9.2.51} \end{equation} from which we have the reciprocity relation \(\pd{C_p}{n_i}{T,p,n_{j \ne i}} = \pd{H_i}{T}{p,\allni}\), or \begin{equation} C_{p,i} = \Pd{H_i}{T}{p,\allni} \tag{9.2.52} \end{equation} | 32,662 | 3,111 |
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In the late 1970’s the lipid hypothesis came in to existences stating that eating saturated fats leads to elevated LDL (Low Density Lipoprotein) which was perceived to be "bad cholesterol." This will result in coronary heart disease which is hardening and narrowing of arteries resulting in heart attack. Fats were eventually classified in to 2 categories: “healthy fats” and “unhealthy fats”. Unhealthy fats where perceived to be fats and healthy fats where perceived to be fats. A meta-analysis of 72 studies with over 103,052 people have found no validity in the lipid hypothesis. The conclusion of the Meta-Analysis was,“In contrast to current recommendations, this systematic review found no evidence that saturated fat increases the risk of coronary disease, or that polyunsaturated fats have a cardio protective effect.”[1] Dietary fats play a critical role in human health. They help keep cells healthy, help with brain development, help with the use of fat soluble vitamins, and they help cushion organs protecting them against blunt trauma. Fats come in multiple forms, saturated, unsaturated and trans fats just to name a few. Saturated fats are solid at room temperature due to their molecular shape. The term saturated is in reference to an sp carbon chain that has its remaining sp orbitals bonded with hydrogen atoms. Thus the term “saturated”. It’s “saturated” with hydrogen. Saturated fats have a chain like structure which allows them to stack very well forming a solid at room temperature. Unsaturated fats are not linear due to double bonded carbons which results in a different molecular shape because the sp carbons are trigonal planar, not tetrahedral (sp carbons) as the carbons are in saturated fats. This change in structure will cause the fat molecules to not stack very well resulting in fats that are liquid at room temperature. Butter is mostly saturated fat, that’s why it’s solid at room temperature. Olive Oil is liquid at room temperature, thus it’s an unsaturated fat. An unsaturated fat can be made in to a saturated fat via hydrogenation reactions. Unsaturated fatty acids may be converted to saturated fatty acids by the relatively simple hydrogenation reaction. Recall that the addition of hydrogen to an alkene (unsaturated) results in an alkane (saturated). A simple hydrogenation reaction is: \[\ce{H_2C=CH_2 + H_2 \rightarrow CH_3CH_3}\] alkene plus hydrogen yields an alkane Vegetable oils are commonly referred to as "polyunsaturated". This simply means that there are several double bonds present. Vegetable oils may be converted from liquids to solids by the hydrogenation reaction. Margarines and shortenings are "hardened" in this way to make them solid or semi-solids. Vegetable oils which have been partially hydrogenated, are now partially saturated so the melting point increases to the point where a solid is present at room temperature. The degree of hydrogenation of unsaturated oils controls the final consistency of the product. What has happened to the healthfulness of the product which has been converted from unsaturated to saturated fats? A major health concern during the hydrogenation process is the production of trans fats. Trans fats are the result of a side reaction with the catalyst of the hydrogenation process. This is the result of an unsaturated fat which is normally found as a cis isomer converts to a trans isomer of the unsaturated fat. Isomers are molecules that have the same molecular formula but are bonded together differently. Focusing on the sp double bonded carbons, a cis isomer has the hydrogens on the same side. Due to the added energy from the hydrogenation process, the activation energy is reached to convert the cis isomers of the unsaturated fat to a trans isomer of the unsaturated fat. The effect is putting one of the hydrogens on the opposite side of one of the carbons. This results in a trans configuration of the double bonded carbons. The human body does not recognize trans fats. Although trans fatty acids are chemically "monounsaturated" or "polyunsaturated," they are considered so different from the cis monounsaturated or polyunsaturated fatty acids that they can not be legally designated as unsaturated for purposes of labeling. Most of the trans fatty acids (although chemically still unsaturated) produced by the partial hydrogenation process are now classified in the same category as saturated fats. The major negative is that trans fat tends to raise "bad" LDL- cholesterol and lower "good" HDL-cholesterol, although not as much as saturated fat. Trans fat are found in margarine, baked goods such as doughnuts and Danish pastry, deep-fried foods like fried chicken and French-fried potatoes, snack chips, imitation cheese, and confectionery fats. | 4,789 | 3,112 |
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A consequence of is that the distribution of electrons in the d orbitals may lead to net stabilization (decrease in energy) of some complexes depending on the specific ligand field geometry and metal d-electron configurations. It is a simple matter to calculate this stabilization since all that is needed is the electron configuration and knowledge of the splitting patterns. The Crystal Field Stabilization Energy is defined as the energy of the electron configuration in the ligand field minus the energy of the electronic configuration in the isotropic field. \[CFSE=\Delta{E}=E_{\text{ligand field}} - E_{\text{isotropic field}} \label{1}\] The CSFE will depend on multiple factors including: For an octahedral complex, an electron in the more stable \(t_{2g}\) subset is treated as contributing \(-2/5\Delta_o\) whereas an electron in the higher energy \(e_g\) subset contributes to a destabilization of \(+3/5\Delta_o\). The final answer is then expressed as a multiple of the crystal field splitting parameter \(\Delta_o\). If any electrons are paired within a single orbital, then the term \(P\) is used to represent the spin pairing energy. What is the Crystal Field Stabilization Energy for a high spin \(d^7\) octahedral complex? The splitting pattern and electron configuration for both isotropic and octahedral ligand fields are compared below. The energy of the isotropic field \((E_{\text{isotropic field}}\)) is \[ E_{\text{isotropic field}}= 7 \times 0 + 2P = 2P \nonumber \] The energy of the octahedral ligand field \(E_{\text{ligand field}}\) is \[E_{\text{ligand field}} = (5 \times -2/5 \Delta_o ) + (2 \times 3/5 \Delta_o) + 2P = -4/5 \Delta_o + 2P \nonumber \] So via Equation \ref{1}, the CFSE is \[\begin{align} CFSE &=E_{\text{ligand field}} - E_{\text{isotropic field}} \nonumber \\[4pt] &=( -4/5\Delta_o + 2P ) - 2P \nonumber \\[4pt] &=-4/5 \Delta_o \nonumber \end{align} \nonumber \] Notice that the Spin pairing Energy falls out in this case (and will when calculating the CFSE of high spin complexes) since the number of paired electrons in the ligand field is the same as that in isotropic field of the free metal ion. What is the Crystal Field Stabilization Energy for a low spin \(d^7\) octahedral complex? The splitting pattern and electron configuration for both isotropic and octahedral ligand fields are compared below. The energy of the isotropic field is the same as calculated for the high spin configuration in Example 1: \[ E_{\text{isotropic field}}= 7 \times 0 + 2P = 2P \nonumber\] The energy of the octahedral ligand\) field \(E_{\text{ligand field}}\) is \[\begin{align} E_{\text{ligand field}} &= (6 \times -2/5 \Delta_o ) + (1 \times 3/5 \Delta_o) + 3P \nonumber \\[4pt] &= -9/5 \Delta_o + 3P \nonumber \end{align} \nonumber \] So via Equation \ref{1}, the CFSE is \[\begin{align} CFSE&=E_{\text{ligand field}} - E_{\text{isotropic field}} \nonumber \\[4pt] &=( -9/5 \Delta_o + 3P ) - 2P \nonumber \\[4pt] &=-9/5 \Delta_o + P \nonumber \end{align} \nonumber \] Adding in the pairing energy since it will require extra energy to pair up one extra group of electrons. This appears more a more stable configuration than the high spin \(d^7\) configuration in Example \(\Page {1}\), but we have then to take into consideration the Pairing energy \(P\) to know definitely, which varies between \(200-400\; kJ\; mol^{-1}\) depending on the metal. \(P\) is the and represents the energy required to pair up electrons within the same orbital. For a given metal ion P (pairing energy) is constant, but it does not vary with ligand and of the metal ion). Similar CFSE values can be constructed for non-octahedral ligand field geometries once the knowledge of the d-orbital splitting is known and the electron configuration within those orbitals known, e.g., the tetrahedral complexes in Table \(\Page {2}\). These energies geoemtries can then be contrasted to the octahedral CFSE to calculate a thermodynamic preference (Enthalpy-wise) for a metal-ligand combination to favor the octahedral geometry. This is quantified via a Octahedral Site Preference Energy defined below. The Octahedral Site Preference Energy (OSPE) is defined as the difference of CFSE energies for a non-octahedral complex and the octahedral complex. For comparing the preference of forming an octahedral ligand field vs. a tetrahedral ligand field, the OSPE is thus: \[OSPE = CFSE_{(oct)} - CFSE_{(tet)} \label{2}\] The OSPE quantifies the preference of a complex to exhibit an octahedral geometry vs. a tetrahedral geometry. Note: the conversion between \(\Delta_o\) and \(\Delta_t\) used for these calculations is: \[\Delta_t \approx \dfrac{4}{9} \Delta_o \label{3}\] which is applicable for comparing octahedral and tetrahedral complexes that involve same ligands only. \(P\) is the and represents the energy required to pair up electrons within the same orbital. Tetrahedral complexes are always high spin since the splitting is appreciably smaller than \(P\) (Equation \ref{3}). After conversion with Equation \ref{3}. The data in Tables \(\Page {1}\) and \(\Page {2}\) are represented graphically by the curves in Figure \(\Page {1}\) below for the high spin complexes only. The low spin complexes require knowledge of \(P\) to graph. The "double-humped" curve in Figure \(\Page {1}\) is found for various properties of the first-row transition metals, including Hydration and of the M(II) ions, ionic radii as well as the stability of M(II) complexes. This suggests that these properties are somehow related to Crystal Field effects. In the case of Hydration Energies describing the of water ligands to a bare metal ion: \[M^{2+} (g) + H_2O \rightarrow [M(OH_2)_6]^{2+} (aq)\] Table \(\Page {3}\) and Figure \(\Page {1}\) shows this type of curve. Note that in any series of this type not all the data are available since a number of ions are not very stable in the M(II) state. | 5,931 | 3,113 |
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A metal ion in solution does not exist in isolation, but in combination with ligands (such as solvent molecules or simple ions) or chelating groups, giving rise to complex ions or coordination compounds. These complexes contain a central atom or ion, often a transition metal, and a cluster of ions or neutral molecules surrounding it. Many complexes are relatively unreactive species remaining unchanged throughout a sequence of chemical or physical operations and can often be isolated as stable solids or liquid compounds. Other complexes have a much more transient existence and may exist only in solution or be highly reactive and easily converted to other species. All metals form complexes, although the extent of formation and nature of these depend very largely on the electronic structure of the metal. The concept of a metal complex originated in the work of Alfred Werner, who in 1913 was awarded the first Nobel Prize in Inorganic chemistry. Complexes may be non-ionic (neutral) or cationic or anionic, depending on the charges carried by the central metal ion and the coordinated groups. The total number of points of attachment to the central element is termed the and this can vary from 2 to greater than 12, but is usually 6. The term come from the latin word , which meaning to bind) was first used by Alfred Stock in 1916 in relation to silicon chemistry. The first use of the term in a British journal was by H. Irving and R.J.P. Williams in , 1948, 162, 746 in their paper describing what is now called the . Ligands can be further characterised as monodentate, bidentate, tridentate etc. where the concept of teeth is introduced, hence the idea of bite angle etc. The term was first applied in 1920 by Sir Gilbert T. Morgan and H.D.K. Drew [ , 1920, 117, 1456], who stated: "The adjective chelate, derived from the great claw or of the lobster or other crustaceans, is suggested for the caliperlike groups which function as two associating units and fasten to the central atom so as to produce heterocyclic rings." Metal complexation is of widespread interest. It is studied not only by inorganic chemists, but by physical and organic chemists, biochemists, pharmacologists, molecular biologists and environmentalists. The "stability of a complex in solution" refers to the degree of association between the two species involved in the state of equilibrium. Qualitatively, the greater the association, the greater the stability of the compound. The magnitude of the (stability or formation) equilibrium constant for the association, quantitatively expresses the stability. Thus, if we have a reaction of the type: \[M + 4L \rightarrow ML_4\] then the larger the stability constant, the higher the proportion of \(ML_4\) that exists in the solution. Free metal ions rarely exist in solution so that \(M\), will usually be surrounded by solvent molecules which will compete with the ligand molecules, \(L\), and be successively replaced by them. For example in water: \[ M(H_2O)_4 + L \rightleftharpoons M(H_2O)_3L + H_2O\] \[ M(H_2O)_3L + L \rightleftharpoons M(H_2O)_2L_2 + H_2O\] However, for simplicity, we generally ignore these solvent molecules and write four stability constants as follows: \(M + L → ML\) with \[K_1 = \dfrac{[ML]}{[M] [L]}\] \(ML + L → ML_2\) with \[K_2 = \dfrac{[ML_2]}{[ML] [L]}\] \(ML_2 + L → ML_3\) with \[K_3 = \dfrac{[ML3]}{[ML_2] [L]}\] \(ML_3 + L → ML_4\) with \[K_4 = \dfrac{[ML_4]}{[ML_3] [L]}\] where \(K_1\), \(K_2\) etc. are referred to as "stepwise stability constants." Alternatively, the "Overall Stability Constant" can be constructed \[ M + 4L \rightarrow ML_4 \beta_4 = \dfrac{[ML_4]}{[M] [L]^4}\] The stepwise and overall stability constants are therefore related as follows: \[\beta_4 =K_1 \times K_2 \times K_3 \times K_4 \] or more generally, \[ \beta_n =K_1 \times K_2 \times K_3 \times K_4 \times ... K_n\] outlines the sequential formation constants for a range of metal-ligand complexes. Consider the four steps involved in the formation of the cuprammonium ion \(Cu(NH_3)_4^{2+}\): \[Cu^{2+} + NH_3 \rightleftharpoons Cu(NH_3)^{2+}\] with \[K_1 = \dfrac{[Cu(NH_3)^{2+}]}{[Cu^{2+}] [NH_3]}\] \[ CuNH_3^{2+} + NH_3 \rightleftharpoons Cu(NH_3)_2^{2+}\] with \[K_2 = \dfrac{ [Cu(NH_3)_2^{2+}]}{[Cu(NH_3)^{2+}] [NH_3]}\] \[ Cu(NH_3)_2^{2+} + NH_3 \rightleftharpoons Cu(NH_3)_3^{2+}\] with \[K_3 = \dfrac{ [Cu(NH_3)_3^{2+}]}{[Cu(NH_3)_2^{2+}] [NH_3]}\] \[ Cu(NH_3)_3^{2+} + NH_3 \rightleftharpoons Cu(NH_3)_4^{2+}\] with \[K_4 = \dfrac{ [Cu(NH_3)_4^{2+}]}{[Cu(NH_3)_3^{2+}] [NH_3]}\] where the \(\{K\}\) constants are the . Also: \[\beta_4 = \dfrac{[Cu(NH_3)_4^{2+}]}{[Cu^{2+}] [NH_3]^4}\] The addition of the four ammine groups to copper shows a pattern found for most formation constants, in that the successive stability constants decrease. In this case, the four constants are: \[\log K_1 =4.0,\] \[\log K_2 =3.2,\] \[\log K_3 =2.7,\] \[\log K_4 =2.0 \; \text{or} \; \log \beta_4 =11.9\] Pay attention to definitions: a number of texts may refer to \(\beta_4\) as the or the of coordination complexes (e.g., termed \(\beta'_4\) below) , which corresponds to the of the formation constant (\(K_4\)), since the reactions referred to are those where fully formed complexes break down to the aqua ion and free ligands. For example, \[\beta'_4 = \dfrac{[Cu^{2+}] [NH_3]^4}{[Cu(NH_3)_4^{2+}]}\] However, this is not definition use here ( ) and should be compared with the equation for the formation constant given earlier. It is usual to represent the metal-binding process by a series of stepwise equilibria which lead to stability constants that may vary numerically from hundreds to enormous values such as 10 and more. That is For this reason, they are commonly reported as logarithms. So log (β) = log (10 ) = 35. It is additionally useful to use logarithms, since log(K) is directly proportional to the free energy of the reaction. \[ΔG^o = -RT \ln(β)\] \[ΔG^o = -2.303 RT \log_{10}(β)\] \[ΔG^o = ΔH^o - TΔS^o\] Below are three tables of the formation constants and thermodynamics properties of example ligand-metal complexes; a more complete table can be found in . The chelate effect can be seen by comparing the reaction of a chelating ligand and a metal ion with the corresponding reaction involving comparable monodentate ligands. For example, comparison of the binding of 2,2'-bipyridine with pyridine or 1,2-diaminoethane (ethylenediamine=en) with ammonia. It has been known for many years that a comparison of this type always shows that the complex resulting from coordination with the chelating ligand is . This can be seen by looking at the values for adding two monodentates compared with adding one bidentate, or adding four monodentates compared to two bidentates, or adding six monodentates compared to three bidentates. A number of points should be highlighted from the formation constants in . In Table \(\Page {1}\), it can be seen that the ΔH° values for the formation steps are almost identical, that is, heat is evolved to about the same extent whether forming a complex involving monodentate ligands or bidentate ligands. What is seen to vary significantly is the ΔS° term which changes from negative (unfavorable) to positive (favorable). Note as well that there is a dramatic increase in the size of the ΔS° term for adding two compared to adding four monodentate ligands. (-5 to -35 JK mol ). What does this imply, if we consider ΔS° to give a measure of disorder? In the case of complex formation of Ni with ammonia or 1,2-diaminoethane, by rewriting the equilibria, the following equations are produced. Using the equilibrium constant for the reaction (3 above) where the bidentate ligands replace the monodentateligands, we find that at a temperature of 25° C: \[\Delta G^° = -2.303 \, RT \, \log_{10} K\] \[= -2.303 \,RT \,(18.28 - 8.61)\] \[= -54 \text{ kJ mol}^{-1}\] Based on measurements made over a range of temperatures, it is possible to break down the \(\Delta G^°\) term into the enthalpy and entropy components. \[\Delta G^° = \Delta H^° - T \Delta S^°\] The result is that: \(\Delta H^° = -29 kJ mol^{-1}\) - TΔS° = -25 kJ mol
and at 25C (298K)
ΔS° = +88 J K mol For many years, these numbers have been in textbooks. For example, the third edition of "Basic Inorganic Chemistry" by F.A. Cotton, G. Wilkinson and P.L. Gaus, John Wiley & Sons, Inc, 1995, on page 186 gives the values as: ΔG° = -67 kJ mol
ΔH° = -12 kJ mol
-TΔS° = -55 kJ mol The conclusion they drew from these incorrect numbers was that the chelate effect was essentially an entropy effect, since the TΔS° contribution was nearly 5 times bigger than ΔH°. In fact, the breakdown of the ΔG° into ΔH° and TΔS° shows that the two terms are nearly equal (-29 and -25 kJ mol , respectively) with the ΔH° term a little bigger! The entropy term found is still much larger than for reactions involving a non-chelating ligand substitution at a metal ion. How can we explain this enhanced contribution from entropy? One explanation is to count the number of species on the left and right hand side of the equation above. It will be seen that on the left-hand-side there are 4 species, whereas on the right-hand-side there are 7 species, that is a net gain of 3 species occurs as the reaction proceeds. This can account for the increase in entropy since it represents an increase in the disorder of the system. An alternative view comes from trying to understand how the reactions might proceed. To form a complex with 6 monodentates requires 6 separate favorable collisions between the metal ion and the ligand molecules. To form the tris-bidentate metal complex requires an initial collision for the first ligand to attach by one arm but remember that the other arm is always going to be nearby and only requires a rotation of the other end to enable the ligand to form the chelate ring. If you consider dissociation steps, then when a monodentate group is displaced, it is lost into the bulk of the solution. On the other hand, if one end of a bidentate group is displaced the other arm is still attached and it is only a matter of the arm rotating around and it can be reattached again. conditions favor the formation of the complex with bidentate groups rather than monodentate groups. Calculate the entropy changes at 25°C, for the following reactions: Zn + 2NH <=> [Zn(NH ) ] ΔH = -28.03 kJ mol and log β = 5.01 Zn + en <=> [Zn(en)] ΔH = -27.6 kJ mol and log β = 6.15 (NB R=8.314 JK mol ) The calculations make use of the equations:
First it is necessary to calculate the ΔG values from the given formation constants and temperature 25C = 298K. For the ammonia complex, ΔG = -8.314 * 298 * 2.303 * 5.01 kJmol , that is -28.6 kJmol . For the 1,2-diaminoethane complex, ΔG= -8.314 * 298 * 2.303 * 6.15 kJmol which corresponds to -35.1 kJmol . This makes use of the relation between Ln and log10 such that ln(x) = 2.303 log10 (x). Then using the second relationship above, ΔS can be found. \[\Delta S = \dfrac{\Delta H - \Delta G}{T}\] For the ammonia complex this gives 1.9 JK mol and for the 1,2-diaminoethane complex 25.2 JK mol , which are the values given as answer 2. | 11,206 | 3,114 |
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The phenomenon of superconductivity was discovered by the Danish physicist H. Kamerlingh Onnes (1853–1926; Nobel Prize in Physics, 1913), who found a way to liquefy helium, which boils at 4.2 K and 1 atm pressure. To exploit the very low temperatures made possible by this new cryogenic fluid, he began a systematic study of the properties of metals, especially their electrical properties. Because the electrical resistance of a sample is technically easier to measure than its conductivity, Onnes measured the of his samples. The resistivity and conductivity of a material are inversely proportional:
\[ conductivity = \dfrac{1}{resistivity} \tag{12.7.1} \] In 1911, Onnes discovered that at about 4 K, the resistivity of metallic mercury (melting point = 234 K) decreased suddenly to essentially zero, rather than continuing to decrease only slowly with decreasing temperature as expected ( ). He called this phenomenon superconductivity because a resistivity of zero means that an electrical current can flow forever. Onnes soon discovered that many other metallic elements exhibit superconductivity at very low temperatures. Each of these superconductors has a characteristic superconducting transition temperature ( ) at which its resistivity drops to zero. At temperatures less than their , superconductors also completely expel a magnetic field from their interior (part (a) in ). This phenomenon is called the Meissner effect after one of its discoverers, the German physicist Walther Meissner, who described the phenomenon in 1933. Due to the Meissner effect, a superconductor will actually “float” over a magnet, as shown in part (b) in . For many years, the phenomenon of superconductivity could not be satisfactorily explained by the laws of conventional physics. In the early 1950s, however, American physicists John Bardeen, Leon Cooper, and John Schrieffer formulated a theory for superconductivity that earned them the Nobel Prize in Physics in 1972. According to the BCS theory (named for the initials of their last names), electrons are able to travel through a solid with zero resistance because of attractive interactions involving two electrons that are at some distance from each other. As one electron moves through the lattice, the surrounding nuclei are attracted to it. The motion of the nuclei can create a transient (short-lived) hole that pulls the second electron in the same direction as the first. The nuclei then return to their original positions to avoid colliding with the second electron as it approaches. The pairs of electrons, called Cooper pairs , migrate through the crystal as a unit. The electrons in Cooper pairs change partners frequently, like dancers in a ballet. According to the BCS theory, as the temperature of the solid increases, the vibrations of the atoms in the lattice increase continuously, until eventually the electrons cannot avoid colliding with them. The collisions result in the loss of superconductivity at higher temperatures. The phenomenon of superconductivity suggested many exciting technological applications. For example, using superconducting wires in power cables would result in zero power losses, even over distances of hundreds of miles. Additionally, because superconductors expel magnetic fields, a combination of magnetic rails and superconducting wheels (or vice versa) could be used to produce of, for example, a train over the track, resulting in friction-free transportation. Unfortunately, for many years the only superconductors known had serious limitations, especially the need for very low temperatures, which required the use of expensive cryogenic fluids such as liquid He. In addition, the superconducting properties of many substances are destroyed by large electrical currents or even moderately large magnetic fields, making them useless for applications in power cables or high-field magnets. The ability of materials such as NbTi, NbSn, Nb Si, and Nb Ge to tolerate rather high magnetic fields, however, has led to a number of commercial applications of superconductors, including high-field magnets for nuclear magnetic resonance (NMR) spectrometers and magnetic resonance imaging (MRI) instruments in medicine, which, unlike x-rays, can detect small changes in soft tissues in the body. Because of these limitations, scientists continued to search for materials that exhibited superconductivity at temperatures greater than 77 K (the temperature of liquid nitrogen, the least expensive cryogenic fluid). In 1986, Johannes G. Bednorz and Karl A. Müller, working for IBM in Zurich, showed that certain mixed-metal oxides containing La, Ba, and Cu exhibited superconductivity above 30 K. These compounds had been prepared by French workers as potential solid catalysts some years earlier, but their electrical properties had never been examined at low temperatures. Although initially the scientific community was extremely skeptical, the compounds were so easy to prepare that the results were confirmed within a few weeks. These high-temperature superconductors earned Bednorz and Müller the Nobel Prize in Physics in 1987. Subsequent research has produced new compounds with related structures that are superconducting at temperatures as high as 135 K. The best known of these was discovered by Paul Chu and Maw-Kuen Wu Jr. and is called the “Chu–Wu phase” or the 1-2-3 superconductor. The formula for the 1-2-3 superconductor is YBa Cu O , where is about 0.1 for samples that superconduct at about 95 K. If ≈ 1.0, giving a formula of YBa Cu O , the material is an electrical insulator. The superconducting phase is thus a nonstoichiometric compound, with a fixed ratio of metal atoms but a variable oxygen content. The overall equation for the synthesis of this material is as follows:
\[ Y_{2}O_{3}\left ( s \right )+4BaCO_{3}\left ( s \right )+6CuO\left ( s \right )+\dfrac{1}{2}O_{2}\left ( g \right )\overset{\Delta }{\rightarrow}2YBa_{2}Cu_{3}O_{7\left ( s \right )}+4CO_{2}\left ( g \right ) \tag{12.7.2} \] If we assume that the superconducting phase is really stoichiometric YBa Cu O , then the average oxidation states of O, Y, Ba, and Cu are −2, +3, +2, and +7/3 respectively. The simplest way to view the average oxidation state of Cu is to assume that two Cu atoms per formula unit are present as Cu and one is present as the rather unusual Cu . In YBa Cu O , the insulating form, the oxidation state of Cu is+5/3 so there are two Cu and one Cu per formula unit. As shown in , the unit cell of the 1-2-3 superconductor is related to the unit cell of the simple perovskite structure (part (b) in ). The only difference between the superconducting and insulating forms of the compound is that an O atom has been removed from between the Cu ions, which destroys the chains of Cu atoms and leaves the Cu in the center of the unit cell as Cu . The chains of Cu atoms are crucial to the formation of the superconducting state. lists the ideal compositions of some of the known high-temperature superconductors that have been discovered in recent years. Engineers have learned how to process the brittle polycrystalline 1-2-3 and related compounds into wires, tapes, and films that can carry enormous electrical currents. Commercial applications include their use in infrared sensors and in analog signal processing and microwave devices. The Composition of Various Superconductors Calculate the average oxidation state of Cu in a sample of YBa Cu O with = 0.5. How do you expect its structure to differ from those shown in for YBa Cu O and YBa Cu O ? stoichiometry average oxidation state and structure Based on the oxidation states of the other component atoms, calculate the average oxidation state of Cu that would make an electrically neutral compound. Compare the stoichiometry of the structures shown in with the stoichiometry of the given compound to predict how the structures differ. The net negative charge from oxygen is (7.0 − 0.5) (−2) = −13, and the sum of the charges on the Y and Ba atoms is [1 × (+3)] + [2 × (+2)] = +7. This leaves a net charge of −6 per unit cell, which must be compensated for by the three Cu atoms, for a net charge of per Cu. The most likely structure would be one in which every other O atom between the Cu atoms in the Cu chains of YBa Cu O has been removed. Exercise Calculate the average oxidation state of Cu in a sample of HgBa Ca Cu O . Assume that Hg is present as Hg . +2 are solids that at low temperatures exhibit zero resistance to the flow of electrical current, a phenomenon known as . The temperature at which the electrical resistance of a substance drops to zero is its . Superconductors also expel a magnetic field from their interior, a phenomenon known as the . Superconductivity can be explained by the , which says that electrons are able to travel through a solid with no resistance because they couple to form pairs of electrons (Cooper pairs). have values greater than 30 K. Why does the BCS theory predict that superconductivity is not possible at temperatures above approximately 30 K? How does the formation of Cooper pairs lead to superconductivity? According to BCS theory, the interactions that lead to formation of Cooper pairs of electrons are so weak that they should be disrupted by thermal vibrations of lattice atoms above about 30 K. | 9,381 | 3,117 |
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If the reaction follows a zeroth order rate law, it can be expressed in terms of the time-rate of change of [A] (which will be negative since A is a reactant): \[-\dfrac{d[A]}{dt} = k \nonumber \] In this case, it is straightforward to separate the variables. Placing time variables on the right and [A] on the left \[ d[A] = - k \,dt \nonumber \] In this form, it is easy to integrate. If the concentration of A is [A] at time t = 0, and the concentration of A is [A] at some arbitrary time later, the form of the integral is \[ \int _{[A]_o}^{[A]} d[A] = - k \int _{t_o}^{t}\,dt \nonumber \] which yields \[ [A] - [A]_o = -kt \nonumber \] or \[ [A] = [A]_o -kt \nonumber \] This suggests that a plot of concentration as a function of time will produce a straight line, the slope of which is –k, and the intercept of which is [A] . If such a plot is linear, then the data are consistent with 0 order kinetics. If they are not, other possibilities must be considered. | 982 | 3,118 |
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As chemists, we are concerned with chemical changes and reactions. The thermodynamics of chemical reactions can be very important in terms of controlling the production of desired products and preventing safety hazards such as explosions. As such, measuring and understanding the thermochemistry of chemical reactions is not only useful, but essential! The techniques of can be used to measure q for a chemical reaction directly. The enthalpy change for a chemical reaction is of significant interest to chemists. An exothermic reaction will release heat (\(q_{reaction} < 0\), \(q_{surroundings} > 0\)) causing the temperature of the surrounding to increase. Conversely, an endothermic reaction (\(q_{reaction} > 0\), \(q_{surroundings} < 0\)) will draw heat from the surroundings, causing the temperature of the surrounding to drop. Measuring the temperature change in the surroundings allows for the determination of how much heat was released or absorbed in the reaction. Bomb calorimetry is used predominantly to measure the heat evolved in combustion reactions, but can be used for a wide variety of reactions. A typical bomb calorimetry set up is shown here. The reaction is contained in a heavy metallic container (the bomb) forcing the reaction to occur at constant volume. As such, the heat evolved (or absorbed) by the reaction is equal to the change in internal energy ( U ). The bomb is then submerged in a reproducible quantity of water, the temperature of which is monitored with a high-precision thermometer. For combustion reactions, the bomb will be loaded with a small sample of the compound to be combusted, and then the bomb is filled with a high pressure (typically about 10 atm) of O . The reaction is initiated by supplying heat using a short piece of resistive wire carrying an electrical current. The calorimeter must be calibrated by carrying out a reaction for which \(\Delta U_{rxn}\) is well known, so that the resulting temperature change can be related to the amount of heat released or absorbed. A commonly used reaction is the combustion of benzoic acid. This makes a good choice since benzoic acid reacts reliably and reproducibly under normal bomb calorimetry conditions. The “water equivalent” of the calorimeter can then be calculated from the temperature change using the following relationship: \[W = \dfrac{n\Delta U_c +e_{wrire}+e_{other}}{\Delta T} \nonumber \] where n is the number of moles of benzoic acid used, \(\Delta U_c\) is the internal energy of combustion for benzoic acid (3225.7 kJ mol at 25 C), \(e_{wire}\) accounts for the energy released in the combustion of the fuse wire, e account for any other corrections (such as heat released due to the combustion of residual nitrogen in the bomb), and T is the measured temperature change in the surrounding water bath. Once the “water equivalent” is determined for a calorimeter, the temperature change can be used to find \(\Delta U_c\) for an unknown compound from the temperature change created upon combustion of a known quantity of the substance. \[ \Delta U_c = \dfrac{W \Delta T - e_{wire} - e_{other}}{n_{sample}} \nonumber \] The experiment above is known as “isothermal bomb calorimetry” as the entire assembly sits in a constant temperature laboratory. Another approach is to employ “adiabatic bomb calorimetry” in which the assembly sits inside of a water jacket, the temperature of which is controlled to match the temperature of the water inside the insulated container. By matching this temperature, there is no thermal gradient, and thus no heat leaks into or out of the assembly during an experiment (and hence the experiment is effectively “adiabatic”). The can be calculated from the internal energy change if the balanced chemical reaction is known. Recall from the definition of enthalpy \[\Delta H = \Delta U + \Delta (pV) \nonumber \] and if the gas-phase reactants and products can be treated as ideal gases (\(pV = nRT\)) \[\Delta H = \Delta U + RT \Delta n_{gas} \nonumber \] at constant temperature. For the combustion of benzoic acid at 25 C \[\ce{C6H5COOH (s) + 15/2 O_2(g) -> 7 CO2(g) + 3 H2O(l)} \nonumber \] it can be seen that \(\Delta n_{gas}\) is -0.5 mol of gas for every mole of benzoic acid reacted. A student burned a 0.7842 g sample of benzoic acid (\(\ce{C7H6O2}\)) in a bomb calorimeter initially at 25.0 C and saw a temperature increase of 2.02 C. She then burned a 0.5348 g sample of naphthalene (\(\ce{C10H8}\)) (again from an initial temperature of 25 C) and saw a temperature increase of 2.24 C. From this data, calculate \(\Delta H_c\) for naphthalene (assuming and are unimportant.) First, the water equivalent: \[W = \dfrac{\left[ (0.7841\,g) \left(\frac{1\,mol}{122.124 \, g} \right)\right] (3225.7 \,kJ/mol)}{2.02 \,°C} = 10.254 \, kJ/°C \nonumber \] Then \(\Delta U_c\) for the sample: \[\Delta U_c = \dfrac{(10.254\, kJ/\,°C)(2.24\,°C )}{(0.5308 \,g)\left(\frac{1\,mol}{128.174 \, g} \right) } = 5546.4 \, kJ/°C \nonumber \] \(\Delta H_c\) is then given by \[\Delta H_c = \Delta U_c + RT \Delta n_{gas} \nonumber \] The reaction for the combustion of naphthalene at 25 C is: \[\ce{ C10H8(s) + 12 O2(g) -> 10 CO2(g) + 4 H2O(l)} \nonumber \] with \(\Delta n_{gas} = -2\). So \[ \Delta H_c = 5546.4 \,kJ/mol + \left( \dfrac{8.314}{1000} kJ/(mol \, K) \right) (298 \,L) (-2) = 5541\, kJ/mol \nonumber \] The literature value (Balcan, Arzik, & Altunata, 1996) is 5150.09 kJ/mol. So that’s not too far off! | 5,487 | 3,120 |
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A first order rate law would take the form \[ \dfrac{d[A]}{dt} = k[A] \nonumber \] Again, separating the variables by placing all of the concentration terms on the left and all of the time terms on the right yields \[ \dfrac{d[A]}{[A]} =-k\,dt \nonumber \] This expression is also easily integrated as before \[ \int_{[A]=0}^{[A]} \dfrac{d[A]}{[A]} =-k \int_{t=0}^{t=t}\,dt \nonumber \] Noting that \[ \dfrac{dx}{x} = d (\ln x) \nonumber \] The form of the integrated rate law becomes \[ \ln [A] - \ln [A]_o = kt \nonumber \] or \[ \ln [A] = \ln [A]_o - kt \label{In1} \] This form implies that a plot of the natural logarithm of the concentration is a linear function of the time. And so a plot of ln[A] as a function of time should produce a linear plot, the slope of which is -k, and the intercept of which is ln[A] . Consider the following kinetic data. Use a graph to demonstrate that the data are consistent with first order kinetics. Also, if the data are first order, determine the value of the rate constant for the reaction. The plot looks as follows: From this plot, it can be seen that the rate constant is 0.0149 s . The concentration at time \(t = 0\) can also be inferred from the intercept. It should also be noted that the integrated rate law (Equation \ref{In1}) can be expressed in exponential form: \[ [A] = [A]_o e^{-kt} \nonumber \] Because of this functional form, 1 order kinetics are sometimes referred to as exponential decay kinetics. Many processes, including radioactive decay of nuclides follow this type of rate law. | 1,563 | 3,121 |
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When cooled, most liquids undergo a simple phase transition to an ordered crystalline solid, a relatively rigid substance that has a fixed shape and volume. In the phase diagrams for these liquids, there are no regions between the liquid and solid phases. Thousands of substances are known, however, that exhibit one or more phases intermediate between the liquid state, in which the molecules are free to tumble and move past one another, and the solid state, in which the molecules or ions are rigidly locked into place. In these intermediate phases, the molecules have an ordered arrangement and yet can still flow like a liquid. Hence they are called liquid crystals , and their unusual properties have found a wide range of commercial applications. They are used, for example, in the liquid crystal displays (LCDs) in digital watches, calculators, and computer and video displays. The first documented example of a liquid crystal was reported by the Austrian Frederick Reinitzer in 1888. Reinitzer was studying the properties of a cholesterol derivative, cholesteryl benzoate, and noticed that it behaved strangely as it melted. The white solid first formed a cloudy white liquid phase at 145°C, which reproducibly transformed into a clear liquid at 179°C. The transitions were completely reversible: cooling molten cholesteryl benzoate below 179°C caused the clear liquid to revert to a milky one, which then crystallized at the melting point of 145°C. In a normal liquid, the molecules possess enough thermal energy to overcome the intermolecular attractive forces and tumble freely. This arrangement of the molecules is described as isotropic , which means that it is equally disordered in all directions. Liquid crystals, in contrast, are anisotropic : their properties depend on the direction in which they are viewed. Hence liquid crystals are not as disordered as a liquid because the molecules have some degree of alignment. Most substances that exhibit the properties of liquid crystals consist of long, rigid rod- or disk-shaped molecules that are easily polarizable and can orient themselves in one of three different ways, as shown in . In the nematic phase , the molecules are not layered but are pointed in the same direction. As a result, the molecules are free to rotate or slide past one another. In the smectic phase , the molecules maintain the general order of the nematic phase but are also aligned in layers. Several variants of the smectic phase are known, depending on the angle formed between the molecular axes and the planes of molecules. The simplest such structure is the so-called smectic A phase, in which the molecules can rotate about their long axes within a given plane, but they cannot readily slide past one another. In the cholesteric phase , the molecules are directionally oriented and stacked in a helical pattern, with each layer rotated at a slight angle to the ones above and below it. As the degree of molecular ordering increases from the nematic phase to the cholesteric phase, the liquid becomes more opaque, although direct comparisons are somewhat difficult because most compounds form only one of these liquid crystal phases when the solid is melted or the liquid is cooled. Molecules that form liquid crystals tend to be rigid molecules with polar groups that exhibit relatively strong dipole–dipole or dipole–induced dipole interactions, hydrogen bonds, or some combination of both. Some examples of substances that form liquid crystals are listed in along with their characteristic phase transition temperature ranges. In most cases, the intermolecular interactions are due to the presence of polar or polarizable groups. Aromatic rings and multiple bonds between carbon and nitrogen or oxygen are especially common. Moreover, many liquid crystals are composed of molecules with two similar halves connected by a unit having a multiple bond. Because of their anisotropic structures, liquid crystals exhibit unusual optical and electrical properties. The intermolecular forces are rather weak and can be perturbed by an applied electric field. Because the molecules are polar, they interact with an electric field, which causes them to change their orientation slightly. Nematic liquid crystals, for example, tend to be relatively translucent, but many of them become opaque when an electric field is applied and the molecular orientation changes. This behavior is ideal for producing dark images on a light or an opalescent background, and it is used in the LCDs in digital watches; handheld calculators; flat-screen monitors; and car, ship, and aircraft instrumentation. Although each application differs in the details of its construction and operation, the basic principles are similar, as illustrated in . Liquid crystals tend to form from long, rigid molecules with polar groups. Changes in molecular orientation that are dependent on temperature result in an alteration of the wavelength of reflected light. Changes in reflected light produce a change in color, which can be customized by using either a single type of liquid crystalline material or mixtures. It is therefore possible to build a liquid crystal thermometer that indicates temperature by color ( ) and to use liquid crystals in heat-sensitive films to detect flaws in electronic board connections where overheating can occur. We also see the effect of liquid crystals in nature. Iridescent green beetles, known as jewel beetles, change color because of the light-reflecting properties of the cells that make up their external skeletons, not because of light absorption from their pigment. The cells form helices with a structure like those found in cholesteric liquid crystals. When the pitch of the helix is close to the wavelength of visible light, the cells reflect light with wavelengths that lead to brilliant metallic colors. Because a color change occurs depending on a person’s angle of view, researchers in New Zealand are studying the beetles to develop a thin material that can be used as a currency security measure. The automobile industry is also interested in exploring such materials for use in paints that would change color at different viewing angles. With only molecular structure as a guide, one cannot precisely predict which of the various liquid crystalline phases a given compound will actually form. One can, however, identify molecules containing the kinds of structural features that tend to result in liquid crystalline behavior, as demonstrated in Example 11. Which molecule is most likely to form a liquid crystalline phase as the isotropic liquid is cooled? compounds liquid crystalline behavior Determine which compounds have a rigid structure and contain polar groups. Those that do are likely to exhibit liquid crystal behavior. Exercise Which compound is least likely to form a liquid crystal phase? (b) Biphenyl; although it is rather long and rigid, it lacks any polar substituents. Many substances exhibit phases that have properties intermediate between those of a crystalline solid and a normal liquid. These substances, which possess long-range molecular order but still flow like liquids, are called . Liquid crystals are typically long, rigid molecules that can interact strongly with one another; they do not have structures, which are completely disordered, but rather have structures, which exhibit different properties when viewed from different directions. In the , only the long axes of the molecules are aligned, whereas in the , the long axes of the molecules are parallel and the molecules are arranged in planes. In the , the molecules are arranged in planes, but each layer is rotated by a certain amount with respect to those above and below it, giving a helical structure. Describe the common structural features of molecules that form liquid crystals. What kind of intermolecular interactions are most likely to result in a long-chain molecule that exhibits liquid crystalline behavior? Does an electrical field affect these interactions? What is the difference between an liquid and an liquid? Which is more anisotropic—a cholesteric liquid crystal or a nematic liquid crystal? | 8,195 | 3,122 |
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Li, Na, K, Rb, and Cs are all group IA elements, also known as the alkali metals. The seventh member of the group, francium (Fr) is radioactive and so rare that only 20 atoms of Fr may exist on Earth at any given moment . The term is derived from an Arabic word meaning “ashes.” Compounds of potassium as well as other alkali metals were obtained from wood ashes by early chemists. All the alkali metals are soft and, except for Cs which is yellow, are silvery-gray in color. Lithium, sodium, potassium, rubidium, and cesium have a great many other properties in common. All are solids at 0°C and melt below 200°C. Each has metallic properties such as good conduction of heat and electricity, malleability (the ability to be hammered into sheets), and ductility (the ability to be drawn into wires). The high thermal (heat) conductivity and the relatively low melting point (for a metal) of sodium make it an ideal heat-transfer fluid. It is used to cool certain types of nuclear reactors (liquid-metal fast breeder reactors, LMFBRs) and to cool the valves of high-powered automobile engines for this reason. Some general properties of the alkali metals are summarized in the table below. All these metal atoms contain a singles electron outside a noble-gas configuration, and so the valence electron is-well shielded from nuclear charge and the atomic radii are relatively large. The large volume of each atom results in a low density—small enough that Li, Na, and K float on water as they react with it. The atoms do not have a strong attraction for the single valence electron, and so it is easily lost (small first ionization energy) to from a +1 ion. Because they readily donate electrons in this way, all the alkali metals are strong reducing agents. They are quite reactive, even reducing water. Weak attraction for the valence electron also results in weak metallic bonding, because it is attraction among nuclei and numerous valence electrons that holds metal atoms together. Weak metallic bonding results in low melting points, especially for the larger atoms toward the bottom of the group. Cs, for example, melts just above room temperature. Weak metallic bonding also accounts for the fact that all these metals are rather soft. That the chemistry of alkali metals is confined to the +1 oxidation state is confirmed by the large second-ionization energies. Removing the first electron from the large, diffuses orbital is easy, but removing a second electron from an octet in an M ion is much too difficult for any oxidizing agent to do. Two other elements are found in group IA. Hydrogen, although many of its compounds have formulas similar to the alkali metals, is a nonmetal and is almost unique in its chemical behavior. Therefore it is not usually included in this group. Francium (Fr) is quite radioactive, and only small quantities are available for study; so it too is usually omitted. Its properties, however, appear to be similar to those of Cs and the other alkali metals. The element lithium combines violently and spectacularly with water. Hydrogen gas is given off, which propels the the lithium metal across the water as it reacts. If the excess water is evaporated, the compound lithium hydroxide (LiOH) remains behind. LiOH is visualized by phenolphthalein indicator, which turns pink as LiOH, a base, is produced. Thus the equation for this reaction is \[\text{2Li}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2LiOH}(aq) + \text{H}_\text{2}(g)\nonumber \] The elements sodium, potassium, rubidium, and cesium also combine violently with water to form hydroxides. The equations for their reactions are \[\text{2Na}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2NaOH}(aq) + \text{H}_\text{2} (g)\nonumber \] \[\text{2K}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2KOH}(aq) + \text{H}_\text{2} (g)\nonumber \] \[\text{2Rb}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2RbOH}(aq) + \text{H}_\text{2} (g)\nonumber \] \[\text{2Cs}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2CsOH}(aq) + \text{H}_\text{2} (g)\nonumber \] Since the alkali metals all react with water in the same way, a may be written: \[\text{2M}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2MOH}(aq) + \text{H}_\text{2} (g)\nonumber \] with M = K, Li, Na, Rb, or Cs. The symbol M represents any one of the five elements. In addition to their behavior when added to water, the alkali metals react directly with many elements. All combine swiftly with oxygen in air to form white oxide: \[\text{4M}(s) + \text{O}_2(g) \rightarrow \text{2M}_2 \text{O}(s)\nonumber \] with M = Li, Na, K, Rb, or Cs (Li O is lithium oxide, Na O is sodium oxide, etc.) All except lithium react further to form yellow peroxides, M O : \[\text{2M}_2 \text{O}(s) + \text{O}_2(g) \rightarrow \text{2M}_2 \text{O}_2(s)\nonumber \] M = Na, K, Rb, or Cs (Na O is sodium peroxide, etc.) Potassium, rubidium, and cesium are sufficiently reactive that yellow superoxides (whose general formula is MO ) can be formed: \[\text{2M}_2 \text{O}_2(s) + \text{O}_2(g) \rightarrow \text{2MO}_2(s)\nonumber \] with M = K, Rb, or Cs Unless the surface of a sample of an alkali metal is scraped clean, it will appear white or gray instead of having a silvery metallic luster. This is due to the oxide, peroxide, or superoxide coating that forms after a few seconds of exposure to air. The following movie shows how a freshly cut piece of lithium is shiny, but dulls to gray when exposed to oxygen in the air. The video also focuses on another important property of alkali metals: they are soft, and easy to cut, compared to other metals. A dull gray oxidized cylinder of lithium metal is cut, revealing a shiny silvery surface. After 1 minute, the surface has dulled, and after 10 minutes, the cut surface has returned to the dull gray of the rest of the lithium metal. Since the alkali metal is lithium, the only reaction with oxygen that occurs is: \[\text{4Li}(s) + \text{O}_2(g) \rightarrow \text{2Li}_2\text{O}(s)\nonumber \] The alkali also combine directly with hydrogen gas to form compounds known as hydrides, MH: \[\text{2M}(s) + \text{H}_2(g) \rightarrow \text{2MH}(s)\nonumber \] with M = Li, Na, K, Rb, or Cs They react with sulfur to form sulfides, M S: \[\text{2M}(s) + \text{S}(g) \rightarrow \text{M}_2\text{S}(s)\nonumber \] with M = Li, Na, K, Rb, or Cs These oxides, hydrides, hydroxides, and sulfides all dissolve in water to give basic solutions, and these compounds are among the strong bases. The peroxides and superoxides formed when the heavier alkali metals react with O also dissolve to give basic solutions: \[\text{2NaO}_2(s) + \text{2H}_2\text{O}(l) \rightarrow \text{4Na}^{+}(aq) + \text{4OH}^{-}(aq) + \text{O}_2(g)\nonumber \] \[\text{4K}_2\text{O}(s) + \text{2H}_2\text{O}(l) \rightarrow \text{4K}^{+}+ \text{4OH}^{-} + \text{3O}_2(g)\nonumber \] Both of the latter equations describe redox as well as acid-base processes, as you can confirm by assigning oxidation numbers. The peroxide and superoxide ions contain O atoms in the unusual (for O) –1 and –½ oxidation states: Therefore (simultaneous oxidation and reduction) of O or O to the more common oxidation states of 0 (in O ) and –2 (in OH ) is possible. The alkali metals also react directly with the halogens, for instance with chlorine, forming chlorides, \[\text{2M}(s) + \text{Cl}_2(g) \rightarrow \text{2MCl}(s)\nonumber \] M = Li, Na, K, Rb, or Cs Below is an example of the reaction of Na with Cl A piece of sodium metal is added to a flask containing chlorine gas. Initially no reaction takes place, but when a drop of water is added, sodium and chlorine react, violently flaring up and producing so much heat that sand is needed in the bottom of the flask to absorb the heat and prevent the glass from cracking. This equation for this reaction is: \[\text{2Na}(s) + \text{Cl}_2(g) \rightarrow \text{2NaCl}(s)\nonumber \] with fluorine to form fluorides, MF: \[\text{2M}(s) + \text{F}_2(g) \rightarrow \text{2MF}(s)\nonumber \] M = Li, Na, K, Rb, or Cs and with bromine to form bromides, MBr: \[\text{2M}(s) + \text{Br}_2(g) \rightarrow \text{2MBr}(s)\nonumber \] M = Li, Na, K, Rb, or Cs Below is an example of K reacting with Br In this video, potassium, which is stored in inert mineral oil due to its high reactivity, is placed in a beaker of liquid bromine after the protective layer of mineral oil has been removed. The potassium reacts explosively with the bromine. The container is covered during the whole process to prevent reactants and products from entering the environment. The chemical equation for this reaction is: \[\text{2K}(s) + \text{Br}_2(g) \rightarrow \text{2KBr}(s)\nonumber \] Sodium and potassium are quite abundant, ranking sixth and seventh among all elements in the earth’s crust, but the other alkali metals are rare. Sodium and potassium ions are components of numerous silicate crystal lattices seen in the Earth's crust, but since most of their compounds are water soluble, they are also important constituents of seawater and underground deposits of brine. Sodium chloride obtained from such brines is the chief commercial source of sodium, while potassium can be obtained from the ores sylvite (KCl) or carnallite (KCl•MgCl •6H O). Both sodium (Na ) and potassium (K ) ions are essential to living systems. Na is the main cation in fluids surrounding the cells, while K is most important inside the cells. Na plays a role in muscle contraction, and both K and Na play a role in transmitting nerve impulses. K is more important than Na in plants, and it is one of three elements (K, P, N) which must be supplied in fertilizer to maintain high crop yields. K is especially abundant in trees—wood ashes from kitchen fires (potash) were the major source of this element as recently as a century ago, and they still make good fertilizer for your garden. Wood ashes contain a mixture of potassium oxide and potassium carbonate, the latter formed by combination of K O with CO produced when C in the wood combines with O : \[\text{K}_2\text{O} + \text{CO}_2 \rightarrow \text{K}_2\text{CO}_3\nonumber \] Na compounds are obtained commercially from brine or from seawater. When an electrical current is passed through an NaCl solution (a process called ), Cl (g), H (g), and a concentrated solution of NaOH (caustic soda or lye) are obtained: \[\text{Na}^{+}(aq) + \text{2Cl}^{-}(aq) + \text{2H}_2\text{O}(l) \xrightarrow{\text{electrolysis}} \text{Cl}_2(g) + \text{H}_2(g) + \text{Na}^{+}(aq) + \text{2OH}^{-}(aq)\nonumber \] This process is described in more detail in the section on electrochemical cells, but you can see from the equation that the electrical current oxidizes Cl to Cl and reduces H O to H . NaOH( ) is used as a strong base in numerous industrial processes to make soap, rayon, cellophane, paper, dyes, and many other products. Lye is also used in home drain cleaners. It must be handled with care because it is strongly basic, highly caustic, and can severely burn the skin. A second important industrial use of brine is the : \[\text{CO}_2(g) + \text{NH}_3(aq) + \text{Na}^{+}(aq) + \text{Cl}^{-}(aq) + \text{H}_2\text{O}(l) \rightarrow \text{NaHCO}_3(s) + \text{NH}_4^{+}(aq) + \text{Cl}^{-}(aq)\nonumber \] The Solvay process is an acid-base reaction combined with a precipitation. The acid anhydride, CO , reacts with H O to produce H CO . This weak acid donates a proton to NH , yielding NH and HCO , and the latter ion precipitates with Na . The weakly basic sodium hydrogen carbonate produced by the Solvay process can be purified for use as an antacid (bicarbonate of soda), but most of it is converted to sodium carbonate (soda ash) by heating: \[\text{2NaHCO}_3(s) \xrightarrow{\Delta } \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)\nonumber \] (The Δ in this equation indicates heating of the reactant.) Sodium carbonate (Na CO ) is used in manufacturing glass and paper, and in some detergents. The carbonate ion is a rather strong base, however, and detergents containing Na CO (washing soda) have resulted in severe chemical burns to some small children who, out of curiosity, have eaten them. | 12,214 | 3,123 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Anti-Markovnikov_Additions_to_Triple_Bonds |
Most Hydrogen halide reactions with alkynes occur in a Markovnikov-manner in which the halide attaches to the most substituted carbon since it is the most positively polarized. A more substituted carbon has more bonds attached to 1) carbons or 2) electron-donating groups such as Fluorine and other halides. However, there are two specific reactions among alkynes where anti-Markovnikov reactions take place: the radical addition of HBr and Hydroboration Oxidation reactions. For alkynes, an anti-Markovnikov addition takes place on a terminal alkyne, an alkyne on the end of a chain. The Br of the Hydrogen Bromide (H-Br) attaches to the less substituted 1-carbon of the terminal alkyne shown below in an anti-Markovnikov manner while the Hydrogen proton attaches to the second carbon. As mentioned above, the first carbon is the less substituted carbon since it has fewer bonds attached to carbons and other substituents. The H-Br reagent must also be reacted with heat or some other radicial initiator such as a peroxide in order for this reaction to proceed in this manner. This presence of the radical or heat leads to the anti-Markovnikov addition since it produces the most stable reaction. For more on Anti-Markovnikov additions: Additions--Anti-Markovnikov Product Formation The product of a terminal alkyne that is reacted with a peroxide (or light) and H-Br is a 1-bromoalkene. : The Bromine can attach in a or manner which means the resulting alkene can be both and . addition is when both Hydrogens attach to the same face or side of the double bond (i.e. ) while the addition is when they attach on opposite sides of the bond ( ). Hydroboration of terminal alkynes reacts in an anti-Markovnikov fashion in which the Boron attacks the less substituted carbon which is the least hindered. It is a stereospecific reaction where addition is observed as the hydroboration occurs on the same side of the alkyne and results in stereochemistry. However, a bulky borane reagent needs to be used to stop at the alkenyl-borane stage. Otherwise, a second hydroboration will occur. In this example, diisoamyl borane or HBsia , a fairly large and sterically-hindered borane reagent, is used. Both of the alkyne's pi bonds will undergo hydroboration if BH (borane) is used by itself. Oxidation is the next step that occurs. The resulting alkylborane is oxidized to a vinyl alcohol due to the reaction with a hydroxide in a basic solution such as aqueous sodium hydroxide. A vinyl alcohol is an alcohol that has both an alkene and an -OH group. After the vinyl alcohol is formed, tautomerization takes place. Tautomerism is the interconversion of isomeric compounds due to the migration of a proton. The alcohol, which in this case is a terminal enol, due to tautomerism to become an aldehyde since the aldehyde form is much more stable than the enol. For additional information on hydroboration oxidation: Hydroboration Oxidation 1. What is the product of this reaction? 2. What process causes the conversion of a vinyl alcohol to an aldeyde and what are some of its distinct features? 3. True or False: Only products are observed in radical H-Br additions to terminal alkynes. 4. Explain why a bulky borane reagent is necessary for reactions. 5. Draw the product. 1. Don't be confused by the borane reagent! Just remember, anytime there is a bulky borane reagent reacting with a terminal alkyne, the hydroboration oxidation reaction will occur and be proceeded by tautomerism which will produce an aldeyde as shown below. 2. The interconversion of an enol or vinyl alcohol to an aldehyde is called tautomerism and it is very distinct since it is a spontaneous reaction that proceeds very quickly. 3. False. Both and products are produced as both and additions are observed. 4. If a small borane reagent is utilized, both pi bonds will be used and a second hydroboration will occur. This will break the double bond of the alkene and the aldehyde product will not be formed. 5. | 4,017 | 3,124 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.01%3A_Prelude_to_Aqueous_Phase_Reactions |
In other sections we emphasized the importance of liquid solutions as a medium for chemical reactions. Water is by far the most important liquid solvent, partly because it is plentiful and partly because of its unique properties. In your body, in other living systems, and in the outside environment a tremendous number of reactions take place in aqueous solutions. Consequently this section, as well as significant portions of many subsequent sections, will be devoted to developing an understanding of reactions which occur in water. Since ionic compounds and polar covalent compounds constitute the main classes which are appreciably soluble in water, reactions in aqueous solutions usually involve these types of substances. There are three important classes of reactions which occur in aqueous solution: , , and . Below are demonstrations of each of the types of reactions we will be investigating throughout the course of this chapter. The following video demonstrates precipitation reactions (carried out in an aqeous solution): Next up is a demonstration of acids and bases reacting with Aluminum (in the form of a soda can): Finally we have a video that shows a few demos of different redox reactions: Ed Vitz (Kutztown University), (University of | 1,283 | 3,125 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.06%3A_Alkanes/8.6.01%3A_Astronomy-_Titan |
The simplest hydrocarbons are , compounds comprised of only carbon and hydrogen linked together by single bonds. Each carbon atom has tetrahedral geometry around it, with angles of 109.5°. Simple diagrams and three-dimensional jmol structures of three alkanes are shown below. Alkanes are very nonpolar due to their symmetry and general lack of dipoles between carbon and hydrogen atoms. By observing the general size and nonpolarity of the molecules shown above, what can you infer about the intermolecular forces between them? Relate the presence of these forces to expected melting and boiling points of these alkanes compared to each other and to a more polar molecule, such as water. The alkanes shown have almost no net dipole and are relatively small molecules. Therefore, the only intermolecular forces involved are London forces. Because propane is larger there will be more London attraction forces holding propane molecules together compared to ethane and methane, so propane will have higher melting and boiling points. However, the forces are still quite weak, and we would expect these alkanes to have lower melting and boiling points than a polar molecule such as water. The intermolecular forces in each molecule are weak enough to the point where methane, ethane, and propane are normally found as gases at room temperature. These hydrocarbons exist in concentrated areas of space, most notably on the Saturnian moon Titan. Titan is unique in that in has a very dense atmosphere rich in hydrocarbons and lakes of simple alkanes . Bodies of methane and ethane play almost the same role on Titan as liquid does on Earth. The heavier propane, used by humans as a fossil fuel, is present in excess of 700 million Earth barrels . Scientists are currently attempting to figure out just what produces these hydrocarbons on Titan, but for now, let's explore the geometry and nomenclature of these three molecules further. Ball-and-stick models of alkane molecules which contain up to three carbon atoms are shown in Figure 1. Methane ( ) has four C—H bonds arranged tetrahedrally around a single carbon atom. Ethane ( ) has a slightly more complicated structure—each carbon is still surrounded tetrahedrally by four bonds, but only three are C―H bonds, while the fourth is a C—C bond. Ethane may be thought of as two methyl groups connected by a single C—C bond. (The group, CH —, has the same structure as methane except that one hydrogen has been removed.) Thus the formula for ethane is CH CH or C H . The third member of the alkane family is propane. As can be seen in part , the three carbon atoms are in a chain. The two methyl groups at the ends of the chain are linked together by a group, —CH —. The formula is CH CH CH or C H . Again the tetrahedral arrangement of C—H or C—C bonds around each carbon atom is maintained. Each of these represents one of the 3-D molecular structures shown in the Jmols in Figure 1. You should compare the projection formulas with the ball-and-stick Jmols, trying to visualize the flat projections in three dimensions. Clearly we could go on to chains of four, five, six, or more carbon atoms by adding more methylene groups to the propane molecule. The first 10 compounds whose structures may be derived in this way are listed in the following table. They are called or , indicating that all contain a single continuous chain of carbon atoms and can be represented by a projection formula whose carbon atoms are in a straight line. \(\Page {1}\) First 10 Note that all the projection formulas in the table have an initial hydrogen atom followed by a number of CH groups. The chain ends with a second single hydrogen atom. The general formula H(CH ) H, or H , may be written, where is the number of CH groups, or the number of C atoms. Propane, for example, has = 3. Its formula is C H and it is referred to as a C hydrocarbon. The designation in front of hydrocarbons butane and above designate a normal straight-chain isomer, with no carbon branches. In addition to the straight chain shown for normal butane, a , in which some carbon atoms are linked to more than two other carbons, is possible. The projection formula for the branched-chain compound 2-methylpropane (also called isobutane because it an isomer of butane) is Unlike normal butane, which has a straight chain of four carbon atoms, the longest chain in isobutane is only three carbon atoms long. The central of these three atoms is bonded to the fourth carbon. Nevertheless, you can verify from the projection formulas or from the ball-and-stick drawings that both normal butane and isobutane have the same molecular formula, C H . The two compounds are isomers, just like ethyl alcohol and dimethyl ether, hence the prefix in the name for one of them. Isomers have have the same structural formula, but they often have dissimilar properties. Isobutane, being more compact, has smaller London Force interactions than n-butane, and thus a lower boiling point. The boiling points can be compared in the table of organic compound boiling points. One should note from this table that of all the organic compounds to consider, alkanes, both straight chained and branched, have the lowest boiling points comparatively. As the number of carbon atoms in an alkane molecule increases, so do the possibilities for isomerism of this kind. There are three isomeric pentanes, all with the formula C H , five isomeric hexanes, C H , and nine isomeric heptanes, C H . The number of possible isomers of tetracontane, C H is larger than 62 million. Thus an inconceivable variety of different molecular structures is possible for compounds containing only carbon and hydrogen atoms connected by single bonds. In crude oil, the most important source of hydrocarbons in the United States, branched-chain and straight-chain alkanes are about equally common. Another aspect of the behavior of alkane molecules (and other molecules containing single bonds) is not apparent from ball-and-stick illustrations or from projection formulas. Like small children, molecules cannot stop wriggling, and most alkane structures are not rigid. Groups on either side of a carbon-carbon single bond are able to rotate freely with respect to each other. Rotation of one methyl group with respect to the other in ethane, CH CH , is shown in Figure 5. While there is free rotation around the carbon bond, certain positions are more stable than others. For the ethane molecule is most stable when the hydrogen atoms on one methyl group are offset from those on the other methyl group(referred to as ) and more energy is needed to pass through the formation, where the hydrogen atoms of both methyl groups line up. Because of this free rotation, and because they collide with other molecules, alkane molecules are constantly flexing and writhing about their C―C bonds, assuming different shapes (different ) all the time. Some feeling for the way in which alkane molecules can adopt a variety of conformations is obtained from the following example. In Figure \(\Page {6}\), five ball-and-stick diagrams labeled ( ) through ( ) are shown. All five correspond to the formula C H (pentane). Since there are only three isomers of pentane, some of these molecules must correspond to different views or different conformations of the same molecule. Decide which of these diagrams correspond to the same isomer, and which isomer each represents. Draw a projection formula for each isomer. When given illustrations of molecules in 3D form, it is often helpful to draw them simply in 2D and compare how different atoms bonds to each other. Molecule ( ) has five carbon atoms in a single continuous sequence. It corresponds to Careful inspection of ( ) reveals that again the five carbon atoms form a single chain. No carbon atom is joined to more than two others. Molecule ( ) is thus also -pentane, but in a different conformation. Molecule ( ) does have a carbon atom joined to three others. The longest chain is four carbon atoms long, and an additional carbon atom is attached to the second carbon atom. The projection formula is accordingly Molecule ( ) is also isopentane, while molecule ( ) corresponds to the third isomer, called neopentane: From ChemPRIME: | 8,290 | 3,126 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/RR._Radical_Reactions/RR1._Introduction_to_Radicals |
Radicals are species that have unpaired electrons. They can be atoms or molecules and they can be neutral species or ions. Frequently, radicals are very reactive. However, their mode of reactivity does not fall neatly within the normal patterns of Lewis acids and bases, or nucleophiles and electrophiles. Radicals play common roles in atmospheric chemistry, including equilibration of the ozone layer. They are also found in a variety of biochemical pathways. In addition, radicals are employed in a number of useful processes, such as the polymerization of methyl methacrylate or vinyl chloride, commonly used to make shatter-resistant "glass" and pipes for plumbing, respectively. Compounds of p-block elements form radicals if one of the atoms has seven electrons in its valence shell rather than the usual eight. Draw structures for the following neutral radicals, making sure to fill in the correct number of electrons. a) Br b) OH c) CH d) CH CH S e) CH CHCH f) NO g) (CH ) NO h) NO A molecule could become a radical in a number of different ways. A bond may break in half via the addition of energy, in the form of either heat or light. Otherwise, it may simply transfer one of its electrons elsewhere. Again, this event may be precipitated by the addition of heat or light energy. Of course, a molecule that receives an additional electron from elsewhere may also become a radical. Draw structures for the following cationic radicals, making sure to fill in the correct number of electrons and the formal charge. a) H C=O b) CH NH c) CH OCH d) CH CH CH Br e) CH CH CHCH Draw structures for the following anionic radicals, making sure to fill in the correct number of electrons. a) O b) H CO c) CH CCCH d) cyclo-C H The compounds above are all simple radicals, containing one unpaired electron. Compounds may also have more than one unpaired electron. For example, elemental oxygen, O , is a diradical. Although its Lewis structure does not suggest anything unusual, its molecular orbital diagram reveals that oxygen actually has two unpaired electrons. Show a molecular orbital interaction diagram illustrating the origins of the molecular orbitals on O from the atomic orbitals of oxygen. A diradical could take two different forms. For example, molecular oxygen has two singly-occupied molecular orbitals. The single electron in each of those orbitals could adopt one of two different spin states. Both could adopt the same spin state (designated with arrow "up", for example), or they could be "spin-paired" (one "up", one "down"). The former situation is called a "triplet state", whereas the latter case is termed a "singlet state". These two situations result in some physical differences, such as different interactions with a magnetic field. Subsequent pages will focus on the reactivity of radicals, with an emphasis on the stages of radical chain reactions. , | 2,898 | 3,128 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.09%3A_Group_VIIIA-_Noble_Gases |
The properties of the noble gases are summarized in the table below. The noble gases have a complete octet of electrons or just for helium, leaving them with little chemical reactivity. Sure enough,the ionization energies of He and Ne are greater than 2000 kJ mol , and it is unlikely that these noble gases will ever be induced to form chemical bonds. The same probably applies to Ar. Kr and especially Xe do form compounds though, which was discussed in the halogens section, and Rn might be expected to be even more reactive. Rn is radioactive, however, and study of its chemistry is difficult. Electron Configuration Usual Oxidation State Radius/pm - Covalent ... ... ... 110 Density/ g cm Electro- negativity Melting Point (in °C) -272 Because of the lack of reactivity of the noble gases, they are often used when an nonreactive atmosphere is needed, such as in welding. Due to their low boiling points, noble gases are also cryogens in their liquid forms. One familiar use of helium is in balloons and blimps, since it is buoyant in the atmosphere, and unlike hydrogen, nonreactive. Another familiar use is as lighting in gas discharge lamps. Referred to popularly as neon lights, they can contain other noble gases, or mixtures of the gases. | 1,265 | 3,129 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.11%3A_Entropy_Randomness_and_Disorder |
A very useful, though somewhat rough, description of the entropy of a substance is as a measure of the of the atoms and molecules which constitute that substance. In these terms the second law of thermodynamics is seen as a tendency for the disorder of the universe to increase. This way of looking at entropy is entirely compatible with the approach presented above. A situation which we intuitively recognize as being orderly is also one which can only be achieved in a limited number of ways. By contrast, situations which we recognize as disordered, random, or chaotic, can he achieved in a whole variety of ways. In other words, , and hence , is small for an ordered situation but large for a disordered situation. There are limits to the lengths one can take this order-disorder approach to entropy, though. It does not lend itself to a quantitative treatment, and it is also difficult to explain some things like the effect of mass in these terms. There is nothing in our intuition about order, for example, which suggests that 1 mol Xe gas is more disordered than 1 mol He gas, even though its entropy is in fact larger. | 1,144 | 3,130 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Other_Reactions_of_Benzene_and_Methylbenzene |
This page gives details of some reactions of benzene and methylbenzene (toluene) not covered elsewhere in this section. It deals with the combustion, hydrogenation and sulfonation of benzene and methylbenzene (toluene), and with the oxidation of side chains attached to benzene rings. Like any other hydrocarbons, benzene and methylbenzene burn in a plentiful supply of oxygen to give carbon dioxide and water. For example: For benzene: \[ 2C_6H_6 + 15O_2 \rightarrow 12CO_2 + 6H_2O\] . . . and methylbenzene: \[ C_6H_5CH_3 + 9O_2 \rightarrow 7CO_2 + 4H_2O\] However, for these hydrocarbons, combustion is hardly ever complete, especially if they are burnt in air. The high proportion of carbon in the molecules means that you need a very high proportion of oxygen to hydrocarbon to get complete combustion. Look at the equations. As a general rule, the hydrogen in a hydrocarbon tends to get what oxygen is available first, leaving the carbon to form carbon itself, or carbon monoxide, if there isn't enough oxygen to go round. The arenes tend to burn in air with extremely smoky flames - full of carbon particles. You almost invariably get incomplete combustion, and the arenes can be recognised by the smokiness of their flames. Hydrogenation is an addition reaction in which hydrogen atoms are added all the way around the benzene ring. A cycloalkane is formed. For example: With benzene: . . . and methylbenzene: These reactions destroy the electron delocalisation in the original benzene ring, because those electrons are being used to form bonds with the new hydrogen atoms. Although the reactions are exothermic overall because of the strengths of all the new carbon-hydrogen bonds being made, there is a high activation barrier to the reaction. The reactions are done using the same finely divided nickel catalyst that is used in hydrogenating alkenes and at similar temperatures (around 150°C), but the pressures used tend to be higher. sulfonation involves replacing one of the hydrogens on a benzene ring by the sulfonic acid group, -SO H. There are two equivalent ways of sulfonating benzene: Or: \[C_6H_6 + H_2SO_4 \rightarrow C_5H_5SO_3H + H_2O\] The product is benzenesulfonic acid. Methylbenzene is more reactive than benzene because of the tendency of the methyl group to "push" electrons towards the ring. The effect of this greater reactivity is that methylbenzene will react with fuming sulfuric acid at 0°C, and with concentrated sulfuric acid if they are heated under reflux for about 5 minutes. As well as the effect on the rate of reaction, with methylbenzene you also have to think about where the sulfonic acid group ends up on the ring relative to the methyl group. Methyl groups have a tendency to "direct" new groups into the 2- and 4- positions on the ring (assuming the methyl group is in the 1- position). Methyl groups are said to be 2,4-directing. So you get a mixture which mainly consists of two isomers. Only about 5 - 10% of the 3- isomer is formed. The main reactions are: and: In the case of sulfonation, the exact proportion of the isomers formed depends on the temperature of the reaction. As the temperature increases, you get increasing proportions of the 4- isomer and less of the 2- isomer. This is because sulfonation is reversible. The sulfonic acid group can fall off the ring again, and reattach somewhere else. This tends to favour the formation of the most thermodynamically stable isomer. This interchange happens more at higher temperatures. The 4- isomer is more stable because there is no cluttering in the molecule as there would be if the methyl group and sulfonic acid group were next door to each other. An alkylbenzene is simply a benzene ring with an alkyl group attached to it. Methylbenzene is the simplest alkylbenzene. Alkyl groups are usually fairly resistant to oxidation. However, when they are attached to a benzene ring, they are easily oxidised by an alkaline solution of potassium manganate(VII) (potassium permanganate). Methylbenzene is heated under reflux with a solution of potassium manganate(VII) made alkaline with sodium carbonate. The purple colour of the potassium manganate(VII) is eventually replaced by a dark brown precipitate of manganese(IV) oxide. The mixture is finally acidified with dilute sulfuric acid. Overall, the methylbenzene is oxidised to benzoic acid. Interestingly, any alkyl group is oxidised back to a -COOH group on the ring under these conditions. So, for example, propylbenzene is also oxidised to benzoic acid. Jim Clark ( ) | 4,548 | 3,131 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/14%3A_Chemical_Kinetics/14.S%3A_Chemical_Kinetics_(Summary) |
Textmap – area of chemistry dealing with speeds/rates of reactions \[\displaystyle \textit{average rate} = \frac{\textit{change #moles B}}{\textit{change in time}}= \frac{\Delta\textit{moles B}}{\Delta t}\textit{ if }A \to B \nonumber \] \[\Delta\textit{moles B} = \textit{moles B at final time}- \textit{moles B at initial time} \nonumber \] \[\displaystyle \textit{average rate} = -\frac{\Delta\textit{moles A}}{\Delta t}\textit{ if }A \to B \nonumber \] \[\displaystyle\textit{rate} = -\frac{1}{a}\frac{\Delta [A]}{\Delta t} = -\frac{1}{b}\frac{\Delta [B]}{\Delta t} = \frac{1}{c}\frac{\Delta [C]}{\Delta t} = \frac{1}{d}\frac{\Delta [D]}{\Delta t} \nonumber \] \[\textit{rate} = -\frac{\Delta [A]}{\Delta t} = k[A] \nonumber \] and in integral form: \[\ln[A]_t - \ln[A]_0 =-kt \nonumber \] or \[\ln\frac{[A]_t}{[A]_0} = -kt \nonumber \] \[\ln[A]_t = - kt + \ln[A]_0 \nonumber \] \[\displaystyle t_{\frac{1}{2}} = -\frac{\ln\frac{1}{2}}{k} = \frac{0.693}{k} \nonumber \] \[\text{Rate} = k[A]^2 \nonumber \] \[\displaystyle\frac{1}{[A]_t} = kt + \frac{1}{[A]_0} \nonumber \] \[\displaystyle\textit{half life} = t_{\frac{1}{2}} = \frac{1}{k[A]_0} \nonumber \] \[\displaystyle k = A e^{\frac{-E_a}{RT}} \nonumber \] \[\displaystyle \ln k = -\frac{E_a}{RT} + \ln A \nonumber \] \[\displaystyle \ln \frac{k_1}{k_2} = \frac{E_a}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right) \nonumber \] | 1,404 | 3,132 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/08%3A_Electrons_in_Atoms/8.07%3A_Quantum_Numbers_and_Electron_Orbitals |
Now that we have introduced the basic concepts of quantum mechanics, we can start to apply these concepts to build up matter, starting from its most elementary constituents, namely atoms, up to molecules, supramolecular complexes (complexes built from weak interactions such as hydrogen bonds and van der Waals interactions), networks, and bulk condensed phases, including liquids, glasses, solids,... As explore the structure of matter, itself, we should stand back and wonder at how the mathematically elegant, yet somehow not quite tangible, structure of quantum theory is able to describe so accurately all phases of matter and types of substances, ranging from metallic and semiconducting crystals to biological macromolecules, to morphologically complex polymeric materials. We begin with the simplest system, the hydrogen atom (and hydrogen-like single-electron cations), which is the atomic system (thus far) for which the Schrödinger equation can be solved exactly for the energy levels and wave functions. To this end, we consider a nucleus of charge \(+Ze\) at the origin and a single electron a distance \(r\) away from it. If we just naïvely start by writing the (total) classical energy \[\underbrace{\dfrac{p_{x}^{2}}{2m}+\dfrac{p_{y}^{2}}{2m}+\dfrac{p_{z}^{2}}{2m}}_{\text{Kinetic Energy}} \underbrace{-\dfrac{Ze^2}{4\pi \epsilon_0 \sqrt{x^2 +y^2 +z^2}}}_{\text{Potential Energy}}=E \label{8.1.1}\] where the constant \(k\) in Coulomb's Law has been rewritten as \(1/4\pi \epsilon_0\), where \(\epsilon_0\) is the permittivity of free space, and where we have written \(r=\sqrt{x^2 +y^2 +z^2}\) in terms of its Cartesian components, then we see immediately that the energy is not a simple sum of terms involving \((x,p_x)\), \((y, p_y)\), and \((z,p_z )\), and therefore, the wave function is not a simple product of a function of \(x\), a function of \(y\) and a function of \(z\). Thus, it would seem that we have already hit a mathematical wall. Fortunately, we are not confined to work in Cartesian coordinates. In fact, there are numerous ways to locate a point \(r\) in space, and for this problem, there is a coordinate system, known as , that is particularly convenient because it uses the distance \(r\) of a point from the origin as one of its explicit coordinates. In Cartesian coordinates, a point in a three-dimensional space requires three numbers to locate it \(r=(x,y,z)\). Thus, if we change to a different coordinate system, we still need three numbers to full locate the point. Figure \(\Page {1}\) shows how to locate a point in the system of spherical coordinates: Thus, we use the distance \(r\) of the point from the origin, which is also the magnitude of the vector \(r\), the angle \(\theta\) the vector \(r\) makes with the positive z -axis, and the angle \(\phi\) between the projection of the vector \(r\) into the \(xy\) plane and the positive x -axis. The angle \(\theta\) is called the and the angle \(\phi\) is called the . These three variables have the following ranges \[r \in [0,\infty ) \;\;\;\; \theta \in [0,\pi ] \;\;\;\; \phi \in [0,2\pi ] \label{8.1.2}\] To map a point from the Cartesian representation \(r=(x,y,z)\) to spherical coordinates \((r, \theta ,\phi )\), a coordinate transformation is needed tells us how points in the two coordinate representations are connected. To determine \(x\) and \(y\), we need the component of \(r\) in the \(xy\) plane, which is \(r\sin\theta \), and its projection into the \(x\) and \(y\) directions, which are \[\begin{align*}x &= r\sin\theta \cos\phi \\ y &= r\sin \theta \sin \phi \end{align*} \label{8.1.3}\] Finally, to determine \(z\), we simply need the component of \(r\) along the z-axis, which is \(r\cos\theta \), so \[z=r\cos\theta \label{8.1.4}\] These three relations tell us how to map the point \((r,\theta ,\phi )\) into the original Cartesian frame \((x,y,z)\). The inverse of this transformation tells us how to map the Cartesian numbers \((x,y,z)\) back into spherical coordinates. It is simple algebra to show that the inverse is \[\begin{align*}r &= \sqrt{x^2 +y^2 +z^2}\\ \phi &= \tan^{-1}\dfrac{y}{x}\\ \theta &= \tan^{-1}\dfrac{\sqrt{x^2+y^2}}{z}\end{align*} \label{8.1.5}\] Consider a two-dimensional vector \(v\) in the \(xy\) plane. The vector can be represented in terms of its two components \(v=(v_x ,v_y)\). That is, just two numbers \(v_x\) and \(v_y\) are needed, and we know everything about the vector. However, if we know the angle \(\theta\) that \(v\) makes with the x-axis, then we can also use the magnitude \(v=|v|\) and the angle as two alternative numbers that can be used to represent the vector. In this case, since \(v_x =v\cos\theta =|v|\cos\theta \) and \(v_y =v\sin \theta =|v|\sin\theta \), then \(v\) is specified as \(v=(v\cos\theta ,v\sin\theta )\). There is still another alternative scheme for representing \(v\). We can use the magnitude \(v=|v|\) and just one of the components of \(v\). Suppose we wish to represent \(v\) in terms of \(v\) and \(v_x\). since \(v=v_{x}^{2}+v_{y}^{2}\), the other component \(v_y\) is given by \(v_{y}^{2}=v^2 -v_{x}^{2}\), then \(v_y =\sqrt{v^2 -v_{x}^{2}}\), we can write \(v=(v_x ,\sqrt{v^2-v_{x}^{2}})\), which means we only need to know the magnitude \(v\) and one of the components to specify \(v\) entirely. This is the scheme that will be used below to represent the angular momentum, as discussed in the next section. In spherical coordinates, the momentum \(p\) of the electron has a radial component \(p_r\), corresponding to motion radially outward from the origin, and an angular component \(L\), corresponding to motion along the surface of a sphere of radius \(r\), i.e. motion perpendicular to the radial direction: \[p=\left ( p_r, \dfrac{L}{r} \right ) \label{8.1.6}\] Note that \(L\) is, itself, still a vector, since motion along the surface of a sphere still has more than one component. Its components include motion along the \(\theta\) and along the \(\phi\) directions. Let us step back briefly into Cartesian components to look at the angular momentum vector \(L\). Its definition is \[L=r \times p \label{8.1.7}\] i.e. the of \(r\) and \(p\), which is why \(L\) is perpendicular to \(r\). In Cartesian components, \(L\) actually has three components, which are given by \[\begin{align*}L_x &= yp_z -zp_y \\ L_y &= zp_x -xp_z \\ L_z &= xp_y -yp_x\end{align*} \label{8.1.8}\] using the definition of the cross product. The important thing about angular momentum is that if the potential energy \(V\) depends only on \(r\), then angular momentum is conserved. For example, suppose \(V(r)\) is the Coulomb potential, which we will write compactly as \[V(r)=-\dfrac{k}{r} \label{8.1.9}\] where \(k=e^2 /(4\pi \epsilon_0 )\). We will show that angular momentum is conserved within the classical mechanical description of the system. If it is true classically, it is also true in quantum mechanics. The force on the classical electron is \[F=-\dfrac{k}{r^3}r \label{8.1.10}\] Therefore, the three force components are \[F_x =-\dfrac{kx}{r^3}\;\;\;\; F_y =-\dfrac{ky}{r^3}\;\;\;\; F_z =-\dfrac{kz}{r^3} \label{8.1.11}\] From Newton's second law: \[F=ma=m\dfrac{dv}{dt}=\dfrac{d(mv)}{dt}=\dfrac{dp}{dt} \label{8.1.12}\] Now consider one of the components of the angular momentum vector, e.g. \(L_z\). Its time derivative is \[\begin{align*}\dfrac{dL_z}{dt} &= \dfrac{dx}{dt}p_y +x\dfrac{dp_y}{dt}-\dfrac{dy}{dt}p_x -y\dfrac{dp_x}{dt}\\ &= \dfrac{p_x}{m}p_y +xF_y -\dfrac{p_y}{m}p_x -yF_x \\ &=\dfrac{p_x p_y }{m} -\dfrac{kxy}{r^3} -\dfrac{p_y p_x}{m}+\dfrac{kyx}{r^3} \\ &= 0\end{align*} \label{8.1.13}\] The same can be shown about \(L_x\) and \(L_y\). Moreover, since \(L^2 =L_{x}^{2}+L_{y}^{2}+L_{z}^{2}\), and the three components \(L_x\), \(L_y\) and \(L_z\) are all constant, \(L^2\) is constant as well. Thus, both the energy and the angular momentum are conserved in classical mechanics, but classically, both quantities can take on any values. In quantum mechanics, the energy can have only certain allowed values, and because angular momentum is conserved, it too can only have certain allowed values, as Bohr predicted in his model of the hydrogen atom. The allowed energies are \[E_n = -\dfrac{Z^2 e^4 m_e}{8\epsilon_{0}^{2}h^2}\dfrac{1}{n^2} \label{8.1.14}\] where \(n=1,2,3,...\) What about angular momentum? In spherical coordinates, the momentum is \(p=(p_r ,L/r)\), and \(L\) only has two components, one along \(\theta\) and one along \(\phi\). We can assign allowed values to these components separately, however, there is a compelling reason to do things a little differently. In spherical coordinates, the classical energy is given by \[\dfrac{p_{r}^{2}}{2m_e}+\dfrac{L^2}{2m_e r^2}-\dfrac{Ze^2}{4\pi \epsilon_0 r}=E \label{8.1.15}\] Thus, \(E\) separates into a purely radial part and an angular part: \[E=\varepsilon_r +\dfrac{L^2}{2m_e r^2}\] Note that the energy depends on \(L^2\). since \(L^2\) and each of the three components \(L_x\), \(L_y\) and \(L_z\) are conserved, we can use the representation of \(L\) in which we choose the magnitude \(L\) and one of the components. Why just one component? Note that even though \(L\) has three Cartesian components, it really is only a two dimensional vector since it describes the motion on a two-dimensional surface. Just like the \(xy\) plane is a two-dimensional surface, the surface of a sphere is also two-dimensional, using the angles \(\theta\) and \(\phi\) to locate points on the surface. The difference is that the surface of a sphere is not flat but curved, so we have to embed it in three-dimensional space. So as a Cartesian representation, three components are needed, but in spherical coordinates, we only need two, i.e. two numbers to represent \(L\). Thus, we can use the magnitude \(L\) and one of its components, which by convention is taken to be the z -component \(L_z\). \(L_z\) describes motion in the \(xy\) plane, i.e. motion in the azimuthal (\(\phi\)) di rection only. However, this is just motion on a ring, so one part of the wave function will just be the particle-on-a-ring wave functions \[\dfrac{1}{\sqrt{2\pi }}e^{im\phi} \ \label{8.1.17}\] where \(m=0,\pm 1,\pm 2,...\), and so the allowed values of \(L_z\) are \(m\hbar\). For \(L^2\), it turns out that the allowed values are \[L^2 \rightarrow l(l+1)\hbar^2 \label{8.1.18}\] where \(l\) is an integer that has the range \(l=0,1,2,...,n-1\) in a given energy level characterized by \(n\). Moreover, because \(L_z \leq |L|\), once the value of \(l\) is specified, \(m\) cannot exceed \(l\), so the range of \(m\) is \(m=-l ,-l+1 ,...,l-1,l\). The Schrödinger equation for single-electron Coulomb systems in spherical coordinates is \[-\dfrac{\hbar^2}{2m_e r^2}\left [ r\dfrac{\partial^2}{\partial r^2}(r\psi (r,\theta ,\phi))+\dfrac{1}{\sin\theta }\dfrac{\partial }{\partial \theta}\sin\theta \dfrac{\partial }{\partial \theta}\psi (r,\theta ,\phi )+\dfrac{1}{\sin^2 \theta}\dfrac{\partial^2}{\partial \phi^2}\psi (r, \theta ,\phi) \right ] -\dfrac{Ze^2}{4\pi \epsilon_0 r}\psi (r, \theta ,\phi ) \\=E\psi (r, \theta ,\phi ) \label{Seq}\] This type of equation is an example of a , which is no simple task to solve. However, solving it gives both the allowed values of the angular momentum discussed above the allowed energies \(E_n \), which agree with the simpler Bohr model. Thus, we do not need to assume anything except the validity of the Schrödinger equation, and the allowed values of energy and angular momentum, together with the corresponding wavefunctions, all emerge from the solution. Obviously, we are not going to go through the solution of the Schrödinger equation, but we can understand something about its mechanics and the solutions from a few simple considerations. Remember that the Schrödinger equation is set up starting from the classical energy, which we said takes the form \[\dfrac{p_{r}^{2}}{2m_e}+\dfrac{L^2}{2m_e r^2}-\dfrac{Ze^2}{4\pi \epsilon_0 r}=E\] which we can write as \[E=\varepsilon_r +\dfrac{L^2}{2m_e r^2}\] where \[\varepsilon_r =\dfrac{p_{r}^{2}}{2m_e}-\dfrac{Ze^2}{4\pi \epsilon_0 r}\] The term \(L^2/(2m_e r^2)\) is actually dependent only on \(\theta \) and \(\phi \), so it is purely angular. Given the separability of the energy into radial and angular terms, the wavefunction can be decomposed into a product of the form \[\psi (r,\theta ,\phi )=R(r)Y(\theta ,\phi )\] Solution of the angular part for the function \(Y(\theta ,\phi )\) yields the allowed values of the angular momentum \(L^2\) and the z -component \(L_z\). The functions \(Y(\theta ,\phi )\) are then characterized by the integers \(l\) and \(m\), and are denoted \(Y_{lm}(\theta ,\phi )\). They are known as . Here we present just a few of them for a few values of \(l\). For \(l=0\), there is just one value of \(m\), \(m=0\), and, therefore, one spherical harmonic, which turns out to be a simple constant: \[Y_{00}(\theta ,\phi )=\dfrac{1}{\sqrt{4\pi}}\] For \(l=1\), there are three values of \(m\), \(m=-1,0,1\), and, therefore, three functions \(Y_{1m}(\theta ,\phi )\). These are given by \[\begin{align*}Y_{11}(\theta ,\phi ) &= -\left ( \dfrac{3}{8\pi} \right )^{1/2}\sin\theta e^{i\phi }\\ Y_{1-1}(\theta ,\phi ) &= \left ( \dfrac{3}{8\pi} \right )^{1/2} \sin\theta e^{-i\phi }\\ Y_{10}(\theta ,\phi ) &= \left ( \dfrac{3}{4\pi } \right )^{1/2}\cos\theta \end{align*}\] Remember that \[e^{i\phi } = \cos\phi +i\sin\phi\] \[e^{-i\phi } = \cos\phi -i\sin\phi\] For \(l=2\), there are five values of \(m\), \(m=-2,-1,0,1,2\), and, therefore, five spherical harmonics, given by \[\begin{align*}Y_{22}(\theta ,\phi ) &=\left ( \dfrac{15}{32\pi } \right )^{1/2}\sin^2 \theta e^{2i\phi }\\ Y_{2-2}(\theta ,\phi ) &= \left ( \dfrac{15}{32\pi} \right )^{1/2}\sin^2\theta e^{-2i\phi }\\ Y_{21}(\theta ,\phi ) &= -\left ( \dfrac{15}{8\pi } \right )^{1/2}\sin\theta \cos\theta e^{i\phi }\\ Y_{2-1}(\theta ,\phi ) &= \left ( \dfrac{15}{8\pi } \right )^{1/2}\sin\theta \cos\theta e^{-i\phi }\\ Y_{20}(\theta ,\phi ) &= \left ( \dfrac{5}{16\pi} \right )^{1/2} (3\cos^2 \theta -1)\end{align*}\] The remaining function \(R(r)\) is characterized by the integers \(n\) and \(l\), as this function satisfies the radial part of the Schrödinger equation, also known as the : \[-\dfrac{\hbar^2}{2m_e}\dfrac{1}{r}\dfrac{d^2}{dr^2}(rR_{nl}(r))-\dfrac{l(l+1)\hbar^2}{2m_e r^2}R_{nl}(r)-\dfrac{Ze^2}{4\pi \epsilon_0 r}R_{nl}(r)=E_n R_{nl}(r)\] Note that, while the functions \(Y_{lm}(\theta ,\phi )\) are not particular to the potential \(V(r)\), the radial functions \(R_{nl}(r)\) are particular for the Coulomb potential. It is the solution of the radial Schrödinger equation that leads to the allowed energy levels. The boundary conditions that lead to the quantized energies are \(rR_{nl}(0)=0\) and \(rR_{nl}(\infty )=0\). The radial parts of the wavefunctions that emerge are given by (for the first few values of \(n\) and \(l\) ): \[\begin{align*}R_{10}(r) &= 2\left ( \dfrac{Z}{a_0} \right )^{3/2}e^{-Zr/a_0}\\ R_{20}(r) &= \dfrac{1}{2\sqrt{2}}\left ( \dfrac{Z}{a_0} \right )^{3/2} \left ( 2-\dfrac{Zr}{a_0} \right ) e^{-Zr/2a_0}\\ R_{21}(r) &= \dfrac{1}{2\sqrt{6}}\left ( \dfrac{Z}{a_0} \right )^{3/2} \dfrac{Zr}{a_0}e^{-Zr/2a_0}\\ R_{30}(r) &= \dfrac{2}{81\sqrt{3}}\left ( \dfrac{Z}{a_0} \right )^{3/2}\left [ 27-18\dfrac{Zr}{a_0}+2 \left ( \dfrac{Zr}{a_0} \right )^2 \right ]e^{-Zr/3a_0}\end{align*}\] where \(a_0\) is the Bohr radius \[a_0 = \dfrac{4\pi \epsilon_0 \hbar^2}{e^2 m_e}=0.529177 \times 10^{-10} \ m\] The full wavefunctions are then composed of products of the radial and angular parts as \[\psi_{nlm}(r,\theta ,\phi)=R_{nl}(r)Y_{lm}(\theta ,\phi )\] At this points, several comments are in order. First, the integers \(n,l,m\) that characterize each state are known as the of the system. Each of them corresponds to a quantity that is classically conserved. The number \(n\) is known as the quantum number, the number \(l\) is known as the quantum number, and the number \(m\) is known as the quantum number. As with any quantum system, the wavefunctions \(\psi_{nlm}(r,\theta ,\phi )\) give the probability amplitude for finding the electron in a particular region of space, and these amplitudes are used to compute actual probabilities associated with measurements of the electron's position. The probability of finding the electron in a small volume element \(dV\) of space around the point \(r=(r,\theta ,\phi )\) is \[\begin{align*}P(electron \ in \ dV \ about \ r,\theta ,\phi ) &= |\psi_{nlm}(r,\theta ,\phi )|^2dV\\ &= R_{nl}^{2}(r)|Y_{lm}(\theta ,\phi)|^2 dV \end{align*}\] What is \(dV\) ? In Cartesian coordinates, \(dV\) is the volume of a small box of dimensions \(dx\), \(dy\), and \(dz\) in the \(x\), \(y\), and \(z\) directions. That is, \[dV=dx\,dy\,dz\] In spherical coordinates, the volume element \(dV\) is a small element of a spherical volume and is given by \[dV=r^2 \sin \theta\, dr\, d\theta \,d\phi \label{dV}\] which is derivable from the transformation equations. If we integrate \(dV\) from Equation \(\ref{dV}\) over a sphere of radius \(R\), we should obtain the volume of the sphere: \[V= \left[\int_0^R r^2 dr\right]\left[\int_0^{\pi}\sin\theta d\theta\right]\left[\int_0^{2\pi}d\phi\right]\] \[ = \left[\left.{r^3 \over 3}\right\vert _0^R\right]\left[\left.-\cos\theta\right\vert _0^{\pi}\right]\left[\left.\phi\right\vert _0^{2\pi}\right]\] \[ = {R^3 \over 3}\times 2\times 2\pi\] \[ = {4 \over 3}\pi R^3\] which is the formula for the volume of a sphere of radius \(R\). The electron in a hydrogen atom \((Z=1)\) is in the state with quantum numbers \(n=1\), \(l=0\) and \(m=0\). What is the probability that a measurement of the electron's position will yield a value \(r\geq 2a_0\)? The wavefunction \(\psi_{100}(r,\theta ,\phi )\) is \[\psi_{100}(r,\theta ,\phi )=\dfrac{2}{a_{0}^{3/2}} \left ( \dfrac{1}{4\pi} \right )^{1/2}e^{-r/a_0}\] Therefore, the probability we seek is \[P(r \geq 2a_0 ) = \int_0^{2\pi}\int_0^{\pi}\int_{2a_0}^{\infty} |\psi_{100}(r,\theta ,\phi)|^2 r^2 \sin\,\theta dr\, d\theta \,d\phi\ \] \[= \dfrac{4}{a_0^{3}}\dfrac{1}{4\pi }\left [ \int_0^{2\pi}d\phi \right ] \left [ \int_0^{\pi} \sin \theta d\theta \right ] \left [ \int_{2a_0}^{\infty}r^2 e^{-2r/a_0}dr \right] \] \[= \dfrac{1}{a_{0}^3\pi}\cdot 2\pi \cdot 2\int_{2a_0}^{\infty}r^2e^{-2r/a_0}dr \] Let \(x=2r/a_0\). Then \[P(r\geq 2a_0)=\dfrac{1}{2}\int_{4}^{\infty}x^2 e^{-x}dx\] After integrating by parts, we find \[P(r\geq 2a_0)=13e^{-4}\approx 0.24=24\%\] which is relatively large given that this is at least two Bohr radii away from the nucleus! The part of the probability involving the product \[P_{nl}(r)=r^2 R_{nl}^{2}(r) \label{radeq}\] is known as the or simply the . \(P_{nl}(r)dr\) is the probability that a measurement of the electron's position yields a value in a radial shell of thickness \(dr\) and radius \(r\) as shown in the Figure \(\Page {2}\): \What the radial probability distribution shows is that the electron cannot be sucked into the nucleus because \(P_{nl}(0)=0\). Hence, as we shrink the radial shell into the nucleus, the probability of finding the electron in that shell goes to 0. Another point concerns the number of allowed states for each allowed energy. Remember that each wavefunction corresponds to a probability distribution in which the electron can be found for each energy. The more possible states there are, the more varied the electronic properties and behavior of the system will be. For \(n=1\), there is one energy \(E_1\) and only one wavefunction \(\psi_{100}\). For \(n=2\), there is one energy \(E_2\) and four possible states, corresponding to the following allowable values of \(l\) and \(m\) \[\begin{align*} l &= 0 \;\;\;\; m=0\\ l &= 1 \;\;\;\; m=-1,0,1\end{align*}\] Thus, there are four wavefunctions \(\psi_{200}\), \(\psi_{21-1}\), \(\psi_{210}\), and \(\psi_{211}\). Whenever there is more than one wavefunction corresponding to a given energy level, then that energy level is said to be . In the above example, the \(n=2\) energy level of the hydrogen atom is . The wavefunctions \(\psi_{nlm}(r,\theta ,\phi)\) of the electron are called , but should not be confused these with trajectories or orbits. These are complete static objects that only give static probabilities when the electron has a well-defined allowable energy. The shapes of the orbitals are largely determined by the values of angular momentum, so we will characterize the orbitals this way. The \(l=0\) orbitals are called \(s\) (\(s\) for "sharp") orbitals. When \(l=0\), \(m=0\) as well, and the wavefunctions are of the form \[\psi_{n00}(r,\theta ,\phi )=R_{n0}(r)Y_{00}(\theta ,\phi )=\left ( \dfrac{1}{4\pi} \right )^{1/2}R_{n0}(r)\] There is no dependence on \(\theta\) or \(\phi\) because \(Y_{00}\) is a constant. Thus, all of these orbitals are spherically symmetric (Figure \(\Page {3}\)). Note several things about these orbitals. First, the density of points dies off exponentially as \(r\) increases, consistent with the exponential dependence of the functions \(R_{n0}(r)\) (Figure \(\Page {4}\)a) . As \(n\) increases, the exponentials decay more slowly as \(e^{-r/a_0}\) for \(n=1\), \(e^{-r/2a_0}\) for \(n=2\) and \(e^{-r/3a_0}\) for \(n=3\). Note, also, that the wavefunctions are peaked at \(r=0\), which would suggest that the amplitude is maximal to find the electron right on top of the nucleus! In fact, we need to be careful about this interpretation, since the radial probability density (Equation \(\ref{radeq}\)) contains an extra \(r^2\) factor from the volume element which goes to \(0\) as \(r \rightarrow 0\) ( Figure \(\Page {4}\)b ). Figure \(\Page {5}\) compares the radial probability densities of the first three \(l=0\) radial wavefunctions. We also see that the wavefunctions have radial nodes where the electron will never be found (Figure \(\Page {5}\)). The number of nodes for \(R_{n0}(r)\) is \(n-1\). The Schrödinger Equation (\(\ref{Seq}\)) is often called the and is limiting case of the more complex that includes an explicit dependence to the wavefunction (where the "wave" part is most applicable). This is demonstrated in the simplified "drum" models below for the first three \(l=0\) orbitals. In all of the modes analogous to orbitals, it can be seen that the center of the \(r=0\) drum vibrates most strongly. The \(l=1\) orbitals are known as \(p\) (\("p"\) for ``principal'') orbitals. The orbitals take the form \[\psi_{nlm}(r,\theta ,\phi )=R_{n1}(r)Y_{1m}(\theta ,\phi )\] where \[\begin{align*}Y_{1\pm 1}(\theta ,\phi ) &= \pm \left ( \dfrac{3}{8\pi} \right )^{1/2} \sin\theta e^{\pm i\phi }\\ Y_{10}(\theta ,\phi ) &= \left ( \dfrac{3}{4\pi } \right )^{1/2} \cos\theta \end{align*}\] Thus, these orbitals are spherically symmetric. The \(m=0\) orbital is known as the \(p_z\) orbital because of the \(\cos\theta \) dependence and lack of \(\phi \) dependence. This resembles the spherical coordinate transformation for \(z\), \(z=r\cos\theta \). Figure \(\Page {6}\) shows the \(2p\) orbitals in more detail The nodal plane in the \(p\) orbital at \(\theta =\pi /2\) arises because \(\cos(\pi /2)=0\) for all \(\phi\), meaning that the entire \(xy\) plane is a nodal plane. The orbitals \(\psi_{n11}(r,\theta ,\phi )\) and \(\psi_{n1-1}(r,\theta ,\phi )\) are not real because of the \(exp(\pm i\phi )\) dependence of \(Y_{1\pm 1}(\theta ,\phi )\). Thus, these orbitals are not entire convenient to work with. Fortunately, because \(\psi_{n11}\) and \(\psi_{n1-1}\) are solutions of the Schrödinger equation with the same energy \(E_n\), we can take any combination of these two functions we wish, and we still have a solution of the Schrödinger equation with the same energy. Thus, it is useful to take two combinations that give us two real orbitals. Consider, for example: \[\begin{align*}\tilde{\psi}_{p_x} &= \dfrac{1}{\sqrt{2}}[\psi_{n1-1}(r,\theta ,\phi )-\psi_{n11}(r,\theta ,\phi )]\\ \tilde{\psi}_{p_y} &= \dfrac{i}{\sqrt{2}}[\psi_{n1-1}(r,\theta ,\phi )+\psi_{n11}(r,\theta ,\phi )]\end{align*}\] which corresponds to defining new spherical harmonics: \[\begin{align*}Y_{p_x}(\theta ,\phi ) &= \dfrac{1}{\sqrt{2}}[Y_{1-1}(\theta ,\phi )-Y_{11}(\theta ,\phi )]=\left ( \dfrac{3}{4\pi} \right )^{1/2}\sin\theta \cos\phi \\ Y_{p_y}(\theta ,\phi ) &= \dfrac{i}{\sqrt{2}}[Y_{1-1}(\theta ,\phi )+Y_{11}(\theta ,\phi )]=\left ( \dfrac{3}{4\pi} \right )^{1/2}\sin\theta \sin\phi \end{align*}\] Again, the notation \(p_x\) and \(p_y\) is used because of the similarity to the spherical coordinate transformations \(x=r\sin\theta \cos\phi \) and \(y=r\sin\theta \sin\phi \). These orbitals have the same shape as the \(p_z\) orbital but are rotated to be oriented along the x -axis for the \(p_x\) orbital and along the y -axis for the \(p_y\) orbital. The \(l=2\) orbitals are known as \(d\) (\("d"\) for "diffuse") orbitals. Again, we seek combinations of the spherical harmonics that give us real orbitals. The combinations we arrive at are known as \(Y_{xy}\), \(Y_{xz}\), \(Y_{yz}\), \(Y_{z^2}\), \(Y_{x^2 -y^2 }\), which gives us the required five orbitals we need for \(m=-2,-1,0,1,2\). The notation again reflects the angular dependence we would have if we took products \(xy\), \(xz\), \(yz\), \(z^2 \), and \(x^2 -y^2 \) using the spherical coordinate transformation equations. Figure \(\Page {7}\) compared the five \(3d\) orbitals. Note the presence of two nodal planes in most of the orbitals. The exception is the \(3d_{z^2}\) orbital which has a . Here, the number of radial nodes is still \(n-l-1\), and, as noted earlier, the overall number of nodes remains the same, so as \(l\) increases, radial nodes are exchanged for angular nodes. For any of the wavefunctions \(\psi_{nlm}(r,\theta ,\phi )\), the result of measuring the distance of the electron from the nucleus many times yields the average value or expectation value of \(r\), which can be shown to be \[\begin{align*}\langle r\rangle &= \int_{0}^{2\pi }\int_{0}^{\pi }\int_{0}^{\infty }|\psi_{nlm}(r,\theta ,\phi )|^2 r^3 \sin\theta drd\theta d\phi \\ &= \dfrac{n^2 a_0}{Z}\left [ 1+\dfrac{1}{2}\left ( 1-\dfrac{l(l+1)}{n^2} \right ) \right ]\end{align*}\] ( ) | 26,215 | 3,133 |
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In all solutions, whether gaseous, liquid, or solid, the substance present in the greatest amount is the solvent, and the substance or substances present in lesser amounts are the solute(s). The solute does not have to be in the same physical state as the solvent, but the physical state of the solvent usually determines the state of the solution. As long as the solute and solvent combine to give a homogeneous solution, the solute is said to be soluble in the solvent. Table \(\Page {1}\) lists some common examples of gaseous, liquid, and solid solutions and identifies the physical states of the solute and solvent in each. The formation of a solution from a solute and a solvent is a physical process, not a chemical one. That is, both solute and solvent can be recovered in chemically unchanged forms using appropriate separation methods. For example, solid zinc nitrate dissolves in water to form an aqueous solution of zinc nitrate: \[\ce{Zn(NO3)2(s) + H2O(l) \rightarrow Zn^{2+}(aq) + 2NO^{-}3(aq)} \label{13.1.1} \] Because \(Zn(NO_3)_2\) can be recovered easily by evaporating the water, this is a physical process. In contrast, metallic zinc appears to dissolve in aqueous hydrochloric acid. In fact, the two substances undergo a chemical reaction to form an aqueous solution of zinc chloride with evolution of hydrogen gas: \[\ce{ Zn(s) + 2H^{+}(aq) + 2Cl^{-}(aq) \rightarrow Zn^{2+}(aq) + 2Cl^{-}(aq) + H2(g)} \label{13.1.2} \] When the solution evaporates, we do not recover metallic zinc, so we cannot say that metallic zinc is soluble in aqueous hydrochloric acid because it is chemically transformed when it dissolves. The dissolution of a solute in a solvent to form a solution does not involve a chemical transformation (that it is a ). Dissolution of a solute in a solvent to form a solution does not involve a chemical transformation. Substances that form a single homogeneous phase in all proportions are said to be completely in one another. Ethanol and water are miscible, just as mixtures of gases are miscible. If two substances are essentially insoluble in each other, such as oil and water, they are . Examples of gaseous solutions that we have already discussed include Earth’s atmosphere. Energy is required to overcome the intermolecular interactions in a solute, which can be supplied only by the new interactions that occur in the solution, when each solute particle is surrounded by particles of the solvent in a process called (or hydration when the solvent is water). Thus all of the solute–solute interactions and many of the solvent–solvent interactions must be disrupted for a solution to form. In this section, we describe the role of enthalpy in this process. Because enthalpy is a , we can use a thermochemical cycle to analyze the energetics of solution formation. The process occurs in three discrete steps, indicated by \(ΔH_1\), \(ΔH_2\), and \(ΔH_3\) in Figure \(\Page {2}\). The overall enthalpy change in the formation of the solution (\( \Delta H_{soln}\)) is the sum of the enthalpy changes in the three steps: \[ \Delta H_{soln} = \Delta H_1 + \Delta H_2 + \Delta H_3 \label{13.1.3} \] When a solvent is added to a solution, steps 1 and 2 are both endothermic because energy is required to overcome the intermolecular interactions in the solvent (\(\Delta H_1\)) and the solute (\(\Delta H_2\)). Because \(ΔH\) is positive for both steps 1 and 2, the solute–solvent interactions (\(\Delta H_3\)) must be stronger than the solute–solute and solvent–solvent interactions they replace in order for the dissolution process to be exothermic (\(\Delta H_{soln} < 0\)). When the solute is an ionic solid, \(ΔH_2\) corresponds to the lattice energy that must be overcome to form a solution. The higher the charge of the ions in an ionic solid, the higher the lattice energy. Consequently, solids that have very high lattice energies, such as \(MgO\) (−3791 kJ/mol), are generally insoluble in all solvents. A positive value for \(ΔH_{soln}\) does not mean that a solution will not form. Whether a given process, including formation of a solution, occurs spontaneously depends on whether the total energy of the system is lowered as a result. Enthalpy is only one of the contributing factors. A high \(ΔH_{soln}\) is usually an indication that the substance is not very soluble. Instant cold packs used to treat athletic injuries, for example, take advantage of the large positive \(ΔH_{soln}\) of ammonium nitrate during dissolution (+25.7 kJ/mol), which produces temperatures less than 0°C (Figure \(\Page {3}\)). The enthalpy change that accompanies a process is important because processes that release substantial amounts of energy tend to occur spontaneously. A second property of any system, its entropy, is also important in helping us determine whether a given process occurs spontaneously. We will discuss entropy in more detail elsewhere, but for now we can state that entropy (\(S\)) is a thermodynamic property of all substances that is proportional to their degree of disorder. A perfect crystal at 0 K, whose atoms are regularly arranged in a perfect lattice and are motionless, has an entropy of zero. In contrast, gases have large positive entropies because their molecules are highly disordered and in constant motion at high speeds. The formation of a solution disperses molecules, atoms, or ions of one kind throughout a second substance, which generally increases the disorder and results in an increase in the entropy of the system. Thus entropic factors almost always favor formation of a solution. In contrast, a change in enthalpy may or may not favor solution formation. The London dispersion forces that hold cyclohexane and n-hexane together in pure liquids, for example, are similar in nature and strength. Consequently, \(ΔH_{soln}\) should be approximately zero, as is observed experimentally. Mixing equal amounts of the two liquids, however, produces a solution in which the n-hexane and cyclohexane molecules are uniformly distributed over approximately twice the initial volume. In this case, the driving force for solution formation is not a negative \(ΔH_{soln}\) but rather the increase in entropy due to the increased disorder in the mixture. All spontaneous processes with \(ΔH \ge 0\) are characterized by an in entropy. In other cases, such as mixing oil with water, salt with gasoline, or sugar with hexane, the enthalpy of solution is large and positive, and the increase in entropy resulting from solution formation is not enough to overcome it. Thus in these cases a solution does not form. All spontaneous processes with ΔH ≥ 0 are characterized by an increase in entropy. Table \(\Page {2}\) summarizes how enthalpic factors affect solution formation for four general cases. The column on the far right uses the relative magnitudes of the enthalpic contributions to predict whether a solution will form from each of the four. Keep in mind that in each case entropy favors solution formation. In two of the cases the enthalpy of solution is expected to be relatively small and can be either positive or negative. Thus the entropic contribution dominates, and we expect a solution to form readily. In the other two cases the enthalpy of solution is expected to be large and positive. The entropic contribution, though favorable, is usually too small to overcome the unfavorable enthalpy term. Hence we expect that a solution will not form readily. In contrast to liquid solutions, the intermolecular interactions in gases are weak (they are considered to be nonexistent in ideal gases). Hence mixing gases is usually a thermally neutral process (\(ΔH_{soln} \approx 0\)), and the entropic factor due to the increase in disorder is dominant (Figure \(\Page {4}\)). Consequently, all gases dissolve readily in one another in all proportions to form solutions. Considering \(\ce{LiCl}\), benzoic acid (\(\ce{C6H5CO2H}\)), and naphthalene, which will be most soluble and which will be least soluble in water? : three compounds relative solubilities in water : Assess the relative magnitude of the enthalpy change for each step in the process shown in Figure \(\Page {2}\). Then use Table \(\Page {2}\) to predict the solubility of each compound in water and arrange them in order of decreasing solubility. The first substance, \(\ce{LiCl}\), is an ionic compound, so a great deal of energy is required to separate its anions and cations and overcome the lattice energy (ΔH is far greater than zero in Equation \(\ref{13.1.1}\)). Because water is a polar substance, the interactions between both Li and Cl ions and water should be favorable and strong. Thus we expect \(ΔH_3\) to be far less than zero, making LiCl soluble in water. In contrast, naphthalene is a nonpolar compound, with only London dispersion forces holding the molecules together in the solid state. We therefore expect \(ΔH_2\) to be small and positive. We also expect the interaction between polar water molecules and nonpolar naphthalene molecules to be weak \(ΔH_3 \approx 0\). Hence we do not expect naphthalene to be very soluble in water, if at all. Benzoic acid has a polar carboxylic acid group and a nonpolar aromatic ring. We therefore expect that the energy required to separate solute molecules (ΔH ) will be greater than for naphthalene and less than for LiCl. The strength of the interaction of benzoic acid with water should also be intermediate between those of LiCl and naphthalene. Hence benzoic acid is expected to be more soluble in water than naphthalene but less soluble than \(\ce{LiCl}\). We thus predict \(\ce{LiCl}\) to be the most soluble in water and naphthalene to be the least soluble. Considering ammonium chloride, cyclohexane, and ethylene glycol (\(HOCH_2CH_2OH\)), which will be most soluble and which will be least soluble in benzene? The most soluble is cyclohexane; the least soluble is ammonium chloride. Solutions are homogeneous mixtures of two or more substances whose components are uniformly distributed on a microscopic scale. The component present in the greatest amount is the solvent, and the components present in lesser amounts are the solute(s). The formation of a solution from a solute and a solvent is a physical process, not a chemical one. Substances that are miscible, such as gases, form a single phase in all proportions when mixed. Substances that form separate phases are immiscible. Solvation is the process in which solute particles are surrounded by solvent molecules. When the solvent is water, the process is called hydration. The overall enthalpy change that accompanies the formation of a solution, \(ΔH_{soln}\), is the sum of the enthalpy change for breaking the intermolecular interactions in both the solvent and the solute and the enthalpy change for the formation of new solute–solvent interactions. Exothermic (\(ΔH_{soln} < 0\)) processes favor solution formation. In addition, the change in entropy, the degree of disorder of the system, must be considered when predicting whether a solution will form. An increase in entropy (a decrease in order) favors dissolution. | 11,128 | 3,134 |
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On a microscopic level we can easily explain why some processes occur of their own accord while others do not. A . A nonspontaneous process, by contrast, corresponds to movement from a probable situation to an improbable one. An example of what probability has to do with spontaneity is provided by expansion of a gas into a vacuum. Let us calculate the probability that the process of gas expansion from flask into a connected flask will reverse itself, that is, the probability that the gas molecules will all collect again in flask . If we choose a particular molecule and label it number 1, we find that it is sometimes in flask and sometimes in flask . Since the molecule’s motion is random and the two flasks contain the same volume, the molecule should spend half its time in each container. The probability of finding molecule 1 in container A is therefore 1/2. Next let us consider the probability that two molecules, labeled 1 and 2, are both in flask . Figure \(\Page {1}\) shows the four possible ways these two molecules can be arranged in the two flasks. All four are equally likely, but only one has both molecules in flask . Thus there is one chance in four that molecules 1 and 2 are both in flask A. This probability of 1/4 equals 1/2 × 1/2; i.e., it is the product of the probability that molecule 1 was in flask times the probability that molecule 2 was in flask . By a similar argument we can show that the probability that three given molecules are all in flask A is 1/2 × 1/2 × 1/2 = (1/2) =1/8, and, in general, that the probability of all gas molecules being in flask at once is (1/2) . If we had 1 mol gas in the flasks, there would be 6.022 × 10 molecules. The probability that all of them would be in flask at the same time would be Similar remarks apply to the probability of reversing other spontaneous processes. In Figure \(\Page {2}\) , some of the atoms in a bar of metal at uniform temperature will be vibrating more than others, but the unusually energetic atoms will be distributed fairly evenly throughout the bar. We will not suddenly find all the energetic metal atoms on the left end of the bar and all the weakly vibrating ones on the right, as in Figure \(\Page {2}\) . The possibility exists that a freak series of collisions between vibrating atoms might produce a high concentration of energetic atoms on the left, but such an occurrence is inconceivably improbable. When a falling book hits the floor, its kinetic energy is converted to heat energy. The floor warms up slightly, and the molecules there start vibrating a little more energetically. For such a process to reverse itself spontaneously, all the floor molecules under the book would suddenly have to become more energetic and vibrate in unison in a vertical direction, flinging the book into the air. As in the previous two examples, this would require a freak series of molecular collisions which is so improbable as never to occur in the entire lifetime of the universe. These principles also apply to the processes considered in the . In the NI reaction, all of the particles distributed in the explosion would all have to return back to the ring stand and reassemble themselves. Even leaving the chemical reaction unconsidered for the moment, this return of all the particles is highly improbable. A similar argument can be put forward for the un-mixing of the dye. The probability that all of the dye molecules will return back to a single uniform drop separate from the water at one moment is another situation so improbable that it will never happen. Simple cases we have described show how spontaneous and nonspontaneous processes can be considered from a microscopic and statistical viewpoint. In any real sample of matter there are a great many molecules jostling each other about, exchanging energy, and sometimes exchanging atoms. This constant jostle is like shuffling a gigantic deck of cards. Because the numbers involved are so large, the laws of probability are inexorable. Some probabilities are large enough to be virtual certainties, while others are small enough to be unthinkable. Invariably the reversal of a spontaneous process turns out to involve movement from an almost certain situation to one which is unimaginably improbable. Conversely, a spontaneous process occurs when a sample of matter finds itself momentarily in a highly improbable situation. As fast as possible, it will adjust on the molecular level until maximum probability is attained. | 4,525 | 3,135 |
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You can experience directly the mass, volume, or temperature of a substance, but you cannot experience its entropy. Consequently you may have the feeling that entropy is somehow less real than other properties of matter. We hope to show in this section that it is quite easy to predict whether the entropy under one set of circumstances will be larger than under another set of circumstances, and also to explain why. With a little practice in making such predictions in simple cases you will acquire an intuitive feel for entropy and it will lose its air of mystery. The entropy of a substance depends on two things: first, the of a substance—its temperature, pressure, and amount; and second, how the substance is . We will discuss how properties affect entropy first. As we saw in the last section, there should be only one way of arranging the energy in a perfect crystal at 0 K. If = 1, then = ln = 0; so that the . This rule, known as the , is obeyed by all solids unless some randomness of arrangement is accidentally “frozen” into the crystal. As energy is fed into the crystal with increasing temperature, we find that an increasing number of alternative ways of dividing the energy between the atoms become possible. increases, and so does . Without exception the entropy of any pure substance . | 1,342 | 3,136 |
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The Lewis electron-pair approach can be used to predict the number and types of bonds between the atoms in a substance, and it indicates which atoms have lone pairs of electrons. This approach gives no information about the actual arrangement of atoms in space, however. We continue our discussion of structure and bonding by introducing the (VSEPR) model (pronounced “vesper”), which can be used to predict the shapes of many molecules and polyatomic ions. Keep in mind, however, that the VSEPR model, like any model, is a limited representation of reality; the model provides no information about bond lengths or the presence of multiple bonds. The VSEPR model can predict the structure of nearly any molecule or polyatomic ion in which the central atom is a nonmetal, as well as the structures of many molecules and polyatomic ions with a central metal atom. The premise of the VSEPR theory is that electron pairs located in bonds and lone pairs repel each other and will therefore adopt the geometry that places electron pairs as far apart from each other as possible. This theory is very simplistic and does not account for the subtleties of orbital interactions that influence molecular shapes; however, the simple VSEPR counting procedure accurately predicts the three-dimensional structures of a large number of compounds, which cannot be predicted using the Lewis electron-pair approach. We can use the VSEPR model to predict the geometry of most polyatomic molecules and ions by focusing only on the number of electron pairs around the , ignoring all other valence electrons present. According to this model, valence electrons in the Lewis structure form , which may consist of a single bond, a double bond, a triple bond, a lone pair of electrons, or even a single unpaired electron, which in the VSEPR model is counted as a lone pair. Because electrons repel each other electrostatically, the most stable arrangement of electron groups (i.e., the one with the lowest energy) is the one that minimizes repulsions. Groups are positioned around the central atom in a way that produces the molecular structure with the lowest energy, as illustrated in Figures \(\Page {1}\) and \(\Page {2}\). In the VSEPR model, the molecule or polyatomic ion is given an AX E designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and and are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the and interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the bond angles. Using this information, we can describe the molecular geometry, the arrangement of the in a molecule or polyatomic ion. This VESPR procedure is summarized as follows: We will illustrate the use of this procedure with several examples, beginning with atoms with two electron groups. In our discussion we will refer to Figure \(\Page {2}\) and Figure \(\Page {3}\), which summarize the common molecular geometries and idealized bond angles of molecules and ions with two to six electron groups. Our first example is a molecule with two bonded atoms and no lone pairs of electrons, \(BeH_2\). 1. The central atom, beryllium, contributes two valence electrons, and each hydrogen atom contributes one. The Lewis electron structure is 3. Both groups around the central atom are bonding pairs (BP). Thus BeH is designated as AX . 4. From Figure \(\Page {3}\) we see that with two bonding pairs, the molecular geometry that minimizes repulsions in BeH is . 1. The central atom, carbon, contributes four valence electrons, and each oxygen atom contributes six. The Lewis electron structure is 2. The carbon atom forms two double bonds. Each double bond is a group, so there are two electron groups around the central atom. Like BeH , the arrangement that minimizes repulsions places the groups 180° apart. 3. Once again, both groups around the central atom are bonding pairs (BP), so CO is designated as AX . 4. VSEPR only recognizes groups around the atom. Thus the lone pairs on the oxygen atoms do not influence the molecular geometry. With two bonding pairs on the central atom and no lone pairs, the molecular geometry of CO is linear (Figure \(\Page {3}\)). The structure of \(\ce{CO2}\) is shown in Figure \(\Page {1}\). 1. The central atom, boron, contributes three valence electrons, and each chlorine atom contributes seven valence electrons. The Lewis electron structure is 3. All electron groups are bonding pairs (BP), so the structure is designated as AX . 4. From Figure \(\Page {3}\) we see that with three bonding pairs around the central atom, the molecular geometry of BCl is , as shown in Figure \(\Page {2}\). 1. The central atom, carbon, has four valence electrons, and each oxygen atom has six valence electrons. As you learned previously, the Lewis electron structure of one of three resonance forms is represented as 3. All electron groups are bonding pairs (BP). With three bonding groups around the central atom, the structure is designated as AX . 4. We see from Figure \(\Page {3}\) that the molecular geometry of CO is trigonal planar with bond angles of 120°. In our next example we encounter the effects of lone pairs and multiple bonds on molecular geometry for the first time. 1. The central atom, sulfur, has 6 valence electrons, as does each oxygen atom. With 18 valence electrons, the Lewis electron structure is shown below. 3. There are two bonding pairs and one lone pair, so the structure is designated as AX E. This designation has a total of three electron pairs, two X and one E. Because a lone pair is not shared by two nuclei, it occupies more space near the central atom than a bonding pair (Figure \(\Page {4}\)). Thus bonding pairs and lone pairs repel each other electrostatically in the order BP–BP < LP–BP < LP–LP. In SO , we have one BP–BP interaction and two LP–BP interactions. 4. The molecular geometry is described only by the positions of the nuclei, by the positions of the lone pairs. Thus with two nuclei and one lone pair the shape is , or , which can be viewed as a trigonal planar arrangement with a missing vertex (Figures \(\Page {2}\) and \(\Page {3}\)). The O-S-O bond angle is expected to be 120° because of the extra space taken up by the lone pair. As with SO , this composite model of electron distribution and negative electrostatic potential in ammonia shows that a lone pair of electrons occupies a larger region of space around the nitrogen atom than does a bonding pair of electrons that is shared with a hydrogen atom. Like lone pairs of electrons, multiple bonds occupy more space around the central atom than a single bond, which can cause other bond angles to be somewhat smaller than expected. This is because a multiple bond has a higher electron density than a single bond, so its electrons occupy more space than those of a single bond. For example, in a molecule such as CH O (AX ), whose structure is shown below, the double bond repels the single bonds more strongly than the single bonds repel each other. This causes a deviation from ideal geometry (an H–C–H bond angle of 116.5° rather than 120°). One of the limitations of Lewis structures is that they depict molecules and ions in only two dimensions. With four electron groups, we must learn to show molecules and ions in three dimensions. 1. The central atom, carbon, contributes four valence electrons, and each hydrogen atom has one valence electron, so the full Lewis electron structure is 2. There are four electron groups around the central atom. As shown in Figure \(\Page {2}\), repulsions are minimized by placing the groups in the corners of a tetrahedron with bond angles of 109.5°. 3. All electron groups are bonding pairs, so the structure is designated as AX . 4. With four bonding pairs, the molecular geometry of methane is (Figure \(\Page {3}\)). 1. In ammonia, the central atom, nitrogen, has five valence electrons and each hydrogen donates one valence electron, producing the Lewis electron structure 2. There are four electron groups around nitrogen, three bonding pairs and one lone pair. Repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron. 3. With three bonding pairs and one lone pair, the structure is designated as AX E. This designation has a total of four electron pairs, three X and one E. We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron. 4. There are three nuclei and one lone pair, so the molecular geometry is . In essence, this is a tetrahedron with a vertex missing (Figure \(\Page {3}\)). However, the H–N–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions (Figure \(\Page {3}\) and Figure \(\Page {4}\)). 1. Oxygen has six valence electrons and each hydrogen has one valence electron, producing the Lewis electron structure 3. With two bonding pairs and two lone pairs, the structure is designated as AX E with a total of four electron pairs. Due to LP–LP, LP–BP, and BP–BP interactions, we expect a significant deviation from idealized tetrahedral angles. 4. With two hydrogen atoms and two lone pairs of electrons, the structure has significant lone pair interactions. There are two nuclei about the central atom, so the molecular shape is , or , with an H–O–H angle that is even less than the H–N–H angles in NH , as we would expect because of the presence of two lone pairs of electrons on the central atom rather than one. This molecular shape is essentially a tetrahedron with two missing vertices. In previous examples it did not matter where we placed the electron groups because all positions were equivalent. In some cases, however, the positions are not equivalent. We encounter this situation for the first time with five electron groups. 1. Phosphorus has five valence electrons and each chlorine has seven valence electrons, so the Lewis electron structure of PCl is 3. All electron groups are bonding pairs, so the structure is designated as AX . There are no lone pair interactions. 4. The molecular geometry of PCl is , as shown in Figure \(\Page {3}\). The molecule has three atoms in a plane in positions and two atoms above and below the plane in positions. The three equatorial positions are separated by 120° from one another, and the two axial positions are at 90° to the equatorial plane. The axial and equatorial positions are not chemically equivalent, as we will see in our next example. 1. The sulfur atom has six valence electrons and each fluorine has seven valence electrons, so the Lewis electron structure is With an expanded valence, this species is an exception to the octet rule. 2. There are five groups around sulfur, four bonding pairs and one lone pair. With five electron groups, the lowest energy arrangement is a trigonal bipyramid, as shown in Figure \(\Page {2}\). 3. We designate SF as AX E; it has a total of five electron pairs. However, because the axial and equatorial positions are not chemically equivalent, where do we place the lone pair? If we place the lone pair in the axial position, we have three LP–BP repulsions at 90°. If we place it in the equatorial position, we have two 90° LP–BP repulsions at 90°. With fewer 90° LP–BP repulsions, we can predict that the structure with the lone pair of electrons in the . We also expect a deviation from ideal geometry because a lone pair of electrons occupies more space than a bonding pair. At 90°, the two electron pairs share a relatively large region of space, which leads to strong repulsive electron–electron interactions. 4. With four nuclei and one lone pair of electrons, the molecular structure is based on a trigonal bipyramid with a missing equatorial vertex; it is described as a . The F –S–F angle is 173° rather than 180° because of the lone pair of electrons in the equatorial plane. 1. The bromine atom has seven valence electrons, and each fluorine has seven valence electrons, so the Lewis electron structure is Once again, we have a compound that is an exception to the octet rule. 2. There are five groups around the central atom, three bonding pairs and two lone pairs. We again direct the groups toward the vertices of a trigonal bipyramid. 3. With three bonding pairs and two lone pairs, the structural designation is AX E with a total of five electron pairs. Because the axial and equatorial positions are not equivalent, we must decide how to arrange the groups to minimize repulsions. If we place both lone pairs in the axial positions, we have six LP–BP repulsions at 90°. If both are in the equatorial positions, we have four LP–BP repulsions at 90°. If one lone pair is axial and the other equatorial, we have one LP–LP repulsion at 90° and three LP–BP repulsions at 90°: Structure (c) can be eliminated because it has a LP–LP interaction at 90°. Structure (b), with fewer LP–BP repulsions at 90° than (a), is lower in energy. However, we predict a deviation in bond angles because of the presence of the two lone pairs of electrons. 4. The three nuclei in BrF determine its molecular structure, which is described as . This is essentially a trigonal bipyramid that is missing two equatorial vertices. The F –Br–F angle is 172°, less than 180° because of LP–BP repulsions (Figure \(\Page {2}\).1). 1. Each iodine atom contributes seven electrons and the negative charge one, so the Lewis electron structure is 2. There are five electron groups about the central atom in I , two bonding pairs and three lone pairs. To minimize repulsions, the groups are directed to the corners of a trigonal bipyramid. 3. With two bonding pairs and three lone pairs, I has a total of five electron pairs and is designated as AX E . We must now decide how to arrange the lone pairs of electrons in a trigonal bipyramid in a way that minimizes repulsions. Placing them in the axial positions eliminates 90° LP–LP repulsions and minimizes the number of 90° LP–BP repulsions. The three lone pairs of electrons have equivalent interactions with the three iodine atoms, so we do not expect any deviations in bonding angles. 4. With three nuclei and three lone pairs of electrons, the molecular geometry of I is linear. This can be described as a trigonal bipyramid with three equatorial vertices missing. The ion has an I–I–I angle of 180°, as expected. Six electron groups form an , a polyhedron made of identical equilateral triangles and six identical vertices (Figure \(\Page {2}\).) 1. The central atom, sulfur, contributes six valence electrons, and each fluorine atom has seven valence electrons, so the Lewis electron structure is With an expanded valence, this species is an exception to the octet rule. 2. There are six electron groups around the central atom, each a bonding pair. We see from Figure \(\Page {2}\) that the geometry that minimizes repulsions is . 3. With only bonding pairs, SF is designated as AX . All positions are chemically equivalent, so all electronic interactions are equivalent. 4. There are six nuclei, so the molecular geometry of SF is octahedral. 1. The central atom, bromine, has seven valence electrons, as does each fluorine, so the Lewis electron structure is With its expanded valence, this species is an exception to the octet rule. 2. There are six electron groups around the Br, five bonding pairs and one lone pair. Placing five F atoms around Br while minimizing BP–BP and LP–BP repulsions gives the following structure: 3. With five bonding pairs and one lone pair, BrF is designated as AX E; it has a total of six electron pairs. The BrF structure has four fluorine atoms in a plane in an equatorial position and one fluorine atom and the lone pair of electrons in the axial positions. We expect all F –Br–F angles to be less than 90° because of the lone pair of electrons, which occupies more space than the bonding electron pairs. 4. With five nuclei surrounding the central atom, the molecular structure is based on an octahedron with a vertex missing. This molecular structure is . The F –B–F angles are 85.1°, less than 90° because of LP–BP repulsions. 1. The central atom, iodine, contributes seven electrons. Each chlorine contributes seven, and there is a single negative charge. The Lewis electron structure is 2. There are six electron groups around the central atom, four bonding pairs and two lone pairs. The structure that minimizes LP–LP, LP–BP, and BP–BP repulsions is 3. ICl is designated as AX E and has a total of six electron pairs. Although there are lone pairs of electrons, with four bonding electron pairs in the equatorial plane and the lone pairs of electrons in the axial positions, all LP–BP repulsions are the same. Therefore, we do not expect any deviation in the Cl–I–Cl bond angles. 4. With five nuclei, the ICl4− ion forms a molecular structure that is , an octahedron with two opposite vertices missing. The relationship between the number of electron groups around a central atom, the number of lone pairs of electrons, and the molecular geometry is summarized in Figure \(\Page {6}\). Using the VSEPR model, predict the molecular geometry of each molecule or ion. two chemical species molecular geometry All electron groups are bonding pairs, so PF is designated as AX . Notice that this gives a total of five electron pairs. With no lone pair repulsions, we do not expect any bond angles to deviate from the ideal. The PF molecule has five nuclei and no lone pairs of electrons, so its molecular geometry is trigonal bipyramidal. There are four electron groups around oxygen, three bonding pairs and one lone pair. Like NH , repulsions are minimized by directing each hydrogen atom and the lone pair to the corners of a tetrahedron. With three bonding pairs and one lone pair, the structure is designated as AX E and has a total of four electron pairs (three X and one E). We expect the LP–BP interactions to cause the bonding pair angles to deviate significantly from the angles of a perfect tetrahedron. There are three nuclei and one lone pair, so the molecular geometry is , in essence a tetrahedron missing a vertex. However, the H–O–H bond angles are less than the ideal angle of 109.5° because of LP–BP repulsions: Using the VSEPR model, predict the molecular geometry of each molecule or ion. trigonal pyramidal octahedral linear Predict the molecular geometry of each molecule. two chemical compounds molecular geometry Use the strategy given in Example\(\Page {1}\). There are five electron groups around the central atom, two bonding pairs and three lone pairs. Repulsions are minimized by placing the groups in the corners of a trigonal bipyramid. From B, XeF is designated as AX E and has a total of five electron pairs (two X and three E). With three lone pairs about the central atom, we can arrange the two F atoms in three possible ways: both F atoms can be axial, one can be axial and one equatorial, or both can be equatorial: The structure with the lowest energy is the one that minimizes LP–LP repulsions. Both (b) and (c) have two 90° LP–LP interactions, whereas structure (a) has none. Thus both F atoms are in the axial positions, like the two iodine atoms around the central iodine in I . All LP–BP interactions are equivalent, so we do not expect a deviation from an ideal 180° in the F–Xe–F bond angle. With two nuclei about the central atom, the molecular geometry of XeF is linear. It is a trigonal bipyramid with three missing equatorial vertices. There are three electron groups around the central atom, two bonding groups and one lone pair of electrons. To minimize repulsions the three groups are initially placed at 120° angles from each other. From B we designate SnCl as AX E. It has a total of three electron pairs, two X and one E. Because the lone pair of electrons occupies more space than the bonding pairs, we expect a decrease in the Cl–Sn–Cl bond angle due to increased LP–BP repulsions. With two nuclei around the central atom and one lone pair of electrons, the molecular geometry of SnCl is bent, like SO , but with a Cl–Sn–Cl bond angle of 95°. The molecular geometry can be described as a trigonal planar arrangement with one vertex missing. Predict the molecular geometry of each molecule. trigonal planar square planar The VSEPR model can be used to predict the structure of somewhat more complex molecules with no single central atom by treating them as linked AX E fragments. We will demonstrate with methyl isocyanate (CH –N=C=O), a volatile and highly toxic molecule that is used to produce the pesticide Sevin. In 1984, large quantities of Sevin were accidentally released in Bhopal, India, when water leaked into storage tanks. The resulting highly exothermic reaction caused a rapid increase in pressure that ruptured the tanks, releasing large amounts of methyl isocyanate that killed approximately 3800 people and wholly or partially disabled about 50,000 others. In addition, there was significant damage to livestock and crops. We can treat methyl isocyanate as linked AX E fragments beginning with the carbon atom at the left, which is connected to three H atoms and one N atom by single bonds. The four bonds around carbon mean that it must be surrounded by four bonding electron pairs in a configuration similar to AX . We can therefore predict the CH –N portion of the molecule to be roughly tetrahedral, similar to methane: The nitrogen atom is connected to one carbon by a single bond and to the other carbon by a double bond, producing a total of three bonds, C–N=C. For nitrogen to have an octet of electrons, it must also have a lone pair: Because multiple bonds are not shown in the VSEPR model, the nitrogen is effectively surrounded by three electron pairs. Thus according to the VSEPR model, the C–N=C fragment should be bent with an angle less than 120°. The carbon in the –N=C=O fragment is doubly bonded to both nitrogen and oxygen, which in the VSEPR model gives carbon a total of two electron pairs. The N=C=O angle should therefore be 180°, or linear. The three fragments combine to give the following structure: Certain patterns are seen in the structures of moderately complex molecules. For example, carbon atoms with four bonds (such as the carbon on the left in methyl isocyanate) are generally tetrahedral. Similarly, the carbon atom on the right has two double bonds that are similar to those in CO , so its geometry, like that of CO , is linear. Recognizing similarities to simpler molecules will help you predict the molecular geometries of more complex molecules. Use the VSEPR model to predict the molecular geometry of propyne (H C–C≡CH), a gas with some anesthetic properties. chemical compound molecular geometry Count the number of electron groups around each carbon, recognizing that in the VSEPR model, a multiple bond counts as a single group. Use Figure \(\Page {3}\) to determine the molecular geometry around each carbon atom and then deduce the structure of the molecule as a whole. Because the carbon atom on the left is bonded to four other atoms, we know that it is approximately tetrahedral. The next two carbon atoms share a triple bond, and each has an additional single bond. Because a multiple bond is counted as a single bond in the VSEPR model, each carbon atom behaves as if it had two electron groups. This means that both of these carbons are linear, with C–C≡C and C≡C–H angles of 180°. Predict the geometry of allene (H C=C=CH ), a compound with narcotic properties that is used to make more complex organic molecules. The terminal carbon atoms are trigonal planar, the central carbon is linear, and the C–C–C angle is 180°. You previously learned how to calculate the of simple diatomic molecules. In more complex molecules with polar covalent bonds, the three-dimensional geometry and the compound’s symmetry determine whether there is a net dipole moment. Mathematically, dipole moments are ; they possess both a and a . The dipole moment of a molecule is therefore the of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. Such is the case for CO , a linear molecule (Figure \(\Page {8a}\)). Each C–O bond in CO is polar, yet experiments show that the CO molecule has no dipole moment. Because the two C–O bond dipoles in CO are equal in magnitude and oriented at 180° to each other, they cancel. As a result, the CO molecule has no dipole moment even though it has a substantial separation of charge. In contrast, the H O molecule is not linear (Figure \(\Page {8b}\)); it is bent in three-dimensional space, so the dipole moments do not cancel each other. Thus a molecule such as H O has a net dipole moment. We expect the concentration of negative charge to be on the oxygen, the more electronegative atom, and positive charge on the two hydrogens. This charge polarization allows H O to hydrogen-bond to other polarized or charged species, including other water molecules. Other examples of molecules with polar bonds are shown in Figure \(\Page {9}\). In molecular geometries that are highly symmetrical (most notably tetrahedral and square planar, trigonal bipyramidal, and octahedral), individual bond dipole moments completely cancel, and there is no net dipole moment. Although a molecule like CHCl is best described as tetrahedral, the atoms bonded to carbon are not identical. Consequently, the bond dipole moments cannot cancel one another, and the molecule has a dipole moment. Due to the arrangement of the bonds in molecules that have V-shaped, trigonal pyramidal, seesaw, T-shaped, and square pyramidal geometries, the bond dipole moments cannot cancel one another. Consequently, molecules with these geometries always have a nonzero dipole moment. Molecules with asymmetrical charge distributions have a net dipole moment. Which molecule(s) has a net dipole moment? three chemical compounds net dipole moment For each three-dimensional molecular geometry, predict whether the bond dipoles cancel. If they do not, then the molecule has a net dipole moment. Which molecule(s) has a net dipole moment? \(\ce{CH3Cl}\) and \(\ce{XeO3}\) Lewis electron structures give no information about , the arrangement of bonded atoms in a molecule or polyatomic ion, which is crucial to understanding the chemistry of a molecule. The allows us to predict which of the possible structures is actually observed in most cases. It is based on the assumption that pairs of electrons occupy space, and the lowest-energy structure is the one that minimizes electron pair–electron pair repulsions. In the VSEPR model, the molecule or polyatomic ion is given an AX E designation, where A is the central atom, X is a bonded atom, E is a nonbonding valence electron group (usually a lone pair of electrons), and and are integers. Each group around the central atom is designated as a bonding pair (BP) or lone (nonbonding) pair (LP). From the BP and LP interactions we can predict both the relative positions of the atoms and the angles between the bonds, called the . From this we can describe the . The VSEPR model can be used to predict the shapes of many molecules and polyatomic ions, but it gives no information about bond lengths and the presence of multiple bonds. A combination of VSEPR and a bonding model, such as Lewis electron structures, is necessary to understand the presence of multiple bonds. Molecules with polar covalent bonds can have a , an asymmetrical distribution of charge that results in a tendency for molecules to align themselves in an applied electric field. Any diatomic molecule with a polar covalent bond has a dipole moment, but in polyatomic molecules, the presence or absence of a net dipole moment depends on the structure. For some highly symmetrical structures, the individual bond dipole moments cancel one another, giving a dipole moment of zero. | 28,146 | 3,138 |
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Generally, there are two types of inorganic compounds that can be formed: ionic compounds and molecular compounds. Nomenclature is the process of naming chemical compounds with different names so that they can be easily identified as separate chemicals. Inorganic compounds are compounds that do not deal with the formation of carbohydrates, or simply all other compounds that do not fit into the description of an organic compound. For example, organic compounds include molecules with carbon rings and/or chains with hydrogen atoms (see picture below). Inorganic compounds, the topic of this section, are every other molecule that does not include these distinctive carbon and hydrogen structures. Compounds made of a metal and nonmetal are commonly known as , where the compound name has an ending of Cations have positive charges while anions have negative charges. The net charge of any ionic compound must be zero which also means it must be electrically neutral. For example, one Na is paired with one Cl ; one Ca is paired with two Br . There are two rules that must be followed through: + = = + + = The may form more than one ion, thus it is needed to be specified which particular ion we are talking about. This is indicated by assigning a Roman numeral after the metal. The Roman numeral denotes the charge and the oxidation state of the transition metal ion. For example, iron can form two common ions, Fe and Fe . To distinguish the difference, Fe would be named iron (II) and Fe would be named iron (III). However, some of the transition metals' charges have specific Latin names. Just like the other nomenclature rules, the ion of the transition metal that has the lower charge has the Latin name ending with and the one with the the higher charge has a Latin name ending with . The most common ones are shown in the table below: Several exceptions apply to the Roman numeral assignment: Aluminum, Zinc, and Silver. Although they belong to the transition metal category, these metals do not have Roman numerals written after their names because these metals only exist in one ion. Instead of using Roman numerals, the different ions can also be presented in plain words. The metal is changed to end in or However, this system is inadequate in some cases, so the Roman numeral system is preferred. This system is used commonly in naming acids, where H SO is commonly known as Sulfuric Acid, and H SO is known as Sulfurous Acid. Compounds that consist of a nonmetal bonded to a nonmetal are commonly known as , where the element with the positive oxidation state is written first. In many cases, nonmetals form more than one binary compound, so are used to distinguish them. CO = carbon oxide BCl = boron chloride CO = carbon oxide N O = nitrogen oxide The prefix is not used for the first element. If there is not a prefix before the first element, it is assumed that there is only one atom of that element. Although HF can be named hydrogen fluoride, it is given a different name for emphasis that it is an acid. An is a substance that dissociates into hydrogen ions (H ) and anions . A quick way to identify acids is to see if there is an H (denoting hydrogen) in front of the molecular formula of the compound. To name acids, the prefix is placed in front of the nonmetal modified to end with . The state of acids is aqueous (aq) because acids are found in water. Some common binary acids include: HF (g) = hydrogen fluor HBr (g) = hydrogen brom -> HCl (g) = hydrogen chlor -> H S (g) = hydrogen sulf -> It is important to include (aq) after the acids because the same compounds can be written in gas phase with hydrogen named first followed by the anion ending with . ____ ____ ____ ClO ClO ClO ClO hypo ite ite ate per ate ----------------> As indicated by the arrow, moving to the right, the following trends occur: (Usage of this example can be seen from the set of compounds containing Cl and O) This occurs because the number of oxygen atoms are increasing from hypochlorite to perchlorate, yet the overall charge of the polyatomic ion is still -1. To correctly specify how many oxygen atoms are in the ion, prefixes and suffixes are again used. In , polyatomic (meaning two or more atoms) are joined together by . Although there may be a element with positive charge like H , it is not joined with another element with an ionic bond. This occurs because if the atoms formed an ionic bond, then it would have already become a compound, thus not needing to gain or loose any electrons. Polyatomic anions are more common than polyatomic cations as shown in the chart below. Polyatomic anions have negative charges while polyatomic cations have positive charges. To indicate different polyatomic ions made up of the same elements, the name of the ion is modified according to the example below: To combine the topic of acids and polyatomic ions, there is nomenclature of aqueous acids. Such acids include sulfuric acid (H SO ) or carbonic acid (H CO ). To name them, follow these quick, simple rules: 1. What is the correct formula for Calcium Carbonate? a. Ca + CO b. CaCO c. CaCO d. 2CaCO 2. What is the correct name for FeO? a. Iron oxide b. Iron dioxide c. Iron(III) oxide d. Iron(II) oxide 3. What is the correct name for Al(NO ) ? a. Aluminum nitrate b. Aluminum(III) nitrate c. Aluminum nitrite d. Aluminum nitrogen trioxide 4. What is the correct formula of phosphorus trichloride? a. P Cl b. PCl c. PCl d. P Cl 5. What is the correct formula of lithium perchlorate? a. Li ClO b. LiClO c. LiClO d. None of these 6. Write the correct name for these compounds. a. BeC O : b. NH MnO : c. CoS O : 7. What is W(HSO ) ? 8. How do you write diphosphorus trioxide? 9. What is H P? 10. By adding oxygens to the molecule in number 9, we now have H PO ? What is the name of this molecule? 1.C 2.D; FeO --> Fe + O --> Iron must have a charge of +2 to make a neutral compound --> Fe + O --> Iron(II) Oxide 3.A 4.B; Phosphorus trichloride --> P + 3Cl --> PCl 5.D, LiClO | 6,078 | 3,140 |
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In we apply a time-dependent potential to an electrochemical cell and measure the resulting current as a function of that potential. We call the resulting plot of current versus applied potential a , and it is the electrochemical equivalent of a spectrum in spectroscopy, providing quantitative and qualitative information about the species involved in the oxidation or reduction reaction [Maloy, J. T. , , 285–289]. The earliest voltammetric technique is polarography, developed by Jaroslav Heyrovsky in the early 1920s—an achievement for which he was awarded the Nobel Prize in Chemistry in 1959. Since then, many different forms of voltammetry have been developed, a few of which are highlighted in . Before examining these techniques and their applications in more detail, we must first consider the basic experimental design for voltammetry and the factors influencing the shape of the resulting voltammogram. For an on-line introduction to much of the material in this section, see by Richard S. Kelly, a resource that is part of the Analytical Sciences Digital Library. Although early voltammetric methods used only two electrodes, a modern voltammeter makes use of a three-electrode potentiostat, such as that shown in In voltammetry we apply a time-dependent potential excitation signal to the working electrode—changing its potential relative to the fixed potential of the reference electrode—and measure the current that flows between the working electrode and the auxiliary electrode. The auxiliary electrode generally is a platinum wire and the reference electrode usually is a SCE or a Ag/AgCl electrode. shows an example of a manual three-electrode potentiostat. Although a modern potentiostat uses very different circuitry, you can use and the accompanying discussion to understand how we can control the potential of working electrode and measure the resulting current. For the working electrode we can choose among several different materials, including mercury, platinum, gold, silver, and carbon. The earliest voltammetric techniques used a mercury working electrode. Because mercury is a liquid, the working electrode usual is a drop suspended from the end of a capillary tube. In the , or HMDE, we extrude the drop of Hg by rotating a micrometer screw that pushes the mercury from a reservoir through a narrow capillary tube (Figure 11.4.1
a). In the , or DME, mercury drops form at the end of the capillary tube as a result of gravity (Figure 11.4.1
b). Unlike the HMDE, the mercury drop of a DME grows continuously—as mercury flows from the reservoir under the influence of gravity—and has a finite lifetime of several seconds. At the end of its lifetime the mercury drop is dislodged, either manually or on its own, and is replaced by a new drop. The , or SMDE, uses a solenoid driven plunger to control the flow of mercury (Figure 11.4.1
c). Activation of the solenoid momentarily lifts the plunger, allowing mercury to flow through the capillary, forming a single, hanging Hg drop. Repeated activation of the solenoid produces a series of Hg drops. In this way the SMDE may be used as either a HMDE or a DME. There is one additional type of mercury electrode: the . A solid electrode—typically carbon, platinum, or gold—is placed in a solution of Hg and held at a potential where the reduction of Hg to Hg is favorable, depositing a thin film of mercury on the solid electrode’s surface. Mercury has several advantages as a working electrode. Perhaps its most important advantage is its high overpotential for the reduction of H O to H , which makes accessible potentials as negative as –1 V versus the SCE in acidic solutions and –2 V versus the SCE in basic solutions (Figure 11.4.2
). A species such as Zn , which is difficult to reduce at other electrodes without simultaneously reducing H O , is easy to reduce at a mercury working electrode. Other advantages include the ability of metals to dissolve in mercury—which results in the formation of an —and the ability to renew the surface of the electrode by extruding a new drop. One limitation to mercury as a working electrode is the ease with which it is oxidized. Depending on the solvent, a mercury electrode can not be used at potentials more positive than approximately –0.3 V to +0.4 V versus the SCE. Solid electrodes constructed using platinum, gold, silver, or carbon may be used over a range of potentials, including potentials that are negative and positive with respect to the SCE (Figure 11.4.2
). For example, the potential window for a Pt electrode extends from approximately +1.2 V to –0.2 V versus the SCE in acidic solutions, and from +0.7 V to –1 V versus the SCE in basic solutions. A solid electrode can replace a mercury electrode for many voltammetric analyses that require negative potentials, and is the electrode of choice at more positive potentials. Except for the carbon paste electrode, a solid electrode is fashioned into a disk and sealed into the end of an inert support with an electrical lead (Figure 11.4.3
). The carbon paste electrode is made by filling the cavity at the end of the inert support with a paste that consists of carbon particles and a viscous oil. Solid electrodes are not without problems, the most important of which is the ease with which the electrode’s surface is altered by the adsorption of a solution species or by the formation of an oxide layer. For this reason a solid electrode needs frequent reconditioning, either by applying an appropriate potential or by polishing. A typical arrangement for a voltammetric electrochemical cell is shown in Figure 11.4.4
. In addition to the working electrode, the reference electrode, and the auxiliary electrode, the cell also includes a N -purge line for removing dissolved O , and an optional stir bar. Electrochemical cells are available in a variety of sizes, allowing the analysis of solution volumes ranging from more than 100 mL to as small as 50 μL. When we oxidize an analyte at the working electrode, the resulting electrons pass through the potentiostat to the auxiliary electrode, reducing the solvent or some other component of the solution matrix. If we reduce the analyte at the working electrode, the current flows from the auxiliary electrode to the cathode. In either case, the current from the redox reactions at the working electrode and the auxiliary electrodes is called a . In this section we consider the factors affecting the magnitude of the faradaic current, as well as the sources of any non-faradaic currents. Because the reaction of interest occurs at the working electrode, we describe the faradaic current using this reaction. A faradaic current due to the analyte’s reduction is a , and its sign is positive. An results from the analyte’s oxidation at the working electrode, and its sign is negative. As an example, let’s consider the faradaic current when we reduce \(\text{Fe(CN)}_6^{3-}\) to \(\text{Fe(CN)}_6^{4-}\) at the working electrode. The relationship between the concentrations of \(\text{Fe(CN)}_6^{3-}\), the concentration of \(\text{Fe(CN)}_6^{4-}\), and the potential is given by the Nernst equation \[E=+0.356 \text{ V}-0.05916 \log \frac{\left[\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\right]_{x=0}}{\left[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\right]_{x=0}} \nonumber\] where +0.356V is the standard-statepotential for the \(\text{Fe(CN)}_6^{3-}\)/\(\text{Fe(CN)}_6^{4-}\) redox couple, and = 0 indicates that the concentrations of \(\text{Fe(CN)}_6^{3-}\)- and \(\text{Fe(CN)}_6^{4-}\) are those at the surface of the working electrode. We use surface concentrations instead of bulk concentrations because the equilibrium position for the redox reaction \[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q)+e^{-}\rightleftharpoons\mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q) \nonumber\] is established at the electrode’s surface. Let’s assume we have a solution for which the initial concentration of \(\text{Fe(CN)}_6^{3-}\) is 1.0 mM and that \(\text{Fe(CN)}_6^{4-}\) is absent. Figure 11.4.5
shows the ladder diagram for this solution. If we apply a potential of +0.530 V to the working electrode, the concentrations of \(\text{Fe(CN)}_6^{3-}\) and \(\text{Fe(CN)}_6^{4-}\) at the surface of the electrode are unaffected, and no faradaic current is observed. If we switch the potential to +0.356 V some of the \(\text{Fe(CN)}_6^{3-}\) at the electrode’s surface is reduced to \(\text{Fe(CN)}_6^{4-}\)until we reach a condition where \[\left[\mathrm{Fe}(\mathrm{CN})_{6}^{3-}\right]_{x=0}=\left[\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\right]_{x=0}=0.50 \text{ mM} \nonumber\] This is the first of the five important principles of electrochemistry outlined in : the electrode’s potential determines the analyte’s form at the electrode’s surface. If this is all that happens after we apply the potential, then there would be a brief surge of faradaic current that quickly returns to zero, which is not the most interesting of results. Although the concentrations of \(\text{Fe(CN)}_6^{3-}\) and \(\text{Fe(CN)}_6^{4-}\) at the electrode surface are 0.50 mM, their concentrations in bulk solution remains unchanged. This is the second of the five important principles of electrochemistry outlined in : the analyte’s concentration at the electrode may not be the same as its concentration in bulk solution. Because of this difference in concentration, there is a concentration gradient between the solution at the electrode’s surface and the bulk solution. This concentration gradient creates a driving force that transports \(\text{Fe(CN)}_6^{4-}\) away from the electrode and that transports \(\text{Fe(CN)}_6^{3-}\) to the electrode (Figure 11.4.6
). As the \(\text{Fe(CN)}_6^{3-}\) arrives at the electrode it, too, is reduced to \(\text{Fe(CN)}_6^{4-}\). A faradaic current continues to flow until there is no difference between the concentrations of \(\text{Fe(CN)}_6^{3-}\) and \(\text{Fe(CN)}_6^{4-}\) at the electrode and their concentrations in bulk solution. Although the potential at the working electrode determines if a faradaic current flows, the magnitude of the current is determined by the rate of the resulting oxidation or reduction reaction. Two factors contribute to the rate of the electrochemical reaction: the rate at which the reactants and products are transported to and from the electrode—what we call —and the rate at which electrons pass between the electrode and the reactants and products in solution. This is the fourth of the five important principles of electrochemistry outlined in : current is a measure of rate. There are three modes of mass transport that affect the rate at which reactants and products move toward or away from the electrode surface: diffusion, migration, and convection. occurs whenever the concentration of an ion or a molecule at the surface of the electrode is different from that in bulk solution. If we apply a potential sufficient to completely reduce \(\text{Fe(CN)}_6^{3-}\) at the electrode surface, the result is a concentration gradient similar to that shown in Figure 11.4.7
. The region of solution over which diffusion occurs is the . In the absence of other modes of mass transport, the width of the diffusion layer, \(\delta\), increases with time as the \(\text{Fe(CN)}_6^{3-}\) must diffuse from an increasingly greater distance. occurs when we mix the solution, which carries reactants toward the electrode and removes products from the electrode. The most common form of convection is stirring the solution with a stir bar; other methods include rotating the electrode and incorporating the electrode into a flow-cell. The final mode of mass transport is , which occurs when a charged particle in solution is attracted to or repelled from an electrode that carries a surface charge. If the electrode carries a positive charge, for example, an anion will move toward the electrode and a cation will move toward the bulk solution. Unlike diffusion and convection, migration affects only the mass transport of charged particles. The movement of material to and from the electrode surface is a complex function of all three modes of mass transport. In the limit where diffusion is the only significant form of mass transport, the current in a voltammetric cell is equal to \[i=\frac{n F A D\left(C_{\text {bulk }}-C_{x=0}\right)}{\delta} \label{11.1}\] where the number of electrons in the redox reaction, is Faraday’s constant, is the area of the electrode, is the diffusion coefficient for the species reacting at the electrode, and are its concentrations in bulk solution and at the electrode surface, and \(\delta\) is the thickness of the diffusion layer. For Equation \ref{11.1} to be valid, convection and migration must not interfere with the formation of a diffusion layer. We can eliminate migration by adding a high concentration of an inert supporting electrolyte. Because ions of similar charge equally are attracted to or repelled from the surface of the electrode, each has an equal probability of undergoing migration. A large excess of an inert electrolyte ensures that few reactants or products experience migration. Although it is easy to eliminate convection by not stirring the solution, there are experimental designs where we cannot avoid convection, either because we must stir the solution or because we are using an electrochemical flow cell. Fortunately, as shown in Figure 11.4.8
, the dynamics of a fluid moving past an electrode results in a small diffusion layer—typically 1–10 μm in thickness—in which the rate of mass transport by convection drops to zero. The rate of mass transport is one factor that influences the current in voltammetry. The ease with which electrons move between the electrode and the species that reacts at the electrode also affects the current. When electron transfer kinetics are fast, the redox reaction is at equilibrium. Under these conditions the redox reaction is and the Nernst equation applies. If the electron transfer kinetics are sufficiently slow, the concentration of reactants and products at the electrode surface—and thus the magnitude of the faradaic current—are not what is predicted by the Nernst equation. In this case the system is . In addition to the faradaic current from a redox reaction, the current in an electrochemical cell includes other, nonfaradaic sources. Suppose the charge on an electrode is zero and we suddenly change its potential so that the electrode’s surface acquires a positive charge. Cations near the electrode’s surface will respond to this positive charge by migrating away from the electrode; anions, on the other hand, will migrate toward the electrode. This migration of ions occurs until the electrode’s positive surface charge and the negative charge of the solution near the electrode are equal. Because the movement of ions and the movement of electrons are indistinguishable, the result is a small, short-lived that we call the . Every time we change the electrode’s potential, a transient charging current flows. The migration of ions in response to the electrode’s surface charge leads to the formation of a structured electrode-solution interface that we call the , or EDL. When we change an electrode’s potential, the charging current is the result of a restructuring of the EDL. The exact structure of the electrical double layer is not important in the context of this text, but you can consult this chapter’s additional resources for additional information. Even in the absence of analyte, a small, measurable current flows through an electrochemical cell. This has two components: a faradaic current due to the oxidation or reduction of trace impurities and a nonfaradaic charging current. Methods for discriminating between the analyte’s faradaic current and the residual current are discussed later in this chapter. The shape of a voltammogram is determined by several experimental factors, the most important of which are how we measure the current and whether convection is included as a means of mass transport. As shown in Figure 11.4.9
, despite an abundance of different voltammetric techniques, several of which are discussed in this chapter, there are only three common shapes for voltammograms. For the voltammogram in Figure 11.4.9
a, the current increases from a background residual current to a , . Because the faradaic current is inversely proportional to \(\delta\) (Equation \ref{11.1}), a limiting current occurs only if the thickness of the diffusion layer remains constant because we are stirring the solution (see ). In the absence of convection the diffusion layer increases with time (see ). As shown in Figure 11.4.9
b, the resulting voltammogram has a instead of a limiting current. For the voltammograms in Figure 11.4.9
a and Figure 11.4.9
b, we measure the current as a function of the applied potential. We also can monitor the change in current, \(\Delta i\), following a change in potential. The resulting voltammogram, shown in Figure 11.4.9
c, also has a peak current. Earlier we described a voltammogram as the electrochemical equivalent of a spectrum in spectroscopy. In this section we consider how we can extract quantitative and qualitative information from a voltammogram. For simplicity we will limit our treatment to voltammograms similar to Figure 11.4.9
a. Let’s assume that the redox reaction at the working electrode is \[O+n e^{-} \rightleftharpoons R \label{11.2}\] where is the analyte’s oxidized form and is its reduced form. Let’s also assume that only initially is present in bulk solution and that we are stirring the solution. When we apply a potential that results in the reduction of to , the current depends on the rate at which diffuses through the fixed diffusion layer shown in . Using Equation \ref{11.1}, the current, , is \[i=K_{O}\left([O]_{\text {bulk }}-[O]_{x=0}\right) \label{11.3}\] where is a constant equal to \(n F A D_O / \delta\). When we reach the limiting current, , the concentration of at the electrode surface is zero and Equation \ref{11.3} simplifies to \[i_{l}=K_{O}[O]_{\mathrm{bulk}} \label{11.4}\] Equation \ref{11.4} shows us that the limiting current is a linear function of the concentration of in bulk solution. To determine the value of we can use any of the standardization methods covered in . Equations similar to Equation \ref{11.4} can be developed for the other two types of voltammograms shown in . To extract the standard-state potential from a voltammogram, we need to rewrite the Nernst equation for reaction \ref{11.2} \[E=E_{O / R}^{\circ}-\frac{0.05916}{n} \log \frac{[R]_{x=0}}{[O]_{x=0}} \label{11.5}\] in terms of current instead of the concentrations of and . We will do this in several steps. First, we substitute Equation \ref{11.4} into Equation \ref{11.3} and rearrange to give \[[O]_{x=0}=\frac{i_{l}-i}{K_{O}} \label{11.6}\] Next, we derive a similar equation for [ ] , by noting that \[i=K_{R}\left([R]_{x=0}-[R]_{\mathrm{bulk}}\right) \nonumber\] Because the concentration of [ ] is zero—remember our assumption that the initial solution contains only —we can simplify this equation \[i=K_{R}[R]_{x=0} \nonumber\] and solve for [ ] . \[[R]_{x=0}=\frac{i}{K_{R}} \label{11.7}\] Now we are ready to finish our derivation. Substituting Equation \ref{11.7} and Equation \ref{11.6} into Equation \ref{11.5} and rearranging leaves us with \[E=E_{O / R}^{\circ}-\frac{0.05916}{n} \log \frac{K_{O}}{K_{R}}-\frac{0.05916}{n} \log \frac{i}{i_{l} - i} \label{11.8}\] When the current, , is half of the limiting current, , \[i=0.5 \times i_{l} \nonumber\] we can simplify Equation \ref{11.8} to \[E_{1 / 2}=E_{O / R}^{\circ}-\frac{0.05916}{n} \log \frac{K_{O}}{K_{R}} \label{11.9}\] where is the half-wave potential (Figure 11.4.10
). If is approximately equal to , which often is the case, then the half-wave potential is equal to the standard-state potential. Note that Equation \ref{11.9} is valid only if the redox reaction is electrochemically reversible. In voltammetry there are three important experimental parameters under our control: how we change the potential applied to the working electrode, when we choose to measure the current, and whether we choose to stir the solution. Not surprisingly, there are many different voltammetric techniques. In this section we consider several important examples. The first important voltammetric technique to be developed— —uses the dropping mercury electrode shown in as the working electrode. As shown in Figure 11.4.11
, the current is measured while applying a linear potential ramp. Although polarography takes place in an unstirred solution, we obtain a limiting current instead of a peak current. When a Hg drop separates from the glass capillary and falls to the bottom of the electrochemical cell, it mixes the solution. Each new Hg drop, therefore, grows into a solution whose composition is identical to the bulk solution. The oscillations in the current are a result of the Hg drop’s growth, which leads to a time-dependent change in the area of the working electrode. The limiting current—which also is called the diffusion current—is measured using either the maximum current, , or from the average current, . The relationship between the analyte’s concentration, , and the limiting current is given by the Ilkovic equations \[i_{\max }=706 n D^{1 / 2} m^{2 / 3} t^{1 / 6} C_{A}=K_{\max } C_{A} \nonumber\] \[i_{avg}=607 n D^{1 / 2} m^{2 / 3} t^{1 / 6} C_{A}=K_{\mathrm{avg}} C_{A} \nonumber\] where is the number of electrons in the redox reaction, is the analyte’s diffusion coefficient, is the flow rate of Hg, is the drop’s lifetime and and are constants. The half-wave potential, , provides qualitative information about the redox reaction. Normal polarography has been replaced by various forms of , several examples of which are shown in Figure 11.4.12
[Osteryoung, J. , , 296–298]. Normal pulse polarography (Figure 11.4.12
a), for example, uses a series of potential pulses characterized by a cycle of time \(\tau\), a pulse-time of , a pulse potential of \(\Delta E_\text{p}\), and a change in potential per cycle of \(\Delta E_\text{s}\). Typical experimental conditions for normal pulse polarography are \(\tau \approx 1 \text{ s}\), ≈ 50 ms, and \(\Delta E_\text{s} \approx 2 \text{ mV}\). The initial value of \(\Delta E_\text{p} \approx 2 \text{ mV}\), and it increases by ≈ 2 mV with each pulse. The current is sampled at the end of each potential pulse for approximately 17 ms before returning the potential to its initial value. The shape of the resulting voltammogram is similar to Figure 11.4.11
, but without the current oscillations. Because we apply the potential for only a small portion of the drop’s lifetime, there is less time for the analyte to undergo oxidation or reduction and a smaller diffusion layer. As a result, the faradaic current in normal pulse polarography is greater than in the polarography, resulting in better sensitivity and smaller detection limits. In differential pulse polarography (Figure 11.4.12
b) the current is measured twice per cycle: for approximately 17 ms before applying the pulse and for approximately 17 ms at the end of the cycle. The difference in the two currents gives rise to the peak-shaped voltammogram. Typical experimental conditions for differential pulse polarography are \(\tau \approx 1 \text{ s}\), ≈ 50 ms, \(\Delta E_\text{p}\) ≈ 50 mV, and \(\Delta E_\text{s}\) ≈ 2 mV. The voltammogram for differential pulse polarography is approximately the first derivative of the voltammogram for normal pulse polarography. To see why this is the case, note that the change in current over a fixed change in potential, \(\Delta i / \Delta E\), approximates the slope of the voltammogram for normal pulse polarography. You may recall that the first derivative of a function returns the slope of the function at each point. The first derivative of a sigmoidal function is a peak-shaped function. Other forms of pulse polarography include staircase polarography (Figure 11.4.12
c) and square-wave polarography (Figure 11.4.12
d). One advantage of square-wave polarography is that we can make \(\tau\) very small—perhaps as small as 5 ms, compared to 1 s for other forms of pulse polarography—which significantly decreases analysis time. For example, suppose we need to scan a potential range of 400 mV. If we use normal pulse polarography with a \(\Delta E_\text{s}\) of 2 mV/cycle and a \(\tau\) of 1 s/cycle, then we need 200 s to complete the scan. If we use square-wave polarography with a \(\Delta E_\text{s}\) of 2 mV/cycle and a \(\tau\) of 5 ms/cycle, we can complete the scan in 1 s. At this rate, we can acquire a complete voltammogram using a single drop of Hg! Polarography is used extensively for the analysis of metal ions and inorganic anions, such as \(\text{IO}_3^-\) and \(\text{NO}_3^-\). We also can use polarography to study organic compounds with easily reducible or oxidizable functional groups, such as carbonyls, carboxylic acids, and carbon-carbon double bonds. In polarography we obtain a limiting current because each drop of mercury mixes the solution as it falls to the bottom of the electrochemical cell. If we replace the DME with a solid electrode (see ), we can still obtain a limiting current if we mechanically stir the solution during the analysis, using either a stir bar or by rotating the electrode. We call this approach . Hydrodynamic voltammetry uses the same potential profiles as in polarography, such as a linear scan ( ) or a differential pulse ( ). The resulting voltammograms are identical to those for polarography, except for the lack of current oscillations from the growth of the mercury drops. Because hydrodynamic voltammetry is not limited to Hg electrodes, it is useful for analytes that undergo oxidation or reduction at more positive potentials. Another important voltammetric technique is , which consists of three related techniques: anodic stripping voltammetry, cathodic stripping voltammetry, and adsorptive stripping voltammetry. Because anodic stripping voltammetry is the more widely used of these techniques, we will consider it in greatest detail. Anodic stripping voltammetry consists of two steps (Figure 11.4.13
). The first step is a controlled potential electrolysis in which we hold the working electrode—usually a hanging mercury drop or a mercury film electrode—at a cathodic potential sufficient to deposit the metal ion on the electrode. For example, when analyzing Cu the deposition reaction is \[\mathrm{Cu}^{2+}+2 e^{-} \rightleftharpoons \mathrm{Cu}(\mathrm{Hg}) \nonumber\] where Cu(Hg) indicates that the copper is amalgamated with the mercury. This step serves as a means of concentrating the analyte by transferring it from the larger volume of the solution to the smaller volume of the electrode. During most of the electrolysis we stir the solution to increase the rate of deposition. Near the end of the deposition time we stop the stirring—eliminating convection as a mode of mass transport—and allow the solution to become quiescent. Typical deposition times of 1–30 min are common, with analytes at lower concentrations requiring longer times. In the second step, we scan the potential anodically—that is, toward a more positive potential. When the working electrode’s potential is sufficiently positive, the analyte is stripped from the electrode, returning to solution in its oxidized form. \[\mathrm{Cu}(\mathrm{Hg})\rightleftharpoons \text{ Cu}^{2+}+2 e^{-} \nonumber\] Monitoring the current during the stripping step gives a peak-shaped voltammogram, as shown in Figure 11.4.13
. The peak current is proportional to the analyte’s concentration in the solution. Because we are concentrating the analyte in the electrode, detection limits are much smaller than other electrochemical techniques. An improvement of three orders of magnitude—the equivalent of parts per billion instead of parts per million—is routine. Anodic stripping voltammetry is very sensitive to experimental conditions, which we must carefully control to obtain results that are accurate and precise. Key variables include the area of the mercury film or the size of the hanging Hg drop, the deposition time, the rest time, the rate of stirring, and the scan rate during the stripping step. Anodic stripping voltammetry is particularly useful for metals that form amalgams with mercury, several examples of which are listed in Table 11.4.1
. Compiled from Peterson, W. M.; Wong, R. V. November 1981, 116–128; Wang, J. May 1985, 41–50. The experimental design for cathodic stripping voltammetry is similar to anodic stripping voltammetry with two exceptions. First, the deposition step involves the oxidation of the Hg electrode to \(\text{Hg}_2^{2+}\), which then reacts with the analyte to form an insoluble film at the surface of the electrode. For example, when Cl is the analyte the deposition step is \[2 \mathrm{Hg}(l)+2 \mathrm{Cl}^{-}(a q) \rightleftharpoons \text{ Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-} \nonumber\] Second, stripping is accomplished by scanning cathodically toward a more negative potential, reducing \(\text{Hg}_2^{2+}\) back to Hg and returning the analyte to solution. \[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)+2 e^{-}\rightleftharpoons 2 \mathrm{Hg}( l)+2 \mathrm{Cl}^{-}(a q) \nonumber\] Table 11.4.1
lists several analytes analyzed successfully by cathodic stripping voltammetry. In adsorptive stripping voltammetry, the deposition step occurs without electrolysis. Instead, the analyte adsorbs to the electrode’s surface. During deposition we maintain the electrode at a potential that enhances adsorption. For example, we can adsorb a neutral molecule on a Hg drop if we apply a potential of –0.4 V versus the SCE, a potential where the surface charge of mercury is approximately zero. When deposition is complete, we scan the potential in an anodic or a cathodic direction, depending on whether we are oxidizing or reducing the analyte. Examples of compounds that have been analyzed by absorptive stripping voltammetry also are listed in Table 11.4.1
. In the voltammetric techniques consider to this point we scan the potential in one direction, either to more positive potentials or to more negative potentials. In we complete a scan in both directions. Figure 11.4.14
a shows a typical potential-excitation signal. In this example, we first scan the potential to more positive values, resulting in the following oxidation reaction for the species . \[R \rightleftharpoons O+n e^{-} \nonumber\] When the potential reaches a predetermined switching potential, we reverse the direction of the scan toward more negative potentials. Because we generated the species on the forward scan, during the reverse scan it reduces back to . \[O+n e^{-} \rightleftharpoons R \nonumber\] Cyclic voltammetry is carried out in an unstirred solution, which, as shown in Figure 11.4.14
b, results in peak currents instead of limiting currents. The voltammogram has separate peaks for the oxidation reaction and for the reduction reaction, each characterized by a peak potential and a peak current. The peak current in cyclic voltammetry is given by the Randles-Sevcik equation \[i_{p}=\left(2.69 \times 10^{5}\right) n^{3 / 2} A D^{1 / 2} \nu^{1 / 2} C_{A} \nonumber\] where is the number of electrons in the redox reaction, is the area of the working electrode, is the diffusion coefficient for the electroactive species, \(\nu\) is the scan rate, and is the concentration of the electroactive species at the electrode. For a well-behaved system, the anodic and the cathodic peak currents are equal, and the ratio / is 1.00. The half-wave potential, , is midway between the anodic and cathodic peak potentials. \[E_{1 / 2}=\frac{E_{p, a}+E_{p, c}}{2} \nonumber\] Scanning the potential in both directions provides an opportunity to explore the electrochemical behavior of species generated at the electrode. This is a distinct advantage of cyclic voltammetry over other voltammetric techniques. Figure 11.4.15
shows the cyclic voltammogram for the same redox couple at both a faster and a slower scan rate. At the faster scan rate, 11.4.15
a, we see two peaks. At the slower scan rate in Figure 11.4.15
b, however, the peak on the reverse scan disappears. One explanation for this is that the products from the reduction of on the forward scan have sufficient time to participate in a chemical reaction whose products are not electroactive. The final voltammetric technique we will consider is , in which we apply a constant potential to the working electrode and measure current as a function of time. Because we do not vary the potential, amperometry does not result in a voltammogram. One important application of amperometry is in the construction of chemical sensors. One of the first amperometric sensors was developed in 1956 by L. C. Clark to measure dissolved O in blood. Figure 11.4.16
shows the sensor’s design, which is similar to a potentiometric membrane electrode. A thin, gas-permeable membrane is stretched across the end of the sensor and is separated from the working electrode and the counter electrode by a thin solution of KCl. The working electrode is a Pt disk cathode, and a Ag ring anode serves as the counter electrode. Although several gases can diffuse across the membrane, including O , N , and CO , only oxygen undergoes reduction at the cathode \[\mathrm{O}_{2}(g)+4 \mathrm{H}_{3} \mathrm{O}^{+}(a q)+4 e^{-}\rightleftharpoons 6 \mathrm{H}_{2} \mathrm{O}(l) \nonumber\] with its concentration at the electrode’s surface quickly reaching zero. The concentration of O at the membrane’s inner surface is fixed by its diffusion through the membrane, which creates a diffusion profile similar to that in . The result is a steady-state current that is proportional to the concentration of dissolved oxygen. Because the electrode consumes oxygen,the sample is stirred to prevent the depletion of O at the membrane’s outer surface. The oxidation of the Ag anode is the other half-reaction. \[\mathrm{Ag}(s)+\text{ Cl}^{-}(a q)\rightleftharpoons \mathrm{AgCl}(s)+e^{-} \nonumber\] Another example of an amperometric sensor is a glucose sensor. In this sensor the single membrane in Figure 11.4.16
is replaced with three membranes. The outermost membrane of polycarbonate is permeable to glucose and O . The second membrane contains an immobilized preparation of glucose oxidase that catalyzes the oxidation of glucose to gluconolactone and hydrogen peroxide. \[\beta-\mathrm{D}-\text {glucose }(a q)+\text{ O}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \text {gluconolactone }(a q)+\text{ H}_{2} \mathrm{O}_{2}(a q) \nonumber\] The hydrogen peroxide diffuses through the innermost membrane of cellulose acetate where it undergoes oxidation at a Pt anode. \[\mathrm{H}_{2} \mathrm{O}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \rightleftharpoons \text{ O}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l)+2 e^{-} \nonumber\] Figure 11.4.17
summarizes the reactions that take place in this amperometric sensor. FAD is the oxidized form of flavin adenine nucleotide—the active site of the enzyme glucose oxidase—and FADH is the active site’s reduced form. Note that O serves a mediator, carrying electrons to the electrode. By changing the enzyme and mediator, it is easy to extend to the amperometric sensor in Figure 11.4.17
to the analysis of other analytes. For example, a CO sensor has been developed using an amperometric O sensor with a two-layer membrane, one of which contains an immobilized preparation of autotrophic bacteria [Karube, I.; Nomura, Y.; Arikawa, Y. , , 295–299]. As CO diffuses through the membranes it is converted to O by the bacteria, increasing the concentration of O at the Pt cathode. Voltammetry has been used for the quantitative analysis of a wide variety of samples, including environmental samples, clinical samples, pharmaceutical formulations, steels, gasoline, and oil. The choice of which voltammetric technique to use depends on the sample’s characteristics, including the analyte’s expected concentration and the sample’s location. For example, amperometry is ideally suited for detecting analytes in flow systems, including the analysis of a patient’s blood or as a selective sensor for the rapid analysis of a single analyte. The portability of amperometric sensors, which are similar to potentiometric sensors, also make them ideal for field studies. Although cyclic voltammetry is used to determine an analyte’s concentration, other methods described in this chapter are better suited for quantitative work. Pulse polarography and stripping voltammetry frequently are interchangeable. The choice of which technique to use often depends on the analyte’s concentration and the desired accuracy and precision. Detection limits for normal pulse polarography generally are on the order of 10 M to 10 M, and those for differential pulse polarography, staircase, and square wave polarography are between 10 M and 10 M. Because we concentrate the analyte in stripping voltammetry, the detection limit for many analytes is as little as 10 M to 10 M. On the other hand, the current in stripping voltammetry is much more sensitive than pulse polarography to changes in experimental conditions, which may lead to poorer precision and accuracy. We also can use pulse polarography to analyze a wider range of inorganic and organic analytes because there is no need to first deposit the analyte at the electrode surface. Stripping voltammetry also suffers from occasional interferences when two metals, such as Cu and Zn, combine to form an intermetallic compound in the mercury amalgam. The deposition potential for Zn . is sufficiently negative that any Cu in the sample also deposits into the mercury drop or film, leading to the formation of intermetallic compounds such as CuZn and CuZn . During the stripping step, zinc in the intermetallic compounds strips at potentials near that of copper, decreasing the current for zinc at its usual potential and increasing the apparent current for copper. It is possible to overcome this problem by adding an element that forms a stronger intermetallic compound with the interfering metal. Thus, adding Ga minimizes the interference of Cu when analyzing for Zn by forming an intermetallic compound of Cu and Ga. In any quantitative analysis we must correct the analyte’s signal for signals that arise from other sources. The total current, , in voltammetry consists of two parts: the current from the analyte’s oxidation or reduction, , and a background or residual current, . \[i_{t o t}=i_{A}+i_{r} \nonumber\] The residual current, in turn, has two sources. One source is a faradaic current from the oxidation or reduction of trace interferents in the sample, . The other source is the charging current, , that accompanies a change in the working electrode’s potential. \[i_{r}=i_{\mathrm{int}}+i_{c h} \nonumber\] We can minimize the faradaic current due to impurities by carefully preparing the sample. For example, one important impurity is dissolved O , which undergoes a two-step reduction: first to H O at a potential of –0.1 V versus the SCE, and then to H O at a potential of –0.9 V versus the SCE. Removing dissolved O by bubbling an inert gas such as N through the sample eliminates this interference. After removing the dissolved O , maintaining a blanket of N over the top of the solution prevents O from reentering the solution. The cell in shows a typical N purge line. There are two methods to compensate for the residual current. One method is to measure the total current at potentials where the analyte’s faradaic current is zero and extrapolate it to other potentials. This is the method shown in . One advantage of extrapolating is that we do not need to acquire additional data. An important disadvantage is that an extrapolation assumes that any change in the residual current with potential is predictable, which may not be the case. A second, and more rigorous approach, is to obtain a voltammogram for an appropriate blank. The blank’s residual current is then subtracted from the sample’s total current. The analysis of a sample with a single analyte is straightforward using any of the standardization methods discussed in . The concentration of As(III) in water is determined by differential pulse polarography in 1 M HCl. The initial potential is set to –0.1 V versus the SCE and is scanned toward more negative potentials at a rate of 5 mV/s. Reduction of As(III) to As(0) occurs at a potential of approximately –0.44 V versus the SCE. The peak currents for a set of standard solutions, corrected for the residual current, are shown in the following table. What is the concentration of As(III) in a sample of water if its peak current is 1.37 μA? Linear regression gives the calibration curve shown in Figure 11.4.18
, with an equation of \[i_{p}=0.0176+3.01 \times[\mathrm{As}(\mathrm{III})] \nonumber\] Substituting the sample’s peak current into the regression equation gives the concentration of As(III) as 4.49 μM. The concentration of copper in a sample of sea water is determined by anodic stripping voltammetry using the method of standard additions. The analysis of a 50.0-mL sample gives a peak current of 0.886 μA. After adding a 5.00-μL spike of 10.0 mg/L Cu , the peak current increases to 2.52 μA. Calculate the μg/L copper in the sample of sea water. For anodic stripping voltammetry, the peak current, , is a linear function of the analyte’s concentration \[i_{p}=K \times C_{\mathrm{Cu}} \nonumber\] where is a constant that accounts for experimental parameters such as the electrode’s area, the diffusion coefficient for Cu , the deposition time, and the rate of stirring. For the analysis of the sample before the standard addition we know that the current is \[i_{p}=0.886 \ \mu \mathrm{A}=K \times C_{\mathrm{Cu}} \nonumber\] and after the standard addition the current is \[i_{p}=2.52 \ \mu \mathrm{A}=K\left\{C_{\mathrm{Cu}} \times \frac{50.00 \ \mathrm{mL}}{50.005 \ \mathrm{mL}}+\frac{10.00 \mathrm{mg} \mathrm{Cu}}{\mathrm{L}} \times \frac{0.005 \ \mathrm{mL}}{50.005 \ \mathrm{mL}}\right\} \nonumber\] where 50.005 mL is the total volume after we add the 5.00 μL spike. Solving each equation for and combining leaves us with the following equation. \[\frac{0.886 \ \mu \mathrm{A}}{C_{\mathrm{Cu}}}=K=\frac{2.52 \ \mu \mathrm{A}}{C_{\mathrm{Cu}} \times \frac{50.00 \ \mathrm{mL}}{50.005 \ \mathrm{mL}}+\frac{10.00 \ \mathrm{mg} \text{ Cu}}{\mathrm{L}} \times \frac{0.005 \ \mathrm{mL}}{50.005 \ \mathrm{mL}}} \nonumber\] Solving this equation for gives its value as \(5.42 \times 10^{-4}\) mg Cu /L, or 0.542 μg Cu /L. Voltammetry is a particularly attractive technique for the analysis of samples that contain two or more analytes. Provided that the analytes behave independently, the voltammogram of a multicomponent mixture is a summation of each analyte’s individual voltammograms. As shown in Figure 11.4.19
, if the separation between the half-wave potentials or between the peak potentials is sufficient, we can determine the presence of each analyte as if it is the only analyte in the sample. The minimum separation between the half-wave potentials or peak potentials for two analytes depends on several factors, including the type of electrode and the potential-excitation signal. For normal polarography the separation is at least ±0.2–0.3 V, and differential pulse voltammetry requires a minimum separation of ±0.04–0.05 V. If the voltammograms for two analytes are not sufficiently separated, a simultaneous analysis may be possible. An example of this approach is outlined the following example. The differential pulse polarographic analysis of a mixture of indium and cadmium in 0.1 M HCl is complicated by the overlap of their respective voltammograms [Lanza P. , , 704–705]. The peak potential for indium is at –0.557 V and that for cadmium is at –0.597 V. When a 0.800-ppm indium standard is analyzed, \(\Delta i_p\) (in arbitrary units) is 200.5 at –0.557 V and 87.5 at –0.597 V relative to a saturated Ag/AgCl reference electorde. A standard solution of 0.793 ppm cadmium has a \(\Delta i_p\) of 58.5 at –0.557 V and 128.5 at –0.597 V. What is the concentration of indium and cadmium in a sample if \(\Delta i_p\) is 167.0 at a potential of –0.557 V and 99.5 at a potential of –0.597V. The change in current, \(\Delta i_p\) in differential pulse polarography is a linear function of the analyte’s concentration \[\Delta i_{p}=k_{A} C_{A} \nonumber\] where is a constant that depends on the analyte and the applied potential, and is the analyte’s concentration. To determine the concentrations of indium and cadmium in the sample we must first find the value of for each analyte at each potential. For simplicity we will identify the potential of –0.557 V as , and that for –0.597 V as . The values of are \[\begin{aligned} k_{\mathrm{In}, E_{1}} &=\frac{200.5}{0.800 \ \mathrm{ppm}}=250.6 \ \mathrm{ppm}^{-1} \\ k_{\mathrm{In}, E_{2}} &=\frac{87.5}{0.800 \ \mathrm{ppm}}=109.4 \ \mathrm{ppm}^{-1} \\ k_{\mathrm{Cd} E_{1}} &=\frac{58.5}{0.793 \ \mathrm{ppm}}=73.8 \ \mathrm{ppm}^{-1} \\ k_{\mathrm{Cd} E_{2}} &=\frac{128.5}{0.793 \ \mathrm{ppm}}=162.0 \ \mathrm{ppm}^{-1} \end{aligned} \nonumber\] Next, we write simultaneous equations for the current at the two potentials. \[\begin{array}{l}{\Delta i_{E_{1}}=167.0=250.6 \ \mathrm{ppm}^{-1} \times C_{\mathrm{In}}+73.8 \ \mathrm{ppm}^{-1} \times C_{\mathrm{Cd}}} \\ {\triangle i_{E_{2}}=99.5=109.4 \ \mathrm{ppm}^{-1} \times C_{\mathrm{In}}+162.0 \ \mathrm{ppm}^{-1} \times C_{\mathrm{Cd}}}\end{array} \nonumber\] Solving the simultaneous equations, which is left as an exercise, gives the concentration of indium as 0.606 ppm and the concentration of cadmium as 0.205 ppm. Voltammetry is one of several important analytical techniques for the analysis of trace metals in environmental samples, including groundwater, lakes, rivers and streams, seawater, rain, and snow. Detection limits at the parts-per-billion level are routine for many trace metals using differential pulse polarography, with anodic stripping voltammetry providing parts-per-trillion detection limits for some trace metals. One interesting environmental application of anodic stripping voltammetry is the determination of a trace metal’s chemical form within a water sample. Speciation is important because a trace metal’s bioavailability, toxicity, and ease of transport through the environment often depends on its chemical form. For example, a trace metal that is strongly bound to colloidal particles generally is not toxic because it is not available to aquatic lifeforms. Unfortunately, anodic stripping voltammetry can not distinguish a trace metal’s exact chemical form because closely related species, such as Pb and PbCl , produce a single stripping peak. Instead, trace metals are divided into “operationally defined” categories that have environmental significance. Operationally defined means that an analyte is divided into categories by the specific methods used to isolate it from the sample. There are many examples of operational definitions in the environmental literature. The distribution of trace metals in soils and sediments, for example, often is defined in terms of the reagents used to extract them; thus, you might find an operational definition for Zn in a lake sediment as that extracted using 1.0 M sodium acetate, or that extracted using 1.0 M HCl. Although there are many speciation schemes in the environmental literature, we will consider one proposed by Batley and Florence [see (a) Batley, G. E.; Florence, T. M. , , 379–388; (b) Batley, G. E.; Florence, T. M. , , 151–158; (c) Batley, G. E.; Florence, T. M. , , 1962–1963; (d) Florence, T. M., Batley, G. E.; , , 219–296]. This scheme, which is outlined in Table 11.4.2
, combines anodic stripping voltammetry with ion-exchange and UV irradiation, dividing soluble trace metals into seven groups. In the first step, anodic stripping voltammetry in a pH 4.8 acetic acid buffer differentiates between labile metals and nonlabile metals. Only labile metals—those present as hydrated ions, weakly bound complexes, or weakly adsorbed on colloidal surfaces—deposit at the electrode and give rise to a signal. Total metal concentration are determined by ASV after digesting the sample in 2 M HNO for 5 min, which converts all metals into an ASV-labile form. A Chelex-100 ion-exchange resin further differentiates between strongly bound metals—usually metals bound to inorganic and organic solids, but also those tightly bound to chelating ligands—and more loosely bound metals. Finally, UV radiation differentiates between metals bound to organic phases and inorganic phases. The analysis of seawater samples, for example, suggests that cadmium, copper, and lead are present primarily as labile organic complexes or as labile adsorbates on organic colloids (Group II in Table 11.4.2
). Differential pulse polarography and stripping voltammetry are used to determine trace metals in airborne particulates, incinerator fly ash, rocks, minerals, and sediments. The trace metals, of course, are first brought into solution using a digestion or an extraction. Amperometric sensors also are used to analyze environmental samples. For example, the dissolved O sensor described earlier is used to determine the level of dissolved oxygen and the biochemical oxygen demand, or BOD, of waters and wastewaters. The latter test—which is a measure of the amount of oxygen required by aquatic bacteria as they decompose organic matter—is important when evaluating the efficiency of a wastewater treatment plant and for monitoring organic pollution in natural waters. A high BOD suggests that the water has a high concentration of organic matter. Decomposition of this organic matter may seriously deplete the level of dissolved oxygen in the water, adversely affecting aquatic life. Other amperometric sensors are available to monitor anionic surfactants in water, and CO , H SO , and NH in atmospheric gases. Differential pulse polarography and stripping voltammetry are used to determine the concentration of trace metals in a variety of clinical samples, including blood, urine, and tissue. The determination of lead in blood is of considerable interest due to concerns about lead poisoning. Because the concentration of lead in blood is so small, anodic stripping voltammetry frequently is the more appropriate technique. The analysis is complicated, however, by the presence of proteins that may adsorb to the mercury electrode, inhibiting either the deposition or stripping of lead. In addition, proteins may prevent the electrodeposition of lead through the formation of stable, nonlabile complexes. Digesting and ashing the blood sample mini- mizes this problem. Differential pulse polarography is useful for the routine quantitative analysis of drugs in biological fluids, at concentrations of less than 10 M [Brooks, M. A. “Application of Electrochemistry to Pharmaceutical Analysis,” Chapter 21 in Kissinger, P. T.; Heinemann, W. R., eds. , Marcel Dekker, Inc.: New York, 1984, pp 539–568.]. Amperometric sensors using enzyme catalysts also have many clinical uses, several examples of which are shown in Table 11.4.3
. Cammann, K.; Lemke, U.; Rohen, A.; Sander, J.; Wilken, H.; Winter, B. , , 516–539. In addition to environmental samples and clinical samples, differential pulse polarography and stripping voltammetry are used for the analysis of trace metals in other sample, including food, steels and other alloys, gasoline, gunpowder residues, and pharmaceuticals. Voltammetry is an important technique for the quantitative analysis of organics, particularly in the pharmaceutical industry where it is used to determine the concentration of drugs and vitamins in formulations. For example, voltammetric methods are available for the quantitative analysis of vitamin A, niacinamide, and riboflavin. When the compound of interest is not electroactive, it often can be derivatized to an electroactive form. One example is the differential pulse polarographic determination of sulfanilamide, which is converted into an electroactive azo dye by coupling with sulfamic acid and 1-napthol. The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of chloropromazine in a pharmaceutical product provides an instructive example of a typical procedure. The description here is based on a method from Pungor, E. , CRC Press: Boca Raton, FL, 1995, pp. 34–37. Chlorpromazine, also is known by its trade name Thorazine, is an antipsychotic drug used in the treatment of schizophrenia. The amount of chlorpromazine in a pharmaceutical product is determined voltammetrically at a graphite working electrode in a unstirred solution, with calibration by the method of standard additions. Add 10.00 mL of an electrolyte solution consisting of 0.01 M HCl and 0.1 M KCl to the electrochemical cell. Place a graphite working electrode, a Pt auxiliary electrode, and a SCE reference electrode in the cell, and record the voltammogram from 0.2 V to 2.0 V at a scan rate of 50 mV/s. Weigh out an appropriate amount of the pharmaceutical product and dissolve it in a small amount of the electrolyte. Transfer the solution to a 100-mL volumetric flask and dilute to volume with the electrolyte. Filter a small amount of the diluted solution and transfer 1.00 mL of the filtrate to the voltammetric cell. Mix the contents of the voltammetric cell and allow the solution to sit for 10 s before recording the voltammogram. Return the potential to 0.2 V, add 1.00 mL of a chlorpromazine standard and record the voltammogram. Report the %w/w chlorpromazine in the formulation. 1. Is chlorpromazine undergoing oxidation or reduction at the graphite working electrode? Because we are scanning toward more positive potentials, we are oxidizing chlorpromazine. 2. Why does this procedure use a graphite electrode instead of a Hg electrode? As shown in , the potential window for a Hg electrode extends from approximately –0.3 V to between –1V and –2 V, de- pending on the pH. Because we are scanning the potential from 0.2 V to 2.0 V, we cannot use a Hg electrode. 3. Many voltammetric procedures require that we first remove dissolved O by bubbling N through the solution. Why is this not necessary for this analysis? Dissolved O is a problem when we scan toward more negative potentials, because its reduction may produce a significant cathodic current. In this procedure we are scanning toward more positive potentials and generating anodic currents; thus, dissolved O is not an interferent and does not need to be removed. 4. What is the purpose of recording a voltammogram in the absence of chlorpromazine? This voltammogram serves as a blank, which provides a measurement of the residual current due to the electrolyte. Because the potential window for a graphite working electrode (see ) does not extend to 2.0 V, there is a measurable anodic residual current due to the solvent’s oxidation. Having measured this residual current, we can subtract it from the total current in the presence of chlorpromazine. 5. Based on the description of this procedure, what is the shape of the resulting voltammogram. You may wish to review the three common shapes shown in . Because the solution is unstirred, the voltammogram will have a peak current similar to that shown in . In the previous section we learned how to use voltammetry to determine an analyte’s concentration in a variety of different samples. We also can use voltammetry to characterize an analyte’s properties, including verifying its electrochemical reversibility, determining the number of electrons transferred during its oxidation or reduction, and determining its equilibrium constant in a coupled chemical reaction. Earlier in this chapter we derived a relationship between and the standard-state potential for a redox couple (Equation \ref{11.9}), noting that a redox reaction must be electrochemically reversible. How can we tell if a redox reaction is reversible by looking at its voltammogram? For a reversible redox reaction Equation \ref{11.8}, which we repeat here, describes the relationship between potential and current for a voltammetric experiment with a limiting current. \[E=E_{O / R}^{\circ}-\frac{0.05916}{n} \log \frac{K_{O}}{K_{R}}-\frac{0.05916}{n} \log \frac{i}{i_{l} - i} \nonumber\] If a reaction is electrochemically reversible, a plot of versus log( / – ) is a straight line with a slope of –0.05916/ . In addition, the slope should yield an integer value for . The following data were obtained from a linear scan hydrodynamic voltammogram of a reversible reduction reaction. The limiting current is 5.15 μA. Show that the reduction reaction is reversible, and determine values for and for . Figure 11.4.20
shows a plot of versus log( / – ). Because the result is a straight-line, we know the reaction is electrochemically reversible under the conditions of the experiment. A linear regression analysis gives the equation for the straight line as \[E=-0.391 \mathrm{V}-0.0300 \log \frac{i}{i_{l}-i} \nonumber\] From Equation \ref{11.8}, the slope is equivalent to –0.05916/ ; solving for gives a value of 1.97, or 2 electrons. From Equation \ref{11.8} and Equation \ref{11.9}, we know that is the -intercept for a plot of versus log( / – ); thus, for the data in this example is –0.391 V versus the SCE. We also can use cyclic voltammetry to evaluate electrochemical reversibility by looking at the difference between the peak potentials for the anodic and the cathodic scans. For an electrochemically reversible reaction, the following equation holds true. \[\Delta E_{p}=E_{p, a}-E_{p, c}=\frac{0.05916 \ \mathrm{V}}{n} \nonumber\] As an example, for a two-electron reduction we expect a \(\Delta E_p\) of approximately 29.6 mV. For an electrochemically irreversible reaction the value of \(\Delta E_p\) is larger than expected. Another important application of voltammetry is determining the equilibrium constant for a solution reaction that is coupled to a redox reaction. The presence of the solution reaction affects the ease of electron transfer in the redox reaction, shifting to a more negative or to a more positive potential. Consider, for example, the reduction of to R \[O+n e^{-} \rightleftharpoons R \nonumber\] the voltammogram for which is shown in Figure 11.4.21
. If we introduce a ligand, , that forms a strong complex with , then we also must consider the reaction \[O+p L\rightleftharpoons O L_{p} \nonumber\] In the presence of the ligand, the overall redox reaction is \[O L_{p}+n e^{-} \rightleftharpoons R+p L \nonumber\] Because of its stability, the reduction of the complex is less favorable than the reduction of . As shown in Figure 11.4.21
, the resulting voltammogram shifts to a potential that is more negative than that for . Furthermore, the shift in the voltammogram increases as we increase the ligand’s concentration. We can use this shift in the value of to determine both the stoichiometry and the formation constant for a metal-ligand complex. To derive a relationship between the relevant variables we begin with two equations: the Nernst equation for the reduction of \[E=E_{O / R}^{\circ}-\frac{0.05916}{n} \log \frac{[R]_{x=0}}{[O]_{x=0}} \label{11.10}\] and the stability constant, \(\beta_p\) for the metal-ligand complex at the electrode surface. \[\beta_{p} = \frac{\left[O L_p\right]_{x = 0}}{[O]_{x = 0}[L]_{x = 0}^p} \label{11.11}\] In the absence of ligand the half-wave potential occurs when [ ] and [ ] are equal; thus, from the Nernst equation we have \[\left(E_{1 / 2}\right)_{n c}=E_{O / R}^{\circ} \label{11.12}\] where the subscript “ ” signifies that the complex is not present.
When ligand is present we must account for its effect on the concentration of . Solving Equation \ref{11.1} for [ ] and substituting into the Equation \ref{11.10} gives \[E=E_{O/R}^{\circ}-\frac{0.05916}{n} \log \frac{[R]_{x=0}[L]_{x=0}^{p} \beta_{p}}{\left[O L_{p}\right]_{x=0}} \label{11.13}\] If the formation constant is sufficiently large, such that essentially all is present as the complex , then [ ] and [ ] are equal at the half-wave potential, and Equation \ref{11.13} simplifies to \[\left(E_{1 / 2}\right)_{c} = E_{O/R}^{\circ} - \frac{0.05916}{n} \log{} [L]_{x=0}^{p} \beta_{p} \label{11.14}\] where the subscript “ ” indicates that the complex is present. Defining \(\Delta E_{1/2}\) as \[\triangle E_{1 / 2}=\left(E_{1 / 2}\right)_{c}-\left(E_{1 / 2}\right)_{n c} \label{11.15}\] and substituting Equation \ref{11.12} and Equation \ref{11.14} and expanding the log term leaves us with the following equation. \[\Delta E_{1 / 2}=-\frac{0.05916}{n} \log \beta_{p}-\frac{0.05916 p}{n} \log {[L]} \label{11.16}\] A plot of \(\Delta E_{1/2}\) versus log[ ] is a straight-line, with a slope that is a function of the metal-ligand complex’s stoichiometric coefficient, , and a -intercept that is a function of its formation constant \(\beta_p\). A voltammogram for the two-electron reduction ( = 2) of a metal, , has a half-wave potential of –0.226 V versus the SCE. In the presence of an excess of ligand, , the following half-wave potentials are recorded. Determine the stoichiometry of the metal-ligand complex and its formation constant. We begin by calculating values of \(\Delta E_{1/2}\) using Equation \ref{11.15}, obtaining the values in the following table. Figure 11.4.22
shows the resulting plot of \(\Delta E_{1/2}\) as a function of log[ ]. A linear regression analysis gives the equation for the straight line as \[\triangle E_{1 / 2}=-0.370 \mathrm{V}-0.0601 \log {[L]} \nonumber\] From Equation \ref{11.16} we know that the slope is equal to –0.05916 / . Using the slope and = 2, we solve for obtaining a value of 2.03 ≈ 2. The complex’s stoichiometry, therefore, is . We also know, from Equation \ref{11.16}, that the -intercept is equivalent to –(0.05916/ )log\(\beta_p\). Solving for \(\beta_2\) gives a formation constant of \(3.2 \times 10^{12}\). The voltammogram for 0.50 mM Cd has an of –0.565 V versus an SCE. After making the solution 0.115 M in ethylenediamine, is –0.845 V, and is –0.873 V when the solution is 0.231 M in ethylenediamine. Determine the stoichiometry of the Cd –ethylenediamine complex and its formation constant. The data in this problem comes from Morinaga, K. “Polarographic Studies of Metal Complexes. V. Ethylenediamine Complexes of Cadmium, Nickel, and Zinc,” , , 793–799. For simplicity, we will use as a shorthand notation for ethylenediamine. From the three half-wave potentials we have a \(\Delta E_{1/2}\) of –0.280 V for 0.115 M en and a \(\Delta E_{1/2}\) of –0.308 V for 0.231 M en. Using Equation \ref{11.16} we write the following two equations. \[\begin{array}{l}{-0.280=-\frac{0.05916}{2} \log \beta_{p}-\frac{0.05916 p}{2} \log (0.115)} \\ {-0.308=-\frac{0.05916}{2} \log \beta_{p}-\frac{0.05916 p}{2} \log (0.231)}\end{array} \nonumber\] To solve for the value of , we first subtract the second equation from the first equation \[0.028=-\frac{0.05916 p}{2} \log (0.115)-\left\{-\frac{0.05916 p}{2} \log (0.231)\right\} \nonumber\] which eliminates the term with \(\beta_p\). Next we solve this equation for \[0.028=\left(2.778 \times 10^{-2}\right) \times p-\left(1.882 \times 10^{-2}\right) \times p =\left(8.96 \times 10^{-3}\right) \times p \nonumber\] obtaining a value of 3.1, or ≈ 3. Thus, the complex is Cd(en) . To find the formation complex, \(\beta_3\), we return to Equation \ref{11.16}, using our value for . Using the data for an en concentration of 0.115 M \[\begin{aligned}-0.280=-& \frac{0.05916}{2} \log \beta_{3}-\frac{0.05916 \times 3}{2} \log (0.115) \\ &-0.363=-\frac{0.05916}{2} \log \beta_{3} \end{aligned} \nonumber\] gives a value for \(\beta_3\) of \(1.92 \times 10^{12}\). Using the data for an en concentration of 0.231 M gives a value of \(2.10 \times 10^{12}\). As suggested by , cyclic voltammetry is one of the most powerful electrochemical techniques for exploring the mechanism of coupled electrochemical and chemical reactions. The treatment of this aspect of cyclic voltammetry is beyond the level of this text, although you can consult this chapter’s additional resources for additional information. Detection levels at the parts-per-million level are routine. For some analytes and for some voltammetric techniques, lower detection limits are possible. Detection limits at the parts-per-billion and the part-per-trillion level are possible with stripping voltammetry. Although most analyses are carried out in conventional electrochemical cells using macro samples, the availability of microelectrodes with diameters as small as 2 μm, allows for the analysis of samples with volumes under 50 μL. For example, the concentration of glucose in 200-μm pond snail neurons was monitored successfully using an amperometric glucose electrode with a 2 mm tip [Abe, T.; Lauw, L. L.; Ewing, A. G. , , 7421–7423]. The accuracy of a voltammetric analysis usually is limited by our ability to correct for residual currents, particularly those due to charging. For an analyte at the parts-per-million level, an accuracy of ±1–3% is routine. Accuracy decreases for samples with significantly smaller concentrations of analyte. Precision generally is limited by the uncertainty in measuring the limiting current or the peak current. Under most conditions, a precision of ±1–3% is reasonable. One exception is the analysis of ultratrace analytes in complex matrices by stripping voltammetry, in which the precision may be as poor as ±25%. In many voltammetric experiments, we can improve the sensitivity by adjusting the experimental conditions. For example, in stripping voltammetry we can improve sensitivity by increasing the deposition time, by increasing the rate of the linear potential scan, or by using a differential-pulse technique. One reason that potential pulse techniques are popular is that they provide an improvement in current relative to a linear potential scan. Selectivity in voltammetry is determined by the difference between half-wave potentials or peak potentials, with a minimum difference of ±0.2–0.3 V for a linear potential scan and ±0.04–0.05 V for differential pulse voltammetry. We often can improve selectivity by adjusting solution conditions. The addition of a complexing ligand, for example, can substantially shift the potential where a species is oxidized or reduced to a potential where it no longer interferes with the determination of an analyte. Other solution parameters, such as pH, also can be used to improve selectivity. Commercial instrumentation for voltammetry ranges from <$1000 for simple instruments to >$20,000 for a more sophisticated instrument. In general, less expensive instrumentation is limited to linear potential scans. More expensive instruments provide for more complex potential-excitation signals using potential pulses. Except for stripping voltammetry, which needs a long deposition time, voltammetric analyses are relatively rapid. | 67,534 | 3,142 |
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Acid rain is a term referring to a mixture of wet and dry deposition (deposited material) from the atmosphere (Figure \(\Page {1}\)) containing higher than normal amounts of nitric and sulfuric acids. The precursors, or chemical forerunners, of acid rain formation result from both natural sources, such as volcanoes and decaying vegetation, and man-made sources, primarily emissions of sulfur dioxide (SO ) and nitrogen oxides (NO ) resulting from fossil fuel combustion. Acid rain occurs when these gases react in the atmosphere with water, oxygen, and other chemicals to form various acidic compounds. The result is a mild solution of sulfuric acid and nitric acid. \[SO_2 + HOH \rightarrow H_2SO_3 \label{1} \] \[2 NO_2 + HOH \rightarrow HNO_2 + HNO_3 \label{2} \] When sulfur dioxide and nitrogen oxides are released from power plants and other sources, blow these compounds across state and national borders, sometimes over hundreds of miles. Acid rain is measured using a scale called “pH.” The lower a substance’s pH, the more acidic it is. Pure water has a pH of 7.0. However, normal rain is slightly acidic because carbon dioxide (CO ) dissolves into it forming weak carbonic acid, giving the resulting mixture a pH of approximately 5.6 at typical atmospheric concentrations of CO . As of 2000, the most acidic rain falling in the U.S. has a pH of about 4.3. Acid rain causes of lakes and streams and contributes to the damage of trees at high elevations (for example, red spruce trees above 2,000 feet) and many sensitive forest soils. In addition, acid rain accelerates the decay of building materials and paints, including irreplaceable buildings, statues, and sculptures that are part of our nation’s cultural heritage. Prior to falling to the earth, sulfur dioxide (SO ) and nitrogen oxide (NO ) gases and their particulate matter derivatives—sulfates and nitrates—contribute to visibility degradation and harm public health. The effects of acid rain are most clearly seen in the aquatic, or water, environments, such as streams, lakes, and marshes. Most lakes and streams have a pH between 6 and 8, although some lakes are naturally acidic even without the effects of acid rain. Acid rain primarily affects sensitive bodies of water, which are located in watersheds whose soils have a limited ability to neutralize acidic compounds (called “buffering capacity”). Lakes and streams become acidic (i.e., the pH value goes down) when the water itself and its surrounding soil cannot buffer the acid rain enough to neutralize it. In areas where buffering capacity is low, acid rain releases aluminum from soils into lakes and streams; aluminum is highly toxic to many species of aquatic organisms. Acid rain causes slower growth, injury, or death of forests as shown in Figure \(\Page {2}\). Of course, acid rain is not the only cause of such conditions. Other factors contribute to the overall stress of these areas, including air pollutants, insects, disease, drought, or very cold weather. In most cases, in fact, the impacts of acid rain on trees are due to the combined effects of acid rain and these other environmental stressors. Acid rain and the dry deposition of acidic particles contribute to the corrosion of (such as bronze). The damage that acid rain does to limestone and marble buildings and sculptures is due to a classic acid–base reaction. Marble and limestone both consist of calcium carbonate (CaCO ), a salt derived from the weak acid H CO . The reaction of a strong acid with a salt of a weak acid goes to completion. Thus we can write the reaction of limestone or marble with dilute sulfuric acid as follows: \[CaCO_{3(s)} + H_2SO_{4(aq)} \rightarrow CaSO_{4(s)} + H_2O_{(l)} + CO_{2(g)} \label{3} \] Because CaSO is sparingly soluble in water, the net result of this reaction is to dissolve the marble or limestone. These effects significantly reduce the societal value of buildings, bridges, cultural objects (such as statues, monuments, and tombstones), and cars (Figure \(\Page {3}\)). Sulfates and nitrates that form in the atmosphere from sulfur dioxide (SO ) and nitrogen oxides (NO ) emissions contribute to meaning we cannot see as far or as clearly through the air. The pollutants that cause acid rain—sulfur dioxide (SO ) and nitrogen oxides (NO )—damage . These gases interact in the atmosphere to form fine sulfate and nitrate particles that can be transported long distances by winds and inhaled deep into people’s lungs. Fine particles can also penetrate indoors. Many scientific studies have identified a relationship between elevated levels of fine particles and increased illness and premature death from heart and lung disorders, such as asthma and bronchitis. ( ) | 4,755 | 3,143 |
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This page explains how adding a catalyst affects the rate of a reaction. It assumes familiarity with basic concepts in the collision theory of reaction rates, and with the of molecular energies in a gas. A catalyst is a substance which speeds up a reaction, but is chemically unchanged at its end. When the reaction has finished, the mass of catalyst is the same as at the beginning. Several examples of catalyzed reactions and their respective catalysts are given below: Collisions only result in a reaction if the particles collide with a certain minimum energy called the activation energy for the reaction. The position of activation energy can be determined from a on a Maxwell-Boltzmann distribution: Only those particles represented by the area to the right of the activation energy will react when they collide. The majority do not have enough energy, and will simply bounce apart. To increase the rate of a reaction, the number of successful collisions must be increased. One possible way of doing this is to provide an alternative way for the reaction to happen which has a lower activation energy. In other words, to move the activation energy to the left on the graph: Adding a catalyst has this effect on activation energy. A catalyst provides an alternative route for the reaction with a lower activation energy. This is illustrated on the following energy profile: Care must be taken when discussing how a catalyst operates. A catalyst provides an route for the reaction with a lower activation energy. It does not "lower the activation energy of the reaction". There is a subtle difference between the two statements that is easily illustrated with a simple analogy. Suppose there is a mountain between two valleys such that the only way for people to get from one valley to the other is over the mountain. Only the most active people will manage to get from one valley to the other. Now suppose a tunnel is cut through the mountain. Many more people will now manage to get from one valley to the other by this easier route. It could be said that the tunnel route has a lower activation energy than going over the mountain, but the mountain itself is not lowered. The tunnel has provided an alternative route but has not lowered the original one. The original mountain is still there, and some people still choose to climb it. In chemical terms, if particles collide with enough energy they can still react in exactly the same way as if the catalyst was not there; it is simply that the majority of particles will react via the easier catalyzed route. Jim Clark ( ) | 2,596 | 3,144 |
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For alkylation reactions of enolate anions to be useful, these intermediates must be generated in high concentration in the absence of other strong nucleophiles and bases. The aqueous base conditions used for the aldol condensation are not suitable because the enolate anions of simple carbonyl compounds are formed in very low concentration, and hydroxide or alkoxide bases induce competing S 2 and E2 reactions of alkyl halides. It is necessary, therefore, to achieve complete conversion of aldehyde or ketone reactants to their enolate conjugate bases by treatment with a very strong base (pK > 25) in a non-hydroxylic solvent before any alkyl halides are added to the reaction system. Some bases that have been used for enolate anion formation are: NaH (sodium hydride, pK > 45), NaNH (sodium amide, pK = 34), and LiN[CH(CH ) ] (lithium diisopropylamide, LDA, pK 36). Ether solvents like tetrahydrofuran (THF) are commonly used for enolate anion formation. With the exception of sodium hydride and sodium amide, most of these bases are soluble in THF. Certain other strong bases, such as alkyl lithium and Grignard reagents, cannot be used to make enolate anions because they rapidly and irreversibly add to carbonyl groups. Nevertheless, these very strong bases are useful in making soluble amide bases. In the preparation of lithium diisopropylamide (LDA), for example, the only other product is the gaseous alkane butane. Because of its solubility in THF, LDA is a widely used base for enolate anion formation. In this application, one equivalent of diisopropylamine is produced along with the lithium enolate, but this normally does not interfere with the enolate reactions and is easily removed from the products by washing with aqueous acid. Although the reaction of carbonyl compounds with sodium hydride is heterogeneous and slow, sodium enolates are formed with the loss of hydrogen, and no other organic compounds are produced. Examples If the formed enolate is stabilized by more than one carbonyl it is possible to use a weaker base such as sodium ethoxide. NaOCH CH = Na OCH CH = NaOEt
Because of the acidity of α hydrogens, carbonyls undergo keto-enol tautomerism. Tautomers are rapidly interconverted constitutional isomers, usually distinguished by a different bonding location for a labile hydrogen atom and a differently located double bond. The equilibrium between tautomers is not only rapid under normal conditions, but it often strongly favors one of the isomers (acetone, for example, is 99.999% keto tautomer). Even in such one-sided equilibria, evidence for the presence of the minor tautomer comes from the chemical behavior of the compound. Tautomeric equilibria are catalyzed by traces of acids or bases that are generally present in most chemical samples. However under acidic and basic conditions the equilibrium can be shifted to the right Acid conditions 1) Protonation of the Carbonyl 2) Enol formation Basic conditions 1) Enolate formation 2) Protonation | 3,022 | 3,145 |
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True chocolate contains cocoa butter. The main types of chocolate, in decreasing order of cocoa liquor content, are: Unsweetened (bitter) chocolate Dark chocolate
Milk chocolate
White chocolate Unsweetened Chocolate Unsweetened chocolate, also known as bitter chocolate, baking chocolate, or cooking chocolate, is pure cocoa liquor mixed with some form of fat to produce a solid substance. The pure ground, roasted cocoa beans impart a strong, deep chocolate flavor. With the addition of sugar in recipes, however, it is used as the base for cakes, brownies, confections, and cookies. Dark (Sweet, Semi-Sweet, Bittersweet) Chocolate Dark chocolate has an ideal balance of cocoa liquor, cocoa butter, and sugar. Thus it has the attractive, rich color and flavor so typical of chocolate, and is also sweet enough to be palatable. It does not contain any milk solids. It can be eaten as is or used in baking. Its flavor does not get lost or overwhelmed, as in many cases when milk chocolate is used. It can be used for fillings, for which more flavorful chocolates with high cocoa percentages ranging from 60% to 99% are often used.Dark is synonymous with semi-sweet, and extra dark with bittersweet, although the ratio of cocoa butter to solids may vary. Sweet chocolate has more sugar, sometimes almost equal to cocoa liquor and butter amounts (45% to 55% range).
Semi-sweet chocolate is frequently used for cooking. It is a dark chocolate with less sugar than sweet chocolate. Bittersweet chocolate has less sugar and more liquor than semi-sweet chocolate, but the two are often interchangeable when baking. Bittersweet and semi-sweet chocolates are sometimes referred to as couverture (see below). The higher the percentage of cocoa, the less sweet the chocolate is. Milk Chocolate Milk chocolate is solid chocolate made with milk, added in the form of milk powder. Milk chocolate contains a higher percentage of fat (the milk contributes to this) and the melting point is slightly lower. It is used mainly as a flavoring and in the production of candies and moulded pieces. White Chocolate The main ingredient in white chocolate is sugar, closely followed by cocoa butter and milk powder. It has no cocoa liquor. It is used mainly as a flavoring in desserts, in the production of candies and, in chunk form in cookies. | 2,335 | 3,147 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Elimination_Reactions/E1_Reactions/Carbocation_Rearrangements |
Carbocation rearrangements are extremely common in organic chemistry reactions are are defined as the movement of a carbocation from an unstable state to a more stable state through the use of various structural reorganizational "shifts" within the molecule. Once the carbocation has shifted over to a different carbon, we can say that there is a structural isomer of the initial molecule. However, this phenomenon is not as simple as it sounds. Whenever alcohols are subject to transformation into various carbocations, the carbocations are subject to a phenomenon known as carbocation rearrangement. A carbocation, in brief, holds the positive charge in the molecule that is attached to three other groups and bears a sextet rather than an octet. However, we do see carbocation rearrangements in reactions that do not contain alcohol as well. Those, on the other hand, require more difficult explanations than the two listed below. There are two types of rearrangements: hydride shift and alkyl shift. These rearrangements usualy occur in many types of carbocations. Once rearranged, the molecules can also undergo further or . Though, most of the time we see either a simple or complex mixture of products. We can expect two products before undergoing carbocation rearrangement, but once undergoing this phenomenon, we see the major product. Whenever a nucleophile attacks some molecules, we typically see products. However, in most cases, we normally see both a major product and a minor product. The major product is typically the rearranged product that is (aka more stable). The minor product, in contract, is typically the normal product that is (aka less stable). The reaction: We see that the formed carbocations can undergo rearrangements called . This means that the two electron hydrogen from the unimolecular substitution moves over to the carbon. We see the phenomenon of hydride shift typically with the reaction of an alcohol and hydrogen halides, which include HBr, HCl, and HI. HF is typically not used because of its instability and its fast reactivity rate. Below is an example of a reaction between an alcohol and hydrogen chloride: The alcohol portion (-OH) has been substituted with the nucleophilic Cl atom. However, it is not a direct substitution of the OH atom as seen in S 2 reactions. In this S 1 reaction, we see that the leaving group, -OH, forms a carbocation on Carbon #3 after receiving a proton from the nucleophile to produce an alkyloxonium ion. Before the Cl atom attacks, the hydrogen atom attached to the Carbon atom directly adjacent to the original Carbon (preferably the more stable Carbon), Carbon #2, can undergo hydride shift. The hydrogen and the carbocation formally switch positions. The Cl atom can now attack the carbocation, in which it forms the more stable structure because of hyperconjugation. The carbocation, in this case, is most stable because it attaches to the tertiary carbon (being attached to 3 different carbons). However, we can still see small amounts of the minor, unstable product. The mechanism for hydride shift occurs in that includes various intermediates and transition states. Below is the mechanism for the given reaction above: In a more complex case, when alkenes undergo hydration, we also observe hydride shift. Below is the reaction of 3-methyl-1-butene with H O that furnishes to make 2-methyl-2-butanol: Once again, we see multiple products. In this case, however, we see two minor products and one major product. We observe the major product because the -OH substitutent is attached to the more substituted carbon. When the reactant undergoes hydration, the proton attaches to carbon #2. The carbocation is therefore on carbon #2. Hydride shift now occurs when the hydrogen on the adjacent carbon formally switch places with the carbocation. The carbocation is now ready to be attacked by H O to furnish an alkyloxonium ion because of stability and hyperconjugation. The final step can be observed by another water molecule attacking the proton on the alkyloxonium ion to furnish an alcohol. We see this mechanism below: Not all carbocations have suitable hydrogen atoms (either secondary or tertiary) that are on adjacent carbon atoms available for rearrangement. In this case, the reaction can undergo a different mode of rearrangement known as (or alkyl group migration). Alkyl Shift acts very similarily to that of hydride shift. Instead of the proton (H) that shifts with the nucleophile, we see an alkyl group that shifts with the nucleophile instead. The shifting group carries its electron pair with it to furnish a bond to the neighboring or adjacent carbocation. The shifted alkyl group and the positive charge of the carbocation switch positions on the moleculeReactions of tertiary carbocations react much faster than that of secondary carbocations. We see alkyl shift from a secondary carbocation to tertiary carbocation in S 1 reactions: We observe slight variations and differences between the two reactions. In reaction #1, we see that we have a secondary substrate. This undergoes alkyl shift because it does not have a suitable hydrogen on the adjacent carbon. Once again, the reaction is similar to hydride shift. The only difference is that we shift an alkyl group rather than shift a proton, while still undergoing various intermediate steps to furnish its final product. With reaction #2, on the other hand, we can say that it undergoes a mechanism. In short, this means that everything happens in one step. This is because primary carbocations be an intermediate and they are relatively difficult processes since they require higher temperatures and longer reaction times. After protonating the alcohol substrate to form the alkyloxonium ion, the water must leave as the alkyl group shifts from the adjacent carbon to skip the formation of the unstable primary carbocation. E1 reactions are also affected by alkyl shift. Once again, we can see both minor and major products. However, we see that the more substituted carbons undergo the effects of E1 reactions and furnish a double bond. See practice problem #4 below for an example as the properties and effects of carbocation rearrangements in E1 reactions are similar to that of alkyl shifts. Typically, hydride shifts can occur at low temperatures. However, by heating the solutionf of a cation, it can easily and readily speed the process of rearrangement. One way to account for a slight barrier is to propose a 1,3-hydride shift interchanging the functionality of two different kinds of methyls. Another possibility is 1,2 hydride shift in which you could yield a secondary carbocation intermediate. Then, a further 1,2 hydride shift would give the more stable rearranged tertiary cation. More distant hydride shifts have been observed, such as 1,4 and 1,5 hydride shifts, but these arrangements are too fast to undergo secondary cation intermediates. Carbocation rearrangements happen very readily and often occur in many organic chemistry reactions. Yet, we typically neglect this step. Dr. Sarah Lievens, a Chemistry professor at the University of California, Davis once said carbocation rearrangements can be observed with various analogies to help her students remember this phenomenon. For hydride shifts: "The new friend (nucleophile) just joined a group (the organic molecule). Because he is new, he only made two new friends. However, the popular kid (the hydrogen) glady gave up his friends to the new friend so that he could have even more friends. Therefore, everyone won't be as lonely and we can all be friends." This analogy works for alkyl shifts in conjunction with hydride shift as well. | 7,715 | 3,148 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Chromatography/Liquid_Chromatography |
Liquid chromatography is a technique used to separate a sample into its individual parts. This separation occurs based on the interactions of the sample with the mobile and stationary phases. Because there are many stationary/mobile phase combinations that can be employed when separating a mixture, there are several different types of chromatography that are classified based on the physical states of those phases. Liquid-solid column chromatography, the most popular chromatography technique and the one discussed here, features a liquid mobile phase which slowly filters down through the solid stationary phase, bringing the separated components with it. Components within a mixture are separated in a column based on each component's affinity for the mobile phase. So, if the components are of different polarities and a mobile phase of a distinct polarity is passed through the column, one component will migrate through the column faster than the other. Because molecules of the same compound will generally move in groups, the compounds are separated into distinct bands within the column. If the components being separated are colored, their corresponding bands can be seen. Otherwise as in high performance liquid chromatography (HPLC), the presence of the bands are detected using other instrumental analysis techniques such as UV-VIS spectroscopy . The following figure shows the migration of two components within a mixture: In the first step, the mixture of components sits atop the wet column. As the mobile phase passes through the column, the two components begin to separate into bands. In this example, the red component has a stronger affinity for the mobile phase while the blue component remains relatively fixed in the stationary phase. As each component is eluted from the column, each can be collected separately and analyzed by whatever method is favored. The relative polarities of these two compounds are determined based on the polarities of the stationary and mobile phases. If this experiment were done as normal phase chromatography, the red component would be less polar than the blue component. On the other hand, this result yielded from reverse phase chromatography would show that the red component is more polar than the blue component. The first known chromatography is traditionally attributed to Russian botanist Mikhail Tswett who used columns of calcium carbonate to separate plant compounds during his research of chlorophyll. This happened in the 20 century (1901). Further development of chromatography occurred when the Nobel Prize was awarded to Archer John Porter Martin and Richard Laurence Millington Synge in 1952. They were able to establish the basics of partition chromatography and also develop Plate theory. The stationary phase in column chromatography is most typically a fine adsorbent solid; a solid that is able hold onto gas or liquid particles on its outer surface. The column typically used in column chromatography looks similar to a Pasteur pipette (Pasteur pipettes are used as columns in small scale column chromatography). The narrow exit of the column is first plugged with glass wool or a porous plate in order to support the column packing material and keep it from escaping the tube. Then the adsorbent solid (usually silica) is tightly packed into the glass tube to make the separating column. The packing of the stationary phase into the glass column must be done carefully to create a uniform distribution of material. A uniform distribution of adsorbent is important to minimize the presence of air bubbles and/or channels within the column. To finish preparing the column, the solvent to be used as the mobile phase is passed through the dry column. Then the column is said to be "wetted" and the column must remain wet throughout the entire experiment. Once the column is correctly prepared, the sample to be separated is placed at the top of the wet column. A photo of a packed separating column can be found in the links. Chromatography is effective because different components within a mixture are attracted to the adsorbent surface of the stationary phase with varying degrees depending on each components polarity and its unique structural characteristics, and also its interaction with the mobile phase. The separation that is achieved using column chromatography is based on factors that are associated with the sample. So, a component that is more attracted to the stationary phase will migrate down the separating column at a slower rate than a component that has a higher affinity for the mobile phase. Also, the efficacy of the separation is dependent on the nature of the adsorbent solid used and the polarity of the mobile phase solvent. The type of adsorbent material used as the stationary phase is vital for efficient separation of components in a mixture. Several different solid may be employed. Adsorbent material can be chosen based on particle size and activity of the solid. The activity of the adsorbent is represented by its activity grade, which is a measure of an adsorbent's attraction for solutes in the sample solution. The solids with the highest activity grading are those that are completely anhydrous. Silica gel and alumina are among the most popular adsorbents used. Alumina caters well to samples that that require specific conditions to adequately separate. However, the use of non-neutral stationary phases should be done with great caution, an increase or decrease of pH in the alumina stationary phase may allow chemical reactions within the components of the mixture. Silica gel, however, is less active than alumina and can generally be used as an all-around adsorbent for most components in solution. Silica is also preferred because of its high sample capacity, making it one of the most popular adsorbent materials. The proper mobile phase must also be chosen for the best separation of the components in an unknown mixture. This eluent will be chosen based on its polarity relative to the sample and the stationary phase. With a strong polar adsorbent stationary phase like alumina, a polar solvent used as the mobile phase will be adsorbed by the stationary phase, which may displace molecules of sample in the mixture and may cause the sample components to elute vary quickly. This will provide little separation of the sample, so it is best to start elution with a solvent of lower polarity to elute the components that are weakly adsorbed to the stationary phase first. The solvent may also be changed during separation in order to change the polarity and therefore elute the various components separately in a more timely manner. This method is very similar to the gradient method of separation used in (HPLC). By using the above apparatus, purchasing expensive air pumps can be avoided. This method is useful to an extent. Since the flow rate of the pressurized gas is controlled manually by the flow rate controller, it is more difficult to quantify the flow rate and keep that flow rate constant. Instruments available for flash chromatography are able to set flow rates digitally and keep flow rate constant. Flash chromatography is similar to HPLC in that the mobile phase is moved through the column by applying pressure to the solvent in order to achieve a quicker result. However, in flash chromatography, only medium pressure is applied to the system within the solution. In HPLC, pressures as high as 5000 psi can be applied in the column by high performance pumps. Plate theory and Rate theory are two theories that are applicable to chromatography. Plate theory describes a chromatography system as being in equilibrium between the stationary and mobile phases. This views the column as divided into a number of imaginary theoretical plates. This is significant because as the number of plates in a column increases or the height equivalent theoretical plates or HETP increases, so does the separation of components. It also provides an equation that describes the elution curve or the chromatogram of a solute it can also be used to find the volume and the column efficiency. \[HETP = \dfrac{L}{N} \] where L= column length and N= number of theoretical plates The Rate theory on the other hand describes the migration of molecules in a column. This included band shape, broadening, and the diffusion of a solute. Rate theory follows the , which is the most appropriate for prediction of dispersion in liquid chromatography columns. It does this by taking into account the various pathways that a sample must travel through a column. Using the Van Deemter equation, it is possible to find the optimum velocity and and a minimum plate height. \[ H=A+\dfrac{B}{u} = Cu \] where \(A\) = Eddy-Diffusion, \(B\) = Longitudinal Diffusion, \(C\) = mass transfer, \(u\) This schematic is of the basic instrumentation of a liquid-solid chromatograph. The solvent inlet brings in the mobile phase which is then pumped through the inline solvent filter and passed through the injection valve. This is where the mobile phase will mix with the injected sample. It then gets passed through another filter and then passed through the column where the sample will be separated into its components. The detector detects the separation of the analytes and the recorder, or usually a computer will record this information. The sample then goes through a backpressure filter and into waste. Liquid-solid column chromatography is an effective separation technique when all appropriate parameters and equipment are used. This method is especially effective when the compounds within the mixture are colored, as this gives the scientist the ability to see the separation of the bands for the components in the sample solution. Even if the bands are not visible, certain components can be observed by other visualization methods. One method that may work for some compounds is irradiation with ultraviolet light. This makes it relatively easy to collect samples one after another. However, if the components within the solution are not visible by any of these methods, it can be difficult to determine the efficacy of the separation that was performed. In this case, separate collections from the column are taken at specified time intervals. Since the human eye is the primary detector for this procedure, it is most effective when the bands of the distinct compounds are visible. Liquid-solid column chromatography is also a less expensive procedure than other methods of separation ( , , etc.). This is because the most basic forms of column chromatography do not require the help of expensive machinery like high pressure solvent pumps used in HPLC. In methods besides flash chromatography, the flow of the mobile phase, the detection of each separation band, and the collection of each component, are all done manually by the scientist. Although this introduces many potential instances of experimental error, this method of separation can be very effective when done correctly. Also, the glass wear used for liquid-solid column chromatography is relatively inexpensive and readily available in many laboratories. Burets are commonly used as the separating column, which in many cases will work just as well as an expensive pre-prepared column. For smaller scale chromatography, Pasteur pipettes are often used. Flash chromatography has the potential to be more costly than the previous methods of separation, especially when sophisticated air pumps and vacuum pumps are needed. When these pieces of machinery are not needed, however, a vacuum line can be instead connected to an aspirator on a water faucet. Also, home-made pressurized air flow controllers can be made as shown previously. | 11,739 | 3,149 |
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One factor that influences the absorbance of a sample is the concentration (c). The expectation would be that, as the concentration goes up, more radiation is absorbed and the absorbance goes up. Therefore, the absorbance is directly proportional to the concentration. A second factor is the path length (b). The longer the path length, the more molecules there are in the path of the beam of radiation, therefore the absorbance goes up. Therefore, the path length is directly proportional to the concentration. When the concentration is reported in moles/liter and the path length is reported in centimeters, the third factor is known as the molar absorptivity (\(\varepsilon\)). In some fields of work, it is more common to refer to this as the extinction coefficient. When we use a spectroscopic method to measure the concentration of a sample, we select out a specific wavelength of radiation to shine on the sample. As you likely know from other experiences, a particular chemical species absorbs some wavelengths of radiation and not others. The molar absorptivity is a measure of how well the species absorbs the particular wavelength of radiation that is being shined on it. The process of absorbance of electromagnetic radiation involves the excitation of a species from the ground state to a higher energy excited state. This process is described as an excitation transition, and excitation transitions have probabilities of occurrences. It is appropriate to talk about the degree to which possible energy transitions within a chemical species are allowed. Some transitions are more allowed, or more favorable, than others. Transitions that are highly favorable or highly allowed have high molar absorptivities. Transitions that are only slightly favorable or slightly allowed have low molar absorptivities. The higher the molar absorptivity, the higher the absorbance. Therefore, the molar absorptivity is directly proportional to the absorbance. If we return to the experiment in which a spectrum (recording the absorbance as a function of wavelength) is recorded for a compound for the purpose of identification, the concentration and path length are constant at every wavelength of the spectrum. The only difference is the molar absorptivities at the different wavelengths, so a spectrum represents a plot of the relative molar absorptivity of a species as a function of wavelength. Since the concentration, path length and molar absorptivity are all directly proportional to the absorbance, we can write the following equation, which is known as the Beer-Lambert law (often referred to as Beer’s Law), to show this relationship. \[\mathrm{A = \varepsilon bc} \nonumber \] Note that Beer’s Law is the equation for a straight line with a y-intercept of zero. Measuring the concentration of a species in a sample involves a multistep process. One important consideration is the wavelength of radiation to use for the measurement. Remember that the higher the molar absorptivity, the higher the absorbance. What this also means is that the higher the molar absorptivity, the lower the concentration of species that still gives a measurable absorbance value. Therefore, the wavelength that has the highest molar absorptivity (\(\lambda\) ) is usually selected for the analysis because it will provide the lowest detection limits. If the species you are measuring is one that has been commonly studied, literature reports or standard analysis methods will provide the \(\lambda\) value. If it is a new species with an unknown \(\lambda\) value, then it is easily measured by recording the spectrum of the species. The wavelength that has the highest absorbance in the spectrum is \(\lambda\) . The second step of the process is to generate a standard curve. The standard curve is generated by preparing a series of solutions (usually 3-5) with known concentrations of the species being measured. Every standard curve is generated using a blank. The blank is some appropriate solution that is assumed to have an absorbance value of zero. It is used to zero the spectrophotometer before measuring the absorbance of the standard and unknown solutions. The absorbance of each standard sample at \(\lambda\) is measured and plotted as a function of concentration. The plot of the data should be linear and should go through the origin as shown in the standard curve in Figure \(\Page {2}\). If the plot is not linear or if the y-intercept deviates substantially from the origin, it indicates that the standards were improperly prepared, the samples deviate in some way from Beer’s Law, or that there is an unknown interference in the sample that is complicating the measurements. Assuming a linear standard curve is obtained, the equation that provides the best linear fit to the data is generated. Note that the slope of the line of the standard curve in Figure \(\Page {2}\) is (\(\varepsilon\)b) in the Beer’s Law equation. If the path length is known, the slope of the line can then be used to calculate the molar absorptivity. The third step is to measure the absorbance in the sample with an unknown concentration. The absorbance of the sample is used with the equation for the standard curve to calculate the concentration. The way to think about this question is to consider the expression we wrote earlier for the absorbance. \[\mathrm{A = \log\left(\dfrac{P_o}{P}\right)} \nonumber \] Since stray radiation always leaks in to the detector and presumably is a fixed or constant quantity, we can rewrite the expression for the absorbance including terms for the stray radiation. It is important to recognize that P the power from the radiation source, is considerably larger than \(P_S\). Also, the numerator (P + P ) is a constant at a particular wavelength. \[\mathrm{A = \log\left(\dfrac{P_o + P_s}{P + P_s}\right)} \nonumber \] Now let’s examine what happens to this expression under the two extremes of low concentration and high concentration. At low concentration, not much of the radiation is absorbed and P is not that much different than P . Since \(P_o \gg P_S\), \(P\) will also be much greater than \(P_S\). If the sample is now made a little more concentrated so that a little more of the radiation is absorbed, P is still much greater than P . Under these conditions the amount of stray radiation is a negligible contribution to the measurements of P and P and has a negligible effect on the linearity of Beer’s Law. As the concentration is raised, P, the radiation reaching the detector, becomes smaller. If the concentration is made high enough, much of the incident radiation is absorbed by the sample and P becomes much smaller. If we consider the denominator (P + P ) at increasing concentrations, P gets small and P remains constant. At its limit, the denominator approaches P , a constant. Since P + P is a constant and the denominator approaches a constant (P ), the absorbance approaches a constant. A plot of what would occur is shown in Figure \(\Page {3}\). The ideal plot is the straight line. The curvature that occurs at higher concentrations that is caused by the presence of stray radiation represents a negative deviation from Beer’s Law. The sample molecules are more likely to interact with each other at higher concentrations, thus the assumption used to derive Beer’s Law breaks down at high concentrations. The effect, which we will not explain in any more detail in this document, also leads to a negative deviation from Beer’s Law at high concentration. Spectroscopic instruments typically have a device known as a monochromator. There are two key features of a monochromator. The first is a device to disperse the radiation into distinct wavelengths. You are likely familiar with the dispersion of radiation that occurs when radiation of different wavelengths is passed through a prism. The second is a slit that blocks the wavelengths that you do not want to shine on your sample and only allows \(\lambda\) to pass through to your sample as shown in Figure \(\Page {4}\). An examination of Figure \(\Page {4}\) shows that the slit has to allow some “packet” of wavelengths through to the sample. The packet is centered on \(\lambda\) , but clearly nearby wavelengths of radiation pass through the slit to the sample. The term defines the packet of wavelengths and it depends on the slit width and the ability of the dispersing element to divide the wavelengths. Reducing the width of the slit reduces the packet of wavelengths that make it through to the sample, meaning that smaller slit widths lead to more monochromatic radiation and less deviation from linearity from Beer’s Law. The important thing to consider is the effect that this has on the power of radiation making it through to the sample (P ). Reducing the slit width will lead to a reduction in P and hence P. An electronic measuring device called a detector is used to monitor the magnitude of P and P. All electronic devices have a background noise associated with them (rather analogous to the static noise you may hear on a speaker and to the discussion of stray radiation from earlier that represents a form of noise). P and P represent measurements of signal over the background noise. As P and P become smaller, the background noise becomes a more significant contribution to the overall measurement. Ultimately the background noise restricts the signal that can be measured and detection limit of the spectrophotometer. Therefore, it is desirable to have a large value of P . Since reducing the slit width reduces the value of P , it also reduces the detection limit of the device. Selecting the appropriate slit width for a spectrophotometer is therefore a balance or tradeoff of the desire for high source power and the desire for high monochromaticity of the radiation. It is not possible to get purely monochromatic radiation using a dispersing element with a slit. Usually the sample has a slightly different molar absorptivity for each wavelength of radiation shining on it. The net effect is that the total absorbance added over all the different wavelengths is no longer linear with concentration. Instead a negative deviation occurs at higher concentrations due to the polychromicity of the radiation. Furthermore, the deviation is more pronounced the greater the difference in the molar absorbtivity. Figure \(\Page {5}\) compares the deviation for two wavelengths of radiation with molar absorptivities that are (a) both 1,000, (b) 500 and 1,500, and (c) 250 and 1,750. As the molar absorptivities become further apart, a greater negative deviation is observed. Therefore, it is preferable to perform the absorbance measurement in a region of the spectrum that is relatively broad and flat. The hypothetical spectrum in Figure \(\Page {6}\) shows a species with two wavelengths that have the same molar absorptivity. The peak at approximately 250 nm is quite sharp whereas the one at 330 nm is rather broad. Given such a choice, the broader peak will have less deviation from the polychromaticity of the radiation and is less prone to errors caused by slight misadjustments of the monochromator. It is important to consider the error that occurs at the two extremes (high concentration and low concentration). Our discussion above about deviations to Beer’s Law showed that several problems ensued at higher concentrations of the sample. Also, the point where only 10% of the radiation is transmitted through the sample corresponds to an absorbance value of 1. Because of the logarithmic relationship between absorbance and transmittance, the absorbance values rise rather rapidly over the last 10% of the radiation that is absorbed by the sample. A relatively small change in the transmittance can lead to a rather large change in the absorbance at high concentrations. Because of the substantial negative deviation to Beer’s law and the lack of precision in measuring absorbance values above 1, it is reasonable to assume that the error in the measurement of absorbance would be high at high concentrations. At very low sample concentrations, we observe that P and P are quite similar in magnitude. If we lower the concentration a bit more, P becomes even more similar to P . The important realization is that, at low concentrations, we are measuring a small difference between two large numbers. For example, suppose we wanted to measure the weight of a captain of an oil tanker. One way to do this is to measure the combined weight of the tanker and the captain, then have the captain leave the ship and measure the weight again. The difference between these two large numbers would be the weight of the captain. If we had a scale that was accurate to many, many significant figures, then we could possibly perform the measurement in this way. But you likely realize that this is an impractical way to accurately measure the weight of the captain and most scales do not have sufficient precision for an accurate measurement. Similarly, trying to measure a small difference between two large signals of radiation is prone to error since the difference in the signals might be on the order of the inherent noise in the measurement. Therefore, the degree of error is expected to be high at low concentrations. The discussion above suggests that it is best to measure the absorbance somewhere in the range of 0.1 to 0.8. Solutions of higher and lower concentrations have higher relative error in the measurement. Low absorbance values (high transmittance) correspond to dilute solutions. Often, other than taking steps to concentrate the sample, we are forced to measure samples that have low concentrations and must accept the increased error in the measurement. It is generally undesirable to record absorbance measurements above 1 for samples. Instead, it is better to dilute such samples and record a value that will be more precise with less relative error. Another question that arises is whether it is acceptable to use a non-linear standard curve. As we observed earlier, standard curves of absorbance versus concentration will show a non-linearity at higher concentrations. Such a non-linear plot can usually be fit using a higher order equation and the equation may predict the shape of the curve quite accurately. Whether or not it is acceptable to use the non-linear portion of the curve depends in part on the absorbance value where the non-linearity starts to appear. If the non-linearity occurs at absorbance values higher than one, it is usually better to dilute the sample into the linear portion of the curve because the absorbance value has a high relative error. If the non-linearity occurs at absorbance values lower than one, using a non-linear higher order equation to calculate the concentration of the analyte in the unknown may be acceptable. One thing that should never be done is to extrapolate a standard curve to higher concentrations. Since non-linearity will occur at some point, and there is no way of knowing in advance when it will occur, the absorbance of any unknown sample must be lower than the absorbance of the highest concentration standard used in the preparation of the standard curve. It is also not desirable to extrapolate a standard curve to lower concentrations. There are occasions when non-linear effects occur at low concentrations. If an unknown has an absorbance that is below that of the lowest concentration standard of the standard curve, it is preferable to prepare a lower concentration standard to ensure that the curve is linear over such a concentration region. Another concern that always exists when using spectroscopic measurements for compound quantification or identification is the potential presence of . The matrix is everything else that is in the sample except for the species being analyzed. A concern can occur when the matrix of the unknown sample has components in it that are not in the blank solution and standards. Components of the matrix can have several undesirable effects. One concern is that a component of the matrix may absorb radiation at the same wavelength as the analyte, giving a false positive signal. Particulate matter in a sample will scatter the radiation, thereby reducing the intensity of the radiation at the detector. Scattered radiation will be confused with absorbed radiation and result in a higher concentration than actually occurs in the sample. Another concern is that some species have the ability to change the value of \(\lambda\) . For some species, the value of \(\lambda\) can show a pronounced dependence on pH. If this is a consideration, then all of the standard and unknown solutions must be appropriately buffered. Species that can hydrogen bond or metal ions that can form donor-acceptor complexes with the analyte may alter the position of \(\lambda\) . Changes in the solvent can affect \(\lambda\) as well. | 16,898 | 3,150 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.S%3A_Solutions_(Summary) |
A is a homogeneous mixture. The major component is the , while the minor component is the . Solutions can have any phase; for example, an is a solid solution. Solutes are or , meaning they dissolve or do not dissolve in a particular solvent. The terms and , instead of soluble and insoluble, are used for liquid solutes and solvents. The statement is a useful guide to predicting whether a solute will dissolve in a given solvent. The amount of solute in a solution is represented by the of the solution. The maximum amount of solute that will dissolve in a given amount of solvent is called the of the solute. Such solutions are . Solutions that have less than the maximum amount are . Most solutions are unsaturated, and there are various ways of stating their concentrations. , , and indicate the percentage of the overall solution that is solute. and are used to describe very small concentrations of a solute. , defined as the number of moles of solute per liter of solution, is a common concentration unit in the chemistry laboratory. express concentrations in terms of moles of charge on ions. When a solution is diluted, we use the fact that the amount of solute remains constant to be able to determine the volume or concentration of the final diluted solution. Dissolving occurs by , the process in which particles of a solvent surround the individual particles of a solute, separating them to make a solution. For water solutions, the word is used. If the solute is molecular, it dissolves into individual molecules. If the solute is ionic, the individual ions separate from each other, forming a solution that conducts electricity. Such solutions are called . If the dissociation of ions is complete, the solution is a . If the dissociation is only partial, the solution is a . Solutions of molecules do not conduct electricity and are called . Solutions have properties that differ from those of the pure solvent. Some of these are properties, which are due to the number of solute particles dissolved, not the chemical identity of the solute. Colligative properties include , , , and . Osmotic pressure is particularly important in biological systems. It is caused by , the passage of solvents through certain membranes like cell walls. The of a solution is the product of a solution’s molarity and the number of particles a solute separates into when it dissolves. Osmosis can be reversed by the application of pressure; this reverse osmosis is used to make fresh water from saltwater in some parts of the world. Because of osmosis, red blood cells placed in hypotonic or hypertonic solutions lose function through either hemolysis or crenation. If they are placed in isotonic solutions, however, the cells are unaffected because osmotic pressure is equal on either side of the cell membrane. | 2,868 | 3,151 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/28%3A_Photochemistry/28.06%3A_The_Sensation_of_Color |
The sensation of color can be achieved in different ways. According to Table 28-1, which relates wavelength to color, we could recognize a given color, say yellow, by direct perception of light encompassing a narrow band of wavelengths around \(580 \: \text{nm}\), or by subtraction of blue light (\(435\)-\(480 \: \text{nm}\)) from white light. A third way of producing color is by an additive process. In fact, a wide range of colors can be achieved by the addition of three colors - red, green, and blue - as indicated in Figure 28-10. Mixing all three so-called , in the right intensities, gives white light; mixing only red and green gives yellow. It is important to recognize that addition of any two primary colors is equivalent to subtracting the third. This point is amplified in Figure 28-10. Subtraction of the three primary additive colors, red, green, and blue, from white light gives, respectively, the three , cyan, magenta, and yellow. Application of additive and subtractive processes in color perception is illustrated in the following sections. | 1,077 | 3,152 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry/Amounts_of_Substances |
Amounts of substances are measured in units of mass (g or kg), volume (L) and mole (mol). Unit interconversions are based on the definitions of the units, and converting amounts from g or kg into mol is based on atomic masses of the elements. are the masses of one mole of elements. A of any element has an Avogadro's number of atoms (= 6.02x10 atoms per mole). The natural units of substances are , which are groups of atoms bonded together, except monatomic molecules of inert gases, \(\ce{He}\), \(\ce{Ne}\), \(\ce{Ar}\), \(\ce{Kr}\), \(\ce{Xe}\), and \(\ce{Rn}\). For example, molecules of oxygen, water, and phosphorous are \(\ce{O2}\), \(\ce{H2O}\), and \(\ce{P4}\) respectively. These molecules have 2, 3, and 4 atoms respectively. Masses of one mole of substances are called . Atomic and molecular weights are called . The above illustrates only a very small number of examples. There are millions of compounds in the world. Please think of some other compounds you know of, and write down their formulas. Then figure out the number of atoms in each mole of your compounds. Different substances have different molecular masses. Thus, equal masses have different numbers of atoms, molecules, or moles. On the other hand, equal numbers of moles of different substances have different masses. The stoichiometric relationships among reactants and products may be complicated in units of g, but much simpler relationships are seen if we deal with units of moles or natural units of atoms and molecules. (mol) represent amounts of substances in the unit of (6.022x10 ) of atoms and molecules. Since empirical formulas such as \(\ce{Fe^2+}\) ions and \(\ce{Fe2O3}\) are used for ionic compounds, a mole represent Avogadro's number of ions or per formula as written. A mole of \(\ce{Fe^2+}\) has 6.022x10 ions, and a mole of \(\ce{Fe2O3}\) has 1.204x10 \(\ce{Fe}\) and 1.8066x10 \(\ce{O}\) atoms, a total of 3.0x10 \(\ce{Fe}\) and \(\ce{O}\) atoms. The mole unit is very important for chemical reactions, as is the skill to convert masses in g to mol. The number of moles of a substance in a sample is the mass in g divided by the , which gives the amount in moles. \(\mathrm{mole = \dfrac{mass\:(g)}{molar\: mass\: (g / mol) }}\) Another common measure of substances is volume. Since density is the mass divided by its volume, conversion between volume and mass is accomplished by the formula: \(\mathrm{density = \dfrac{mass\:(g)}{volume\: (cm^3)}}\) \(\mathrm{mass = density\: (g\: cm^{-3}) \times volume\: (cm^3)}\) These fundamental formulas are results of the definition of these terms. Know where to find: 1.0, 16.0, 55.9, 197 know where to find molar masses of elements. 1000/197 = ? to convert mass in g into moles. 1000 g / 18 g = 55.6 mol calculate molar masses of molecular compounds 0.20 mol / 22.4 L = 8.93e-3 mol/L express amount in volume. 32 g/mol * 8.93e-3 mol = 0.286 g convert amounts in moles to mass in g or kg. \(\ce{Fe2O3}\) \(\ce{Fe}\) \(\ce{O}\) 1000000 g/(159.6 g/mol) = 6265 mol convert amounts between moles and kg of compounds represented by chemical formulas. | 3,132 | 3,154 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/11%3A_Fluids/11.07%3A_Phase_Diagrams |
The state exhibited by a given sample of matter depends on the identity, temperature, and pressure of the sample. A phase diagram is a graphic summary of the physical state of a substance as a function of temperature and pressure in a closed system. A typical phase diagram consists of discrete regions that represent the different phases exhibited by a substance ( ). Each region corresponds to the range of combinations of temperature and pressure over which that phase is stable. The combination of high pressure and low temperature (upper left of ) corresponds to the solid phase, whereas the gas phase is favored at high temperature and low pressure (lower right). The combination of high temperature and high pressure (upper right) corresponds to a supercritical fluid. The solid phase is favored at low temperature and high pressure; the gas phase is favored at high temperature and low pressure. The lines in a phase diagram correspond to the combinations of temperature and pressure at which two phases can coexist in equilibrium. In the line that connects points A and D separates the solid and liquid phases and shows how the melting point of a solid varies with pressure. The solid and liquid phases are in equilibrium all along this line; crossing the line horizontally corresponds to melting or freezing. The line that connects points A and B is the vapor pressure curve of the liquid, which we discussed in . It ends at the critical point, beyond which the substance exists as a supercritical fluid. The line that connects points A and C is the vapor pressure curve of the phase. Along this line, the solid is in equilibrium with the vapor phase through sublimation and deposition. Finally, point A, where the solid/liquid, liquid/gas, and solid/gas lines intersect, is the triple point ; it is the combination of temperature and pressure at which all three phases (solid, liquid, and gas) are in equilibrium and can therefore exist simultaneously. Because no more than three phases can ever coexist, a phase diagram can never have more than three lines intersecting at a single point. Remember that a phase diagram, such as the one in , is for a single pure substance in a closed system, not for a liquid in an open beaker in contact with air at 1 atm pressure. In practice, however, the conclusions reached about the behavior of a substance in a closed system can usually be extrapolated to an open system without a great deal of error. shows the phase diagram of water and illustrates that the triple point of water occurs at 0.01°C and 0.00604 atm (4.59 mmHg). Far more reproducible than the melting point of ice, which depends on the amount of dissolved air and the atmospheric pressure, the triple point (273.16 K) is used to define the absolute (Kelvin) temperature scale. The triple point also represents the lowest pressure at which a liquid phase can exist in equilibrium with the solid or vapor. At pressures less than 0.00604 atm, therefore, ice does not melt to a liquid as the temperature increases; the solid sublimes directly to water vapor. Sublimation of water at low temperature and pressure can be used to “freeze-dry” foods and beverages. The food or beverage is first cooled to subzero temperatures and placed in a container in which the pressure is maintained below 0.00604 atm. Then, as the temperature is increased, the water sublimes, leaving the dehydrated food (such as that used by backpackers or astronauts) or the powdered beverage (as with freeze-dried coffee). The phase diagram for water illustrated in part (b) in shows the boundary between ice and water on an expanded scale. The melting curve of ice slopes up and slightly to the left rather than up and to the right as in ; that is, the melting point of ice with increasing pressure; at 100 MPa (987 atm), ice melts at −9°C. Water behaves this way because it is one of the few known substances for which the crystalline solid is than the liquid (others include antimony and bismuth). Increasing the pressure of ice that is in equilibrium with water at 0°C and 1 atm tends to push some of the molecules closer together, thus decreasing the volume of the sample. The decrease in volume (and corresponding increase in density) is smaller for a solid or a liquid than for a gas, but it is sufficient to melt some of the ice. In part (b) in , point A is located at = 1 atm and = −1.0°C, within the solid (ice) region of the phase diagram. As the pressure increases to 150 atm while the temperature remains the same, the line from point A crosses the ice/water boundary to point B, which lies in the liquid water region. Consequently, applying a pressure of 150 atm will melt ice at −1.0°C. We have already indicated that the pressure dependence of the melting point of water is of vital importance. If the solid/liquid boundary in the phase diagram of water were to slant up and to the right rather than to the left, ice would be denser than water, ice cubes would sink, water pipes would not burst when they freeze, and antifreeze would be unnecessary in automobile engines. In contrast to the phase diagram of water, the phase diagram of CO ( ) has a more typical melting curve, sloping up and to the right. The triple point is −56.6°C and 5.11 atm, which means that liquid CO cannot exist at pressures lower than 5.11 atm. At 1 atm, therefore, solid CO sublimes directly to the vapor while maintaining a temperature of −78.5°C, the normal sublimation temperature. Solid CO is generally known as dry ice because it is a cold solid with no liquid phase observed when it is warmed. Also notice the critical point at 30.98°C and 72.79 atm. In addition to the uses discussed in , supercritical carbon dioxide is emerging as a natural refrigerant, making it a low carbon (and thus a more environmentally friendly) solution for domestic heat pumps. Referring to the phase diagram of water in , phase diagram, temperature, and pressure physical form and physical changes Identify the region of the phase diagram corresponding to the initial conditions and identify the phase that exists in this region. Draw a line corresponding to the given pressure. Move along that line in the appropriate direction (in this case cooling) and describe the phase changes. Exercise Referring to the phase diagram of water in , predict the physical form of a sample of water at −0.0050°C as the pressure is gradually increased from 1.0 mmHg to 218 atm. The sample is initially a gas, condenses to a solid as the pressure increases, and then melts when the pressure is increased further to give a liquid. The states of matter exhibited by a substance under different temperatures and pressures can be summarized graphically in a , which is a plot of pressure versus temperature. Phase diagrams contain discrete regions corresponding to the solid, liquid, and gas phases. The solid and liquid regions are separated by the melting curve of the substance, and the liquid and gas regions are separated by its vapor pressure curve, which ends at the critical point. Within a given region, only a single phase is stable, but along the lines that separate the regions, two phases are in equilibrium at a given temperature and pressure. The lines separating the three phases intersect at a single point, the , which is the only combination of temperature and pressure at which all three phases can coexist in equilibrium. Water has an unusual phase diagram: its melting point decreases with increasing pressure because ice is less dense than liquid water. The phase diagram of carbon dioxide shows that liquid carbon dioxide cannot exist at atmospheric pressure. Consequently, solid carbon dioxide sublimes directly to a gas. A phase diagram is a graphic representation of the stable phase of a substance at any combination of temperature and pressure. What do the lines separating different regions in a phase diagram indicate? What information does the slope of a line in a phase diagram convey about the physical properties of the phases it separates? Can a phase diagram have more than one point where three lines intersect? If the slope of the line corresponding to the solid/liquid boundary in the phase diagram of water were positive rather than negative, what would be the effect on aquatic life during periods of subzero temperatures? Explain your answer. The lines in a phase diagram represent boundaries between different phases; at any combination of temperature and pressure that lies on a line, two phases are in equilibrium. It is physically impossible for more than three phases to coexist at any combination of temperature and pressure, but in principle there can be more than one triple point in a phase diagram. The slope of the line separating two phases depends upon their relative densities. For example, if the solid–liquid line slopes up and to the , the liquid is less dense than the solid, while if it slopes up and to the , the liquid is denser than the solid. Naphthalene (C H ) is the key ingredient in mothballs. It has normal melting and boiling points of 81°C and 218°C, respectively. The triple point of naphthalene is 80°C at 1000 Pa. Use these data to construct a phase diagram for naphthalene and label all the regions of your diagram. Argon is an inert gas used in welding. It has normal boiling and freezing points of 87.3 K and 83.8 K, respectively. The triple point of argon is 83.8 K at 0.68 atm. Use these data to construct a phase diagram for argon and label all the regions of your diagram. | 9,549 | 3,155 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Photoreceptors/Chemistry_of_Vision/Photochemical_Changes_in_Opsin |
Photochemical events in vision involve the protein opsin and the cis/trans isomers of retinal. Opsin does not absorb visible light, but when it is bonded with 11-cis-retinal to form rhodopsin, which has a very broad absorption band in the visible region of the spectrum. The peak of the absorption is around 500 nm, which matches the output of the sun closely. Upon absorption of a photon of light in the visible range, cis-retinal can isomerize to all-trans-retinal. The shape of the molecule changes as a result of this isomerization. The molecule changes from an overall bent structure to one that is more or less linear. All of this is the result of trigonal planar bonding (120 bond angles) about the double bonds. As we shall see below, the isomerization of retinal has an important effect on special proteins in the rod cell: the isomerization event actually causes the proteins to change their shape. This shape change ultimately leads to the generation of a nerve impulse. Hence, the next step in understanding the vision process for monochrome vision is to describe these proteins, and how they change their shape after retinal isomerizes. Opsin consists of 348 amino acids, covalently linked together to form a single chain. This chain has seven hydrophobic, or water-repelling, alpha-helical regions that pass through the lipid membrane of the pigment-containing discs. This region consists primarily of nonpolar amino acids, which do not attract the polar water molecule. The cis-retinal is situated among these alpha helixes in the hydrophobic region. It is covalently linked to Lysine 296, one of the amino acids in the opsin peptide chain. The linkage is as a Schiff base reaction. When the cis-retinal absorbs a photon it isomerizes to the all-trans configuration without (at first) any accompanying change in the structure of the protein. Rhodopsin containing the all-trans isomer of retinal is known as bathorhodopsin. However, the trans isomer does not fit well into the protein, due to its rigid, elongated shape. While it is contained in the protein, the all-trans-retinal adopts a twisted conformation, which is energetically unfavorable. The molecule undergoes a series of shape changes to try and better fit the binding site. Therefore, a series of changes in the protein occurs to expel the trans-retinal from the protein. These rapid movements of the retinal are transferred to the protein, and from there into the lipid membrane and nerve cells to which it is attached. This generates nerve impulses which travel along the optic nerve to the brain, and we perceive them as visual signals - sight. The free all-trans-retinal is then converted back into the cis form by a series of enzyme-catalyzed reactions, whereupon is reattaches to another opsin ready for the next photon to begin the process again. | 2,854 | 3,156 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/09%3A_Chemical_Equilibria/9.02%3A_Chemical_Potential |
Equilibrium can be understood as accruing at the composition of a reaction mixture at which the aggregate chemical potential of the products is equal to that of the reactants. Consider the simple reaction \[A(g) \rightleftharpoons B(g) \nonumber \] The criterion for equilibrium will be \[ \mu_A=\mu_B \nonumber \] If the gases behave ideally, the chemical potentials can be described in terms of the mole fractions of \(A\) and \(B\) \[ \mu_A^o + RT \ln\left( \dfrac{p_A}{p_{tot}} \right) = \mu_B^o + RT \ln\left( \dfrac{p_B}{p_{tot}} \right) \label{eq2} \] where Dalton’s Law has been used to express the mole fractions. \[ \chi_i = \dfrac{p_i}{p_{tot}} \nonumber \] Equation \ref{eq2} can be simplified by collecting all chemical potentials terms on the left \[ \mu_A^o - \mu_B^o = RT \ln \left( \dfrac{p_B}{p_{tot}} \right) - RT \ln\left( \dfrac{p_A}{p_{tot}} \right) \label{eq3} \] Combining the logarithms terms and recognizing that \[\mu_A^o - \mu_B^o –\Delta G^o \nonumber \] for the reaction, one obtains \[–\Delta G^o = RT \ln \left( \dfrac{p_B}{p_{A}} \right) \nonumber \] And since \(p_A/p_B = K_p\) for this reaction (assuming perfectly ideal behavior), one can write \[ \Delta G^o = RT \ln K_p \nonumber \] Another way to achieve this result is to consider the Gibbs function change for a reaction mixture in terms of the . The reaction quotient can be expressed as \[ Q_p = \dfrac{\prod_i p_i^{\nu_i}}{\prod_j p_j^{\nu_j}} \nonumber \] where \(\nu_i\) are the stoichiometric coefficients for the products, and \(\nu_j\) are those for the reactants. Or if the stoichiometric coefficients are defined by expressing the reaction as a sum \[ 0 =\sum_i \nu_i X_i \nonumber \] where \(X_i\) refers to one of the species in the reaction, and \(\nu_i\) is then the stoichiometric coefficient for that species, it is clear that \(\nu_i\) will be negative for a reactant (since its concentration or partial pressure will reduce as the reaction moves forward) and positive for a product (since the concentration or partial pressure will be increasing.) If the stoichiometric coefficients are expressed in this way, the expression for the reaction quotient becomes \[Q_p = \prod_i p_i^{\nu_i} \nonumber \] Using this expression, the Gibbs function change for the system can be calculated from \[ \Delta G =\Delta G^o + RT \ln Q_p \nonumber \] And since at equilibrium \[\Delta G = 0 \nonumber \] and \[Q_p=K_p \nonumber \] It is evident that \[ \Delta G_{rxn}^o = -RT \ln K_p \label{triangle} \] It is in this simple way that \(K_p\) and \(\Delta G^o\) are related. It is also of value to note that the criterion for a spontaneous chemical process is that \(\Delta G_{rxn}\ < 0\), rather than \(\Delta G_{rxn}^o\), as is stated in many texts! Recall that \(\Delta G_{rxn}^o\) is a function of all of the reactants and products being in their standard states of unit fugacity or activity. However, the direction of spontaneous change for a chemical reaction is dependent on the composition of the reaction mixture. Similarly, the magnitude of the equilibrium constant is insufficient to determine whether a reaction will spontaneously form reactants or products, as the direction the reaction will shift is also a function of not just the equilibrium constant, but also the composition of the reaction mixture! Based on the data below at 298 K, calculate the value of the equilibrium constant (\(K_p\)) for the reaction \[2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g) \nonumber \] First calculate the value of \(\Delta G_{rxn}^o\) from the \(\Delta G_{f}^o\) data. \[ \Delta G_{rxn}^o = 2 \times (51.53 \,kJ/mol) - 2 \times (86.55 \,kJ/mol) = -70.04 \,kJ/mol \nonumber \] And now use the value to calculate \(K_p\) using Equation \ref{triangle}. \[ -70040\, J/mol = -(8.314 J/(mol\, K) (298 \, K) \ln K_p \nonumber \] \[ K_p = 1.89 \times 10^{12} \nonumber \] : as expected for a reaction with a very large negative \(\Delta G_{rxn}^o\), the equilibrium constant is large, favoring the formation of the products. | 4,037 | 3,157 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.16%3A_Equilibrium_Constants_Revisited |
In discussing entropy and spontaneity, we have tended to treat chemical reactions as though they had no other alternatives than either going to completion or not occurring at all. It is only when both reactants and products are pure solids or pure liquids, however, that we find this all-or-nothing-at-all type of behavior. Reactions which involve gases or solutions are governed by an equilibrium constant, and as we saw in the , this means that there is always some reactant and some product in the equilibrium mixture. Consequently such reactions must always occur to some extent, no matter how minute, and can never go quite to completion. Figure \(\Page {1}\) illustrates how the free energy varies as the reaction proceeds in the two cases. If only pure solids and liquids are involved, then a plot of against the extent of the reaction is a straight line, as shown in part of the figure for the reaction \[\ce{Hg(l) + HgBr2(s)→Hg2Br2(s)}\qquad 1 \text{ atm, 298K} \nonumber \] In such a case, if Δ °, the free-energy difference between reactants and products, is , then the reaction will attain the lowest value of possible by going to completion. When gases and solutions are involved in the reaction, a plot of against the extent of the reaction is no longer a straight line but exhibits a “sag,“ as shown in Figure \(\Page {1}\) for the reaction \[\ce{2NO2(g)→N2O4(g)}\qquad 1 \text{ atm}\ \label{2} \] \[{\Delta G_{m}^{ o}}={-RT}{\text{ ln } K^{o}} \label{3} \] or, if base 10 logarithms are used, \[{\Delta G_{m}^{ o}}={-2.303 RT}{\text{ ln } K^{o}} \label{4} \] Where ° ( standard) is called the standard . ° is closely related to and differs from it only by being a dimensionless number. Recall from the section on that is defined for the general chemical reaction \[ a \text{A} + b \text{B} + \cdots \rightleftharpoons c \text{C} + d \text{D} + \cdots \nonumber \] by the equation \[K_{p}=\dfrac{p_{\text{C}}^{c}\text{ }\times \text{ }p_{\text{D}}^{d}\text{ }\times \text{ }\cdots }{p_{\text{A}}^{a}\text{ }\times \text{ }p_{\text{B}}^{b}\text{ }\times \text{ }\cdots } \nonumber \] etc., are partial pressures. The definition of ° is identical except that it involves pressure (i.e., pure numbers) rather than pressures: \[K\text{ }\!\!{}^\circ\!\!\text{ }=\dfrac{\left( \dfrac{p_{\text{C}}}{p\text{ }\!\!{}^\circ\!\!\text{ }} \right)^{c}\text{ }\times \text{ }\left( \dfrac{p_{\text{D}}}{p\text{ }\!\!{}^\circ\!\!\text{ }} \right)^{d}\text{ }\times \text{ }\cdots }{\left( \dfrac{p_{\text{A}}}{p\text{ }\!\!{}^\circ\!\!\text{ }} \right)^{a}\text{ }\times \text{ }\left( \dfrac{p_{\text{B}}}{p\text{ }\!\!{}^\circ\!\!\text{ }} \right)^{b}\text{ }\times \text{ }\cdots } \nonumber \] Where ° is a standard pressure—almost always 1 atm (101.325 kPa). Thus if the partial pressures , , etc., are expressed in atmospheres, ° involves the same number as . The equilibrium constant for the reaction \[ \ce{2NO2 <=> N2O4} \nonumber \] is 0.0694 kPa at 298 K. Use this value to find Δ °(298 K) for the reaction. We must first express in atmospheres: \[K_{p}=\text{0}\text{.0694 kPa}^{-\text{1}}=\dfrac{\text{0}\text{.0694}}{\text{kPa}}\text{ }\times \text{ }\dfrac{\text{101}\text{.3 kPa}}{\text{1 atm}}=\text{7}\text{.03 atm}^{-\text{1}} \nonumber \] Thus \( K^\circ = 7.03 \qquad \text{ a pure number} \) From Eq. \(\ref{4}\) we now have \(\begin{align} \Delta G_m^{\circ} & = –2.303RT \text{ log } K^{\circ} \\ &= – 8.3143 \text{ J K}^{-1} \text{ mol}^{-1} \times 298 \text{ K} \times 2.303 \times \text{ log } 7.03 \\ & = –4833 \text{ J mol}^{-1} = –4.833 \text{ kJ mol}^{-1} \end{align} \) : You may also use ln as well as log as long as you use Eq. \(\ref{3}\) directly: \[ \Delta G_m^{\circ} = –RT \text{ ln } K^{\circ} = – RT \times \text{ ln } 7.03 = – RT \times 1.950 = –4.833 \nonumber \] In , we argued that the value of an equilibrium constant is the product of two factors: \[{K}=\text{energy factor}\times\text{probability factor} \label{14} \] The energy factor takes account of the fact that a higher-energy species is less likely to occur in an equilibrium mixture than a lower-energy species, especially at low temperatures. The probability factor reflects the fact that if there are a larger number of ways in which a molecule can arrange itself in space, a molecule is more likely to occur in that state than in one for which a smaller number of spatial arrangements is possible. We are now in a position to make quantitative the qualitative argument presented. Combining \[{\Delta G_{m}^{o}}={\Delta H_{m}^{ o}}-{T\Delta S_{m}^{ o}} \nonumber \] with \[{\Delta G_{m}^{o}}=\text{RT}{\text{ ln }{K^o}} \nonumber \] we find \[{-}\text{RT}{\text{ ln }{K^o}}={\Delta H_{m}^{ o}}-{T\Delta S_{m}^{ o}} \nonumber \] giving us \[{ \text{ ln }{K^o}}=\dfrac{-\Delta H_{m}^{ o}}{RT} + \dfrac{\Delta S_{m}^{ o}}{R} \label{18} \] or, in base 10 logarithms \[{ \text{ log }{K^o}}=\dfrac{-\Delta H_{m}^{ o}}{\text{2}\text{.303}RT} + \dfrac{\Delta S_{m}^{ o}}{\text{2}\text{.303}R} \label{19} \] If we now take the logarithm of each side of Eq. \(\ref{14}\), we have \[\text{ log }{K}=\text{ log }(\text{energy factor})+{\text{ log }(\text{probability factor})}\, \nonumber \] This equation has the same form as Eq. \(\ref{19}\), and if we confine ourselves to the standard equilibrium constant °, we can say that \[ \text{ log } (\text{energy factor})=\dfrac{-\Delta H_{m}^{ o}}{\text{2}\text{.303}RT} \label{21} \] and \[ \text{ log } (\text{probability factor})=\dfrac{\Delta S_{m}^{ o}}{\text{2}\text{.303}R} \label{22} \] Although we did not formally derive Eq. \(\ref{3}\), on which these results are based, we can examine the last two equalities [Eqs. \(\ref{21}\) and \(\ref{22}\)] to see that they agree with our qualitative expectations. If Δ ° is negative, we know that the products of a reaction have a lower enthalpy (and hence lower energy) than the reactants. Thus we would predict that products would he favored by the energy factor, and this is what Eq. \(\ref{21}\) says—the more negative Δ °, the more positive the logarithm of the energy factor and the larger the standard equilibrium constant. Also, since appears in the denominator of the right-hand side of Eq. \(\ref{21}\), the smaller the value of , the larger (and hence more important) the energy factor for a given value of Δ °. We have also seen in that an increase in entropy of a system corresponds to an increase in thermodynamic probability. This is precisely what Eq. \(\ref{22}\) says. The larger the value of Δ °, the larger the probability factor and hence the larger the standard equilibrium constant. Thus our qualitative description of the two factors affecting the equilibrium constant has been refined to the point where macroscopic quantities, Δ ° and Δ °, can be related to what is happening on the microscopic level. In addition to the feature we have just mentioned, Eq. \(\ref{19}\) is useful in another way. If we can measure or estimate Δ ° and Δ ° at temperatures other than the usual 298.15 K, we can obtain ° and calculate the extent of reaction as shown in the next example. Find the concentration of NO in equilibrium with air at 1000 K and 1 atm pressure. \[ \ce{N2 (g) + O2 (g) -> 2NO (g) } \Delta H_m^{\circ} = 181 \text{kJ mol}^{-1} \nonumber \] \[ \Delta S_m^{\circ} = 35.6 \text{J K}^{-1} \text{ mol}^{-1} \nonumber \] We first find ° : \[{\text{ log }{K^o}}=\dfrac{-\Delta H_{m}^{ o}}{\text{2}\text{.303}RT} + \dfrac{\Delta S_{m}^{ o}}{\text{2}\text{.303}R}= \text{–9.45 + 1.34}= \text{–8.11} \nonumber \] Thus \[ K^{\circ} = 10^{-8.11} = 7.71 \times 10^{-9} \nonumber \] However, since Δ = 0, \(K^{\circ} = K_p = 7.71 \times 10^{-9}\) Assuming the air to be 80% N and 20% O , we can write \[ p_{\text{N2}} = 0.8 \text{ atm} ~~~~~\text{and}~~~~~ p_{\text{O2}} = 0.2 \text{ atm} \nonumber \] Thus \[{K}_{p}=\dfrac{p_{\text{NO}_{\text{2}}}^{\text{2}}}{\text{(}p_{\text{N}_{\text{2}}}\text{)(}p_{\text{O}_{\text{2}}}\text{)}} = \dfrac{p_{\text{NO}_{\text{2}}}^{\text{2}}}{\text{0}\text{.8 atm }\times \text{ 0}\text{.2 atm}} = {\text{7.71}}\times{\text{10}}^{-\text{9}} \nonumber \] Giving \(p_{\text{NO}_{\text{2}}}^{\text{2}} = {\text{7.71}}\times{\text{10}}^{-\text{9}}\times {\text{0.16 atm}}^{\text{2}}={\text{1.23}}\times{\text{10}}^{-\text{9}}{\text{atm}}^{\text{2}}\) \({p}_{\text{NO}_2}=\sqrt{\text{1}\text{.23 }\times \text{ 10}^{-9}\text{ atm}^{\text{2}}}= {\text{3.5}}\times{\text{10}}^{- \text{5}}{\text{atm}} ={\text{3.5}}\times{\text{10}}^{- \text{3}}{\text{kPa}}\) But \({c}_{\text{NO}_2}=\dfrac{c_{\text{NO}_{\text{2}}}}{V} = \dfrac{p_{\text{NO}_{\text{2}}}}{RT}\) Thus \[{c}_{\text{NO}_2}=\dfrac{\text{3}\text{.5 }\times \text{ 10}^{-\text{3}}\text{ kPa}}{\text{8}\text{.3143 J K}^{-\text{1}}\text{ mol}^{-\text{1}}}\text{ }\times \text{ }\dfrac{\text{1}}{\text{1000 K}} = {\text{4.3}}\times{\text{10}}^{- \text{7}}{\text{dm}}^{- \text{3}} \nonumber \] Almost all combustion processes heat N and O to high temperatures, producing small concentrations of NO. Under most circumstances this is not of great consequence, but in the presence of sunlight and partially burned gasoline, NO can initiate a form of air pollution called photochemical smog. The presence of minute concentrations of NO in the upper atmosphere from high-flying supersonic jet airplanes can also deplete the ozone layer there. | 9,529 | 3,159 |
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