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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/08%3A_Phase_Equilibrium/8.01%3A_Prelude_to_Phase_Equilibrium |
From the very elementary stages of our journey to describe the physical nature of matter, we learn to classify mater into three (or more) phases: solid, liquid, and gas. This is a fairly easy classification system that can be based on such simple ideas as shape and volume. As we have progressed, we have seen that solids and liquids are not completely incompressible as they may have non-zero values of \(\kappa_T\). And we learn that there are a number of finer points to describing the nature of the phases about which we all learn in grade school. In this chapter, we will employ some of the tools of thermodynamics to explore the nature of phase boundaries and see what we can conclude about them. | 714 | 2,795 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/05%3A_The_Second_Law/5.07%3A_The_Third_Law_of_Thermodynamics |
One important consequence of Botlzmann’s proposal is that a perfectly ordered crystal (i.e. one that has only one energetic arrangement in its lowest energy state) will have an entropy of 0. This makes entropy qualitatively different than other thermodynamic functions. For example, in the case of enthalpy, it is impossible have a zero to the scale without setting an arbitrary reference (which is that the enthalpy of formation of elements in their standard states is zero.) But entropy has a natural zero! It is the state at which a system has perfect order. This also has another important consequence, in that it suggests that there must also be a zero to the temperature scale. These consequences are summed up in the . The entropy of a perfectly ordered crystal at 0 K is zero. This also suggests that absolute molar entropies can be calculated by \[S = \int_o^{T} \dfrac{C}{T} dT \nonumber \] where \(C\) is the heat capacity. An entropy value determined in this manner is called a . Naturally, the heat capacity will have some temperature dependence. It will also change abruptly if the substance undergoes a phase change. Unfortunately, it is exceedingly difficult to measure heat capacities very near zero K. Fortunately, many substances follow the in that at very low temperatures, their heat capacities are proportional to T . Using this assumption, we have a temperature dependence model that allows us to extrapolate absolute zero based on the heat capacity measured at as low a temperature as can be found. SiO is found to have a molar heat capacity of 0.777 J mol K at 15 K (Yamashita, et al., 2001). Calculate the molar entropy of SiO at 15 K. Using the Debye model, the heat capacity is given by The value of a can be determined by The entropy is then calculated by Start at 0 K, and go from there! | 1,837 | 2,796 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.08%3A_Dependence_of_S_on_Molecular_Structure |
Since they all refer to the same temperature and pressure, and also to 1 mol of substance, any differences in the entropy values listed in the table of standard entropies must be due to of the various substances listed. There are two aspects of the molecular structure of a substance which affect the value of its entropy: (1) The degree to which the movement of the atoms and molecules in the structure is restricted—the less restricted this movement, the greater the entropy. (2) The mass of the atoms and molecules which are moving—the greater the mass, the larger the entropy. We will consider each of these factors in turn. \(\Page {1}\) Entropy Values We have already encountered two examples of how the removal of a restriction on the motion of molecules allows an increase in the number of alternative ways in which the molecules can arrange themselves in space. The first of these is the case of the ring and chain forms of 1,4-butanediol discussed in . When the two ends of this molecule hydrogen bond to form a ring, the result is a fairly rigid structure. However, if the restricting influence of the hydrogen bond is removed, the more flexible chain form of the molecule is capable of many more alternative configurations. As a result, we find that for a mole of molecules in the chain form is very much larger than for a mole of molecules in the ring form. The chain form turns out to have a molar entropy which is 41 J K mol higher than that for the ring form. A second example of the effect of relaxing a restriction on molecular motion is the expansion of a gas into a vacuum. When the molecules have more freedom of movement, many more alternative arrangements are possible, and hence is larger. As we have already calculated, doubling the volume of a gas increases its molar entropy by 5.76 J K mol . The effect of closeness of energy levels on the entropy is shown in Figure \(\Page {1}\). The more closely spaced the levels, the more different ways the same quantity of energy can be distributed among them. This applies in general for any number of particles and any quantity of energy. Therefore, . This effect is quite obvious among the noble gases—their molar entropies increase steadily with molar (and hence molecular) mass. | 2,284 | 2,797 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Reducing_Alkynes-The_Reactivity_of_the_Two__Bonds |
Reactions between alkynes and catalysts are a common source of alkene formation. Because alkynes differ from alkenes on account of their two procurable bonds, alkynes are more susceptible to additions. Aside from turning them into alkenes, these catalysts affect the arrangement of substituents on the newly formed alkene molecule. Depending on which catalyst is used, the catalysts cause anti- or syn-addition of hydrogens. Alkynes can readily undergo additions because of their availability of two bonds. Alkynes can be fully hydrogenated into alkanes with the help of a platinum catalyst. However, the use of two other catalysts can be used to hydrogenate alkynes to alkanes. These catalysts are: Palladium dispersed on carbon (Pd/C) and finely dispersed nickel (Raney-Ni). Because hydrogenation is an interruptible process involving a series of steps, hydrogenation can be stopped, using modified catalysts (e.g., Lindlar’s Catalyst) at the transitional alkene stage. Lindar’s catalyst has three components: Palladium-Calcium Carbonate, lead acetate and quinoline. The quinoline serves to prevent complete hydrogenation of the alkyne to an alkane. Lindlar’s Catalyst transforms an alkyne to a cis-alkene. Alkynes can be reduced to trans-alkenes with the use of sodium dissolved in an ammonia solvent. An Na radical donates an electron to one of the bonds in a carbon-carbon triple bond. This forms an anion, which can be protonated by a hydrogen in an ammonia solvent. This prompts another Na radical to donate an electron to the second orbital. Soon after this anion is also protonated by a hydrogen from the ammonia solvent, resulting in a trans-alkene. | 1,685 | 2,798 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/17%3A_Additional_Aspects_of_Acid-Base_Equilibria/17.3%3A_Acid-Base_Indicators |
Certain organic substances change color in dilute solution when the hydronium ion concentration reaches a particular value. For example, phenolphthalein is a colorless substance in any aqueous solution with a hydronium ion concentration greater than 5.0 × 10 (pH < 8.3). In more basic solutions where the hydronium ion concentration is less than 5.0 × 10 (pH > 8.3), it is red or pink. Substances such as phenolphthalein, which can be used to determine the pH of a solution, are called . Acid-base indicators are either weak organic acids or weak organic bases. The equilibrium in a solution of the acid-base indicator methyl orange, a weak acid, can be represented by an equation in which we use HIn as a simple representation for the complex methyl orange molecule: \[\underbrace{\ce{HIn}_{(aq)}}_{\ce{red}}+\ce{H2O}_{(l)}⇌\ce{H3O+}_{(aq)}+\underbrace{\ce{In-}_{(aq)}}_{\ce{yellow}}\] \[K_\ce{a}=\ce{\dfrac{[H3O+,In- ]}{[HIn]}}=4.0×10^{−4}\] The anion of methyl orange, In , is yellow, and the nonionized form, HIn, is red. When we add acid to a solution of methyl orange, the increased hydronium ion concentration shifts the equilibrium toward the nonionized red form, in accordance with Le Chatelier’s principle. If we add base, we shift the equilibrium towards the yellow form. This behavior is completely analogous to the action of buffers. An indicator’s color is the visible result of the ratio of the concentrations of the two species In and HIn. If most of the indicator (typically about 60−90% or more) is present as In , then we see the color of the In ion, which would be yellow for methyl orange. If most is present as HIn, then we see the color of the HIn molecule: red for methyl orange. For methyl orange, we can rearrange the equation for and write: \[\mathrm{\dfrac{[In^-]}{[HIn]}=\dfrac{[substance\: with\: yellow\: color]}{[substance\: with\: red\: color]}=\dfrac{\mathit{K}_a}{[H_3O^+]}}\] This shows us how the ratio of \(\ce{\dfrac{[In- ]}{[HIn]}}\) varies with the concentration of hydronium ion. The above expression describing the indicator equilibrium can be rearranged: \[\mathrm{\dfrac{[H_3O^+]}{\mathit{K}_a}=\dfrac{[HIn]}{[In^- ]}}\] \[\mathrm{log\left(\dfrac{[H_3O^+]}{\mathit{K}_a}\right)=log\left(\dfrac{[HIn]}{[In^- ]}\right)}\] \[\mathrm{log([H_3O^+])-log(\mathit{K}_a)=-log\left(\dfrac{[In^-]}{[HIn]}\right)}\] \[\mathrm{-pH+p\mathit{K}_a=-log\left(\dfrac{[In^-]}{[HIn]}\right)}\] \[\mathrm{pH=p\mathit{K}_a+log\left(\dfrac{[In^-]}{[HIn]}\right)\:or\:pH=p\mathit{K}_a+log\left(\dfrac{[base]}{[acid]}\right)}\] The last formula is the same as the Henderson-Hasselbalch equation, which can be used to describe the equilibrium of indicators. When [H O ] has the same numerical value as , the ratio of [In ] to [HIn] is equal to 1, meaning that 50% of the indicator is present in the red form (HIn) and 50% is in the yellow ionic form (In ), and the solution appears orange in color. When the hydronium ion concentration increases to 8 × 10 (a pH of 3.1), the solution turns red. No change in color is visible for any further increase in the hydronium ion concentration (decrease in pH). At a hydronium ion concentration of 4 × 10 (a pH of 4.4), most of the indicator is in the yellow ionic form, and a further decrease in the hydronium ion concentration (increase in pH) does not produce a visible color change. The pH range between 3.1 (red) and 4.4 (yellow) is the of methyl orange; the pronounced color change takes place between these pH values. Many different substances can be used as indicators, depending on the particular reaction to be monitored. For example, red cabbage juice contains a mixture of colored substances that change from deep red at low pH to light blue at intermediate pH to yellow at high pH (Figure \(\Page {1}\)). In all cases, though, a good indicator must have the following properties: Red cabbage juice contains a mixture of substances whose color depends on the pH. Each test tube contains a solution of red cabbage juice in water, but the pH of the solutions varies from pH = 2.0 (far left) to pH = 11.0 (far right). At pH = 7.0, the solution is blue. Synthetic indicators have been developed that meet these criteria and cover virtually the entire pH range. Figure \(\Page {2}\) shows the approximate pH range over which some common indicators change color and their change in color. In addition, some indicators (such as thymol blue) are polyprotic acids or bases, which change color twice at widely separated pH values. It is important to be aware that an indicator does not change color abruptly at a particular pH value; instead, it actually undergoes a pH titration just like any other acid or base. As the concentration of HIn decreases and the concentration of In− increases, the color of the solution slowly changes from the characteristic color of HIn to that of In−. As we will see in Section 16, the [In−]/[HIn] ratio changes from 0.1 at a pH one unit below pKin to 10 at a pH one unit above pKin. Thus most indicators change color over a pH range of about two pH units. We have stated that a good indicator should have a pKin value that is close to the expected pH at the equivalence point. For a strong acid–strong base titration, the choice of the indicator is not especially critical due to the very large change in pH that occurs around the equivalence point. In contrast, using the wrong indicator for a titration of a weak acid or a weak base can result in relatively large errors, as illustrated in Figure \(\Page {3}\). This figure shows plots of pH versus volume of base added for the titration of 50.0 mL of a 0.100 M solution of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M \(NaOH\). The pH ranges over which two common indicators (methyl red, \(pK_{in} = 5.0\), and phenolphthalein, \(pK_{in} = 9.5\)) change color are also shown. The horizontal bars indicate the pH ranges over which both indicators change color cross the HCl titration curve, where it is almost vertical. Hence both indicators change color when essentially the same volume of \(NaOH\) has been added (about 50 mL), which corresponds to the equivalence point. In contrast, the titration of acetic acid will give very different results depending on whether methyl red or phenolphthalein is used as the indicator. Although the pH range over which phenolphthalein changes color is slightly greater than the pH at the equivalence point of the strong acid titration, the error will be negligible due to the slope of this portion of the titration curve. Just as with the HCl titration, the phenolphthalein indicator will turn pink when about 50 mL of \(NaOH\) has been added to the acetic acid solution. In contrast, methyl red begins to change from red to yellow around pH 5, which is near the midpoint of the acetic acid titration, not the equivalence point. Adding only about 25–30 mL of \(NaOH\) will therefore cause the methyl red indicator to change color, resulting in a huge error. The graph shows the results obtained using two indicators (methyl red and phenolphthalein) for the titration of 0.100 M solutions of a strong acid (HCl) and a weak acid (acetic acid) with 0.100 M \(NaOH\). Due to the steepness of the titration curve of a strong acid around the equivalence point, either indicator will rapidly change color at the equivalence point for the titration of the strong acid. In contrast, the pKin for methyl red (5.0) is very close to the pKa of acetic acid (4.76); the midpoint of the color change for methyl red occurs near the midpoint of the titration, rather than at the equivalence point. In general, for titrations of strong acids with strong bases (and vice versa), any indicator with a pK in between about 4.0 and 10.0 will do. For the titration of a weak acid, however, the pH at the equivalence point is greater than 7.0, so an indicator such as phenolphthalein or thymol blue, with pKin > 7.0, should be used. Conversely, for the titration of a weak base, where the pH at the equivalence point is less than 7.0, an indicator such as methyl red or bromocresol blue, with pKin < 7.0, should be used. In the titration of a weak acid with a strong base, which indicator would be the best choice? The correct answer is C. In the titration of a weak acid with a strong base, the conjugate base of the weak acid will make the pH at the equivalence point greater than 7. Therefore, you would want an indicator to change in that pH range. Both methyl orange and bromocresol green change color in an acidic pH range, while phenolphtalein changes in a basic pH. The existence of many different indicators with different colors and pKin values also provides a convenient way to estimate the pH of a solution without using an expensive electronic pH meter and a fragile pH electrode. Paper or plastic strips impregnated with combinations of indicators are used as “pH paper,” which allows you to estimate the pH of a solution by simply dipping a piece of pH paper into it and comparing the resulting color with the standards printed on the container (Figure \(\Page {4}\)). pH Indicators: Acid–base indicators are compounds that change color at a particular pH. They are typically weak acids or bases whose changes in color correspond to deprotonation or protonation of the indicator itself. ). | 9,331 | 2,800 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.16%3A_Examples_of_Lewis_Structures |
Here are a set of examples showing 3D twirlymol molecular models of from other sections in the text, along with Lewis structures of these molecules as well. It is useful to see what can be conveyed via 3D models, molecular geometry for instance, and what is conveyed by Lewis structure models, an example being the assignment of electrons around atoms and assignment of electrons for bonding. An important clarification, is that the twirlymols display , not bonds, so double and triple bonds do not show up. Double and triple bonds are shown in the Lewis structures. Molecules from a common page are grouped together, with a link back to the original page. HOCl CH O CO CH OH C H CO HClO N O O CO SO OH H O NH SO CH COOH NH Ed Vitz (Kutztown University), (University of | 809 | 2,802 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.10%3A_Ions_and_Electrolytes/8.10.9B%3A_8.10.9B%3A_The_nature_of_ions_in_aqueous_solution |
The kinds of ions we will consider in this lesson are mostly those found in solutions of common acids or salts. As is evident from the image below, most of the metallic elements form monatomic cations, but the number of monatomic anions is much smaller. This reflects the fact that many single-atom anions such as hydride H , oxide O , sulfide S and those in and , are unstable in ( , react with) water, and their major forms are those in which they are combined with other elements, particularly oxygen. Some of the more familiar are hydroxide OH , carbonate CO , nitrate NO , sulfate SO , chlorate ClO , and arsenate AsO . Conductivity water" with κ = 0.043 × 10 S cm at 18°C. Ordinary distilled water in equilibrium with atmospheric CO has a conductivity that is 16 times greater. It is now known that ordinary distillation cannot entirely remove all impurities from water. Ionic impurities get entrained into the fog created by breaking bubbles and are carried over into the distillate by capillary flow along the walls of the apparatus. Organic materials tend to be steam-volatile ("steam-distilled"). The best current practice is to employ a special still made of fused silica in which the water is volatilized from its surface without boiling. Complete removal of organic materials is accomplished by passing the water vapor through a column packed with platinum gauze heated to around 800°C through which pure oxygen gas is passed to ensure complete oxidation of carbon compounds. Conductance measurements are widely used to gauge water quality, especially in industrial settings in which concentrations of dissolved solids must be monitored in order to schedule maintenance of boilers and cooling towers. The conductance of a solution depends on the concentration of the ions it contains, on the number of charges carried by each ion, and on the of these ions. The latter term refers to the ability of the ion to make its way through the solution, either by ordinary thermal diffusion or in response to an electric potential gradient. The first step in comparing the conductances of different solutes is to reduce them to a common concentration. For this, we define the which is known as the , denoted by the upper-case Greek : \[Λ = \dfrac{κ}{c}\] When κ is expressed in S cm , should be in mol cm , so Λ will have the units S cm . This is best visualized as the conductance of a cell having 1-cm electrodes spaced 1 cm apart — that is, of a 1 cm cube of solution. But because chemists generally prefer to express concentrations in mol L or mol dm (mol/1000 cm ) , it is common to write the expression for molar conductivity as \[Λ = \dfrac{1000κ}{c}\] But if is the concentration in moles per liter, this will still not fairly compare two salts having different stoichiometries, such as AgNO and FeCl , for example. If we assume that both salts dissociate completely in solution, each mole of AgNO yields two moles of charges, while FeCl releases six (i.e., one Fe ion, and three Cl ions.) So if one neglects the [rather small] differences in the ionic mobilities, the molar conductivity of FeCl would be three times that of AgNO . The most obvious way of getting around this is to note that one mole of a 1:1 salt such as AgNO is "equivalent" (in this sense) to 1/3 of a mole of FeCl , and of ½ a mole of MgBr . To find the number of equivalents that correspond a given quantity of a salt, just divide the number of moles by the total number of positive charges in the formula unit. (If you like, you can divide by the number of negative charges instead; because these substances are electrically neutral, the numbers will be identical.) We can refer to equivalent concentrations of individual ions as well as of neutral salts. Also, since acids can be regarded as salts of H , we can apply the concept to them; thus a 1M L solution of sulfuric acid H SO has a concentration of 2 eq L . The following diagram summarizes the relation between moles and equivalents for CuCl : What is the concentration, in equivalents per liter, of a solution made by dissolving 4.2 g of chromium(III) sulfate pentahydrate Cr (SO ) ·5H O in sufficient water to give a total volume of 500 mL? (The molar mass of the hydrate is 482 g) Assume that the salt dissociates into 6 positive charges and 6 negative charges. The concept of equivalent concentration allows us to compare the conductances of different salts in a meaningful way. is defined similarly to molar conductivity \[Λ = \dfrac{κ}{c}\] except that the concentration term is now expressed in equivalents per liter instead of moles per liter. (In other words, the equivalent conductivity of an electrolyte is the conductance per equivalent per liter.) | 4,750 | 2,803 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/23%3A_The_Transition_Elements/23.1%3A_General_Properties_of_Transition_Metals |
We have daily contact with many transition metals. Iron occurs everywhere—from the rings in your spiral notebook and the cutlery in your kitchen to automobiles, ships, buildings, and in the hemoglobin in your blood. Titanium is useful in the manufacture of lightweight, durable products such as bicycle frames, artificial hips, and jewelry. Chromium is useful as a protective plating on plumbing fixtures and automotive detailing. Transition metals are defined as those elements that have (or readily form) partially filled orbitals. As shown in Figure \(\Page {2}\), the in groups 3–11 are transition elements. The , also called (the lanthanides and actinides), also meet this criterion because the orbital is partially occupied before the orbitals. The orbitals fill with the copper family (group 11); for this reason, the next family (group 12) are technically not transition elements. However, the group 12 elements do display some of the same chemical properties and are commonly included in discussions of transition metals. Some chemists do treat the group 12 elements as transition metals. The -block elements are divided into the (the elements Sc through Cu), the (the elements Y through Ag), and the (the element La and the elements Hf through Au). Actinium, Ac, is the first member of the , which also includes Rf through Rg. The -block elements are the elements Ce through Lu, which constitute the (or ), and the elements Th through Lr, which constitute the (or ). Because lanthanum behaves very much like the lanthanide elements, it is considered a lanthanide element, even though its electron configuration makes it the first member of the third transition series. Similarly, the behavior of actinium means it is part of the actinide series, although its electron configuration makes it the first member of the fourth transition series The elements of the second and third rows of the show gradual changes in properties across the table from left to right as expected. Electrons in the outer shells of the atoms of these elements have little shielding effects resulting in an increase in effective nuclear charge due to the addition of protons in the nucleus. Consequently, the effects on atomic properties are: smaller atomic radius, increased first ionization energy, enhanced electronegativity and more nonmetallic character. This trend continues until one reaches calcium (Z=20). There is an abrupt break at this point. The next ten elements called the first transition series are remarkably similar in their physical and chemical properties. This general similarity in properties has been explained in terms of their relatively small difference in effective nuclear charge over the series. This occurs because each additional electron enters the penultimate 3d shell providing an effective shield between the nucleus and the outer 4s shell. Thus, the transition elements can be defined as those in which the d electron shells are being filled and so we generally ignore Sc and Zn where Sc(III) is d and Zn(II) is d . It is useful, at the beginning, to identify the physical and chemical properties of transition elements which differ from main group elements (s-block). Properties of transition elements include: The transition elements are much denser than the s-block elements and show a gradual increase in density from scandium to copper. This trend in density can be explained by the small and irregular decrease in metallic radii coupled with the relative increase in atomic mass. The melting points and the molar enthalpies of fusion of the transition metals are both high in comparison to main group elements. This arises from strong metallic bonding in transition metals which occurs due to delocalization of electrons facilitated by the availability of both d and s electrons. In moving across the series of metals from scandium to zinc a small change in the values of the first and second ionization energies is observed. This is due to the build-up of electrons in the immediately underlying d-sub-shells that efficiently shields the 4s electrons from the nucleus and minimizing the increase in effective nuclear charge \(Z_{eff}\) from element to element. The increases in third and fourth ionization energy values are more rapid. However, the trends in these values show the usual discontinuity half way along the series. The reason is that the five d electrons are all unpaired, in singly occupied orbitals. When the sixth and subsequent electrons enter, the electrons have to share the already occupied orbitals resulting in inter-electron repulsions, which would require less energy to remove an electron. Hence, the third ionization energy curve for the last five elements is identical in shape to the curve for the first five elements, but displaced upwards by 580 kJ mol . Transition metals can form compounds with a wide range of oxidation states. Some of the observed oxidation states of the elements of the first transition series are shown in Figure \(\Page {5}\). As we move from left to right across the first transition series, we see that the number of common oxidation states increases at first to a maximum towards the middle of the table, then decreases. The values in the table are typical values; there are other known values, and it is possible to synthesize new additions. For example, in 2014, researchers were successful in synthesizing a new oxidation state of iridium (9+). For the elements scandium through manganese (the first half of the first transition series), the highest oxidation state corresponds to the loss of all of the electrons in both the and orbitals of their valence shells. The titanium(IV) ion, for example, is formed when the titanium atom loses its two 3 and two 4 electrons. These highest oxidation states are the most stable forms of scandium, titanium, and vanadium. However, it is not possible to continue to remove all of the valence electrons from metals as we continue through the series. Iron is known to form oxidation states from 2+ to 6+, with iron(II) and iron(III) being the most common. Most of the elements of the first transition series form ions with a charge of 2+ or 3+ that are stable in water, although those of the early members of the series can be readily oxidized by air. The elements of the second and third transition series generally are more stable in higher oxidation states than are the elements of the first series. In general, the atomic radius increases down a group, which leads to the ions of the second and third series being larger than are those in the first series. Removing electrons from orbitals that are located farther from the nucleus is easier than removing electrons close to the nucleus. For example, molybdenum and tungsten, members of group 6, are limited mostly to an oxidation state of 6+ in aqueous solution. Chromium, the lightest member of the group, forms stable Cr ions in water and, in the absence of air, less stable Cr ions. The sulfide with the highest oxidation state for chromium is Cr S , which contains the Cr ion. Molybdenum and tungsten form sulfides in which the metals exhibit oxidation states of 4+ and 6+. The electronic configuration of the atoms of the first row transition elements are basically the same. It can be seen in the Table above that there is a gradual filling of the 3d orbitals across the series starting from scandium. This filling is, however, not regular, since at chromium and copper the population of 3d orbitals increase by the acquisition of an electron from the 4s shell. This illustrates an important generalization about orbital energies of the first row transition series. At chromium, both the 3d and 4s orbitals are occupied, but neither is completely filled in preference to the other. This suggests that the energies of the 3d and 4s orbitals are relatively close for atoms in this row. In the case of copper, the 3d level is full, but only one electron occupies the 4s orbital. This suggests that in copper the 3d orbital energy is lower than the 4s orbital. Thus the 3d orbital energy has passed from higher to lower as we move across the period from potassium to zinc. However, the whole question of preference of an atom to adopt a particular electronic configuration is not determined by orbital energy alone. In chromium it can be shown that the 4s orbital energy is still below the 3d which suggests a configuration [Ar] 3d 4s . However due to the effect of electronic repulsion between the outer electrons the actual configuration becomes [Ar]3d 4s where all the electrons in the outer orbitals are unpaired. It should be remembered that the factors that determine electronic configuration in this period are indeed delicately balanced. This shows that elemental Mn is a stronger reductant than molecular hydrogen and hence should be able to displace hydrogen gas from 1 mol dm hydrochloric acid. Review how to write electron configurations, covered in the chapter on electronic structure and periodic properties of elements. Recall that for the transition and inner transition metals, it is necessary to remove the electrons before the or electrons. Then, for each ion, give the electron configuration: For the examples that are transition metals, determine to which series they belong. For ions, the -valence electrons are lost prior to the or electrons. Give an example of an ion from the first transition series with no electrons. V is one possibility. Other examples include Sc , Ti , Cr , and Mn . Transition metals demonstrate a wide range of chemical behaviors. As can be seen from their reduction potentials ( ), some transition metals are strong reducing agents, whereas others have very low reactivity. For example, the lanthanides all form stable 3+ aqueous cations. The driving force for such oxidations is similar to that of alkaline earth metals such as Be or Mg, forming Be and Mg . On the other hand, materials like platinum and gold have much higher reduction potentials. Their ability to resist oxidation makes them useful materials for constructing circuits and jewelry. Ions of the lighter -block elements, such as Cr , Fe , and Co , form colorful hydrated ions that are stable in water. However, ions in the period just below these (Mo , Ru , and Ir ) are unstable and react readily with oxygen from the air. The majority of simple, water-stable ions formed by the heavier -block elements are oxyanions such as \(\ce{MoO4^2-}\) and \(\ce{ReO4-}\). Ruthenium, osmium, rhodium, iridium, palladium, and platinum are the . With difficulty, they form simple cations that are stable in water, and, unlike the earlier elements in the second and third transition series, they do not form stable oxyanions. Both the - and -block elements react with nonmetals to form binary compounds; heating is often required. These elements react with halogens to form a variety of halides ranging in oxidation state from 1+ to 6+. On heating, oxygen reacts with all of the transition elements except palladium, platinum, silver, and gold. The oxides of these latter metals can be formed using other reactants, but they decompose upon heating. The -block elements, the elements of group 3, and the elements of the first transition series except copper react with aqueous solutions of acids, forming hydrogen gas and solutions of the corresponding salts. Which is the strongest oxidizing agent in acidic solution: dichromate ion, which contains chromium(VI), permanganate ion, which contains manganese(VII), or titanium dioxide, which contains titanium(IV)? First, we need to look up the reduction half reactions ( ) for each oxide in the specified oxidation state: \[\ce{Cr2O7^2- + 14H+ + 6e- ⟶ 2Cr^3+ + 7H2O} \hspace{20px} \mathrm{+1.33\: V}\] \[\ce{MnO4- + 8H+ + 5e- ⟶ Mn^2+ + H2O} \hspace{20px} \mathrm{+1.51\: V}\] \[\ce{TiO2 + 4H+ + 2e- ⟶ Ti^2+ + 2H2O} \hspace{20px} \mathrm{−0.50\: V}\] A larger reduction potential means that it is easier to reduce the reactant. Permanganate, with the largest reduction potential, is the strongest oxidizer under these conditions. Dichromate is next, followed by titanium dioxide as the weakest oxidizing agent (the hardest to reduce) of this set. Predict what reaction (if any) will occur between HCl and Co( ), and between HBr and Pt( ). You will need to use the standard reduction potentials from ( ). \(\ce{Co}(s)+\ce{2HCl}⟶\ce{H2}+\ce{CoCl2}(aq)\); no reaction because Pt( ) will not be oxidized by H ). | 12,529 | 2,804 |
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In addition to exerting an effect on the speed of reaction, substituents on the benzene ring also influence the regiochemistry of the reaction. That is, they control the new substituent appears in the product. Remember, there are three different position on the bezene ring where a new substituent can attach, relative to the original substituent. Substitution could actually occur on five positions around the ring, but two pairs are related by symmetry. Isomerism in disubstituted benzenes can be described by numbering the substituents (1,2- etc) or by the relationships -, - and -. There are two positions - to the initial substituent and two positions - to it. Ingold and colleagues investigated the question of regiochemistry in nitration. They reported the following observations: In looking at the table, you might see that there are two groups of substituents. One group reacts to make mixtures of - and - products. There may be different ratios of - to - and there may be small amounts of -, but don't get bogged down in the details. Focus on the bigger picture. Some groups are " -/ -directors". The other group reacts to makemostly -substituted products. here may be small amounts of - and - products, but don't worry about that. Focus on the bigger picture. Some groups are " -directors". These regiochemical effects are very closely related to the activating and directing effects we have already seen. If we want to understand this data, we need to think about things like π-donation, π-acceptance, inductive effects and cation stability. Show resonance structures for the cationic intermediate that results during nitration of toluene (methylbenzene). Explain why a mixture of and substitution results. Show resonance structures for the cationic intermediate that results during nitration of chlorobenzene. Explain why a mixture of and substitution results. Show resonance structures for the cationic intermediate that results during nitration of acetophenone (C H COCH ). Explain why mostly substitution results. Show resonance structures for the cationic intermediate that results during nitration of acetanilide (C H NH(CO)CH ). Explain why a mixture of and substitution results. In general, we can divide these substituents into three groups: Note that, once again, we may have two competing effects in one substituent, such as a halogen. In halogens, although the net effect may be to slow the reaction down, that weak π-donation is still enough to tilt the balance of products in favour of ortho- and para- substitution. Fill in the major organic products of the following reactions. Fill in the starting materials and reagents needed to obtain the major product shown via electrophilic aromatic substitution. Given two different substituents on a benzene, there can sometimes be a conflict in predicting which substitution pattern will result. Generally, the group with the stronger activating effect wins out. Predict the major products of the following reactions. , | 3,031 | 2,805 |
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Crystalline solids fall into one of four categories. All four categories involve packing discrete molecules or atoms into a lattice or repeating array, though are a special case. The categories are distinguished by the nature of the interactions holding the discrete molecules or atoms together. Based on the nature of the forces that hold the component atoms, molecules, or ions together, solids may be formally classified as ionic, molecular, covalent (network), or metallic. The variation in the relative strengths of these four types of interactions correlates nicely with their wide variation in properties. In ionic and molecular solids, there are no chemical bonds between the molecules, atoms, or ions. The solid consists of discrete chemical species held together by intermolecular forces that are or in nature. This behavior is most obvious for an ionic solid such as \(NaCl\), where the positively charged Na ions are attracted to the negatively charged \(Cl^-\) ions. Even in the absence of ions, however, electrostatic forces are operational. For polar molecules such as \(CH_2Cl_2\), the positively charged region of one molecular is attracted to the negatively charged region of another molecule ( ). For a nonpolar molecule such as \(CO_2\), which has no permanent dipole moment, the random motion of electrons gives rise to temporary polarity (a temporary dipole moment). Electrostatic attractions between two temporarily polarized molecules are called . is a term describing an attractive interaction between a hydrogen atom from a molecule or a molecular fragment X–H in which X is more electronegative than H, and an atom or a group of atoms in the same or a different molecule, in which there is evidence of bond formation. (See the Provisional Recommendation on the definition of a hydrogen bond.) Dots are employed to indicate the presence of a hydrogen bond: X–H•••Y. The attractive interaction in a hydrogen bond typically has a strong electrostatic contribution, but dispersion forces and weak covalent bonding are also present. In metallic solids and network solids, however, chemical bonds hold the individual chemical subunits together. The crystal is essential a single, macroscopic molecule with continuous chemical bonding throughout the entire structure. In metallic solids, the valence electrons are no longer exclusively associated with a single atom. Instead these electrons exist in molecular orbitals that are delocalized over many atoms, producing an electronic band structure. The metallic crystal essentially consists of a set of metal cations in a sea of electrons. This type of chemical bonding is called . You learned previously that an ionic solid consists of positively and negatively charged ions held together by electrostatic forces. The strength of the attractive forces depends on the charge and size of the ions that compose the lattice and determines many of the physical properties of the crystal. The (i.e., the energy required to separate 1 mol of a crystalline ionic solid into its component ions in the gas phase) is directly proportional to the product of the ionic charges and inversely proportional to the sum of the radii of the ions. For example, NaF and CaO both crystallize in the face-centered cubic (fcc) sodium chloride structure, and the sizes of their component ions are about the same: Na (102 pm) versus Ca (100 pm), and F (133 pm) versus O (140 pm). Because of the higher charge on the ions in CaO, however, the lattice energy of CaO is almost four times greater than that of NaF (3401 kJ/mol versus 923 kJ/mol). The forces that hold Ca and O together in CaO are much stronger than those that hold Na and F together in NaF, so the heat of fusion of CaO is almost twice that of NaF (59 kJ/mol versus 33.4 kJ/mol), and the melting point of CaO is 2927°C versus 996°C for NaF. In both cases, however, the values are large; that is, simple ionic compounds have high melting points and are relatively hard (and brittle) solids. Molecular solids consist of atoms or molecules held to each other by dipole–dipole interactions, London dispersion forces, or hydrogen bonds, or any combination of these. The arrangement of the molecules in solid benzene is as follows: Because the intermolecular interactions in a molecular solid are relatively weak compared with ionic and covalent bonds, molecular solids tend to be soft, low melting, and easily vaporized (\(ΔH_{fus}\) and \(ΔH_{vap}\) are low). For similar substances, the strength of the London dispersion forces increases smoothly with increasing molecular mass. For example, the melting points of benzene (C H ), naphthalene (C H ), and anthracene (C H ), with one, two, and three fused aromatic rings, are 5.5°C, 80.2°C, and 215°C, respectively. The enthalpies of fusion also increase smoothly within the series: benzene (9.95 kJ/mol) < naphthalene (19.1 kJ/mol) < anthracene (28.8 kJ/mol). If the molecules have shapes that cannot pack together efficiently in the crystal, however, then the melting points and the enthalpies of fusion tend to be unexpectedly low because the molecules are unable to arrange themselves to optimize intermolecular interactions. Thus toluene (C H CH ) and m-xylene [m-C H (CH ) ] have melting points of −95°C and −48°C, respectively, which are significantly lower than the melting point of the lighter but more symmetrical analog, benzene. Self-healing rubber is an example of a molecular solid with the potential for significant commercial applications. The material can stretch, but when snapped into pieces it can bond back together again through reestablishment of its hydrogen-bonding network without showing any sign of weakness. Among other applications, it is being studied for its use in adhesives and bicycle tires that will self-heal. Covalent solids are formed by networks or chains of atoms or molecules held together by covalent bonds. A perfect single crystal of a covalent solid is therefore a single giant molecule. For example, the structure of diamond, shown in part (a) in Figure \(\Page {1}\), consists of sp3 hybridized carbon atoms, each bonded to four other carbon atoms in a tetrahedral array to create a giant network. The carbon atoms form six-membered rings. The unit cell of diamond can be described as an fcc array of carbon atoms with four additional carbon atoms inserted into four of the tetrahedral holes. It thus has the zinc blende structure described in Section 12.3, except that in zinc blende the atoms that compose the fcc array are sulfur and the atoms in the tetrahedral holes are zinc. Elemental silicon has the same structure, as does silicon carbide (SiC), which has alternating C and Si atoms. The structure of crystalline quartz (SiO ), shown in Section 12.1, can be viewed as being derived from the structure of silicon by inserting an oxygen atom between each pair of silicon atoms. All compounds with the diamond and related structures are hard, high-melting-point solids that are not easily deformed. Instead, they tend to shatter when subjected to large stresses, and they usually do not conduct electricity very well. In fact, diamond (melting point = 3500°C at 63.5 atm) is one of the hardest substances known, and silicon carbide (melting point = 2986°C) is used commercially as an abrasive in sandpaper and grinding wheels. It is difficult to deform or melt these and related compounds because strong covalent (C–C or Si–Si) or polar covalent (Si–C or Si–O) bonds must be broken, which requires a large input of energy. Other covalent solids have very different structures. For example, graphite, the other common allotrope of carbon, has the structure shown in part (b) in Figure \(\Page {1}\). It contains planar networks of six-membered rings of sp2 hybridized carbon atoms in which each carbon is bonded to three others. This leaves a single electron in an unhybridized 2pz orbital that can be used to form C=C double bonds, resulting in a ring with alternating double and single bonds. Because of its resonance structures, the bonding in graphite is best viewed as consisting of a network of C–C single bonds with one-third of a π bond holding the carbons together, similar to the bonding in benzene. To completely describe the bonding in graphite, we need a molecular orbital approach similar to the one used for benzene in Chapter 9. In fact, the C–C distance in graphite (141.5 pm) is slightly longer than the distance in benzene (139.5 pm), consistent with a net carbon–carbon bond order of 1.33. In graphite, the two-dimensional planes of carbon atoms are stacked to form a three-dimensional solid; only London dispersion forces hold the layers together. As a result, graphite exhibits properties typical of both covalent and molecular solids. Due to strong covalent bonding within the layers, graphite has a very high melting point, as expected for a covalent solid (it actually sublimes at about 3915°C). It is also very soft; the layers can easily slide past one another because of the weak interlayer interactions. Consequently, graphite is used as a lubricant and as the “lead” in pencils; the friction between graphite and a piece of paper is sufficient to leave a thin layer of carbon on the paper. Graphite is unusual among covalent solids in that its electrical conductivity is very high parallel to the planes of carbon atoms because of delocalized C–C π bonding. Finally, graphite is black because it contains an immense number of alternating double bonds, which results in a very small energy difference between the individual molecular orbitals. Thus light of virtually all wavelengths is absorbed. Diamond, on the other hand, is colorless when pure because it has no delocalized electrons. Table \(\Page {2}\) compares the strengths of the intermolecular and intramolecular interactions for three covalent solids, showing the comparative weakness of the interlayer interactions. In network solids, conventional chemical bonds hold the chemical subunits together. The bonding between chemical subunits, however, is identical to that within the subunits, resulting in a continuous network of chemical bonds. One common examples of network solids are diamond (a form of pure carbon) Carbon exists as a pure element at room temperature in three different forms: graphite (the most stable form), diamond, and fullerene. The structure of diamond is shown at the right in a "ball-and-stick" format. The balls represent the carbon atoms and the sticks represent a covalent bond. Be aware that in the "ball-and-stick" representation the size of the balls do not accurately represent the size of carbon atoms. In addition, a single stick is drawn to represent a covalent bond irrespective of whether the bond is a single, double, or triple bond or requires resonance structures to represent. In the diamond structure, all bonds are single covalent bonds (\(\sigma\) bonds). The "space-filling" format is an alternate representation that displays atoms as spheres with a radius equal to the van der Waals radius, thus providing a better sense of the size of the atoms. Notice that diamond is a network solid. The entire solid is an "endless" repetition of carbon atoms bonded to each other by covalent bonds. (In the display at the right, the structure is truncated to fit in the display area.) The most stable form of carbon is graphite. Graphite consists of sheets of carbon atoms covalently bonded together. These sheets are then stacked to form graphite. Figure \(\Page {3}\) shows a ball-and-stick representation of graphite with sheets that extended "indefinitely" in the xy plane, but the structure has been truncated for display purposed. Graphite may also be regarded as a network solid, even though there is no bonding in the z direction. Each layer, however, is an "endless" bonded network of carbon atoms. Until the mid 1980's, pure carbon was thought to exist in two forms: graphite and diamond. The discovery of C molecules in interstellar dust in 1985 added a third form to this list. The existence of C , which resembles a soccer ball, had been hypothesized by theoreticians for many years. In the late 1980's synthetic methods were developed for the synthesis of C , and the ready availability of this form of carbon led to extensive research into its properties. The C molecule (Figure \(\Page {4}\); left), is called buckminsterfullerene, though the shorter name fullerene is often used. The name is a tribute to the American architect R. Buckminster Fuller, who is famous for designing and constructing geodesic domes which bear a close similarity to the structure of C . As is evident from the display, C is a sphere composed of six-member and five-member carbon rings. These balls are sometimes fondly referred to as "Bucky balls". It should be noted that fullerenes are an entire class of pure carbon compounds rather than a single compound. A distorted sphere containing more than 60 carbon atoms have also been found, and it is also possible to create long tubes (Figure \(\Page {4}\); right). All of these substances are pure carbon. such as crystals of copper, aluminum, and iron are formed by metal atoms Figure \(\Page {5}\). The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a “sea” of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their malleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials. Metals are characterized by their ability to reflect light, called luster, their high electrical and thermal conductivity, their high heat capacity, and their malleability and ductility. Every lattice point in a pure metallic element is occupied by an atom of the same metal. The packing efficiency in metallic crystals tends to be high, so the resulting metallic solids are dense, with each atom having as many as 12 nearest neighbors. Bonding in metallic solids is quite different from the bonding in the other kinds of solids we have discussed. Because all the atoms are the same, there can be no ionic bonding, yet metals always contain too few electrons or valence orbitals to form covalent bonds with each of their neighbors. Instead, the valence electrons are delocalized throughout the crystal, providing a strong cohesive force that holds the metal atoms together. Valence electrons in a metallic solid are delocalized, providing a strong cohesive force that holds the atoms together. The strength of metallic bonds varies dramatically. For example, cesium melts at 28.4°C, and mercury is a liquid at room temperature, whereas tungsten melts at 3680°C. Metallic bonds tend to be weakest for elements that have nearly empty (as in Cs) or nearly full (Hg) valence subshells, and strongest for elements with approximately half-filled valence shells (as in W). As a result, the melting points of the metals increase to a maximum around group 6 and then decrease again from left to right across the d block. Other properties related to the strength of metallic bonds, such as enthalpies of fusion, boiling points, and hardness, have similar periodic trends. A somewhat oversimplified way to describe the bonding in a metallic crystal is to depict the crystal as consisting of positively charged nuclei in an electron sea (Figure \(\Page {6}\)). In this model, the valence electrons are not tightly bound to any one atom but are distributed uniformly throughout the structure. Very little energy is needed to remove electrons from a solid metal because they are not bound to a single nucleus. When an electrical potential is applied, the electrons can migrate through the solid toward the positive electrode, thus producing high electrical conductivity. The ease with which metals can be deformed under pressure is attributed to the ability of the metal ions to change positions within the electron sea without breaking any specific bonds. The transfer of energy through the solid by successive collisions between the metal ions also explains the high thermal conductivity of metals. This model does not, however, explain many of the other properties of metals, such as their metallic luster and the observed trends in bond strength as reflected in melting points or enthalpies of fusion. Some general properties of the four major classes of solids are summarized in Table \(\Page {2}\). The general order of increasing strength of interactions in a solid is: Classify Ge, RbI, C (CH ) , and Zn as ionic, molecular, covalent, or metallic solids and arrange them in order of increasing melting points. : compounds : classification and order of melting points : : Germanium lies in the p block just under Si, along the diagonal line of semimetallic elements, which suggests that elemental Ge is likely to have the same structure as Si (the diamond structure). Thus Ge is probably a covalent solid. RbI contains a metal from group 1 and a nonmetal from group 17, so it is an ionic solid containing Rb and I ions. The compound C6(CH3)6 is a hydrocarbon (hexamethylbenzene), which consists of isolated molecules that stack to form a molecular solid with no covalent bonds between them. Zn is a d-block element, so it is a metallic solid. Arranging these substances in order of increasing melting points is straightforward, with one exception. We expect C (CH ) to have the lowest melting point and Ge to have the highest melting point, with RbI somewhere in between. The melting points of metals, however, are difficult to predict based on the models presented thus far. Because Zn has a filled valence shell, it should not have a particularly high melting point, so a reasonable guess is C (CH ) < Zn ~ RbI < Ge. The actual melting points are C (CH ) , 166°C; Zn, 419°C; RbI, 642°C; and Ge, 938°C. This agrees with our prediction. Classify C , BaBr , GaAs, and AgZn as ionic, covalent, molecular, or metallic solids and then arrange them in order of increasing melting points. C (molecular) < AgZn (metallic) ~ BaBr (ionic) < GaAs (covalent). The actual melting points are C60, about 300°C; AgZn, about 700°C; BaBr , 856°C; and GaAs, 1238°C. The major types of solids are ionic, molecular, covalent, and metallic. Ionic solids consist of positively and negatively charged ions held together by electrostatic forces; the strength of the bonding is reflected in the lattice energy. Ionic solids tend to have high melting points and are rather hard. Molecular solids are held together by relatively weak forces, such as dipole–dipole interactions, hydrogen bonds, and London dispersion forces. As a result, they tend to be rather soft and have low melting points, which depend on their molecular structure. Covalent solids consist of two- or three-dimensional networks of atoms held together by covalent bonds; they tend to be very hard and have high melting points. Metallic solids have unusual properties: in addition to having high thermal and electrical conductivity and being malleable and ductile, they exhibit luster, a shiny surface that reflects light. An alloy is a mixture of metals that has bulk metallic properties different from those of its constituent elements. Alloys can be formed by substituting one metal atom for another of similar size in the lattice (substitutional alloys), by inserting smaller atoms into holes in the metal lattice (interstitial alloys), or by a combination of both. Although the elemental composition of most alloys can vary over wide ranges, certain metals combine in only fixed proportions to form intermetallic compounds with unique properties. ( ) ( ) ). | 19,893 | 2,806 |
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The strongest activating and ortho/para-directing substituents are the amino (-NH ) and hydroxyl (-OH) groups. Direct nitration of phenol (hydroxybenzene) by dilute nitric acid gives modest yields of nitrated phenols and considerable oxidative decomposition to tarry materials; aniline (aminobenzene) is largely destroyed. Bromination of both phenol and aniline is difficult to control, with di- and tri-bromo products forming readily. Because of their high nucleophilic reactivity, aniline and phenol undergo substitution reactions with iodine, a halogen that is normally unreactive with benzene derivatives. The mixed halogen iodine chloride (ICl) provides a more electrophilic iodine moiety, and is effective in iodinating aromatic rings having less powerful activating substituents. By acetylating the heteroatom substituent on phenol and aniline, its activating influence can be substantially attenuated. For example, acetylation of aniline gives acetanilide (first step in the following equation), which undergoes nitration at low temperature, yielding the para-nitro product in high yield. The modifying acetyl group can then be removed by acid-catalyzed hydrolysis (last step), to yield para-nitroaniline. Although the activating influence of the amino group has been reduced by this procedure, the acetyl derivative remains an ortho/para-directing and activating substituent. | 1,396 | 2,807 |
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Alkenes can be treated with aqueous acids or, much more efficiently, with aqueous mercuric salts, followed by sodium borohydride, to produce alcohols. In those cases, the hydroxyl group is found in different places. Treatment with acid often results in a mixture of alcohols in which the OH groups are found in the most substituted positions in the structure, regardless of the position of the original alkene. Oxymercuration - demercuration results in the hydroxyl being fixed at the most substituted end of the former C=C bond. Those reactions are generally called hydration reactions because they result in the overall addition of H-OH across the double bond. Hydroboration - oxidation is a two-step sequence of reactions that also results in hydration of a double bond. However, this reaction is complementary to oxymercuration - demercuration. Instead of leaving an OH group at the most substituted end of the double bond, the hydroxy group is placed at the least substituted end of the double bond. Let's modify that statement a little bit. In reality, the reaction scheme above just shows the major product. The minor product has the hydroxy group at the more substituted position of the double bond. These two products might be found in different ratios, maybe even as close as 55:45, but the least substituted product always predominates. We will see more efficient hydroboration methods soon, leading to ratios above 95:5, or almost entirely the least substituted product. The product of non-hydrogen addition (i.e. OH group addition) at the most substituted end of the alkene is called a Markovnikov addition product. The product of non-hydrogen addition at the least substituted end of the alkene is called an anti-Markovnikov addition product. This selectivity is important in synthetic applications. We use natural products all the time as pharmaceuticals, vitamins and other health and beauty applications, but we can't always obtain these compounds directly from nature, for a number of reasons. It could be that the organism needs to be killed in order to harvest its products, or that there isn't enough of the source in nature ro meet demand. Frequently it is more economical to produce commercially useful compounds from convenient chemical feedstocks. Over the last century and a half, those feedstocks have come from coal tar and, later, petroleum. Currently, there is rapid progress underway to develop chemical feedstocks from sources such as vegetable and algal oil (i.e. oil from seaweed). These feedstocks are just compounds that can be converted synthetically into pharmaceuticals as well as in plastics, paints, coatings and other materials. Frequently, the starting materials for these processes contain C=C bonds that can be functionalized through electrophilic addition. Thus, electrophilic addition and related reactions are among the most important in the world, economically speaking. It's very valuable to be able to control the outcome of these reactions in order to make processes more efficient, producing fewer wasteful by-products. List advantages and disadvantages of producing materials based on Again, we are going to focus on the first of the two reactions in this sequence. That part is where the placement of the new substituent is decided. After the addition of the borane, an alkylborane is formed. The major isomer results from anti-Markovnikov addition. It seems pretty clear at this point that this reaction must proceed like other electrophilic additions to alkenes. The π electrons donate to the electrophile. In this case that's boron, which is strongly Lewis acidic because it lacks an octet. The boron ends up at the least substituted end of the double bond. That outcome would certainly be favoured over this one: That boron is beginning to look less electrophilic and more nucleophilic. We can easily imagine a hydride nucleophile being delivered to the carbocation. So far, the picture of how the alkylboration reaction works fits pretty well within our electrophilic addition framework. Unfortunately, there are some problems with this model. First of all, maybe 55% of the boration takes place in a Markovnikov sense, but the other 45% is added to form an anti-Markovnikov product. Certainly the secondary cation is favoured over the primary one, but if the reaction is proceeding through a carbocation, then the primary one shouldn't happen at all. Something is wrong with our model. Another hole is torn in the argument when we look at the results of stereochemical studies. We could, for example, take the following deuterium-labelled hexene and treat it with borane. We could look at the products via H NMR spectroscopy, and if we could see the coupling constant between the two protons shown in the structure, then we would know their relative arrangement in space. We would know their stereochemistry. If we did that experiment, then we would see that the hydrogen and the boron from the borane are added to the same face of the alkene. We don't get addition of boron to one face and hydrogen to the other. This type of addition is called addition; it is the opposite of addition. Show how a cationic intermediate and conformational changes would allow both and addition of borane to propene. That result means that, although some elements of our mechanism may reflect reality, we at least have a problem with timing. How can the hydride be delivered before the conformation has a chance to change? It has to happen pretty quickly. What if it happens at the same time as π bond donation to the boron? This reaction would best be described as a concerted addition. Two groups are added to the two ends of the double bond at the same time. During the transition state, two bonds would be breaking and two would be forming at the same time. We can further improve our model of how the alkylboration works if we consider that disiamylborane and 9-BBN are much more effecient than borane in terms of regioselectivity. These reagents can produce close to 100% anti-Markovnikov addition. Explain how with the help of drawings. The subsequent reaction in this series involves removal of the boron and replacement with a hydroxyl group. The mechanism of this reaction may not be worth memorizing becuase it doesn't fit well within categories we have looked at so far. The important thing to know is that the oxygen ends up in exactly the same place as the boron. There is no change in stereochemistry at that position. Overall, the hydrogen and the hydroxy effectively group undergo addition, although they are added in different steps. Borane is frequently used in THF because borane alone is not very stable; it is quite pyrophoric, bursting into flame upon contact with air. In THF, borane forms an equilibrium with a Lewis-acid-base complex. Show this equilibrium reaction. Show products of the following reactions. Provide reagents for the following reactions , | 6,949 | 2,808 |
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There are several different ways to quantitatively describe the concentration of a solution. For example, molarity is a useful way to describe solution concentrations for reactions that are carried out in solution. Mole fractions are used not only to describe gas concentrations but also to determine the vapor pressures of mixtures of similar liquids. Example \(\Page {1}\) reviews the methods for calculating the molarity and mole fraction of a solution when the masses of its components are known. Commercial vinegar is essentially a solution of acetic acid in water. A bottle of vinegar has 3.78 g of acetic acid per 100.0 g of solution. Assume that the density of the solution is 1.00 g/mL. : mass of substance and mass and density of solution : molarity and mole fraction : : The molarity is the number of moles of acetic acid per liter of solution. We can calculate the number of moles of acetic acid as its mass divided by its molar mass. \[ \begin{align*} \text{moles } \ce{CH_3CO_2H} &=\dfrac{3.78\; \cancel{\ce{g}}\; \ce{CH_3CO_2H}}{60.05\; \cancel{\ce{g}}/\ce{mol}} \\[4pt] &=0.0629 \; \ce{mol} \end{align*} \nonumber \] The volume of the solution equals its mass divided by its density. \[ \begin{align*} \text{volume} &=\dfrac{\text{mass}}{\text{density}} \\[4pt] &=\dfrac{100.0\; \cancel{\ce{g}}\; \text{solution}}{1.00\; \cancel{\ce{g}}/\ce{mL}}=100\; mL\nonumber \end{align*} \nonumber \] Then calculate the molarity directly. \[ \begin{align*} \text{molarity of } \ce{CH_3CO_2H} &=\dfrac{\text{moles } \ce{CH3CO2H} }{\text{liter solution}} \\[4pt] &=\dfrac{0.0629\; mol\; \ce{CH_3CO_2H}}{(100\; \cancel{\ce{mL}})(1\; L/1000\; \cancel{\ce{mL}})}=0.629\; M \; \ce{CH_3CO_2H} \end{align*} \nonumber \] This result makes intuitive sense. If 100.0 g of aqueous solution (equal to 100 mL) contains 3.78 g of acetic acid, then 1 L of solution will contain 37.8 g of acetic acid, which is a little more than \(\ce{ 1/2}\) mole. Keep in mind, though, that the mass and volume of a solution are related by its density; concentrated aqueous solutions often have densities greater than 1.00 g/mL. To calculate the mole fraction of acetic acid in the solution, we need to know the number of moles of both acetic acid and water. The number of moles of acetic acid is 0.0629 mol, as calculated in part (a). We know that 100.0 g of vinegar contains 3.78 g of acetic acid; hence the solution also contains (100.0 g − 3.78 g) = 96.2 g of water. We have \[moles\; \ce{H_2O}=\dfrac{96.2\; \cancel{\ce{g}}\; \ce{H_2O}}{18.02\; \cancel{\ce{g}}/mol}=5.34\; mol\; \ce{H_2O}\nonumber \] The mole fraction \(\chi\) of acetic acid is the ratio of the number of moles of acetic acid to the total number of moles of substances present: \[ \begin{align*} \chi_{\ce{CH3CO2H}} &=\dfrac{moles\; \ce{CH_3CO_2H}}{moles \; \ce{CH_3CO_2H} + moles\; \ce{H_2O}} \\[4pt] &=\dfrac{0.0629\; mol}{0.0629 \;mol + 5.34\; mol} \\[4pt] &=0.0116=1.16 \times 10^{−2} \end{align*} \nonumber \] This answer makes sense, too. There are approximately 100 times as many moles of water as moles of acetic acid, so the ratio should be approximately 0.01. A solution of \(\ce{HCl}\) gas dissolved in water (sold commercially as “muriatic acid,” a solution used to clean masonry surfaces) has 20.22 g of \(\ce{HCl}\) per 100.0 g of solution, and its density is 1.10 g/mL. 6.10 M HCl \(\chi_{HCl} = 0.111\) The concentration of a solution can also be described by its molality (m), the number of moles of solute per kilogram of solvent: \[ \text{molality (m)} =\dfrac{\text{moles solute}}{\text{kilogram solvent}} \label{Eq1} \] Molality, therefore, has the same numerator as molarity (the number of moles of solute) but a different denominator (kilogram of solvent rather than liter of solution). For dilute aqueous solutions, the molality and molarity are nearly the same because dilute solutions are mostly solvent. Thus because the density of water under standard conditions is very close to 1.0 g/mL, the volume of 1.0 kg of \(H_2O\) under these conditions is very close to 1.0 L, and a 0.50 M solution of \(KBr\) in water, for example, has approximately the same concentration as a 0.50 m solution. Another common way of describing concentration is as the ratio of the mass of the solute to the total mass of the solution. The result can be expressed as mass percentage, parts per million (ppm), or parts per billion (ppb): \[ \begin{align} \text{mass percentage}&=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 100 \label{Eq2} \\[4pt] \text{parts per million (ppm)} &=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{6} \label{Eq3} \\[4pt] \text{parts per billion (ppb)}&=\dfrac{\text{mass of solute}}{\text{mass of solution}} \times 10^{9} \label{Eq4} \end{align} \] In the health sciences, the concentration of a solution is often expressed as parts per thousand (ppt), indicated as a proportion. For example, adrenalin, the hormone produced in high-stress situations, is available in a 1:1000 solution, or one gram of adrenalin per 1000 g of solution. The labels on bottles of commercial reagents often describe the contents in terms of mass percentage. Sulfuric acid, for example, is sold as a 95% aqueous solution, or 95 g of \(\ce{H_2SO_4}\) per 100 g of solution. Parts per million and parts per billion are used to describe concentrations of highly dilute solutions. These measurements correspond to milligrams and micrograms of solute per kilogram of solution, respectively. For dilute aqueous solutions, this is equal to milligrams and micrograms of solute per liter of solution (assuming a density of 1.0 g/mL). Several years ago, millions of bottles of mineral water were contaminated with benzene at ppm levels. This incident received a great deal of attention because the lethal concentration of benzene in rats is 3.8 ppm. A 250 mL sample of mineral water has 12.7 ppm of benzene. Because the contaminated mineral water is a very dilute aqueous solution, we can assume that its density is approximately 1.00 g/mL. : volume of sample, solute concentration, and density of solution : molarity of solute and mass of solute in 250 mL : : a. A To calculate the molarity of benzene, we need to determine the number of moles of benzene in 1 L of solution. We know that the solution contains 12.7 ppm of benzene. Because 12.7 ppm is equivalent to 12.7 mg/1000 g of solution and the density of the solution is 1.00 g/mL, the solution contains 12.7 mg of benzene per liter (1000 mL). The molarity is therefore \[\begin{align*} \text{molarity}&=\dfrac{\text{moles}}{\text{liter solution}} \\[4pt] &=\dfrac{(12.7\; \cancel{mg}) \left(\frac{1\; \cancel{g}}{1000\; \cancel{mg}}\right)\left(\frac{1\; mol}{78.114\; \cancel{g}}\right)}{1.00\; L} \\[4pt] &=1.63 \times 10^{-4} M\end{align*} \nonumber \] b. B We are given that there are 12.7 mg of benzene per 1000 g of solution, which is equal to 12.7 mg/L of solution. Hence the mass of benzene in 250 mL (250 g) of solution is \[\begin{align*} \text{mass of benzene} &=\dfrac{(12.7\; mg\; \text{benzene})(250\; \cancel{mL})}{1000\; \cancel{mL}} \\[4pt] &=3.18\; mg \\[4pt] &=3.18 \times 10^{-3}\; g\; \text{benzene} \end{align*} \nonumber \] The maximum allowable concentration of lead in drinking water is 9.0 ppb. 4.3 × 10 M 2 × 10 g How do chemists decide which units of concentration to use for a particular application? Although molarity is commonly used to express concentrations for reactions in solution or for titrations, it does have one drawback—molarity is the number of moles of solute divided by the volume of the solution, and the volume of a solution depends on its density, which is a function of temperature. Because volumetric glassware is calibrated at a particular temperature, typically 20°C, the molarity may differ from the original value by several percent if a solution is prepared or used at a significantly different temperature, such as 40°C or 0°C. For many applications this may not be a problem, but for precise work these errors can become important. In contrast, mole fraction, molality, and mass percentage depend on only the masses of the solute and solvent, which are independent of temperature. Mole fraction is not very useful for experiments that involve quantitative reactions, but it is convenient for calculating the partial pressure of gases in mixtures, as discussed previously. Mole fractions are also useful for calculating the vapor pressures of certain types of solutions. Molality is particularly useful for determining how properties such as the freezing or boiling point of a solution vary with solute concentration. Because mass percentage and parts per million or billion are simply different ways of expressing the ratio of the mass of a solute to the mass of the solution, they enable us to express the concentration of a substance even when the molecular mass of the substance is unknown. Units of ppb or ppm are also used to express very low concentrations, such as those of residual impurities in foods or of pollutants in environmental studies. Table \(\Page {1}\) summarizes the different units of concentration and typical applications for each. When the molar mass of the solute and the density of the solution are known, it becomes relatively easy with practice to convert among the units of concentration we have discussed, as illustrated in Example \(\Page {3}\). Vodka is essentially a solution of ethanol in water. Typical vodka is sold as “80 proof,” which means that it contains 40.0% ethanol by volume. The density of pure ethanol is 0.789 g/mL at 20°C. If we assume that the volume of the solution is the sum of the volumes of the components (which is not strictly correct), calculate the following for the ethanol in 80-proof vodka. : volume percent and density : mass percentage, mole fraction, molarity, and molality : : The key to this problem is to use the density of pure ethanol to determine the mass of ethanol (\(CH_3CH_2OH\)), abbreviated as EtOH, in a given volume of solution. We can then calculate the number of moles of ethanol and the concentration of ethanol in any of the required units. A Because we are given a percentage by volume, we assume that we have 100.0 mL of solution. The volume of ethanol will thus be 40.0% of 100.0 mL, or 40.0 mL of ethanol, and the volume of water will be 60.0% of 100.0 mL, or 60.0 mL of water. The mass of ethanol is obtained from its density: \[mass\; of\; EtOH=(40.0\; \cancel{mL})\left(\dfrac{0.789\; g}{\cancel{mL}}\right)=31.6\; g\; EtOH\nonumber \] If we assume the density of water is 1.00 g/mL, the mass of water is 60.0 g. We now have all the information we need to calculate the concentration of ethanol in the solution. B The mass percentage of ethanol is the ratio of the mass of ethanol to the total mass of the solution, expressed as a percentage: \[ \begin{align*} \%EtOH &=\left(\dfrac{mass\; of\; EtOH}{mass\; of\; solution}\right)(100) \\[4pt] &=\left(\dfrac{31.6\; \cancel{g}\; EtOH}{31.6\; \cancel{g} \;EtOH +60.0\; \cancel{g} \; H_2O} \right)(100) \\[4pt]&= 34.5\%\end{align*} \nonumber \] C The mole fraction of ethanol is the ratio of the number of moles of ethanol to the total number of moles of substances in the solution. Because 40.0 mL of ethanol has a mass of 31.6 g, we can use the molar mass of ethanol (46.07 g/mol) to determine the number of moles of ethanol in 40.0 mL: \[ \begin{align*} moles\; \ce{CH_3CH_2OH}&=(31.6\; \cancel{g\; \ce{CH_3CH_2OH}}) \left(\dfrac{1\; mol}{46.07\; \cancel{g\; \ce{CH_3CH_2OH}}}\right) \\[4pt] &=0.686 \;mol\; \ce{CH_3CH_2OH} \end{align*} \nonumber \] Similarly, the number of moles of water is \[ moles \;\ce{H_2O}=(60.0\; \cancel{g \; \ce{H_2O}}) \left(\dfrac{1 \;mol\; \ce{H_2O}}{18.02\; \cancel{g\; \ce{H_2O}}}\right)=3.33\; mol\; \ce{H_2O}\nonumber \] The mole fraction of ethanol is thus \[ \chi_{\ce{CH_3CH_2OH}}=\dfrac{0.686\; \cancel{mol}}{0.686\; \cancel{mol} + 3.33\;\cancel{ mol}}=0.171\nonumber \] D The molarity of the solution is the number of moles of ethanol per liter of solution. We already know the number of moles of ethanol per 100.0 mL of solution, so the molarity is The molality of the solution is the number of moles of ethanol per kilogram of solvent. Because we know the number of moles of ethanol in 60.0 g of water, the calculation is again straightforward: \[ m_{\ce{CH_3CH_2OH}}=\left(\dfrac{0.686\; mol\; EtOH}{60.0\; \cancel{g}\; H_2O } \right) \left(\dfrac{1000\; \cancel{g}}{kg}\right)=\dfrac{11.4\; mol\; EtOH}{kg\; H_2O}=11.4 \;m\nonumber \] A solution is prepared by mixing 100.0 mL of toluene with 300.0 mL of benzene. The densities of toluene and benzene are 0.867 g/mL and 0.874 g/mL, respectively. Assume that the volume of the solution is the sum of the volumes of the components. Calculate the following for toluene. mass percentage toluene = 24.8% \(\chi_{toluene} = 0.219\) 2.35 M toluene 3.59 m toluene A Discussing Different Measures of Concentration. Link: A Discussing how to Convert Measures of Concentration. Link: Different units are used to express the concentrations of a solution depending on the application. The concentration of a solution is the quantity of solute in a given quantity of solution. It can be expressed in several ways: molarity (moles of solute per liter of solution); mole fraction, the ratio of the number of moles of solute to the total number of moles of substances present; mass percentage, the ratio of the mass of the solute to the mass of the solution times 100; parts per thousand (ppt), grams of solute per kilogram of solution; parts per million (ppm), milligrams of solute per kilogram of solution; parts per billion (ppb), micrograms of solute per kilogram of solution; and molality (m), the number of moles of solute per kilogram of solvent. | 13,802 | 2,809 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/20%3A_Electrochemistry/20.4%3A_Cell_Potential_as_a_Function_of_Concentrations |
The enables the determination of cell potential under non-standard conditions. It relates the measured cell potential to the and allows the accurate determination of equilibrium constants (including solubility constants). Recall that the actual free-energy change for a reaction under nonstandard conditions, \(\Delta{G}\), is given as follows: \[\Delta{G} = \Delta{G°} + \ln Q \label{Eq1} \] We also know that \(ΔG = −nFE_{cell}\) (under non-standard confitions) and \(ΔG^o = −nFE^o_{cell}\) (under standard conditions). Substituting these expressions into Equation \(\ref{Eq1}\), we obtain \[−nFE_{cell} = −nFE^o_{cell} + RT \ln Q \label{Eq2} \] Dividing both sides of this equation by \(−nF\), \[E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln Q \label{Eq3} \] Equation \(\ref{Eq3}\) is called the , after the German physicist and chemist Walter Nernst (1864–1941), who first derived it. The Nernst equation is arguably the most important relationship in electrochemistry. When a redox reaction is at equilibrium (\(ΔG = 0\)), then Equation \(\ref{Eq3}\) reduces to Equation \(\ref{Eq31}\) and \(\ref{Eq32}\) because \(Q = K\), and there is no net transfer of electrons (i.e., E = 0). \[E_\textrm{cell}=E^\circ_\textrm{cell}-\left(\dfrac{RT}{nF}\right)\ln K = 0 \label{Eq31} \] since \[E^\circ_\textrm{cell}= \left(\dfrac{RT}{nF}\right)\ln K \label{Eq32} \] \(\ref{Eq3}\) \[E_{\textrm{cell}}=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \label{Eq4} \] The Nernst Equation (\(\ref{Eq3}\)) can be used to determine the value of E , and thus the direction of spontaneous reaction, for any redox reaction under any conditions. Equation \(\ref{Eq4}\) allows us to calculate the potential associated with any electrochemical cell at 298 K for any combination of reactant and product concentrations under any conditions. We can therefore determine the spontaneous direction of any redox reaction under any conditions, as long as we have tabulated values for the relevant standard electrode potentials. Notice in Equation \(\ref{Eq4}\) that the cell potential changes by 0.0591/n V for each 10-fold change in the value of \(Q\) because log 10 = 1. The following reaction proceeds spontaneously under standard conditions because E° > 0 (which means that ΔG° < 0): \[\ce{2Ce^{4+}(aq) + 2Cl^{–}(aq) -> 2Ce^{3+}(aq) + Cl2(g)}\;\; E^°_{cell} = 0.25\, V \nonumber \] Calculate \(E_{cell}\) for this reaction under the following nonstandard conditions and determine whether it will occur spontaneously: [Ce ] = 0.013 M, [Ce ] = 0.60 M, [Cl ] = 0.0030 M, \(P_\mathrm{Cl_2}\) = 1.0 atm, and T = 25°C. balanced redox reaction, standard cell potential, and nonstandard conditions cell potential Determine the number of electrons transferred during the redox process. Then use the Nernst equation to find the cell potential under the nonstandard conditions. We can use the information given and the Nernst equation to calculate E . Moreover, because the temperature is 25°C (298 K), we can use Equation \(\ref{Eq4}\) instead of Equation \(\ref{Eq3}\). The overall reaction involves the net transfer of two electrons: \[2Ce^{4+}_{(aq)} + 2e^− \rightarrow 2Ce^{3+}_{(aq)}\nonumber \] \[2Cl^−_{(aq)} \rightarrow Cl_{2(g)} + 2e^−\nonumber \] so n = 2. Substituting the concentrations given in the problem, the partial pressure of Cl , and the value of E° into Equation \(\ref{Eq4}\), \[\begin{align*}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \\
& =\textrm{0.25 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Ce^{3+}}]^2P_\mathrm{Cl_2}}{[\mathrm{Ce^{4+}}]^2[\mathrm{Cl^-}]^2}\right) \\
& =\textrm{0.25 V}-[(\textrm{0.0296 V})(8.37)]=\textrm{0.00 V}\end{align*} \nonumber \] Thus the reaction will not occur spontaneously under these conditions (because E = 0 V and ΔG = 0). The composition specified is that of an equilibrium mixture Molecular oxygen will not oxidize \(MnO_2\) to permanganate via the reaction \[\ce{4MnO2(s) + 3O2(g) + 4OH^{−} (aq) -> 4MnO^{−}4(aq) + 2H2O(l)} \;\;\; E°_{cell} = −0.20\; V\nonumber \] Calculate \(E_{cell}\) for the reaction under the following nonstandard conditions and decide whether the reaction will occur spontaneously: pH 10, \(P_\mathrm{O_2}\)= 0.20 atm, [MNO ] = 1.0 × 10 M, and T = 25°C. E = −0.22 V; the reaction will not occur spontaneously. Applying the Nernst equation to a simple electrochemical cell such as the Zn/Cu cell allows us to see how the cell voltage varies as the reaction progresses and the concentrations of the dissolved ions change. Recall that the overall reaction for this cell is as follows: \[Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)\;\;\;E°cell = 1.10 V \label{Eq5} \] The reaction quotient is therefore \(Q = [Zn^{2+}]/[Cu^{2+}]\). Suppose that the cell initially contains 1.0 M Cu and 1.0 × 10 M Zn . The initial voltage measured when the cell is connected can then be calculated from Equation \(\ref{Eq4}\): \[\begin{align}E_\textrm{cell} & =E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log\dfrac{[\mathrm{Zn^{2+}}]}{[\mathrm{Cu^{2+}}]}\\
& =\textrm{1.10 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{1.0\times10^{-6}}{1.0}\right)=\textrm{1.28 V}\end{align} \label{Eq6} \] Thus the initial voltage is greater than E° because \(Q<1\). As the reaction proceeds, [Zn ] in the anode compartment increases as the zinc electrode dissolves, while [Cu ] in the cathode compartment decreases as metallic copper is deposited on the electrode. During this process, the ratio Q = [Zn ]/[Cu ] steadily increases, and the cell voltage therefore steadily decreases. Eventually, [Zn ] = [Cu ], so Q = 1 and E = E° . Beyond this point, [Zn ] will continue to increase in the anode compartment, and [Cu ] will continue to decrease in the cathode compartment. Thus the value of Q will increase further, leading to a further decrease in E . When the concentrations in the two compartments are the opposite of the initial concentrations (i.e., 1.0 M Zn and 1.0 × 10 M Cu ), Q = 1.0 × 10 , and the cell potential will be reduced to 0.92 V. The variation of E with \(\log{Q}\) over this range is linear with a slope of −0.0591/n, as illustrated in Figure \(\Page {1}\). As the reaction proceeds still further, \(Q\) continues to increase, and E continues to decrease. If neither of the electrodes dissolves completely, thereby breaking the electrical circuit, the cell voltage will eventually reach zero. This is the situation that occurs when a battery is “dead.” The value of \(Q\) when E = 0 is calculated as follows: Recall that at equilibrium, \(Q = K\). Thus the equilibrium constant for the reaction of Zn metal with Cu to give Cu metal and Zn is 1.7 × 10 at 25°C. The Nernst Equation: A voltage can also be generated by constructing an electrochemical cell in which each compartment contains the same redox active solution but at different concentrations. The voltage is produced as the concentrations equilibrate. Suppose, for example, we have a cell with 0.010 M AgNO in one compartment and 1.0 M AgNO in the other. The cell diagram and corresponding half-reactions are as follows: \[\ce{Ag(s)\,|\,Ag^{+}}(aq, 0.010 \;M)\,||\,\ce{Ag^{+}}(aq, 1.0 \;M)\,|\,\ce{Ag(s)} \label{Eq8} \] cathode: \[\ce{Ag^{+}} (aq, 1.0\; M) + \ce{e^{−}} \rightarrow \ce{Ag(s)} \label{Eq9} \] anode: \[\ce{Ag(s)} \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) + \ce{e^{−}} \label{Eq10} \] Overall \[\ce{Ag^{+}}(aq, 1.0 \;M) \rightarrow \ce{Ag^{+}}(aq, 0.010\; M) \label{Eq11} \] As the reaction progresses, the concentration of \(Ag^+\) will increase in the left (oxidation) compartment as the silver electrode dissolves, while the \(Ag^+\) concentration in the right (reduction) compartment decreases as the electrode in that compartment gains mass. The total mass of \(Ag(s)\) in the cell will remain constant, however. We can calculate the potential of the cell using the Nernst equation, inserting 0 for E° because E° = −E° : \[\begin{align*} E_\textrm{cell}&=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \\[4pt] &=0-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{0.010}{1.0}\right) \\[4pt] &=\textrm{0.12 V} \end{align*} \nonumber \] An electrochemical cell of this type, in which the anode and cathode compartments are identical except for the concentration of a reactant, is called a . As the reaction proceeds, the difference between the concentrations of Ag in the two compartments will decrease, as will E . Finally, when the concentration of Ag is the same in both compartments, equilibrium will have been reached, and the measured potential difference between the two compartments will be zero (E = 0). Calculate the voltage in a galvanic cell that contains a manganese electrode immersed in a 2.0 M solution of MnCl as the cathode, and a manganese electrode immersed in a 5.2 × 10 M solution of MnSO as the anode (T = 25°C). galvanic cell, identities of the electrodes, and solution concentrations voltage This is a concentration cell, in which the electrode compartments contain the same redox active substance but at different concentrations. The anions (Cl and SO ) do not participate in the reaction, so their identity is not important. The overall reaction is as follows: \[\ce{ Mn^{2+}}(aq, 2.0\, M) \rightarrow \ce{Mn^{2+}} (aq, 5.2 \times 10^{−2}\, M)\nonumber \] For the reduction of Mn (aq) to Mn(s), n = 2. We substitute this value and the given Mn concentrations into Equation \(\ref{Eq4}\): \[ \begin{align*} E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}{n}\right)\log Q \\[4pt] &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{5.2\times10^{-2}}{2.0}\right) \\[4pt] &=\textrm{0.047 V}\end{align*} \nonumber \] Thus manganese will dissolve from the electrode in the compartment that contains the more dilute solution and will be deposited on the electrode in the compartment that contains the more concentrated solution. Suppose we construct a galvanic cell by placing two identical platinum electrodes in two beakers that are connected by a salt bridge. One beaker contains 1.0 M HCl, and the other a 0.010 M solution of Na SO at pH 7.00. Both cells are in contact with the atmosphere, with \(P_\mathrm{O_2}\) = 0.20 atm. If the relevant electrochemical reaction in both compartments is the four-electron reduction of oxygen to water: \[\ce{O2(g) + 4H^{+}(aq) + 4e^{−} \rightarrow 2H2O(l)} \nonumber \] What will be the potential when the circuit is closed? 0.41 V Because voltages are relatively easy to measure accurately using a voltmeter, electrochemical methods provide a convenient way to determine the concentrations of very dilute solutions and the solubility products (\(K_{sp}\)) of sparingly soluble substances. As you learned previously, solubility products can be very small, with values of less than or equal to 10 . Equilibrium constants of this magnitude are virtually impossible to measure accurately by direct methods, so we must use alternative methods that are more sensitive, such as electrochemical methods. To understand how an electrochemical cell is used to measure a solubility product, consider the cell shown in Figure \(\Page {1}\), which is designed to measure the solubility product of silver chloride: \[K_{sp} = [\ce{Ag^{+}},\ce{Cl^{−}}]. \nonumber \] In one compartment, the cell contains a silver wire inserted into a 1.0 M solution of Ag ; the other compartment contains a silver wire inserted into a 1.0 M Cl solution saturated with AgCl. In this system, the Ag ion concentration in the first compartment equals K . We can see this by dividing both sides of the equation for K by [Cl ] and substituting: \[\begin{align*}[\ce{Ag^{+}}] &= \dfrac{K_{sp}}{[\ce{Cl^{−}}]} \\[4pt] &= \dfrac{K_{sp}}{1.0} = K_{sp}. \end{align*} \nonumber \] The overall cell reaction is as follows: Ag (aq, concentrated) → Ag (aq, dilute) Thus the voltage of the concentration cell due to the difference in [Ag ] between the two cells is as follows: \[\begin{align} E_\textrm{cell} &=\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{1}\right)\log\left(\dfrac{[\mathrm{Ag^+}]_\textrm{dilute}}{[\mathrm{Ag^+}]_\textrm{concentrated}}\right) \nonumber \\[4pt] &= -\textrm{0.0591 V } \log\left(\dfrac{K_{\textrm{sp}}}{1.0}\right) \nonumber \\[4pt] &=-\textrm{0.0591 V }\log K_{\textrm{sp}} \label{Eq122} \end{align} \] By closing the circuit, we can measure the potential caused by the difference in [Ag+] in the two cells. In this case, the experimentally measured voltage of the concentration cell at 25°C is 0.580 V. Solving Equation \(\ref{Eq122}\) for \(K_{sp}\), \[\begin{align*}\log K_\textrm{sp} & =\dfrac{-E_\textrm{cell}}{\textrm{0.0591 V}}=\dfrac{-\textrm{0.580 V}}{\textrm{0.0591 V}}=-9.81 \\[4pt]
K_\textrm{sp} & =1.5\times10^{-10}\end{align*} \nonumber \] Thus a single potential measurement can provide the information we need to determine the value of the solubility product of a sparingly soluble salt. To measure the solubility product of lead(II) sulfate (PbSO ) at 25°C, you construct a galvanic cell like the one shown in Figure \(\Page {1}\), which contains a 1.0 M solution of a very soluble Pb salt [lead(II) acetate trihydrate] in one compartment that is connected by a salt bridge to a 1.0 M solution of Na SO saturated with PbSO in the other. You then insert a Pb electrode into each compartment and close the circuit. Your voltmeter shows a voltage of 230 mV. What is K for PbSO ? Report your answer to two significant figures. galvanic cell, solution concentrations, electrodes, and voltage K You have constructed a concentration cell, with one compartment containing a 1.0 M solution of \(\ce{Pb^{2+}}\) and the other containing a dilute solution of Pb in 1.0 M Na SO . As for any concentration cell, the voltage between the two compartments can be calculated using the Nernst equation. The first step is to relate the concentration of Pb in the dilute solution to K : \[\begin{align*}[\mathrm{Pb^{2+}},\mathrm{SO_4^{2-}}] & =K_\textrm{sp} \\
[\mathrm{Pb^{2+}}] &=\dfrac{K_\textrm{sp}}{[\mathrm{SO_4^{2-}}]}=\dfrac{K_\textrm{sp}}{\textrm{1.0 M}}=K_\textrm{sp}\end{align*} \nonumber \] The reduction of Pb to Pb is a two-electron process and proceeds according to the following reaction: Pb (aq, concentrated) → Pb (aq, dilute) so \[\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{0.0591}{n}\right)\log Q \\
\textrm{0.230 V} & =\textrm{0 V}-\left(\dfrac{\textrm{0.0591 V}}{2}\right)\log\left(\dfrac{[\mathrm{Pb^{2+}}]_\textrm{dilute}}{[\mathrm{Pb^{2+}}]_\textrm{concentrated}}\right)=-\textrm{0.0296 V}\log\left(\dfrac{K_\textrm{sp}}{1.0}\right) \\
-7.77 & =\log K_\textrm{sp} \\
1.7\times10^{-8} & =K_\textrm{sp}\end{align*} \nonumber \] A concentration cell similar to the one described in Example \(\Page {3}\) contains a 1.0 M solution of lanthanum nitrate [La(NO ) ] in one compartment and a 1.0 M solution of sodium fluoride saturated with LaF in the other. A metallic La strip is inserted into each compartment, and the circuit is closed. The measured potential is 0.32 V. What is the K for LaF ? Report your answer to two significant figures. 5.7 × 10 Another use for the Nernst equation is to calculate the concentration of a species given a measured potential and the concentrations of all the other species. We saw an example of this in Example \(\Page {3}\), in which the experimental conditions were defined in such a way that the concentration of the metal ion was equal to K . Potential measurements can be used to obtain the concentrations of dissolved species under other conditions as well, which explains the widespread use of electrochemical cells in many analytical devices. Perhaps the most common application is in the determination of [H ] using a pH meter, as illustrated below. Suppose a galvanic cell is constructed with a standard Zn/Zn couple in one compartment and a modified hydrogen electrode in the second compartment. The pressure of hydrogen gas is 1.0 atm, but [H ] in the second compartment is unknown. The cell diagram is as follows: \[\ce{Zn(s)}|\ce{Zn^{2+}}(aq, 1.0\, M) || \ce{H^{+}} (aq, ?\, M)| \ce{H2} (g, 1.0\, atm)| Pt(s) \nonumber \] What is the pH of the solution in the second compartment if the measured potential in the cell is 0.26 V at 25°C? galvanic cell, cell diagram, and cell potential pH of the solution Under standard conditions, the overall reaction that occurs is the reduction of protons by zinc to give H (note that Zn lies below H in ): By substituting the given values into the simplified Nernst equation (Equation \(\ref{Eq4}\)), we can calculate [H ] under nonstandard conditions: \[\begin{align*}E_\textrm{cell} &=E^\circ_\textrm{cell}-\left(\dfrac{\textrm{0.0591 V}}n\right)\log\left(\dfrac{[\mathrm{Zn^{2+}}]P_\mathrm{H_2}}{[\mathrm{H^+}]^2}\right) \\
\textrm{0.26 V} &=\textrm{0.76 V}-\left(\dfrac{\textrm{0.0591 V}}2\right)\log\left(\dfrac{(1.0)(1.0)}{[\mathrm{H^+}]^2}\right) \\
16.9 &=\log\left(\dfrac{1}{[\mathrm{H^+}]^2}\right)=\log[\mathrm{H^+}]^{-2}=(-2)\log[\mathrm{H^+}] \\
8.46 &=-\log[\mathrm{H^+}] \\
8.5 &=\mathrm{pH}\end{align*} \nonumber \] Thus the potential of a galvanic cell can be used to measure the pH of a solution. Suppose you work for an environmental laboratory and you want to use an electrochemical method to measure the concentration of Pb in groundwater. You construct a galvanic cell using a standard oxygen electrode in one compartment (E° = 1.23 V). The other compartment contains a strip of lead in a sample of groundwater to which you have added sufficient acetic acid, a weak organic acid, to ensure electrical conductivity. The cell diagram is as follows: \[Pb_{(s)} ∣Pb^{2+}(aq, ? M)∥H^+(aq), 1.0 M∣O_2(g, 1.0 atm)∣Pt_{(s)}\nonumber \] When the circuit is closed, the cell has a measured potential of 1.62 V. Use to determine the concentration of Pb in the groundwater. \(1.2 \times 10^{−9}\; M\) The Nernst equation can be used to determine the direction of spontaneous reaction for any redox reaction in aqueous solution. The Nernst equation allows us to determine the spontaneous direction of any redox reaction under any reaction conditions from values of the relevant standard electrode potentials. Concentration cells consist of anode and cathode compartments that are identical except for the concentrations of the reactant. Because ΔG = 0 at equilibrium, the measured potential of a concentration cell is zero at equilibrium (the concentrations are equal). A galvanic cell can also be used to measure the solubility product of a sparingly soluble substance and calculate the concentration of a species given a measured potential and the concentrations of all the other species. | 18,829 | 2,811 |
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Einstein’s photons of light were individual packets of energy having many of the characteristics of particles. Recall that the collision of an electron (a particle) with a sufficiently energetic photon can eject a from the surface of a metal. Any excess energy is transferred to the electron and is converted to the kinetic energy of the ejected electron. Einstein’s hypothesis that energy is concentrated in localized bundles, however, was in sharp contrast to the classical notion that energy is spread out uniformly in a wave. We now describe Einstein’s theory of the relationship between energy and mass, a theory that others built on to develop our current model of the atom. Einstein initially assumed that photons had zero mass, which made them a peculiar sort of particle indeed. In 1905, however, he published his special theory of relativity, which related energy and mass according to the following equation: \[ E=h\nu=h\dfrac{c}{\lambda }=mc^{2} \label{8.5.1} \] According to this theory, a photon of wavelength λ and frequency ν has a nonzero mass, which is given as follows: \[ m=\dfrac{E}{c^{2}}=\dfrac{h\nu }{c^{2}}=\dfrac{h}{\lambda c} \label{8.5.2}\] That is, light, which had always been regarded as a wave, also has properties typical of particles, a condition known as wave–particle duality (a principle that matter and energy have properties typical of both waves and particles). Depending on conditions, light could be viewed as either a wave or a particle. In 1922, the American physicist Arthur Compton (1892–1962) reported the results of experiments involving the collision of x-rays and electrons that supported the particle nature of light. At about the same time, a young French physics student, Louis de Broglie (1892–1972), began to wonder whether the converse was true: Could particles exhibit the properties of waves? In his PhD dissertation submitted to the Sorbonne in 1924, de Broglie proposed that a particle such as an electron could be described by a wave whose wavelength is given by \[\lambda =\dfrac{h}{mv} \label{8.5.3}\] where This revolutionary idea was quickly confirmed by American physicists Clinton Davisson (1881–1958) and Lester Germer (1896–1971), who showed that beams of electrons, regarded as particles, were diffracted by a sodium chloride crystal in the same manner as x-rays, which were regarded as waves. It was proven experimentally that electrons do exhibit the properties of waves. For his work, de Broglie received the Nobel Prize in Physics in 1929. If particles exhibit the properties of waves, why had no one observed them before? The answer lies in the numerator of de Broglie’s equation, which is an extremely small number. As you will calculate in Example \(\Page {1}\), Planck’s constant (6.63 × 10 J•s) is so small that the wavelength of a particle with a large mass is too short (less than the diameter of an atomic nucleus) to be noticeable. Calculate the wavelength of a baseball, which has a mass of 149 g and a speed of 100 mi/h. mass and speed of object wavelength The wavelength of a particle is given by λ = / . We know that = 0.149 kg, so all we need to find is the speed of the baseball: \( v=\left ( \dfrac{100\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right ) \) Recall that the joule is a derived unit, whose units are (kg•m )/s . Thus the wavelength of the baseball is \[ \lambda =\dfrac{6.626\times 10^{-34}\; J\cdot s}{\left ( 0.149\; kg \right )\left ( 44.69\; m\cdot s \right )}= \dfrac{6.626\times 10^{-34}\; \cancel{kg}\cdot m{^\cancel{2}\cdot \cancel{s}{\cancel{^{-2}}\cdot \cancel{s}}}}{\left ( 0.149\; \cancel{kg} \right )\left ( 44.69\; \cancel{m}\cdot \cancel{s^{-1}} \right )}=9.95\times 10^{-35}\; m \] (You should verify that the units cancel to give the wavelength in meters.) Given that the diameter of the nucleus of an atom is approximately 10 m, the wavelength of the baseball is almost unimaginably small. Calculate the wavelength of a neutron that is moving at 3.00 × 10 m/s. 1.32 Å, or 132 pm As you calculated in Example \(\Page {1}\), objects such as a baseball or a neutron have such short wavelengths that they are best regarded primarily as particles. In contrast, objects with very small masses (such as photons) have large wavelengths and can be viewed primarily as waves. Objects with intermediate masses, such as electrons, exhibit the properties of both particles and waves. Although we still usually think of electrons as particles, the wave nature of electrons is employed in an , which has revealed most of what we know about the microscopic structure of living organisms and materials. Because the wavelength of an electron beam is much shorter than the wavelength of a beam of visible light, this instrument can resolve smaller details than a light microscope can (Figure \(\Page {1}\)). A wave is a disturbance that travels in space. The magnitude of the wave at any point in space and time varies sinusoidally. While the absolute value of the magnitude of one wave at any point is not very important, the displacement of two waves called the phase difference, is vitally important because it determines whether the waves reinforce or interfere with each other. Figure \(\Page {2}\)a, on the right shows an arbitrary phase difference between two wave. Figure \(\Page {2}\)b shows what happens when the two waves are 180 degrees out of phase. The green line is their sum. Figure \(\Page {2}\)c shows what happens when the two lines are in phase, exactly superimposed on each other. Again, the green line is the sum of the intensities. De Broglie also investigated why only certain orbits were allowed in Bohr’s model of the hydrogen atom. He hypothesized that the electron behaves like a standing wave . An example of a standing wave is the motion of a string of a violin or guitar. When the string is plucked, it vibrates at certain fixed frequencies because it is fastened at both ends (Figure \(\Page {3}\) ). If the length of the string is , then the lowest-energy vibration (the fundamental ) has wavelength \[ \begin{array}{ll} \dfrac{\lambda }{2}=L \\ \lambda =2L \end{array} \label{8.5.4}\] Higher-energy vibrations are called overtones ( ) and are produced when the string is plucked more strongly; they have wavelengths given by \[ \lambda=\dfrac{2L}{n} \label{8.5.5}\] where has any integral value. Thus the resonant vibrational energies of the string are quantized. When plucked, all other frequencies die out immediately. Only the resonant frequencies survive and are heard. By analogy we can think of the resonant frequencies as being quantized. Notice in Figure \(\Page {3}\) that all overtones have one or more nodes , points where the string does not move. The amplitude of the wave at a node is zero. Quantized vibrations and overtones containing nodes are not restricted to one-dimensional systems, such as strings. A two-dimensional surface, such as a drumhead, also has quantized vibrations. Similarly, when the ends of a string are joined to form a circle, the only allowed vibrations are those with wavelength \[2πr = nλ \label{8.5.6}\] where is the radius of the circle. De Broglie argued that Bohr’s allowed orbits could be understood if the electron behaved like a (Figure \(\Page {4}\)). The standing wave could exist only if the circumference of the circle was an integral multiple of the wavelength such that the propagated waves were all in phase, thereby increasing the net amplitudes and causing . Otherwise, the propagated waves would be out of phase, resulting in a net decrease in amplitude and causing The non resonant waves interfere with themselves! De Broglie’s idea explained Bohr’s allowed orbits and energy levels nicely: in the lowest energy level, corresponding to = 1 in Equation 8.5.2, one complete wavelength would close the circle. Higher energy levels would have successively higher values of with a corresponding number of nodes. Standing waves are often observed on rivers, reservoirs, ponds, and lakes when seismic waves from an earthquake travel through the area. The waves are called , a term first used in 1955 when lake levels in England and Norway oscillated from side to side as a result of the Assam earthquake of 1950 in Tibet. They were first described in the Proceedings of the Royal Society in 1755 when they were seen in English harbors and ponds after a large earthquake in Lisbon, Portugal. Seismic seiches were also observed in many places in North America after the Alaska earthquake of March 28, 1964. Those occurring in western reservoirs lasted for two hours or longer, and amplitudes reached as high as nearly 6 ft along the Gulf Coast. The height of seiches is approximately proportional to the thickness of surface sediments; a deeper channel will produce a higher seiche. Still, as all analogies, although the standing wave model helps us understand much about why Bohr's theory worked, it also, if pushed too far can mislead. As you will see, some of de Broglie’s ideas are retained in the modern theory of the electronic structure of the atom: the wave behavior of the electron and the presence of nodes that increase in number as the energy level increases. Unfortunately, his (and Bohr's) explanation also contains one major feature that we know to be incorrect: in the currently accepted model, the electron in a given orbit is always at the same distance from the nucleus. The de Broglie Equation: Because a wave is a disturbance that travels in space, it has no fixed position. One might therefore expect that it would also be hard to specify the exact position of a that exhibits wavelike behavior. A characteristic of light is that is can be bent or spread out by passing through a narrow slit as shown in the video below. You can literally see this by half closing your eyes and looking through your eye lashes. This reduces the brightness of what you are seeing and somewhat fuzzes out the image, but the light bends around your lashes to provide a complete image rather than a bunch of bars across the image. This is called diffraction. This behavior of waves is captured in Maxwell's equations (1870 or so) for electromagnetic waves and was and is well understood. Heisenberg's uncertainty principle for light is, if you will, merely a conclusion about the nature of electromagnetic waves and nothing new. DeBroglie's idea of wave particle duality means that particles such as electrons which all exhibit wave like characteristics, will also undergo diffraction from slits whose size is of the order of the electron wavelength. This situation was described mathematically by the German physicist Werner Heisenberg (1901–1976; Nobel Prize in Physics, 1932), who related the position of a particle to its momentum. Referring to the electron, Heisenberg stated that “at every moment the electron has only an inaccurate position and an inaccurate velocity, and between these two inaccuracies there is this uncertainty relation.” Mathematically, the Heisenberg uncertainty principle states that the uncertainty in the position of a particle (Δ ) multiplied by the uncertainty in its momentum [Δ( )] is greater than or equal to Planck’s constant divided by 4π: \[ \left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\geqslant \dfrac{h}{4\pi } \label{8.5.7} \] Because Planck’s constant is a very small number, the Heisenberg uncertainty principle is important only for particles such as electrons that have very low masses. These are the same particles predicted by de Broglie’s equation to have measurable wavelengths. If the precise position of a particle is known absolutely (Δ = 0), then the uncertainty in its momentum must be infinite: \[ \left ( \Delta \left [ mv \right ] \right )= \dfrac{h}{4\pi \left ( \Delta x \right ) }=\dfrac{h}{4\pi \left ( 0 \right ) }=\infty \label{8.5.8} \] Because the mass of the electron at rest ( ) is both constant and accurately known, the uncertainty in Δ( ) must be due to the Δ term, which would have to be infinitely large for Δ( ) to equal infinity. That is, according to Equation 8.5.8, the more accurately we know the exact position of the electron (as Δ → 0), the less accurately we know the speed and the kinetic energy of the electron (1/2 ) because Δ( ) → ∞. Conversely, the more accurately we know the precise momentum (and the energy) of the electron [as Δ( ) → 0], then Δ → ∞ and we have no idea where the electron is. Bohr’s model of the hydrogen atom violated the Heisenberg uncertainty principle by trying to specify both the position (an orbit of a particular radius) and the energy (a quantity related to the momentum) of the electron. Moreover, given its mass and wavelike nature, the electron in the hydrogen atom could not possibly orbit the nucleus in a well-defined circular path as predicted by Bohr. You will see, however, that the of the electron in the hydrogen atom is exactly the one predicted by Bohr’s model. Calculate the minimum uncertainty in the position of the pitched baseball from Example \(\Page {1}\) that has a mass of exactly 149 g and a speed of 100 ± 1 mi/h. mass and speed of object minimum uncertainty in its position The Heisenberg uncertainty principle tells us that (Δ )[Δ( )] = /4π. Rearranging the inequality gives We know that = 6.626 × 10 J•s and = 0.149 kg. Because there is no uncertainty in the mass of the baseball, Δ( ) = Δ and Δ = ±1 mi/h. We have \( \Delta \nu =\left ( \dfrac{1\; \cancel{mi}}{\cancel{h}} \right )\left ( \dfrac{1\; \cancel{h}}{60\; \cancel{min}} \right )\left ( \dfrac{1\; \cancel{min}}{60\; s} \right )\left ( \dfrac{1.609\; \cancel{km}}{\cancel{mi}} \right )\left ( \dfrac{1000\; m}{\cancel{km}} \right )=0.4469\; m/s \) \( \Delta x \geqslant \left ( \dfrac{6.626\times 10^{-34}\; J\cdot s}{4\left ( 3.1416 \right )} \right ) \left ( \dfrac{1}{\left ( 0.149\; kg \right )\left ( 0.4469\; m\cdot s^{-1} \right )} \right ) \) Inserting the definition of a joule (1 J = 1 kg•m /s ) gives \( \Delta x \geqslant \left ( \dfrac{6.626\times 10^{-34}\; \cancel{kg} \cdot m^{\cancel{2}} \cdot s}{4\left ( 3.1416 \right )\left ( \cancel{s^{2}} \right )} \right ) \left ( \dfrac{1\; \cancel{s}}{\left ( 0.149\; \cancel{kg} \right )\left ( 0.4469\; \cancel{m} \right )} \right ) \) \( \Delta x \geqslant 7.92 \pm \times 10^{-34}\; m \) This is equal to 3.12 × 10 inches. We can safely say that if a batter misjudges the speed of a fastball by 1 mi/h (about 1%), he will not be able to blame Heisenberg’s uncertainty principle for striking out. Calculate the minimum uncertainty in the position of an electron traveling at one-third the speed of light, if the uncertainty in its speed is ±0.1%. Assume its mass to be equal to its mass at rest. 6 × 10 m, or 0.6 nm (about the diameter of a benzene molecule) Answers for these quizzes are included. The modern model for the electronic structure of the atom is based on recognizing that an electron possesses particle and wave properties, the so-called . Louis de Broglie showed that the wavelength of a particle is equal to Planck’s constant divided by the mass times the velocity of the particle. \[ \lambda =\dfrac{h}{mv} \tag{8.5.3} \] The electron in Bohr’s circular orbits could thus be described as a , one that does not move through space. Standing waves are familiar from music: the lowest-energy standing wave is the vibration, and higher-energy vibrations are and have successively more , points where the amplitude of the wave is always zero. Werner Heisenberg’s states that it is impossible to precisely describe both the location and the speed of particles that exhibit wavelike behavior. \[ \left ( \Delta x \right )\left ( \Delta \left [ mv \right ] \right )\geqslant \dfrac{h}{4\pi } \tag{8.5.7} \] ( ) | 15,929 | 2,812 |
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The same underlying principles of thin layer chromatography (TLC) apply to column chromatography. In fact, a TLC is always run before performing a column to assess the situation and determine the proper solvent ratio. In order to get good separation, it is ideal if the desired component has an and is separated from other components by at least 0.2 \(R_f\) units.\(^8\) If the spots to be separated are very close (if the difference in \(R_f\) is < 0.2), it's best if the middle of the spots has an \(R_f\) of 0.35. An \(R_f\) near 0.35 is ideal because it is slow enough that stationary-mobile phase equilibration can occur, but fast enough to minimize band widening from diffusion. There are a few variables which aren't applicable to TLC, but which affect the separation of components in column chromatography. These include the column diameter, quantity of adsorbent used, and solvent flow rate. Table 2.5 summarizes variable recommendations based on the sample size and degree of separation between components. In all scenarios, the columns are to be prepped between 5-6 inches high. For example, a column of one-inch diameter (\(1 \: \text{in}\) is \(25.4 \: \text{mm}\)) should be able to purify around \(400 \: \text{mg}\) of material if the separation is good (\(\Delta R_f\) > 0.2, third column in Table 2.5), or around \(160 \: \text{mg}\) if the separation is difficult (\(\Delta R_f\) > 0.1). The column should be able to be prepared and eluted using around \(200 \: \text{mL}\) of solvent, and the fractions can be collected with approximately \(10 \: \text{mL}\) of solution each.\(^9\) There are multiple variations on how to physically run a column, and your instructor may prefer a certain method. One large difference in methods is how the column is prepared. In the "dry packing" method, dry silica or alumina is added directly to a column, and solvent is allowed to trickle through in portions, then with pressure. In the "wet packing" method, the column is filled with solvent first, then dry silica or alumina is lightly shaken in, then packed with pressure. In the "slurry" method, solvent is added to the silica or alumina in an Erlenmeyer flask, poured onto the column as a sludgy material, then packed with pressure. It is important to know that heat is liberated when solvent is added to silica or alumina (they have an exothermic heat of solvation). The slurry method is presented in this section, with the main reason being that it allows this exothermic step to happen in an Erlenmeyer flask instead of on the column. If heat is liberated during the packing of the column, it may generate bubbles from the boiling of solvent. These can interfere with the separation of the column if they are not adequately removed, and can crack the adsorbent material in the column. The column pictured in this section shows purification of a \(0.20 \: \text{g}\) sample containing a mixture of ferrocene and acetylferrocene (crude TLC is in Figure 2.51a). Roughly \(8 \: \text{mL}\) fractions were collected into small test tubes, and roughly \(400 \: \text{mL}\) of eluent was used. Once the sample is applied to the column, there is a race against time, as diffusion will start to broaden the material. A sample should not be applied until you are ready to complete the column immediately and in its entirety. This process may take between 15 and 90 minutes! If using test tubes to collect fractions, the test tubes should be arranged in a rack prior to adding the sample, and the column height should be adjusted so that the test tube rack can slide underneath. Make sure there is a frit or cotton plug in the bottom of the column. Fill the column with silica or alumina to 5-6 inches in the fume hood. Pour the adsorbent into an Erlenmeyer flask and add eluent to make a pourable slurry. The eluent should give the desired component an by TLC. Pour in the slurry and immediately rinse the sides of the column with eluent and a pipette. Jostle the column to remove air bubbles. Use air pressure to pack the column. Keep it perfectly vertical. Add \(0.5 \: \text{cm}\) of sand. Rinse the sand off the sides carefully with a swirling motion. Don't disrupt the top surface of the silica or alumina with rinsing. Adjust the eluent level to the sand layer, and then add the sample (pure liquid, dissolved in \(\ce{CH_2Cl_2}\), or solid adsorbed onto a portion of silica). Rinse the sides and use air pressure to force the eluent down onto the silica/alumina layer. Fill the reservoir with eluent (carefully to not disrupt the top surface). Use steady air pressure to elute the column. Collect fractions in test tubes in a rack (keep in order). Rinse the column tip if a component has finished coming off the column. Possibly increase the eluent polarity to make components elute faster. Use TLC to determine the purity of the fractions, and combine appropriate fractions. Remove the solvent with the rotary evaporator. It is quite common to break the tip of a Pasteur pipette while rinsing the column, which often falls and wedges itself into the delicate column. Unfortunately, a broken pipette in a column can cause problems with the separation of components. For example, Figure 2.65 shows the effect of a broken pipette on the separation of two components. A broken pipette is embedded in the column and is the nearly vertical line of orange seen between the two bands on the column. Since stationary-mobile phase equilibration does not occur on the glass surface of the pipette, compounds stream down quickly and move faster than they should based on their \(R_f\). The orange vertical line is the top component draining into the band of the bottom component, contaminating it. If a broken pipette pierces the column, and the sand or sample has yet been applied, attempt to remove the pipette with long forceps. After removal, vigorously jostle the column to pack it again and continue on with the column. If the pipette cannot be removed with forceps, you may consider redoing the column. If the sand already been applied, removal of the pipette and jostling the column will often ruin the horizontal surface of the adsorbent, so it will be necessary to re-pack the column. If the sample has already been applied, there isn't anything to do but continue on and hope for the best. If the column results in an unsuccessful separation, all fractions containing the compound of interest can be combined, solvent evaporated by the rotary evaporator, and another purification method (or a second column) can be attempted. Like a broken pipette, an air bubble is an empty pocket where stationary-mobile phase equilibration does not happen, so components move quicker around an air bubble than they should. This may lead to uneven eluting bands, which can cause overlap if the separation of the mixture is difficult (if the components have very close \(R_f\) values, as in Figure 2.66). If air bubbles are seen in the column and the sand or sample has yet been applied, give the column a good jostling during packing to remove all air bubbles. See your instructor if the bubbles are not budging as you may be approaching the task too delicately. If the sand or sample has already been applied, it's best to leave the column as is and hope the air bubbles do not affect the separation. If the components of a mixture are colored, it may be obvious when the bands elute in a crooked manner. This is most likely due to the column being clamped at a slight diagonal. If the column is clamped in a slanted manner, components will travel in a slanted manner (Figure 2.67). This may cause separation problems if the components have a similar \(R_f\). There is no way to fix this problem midway through a column, but if the components have very different \(R_f\) values, the slanted bands may have no effect on the separation. In the future, be sure to check that the column is perfectly vertical in both the side-to-side and front-back directions. \(^8\)W.C. Still, M. Kahn, A. Mitra, Vol. 43, No. 14, . \(^9\)A typical small test tube \(\left( 13 x 100 \: \text{mm} \right)\) has a capacity of \(9 \: \text{mL}\), and a typical medium test tube \(\left( 18 x 150 \: \text{mm} \right)\) has a capacity of \(27 \: \text{mL}\). \(^{10}\)W.C. Still, M. Kahn, A. Mitra, , Vol. 43, No. 14, . | 8,316 | 2,813 |
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Every discipline has its own vocabulary and your success in studying ana- lytical chemistry will improve if you master this vocabulary. Be sure you understand the difference between an analyte and its matrix, between a technique and a method, between a procedure and a protocol, and between a total analysis technique and a concentration technique. In selecting an analytical method we consider criteria such as accu- racy, precision, sensitivity, selectivity, robustness, ruggedness, the amount of available sample, the amount of analyte in the sample, time, cost, and the availability of equipment. These criteria are not mutually independent, and often it is necessary to find an acceptable balance between them. In developing a procedure or protocol, we give consideration to compensating for interferences, calibrating the method, obtaining an appropriate sample, and validating the analysis. Poorly designed procedures and protocols produce results that are insufficient to meet the needs of the analysis. accuracy calibration detection limit matrix method blank protocol rugged sensitivity technique analysis calibration curve determination measurement precision QA/QC selectivity signal total analysis technique analyte concentration technique interferent method procedure robust selectivity coefficient specificity validation | 1,347 | 2,814 |
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An atom consists of a positively charged nucleus, surrounded by one or more negatively charged particles called electrons. The positive charges equal the negative charges, so the atom has no overall charge; it is electrically neutral. Most of an atom’s mass is in its nucleus; the mass of an electron is only 1/1836 the mass of the lightest nucleus, that of hydrogen. Although the nucleus is heavy, it is quite small compared with the overall size of an atom. The radius of a typical atom is around 1 to 2.5 angstroms (Å), whereas the radius of a nucleus is about 10 Å. If an atom were enlarged to the size of the earth, its nucleus would be only 200 feet in diameter and could easily rest inside a small football stadium. The nucleus of an atom contains protons and neutrons. Protons and neutrons have nearly equal masses, but they differ in charge. A neutron has no charge, whereas a proton has a positive charge that exactly balances the negative charge on an electron. Table \(\Page {1}\) lists the charges of these three fundamental particles, and gives their masses expressed in atomic mass units. The atomic mass unit (amu) is defined as exactly one-twelfth the mass of a carbon atom that has six protons and six neutrons in its nucleus. With this scale, protons and neutrons have masses that are close to, but not precisely, 1 u each (there are 6.022 x 10 u in 1 gram This number is known as , N, and one of the ways this number can be calculated is discussed below). The number of protons in the nucleus of an atom is known as the atomic number, Z. It is equal to the number of electrons around the nucleus, because an atom is electrically neutral. The mass number of an atom is equal to the total number of heavy particles: protons and neutrons. When two atoms are close enough to combine chemically to All atoms with the same atomic number behave in much the same way chemically, and are classified as the same chemical element. Each element has its own name and a one- or two-letter symbol (usually derived from the element’s English or Latin name). For example, the symbol for carbon is C, and the symbol for calcium is Ca. The symbol for sodium is Na-the first two letters of its Latin (and German) name, , to distinguish it from nitrogen, N, and sulfur, S. What is the atomic symbol for bromine, and what is its atomic number? Why isn’t the symbol for bromine just the first letter of its name? What other element preempts the symbol B? (Refer to the ) Bromine’s atomic number is 35, and its symbol is Br; B is the symbol for boron | 2,569 | 2,816 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Fundamentals/WKB_Approximation |
The WKB Approximation, named after scientists Wentzel–Kramers–Brillouin, is a method to approximate solutions to a time-independent linear differential equation or in this case, the Schrödinger Equation. Its principal applications are for calculating bound-state energies and tunneling rates through potential barriers. The WKB Approximation is most often applied to 1D problems, but also works for 3D spherically symmetric problems. As a general overview, the wavefunction is assumed to be an exponential function with either amplitude or phase taken to be slowly changing relative to the \(λ\). It is then semi-classically expanded. ö \[\psi(x)=Ae^{\pm ikx} \nonumber \] where \[k=\dfrac{2\pi}{\lambda}=\sqrt{\dfrac{2m(E-U)}{\hbar^2}}=constant\label{56.1} \] \[\psi(x)=Ae^{\pm i\phi(x)}\label{56.2} \] where \(\phi (x)=xk(x)\). \(\phi (x)=\pm kx\) \(\phi (x)\) \(\pm kx\) This occurs because the mass stops and reverses its velocity is zero. two wavefunctions must be properly matched \[\psi(x)=Ae^{\pm ikx},\label{56.3} \] λ \[k(x)=\sqrt{\dfrac{2m(E-U(x))}{\hbar^2}}\label{56.4} \] If \(E < U\), the solution to the Schrödinger equation for a constant \(U\) is: \[\psi(x)=Ae^{\pm\kappa x}. \label{56.5} \] In these regions, a classical particle would not be allowed, but there is a finite probability that a particle can pass through a potential energy barrier in quantum mechanics. The quantum particle is described as ‘tunnelling’, which is important in determining the rates of chemical reactions, particularly at lower temperatures. \[k(x)=-i\sqrt{\dfrac{2m(U(x)-E)}{\hbar^2}} = -i\kappa(x).\label{56.6} \] If \(U(x)\) is not a constant, but instead varies very slowly on a distance scale of \(λ\), then it is reasonable to suppose that ψ remains practically sinusoidal, except that the wavelength and amplitude change slowly with \(x\). Substituting in the normalized version of Equation \ref{56.2}, the Schrödinger Equation: \[\dfrac{-\hbar^2}{2m}\dfrac{\partial^2}{\partial x^2}\psi(x)+U(x)\psi(x)=E\psi(x)\label{56.7} \] becomes \[i\dfrac{\partial^2\phi}{\partial x^2}-\left(\dfrac{\partial\phi}{\partial x}\right)^2+(k(x))^2=0.\label{56.8} \] \(\phi (x)\) \[\dfrac{\partial^2\phi}{\partial x^2} \ge 0.\label{56.9}\] Thus, \[\left(\dfrac{\partial\phi_0}{\partial x}\right)^2=(k(x))^2\label{56.10} \] \[\phi_0(x)=\pm\int k(x)dx+C_0\label{56.11} \] \(\phi_0(x)\) \[\psi(x)=e^{i(\pm\int k(x)dx+C_0)}.\label{56.12} \] To obtain a more accurate solution, we manipulate Equation \ref{56.8} to solve for \(\phi(x)\). \[i\dfrac{\partial^2\phi}{\partial x^2}-\left(\dfrac{\partial\phi}{\partial x}\right)^2+(k(x))^2=0 \nonumber \] \[\left(\dfrac{\partial\phi}{\partial x}\right)^2=(k(x))^2+i\dfrac{\partial^2\phi}{\partial x^2} \nonumber \] \[\dfrac{\partial\phi}{\partial x}=\sqrt{(k(x))^2+i\dfrac{\partial^2\phi}{\partial x^2}} \nonumber \] So, \[\phi(x)=\pm\int\sqrt{(k(x))^2+i\dfrac{\partial^2\phi}{\partial x^2}}dx+C_1\label{56.13} \] \[\dfrac{\partial\phi_0}{\partial x}=\pm k(x).\label{56.14} \] \[\dfrac{\partial^2\phi_0}{\partial x^2}=\pm \dfrac{\partial}{\partial x}k(x).\label{56.15} \] \[\phi_1(x)=\pm\int\sqrt{(k(x))^2\pm i\dfrac{\partial}{\partial x}k(x)}dx+C_1\label{56.16} \] \(\phi_1(x)\) \[\psi(x)=e^{i\left( \pm\int\sqrt{(k(x))^2\pm i\dfrac{\partial}{\partial x}k(x)}dx+C_1\right)}.\label{56.17} \] Determine the tunneling probability \(T\) at a finite width potential barrier. Given: \[T = \dfrac{\psi^*(L) \psi(L)}{\psi^*(0)\psi(0)} \nonumber \] where \[\psi(x)=\psi(0)e^{i\pm(\int k(x)dx+C_1)} \nonumber \] \(k(x)=-i\sqrt{\dfrac{2m(U(x)-E)}{\hbar^2}}\) for \(E < U(x)\), assuming \(U(x) = U\). Plugging in \(k(x)\) to solve for the wavefunction, \[ \begin{align*} \psi(x) &=\psi(0)e^{i\left(\pm\int_{0}^{x}-i\sqrt{\dfrac{2m(U(x)-E)}{\hbar^2}}dx\right)} \\[4pt] &=\psi(0)e^{\left(-\int_{0}^{x}\sqrt{\dfrac{2m(U-E)}{\hbar^2}}dx\right)} \\[4pt] &=\psi(0)e^{\left(-\left(\sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right)x\right)} \end{align*} \] Thus solving for the the tunneling probability \(T\), \[ \begin{align*} T &= \dfrac{\psi^*(L) \psi(L)}{\psi^*(0)\psi(0)} \\[4pt] &= \dfrac{\psi(0)e^{\left( - \left( \sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right) L \right)} \psi(0)e^{\left( - \left( \sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right) L \right)}}{\psi(0)e^{\left( - \left( \sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right) 0 \right)} \psi(0)e^{\left( - \left( \sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right) 0 \right)}} \\[4pt] &= e^{\left( -2 \left( \sqrt{\dfrac{2m(U-E)}{\hbar^2}}\right) L \right)} \end{align*} \] | 4,520 | 2,817 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.02%3A_Digestion_of_food |
Food digestion, stage 1 of food catabolism, happens in the digestive system shown in the figure on the right (Copyright: public domain, via Wikimedia Common). It is primarily the process of hydrolysis of biopolymers or larger molecules in food into smaller molecules that can transfer from the digestive system to the bloodstream and transport to cells. Carbohydrates are hydrolyzed into monosaccharides, fats into fatty acids and mono- or di-glycerides, proteins into amino acids, and DNA and RNA into mononucleotides by different enzymes. The carbohydrates in human food are starch, sucrose, and lactose. Their digestion begins in the mouth through enzymes in the saliva and continues in the small intestine by pancreatic amylase. The enzymes degrade the starch into smaller and smaller fragments that ultimately result in glucose and maltose which can be absorbed by the small intestine. Sucrose (table sugar) is hydrolyzed into glucose and fructose by the enzyme sucrase. Lactose which is found in milk is hydrolyzed into glucose and galactose by the enzyme lactase. The majority of the adult population have problems digesting unfermented dairy because they can not produce sufficient amounts of the enzyme lactase -a condition called lactose intolerance. Digestion of fats begins in the mouth through lingual lipase but it mainly happens in the small intestine. Bile acids emulsify the fats and the enzyme pancreatic lipase hydrolysis them into free fatty acids along with mono- and di-glycerides. Digestion of proteins happens in the stomach and in the duodenum i.e., the first portion of the small intestine. Three enzymes: pepsin secreted by the stomach; and trypsin and chymotrypsin secreted by the pancreas hydrolyze proteins into smaller peptides which are then hydrolyzed into amino acids by various exopeptidases and dipeptidases. | 1,858 | 2,818 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Vitamins_Cofactors_and_Coenzymes/Nicotinamide_Adenine_Dinucleotide_(NAD) |
Nicotinamide is from the niacin vitamin. The NAD coenzyme is involved with many types of oxidation reactions where alcohols are converted to ketones or aldehydes. It is also involved in the first enzyme complex 1 of the electron transport chain.The structure for the coenzyme, NAD , Nicotinamide Adenine Dinucleotide is shown in Figure \(\Page {1}\). One role of \(NAD^+\) is to initiate the electron transport chain by the reaction with an organic (intermediate in metabolic reactions). This is an oxidation reaction where 2 hydrogen atoms (or 2 hydrogen ions and 2 electrons) are removed from the organic metabolite. (The organic metabolites are usually from the citric acid cycle and the oxidation of fatty acids--details in following pages.) The reaction can be represented simply where M = any metabolite. \[ MH_2 + NAD^+ \rightarrow NADH + H^+ + M: + \text{energy}\] One hydrogen is removed with 2 electrons as a hydride ion (\(H^-\)) while the other is removed as the positive ion (\(H+\)). Usually the metabolite is some type of alcohol which is oxidized to a ketone. The NAD is represented as cyan in Figure \(\Page {2}\). The alcohol is represented by the space filling red, gray, and white atoms. The reaction is to convert the alcohol, ethanol, into ethanal, an aldehyde. \[ CH_3CH_2OH + NAD^+ \rightarrow CH_3CH=O + NADH + H^+ \] This is an oxidation reaction and results in the removal of two hydrogen ions and two electrons which are added to the NAD , converting it to NADH and H . This is the first reaction in the metabolism of alcohol. The active site of ADH has two binding regions. The coenzyme binding site, where NAD binds, and the substrate binding site, where the alcohol binds. Most of the binding site for the NAD is hydrophobic as represented in green. Three key amino acids involved in the catalytic oxidation of alcohols to aldehydes and ketones. They are ser-48, phe 140, and phe 93. | 1,944 | 2,820 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al)/10%3A_Chemical_Bonding_I%3A_Basic_Concepts/108%3A_Bond_Order_and_Bond_Lengths |
Bond order is the number of chemical bonds between a pair of atoms and indicates the stability of a bond. For example, in diatomic nitrogen, N≡N, the bond order is 3; in acetylene, H−C≡C−H, the carbon-carbon bond order is also 3, and the C−H bond order is 1. Bond order and bond length indicate the type and strength of between atoms. Bond order and length are inversely proportional to each other: when bond order is increased, bond length is decreased. Chemistry deals with the way in which subatomic particles bond together to form atoms. Chemistry also focuses on the way in which atoms bond together to form molecules. In the atomic structure, electrons surround the atomic nucleus in regions called orbitals. Each orbital shell can hold a certain number of electrons. When the nearest orbital shell is full, new electrons start to gather in the next orbital shell out from the nucleus, and continue until that shell is also full. The collection of electrons continues in ever widening orbital shells as larger atoms have more electrons than smaller atoms. When two atoms bond to form a molecule, their electrons bond them together by mixing into openings in each others' orbital shells. As with the collection of electrons by the atom, the formation of bonds by the molecule starts at the nearest available orbital shell opening and expand outward. Bond order is the number of bonding pairs of electrons between two atoms. In a between two atoms, a single bond has a bond order of one, a double bond has a bond order of two, a triple bond has a bond order of three, and so on. To determine the bond order between two covalently bonded atoms, follow these steps: If the bond order is zero, the molecule cannot form. The higher bond orders indicate greater stability for the new molecule. In molecules that have , the bond order does not need to be an integer. Determine the bond order for cyanide, CN . 1) Draw the Lewis structure. 2) Determine the type of bond between the two atoms. Because there are 3 dashes, the bond is a triple bond. A triple bond corresponds to a bond order of 3. Determine the bond order for hydrogen gas, H . 1) Draw the Lewis structure. 2) Determine the type of bond between the two atoms. There is only one pair of shared electrons (or dash), indicating is a single bond, with a bond order of 1. If there are more than two atoms in the molecule, follow these steps to determine the bond order: Determine the bond order for nitrate, \(NO_3^-\). 1) Draw the Lewis structure. 2) Count the total number of bonds. 4 The total number of bonds is 4. 3) Count the number of bond groups between individual atoms. 3 The number of bond groups between individual atoms is 3. 4) Divide the number of bonds between individual atoms by the total number of bonds. \[\dfrac{4}{3}= 1.33 \] The bond order is 1.33 Determine the bond order for nitronium ion: \(NO_2^+\). 1) Draw the Lewis Structure. 2) Count the total number of bonds. 4 The total number of bonds is 4. 3) Count the number of bond groups between individual atoms. 2 The number of bond groups between atoms is 2. 4) Divide the bond groups between individual atoms by the total number of bonds. \[\frac{4}{2} = 2\] The bond order is 2. A high bond order indicates more attraction between electrons. A higher bond order also means that the atoms are held together more tightly. With a lower bond order, there is less attraction between electrons and this causes the atoms to be held together more loosely. Bond order also indicates the stability of the bond. The higher the bond order, the more electrons holding the atoms together, and therefore the greater the stability. Bond order increases across a period and decreases down a group. Bond length is defined as the distance between the centers of two covalently bonded atoms. The length of the bond is determined by the number of bonded electrons (the bond order). The higher the bond order, the stronger the pull between the two atoms and the shorter the bond length. Generally, the length of the bond between two atoms is approximately the sum of the covalent radii of the two atoms. Bond length is reported in picometers. Therefore, bond length increases in the following order: triple bond < double bond < single bond. To find the bond length, follow these steps: Determine the carbon-to-chlorine bond length in CCl . Using , a C single bond has a length of 75 picometers and that a Cl single bond has a length of 99 picometers. When added together, the bond length of a C-Cl bond is approximately 174 picometers. Determine the carbon-oxygen bond length in CO . Using , we see that a C double bond has a length of 67 picometers and that an O double bond has a length of 57 picometers. When added together, the bond length of a C=O bond is approximately 124 picometers. Because the bond length is proportional to the , the bond length trends in the periodic table follow the same trends as atomic radii: bond length decreases across a period and increases down a group. 1. First, write the Lewis structure for \(O_2\). There is a double bond between the two oxygen atoms; therefore, the bond order of the molecule is 2. 2. The Lewis structure for NO is given below: To find the bond order of this molecule, take the average of the bond orders. N=O has a bond order of two, and both N-O bonds have a bond order of one. Adding these together and dividing by the number of bonds (3) reveals that the bond order of nitrate is 1.33. 3. To find the carbon-nitrogen bond length in HCN, draw the Lewis structure of HCN. The bond between carbon and nitrogen is a triple bond, and a triple bond between carbon and nitrogen has a bond length of approximately 60 + 54 =114 pm. 4. From the Lewis structures for CO and CO, there is a double bond between the carbon and oxygen in CO and a triple bond between the carbon and oxygen in CO. Referring to the table above, a double bond between carbon and oxygen has a bond length of approximately 67 + 57 = 124 pm and a triple bond between carbon and oxygen has a bond length of approximately 60 + 53 =113 pm. Therefore, the bond length is greater in CO . Another method makes use of the fact that the more electron bonds between the atoms, the tighter the electrons are pulling the atoms together. Therefore, the bond length is greater in CO . 5. To find the nitrogen-to-fluorine bond length in NF , draw the Lewis structure. The bond between fluorine and nitrogen is a single bond. From the table above, a single bond between fluorine and nitrogen has a bond length of approximately 64 + 71 =135 pm. | 6,594 | 2,821 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Hydrides |
The term hydride is commonly named after binary compounds that hydrogen forms with other elements of the periodic table. Hydride compounds in general form with almost any element, except a few noble gases. The trends and properties vary according to the type of intermolecular force that bonds the elements together, the temperature, its molecular masses, and other components. Hydrides are classified into three major groups, depending on what elements the hydrogen bonds to. The three major groups are covalent, ionic, and metallic hydrides. Formally, hydride is known as the negative ion of a hydrogen, H , also called a . Because of this negative charge, hydrides have reducing, or basic properties. Its special characteristics will be further discussed. The first major group is covalent hydrides, which is when a hydrogen atom and one or more non-metals form compounds. This occurs when hydrogen covalently bonds to a more electropositive element by sharing electron pairs. These hydrides can be volatile or non-volatile. Volatile simply means being readily able to be vaporized at low temperatures. One such example of a covalent hydride is when hydrogen bonds with chlorine and forms hydrochloric acid (\(HCl\)). Examples are listed below: \[\ce{H2(g) + Cl2(g) -> 2HCl(g)} \label{1}\] \[\ce{3H2(g) + N2(g) -> 2NH3(g)} \label{2}\] The hydrides of nonmetals on the periodic table become more electronegative as you move from group 13 to 17. This means that they are less capable of donating an electron, and want to keep them because their electron orbital becomes fuller. Instead of donating a \(H^-\), they would instead donate a \(H^+\) because they are more acidic. Boron can form many different types of hydrides; one of them is borane (\(\ce{BH3}\)), which reacts violently with air and is easily oxidized. Borane occurs as a gaseous substance, and can form \(\ce{B2H6}\) by two borane molecules combined with each other. Borane is not a stable compound because it does not follow a complete octet rule since it has only six valence electrons. Ammonia is an important nitrogen hydride that is possible due to the synthesis of nitrogen and water which is called the . The chemical equation for this reaction is: \[\ce{N2(g) + 3H2(g) <=> 2NH3(g)} \nonumber\] To yield ammonia, there needs to be a catalyst to speed up the reaction, a high temperature and a high pressure. Ammonia is a reagent used in many chemistry experiments and is used as fertilizer. Ammonia can react with sulfuric acid to produce ammonium sulfate, which is also an important fertilizer. In this reaction, ammonia acts as a base since it receives electrons while sulfuric acid gives off electrons. \[\ce{2NH3(aq) + H2SO4(aq) <=> (NH4)2SO4(aq)} \nonumber\] Other hydrides of nitrogen include ammonium chloride, hydrazine and hydroxylamine. Ammonium chloride is widely used in dry-cell batteries and clean metals. The second category of hydrides are (also known as or ). These compounds form between hydrogen and the most active metals, especially with the alkali and alkaline-earth metals of group one and two elements. In this group, the hydrogen acts as the hydride ion (\(H^-\)). They bond with more electropositive metal atoms. Ionic hydrides are usually binary compounds (i.e., only two elements in the compound) and are also insoluble in solutions. \[\ce{A(s) + H2(g) -> 2AH(s)} \label{3}\] with \(A\) as any metal. \[\ce{A(s) + H2(g) -> AH2(s)} \label{4}\] with \(A\) as any metal. Ionic hydrides combine vigorously with water to produce hydrogen gas. As ionic hydrides, alkali metal hydrides contain the hydride ion \(H^-\) as well. They are all very reactive and readily react with various compounds. For example, when an alkali metal reacts with hydrogen gas under heat, an ionic hydride is produced. Alkali metal hydrides also react with water to produce hydrogen gas and a hydroxide salt: \[\ce{MH(s) + H2O(l) -> MOH(aq) + H2(g)} \nonumber\] The third category of hydrides are metallic hydrides, also known as interstitial hydrides. Hydrogen bonds with transition metals. One interesting and unique characteristic of these hydrides are that they can be nonstoichiometric, meaning basically that the fraction of H atoms to the metals are not fixed. Nonstoichiometric compounds have a variable composition. The idea and basis for this is that with metal and hydrogen bonding there is a crystal lattice that H atoms can and may fill in between the lattice while some might, and is not a definite ordered filling. Thus it is not a fixed ratio of H atoms to the metals. Even so, metallic hydrides consist of more basic stoichiometric compounds as well. You may think that hydrides are all intact through hydrogen bonding because of the presence of at least a hydrogen atom, but that is false. Only some hydrides are connected with . Hydrogen bonds have energies of the order of 15-40 kJ/mol, which are fairly strong but in comparison with covalent bonds at energies greater than 150 kJ/mol, they are still much weaker. Some hydrogen bonding can be weak if they are mildly encountered with neighboring molecules. Specifically fluorine, oxygen, and nitrogen are more vulnerable to hydrogen bonding. In hydrides, hydrogen is bonded with a highly electronegative atom so their properties are more distinguished. Such that in the chart below comparing boiling points of groups 14-17 hydrides, the values of ammonia (NH ), water (H O), and hydrogen fluoride (HF) break the increasing boiling point trend. Supposedly, as the molecular mass increases, the boiling points increase as well. Due to the hydrogen bonds of the three following hydrides, they distinctly have high boiling points instead of the initial assumption of having the lowest boiling points. What occur in these hydrogen bonds are strong dipole-dipole attractions because of the high ionic character of the compounds. | 5,896 | 2,822 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.04%3A_Citric_acid_cycle |
The is the third stage of the catabolism of food, including carbohydrates, fats, and proteins. It is called the citric acid cycle because a \(\ce{2C}\) acetyl-\(\ce{CoA}\) produced in the 2nd stage of catabolism of foods reacts with a \(\ce{4C}\) oxaloacetate and produces \(\ce{6C}\)-citrate in the first reaction. The citrate is oxidized through a series of eight reactions producing two carbon dioxide (\(\ce{CO2}\) and re-generates the oxaloacetate to repeat the next round of the cycle, as shown in Figure \(\Page {1}\). It is also called the because citrate has three carboxylates (\(\ce{-COO^{-}}\)) groups, i.e., triacid. Another name for it is the in honor of Hans Krebs who discovered this metabolic pathway. The oxidation reactions in the citric acid cycle release energy which is coupled to the reduction of \(\ce{NAD^{+}}\) to \(\ce{NADH}\), \(\ce{FAD}\) to \(\ce{FADH}\), or conversions of \(\ce{ADP}\) to high-energy \(\ce{ATP}\). Guanosine triphosphate (\(\ce{GTP}\)) -another high-energy molecule, is also produced from guanosine diphosphate (\(\ce{GDP}\)) but this reaction ultimately reverses to transfer its energy to produce \(\ce{ATP}\). The following equations of the reactions are modified from a public domain at . To understand the firs reaction, recall that the \(\ce{-CH3}\) group of acetyl-\(\ce{CoA}\) has relatively acidic protons as it is \(\alpha\ce{-C}\) to a carbonyl \(\ce{C=O}\) group. An enzyme, acting as a base, removes a proton from the \(\ce{-CH3}\) group making it a carbanion which, being a strong nucleophile, attacks the ketone-\(\ce{C}\) of oxaloacetate, which is an electrophile. This nucleophilic reaction converts the \(\ce{C=O}\) into an \(\ce{-OH}\) group. It is followed by the hydrolysis of the thioester by \(\ce{H2O}\) through nucleophilic acyl substitution mechanism producing citrate and \(\ce{HS-CoA}\), as shown in the following overall reaction. To understand the second reaction, recall that tertiary \(\ce{-OH}\) do not oxidize, but they are easy to eliminate through E2-dehydration mechanics. The tertiary \(\ce{-OH}\) of citrate is eliminated and then \(\ce{H2O}\) adds to the alkene intermediate, but it installs a secondary \(\ce{-OH}\) group in isocitrate product, as shown below. The third reaction has three steps. In the first step, the secondary \(\ce{-OH}\) in isocitrate is oxidized to a ketone group at the expense of reduction of a \(\ce{NAD^{+}}\) into \(\ce{NADH}\). In this step, one \(\ce{-H}\) is picked up by an enzyme (\(\ce{A^{-}}\) and the second by the nicotinamide ring (shown in blue color) of the coenzyme \(\ce{NAD^{+}}\). The \(\ce{COO^{-}}\) group in the intermediate-I, being \(\beta\) to the \(\ce{C=O}\) group, eliminates (decarboxylate) easily as \(\ce{CO2}\). The decarboxylation is facilitated by cofactor \(\ce{Mg^{2+}}\) that binds with and draws electrons from the \(\ce{O's}\). The enzyme \(\ce{H-A}\) protonates the enole \(\ce{C=C}\) bond of the intermediate-II that produces \(\alpha\)-ketoglutarate, as shown below. The fourth reaction is similar to the previous link reaction's oxidative decarboxylation of pyruvate to acetyl-\(\ce{CoA}\). In this reaction \(\ce{NAD^{+}}\) is reduced into \(\ce{NADH}\) at the expense of oxidative decarboxylation of \(\alpha\)-ketoglutarate and the resulting acyl-group is transferred to \(\ce{HS-CoA}\) resulting in succinyl-\(\ce{CoA}\) product as shown in the reaction below. The fifth reaction has two steps. In the first step, a phosphate (\(\ce{Pi}\)) group substitutes \(\ce{-S-CoA}\) group from succinyl-\(\ce{CoAi}\). In the second step, the phosphate group is transferred to \(\ce{GDP}\) that converts to a \(\ce{GTP}\) by the same mechanism as in steps #7 and step#10 of glycolysis. \(\ce{GTP}\) reverts back to \(\ce{GDP}\) and transfers its phosphate group to \(\ce{ADP}\) that converts to one \(\ce{ATP}\). So, production of one \(\ce{GTP}\) is equal to production of one \(\ce{ATP}\). The overall reaction five is shown below. In the sixth creation, one \(\ce{FAD}\) is reduced into \(\ce{FADH_{2}}\) at the expense of oxidation of succinate to fumarate, by the following overall reaction. In the seventh reaction \(\ce{H2O}\) hydrates the \(\ce{C=C}\) of fumarate producing malate, as shown below. Recall that all of these reactions are catalyzed by enzymes and the enzymes are stereospecific. In this case, only fumarate (not its cis-isomer) reacts with the enzyme and only (S)-malate (not its enantiomers (R)-malate) is produced. In the eighth step, one more \(\ce{NAD^{+}}\) is reduced into \(\ce{NADH}\) at the expense of oxidation of malate into oxaloacetate, as shown in the reaction below. The oxaloacetate is the reactant of step#1 and begins the cycle again by reacting with another acetyl-. One citric acid cycle has the following overall reaction with one acetyl-\(\ce{CoA}\). \(\ce{acetyl-CoA + 3NAD^{+} + FAD + GDP + P_{i} + 2H2O -> HS-CoA + 3NADH + 3H^{+} + FADH_{2} + GTP + 2CO2}\) Recall that the catabolism of one glucose molecule produces two pyruvates that, intern, produce two acetyl-\(\ce{CoA}\) in the oxidative decarboxylation of pyruvate under aerobic conditions. So, the citric acid cycle runs twice for the catabolism of one glucose molecule. \(\ce{GTP}\) converts to \(\ce{ATP}\). \(\ce{NADH}\) and \(\ce{FADH}\) are also high energy molecules which are oxidized by \(\ce{O2}\) in the fourth stage of catabolism of food. The energy released from the oxidation of \(\ce{NADH}\) and \(\ce{FADH}\) in the fourth stage is used to produce more \(\ce{ATPs}\) which is described in the next section. The following video provides a summary of the citric acid cycle, also called . | 5,692 | 2,823 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Diazomethane_Carbenes_and_Cyclopropane_Synthesis |
The highly strained nature of cyclopropane compounds makes them very reactive and interesting synthetic targets. Additionally cyclopropanes are present in numerous biological compounds. One common method of cyclopropane synthesis is the reaction of carbenes with the double bond in alkenes or cycloalkenes. Methylene, H C, is simplest carbene, and in general carbenes have the formula R C. Other species that will also react with alkenes to form cyclopropanes but do not follow the formula of carbenes are referred to as carbenoids. Carbenes were once only thought of as short lived intermediates. The reactions of this section only deal with these short lived carbenes which are mostly prepared in situ, in conjunction with the main reaction. However, there do exist so called persistent carbenes. These persistent carbenes are stabilized by a variety of methods often including aromatic rings or transition metals. In general a carbene is neutral and has 6 valence electrons, 2 of which are non bonding. These electrons can either occupy the same sp hybridized orbital to form a singlet carbene (with paired electrons), or two different sp orbitals to from a triplet carbene (with unpaired electrons). The chemistry of triplet and singlet carbenes is quite different but can be oversimplified to the statement: singlet carbenes usually retain stereochemistry while triplet carbenes do not. The carbenes discussed in this section are singlet and thus retain stereochemistry. The reactivity of a singlet carbene is concerted and similar to that of electrophilic or nucleophilic addition (although, triplet carbenes react like biradicals, explaining why sterochemistry is not retained). The highly reactive nature of carbenes leads to very fast reactions in which the rate determining step is generally carbene formation. The preparation of methylene starts with the yellow gas diazomethane, CH N . Diazomethane can be exposed to light, heat or copper to facilitate the loss of nitrogen gas and the formation of the simplest carbene methylene. The process is driven by the formation of the nitrogen gas which is a very stable molecule. A carbene such as methlyene will react with an alkene which will break the double bond and result with a cyclopropane. The reaction will usually leave stereochemistry of the double bond unchanged. As stated before, carbenes are generally formed along with the main reaction; hence the starting material is diazomethane not methylene. In the above case 2-butene is converted to -1,2-dimethylcyclopropane. Likewise, below the configuration is maintained. 1. Knowing that cycloalkenes react much the same as regular alkenes what would be the expected structure of the product of cyclohexene and diazomethane facilitated by copper metal? 2. What would be the result of a Simmons-Smith reaction that used -3-pentene as a reagent? 3. What starting material could be used to form -1,2-diethylcyclopropane? 4. What would the following reaction yield? 5. Draw the product of this reaction. What type of reaction is this? 1. The product will be a bicyclic ring, Bicyclo[4.1.0]heptane. 5. This is a Simmons-Smith reaction which uses the carbenoid formed by the CH I and Zu-Cu. The reaction results in the same product as if methylene was used and retains stereospecificity. Iodine metal and the Zn-Cu are not part of the product. The product is -1,2-ethyl-methylcyclopropane. | 3,427 | 2,824 |
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How do alkali metals differ from alkaline earth metals in atomic structure and general properties? The alkali metals all have a single electron in their outermost shell. In contrast, the alkaline earth metals have a completed subshell in their outermost shell. In general, the alkali metals react faster and are more reactive than the corresponding alkaline earth metals in the same period. Why does the reactivity of the alkali metals decrease from cesium to lithium? Predict the formulas for the nine compounds that may form when each species in column 1 of reacts with each species in column 2. \[\ce{Na + I2 ⟶ 2NaI\\
2Na + Se ⟶ Na2Se\\
2Na + O2 ⟶ Na2O2}\] \[\ce{Sr + I2⟶SrI2\\
Sr + Se⟶SeSe\\
2Sr + O2⟶2SrO}\] \[\ce{2Al + 3I2⟶2AlI3\\
2Al + 3Se⟶Al2Se3\\
4Al + 3O2⟶2Al2O3}\] Predict the best choice in each of the following. You may wish to review the chapter on electronic structure for relevant examples. Sodium chloride and strontium chloride are both white solids. How could you distinguish one from the other? The possible ways of distinguishing between the two include infrared spectroscopy by comparison of known compounds, a flame test that gives the characteristic yellow color for sodium (strontium has a red flame), or comparison of their solubilities in water. At 20 °C, NaCl dissolves to the extent of \(\mathrm{\dfrac{35.7\: g}{100\: mL}}\) compared with \(\mathrm{\dfrac{53.8\: g}{100\: mL}}\) for SrCl . Heating to 100 °C provides an easy test, since the solubility of NaCl is \(\mathrm{\dfrac{39.12\: g}{100\: mL}}\), but that of SrCl is \(\mathrm{\dfrac{100.8\: g}{100\: mL}}\). Density determination on a solid is sometimes difficult, but there is enough difference (2.165 g/mL NaCl and 3.052 g/mL SrCl ) that this method would be viable and perhaps the easiest and least expensive test to perform. The reaction of quicklime, CaO, with water produces slaked lime, Ca(OH) , which is widely used in the construction industry to make mortar and plaster. The reaction of quicklime and water is highly exothermic: \[\ce{CaO}(s)+\ce{H2O}(l)⟶\ce{Ca(OH)2}(s) \hspace{20px} ΔH=\mathrm{−350\: kJ\:mol^{−1}}\] Write a balanced equation for the reaction of elemental strontium with each of the following: (a) \(\ce{2Sr}(s)+\ce{O2}(g)⟶\ce{2SrO}(s)\); (b) \(\ce{Sr}(s)+\ce{2HBr}(g)⟶\ce{SrBr2}(s)+\ce{H2}(g)\); (c) \(\ce{Sr}(s)+\ce{H2}(g)⟶\ce{SrH2}(s)\); (d) \(\ce{6Sr}(s)+\ce{P4}(s)⟶\ce{2Sr3P2}(s)\); (e) \(\ce{Sr}(s)+\ce{2H2O}(l)⟶\ce{Sr(OH)2}(aq)+\ce{H2}(g)\) How many moles of ionic species are present in 1.0 L of a solution marked 1.0 mercury(I) nitrate? What is the mass of fish, in kilograms, that one would have to consume to obtain a fatal dose of mercury, if the fish contains 30 parts per million of mercury by weight? (Assume that all the mercury from the fish ends up as mercury(II) chloride in the body and that a fatal dose is 0.20 g of HgCl .) How many pounds of fish is this? 11 lb The elements sodium, aluminum, and chlorine are in the same period. Does metallic tin react with HCl? Yes, tin reacts with hydrochloric acid to produce hydrogen gas. What is tin pest, also known as tin disease? Compare the nature of the bonds in PbCl to that of the bonds in PbCl . In PbCl , the bonding is ionic, as indicated by its melting point of 501 °C. In PbCl , the bonding is covalent, as evidenced by it being an unstable liquid at room temperature. Is the reaction of rubidium with water more or less vigorous than that of sodium? How does the rate of reaction of magnesium compare? Write an equation for the reduction of cesium chloride by elemental calcium at high temperature. \[\ce{2CsCl}(l)+\ce{Ca}(g)\:\mathrm{\overset{countercurrent \\ fractionating \\ tower}{\xrightarrow{\hspace{40px}}}}\:\ce{2Cs}(g)+\ce{CaCl2}(l)\] Why is it necessary to keep the chlorine and sodium, resulting from the electrolysis of sodium chloride, separate during the production of sodium metal? Give balanced equations for the overall reaction in the electrolysis of molten lithium chloride and for the reactions occurring at the electrodes. You may wish to review the chapter on electrochemistry for relevant examples. Cathode (reduction): \(\ce{2Li+} + \ce{2e-}⟶\ce{2Li}(l)\); Anode (oxidation): \(\ce{2Cl-}⟶\ce{Cl2}(g)+\ce{2e-}\); Overall reaction: \(\ce{2Li+}+\ce{2Cl-}⟶\ce{2Li}(l)+\ce{Cl2}(g)\) The electrolysis of molten sodium chloride or of aqueous sodium chloride produces chlorine. Calculate the mass of chlorine produced from 3.00 kg sodium chloride in each case. You may wish to review the chapter on electrochemistry for relevant examples. What mass, in grams, of hydrogen gas forms during the complete reaction of 10.01 g of calcium with water? 0.5035 g H How many grams of oxygen gas are necessary to react completely with 3.01 × 10 atoms of magnesium to yield magnesium oxide? Magnesium is an active metal; it burns in the form of powder, ribbons, and filaments to provide flashes of brilliant light. Why is it possible to use magnesium in construction? Despite its reactivity, magnesium can be used in construction even when the magnesium is going to come in contact with a flame because a protective oxide coating is formed, preventing gross oxidation. Only if the metal is finely subdivided or present in a thin sheet will a high-intensity flame cause its rapid burning. Why is it possible for an active metal like aluminum to be useful as a structural metal? Describe the production of metallic aluminum by electrolytic reduction. Extract from ore: \(\ce{AlO(OH)}(s)+\ce{NaOH}(aq)+\ce{H2O}(l)⟶\ce{Na[Al(OH)4]}(aq)\) Recover: \(\ce{2Na[Al(OH)4]}(s)+\ce{H2SO4}(aq)⟶\ce{2Al(OH)3}(s)+\ce{Na2SO4}(aq)+\ce{2H2O}(l)\) Sinter: \(\ce{2Al(OH)3}(s)⟶\ce{Al2O3}(s)+\ce{3H2O}(g)\) Dissolve in Na AlF ( ) and electrolyze: \(\ce{Al^3+}+\ce{3e-}⟶\ce{Al}(s)\) What is the common ore of tin and how is tin separated from it? A chemist dissolves a 1.497-g sample of a type of metal (an alloy of Sn, Pb, Sb, and Cu) in nitric acid, and metastannic acid, H SnO , is precipitated. She heats the precipitate to drive off the water, which leaves 0.4909 g of tin(IV) oxide. What was the percentage of tin in the original sample? 25.83% Consider the production of 100 kg of sodium metal using a current of 50,000 A, assuming a 100% yield. (a) How long will it take to produce the 100 kg of sodium metal? (b) What volume of chlorine at 25 °C and 1.00 atm forms? What mass of magnesium forms when 100,000 A is passed through a MgCl melt for 1.00 h if the yield of magnesium is 85% of the theoretical yield? 39 kg Give the hybridization of the metalloid and the molecular geometry for each of the following compounds or ions. You may wish to review the chapters on chemical bonding and advanced covalent bonding for relevant examples. Write a Lewis structure for each of the following molecules or ions. You may wish to review the chapter on chemical bonding. (a) H BPH : ; (b) \(\ce{BF4-}\): ; (c) BBr : ; (d) B(CH ) : ; (e) B(OH) : Describe the hybridization of boron and the molecular structure about the boron in each of the following: Using only the periodic table, write the complete electron configuration for silicon, including any empty orbitals in the valence shell. You may wish to review the chapter on electronic structure. 1 2 2 3 3 3 . Write a Lewis structure for each of the following molecules and ions: Describe the hybridization of silicon and the molecular structure of the following molecules and ions: (a) (CH ) SiH: bonding about Si; the structure is tetrahedral; (b) \(\ce{SiO4^4-}\): sp bonding about Si; the structure is tetrahedral; (c) Si H : bonding about each Si; the structure is linear along the Si-Si bond; (d) Si(OH) : bonding about Si; the structure is tetrahedral; (e) \(\ce{SiF6^2-}\): sp bonding about Si; the structure is octahedral Describe the hybridization and the bonding of a silicon atom in elemental silicon. Classify each of the following molecules as polar or nonpolar. You may wish to review the chapter on chemical bonding. (a) SiH (b) Si H (c) SiCl H (d) SiF (e) SiCl F (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar Silicon reacts with sulfur at elevated temperatures. If 0.0923 g of silicon reacts with sulfur to give 0.3030 g of silicon sulfide, determine the empirical formula of silicon sulfide. Name each of the following compounds: (a) tellurium dioxide or tellurium(IV) oxide; (b) antimony(III) sulfide; (c) germanium(IV) fluoride; (d) silane or silicon(IV) hydride; (e) germanium(IV) hydride Write a balanced equation for the reaction of elemental boron with each of the following (most of these reactions require high temperature): Why is boron limited to a maximum coordination number of four in its compounds? Boron has only and orbitals available, which can accommodate a maximum of four electron pairs. Unlike silicon, no orbitals are available in boron. Write a formula for each of the following compounds: From the data given in , determine the standard enthalpy change and the standard free energy change for each of the following reactions: (a) Δ ° = 87 kJ; Δ ° = 44 kJ; (b) Δ ° = −109.9 kJ; Δ = −154.7 kJ; (c) Δ ° = −510 kJ; Δ ° = −601.5 kJ A hydride of silicon prepared by the reaction of Mg Si with acid exerted a pressure of 306 torr at 26 °C in a bulb with a volume of 57.0 mL. If the mass of the hydride was 0.0861 g, what is its molecular mass? What is the molecular formula for the hydride? Suppose you discovered a diamond completely encased in a silicate rock. How would you chemically free the diamond without harming it? A mild solution of hydrofluoric acid would dissolve the silicate and would not harm the diamond. Carbon forms a number of allotropes, two of which are graphite and diamond. Silicon has a diamond structure. Why is there no allotrope of silicon with a graphite structure? Nitrogen in the atmosphere exists as very stable diatomic molecules. Why does phosphorus form less stable P molecules instead of P molecules? In the N molecule, the nitrogen atoms have an σ bond and two π bonds holding the two atoms together. The presence of three strong bonds makes N a very stable molecule. Phosphorus is a third-period element, and as such, does not form π bonds efficiently; therefore, it must fulfill its bonding requirement by forming three σ bonds. Write balanced chemical equations for the reaction of the following acid anhydrides with water: Determine the oxidation number of each element in each of the following compounds: (a) H = 1+, C = 2+, and N = 3−; (b) O = 2+ and F = 1−; (c) As = 3+ and Cl = 1− Determine the oxidation state of sulfur in each of the following: Arrange the following in order of increasing electronegativity: F; Cl; O; and S. S < Cl < O < F Why does white phosphorus consist of tetrahedral P molecules while nitrogen consists of diatomic N molecules? Why does hydrogen not exhibit an oxidation state of 1− when bonded to nonmetals? The electronegativity of the nonmetals is greater than that of hydrogen. Thus, the negative charge is better represented on the nonmetal, which has the greater tendency to attract electrons in the bond to itself. The reaction of calcium hydride, CaH , with water can be characterized as a Lewis acid-base reaction: \[\ce{CaH2}(s)+\ce{2H2O}(l)⟶\ce{Ca(OH)2}(aq)+\ce{2H2}(g)\] Identify the Lewis acid and the Lewis base among the reactants. The reaction is also an oxidation-reduction reaction. Identify the oxidizing agent, the reducing agent, and the changes in oxidation number that occur in the reaction. In drawing Lewis structures, we learn that a hydrogen atom forms only one bond in a covalent compound. Why? Hydrogen has only one orbital with which to bond to other atoms. Consequently, only one two-electron bond can form. What mass of CaH is necessary to react with water to provide enough hydrogen gas to fill a balloon at 20 °C and 0.8 atm pressure with a volume of 4.5 L? The balanced equation is: \[\ce{CaH2}(s)+\ce{2H2O}(l)⟶\ce{Ca(OH)2}(aq)+\ce{2H2}(g)\] What mass of hydrogen gas results from the reaction of 8.5 g of KH with water? \[\ce{KH + H2O ⟶ KOH + H2}\] 0.43 g H Carbon forms the \(\ce{CO3^2-}\) ion, yet silicon does not form an analogous \(\ce{SiO3^2-}\) ion. Why? Complete and balance the following chemical equations: (a) hardening of plaster containing slaked lime \[\ce{Ca(OH)2 + CO2 ⟶}\] (b) removal of sulfur dioxide from the flue gas of power plants \[\ce{CaO + SO2 ⟶}\] (c) the reaction of baking powder that produces carbon dioxide gas and causes bread to rise \[\ce{NaHCO3 + NaH2PO4 ⟶}\] (a) \(\ce{Ca(OH)2}(aq)+\ce{CO2}(g)⟶\ce{CaCO3}(s)+\ce{H2O}(l)\); (b) \(\ce{CaO}(s)+\ce{SO2}(g)⟶\ce{CaSO3}(s)\); (c) \(\ce{2NaHCO3}(s)+\ce{NaH2PO4}(aq)⟶\ce{Na3PO4}(aq)+\ce{2CO2}(g)+\ce{2H2O}(l)\) Heating a sample of Na CO ⋅ H O weighing 4.640 g until the removal of the water of hydration leaves 1.720 g of anhydrous Na CO . What is the formula of the hydrated compound? Write the Lewis structures for each of the following: (a) NH : For each of the following, indicate the hybridization of the nitrogen atom (for \(\ce{N3-}\), the central nitrogen). Explain how ammonia can function both as a Brønsted base and as a Lewis base. Ammonia acts as a Brønsted base because it readily accepts protons and as a Lewis base in that it has an electron pair to donate. Determine the oxidation state of nitrogen in each of the following. You may wish to review the chapter on chemical bonding for relevant examples. For each of the following, draw the Lewis structure, predict the ONO bond angle, and give the hybridization of the nitrogen. You may wish to review the chapters on chemical bonding and advanced theories of covalent bonding for relevant examples. (a) NO (b) \(\ce{NO2-}\) (c) \(\ce{NO2+}\) (a) NO : Nitrogen is hybridized. The molecule has a linear geometry with an ONO bond angle of 180°. How many grams of gaseous ammonia will the reaction of 3.0 g hydrogen gas and 3.0 g of nitrogen gas produce? Although PF and AsF are stable, nitrogen does not form NF molecules. Explain this difference among members of the same group. Nitrogen cannot form a NF molecule because it does not have orbitals to bond with the additional two fluorine atoms. The equivalence point for the titration of a 25.00-mL sample of CsOH solution with 0.1062 HNO is at 35.27 mL. What is the concentration of the CsOH solution? Write the Lewis structure for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry. (a) ; (b) ; (c) ; (d) ; (e) Describe the molecular structure of each of the following molecules or ions listed. You may wish to review the chapter on chemical bonding and molecular geometry. Complete and balance each of the following chemical equations. (In some cases, there may be more than one correct answer.) (a) \(\ce{P4}(s)+\ce{4Al}(s)⟶\ce{4AlP}(s)\); (b) \(\ce{P4}(s)+\ce{12Na}(s)⟶\ce{4Na3P}(s)\); (c) \(\ce{P4}(s)+\ce{10F2}(g)⟶\ce{4PF5}(l)\); (d) \(\ce{P4}(s)+\ce{6Cl2}(g)⟶\ce{4PCl3}(l)\) or \(\ce{P4}(s)+\ce{10Cl2}(g)⟶\ce{4PCl5}(l)\); (e) \(\ce{P4}(s)+\ce{3O2}(g)⟶\ce{P4O6}(s)\) or \(\ce{P4}(s)+\ce{5O2}(g)⟶\ce{P4O10}(s)\); (f) \(\ce{P4O6}(s)+\ce{2O2}(g)⟶\ce{P4O10}(s)\) Describe the hybridization of phosphorus in each of the following compounds: P O , P O , PH I (an ionic compound), PBr , H PO , H PO , PH , and P H . You may wish to review the chapter on advanced theories of covalent bonding. What volume of 0.200 NaOH is necessary to neutralize the solution produced by dissolving 2.00 g of PCl is an excess of water? Note that when H PO is titrated under these conditions, only one proton of the acid molecule reacts. 291 mL How much POCl can form from 25.0 g of PCl and the appropriate amount of H O? How many tons of Ca (PO ) are necessary to prepare 5.0 tons of phosphorus if the yield is 90%? 28 tons Write equations showing the stepwise ionization of phosphorous acid. Draw the Lewis structures and describe the geometry for the following: (a) ; (b) ; (c) ; (d) Why does phosphorous acid form only two series of salts, even though the molecule contains three hydrogen atoms? Assign an oxidation state to phosphorus in each of the following: (a) P = 3+; (b) P = 5+; (c) P = 3+; (d) P = 5+; (e) P = 3−; (f) P = 5+ Phosphoric acid, one of the acids used in some cola drinks, is produced by the reaction of phosphorus(V) oxide, an acidic oxide, with water. Phosphorus(V) oxide is prepared by the combustion of phosphorus. Predict the product of burning francium in air. FrO Using equations, describe the reaction of water with potassium and with potassium oxide. Write balanced chemical equations for the following reactions: (a) \(\ce{2Zn}(s)+\ce{O2}(g)⟶\ce{2ZnO}(s)\); (b) \(\ce{ZnCO3}(s)⟶\ce{ZnO}(s)+\ce{CO2}(g)\); (c) \(\ce{ZnCO3}(s)+\ce{2CH3COOH}(aq)⟶\ce{Zn(CH3COO)2}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\); (d) \(\ce{Zn}(s)+\ce{2HBr}(aq)⟶\ce{ZnBr2}(aq)+\ce{H2}(g)\) Write balanced chemical equations for the following reactions: Illustrate the amphoteric nature of aluminum hydroxide by citing suitable equations. \(\ce{Al(OH)3}(s)+\ce{3H+}(aq)⟶\ce{Al^3+}+\ce{3H2O}(l)\); \(\ce{Al(OH)3}(s)+\ce{OH-}⟶\ce{[Al(OH)4]-}(aq)\) Write balanced chemical equations for the following reactions: Write balanced chemical equations for the following reactions: (a) \(\ce{Na2O}(s)+\ce{H2O}(l)⟶\ce{2NaOH}(aq)\); (b) \(\ce{Cs2CO3}(s)+\ce{2HF}(aq)⟶\ce{2CsF}(aq)+\ce{CO2}(g)+\ce{H2O}(l)\); (c) \(\ce{Al2O3}(s)+\ce{6HClO4}(aq)⟶\ce{2Al(ClO4)3}(aq)+\ce{3H2O}(l)\); (d) \(\ce{Na2CO3}(aq)+\ce{Ba(NO3)2}(aq)⟶\ce{2NaNO3}(aq)+\ce{BaCO3}(s)\); (e) \(\ce{TiCl4}(l)+\ce{4Na}(s)⟶\ce{Ti}(s)+\ce{4NaCl}(s)\) What volume of 0.250 H SO solution is required to neutralize a solution that contains 5.00 g of CaCO ? Which is the stronger acid, HClO or HBrO ? Why? HClO is the stronger acid because, in a series of oxyacids with similar formulas, the higher the electronegativity of the central atom, the stronger is the attraction of the central atom for the electrons of the oxygen(s). The stronger attraction of the oxygen electron results in a stronger attraction of oxygen for the electrons in the O-H bond, making the hydrogen more easily released. The weaker this bond, the stronger the acid. Write a balanced chemical equation for the reaction of an excess of oxygen with each of the following. Remember that oxygen is a strong oxidizing agent and tends to oxidize an element to its maximum oxidation state. Which is the stronger acid, H SO or H SeO ? Why? You may wish to review the chapter on acid-base equilibria. As H SO and H SeO are both oxyacids and their central atoms both have the same oxidation number, the acid strength depends on the relative electronegativity of the central atom. As sulfur is more electronegative than selenium, H SO is the stronger acid. Explain why hydrogen sulfide is a gas at room temperature, whereas water, which has a lower molecular mass, is a liquid. Give the hybridization and oxidation state for sulfur in SO , in SO , and in H SO . SO , 4+; SO , , 6+; H SO , , 6+ Which is the stronger acid, NaHSO or NaHSO ? Determine the oxidation state of sulfur in SF , SO F , and KHS. SF : S = 6+; SO F : S = 6+; KHS: S = 2− Which is a stronger acid, sulfurous acid or sulfuric acid? Why? Oxygen forms double bonds in O , but sulfur forms single bonds in S . Why? Sulfur is able to form double bonds only at high temperatures (substantially endothermic conditions), which is not the case for oxygen. Give the Lewis structure of each of the following: Write two balanced chemical equations in which sulfuric acid acts as an oxidizing agent. There are many possible answers including: \[\ce{Cu}(s)+\ce{2H2SO4}(l)⟶\ce{CuSO4}(aq)+\ce{SO2}(g)+\ce{2H2O}(l)\] \[\ce{C}(s)+\ce{2H2SO4}(l)⟶\ce{CO2}(g)+\ce{2SO2}(g)+\ce{2H2O}(l)\] Explain why sulfuric acid, H SO , which is a covalent molecule, dissolves in water and produces a solution that contains ions. How many grams of Epsom salts (MgSO ⋅7H O) will form from 5.0 kg of magnesium? 5.1 × 10 g What does it mean to say that mercury(II) halides are weak electrolytes? Why is SnCl4 not classified as a salt? SnCl4 is not a salt because it is covalently bonded. A salt must have ionic bonds. The following reactions are all similar to those of the industrial chemicals. Complete and balance the equations for these reactions: (a) reaction of a weak base and a strong acid \[\ce{NH3 + HClO4⟶}\] (b) preparation of a soluble silver salt for silver plating \[\ce{Ag2CO3 + HNO3⟶}\] (c) preparation of strontium hydroxide by electrolysis of a solution of strontium chloride \[\ce{SrCl2}(aq)+\ce{H2O}(l)\xrightarrow{\ce{electrolysis}}\] Which is the stronger acid, HClO3 or HBrO3? Why? In oxyacids with similar formulas, the acid strength increases as the electronegativity of the central atom increases. HClO3 is stronger than HBrO3; Cl is more electronegative than Br. What is the hybridization of iodine in IF3 and IF5? Predict the molecular geometries and draw Lewis structures for each of the following. You may wish to review the chapter on chemical bonding and molecular geometry. (a) IF (b) \(\ce{I3-}\) (c) PCl (d) SeF (e) ClF (a) ; (b) ; (c) ; (d) ; (e) Which halogen has the highest ionization energy? Is this what you would predict based on what you have learned about periodic properties? Name each of the following compounds: (a) BrF (b) NaBrO (c) PBr (d) NaClO (e) KClO (a) bromine trifluoride; (b) sodium bromate; (c) phosphorus pentabromide; (d) sodium perchlorate; (e) potassium hypochlorite Explain why, at room temperature, fluorine and chlorine are gases, bromine is a liquid, and iodine is a solid. What is the oxidation state of the halogen in each of the following? (a) H IO (b) \(\ce{IO4-}\) (c) ClO (d) ICl (e) F (a) I: 7+; (b) I: 7+; (c) Cl: 4+; (d) I: 3+; Cl: 1−; (e) F: 0 Physiological saline concentration—that is, the sodium chloride concentration in our bodies—is approximately 0.16 . A saline solution for contact lenses is prepared to match the physiological concentration. If you purchase 25 mL of contact lens saline solution, how many grams of sodium chloride have you bought? Give the hybridization of xenon in each of the following. You may wish to review the chapter on the advanced theories of covalent bonding. (a) hybridized; (b) hybridized; (c) hybridized; (d) hybridized; (e) hybridized; What is the molecular structure of each of the following molecules? You may wish to review the chapter on chemical bonding and molecular geometry. Indicate whether each of the following molecules is polar or nonpolar. You may wish to review the chapter on chemical bonding and molecular geometry. (a) nonpolar; (b) nonpolar; (c) polar; (d) nonpolar; (e) polar What is the oxidation state of the noble gas in each of the following? You may wish to review the chapter on chemical bonding and molecular geometry. A mixture of xenon and fluorine was heated. A sample of the white solid that formed reacted with hydrogen to yield 81 mL of xenon (at STP) and hydrogen fluoride, which was collected in water, giving a solution of hydrofluoric acid. The hydrofluoric acid solution was titrated, and 68.43 mL of 0.3172 sodium hydroxide was required to reach the equivalence point. Determine the empirical formula for the white solid and write balanced chemical equations for the reactions involving xenon. The empirical formula is XeF , and the balanced reactions are: \[\ce{Xe}(g)+\ce{3F2}(g)\xrightarrow{Δ}\ce{XeF6}(s)\]
\[\ce{XeF6}(s)+\ce{3H2}(g)⟶\ce{6HF}(g)+\ce{Xe}(g)\] Basic solutions of Na XeO are powerful oxidants. What mass of Mn(NO ) •6H O reacts with 125.0 mL of a 0.1717 basic solution of Na XeO that contains an excess of sodium hydroxide if the products include Xe and solution of sodium permanganate? | 23,711 | 2,825 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/Clemmensen_Reduction |
The reaction of aldehydes and ketones with zinc amalgam (Zn/Hg alloy) in concentrated hydrochloric acid, which reduces the aldehyde or ketone to a hydrocarbon, is called Clemmensen reduction. This alternative reduction involves heating a carbonyl compound with finely divided, amalgamated zinc in a hydroxylic solvent (often an aqueous mixture) containing a mineral acid such as HCl. The mercury alloyed with the zinc does not participate in the reaction, it serves only to provide a clean active metal surface. The mechanism of Clemmensen reduction is not fully understood; intermediacy of radicals are implicated. Clemmensen reduction is complementary to , which also converts aldehydes and ketones to hydrocarbons, in that the former is carried out in strongly acidic conditions and the latter in strongly basic conditions. ) ), | 855 | 2,826 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/Carbonyls_are_Electrophiles |
Carbonyl compounds are very common in biological chemistry. Classes of compounds that contain the C=O bond include amides (found in proteins & peptides, used as signaling molecules and to help catalyze and guide reactions), aldehydes and ketones (found in carbohydrates, which play structural roles in cellulose, starch and DNA, for example) and esters (found in fats that form cell membranes, among other things). Understanding the reactivity of these bonds will help you to learn about many biological processes, as well as other transformations that are important in human society. Figure CO1.1. A tripeptide, composed of three amino acids. See if you can identify the three separate sub-units. Figure CO1.2. A simple carbohydrate, glyceraldehyde. Figure CO1.3. A triglyceride. The carbonyl bond is very polar. There is a partial positive charge on the carbon and a partial negative charge on the oxygen, because oxygen is more electronegative than carbon. This charge separation is intensified because of the double bond between the carbon and oxygen. Rather than just pulling one pair of bonding electrons towards itself, the oxygen pulls two pairs of electrons towards itself. Sometimes, a resonance structure is drawn to emphasize the charge separation in the carbonyl. The structure has only one bond between the carbon and oxygen. In this structure, oxygen has an octet but carbon does not. This is not really a good Lewis structure, because the other resonance structure satisfies octets on all the atoms. However, this Lewis structure emphasizes the polarity of the bond and is sometimes drawn to reinforce that idea. Figure CO1.4. Thinking about charge distribution in a carbonyl group. Because of the positive charge on the carbonyl carbon, the most important theme in carbonyl chemistry is reaction of the carbonyl as a Lewis acid. Reactions of carbonyls almost always involve addition of an electron donor to the carbonyl carbon. The electrophilicity of carbonyls is very important in their reactivity. The goal of this chapter is to develop an understanding of how carbonyls react. We will learn about a few key factors that will be used in different combinations under different circumstances. Eventually, you will build an understanding that will allow you to follow both biological reactions and modern synthetic reactions. Figure CO1.5. Reactivity in carbonyl compounds. The carbonyl in the lower sugar on the left has reacted with the neighboring molecule. It is important to realize that biological reactions, such as carbohydrate synthesis, are very complex and can involve many, many steps. For example, the carbohydrate synthesis shown above involves additional acid-base steps as well as a reaction of a carbonyl. The additional acid base steps may involve proton donors and acceptors as well as more general Lewis acids. Locate the carbonyls in the following biological molecules and identify the functional groups that contain carbonyls in each case. Ginkgolides are biologically active terpenoids from Ginkgo trees. They are thought to have medicinal properties. Okundoperoxide is isolated from a type of sedge in Cameroon. It has modest anti-malarial properties. D-erythrose is a typical carbohydrate. Frequently, the carbonyls in carbohydrates are "masked," as in deoxyribose (below). By moving one proton from one position to another, and then breaking a single C-O bond, discover where the carbonyl is hiding. , | 3,457 | 2,827 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/11%3A_Chemical_Kinetics_I/11.08%3A_The_Method_of_Half-Lives |
Another method for determining the order of a reaction is to examine the behavior of the as the reaction progresses. The half-life can be defined as the time it takes for the concentration of a reactant to fall to half of its original value. The method of half-lives involved measuring the half-life’s dependence on concentration. The expected behavior can be predicted using the integrated rate laws we derived earlier. Using the definition of the half-life, at time \(t_{1/2}\) the concentration \([A]\) drops to half of its original value, \([A]_0\). \[ [A] = \dfrac{1}{2} [A]_o \nonumber \] at \(t=t_{1/2}\). So if the reaction is 0 order in \(A\), after one half-life \[\dfrac{1}{2} [A]_o = [A]_o - k t_{1/2} \nonumber \] Solving for \(t_{1/2}\) reveals the dependence of the half-life on the initial concentration. \[\dfrac{[A]_o}{2k} = t_{1/2} \nonumber \] So as the original concentration is , the half-life of a 0 order reaction will also . Similarly, for a first order reaction, \[\dfrac{1}{2} [A]_o = [A]_o e^{- k t_{1/2}} \nonumber \] and solving for \(t_{1/2}\) results in a concentration independent expression \[ \dfrac{\ln 2}{k} = t_{1/2} \nonumber \] It is because the half-life of a 1 order reaction is of concentration that it is oftentimes used to describe the rate of first order processes (such as radioactive decay.) For a 2 order reaction, the half-life can be expressed based on the integrated rate law. \[\dfrac{1}{\dfrac{1}{2} [A]_o} = \dfrac{1}{[A]_o} + k t_{1/2} \nonumber \] solving for \(t_{1/2}\) yields \[\dfrac{1}{ t_{1/2} } = t_{1/2} \nonumber \] In the case of a second order reaction, the half-life with initial concentration. For reactions in which the rate law depends on the concentration of more than one species, the half-life can take a much more complex form that may depend on the initial concentrations of multiple reactants, or for that matter, products! Carbon-14 decays into nitrogen-14 with first order kinetics and with a half-life of 5730 years. \[ \ce{^{14}C} \rightarrow \ce{^{14}N} \nonumber \] What is the rate constant for the decay process? What percentage of carbon-14 will remain after a biological sample has stopped ingesting carbon-14 for 1482 years? The rate constant is fairly easy to calculate: \[ t_{1/2} = \dfrac{\ln 2}{k} = \dfrac{\ln 2}{5730\, yr} = 1.21 \times 10^{-4} \,yr^{-1} \nonumber \] Now the integrated rate law can be used to solve the second part of the problem. \[ [\ce{^{14}C} ] = [\ce{^{14}C}]_o e^{-k t} \nonumber \] this can be rewritten in term of relative loss of \([\ce{^{14}C} ]\). \[ \dfrac{[\ce{^{14}C} ] }{[\ce{^{14}C}]_o} = e^{-k t} \nonumber \] so \[ \dfrac{[\ce{^{14}C} ] }{[\ce{^{14}C}]_o} = e^{- (1.21 \times 10^{-4} \,yr^{-1} )(1482 \,ys)} = 0.836 \nonumber \] So after 1482 years, there is 83.6 % of \(\ce{^{14}C}\) still left. Based on the following concentration data as a function of time, determine the behavior of the half-life as the reaction progresses. Use this information to determine if the following reaction is 0 order, 1 order, or 2 order in A. Also, use the data to estimate the rate constant for the reaction. If the original concentration is taken as 1.200 M, half of the original concentration is 0.600 M. The reaction takes 20 seconds to reduce the concentration to half of its original value. If the original concentration is taken as 0.800 M, it clearly takes 30 seconds for the concentration to reach half of that value. Based on this methodology, the following table is easy to generate: The rate constant can be calculated using any of these values: \[ \begin{align} k &=\dfrac{1}{[A]t_{1/2}} \\[4pt] &= \dfrac{1}{(0.8\,M)(30\,s)} \\[4pt] &= 0.0417 \, M^{-1}s^{-1} \end{align} \nonumber \] | 3,740 | 2,828 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/11%3A_Chemical_Kinetics_I/11.02%3A_Measuring_Reaction_Rates |
There are several methods that can be used to measure chemical reactions rates. A common method is to use spectrophotometry to monitor the concentration of a species that will absorb light. If it is possible, it is preferable to measure the appearance of a product rather than the disappearance of a reactant, due to the low background interference of the measurement. However, high-quality kinetic data can be obtained either way. The involves using flow control (which can be provided by syringes or other valves) to control the flow of reactants into a mixing chamber where the reaction takes place. The reaction mixture can then be probed spectrophotometrically. Stopped-flow methods are commonly used in physical chemistry laboratory courses (Progodich, 2014). Some methods depend on measuring the initial rate of a reaction, which can be subject to a great deal of experimental uncertainty due to fluctuations in instrumentation or conditions. Other methods require a broad range of time and concentration data. These methods tend to produce more reliable results as they can make use of the broad range of data to smooth over random fluctuations that may affect measurements. Both approaches (initial rates and full concentration profile data methods) will be discussed below. | 1,297 | 2,829 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/Pyranose_and_Furanose_Forms |
Carbohydrates are an important class of biological molecules. Although their best-known role is in energy storage in the form of glucose and starch, carbohydrates play a number of other roles. For example, they lend structural support in the backbone of DNA. They aid in homebody security and defense operations by forming molecular codes on the surfaces of cells that identify whether the cell is one of our own or an intruder. They carry specialized chemical reagents into enzymes where reactions essential to life are carried out. Some of them are sweet. The term "sugar" is often applied to any simple carbohydrate. One strict definition of a carbohydrate is a polyhydroxylated aldehyde or ketone. In general, a sugar is a molecule that contains an aldehyde or a ketone, and every carbon other than the carbonyl carbon has a hydroxyl group attached to it. This sort of structure presents lots of possibilities for reactions. In just one molecule, we have both an electrophile (the carbonyl) and a number of nucleophiles (the hydroxyls). Can sugars react with each other? Yes. Can a sugar react with itself? Of course. In fact, if you have seen drawings of sugars before, you might not have noticed the carbonyl. That's because the carbonyl is usually "masked" as a hemiacetal. The hemiacetal forms when a hydroxyl group along the carbon chain reaches back and bonds to the electrophilic carbonyl carbon. As a result, five- and six-membered rings are very common in sugars. Five-membered rings are called "furanoses" and six-membered rings are called "pyranoses". The most common way of drawing these rings are in "Haworth projections." Haworth projections don't reflect the real shape of the ring. For example, in a six-membered ring, the atoms in the ring adopt a zig-zag, up-and-down pattern in order to optimize bond angles. The chair drawing shows that relationship, but in a Haworth projection, the ring is drawn as though it were flat. Also, substituents on the atoms in the ring can be found above the ring, below the ring, or sticking out around the edge of the ring. The chair drawing or "diamond lattice projection" shows these relationships pretty well. A Haworth projection doesn't try to do that. Instead, it tries to depict stereochemical relationships: whether two substituents are on the same face of the ring or opposite faces of the ring. In a diamond lattice projection, we have to keep track of whether a given substituent is in the upper or lower position at its particular site on the ring, and that requires careful attention. In a Haworth projection, substituents are simply drawn straight up or straight down. Because there are many chiral canters in sugars, and becuase two sugars can differ by just one chiral center, Haworth projections make it easier to tell different sugars apart. When an open-chain sugar cyclizes by forming a hemiacetal, it forms a new stereocenter. Because the carbonyl carbon is trigonal planar, the hydroxyl group can approach it from either face. There is nothing to distinguish one face from the other, and so approach from either face is equally likely. That means that two different stereochemical configurations can form at the hemiacetal carbon: R and S. In sugar chemistry, these two isomers are named a different way: alpha and beta. To distinguish these two designations, you need to look at the . In a Haworth projection, the lower edge of the ring is read as being nearer to you. The upper edge is read as being farther away. Remember, that's how we usually read a chair structure, too. However, in a Haworth projection, we have to orient the ring in a specific way. The hemiacetal carbon is always placed at the right edge of the drawing. In addition, we always keep the oxygen atom on the back edge of the ring (i.e. the upper edge of the drawing). That means the ring oxygen in a Haworth projection is always found in the upper or upper right part of the drawing, with the hemiacetal carbon directly beside it to the right. If we have the Haworth projection, we can designate whether we have the alpha or beta from by seeing whether the hydroxy part of the hemiacetal points up or down. If it is down like the ants, we have an alpha isomer. If it is up like the butterflies, we have a beta isomer. Because the hemiacetal carbon can adopt either of two configurations in a ring, it is given a special name. It is called the anomeric position. , | 4,428 | 2,830 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Lasers/Semiconductor_and_Solid-state_lasers |
In both solid-state and semiconductor lasers the lasing medium is a solid. Aside from this similarity, however, these two laser types are very different from each other. In the case of the solid-state lasers the lasing species is typically an impurity that resides in a solid host, a crystal of some sort. The crystal modifies some of the quantized energy levels of the impurity, but still the lasing is almost atomic - similar to gas lasers. The physics of the quantization, in the case of semiconductor lasers, is very different. In addition, solid-state lasers are always optically pumped, whereas semiconductor lasers are excited by the passage of electric current through them. We have already seen how atomic energy levels become modified when two or more atoms bind to form a molecule. The energy quantization picture becomes more complex when atoms bind together to form a solid. Below, we will first examine energy quantization in solids before we begin a more detailed examination of solid-state and semiconductor lasers. You may recall from high school chemistry that a model that is very successful in explaining the electronic composition of stable atomic species is called the shell model. In this model, very similar to the planetary model, the atom is thought to be made of shells of different types. Each shell-type can accommodate a certain number of electrons. For example, an "s-shell" can be occupied by no more than two electrons, whereas p-shells can have up to six electrons in them, etc. To "make stable atoms" we start by filling electrons in these shells starting in "level" 1, then level-2, etc., following the specified hierarchy: 1s; 2s, 2p; 3s, 3p; 4s, 3d, 4p; 5s, 4d, 5p; etc. (Remember that the chemistry of materials is due to electromagnetic interactions alone, so for this we only need to concern ourselves with electrons.) We start with the innermost shell, an s-shell. Our first atom-type has one electron in its s-shell; that's of course the hydrogen atom, H. The next atom has two electrons in its s-shell; that's He. The third atom on our list (see the Periodic Table) has three electrons, two in its 1 s-shell and then one in its 2 s-shell; and so on. According to this model, which incidentally is very well supported by the more rigorous and fundamental laws of Quantum Mechanics, atoms prefer to fill their outer most shell. Inert gases, for example, all (except for He) have a completely filled outer p-shell. Those atoms that manage this are energetically very stable. Those that don't would then prefer to interact with another atom of the correct type to allow an exchange of electrons so that they come closer to fulfilling this desire for closed shells in order to reduce their overall energy. A bond between two types of atoms by this give-and-take is strongest when both atoms get the most fulfillment - both end up with closed shells. A good example of this is the case of table-salt: NaCl. Sodium atoms, Na, are hydrogenic (hydrogen-like); i.e. they have one electron in their outermost shell, which is an s-shell just outside an inert gas shell. So, to emulate the rare gases, all they need to do is to get rid of this electron. They cannot do this on their own (why?), but they can bind with an atom that wants an additional electron to complete its outer most shell. This happens to be true for the chlorine atom that has 5 electrons in its outermost shell, a p-shell. By sharing this electron with chlorine, the sodium atom still remains neutral, but now it, as well as the Cl, both end up with "filled" outer shells. In more ways than one, this is a happy marriage. Of course, in a grain of salt there are many Na and Cl atoms. (One grain of salt has roughly about 1 mg of mass. One proton is about 10-27 kg. How many atoms are then in one grain?) These atoms arrange themselves in a very organized pattern, called a crystal. These are rows and columns of atoms, as in a three-dimensional array, with interchanging Na and Cl species. In a salt crystal, a Na atom shares its electron with several Cl atoms, and yet because of the geometry of the crystal each Cl atom ends up with a net total of one extra and each Na atom with only one fewer electron (how?). The electric attraction of the electrons to the atom's positively charged nucleus is the primary force that holds the atom together as a whole. So, in our shell model of the atom we must also account for the difference in attraction of electrons in different shells as well as the repulsion among the electrons themselves. Electrons in outer shells are further away from the nucleus, so their force of attraction is weaker than the inner shell electrons. In this respect, those electrons that belong to the unfilled outermost shell are the most loosely bound electrons in the atom. These electrons are the ones that are mostly responsible for the atom's interaction with another atom, so they are given a special name: the valence electrons. How is the shell model related to our picture of atomic energy levels? It is primarily the valence electrons that connect these two pictures. Each atom species has its own unique energy levels, but its shell structure is not unique. When it is in its ground state, then its shells are filled according to the shell model's recipe, mentioned above. But once the atom is excited, then its electron configuration will change. The first excited state, for example, could require the valence electron to jump into the next higher shell. Further excitation of the atom could lead to even more drastic changes in its shell structure. But each of the excited energy levels corresponds to a different occupation of the accessible shells by the atom's electrons. If atoms were made of (stationary) electrostatic charges, then most atomic species would have no reason to get together to make molecules or solids. But fortunately for us, this is not the case. Two neutral atoms can attract each other simply by changing their charge distribution so that they each become polar, i.e. the electron cloud of charges shifts over so that its center is not at the positive nucleus (see the figure below). In this sense even atoms of the same species find it energetically favorable to form a molecule. Because of this two totally neutral atoms of nitrogen prefer to form an N2 molecule even in the gas form. See the model drawing, below, which shows three time frames. The top frame shows two atoms with spherically symmetric charge distributions. These atoms do not initially interact with each other. But a momentary polarization of one atom can cause the other atom to polarize as well. This is the picture in the middle frame. The bottom frame the shows that two polarized atoms attract and form a molecule in which the electronic clouds of the two atoms overlaps. When we cool a gas, the ease with which it makes a liquid has a great deal to do with how easily the atoms in the gas form a molecule. Things get even more interesting when the liquid is further cooled to make a solid. The form, i.e. atomic structure, in the solid has not only to do with the atoms wanting to share electrons, but also with a totally different aspect of energy for a large collection of atoms. This relates to the probability of arrangements, or the so called entropy. What concerns us here is that for some species of atoms the lowest possible energy is reached when the atoms arrange themselves is a very regular and repeatable array of positions, called a crystal. In every-day terminology the word "crystal" usually refers to a glassy and transparent object. But in our sense of the "regular network" a crystal may be totally opaque. In fact, most metals are crystalline in structure and are not at all transparent (at least in the visible range). Why should atoms form a regular array and make up a crystal? Why shouldn't they just clump together? In most substances atoms clump randomly, forming a solid. But in some special situations atoms crystallize to make very regular network structures. These networks can have different geometrical regularities. For example, in some crystals, the network structure is a simple cube that repeats in all directions. One atom sits at each corner of the cube, making up a simple cubic crystal. In a variation of this, in addition to the simple cubic structure each face of the cube has an atom at its center; this is called a face centered cubic crystal, etc. The salt crystal of NaCl has the simple cubic structure in which Cl and Na atoms occupy alternate corners of the cube, shown below: This network structure in the crystal is called a lattice. The atoms in the lattice are not motionless. They jiggle and oscillate, but on average remain at a fixed site. The higher the temperature of the solid, the more vigorous is the jiggling oscillation of the lattice atoms. In the case of metals, which are electric conductors, the electron clouds of the lattice atoms overlap allowing the valence electrons to jump from lattice site to site when an electric potential (or voltage) is applied to the metal. So conductors allow these electrons to move freely - they are called "free electrons" - and make up an electric current, or flow of electric charge. Insulators, on the other hand, do not share their electrons and they are not free to migrate around. These materials, such as glass, wood, rubber, most plastic, and pure water, do not support electric currents under normal circumstances. What happens to the quantized energy levels of an atom when it is influenced by the presence of all the other lattice atoms in the crystal? It turns out that the idea of quantized energy levels still remains valid, but in a totally different picture. Instead of each atom having its own set of energy levels the solid as a whole can be described by a set of so called energy bands. In this picture the bands represent the quantized energy of the whole crystal. This is represented by the number of the energy bands, their energy widths, the gap between the bands, and whether the band is full, partially full, or empty. So, in a way these bands in the crystal play a similar role to the shells in the case of individual atoms. Just as in the atomic shell model we "fill" energy bands in a solid with electrons. Depending on the atomic species that make up the crystal, these bands get filled or not. For example, in the case of the sodium atom that we examined above, the corresponding energy band picture for its solid form is shown below: Again it should be stressed that this band picture is really a combination of the energy-level-diagram and the occupancy picture of the shell model. So, here the large gap between the 1s and 2s bands represents the relatively large amount of energy that is necessary to promote an electron in a single atom of sodium from the 1s to the 2s shell. Also the filled 1s, 2s, and 2p bands indicate Na's shell structure. At the same time, the unfilled 3s band shows that these valence electrons in solid sodium can continuously change their energy within this band. This is totally due to the huge number of atoms in the solid structure and was not allowed in the quantized single atom. The electric conductivity of different materials is a consequence of their individual band structure. Metals, such as sodium, copper, aluminum, etc. have a crystalline band structure in which the upper most energy band, the valence band, is partially filled (or better yet, partially empty!). Electrons in this band can gain energy from an applied external electric potential, such as a battery, without the restriction of quantization. Insulators, on the other hand, have full valence bands, so for an electron to be promoted to its next higher available energy it has to leave this band and jump into the upper empty band. This is possible, but it takes a huge electric potential to make it happen (that's why lightning can pass through walls and kill people, but typical wiring is safe with just a thin plastic insulation over it). Semiconductors are intermediate between insulators and conductors. Their band structure is the same as an insulators' but the separating gap between their valance band and the next empty band, the conduction band, is very narrow. Because of this narrow so called band gap, semiconductors can use the thermal energy of their lattice (collisions between the valence electrons and the oscillating lattice atoms) to promote a few of their valence electrons to the conduction band. In this way these material can conduct electricity, but because of the limited number of electrons that get promoted, they are not very good conductors. Examples of crystalline materials with narrow band gaps are silicon (Si) and germanium (Ge). In both of these materials the outermost shell is a partially filled p-shell with only two (out of a maximum of 6) electrons in it. In these crystals the two valence p-shell electrons are shared by the other neighboring atoms in the lattice. To make a better conductor out of these semiconductors, atom impurities are introduced in the crystal formation through a process called doping. If one of the silicon atoms in the lattice is replaced by an arsenic (As) atom, which has three p-shell valence electrons, then the extra electron becomes loosely attached to its mother atom. This loose electron can take part in the electric conduction and therefore increases the conductivity of the crystal. Strangely enough, a similar process can occur if one of the silicon atoms is replaced by a gallium (Ga) atom, which has only one p-shell valence electron. In this case, the Ga impurity strongly attracts one of the p-shell electrons from one of its Si neighbors. This leads to the production of a so called "electron hole". The resulting electron vacancy (hole) can travel in the crystal just as an electron would - but in the opposite direction - and results in an increased conductivity. Doped Si crystals with extra electrons, negative carriers, are referred to as n-type; those with holes are called p-type. These impurities modify the crystals' energy band structure by introducing atomic-type energy levels between the valance and conduction bands of the crystal and thus further reduce the band gap. This is shown below: Diodes are the most basic of semiconductor devices. They are simply one-way conductors. By suitably doping a Si crystal, its conductivity can be altered to make it behave as a conductor in one direction and an insulator in the other direction! This is commonly done in a p-n junction. When the crystal is being formed it is first introduced to a donor-type impurity (say As for an Si crystal). As the crystal grows the concentration of the donar impurity is reduced and replaced with a growing concentration of an acceptor impurity (say Ga for an Si crystal). So, the resulting crystal begins at one end as an n-type and eventually becomes a p-type at the other end. The n-type portion has a large concentration of electrons and the p-type end a large concentration of holes. This variation causes this device to act as an effective insulator for the flow of electricity in it from the n to the p-side, but a very good conductor in the opposite direction. So, unlike a regular conductor that allows internal flow of electricity in both directions, this device acts like a unidirectional valve. An electronic component that allows electricity to flow in one way (the forward bias), but not the opposite way (reverse bias) is called a diode. Transistors are the most common of semiconductor devices. They are controllable switches made from either p-n-p or n-p-n arrays of doped silicon. Today's technology can put millions of transistors together into single integrated modules, or chips, to make a large variety of different types of devices such as computer memory chips or microprocessors. Another semiconductor device is the familiar solar cell. This is also a p-n junction with variations in the hole and electron concentrations along the crystal. When light strikes this device it excites bound electrons from the valence band into the conduction band. This creates not only a mobile electron, but a hole in the valence band. Both the electron and the oppositely traveling hole produce electric current. So, when light strikes this device it behaves as a source of electricity, just like a battery. In the case of a battery chemical energy is used to produce free electrons and thereby the electric current. In solar cells photon energy is converted into the production of electron-hole pairs and the resulting electric current in the junction. Light Emitting Diodes (LEDs) behave just the opposite of solar cells. In these devices, also having a p-n junction, the combination of hole-electron pairs, created by an applied source of electricity, give off light. So, light is produced when a free electron fills a hole, i.e. a vacancy, in the crystal that is created by the presence of a p-type impurity. The light generated by LEDs is very broad-band and not of a single wavelength. It is also generated in all directions, so these devices do not make laser light. But they are useful light sources in that they are very compact and consume little electric power. To make them more efficient and reduce absorption of the generated light by the crystal's lattice, typically the junction is made near the surface. These light sources were first made in the early 1960's. Their creation soon lead to the discovery of semiconductor lasers. Also known as diode-lasers, these are by far the most inexpensive and commonly used lasers in the world. The first diode-laser was invented in 1962 at the General Electric Research and Development Center, in Niskayuna, New York - just a few miles from the Union College campus. But it was not until the early 1980's, with the development of innovative semiconductor chip manufacturing techniques, that their mass production reached the consumer market. Similar to their LED cousins, these semiconductor devices generate light from the energy extracted when electron-hole pairs recombine. Also, as in the case of LEDs, the electron-hole pairs in the semiconductor lasers are produced by the flow of electric current in the junction - this is called the injection current. The primary difference between these lasers and photodiodes is that at high injection currents the electron-hole pair densities increase to produce population inversion, which leads to lasing. These lasers, in fact, behave very much like LEDs until the critical injection current, called the threshold current, is reached. A typical semiconductor laser, shown above, is about a few hundred microns in dimension, much smaller than a grain of table salt. (Recall that a human hair is about 100 microns thick.) It produces light from an active area which is still smaller, a few tenths of one micron, perpendicular to the flow of electric current. Therefore (recall that diffraction from a very small exit aperture leads to a large angular spread of the beam) its laser light is very divergent and requires the use of a lens to collimate it. Most commonly used diode lasers are made of several different layers of compound semiconductors, often GaAlAs. Both the doping and thickness designs of these layers are used to control the confinement of laser light in the active lasing region. For some applications reflective coatings are deposited on the front and back facets to increase the efficiency of the optical amplification by acting as mirrors. But in most cases simple cleaving of the output facets leads to sufficient light amplification for lasing action. One of the drawbacks of the diode laser's small size is that it has a short coherence length. This limits its use in applications that require large coherence lengths, such as interferometry and holography. But the portability of its size makes it far more useful in many other applications. Below are two photographs of typical diode laser modules, shown in comparison with a typical writing pen or with a quarter. Notice that the laser's packaging "can" has a glass window to protect the fragile diode laser housed inside it. This package also includes a diagnostic photodiode detector that measures the output laser light leaking from the back facet of the laser chip. Diode lasers operate in several modes. Their wavelength depends primarily on the size of the band gap of the semiconductor. But the design of the semiconductor layers, as well as active feedback into the laser can generate band-widths of a few hundred kHz. Since changing the temperature of the diode expands or contracts the crystal and in this way changes the dimensions of the laser's optical cavity, the laser's wavelength can be changed either by adjusting the injection current or by directly changing diode's temperature with another device. Wavelength tunability of a typical diode laser is about a few nm over a temperature range of about 10 oC. (This is very small compared to tunable dye lasers, for example.) Room temperature diode lasers have been manufactured with wavelength outputs from a few microns in the IR all the way to the green in the visible. Researchers are currently producing blue diode lasers, but these are not yet mass produced. These lasers have a very high efficiency and are manufactured to produce laser light from a few mW to several tenths of a Watt. They can be made to operate in cw as well as in a pulsed mode. Because of their compact size an array of these can be manufactured on single chips, called laser diode arrays, that can generate output powers in the Watt range. But the single property that makes these lasers extremely useful is that their output light can be modulated at very fast rates. Because of this, and the very narrow band-width light that they can produce, diode lasers have overtaken all other lasers used in the communication industry. Measurement of the speed of light using a GHz pulsed diode laser (center) with its beam split by the glass slide to either directly enter the fast photodiode detector on the right or to travel several meters distance before being steered to enter the second photodiode detector on the left. By measuring the total path traveled and the time between the arrival of the two pulses, the speed of light can be measured on a table top to better than 1% accuracy. The first laser ever made, the ruby laser, was a solid-state laser. In this laser the lasing is a result of atomic transitions of an impurity atom in a crystalline host. Even though the atomic levels of the lasing species are often modified because of the host crystal, the lasing process is atomic and very different from that in semiconductor lasers. All solid-state lasers are pumped optically. So, in these lasers the host must be transparent to the pumping radiation. (Why?) Also, the host must be a good heat conductor in order for it to effectively dissipate waste energy that is not used for lasing. Below we will review two of the most commonly used solid-state lasers: the ruby and YAG lasers. Ruby is an aluminum oxide (\(Al_2O_3\)) crystal, called sapphire, with a small amount of chromium oxide (\(Cr_2O_3\)) added to it. Sapphire is colorless and transparent, but the chromium doped crystal is pinkish-red in color because it strongly absorbs both in the green and in the blue. When this crystal is excited through the absorption of blue and/or green light it soon causes excitation of a meta-stable energy state of the chromium ion (\(Cr^{3+}\)). After a typical lifetime of a few milliseconds this state de-excites to the ground state with the emission of a 694.3 nm photon, which is visibly red in color. So, the primary role of the aluminum oxide crystal, aside from hosting the chromium ion, is to absorb the pump energy and to excite the ion through collisions. The above simplified energy-level diagram shows the two optically pumped broad levels that quickly decay into the upper metastable lasing levels via non-radiative transitions (shown with dashed arrows). The energy of these non-radiative transitions is lost to the crystal as heat. Finally, the chromium ion de-excites back to its ground state by emitting the laser's 694.3 nm photon. Note that Ruby is a 3-level laser and as such it requires a good deal of pump energy to achieve a population inversion. It can not produce a cw beam but produces a series of bright laser pulses. To achieve population inversion, a strong pulse of broadband light is used to excite most chromium ions to their excited state. This is typically accomplished with the use of a helical flashlamp that surrounds a cylindrical ruby crystal. The two ends of this cylinder are coated (using evaporation techniques) to reflect the red 694.3 nm light. Because the wavelength output of the ruby laser is narrow-band and the pulse of light can be strong, this laser is one of the most preferred for holography. The following diagram shows a ruby rod surrounded with a helical flashtube. The cylindrical ends are coated for reflectivity in the red to form the optical cavity of the laser. Typical ruby rods range in length from about 10 cm to 1/4 m and in diameter from a few to about 25 mm. The standard ruby lasers produce a few ms long duration pulses that range in energy from 10 to 100 J. But because ruby rods degrade quickly from excessive heating, these lasers are often operated at rather slow repetition rates (rep-rate) of few pulses per second (few Hz). Applications that require large numbers of photons in a short period of time, benefit from laser pulses that are short in duration, but have a high peak. The following graph shows two pulses of light that carry equal number of photons (equal areas under their graphs), but have very different peak heights. The one that has the higher peak, then has the shortest duration. (Note that for simplicity these pulses are depicted as rectangular in shape suggesting that lasing turns on and off at precise instants. In reality, however, laser pulses look more like hills; gradually increasing, reaching a peak, and then gradually decreasing before fully turning off.) Since the energy of each photon is fixed depending on the value of its wavelength (or its frequency), then both of these pulses carry the same amount of energy. For example, if we are told that the above pulses each have N photons emitted by a ruby laser, then we could easily calculate the pulse energies: \[\text{pulse energy} = \text{(number of photons in the pulse)} \times \text{(energy of a single photon)}\] i.e. pulse energy = N x (hn) = N x ( hc / l ) = N x { (6.63 x 10-34) (3.00 x 108) / (694.3 x 10-9) } \[ = N \times ( 2.87 \times 10^{-19}) Joules.\] Alternatively, if we were told that each of the above ruby laser pulses is a 10 J pulse, then we could easily calculate the number of photons in each pulse: number of photons = N = (pulse energy ) / (energy of a single photon) or, \[N = \dfrac{10}{ 2.87 \times 10^{-19}} = 3.5 \times 10^{19}\] Devices that measure laser light energies do not count the number of photons. Instead, they measure the energy that the photons produce in the device. (For example, photodiodes, which work like solar cells, absorb the photon energy and create electron-hole pairs, which produce a voltage across the p-n junction. This voltage is proportional to the absorbed energy and is displayed by the device indicating the pulse energy.) The following graphs are more realistic depictions of pulse energy measurements: Here again we see two graphs of roughly equal areas and therefore equal pulse energies. The top pulse creates "less energy for a longer time" while the lower one creates "more energy for a shorter time". Since power is defined as: power = ( energy ) / ( time ) (with units of 1 Watt = 1 Joule / 1 second ) Then the pulse shown on the lower graph has much more power than the top one. (Why?) This is called the pulse power, or the peak power. For comparison with a cw laser it is convenient to average the energy of the pulse over not just its duration, but also over the time that the laser is on but there is not light emitted from it. To do this, all we need do is divide the total pulse energy by the time interval from when a pulse begins until the time that the next pulse begins; i.e. the "pulse-to-pulse time": average power = ( pulse energy ) / ( pulse-to-pulse time ) In pulsed lasers there are several ways to achieve a large peak power. One of these is called Q-switching, which refers to an increase in the efficiency of the optical cavity. The Q of an amplifier is just a measure of its efficiency. An amplifier that uses little energy for its amplification is said to have a large-Q. In pulsed lasers a method of increasing efficiency is to simply increase peak power. This can happen when instead of producing a long pulse one makes the laser produce a shorter pulse with a larger peak energy. One method for doing this is, instead of causing stimulated emission, to let the medium buildup more and more of a population inversion in its metastable state. This can be done rather easily by "blocking" one of the optical cavity's reflective mirrors for long enough that the maximum number of the atoms are in the upper laser level. Then to cause all of the atoms in the upper lasing level to undergo stimulated emission "all at once" a very quick "unblocking" of the reflective mirror results in many atoms to undergo stimulated emission simultaneously; therefore the resulting laser pulse has a high peak power. In this way Q-switched lasers can produce very high peak powers, in the hundreds of megawatts! Yttrium-Aluminum-Garnet (\(Y_3Al_5O_{12}\)), YAG, is a clear and transparent crystal that is most commonly used as the host crystal for neodymium impurity atoms in Nd:YAG lasers. Even though the lasing transition occurs in the Neodymium (Nd) ions, these lasers are often called YAG lasers. Typically 1 - 2% of the Y is replaced by Nd in these lasers. Similar to the ruby laser, here again the crystal not only hosts the atom, but also its broad absorption bands (in the presence of the Nd-impurity) effectively absorb optical radiation - mostly around 700 nm and 800 nm. This energy is then transferred to the impurity ions through non-radiative processes. Similar to the level diagram for the ruby laser, the above diagram is a simplified representation of the transitions for the neodymium ion, Nd+++, that take place in the YAG laser. Again, the dashed arrows indicate non-radiative transitions, where the energy lost in them is transferred as heat to the crystal. But unlike the ruby laser, the Nd:YAG is a four-level laser and has much more efficient lasing transitions than the ruby laser. Because of this efficiency in population inversion it can be pumped to produce a wide variety of laser output energies and can be operated in the cw as well as pulsed modes. The lasing photon has a wavelength of 1064 nm, well in the IR range. Another host used for Nd is plain glass, but its optical and thermal properties are not as desirable as YAG's. The trade off is that growing large YAG crystals is not easy. A typical YAG rod, which is drilled out of a crystal block, is smaller than 1 cm in diameter and from a few to 10 cm in length. Still, these lasers can produce high output powers by using several rods in tandem. Pulsed YAG lasers can produce high peak powers by Q-switching and/or by using several rods as amplifiers. When used as amplifiers, the rods are not coated for reflectivity to produce optical amplification in the rod. Instead, the output of the first rod (the oscillator) is used as a Q-switch to generate stimulated emission in the next rod (the amplifier), and so on. In the cw mode YAG lasers are pumped either by an arc lamp or a semiconductor laser. Pumping with a diode laser is more efficient because the diode's wavelength can be chosen to closely match the absorption of the YAG. Flashlamps are often used for pulsed operation of these lasers. In this mode Q-switching the laser can produce ns pulses with relatively high peak powers. But, to reach very high powers, like that generated by the NOVA laser, many rod amplifications are necessary. YAG lasers are extremely versatile. They are used in applications from welding and drilling to range finding. Because their output wavelength is not visible, YAG lasers are used in many applications that require secrecy; from military to security applications. It is now possible to convert IR radiation into the visible range using so called non-linear crystals (also known as second-harmonic-generators). When high density photons impinge on such a crystal, the non-linear properties of its index of refraction causes the crystal to absorb two long wavelength photons and generate, in their place, one photon with twice their energy and therefore half their wavelength. Using these crystals, the YAG laser wavelength can be shortened to 532 nm (1064 nm/2), which is seen by the eye as green light. Today's green-colored laser pointers are diode-pumped YAG lasers with doubling crystals. Although we can change the wavelength of the YAG laser from IR to green (and even further into UV, via another second harmonic generation), this is not a continuous tuning. Other interesting solid-state lasers, called vibronic lasers, use effective bands as their lower lasing level. In this way these lasers can produce a tunable output; i.e. one with a continually changeable wavelength. The most commonly used laser of this type is the Ti:sapphire laser, which is an aluminum oxide crystal, the same as ruby's host, but doped with titanium instead of chromium atoms. Jay Newman and Seyffie Maleki ( ) | 33,732 | 2,833 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/04%3A_Transport/16%3A_Targeted_Diffusion/16.01%3A_Diffusion_to_Capture |
In this section we will discuss the kinetics of association of a diffusing particle with a target. What is the rate at which a diffusing molecule reaches its target? These diffusion-to-capture problems show up in many contexts. For instance: We will consider two approaches to dealing with these problems: What is the rate of encounter of a diffusing species with a spherical target? We can find a steady-state solution by determining the steady-state radial concentration profile \(C(r)\). Assume that reaction is immediate on encounter at a radius \(a\). This sets the boundary condition, \(C(a) = 0\). We also know the bulk concentration \(C_0 = C(∞)\). From our earlier discussion, the steady state solution to this problem is \[ C(r) = C_0\left( 1- \dfrac{a}{r} \right) \nonumber \] Next, to calculate the rate of collisions with the sphere, we first calculate the flux density of molecules incident on the surface of the sphere (\(r = a\)): \[ J(a) = -D \dfrac{\partial C}{\partial r}\Bigg{|}_{r=a} = - \dfrac{DC_0}{a} \] \(J\) is expressed as (molec area sec ) or [(mol/L) area sec ]. We then calculate the flux, or rate of collisions of molecules with the sphere (molec sec ), by multiplying the flux density by the surface area of the sphere (A = 4πa ): \[ \begin{aligned} j &= \dfrac{dN}{dt} = JA = \left( \dfrac{DC_0}{a} \right) (4\pi a^2) \\ &=4\pi DaC_0 \\ &\equiv kC_0 \end{aligned} \] We associate the constant or proportionality between rate of collisions and concentration with the pseudo first-order association rate constant, \(k = 4πDa\), which is proportional to the rate of diffusion to the target and the size of the target. The discussion above describes the rate of collisions of solutes with an absorbing sphere, which are applicable if the absorbing sphere is fixed. For problems involving the encounter between two species that are both diffusing in solution \( (A+B \rightarrow X) \), you can extend this treatment to the encounter of two types of particles A and B, which are characterized by two bulk concentrations C and C , two radii R and R , and two diffusion constants D and D . To describe the rate of reaction, we need to calculate the total rate of collisions between A and B molecules. Rather than describing the diffusion of both A and B molecules, it is simpler to fix the frame of reference on B and recognize that we want to describe the diffusion of A with respect to B. In that case, the effective diffusion constant is \(D=D_a+D_b \) Furthermore, we expand our encounter radius to the sum of the radii of the two spheres (R = r + r ). The flux density of A molecules incident on a single B at an encounter radius of R is given by eq. (1) \[J_{a\rightarrow b} = \dfrac{DC_A}{R_{AB}} \nonumber \] Here J describes the number of molecules of A incident per unit area at a radius R from B molecules per unit time, [molec A] [area of B] sec . If we treat the motion of B to be uncorrelated with A, then the total rate of collisions between A and B can be obtained from the product of J with the area of a sphere of radius R and the total concentration of B: \[ \begin{aligned} \dfrac{dN_{A \leftrightarrow B}}{dt} &= J_{A\rightarrow B} A_{AB}C_B \\ &=J_{A\rightarrow B}(4\pi R^2_{AB}) C_B \\ &=4\pi DR_{AB}C_AC_B \end{aligned} \] The same result is obtained if we begin with the flux density of B incident on A, J , using the same encounter radius and diffusion constant. Now comparing this with expected second order rate law for a bimolecular reaction \[ \dfrac{dN_{A \leftrightarrow B}}{dt} = k_aC_AC_B \nonumber \] we see \[ k_a = 4\pi (D_A+D_B) R_{AB} \nonumber \] k is the rate constant for a diffusion limited reaction (association). It has units of cm s , which can be converted to (L mol s ) by multiplying by Avagadro’s number. If you modify these expressions so that only part of the sphere is reactive, then similar results ensue, in which one recovers the same diffusion limited association rate (k ) multiplied by an additional factor that depends on the geometry of the surface area that is active: k =k ∙[constant]. For instance if we consider a small circular patch on a sphere that subtends a half angle θ, the geometric factor should scale as sinθ. For small θ, sinθ≈θ. If you have small patches on two spheres, which must diffusively encounter each other, the slowing of the association rate relative to the case with the fully accessible spherical surface area is \( k_a /k_{a,0} = \theta_A\theta_B(\theta_A+\theta_B)/8 \) For the association rate of molecules with a sphere of radius R covered with n absorbing spots of radius b: \(k_a = k_{a,0} = \left( 1 + \dfrac{\pi R}{nb} \right)^{-1} \) Additional configurations are explored in Berg. | 4,750 | 2,834 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/07%3A_Mixtures_and_Solutions/7.06%3A_Colligative_Properties |
Colligative properties are important properties of as they describe how the properties of the will change as (or solutes) is (are) added. Before discussing these important properties, let us first review some definitions. Solutions can exist in solid (alloys of metals are an example of solid-phase solutions), liquid, or gaseous (aerosols are examples of gas-phase solutions) forms. For the most part, this discussion will focus on liquid-phase solutions. In general (and as will be discussed in Chapter 8 in more detail) a liquid will freeze when \[ \mu_{solid} \le \mu_{liquid} \nonumber \] As such, the freezing point of the solvent in a solution will be affected by anything that changes the chemical potential of the solvent. As it turns out, the chemical potential of the solvent is reduced by the presence of a solute. In a mixture, the chemical potential of component \(A\) can be calculated by \[\mu_A=\mu_A^o + RT \ln \chi_A \label{chemp} \] And because \(\chi_A\) is always less than (or equal to) 1, the chemical potential is always reduced by the addition of another component. The condition under which the solvent will freeze is \[ \mu_{A,solid} = \mu_{A,liquid} \nonumber \] where the chemical potential of the liquid is given by Equation \ref{chemp}, which rearrangement to \[\dfrac{ \mu_A -\mu_A^o}{RT} = \ln \chi_A \nonumber \] To evaluate the temperature dependence of the chemical potential, it is useful to consider the temperature derivative at constant pressure. \[ \left[ \dfrac{\partial}{\partial T} \left( \dfrac{\mu_A-\mu_A^o}{RT} \right) \right]_{p} = \left( \dfrac{\partial \ln \chi_A}{\partial T} \right)_p \nonumber \] \[- \dfrac{\mu_A - \mu_A^o}{RT^2} + \dfrac{1}{RT} \left[ \left( \dfrac{\partial \mu_A}{\partial T} \right)_p -\left( \dfrac{\partial \mu_A^o}{\partial T} \right)_p \right] =\left( \dfrac{\partial \ln \chi_A}{\partial T} \right)_p \label{bigeq} \] Recalling that \[ \mu = H = TS \nonumber \] and \[\left( \dfrac{\partial \mu}{\partial T} \right)_p =-S \nonumber \] Equation \ref{bigeq} becomes \[ - \dfrac{(H_A -TS_A - H_A^o + TS^o_A)}{RT^2} + \dfrac{1}{RT} \left[ -S_A + S_A^o\right] =\left( \dfrac{\partial \ln \chi_A}{\partial T} \right)_p \label{bigeq2} \] And noting that in the case of the solvent freezing, \(H_A^o\) is the enthalpy of the pure solvent in solid form, and \(H_A\) is the enthalpy of the solvent in the liquid solution. So \[ H_A^o - H_a = \Delta H_{fus} \nonumber \] Equation \ref{bigeq2} then becomes \[ \dfrac{\Delta H_{fus}}{RT^2} - \cancel{ \dfrac{-S_A + S_A^o}{RT}} + \cancel{\dfrac{-S_A + S_A^o}{RT}}=\left( \dfrac{\partial \ln \chi_A}{\partial T} \right)_p \nonumber \] or \[ \dfrac{\Delta H_{fus}}{RT^2} = \left( \dfrac{\partial \ln \chi_A}{\partial T} \right)_p \nonumber \] Separating the variables puts the equation into an integrable form. \[ \int_{T^o}^T \dfrac{\Delta H_{fus}}{RT^2} dT = \int d \ln \chi_A \label{int1} \] where \(T^{o}\) is the freezing point of the pure solvent and \(T\) is the temperature at which the solvent will begin to solidify in the solution. After integration of Equation \ref{int1}: \[ - \dfrac{\Delta H_{fus}}{R} \left( \dfrac{1}{T} - \dfrac{1}{T^{o}} \right) = \ln \chi_A \label{int3} \] This can be simplified further by noting that \[ \dfrac{1}{T} - \dfrac{1}{T^o} = \dfrac{T^o - T}{TT^o} = \dfrac{\Delta T}{TT^o} \nonumber \] where \(\Delta T\) is the difference between the freezing temperature of the pure solvent and that of the solvent in the solution. Also, for small deviations from the pure freezing point, \(TT^o\) can be replaced by the approximate value \((T^o)^2\). So the Equation \ref{int3} becomes \[ - \dfrac{\Delta H_{fus}}{R(T^o)^2} \Delta T = \ln \chi_A \label{int4} \] Further, for dilute solutions, for which \(\chi_A\) , the mole fraction of the solvent is very nearly 1, then \[ \ln \chi_A \approx -(1 -\chi_A) = -\chi_B \nonumber \] where \(\chi_B\) is the mole fraction of the solute. After a small bit of rearrangement, this results in an expression for freezing point depression of \[ \Delta T = \left( \dfrac{R(T^o)^2}{\Delta H_{fus}} \right) \chi_B \nonumber \] The first factor can be replaced by \(K_f\): \[\dfrac{R(T^o)^2}{\Delta H_{fus}} = K_f \nonumber \] which is the for the solvent. \(\Delta T\) gives the magnitude of the reduction of freezing point for the solution. Since \(\Delta H_{fus}\) and \(T^o\) are properties of the solvent, the freezing point depression property is independent of the solute and is a property based solely on the nature of the solvent. Further, since \(\chi_B\) was introduced as \((1 - \chi_A)\), it represents the sum of the mole fractions of all solutes present in the solution. It is important to keep in mind that for a real solution, freezing of the solvent changes the composition of the solution by decreasing the mole fraction of the solvent and increasing that of the solute. As such, the magnitude of \(\Delta T\) will change as the freezing process continually removes solvent from the liquid phase of the solution. The derivation of an expression describing boiling point elevation is similar to that for freezing point depression. In short, the introduction of a solute into a liquid solvent lowers the chemical potential of the solvent, cause it to favor the liquid phase over the vapor phase. As sch, the temperature must be increased to increase the chemical potential of the solvent in the liquid solution until it is equal to that of the vapor-phase solvent. The increase in the boiling point can be expressed as \[ \Delta T = K_b \chi_B \nonumber \] where \[\dfrac{R(T^o)^2}{\Delta H_{vap}} = K_b \nonumber \] is called the and, like the cryoscopic constant, is a property of the solvent that is independent of the solute or solutes. A very elegant derivation of the form of the models for freezing point depression and boiling point elevation has been shared by F. E. Schubert (Schubert, 1983). Cryoscopic and ebullioscopic constants are generally tabulated using molality as the unit of solute concentration rather than mole fraction. In this form, the equation for calculating the magnitude of the freezing point decrease or the boiling point increase is \[ \Delta T = K_f \,m \nonumber \] or \[ \Delta T = K_b \,m \nonumber \] where \(m\) is the concentration of the solute in moles per kg of solvent. Some values of \(K_f\) and \(K_b\) are shown in the table below. The boiling point of a solution of 3.00 g of an unknown compound in 25.0 g of CCl raises the boiling point to 81.5 °C. What is the molar mass of the compound? The approach here is to find the number of moles of solute in the solution. First, find the concentration of the solution: \[(85.5\, °C- 76.8\, °C) = \left( 5.02\, °C\,Kg/mol \right) m \nonumber \] \[ m= 0.936\, mol/kg \nonumber \] Using the number of kg of solvent, one finds the number for moles of solute: \[ \left( 0.936 \,mol/\cancel{kg} \right) (0.02\,\cancel{kg}) =0.0234 \, mol \nonumber \] The ratio of mass to moles yields the final answer: \[\dfrac{3.00 \,g}{0.0234} = 128 g/mol \nonumber \] For much the same reason as the lowering of freezing points and the elevation of boiling points for solvents into which a solute has been introduced, the vapor pressure of a volatile solvent will be decreased due to the introduction of a solute. The magnitude of this decrease can be quantified by examining the effect the solute has on the chemical potential of the solvent. In order to establish equilibrium between the solvent in the solution and the solvent in the vapor phase above the solution, the chemical potentials of the two phases must be equal. \[\mu_{vapor} = \mu_{solvent} \nonumber \] If the solute is not volatile, the vapor will be pure, so (assuming ideal behavior) \[\mu_{vap}^o + RT \ln \dfrac{p'}{p^o} = \mu_A^o + RT \ln \chi_A \label{eq3} \] Where \(p’\) is the vapor pressure of the solvent over the solution. Similarly, for the pure solvent in equilibrium with its vapor \[ \mu_A^o = \mu_{vap}^o + RT \ln \dfrac{p_A}{p^o} \label{eq4} \] where \(p^o\) is the standard pressure of 1 atm, and \(p_A\) is the vapor pressure of the pure solvent. Substituting Equation \ref{eq4} into Equation \ref{eq3} yields \[ \cancel{\mu_{vap}^o} + RT \ln \dfrac{p'}{p^o}= \left ( \cancel{\mu_{vap}^o} + RT \ln \dfrac{p_A}{p^o} \right) + RT \ln \chi_A \nonumber \] The terms for \(\mu_{vap}^o\) cancel, leaving \[ RT \ln \dfrac{p'}{p^o}= RT \ln \dfrac{p_A}{p^o} + RT \ln \chi_A \nonumber \] Subtracting \(RT \ln(P_A/P^o)\) from both side produces \[ RT \ln \dfrac{p'}{p^o} - RT \ln \dfrac{p_A}{p^o} = RT \ln \chi_A \nonumber \] which rearranges to \[RT \ln \dfrac{p'}{p_A} = RT \ln \chi_A \nonumber \] Dividing both sides by \(RT\) and then exponentiating yields \[ \dfrac{p'}{p_A} = \chi_A \nonumber \] or \[ p'=\chi_Ap_A \label{RL} \] This last result is Raoult’s Law. A more formal derivation would use the fugacities of the vapor phases, but would look essentially the same. Also, as in the case of freezing point depression and boiling point elevations, this derivation did not rely on the nature of the solute! However, unlike freezing point depression and boiling point elevation, this derivation did not rely on the solute being dilute, so the result should apply the entire range of concentrations of the solution. Consider a mixture of two volatile liquids A and B. The vapor pressure of pure A is 150 Torr at some temperature, and that of pure B is 300 Torr at the same temperature. What is the total vapor pressure above a mixture of these compounds with the mole fraction of B of 0.600. What is the mole fraction of B in the vapor that is in equilibrium with the liquid mixture? Using Raoult’s Law (Equation \ref{RL}) \[ p_A = (0.400)(150\, Toor) =60.0 \,Torr \nonumber \] \[ p_B = (0.600)(300\, Toor) =180.0 \,Torr \nonumber \] \[ p_{tot} = p_A + p_B = 240 \,Torr \nonumber \] To get the mole fractions in the gas phase, one can use Dalton’s Law of partial pressures. \[ \chi_A = \dfrac{ p_A}{p_{tot}} = \dfrac{60.0 \,Torr}{240\,Torr} = 0.250 \nonumber \] \[ \chi_B = \dfrac{ p_B}{p_{tot}} = \dfrac{180.0 \,Torr}{240\,Torr} = 0.750 \nonumber \] And, of course, it is also useful to note that the sum of the mole fractions is 1 (as it must be!) \[ \chi_A+\chi_B =1 \nonumber \] is a process by which solvent can pass through a semi-permeable membrane (a membrane through which solvent can pass, but not solute) from an area of low solute concentration to a region of high solute concentration. The is the pressure that when exerted on the region of high solute concentration will halt the process of osmosis. The nature of osmosis and the magnitude of the osmotic pressure can be understood by examining the chemical potential of a pure solvent and that of the solvent in a solution. The chemical potential of the solvent in the solution (before any extra pressure is applied) is given by \[ \mu_A = \mu_A^o + RT \ln \chi_A \nonumber \] And since x < 1, the chemical potential is of the solvent in a solution is always lower than that of the pure solvent. So, to prevent osmosis from occurring, something needs to be done to raise the chemical potential of the solvent in the solution. This can be accomplished by applying pressure to the solution. Specifically, the process of osmosis will stop when the chemical potential solvent in the solution is increased to the point of being equal to that of the pure solvent. The criterion, therefore, for osmosis to cease is \[ \mu_A^o(p) = \mu_A(\chi_b, +\pi) \nonumber \] To solve the problem to determine the magnitude of , the pressure dependence of the chemical potential is needed in addition to understanding the effect the solute has on lowering the chemical potential of the solvent in the solution. The magnitude, therefore, of the increase in chemical potential due to the application of excess pressure must be equal to the magnitude of the reduction of chemical potential by the reduced mole fraction of the solvent in the solution. We already know that the chemical potential of the solvent in the solution is reduced by an amount given by \[ \mu^o_A - \mu_A = RT \ln \chi_A \nonumber \] And the increase in chemical potential due to the application of excess pressure is given by \[ \mu(p+\pi) = \mu(p) + \int _{p}^{\pi} \left( \dfrac{\partial \mu}{\partial p} \right)_T dp \nonumber \] The integrals on the right can be evaluated by recognizing \[\left( \dfrac{\partial \mu}{\partial p} \right)_T = V \nonumber \] where \(V\) is the molar volume of the substance. Combining these expressions results in \[ -RT \ln \chi_A = \int_{p}^{p+\pi} V\,dp \nonumber \] If the molar volume of the solvent is independent of pressure (has a very small value of \(\kappa_T\) – which is the case for most liquids) the term on the right becomes. \[ \int_{p}^{\pi} V\,dP = \left. V p \right |_{p}^{p+\pi} = V\pi \nonumber \] Also, for values of \(\chi_A\) very close to 1 \[ \ln \chi_A \approx -(1- \chi_A) = - \chi_B \nonumber \] So, for dilute solutions \[ \chi_B RT = V\pi \nonumber \] Or after rearrangement \[ \pi \dfrac{\chi_B RT}{V} \nonumber \] again, where \(V\) is the molar volume of the solvent. And finally, since \(\chi_B/V\) is the concentration of the solute \(B\) for cases where \(n_B \ll n_A\). This allows one to write a simplified version of the expression which can be used in the case of very dilute solutions \[ \pi = [B]RT \nonumber \] When a pressure exceeding the osmotic pressure \(\pi\) is applied to the solution, the chemical potential of the solvent in the solution can be made to exceed that of the pure solvent on the other side of the membrane, causing reverse osmosis to occur. This is a very effective method, for example, for recovering pure water from a mixture such as a salt/water solution. | 13,780 | 2,835 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/04%3A_Transport/15%3A_Passive_Transport/15.02%3A_Facilitated_Diffusion |
Facilitated diffusion is a type of dimensionality reduction that has been used to describe the motion of transcription factors and regulatory proteins looking for their binding target on DNA. Experiments by Riggs et al. showed that Lac repressor finds its binding site about one hundred times faster than expected by 3D diffusion. They measured k =7×10 M s , which is 100–1000 times faster than typical rates. The calculated diffusion-limited association rate from the Smoluchowski equation is k ≈10 M s using estimated values of D≈5×10 cm s and R≈5×10 cm. Berg and von Hippel theoretically described the possible ways in which nonspecific binding to DNA enabled more efficient one-dimensional motion coupled to three-dimensional transport. The transcription factor diffuses in 1D along DNA with the objective of locating a specific binding site. The association of the protein and DNA at all points is governed by a nonspecific interaction. Sliding requires a balance of nonspecific attractive forces that are not too strong (or the protein will not move) or too weak (or it will not stay bound). The nonspecific interaction is governed by an equilibrium constant and exchange rates between the bound and free forms: \[ F \overset{k_a}{\underset{k_d}{\rightleftharpoons}} B \qquad \qquad \qquad K = \dfrac{k_a}{k_d} = \dfrac{\overline{\tau}_{1D}}{\overline{\tau}_{3D}} \nonumber \] We can also think of this equilibrium constant in terms of the average times spent diffusing in 1D or 3D. The protein stays bound for a period of time dictated by the dissociation rate kd. It can then diffuse in 3D until reaching a contact with DNA again, at a point which may be short range in distance but widely separated in sequence. The target for the transcription factor search can be much larger that the physical size of the binding sequence. Since the 1D sliding is the efficient route to finding the binding site, the target size is effectively covered by the mean 1D diffusion length of the protein, that is, the average distance over which the protein will diffuse in 1D before it dissociates. Since one can express the average time that a protein remains bound as \(\overline{\tau}_{1D}=k^{-1}_d\), the target will have DNA contour length of \[ R^*=\left( \dfrac{4D_1}{k_d} \right)^{1/2} \nonumber \] If the DNA is treated as an infinitely long cylinder with radius b, and the protein is considered to have a uniform probability of nonspecifically associating with the entire surface of the DNA, then one can solve for the steady-state solution for the diffusion equation, assuming a completely absorbing target. The rate constant for specific binding to the target has been determined as \[ \eta = \dfrac{D_1K'}{D_3b} \nonumber \] where K' is the equilibrium constant for nonspecific binding per unit surface area of the cylinder (M cm or cm). We can express the equilibrium constant per base-pair as \(K=2\pi \ell bK'\), where \(\ell \) is the length of a base pair along the contour of the DNA. The association rate will be given by the product of k and the concentration of protein. _________________________________ | 3,144 | 2,836 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/05%3A_The_Second_Law/5.01%3A_Introduction_to_the_Second_Law |
Rudolph Clausius is kind enough in his 1879 work “The Mechanical Theory of Heat” (Clausius, 1879) to indicate where we have been in our discussion of thermodynamics, as well as where we are going. “The fundamental laws of the universe which correspond to the two fundamental theorems of the mechanical theory of heat:
1. The energy of the universe is constant.
2. The entropy of the universe tends to a maximum.”
― Rudolf Clausius, The , which introduces us to the topic of entropy, is amazing in how it constrains what we can experience and what we can do in the universe. As Sean M. Carroll, a CalTech Theoretical physicist, suggests in a 2010 interview with Wired Magazine (Biba, 2010), I’m trying to understand how time works. And that’s a huge question that has lots of different aspects to it. A lot of them go back to Einstein and spacetime and how we measure time using clocks. But the particular aspect of time that I’m interested in is the arrow of time: the fact that the past is different from the future. We remember the past but we don’t remember the future. There are irreversible processes. There are things that happen, like you turn an egg into an omelet, but you can’t turn an omelet into an egg. We, as observers of nature, are time travelers. And the constraints on what we can observe as we move through time step from the . But more than just understanding what the second law says, we are interested in what sorts of processes are possible. And even more to the point, what sorts of processes are . A spontaneous process is one that will occur without external forces pushing it. A process can be spontaneous even if it happens very slowly. Unfortunately, Thermodynamics is silent on the topic of how fast processes will occur, but is provides us with a powerful toolbox for predicting which processes will be spontaneous. But in order to make these predictions, a new thermodynamic law and variable is needed since the first law (which defined \(\Delta U\) and \(\Delta H\)) is insufficient. Consider the following processes: \[NaOH(s) \rightarrow Na^+(aq) + OH^-(aq) \nonumber \] with \(\Delta H < 0\) \[NaHCO_3(s) \rightarrow Na+(aq) + HCO_3^-(aq) \nonumber \] with \(\Delta H > 0\) Both reactions will occur spontaneously, but one is exothermic and the other endothermic. So while it is intuitive to think that an exothermic process will be spontaneous, there is clearly more to the picture than simply the release of energy as heat when it comes to making a process spontaneous. The Carnot cycle because a useful thought experiment to explore to help to answer the question of why a process is spontaneous. | 2,665 | 2,837 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Fundamentals/Quantum_Tunneling |
\( A_0 e^{-\alpha x}\) \(\alpha \) \[ -{\frac{\hbar^2}{2m}} {\frac{\partial^2 \psi}{\partial x^2}} + V(r) \psi = E \psi \label{eq1}\] \(V (x)\) \(a\) \(V_0\) \[
V =
\begin{cases}
0 & \text{if } -\infty<x\leq 0\\[3pt]
V_0 & \text{if } 0<x<a\\[3pt]
0 & \text{if } a\leq x<\infty
\end{cases}
\label{eq2}\] \(a\) \(\psi (x)\) \(\psi (x)\) \[
\psi =
\begin{cases}
A\sin kx + B\cos kx & \text{if } -\infty<x\leq 0\\[3pt]
Ce^{-\alpha x} + De^{\alpha x} & \text{if } 0<x<a\\[3pt]
E\sin kx + F\cos kx & \text{if } a\leq x<\infty
\end{cases}
\] \(k = \frac{\sqrt{2mE}}{\hbar} \) \(\alpha = \frac{\sqrt{2m(V_o -E)}}{\hbar} \) \(\psi_1 (0) = \psi_2 (0) \) \(\psi_1 (a) = \psi_2 (a) \) \[A\sin 0 + B\cos 0 = Ce^{0} + De^{0}\] \( A=0 \) \( B=C+D \) \[A\sin ka + B\cos ka = Ce^{-\alpha a} + De^{\alpha a}\] \( a \) \( D \) \( E \) \( F \) \( a \) \( a \) \( \infty \) \( Ce^{-\alpha a}\cos(k(x-a)) \) \( a \) and setting the amplitude to the boundary value \[
\psi =
\begin{cases}
B\cos kx & \text{if } -\infty<x\leq 0\\[3pt]
Be^{-\alpha x} & \text{if } 0<x<a\\[3pt]
Be^{-\alpha a}\cos(k(x-a)) & \text{if } a\leq x<\infty
\end{cases}
\] \( e^{-a\frac{\sqrt{2m(V_o -E)}}{\hbar}} \) \( a \) \( (V_o -E) \) \( e^{-2a\frac{\sqrt{2m(V_o -E)}}{\hbar}} \) \( k \) | 1,282 | 2,838 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Chemical_Reactions/Stoichiometry/Excess_and_Limiting_Reagents |
Chemical reaction equations give the ideal stoichiometric relationship among reactants and products. However, the reactants for a reaction in an experiment are not necessarily a stoichiometric mixture. In a chemical reaction, reactants that are not used up when the reaction is finished are called . The reagent that is completely used up or reacted is called the , because its quantity limits the amount of products formed. Let us consider the reaction between solid sodium and chlorine gas. The reaction can be represented by the equation: \[\mathrm{2 Na_{\large{(s)}} + Cl_{2\large{(g)}} \rightarrow 2 NaCl_{\large{(s)}}}\] It represents a reaction of a metal and a diatomic gas chlorine. This balanced reaction equation indicates that two \(\ce{Na}\) atoms would react with two \(\ce{Cl}\) atoms or one \(\ce{Cl2}\) molecule. Thus, if you have 6 \(\ce{Na}\) atoms, 3 \(\ce{Cl2}\) molecules will be required. If there is an excess number of \(\ce{Cl2}\) molecules, they will remain unreacted. We can also state that 6 moles of sodium will require 3 moles of \(\ce{Cl2}\) gas. If there are more than 3 moles of \(\ce{Cl2}\) gas, some will remain as an excess reagent, and the sodium is a limiting reagent. It limits the amount of the product that can be formed. Chemical reactions with stoichiometric amounts of reactants have limiting or excess reagents. Calculate the number of moles of \(\ce{CO2}\) formed in the combustion of ethane \(\ce{C2H6}\) in a process when 35.0 mol of \(\ce{O2}\) is consumed. The reaction is \[\ce{2 C2H6 + 7 O2 \rightarrow 4 CO2 + 6 H2O}\] \[\mathrm{35.0\: mol\: O_2 \times\dfrac{4\: mol\: CO_2}{7\: mol\: O_2} = 20.0\: mol\: CO_2}\] A balanced equation for the reaction is a basic requirement for identifying the limiting reagent even if amounts of reactants are known. Two moles of \(\ce{Mg}\) and five moles of \(\ce{O2}\) are placed in a reaction vessel, and then the \(\ce{Mg}\) is ignited according to the reaction \(\mathrm{Mg + O_2 \rightarrow MgO}\). Identify the limiting reagent in this experiment. Before a limiting reagent is identified, the reaction be balanced. The balanced reaction is \[\mathrm{2 Mg + O_2 \rightarrow 2 MgO}\] Thus, two moles of \(\ce{Mg}\) require only ONE mole of \(\ce{O2}\). Four moles of oxygen will remain unreacted. Therefore, oxygen is the excess reagent, and \(\ce{Mg}\) is the limiting reagent. Answer these questions:
How many moles of \(\ce{MgO}\) is formed?
What is the weight of \(\ce{MgO}\) formed? \(\ce{Li}\) \(\ce{Na}\) \(\ce{K}\) \(\ce{Fe}\) \(\ce{Al}\) \(\ce{Mg}\) \(\ce{Ar}\) Hint: Potassium, \(\ce{K}\)
Predict reactivity from location of element in the periodic table. Hint: Two moles
Write a reaction equation for a chemical reaction. Hint: Solid
Know where to look up properties of elements or substances. Hint: Sodium metal Hint: sodium ions Hint: chlorine is the limiting reagent
Identify excess and limiting reagent. Hint: hydrogen \(\ce{H2}\) \(\ce{F2}\) Hint: Fluorine Identify the excess and limiting reagent.
\(\mathrm{2 Al_{\large{(s)}} + Fe_2O_3_{\large{(s)}} \rightarrow 2 Fe_{\large{(liquid)}} + Al_2O_3}\)
\(\ce{Al}\) \(\ce{Fe2O3}\) Hint: Iron oxide
A stoichiometric mixture has a mass ratio of 54:160 (nearly 1:3) for \(\ce{Al:Fe2O3}\). | 3,285 | 2,839 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/Cycloalkanes/Physical_Properties_of_Cycloalkanes |
Cycloalkanes are types of alkanes that have one or more rings of carbon atoms in their structure. The physical properties of cycloalkanes are similar to those of alkanes, but they have higher boiling points, melting points and higher densities due to the greater number of London forces that they contain. Cycloalkanes consist of carbon and hydrogen atoms that are saturated because of the single carbon-carbon bond (meaning that no more hydrogen atoms can be added). Cycloalkanes are also non polar and do not have intermolecular hydrogen bonding; they are usually hydrophobic (meaning they do not dissolve in water) and are less dense than water. Cycloalkanes can also be used for many different purposes. These uses are typically classified by the number of carbons in the cycloalkane ring. Many cycloalkanes are used in motor fuel, natural gas, petroleum gas, kerosene, diesel, and many other heavy oils. There are 4 general groups of cycloalkanes: Cycloalkanes can be substituted and named as cycloalkyl derivatives, and disubstituted cycloalkanes can be stereoisomers. In one isomer, if two substituents are placed on the same face or side of the ring, they are called . If the two substituents are on opposite faces, they are called . Substituents can also be either equitorial or axial on certain cycloalkanes, such as cyclohexane. Generally, the melting point, the boiling point and the density of cycloalkanes increase as the number of carbons increases. This trend occurs because of the greater number of bonds that are in higher membered rings, thus making the bonds harder to break. Although alkanes are similar to cycloalkanes, they have higher London Dispersion forces because the ring shape allows for a greater area of contact. Ring strain also causes certain cylcoalkanes to be more reactive. are the attractive or repulsive forces between molecules or between parts of the same molecule. For cycloalkanes, London dispersion forces refer to the repulsive forces between the molecules that cause ring strain. occurs because the carbons in cycloalkanes are sp hybridized, which means that they do not have the expected ideal bond angle of 109.5 ; this causes an increase in the potential energy because of the desire for the carbons to be at an ideal 109.5 . An example of ring strain can be seen in the diagram of cyclopropane below in which the bond angle is 60 between the carbons. The reason for ring strain can be seen through the tetrahedral carbon model. The C-C-C bond angles in cyclopropane (diagram above) (60 ) and cyclobutane (90 ) are much different than the ideal bond angle of 109.5 . This bond angle causes cyclopropane and cyclobutane to have a high ring strain. However, molecules, such as cyclohexane and cyclopentane, would have a much lower ring strain because the bond angle between the carbons is much closer to 109.5 . Below are some examples of cycloalkanes. Ring strain can be seen more prevalently in the cyclopropane and cyclobutane models. Below is a chart of cycloalkanes and their respective heats of combustion ( Δ ). The Δ value increases as the number of carbons in the cycloalkane increases (higher membered ring), and the Δ /CH ratio decreases. The increase in Δ can be attributed to the greater amount of London Dispersion forces. However, the decrease in Δ /CH can be attributed to a decrease in the ring strain. How do cycloalkanes deal with ring strain? Certain cycloalkanes, such as cyclohexane, deal with ring strain by forming . A conformer is a stereoisomer in which molecules of the same connectivity and formula exist as different isomers, in this case, to reduce ring strain. The ring strain is reduced in conformers due to the rotations around the sigma bonds. More about cyclohexane and its conformers can be seen . There are many different types of strain that occur with cycloalkanes. In addition to ring strain, there is also transannular strain, eclipsing, or torsional strain and bond angle strain.Transannular strain exists when there is steric repulsion between atoms. Eclipsing (torsional) strain exists when a cycloalkane is unable to adopt a staggered conformation around a C-C bond, and bond angle strain is the energy needed to distort the tetrahedral carbons enough to close the ring. The presence of angle strain in a molecule indicates that there are bond angles in that particular molecule that deviate from the ideal bond angles required (i.e., that molecule has conformers). | 4,485 | 2,840 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/12%3A_Chemical_Kinetics_II/12.07%3A_The_Lindemann_Mechanism |
The ( ) is a useful one to demonstrate some of the techniques we use for relating chemical mechanisms to rate laws. In this mechanism, a reactant is collisionally activated to a highly energetic form that can then go on to react to form products. \[ A + A \xrightleftharpoons [k_1]{k_{-1}} A^* \nonumber \] \[ A^* \xrightarrow{k_2} P \nonumber \] If the steady state approximation is applied to the intermediate \(A^*\) \[ \dfrac{d[A^*]}{dt} = k_1[A]^2 - k_{-1}[A^*,A] - k_2[A^*] \approx 0 \nonumber \] an expression can be derived for \([A^*]\). \[A^*]= \dfrac{ k_1[A]^2 }{k_{-1}[A] + k_2} \nonumber \] Substituting this into an expression for the rate of the production of the product \(P\) \[\dfrac{d[P]}{dt} = k_2[A^*] \nonumber \] yields \[\dfrac{d[P]}{dt} = \dfrac{ k_2 k_1[A]^2 }{k_{-1}[A] + k_2} \nonumber \] In the limit that \(k_{-1}[A] \ll k_2\), the rate law becomes first order in \([A]\) since \(k_{-1}[A] + k_2 \approx k_{-1}[A]\). \[\dfrac{d[P]}{dt} = \dfrac{ k_2 k_1 }{k_{-1}} [A] \nonumber \] This will happen if the second step is very slow (and is the rate determining step), such that the reverse of the first step “wins” in the competition for [A*]. However, in the other limit, that \(k_2 \gg k_{-1}[A]\), the reaction becomes second order in \([A]\) since \(k_{-1}[A] + k_2 \approx k_2\). \[\dfrac{d[P]}{dt} = k_1[A]^2 \nonumber \] which is consistent with the forward reaction of the first step being the rate determining step, since \(A^*\) is removed from the reaction (through the formation of products) very quickly as soon as it is formed. Sometimes, the is provided by an inert species \(M\), perhaps by filling the reaction chamber with a heavy non-reactive species, such as Ar. In this case, the mechanism becomes \[ A + M \xrightleftharpoons [k_1]{k_{-1}} A^* + M \nonumber \] \[ A^* \xrightarrow{k_2} P \nonumber \] And in the limit that \([A^*]\) can be treated using the steady state approximation, the rate of production of the product becomes \[\dfrac{d[P]}{dt} = \dfrac{ k_2 k_1[M] }{k_{-1}[M] + k_2} \nonumber \] And if the concentration of the third body collider is constant, it is convenient to define an , \(k_{uni}\). \[k_{uni} = \dfrac{ k_2 k_1[M] }{k_{-1}[M] + k_2 } \nonumber \] The utility is that important information about the individual step rate constants can be extracted by plotting \(1/k_{uni}\) as a function of \(1/[M]\). \[ \dfrac{1}{k_{uni}} = \dfrac{k_{-1}}{ k_2 k_1 } + k_2 \left( \dfrac{1}{[M]} \right) \nonumber \] The plot should yield a straight line, the slope of which gives the value of \(k_2\), and the intercept gives \((k_{-1}/k_2k_1)\). | 2,632 | 2,841 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/09%3A_Chemical_Equilibria/9.03%3A_Activities_and_Fugacities |
To this point, we have mostly ignored deviations from ideal behavior. But it should be noted that thermodynamic equilibrium constants are not expressed in terms of concentrations or pressures, but rather in terms of activities and fugacities (both being discussed in Chapter 7). Based on these quantities, \[ K_p = \prod_i f_i^{\nu_i} \label{eq1} \] and \[ K_c = \prod_i a_i^{\nu_i} \nonumber \] And since activities and fugacities are unitless, thermodynamic equilibrium constants are unitless as well. Further, it can be noted that the activities of solids and pure liquids are unity (assuming ideal behavior) since they are in their standard states at the given temperature. As such, these species never change the magnitude of the equilibrium constant and are generally omitted from the equilibrium constant expression. Thermodynamic equilibrium constants are unitless. Oftentimes it is desirable to express the equilibrium constant in terms of concentrations (or activities for systems that deviate from ideal behavior.) To make this conversion, the relationship between pressure and concentration from the ideal gas law can be used. \[p= RT \left( \dfrac{n}{V}\right) \nonumber \] And noting that the concentration is given by (\(n/V\)), the expression for the equilibrium constant (Equation \ref{eq1}) becomes \[ K_p = \prod_i (RT[X_i])^{\nu_i} \label{eq} \] And since for a given temperature, \(RT\) is a constant and can be factored out of the expression, leaving \[ \begin{align} K_p &=\left( \prod_i(RT)^{\nu_i} \right) \left( \prod_i [X_i]^{\nu_i}\right) \\[10pt] &= (RT)^{\sum \nu_i} \prod [X_i]^{\nu_i} \\[10pt] &= (RT)^{\sum \nu_i} K_c \end{align} \nonumber \] This conversion works for reactions in which all reactants and products are in the gas phase. Care must be used when applying this relationship to heterogeneous equilibria. | 1,864 | 2,843 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/02%3A_Gas_Laws/2.16%3A_Problems |
\[ \begin{array}{|c|c|c|c|} \hline \text{ Compound } & \text{ Mol Mass, g mol}^{–1} ~ & \text{ Density, g mL}^{–1} & \text{ Van der Waals } b, \text{ L mol}^{–1} \\ \hline \text{Acetic acid} & 60.05 & 1.0491 & 0.10680 \\ \hline \text{Acetone} & 58.08 & 0.7908 & 0.09940 \\ \hline \text{Acetonitrile} & 41.05 & 0.7856 & 0.11680 \\ \hline \text{Ammonia} & 17.03 & 0.7710 & 0.03707 \\ \hline \text{Aniline} & 93.13 & 1.0216 & 0.13690 \\ \hline \text{Benzene} & 78.11 & 0.8787 & 0.11540 \\ \hline \text{Benzonitrile} & 103.12 & 1.0102 & 0.17240 \\ \hline \text{iso-Butylbenzene } & 134.21 & 0.8621 & 0.21440 \\ \hline \text{Chlorine} & 70.91 & 3.2140 & 0.05622 \\ \hline \text{Durene} & 134.21 & 0.8380 & 0.24240 \\ \hline \text{Ethane} & 30.07 & 0.5720 & 0.06380 \\ \hline \text{Hydrogen chloride} & 36.46 & 1.1870 & 0.04081 \\ \hline \text{Mercury} & 200.59 & 13.5939 & 0.01696 \\ \hline \text{Methane} & 16.04 & 0.4150 & 0.04278 \\ \hline \text{Nitrogen dioxide} & 46.01 & 1.4494 & 0.04424 \\ \hline \text{Silicon tetrafluoride} & 104.08 & 1.6600 & 0.05571 \\ \hline \text{Water} & 18.02 & 1.0000 & 0.03049 \\ \hline \end{array}\] Notes We use the over-bar to indicate that the quantity is per mole of substance. Thus, we write \(\overline{N}\) to indicate the number of particles per mole. We write \(\overline{M}\) to represent the gram molar mass. In Chapter 14, we introduce the use of the over-bar to denote a partial molar quantity; this is consistent with the usage introduced here, but carries the further qualification that temperature and pressure are constant at specified values. We also use the over-bar to indicate the arithmetic average; such instances will be clear from the context. The unit of temperature is named the kelvin, which is abbreviated as K. A redefinition of the size of the unit of temperature, the kelvin, is under consideration. The practical effect will be inconsequential for any but the most exacting of measurements. For a thorough discussion of the development of the concept of temperature, the evolution of our means to measure it, and the philosophical considerations involved, see Hasok Chang, Inventing Temperature, Oxford University Press, 2004. See T. L. Hill, , Addison-Wesley Publishing Company, 1960, p 286. See S. M. Blinder, , The Macmillan Company, Collier-Macmillan Canada, Ltd., Toronto, 1969, pp 185-189 | 2,371 | 2,845 |
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The Helmholtz energy \(A\) is developed for isochoric changes and as we have often said before it is much easier to deal with isobaric ones where \(P=\) constant. We can therefore repeat the above treatment for the enthalpy and introduce another state function the Gibbs energy \[\begin{align*} G &≡ H -TS \\[4pt] &= U + PV - TS \\[4pt] &= A + PV \end{align*}\] If we take both \(T\) and \(P\) constant we get \[dU-TdS + PdV \le 0 \nonumber \] \[dG \le 0 \nonumber \] \(G\) either decreases (spontaneously) or is constant (at equilibrium). Calculating the state function between two end points we get: \[ΔG = ΔH - TΔS ≤ 0 (T,P\text{ constant}) \nonumber \] This quantity is key to the question of spontaneity under the conditions we usually work under. If for a process is positive it does not occur spontaneous and can only be made to occur if it is 'pumped', i.e. coupled with a process that has a negative . The latter is spontaneous. If . Because the \(\Delta S\) term contains the temperature \(T\) as coefficient the spontaneous direction of a process, e.g. a chemical reaction can with temperature depending on the values of the enthalpy and the entropy change and . This is true for the melting process, e.g. for water below 0 C we get water=>ice, above this temperature ice melts to water, but it also goes for chemical reactions. Consider \[\ce{NH3(g) + HCl(g) <=> NH4Cl(s)} \nonumber \] at 298K / 1 bar is -176.2 kJ. The change in entropy is -0.285 kJ/K so that at 298K is -91.21 kJ. Clearly this is a reaction that will proceed to the depletion of whatever is the limiting reagent on the left. However at 618 K this is a different story. Above this temperature is positive! (assuming enthalpy and entropy have remained the same, which is almost but not completely true) The reaction will not proceed. Instead the reaction would proceed spontaneously. The salt on the right would decompose in the two gases -base and acid- on the left. As we have seen, can be related to the maximal amount of work that a system can perform at constant \(V\) and \(T\). We can hold an analogous argument for \(\Delta G\) except that \(V\) is not constant so that we have to consider volume work (zero at constant volume) . \[dG = d(U+PV-TS) = dU -TdS - SdT - PdV +VdP \nonumber \] As \(dU = TdS + δw_{rev}\) \[dG = δw_{rev} -SdT + VdP + PdV \nonumber \] As the later term is \(-δw_{volume}\) \[dG = δw_{rev} -SdT + VdP - δw_{volume} \nonumber \] At constant \(T\) and \(P\) the two middle terms drop out \[dG = δw_{rev} - δw_{volume} = δw_{other useful work} \nonumber \] stands for the (maximal) reversible, isobaric isothermal non-\(PV\) work that a certain spontaneous change can perform. The volume work may not be zero, but is corrected for. Because \(G ≡ H-TS\), we can write \[dG = dH -TdS -SdT \nonumber \] \[dG = TdS +VdP -TdS -SdT = VdP - SdT \nonumber \] The natural variables of \(G\) are pressure \(P\) and temperature \(T\). This is what makes this function the most useful of the four \(U\), \(H\), \(A\), and \(G\): these are the natural variables of most of your laboratory experiments! We now have developed the basic set of concepts and functions that together form the framework of thermodynamics. Let's summarize four very basic state functions: Note: We are now ready to begin applying thermodynamics to a number of very diverse situations, but we will first develop some useful partial differential machinery. | 3,463 | 2,846 |
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The properties of liquids are intermediate between those of gases and solids but are more similar to solids. In contrast to molecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, molecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. For example, it requires 927 kJ to overcome the intramolecular forces and break both O–H bonds in 1 mol of water, but it takes only about 41 kJ to overcome the intermolecular attractions and convert 1 mol of liquid water to water vapor at 100°C. (Despite this seemingly low value, the intermolecular forces in liquid water are among the strongest such forces known!) Given the large difference in the strengths of intra- and intermolecular forces, changes between the solid, liquid, and gaseous states almost invariably occur for molecular substances . The properties of liquids are intermediate between those of gases and solids but are more similar to solids. Intermolecular forces determine bulk properties such as the melting points of solids and the boiling points of liquids. Liquids boil when the molecules have enough thermal energy to overcome the intermolecular attractive forces that hold them together, thereby forming bubbles of vapor within the liquid. Similarly, solids melt when the molecules acquire enough thermal energy to overcome the intermolecular forces that lock them into place in the solid. Intermolecular forces are electrostatic in nature; that is, they arise from the interaction between positively and negatively charged species. Like covalent and ionic bonds, intermolecular interactions are the sum of both attractive and repulsive components. Because electrostatic interactions fall off rapidly with increasing distance between molecules, intermolecular interactions are most important for solids and liquids, where the molecules are close together. These interactions become important for gases only at very high pressures, where they are responsible for the observed deviations from the ideal gas law at high pressures. (For more information on the behavior of real gases and deviations from the ideal gas law, see Section 7.1.) In this section, we explicitly consider three kinds of intermolecular interactions: , , and . The first two are often described collectively as van der Waals forces . Recall from that polar covalent bonds behave as if the bonded atoms have localized fractional charges that are equal but opposite (i.e., the two bonded atoms generate a ). If the structure of a molecule is such that the individual bond dipoles do not cancel one another, then the molecule has a net dipole moment. Molecules with net dipole moments tend to align themselves so that the positive end of one dipole is near the negative end of another and vice versa, as shown in part (a) in . These arrangements are more stable than arrangements in which two positive or two negative ends are adjacent (part (c) in ). Hence dipole–dipole interactions , such as those in part (b) in , are , whereas those in part (d) in are . Because molecules in a liquid move freely and continuously, molecules always experience both attractive and repulsive dipole–dipole interactions simultaneously, as shown in . On average, however, the attractive interactions dominate. Because each end of a dipole possesses only a fraction of the charge of an electron, dipole–dipole interactions are substantially weaker than the interactions between two ions, each of which has a charge of at least ±1, or between a dipole and an ion, in which one of the species has at least a full positive or negative charge. In addition, the attractive interaction between dipoles falls off much more rapidly with increasing distance than do the ion–ion interactions we considered in . Recall that the attractive energy between two ions is proportional to 1/ , where is the distance between the ions. Doubling the distance ( → 2 ) decreases the attractive energy by one-half. In contrast, the energy of the interaction of two dipoles is proportional to 1/ , so doubling the distance between the dipoles decreases the strength of the interaction by 2 , or 64-fold. Thus a substance such as HCl, which is partially held together by dipole–dipole interactions, is a gas at room temperature and 1 atm pressure, whereas NaCl, which is held together by interionic interactions, is a high-melting-point solid. Within a series of compounds of similar molar mass, the strength of the intermolecular interactions increases as the dipole moment of the molecules increases, as shown in . Using what we learned in about predicting relative bond polarities from the electronegativities of the bonded atoms, we can make educated guesses about the relative boiling points of similar molecules. The attractive energy between two ions is proportional to 1/ , whereas the attractive energy between two dipoles is proportional to 1/ . Arrange ethyl methyl ether (CH OCH CH ), 2-methylpropane [isobutane, (CH ) CHCH ], and acetone (CH COCH ) in order of increasing boiling points. Their structures are as follows: compounds order of increasing boiling points Compare the molar masses and the polarities of the compounds. Compounds with higher molar masses and that are polar will have the highest boiling points. The three compounds have essentially the same molar mass (58–60 g/mol), so we must look at differences in polarity to predict the strength of the intermolecular dipole–dipole interactions and thus the boiling points of the compounds. The first compound, 2-methylpropane, contains only C–H bonds, which are not very polar because C and H have similar electronegativities. It should therefore have a very small (but nonzero) dipole moment and a very low boiling point. Ethyl methyl ether has a structure similar to H O; it contains two polar C–O single bonds oriented at about a 109° angle to each other, in addition to relatively nonpolar C–H bonds. As a result, the C–O bond dipoles partially reinforce one another and generate a significant dipole moment that should give a moderately high boiling point. Acetone contains a polar C=O double bond oriented at about 120° to two methyl groups with nonpolar C–H bonds. The C–O bond dipole therefore corresponds to the molecular dipole, which should result in both a rather large dipole moment and a high boiling point. Thus we predict the following order of boiling points: 2-methylpropane < ethyl methyl ether < acetone. This result is in good agreement with the actual data: 2-methylpropane, boiling point = −11.7°C, and the dipole moment (μ) = 0.13 D; methyl ethyl ether, boiling point = 7.4°C and μ = 1.17 D; acetone, boiling point = 56.1°C and μ = 2.88 D. Exercise Arrange carbon tetrafluoride (CF ), ethyl methyl sulfide (CH SC H ), dimethyl sulfoxide [(CH ) S=O], and 2-methylbutane [isopentane, (CH ) CHCH CH ] in order of decreasing boiling points. dimethyl sulfoxide (boiling point = 189.9°C) > ethyl methyl sulfide (boiling point = 67°C) > 2-methylbutane (boiling point = 27.8°C) > carbon tetrafluoride (boiling point = −128°C) Thus far we have considered only interactions between polar molecules, but other factors must be considered to explain why many nonpolar molecules, such as bromine, benzene, and hexane, are liquids at room temperature, and others, such as iodine and naphthalene, are solids. Even the noble gases can be liquefied or solidified at low temperatures, high pressures, or both ( ) What kind of attractive forces can exist between nonpolar molecules or atoms? This question was answered by Fritz London (1900–1954), a German physicist who later worked in the United States. In 1930, London proposed that temporary fluctuations in the electron distributions within atoms and nonpolar molecules could result in the formation of short-lived instantaneous dipole moments , which produce attractive forces called London dispersion forces between otherwise nonpolar substances. Consider a pair of adjacent He atoms, for example. On average, the two electrons in each He atom are uniformly distributed around the nucleus. Because the electrons are in constant motion, however, their distribution in one atom is likely to be asymmetrical at any given instant, resulting in an instantaneous dipole moment. As shown in part (a) in , the instantaneous dipole moment on one atom can interact with the electrons in an adjacent atom, pulling them toward the positive end of the instantaneous dipole or repelling them from the negative end. The net effect is that the first atom causes the temporary formation of a dipole, called an induced dipole , in the second. Interactions between these temporary dipoles cause atoms to be attracted to one another. These attractive interactions are weak and fall off rapidly with increasing distance. London was able to show with quantum mechanics that the attractive energy between molecules due to temporary dipole–induced dipole interactions falls off as 1/ . Doubling the distance therefore decreases the attractive energy by 2 , or 64-fold. Instantaneous dipole–induced dipole interactions between nonpolar molecules can produce intermolecular attractions just as they produce interatomic attractions in monatomic substances like Xe. This effect, illustrated for two H molecules in part (b) in tends to become more pronounced as atomic and molecular masses increase ( ). For example, Xe boils at −108.1°C, whereas He boils at −269°C. The reason for this trend is that the strength of London dispersion forces is related to the ease with which the electron distribution in a given atom can be perturbed. In small atoms such as He, the two 1 electrons are held close to the nucleus in a very small volume, and electron–electron repulsions are strong enough to prevent significant asymmetry in their distribution. In larger atoms such as Xe, however, the outer electrons are much less strongly attracted to the nucleus because of filled intervening shells. (For more information on shielding, see ) As a result, it is relatively easy to temporarily deform the electron distribution to generate an instantaneous or induced dipole. The ease of deformation of the electron distribution in an atom or molecule is called its polarizability . Because the electron distribution is more easily perturbed in large, heavy species than in small, light species, we say that heavier substances tend to be much more than lighter ones. For similar substances, London dispersion forces get stronger with increasing molecular size. The polarizability of a substance also determines how it interacts with ions and species that possess permanent dipoles, as we shall see when we discuss solutions in . Thus London dispersion forces are responsible for the general trend toward higher boiling points with increased molecular mass and greater surface area in a homologous series of compounds, such as the alkanes (part (a) in ). The strengths of London dispersion forces also depend significantly on molecular shape because shape determines how much of one molecule can interact with its neighboring molecules at any given time. For example, part (b) in shows 2,2-dimethylpropane (neopentane) and -pentane, both of which have the empirical formula C H . Neopentane is almost spherical, with a small surface area for intermolecular interactions, whereas -pentane has an extended conformation that enables it to come into close contact with other -pentane molecules. As a result, the boiling point of neopentane (9.5°C) is more than 25°C lower than the boiling point of -pentane (36.1°C). All molecules, whether polar or nonpolar, are attracted to one another by London dispersion forces in addition to any other attractive forces that may be present. In general, however, dipole–dipole interactions in small polar molecules are significantly stronger than London dispersion forces, so the former predominate. Arrange -butane, propane, 2-methylpropane [isobutene, (CH ) CHCH ], and -pentane in order of increasing boiling points. compounds order of increasing boiling points Determine the intermolecular forces in the compounds and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. The four compounds are alkanes and nonpolar, so London dispersion forces are the only important intermolecular forces. These forces are generally stronger with increasing molecular mass, so propane should have the lowest boiling point and -pentane should have the highest, with the two butane isomers falling in between. Of the two butane isomers, 2-methylpropane is more compact, and -butane has the more extended shape. Consequently, we expect intermolecular interactions for -butane to be stronger due to its larger surface area, resulting in a higher boiling point. The overall order is thus as follows, with actual boiling points in parentheses: propane (−42.1°C) < 2-methylpropane (−11.7°C) < -butane (−0.5°C) < -pentane (36.1°C). Exercise Arrange GeH , SiCl , SiH , CH , and GeCl in order of decreasing boiling points. GeCl (87°C) > SiCl (57.6°C) > GeH (−88.5°C) > SiH (−111.8°C) > CH (−161°C) Molecules with hydrogen atoms bonded to electronegative atoms such as O, N, and F (and to a much lesser extent Cl and S) tend to exhibit unusually strong intermolecular interactions. These result in much higher boiling points than are observed for substances in which London dispersion forces dominate, as illustrated for the covalent hydrides of elements of groups 14–17 in . Methane and its heavier congeners in group 14 form a series whose boiling points increase smoothly with increasing molar mass. This is the expected trend in nonpolar molecules, for which London dispersion forces are the exclusive intermolecular forces. In contrast, the hydrides of the lightest members of groups 15–17 have boiling points that are more than 100°C greater than predicted on the basis of their molar masses. The effect is most dramatic for water: if we extend the straight line connecting the points for H Te and H Se to the line for period 2, we obtain an estimated boiling point of −130°C for water! Imagine the implications for life on Earth if water boiled at −130°C rather than 100°C. Why do strong intermolecular forces produce such anomalously high boiling points and other unusual properties, such as high enthalpies of vaporization and high melting points? The answer lies in the highly polar nature of the bonds between hydrogen and very electronegative elements such as O, N, and F. The large difference in electronegativity results in a large partial positive charge on hydrogen and a correspondingly large partial negative charge on the O, N, or F atom. Consequently, H–O, H–N, and H–F bonds have very large bond dipoles that can interact strongly with one another. Because a hydrogen atom is so small, these dipoles can also approach one another more closely than most other dipoles. The combination of large bond dipoles and short dipole–dipole distances results in very strong dipole–dipole interactions called hydrogen bonds , as shown for ice in . A hydrogen bond is usually indicated by a dotted line between the hydrogen atom attached to O, N, or F (the ) and the atom that has the lone pair of electrons (the ). Because each water molecule contains two hydrogen atoms and two lone pairs, a tetrahedral arrangement maximizes the number of hydrogen bonds that can be formed. In the structure of ice, each oxygen atom is surrounded by a distorted tetrahedron of hydrogen atoms that form bridges to the oxygen atoms of adjacent water molecules. The bridging hydrogen atoms are equidistant from the two oxygen atoms they connect, however. Instead, each hydrogen atom is 101 pm from one oxygen and 174 pm from the other. In contrast, each oxygen atom is bonded to two H atoms at the shorter distance and two at the longer distance, corresponding to two O–H covalent bonds and two hydrogen bonds from adjacent water molecules, respectively. The resulting open, cagelike structure of ice means that the solid is actually slightly less dense than the liquid, which explains why ice floats on water rather than sinks. Hydrogen bond formation requires a hydrogen bond donor a hydrogen bond acceptor. Because ice is less dense than liquid water, rivers, lakes, and oceans freeze from the top down. In fact, the ice forms a protective surface layer that insulates the rest of the water, allowing fish and other organisms to survive in the lower levels of a frozen lake or sea. If ice were denser than the liquid, the ice formed at the surface in cold weather would sink as fast as it formed. Bodies of water would freeze from the bottom up, which would be lethal for most aquatic creatures. The expansion of water when freezing also explains why automobile or boat engines must be protected by “antifreeze” (we will discuss how antifreeze works in ) and why unprotected pipes in houses break if they are allowed to freeze. Although hydrogen bonds are significantly weaker than covalent bonds, with typical dissociation energies of only 15–25 kJ/mol, they have a significant influence on the physical properties of a compound. Compounds such as HF can form only two hydrogen bonds at a time as can, on average, pure liquid NH . Consequently, even though their molecular masses are similar to that of water, their boiling points are significantly lower than the boiling point of water, which forms hydrogen bonds at a time. Considering CH OH, C H , Xe, and (CH ) N, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures. compounds formation of hydrogen bonds and structure Identify the compounds with a hydrogen atom attached to O, N, or F. These are likely to be able to act as hydrogen bond donors. Of the compounds that can act as hydrogen bond donors, identify those that also contain lone pairs of electrons, which allow them to be hydrogen bond acceptors. If a substance is both a hydrogen donor and a hydrogen bond acceptor, draw a structure showing the hydrogen bonding. Of the species listed, xenon (Xe), ethane (C H ), and trimethylamine [(CH ) N] do not contain a hydrogen atom attached to O, N, or F; hence they cannot act as hydrogen bond donors. The one compound that can act as a hydrogen bond donor, methanol (CH OH), contains both a hydrogen atom attached to O (making it a hydrogen bond donor) and two lone pairs of electrons on O (making it a hydrogen bond acceptor); methanol can thus form hydrogen bonds by acting as either a hydrogen bond donor or a hydrogen bond acceptor. The hydrogen-bonded structure of methanol is as follows: Exercise Considering CH CO H, (CH ) N, NH , and CH F, which can form hydrogen bonds with themselves? Draw the hydrogen-bonded structures. CH CO H and NH ; Arrange C (buckminsterfullerene, which has a cage structure), NaCl, He, Ar, and N O in order of increasing boiling points. compounds order of increasing boiling points Identify the intermolecular forces in each compound and then arrange the compounds according to the strength of those forces. The substance with the weakest forces will have the lowest boiling point. Electrostatic interactions are strongest for an ionic compound, so we expect NaCl to have the highest boiling point. To predict the relative boiling points of the other compounds, we must consider their polarity (for dipole–dipole interactions), their ability to form hydrogen bonds, and their molar mass (for London dispersion forces). Helium is nonpolar and by far the lightest, so it should have the lowest boiling point. Argon and N O have very similar molar masses (40 and 44 g/mol, respectively), but N O is polar while Ar is not. Consequently, N O should have a higher boiling point. A C molecule is nonpolar, but its molar mass is 720 g/mol, much greater than that of Ar or N O. Because the boiling points of nonpolar substances increase rapidly with molecular mass, C should boil at a higher temperature than the other nonionic substances. The predicted order is thus as follows, with actual boiling points in parentheses: He (−269°C) < Ar (−185.7°C) < N O (−88.5°C) < C (>280°C) < NaCl (1465°C). Exercise Arrange 2,4-dimethylheptane, Ne, CS , Cl , and KBr in order of decreasing boiling points. KBr (1435°C) > 2,4-dimethylheptane (132.9°C) > CS (46.6°C) > Cl (−34.6°C) > Ne (−246°C) Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions. Transitions between the solid and liquid or the liquid and gas phases are due to changes in intermolecular interactions but do not affect intramolecular interactions. The three major types of intermolecular interactions are dipole–dipole interactions, London dispersion forces (these two are often referred to collectively as ), and hydrogen bonds. arise from the electrostatic interactions of the positive and negative ends of molecules with permanent dipole moments; their strength is proportional to the magnitude of the dipole moment and to 1/ , where is the distance between dipoles. are due to the formation of in polar or nonpolar molecules as a result of short-lived fluctuations of electron charge distribution, which in turn cause the temporary formation of an in adjacent molecules. Like dipole–dipole interactions, their energy falls off as 1/ . Larger atoms tend to be more than smaller ones because their outer electrons are less tightly bound and are therefore more easily perturbed. are especially strong dipole–dipole interactions between molecules that have hydrogen bonded to a highly electronegative atom, such as O, N, or F. The resulting partially positively charged H atom on one molecule (the ) can interact strongly with a lone pair of electrons of a partially negatively charged O, N, or F atom on adjacent molecules (the ). Because of strong hydrogen bonding between water molecules, water has an unusually high boiling point, and ice has an open, cagelike structure that is less dense than liquid water. What is the main difference between intramolecular interactions and intermolecular interactions? Which is typically stronger? How are changes of state affected by these different kinds of interactions? Describe the three major kinds of intermolecular interactions discussed in this chapter and their major features. The hydrogen bond is actually an example of one of the other two types of interaction. Identify the kind of interaction that includes hydrogen bonds and explain why hydrogen bonds fall into this category. Which are stronger—dipole–dipole interactions or London dispersion forces? Which are likely to be more important in a molecule with heavy atoms? Explain your answers. Explain why hydrogen bonds are unusually strong compared to other dipole–dipole interactions. How does the strength of hydrogen bonds compare with the strength of covalent bonds? Liquid water is essential for life as we know it, but based on its molecular mass, water should be a gas under standard conditions. Why is water a liquid rather than a gas under standard conditions? Describe the effect of polarity, molecular mass, and hydrogen bonding on the melting point and boiling point of a substance. Why are intermolecular interactions more important for liquids and solids than for gases? Under what conditions must these interactions be considered for gases? Using acetic acid as an example, illustrate both attractive and repulsive intermolecular interactions. How does the boiling point of a substance depend on the magnitude of the repulsive intermolecular interactions? In group 17, elemental fluorine and chlorine are gases, whereas bromine is a liquid and iodine is a solid. Why? The boiling points of the anhydrous hydrogen halides are as follows: HF, 19°C; HCl, −85°C; HBr, −67°C; and HI, −34°C. Explain any trends in the data, as well as any deviations from that trend. Identify the most important intermolecular interaction in each of the following. Identify the most important intermolecular interaction in each of the following. Would you expect London dispersion forces to be more important for Xe or Ne? Why? (The atomic radius of Ne is 38 pm, whereas that of Xe is 108 pm.) Arrange Kr, Cl , H , N , Ne, and O in order of increasing polarizability. Explain your reasoning. Both water and methanol have anomalously high boiling points due to hydrogen bonding, but the boiling point of water is greater than that of methanol despite its lower molecular mass. Why? Draw the structures of these two compounds, including any lone pairs, and indicate potential hydrogen bonds. The structures of ethanol, ethylene glycol, and glycerin are as follows: Arrange these compounds in order of increasing boiling point. Explain your rationale. Do you expect the boiling point of H S to be higher or lower than that of H O? Justify your answer. Ammonia (NH ), methylamine (CH NH ), and ethylamine (CH CH NH ) are gases at room temperature, while propylamine (CH CH CH NH ) is a liquid at room temperature. Explain these observations. Why is it not advisable to freeze a sealed glass bottle that is completely filled with water? Use both macroscopic and microscopic models to explain your answer. Is a similar consideration required for a bottle containing pure ethanol? Why or why not? Which compound in the following pairs will have the higher boiling point? Explain your reasoning. Some recipes call for vigorous boiling, while others call for gentle simmering. What is the difference in the temperature of the cooking liquid between boiling and simmering? What is the difference in energy input? Use the melting of a metal such as lead to explain the process of melting in terms of what is happening at the molecular level. As a piece of lead melts, the temperature of the metal remains constant, even though energy is being added continuously. Why? How does the O–H distance in a hydrogen bond in liquid water compare with the O–H distance in the covalent O–H bond in the H O molecule? What effect does this have on the structure and density of ice? Water is a liquid under standard conditions because of its unique ability to form four strong hydrogen bonds per molecule. As the atomic mass of the halogens increases, so does the number of electrons and the average distance of those electrons from the nucleus. Larger atoms with more electrons are more easily polarized than smaller atoms, and the increase in polarizability with atomic number increases the strength of London dispersion forces. These intermolecular interactions are strong enough to favor the condensed states for bromine and iodine under normal conditions of temperature and pressure. Water has two polar O–H bonds with H atoms that can act as hydrogen bond donors, plus two lone pairs of electrons that can act as hydrogen bond acceptors, giving a net of hydrogen bonds per H O molecule. Although methanol also has two lone pairs of electrons on oxygen that can act as hydrogen bond acceptors, it only has one O–H bond with an H atom that can act as a hydrogen bond donor. Consequently, methanol can only form hydrogen bonds per molecule on average, versus four for water. Hydrogen bonding therefore has a much greater effect on the boiling point of water. Vigorous boiling causes more water molecule to escape into the vapor phase, but does not affect the temperature of the liquid. Vigorous boiling requires a higher energy input than does gentle simmering. , Scott Sinex and Scott Johnson | 27,707 | 2,847 |
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Electrophilic addition to alkenes generally takes place via donation of the π-bonding electron pair from the alkene to an electrophile. So far, we have only looked at protic electrophiles, but the reaction proceeds with others, as well. For instance, alkenes react quite easily with bromine. Dripping a solution of bromine into a solution of alkene provides a clear sign of reaction. The red-brown colour of bromine disappears almost instantly. Although bromine isn't an obvious electrophile, most of the common diatomic elements can behave that way; the exception is dinitrogen. A fleeting asymmetry of electrons can polarize the molecules to one end. That event leaves one atom partially positive and the other end partially negative. Because these elements tend to form somewhat stable anions, the partially negative atom can be displaced failry easily. As before, a nucleophile connects with the cation in a second step. In this case, a dibromide compound is formed. However, it isn't formed in quite the way that is shown below. We know that the mechanism shown above does not convey the whole picture because it isn't consistent with the stereochemistry of the reaction. The stereochemical outcome is shown below. The enantiomer is formed as well. Assign configurations R or S to the product shown in the above mechanism. However, although two enantiomers are formed in the reaction, the corresponding diastereomer is not. The following step does not occur. Assign configurations R or S to the product shown in the above mechanism and explain why it is not an enantiomer to the compound in problem EA4.1. Instead, the cation that forms in the reaction appears to be stabilized by lone pair donation from bromine. The intermediate species below is called a cyclic bromnium ion. The bromine prevents approach of the nucleophilic bromide from one side, ensuring formation of product through addition only. The product forms as a result. Additional evidence of the stabilized bromonium comes from the observation of just one product in the bromination of 1-hexene. How many products would be expected in the absence of a stabilized cation? Explain with a mechanism. Bromonium ions like the one shown below have been isolated and characterized by X-ray crystallography in at least one case. Explain why the intermediate was isolated in this case, rather than a dibromo product. In some cases, the reaction of an alkene with bromine does not provide dibromo products. Show products of the reaction of cyclohexene under the following conditions and justify your choices with mechanisms. a) Br in water b) Br and NH Cl in THF Provide products for the reaction of bromine with each of the following compounds. , | 2,728 | 2,848 |
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Question 1) How are these five molecules different?
Question 2) If applicable, how are these five molecules similar?
Question 3) What does the red sphere represent? Question 4) Choose the molecule that you think will be the most viscous. Write down your answer on a piece of paper and briefly write why you chose that molecule.
Question 5) Choose the molecule that you think will be the least viscous. Write down your answer on a piece of paper and briefly write why you chose that molecule.
Question 6) How does the red sphere affect how fast the liquid falls?
Question 7) How does the time the liquid falls relate to viscosity? Used with permission. | 678 | 2,849 |
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We have covered the basic ideas of atomic structure, but it is worth realizing that of all the elements only the noble gases are found naturally in a form such that their atoms occur as individuals, widely separated from all other atoms. Under the conditions that prevail on the surface of the earth, almost all atoms are linked by chemical bonds to other atoms. Oxygen, for example, is the most common element on earth. It is found in combination with metals in rocks, with hydrogen in water, with carbon and hydrogen in living organisms, or as the diatomic molecule O in the atmosphere, but individual oxygen atoms are quite rare. Most other elements behave in a similar way. Thus, if we want to understand the chemistry of everyday matter, we need to understand the nature of the chemical bonds which hold atoms together. Theories of chemical bonding invariably involve electrons. When one atom approaches another, the valence electrons, found in the outermost regions of the atoms, interact long before the nuclei can come close together. Electrons are the least massive components of an atom, and so they can relocate to produce electrostatic forces which hold atoms together. According to Coulomb’s law, such electrostatic or Coulombic forces are quite large when charges are separated by distances of a few hundred picometers—the size of an atom. Coulombic forces, then, are quite capable of explaining the strengths of the bonds by which atoms are held together. An important piece of evidence relating electrons and chemical bonding was noted by G. N. Lewis shortly after the discovery that the atomic number indicated how many electrons were present in each kind of atom. . Thus H O has 2 electrons from the 2 H's and 8 from O for a total of 10, NCl has 7 + (3 × 17) = 58 electrons, and so on. This is a bit surprising when you consider that half the elements have odd atomic numbers so that their atoms have an odd number of electrons. Lewis suggested that , thus accounting for the predominance of even numbers of electrons in chemical formulas. These pairs often leave bonded atoms with eight electrons in the outer most shell, known as the . The orbital shown below gives a visual example of the octet rule, as it has a full octet, or 8 valence electrons. There are two important ways in which the valence shells of different atoms can interact to produce electron pairs and chemical bonds. When two atoms have quite different degrees of attraction for their outermost electrons, one or more electrons may transfer their allegiance from one atom to another, pairing with electrons already present on the second atom. The atom to which electrons are transferred will acquire excess negative charge, becoming a negative ion, while the atom which loses electrons will become a positive ion. These oppositely charged ions will be held together by the coulombic forces of attraction between them, forming an . Since electrons are shifted so that one neutral atom becomes positively charged, and the other becomes negatively charged, substances formed from ionic bonding are often in pairs and are then referred to as . Binary ionic compounds are not too common, but the existence of polyatomic ions greatly extends the number of ionic substances. Oppositely charged ions are held in by strong coulombic forces, an example of which is pictured below. This crystal lattice structure means the include hardness, brittleness, and having high melting and boiling points. The majority of them dissolve in water, and in solution each ion exhibits its own chemical properties. Ionic compounds obey the octet rule, which explains why ions are generally in . An everyday example of an ionic compound, salt, can be seen below. On the other hand, when two atoms have the same degree of attraction for their valence electrons, it is possible for them to pairs of electrons in the region between their nuclei. Such shared pairs of electrons attract nuclei, holding them together with a . Sharing one or more pairs of electrons between two atoms attracts the nuclei together and usually around each atom. This process of sharing electrons is demonstrated below, where two Hydrogen atoms share their only electrons with each other, thus forming a covalent bond.
Figure 6.0.4 Covalent bonding often produces individual molecules, like CO or CH CH OH, which have no net electrical charge and little attraction for each other. Thus covalent substances often have low melting and boiling points and are liquids or gases at room temperature. Occasionally, as in the case of SiO , an extended network of covalent bonds is required to satisfy the octet rule. Such giant molecules result in solid compounds with high melting points. A common (if ) covalent solid, known for its hardness, is the diamond, pictured below. A number of atomic properties, such as , , , and , are important in determining whether certain elements will form covalent or ionic compounds and what properties those compounds will have. In the next sections we will consider the formation of covalent and ionic bonds and the properties of some substances containing each type of bond. The figure below previews how each atomic property important to understanding bonding varies according to an atom’s position in the periodic table. The figure also includes one atomic property, electronegativity, which will be covered when we explore in more depth. | 5,451 | 2,850 |
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Wöhler synthesis of Urea in 1828 heralded the birth of modern chemistry. The Art of synthesis is as old as Organic chemistry itself. Natural product chemistry is firmly rooted in the science of degrading a molecule to known smaller molecules using known chemical reactions and conforming the assigned structure by chemical synthesis from small, well known molecules using well established synthetic chemistry techniques. Once this art of synthesizing a molecule was mastered, chemists attempted to modify bioactive molecules in an attempt to develop new drugs and also to unravel the mystery of biomolecular interactions. Until the middle of the 20th Century, organic chemists approached the task of synthesis of molecules as independent tailor made projects, guided mainly by chemical intuition and a sound knowledge of chemical reactions. During this period, a strong foundation was laid for the development of mechanistic principles of organic reactions, new reactions and reagents. More than a century of such intensive studies on the chemistry of carbohydrates, alkaloids, terpenes and steroids laid the foundation for the development of logical approaches for the synthesis of molecules. The job of a synthetic chemist is akin to that of an architect (or civil engineer). While the architect could actually see the building he is constructing, a molecular architect called Chemist is handicapped by the fact that the molecule he is synthesizing is too small to be seen even through the most powerful microscope developed to date. With such a limitation, how does he ‘see’ the developing structure? For this purpose, a chemist makes use of spectroscopic tools. How does he cut, tailor and glue the components on a molecule that he cannot see? For this purpose chemists have developed molecular level tools called Reagents and Reactions. How does he clean the debris and produce pure molecules? This feat is achieved by crystallization, distillation and extensive use of Chromatography techniques. A mastery over several such techniques enables the molecular architect (popularly known as organic chemist) to achieve the challenging task of synthesizing the mirade molecular structures encountered in Natural Products Chemistry, Drug Chemistry and modern Molecular Materials. In this task, he is further guided by several ‘thumb rules’ that chemists have evolved over the past two centuries. The discussions on the topics for synthesis, and are beyond the scope of this write-up. Let us begin with a brief look at some of the important We would then discuss Protection and Deprotection of some important functional groups. We could then move on to the Logic of planning Organic Synthesis. A multi-step synthesis of any organic compound requires the chemist to accomplish three related tasks: Other factors must also be considered, always in context with those listed above: ), | 2,911 | 2,851 |
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A mixture of metal ions in a solution can be separated by precipitation with anions such as \(\ce{Cl-}\), \(\ce{Br-}\), \(\ce{SO4^2-}\), \(\ce{CO3^2-}\), \(\ce{S^2-}\), \(\ce{Cr2O4^2-}\), \(\ce{PO4^2-}\), \(\ce{OH-}\) etc. When a metal ion or a group of metal ions form insoluble salts with a particular anion, they can be separated from others by precipitation. We can also separate the anions by precipitating them with appropriate metal ions. There are no definite dividing lines between , and , but concentrations of their saturated solutions are small, medium, and large. Solubility products are usually listed for insoluble and sparingly soluble salts, but they are not given for soluble salts. Solubility products for soluble salts are very large. What type of salts are usually soluble, sparingly soluble and insoluble? The following are some general guidelines, but these are not precise laws. These are handy rules for us to have if we deal with salts often. On the other hand, solubility is an important physical property of a substance, and these properties are listed in handbooks. Formation of crystals from a saturated solution is a phenomenon, and it can be applied to separate various chemicals or ions in a solution. When solubilities of two metal salts are very different, they can be separated by precipitation. The values for various salts are valuable information, and some data are given in Table E3. In the first two examples, we show how barium and strontium can be separated as chromate. The for strontium chromate is \(3.6 \times 10^{-5}\) and the for barium chromate is \(1.2 \times10^{-10}\). What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.30 M solution of barium and strontium ions without precipitating the other?
Since the for barium chromate is smaller, we know that \(\ce{BaCrO4}\) will form a precipitate first as \(\ce{[CrO4^2- ]}\) increases so that for \(\ce{BaCrO4}\) also increases from zero to of \(\ce{BaCrO4}\), at which point, \(\ce{BaCrO4}\) precipitates. As \(\ce{[CrO4^2- ]}\) increases, \(\ce{[Ba^2+]}\) decreases. Further increase of \(\ce{[CrO4^2- ]}\) till for \(\ce{SrCrO4}\) increases to of \(\ce{SrCrO4}\); it then precipitates. Let us write the equilibrium equations and data down to help us think. Let \(x\) be the concentration of chromate to precipitate \(\ce{Sr^2+}\), and \(y\) be that to precipitate \(\ce{Ba^2+}\): \[\ce{SrCrO4(s) \rightarrow Sr^{2+}(aq) + CrO4^{2-}(aq)} \nonumber \] According to the definition of we have we have \(K_{\ce{sp}} = (0.30)(x) = 3.6 \times 10^{-5} \). Solving for \(x\) gives \[x = \dfrac{3.6 \times 10^{5}}{0.30} = 1.2 \times 10^{-4} M \nonumber \] Further, let \(y\) be the concentration of chromate to precipitate precipitate \(\ce{Ba^2+}\): \[\ce{BaCrO4(s) \rightarrow Ba^{2+}(aq) + CrO4^{2-}(aq)} \nonumber \] with \(K_{\ce{sp}} = (0.30)(y) = 1.2 \times 10^{-10}\). Solving for \(y\) gives \[y = \dfrac{1.2 \times 10^{-10}}{0.30} = 4.0 \times 10^{-10} \;M \nonumber \] The 's for the two salts indicate \(\ce{BaCrO4}\) to be much less soluble, and it will precipitate before any \(\ce{SrCrO4}\) precipitates. If chromate concentration is maintained less than \(1.2 \times 10^{-4} M\), then all \(\ce{Sr^2+}\) ions will remain in the solution. In reality, controling the increase of \(\ce{[CrO4^2- ]}\) is very difficult. The for strontium chromate is \(3.6\times 10^{-5}\) and the for barium chromate is \(1.2\times 10^{-10}\). Potassium chromate is added a small amount at a time to first precipitate \(\ce{BaCrO4}\). Calculate \(\ce{[Ba^2+]}\) when the first trace of \(\ce{SrCrO4}\) precipitate starts to form in a solution that contains 0.30 M each of \(\ce{Ba^2+}\) and \(\ce{Sr^2+}\) ions. From the solution given in Example \(\Page {1}\), \(\ce{[CrO4^2- ]} = 3.6\times 10^{-4}\; M\) when \(\ce{SrCrO4}\) starts to form. At this concentration, the \(\ce{[Ba^2+]}\) is estimated at \(3.6 \times 10^{-4} = 1.2\times 10^{-10}\). The of \(\ce{BaCrO4}\). Thus, \[\ce{[Ba^2+]} = 3.33 \times 10^{-7}\, M \nonumber \] Very small indeed, compared to 0.30. In the fresh precipitate of \(\ce{SrCrO4}\), the molar ratio of \(\ce{SrCrO4}\) to \(\ce{BaCrO4}\) is \[\dfrac{0.30}{3.33 \times 10^{-7}} = 9.0 \times 10^{5}. \nonumber \] Hence, the amount of \(\ce{Ba^2+}\) ion in the solid is only \(1 \times 10^{-6}\) (i.e., 1 ppm) of all metal ions, providing that all the solid was removed when \[\ce{[CrO4^{2-}]} = 3.6 \times 10^{-4} M. \nonumber \] The calculation shown here indicates that the separation of \(\ce{Sr}\) and \(\ce{Ba}\) is pretty good. In practice, an impurity level of 1 ppm is a very small value. What reagent should you use to separate silver and lead ions that are present in a solution? What data or information will be required for this task? The 's for salts of silver and lead are required. We list the 's for chlorides and sulfates in a table here. These value are found in the Handbook Menu of our website as Salts . Because the 's \(\ce{AgCl}\) and \(\ce{PbCl2}\) are very different, chloride, \(\ce{Cl-}\), apppears a good choice of negative ions for their separation. The literature also indicates that \(\ce{PbCl2}\) is rather soluble in warm water, and by heating the solution to 350 K (80 C), you can keep \(\ce{Pb^2+}\) ions in solution and precipitate \(\ce{AgCl}\) as a solid. The solubility of \(\ce{AgCl}\) is very small even at high temperatures. Find more detailed information about the solubility of lead chloride as a function of temperature. Can sulfate be used to separate silver and lead ions? Which one will form a precipitate first as the sulfate ion concentration increases? What is the \(\ce{[Pb^2+]}\) when \(\ce{Ag2SO4}\) begins to precipitate in a solution that contains 0.10 M \(\ce{Ag+}\)? The Separation of Two Ions by a Difference in Solubility: | 5,942 | 2,852 |
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The chemical structures of the solute and solvent dictate the types of forces possible and, consequently, are important factors in determining solubility. For example, under similar conditions, the water solubility of oxygen is approximately three times greater than that of helium, but 100 times less than the solubility of chloromethane, CHCl . Considering the role of the solvent’s chemical structure, note that the solubility of oxygen in the liquid hydrocarbon hexane, C H , is approximately 20 times greater than it is in water. Other factors also affect the solubility of a given substance in a given solvent. Temperature is one such factor, with gas solubility typically decreasing as temperature increases (Figure \(\Page {1}\)). This is one of the major impacts resulting from the thermal pollution of natural bodies of water. When the temperature of a river, lake, or stream is raised abnormally high, usually due to the discharge of hot water from some industrial process, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water’s ecosystems and, in severe cases, can result in large-scale fish kills (Figure \(\Page {2}\)). The solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the solution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with CO at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (Figure \(\Page {3}\)). At this point, the beverage is with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become “flat.” For many gaseous solutes, the relation between solubility, , and partial pressure, , is a proportional one: where is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature. This is a mathematical statement of : At 20 °C, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa (760 torr) is 1.38 × 10 mol L . Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa (155 torr), the approximate pressure of oxygen in earth’s atmosphere. According to Henry’s law, for an ideal solution the solubility, , of a gas (1.38 × 10 mol L , in this case) is directly proportional to the pressure, , of the undissolved gas above the solution (101.3 kPa, or 760 torr, in this case). Because we know both and , we can rearrange this expression to solve for . \[\begin{align*}
C_\ce{g}&=kP_\ce{g}\\[4pt]
k&=\dfrac{C_\ce{g}}{P_\ce{g}}\\[4pt]
&=\mathrm{\dfrac{1.38×10^{−3}\:mol\:L^{−1}}{101.3\:kPa}}\\[4pt]
&=\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}}\\[4pt]
&\hspace{15px}\mathrm{(1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1})}
\end{align*}\] Now we can use to find the solubility at the lower pressure. \[C_\ce{g}=kP_\ce{g}\] \[\mathrm{1.36×10^{−5}\:mol\:L^{−1}\:kPa^{−1}×20.7\:kPa\\[4pt]
(or\:1.82×10^{−6}\:mol\:L^{−1}\:torr^{−1}×155\:torr)\\[4pt]
=2.82×10^{−4}\:mol\:L^{−1}}\] Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable. Exposing a 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 20.26 kPa (152 torr) resulted in the dissolution of 1.45 × 10 g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 101.3 kPa (760 torr). 7.25 × 10 g in 100.0 mL or 0.0725 g/L Decompression sickness (DCS), or “the bends,” is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver’s blood are proportionally higher per Henry’s law. As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver’s blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds (10 or 20 m/min) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventive measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure \(\Page {4}\)). Deviations from are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water does not increase as rapidly with increasing pressure as predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions. Gases can form supersaturated solutions. If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (Figure \(\Page {5}\)), a deep lake in a volcanic crater. The water at the bottom of Lake Nyos is saturated with carbon dioxide by volcanic activity beneath the lake. It is believed that the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved CO were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley. Henry's Law (The Solubility of Gases in Solvents): ). | 6,747 | 2,853 |
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We have seen that when elements react, they often gain or lose enough electrons to achieve the valence electron configuration of the nearest noble gas. In this section, we develop a more quantitative approach to predicting such reactions by examining periodic trends in the energy changes that accompany ion formation. Because atoms do not spontaneously lose electrons, energy is required to remove an electron from an atom to form a cation. Chemists define the ionization energy (\(I\)) of an element as the amount of energy needed to remove an electron from the gaseous atom \(E\) in its ground state. \(I\) is therefore the energy required for the reaction \[ E_{(g)} \rightarrow E^+_{(g)} +e^- \;\;\ \text{energy required=I } \label{9.4.1}\] Because an input of energy is required, the ionization energy is always positive (\(I > 0\)) for the reaction as written in Equation 9.4.1. Larger values of mean that the electron is more tightly bound to the atom and harder to remove. Typical units for ionization energies are kilojoules/mole (kJ/mol) or electron volts (eV): If an atom possesses more than one electron, the amount of energy needed to remove successive electrons increases steadily. We can define a first ionization energy ( ), a second ionization energy ( ), and in general an th ionization energy ( ) according to the following reactions: \[ E_{(g)} \rightarrow E^+_{(g)} +e^- \;\;\ I_1=\text{1st ionization energy} \label{9.4.2}\] \[ E_{(g)} \rightarrow E^+_{(g)} +e^- \;\;\ I_2=\text{2nd ionization energy} \label{9.4.3}\] \[ E^+_{(g)} \rightarrow E^{2+}_{(g)} +e^- \;\;\ I_3=\text{3rd ionization energy} \label{9.4.4}\] Values for the ionization energies of \(Li\) and \(Be\) listed in Table \(\Page {1}\) show that successive ionization energies for an element increase steadily; that is, it takes more energy to remove the second electron from an atom than the first, and so forth. There are two reasons for this trend. First, the second electron is being removed from a positively charged species rather than a neutral one, so in accordance with Coulomb’s law, more energy is required. Second, removing the first electron reduces the repulsive forces among the remaining electrons, so the attraction of the remaining electrons to the nucleus is stronger. Successive ionization energies for an element steadily. The most important consequence of the values listed in Table \(\Page {1}\) is that the chemistry of \(Li\) is dominated by the \(Li^+\) ion, while the chemistry of \(Be\) is dominated by the +2 oxidation state. The energy required to remove the electron from \(Li\) \[Li^+_{(g)} \rightarrow Li^{2+}_{(g)} + e^− \label{9.4.5}\] is more than 10 times greater than the energy needed to remove the first electron. Similarly, the energy required to remove the electron from \(Be\) \[Be^{2+}_{(g)} \rightarrow Be^{3+}_{(g)} + e^− \label{9.4.6}\] is about 15 times greater than the energy needed to remove the first electron and around 8 times greater than the energy required to remove the second electron. Both \(Li^+\) and \(Be^{2+}\) have 1 closed-shell configurations, and much more energy is required to remove an electron from the 1 core than from the 2 valence orbital of the same element. The chemical consequences are enormous: lithium (and all the alkali metals) forms compounds with the 1+ ion but not the 2+ or 3+ ions. Similarly, beryllium (and all the alkaline earth metals) forms compounds with the 2+ ion but not the 3+ or 4+ ions. The energy required to remove electrons from a filled core is prohibitively large under normal reaction conditions. Ionization energies of the elements in the third row of the periodic table exhibit the same pattern as those of \(Li\) and \(Be\) (Table \(\Page {2}\)): successive ionization energies increase steadily as electrons are removed from the valence orbitals (3 or 3 , in this case), followed by an especially large increase in ionization energy when electrons are removed from filled core levels as indicated by the bold diagonal line in Table \(\Page {2}\). Thus in the third row of the periodic table, the largest increase in ionization energy corresponds to removing the fourth electron from \(Al\), the fifth electron from Si, and so forth—that is, removing an electron from an ion that has the valence electron configuration of the preceding noble gas. This pattern explains why the chemistry of the elements normally involves only valence electrons. Too much energy is required to either remove or share the inner electrons. From their locations in the periodic table, predict which of these elements has the highest fourth ionization energy: B, C, or N. three elements element with highest fourth ionization energy These elements all lie in the second row of the periodic table and have the following electron configurations: The fourth ionization energy of an element (\(I_4\)) is defined as the energy required to remove the fourth electron: \[E^{3+}_{(g)} \rightarrow E^{4+}_{(g)} + e^-\] Because carbon and nitrogen have four and five valence electrons, respectively, their fourth ionization energies correspond to removing an electron from a partially filled valence shell. The fourth ionization energy for boron, however, corresponds to removing an electron from the filled 1 subshell. This should require much more energy. The actual values are as follows: B, 25,026 kJ/mol; C, 6223 kJ/mol; and N, 7475 kJ/mol. From their locations in the periodic table, predict which of these elements has the lowest second ionization energy: Sr, Rb, or Ar. Sr The first column of data in Table \(\Page {2}\) shows that first ionization energies tend to increase across the third row of the periodic table. This is because the valence electrons do not screen each other very well, allowing the effective nuclear charge to increase steadily across the row. The valence electrons are therefore attracted more strongly to the nucleus, so atomic sizes decrease and ionization energies increase. These effects represent two sides of the same coin: stronger electrostatic interactions between the electrons and the nucleus further increase the energy required to remove the electrons. However, the first ionization energy decreases at Al ([Ne]3 3 ) and at S ([Ne]3 3 ). The electrons in aluminum’s filled 3 subshell are better at screening the 3 electron than they are at screening each other from the nuclear charge, so the electrons penetrate closer to the nucleus than the electron does. The decrease at S occurs because the two electrons in the same orbital repel each other. This makes the S atom slightly less stable than would otherwise be expected, as is true of all the group 16 elements. The first ionization energies of the elements in the first six rows of the periodic table are plotted in Figure \(\Page {1}\) and are presented numerically and graphically in Figure \(\Page {2}\). These figures illustrate three important trends: Generally, \(I_1\) increases diagonally from the lower left of the periodic table to the upper right. Gallium (Ga), which is the first element following the first row of transition metals, has the following electron configuration: [Ar]4 3 4 . Its first ionization energy is significantly lower than that of the immediately preceding element, zinc, because the filled 3 subshell of gallium lies inside the 4 subshell, screening the single 4 electron from the nucleus. Experiments have revealed something of even greater interest: the second and third electrons that are removed when gallium is ionized come from the 4 orbital, the 3 subshell. The chemistry of gallium is dominated by the resulting Ga ion, with its [Ar]3 electron configuration. This and similar electron configurations are particularly stable and are often encountered in the heavier -block elements. They are sometimes referred to as pseudo noble gas configurations. In fact, for elements that exhibit these configurations, . As we noted, the first ionization energies of the transition metals and the lanthanides change very little across each row. Differences in their second and third ionization energies are also rather small, in sharp contrast to the pattern seen with the - and -block elements. The reason for these similarities is that the transition metals and the lanthanides form cations by losing the electrons before the ( − 1) or ( − 2) electrons, respectively. This means that transition metal cations have ( − 1) valence electron configurations, and lanthanide cations have ( − 2) valence electron configurations. Because the ( − 1) and ( − 2) shells are closer to the nucleus than the shell, the ( − 1) and ( − 2) electrons screen the electrons quite effectively, reducing the effective nuclear charge felt by the electrons. As increases, the increasing positive charge is largely canceled by the electrons added to the ( − 1) or ( − 2) orbitals. That the electrons are removed before the ( − 1) or ( − 2) electrons may surprise you because the orbitals were filled in the reverse order. In fact, the , the ( − 1) , and the ( − 2) orbitals are so close to one another in energy, and interpenetrate one another so extensively, that very small changes in the effective nuclear charge can change the order of their energy levels. As the orbitals are filled, the effective nuclear charge causes the 3 orbitals to be slightly lower in energy than the 4 orbitals. The [Ar]3 electron configuration of Ti tells us that the 4 electrons of titanium are lost before the 3 electrons; this is confirmed by experiment. A similar pattern is seen with the lanthanides, producing cations with an ( − 2) valence electron configuration. Because their first, second, and third ionization energies change so little across a row, these elements have important similarities in chemical properties in addition to the expected vertical similarities. For example, all the first-row transition metals except scandium form stable compounds as M ions, whereas the lanthanides primarily form compounds in which they exist as M ions. Use their locations in the periodic table to predict which element has the lowest first ionization energy: Ca, K, Mg, Na, Rb, or Sr. six elements element with lowest first ionization energy Locate the elements in the periodic table. Based on trends in ionization energies across a row and down a column, identify the element with the lowest first ionization energy. These six elements form a rectangle in the two far-left columns of the periodic table. Because we know that ionization energies increase from left to right in a row and from bottom to top of a column, we can predict that the element at the bottom left of the rectangle will have the lowest first ionization energy: Rb. Use their locations in the periodic table to predict which element has the highest first ionization energy: As, Bi, Ge, Pb, Sb, or Sn. As Ionization Energy: The tendency of an element to lose is one of the most important factors in determining the kind of compounds it forms. Periodic behavior is most evident for , the energy required to remove an electron from a gaseous atom. The energy required to remove successive electrons from an atom increases steadily, with a substantial increase occurring with the removal of an electron from a filled inner shell. Consequently, only valence electrons can be removed in chemical reactions, leaving the filled inner shell intact. Ionization energies explain the common oxidation states observed for the elements. Ionization energies increase diagonally from the lower left of the periodic table to the upper right. Minor deviations from this trend can be explained in terms of particularly stable electronic configurations, called , in either the parent atom or the resulting ion. | 11,834 | 2,854 |
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Bromine will not add across the double bond of benzene. Instead, a bromine atom can replace one of the hydrogen atoms on the benzene. This reaction is especially easy in the presence of a catalyst. How does that outcome happen? Why does that outcome happen? There has been a good deal of study of these reactions and there is strong evidence of the steps through which they occur. As expected, the reaction involves donation of π electrons from the benzene. For the moment, we'll assume the electrophile is a bromine cation; we will deal with its exact structure later. The problem is, that initial step results in the loss of aromaticity. The aromatic system confers a little extra stability on the π system, so the molecule is motivated to restore the aromaticity. The easiest way to do that, and get rid of a positive charge at the same time, would be to deprotonate the cation. Some base will pick up the proton; it is likely a bromide ion in this case. We will see later where that bromide comes from. How do we know that the mechanism unfolds this way? There are three basic steps that are clearly accomplished during the course of the reaction: the C-H bond is broken, the C-Br bond is formed, and the Br-Br bond is broken. When is the C-H bond broken? That question can be answered by looking for what is called an "isotope effect." The most common isotope of hydrogen is H, or protium, but H is also available; it is called "deuterium." Deuterium is often represented by the symbol D and protium by the symbol H. Deuterium is twice as heavy as the common protium. That mass difference leads to a lower vibrational frequency of a C-D bond than a C-H bond. The C-H bond vibrates more rapidly and energetically than a C-D bond; as a consequence, the C-H bond is more easily broken than the C-D bond. If we take a sample of ordinary benzene, C H , and a sample of deuterated benzene, C D , we can measure how quickly they each undergo a bromination reaction. Very often, a reaction that involves C-H bond cleavage will slow down if a C-D bond is involved. This outcome is observed in E2 eliminations, for instance. This slowing of the reaction with the heavier isotope is called the deuterium isotope effect. However, no deuterium isotope effect is observed during bromination, or other aromatic electrophilic substitution reactions. That absence of an isotope effect usually means the C-H bond cleavage is a sort of an afterthought. The hard part of the reaction is already done. Both the C-H and C-D bonds are broken so quickly and easily, by comparison, that we don't really notice the difference between them. There is even more evidence. In a few exceptional cases, the cationic intermediate in this reaction is stable enough to be isolated and crystallized. X-ray diffraction shows that there is a tetrahedral carbon in the ring, indicating that the C-H bond has not broken yet. The C-H bond is broken at the end of the reaction. When is the Br-Br bond broken? That question is a little harder to answer. We can't use the same isotope strategy that we used with the C-H bond. Although deuterium is twice as heavy as protium, producing a substantial isotope effect, Br is only 2.5% more massive than Br. Any difference in rates involving these isotopes is undetectable. The exact nature of the bromine species in the reaction is complicated, and may even be different under different conditions. In the case of uncatalyzed bromination reactions, the addition of salts such as NaBr has no effect on the reaction rate, indicating that the arene reacts directly with Br rather than Br . Explain this line of reasoning. Given each of the following electrophiles, provide a mechanism for electrophilic aromatic substitution. a) NO b) CH CH c) SO H d) CH CO , | 3,784 | 2,855 |
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Alkyl halides (also known as haloalkanes) are compounds in which one or more hydrogen atoms in an alkane have been replaced by halogen atoms (fluorine, chlorine, bromine or iodine). We will only look at compounds containing one halogen atom. For example: alkyl halides fall into different classes depending on how the halogen atom is positioned on the chain of carbon atoms. There are some chemical differences between the various types. The chart shows the boiling points of some simple alkyl halides. Notice that three of these have boiling points below room temperature (taken as being about 20°C). These will be gases at room temperature. All the others you are likely to come across are liquids. Remember: The patterns in boiling point reflect the patterns in . These attractions get stronger as the molecules get longer and have more electrons. That increases the sizes of the temporary dipoles that are set up. This is why the boiling points increase as the number of carbon atoms in the chains increases. Look at the chart for a particular type of halide (a chloride, for example). Dispersion forces get stronger as you go from 1 to 2 to 3 carbons in the chain. It takes more energy to overcome them, and so the boiling points rise. The increase in boiling point as you go from a chloride to a bromide to an iodide (for a given number of carbon atoms) is also because of the increase in number of electrons leading to larger dispersion forces. There are lots more electrons in, for example, iodomethane than there are in chloromethane - count them! The carbon-halogen bonds (apart from the carbon-iodine bond) are polar, because the electron pair is pulled closer to the halogen atom than the carbon. This is because (apart from iodine) the halogens are more electronegative than carbon. The electronegativity values are: This means that in addition to the dispersion forces there will be forces due to the attractions between the permanent dipoles (except in the iodide case). The size of those dipole-dipole attractions will fall as the bonds get less polar (as you go from chloride to bromide to iodide, for example). Nevertheless, the boiling points rise! This shows that the effect of the permanent dipole-dipole attractions is much less important than that of the temporary dipoles which cause the dispersion forces. The large increase in number of electrons by the time you get to the iodide completely outweighs the loss of any permanent dipoles in the molecules. The examples show that the boiling points fall as the isomers go from a primary to a secondary to a tertiary halogenoalkane. This is a simple result of the fall in the effectiveness of the dispersion forces. The temporary dipoles are greatest for the longest molecule. The attractions are also stronger if the molecules can lie closely together. The tertiary halogenoalkane is very short and fat, and won't have much close contact with its neighbours. The alkyl halides are at best only slightly soluble in water. For a halogenoalkane to dissolve in water you have to break attractions between the halogenoalkane molecules (van der Waals dispersion and dipole-dipole interactions) and break the hydrogen bonds between water molecules. Both of these cost energy. Energy is released when new attractions are set up between the halogenoalkane and the water molecules. These will only be dispersion forces and dipole-dipole interactions. These aren't as strong as the original hydrogen bonds in the water, and so not as much energy is released as was used to separate the water molecules. The energetics of the change are sufficiently "unprofitable" that very little dissolves. This energetic picture of solvation is not the whole story and entropic factors must also be considered to understand solvation properly. alkyl halides tend to dissolve in organic solvents because the new intermolecular attractions have much the same strength as the ones being broken in the separate halogenoalkane and solvent. The pattern in strengths of the four carbon-halogen bonds are: Notice that bond strength falls as you go from C-F to C-I, and notice how much stronger the carbon-fluorine bond is than the rest. To react with the alkyl halides, the carbon-halogen bond has got to be broken. Because that gets easier as you go from fluoride to chloride to bromide to iodide, the compounds get more reactive in that order. Iodoalkanes are the most reactive and fluoroalkanes are the least. In fact, fluoroalkanes are so unreactive that we shall pretty well ignore them completely from now on in this section! Of the four halogens, fluorine is the most electronegative and iodine the least. That means that the electron pair in the carbon-fluorine bond will be dragged most towards the halogen end. Looking at the methyl halides as simple examples: The electronegativities of carbon and iodine are equal and so there will be no separation of charge on the bond. One of the important set of reactions of alkyl halides involves replacing the halogen by something else - . These reactions involve either: You might have thought that either of these would be more effective in the case of the carbon-fluorine bond with the quite large amounts of positive and negative charge already present. But that's not so - quite the opposite is true! The thing that governs the reactivity is the strength of the bonds which have to be broken. If is difficult to break a carbon-fluorine bond, but easy to break a carbon-iodine one. Jim Clark ( ) | 5,507 | 2,856 |
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Proton transfer is rapid, especially if it is transferred from a very acidic position. For example, a proton can easily be transferred from a positively charged oxygen atom to a neutral oxygen (resulting in a new, neutral oxygen and a new, positive oxygen). These species would be in equilibrium with each other. It would not be as easy to transfer a proton from a neutral oxygen to another neutral oxygen. Sometimes, a neutral oxygen can transfer a proton to a negatively charged one, but the equilibrium will depend on the relative pKa values of the two species. In the case below, the tert-butoxide is a less stable anion than the hydroxide because the tert-butoxide is larger and requires more organization of solvent molecules around it. It is tempting to think that a proton could be transferred directly from a cationic position to an anionic position in the same molecule. That might not occur, however. In terms of conformational analysis, the two positions might not be able to twist around and reach each other. The usual rule applies: two atoms may need to be greater than five atoms away from each other along a chain before they can reach around and make contact. If the solvent has a lone pair, it may pick up the proton from the acidic position and drop it off on the basic position. These events are made easy by the fact that the reacting molecules are usually surrounded by many solvent molecules. , | 1,431 | 2,857 |
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Draw 4 mL of any gas (A) into a 10 cc syringe (a colored gas such as nitrogen dioxide is ideal, but not necessary). Draw 6 mL of a second gas (B) into the syringe. Now, for the two gases in the syringe, 1. What is the volume of A? 2. What is the volume of B? The amounts of the two gases cannot be distinguished by their volumes! 3. What is the pressure of A, since it expanded from 6 to 10 mL? Boyle's law tells us: (1Atm)(6 mL) = P (10 mL) 4. What is the pressure of B? Dalton's Law tells us: If the pressure of A is 0.6 Atm, and the total pressure is 1.0 Atm, the pressure of B must be 0.4 Atm. 5. Use Avogadro's Law to derive other useful forms of Dalton's Law. P / P = n / n | 695 | 2,858 |
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The electron affinity (\(EA\)) of an element \(E\) is defined as the energy change that occurs when an electron is added to a gaseous atom: \[ E_{(g)}+e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA \label{7.5.1}\] Unlike ionization energies, which are always positive for a neutral atom because energy is required to remove an electron, electron affinities can be negative (energy is released when an electron is added), positive (energy must be added to the system to produce an anion), or zero (the process is energetically neutral). This sign convention is consistent with a negative value corresponded to the energy change for an , which is one in which heat is released. The chlorine atom has the most negative electron affinity of any element, which means that more energy is released when an electron is added to a gaseous chlorine atom than to an atom of any other element: \[ Cl_{(g)}+e^- \rightarrow Cl^-_{(g)} \;\;\; EA=-346\; kJ/mol \label{7.5.2}\] In contrast, beryllium does not form a stable anion, so its effective electron affinity is \[ Be_{(g)}+e^- \rightarrow Be^-_{(g)} \;\;\; EA \ge 0 \label{7.5.3}\] Nitrogen is unique in that it has an electron affinity of approximately zero. Adding an electron neither releases nor requires a significant amount of energy: \[ N_{(g)}+e^- \rightarrow N^-_{(g)} \;\;\; EA \approx 0 \label{7.5.4}\] Generally, electron affinities become more negative across a row of the periodic table. In general, electron affinities of the main-group elements become less negative as we proceed down a column. This is because as increases, the extra electrons enter orbitals that are increasingly far from the nucleus. Atoms with the largest radii, which have the lowest ionization energies (affinity for their own valence electrons), also have the lowest affinity for an added electron. There are, however, two major exceptions to this trend: In general, electron affinities become more negative across a row and less negative down a column. The equations for second and higher electron affinities are analogous to those for second and higher ionization energies: \[E_{(g)} + e^- \rightarrow E^-_{(g)} \;\;\; \text{energy change=}EA_1 \label{7.5.5}\] \[E^-_{(g)} + e^- \rightarrow E^{2-}_{(g)} \;\;\; \text{energy change=}EA_2 \label{7.5.6}\] As we have seen, the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons. For example, the first electron affinity of oxygen is −141 kJ/mol, but the second electron affinity is +744 kJ/mol: \[O_{(g)} + e^- \rightarrow O^-_{(g)} \;\;\; EA_1=-141 \;kJ/mol \label{7.5.7}\] Thus the formation of a gaseous oxide (\(O^{2−}\)) ion is energetically quite (estimated by adding both steps): \[O_{(g)} + 2e^- \rightarrow O^{2-}_{(g)} \;\;\; EA=+603 \;kJ/mol \label{7.5.9}\] Similarly, the formation of all common dianions (such as \(S^{2−}\)) or trianions (such as \(P^{3−}\)) is energetically unfavorable in the gas phase. While first electron affinities can be negative, positive, or zero, second electron affinities are always positive. If energy is required to form both positively charged ions and monatomic polyanions, why do ionic compounds such as \(MgO\), \(Na_2S\), and \(Na_3P\) form at all? The key factor in the formation of stable ionic compounds is the favorable electrostatic interactions between the cations and the anions . Based on their positions in the periodic table, which of Sb, Se, or Te would you predict to have the most negative electron affinity? three elements element with most negative electron affinity We know that electron affinities become less negative going down a column (except for the anomalously low electron affinities of the elements of the second row), so we can predict that the electron affinity of Se is more negative than that of Te. We also know that electron affinities become more negative from left to right across a row, and that the group 15 elements tend to have values that are less negative than expected. Because Sb is located to the left of Te and belongs to group 15, we predict that the electron affinity of Te is more negative than that of Sb. The overall order is Se < Te < Sb, so Se has the most negative electron affinity among the three elements. Based on their positions in the periodic table, which of Rb, Sr, or Xe would you predict to most likely form a gaseous anion? Rb Electron Affinity: The ( of an element is the energy change that occurs when an electron is added to a gaseous atom to give an anion. In general, elements with the most negative electron affinities (the highest affinity for an added electron) are those with the smallest size and highest ionization energies and are located in the upper right corner of the periodic table. | 5,005 | 2,859 |
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Crystalline solids have regular ordered arrays of components held together by uniform intermolecular forces, whereas the components of amorphous solids are not arranged in regular arrays. With few exceptions, the particles that compose a solid material, whether ionic, molecular, covalent, or metallic, are held in place by strong attractive forces between them. When we discuss solids, therefore, we consider the positions of the atoms, molecules, or ions, which are essentially fixed in space, rather than their motions (which are more important in liquids and gases). The constituents of a solid can be arranged in two general ways: they can form a regular repeating three-dimensional structure called a crystal lattice, thus producing a crystalline solid, or they can aggregate with no particular order, in which case they form an amorphous solid (from the Greek ámorphos, meaning “shapeless”). Crystalline solids, or crystals, have distinctive internal structures that in turn lead to distinctive flat surfaces, or faces. The faces intersect at angles that are characteristic of the substance. When exposed to x-rays, each structure also produces a distinctive pattern that can be used to identify the material. The characteristic angles do not depend on the size of the crystal; they reflect the regular repeating arrangement of the component atoms, molecules, or ions in space. When an ionic crystal is cleaved (Figure \(\Page {2}\), for example, repulsive interactions cause it to break along fixed planes to produce new faces that intersect at the same angles as those in the original crystal. In a covalent solid such as a cut diamond, the angles at which the faces meet are also not arbitrary but are determined by the arrangement of the carbon atoms in the crystal. Crystals tend to have relatively sharp, well-defined melting points because all the component atoms, molecules, or ions are the same distance from the same number and type of neighbors; that is, the regularity of the crystalline lattice creates local environments that are the same. Thus the intermolecular forces holding the solid together are uniform, and the same amount of thermal energy is needed to break every interaction simultaneously. Amorphous solids have two characteristic properties. When cleaved or broken, they produce fragments with irregular, often curved surfaces; and they have poorly defined patterns when exposed to x-rays because their components are not arranged in a regular array. An amorphous, translucent solid is called a glass. Almost any substance can solidify in amorphous form if the liquid phase is cooled rapidly enough. Some solids, however, are intrinsically amorphous, because either their components cannot fit together well enough to form a stable crystalline lattice or they contain impurities that disrupt the lattice. For example, although the chemical composition and the basic structural units of a quartz crystal and quartz glass are the same—both are SiO and both consist of linked SiO tetrahedra—the arrangements of the atoms in space are not. Crystalline quartz contains a highly ordered arrangement of silicon and oxygen atoms, but in quartz glass the atoms are arranged almost randomly. When molten SiO is cooled rapidly (4 K/min), it forms quartz glass, whereas the large, perfect quartz crystals sold in mineral shops have had cooling times of thousands of years. In contrast, aluminum crystallizes much more rapidly. Amorphous aluminum forms only when the liquid is cooled at the extraordinary rate of 4 × 10 K/s, which prevents the atoms from arranging themselves into a regular array. In an amorphous solid, the local environment, including both the distances to neighboring units and the numbers of neighbors, varies throughout the material. Different amounts of thermal energy are needed to overcome these different interactions. Consequently, amorphous solids tend to soften slowly over a wide temperature range rather than having a well-defined melting point like a crystalline solid. If an amorphous solid is maintained at a temperature just below its melting point for long periods of time, the component molecules, atoms, or ions can gradually rearrange into a more highly ordered crystalline form. Crystals have sharp, well-defined melting points; amorphous solids do not. Because a crystalline solid consists of repeating patterns of its components in three dimensions (a crystal lattice), we can represent the entire crystal by drawing the structure of the smallest identical units that, when stacked together, form the crystal. This basic repeating unit is called a unit cell. For example, the unit cell of a sheet of identical postage stamps is a single stamp, and the unit cell of a stack of bricks is a single brick. In this section, we describe the arrangements of atoms in various unit cells. Unit cells are easiest to visualize in two dimensions. In many cases, more than one unit cell can be used to represent a given structure, as shown for the Escher drawing in the chapter opener and for a two-dimensional crystal lattice in Figure \(\Page {4}\). Usually the smallest unit cell that completely describes the order is chosen. The only requirement for a valid unit cell is that repeating it in space must produce the regular lattice. Thus the unit cell in Figure \(\Page {4d}\) is not a valid choice because repeating it in space does not produce the desired lattice (there are triangular holes). The concept of unit cells is extended to a three-dimensional lattice in the schematic drawing in Figure \(\Page {5}\). There are seven fundamentally different kinds of unit cells, which differ in the relative lengths of the edges and the angles between them (Figure \(\Page {6}\)). Each unit cell has six sides, and each side is a parallelogram. We focus primarily on the cubic unit cells, in which all sides have the same length and all angles are 90°, but the concepts that we introduce also apply to substances whose unit cells are not cubic. If the cubic unit cell consists of eight component atoms, molecules, or ions located at the corners of the cube, then it is called simple cubic (Figure \(\Page {7a}\)). If the unit cell also contains an identical component in the center of the cube, then it is body-centered cubic (bcc) (\(\Page {7b}\)). If there are components in the center of each face in addition to those at the corners of the cube, then the unit cell is face-centered cubic (fcc) (Figure \(\Page {7c}\)). As indicated in Figure \(\Page {7}\), a solid consists of a large number of unit cells arrayed in three dimensions. Any intensive property of the bulk material, such as its density, must therefore also be related to its unit cell. Because density is the mass of substance per unit volume, we can calculate the density of the bulk material from the density of a single unit cell. To do this, we need to know the size of the unit cell (to obtain its volume), the molar mass of its components, and the number of components per unit cell. When we count atoms or ions in a unit cell, however, those lying on a face, an edge, or a corner contribute to more than one unit cell, as shown in Figure \(\Page {7}\). For example, an atom that lies on a face of a unit cell is shared by two adjacent unit cells and is therefore counted as \({1\over 2}\) atom per unit cell. Similarly, an atom that lies on the edge of a unit cell is shared by four adjacent unit cells, so it contributes \({1 \over 4}\) atom to each. An atom at a corner of a unit cell is shared by all eight adjacent unit cells and therefore contributes \({1 \over 8}\) atom to each. The statement that atoms lying on an edge or a corner of a unit cell count as \({1 \over 4}\) or \({1 \over 8}\) atom per unit cell, respectively, is true for all unit cells except the hexagonal one, in which three unit cells share each vertical edge and six share each corner (Figure \(\Page {7}\):), leading to values of \({1 \over 3}\) and \({1 \over 6}\) atom per unit cell, respectively, for atoms in these positions. In contrast, atoms that lie entirely within a unit cell, such as the atom in the center of a body-centered cubic unit cell, belong to only that one unit cell. For all unit cells except hexagonal, atoms on the faces contribute \({1\over 2}\) atom to each unit cell, atoms on the edges contribute \({1 \over 4}\) atom to each unit cell, and atoms on the corners contribute \({1 \over 8}\) atom to each unit cell. Metallic gold has a face-centered cubic unit cell (\(\Page {7c}\)). How many Au atoms are in each unit cell? : unit cell : number of atoms per unit cell Using Figure \(\Page {7}\), identify the positions of the Au atoms in a face-centered cubic unit cell and then determine how much each Au atom contributes to the unit cell. Add the contributions of all the Au atoms to obtain the total number of Au atoms in a unit cell. As shown in Figure \(\Page {7}\), a face-centered cubic unit cell has eight atoms at the corners of the cube and six atoms on the faces. Because atoms on a face are shared by two unit cells, each counts as \({1 \over 2}\) atom per unit cell, giving \(6\times{1 \over 2}=3\) Au atoms per unit cell. Atoms on a corner are shared by eight unit cells and hence contribute only \({1\over 8}\) atom per unit cell, giving \(8\times{1\over 8}=1\) Au atom per unit cell. The total number of Au atoms in each unit cell is thus \(3 + 1 = 4\). Metallic iron has a body-centered cubic unit cell (Figure \(\Page {7b}\)). How many Fe atoms are in each unit cell? two Now that we know how to count atoms in unit cells, we can use unit cells to calculate the densities of simple compounds. Note, however, that we are assuming a solid consists of a perfect regular array of unit cells, whereas real substances contain impurities and defects that affect many of their bulk properties, including density. Consequently, the results of our calculations will be close but not necessarily identical to the experimentally obtained values. Calculate the density of metallic iron, which has a body-centered cubic unit cell (Figure \(\Page {7b}\)) with an edge length of 286.6 pm. : unit cell and edge length : density : : We know from Example \(\Page {1}\) that each unit cell of metallic iron contains two Fe atoms. The molar mass of iron is 55.85 g/mol. Because density is mass per unit volume, we need to calculate the mass of the iron atoms in the unit cell from the molar mass and Avogadro’s number and then divide the mass by the volume of the cell (making sure to use suitable units to get density in g/cm ): \[ \textit{mass of Fe} =\left ( 2 \; \cancel{atoms} \; Fe \right )\left ( \dfrac{ 1 \; \cancel{mol}}{6.022\times 10^{23} \; \cancel{atoms}} \right )\left ( \dfrac{55.85 \; g}{\cancel{mol}} \right ) =1.855\times 10^{-22} \; g \nonumber \] \[ volume=\left [ \left ( 286.6 \; pm \right )\left ( \dfrac{10^{-12 }\; \cancel{m}}{\cancel{pm}} \right )\left ( \dfrac{10^{2} \; cm}{\cancel{m}} \right ) \right ] =2.345\times 10^{-23} \; cm^{3} \nonumber \] \[ density = \dfrac{1.855\times 10^{-22} \; g}{2.345\times 10^{-23} \; cm^{3}} = 7.880 g/cm^{3} \nonumber \] This result compares well with the tabulated experimental value of 7.874 g/cm . Calculate the density of gold, which has a face-centered cubic unit cell (Figure \(\Page {7c}\)) with an edge length of 407.8 pm. 19.29 g/cm Our discussion of the three-dimensional structures of solids has considered only substances in which all the components are identical. As we shall see, such substances can be viewed as consisting of identical spheres packed together in space; the way the components are packed together produces the different unit cells. Most of the substances with structures of this type are metals. The arrangement of the atoms in a solid that has a simple cubic unit cell was shown in Figure \(\Page {5a}\). Each atom in the lattice has only six nearest neighbors in an octahedral arrangement. Consequently, the simple cubic lattice is an inefficient way to pack atoms together in space: only 52% of the total space is filled by the atoms. The only element that crystallizes in a simple cubic unit cell is polonium. Simple cubic unit cells are, however, common among binary ionic compounds, where each cation is surrounded by six anions and vice versa (Figure \(\Page {8}\)). The body-centered cubic unit cell is a more efficient way to pack spheres together and is much more common among pure elements. Each atom has eight nearest neighbors in the unit cell, and 68% of the volume is occupied by the atoms. As shown in Figure \(\Page {8}\), the body-centered cubic structure consists of a single layer of spheres in contact with each other and aligned so that their centers are at the corners of a square; a second layer of spheres occupies the square-shaped “holes” above the spheres in the first layer. The third layer of spheres occupies the square holes formed by the second layer, so that each lies directly above a sphere in the first layer, and so forth. All the alkali metals, barium, radium, and several of the transition metals have body-centered cubic structures. The most efficient way to pack spheres is the close-packed arrangement, which has two variants. A single layer of close-packed spheres is shown in Figure \(\Page {6a}\). Each sphere is surrounded by six others in the same plane to produce a hexagonal arrangement. Above any set of seven spheres are six depressions arranged in a hexagon. In principle, all six sites are the same, and any one of them could be occupied by an atom in the next layer. Actually, however, these six sites can be divided into two sets, labeled B and C in Figure \(\Page {9a}\). Sites B and C differ because as soon as we place a sphere at a B position, we can no longer place a sphere in any of the three C positions adjacent to A and vice versa. If we place the second layer of spheres at the B positions in Figure \(\Page {9a}\), we obtain the two-layered structure shown in Figure \(\Page {9b}\). There are now two alternatives for placing the first atom of the third layer: we can place it directly over one of the atoms in the first layer (an A position) or at one of the C positions, corresponding to the positions that we did not use for the atoms in the first or second layers (Figure \(\Page {9c}\)). If we choose the first arrangement and repeat the pattern in succeeding layers, the positions of the atoms alternate from layer to layer in the pattern ABABAB…, resulting in a hexagonal close-packed (hcp) structure (Figure \(\Page {9a}\)). If we choose the second arrangement and repeat the pattern indefinitely, the positions of the atoms alternate as ABCABC…, giving a cubic close-packed (ccp) structure (Figure \(\Page {9b}\)). Because the ccp structure contains hexagonally packed layers, it does not look particularly cubic. As shown in Figure \(\Page {9b}\), however, simply rotating the structure reveals its cubic nature, which is identical to a fcc structure. The hcp and ccp structures differ only in the way their layers are stacked. Both structures have an overall packing efficiency of 74%, and in both each atom has 12 nearest neighbors (6 in the same plane plus 3 in each of the planes immediately above and below). Table \(\Page {1}\) compares the packing efficiency and the number of nearest neighbors for the different cubic and close-packed structures; the number of nearest neighbors is called the coordination number. Most metals have hcp, ccp, or bcc structures, although several metals exhibit both hcp and ccp structures, depending on temperature and pressure. A crystalline solid can be represented by its unit cell, which is the smallest identical unit that when stacked together produces the characteristic three-dimensional structure. Solids are characterized by an extended three-dimensional arrangement of atoms, ions, or molecules in which the components are generally locked into their positions. The components can be arranged in a regular repeating three-dimensional array (a crystal lattice), which results in a crystalline solid, or more or less randomly to produce an amorphous solid. Crystalline solids have well-defined edges and faces, diffract x-rays, and tend to have sharp melting points. In contrast, amorphous solids have irregular or curved surfaces, do not give well-resolved x-ray diffraction patterns, and melt over a wide range of temperatures. The smallest repeating unit of a crystal lattice is the unit cell. The simple cubic unit cell contains only eight atoms, molecules, or ions at the corners of a cube. A body-centered cubic (bcc) unit cell contains one additional component in the center of the cube. A face-centered cubic (fcc) unit cell contains a component in the center of each face in addition to those at the corners of the cube. Simple cubic and bcc arrangements fill only 52% and 68% of the available space with atoms, respectively. The hexagonal close-packed (hcp) structure has an ABABAB… repeating arrangement, and the cubic close-packed (ccp) structure has an ABCABC… repeating pattern; the latter is identical to an fcc lattice. The hcp and ccp arrangements fill 74% of the available space and have a coordination number of 12 for each atom in the lattice, the number of nearest neighbors. The simple cubic and bcc lattices have coordination numbers of 6 and 8, respectively. | 17,359 | 2,860 |
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When any weak acid, which we will denote by the general formula HA, is dissolved in water, the reaction \[\text{HA} + \text{H}_{2}\text{O}\rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{ A}^{-}\nonumber \] proceeds to only a limited extent, and we must allow for this in calculating the hydronium-ion concentration and hence the pH of such a solution. In general the pH of a solution of a weak acid depends on only two factors, the concentration of the acid, , and the magnitude of an equilibrium constant , called the acid constant, which measures the strength of the acid. The acid constant is defined by the relationship: \[K_{c} \times \text{ 1}=K_{a}=\frac{[\text{ H}_{3}\text{O}^{+},\text{ A}^{-}]}{[\text{ HA }] }\nonumber \] (The acid constant is \(K_c\) multiplied by the activity of the water, which has a value of 1 because water is the solvent as already defined in the section on .) In we showed how measurements of the conductivity of acetic acid solutions could be used to find the acid constant for acetic acid, (CH COOH). You will recall from that example that was only approximately a constant, varying from a value of 1.81 × 10 mol/L in very dilute solutions to a value of 1.41 × 10 mol/L in a 1 solution. A similar variation is found for other weak acids, so that most of the calculations we do using are only approximate. Only two or possibly three significant figures should be retained. The table below gives the values for a few selected acids arranged in order of their strength. This table is part of our larger collection of acid and base resources. It is at once apparent from this table that the larger the value, the stronger the acid. The strongest acids, like HCl and H SO have values which are too large to measure, while another strong acid, HNO , has value close to 20 mol/L. Typical weak acids such as HF and CH COOH have acid constants with a value of 10 or 10 mol/L. Acids like the ammonium ion, NH , and hydrogen cyanide, HCN, for which is less than 10 mol/L are very weakly acidic. You may notice that some acids have multiple K values. This is due to the fact they are polyprotic, which means they are able to give up more than one proton. However, due to molecular forces, the ease with which each proton gives up its protons varies, giving different K values for each proton given up. \(\ce{CH3COOH + H2O <=> H3O^+ +CH3COO^-}\) \(\ce{H3PO4 + H2O <=> H3O^+ + H2PO4^-}\) \(\ce{H2PO4^- + H2O <=> H3O^+ + HPO4^{2-}}\) \(\ce{HPO4^{2-} + H2O <=> H3O^+ + PO4^{3-}}\) \(K_1 = 7.2 \times10^{-3}\) \(K_2 = 6.3 \times10^{-8}\) \(K_3 = 4.6 \times10^{-13}\) \(\ce{H2SO4 + H20 <=> H3O^+ + HSO4^-}\) \(\ce{HSO4^- + H20 <=> H3O^+ +SO4^{2-}}\) \(K_1 = \text{very large}\) \(K_2 = 1.1\times 10^{-2}\) Before we can go on to discuss how the hydronium-ion concentration and the pH of a solution of a weak acid depend on the concentration of the acid, we need to clarify a point of terminology. In order to do this let us take as an example a 0.0010 solution of acetic acid. Conductivity measurements discussed in the section on show that only about 10 percent of the acid molecules have donated protons to water at any given time. We thus have a situation which can be summarized schematically in the following way: \[\underset{\text{0.0009 mol L}^{-1}}{\overset{\text{90 }\% }{\mathop{\text{CH}_{3}\text{COOH}}}}\,\text{ + H}_{2}\text{O } \rightleftharpoons \underset{\text{0.0001 mol L}^{-1}}{\overset{\text{10 }\% }{\mathop{\text{CH}_{3}\text{COO}^{-}}}}\,\text{ + }\underset{\text{0.0001 mol L}^{-1}}{\mathop{\text{H}_{3}\text{O}^{\text{+}}}} ~~~~~~~ \nonumber \] In such a solution there is some ambiguity as to what we mean by the phrase . Do we mean 0.0010 mol L , or do we mean 90 percent of this value, namely, 0.0009 mol L ? In order to resolve this difficulty, we will use the term of acid and the symbol to indicate the quantity 0.0010 mol L , that is, to indicate the total amount of acetic acid originally added per unit volume of solution. On the other hand we will use the term and the symbol [CH COOH] indicate the quantity 0.0009 mol L that is, the final concentration of this species in the equilibrium mixture. Let us now consider the general problem of finding [H O ]in a solution of a weak acid HA whose acid constant is and whose stoichiometric concentration is . According to the equation for the equilibrium, \[\text{HA} + \text{H}_{2}\text{O}\rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{ A}^{-}\nonumber \] for every mole of H O produced, there must also be a mole of A produced. At the same time a mole of HA and a mole of H O must be consumed. Since the volume which all these species occupy is the same, any increase in [H O ] must be accompanied by an equal increase in [A ] and an equal decrease in [HA]. Consequently we can draw up the following table (in which equilibrium concentrations of all species have been expressed in terms of [H O ]: The hydronium-ion concentration actually increases from 10 mol/L to the equilibrium concentration, and so the change in each of the concentrations is ± ([H O ] –10 mol/L). However, the concentration of hydronium ions produced by the weak acid is usually so much larger than –10 mol L that the latter quantity can be ignored. In the case of 0.0010 acetic acid, for example, [H O ] ≈ 1 × 10 mol/L. Subtracting gives: \[\begin{align*} & \text{ 0.000 100 0 mol L}^{-1} \\ - & \underline{\text{0.000 000 1 mol L}^{-1}} \\ & \text{0.000 099 9 mol L}^{-1} \end{align*} \nonumber \] which is very close to 1 × 10 mol/L. We can now substitute the equilibrium concentrations into the expression \[K_{a}=\frac{ [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] [ \text{ A}^{-} ] }{ [ \text{ HA } ] }=\frac{ [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] ^{\text{2}}}{c_{a}- [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] } \label{8} \] This could be solved for [H O ] by means of the quadratic formula, but in most cases a quicker approximate method is available. Since the acid is weak, only a small fraction of the HA molecules will have donated protons to form H O ions. Therefore [H O ] is only a small fraction of and can be ignored when we calculate – [H O ]. That is, \[ c_{a} – \text{[H}_{3} \text{O}^{+} ] \approx c_{a} ~~~~~~~~~ \label{5} \] Equation \(\ref{8}\) then becomes \[K_{a}= \frac{\text{ [} \text{ H}_{\text{3}}\text{O}^{\text{+}} ] ^{\text{2}}}{c_{a}}\nonumber \] which rearranges to \[ \text{[H}_{3} \text{O}^{+}]^{2} \approx K_{a}c_{a} \nonumber \] Taking the square root of both sides gives an important approximate formula: \[\text{[H}_{3} \text{O}^{+} ] \approx \sqrt{K_{a}c_{a}} ~~~~~~~\label{12} \] Use Equation \(\ref{12}\) to calculate the pH of a 0.0200- solution of acetic acid. Compare this with the pH obtained using the [H O ] of 5.92 × 10 mol L derived from accurate conductivity measurements. From first table above, = 1.8 × 10 mol L . Since = 2.00 × 10 mol L , we have \[\begin{align*} [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] &=\sqrt{K_{a}c_{a}} \\[4pt] &=\sqrt{\text{(1}\text{.8 }\times \text{ 10}^{-\text{5}}\text{ mol L}^{-\text{1}}\text{)(2}\text{.00 }\times \text{ 10}^{-\text{2}}\text{ mol L}^{-\text{1}}\text{)}} \\[4pt] &=\sqrt{\text{3}\text{.6 }\times \text{ 10}^{-\text{7}}\text{ mol}^{\text{2}}\text{ L}^{-2}} \\[4pt] &=\sqrt{\text{36 }\times \text{ 10}^{-8}\text{ mol L}^{-\text{1}}} \\[4pt] [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] &=\text{6}\text{.0 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}} \\\text{ pH} &=-\text{log(6}\text{.0 }\times \text{ 10}^{-\text{4}}\text{)} \\[4pt] &=-\text{(0}\text{0.78}-\text{4)} \\[4pt] &=\text{3}\text{.22} \end{align*} \nonumber \] Using the accurate [H O ] from conductivity measurements, \[\begin{align*} \text{pH} &= – \text{log} (5.92 \times 10^{-4}) \\[4pt] &= –(0.772 – 4) \\[4pt] &= 3.228 \end{align*} \nonumber \] Note that the approximate equation gives an [H O ] which differs by 1 in the second significant digit from the accurate value. The calculated pH differs by 1 in the second place to the right of the decimal―roughly the same as the accuracy of simple pH measurements. In a few cases, if the acid is quite strong or the solution very dilute, it turns out that Equation \(\ref{12}\) is too gross an approximation. A convenient rule of thumb for determining when this is the case is to take the ratio [H O ]/ If this is larger than 5% or so, we need to make a second approximation, and then the rules for can be applied. Equation \(\ref{8}\) can be converted to a convenient form for successive approximations by multiplying both sides by – [H O ]: \[ \text{[H}_{3} \text{O}^{+}]^{2} = K_{a}(c_{a} - \text{[H}_{3} \text{O}^{+}]) \nonumber \] or \[\text{[H}_{3} \text{O}^{+} ] = \sqrt{K_{\text{a}}\text{(}c_{\text{a}}- [\text{ H}_{\text{3}}\text{O}^{\text{+}} ]\text{ )}} \nonumber \] To get a second approximation for [H O ] we can feed the first approximation into the right side of this equation. The exact procedure is detailed in the following example. Find the pH of 0.0200 \(\ce{HF}\) using the table of Acid Constants on this page. Using Equation \(\ref{12}\) we find \[\begin{align*} [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] =\sqrt{K_{a}c_{a}} =\sqrt{\text{6}\text{.8 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\times \text{ 0}\text{.0200 mol L}^{-\text{1}}} =\text{3}\text{.69 }\times \text{ 10}^{-3}\text{ mol L}^{-\text{1}} \\\end{align*} \nonumber \] Checking we find \[\frac{ [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] }{c_{a}}=\frac{\text{0}\text{.003 69}}{\text{0}\text{0.0200}}= 0.185 \nonumber \] that is 18.5 percent. A second approximation is thus necessary. We feed our first approximation of [H O ]= 3.69 × 10 mol L into Equation \ref{12}: \[\begin{align*} [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] &=\sqrt{K_{a}c_{a}} \\[5pt] &=\sqrt{\text{6}\text{.8 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\text{(0}\text{0.02}-\text{0}\text{0.003 69) mol L}^{-\text{1}}} \\[4pt] &=\text{3}\text{.33 }\times \text{ 10}^{-3}\text{ mol L}^{-\text{1}} \end{align*} \nonumber \] Taking a third approximation we find \[\begin{align*}[ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] &=\sqrt{\text{6}\text{.8 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\text{(0}\text{.02}-\text{0}\text{.003 33) mol L}^{-\text{1}}} \\[4pt] &=\text{3}\text{.37 }\times \text{ 10}^{-3}\text{ mol L}^{-\text{1}} \end{align*} \nonumber \] Since this differs from the second approximation by less than 5 percent, we take it as the final result. The pH is \[\mathrm{pH}=-\log \left(3.37 \times 10^{-3}\right)=2.47 \nonumber \] : Since \(\ce{HF}\) is a stronger acid than acetic acid, we expect this solution to have a lower pH than 0.02 \(\ce{CH3COOH}\) and indeed it does. | 10,791 | 2,861 |
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The pH of a solution of a weak base can be calculated in a way which is very similar to that used for a weak acid. Instead of an acid constant , a base constant must be used. If a weak base B accepts protons from water according to the equation \[\text{B} + \text{ H}_{\text{2}}\text{O}\rightleftharpoons\text{BH}^{+} + \text{OH}^{-} \label{1} \] then the base constant is defined by the expression \[K_{b}=\dfrac{ \text{ BH}^{\text{+}} \text{ OH}^{-} }{ \text{ B } } \label{2} \] A list of values for selected bases arranged in order of strength is given in the table below. This table is part of our larger collection of acid-base resources. To find the pH we follow the same general procedure as in the case of a weak acid. If the stoichiometric concentration of the base is indicated by , the result is entirely analogous to ; namely, \[K_{b}=\dfrac{ [\text{OH}^{-}]^2}{c_{b}- [\text{ OH}^{-}] } \label{3} \] Under most circumstances we can make the approximation \[c_b – [OH^–] \approx c_b \nonumber \] in which case Equation \ref{3} reduces to the approximation \[[OH^–] ≈ \sqrt{K_{b}c_{b}} \label{4} \] which is identical to the expression obtained in the acid case (approximation shown in ) except that OH replaces H O and replaces . Once we have found the hydroxide-ion concentration from this approximation, we can then easily find the pOH, and from it the pH. Using the value for listed in the table, find the pH of 0.100 NH . It is not a bad idea to guess an approximate pH before embarking on the calculation. Since we have a dilute solution of a weak base, we expect the solution to be only mildly basic. A pH of 13 or 14 would be too basic, while a pH of 8 or 9 is too close to neutral. A pH of 10 or 11 seems reasonable. Using Equation \ref{4} we have \[\begin{align*} [\text{ OH}^{-}] &=\sqrt{K_{b}c_{b}} \\[4pt] & =\sqrt{\text{1.8 }\times \text{ 10}^{-\text{5}}\text{ mol L}^{-\text{1}} \times \text{ 0.100 mol L}^{-\text{1}}} \\[4pt] &=\sqrt{\text{1.8 }\times \text{ 10}^{-\text{6}}\text{ mol}^{\text{2}}\text{ L}^{-2}} \\[4pt] &=\text{1.34 }\times \text{ 10}^{-\text{3}}\text{ mol L}^{-\text{1}} \end{align*} \nonumber \] Checking the accuracy of the approximation, we find \(\dfrac{ [\text{ OH}^{-} ]}{c_{\text{b}}}=\dfrac{\text{1.34 }\times \text{ 10}^{-\text{3}}}{\text{0.1}}\approx \text{1 percent}\) The approximation is valid, and we thus proceed to find the pOH. \(\text{pOH}=-\text{log}\dfrac{ [\text{ OH}^{-} ]}{\text{mol L}^{-\text{1}}}=-\text{log(1.34 }\times \text{ 10}^{-\text{3}}\text{)}=\text{2.87}\) From which \[pH = 14.00 – pOH = 14.00 – 2.87 = 11.13 \nonumber \] This calculated value checks well with our initial guess. Occasionally we will find that the approximation \[c_b – [OH^{–}] ≈ c_b \nonumber \] is not valid, in which case we must use a series of successive approximations similar to that outlined above for acids. The appropriate formula can be derived from Equation \ref{3} and reads \[[OH^{-}] \approx \sqrt{K_{b} ( c_b - [OH^{-}] )} \nonumber \] | 3,031 | 2,862 |
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“A microscopic cell of photosynthetic cyanobacteria constitutes a complex of chemical factories that carry out a multitude of biochemical processes. Powered by solar energy and operating under ambient conditions, these organisms take carbon dioxide and nitrogen from air and simple inorganic ions dissolved in water and make all the life molecules they need for their metabolism and reproduction. In eons past these kinds of organisms generated all of the oxygen that is in Earth’s atmosphere. For all their knowledge of chemistry it would be impossible for humans to reproduce the chemical processes of these remarkable bacteria”
| 645 | 2,863 |
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It is possible to use kinetics studies of a chemical system, such as the effect of changes in reactant concentrations, to deduce events that occur on a microscopic scale, such as collisions between individual particles. Such studies have led to the collision model of chemical kinetics, which is a useful tool for understanding the behavior of reacting chemical species. The collision model explains why chemical reactions often occur more rapidly at higher temperatures. For example, the reaction rates of many reactions that occur at room temperature approximately double with a temperature increase of only 10°C. In this section, we will use the collision model to analyze this relationship between temperature and reaction rates. Before delving into the relationship between temperature and reaction rate, we must discuss three microscopic factors that influence the observed macroscopic reaction rates. Central to collision model is that a chemical reaction can occur only when the reactant molecules, atoms, or ions collide. Hence, the observed rate is influence by the frequency of collisions between the reactants. The is the average rate in which two reactants collide for a given system and is used to express the average number of collisions per unit of time in a defined system. While deriving the collisional frequency (\(Z_{AB}\)) between two species in a gas is , it is beyond the scope of this text and the equation for collisional frequency of \(A\) and \(B\) is the following: \[Z_{AB} = N_{A}N_{B}\left(r_{A} + r_{B}\right)^2\sqrt{ \dfrac{8\pi k_{B}T}{\mu_{AB}}} \label{freq} \] with The specifics of Equation \ref{freq} are not important for this conversation, but it is important to identify that \(Z_{AB}\) increases with increasing density (i.e., increasing \(N_A\) and \(N_B\)), with increasing reactant size ( . A Discussing Collision Theory of Kinetics: Previously, we discussed the kinetic molecular theory of gases, which showed that the average kinetic energy of the particles of a gas increases with increasing temperature. Because the speed of a particle is proportional to the square root of its kinetic energy, increasing the temperature will also increase the number of collisions between molecules per unit time. What the kinetic molecular theory of gases does not explain is why the reaction rate of most reactions approximately doubles with a 10°C temperature increase. This result is surprisingly large considering that a 10°C increase in the temperature of a gas from 300 K to 310 K increases the kinetic energy of the particles by only about 4%, leading to an increase in molecular speed of only about 2% and a correspondingly small increase in the number of bimolecular collisions per unit time. The collision model of chemical kinetics explains this behavior by introducing the concept of (\(E_a\)). We will define this concept using the reaction of \(\ce{NO}\) with ozone, which plays an important role in the depletion of ozone in the ozone layer: \[\ce{NO(g) + O_3(g) \rightarrow NO_2(g) + O_2(g)} \nonumber \] Increasing the temperature from 200 K to 350 K causes the rate constant for this particular reaction to increase by a factor of more than 10, whereas the increase in the frequency of bimolecular collisions over this temperature range is only 30%. Thus something other than an increase in the collision rate must be affecting the reaction rate. Experimental rate law for this reaction is \[\text{rate} = k [\ce{NO},\ce{O3}] \nonumber \] and is used to identify how the reaction rate (not the rate constant) vares with concentration. The rate constant, however, does vary with temperature. gure \(\Page {1}\) shows a plot of the rate constant of the reaction of \(\ce{NO}\) with \(\ce{O3}\) at various temperatures. The relationship is not linear but instead resembles the relationships seen in graphs of vapor pressure versus temperature (e.g, the ). In all three cases, the shape of the plots results from a distribution of kinetic energy over a population of particles (electrons in the case of conductivity; molecules in the case of vapor pressure; and molecules, atoms, or ions in the case of reaction rates). Only a fraction of the particles have sufficient energy to overcome an energy barrier. In the case of vapor pressure, particles must overcome an energy barrier to escape from the liquid phase to the gas phase. This barrier corresponds to the energy of the intermolecular forces that hold the molecules together in the liquid. In conductivity, the barrier is the energy gap between the filled and empty bands. In chemical reactions, the energy barrier corresponds to the amount of energy the particles must have to react when they collide. This energy threshold, called the , was first postulated in 1888 by the Swedish chemist Svante Arrhenius (1859–1927; Nobel Prize in Chemistry 1903). It is the minimum amount of energy needed for a reaction to occur. Reacting molecules must have enough energy to overcome electrostatic repulsion, and a minimum amount of energy is required to break chemical bonds so that new ones may be formed. Molecules that collide with less than the threshold energy bounce off one another chemically unchanged, with only their direction of travel and their speed altered by the collision. Molecules that are able to overcome the energy barrier are able to react and form an arrangement of atoms called the or the of the reaction. The activated complex is not a reaction intermediate; it does not last long enough to be detected readily. Any phenomenon that depends on the distribution of thermal energy in a population of particles has a nonlinear temperature dependence. We can graph the energy of a reaction by plotting the potential energy of the system as the reaction progresses. shows a plot for the NO–O system, in which the vertical axis is potential energy and the horizontal axis is the reaction coordinate, which indicates the progress of the reaction with time. The activated complex is shown in brackets with an asterisk. The overall change in potential energy for the reaction (\(ΔE\)) is negative, which means that the reaction releases energy. (In this case, \(ΔE\) is −200.8 kJ/mol.) To react, however, the molecules must overcome the energy barrier to reaction (\(E_a\) is 9.6 kJ/mol). That is, 9.6 kJ/mol must be put into the system as the activation energy. Below this threshold, the particles do not have enough energy for the reaction to occur. re \(\Page {3a}\) illustrates the general situation in which the products have a lower potential energy than the reactants. In contrast, F re \(\Page {3b}\) illustrates the case in which the products have a higher potential energy than the reactants, so the overall reaction requires an input of energy; that is, it is energetically uphill, and \(Δ > 0\). Although the energy changes that result from a reaction can be positive, negative, or even zero, in most cases an energy barrier must be overcome before a reaction can occur. This means that the activation energy is almost always positive; there is a class of reactions called barrierless reactions, but those are discussed elsewhere. For similar reactions under comparable conditions, the one with the smallest will occur most rapidly. Whereas \(ΔE\) is related to the tendency of a reaction to occur spontaneously, \(E_a\) gives us information about the reaction rate and how rapidly the reaction rate changes with temperature. For two similar reactions under comparable conditions, the reaction with the smallest \(E_a\) will occur more rapidly. Figure \(\Page {4}\) shows both the kinetic energy distributions and a potential energy diagram for a reaction. The shaded areas show that at the lower temperature (300 K), only a small fraction of molecules collide with kinetic energy greater than ; however, at the higher temperature (500 K) a much larger fraction of molecules collide with kinetic energy greater than . Consequently, the reaction rate is much slower at the lower temperature because only a relatively few molecules collide with enough energy to overcome the potential energy barrier. Discussing Transition State Theory: Even when the energy of collisions between two reactant species is greater than \(E_a\), most collisions do not produce a reaction. The probability of a reaction occurring depends not only on the collision energy but also on the spatial orientation of the molecules when they collide. For \(\ce{NO}\) and \(\ce{O3}\) to produce \(\ce{NO2}\) and \(\ce{O2}\), a terminal oxygen atom of \(\ce{O3}\) must collide with the nitrogen atom of \(\ce{NO}\) at an angle that allows \(\ce{O3}\) to transfer an oxygen atom to \(\ce{NO}\) to produce \(\ce{NO2}\) ( re \(\Page {4}\)). All other collisions produce no reaction. Because fewer than 1% of all possible orientations of \(\ce{NO}\) and \(\ce{O3}\) result in a reaction at kinetic energies greater than \(E_a\), most collisions of \(\ce{NO}\) and \(\ce{O3}\) are unproductive. The fraction of orientations that result in a reaction is called the (\(\rho\)) and its value can range from \(\rho=0\) (no orientations of molecules result in reaction) to \(\rho=1\) (all orientations result in reaction). The collision model explains why most collisions between molecules do not result in a chemical reaction. For example, nitrogen and oxygen molecules in a single liter of air at room temperature and 1 atm of pressure collide about 10 times per second. If every collision produced two molecules of \(\ce{NO}\), the atmosphere would have been converted to \(\ce{NO}\) and then \(\ce{NO2}\) a long time ago. Instead, in most collisions, the molecules simply bounce off one another without reacting, much as marbles bounce off each other when they collide. For an \(A + B\) elementary reaction, all three microscopic factors discussed above that affect the reaction rate can be summarized in a single relationship: \[\text{rate} = (\text{collision frequency}) \times (\text{steric factor}) \times (\text{fraction of collisions with } E > E_a ) \nonumber \] where \[\text{rate} = k[A,B] \label{14.5.2} \] Arrhenius used these relationships to arrive at an equation that relates the magnitude of the rate constant for a reaction to the temperature, the activation energy, and the constant, \(A\), called the : \[k=Ae^{-E_{\Large a}/RT} \label{14.5.3} \] The frequency factor is used to convert concentrations to collisions per second (scaled by the steric factor). Because the frequency of collisions depends on the temperature, \(A\) is actually not constant (Equation \ref{freq}). Instead, \(A\) increases slightly with temperature as the increased kinetic energy of molecules at higher temperatures causes them to move slightly faster and thus undergo more collisions per unit time. is known as the and summarizes the collision model of chemical kinetics, where \(T\) is the absolute temperature (in K) and is the ideal gas constant [8.314 J/(K·mol)]. \(E_a\) indicates the sensitivity of the reaction to changes in temperature. The reaction rate with a large \(E_a\) increases rapidly with increasing temperature, whereas the reaction rate with a smaller \(E_a\) increases much more slowly with increasing temperature. If we know the reaction rate at various temperatures, we can use the Arrhenius equation to calculate the activation energy. Taking the natural logarithm of both sides of , \[\begin{align} \ln k &=\ln A+\left(-\dfrac{E_{\textrm a}}{RT}\right) \\[4pt] &=\ln A+\left[\left(-\dfrac{E_{\textrm a}}{R}\right)\left(\dfrac{1}{T}\right)\right] \label{14.5.4} \end{align} \] is the equation of a straight line, \[y = mx + b \nonumber \] where \(y = \ln k\) and \(x = 1/T\). This means that a plot of \(\ln k\) versus \(1/T\) is a straight line with a slope of \(−E_a/R\) and an intercept of \(\ln A\). In fact, we need to measure the reaction rate at only two temperatures to estimate \(E_a\). Knowing the \(E_a\) at one temperature allows us to predict the reaction rate at other temperatures. This is important in cooking and food preservation, for example, as well as in controlling industrial reactions to prevent potential disasters. The procedure for determining \(E_a\) from reaction rates measured at several temperatures is illustrated in Example \(\Page {1}\). A Discussing The Arrhenius Equation: Many people believe that the rate of a tree cricket’s chirping is related to temperature. To see whether this is true, biologists have carried out accurate measurements of the rate of tree cricket chirping (\(f\)) as a function of temperature (\(T\)). Use the data in the following table, along with the graph of ln[chirping rate] versus \(1/T\) to calculate \(E_a\) for the biochemical reaction that controls cricket chirping. Then predict the chirping rate on a very hot evening, when the temperature is 308 K (35°C, or 95°F). chirping rate at various temperatures activation energy and chirping rate at specified temperature If cricket chirping is controlled by a reaction that obeys the Arrhenius equation, then a plot of \(\ln f\) versus \(1/T\) should give a straight line ( ). Also, the slope of the plot of \(\ln f\) versus \(1/T\) should be equal to \(−E_a/R\). We can use the two endpoints in to estimate the slope: \[\begin{align*}\textrm{slope}&=\dfrac{\Delta\ln f}{\Delta(1/T)} \\[4pt] &=\dfrac{5.30-4.37}{3.34\times10^{-3}\textrm{ K}^{-1}-3.48\times10^{-3}\textrm{ K}^{-1}} \\[4pt] &=\dfrac{0.93}{-0.14\times10^{-3}\textrm{ K}^{-1}} \\[4pt] &=-6.6\times10^3\textrm{ K}\end{align*} \nonumber \] A computer best-fit line through all the points has a slope of −6.67 × 10 K, so our estimate is very close. We now use it to solve for the activation energy: \[\begin{align*} E_{\textrm a} &=-(\textrm{slope})(R) \\[4pt] &=-(-6.6\times10^3\textrm{ K})\left(\dfrac{8.314 \textrm{ J}}{\mathrm{K\cdot mol}}\right)\left(\dfrac{\textrm{1 KJ}}{\textrm{1000 J}}\right) \\[4pt] &=\dfrac{\textrm{55 kJ}}{\textrm{mol}} \end{align*} \nonumber \] If the activation energy of a reaction and the rate constant at one temperature are known, then we can calculate the reaction rate at any other temperature. We can use to express the known rate constant (\(k_1\)) at the first temperature (\(T_1\)) as follows: \[\ln k_1=\ln A-\dfrac{E_{\textrm a}}{RT_1} \nonumber \] Similarly, we can express the unknown rate constant (\(k_2\)) at the second temperature (\(T_2\)) as follows: \[\ln k_2=\ln A-\dfrac{E_{\textrm a}}{RT_2} \nonumber \] These two equations contain four known quantities ( , , , and ) and two unknowns ( and ). We can eliminate by subtracting the first equation from the second: \[\begin{align*} \ln k_2-\ln k_1 &=\left(\ln A-\dfrac{E_{\textrm a}}{RT_2}\right)-\left(\ln A-\dfrac{E_{\textrm a}}{RT_1}\right) \\[4pt] &=-\dfrac{E_{\textrm a}}{RT_2}+\dfrac{E_{\textrm a}}{RT_1} \end{align*} \nonumber \] Then \[\ln \dfrac{k_2}{k_1}=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \nonumber \] To obtain the best prediction of chirping rate at 308 K ( ), we try to choose for and the measured rate constant and corresponding temperature in the data table that is closest to the best-fit line in the graph. Choosing data for = 296 K, where = 158, and using the \(E_a\) calculated previously, \[\begin{align*} \ln\dfrac{k_{T_2}}{k_{T_1}} &=\dfrac{E_{\textrm a}}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \\[4pt] &=\dfrac{55\textrm{ kJ/mol}}{8.314\textrm{ J}/(\mathrm{K\cdot mol})}\left(\dfrac{1000\textrm{ J}}{\textrm{1 kJ}}\right)\left(\dfrac{1}{296 \textrm{ K}}-\dfrac{1}{\textrm{308 K}}\right) \\[4pt] &=0.87 \end{align*} \nonumber \] Thus / = 2.4 and = (2.4)(158) = 380, and the chirping rate on a night when the temperature is 308 K is predicted to be 380 chirps per minute. The equation for the decomposition of \(\ce{NO2}\) to \(\ce{NO}\) and \(\ce{O2}\) is second order in \(\ce{NO2}\): \[\ce{2NO2(g) → 2NO(g) + O2(g)} \nonumber \] Data for the reaction rate as a function of temperature are listed in the following table. Calculate \(E_a\) for the reaction and the rate constant at 700 K. \(E_a\) = 114 kJ/mol; = 18,600 M ·s = 1.86 × 10 M ·s . What \(E_a\) results in a doubling of the reaction rate with a 10°C increase in temperature from 20° to 30°C? about 51 kJ/mol A Discussing Graphing Using the Arrhenius Equation: For a chemical reaction to occur, an energy threshold must be overcome, and the reacting species must also have the correct spatial orientation. The Arrhenius equation is \(k=Ae^{-E_{\Large a}/RT}\). A minimum energy (activation energy,v\(E_a\)) is required for a collision between molecules to result in a chemical reaction. Plots of potential energy for a system versus the reaction coordinate show an energy barrier that must be overcome for the reaction to occur. The arrangement of atoms at the highest point of this barrier is the activated complex, or transition state, of the reaction. At a given temperature, the higher the , the slower the reaction. The fraction of orientations that result in a reaction is the steric factor. The frequency factor, steric factor, and activation energy are related to the rate constant in the Arrhenius equation: \(k=Ae^{-E_{\Large a}/RT}\). A plot of the natural logarithm of versus 1/ is a straight line with a slope of − / . | 17,336 | 2,864 |
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Chemists study the structures, physical properties, and chemical properties of material substances. These consist of , which is anything that occupies space and has mass. Gold and iridium are matter, as are peanuts, people, and postage stamps. Smoke, smog, and laughing gas are matter. Energy, light, and sound, however, are not matter; ideas and emotions are also not matter. The of an object is the quantity of matter it contains. Do not confuse an object’s mass with its , which is a force caused by the gravitational attraction that operates on the object. Mass is a fundamental property of an object that does not depend on its location.In physical terms, the mass of an object is directly proportional to the force required to change its speed or direction. A more detailed discussion of the differences between weight and mass and the units used to measure them is included in Essential Skills 1 ( ). Weight, on the other hand, depends on the location of an object. An astronaut whose mass is 95 kg weighs about 210 lb on Earth but only about 35 lb on the moon because the gravitational force he or she experiences on the moon is approximately one-sixth the force experienced on Earth. For practical purposes, weight and mass are often used interchangeably in laboratories. Because the force of gravity is considered to be the same everywhere on Earth’s surface, 2.2 lb (a weight) equals 1.0 kg (a mass), regardless of the location of the laboratory on Earth. Under normal conditions, there are three distinct states of matter: solids, liquids, and gases. are relatively rigid and have fixed shapes and volumes. A rock, for example, is a solid. In contrast, have fixed volumes but flow to assume the shape of their containers, such as a beverage in a can. , such as air in an automobile tire, have neither fixed shapes nor fixed volumes and expand to completely fill their containers. Whereas the volume of gases strongly depends on their temperature and (the amount of force exerted on a given area), the volumes of liquids and solids are virtually independent of temperature and pressure. Matter can often change from one physical state to another in a process called a . For example, liquid water can be heated to form a gas called steam, or steam can be cooled to form liquid water. However, such changes of state do not affect the chemical composition of the substance. A pure chemical substance is any matter that has a fixed chemical composition and characteristic properties. Oxygen, for example, is a pure chemical substance that is a colorless, odorless gas at 25°C. Very few samples of matter consist of pure substances; instead, most are mixtures, which are combinations of two or more pure substances in variable proportions in which the individual substances retain their identity. Air, tap water, milk, blue cheese, bread, and dirt are all mixtures. If all portions of a material are in the same state, have no visible boundaries, and are uniform throughout, then the material is . Examples of homogeneous mixtures are the air we breathe and the tap water we drink. Homogeneous mixtures are also called solutions. Thus air is a solution of nitrogen, oxygen, water vapor, carbon dioxide, and several other gases; tap water is a solution of small amounts of several substances in water. The specific compositions of both of these solutions are not fixed, however, but depend on both source and location; for example, the composition of tap water in Boise, Idaho, is not the same as the composition of tap water in Buffalo, New York. Although most solutions we encounter are liquid, solutions can also be solid. The gray substance still used by some dentists to fill tooth cavities is a complex solid solution that contains 50% mercury and 50% of a powder that contains mostly silver, tin, and copper, with small amounts of zinc and mercury. Solid solutions of two or more metals are commonly called alloys. If the composition of a material is not completely uniform, then it is (e.g., chocolate chip cookie dough, blue cheese, and dirt). Mixtures that appear to be homogeneous are often found to be heterogeneous after microscopic examination. Milk, for example, appears to be homogeneous, but when examined under a microscope, it clearly consists of tiny globules of fat and protein dispersed in water. The components of heterogeneous mixtures can usually be separated by simple means. Solid-liquid mixtures such as sand in water or tea leaves in tea are readily separated by filtration, which consists of passing the mixture through a barrier, such as a strainer, with holes or pores that are smaller than the solid particles. In principle, mixtures of two or more solids, such as sugar and salt, can be separated by microscopic inspection and sorting. More complex operations are usually necessary, though, such as when separating gold nuggets from river gravel by panning. First solid material is filtered from river water; then the solids are separated by inspection. If gold is embedded in rock, it may have to be isolated using chemical methods. Homogeneous mixtures (solutions) can be separated into their component substances by physical processes that rely on differences in some physical property, such as differences in their boiling points. Two of these separation methods are distillation and crystallization. makes use of differences in volatility, a measure of how easily a substance is converted to a gas at a given temperature. A simple distillation apparatus for separating a mixture of substances, at least one of which is a liquid. The most volatile component boils first and is condensed back to a liquid in the water-cooled condenser, from which it flows into the receiving flask. If a solution of salt and water is distilled, for example, the more volatile component, pure water, collects in the receiving flask, while the salt remains in the distillation flask. Mixtures of two or more liquids with different boiling points can be separated with a more complex distillation apparatus. One example is the refining of crude petroleum into a range of useful products: aviation fuel, gasoline, kerosene, diesel fuel, and lubricating oil (in the approximate order of decreasing volatility). Another example is the distillation of alcoholic spirits such as brandy or whiskey. This relatively simple procedure caused more than a few headaches for federal authorities in the 1920s during the era of Prohibition, when illegal stills proliferated in remote regions of the United States. separates mixtures based on differences in solubility, a measure of how much solid substance remains dissolved in a given amount of a specified liquid. Most substances are more soluble at higher temperatures, so a mixture of two or more substances can be dissolved at an elevated temperature and then allowed to cool slowly. Alternatively, the liquid, called the solvent, may be allowed to evaporate. In either case, the least soluble of the dissolved substances, the one that is least likely to remain in solution, usually forms crystals first, and these crystals can be removed from the remaining solution by filtration. Most mixtures can be separated into pure substances, which may be either elements or compounds. An , such as gray, metallic sodium, is a substance that cannot be broken down into simpler ones by chemical changes; a , such as white, crystalline sodium chloride, contains two or more elements and has chemical and physical properties that are usually different from those of the elements of which it is composed. With only a few exceptions, a particular compound has the same elemental composition (the same elements in the same proportions) regardless of its source or history. The chemical composition of a substance is altered in a process called a . The conversion of two or more elements, such as sodium and chlorine, to a chemical compound, sodium chloride, is an example of a chemical change, often called a chemical reaction. Currently, about 115 elements are known, but millions of chemical compounds have been prepared from these 115 elements. The known elements are listed in . In general, a reverse chemical process breaks down compounds into their elements. For example, water (a compound) can be decomposed into hydrogen and oxygen (both elements) by a process calledelectrolysis. In electrolysis, electricity provides the energy needed to separate a compound into its constituent elements ( ). A similar technique is used on a vast scale to obtain pure aluminum, an element, from its ores, which are mixtures of compounds. Because a great deal of energy is required for electrolysis, the cost of electricity is by far the greatest expense incurred in manufacturing pure aluminum. Thus recycling aluminum is both cost-effective and ecologically sound. The overall organization of matter and the methods used to separate mixtures are summarized in . Figure \(\Page {6}\): Relationships between the Types of Matter and the Methods Used to Separate Mixtures Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution). : a chemical substance : its classification Identify each substance as a compound, an element, a heterogeneous mixture, or a homogeneous mixture (solution). Different Definitions of Matter: Matter can be classified according to physical and chemical properties. Matter is anything that occupies space and has mass. The three states of matter are solid, liquid, and gas. A physical change involves the conversion of a substance from one state of matter to another, without changing its chemical composition. Most matter consists of mixtures of pure substances, which can be homogeneous (uniform in composition) or heterogeneous (different regions possess different compositions and properties). Pure substances can be either chemical compounds or elements. Compounds can be broken down into elements by chemical reactions, but elements cannot be separated into simpler substances by chemical means. The properties of substances can be classified as either physical or chemical. Scientists can observe physical properties without changing the composition of the substance, whereas chemical properties describe the tendency of a substance to undergo chemical changes (chemical reactions) that change its chemical composition. Physical properties can be intensive or extensive. Intensive properties are the same for all samples; do not depend on sample size; and include, for example, color, physical state, and melting and boiling points. Extensive properties depend on the amount of material and include mass and volume. The ratio of two extensive properties, mass and volume, is an important intensive property called density. ( ) | 10,807 | 2,865 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Definitions_of_Oxidation_and_Reduction |
This page discusses redox The terms and can be defined in terms of the adding or removing oxygen to a compound. while this is not the most robust definition, as discussed below, it is the easiest to remember. For example, in the extraction of iron from its ore: Because both reduction and oxidation are occurring simultaneously, this is known as a redox reaction. An oxidizing agent is substance which oxidizes something else. In the above example, the iron(III) oxide is the oxidizing agent. A reducing agent reduces something else. In the equation, the carbon monoxide is the reducing agent. These are old definitions which are no longer used, except occasionally in organic chemistry. Notice that these are exactly the opposite of the oxygen definitions (#1). For example, ethanol can be oxidized to ethanal: An oxidizing agent is required to remove the hydrogen from the ethanol. A commonly used oxidizing agent is potassium dichromate(VI) solution acidified with dilute sulfuric acid. Ethanal can also be reduced back to ethanol by adding hydrogen. A possible reducing agent is sodium tetrahydridoborate, NaBH . Again the equation is too complicated to consider at this point. More precise definitionsof oxidizing and reducing agents are Remembering these definitions is essential, and easily done using this convenient acronym: The equation below shows an obvious example of oxygen transfer in a simple redox reaction: \[ \ce{CuO + Mg \rightarrow Cu + MgO} \nonumber\] Copper(II) oxide and magnesium oxide are both ionic compounds. If the above is written as an ionic equation, it becomes apparent that the oxide ions are spectator ions. Omitting them gives: In the above reaction, magnesium reduces the copper(II) ion by transferring electrons to the ion and neutralizing its charge. Therefore, magnesium is a reducing agent. Another way of putting this is that the copper(II) ion is removing electrons from the magnesium to create a magnesium ion. The copper(II) ion is acting as an oxidizing agent. Confusion can result from trying to learn both the definitions of oxidation and reduction in terms of electron transfer and the definitions of oxidizing and reducing agents in the same terms. The following thought pattern can be helpful: Here is another mental exercise: | 2,293 | 2,868 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/15%3A_Alcohols_and_Ethers/15.04%3A_Preparation_of_Alcohols |
Many of the common laboratory methods for the preparation of alcohols have been discussed in previous post or will be considered later; thus to avoid undue repetition we shall not consider them in detail at this time. Included among these methods are hydration ( ) and hydroboration ( ), addition of hypohalous acids to alkenes ( ), \(S_\text{N}1\) and \(S_\text{N}2\) hydrolysis of alkyl halides ( to 8-7) and of allylic and benzylic halides ( and ), addition of Grignard reagents to carbonyl compounds ( ), and the reduction of carbonyl compounds ( and ). These methods are summarized in Table 15-2. Some of the reactions we have mentioned are used for large-scale industrial production. For example, ethanol is made in quantity by the hydration of ethene, using an excess of steam under pressure at temperatures around \(300^\text{o}\) in the presence of phosphoric acid: A dilute solution of ethanol is obtained, which can be concentrated by distillation to a constant-boiling point mixture that contains \(95.6\%\) ethanol by weight. Dehydration of the remaining few percent of water to give “absolute alcohol” is achieved either by chemical means or by distillation with benzene, which results in preferential separation of the water. Ethanol also is made in large quantities by fermentation, but this route is not competitive for industrial uses with the hydration of ethene. Isopropyl alcohol and -butyl alcohol also are manufactured by hydration of the corresponding alkenes. The industrial synthesis of methyl alcohol involves hydrogenation of carbon monoxide. Although this reaction has the favorable \(\Delta H^0\) value of \(-28.4 \: \text{kcal mol}^{-1}\), it requires high pressures and high temperatures and a suitable catalyst; excellent conversions are achieved using zinc oxide-chromic oxide as a catalyst: Various methods of synthesis of other alcohols by reduction of carbonyl compounds will be discussed in . and (1977) | 1,968 | 2,869 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_(Wade)/21%3A_Carboxylic_Acids |
learning objectives After reading this chapter and completing ALL the exercises, a student can Please note: IUPAC nomenclature and important common names of carboxylic acids were explained in Chapter 3. It can useful is often required to memorize the structures for the following common names: formic acid, acetic acid, acetyl chloride, acetic anhydride, acetic formic anhydride, ethyl acetate, sodium and potassium salts of formate, acetate, and benzoate, acetamide, benzamide, acetonitrile, benzonitrile, carbonic acid, oxalic acid, malonic acid, succinic acid, glutaric acid, adipic acid, and phthalic acid
| 631 | 2,870 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/The_Wittig_Reaction |
Organophosphorus react with aldehydes or ketones to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration. It has been noted that dipolar phosphorus compounds are stabilized by p-d bonding. This bonding stabilization extends to carbanions adjacent to phosphonium centers, and the zwitterionic conjugate bases derived from such cations are known as ylides. An is defined as a compound with opposite charges on adjacent atoms both of which have complete octets. For the Wittig reaction discussed below an organophosphorus ylide, also called Wittig reagents, will be used. The ability of phosphorus to hold more than eight valence electrons allows for a resonance structure to be drawn forming a double bonded structure. The stabilization of the carbanion provided by the phosphorus causes an increase in acidity (pKa ~35). Very strong bases, such as butyl lithium, are required for complete formation of ylides. The ylides shown here are all strong bases. Like other strongly basic organic reagents, they are protonated by water and alcohols, and are sensitive to oxygen. Water decomposes phosphorous ylides to hydrocarbons and phosphine oxides, as shown. Although many ylides are commercially available it is often necessary to create them synthetically. Ylides can be synthesized from an alkyl halide and a trialkyl phosphine. Typically triphenyl phosphine is used to synthesize ylides. Because a S 2 reaction is used in the ylide synthesis methyl and primary halides perform the best. Secondary halides can also be used but the yields are generally lower. This should be considered when planning out a synthesis which involves a synthesized Wittig reagent. 1) S 2 reaction 2) Deprotonation The most important use of ylides in synthesis comes from their reactions with aldehydes and ketones, which are initiated in every case by a covalent bonding of the nucleophilic alpha-carbon to the electrophilic carbonyl carbon. Ylides react to give substituted alkenes in a transformation called the Wittig reaction. This reaction is named for George Wittig who was awarded the Nobel prize for this work in 1979. A principal advantage of alkene synthesis by the Wittig reaction is that the location of the double bond is absolutely fixed, in contrast to the mixtures often produced by alcohol dehydration. Going from reactants to products simplified Following the initial carbon-carbon bond formation, two intermediates have been identified for the Wittig reaction, a dipolar charge-separated species called a betaine and a four-membered heterocyclic structure referred to as an oxaphosphatane. Cleavage of the oxaphosphatane to alkene and phosphine oxide products is exothermic and irreversible. 1) Nucleophillic attack on the carbonyl 2) Formation of a 4 membered ring 3) Formation of the alkene If possible both E and Z isomer of the double bond will be formed. This should be considered when planning a synthesis involving a Wittig Reaction. 1) Please write the product of the following reactions. 2) Please indicate the starting material required to produce the product. 3) Please draw the structure of the oxaphosphetane which is made during the mechanism of the reaction given that produces product . 4) Please draw the structure of the betaine which is made during the mechanism of the reaction given that produces product . 5) Please give a detailed mechanism and the final product of this reaction 6) It has been shown that reacting and epoxide with triphenylphosphine forms an alkene. Please propose a mechanism for this reaction. Review the section on epoxide reactions if you need help. 1) 2) 3) 4) 5) Nucleophillic attack on the carbonyl Formation of a 4 membered ring Formation of the alkene 6) Nucleophillic attack on the epoxide Formation of a 4 membered ring Formation of the alkene ) ), | 4,130 | 2,871 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Analytical_Chemiluminescence/4%3A_Instrumentation/4.04%3A_Lab_on_a_Chip |
Micro total analytical systems (also called “chips”) are miniaturized microfluidic devices, fabricated from a variety of materials within which channels are constructed for the transport of samples and reagents. The small size minimizes the consumption of reagents, reduces manufacturing costs and increases the possibilities for automation. Miniaturization of detectors, however, leads to problems due to the reduced volume of liquid in the detector and to difficulties inherent in scaling down the size of the particular detector. One solution is to interface the chip with a macro-scale detector such as a photomultiplier tube; this is called the “off-chip” approach. This can be achieved, for example, by using optical fibres to carry light from the chip to the detector. An alternative solution – the “on-chip” approach - is to assemble a compact version of the detector and integrate this on the chip with the rest of the analytical system . Chemiluminescence detection offers high sensitivity, low detection limits and instrumental simplicity but requires a relatively complex manifold on the microchip, the details depending on the chemiluminescence reaction system being used; for example, a Y-shaped channel junction works best when using peroxide-luminol chemiluminescence. Reagent is delivered by a micropump. The chip design must ensure that a high proportion of the emitted light enters the off-chip photomultiplier; this frequently involves coupling with an optical fibre. Such an arrangement typically achieves micromolar detection limits and has been used for a range of analytes including catechol, dopamine, amino-acids, cytochrome c and myoglobin as well as the determination of chip-separated chromium(III), cobalt(II) and copper(II). Horseradish peroxidise can be determined at sub-nanomolar levels. Micromolar concentrations of ATP (adenosine triphosphate) can be measured by means of luciferin-luciferase bioluminescence. The effect of antioxidants has been measured using a microfluidic system incorporating peroxy-oxalate chemiluminescence, by injecting the antioxidants into the hydrogen peroxide stream. The method is simple and rapid and excellent analytical performance is obtained in terms of sensitivity, dynamic range and precision. Electrochemiluminescence detection has been applied for microchip separations using electrodes installed during fabrication. Photodiodes have been fabricated into chips at the bottoms of the microfluidic channels and have been used for on-chip chemiluminescence detection of DNA produced by the polymerase chain reaction and separated on the same chip by capillary electrophoresis. These devices have been used also to detect luminol chemiluminescence for the micromolar determination of hydrogen peroxide generated by the oxidation of glucose with glucose oxidase. Thin-film organic photodiodes can be fabricated by vacuum deposition and integrated into chips. Copper-phthalocyanine-fullerene small molecule diodes have high quantum efficiency and have been used to determine hydrogen peroxide by peroxy-oxalate chemiluminescence. Another example has been used for hydrogen peroxide determination by luminol chemiluminescence. | 3,211 | 2,872 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/11%3A_Chemical_Kinetics_I/11.11%3A_Transition_State_Theory |
was proposed in 1935 by Henry Erying, and further developed by Merrideth G. Evans and Michael Polanyi (Laidler & King, 1983), as another means of accounting for chemical reaction rates. It is based on the idea that a molecular collision that leads to reaction must pass through an intermediate state known as the transition state. For example, the reaction \[ A + BC \rightarrow AB + C \nonumber \] would have an intermediate (\(ABC\)) where the \(B-C\) bond is partially broken, and the \(A-B\) bond is partially formed. \[A + B-C \rightarrow (A-B-C)^‡ \rightarrow A-B + C \nonumber \] So the reaction is mediated by the formation of an activated complex (denoted with the double-dagger symbol ‡) and the decomposition of that complex into the reaction products. Using this theory, the rate of reaction can be expressed as the product of two factors \[\text{rate} = (\text{transition state concentration}) \times (\text{decomposition frequency}) \nonumber \] If the formation of the activated complex is considered to reach an equilibrium, \[ K^‡ = \dfrac{[ABC]^‡}{[A,BC]} \nonumber \] So the concentration of the transition state complex can be expressed by \[[ABC]^‡ = K^‡ [A,BC] \nonumber \] Using the relationship from Chapter 9 for the equilibrium constant, \(K^‡\) can be expressed in terms of the free energy of formation of the complex (\(\ ^‡\)) \[K^‡ = e^{-\Delta G^‡ /RT} \nonumber \] And so the reaction rate is given by \[\text{rate} = (\text{frequency}) [A,BC]e^{-\Delta G^‡ /RT} \nonumber \] and the remaining task is to derive an expression for the frequency factor. If the frequency is taken to be equal to the vibrational frequency for the vibration of the bond being broken in the activated complex in order to form the reaction products, it can be expressed in terms of the energy of the oscillation of the bond as the complex vibrates. \[ E = h\nu = k_BT \nonumber \] or \[\nu = \dfrac{k_BT}{h} \nonumber \] The reaction rate is then predicted to be \[\text{rate} = \dfrac{k_BT}{h} [A,BC]e^{-\Delta G^‡ /RT} \nonumber \] And the rate constant is thus given by \[ k = \dfrac{k_BT}{h} e^{-\Delta G^‡ /RT} \nonumber \] An alternative description gives the transition state formation equilibrium constant in terms of the partition functions describing the reactants and the transition state: \[K^‡ = \dfrac{Q^‡}{Q_A^‡Q_{BC}^‡} e^{-\Delta G^‡ /RT} \nonumber \] where Q is the describing the i species. If the partition function of the transition state is expressed as a product of the partition function excluding and contribution from the vibration leading to the bond cleavage that forms the products and the partition function of that specific vibrational mode \[ Q^‡ = Q^{‡'}q_v^‡ \nonumber \] In this case, \(q_v^‡\) can be expressed by \[q_v^‡ = \dfrac{1}{1-e^{-h\nu^‡/RT}} \approx \dfrac{k_BT}{h\nu^‡} \nonumber \] So the equilibrium constant can be expressed \[K^‡ = \dfrac{k_BT}{h} \dfrac{Q^‡}{Q_A^‡Q_{BC}^‡} e^{-\Delta G^‡ /RT} \nonumber \] And so the rate constant, which is the product of \(n^‡\) and \(K^‡\), is given by \[ k = \dfrac{k_BT}{h} \dfrac{Q^‡}{Q_A^‡Q_{BC}^‡} e^{-\Delta G^‡ /RT} \nonumber \] which looks very much like the Arrhenius equation proposed quite a few years earlier! Thus, if one understands the vibrational dynamics of the activated complex, and can calculate the partition functions describing the reactants and the transition state, one can, at least in theory, predict the rate constant for the reaction. In the next chapter, we will take a look at how kinetics studies can shed some light on chemical reaction mechanisms. | 3,600 | 2,873 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Properties_of_Alkyl_Halides/Haloalkanes/Alkyl_Halide_Occurrence |
Halogen containing organic compounds are relatively rare in terrestrial plants and animals. The thyroid hormones T and T are exceptions; as is fluoroacetate, the toxic agent in the South African shrub , known as "gifblaar". However, the halogen rich environment of the ocean has produced many interesting natural products incorporating large amounts of halogen. Some examples are shown below. The ocean is the largest known source for atmospheric methyl bromide and methyl iodide. Furthermore, the ocean is also estimated to supply 10-20% of atmospheric methyl chloride, with other significant contributions coming from biomass burning, salt marshes and wood-rotting fungi. Many subsequent chemical and biological processes produce poly-halogenated methanes.
Synthetic organic halogen compounds are readily available by direct halogenation of hydrocarbons and by addition reactions to alkenes and alkynes. Many of these have proven useful as intermediates in traditional synthetic processes. Some halogen compounds, shown in the box. have been used as pesticides, but their persistence in the environment, once applied, has led to restrictions, including banning, of their use in developed countries. Because DDT is a cheap and effective mosquito control agent, underdeveloped countries in Africa and Latin America have experienced a dramatic increase in malaria deaths following its removal, and arguments are made for returning it to limited use. 2,4,5-T and 2,4-D are common herbicides that are sold by most garden stores. Other organic halogen compounds that have been implicated in environmental damage include the polychloro- and polybromo-biphenyls (PCBs and PBBs), used as heat transfer fluids and fire retardants; and freons (e.g. CCl F and other chlorofluorocarbons) used as refrigeration gases and fire extinguishing agents. | 1,853 | 2,874 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Reactivity_of_Arenes/Friedel-Crafts_Acylation/Friedel-Crafts_Reactions |
This page gives details of the Friedel-Crafts reactions of benzene and methylbenzene (toluene). An acyl group is an alkyl group attached to a carbon-oxygen double bond. If "R" represents any alkyl group, then an acyl group has the formula RCO-. Acylation means substituting an acyl group into something - in this case, into a benzene ring. The most commonly used acyl group is CH CO-. This is called the ethanoyl group, and in this case the reaction is sometimes called "ethanoylation". In the example which follows we are substituting a CH CO- group into the ring, but you could equally well use any other acyl group. The most reactive substance containing an acyl group is an acyl chloride (also known as an acid chloride). These have the general formula RCOCl. Benzene is treated with a mixture of ethanoyl chloride, CH COCl, and aluminium chloride as the catalyst. The mixture is heated to about 60°C for about 30 minutes. A ketone called phenylethanone (old name: acetophenone) is formed. or, if you want a more compact form: \[ C_6H_6 + CH_3COCl \longrightarrow C_6H_5COCH_3 + HCl\] The aluminium chloride isn't written into these equations because it is acting as a catalyst. If you wanted to include it, you could write AlCl over the top of the arrow (see below). The reaction is just the same with methylbenzene except that you have to worry about where the acyl group attaches to the ring relative to the methyl group. Normally, the methyl group in methylbenzene directs new groups into the 2- and 4- positions (assuming the methyl group is in the 1- position). In acylation, though, virtually all the substitution happens in the 4- position. Alkylation means substituting an alkyl group into something - in this case into a benzene ring. A hydrogen on the ring is replaced by a group like methyl or ethyl and so on. Friedel-Crafts alkylation of benzene: Benzene reacts at room temperature with a chloroalkane (for example, chloromethane or chloroethane) in the presence of aluminium chloride as a catalyst. On this page, we will look at substituting a methyl group, but any other alkyl group could be used in the same way. Substituting a methyl group gives methylbenzene. or: \[ C_6H_6 + CH_3Cl \rightarrow C_6H_5CH_3 + HCl\] Again, the reaction is just the same with methylbenzene except that you have to worry about where the alkyl group attaches to the ring relative to the methyl group. Unfortunately this time there is a problem! Where the incoming alkyl group ends up depends to a large extent on the temperature of the reaction. At 0°C, substituting methyl groups into methylbenzene, you get a mixture of the 2-.3- and 4- isomers in the proportion 54% / 17% / 29%. That's a higher proportion of the 3- isomer than you might expect. At 25°C, the proportions change to 3% / 69% / 28%. In other words the proportion of the 3- isomer has increased even more. Raise the temperature some more and the trend continues. The manufacture of ethylbenzene: Ethylbenzene is an important industrial chemical used to make styrene (phenylethene), which in turn is used to make polystyrene - poly(phenylethene). It is manufactured from benzene and ethene. There are several ways of doing this, some of which use a variation on Friedel-Crafts alkylation. The reaction is done in the liquid state. Ethene is passed through a liquid mixture of benzene, aluminium chloride and a catalyst promoter which might be chloroethane or hydrogen chloride. We are going to assume it is HCl. Promoters are used to make catalysts work better. There are two variants on the process. One (the Union Carbide / Badger process) uses a temperature no higher than 130°C and a pressure just high enough to keep everything liquid. The other (the Monsanto process) uses a slightly higher temperature of 160°C which needs less catalyst. (Presumably - although I haven't been able to confirm this - the pressure would also need to be higher to keep everything liquid at the higher temperature.) or \[C_6H_6 + \ce{CH_2=CH_2} \rightarrow C_6H_5CH_2CH_3\] Again, the aluminium chloride and HCl aren't written into these equations because they are acting as catalysts. If you wanted to include them, you could write AlCl and HCl over the top of the arrow. Jim Clark ( ) | 4,252 | 2,875 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Field_Theory/Tetrahedral_vs._Square_Planar_Complexes |
Tetrahedral geometry is a bit harder to visualize than square planar geometry. Tetrahedral geometry is analogous to a pyramid, where each of corners of the pyramid corresponds to a ligand, and the central molecule is in the middle of the pyramid. This geometry also has a coordination number of 4 because it has 4 ligands bound to it. Finally, the bond angle between the ligands is 109.5 . An example of the tetrahedral molecule \(\ce{CH4}\), or methane. In a tetrahedral complex, \(Δ_t\) is relatively small even with strong-field ligands as there are fewer ligands to bond with. It is rare for the \(Δ_t\) of tetrahedral complexes to exceed the pairing energy. Usually, electrons will move up to the higher energy orbitals rather than pair. Because of this, In square planar molecular geometry, a central atom is surrounded by constituent atoms, which form the corners of a square on the same plane. The geometry is prevalent for transition metal complexes with d configuration. This includes Rh(I), Ir(I), Pd(II), Pt(II), and Au(III). Notable examples include the anticancer drugs (\(\ce{PtCl2(NH3)2}\)). A square planar complex also has a coordination number of 4. The structure of the complex differs from tetrahedral because the ligands form a simple square on the x and y axes. Because of this, the crystal field splitting is also different (Figure \(\Page {1}\)). Since there are no ligands along the z-axis in a square planar complex, the repulsion of electrons in the \(d_{xz}\), \(d_{yz}\), and the \(d_{z^2}\) orbitals are considerably lower than that of the octahedral complex (the \(d_{z^2}\) orbital is slightly higher in energy to the "doughnut" that lies on the x,y axis). The \(d_{x^2-y^2}\) orbital has the most energy, followed by the \(d_{xy}\) orbital, which is followed by the remaining orbtails (although \(d_{z^2}\) has slightly more energy than the \(d_{xz}\) and \(d_{yz}\) orbital). This pattern of orbital splitting remains constant throughout all geometries. Whichever orbitals come in direct contact with the ligand fields will have higher energies than orbitals that slide past the ligand field and have more of indirect contact with the ligand fields. So when confused about which geometry leads to which splitting, think about the way the ligand fields interact with the electron orbitals of the central atom. In square planar complexes \(Δ\) will almost always be large (Figure \(\Page {1}\)), even with a weak-field ligand. Electrons tend to be paired rather than unpaired because paring energy is usually much less than \(Δ\). Therefore, The molecule \(\ce{[PdCl4]^{2−}}\) is diamagnetic, which indicates a square planar geometry as all eight d-electrons are paired in the lower-energy orbitals. However, \(\ce{[NiCl4]^{2−}}\) is also d but has two unpaired electrons, indicating a tetrahedral geometry. Why is \(\ce{[PdCl4]^{2−}}\) square planar if \(\ce{Cl^{-}}\) is not a strong-field ligand? The geometry of the complex changes going from \(\ce{[NiCl4]^{2−}}\) to \(\ce{[PdCl4]^{2−}}\). Clearly this cannot be due to any change in the ligand since it is the same in both cases. It is the other factor, the metal, that leads to the difference. Consider the splitting of the d-orbitals in a generic d complex. If it were to adopt a square planar geometry, the electrons will be stabilized (with respect to a tetrahedral complex) as they are placed in orbitals of lower energy. However, this comes at a cost: two of the electrons, which were originally unpaired in the tetrahedral structure, are now paired in the square-planer structure: We can label these two factors as \(ΔE\) (stabilization derived from occupation of lower-energy orbitals) and \(P\) (spin pairing energy) respectively. One can see that: This is analogous to deciding whether an octahedral complex adopts a high- or low-spin configuration; where the crystal field splitting parameter \(Δ_o\) \(ΔE\) does above. Unfortunately, unlike \(Δ_o\) in octahedral complexes, there is no simple graphical way to represent \(ΔE\) on the diagram above since multiple orbitals are changed in energy between the two geometries. Interpreting the origin of metal-dependent stabilization energies can be tricky. However, we know experimentally that \(\ce{Pd^{2+}}\) has a larger splitting of the d-orbitals and hence a larger \(\Delta E\) than \(\ce{Ni^{2+}}\) (moreover \(P\) is also smaller). Practically all 4d and 5d d \(\ce{ML4}\) complexes adopt a square planar geometry, irrespective if the ligands are strong-field ligand or not. Other examples of such square planar complexes are \(\ce{[PtCl4]^{2−}}\) and \(\ce{[AuCl4]^{-}}\). | 4,653 | 2,876 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Fundamentals/Structure_of_Organic_Molecules/The_E-Z_system_for_naming_alkenes |
The traditional system for naming the geometric isomers of an alkene, in which the same groups are arranged differently, is to name them as cis or trans. However, it is easy to find examples where the cis-trans system is not easily applied. IUPAC has a more complete system for naming alkene isomers. The R-S system is based on a set of "priority rules", which allow you to rank any groups. The rigorous IUPAC system for naming alkene isomers, called the E-Z system, is based on the same priority rules.These priority rules are often called the Cahn-Ingold-Prelog (CIP) rules, after the chemists who developed the system The general strategy of the E-Z system is to analyze the two groups at each end of the double bond. At each end, rank the two groups, using the CIP priority rules, discussed in Ch 15. Then, see whether the higher priority group at one end of the double bond and the higher priority group at the other end of the double bond are on the side (Z, from German = together) or on sides (E, from German = opposite) of the double bond. The Figure below shows the two isomers of 2-butene. You should recognize them as cis and trans. Let's analyze them to see whether they are E or Z. Start with the left hand structure (the cis isomer). On C2 (the left end of the double bond), the two atoms attached to the double bond are C and H. By the CIP priority rules, C is higher priority than H (higher atomic number). Now look at C3 (the right end of the double bond). Similarly, the atoms are C and H, with C being higher priority. We see that the higher priority group is "down" at C2 and "down" at C3. Since the two priority groups are both on the side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is (Z)-2-butene. Now look at the right hand structure (the trans isomer). In this case, the priority group is "down" on the left end of the double bond and "up" on the right end of the double bond. Since the two priority groups are on sides of the double bond, they are entgegen = opposite. Therefore, this is (E)-2-butene. In simple cases, such as 2-butene, Z corresponds to cis and E to trans. However, that is a rule. This section and the following one illustrate some idiosyncrasies that happen when you try to compare the two systems. The real advantage of the E-Z system is that it will always work. In contrast, the cis-trans system breaks down with many ambiguous cases. The following figure shows two isomers of an alkene with four different groups on the double bond, 1-bromo-2-chloro-2-fluoro-1-iodoethene. It should be apparent that the two structures shown are distinct chemicals. However, it is impossible to name them as cis or trans. On the other hand, the E-Z system works fine... Consider the left hand structure. On C1 (the left end of the double bond), the two atoms attached to the double bond are Br and I. By the CIP priority rules, I is higher priority than Br (higher atomic number). Now look at C2. The atoms are Cl and F, with Cl being higher priority. We see that the higher priority group is "down" at C1 and "down" at C2. Since the two priority groups are both on the side of the double bond ("down", in this case), they are zusammen = together. Therefore, this is the (Z) isomer. Similarly, the right hand structure is (E). Consider the molecule shown at the left. This is 2-bromo-2-butene -- ignoring the geometric isomerism for now. Cis or trans? This molecule is clearly cis. The two methyl groups are on the same side. More rigorously, the "parent chain" is cis. E or Z? There is a methyl at each end of the double bond. On the left, the methyl is the high priority group -- because the other group is -H. On the right, the methyl is the low priority group -- because the other group is -Br. That is, the high priority groups are -CH (left) and -Br (right). Thus the two priority groups are on opposite sides = entgegen = E. This example should convince you that cis and Z are not synonyms. Cis/trans and E,Z are determined by distinct criteria. There may seem to be a simple correspondence, but it is not a rule. Be sure to determine cis,trans or E,Z separately, as needed. If the compound contains more than one double bond, then each one is analyzed and declared to be E or Z. The configuration at the left hand double bond is E; at the right hand double bond it is Z. Thus this compound is (1E,4Z)-1,5-dichloro-1,4-hexadiene. Consider the compound below: This is 1-chloro-2-ethyl-1,3-butadiene -- ignoring, for the moment, the geometric isomerism. There is no geometric isomerism at the second double bond, at 3-4, because it has 2 H at its far end. What about the first double bond, at 1-2? On the left hand end, there is H and Cl; Cl is higher priority (by atomic number). On the right hand end, there is -CH -CH (an ethyl group) and -CH=CH (a vinyl or ethenyl group). Both of these groups have C as the first atom, so we have a tie so far and must look further. What is attached to this first C? For the ethyl group, the first C is attached to C, H, and H. For the ethenyl group, the first C is attached to a C twice, so we count it twice; therefore that C is attached to C, C, H. CCH is higher than CHH; therefore, the ethenyl group is higher priority. Since the priority groups, Cl and ethenyl, are on the same side of the double bond, this is the Z-isomer; the compound is (Z)-1-chloro-2-ethyl-1,3-butadiene. Which is higher priority, by the CIP rules: a C with an O and 2 H attached to it or a C with three C? The first C has one atom of high priority but also two atoms of low priority. How do these "balance out"? Answering this requires a clear understanding of how the ranking is done. The simple answer is that the first point of difference is what matters; the O wins. To illustrate this, consider the molecule at the left. Is the double bond here E or Z? At the left end of the double bond, Br > H. But the right end of the double bond requires a careful analysis. At the right hand end, the first atom attached to the double bond is a C at each position. A tie, so we look at what is attached to this first C. For the upper C, it is CCC (since the triple bond counts three times). For the lower C, it is OHH -- listed in order from high priority atom to low. OHH is higher priority than CCC, because of the first atom in the list. That is, the O of the lower group beats the C of the upper group. In other words, the O is the highest priority atom of any in this comparison; thus the O "wins". Therefore, the high priority groups are "up" on the left end (the -Br) and "down" on the right end (the -CH -O-CH ). This means that the isomer shown is opposite = entgegen = E. And what is the name? The "name" feature of ChemSketch says it is (2E)-2-(1-bromoethylidene)pent-3-ynyl methyl ether. >Robert Bruner ( ) | 6,847 | 2,878 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.06%3A_Le_Chatelier's_Principle |
Chemical equilibrium was studied by the French chemist Henri Le Chatelier (1850 - 1936) and his description of how a system responds to a stress to equilibrium has become known as : When a chemical system that is at equilibrium is disturbed by a stress, the system will respond in order to relieve the stress. Stresses to a chemical system involve changes in the concentrations of reactants or products, changes in the temperature of the system, or changes in the pressure of the system. We will discuss each of these stresses separately. The change to the equilibrium position in every case is either a favoring of the forward reaction or a favoring of the reverse reaction. When the forward reaction is favored, the concentrations of products increase, while the concentrations of reactants decrease. When the reverse reaction is favored, the concentrations of the products decrease, while the concentrations of reactants increase. \[\begin{array}{lll} \textbf{Original Equilibrium} & \textbf{Favored Reaction} & \textbf{Result} \\ \ce{A} \rightleftharpoons \ce{B} & \text{Forward:} \: \ce{A} \rightarrow \ce{B} & \left[ \ce{A} \right] \: \text{decreases}; \: \left[ \ce{B} \right] \: \text{increases} \\ \ce{A} \rightleftharpoons \ce{B} & \text{Reverse:} \: \ce{A} \leftarrow \ce{B} & \left[ \ce{A} \right] \: \text{increases}; \: \left[ \ce{B} \right] \: \text{decreases} \end{array} \nonumber \] A change in concentration of one of the substances in an equilibrium system typically involves either the addition or the removal of one of the reactants or products. Consider the Haber-Bosch process for the industrial production of ammonia from nitrogen and hydrogen gases. \[\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right) \nonumber \] If the concentration of one substance in a system is increased, the system will respond by favoring the reaction that removes that substance. When more \(\ce{N_2}\) is added, the forward reaction will be favored because the forward reaction uses up \(\ce{N_2}\) and converts it to \(\ce{NH_3}\). The forward reaction speeds up temporarily as a result of the addition of a reactant. The position of equilibrium shifts as more \(\ce{NH_3}\) is produced. The concentration of \(\ce{NH_3}\) increases, while the concentrations of \(\ce{N_2}\) and \(\ce{H_2}\) decrease. After some time passes, equilibrium is reestablished with new concentrations of all three substance. As can be seen in the figure below, if more \(\ce{N_2}\) is added, a new equilibrium is achieved by the system. The new concentration of \(\ce{NH_3}\) is higher because of the favoring of the forward reaction. The new concentration of the \(\ce{H_2}\) is lower .The concentration of \(\ce{N_2}\) is higher than in the original equilibrium, but went down slightly following the addition of the \(\ce{N_2}\) that disturbed the original equilibrium. By responding in this way, the value of the equilibrium constant for the reaction, \(K_\text{eq}\), does not change as a result of the stress to the system. In other words, the amount of each substance is different but the ratio of the amount of each remains the same. If more \(\ce{NH_3}\) were added, the reverse reaction would be favored. This "favoring" of a reaction means temporarily speeding up the reaction in that direction until equilibrium is reestablished. Recall that once equilibrium is reestablished, the rates of the forward and reverse reactions are again equal. The addition of \(\ce{NH_3}\) would result in increased formation of the reactants, \(\ce{N_2}\) and \(\ce{H_2}\). An equilibrium can also be disrupted by the removal of one of the substances. If the concentration of a substance is decreased, the system will respond by favoring the reaction that replaces that substance. In the industrial Haber-Bosch process, \(\ce{NH_3}\) is removed from the equilibrium system as the reaction proceeds. As a result, the forward reaction is favored so that more \(\ce{NH_3}\) is produced. The concentrations of \(\ce{N_2}\) and \(\ce{H_2}\) decrease. Continued removal of \(\ce{NH_3}\) will eventually force the reaction to go to completion until all of the reactants are used up. If either \(\ce{N_2}\) or \(\ce{H_2}\) were removed from the equilibrium system, the reverse reaction would be favored and the concentration of \(\ce{NH_3}\) would decrease. The effect of changes in concentration on an equilibrium system according to Le Chatelier's principle is summarized in the table below. Given this reaction at equilibrium: \[N_{2}+3H_{2}\rightleftharpoons 2NH_{3} \nonumber \] How will it affect the reaction if the equilibrium is stressed by each change? Given this reaction at equilibrium: \[CO(g)+Br_{2}(g)\rightleftharpoons COBr_{2}(g) \nonumber \] How will it affect the reaction if the equilibrium is stressed by each change? 1. shift to the left (toward reactants) 2. shift to the left (toward reactants) Increasing or decreasing the temperature of a system at equilibrium is also a stress to the system. The equation for the Haber-Bosch process is written again below, as a thermochemical equation (i.e. it contains information about the energy gained or lost when the reaction occurs). \[\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightleftharpoons 2 \ce{NH_3} \left( g \right) + 91 \: \text{kJ} \nonumber \] The forward reaction is the exothermic direction: the formation of \(\ce{NH_3}\) releases heat which is why that is shown as a product. The reverse reaction is the endothermic direction: as \(\ce{NH_3}\) decomposes to \(\ce{N_2}\) and \(\ce{H_2}\), heat is absorbed. An increase in the temperature for this is like adding a product because heat is being released by the reaction. If we add a product then the reaction proceeds towards the formation of more reactants. Reducing the temperature for this system would be similar to removing a product which would favor the formation of more products. The amount of \(\ce{NH_3}\) will increase and the amount of \(\ce{N_2}\) and \(\ce{H_2}\) will decrease. For changes in concentration, the system responds in such a way that the value of the equilibrium constant, \(K_\text{eq}\) is unchanged. However, a change in temperature shifts the equilibrium and the \(K_\text{eq}\) value either increases or decreases. As discussed in the previous section, values of \(K_\text{eq}\) are dependent on the temperature. When the temperature of the system for the Haber-Bosch process is increased, the resultant shift in equilibrium towards the reactants means that the \(K_\text{eq}\) value decreases. When the temperature is decreased, the shift in equilibrium towards the products means that the \(K_\text{eq}\) value increases. Le Chatelier's principle as related to temperature changes can be illustrated easily be the reaction in which dinitrogen tetroxide is in equilibrium with nitrogen dioxide. \[\ce{N_2O_4} \left( g \right) + \text{heat} \rightleftharpoons 2 \ce{NO_2} \left( g \right) \nonumber \] Dinitrogen tetroxide \(\left( \ce{N_2O_4} \right)\) is colorless, while nitrogen dioxide \(\left( \ce{NO_2} \right)\) is dark brown in color. When \(\ce{N_2O_4}\) breaks down into \(\ce{NO_2}\), heat is absorbed (endothermic) according to the forward reaction above. Therefore, an increase in temperature (adding heat) of the system will favor the forward reaction. Conversely, a decrease in temperature (removing heat) will favor the reverse reaction. Predict the effect of increasing the temperature on this equilibrium. \[PCl_{3}+Cl_{2}\rightleftharpoons PCl_{5}+60kJ \nonumber \] Because energy is listed as a product, it is being produced, so the reaction is exothermic. If the temperature is increasing, a product is being added to the equilibrium, so the equilibrium shifts to minimize the addition of extra product: it shifts to the left (back toward reactants). Predict the effect of decreasing the temperature on this equilibrium. \[N_{2}O_{4}+57kJ\rightleftharpoons 2NO_{2} \nonumber \] Equilibrium shifts to the left (toward reactants). Changing the pressure of an equilibrium system in which gases are involved is also a stress to the system. A change in the pressure on a liquid or a solid has a negligible effect. We will return again the equilibrium for the Haber-Bosch process. Imagine the gases are contained in a closed system in which the volume of the system is controlled by an adjustable piston as shown in the figure below. On the far left, the reaction system contains primarily \(\ce{N_2}\) and \(\ce{H_2}\), with only one molecule of \(\ce{NH_3}\) present. As the piston is pushed inwards, the pressure of the system increases according to Boyle's law. This is a stress to the equilibrium. In the middle image, the same number of molecules is now confined in a smaller space and so the pressure has increased. According to Le Chatelier's principle, the system responds in order to relieve the stress. In the image on the right, the forward reaction has been favored and more \(\ce{NH_3}\) is produced. The overall result is a decrease in the number of gas molecules in the entire system. This in turn decreases the pressure and provides a relief to the original stress of a pressure increase. An increase in pressure on an equilibrium system favors the reaction which products fewer total moles of gas. In this case, it is the forward reaction that is favored. A decrease in pressure on the above system could be achieved by pulling the piston outward, increasing the container volume. The equilibrium would respond by favoring the reverse reaction in which \(\ce{NH_3}\) decomposes to \(\ce{N_2}\) and \(\ce{H_2}\). This is because the overall number of gas molecules would increase and so would the pressure. A decrease in pressure on an equilibrium system favors the reaction which produces more total moles of gas. This is summarized in the table below. Like changes in concentration, the \(K_\text{eq}\) value for a given reaction is unchanged by a change in pressure. The amounts of each substance will change but the ratio will not. It is important to remember when analyzing the effect of a pressure change on equilibrium that only gases are affected. If a certain reaction involves liquids or solids, they should be ignored. For example, calcium carbonate decomposes according to the equilibrium reaction: \[\ce{CaCO_3} \left( s \right) \rightleftharpoons \ce{CaO} \left( s \right) + \ce{O_2} \left( g \right) \nonumber \] Oxygen is the only gas in the system. An increase in the pressure of the system slows the rate of decomposition of \(\ce{CaCO_3}\) because the reverse reaction is favored. When a system contains equal moles of gas on both sides of the equation, pressure has no effect on the equilibrium position, as in the formation of \(\ce{HCl}\) from \(\ce{H_2}\) and \(\ce{Cl_2}\). \[\ce{H_2} \left( g \right) + \ce{Cl_2} \left( g \right) \rightleftharpoons 2 \ce{HCl} \left( g \right) \nonumber \] What is the effect on this equilibrium if pressure is increased? \[N_{2}(g)+3H_{2}(g)\rightleftharpoons 2NH_{3}(g) \nonumber \] According to Le Chatelier's principle, if pressure is increased, then the equilibrium shifts to the side with the fewer number of moles of gas. This particular reaction shows a total of 4 mol of gas as reactants and 2 mol of gas as products, so the reaction shifts to the right (toward the products side). What is the effect on this equilibrium if pressure is decreased? \[3O_{2}(g)\rightleftharpoons 2O_{3}(g) \nonumber \] Reaction shifts to the left (toward reactants). In aerobic respiration, oxygen is transported to the cells where it is combined with glucose and metabolized to carbon dioxide, which then moves back to the lungs from which it is expelled. hemoglobin + O oxyhemoglobin The partial pressure of O in the air is 0.2 atm, sufficient to allow these molecules to be taken up by hemoglobin (the red pigment of blood) in which it becomes loosely bound in a complex known as oxyhemoglobin. At the ends of the capillaries which deliver the blood to the tissues, the O concentration is reduced by about 50% owing to its consumption by the cells. This shifts the equilibrium to the left, releasing the oxygen so it can diffuse into the cells. , Ph.D. (Department of Chemistry, University of Kentucky) | 12,316 | 2,879 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Nucleophilic_Reactivity_of_Deprotonated_Alkynes |
Conjugate base anions of terminal alkynes acetylide nucleophiles nucleophilic nucleophilic Terminal are much more acidic than most other hydrocarbons. Removal of the proton leads to the formation of an acetylide anion. The origin of the enhanced acidity can be attributed to the stability of the acetylide anion, which has the unpaired electrons in an sp hybridized orbital. The stability results from occupying an orbital with a high degree of s-orbital character. There is a strong correlation between s-character in the orbital containing the non-bonding electrons in the anion and the acidity of hydrocarbons. The enhanced acidity with greater s-character occurs despite the fact that the is larger. Consequently, acetylide anions can be readily formed by deprotonation using a sufficiently strong base. Amide anion (NH ), in the form of NaNH is commonly used for the formation of acetylide anions. Acetylide anions are strong bases and strong nucleophiles. Therefore, they are able to displace halides and other leaving groups in substitution reactions. The product is a substituted alkyne. Because the ion is a very strong base, the substitution reaction is most efficient with methyl or without substitution near the reaction center, Secondary, tertiary or even bulky primary substrates will give elimination by the E2 mechanism. Acetylide anions will add to to form alkoxides, which, upon protonation, give propargyl alcohols. With aldehydes and non-symmetric ketones, in the absence of chiral catalyst, the product will be a racemic mixture of the two enantiomers. The triple bond in the propargyl alcohol can be modified by using the reactivity of the alkyne. For example, Markovnikov and anti-Markovnikov hydration of the triple bond leads to formation of the hydroxy-substituted ketone and aldehyde, respectively, after enol-keto tautomerization. 1. The pK of ammonia is 35. Estimate the equilibrium constant for the deprotonation of pent-1-yne by amide, as shown above. 1. Assuming the pK of pent-1-yne is about 25, then the difference in pK s is 10. Since pentyne is more acidic, the formation of the acetylide will be favored at equilibrium, so the equilibrium constant for the reaction is about 10 | 2,241 | 2,880 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Addition_to_Carbonyls/CO16._Biological_Reduction |
Addition to a carbonyl by a hydride, such as NaBH , results in conversion of the carbonyl compound to an alcohol. The hydride from the BH anion acts as a nucleophile, adding H to the carbonyl carbon. A proton source can then protonate the oxygen of the resulting alkoxide ion, forming an alcohol. Formally, that process is referred to as a reduction. Reduction generally means a reaction in which electrons are added to a compound; the compound that gains electrons is said to be reduced. Because hydride can be thought of as a proton plus two electrons, we can think of conversion of a ketone or an aldehyde to an alcohol as a two-electron reduction. An aldehyde plus two electrons and two protons becomes an alcohol. Aldehydes, ketones and alcohols are very common features in biological molecules. Converting between these compounds is a frequent event in many biological pathways. However, semi-anionic compounds like sodium borohydride don't exist in the cell. Instead, a number of biological hydride donors play a similar role. NADH is a common biological reducing agent. NADH is an acronym for nicotinamide adenine dinucleotide hydride. Insetad of an anionic donor that provides a hydride to a carbonyl, NADH is actually a neutral donor. It supplies a hydride to the carbonyl under very specific circumstances. In doing so, it forms a cation, NAD . However, NAD is stabilized by the fact that its nicotinamide ring is aromatic; it was not aromatic in NADH. , | 1,483 | 2,881 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/05%3A_Cooperativity/18%3A_Cooperativity/18.01%3A_HelixCoil_Transition |
Cooperativity plays an important role in the description of the helix–coil transition, which refers to the reversible transition of macromolecules between coil and extended helical structures. This phenomenon was observed by Paul Doty in the 1950s for the conversion of polypeptides between a coil and α-helical form, and for the melting and hybridization of DNA. Bruno Zimm developed a statistical theory with J. Bragg that described the helix–coil transition, which forms the basis of our discussion. One of the observations that motivated this work is shown in the figure below. The fraction of helical structure observed in the polypeptide poly-benzylglutamate showed a temperature-dependent melting behavior in which the steepness of the transition increased with polymer chain length. This length dependence indicates a higher probability of forming helices when more residues are present, and that the linkages do not act independently. This suggests a two-step mechanism. The rate-limiting step of forming an \(α\) helix is the nucleation of a single hydrogen bonded residue \(i → i+4\) loop. Once this occurs, the addition of further hydrogen bonds to extend this helix is much easier and occurs in rapid succession. To model this behavior, we imagine that the polypeptide consists of a chain of segments that can take on two configurations, H or C. H: helix (decreases entropy but also lower enthalpy) C: coil (raises entropy) To specify the state of a conformation through a sequence, i.e., ...HCHHHHCCCCHHH... Remember to not take this too literally, and be flexible in the interpretation of your model. Although this model was derived with an \(α\)-helix formation in polypeptides in mind, in a more general sense \(H\) and \(C\) do not necessarily refer explicitly to residues of a sequence, but just for independently interacting regions. If there are \(n\) segments, these can be divided into \(n_H\) helical and \(n_C\) coil segments. \[n_H + n_C = n \nonumber \] The segments need not correspond directly to amino acids, but structurally and energetically distinct regions. Our goal will be to calculate the fractional helicity of this system \(\theta_H\) as a function of temperature, by calculating the conformational partition function, q , by an explicit summation over i microstates, Boltzmann weighed by the microstate energy E : \[ q_{\mathrm{conf}} (n) = \sum_{i\, \mathrm{config.}} e^{-E_i/k_BT} \] We start our analysis by discussing a non-cooperative model. We assume: \[ E_i = E(n_H) = n_H\Delta \epsilon \nonumber \] Then, we can calculate q using g(n,n ), the degeneracy of distinguishable states for a polymer of length n with n helical segments. The conformational partition function is obtained by \[q_{\mathrm{conf}}(n) = \sum_{n_H=0}^n g(n,n_H) e^{-n_H \Delta \epsilon / k_BT} \] In evaluating the partition functions in helix–coil transition models, it is particularly useful to define a “statistical weight” for the helical configuration. It describes the influence of having an H on the probability of observing a particular configuration at kBT: \[s = e^{-\Delta \epsilon / k_BT} \] For the present model, we can think of s as an equilibrium constant for the process of adding a helical residue to a sequence: \[s = \dfrac{P(n_H+1)}{P(n_H)} \nonumber \] This equilibrium constant is related to the free energy change for adding a helical residue to the growing chain. Then we can write eq. (18.1.2) as \[q_{\mathrm{conf}}(n) = \sum_{n_H=0}^n g(n,n_H) s^{n_H} \nonumber \] Since there are only two possible configurations (H and C), the degeneracy of configurations with \(n_H\) helical segments in a chain of length n is given by the binomial coefficients: \[ g(n,n_H) = \dfrac{n!}{n_H!n_C!} = \begin{pmatrix} n \\ n_H \end{pmatrix} \] since \(n_C=n-n_H\). Then using the binomial theorem, we obtain \[q_{\mathrm{conf}}(n) = (1+s)^n \] Also, the probability of a chain with n segments having n helical linkages is \[P(n,n_H) = \dfrac{g(n,n_H)e^{-E(n_H)/k_BT}}{q_{\mathrm{conf}}} = \begin{pmatrix} n \\ n_H \end{pmatrix} \dfrac{s^{n_H}}{(1+s)^n} \] Example: n = 4 The conformations available are at right. The molecular conformational partition function is \(\begin{aligned} q_{\mathrm{conf}} &= 1+4e^{-\Delta \epsilon /k_BT} +6e^{-2\Delta \epsilon /k_BT} +4e^{-3\Delta \epsilon /k_BT} +e^{-4\Delta \epsilon /k_BT} \\ &= 1+4s+6s^2+4s^3+s^4 \\ &= (1+s)^4 \end{aligned} \) The last step follows from Pascal’s Rule for binomial coefficients. From eq. (18.1.6), the probability of having two helical residues in a four-residue sequence is: \[P(4,2) = \dfrac{6s^2}{(1+s)^4} \nonumber \] To relate this to an observable quantity, we define the fractional helicity, the average fraction of residues that are in the H form. \[ \theta_H = \dfrac{\langle n_H \rangle}{n} \] \[ \langle n_H \rangle = \sum_{n_H = 0}^{n} n_HP(n,n_H) \] Using this amazing little identity, which we derive below, \[ \langle n_H \rangle = \dfrac{s}{q} \dfrac{\partial q}{\partial s} \] You can use eq. (18.1.5) to show: \[ \langle n_H \rangle = \dfrac{ns}{1+s} \] and \[ \theta_H = \dfrac{s}{1+s} \] This takes the same form as one would expect for the simple chemical equilibrium of an \(C \rightleftharpoons H\) molecular reaction. If we define the equilibrium constant K = [H]/[C], then the fraction of molecules in the H state is \(\theta_H = [H]/([C]+[H]) = K_{HC}/(1+K_{HC})\). In this limit s = K . Below we plot eq. (18.1.11), choosing Δε to be independent of temperature. θ is a smooth and slowly varying function of T and does not show cooperative behavior. Its high temperature limit is θ = 0.5, reflecting the fact that in the absence of barriers, the H and C configurations are equally probable for every residue. We can look a bit deeper at what is happening with the structures present by plotting the probability distribution function for finding nH helical segments within a chain of length n, eq. (18.1.6), and the associated energy landscape (a potential of mean force): \[F(n,n_H) = -Nk_BT\ln{[P(n,n_H)]} \approx -Nk_BT \ln{[g(n,n_H)s^{n_H}]} \nonumber \] The maximum probability and free-energy minimum is located at full helix content at the lowest temperature, and gradually shifts toward nH/n = 0.5 with increasing temperature. The probability density appears Gaussian, and the corresponding free energy appears parabolic. Using similar methods to that described above, we can show that the variance in this distribution scales as n−1/2.The presence of a single shifting minimum is referred to as a transition in a one-state system, rather than two-state behavior expected for phase transitions. Here nH is the order parameter that characterizes the extend of folding of the helix. Where does eq. (18.1.9) come from? For the moment, we will drop the “conf” and “H” subscripts, mainly to write things more compactly, but also to emphasize the generality of this method to all polynomial expansions. Using eq. (18.1.2), \(q= \sum_n gs^n \), and recognizing that g is not a function of s: \[ \begin{aligned} \dfrac{\partial q}{\partial s} &= \sum_n ngs^{n-1} \\ &= \dfrac{1}{s}\sum_nngs^n \end{aligned} \] \[\] From eq. (18.1.6), \(P_n = gs^n/1 \), we can write this in terms of the helical segment probability \[ \dfrac{1}{q} \dfrac{\partial q}{\partial s} = \dfrac{1}{s}\sum_nnP_n \] Comparing eq. (18.1.13) with eq. (18.1.12), \( \boldsymbol{\langle} n \boldsymbol{\rangle} = \sum_nnP_n \), we see that \[\dfrac{s}{q} \dfrac{\partial q}{\partial s} = \langle n \rangle \quad \mathrm{or} \quad \dfrac{\partial \ln{q}}{\partial \ln{s}} = \boldsymbol{\langle} n \boldsymbol{\rangle} \] This method of obtaining averages from derivatives of a polynomial appears regularly in statistical mechanics. Let’s modify the model to add an element of cooperativity to the segments in the chain. In order to form a helix, you need to nucleate a helical turn and then adding adjacent helical segments is easier. The probability of forming a turn is relatively low, meaning the free energy barrier for nucleation of one H in a sequence of C is relatively high: \(\Delta G_{nuc}>0\). However the free-energy change per residue for forming H from C within a helical stretch, \(\Delta G_{HC} \), stabilizes the growing helix. Based on these free energies, we define statistical weights: \[ s = e^{-\Delta G_{HC}/k_BT} \nonumber \] \[\sigma = e^{-\Delta{nuc}/k_BT}\nonumber \] s and σ are also known as the Zimm–Bragg parameters. Here, s is the statistical weight to add one helical segment to an existing continuous sequence (or stretch) of H, which we interpret as an equilibrium constant: \[ s = \dfrac{[...CHHHHCC...]}{[...CHHHCCC...]}= \dfrac{P_H(n_H+1)}{P_H(n_H)} \nonumber \] σ is the statistical weight for each stretch of H. This is purely to reflect the probability of forming a new helical segment within a stretch of C. The energy benefit of making the helical form is additional: \[ \sigma s = \dfrac{[...CCCHCC...]}{[...CCCCCC...]}= \dfrac{P_H(\nu_H+1)}{P_H(\nu_H)} \nonumber \] \(\nu \) is the number of helical stretch segments in a chain. Note that the formation of the first helical segment has a contribution from both the nucleation barrier (σ) and the formation of the first stabilizing interaction (s). The statistical weight for a particular microstate is then \(e^{-E_i/k_BT } = s^{n_H}\sigma^{\nu_H}\) Since \(\Delta G_{nucl} will be large and positive, σ≪ 1. Also, we take s > 1, and the presence of cooperativity will mainly hinge on σ ≪ s. A 35 segment chain has 235 = 3.4×1010 possible configurations. This particular microstate has fifteen helical segments (nH = 16) partitioned into three helical stretches (νH = 3): \[ CCCCCC \underbrace{HHHHH}_5 CCC \underbrace{H}_1 CCCCCCCC \underbrace{HHHHHHHHHH}_{10}CC \nonumber \] We ignore all Cs since the C state is the ground state and their statistical weight is 1. \[ e^{-E_i/k_BT} = s^{n_H}\sigma^{\nu_H} = s^{16}\sigma^3 \nonumber \] Now the partition function involves a sum over all possible helical segments and stretches: \[q_{conf}(n) = \sum_{n_H=0}^n \sum_{\nu_H = 0}^{\nu_{max}} g(n,n_H,\nu_H )s^{n_H}\sigma^{\nu_H} \] Since the all-coil state (n = 0) is the reference state, it contributes a value of 1 to the partition function (the leading term in the summation). Therefore, the probability of observing the all-coil state is \[P(n,n_H = 0) = q_{conf}^{-1} \] From eq. (18.1.15), the mean number of helical residues is \[ \langle n_H \rangle =\dfrac{1}{q_{conf}} \sum_{n_H=0}^n \sum_{\nu_H = 0}^{\nu_{max}} n_H g(n,n_H,\nu_H) s^{n_H}\sigma^{\nu_H} \nonumber \] In these equations, ν refers to the maximum number of helical stretches for a given n , n /2 for even n and (n /2)+1 for odd n . As a next step, we examine what happens with the simplifying assumption that one helical stretch is allowed. This is the single stretch approximation or the zipper model, in which conversion to a helix proceeds quickly once a single turn has been nucleated. This is reasonable for short chains in which two stretches are unlikely due to steric constraints. For the single stretch case, we only need to account for ν = 0 and 1. For ν = 0 the system is all coil (n = 0) and there is only one microstate to count, g(n,0,0) = 1. For a single helical stretch we need to accounts for the number of ways of positioning a single helical stretch of n residues on a chain of length n: g(n,n ,1) = n-n +1. Then the partition function, eq. (18.1.15), is \[q_{zip}(n) = 1+\sigma \sum_{n_H=1}^n (n-n_H+1)s^{n_H} \] We can evaluate these sums using the relations \[ \begin{aligned} \sum_{n_H=1}^n &= \dfrac{s^{n+1}-s}{s-1} \\ \sum_{n_H=1}^n n_H s^{n_H} &= \dfrac{s}{(s-1)^2} \left[ ns^{n+1}-(n+1)s^n +1 \right] \end{aligned} \] which leads to \[ q_{zip}(n) = 1+ \dfrac{\sigma s^2}{(s-1)^2} \left( s^n + \dfrac{n}{s}-(n+1) \right) \nonumber \] Following the general expression in eq. (18.1.6), and counting the degeneracy of ways to place a stretch of n segments, the probability distribution of helical segments is \[ P_H(n,n_H) = \dfrac{(n-n_H+1)\sigma s^{n_H}}{q_{conf}} \qquad \qquad 1\leq n_H \leq n\] This expression does not apply to the case n = 0, for which we turn to eq. (18.1.16). The helical fraction is obtained from \( \theta_H = \frac{s}{n}(\partial \ln {q_{zip}}/\partial s) \) : \[ \theta_H = \dfrac{\sigma s}{(s-1)^3} \left( \dfrac{ns^{n+2}-(n+2)s^{n+1}+(n+2)s - n}{n \{ 1+\left( \sigma s/(s-1)^2 \right) \left( s^{n+1} + n - (n+1)s \right) \}} \right) \nonumber \] Multiple stretches Expressions for the full partition function of chains with length n, eq. (18.1.15), can be evaluated for one-dimensional models that account for nearest neighbor interactions (Ising model) using an approach based on a statistical weight matrix, . You can show that the Zimm–Bragg partition function can be written as a product of matrices of the form \[ \begin{aligned} q_{conf} (n) &= \begin{pmatrix} 1 &0 \end{pmatrix} \bf{M}^n \begin{pmatrix} 1 \\ 1 \end{pmatrix} \\ \bf{M} &= \begin{pmatrix} 1&\sigma s \\ 1&s \end{pmatrix} \end{aligned} \] Each matrix represents possible configurations of two adjoining partners, and raised to the nth power gives all configurations for a chain of length n. This form also indicates that we can obtain a closed form for q from the eigenvalues of raised to the n power. If T is the transformation that diagonalizes , = , then = . This approach allows us to write \[ q_{conf} = \underset{ \sim }{\lambda}^{-1} \left( \lambda^{n+1}_+ (1-\lambda_-)-\lambda^{n+1}_-(1-\lambda_+)\right) \nonumber \] \(\mathrm{with} \begin{aligned} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad &\lambda_{\pm} = \frac{1}{2} \left( (1-s) \pm \underset{ \sim }{\lambda} \right) \\ &\underset{ \sim }{\lambda} = \lambda_+ - \lambda_- = \left( (1-s)^2+4\sigma s\right)^{-1/2} \end{aligned} \) and the fractional helicity is obtained from \[\theta_H = \dfrac{\langle n_H \rangle}{n}= \dfrac{s}{n} \dfrac{\partial \ln{q_{conf}}}{\partial s} \] Simplifying these expressions for the limit of long chains \( (n \rightarrow \infty , \lambda^{n+1}_+ \gg \lambda_-^{n+1} ) \), one finds \[ q_{conf} \approx \left( \dfrac{1+s+\underset{ \sim }{\lambda}}{2} \right)^n \nonumber \] and \[ \theta_H = s \left( \dfrac{1+\frac{1}{\underset{ \sim }{\lambda}}(s-1+2\sigma }{1+s+\underset{ \sim }{\lambda}} \right) \] Note that when you set σ =1, you recover the noncooperative expression, eq. (18.1.11). When s→1, θ →0.5. Below, we examine the transition behavior in the large n limit from eq. (18.1.20) as a function of the cooperativity parameter σ. We note that a sharp transition between an ensemble that is mostly coil to one that is mostly helix occurs near s = 1, the point where these states exist with equal probability. When the \(C\rightleftharpoons H\) equilibrium shifts slightly to favor H (s slightly greater than 1), most of the sample quickly converts to helical form. When the equilibrium shifts slightly toward C, most of the sample follows. As σ decreases, the steepness of this transition grows as \( (d\theta / ds)_{s=1}=1/4\sigma^{1/2}\). Therefore, we conclude that highly cooperative transitions will have s ≈ 1 and σ≪ s. In practice for polypeptides, we find that σ/s lies between 5×10 and 5×10 . Next, we explore the chain-length dependence for finite chains. We find that the cooperativity of this transition, observed through the steepness of the curve at θ = 0.5 increases with n. We also observe that the observed midpoint (θ = 0.5) lies at s > 1, where the single linkage equilibrium favors the H form. This reflects the constraints on the length of helical stretches available a given chain. Now let’s describe the temperature dependence of the cooperative model. The helix–coil transition shows a cooperative melting transition, where heating the sample a few degrees causes a dramatic change from a sample that is primarily in the C form to one that is primarily H. Multiple temperature-dependent factors make this a bit difficult to deal with analytically, therefore we focus on the behavior at the melting temperature T , which we define as the point where θ (T ) = 0.5. Look at the slope of θ at T . From chain rule: \[ \dfrac{d\theta }{dT} = \dfrac{d\theta}{ds} \cdot \dfrac{ds}{dT} = \dfrac{d\theta}{ds} \cdot s\dfrac{d\ln{s}}{dT} \nonumber \] Since we interpret s as an equilibrium constant for the addition of one helical residue to a stretch, we can write a van’t Hoff relation \[ \dfrac{d\ln{s}}{dT} = \dfrac{\Delta H^0_{HC}}{k_BT^2} \nonumber \] Note that this relation assumes that ΔH is independent of temperature, which generally is a concern, but we will not worry too much since we are just evaluating this at T . Next we focus our discussion on the high n limit. From the Zimm–Bragg model: \[ \left( \dfrac{d\theta}{ds} \right)_{s=1} = \dfrac{1}{4\sigma^{1/2}} \nonumber \] Then, we set s(T ) = 1, and combine these results to give the slope of the melting curve at T : \[ \left( \dfrac{d\theta }{dT} \right)_{T=T_m} = \dfrac{\Delta H^0_{HC}}{4\sigma^{1/2}k_BT^2_m} \nonumber \] The slope of \(\theta \mathrm{at} T_m\) has units of inverse temperature, so we can also express this as a transition width: \(\Delta T_m = (d\theta / dT)^{-1}_{T_m}\). Keep in mind this van’t Hoff analysis comes with some real limitations when applied to experimental data. It does not account for the finite size of the system, which we have seen shifts s(T ) to be >1, and the knowledge of parameters at T does not necessarily translate to other temperatures. To the extent that you can apply the assumptions, the van’t Hoff expression can also be used to predict the helical fraction as a function of temperature in the vicinity of T using \(\ln{s} = \dfrac{\Delta H^0_{HC}}{k_B}\left( \dfrac{1}{T_M}-\dfrac{1}{T} \right) \) and assuming that σ is independent of temperature. Below we show the length dependence of the melting temperature. As the length of the chain approaches infinite, the helix/coil transition becomes a step function in temperature. This trend matches the expectations for a phase transition: in the thermodynamic limit, the infinite system, will show discontinuous behavior. For finite lengths, the melting temperature Tm is lower that for the infinite chain (T ,∞), but approaches this value for n>300. Side-chain only has a small effect on the helix–coil propagation parameter: \(\Delta H_{HC}^0\) (kcal mol residue ) Alanine-rich peptides Ac-Y(AEAAKA)8F-NH2 Ac-(AAKAA)kY-NH2 Poly(L-lysine) Poly(L-glutamate) Finally, we investigate the free-energy landscape for the Zimm–Bragg model of the helix–coil transition. The figure below shows the helical probability distribution and corresponding energy landscape for different values of the reduced temperature kBT/Δε for a chain length of n=40 and σ=10-3. Note that P(nH) is calculated from eq. (18.1.18) for all but the all-coil state, which comes from eq. (18.1.16). The cooperative model shows two-state behavior. At low temperature and high temperature, the system is almost entirely in the all-helix or all-coil configuration, respectively; however, at intermediate temperatures, the distribution of helical configurations can be very broad. The least probable configuration is a chain with only one helical segment. This behavior looks much closer to the two-state behavior expected from phase-transition behavior. The free energy has minima for n = 0 and for n > 1, and the free energy difference between these states shifts with temperature to favor one or the other minimum. ___________________________________________________________________ | 19,665 | 2,883 |
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Natural diamonds are mined from sites around the world. However, the price of natural diamonds is carefully controlled, so other sources of diamonds are being explored. Several different methods for producing synthetic diamonds are available, usually involving treating carbon at very high temperatures and pressures. The diamonds produced are now of high quality, but are primarily used in industrial applications. Diamonds are one of the hardest materials available and are widely used for cutting and grinding tools. An application of Hess's law allows us to use standard heats of formation to indirectly calculate the heat of reaction for any reaction that occurs at standard conditions. An enthalpy change that occurs specifically under standard conditions is called the and is given the symbol \(\Delta H^\text{o}\). The standard heat of reaction can be calculated by using the following equation. \[\Delta H^\text{o} = \sum n \Delta H^\text{o}_\text{f} \: \text{(products)} - \sum n \Delta H^\text{o}_\text{f} \: \text{(reactants)}\nonumber \] The symbol \(\Sigma\) is the Greek letter sigma and means "the sum of". The standard heat of reaction is equal to the sum of all the standard heats of formation of the products minus the sum of all the standard heats of formation of the reactants. The symbol "\(n\)" signifies that each heat of formation must first be multiplied by its coefficient in the balanced equation. Calculate the standard heat of reaction \(\left( \Delta H^\text{o} \right)\) for the reaction of nitrogen monoxide gas with oxygen to form nitrogen dioxide gas. First write the balanced equation for the reaction. Then apply the equation to calculate the standard heat of reaction from the standard heats of formation. The balanced equation is: \(2 \ce{NO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)\) Applying the equation from the text: \[\begin{align*} \Delta H^\text{o} &= \left[ 2 \: \text{mol} \: \ce{NO_2} \left( 33.85 \: \text{kJ/mol} \right) \right] - \left[ 2 \: \text{mol} \: \ce{NO} \left( 90.4 \: \text{kJ/mol} \right) + 1 \: \text{mol} \: \ce{O_2} \left( 0 \: \text{kJ/mol} \right) \right] \\ &= -113 \: \text{kJ} \end{align*}\nonumber \] The standard heat of reaction is \(-113 \: \text{kJ}\nonumber \] The reaction is exothermic, which makes sense because it is a combustion reaction and combustion reactions always release heat. | 2,436 | 2,885 |
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The development of the atomic theory owes much to the work of two men: Antoine Lavoisier, who did not himself think of matter in terms of atoms but whose work laid organization groundwork for thinking about elements, and John Dalton, to whom the atomic theory is attributed. Much of Lavoisier’s work as a chemist was devoted to the study of combustion. He became convinced that when a substance is burned in air, it combines with some component of the air. Eventually he realized that this component was the which had been discovered by Joseph Priestly (1733 to 1804) a few years earlier. Lavoisier renamed this substance . In an important series of experiments he showed that when mercury is heated in oxygen at a moderate temperature, a red substance, , is obtained. (A is the ash left when a substance burns in air.) At a higher temperature this calx decomposes into mercury and oxygen. Lavoisier’s careful experiments also revealed that the combined masses of mercury and oxygen were exactly equal to the mass of calx of mercury. That is, there was no change in mass upon formation or decomposition of the calx. Lavoisier hypothesized that this should be true of all chemical changes, and further experiments showed that he was right. This principle is now called the . As Lavoisier continued his experiments with oxygen, he noticed something else. Although oxygen combined with many other substances, it never behaved as though it were itself a combination of other substances. Lavoisier was able to decompose the red calx into mercury and oxygen, but he could find no way to break down oxygen into two or more new substances. Because of this he suggested that oxygen must be an —an ultimately simple substance which could not be decomposed by chemical changes. Lavoisier did not originate the idea that certain substances (elements) were fundamental and all others could be derived from them. This had first been proposed in Greece during the fifth century B.C. by Empedocles, who speculated that all matter consisted of combinations of earth, air, fire, and water. These ideas were further developed and taught by Aristotle and remained influential for 2000 years. Lavoisier did produce the first table of the elements which contained a large number of substances that modern chemists would agree should be classifies as elements. He published it with the knowledge that further research might succeed decomposing some of the substances listed, thus showing them not to be elements. One of his objectives was to prod his contemporaries into just that kind of research. Sure enough the “earth substances” listed at the bottom were eventually shown to be combinations of certain metals with oxygen. It is also interesting to note that not even Lavoisier could entirely escape from Aristotle’s influence. The second element in his list is Aristotle’s “fire,” which Lavoisier called “caloric,” and which we now call “heat.” Both heat and light, the first two items in the table, are now regarded as forms of energy rather than of matter. Although his table of elements was incomplete, and even incorrect in some instances, Lavoisier’s work represented a major step forward. By classifying certain substances as elements, he stimulated much additional chemical research and brought order and structure to the subject where none had existed before. His contemporaries accepted his ideas very readily, and he became known as the father of chemistry. John Dalton (1766 to 1844) was a generation younger than Lavoisier and different from him in almost every respect. Dalton came from a working class family and only attended elementary school. Apart from this, he was entirely self-taught. Even after he became famous, he never aspired beyond a modest bachelor’s existence in which he supported himself by teaching mathematics to private pupils. Dalton made many contributions to science, and he seems not to have realized that his atomic theory was the most important of them. In his “New System of Chemical Philosophy” published in 1808, only the last seven pages out of a total of 168 are devoted to it! The postulates of the atomic theory are given below. The first is no advance on the ancient Greek philosopher Democritus who had theorized almost 2000 years earlier that matter consists of very small particles. The second postulate, however, shows the mark of an original genius; here Dalton links the idea of to the idea of . Lavoisier’s criterion for an element had been essentially a macroscopic, experimental one. If a substance could not be decomposed chemically, then it was probably an element. By contrast, Dalton defines an element in theoretical, sub-microscopic terms. . Different elements have different atoms. There are just as many different kinds of elements as there are different kinds of atoms. Now look back a moment to the physical states of mercury, where sub-microscopic pictures of solid, liquid, and gaseous mercury were given. Applying Dalton’s second postulate to this figure, you can immediately conclude that mercury is an element, because only one kind of atom appears. Although mercury atoms are drawn as spheres in the figure, it would be more common today to represent them using chemical symbols. The chemical symbol for an element (or an atom of that element) is a one- or two-letter abbreviation of its name. Usually, but not always, the first two letters are used. To complicate matters further, chemical symbols are sometimes derived from a language other than English. For example the symbol for Hg for mercury comes from the first and seventh letters of the element’s Latin name, The chemical symbols for all the currently known elements are listed above in the table, which also includes atomic weights. These symbols are the basic vocabulary of chemistry because the atoms they represent make up all matter. You will see symbols for the more important elements over and over again, and the sooner you know what element they stand for, the easier it will be for you to learn chemistry. These more important element have been indicated in the above table by colored shading around their names. Dalton’s fourth postulate states that atoms may combine to form molecules. An example of this is provided by bromine, the only element other than mercury which is a liquid at ordinary room temperature (20°C). Macroscopically, bromine consists of dark-colored crystals below –7.2°C and a reddish brown gas above 58.8°C. The liquid is dark red-brown and has a pungent odor similar to the chlorine used in swimming pools. It can cause severe burns on human skin and should not be handled without the protection of rubber gloves. Figure \(\Page {1}\) Sub-microscopic view of the diatomic molecules of the element bromine (a) in the gaseous state (above 58°C); (b) in liquid form (between -7.2 and 58.8°C); and (c) in solid form (below -7.2°C). The sub-microscopic view of bromine in the following figure is in agreement with its designation as an element—only one kind of atom is present. Except at very high temperatures, though, bromine atoms always double up. Whether in solid, liquid, or gas, they go around in pairs. Such a tightly held combination of two or more atoms is called a . The composition of a molecule is indicated by a . A subscript to the right of the symbol for each element tells how many atoms of that element are in the molecule. For example, the atomic weights table gives the chemical Br for bromine, but each molecule contains two bromine atoms, and so the chemical is Br . According to Dalton’s fourth postulate, atoms combine in the ratio of small whole numbers, and so the subscripts in a formula should be small whole numbers. | 7,722 | 2,886 |
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A setup (Figure 1.58) allows for liquid to boil and condense, with the condensed liquid returning to the original flask. A reflux setup is analogous to a distillation, with the main difference being the vertical placement of the condenser. The liquid remains at the boiling point of the solvent (or solution) during active reflux. A reflux apparatus allows for facile heating of a solution, but without the loss of solvent that would result from heating in an open vessel. In a reflux setup, solvent vapors are trapped by the condenser, and the concentration of reactants remains constant throughout the process. The main purpose of refluxing a solution is to heat a solution in a controlled manner at a constant temperature. For example, imagine that you want to heat a solution to \(60^\text{o} \text{C}\) for one hour in order to conduct a chemical reaction. It would be difficult to maintain a warm water bath at \(60^\text{o} \text{C}\) without special equipment, and it would require regular monitoring. However, if methanol was the solvent, the solution could be heated to reflux, and it would maintain its temperature without regular maintenance at the boiling point of methanol \(\left( 65^\text{o} \text{C} \right)\). True, \(65^\text{o} \text{C}\) is not \(60^\text{o} \text{C}\) and if the specific temperature were crucial to the reaction, then specialized heating equipment would be necessary. But often the boiling point of the solvent is chosen as the reaction temperature because of its practicality.
Do not turn off the water flowing through the condenser until the solution is only warm to the touch. After a few minutes of air cooling, the round bottomed flask can be immersed in a tap water bath to accelerate the cooling process (Figure 1.65b). Pour liquid into the flask along with a stir bar or boiling stones. Use an extension clamp on the round bottomed flask to connect to the ring stand or latticework. Attach the condenser, and connect the hoses so that water travels against gravity (cooling water comes into the bottom and drains out the top). Be sure there is a secure connection between the round bottomed flask and condenser, as vapors escaping this joint have the potential to catch on fire. Circulate water through the condenser, then begin heating the flask (by using a heating mantle, sand, water, or oil bath). Use an adjustable platform so the heat can be lowered and removed at the end of the reflux, or if something unexpected occurs. Heat so that the "reflux ring" is seen in the lower third of the condenser. Turn down the heat if the refluxing vapors reach higher than halfway up the condenser. At the end of the reflux period, lower the heat source from the flask or raise the apparatus. Keep circulating water in the condenser until the flask is just warm to the touch. After air cooling somewhat, the flask can be quickly cooled by immersing in a container of tap water. | 2,936 | 2,887 |
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The development of the atomic theory owes much to the work of two men: Antoine Lavoisier, who did not himself think of matter in terms of atoms but whose work laid organization groundwork for thinking about elements, and John Dalton, to whom the atomic theory is attributed. Much of Lavoisier’s work as a chemist was devoted to the study of combustion. He became convinced that when a substance is burned in air, it combines with some component of the air. Eventually he realized that this component was the which had been discovered by Joseph Priestly (1733 to 1804) a few years earlier. Lavoisier renamed this substance . In an important series of experiments he showed that when mercury is heated in oxygen at a moderate temperature, a red substance, , is obtained. (A is the ash left when a substance burns in air.) At a higher temperature this calx decomposes into mercury and oxygen. Lavoisier’s careful experiments also revealed that the combined masses of mercury and oxygen were exactly equal to the mass of calx of mercury. That is, there was no change in mass upon formation or decomposition of the calx. Lavoisier hypothesized that this should be true of all chemical changes, and further experiments showed that he was right. This principle is now called the . As Lavoisier continued his experiments with oxygen, he noticed something else. Although oxygen combined with many other substances, it never behaved as though it were itself a combination of other substances. Lavoisier was able to decompose the red calx into mercury and oxygen, but he could find no way to break down oxygen into two or more new substances. Because of this he suggested that oxygen must be an —an ultimately simple substance which could not be decomposed by chemical changes. Lavoisier did not originate the idea that certain substances (elements) were fundamental and all others could be derived from them. This had first been proposed in Greece during the fifth century B.C. by Empedocles, who speculated that all matter consisted of combinations of earth, air, fire, and water. These ideas were further developed and taught by Aristotle and remained influential for 2000 years. Lavoisier did produce the first table of the elements which contained a large number of substances that modern chemists would agree should be classifies as elements. He published it with the knowledge that further research might succeed decomposing some of the substances listed, thus showing them not to be elements. One of his objectives was to prod his contemporaries into just that kind of research. Sure enough the “earth substances” listed at the bottom were eventually shown to be combinations of certain metals with oxygen. It is also interesting to note that not even Lavoisier could entirely escape from Aristotle’s influence. The second element in his list is Aristotle’s “fire,” which Lavoisier called “caloric,” and which we now call “heat.” Both heat and light, the first two items in the table, are now regarded as forms of energy rather than of matter. Although his table of elements was incomplete, and even incorrect in some instances, Lavoisier’s work represented a major step forward. By classifying certain substances as elements, he stimulated much additional chemical research and brought order and structure to the subject where none had existed before. His contemporaries accepted his ideas very readily, and he became known as the father of chemistry. John Dalton (1766 to 1844) was a generation younger than Lavoisier and different from him in almost every respect. Dalton came from a working class family and only attended elementary school. Apart from this, he was entirely self-taught. Even after he became famous, he never aspired beyond a modest bachelor’s existence in which he supported himself by teaching mathematics to private pupils. Dalton made many contributions to science, and he seems not to have realized that his atomic theory was the most important of them. In his “New System of Chemical Philosophy” published in 1808, only the last seven pages out of a total of 168 are devoted to it! The postulates of the atomic theory are given below. The first is no advance on the ancient Greek philosopher Democritus who had theorized almost 2000 years earlier that matter consists of very small particles. The second postulate, however, shows the mark of an original genius; here Dalton links the idea of to the idea of . Lavoisier’s criterion for an element had been essentially a macroscopic, experimental one. If a substance could not be decomposed chemically, then it was probably an element. By contrast, Dalton defines an element in theoretical, sub-microscopic terms. . Different elements have different atoms. There are just as many different kinds of elements as there are different kinds of atoms. Now look back a moment to the physical states of mercury, where sub-microscopic pictures of solid, liquid, and gaseous mercury were given. Applying Dalton’s second postulate to this figure, you can immediately conclude that mercury is an element, because only one kind of atom appears. Although mercury atoms are drawn as spheres in the figure, it would be more common today to represent them using chemical symbols. The chemical symbol for an element (or an atom of that element) is a one- or two-letter abbreviation of its name. Usually, but not always, the first two letters are used. To complicate matters further, chemical symbols are sometimes derived from a language other than English. For example the symbol for Hg for mercury comes from the first and seventh letters of the element’s Latin name, The chemical symbols for all the currently known elements are listed above in the table, which also includes atomic weights. These symbols are the basic vocabulary of chemistry because the atoms they represent make up all matter. You will see symbols for the more important elements over and over again, and the sooner you know what element they stand for, the easier it will be for you to learn chemistry. These more important element have been indicated in the above table by colored shading around their names. Dalton’s fourth postulate states that atoms may combine to form molecules. An example of this is provided by bromine, the only element other than mercury which is a liquid at ordinary room temperature (20°C). Macroscopically, bromine consists of dark-colored crystals below –7.2°C and a reddish brown gas above 58.8°C. The liquid is dark red-brown and has a pungent odor similar to the chlorine used in swimming pools. It can cause severe burns on human skin and should not be handled without the protection of rubber gloves. Figure \(\Page {1}\) Sub-microscopic view of the diatomic molecules of the element bromine (a) in the gaseous state (above 58°C); (b) in liquid form (between -7.2 and 58.8°C); and (c) in solid form (below -7.2°C). The sub-microscopic view of bromine in the following figure is in agreement with its designation as an element—only one kind of atom is present. Except at very high temperatures, though, bromine atoms always double up. Whether in solid, liquid, or gas, they go around in pairs. Such a tightly held combination of two or more atoms is called a . The composition of a molecule is indicated by a . A subscript to the right of the symbol for each element tells how many atoms of that element are in the molecule. For example, the atomic weights table gives the chemical Br for bromine, but each molecule contains two bromine atoms, and so the chemical is Br . According to Dalton’s fourth postulate, atoms combine in the ratio of small whole numbers, and so the subscripts in a formula should be small whole numbers. | 7,722 | 2,888 |
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Before we consider the enzymatically controlled reactions of dioxygen in living systems, it is instructive to consider the uncontrolled and deleterious reactions that must also occur in aerobic organisms. Life originally appeared on Earth at a time when the atmosphere contained only low concentrations of dioxygen, and was reducing rather than oxidizing, as it is today. With the appearance of photosynthetic organisms approximately 2.5 billion years ago, however, the conversion to an aerobic, oxidizing atmosphere exposed the existing anaerobic organisms to a gradually increasing level of oxidative stress. Modern-day anaerobic bacteria, the descendants of the original primitive anaerobic organisms, evolved in ways that enabled them to avoid contact with normal atmospheric concentrations of dioxygen. Modern-day aerobic organisms, by contrast, evolved by developing aerobic metabolism to harness the oxidizing power of dioxygen and thus to obtain usable metabolic energy. This remarkably successful adaptation enabled life to survive and flourish as the atmosphere became aerobic, and also allowed larger, multicellular organisms to evolve. An important aspect of dioxygen chemistry that enabled the development of aerobic metabolism is the relatively slow rate of dioxygen reactions in the absence of catalysts. Thus, enzymes could be used to direct and control the oxidation of substrates either for energy generation or for biosynthesis. Nevertheless, the balance achieved between constructive and destructive oxidation is a delicate one, maintained in aerobic organisms by several means, e.g.: compartmentalization of oxidative reactions in mitochondria, peroxisomes, and chloroplasts; scavenging or detoxification of toxic byproducts of dioxygen reactions; repair of some types of oxidatively damaged species; and degradation and replacement of other species. The classification "anaerobic" actually includes organisms with varying degrees of tolerance for dioxygen: strict anaerobes, for which even small concentrations of O are toxic; moderate anaerobes, which can tolerate low levels of dioxygen; and microaerophiles, which require low concentrations of O for growth, but cannot tolerate normal atmospheric concentrations, i.e., 21 percent O , 1 atm pressure. Anaerobic organisms thrive in places protected from the atmosphere, for example, in rotting organic material, decaying teeth, the colon, and gangrenous wounds. Dioxygen appears to be toxic to anaerobic organisms largely because it depletes the reducing equivalents in the cell that are needed for normal biosynthetic reactions. Aerobic organisms can, of course, live in environments in which they are exposed to normal atmospheric concentrations of O . Nevertheless, there is much evidence that O is toxic to these organisms as well. For example, plants grown in varying concentrations of O have been observed to grow faster in lower than normal concentrations of O . grown under 5 atm of O ceased to grow unless the growth medium was supplemented with branched-chain amino acids or precursors. High concentrations of O damaged the enzyme dihydroxy acid dehydratase, an important component in the biosynthetic pathway for those amino acids. In mammals, elevated levels of O are clearly toxic, leading first to coughing and soreness of the throat, and then to convulsions when the level of 5 atm of 100 percent O is reached. Eventually, elevated concentrations of O lead to pulmonary edema and irreversible lung damage, with obvious damage to other tissues as well. The effects of high concentrations of O on humans is of some medical interest, since dioxygen is used therapeutically for patients experiencing difficulty breathing, or for those suffering from infection by anaerobic organisms. The major biochemical targets of O toxicity appear to be lipids, DNA, and proteins. The chemical reactions accounting for the damage to each type of target are probably different, not only because of the different reactivities of these three classes of molecules, but also because of the different environment for each one inside the cell. Lipids, for example, are essential components of membranes and are extremely hydrophobic. The oxidative damage that is observed is due to free-radical autoxidation (see Reactions 5.16 to 5.21), and the products observed are lipid hydroperoxides (see Reaction 5.23). The introduction of the hydroperoxide group into the interior of the lipid bilayer apparently causes that structure to be disrupted, as the configuration of the lipid rearranges in order to bring that polar group out of the hydrophobic membrane interior and up to the membrane-water interface. DNA, by contrast, is in the interior of the cell, and its exposed portions are surrounded by an aqueous medium. It is particularly vulnerable to oxidative attack at the base or at the sugar, and multiple products are formed when samples are exposed to oxidants . Since oxidation of DNA may lead to mutations, this type of damage is potentially very serious. Proteins also suffer oxidative damage, with amino-acid side chains, particularly the sulfur-containing residues cysteine and methionine, appearing to be the most vulnerable sites. The biological defense systems protecting against oxidative damage and its consequences are summarized below. Some examples of small-molecule antioxidants are \(\alpha\)-tocopherol (vitamin E; 5.24), which is found dissolved in cell membranes and protects them against lipid peroxidation, and ascorbate (vitamin C; 5.25) and glutathione (5.26), which are found in the cytosol of many cells. Several others are known as well. \(\tag{5.24}\) \(\tag{5.25}\) \(\tag{5.26}\) The enzymatic antioxidants are (a) catalase and the various peroxidases, whose presence lowers the concentration of hydrogen peroxide, thereby preventing it from entering into potentially damaging reactions with various cell components (see Section VI and Reactions 5.82 and 5.83), and (b) the superoxide dismutases, whose presence provides protection against dioxygen toxicity that is believed to be mediated by the superoxide anion, O (see Section VII and Reaction 5.95). Some of the enzymatic and nonenzymatic antioxidants in the cell are illustrated in Figure 5.1. Redox-active metal ions are present in the cell in their free, uncomplexed state only in extremely low concentrations. They are instead sequestered by metal-ion storage and transport proteins, such as ferritin and transferrin for iron (see Chapter 1) and ceruloplasmin for copper. This arrangement prevents such metal ions from catalyzing deleterious oxidative reactions, but makes them available for incorporation into metalloenzymes as they are needed. In vitro experiments have shown quite clearly that redox-active metal ions such as Fe or Cu are extremely good catalysts for oxidation of sulfhydryl groups by O (Reaction 5.27). \[4RSH + O_{2} \xrightarrow{M^{n+}} 2RSSR + 2H_{2}O \tag{5.27}\] In addition, in the reducing environment of the cell, redox-active metal ions catalyze a very efficient one-electron reduction of hydrogen peroxide to produce hydroxyl radical, one of the most potent and reactive oxidants known (Reactions 5.28 to 5.30). \[M^{n+} + Red^{-} \rightarrow M^{(n-1)+} + Red \tag{5.28}\] \[M^{(n-1)+} + H_{2}O_{2} \rightarrow M^{n+} + OH^{-} + HO \cdotp \tag{5.29}\] \[Red^{-} + H_{2}O_{2} \rightarrow Red + OH^{-} + HO \cdotp \tag{5.30}\] \[(Red^{-} = reducing\; agent)\] Binding those metal ions in a metalloprotein usually prevents them from entering into these types of reactions. For example, transferrin, the iron-transport enzyme in serum, is normally only 30 percent saturated with iron. Under conditions of increasing iron overload, the empty iron-binding sites on transferrin are observed to fill, and symptoms of iron poisoning are not observed until after transferrin has been totally saturated with iron. Ceruloplasmin and metallothionein may playa similar role in preventing copper toxicity. It is very likely that both iron and copper toxicity are largely due to catalysis of oxidation reactions by those metal ions. Repair of oxidative damage must go on constantly, even under normal conditions of aerobic metabolism. For lipids, repair of peroxidized fatty-acid chains is catalyzed by phospholipase A , which recognizes the structural changes at the lipid-water interface caused by the fatty-acid hydroperoxide, and catalyzes removal of the fatty acid at that site. The repair is then completed by enzymatic reacylation. Although some oxidatively damaged proteins are repaired, more commonly such proteins are recognized, degraded at accelerated rates, and then replaced. For DNA, several multi-enzyme systems exist whose function is to repair oxidatively damaged DNA. For example, one such system catalyzes recognition and removal of damaged bases, removal of the damaged part of the strand, synthesis of new DNA to fill in the gaps, and religation to restore the DNA to its original, undamaged state. Mutant organisms that lack these repair enzymes are found to be hypersensitive to O , H O , or other oxidants. One particularly interesting aspect of oxidant stress is that most aerobic organisms can survive in the presence of normally lethal levels of oxidants if they have first been exposed to lower, nontoxic levels of oxidants. This phenomenon has been observed in animals, plants, yeast, and bacteria, and suggests that low levels of oxidants cause antioxidant systems to be induced . In certain bacteria, the mechanism of this induction is at least partially understood. A DNA-binding regulatory protein named OxyR that exists in two redox states has been identified in these systems. Increased oxidant stress presumably increases concentration of the oxidized form, which then acts to turn on the transcription of the genes for some of the antioxidant enzymes. A related phenomenon may occur when bacteria and yeast switch from anaerobic to aerobic metabolism. When dioxygen is absent, these microorganisms live by fermentation, and do not waste energy by synthesizing the enzymes and other proteins needed for aerobic metabolism. However, when they are exposed to dioxygen, the synthesis of the respiratory apparatus is turned on. The details of this induction are not known completely, but some steps at least depend on the presence of heme, the prosthetic group of hemoglobin and other heme proteins, whose synthesis requires the presence of dioxygen. What has been left out of the preceding discussion is the identification of the species responsible for oxidative damage, i.e., the agents that directly attack the various vulnerable targets in the cell. They were left out because the details of the chemistry responsible for dioxygen toxicity are largely unknown. In 1954, Rebeca Gerschman formulated the "free-radical theory of oxygen toxicity" after noting that tissues subjected to ionizing radiation resemble those exposed to elevated levels of dioxygen. Fourteen years later, Irwin Fridovich proposed that the free radical responsible for dioxygen toxicity was superoxide, O , based on his identification of the first of the superoxide dismutase enzymes. Today it is still not known if superoxide is the principal agent of dioxygen toxicity, and, if so, what the chemistry responsible for that toxicity is. There is no question that superoxide is formed during the normal course of aerobic metabolism, although it is difficult to obtain estimates of the amount under varying conditions, because, even in the absence of a catalyst, superoxide disproportionates quite rapidly to dioxygen and hydrogen peroxide (Reaction 5.4) and therefore never accumulates to any great extent in the cell under normal conditions of pH. One major problem in this area is that a satisfactory chemical explanation for the purported toxicity of superoxide has never been found, despite much indirect evidence from experiments that the presence of superoxide can lead to undesirable oxidation of various cell components and that such oxidation can be inhibited by superoxide dismutase. The mechanism most commonly proposed is production of hydroxyl radicals via Reactions (5.28) to (5.30) with Red = O , which is referred to as the "Metal-Catalyzed Haber-Weiss Reaction". The role of superoxide in this mechanism is to reduce oxidized metal ions, such as Cu or Fe , present in the cell in trace amounts, to a lower oxidation state. Hydroxyl radical is an extremely powerful and indiscriminate oxidant. It can abstract hydrogen atoms from organic substrates, and oxidize most reducing agents very rapidly. It is also a very effective initiator of free-radical autoxidation reactions (see Section II.C above). Therefore, reactions that produce hydroxyl radical in a living cell will probably be very deleterious. The problem with this explanation for superoxide toxicity is that the only role played by superoxide here is that of a reducing agent of trace metal ions. The interior of a cell is a highly reducing environment, however, and other reducing agents naturally present in the cell such as, for example, ascorbate anion can also act as Red in Reaction (5.28), and the resulting oxidation reactions due to hydroxyl radical are therefore no longer inhibitable by SOD. Other possible explanations for superoxide toxicity exist, of course, but none has ever been demonstrated experimentally. Superoxide might bind to a specific enzyme and inhibit it, much as cytochrome oxidase is inhibited by cyanide or hemoglobin by carbon monoxide. Certain enzymes may be extraordinarily sensitive to direct oxidation by superoxide, as has been suggested for the enzyme aconitase, an iron-sulfur enzyme that contains an exposed iron atom. Another possibility is that the protonated and therefore neutral form of superoxide, HO , dissolves in membranes and acts as an initiator of lipid peroxidation. It has also been suggested that superoxide may react with nitric oxide, NO, in the cell producing peroxynitrite, a very potent oxidant. One particularly appealing mechanism for superoxide toxicity that has gained favor in recent years is the "Site-Specific Haber-Weiss Mechanism." The idea here is that traces of redox-active metal ions such as copper and iron are bound to macromolecules under normal conditions in the cell. Most reducing agents in the cell are too bulky to come into close proximity to these sequestered metal ions. Superoxide, however, in addition to being an excellent reducing agent, is very small, and could penetrate to these metal ions and reduce them. The reduced metal ions could then react with hydrogen peroxide, generating hydroxyl radical, which would immediately attack at a site near the location of the bound metal ion. This mechanism is very similar to that of the metal complexes that cause DNA cleavage; by reacting with hydrogen peroxide while bound to DNA, they generate powerful oxidants that react with DNA with high efficiency because of their proximity to it (see Chapter 8). Although we are unsure what specific chemical reactions superoxide might undergo inside of the cell, there nevertheless does exist strong evidence that the superoxide dismutases play an important role in protection against dioxygen-induced damage. Mutant strains of bacteria and yeast that lack superoxide dismutases are killed by elevated concentrations of dioxygen that have no effect on the wild-type cells. This extreme sensitivity to dioxygen is alleviated when the gene coding for a superoxide dismutase is reinserted into the cell, even if the new SOD is of another type and from a different organism. In summary, we know a great deal about the sites that are vulnerable to oxidative damage in biological systems, about the agents that protect against such damage, and about the mechanisms that repair such damage. Metal ions are involved in all this chemistry, both as catalysts of deleterious oxidative reactions and as cofactors in the enzymes that protect against and repair such damage. What we still do not know at this time, however, is how dioxygen initiates the sequence of chemical reactions that produce the agents that attack the vulnerable biological targets | 16,226 | 2,889 |
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In a double bond, two electron pairs are shared between a pair of atomic nuclei. Despite the fact that the two electron pairs repel each other, they must remain between the nuclei, and so they cannot avoid each other. Therefore, for purposes of predicting molecular geometry, the two electron pairs in a double bond behave as one. They will, however, be somewhat “fatter” than a single electron-pair bond. For the same reason the three electron pairs in a triple bond behave as an “extra-fat” bond. As an example of the multiple-bond rules, consider hydrogen cyanide, HCN. The is Treating the triple bond as if it were a single “fat” electron pair, we predict a linear molecule with an H―C―H angle of 180°. This is confirmed experimentally. Another example is formaldehyde, CH O, whose Lewis structure is Since no lone pairs are present on C, the two H’s and the O should be arranged trigonally, with all four atoms in the same plane. Also, because of the “fatness” of the double bond, squeezing the C—H bond pairs together, we expect the H―C―H angle to be slightly less than 120°. Experimentally it is found to have the value of 117°. Predict the shape of the two molecules (a) nitrosyl chloride, NOCl, and (b) carbon dioxide, CO . We must first construct a skeleton structure and then a Lewis diagram. Since N has a valence of 3, O a valence of 2, and Cl is monovalent, a probable structure for NOCl is Completing the Lewis diagram, we find Since N has two bonds and one lone pair, the molecule must be . The O—N—Cl angle should be about 120°. Since the “fat” lone pair would act to reduce this angle while the “fat” double bond would tend to increase it, it is impossible to predict at this level of argument whether the angle will be slightly larger or smaller than 120°. The Lewis structure of CO was considered in the previous chapter and found to be Since C has no lone pairs in its valence shell and each double bond acts as a fat bond pair, we conclude that the two O atoms are separated by 180° and that the molecule is . | 2,054 | 2,890 |
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described catalysts as substances that increase the reaction rate of a chemical reaction without being consumed in the process. A catalyst, therefore, does not appear in the overall stoichiometry of the reaction it catalyzes, but it appear in at least one of the elementary reactions in the mechanism for the catalyzed reaction. The catalyzed pathway has a lower , but the change in energy that results from the reaction (the difference between the energy of the reactants and the energy of the products) is affected by the presence of a catalyst ( ). Nevertheless, because of its lower , the reaction rate of a catalyzed reaction is faster than the reaction rate of the uncatalyzed reaction at the same temperature. Because a catalyst decreases the height of the energy barrier, its presence increases the reaction rates of the forward the reverse reactions by the same amount. In this section, we will examine the three major classes of catalysts: heterogeneous catalysts, , and . A catalyst affects , not Δ . In heterogeneous catalysis , the catalyst is in a different phase from the reactants. At least one of the reactants interacts with the solid surface in a physical process called in such a way that a chemical bond in the reactant becomes weak and then breaks. are substances that bind irreversibly to catalysts, preventing reactants from adsorbing and thus reducing or destroying the catalyst’s efficiency. An example of heterogeneous catalysis is the interaction of hydrogen gas with the surface of a metal, such as Ni, Pd, or Pt. As shown in part (a) in , the hydrogen–hydrogen bonds break and produce individual adsorbed hydrogen atoms on the surface of the metal. Because the adsorbed atoms can move around on the surface, two hydrogen atoms can collide and form a molecule of hydrogen gas that can then leave the surface in the reverse process, called . Adsorbed H atoms on a metal surface are substantially more reactive than a hydrogen molecule. Because the relatively strong H–H bond (dissociation energy = 432 kJ/mol) has already been broken, the energy barrier for most reactions of H is substantially lower on the catalyst surface. shows a process called , in which hydrogen atoms are added to the double bond of an alkene, such as ethylene, to give a product that contains C–C single bonds, in this case ethane. Hydrogenation is used in the food industry to convert vegetable oils, which consist of long chains of alkenes, to more commercially valuable solid derivatives that contain alkyl chains. Hydrogenation of some of the double bonds in polyunsaturated vegetable oils, for example, produces margarine, a product with a melting point, texture, and other physical properties similar to those of butter. Several important examples of industrial heterogeneous catalytic reactions are in . Although the mechanisms of these reactions are considerably more complex than the simple hydrogenation reaction described here, they all involve adsorption of the reactants onto a solid catalytic surface, chemical reaction of the adsorbed species (sometimes via a number of intermediate species), and finally desorption of the products from the surface. In homogeneous catalysis , the catalyst is in the same phase as the reactant(s). The number of collisions between reactants and catalyst is at a maximum because the catalyst is uniformly dispersed throughout the reaction mixture. Many homogeneous catalysts in industry are transition metal compounds ( ), but recovering these expensive catalysts from solution has been a major challenge. As an added barrier to their widespread commercial use, many homogeneous catalysts can be used only at relatively low temperatures, and even then they tend to decompose slowly in solution. Despite these problems, a number of commercially viable processes have been developed in recent years. High-density polyethylene and polypropylene are produced by homogeneous catalysis. The scalding, foul-smelling spray emitted by this bombardier beetle is produced by the catalytic decomposition of H O . , catalysts that occur naturally in living organisms, are almost all protein molecules with typical molecular masses of 20,000–100,000 amu. Some are homogeneous catalysts that react in aqueous solution within a cellular compartment of an organism. Others are heterogeneous catalysts embedded within the membranes that separate cells and cellular compartments from their surroundings. The reactant in an enzyme-catalyzed reaction is called a substrate . Because enzymes can increase reaction rates by enormous factors (up to 10 times the uncatalyzed rate) and tend to be very specific, typically producing only a single product in quantitative yield, they are the focus of active research. At the same time, enzymes are usually expensive to obtain, they often cease functioning at temperatures greater than 37°C, have limited stability in solution, and have such high specificity that they are confined to turning one particular set of reactants into one particular product. This means that separate processes using different enzymes must be developed for chemically similar reactions, which is time-consuming and expensive. Thus far, enzymes have found only limited industrial applications, although they are used as ingredients in laundry detergents, contact lens cleaners, and meat tenderizers. The enzymes in these applications tend to be , which are able to cleave the amide bonds that hold amino acids together in proteins. Meat tenderizers, for example, contain a protease called papain, which is isolated from papaya juice. It cleaves some of the long, fibrous protein molecules that make inexpensive cuts of beef tough, producing a piece of meat that is more tender. Some insects, like the bombadier beetle, carry an enzyme capable of catalyzing the decomposition of hydrogen peroxide to water ( ). Enzyme inhibitors cause a decrease in the reaction rate of an enzyme-catalyzed reaction by binding to a specific portion of an enzyme and thus slowing or preventing a reaction from occurring. Irreversible inhibitors are therefore the equivalent of poisons in heterogeneous catalysis. One of the oldest and most widely used commercial enzyme inhibitors is aspirin, which selectively inhibits one of the enzymes involved in the synthesis of molecules that trigger inflammation. The design and synthesis of related molecules that are more effective, more selective, and less toxic than aspirin are important objectives of biomedical research. participate in a chemical reaction and increase its rate. They do not appear in the reaction’s net equation and are not consumed during the reaction. Catalysts allow a reaction to proceed via a pathway that has a lower activation energy than the uncatalyzed reaction. In , catalysts provide a surface to which reactants bind in a process of adsorption. In , catalysts are in the same phase as the reactants. are biological catalysts that produce large increases in reaction rates and tend to be specific for certain reactants and products. The reactant in an enzyme-catalyzed reaction is called a . cause a decrease in the reaction rate of an enzyme-catalyzed reaction. What effect does a catalyst have on the activation energy of a reaction? What effect does it have on the frequency factor ( )? What effect does it have on the change in potential energy for the reaction? How is it possible to affect the product distribution of a reaction by using a catalyst? A heterogeneous catalyst works by interacting with a reactant in a process called . What occurs during this process? Explain how this can lower the activation energy. What effect does increasing the surface area of a heterogeneous catalyst have on a reaction? Does increasing the surface area affect the activation energy? Explain your answer. Identify the differences between a heterogeneous catalyst and a homogeneous catalyst in terms of the following. An area of intensive chemical research involves the development of homogeneous catalysts, even though homogeneous catalysts generally have a number of operational difficulties. Propose one or two reasons why a homogenous catalyst may be preferred. Consider the following reaction between cerium(IV) and thallium(I) ions: This reaction is slow, but Mn catalyzes it, as shown in the following mechanism: In what way does Mn increase the reaction rate? The text identifies several factors that limit the industrial applications of enzymes. Still, there is keen interest in understanding how enzymes work for designing catalysts for industrial applications. Why? Most enzymes have an optimal pH range; however, care must be taken when determining pH effects on enzyme activity. A decrease in activity could be due to the effects of changes in pH on groups at the catalytic center or to the effects on groups located elsewhere in the enzyme. Both examples are observed in chymotrypsin, a digestive enzyme that is a protease that hydrolyzes polypeptide chains. Explain how a change in pH could affect the catalytic activity due to (a) effects at the catalytic center and (b) effects elsewhere in the enzyme. ( : remember that enzymes are composed of functional amino acids.) A catalyst lowers the activation energy of a reaction. Some catalysts can also orient the reactants and thereby increase the frequency factor. Catalysts have no effect on the change in potential energy for a reaction. In adsorption, a reactant binds tightly to a surface. Because intermolecular interactions between the surface and the reactant weaken or break bonds in the reactant, its reactivity is increased, and the activation energy for a reaction is often decreased. The Mn ion donates two electrons to Ce , one at a time, and then accepts two electrons from Tl . Because Mn can exist in three oxidation states separated by one electron, it is able to couple one-electron and two-electron transfer reactions. At some point during an enzymatic reaction, the concentration of the activated complex, called an enzyme–substrate complex (ES), and other intermediates involved in the reaction is nearly constant. When a single substrate is involved, the reaction can be represented by the following sequence of equations: \( \begin{matrix}
enzyme\;\left ( E \right ) + substrate\;\left ( S \right ) \rightleftharpoons\\
enzyme-substrate\;complex\;\left ( ES \right ) \rightleftharpoons\\
enzyme\;\left ( E \right ) + product\;\left ( P \right )
\end{matrix} \) This can also be shown as follows: \( E + S \xrightarrow[k_{-1}]{k_{1}} ES \xleftarrow[k_{-2}]{k_{2}} E + P\) Using molar concentrations and rate constants, write an expression for the rate of disappearance of the enzyme–substrate complex. Typically, enzyme concentrations are small, and substrate concentrations are high. If you were determining the rate law by varying the substrate concentrations under these conditions, what would be your apparent reaction order? A particular reaction was found to proceed via the following mechanism: What is the overall reaction? Is this reaction catalytic, and if so, what species is the catalyst? Identify the intermediates. A particular reaction has two accessible pathways (A and B), each of which favors conversion of to a different product ( and , respectively). Under uncatalyzed conditions pathway A is favored, but in the presence of a catalyst pathway B is favored. Pathway B is reversible, whereas pathway A is not. Which product is favored in the presence of a catalyst? without a catalyst? Draw a diagram illustrating what is occurring with and without the catalyst. The kinetics of an enzyme-catalyzed reaction can be analyzed by plotting the reaction rate versus the substrate concentration. This type of analysis is referred to as a Michaelis–Menten treatment. At low substrate concentrations, the plot shows behavior characteristic of first-order kinetics, but at very high substrate concentrations, the behavior shows zeroth-order kinetics. Explain this phenomenon. \( \dfrac{\Delta \left ( ES \right )}{\Delta t} =-\left ( k_{2}+k_{-1} \right ) \left [ES \right ] + k_{1} \left [ E \right ]\left [ S \right ] + k_{-2}\left [ E \right ]\left [ P \right ] \approx 0 \); zeroth order in substrate. In both cases, the product of pathway A is favored. All of the produced in the catalyzed reversible pathway B will eventually be converted to as is converted irreversibly to by pathway A. | 12,478 | 2,891 |
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On the page discussing , it is shown that the density of electrons in a covalent bond is shared between both atoms. When drawing Lewis Structures it is sometimes useful to see which structure can be deemed the best. The is a somewhat artificial device that exists in the minds of chemists (not within the molecules, themselves) to help keep track of electrons in their bonding configurations. The Here are some rules for determining the Formal Charge on each atom in a molecule or polyatomic ion: The Formal Charge is defined by the relationship: With the definitions above, we can calculate the Formal Charge on the thiocyanate Ion, SCN : Notice how the sum of all of the formal charges adds up to the charge of the thiocyanate ion (-1). When drawing Lewis Structures, we use this information to determine which structure would be the most likely. The following rules apply: We have also discussed electronegativity, which gives rise to polarity in bonds and molecules. Thus, sometimes it is helpful for us to define another somewhat artificial device - invented by chemists, not by molecules - which enables us to keep track of electrons in complicated reactions where electrons rearrange into new bonds. We can obtain oxidation numbers by arbitrarily assigning the electrons of each covalent bond to the more electronegative atom in the bond. This is in contrast to the Formal Charge which divides each bonding pair equally without concern for which atom may be more electronegative. When this division has been done for all bonds, the charge remaining on each atom is said to be its oxidation number. If two like atoms are joined, each atom is assigned half the bonding electrons. Determine the oxidation number of each atom in each of the following formulas: (a) Cl ; (b) CH ; (c) NaCl; (d) OF ; (e) H O . In each case we begin by drawing a Lewis diagram: In each Lewis diagram, electrons have been color coded to indicate the atom from which they came originally. The boxes enclose electrons assigned to a given atom by the rules for determining oxidation number. Since the bond in Cl is purely covalent and the electrons are shared equally, one electron from the bond is assigned to each Cl, giving the same number of valence electrons (7) as a neutral Cl atom. Thus neither atom has lost any electrons, and the oxidation number is 0. This is indicated by writing a 0 above the symbol for chlorine in the formula \[\overset{0}{\mathop{\text{Cl}}}\,_{\text{2}} \nonumber \] Since C is more electronegative than H, the pair of electrons in each C―H bond is assigned to C. Therefore each H has the one valence electron it originally had, giving an oxidation number of +1. The C atom has four electrons, giving it a negative charge and hence an oxidation number of – 4: \[\overset{-\text{4}}{\mathop{\text{C}}}\,\overset{\text{+1}}{\mathop{\text{H}}}\,_{\text{4}} \nonumber \] In NaCl each Na atom has lost an electron to form an Na ion, and each Cl atom has gained an electron to form Cl . The oxidation numbers therefore correspond to the ionic charges: \[\overset{\text{+1}}{\mathop{\text{Na}}}\,\overset{-\text{1}}{\mathop{\text{Cl}}}\, \nonumber \] Since F is more electronegative than O, the bonding pairs are assigned to F in oxygen difluoride (OF ). The O is left with four valence electrons, and each F has eight. The oxidation numbers are \[\overset{\text{+2}}{\mathop{\text{O}}}\,\overset{-\text{1}}{\mathop{\text{F}_{\text{2}}}}\, \nonumber \] In Hydrogen peroxide (H O ) the O—H bond pairs are assigned to the more electronegative O’s, but the O―O bond is purely covalent, and the electron pair is divided equally. This gives each O seven electrons, a gain of 1 over the neutral atom. The oxidation numbers are \[\overset{\text{+1}}{\mathop{\text{H}_{\text{2}}}}\,\overset{-\text{1}}{\mathop{\text{O}_{\text{2}}}}\, \nonumber \] Although one could always work out Lewis diagrams to obtain oxidation numbers as shown in Example \(\Page {1}\), it is often easier to use a few simple rules to obtain them. The rules summarize the properties of oxidation numbers illustrated in Example \(\Page {2}\). As an illustration of these rules, let us consider a few more examples. Determine the oxidation number of each element in each of the following formulas: (a) NaClO; (b) ClO ; and (c) MgH . Since Na is a group IA element, its oxidation number is +1 (rule 3 ). The oxidation number of O is usually –2 (rule 3 ). Therefore (rule 4), +1 + oxidation number of Cl + (–2) = 0. \[ \text{Oxidation number of Cl} = 2 – 1 = +1 \nonumber \] Thus we write the formula \[\overset{\text{+1}}{\mathop{\text{Na}}}\,\overset{+\text{1}}{\mathop{\text{Cl}}}\,\overset{-\text{2}}{\mathop{\text{O}}}\, \nonumber \] if oxidation numbers are to be included. In this case the oxidation numbers must add to –1, the charge on the polyatomic ion. Since O is usually –2, we have \[ \text{Oxidation number of Cl} + 4(–2) = –1 \nonumber \] \[ \text{Oxidation number of Cl} = –1 + 8 = +7 \nonumber \] In MgH , H is combined with an element more electropositive than itself, and so its oxidation number is –1. Mg is in group IIA,and so its oxidation number is +2: \[\overset{\text{+2}}{\mathop{\text{Mg}}}\,\overset{-\text{1}}{\mathop{\text{H}}}\,_{\text{2}} \nonumber \] As a check on these assignments, it is wise to make sure that the oxidation numbers sum to 0: \[ +2 + 2(–1) = 0 \qquad \text{OK} \nonumber \] Oxidation numbers are mainly used by chemists to identify and handle a type of chemical reaction called a , or an . This type of reaction can be recognized because it involves a of at least one element. More information on these reactions is found in the section on redox reactions. Oxidation numbers are also used in the names of compounds. The internationally recommended rules of nomenclature involve roman numerals which represent oxidation numbers. For example, the two bromides of mercury, Hg Br and HgBr , are called mercury(I) bromide and mercury(II) bromide, respectively. Here the numeral I refers to an oxidation number of +1 for mercury, and II to an oxidation number of +2. Oxidation numbers can sometimes also be useful in writing Lewis structures, particularly for oxyanions. In the sulfite ion, SO for example, the oxidation number of sulfur is +4, suggesting that only sulfur electrons are involved in the bonding. Since sulfur has six valence electrons, we conclude that two electrons are not involved in the bonding, i.e., that there is a pair. With this clue, a plausible Lewis structure is much easier to draw: | 6,567 | 2,892 |
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Many enzymes only operate in combination with organic molecules that are actually reagents for the reaction. These substances are called or . Some coenzymes function with more than one enzyme and are involved in reactions with a number of different substrates. Several of the B vitamins function as coenzymes or as precursors of coenzymes; some of these have been mentioned previously. Nicotinamide adenine dinucleotide \(\left( \ce{NAD}^\oplus \right)\) which, in conjunction with the enzyme , oxidizes ethanol to ethanal ( ), also is the oxidant in the citric acid cycle ( ). The precursor to \(\ce{NAD}^\oplus\) is the B vitamin niacin or nicotinic acid ( ). Riboflavin (vitamin B\(_2\)) is a precursor of flavin adenine nucleotide \(\left( \ce{FAD} \right)\), a coenzyme in redox processes rather like \(\ce{NAD}^\oplus\) ( ). Another example of a coenzyme is pyridoxal (vitamin B\(_6\)), mentioned in connection with the deamination and decarboxylation of amino acids ( ). Yet another is coenzyme A \(\left( \textbf{CoA} \ce{SH} \right)\), which is essential for metabolism and biosynthesis ( , , and ). An especially interesting coenzyme is thiamine pyrophosphate (vitamin B\(_1\)) which, in conjunction with the appropriate enzyme, decarboxylates 2-oxopropanoic acid (pyruvic acid; ). We can write the overall reaction as follows: Although we do not know just how thiamine binds to the enzyme, the essential features of the reaction are quite well understood. Thiamine has an acidic hydrogen at the 2-position of the azathiacyclopentadiene ring, and you should recognize that the conjugate base, \(10\), is both a nitrogen ylide and a sulfur ylide ( ): The acidity of the ring proton of the thiamine ring is a consequence of the adjacent positive nitrogen and the known ability of sulfur to stabilize an adjacent carbanion. Nucleophilic attack of the anionic carbon of \(10\) on \(\ce{C_2}\) of 2-oxopropanoic acid is followed by decarboxylation: The overall reaction introduces a two-carbon chain at the \(\ce{C_2}\) position of the thiamine ring and the resulting modified coenzyme, \(11\), functions in subsequent biological reactions as a carrier of a \(\ce{CH_3-CHOH}-\) group and a potential source of a \(\ce{CH_3CO}-\) group. The metabolism of glucose ( ) requires the conversion of pyruvate to ethanoyl by way of \(11\); and, in fermentation, the hydroxylethyl group of \(11\) is released as ethanal, which is reduced to ethanol by \(\ce{NADH}\) (see for discussion of the reverse reaction). Thiamine pyrophosphate also plays a key role in the biosynthetic reactions that build (or degrade) pentoses from hexoses. We have mentioned these reactions previously in connection with the Calvin cycle ( ) and the pentose-phosphate pathway ( ). and (1977) | 2,794 | 2,894 |
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The term oxidation was first used to describe reactions in which metals react with oxygen in air to produce metal oxides. When iron is exposed to air in the presence of water, for example, the iron turns to rust—an iron oxide. When exposed to air, aluminum metal develops a continuous, transparent layer of aluminum oxide on its surface. In both cases, the metal acquires a positive charge by transferring electrons to the neutral oxygen atoms of an oxygen molecule. As a result, the oxygen atoms acquire a negative charge and form oxide ions (O ). Because the metals have lost electrons to oxygen, they have been oxidized; oxidation is therefore the loss of electrons. Conversely, because the oxygen atoms have gained electrons, they have been reduced, so reduction is the gain of electrons. For every oxidation, there must be an associated reduction. Therefore, these reactions are known as oxidation-reduction reactions, or "redox" reactions for short. Any oxidation must be accompanied by a reduction and vice versa. Originally, the term reduction referred to the decrease in mass observed when a metal oxide was heated with carbon monoxide, a reaction that was widely used to extract metals from their ores. When solid copper(I) oxide is heated with hydrogen, for example, its mass decreases because the formation of pure copper is accompanied by the loss of oxygen atoms as a volatile product (water vapor). The reaction is as follows: \[ \ce{Cu_2O (s) + H_2 (g) \rightarrow 2Cu (s) + H_2O (g)} \label{4.4.1} \] Oxidation-reduction reactions are now defined as reactions that exhibit a change in the oxidation states of one or more elements in the reactants by a transfer of electrons, which follows the mnemonic "oxidation is loss, reduction is gain", or . The of each atom in a compound is the charge an atom would have if all its bonding electrons were transferred to the atom with the greater attraction for electrons. Atoms in their elemental form, such as O or H , are assigned an oxidation state of zero. For example, the reaction of aluminum with oxygen to produce aluminum oxide is \[\ce{ 4 Al (s) + 3O_2 \rightarrow 2Al_2O_3 (s)} \label{4.4.2} \] Each neutral oxygen atom gains two electrons and becomes negatively charged, forming an oxide ion; thus, oxygen has an oxidation state of −2 in the product and has been reduced. Each neutral aluminum atom loses three electrons to produce an aluminum ion with an oxidation state of +3 in the product, so aluminum has been oxidized. In the formation of Al O , electrons are transferred as follows (the small overset number emphasizes the oxidation state of the elements): \[ 4 \overset{0}{\ce{Al}} + 3 \overset{0}{\ce{O2}} \rightarrow \ce{4 Al^{3+} + 6 O^{2-} }\label{4.4.3} \] and are examples of oxidation–reduction (redox) reactions. In redox reactions, there is a net transfer of electrons from one reactant to another. In any redox reaction, the total number of electrons lost must equal the total of electrons gained to preserve electrical neutrality. In , for example, the total number of electrons lost by aluminum is equal to the total number gained by oxygen: \[ \begin{align*} \text{electrons lost} &= \ce{4 Al} \, \text{atoms} \times {3 \, e^- \, \text{lost} \over \ce{Al} \, \text{atom} } \\[4pt] &= 12 \, e^- \, \text{lost} \label{4.4.4a} \end{align*} \] \[ \begin{align*} \text{electrons gained} &= \ce{6 O} \, \text{atoms} \times {2 \, e^- \, \text{gained} \over \ce{O} \, \text{atom}} \\[4pt] &= 12 \, e^- \, \text{gained} \label{4.4.4b}\end{align*} \] The same pattern is seen in all oxidation–reduction reactions: the number of electrons lost must equal the number of electrons gained. An additional example of a redox reaction, the reaction of sodium metal with chlorine is illustrated in . In all oxidation–reduction (redox) reactions, the number of electrons lost equals the number of electrons gained. Assigning oxidation states to the elements in binary ionic compounds is straightforward: the oxidation states of the elements are identical to the charges on the monatomic ions. Previously, you learned how to predict the formulas of simple ionic compounds based on the sign and magnitude of the charge on monatomic ions formed by the neutral elements. Examples of such compounds are sodium chloride (NaCl; ), magnesium oxide (MgO), and calcium chloride (CaCl ). In covalent compounds, in contrast, atoms share electrons. However, we can still assign oxidation states to the elements involved by treating them as if they were ionic (that is, as if all the bonding electrons were transferred to the more attractive element). Oxidation states in covalent compounds are somewhat arbitrary, but they are useful bookkeeping devices to help you understand and predict many reactions. A set of rules for assigning oxidation states to atoms in chemical compounds follows. Nonintegral (fractional) oxidation states are encountered occasionally. They are usually due to the presence of two or more atoms of the same element with different oxidation states. In any chemical reaction, the net charge must be conserved; that is, in a chemical reaction, the total number of electrons is constant, just like the total number of atoms. Consistent with this, rule 1 states that the sum of the individual oxidation states of the atoms in a molecule or ion must equal the net charge on that molecule or ion. In NaCl, for example, Na has an oxidation state of +1 and Cl is −1. The net charge is zero, as it must be for any compound. Rule 3 is required because fluorine attracts electrons more strongly than any other element, for reasons you will discover in . Hence fluorine provides a reference for calculating the oxidation states of other atoms in chemical compounds. Rule 4 reflects the difference in chemistry observed for compounds of hydrogen with nonmetals (such as chlorine) as opposed to compounds of hydrogen with metals (such as sodium). For example, NaH contains the H ion, whereas HCl forms H and Cl ions when dissolved in water. Rule 5 is necessary because fluorine has a greater attraction for electrons than oxygen does; this rule also prevents violations of rule 2. So the oxidation state of oxygen is +2 in OF but −½ in KO . Note that an oxidation state of −½ for O in KO is perfectly acceptable. The reduction of copper(I) oxide shown in Equation \(\ref{4.4.5}\) demonstrates how to apply these rules. Rule 1 states that atoms in their elemental form have an oxidation state of zero, which applies to H and Cu. From rule 4, hydrogen in H O has an oxidation state of +1, and from rule 5, oxygen in both Cu O and H O has an oxidation state of −2. Rule 6 states that the sum of the oxidation states in a molecule or formula unit must equal the net charge on that compound. This means that each Cu atom in Cu O must have a charge of +1: 2(+1) + (−2) = 0. So the oxidation states are as follows: \[ \overset {\color{ref}{+1}}{\ce{Cu_2}} \overset {\color{ref}-2}{\ce{O}} (s) + \overset {\color{ref}0}{\ce{H_2}} (g) \rightarrow 2 \overset {\color{ref}0}{\ce{Cu}} (s) + \overset {\color{ref}+1}{\ce{H}}_2 \overset {\color{ref}-2}{\ce{O}} (g) \label{4.4.5} \] Assigning oxidation states allows us to see that there has been a net transfer of electrons from hydrogen (0 → +1) to copper (+1 → 0). Thus, this is a redox reaction. Once again, the number of electrons lost equals the number of electrons gained, and there is a net conservation of charge: \[ \text{electrons lost} = 2 \, H \, \text{atoms} \times {1 \, e^- \, \text{lost} \over H \, \text{atom} } = 2 \, e^- \, \text{lost} \label{4.4.6a} \] \[ \text{electrons gained} = 2 \, Cu \, \text{atoms} \times {1 \, e^- \, \text{gained} \over Cu \, \text{atom}} = 2 \, e^- \, \text{gained} \label{4.4.6b} \] Remember that oxidation states are useful for visualizing the transfer of electrons in oxidation–reduction reactions, but the oxidation state of an atom and its actual charge are the same only for simple ionic compounds. Oxidation states are a convenient way of assigning electrons to atoms, and they are useful for predicting the types of reactions that substances undergo. Assign oxidation states to all atoms in each compound. : molecular or empirical formula : oxidation states : Begin with atoms whose oxidation states can be determined unambiguously from the rules presented (such as fluorine, other halogens, oxygen, and monatomic ions). Then determine the oxidation states of other atoms present according to rule 1. : a. We know from rule 3 that fluorine always has an oxidation state of −1 in its compounds. The six fluorine atoms in sulfur hexafluoride give a total negative charge of −6. Because rule 1 requires that the sum of the oxidation states of all atoms be zero in a neutral molecule (here SF ), the oxidation state of sulfur must be +6: [(6 F atoms)(−1)] + [(1 S atom) (+6)] = 0 b. According to rules 4 and 5, hydrogen and oxygen have oxidation states of +1 and −2, respectively. Because methanol has no net charge, carbon must have an oxidation state of −2: [(4 H atoms)(+1)] + [(1 O atom)(−2)] + [(1 C atom)(−2)] = 0 c. Note that (NH ) SO is an ionic compound that consists of both a polyatomic cation (NH ) and a polyatomic anion (SO ) (see ). We assign oxidation states to the atoms in each polyatomic ion separately. For NH , hydrogen has an oxidation state of +1 (rule 4), so nitrogen must have an oxidation state of −3: [(4 H atoms)(+1)] + [(1 N atom)(−3)] = +1, the charge on the NH ion For SO42−, oxygen has an oxidation state of −2 (rule 5), so sulfur must have an oxidation state of +6: [(4 O atoms) (−2)] + [(1 S atom)(+6)] = −2, the charge on the sulfate ion d. Oxygen has an oxidation state of −2 (rule 5), giving an overall charge of −8 per formula unit. This must be balanced by the positive charge on three iron atoms, giving an oxidation state of +8/3 for iron: Fractional oxidation states are allowed because oxidation states are a somewhat arbitrary way of keeping track of electrons. In fact, Fe O can be viewed as having two Fe ions and one Fe ion per formula unit, giving a net positive charge of +8 per formula unit. Fe O is a magnetic iron ore commonly called magnetite. In ancient times, magnetite was known as lodestone because it could be used to make primitive compasses that pointed toward Polaris (the North Star), which was called the “lodestar.” e. Initially, we assign oxidation states to the components of CH CO H in the same way as any other compound. Hydrogen and oxygen have oxidation states of +1 and −2 (rules 4 and 5, respectively), resulting in a total charge for hydrogen and oxygen of [(4 H atoms)(+1)] + [(2 O atoms)(−2)] = 0 So the oxidation state of carbon must also be zero (rule 6). This is, however, an average oxidation state for the two carbon atoms present. Because each carbon atom has a different set of atoms bonded to it, they are likely to have different oxidation states. To determine the oxidation states of the individual carbon atoms, we use the same rules as before but with the additional assumption that bonds between atoms of the same element do not affect the oxidation states of those atoms. The carbon atom of the methyl group (−CH ) is bonded to three hydrogen atoms and one carbon atom. We know from rule 4 that hydrogen has an oxidation state of +1, and we have just said that the carbon–carbon bond can be ignored in calculating the oxidation state of the carbon atom. For the methyl group to be electrically neutral, its carbon atom must have an oxidation state of −3. Similarly, the carbon atom of the carboxylic acid group (−CO H) is bonded to one carbon atom and two oxygen atoms. Again ignoring the bonded carbon atom, we assign oxidation states of −2 and +1 to the oxygen and hydrogen atoms, respectively, leading to a net charge of [(2 O atoms)(−2)] + [(1 H atom)(+1)] = −3 To obtain an electrically neutral carboxylic acid group, the charge on this carbon must be +3. The oxidation states of the individual atoms in acetic acid are thus \[ \underset {-3}{C} \overset {+1}{H_3} \overset {+3}{C} \underset {-2}{O_2} \overset {+1}{H} \nonumber \] Thus the sum of the oxidation states of the two carbon atoms is indeed zero. Assign oxidation states to all atoms in each compound. Ba, +2; F, −1 C, 0; H, +1; O, −2 K, +1; Cr, +6; O, −2 Cs, +1; O, −½ C, −3; H, +1; C, −1; H, +1; O, −2; H, +1 Many types of chemical reactions are classified as redox reactions, and it would be impossible to memorize all of them. However, there are a few important types of redox reactions that you are likely to encounter and should be familiar with. These include: The following sections describe another important class of redox reactions: single-displacement reactions of metals in solution. \[\ce{ Fe(s) + 2HCl(aq) \rightarrow FeCl_2(aq) + H_2(g)} \label{4.4.81} \] In subsequent steps, \(\ce{FeCl2}\) undergoes oxidation to form a reddish-brown precipitate of \(\ce{Fe(OH)3}\). Many metals dissolve through reactions of this type, which have the general form \[\text{metal} + \text{acid} \rightarrow \text{salt} + \text{hydrogen} \label{4.4.82} \] Some of these reactions have important consequences. For example, it has been proposed that one factor that contributed to the fall of the Roman Empire was the widespread use of lead in cooking utensils and pipes that carried water. Rainwater, as we have seen, is slightly acidic, and foods such as fruits, wine, and vinegar contain organic acids. In the presence of these acids, lead dissolves: \[ \ce{Pb(s) + 2H^+(aq) \rightarrow Pb^{2+}(aq) + H_2(g) } \label{4.4.83} \] Consequently, it has been speculated that both the water and the food consumed by Romans contained toxic levels of lead, which resulted in widespread lead poisoning and eventual madness. Perhaps this explains why the Roman Emperor Caligula appointed his favorite horse as consul! Certain metals are oxidized by aqueous acid, whereas others are oxidized by aqueous solutions of various metal salts. Both types of reactions are called single-displacement reactions, in which the ion in solution is displaced through oxidation of the metal. Two examples of single-displacement reactions are the reduction of iron salts by zinc (Equation \(\ref{4.4.84}\)) and the reduction of silver salts by copper (Equation \(\ref{4.4.85}\) and Figure \(\Page {3}\)): \[ \ce{Zn(s) + Fe^{2+}(aq) \rightarrow Zn^{2+}(aq) + Fe(s)} \label{4.4.84} \] \[ \ce{ Cu(s) + 2Ag^+(aq) \rightarrow Cu^{2+}(aq) + 2Ag(s)} \label{4.4.85} \] The reaction in Equation \(\ref{4.4.84}\) is widely used to prevent (or at least postpone) the corrosion of iron or steel objects, such as nails and sheet metal. The process of “galvanizing” consists of applying a thin coating of zinc to the iron or steel, thus protecting it from oxidation as long as zinc remains on the object. By observing what happens when samples of various metals are placed in contact with solutions of other metals, chemists have arranged the metals according to the relative ease or difficulty with which they can be oxidized in a single-displacement reaction. For example, metallic zinc reacts with iron salts, and metallic copper reacts with silver salts. Experimentally, it is found that zinc reacts with both copper salts and silver salts, producing \(\ce{Zn2+}\). Zinc therefore has a greater tendency to be oxidized than does iron, copper, or silver. Although zinc will not react with magnesium salts to give magnesium metal, magnesium metal will react with zinc salts to give zinc metal: \[ \ce{Zn(s) + Mg^{2+}(aq) \xcancel{\rightarrow} Zn^{2+}(aq) + Mg(s)} \label{4.4.10} \] \[ \ce{Mg(s) + Zn^{2+}(aq) \rightarrow Mg^{2+}(aq) + Zn(s)} \label{4.4.11} \] Magnesium has a greater tendency to be oxidized than zinc does. Pairwise reactions of this sort are the basis of the activity series (Figure \(\Page {4}\)), which lists metals and hydrogen in order of their relative tendency to be oxidized. The metals at the top of the series, which have the greatest tendency to lose electrons, are the alkali metals (group 1), the alkaline earth metals (group 2), and Al (group 13). In contrast, the metals at the bottom of the series, which have the lowest tendency to be oxidized, are the precious metals or coinage metals—platinum, gold, silver, and copper, and mercury, which are located in the lower right portion of the metals in the periodic table. You should be generally familiar with which kinds of metals are active metals, which have the greatest tendency to be oxidized. (located at the top of the series) and which are inert metals, which have the least tendency to be oxidized. (at the bottom of the series). When using the activity series to predict the outcome of a reaction, keep in mind that . Because magnesium is above zinc in Figure \(\Page {4}\), magnesium metal will reduce zinc salts but not vice versa. Similarly, the precious metals are at the bottom of the activity series, so virtually any other metal will reduce precious metal salts to the pure precious metals. Hydrogen is included in the series, and the tendency of a metal to react with an acid is indicated by its position relative to hydrogen in the activity series. Because the precious metals lie below hydrogen, they do not dissolve in dilute acid and therefore do not corrode readily. Example \(\Page {2}\) demonstrates how a familiarity with the activity series allows you to predict the products of many single-displacement reactions. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation. reactants overall reaction and net ionic equation \[ \ce{ Al(s) + 3Ag^+(aq) \rightarrow Al^{3+}(aq) + 3Ag(s)} \nonumber \] Recall from our discussion of solubilities that most nitrate salts are soluble. In this case, the nitrate ions are spectator ions and are not involved in the reaction. \[ \ce{Pb(s) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + H_2(g) } \nonumber \] Lead(II) sulfate is the white solid that forms on corroded battery terminals. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation. \(no\: reaction\) \(3Zn(s) + 2Cr^{3+}(aq) \rightarrow 3Zn^{2+}(aq) + 2Cr(s)\) \(2Al(s) + 6CH_3CO_2H(aq) \rightarrow 2Al^{3+}(aq) + 6CH_3CO_2^-(aq) + 3H_2(g)\ Oxidation–reduction reactions are balanced by separating the overall chemical equation into an oxidation equation and a reduction equation. In oxidation–reduction reactions, electrons are transferred from one substance or atom to another. We can balance oxidation–reduction reactions in solution using the (Table \(\Page {1}\)), in which the overall reaction is separated into an oxidation equation and a reduction equation. There are many types of redox reactions. are reactions of metals with either acids or another metal salt that result in dissolution of the first metal and precipitation of a second (or evolution of hydrogen gas). The outcome of these reactions can be predicted using the (Figure \(\Page {4}\)), which arranges metals and H in decreasing order of their tendency to be oxidized. Any metal will reduce metal ions below it in the activity series. lie at the top of the activity series, whereas are at the bottom of the activity series. | 19,266 | 2,895 |
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The postulates of the kinetic molecular theory of gases ignore both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive. In reality, however, all gases have nonzero molecular volumes. Furthermore, the molecules of real gases interact with one another in ways that depend on the structure of the molecules and therefore differ for each gaseous substance. In this section, we consider the properties of real gases and how and why they differ from the predictions of the ideal gas law. We also examine liquefaction, a key property of real gases that is not predicted by the kinetic molecular theory of gases. For an ideal gas, a plot of \(PV/nRT\) versus \(P\) gives a horizontal line with an intercept of 1 on the \(PV/nRT\) axis. Real gases, however, show significant deviations from the behavior expected for an ideal gas, particularly at high pressures (Figure \(\Page {1a}\)). Only at relatively low pressures (less than 1 atm) do real gases approximate ideal gas behavior (Figure \(\Page {1b}\)). Real gases also approach ideal gas behavior more closely at higher temperatures, as shown in Figure \(\Page {2}\) for \(N_2\). Why do real gases behave so differently from ideal gases at high pressures and low temperatures? Under these conditions, the two basic assumptions behind the ideal gas law—namely, that gas molecules have negligible volume and that intermolecular interactions are negligible—are no longer valid. Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container. In contrast, the molecules of a real gas have small but measurable volumes. At low pressures, the gaseous molecules are relatively far apart, but as the pressure of the gas increases, the intermolecular distances become smaller and smaller (Figure \(\Page {3}\)). As a result, the volume occupied by the molecules becomes significant compared with the volume of the container. Consequently, the total volume occupied by the gas is greater than the volume predicted by the ideal gas law. Thus at very high pressures, the experimentally measured value of / is greater than the value predicted by the ideal gas law. Moreover, all molecules are attracted to one another by a combination of forces. These forces become particularly important for gases at low temperatures and high pressures, where intermolecular distances are shorter. Attractions between molecules reduce the number of collisions with the container wall, an effect that becomes more pronounced as the number of attractive interactions increases. Because the average distance between molecules decreases, the pressure exerted by the gas on the container wall decreases, and the observed pressure is than expected (Figure \(\Page {4}\)). Thus as shown in Figure \(\Page {2}\), at low temperatures, the ratio of (PV/nRT\) is lower than predicted for an ideal gas, an effect that becomes particularly evident for complex gases and for simple gases at low temperatures. At very high pressures, the effect of nonzero molecular volume predominates. The competition between these effects is responsible for the minimum observed in the \(PV/nRT\) versus \(P\) plot for many gases. Nonzero molecular volume makes the actual volume greater than predicted at high pressures; intermolecular attractions make the pressure less than predicted. At high temperatures, the molecules have sufficient kinetic energy to overcome intermolecular attractive forces, and the effects of nonzero molecular volume predominate. Conversely, as the temperature is lowered, the kinetic energy of the gas molecules decreases. Eventually, a point is reached where the molecules can no longer overcome the intermolecular attractive forces, and the gas liquefies (condenses to a liquid). The Dutch physicist Johannes van der Waals (1837–1923; Nobel Prize in Physics, 1910) modified the ideal gas law to describe the behavior of real gases by explicitly including the effects of molecular size and intermolecular forces. In his description of gas behavior, the so-called equation, \[ \underbrace{ \left(P + \dfrac{an^2}{V^2}\right)}_{\text{Pressure Term}} \overbrace{(V − nb)}^{\text{Pressure Term}} =nRT \label{10.9.1} \] and are empirical constants that are different for each gas. The values of \(a\) and \(b\) are listed in Table \(\Page {1}\) for several common gases. The pressure term in Equation \(\ref{10.9.1}\) corrects for intermolecular attractive forces that tend to reduce the pressure from that predicted by the ideal gas law. Here, \(n^2/V^2\) represents the concentration of the gas (\(n/V\)) squared because it takes two particles to engage in the pairwise intermolecular interactions of the type shown in Figure \(\Page {4}\). The volume term corrects for the volume occupied by the gaseous molecules. The correction for volume is negative, but the correction for pressure is positive to reflect the effect of each factor on and , respectively. Because nonzero molecular volumes produce a measured volume that is than that predicted by the ideal gas law, we must subtract the molecular volumes to obtain the actual volume available. Conversely, attractive intermolecular forces produce a pressure that is than that expected based on the ideal gas law, so the \(an^2/V^2\) term must be added to the measured pressure to correct for these effects. You are in charge of the manufacture of cylinders of compressed gas at a small company. Your company president would like to offer a 4.00 L cylinder containing 500 g of chlorine in the new catalog. The cylinders you have on hand have a rupture pressure of 40 atm. Use both the ideal gas law and the van der Waals equation to calculate the pressure in a cylinder at 25°C. Is this cylinder likely to be safe against sudden rupture (which would be disastrous and certainly result in lawsuits because chlorine gas is highly toxic)? volume of cylinder, mass of compound, pressure, and temperature safety Use the molar mass of chlorine to calculate the amount of chlorine in the cylinder. Then calculate the pressure of the gas using the ideal gas law. Obtain and values for Cl from Table \(\Page {1}\). Use the van der Waals equation (\(\ref{10.9.1}\)) to solve for the pressure of the gas. Based on the value obtained, predict whether the cylinder is likely to be safe against sudden rupture. We begin by calculating the amount of chlorine in the cylinder using the molar mass of chlorine (70.906 g/mol): \[\begin{align} n &=\dfrac{m}{M} \\[4pt] &= \rm\dfrac{500\;g}{70.906\;g/mol} \\[4pt] &=7.052\;mol\nonumber \end{align} \nonumber \] Using the ideal gas law and the temperature in kelvin (298 K), we calculate the pressure: \[\begin{align} P &=\dfrac{nRT}{V} \\[4pt] &=\rm\dfrac{7.052\;mol\times 0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L} \\[4pt] &= 43.1\;atm \end{align} \nonumber \] If chlorine behaves like an ideal gas, you have a real problem! Now let’s use the van der Waals equation with the and values for Cl from Table \(\Page {1}\). Solving for \(P\) gives \[\begin{align}P&=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\\&=\rm\dfrac{7.052\;mol\times0.08206\dfrac{L\cdot atm}{mol\cdot K}\times298\;K}{4.00\;L-7.052\;mol\times0.0542\dfrac{L}{mol}}-\dfrac{6.260\dfrac{L^2atm}{mol^2}\times(7.052\;mol)^2}{(4.00\;L)^2}\\&=\rm28.2\;atm\end{align} \nonumber \] This pressure is well within the safety limits of the cylinder. The ideal gas law predicts a pressure 15 atm higher than that of the van der Waals equation. A 10.0 L cylinder contains 500 g of methane. Calculate its pressure to two significant figures at 27°C using the 77 atm 67 atm Liquefaction of gases is the condensation of gases into a liquid form, which is neither anticipated nor explained by the kinetic molecular theory of gases. Both the theory and the ideal gas law predict that gases compressed to very high pressures and cooled to very low temperatures should still behave like gases, albeit cold, dense ones. As gases are compressed and cooled, however, they invariably condense to form liquids, although very low temperatures are needed to liquefy light elements such as helium (for He, 4.2 K at 1 atm pressure). Liquefaction can be viewed as an extreme deviation from ideal gas behavior. It occurs when the molecules of a gas are cooled to the point where they no longer possess sufficient kinetic energy to overcome intermolecular attractive forces. The precise combination of temperature and pressure needed to liquefy a gas depends strongly on its molar mass and structure, with heavier and more complex molecules usually liquefying at higher temperatures. In general, substances with large van der Waals \(a\) coefficients are relatively easy to liquefy because large coefficients indicate relatively strong intermolecular attractive interactions. Conversely, small molecules with only light elements have small coefficients, indicating weak intermolecular interactions, and they are relatively difficult to liquefy. Gas liquefaction is used on a massive scale to separate O , N , Ar, Ne, Kr, and Xe. After a sample of air is liquefied, the mixture is warmed, and the gases are separated according to their boiling points. A large value of a in the van der Waals equation indicates the presence of relatively strong intermolecular attractive interactions. The ultracold liquids formed from the liquefaction of gases are called cryogenic liquids, from the Greek , meaning “cold,” and , meaning “producing.” They have applications as refrigerants in both industry and biology. For example, under carefully controlled conditions, the very cold temperatures afforded by liquefied gases such as nitrogen (boiling point = 77 K at 1 atm) can preserve biological materials, such as semen for the artificial insemination of cows and other farm animals. These liquids can also be used in a specialized type of surgery called , which selectively destroys tissues with a minimal loss of blood by the use of extreme cold. Moreover, the liquefaction of gases is tremendously important in the storage and shipment of fossil fuels (Figure \(\Page {5}\)). Liquefied natural gas (LNG) and liquefied petroleum gas (LPG) are liquefied forms of hydrocarbons produced from natural gas or petroleum reserves. consists mostly of methane, with small amounts of heavier hydrocarbons; it is prepared by cooling natural gas to below about −162°C. It can be stored in double-walled, vacuum-insulated containers at or slightly above atmospheric pressure. Because LNG occupies only about 1/600 the volume of natural gas, it is easier and more economical to transport. is typically a mixture of propane, propene, butane, and butenes and is primarily used as a fuel for home heating. It is also used as a feedstock for chemical plants and as an inexpensive and relatively nonpolluting fuel for some automobiles. No real gas exhibits ideal gas behavior, although many real gases approximate it over a range of conditions. Deviations from ideal gas behavior can be seen in plots of / versus at a given temperature; for an ideal gas, / versus = 1 under all conditions. At high pressures, most real gases exhibit larger / values than predicted by the ideal gas law, whereas at low pressures, most real gases exhibit / values close to those predicted by the ideal gas law. Gases most closely approximate ideal gas behavior at high temperatures and low pressures. Deviations from ideal gas law behavior can be described by the , which includes empirical constants to correct for the actual volume of the gaseous molecules and quantify the reduction in pressure due to intermolecular attractive forces. If the temperature of a gas is decreased sufficiently, occurs, in which the gas condenses into a liquid form. Liquefied gases have many commercial applications, including the transport of large amounts of gases in small volumes and the uses of ultracold . | 12,027 | 2,896 |
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Enzymes called catalyze the transfer of phosphoryl groups to organic molecules. The source of the phosphoryl group in most phosphorylation reactions is a molecule called adenosine triphosphate, abbreviated ATP. Notice that there are essentially three parts to the molecule: an adenine nucleoside base, a five-carbon sugar (ribose), and a triphosphate group. The three phosphates are designated by Greek letters a, b, and g. Adenosine diphosphate (ADP) and adenosine monophosphate (AMP) are also important players in the reactions of this chapter. You will see ATP, ADP, and AMP abbreviated in many different ways in this text and throughout the biochemical literature. For example, the three structures below are all abbreviated depictions of ATP: The following exercise will give you some practice in recognizing different abbreviations for biological molecules that contain phosphate groups. Below are a number of representations, labeled A-S, of molecules that contain phosphoryl groups. Different abbreviations are used. Arrange A-S into groups of drawings that depict the same molecule, using different abbreviations (or no abbreviation at all). When we speak of the energy of an ATP molecule being 'spent', what we mean is that a phosphoryl group is being transferred from ATP to some other acceptor molecule, making the acceptor molecule more reactive. For example, in many phosphoryl transfer reactions (such as the phosphorylation of glucose, which we used as an example in ) the gamma ( ) phosphate of ATP is transferred to an organic acceptor, releasing ADP. In other reactions, the base, ribose, and the alpha phosphate is transferred to the organic molecule to form an organic-AMP adduct, while inorganic pyrophosphate (PP ) is released. Occasionally, the beta and gamma phosphate groups are transferred together, with the release of AMP. In all of these reactions, a relatively stable organic molecule is being transformed into a higher energy phosphorylated product. This activated product can then go on to react in ways that its more stable, non-phosphorylated counterpart could not - phosphoryl groups, as we know, are much better leaving groups in nucleophilic substitution reactions than the hydroxyl group of alcohols. Even though the conversion of a lower-energy starting compound into a higher energy product is, by itself, a thermodynamically process, the overall phosphoryl transfer reaction is thermodynamically , because the . In other words, the energy stored in the phosphate anhydride bond of ATP has been 'spent' to create an activated (higher energy) molecule. When AMP or ADP is converted back to ATP, energy from fuel molecules (or from sunlight) is required to re-form the high energy anhydride bond (this process is the subject of discussion later in this section) The explanation for why the phosphate anhydrides linkages in ATP are so energetic lies primarily in the concept of charge separation. Recall from that ATP, at the physiological pH of ~7, is almost completely ionized with a total charge of close to -4. When one of the two anhydride bonds is broken, the negative charges on the phosphate groups are able to separate, eliminating some of the same-charge repulsion that existed in ATP. One way to picture this is as a coil springing open. Another reason has to do with the energy of solvation by water. When the gamma phosphate of ATP is transferred to an alcohol, for example, surrounding water molecules are able to form more hydrogen-bonding interactions with the products (ADP and the organic phosphate) than was possible with ATP and alcohol. These additional solvation interactions stabilize the products of the phosphorylation reaction relative to the starting compounds. You will learn more about the thermodynamic role of ATP in metabolic pathways if you take a class in biochemistry - what is most important to understand at this point is that, because of the energy stored in its phosphate anhydride bonds, ATP is a powerful phosphoryl group donor, and is used as such in many important biochemical reactions. Some examples are discussed in the remainder of this section. Recall that almost all biomolecules are charged species, which 1) keeps them water soluble, and 2) prevents them from diffusing across lipid bilayer membranes. Although many biomolecules are ionized by virtue of negatively charged carboxylate and positively charged amino groups, the most common ionic group in biologically important organic compounds is phosphate - thus the phosphorylation of alcohol groups is a critical metabolic step. In alcohol phosphorylations, ATP is almost always the phosphate donor, and the mechanism is very consistent: the alcohol oxygen acts as a nucleophile, attacking the gamma-phosphorus of ATP and expelling ADP (look again, for example, at the glucose kinase reaction that we first saw in ). + B-H The glucose kinase reaction is the first step in glycolysis, a metabolic pathway in which the 6-carbon sugar glucose is broken down into two 3-carbon fragments called pyruvate. The third step of glycolysis also a kinase reaction: this time, it is the hydroxyl group on carbon #1 of fructose-6-phosphate that is phosphorylated (step 2 of glycolysis is the isomerization of glucose-6-phosphate to fructose-6-phosphate, a reaction we will study in ) Once again, ATP is the phosphate donor in the fructose-6-phosphate kinase reaction: Now, when the 6-carbon sugar breaks into two 3-carbon pieces, each piece has its own phosphate group (the carbon-carbon bond-breaking step is a reaction that we will learn about in ). The biological activity of many proteins is regulated by means of a very similar phosphorylation reaction catalyzed by . In these reactions, the side chain hydroxyl groups on serine, threonine, and tyrosine residues of certain proteins are modified with the gamma phosphate from ATP. Notice the new " ATP in, ADP out" notation used in this figure, showing that ATP is converted to ADP in the course of the reaction. From here on, we will frequently use this shorthand convention to indicate when common molecules such as ATP, water, or phosphate are participants in a reaction, either as reactants or products. We have just seen how alcohol groups can be converted to monophosphates using ATP as the phosphate donor. In some biochemical pathways, the next step is the addition of a second phosphate group to form a diphosphate. In the early stages of the biosynthesis of ‘isoprenoid’ compounds such as cholesterol, for example, two phosphates are added sequentially to a primary alcohol group on an intermediate compound called mevalonate. The first phosphorylation is essentially the same as the reactions described in part B of this section. In the second phosphorylation reaction, the gamma phosphate of a second ATP molecule is transferred to an oxygen atom on the first phosphate, forming a new phosphate anhydride linkage. Another example of a diphosphorylation reaction takes place in a single step, rather than sequentially. Phosphoribosyl diphosphate (PRPP) is a very important intermediate compound in the biosynthesis of nucleotides and some amino acids, and is the product of the diphosphorylation of ribose-5-phosphate. In this reaction, two phosphate groups (the beta and the gamma) are transferred from ATP to a hydroxyl on ribose-5-phosphate. Notice that the beta phosphorus of ATP is the electrophile in this case, rather then the alpha phosphorus. Consequentially, the reaction results in the conversion on one ATP to one AMP, rather than two ATPs to two ADPs. Are we getting more for our 'ATP money' in this one-step diphosphorylation? Not really - in order to convert the energy-poor AMP back up to the energy-rich ATP, the cell first has to transfer a phosphate from a second ATP molecule in a reaction catalyzed by an enzyme called adenylate kinase. So in the end, the diphosphorylation reaction still costs the cell two ATP-to-ADP conversions. It is worth noting that both of the diphosphate groups produced in these reactions end up as leaving groups in subsequent nucleophilic substitution reactions. Mevalonate diphosphate is eventually converted to isoprenyl diphosphate, the substrate for protein prenyltransferase ( ). PRPP is the starting point for the biosynthesis of both pyrimidine (C and U/T) and purine (G, and A) nucleotides. The S 1 displacement of pyrophosphate in pyrimidine biosynthesis is shown below. Thus far we have seen alcohol oxygens and phosphate oxygens acting as nucleophilic accepting groups in phosphoryl transfer reactions. Consider next the first step of the reaction catalyzed by the enzyme glutamine synthase: Once again in this reaction, the gamma-phosphate of ATP is transferred to an oxygen acceptor - however in this case the acceptor is a carboxylate oxygen, and the product is an acyl phosphate. As we shall see in , acyl phosphates are commonly referred to as 'activated carboxylates', and are primed to undergo reactions called 'nucleophilic acyl substitutions'. Activation of the side chain carboxylate of aspartate is somewhat different from the parallel activation of glutamate shown above. While the carboxylate group in glutamate accepts a simple phosphate group from ATP, the carboxylate in aspartate attacks the alpha-phosphate of ATP, displacing inorganic pyrophosphate and accepting an entire AMP group. The resulting 'acyl adenosine phosphate', which is technically a phosphate diester, is another form of 'activated carboxylate' that we will learn more about in . For some interesting variations on the phosphoryl transfer reaction, consider the early steps of isoprenoid biosynthesis in bacteria (this is a completely different pathway than that mentioned in , which is operative in animals). In the first step, the oxygen of a monophosphate ester attacks the a phosphate of CTP ( ATP!) to expel inorganic pyrophosphate. In step 2, a second hydroxyl group is phosphorylated in the normal way by an ATP-dependent kinase, and in step 3 that phosphate proceeds to attack the first of the two phosphates on the nucleotide diphosphate diester, expelling CMP and forming a cyclic diphosphate. Several more steps lead to the formation of isopentenyl diphosphate, the building block molecule for all isoprenoid compounds. Throughout this section we have seen reactions in which the energy contained in an ATP anhydride bond is 'spent', and ADP or AMP is formed as a result. In order to regenerate ATP, a phosphate group must be transferred to ADP, which is of course a thermodynamically uphill reaction requiring the input of energy from the breakdown of fuel molecules or, in the case of plants, from sunlight. By far the most important source of ATP regeneration is the enzyme ATP synthase, which catalyzes the direct condensation between inorganic phosphate and ADP. Despite the apparent simplicity of the chemistry going on here, ATP synthase is an extremely large, complex, and fascinating enzyme, with multiple protein subunits and a intricate 'molecular motor' design. The reaction must be 'driven' uphill by using the energy from a proton gradient that is set up across the inner mitochondrial membrane. You will learn much more about this amazing biochemical machine if you take a course in biochemistry. Two other reactions in the gycolytic (sugar breakdown) pathway also result in the generation of ATP from ADP, but these are minor physiological sources of ATP compared to ATP synthase. Phosphoglycerate kinase (named according to the reverse of the reaction shown below) transfers a phosphate from an acyl phosphate to ADP. Note that here the leaving group is a carboxylate group. Pyruvate kinase (again the name refers to the reverse reaction) catalyzes a less familiar-looking phosphate transfer. We will revisit this reaction in . | 11,907 | 2,897 |
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Suppose we dispense 20 mL of a reagent using the Class A 10-mL pipet whose calibration information is given in . If the volume and uncertainty for one use of the pipet is 9.992 ± 0.006 mL, what is the volume and uncertainty if we use the pipet twice? As a first guess, we might simply add together the volume and the maximum uncertainty for each delivery; thus (9.992 mL + 9.992 mL) ± (0.006 mL + 0.006 mL) = 19.984 ± 0.012 mL It is easy to appreciate that combining uncertainties in this way overestimates the total uncertainty. Adding the uncertainty for the first delivery to that of the second delivery assumes that with each use the indeterminate error is in the same direction and is as large as possible. At the other extreme, we might assume that the uncertainty for one delivery is positive and the other is negative. If we subtract the maximum uncertainties for each delivery, (9.992 mL + 9.992 mL) ± (0.006 mL – 0.006 mL) = 19.984 ± 0.000 mL we clearly underestimate the total uncertainty. So what is the total uncertainty? From the discussion above, we reasonably expect that the total uncertainty is greater than ±0.000 mL and that it is less than ±0.012 mL. To estimate the uncertainty we use a mathematical technique known as the propagation of uncertainty. Our treatment of the propagation of uncertainty is based on a few simple rules. A allows us to estimate the uncertainty in a result from the uncertainties in the measurements used to calculate that result. For the equations in this section we represent the result with the symbol , and we represent the measurements with the symbols , , and . The corresponding uncertainties are , , , and . We can define the uncertainties for , , and using standard deviations, ranges, or tolerances (or any other measure of uncertainty), as long as we use the same form for all measurements. The requirement that we express each uncertainty in the same way is a critically important point. Suppose you have a range for one measurement, such as a pipet’s tolerance, and standard deviations for the other measurements. All is not lost. There are ways to convert a range to an estimate of the standard deviation. See for more details. When we add or subtract measurements we propagate their absolute uncertainties. For example, if the result is given by the equation \[R = A + B - C \nonumber\] the the absolute uncertainty in is \[u_R = \sqrt{u_A^2 + u_B^2 + u_C^2} \label{4.1}\] If we dispense 20 mL using a 10-mL Class A pipet, what is the total volume dispensed and what is the uncertainty in this volume? First, complete the calculation using the manufacturer’s tolerance of 10.00 mL±0.02 mL, and then using the calibration data from . To calculate the total volume we add the volumes for each use of the pipet. When using the manufacturer’s values, the total volume is \[V = 10.00 \text{ mL} + 10.00 \text{ mL} = 20.00 \text{ mL} \nonumber\] and when using the calibration data, the total volume is \[V = 9.992 \text{ mL} + 9.992 \text{ mL} = 19.984 \text{ mL} \nonumber\] Using the pipet’s tolerance as an estimate of its uncertainty gives the uncertainty in the total volume as \[u_R = (0.02)^2 + (0.02)^2 = 0.028 \text{ mL} = 0.028 \text{ mL} \nonumber\] and using the standard deviation for the data in gives an uncertainty of \[u_R = (0.006)^2 + (0.006)^2 = 0.0085 \text{ mL} \nonumber\] Rounding the volumes to four significant figures gives 20.00 mL ± 0.03 mL when we use the tolerance values, and 19.98 ± 0.01 mL when we use the calibration data. When we multiple or divide measurements we propagate their relative uncertainties. For example, if the result is given by the equation \[R = \frac {A \times B} {C} \nonumber\] then the relative uncertainty in is \[\frac {u_R} {R}= \sqrt{\left( \frac {u_A} {A} \right)^2 + \left( \frac {u_B} {B} \right)^2 + \left( \frac {u_C} {C} \right)^2} \label{4.2}\] The quantity of charge, , in coulombs that passes through an electrical circuit is \[Q = i \times t \nonumber\] where is the current in amperes and is the time in seconds. When a current of 0.15 A ± 0.01 A passes through the circuit for 120 s ± 1 s, what is the total charge and its uncertainty? The total charge is \[Q = (0.15 \text{ A}) \times (120 \text{ s}) = 18 \text{ C} \nonumber\] Since charge is the product of current and time, the relative uncertainty in the charge is \[\frac {u_R} {R} = \sqrt{\left( \frac {0.01} {0.15} \right)^2 + \left( \frac {1} {120} \right)^2} = 0.0672 \nonumber\] and the charge’s absolute uncertainty is \[u_R = R \times 0.0672 = (18 \text{ C}) \times (0.0672) = 1.2 \text{ C} \nonumber\] Thus, we report the total charge as 18 C ± 1 C. Many chemical calculations involve a combination of adding and subtracting, and of multiply and dividing. As shown in the following example, we can calculate the uncertainty by separately treating each operation using Equation \ref{4.1} and Equation \ref{4.2} as needed. For a concentration technique, the relationship between the signal and the an analyte’s concentration is \[S_{total} = k_A C_A + S_{mb} \nonumber\] What is the analyte’s concentration, , and its uncertainty if is 24.37 ± 0.02, is 0.96 ± 0.02, and is \(0.186 \pm 0.003 \text{ ppm}^{-1}\)? Rearranging the equation and solving for \[C_A = \frac {S_{total} - S_{mb}} {k_A} = \frac {24.37 - 0.96} {0.186 \text{ ppm}^{-1}} = \frac {23.41} {0.186 \text{ ppm}^{-1}} = 125.9 \text{ ppm} \nonumber\] gives the analyte’s concentration as 126 ppm. To estimate the uncertainty in , we first use Equation \ref{4.1} to determine the uncertainty for the numerator. \[u_R = \sqrt{(0.02)^2 + (0.02)^2} = 0.028 \nonumber\] The numerator, therefore, is 23.41 ± 0.028. To complete the calculation we use Equation \ref{4.2} to estimate the relative uncertainty in . \[\frac {u_R} {R} = \sqrt{\left( \frac {0.028} {23.41} \right)^2 + \left( \frac {0.003} {0.186} \right)^2} = 0.0162 \nonumber\] The absolute uncertainty in the analyte’s concentration is \[u_R = (125.9 \text{ ppm}) \times (0.0162) = 2.0 \text{ ppm} \nonumber\] Thus, we report the analyte’s concentration as 126 ppm ± 2 ppm. To prepare a standard solution of Cu you obtain a piece of copper from a spool of wire. The spool’s initial weight is 74.2991 g and its final weight is 73.3216 g. You place the sample of wire in a 500-mL volumetric flask, dissolve it in 10 mL of HNO , and dilute to volume. Next, you pipet a 1 mL portion to a 250-mL volumetric flask and dilute to volume. What is the final concentration of Cu in mg/L, and its uncertainty? Assume that the uncertainty in the balance is ±0.1 mg and that you are using Class A glassware. The first step is to determine the concentration of Cu in the final solution. The mass of copper is \[74.2991 \text{ g} - 73.3216 \text{ g} = 0.9775 \text{ g Cu} \nonumber\] The 10 mL of HNO used to dissolve the copper does not factor into our calculation. The concentration of Cu is \[\frac {0.9775 \text{ g Cu}} {0.5000 \text{ L}} \times \frac {1.000 \text{ mL}} {250.0 \text{ mL}} \times \frac {1000 \text{ mg}} {\text{g}} = 7.820 \text{ mg } \ce{Cu^{2+}} \text{/L} \nonumber\] Having found the concentration of Cu , we continue with the propagation of uncertainty. The absolute uncertainty in the mass of Cu wire is \[u_\text{g Cu} = \sqrt{(0.0001)^2 + (0.0001)^2} = 0.00014 \text{ g} \nonumber\] The relative uncertainty in the concentration of Cu is \[\frac {u_\text{mg/L}} {7.820 \text{ mg/L}} = \sqrt{\left( \frac {0.00014} {0.9775} \right)^2 + \left( \frac {0.20} {500.0} \right)^2 + \left( \frac {0.006} {1.000} \right)^2 + \left( \frac {0.12} {250.0} \right)^2} = 0.00603 \nonumber\] Solving for gives the uncertainty as 0.0472. The concentration and uncertainty for Cu is 7.820 mg/L ± 0.047 mg/L. Many other mathematical operations are common in analytical chemistry, including the use of powers, roots, and logarithms. Table 4.3.1
provides equations for propagating uncertainty for some of these function where and are independent measurements and where is a constant whose value has no uncertainty. If the pH of a solution is 3.72 with an absolute uncertainty of ±0.03, what is the [H ] and its uncertainty? The concentration of H is \[[\ce{H+}] = 10^{-\text{pH}} = 10^{-3.72} = 1.91 \times 10^{-4} \text{ M} \nonumber\] or \(1.9 \times 10^{-4}\) M to two significant figures. From Table 4.3.1
the relative uncertainty in [H ] is \[\frac {u_R} {R} = 2.303 \times u_A = 2.303 \times 0.03 = 0.069 \nonumber\] The uncertainty in the concentration, therefore, is \[(1.91 \times 10^{-4} \text{ M}) \times (0.069) = 1.3 \times 10^{-5} \text{ M} \nonumber\] We report the [H ] as \(1.9 (\pm 0.1) \times 10^{-4}\) M, which is equivalent to \(1.9 \times 10^{-4} \text{ M } \pm 0.1 \times 10^{-4} \text{ M}\). A solution of copper ions is blue because it absorbs yellow and orange light. Absorbance, , is defined as \[A = - \log T = - \log \left( \frac {P} {P_\text{o}} \right) \nonumber\] where, is the transmittance, is the power of radiation as emitted from the light source and is its power after it passes through the solution. What is the absorbance if is \(3.80 \times 10^2\) and is \(1.50 \times 10^2\)? If the uncertainty in measuring and is 15, what is the uncertainty in the absorbance? The first step is to calculate the absorbance, which is \[A = - \log T = -\log \frac {P} {P_\text{o}} = - \log \frac {1.50 \times 10^2} {3.80 \times 10^2} = 0.4037 \approx 0.404 \nonumber\] Having found the absorbance, we continue with the propagation of uncertainty. First, we find the uncertainty for the ratio / , which is the transmittance, T. \[\frac {u_{T}} {T} = \sqrt{\left( \frac {15} {3.80 \times 10^2} \right)^2 + \left( \frac {15} {1.50 \times 10^2} \right)^2 } = 0.1075 \nonumber\] Finally, from the uncertainty in the absorbance is \[u_A = 0.4343 \times \frac {u_{T}} {T} = (0.4343) \times (0.1075) = 4.669 \times 10^{-2} \nonumber\] The absorbance and uncertainty is 0.40 ± 0.05 absorbance units. Given the effort it takes to calculate uncertainty, it is worth asking whether such calculations are useful. The short answer is, yes. Let’s consider three examples of how we can use a propagation of uncertainty to help guide the development of an analytical method. One reason to complete a propagation of uncertainty is that we can compare our estimate of the uncertainty to that obtained experimentally. For example, to determine the mass of a penny we measure its mass twice—once to tare the balance at 0.000 g and once to measure the penny’s mass. If the uncertainty in each measurement of mass is ±0.001 g, then we estimate the total uncertainty in the penny’s mass as \[u_R = \sqrt{(0.001)^2 + (0.001)^2} = 0.0014 \text{ g} \nonumber\] If we measure a single penny’s mass several times and obtain a standard deviation of ±0.050 g, then we have evidence that the measurement process is out of control. Knowing this, we can identify and correct the problem. We also can use a propagation of uncertainty to help us decide how to improve an analytical method’s uncertainty. In , for instance, we calculated an analyte’s concentration as 126 ppm ± 2 ppm, which is a percent uncertainty of 1.6%. Suppose we want to decrease the percent uncertainty to no more than 0.8%. How might we accomplish this? Looking back at the calculation, we see that the concentration’s relative uncertainty is determined by the relative uncertainty in the measured signal (corrected for the reagent blank) \[\frac {0.028} {23.41} = 0.0012 \text{ or } 0.12\% \nonumber\] and the relative uncertainty in the method’s sensitivity, , \[\frac {0.003 \text{ ppm}^{-1}} {0.186 \text{ ppm}^{-1}} = 0.016 \text{ or } 1.6\% \nonumber\] Of these two terms, the uncertainty in the method’s sensitivity dominates the overall uncertainty. Improving the signal’s uncertainty will not improve the overall uncertainty of the analysis. To achieve an overall uncertainty of 0.8% we must improve the uncertainty in to ±0.0015 ppm . Verify that an uncertainty of ±0.0015 ppm for is the correct result. An uncertainty of 0.8% is a relative uncertainty in the concentration of 0.008; thus, letting be the uncertainty in \[0.008 = \sqrt{\left( \frac {0.028} {23.41} \right)^2 + \left( \frac {u} {0.186} \right)^2} \nonumber\] Squaring both sides of the equation gives \[6.4 \times 10^{-5} = \left( \frac {0.028} {23.41} \right)^2 + \left( \frac {u} {0.186} \right)^2 \nonumber\] Solving for the uncertainty in gives its value as \(1.47 \times 10^{-3}\) or ±0.0015 ppm . Finally, we can use a propagation of uncertainty to determine which of several procedures provides the smallest uncertainty. When we dilute a stock solution usually there are several combinations of volumetric glassware that will give the same final concentration. For instance, we can dilute a stock solution by a factor of 10 using a 10-mL pipet and a 100-mL volumetric flask, or using a 25-mL pipet and a 250-mL volumetric flask. We also can accomplish the same dilution in two steps using a 50-mL pipet and 100-mL volumetric flask for the first dilution, and a 10-mL pipet and a 50-mL volumetric flask for the second dilution. The overall uncertainty in the final concentration—and, therefore, the best option for the dilution—depends on the uncertainty of the volumetric pipets and volumetric flasks. As shown in the following example, we can use the tolerance values for volumetric glassware to determine the optimum dilution strategy [Lam, R. B.; Isenhour, T. L. , , 1158–1161]. Which of the following methods for preparing a 0.0010 M solution from a 1.0 M stock solution provides the smallest overall uncertainty?
(a) A one-step dilution that uses a 1-mL pipet and a 1000-mL volumetric flask.
(b) A two-step dilution that uses a 20-mL pipet and a 1000-mL volumetric flask for the first dilution, and a 25-mL pipet and a 500-mL volumetric flask for the second dilution. The dilution calculations for case (a) and case (b) are \[\text{case (a): 1.0 M } \times \frac {1.000 \text { mL}} {1000.0 \text { mL}} = 0.0010 \text{ M} \nonumber\] \[\text{case (b): 1.0 M } \times \frac {20.00 \text { mL}} {1000.0 \text { mL}} \times \frac {25.00 \text{ mL}} {500.0 \text{mL}} = 0.0010 \text{ M} \nonumber\] Using tolerance values from , the relative uncertainty for case (a) is \[u_R = \sqrt{\left( \frac {0.006} {1.000} \right)^2 + \left( \frac {0.3} {1000.0} \right)^2} = 0.006 \nonumber\] and for case (b) the relative uncertainty is \[u_R = \sqrt{\left( \frac {0.03} {20.00} \right)^2 + \left( \frac {0.3} {1000} \right)^2 + \left( \frac {0.03} {25.00} \right)^2 + \left( \frac {0.2} {500.0} \right)^2} = 0.002 \nonumber\] Since the relative uncertainty for case (b) is less than that for case (a), the two-step dilution provides the smallest overall uncertainty. Of course we must balance the smaller uncertainty for case (b) against the increased opportunity for introducing a determinate error when making two dilutions instead of just one dilution, as in case (a). | 15,013 | 2,898 |
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When we turn our attention from the potential energy of charged macroscopic particles which have a definite location in space to microscopic particles like the electron, we immediately encounter a difficulty. The electron in an atom is not a fixed distance from the nucleus but is “smeared out” in space in a wave pattern over a large range of distances. Nevertheless it is still meaningful to talk about the potential energy of such an electron cloud. Consider the 1 electron illustrated by the dot-density diagram in , for example. If the electron were actually positioned at one of these dots momentarily, it would have a definite potential energy at that moment. If we now add up the potential energy for each dot and divide by the number of dots, we obtain an average potential energy, which is a good approximation to the potential energy of the electron cloud. The more dots we have, the closer such an approximation is to the exact answer. In practice we can often decide which of two electron clouds has the higher potential energy by looking at them. In , for example, it is easy to see that the potential energy of an electron in a 1 orbital is lower than that of a 2 electron. An electron in a 1 orbital is almost always closer than 200 pm to the nucleus, while in a 2 orbital it is usually farther away. In the same way we have no difficulty in estimating that a 3 electron is on average farther from the nucleus and hence higher in potential energy than a 2 electron. It is also easy to see that electron clouds which differ only in their orientation in space must have the potential energy. An example would be the 2p , 2p , and 2p clouds. When we compare orbitals with different basic shapes, mere inspection of the dot-density diagrams is often insufficient to tell us about the relative potential energies. It is not apparent from , for instance, whether the 2 or 2 orbital has the higher potential energy. Actually both have the same energy in a hydrogen atom, though not in other atoms. In the same way the 3 , the three 3 , and the five 3 orbitals are all found to have the same energy in the hydrogen atom. Although dot-density diagrams are very informative about the potential energy of an electron in an orbital, they tell us nothing at all about its kinetic energy. It is impossible, for example, to decide from Figure 5.6 whether the electron in a 1 orbital is moving faster on the whole than an electron in a 2s orbital, or even whether it is moving at all! Fortunately it turns out that this difficulty is unimportant. The energy (kinetic + potential) of an electron in an atom or a molecule is always one-half its potential energy. Thus, for example, when an electron is shifted from a 1 to a 2 orbital, its potential energy increases by 3.27 aJ. At the same time the electron slows down and its kinetic energy drops by half this quantity, namely, 1.635 aJ. The net result is that the total energy (kinetic + potential) increases by exactly half the increase in potential energy alone; i.e., it increases by 1.635 aJ. A similar statement can he made for change inflicted on any electron in any atomic or molecular system. This result is known as the . Because of this theorem we can, if we want, ignore the kinetic energy of an electron and concentrate exclusively on its potential energy. | 3,352 | 2,899 |
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The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl , they can each complete their valence shell: Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pair; the other three pairs of electrons on each chlorine atom are called lone pairs. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond. Examples of this type of bonding are presented in Section 8.6 when we discuss atoms with less than an octet of electrons. We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols: The structure on the right is the , or , for H O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown here for our two examples: The following procedure can be used to construct Lewis electron structures for more complex molecules and ions: Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed. The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal. This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6. Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for an ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant. 1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows: 2. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons. 3. Placing a bonding pair of electrons between each pair of bonded atoms gives the following: Six electrons are used, and 6 are left over. 4. Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following: Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons. 5. There are no electrons left to place on the central atom. 6. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond: Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid. An alternative structure can be drawn with one H bonded to O. , discussed later in this section, suggest that such a structure is less stable than that shown previously. Write the Lewis electron structure for each species. chemical species Lewis electron structures Use the six-step procedure to write the Lewis electron structure for each species. Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States. Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen: Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following: All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas. Write Lewis electron structures for CO and SCl , a vile-smelling, unstable red liquid that is used in the manufacture of rubber. Lewis Structure of Molecules: Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. In the Lewis model, the number of bonds formed by an element in a neutral compound is the same as the number of unpaired electrons it must share with other atoms to complete its octet of electrons. For the elements of , this number is one; for the elements of , it is two; for , three; and for four. These requirements are illustrated by the following Lewis structures for the hydrides of the lightest members of each group: Elements may form multiple bonds to complete an octet. In ethylene, for example, each carbon contributes two electrons to the double bond, giving each carbon an octet (two electrons/bond × four bonds = eight electrons). Neutral structures with fewer or more bonds exist, but they are unusual and violate the octet rule. Allotropes of an element can have very different physical and chemical properties because of different three-dimensional arrangements of the atoms; the number of bonds formed by the component atoms, however, is always the same. As noted at the beginning of the chapter, diamond is a hard, transparent solid; graphite is a soft, black solid; and the fullerenes have open cage structures. Despite these differences, the carbon atoms in all three allotropes form four bonds, in accordance with the octet rule. Lewis structures explain why the elements of groups 14–17 form neutral compounds with four, three, two, and one bonded atom(s), respectively. Elemental phosphorus also exists in three forms: white phosphorus, a toxic, waxy substance that initially glows and then spontaneously ignites on contact with air; red phosphorus, an amorphous substance that is used commercially in safety matches, fireworks, and smoke bombs; and black phosphorus, an unreactive crystalline solid with a texture similar to graphite (Figure \(\Page {3}\)). Nonetheless, the phosphorus atoms in all three forms obey the octet rule and form three bonds per phosphorus atom. It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal charge on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure. To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules: For each atom, we then compute a formal charge: \( \begin{matrix}
formal\; charge= & valence\; e^{-}- & \left ( non-bonding\; e^{-}+\frac{bonding\;e^{-}}{2} \right )\\
& ^{\left ( free\; atom \right )} & ^{\left ( atom\; in\; Lewis\; structure \right )}
\end{matrix} \label{8.5.1} \) (atom in Lewis structure) To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH ) whose Lewis electron structure is as follows: A neutral nitrogen atom has five valence electrons (it is in group 15). From its Lewis electron structure, the nitrogen atom in ammonia has one lone pair and shares three bonding pairs with hydrogen atoms, so nitrogen itself is assigned a total of five electrons [2 nonbonding e + (6 bonding e /2)]. Substituting into Equation 8.5.2, we obtain \[ formal\; charge\left ( N \right )=5\; valence\; e^{-}-\left ( 2\; non-bonding\; e^{-} +\frac{6\; bonding\; e^{-}}{2} \right )=0 \label{8.5.2}\] A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule shares one pair of bonding electrons and is therefore assigned one electron [0 nonbonding e + (2 bonding e /2)]. Using Equation 8.5.2 to calculate the formal charge on hydrogen, we obtain \[ formal\; charge\left ( H \right )=1\; valence\; e^{-}-\left ( 0\; non-bonding\; e^{-} +\frac{2\; bonding\; e^{-}}{2} \right )=0 \label{8.5.3}\] The hydrogen atoms in ammonia have the same number of electrons as neutral hydrogen atoms, and so their formal charge is also zero. Adding together the formal charges should give us the overall charge on the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall charge on the NH molecule. An atom, molecule, or ion has a formal charge of zero if it has the number of bonds that is typical for that species. Typically, the structure with the most charges on the atoms closest to zero is the more stable Lewis structure. In cases where there are positive or negative formal charges on various atoms, stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges. Lewis Structure of Charged Molecules: Calculate the formal charges on each atom in the NH ion. chemical species formal charges Identify the number of valence electrons in each atom in the NH ion. Use the Lewis electron structure of NH to identify the number of bonding and nonbonding electrons associated with each atom and then use Equation 8.5.2 to calculate the formal charge on each atom. The Lewis electron structure for the NH ion is as follows: The nitrogen atom shares four bonding pairs of electrons, and a neutral nitrogen atom has five valence electrons. Using Equation 8.5.1, the formal charge on the nitrogen atom is therefore \[ formal\; charge\left ( N \right )=5-\left ( 0+\frac{8}{2} \right )=0 \] Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore \[ formal\; charge\left ( H \right )=1-\left ( 0+\frac{2}{2} \right )=0 \] The formal charges on the atoms in the NH ion are thus Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1. Write the formal charges on all atoms in BH . If an atom in a molecule or ion has the number of bonds that is typical for that atom (e.g., four bonds for carbon), its formal charge is zero. As an example of how formal charges can be used to determine the most stable Lewis structure for a substance, we can compare two possible structures for CO . Both structures conform to the rules for Lewis electron structures. This structure has an octet of electrons around each O atom but only 4 electrons around the C atom. Both Lewis electron structures give all three atoms an octet. How do we decide between these two possibilities? The formal charges for the two Lewis electron structures of CO are as follows: Both Lewis structures have a net formal charge of zero, but the structure on the right has a +1 charge on the more electronegative atom (O). Thus the symmetrical Lewis structure on the left is predicted to be more stable, and it is, in fact, the structure observed experimentally. Remember, though, that formal charges do represent the actual charges on atoms in a molecule or ion. They are used simply as a bookkeeping method for predicting the most stable Lewis structure for a compound. The Lewis structure with the set of formal charges closest to zero is usually the most stable. The thiocyanate ion (SCN ), which is used in printing and as a corrosion inhibitor against acidic gases, has at least two possible Lewis electron structures. Draw two possible structures, assign formal charges on all atoms in both, and decide which is the preferred arrangement of electrons. chemical species Lewis electron structures, formal charges, and preferred arrangement Possible Lewis structures for the SCN ion are as follows: We must calculate the formal charges on each atom to identify the more stable structure. If we begin with carbon, we notice that the carbon atom in each of these structures shares four bonding pairs, the number of bonds typical for carbon, so it has a formal charge of zero. Continuing with sulfur, we observe that in (a) the sulfur atom shares one bonding pair and has three lone pairs and has a total of six valence electrons. The formal charge on the sulfur atom is therefore \( 6-\left ( 6+\frac{2}{2} \right )=-1.5-\left ( 4+\frac{4}{2} \right )=-1 \) In (c), nitrogen has a formal charge of −2. Which structure is preferred? Structure (b) is preferred because the negative charge is on the more electronegative atom (N), and it has lower formal charges on each atom as compared to structure (c): 0, −1 versus +1, −2. Salts containing the fulminate ion (CNO ) are used in explosive detonators. Draw three Lewis electron structures for CNO and use formal charges to predict which is more stable. (Note: N is the central atom.) The second structure is predicted to be more stable. A plot of the overall energy of a covalent bond as a function of internuclear distance is identical to a plot of an ionic pair because both result from attractive and repulsive forces between charged entities. In Lewis electron structures, we encounter , which are shared by two atoms, and , which are not shared between atoms. If both electrons in a covalent bond come from the same atom, the bond is called a . Lewis structures are an attempt to rationalize why certain stoichiometries are commonly observed for the elements of particular families. Neutral compounds of group 14 elements typically contain four bonds around each atom (a double bond counts as two, a triple bond as three), whereas neutral compounds of group 15 elements typically contain three bonds. In cases where it is possible to write more than one Lewis electron structure with octets around all the nonhydrogen atoms of a compound, the on each atom in alternative structures must be considered to decide which of the valid structures can be excluded and which is the most reasonable. The formal charge is the difference between the number of valence electrons of the free atom and the number of electrons assigned to it in the compound, where bonding electrons are divided equally between the bonded atoms. The Lewis structure with the lowest formal charges on the atoms is almost always the most stable one. ( ) | 15,293 | 2,900 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Carboxylic_Acids/Reactivity_of_Carboxylic_Acids/Reactions_of_Carboxylic_Acids |
Because of their enhanced acidity, carboxylic acids react with bases to form ionic salts, as shown in the following equations. In the case of alkali metal hydroxides and simple amines (or ammonia) the resulting salts have pronounced ionic character and are usually soluble in water. Heavy metals such as silver, mercury and lead form salts having more covalent character (3rd example), and the water solubility is reduced, especially for acids composed of four or more carbon atoms. Carboxylic acids and salts having alkyl chains longer than six carbons exhibit unusual behavior in water due to the presence of both hydrophilic (CO ) and hydrophobic (alkyl) regions in the same molecule. Such molecules are termed (Gk. amphi = both) or . Depending on the nature of the hydrophilic portion these compounds may form monolayers on the water surface or sphere-like clusters, called micelles, in solution. This reaction class could be termed , and is defined as follows ( is an electrophile). Some examples of this substitution are provided in equations (1) through (4). If is a strong electrophile, as in the first equation, it will attack the nucleophilic oxygen of the carboxylic acid directly, giving a positively charged intermediate which then loses a proton. If is a weak electrophile, such as an alkyl halide, it is necessary to convert the carboxylic acid to the more nucleophilic carboxylate anion to facilitate the substitution. This is the procedure used in reactions 2 and 3. Equation 4 illustrates the use of the reagent diazomethane (CH N ) for the preparation of methyl esters. This toxic and explosive gas is always used as an ether solution (bright yellow in color). The reaction is easily followed by the evolution of nitrogen gas and the disappearance of the reagent's color. This reaction is believed to proceed by the rapid bonding of a strong electrophile to a carboxylate anion. The nature of S 2 reactions, as in equations 2 & 3, has been described elsewhere. The mechanisms of reactions 1 & 4 will be displayed by clicking the " " button below the diagram. Alkynes may also serve as electrophiles in substitution reactions of this kind, as illustrated by the synthesis of vinyl acetate from acetylene. Intramolecular carboxyl group additions to alkenes generate cyclic esters known as . Five-membered (gamma) and six-membered (delta) lactones are most commonly formed. Electrophilic species such as acids or halogens are necessary initiators of lactonizations. Even the weak electrophile iodine initiates iodolactonization of γ,δ- and δ,ε-unsaturated acids. Examples of these reactions will be displayed by clicking the " " button. Reactions in which the hydroxyl group of a carboxylic acid is replaced by another nucleophilic group are important for preparing . The alcohols provide a useful against which this class of transformations may be evaluated. In general, the hydroxyl group proved to be a poor leaving group, and virtually all alcohol reactions in which it was lost involved a prior conversion of –OH to a better leaving group. This has proven to be true for the carboxylic acids as well. Four examples of these hydroxyl substitution reactions are presented by the following equations. In each example, the new bond to the carbonyl group is colored magenta and the nucleophilic atom that has replaced the hydroxyl oxygen is colored green. The hydroxyl moiety is often lost as water, but in reaction #1 the hydrogen is lost as HCl and the oxygen as SO . This reaction parallels a similar transformation of alcohols to alkyl chlorides, although its is different. Other reagents that produce a similar conversion to acyl halides are PCl and SOBr . The amide and anhydride formations shown in equations #2 & 3 require strong heating, and milder procedures that accomplish these transformations will be described in the next chapter. A thoughtful examination of this reaction (#4) leads one to question why it is classified as a hydroxyl substitution rather than a hydrogen substitution. The following equations, in which the hydroxyl oxygen atom of the carboxylic acid is colored red and that of the alcohol is colored blue, illustrate this distinction (note that the starting compounds are in the center). In order to classify this reaction correctly and establish a plausible mechanism, the oxygen atom of the alcohol was isotopically labeled as O (colored blue in our equation). Since this oxygen is found in the ester product and not the water, the hydroxyl group of the acid must have been replaced in the substitution. A mechanism for this general esterification reaction will be displayed on clicking the " " button; also, once the mechanism diagram is displayed, a reaction coordinate for it can be seen by clicking the head of the green " " arrow. Addition-elimination mechanisms of this kind proceed by way of tetrahedral intermediates (such as and in the mechanism diagram) and are common in acyl substitution reactions. Acid catalysis is necessary to increase the electrophilic character of the carboxyl carbon atom, so it will bond more rapidly to the nucleophilic oxygen of the alcohol. Base catalysis is not useful because base converts the acid to its carboxylate anion conjugate base, a species in which the electrophilic character of the carbon is reduced. Since a tetrahedral intermediate occupies more space than a planar carbonyl group, we would expect the rate of this reaction to be retarded when bulky reactants are used. To test this prediction the esterification of acetic acid was compared with that of 2,2-dimethylpropanoic acid, (CH ) CO H. Here the relatively small methyl group of acetic acid is replaced by a larger tert-butyl group, and the bulkier acid reacted fifty times slower than acetic acid. Increasing the bulk of the alcohol reactant results in a similar rate reduction. | 5,875 | 2,901 |
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The Earth's atmosphere is composed of several layers (Figure \(\Page {1}\)). The lowest layer, the , extends from the Earth's surface up to about 6 miles or 10 kilometers (km) in altitude. Virtually all human activities occur in the troposphere. Mt. Everest, the tallest mountain on the planet, is only about 5.6 miles (9 km) high. The next layer, the , continues from 6 miles (10 km) to about 31 miles (50 km). Most commercial airplanes fly in the lower part of the stratosphere. Most atmospheric ozone is concentrated in a layer in the stratosphere, about 9 to 18 miles (15 to 30 km) above the Earth's surface (see the figure below). Ozone is a molecule that contains three oxygen atoms. At any given time, ozone molecules are constantly formed and destroyed in the stratosphere. The total amount has remained relatively stable during the decades that it has been measured. The ozone layer in the stratosphere absorbs a portion of the radiation from the sun, preventing it from reaching the planet's surface. Most importantly, it absorbs the portion of UV light called . UVB has been linked to many , including skin cancers, cataracts, xand harm to some crops and marine life. The mesosphere starts just above the stratosphere and extends to 85 kilometers (53 miles) high. Meteors burn up in this layer
The thermosphere starts just above the mesosphere and extends to 600 kilometers (372 miles) high. Aurora and satellites occur in this layer. The ionosphere is an abundant layer of electrons and ionized atoms and molecules that stretches from about 48 kilometers (30 miles) above the surface to the edge of space at about 965 km (600 mi), overlapping into the mesosphere and thermosphere. This dynamic region grows and shrinks based on solar conditions and divides further into the sub-regions: D, E and F; based on what wavelength of solar vvradiation is absorbed. The ionosphere is a critical link in the chain of Sun-Earth interactions. This region is what makes radio communications possible. The exosphere is is the upper limit of the atmosphere. It extends from the top of the thermosphere up to 10,000 km (6,200 mi). Except for water vapor, whose atmospheric abundance varies from practically zero up to 4%, the fractions of the major atmospheric components N , O , and Ar are remarkably uniform below about 100 km (Table \(\Page {1}\)). At greater heights, diffusion becomes the principal transport process, and the lighter gases become relatively more abundant. In addition, photochemical processes result in the formation of new species whose high reactivities would preclude their existence in significant concentrations at the higher pressures found at lower elevations. The atmospheric gases fall into three abundance categories: major, minor, and trace. Nitrogen, the most abundant component, has accumulated over time as a result of its geochemical inertness; a very small fraction of it passes into the other phases as a result of biological activity and natural fixation by lightning. It is believed that denitrifying bacteria in marine sediments may provide the major route for the return of N to the atmosphere. Oxygen is almost entirely of biological origin, and cycles through the hydrosphere, the biosphere, and sedimentary rocks. Argon consists mainly of Ar which is a decay product of K in the mantle and crust. The most abundant of the minor gases aside from water vapor is carbon dioxide (Table \(\Page {2}\)). Next in abundance are neon and helium. Helium is a decay product of radioactive elements in the earth, but neon and the other inert gases are primordial, and have probably been present in their present relative abundances since the earth’s formation. Two of the minor gases, ozone and carbon monoxide, have abundances that vary with time and location. A variable abundance implies an imbalance between the rates of formation and removal. In the case of carbon monoxide, whose major source is anthropogenic (a small amount is produced by biological action), the variance is probably due largely to localized differences in fuel consumption, particularly in internal combustion engines. The nature of the carbon monoxide sink (removal mechanism) is not entirely clear; it may be partly microbial. ) ( ) | 4,272 | 2,902 |
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Rutherford’s nuclear model of the atom helped explain why atoms of different elements exhibit different chemical behavior. The identity of an element is defined by its , the number of protons in the nucleus of an atom of the element. The atomic number is therefore different for each element. The known elements are arranged in order of increasing Z in the ( ). The rationale for the peculiar format of the periodic table is explained later. ach element is assigned a unique one-, two-, or three-letter symbol. The names of the elements are listed in the periodic table, along with their symbols, atomic numbers, and atomic masses. The chemistry of each element is determined by its number of protons and electrons. In a neutral atom, the number of electrons equals the number of protons. The elements are arranged in a , which is probably the single most important learning aid in chemistry. It summarizes huge amounts of information about the elements in a way that facilitates the prediction of many of their properties and chemical reactions. The elements are arranged in seven horizontal rows, in order of increasing atomic number from left to right and top to bottom. The rows are called periods, and they are numbered from 1 to 7. The elements are stacked in such a way that elements with similar chemical properties form vertical columns, called groups, numbered from 1 to 18 (older periodic tables use a system based on roman numerals). Groups 1, 2, and 13–18 are the main group elements, listed as A in older tables. Groups 3–12 are in the middle of the periodic table and are the transition elements, listed as B in older tables. The two rows of 14 elements at the bottom of the periodic table are the lanthanides and the actinides, whose positions in the periodic table are indicated in group 3. The heavy orange zigzag line running diagonally from the upper left to the lower right through groups 13–16 in divides the elements into metals (in blue, below and to the left of the line) and nonmetals (in bronze, above and to the right of the line). Gold-colored lements that lie along the diagonal line exhibit properties intermediate between metals and nonmetals; they are called semimetals. The distinction between metals and nonmetals is one of the most fundamental in chemistry. Metals—such as copper or gold—are good conductors of electricity and heat; they can be pulled into wires because they are ductile; they can be hammered or pressed into thin sheets or foils because they are malleable; and most have a shiny appearance, so they are lustrous. The vast majority of the known elements are metals. Of the metals, only mercury is a liquid at room temperature and pressure; all the rest are solids. Nonmetals, in contrast, are generally poor conductors of heat and electricity and are not lustrous. Nonmetals can be gases (such as chlorine), liquids (such as bromine), or solids (such as iodine) at room temperature and pressure. Most solid nonmetals are brittle, so they break into small pieces when hit with a hammer or pulled into a wire. As expected, semimetals exhibit properties intermediate between metals and nonmetals. Based on its position in the periodic table, do you expect selenium to be a metal, a nonmetal, or a semimetal? : element : classification : Find selenium in the periodic table shown in and then classify the element according to its location. : The atomic number of selenium is 34, which places it in period 4 and group 16. In , selenium lies above and to the right of the diagonal line marking the boundary between metals and nonmetals, so it should be a nonmetal. Note, however, that because selenium is close to the metal-nonmetal dividing line, it would not be surprising if selenium were similar to a semimetal in some of its properties. Based on its location in the periodic table, do you expect indium to be a nonmetal, a metal, or a semimetal? : metal As previously noted, the periodic table is arranged so that elements with similar chemical behaviors are in the same group. Chemists often make general statements about the properties of the elements in a group using descriptive names with historical origins. For example, the elements of Group 1 are known as the alkali metals, Group 2 are the alkaline earth metals, Group 17 are the halogens, and Group 18 are the noble gases. The alkali metals are lithium, sodium, potassium, rubidium, cesium, and francium. Hydrogen is unique in that it is generally placed in Group 1, but it is not a metal. The compounds of the alkali metals are common in nature and daily life. One example is table salt (sodium chloride); lithium compounds are used in greases, in batteries, and as drugs to treat patients who exhibit manic-depressive, or bipolar, behavior. Although lithium, rubidium, and cesium are relatively rare in nature, and francium is so unstable and highly radioactive that it exists in only trace amounts, sodium and potassium are the seventh and eighth most abundant elements in Earth’s crust, respectively. are beryllium, magnesium, calcium, strontium, barium, and radium. Beryllium, strontium, and barium are rare, and radium is unstable and highly radioactive. In contrast, calcium and magnesium are the fifth and sixth most abundant elements on Earth, respectively; they are found in huge deposits of limestone and other minerals. are fluorine, chlorine, bromine, iodine, and astatine. The name halogen is derived from the Greek words for “salt forming,” which reflects that all the halogens react readily with metals to form compounds, such as sodium chloride and calcium chloride (used in some areas as road salt). Compounds that contain the fluoride ion are added to toothpaste and the water supply to prevent dental cavities. Fluorine is also found in Teflon coatings on kitchen utensils. Although chlorofluorocarbon propellants and refrigerants are believed to lead to the depletion of Earth’s ozone layer and contain both fluorine and chlorine, the latter is responsible for the adverse effect on the ozone layer. Bromine and iodine are less abundant than chlorine, and astatine is so radioactive that it exists in only negligible amounts in nature. The noble gases are helium, neon, argon, krypton, xenon, and radon. Because the noble gases are composed of only single atoms, they are called monatomic. At room temperature and pressure, they are unreactive gases. Because of their lack of reactivity, for many years they were called inert gases or rare gases. However, the first chemical compounds containing the noble gases were prepared in 1962. Although the noble gases are relatively minor constituents of the atmosphere, natural gas contains substantial amounts of helium. Because of its low reactivity, argon is often used as an unreactive (inert) atmosphere for welding and in light bulbs. The red light emitted by neon in a gas discharge tube is used in neon lights. The noble gases are unreactive at room temperature and pressure. Periodic Law in the Periodic Table: The periodic table is an arrangement of the elements in order of increasing atomic number. Elements that exhibit similar chemistry appear in vertical columns called groups (numbered 1–18 from left to right); the seven horizontal rows are called periods. Some of the groups have widely-used common names, including the alkali metals (Group 1) and the alkaline earth metals (Group 2) on the far left, and the halogens (Group 17) and the noble gases (Group 18) on the far right. The elements can be broadly divided into metals, nonmetals, and semimetals. Semimetals exhibit properties intermediate between those of metals and nonmetals. Metals are located on the left of the periodic table, and nonmetals are located on the upper right. They are separated by a diagonal band of semimetals. Metals are lustrous, good conductors of electricity, and readily shaped (they are ductile and malleable), whereas solid nonmetals are generally brittle and poor electrical conductors. Other important groupings of elements in the periodic table are the main group elements, the transition metals, the lanthanides, and the actinides. | 8,143 | 2,904 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/Writing_Equations_for_Redox_Reactions |
This page explains how to work out electron-half-reactions for oxidation and reduction processes, and then how to combine them to give the overall ionic equation for a redox reaction. This is an important skill in inorganic chemistry. The ionic equation for the magnesium-aided reduction of hot copper(II) oxide to elemental copper is given below : \[\ce{Cu^{2+} + Mg \rightarrow Cu + Mg^{2+}}\nonumber \] The equation can be split into two parts and considered from the separate perspectives of the elemental magnesium and of the copper(II) ions. This arrangement clearly indicates that the magnesium has lost two electrons, and the copper(II) ion has gained them. \[ \ce{ Mg \rightarrow Mg^{2+} + 2e^-}\nonumber \] \[\ce{Cu^{2+} + 2e^-} \rightarrow Cu\nonumber \] These two equations are described as "electron-half-equations," "half-equations," or "ionic-half-equations," or "half-reactions." Every redox reaction is made up of two half-reactions: in one, electrons are lost (an oxidation process); in the other, those electrons are gained (a reduction process). In the example above, the electron-half-equations were obtained by extracting them from the overall ionic equation. In practice, the reverse process is often more useful: starting with the electron-half-equations and using them to build the overall ionic equation. From this information, the overall reaction can be obtained. The chlorine reaction, in which chlorine gas is reduced to chloride ions, is considered first: \[\ce{ Cl_2 \rightarrow Cl^{-}}\nonumber \] The atoms in the equation must be balanced: \[\ce{ Cl_2 \rightarrow 2Cl^{-}}\nonumber \] This step is crucial. If any atoms are unbalanced, problems will arise later. To completely balance a half-equation, all charges and extra atoms must be equal on the reactant and product sides. In order to accomplish this, the following can be added to the equation: In the chlorine case, the only problem is a charge imbalance. The left-hand side of the equation has no charge, but the right-hand side carries 2 negative charges. This is easily resolved by adding two electrons to the left-hand side. The fully balanced half-reaction is: \[\ce{ Cl_2 +2 e^- \rightarrow 2Cl^{-}}\nonumber \] Next the iron half-reaction is considered. Iron(II) ions are oxidized to iron(III) ions as shown: \[ \ce{Fe^{2+} \rightarrow Fe^{3+}}\nonumber \] The atoms balance, but the charges do not. There are 3 positive charges on the right-hand side, but only 2 on the left. To reduce the number of positive charges on the right-hand side, an electron is added to that side: \[ \ce{Fe^{2+} \rightarrow Fe^{3+} } + e-\nonumber \] It is obvious that the iron reaction will have to happen twice for every chlorine reaction. This is accounted for in the following way: each equation is multiplied by the value that will give equal numbers of electrons, and the two resulting equations are added together such that the electrons cancel out: At this point, it is important to check once more for atom and charge balance. In this case, no further work is required. The first example concerned a very simple and familiar chemical equation, but the technique works just as well for more complicated (and perhaps unfamiliar) chemistry. Manganate(VII) ions, MnO , oxidize hydrogen peroxide, H O , to oxygen gas. The reaction is carried out with potassium manganate(VII) solution and hydrogen peroxide solution acidified with dilute sulfuric acid. As the oxidizing agent, M anganate (VII) is reduced to manganese(II). The hydrogen peroxide reaction is written first according to the information given: \[ \ce{H_2O_2 \rightarrow O_2} \nonumber \] The oxygen is already balanced, but the right-hand side has no hydrogen. All you are allowed to add to this equation are water, hydrogen ions and electrons. Adding water is obviously unhelpful: if water is added to the right-hand side to supply extra hydrogen atoms, an additional oxygen atom is needed on the left. Hydrogen ions are a better choice. Adding two hydrogen ions to the right-hand side gives: \[ \ce{ H_2O_2 \rightarrow O_2 + 2H^{+}} \nonumber \] Next the charges are balanced by adding two electrons to the right, making the overall charge on both sides zero: \[ \ce{ H_2O_2 \rightarrow O_2 + 2H^{+} + 2e^{-}}\nonumber \] Next the manganate(VII) half-equation is considered: \[MnO_4^- \rightarrow Mn^{2+}\nonumber \] The manganese atoms are balanced, but the right needs four extra oxygen atoms. These can only come from water, so four water molecules are added to the right: \[ MnO_4^- \rightarrow Mn^{2+} + 4H_2O\nonumber \] The water introduces eight hydrogen atoms on the right. To balance these, eight hydrogen ions are added to the left: \[ MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O\nonumber \] Now that all the atoms are balanced, only the charges are left. There is a net +7 charge on the left-hand side (1- and 8+), but only a charge of +2 on the right. 5 electrons are added to the left-hand side to reduce the +7 to +2: \[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} _ 4H_2O\nonumber \] This illustrates the strategy for balancing half-equations, summarized as followed: Now the half-equations are combined to make the ionic equation for the reaction. As before, the equations are multiplied such that both have the same number of electrons. In this case, the least common multiple of electrons is ten: This often occurs with hydrogen ions and water molecules in more complicated redox reactions. Subtracting 10 hydrogen ions from both sides leaves the simplified ionic equation. \[ 2MnO_4^- + 6H^+ + 5H_2O_2 \rightarrow 2Mn^{2+} + 8H_2O + 5O_2\nonumber \] This technique can be used just as well in examples involving organic chemicals. Potassium dichromate(VI) solution acidified with dilute sulfuric acid is used to oxidize ethanol, CH CH OH, to ethanoic acid, CH COOH. The oxidizing agent is the dichromate(VI) ion, Cr O , which is reduced to chromium(III) ions, Cr . The ethanol to ethanoic acid half-equation is considered first: \[ CH_3CH_2OH \rightarrow CH_3COOH\nonumber \] The oxygen atoms are balanced by adding a water molecule to the left-hand side: \[ CH_3CH_2OH + H_2O \rightarrow CH_3COOH\nonumber \] Four hydrogen ions to the right-hand side to balance the hydrogen atoms: \[ CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+\nonumber \] The charges are balanced by adding 4 electrons to the right-hand side to give an overall zero charge on each side: \[ CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+ + 4e^-\nonumber \] The unbalanced dichromate (VI) half reaction is written as given: \[ Cr_2O_7^{2-} \rightarrow Cr^{3+}\nonumber \] At this stage, students often forget to balance the chromium atoms, making it impossible to obtain the overall equation. To avoid this, the chromium ion on the right is multiplied by two: \[ Cr_2O_7^{2-} \rightarrow 2Cr^{3+}\nonumber \] The oxygen atoms are balanced by adding seven water molecules to the right: \[ Cr_2O_7^{2-} \rightarrow 2Cr^{3+} + 7H_2O\nonumber \] The resulting hydrogen atoms are balanced by adding fourteen hydrogen ions to the left: \[ Cr_2O_7^{2-} + 14H^+ \rightarrow 2Cr^{3+} + 7H_2O\nonumber \] Six electrons are added to the left to give a net +6 charge on each side. \[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\nonumber \] \[ CH_3CH_2OH + H_2O \rightarrow CH_3COOH + 4H^+ + 4e^-\nonumber \] \[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O\nonumber \] The least common multiple of 4 and 6 is 12. Therefore, the first equation is multiplied by 3 and the second by 2, giving 12 electrons in each equation: Simplifying the water molecules and hydrogen ions gives final equation: Working out half-equations for reactions in alkaline solution is decidedly more tricky than the examples above. As some curricula do not include this type of problem, the process for balancing alkaline redox reactions is covered on a separate page. | 7,904 | 2,907 |
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. In addition to these individual basis; please contact 1. Explain why buffers are crucial for the proper functioning of biological systems. 2. What is the role of a buffer in chemistry and biology? Is it correct to say that buffers prevent a change in \([H_3O^+]\)? Explain your reasoning. 3. Explain why the most effective buffers are those that contain approximately equal amounts of the weak acid and its conjugate base. 4. Which region of the titration curve of a weak acid or a weak base corresponds to the region of the smallest change in pH per amount of added strong acid or strong base? 5. If you were given a solution of sodium acetate, describe two ways you could convert the solution to a buffer. 6. Why are buffers usually used only within approximately one pH unit of the \(pK_a\) or \(pK_b\) of the parent weak acid or base? 7. The titration curve for a monoprotic acid can be divided into four regions: the starting point, the region around the midpoint of the titration, the equivalence point, and the region after the equivalence point. For which region would you use each approach to describe the behavior of the solution? 8. Which of the following will produce a buffer solution? Explain your reasoning in each case. 9. Which of the following will produce a buffer solution? Explain your reasoning in each case. 10. Use the definition of Kb for a weak base to derive the following expression, which is analogous to the Henderson-Hasselbalch approximation but for a weak base (B) rather than a weak acid (HA): 11. Why do biological systems use overlapping buffer systems to maintain a constant pH? 12. The \(CO_2/HCO_3^−\) buffer system of blood has an effective \(pK_a\) of approximately 6.1, yet the normal pH of blood is 7.4. Why is \(CO_2/HCO_3^−\) an effective buffer when the \(pK_a\) is more than 1 unit below the pH of blood? What happens to the pH of blood when the \(CO_2\) pressure increases? when the \(O_2\) pressure increases? 13. Carbon dioxide produced during respiration is converted to carbonic acid (\(H_2CO_3\)). The \(pK_a1\) of carbonic acid is 6.35, and its \(pK_a2\) is 10.33. Write the equations corresponding to each pK value and predict the equilibrium position for each reaction. 1. Buffers are crucial for the proper function of biological systems because they help maintain a relatively constant \(pH\) that maintains homeostasis and this is required for bodily function. For example, buffers can prevent stomach acid from being too acidic in which tissues and other parts of the body would be harmed. 2. The role of a buffer in chemistry and biology is to resist pH changes upon the addition of an acid or base. It is correct to say that buffers prevent a change in hydronium ion concentration because the pH is a measure of the acidity or alkalinity of a solution. 3. The most effective buffers are those that contain approximately equal amounts of the weak acid and its conjugate base because when placed in the Henderson-Hasselbalch equation the \(pH\) approximately equals the \(pK_{a}\). This is the goal because the more the ratio needs to differ to achieve the desired pH, the less effective the buffer and it should not differ by more than tenfold. 4. The region after the equivalence point on a titration curve of a weak acid or a weak base corresponds to the smallest change in pH per amount of added strong acid or strong base. This is because the strong acid or strong base completely neutralized the weak acid or weak base in which the pH remains constant relative to the strong acid or strong base. 5. Sodium acetate is a salt, therefore it can’t be regarded as an acid or a base. One way to convert the solution to a buffer is to combine sodium acetate with a weak acid such as acetic acid. The sodium acetate would react with the hydronium ions and the acetic acid would react with hydroxide ions to prevent drastic changes in pH. Another way to convert the solution to a buffer is by titrating a weak acid and strong base or strong base and weak acid. For example, combining acetic acid to sodium hydroxide to produce sodium acetate (a neutralization reaction). 6. Buffers are usually used only within approximately one \(pH\) unit of the \(pK_a\) or \(pK_b\) of the parent weak acid or base because the ability of a buffer solution to maintain a nearly constant pH due to a small amount of acid or base is greatest at the \(pK_a\) and decreases as the pH of the solution goes above or below the \(pK_a\). \(pH=pK_a+log(\frac{[B]}{[A]})\) When \(\frac{[B]}{[A]}=10\): \(pH=pK_a+1\) When \(\frac{[B]}{[A]}=\frac{1}{10}\): \(pH=pK_a-1\) 7. a. I would use the region around the midpoint of the titration to describe the behavior of a buffer because the midpoint of the titration is defined as the point at which exactly enough acid or base has been added to neutralize one-half of the acid or the base originally present while maintaining a relatively constant pH. b. I would use the equivalence point region to describe the behavior of a solution of a salt of a weak base because at this point the amount of titrant added is enough to completely neutralize the solution. c. I would use the starting point region of the titration curve to describe the behavior of the solution of a weak acid because it is expected to be at a higher pH than a strong acid but lower than \(pH\) 7. d. I would use the region after the equivalence point to describe the dilution of a strong base because in this region one can predict the pH by simply taking into account the amount of excess base. 8. a. This will not produce a buffer solution because the hydrochloric acid completely neutralizes the sodium fluoride to give sodium chloride. b. This will produce a buffer solution because the hydrochloric acid neutralizes only half of the acetic acid to give a solution containing equal amounts of hydrofluoric acid and sodium chloride. c. This will not produce a buffer solution because hydrofluoric acid is a weak acid and hydrochloric acid is a strong acid. d. This will produce a buffer solution because the solution will contain a 2:1 ratio of hydrofluoric acid and sodium hydroxide. e. This will not produce a buffer solution solution because sodium fluoride is a weak base and sodium hydroxide is a strong base. 9. a. This will not produce a buffer because the hydrochloric acid completely neutralizes the sodium acetate to give acetic acid and sodium chloride. b. This will produce a buffer because the HCl neutralizes only half of the sodium acetate to give a solution containing equal amounts of acetic acid and sodium acetate. c. This will not produce a buffer because the sodium hydroxide completely neutralizes the acetic acid to give sodium acetate. d. This will produce a buffer because the sodium hydroxide neutralizes only half of the acetic acid to give a solution containing equal amounts of acetic acid and sodium acetate. e. This will produce a buffer because the solution will contain a 2:1 ratio of sodium acetate and acetic acid. 10. For \(pH\): \(HA \overset{K_a}{\rightleftharpoons} A^{-}+H^{+}\) \(K_a= \frac{[H^{+},A^{-}]}{[HA]}\) \(log(K_a)=log([H^{+}])+log(\frac{[A^{-}]}{[HA]})\) \(-pK_a=-pH+log(\frac{[A^{-}]}{[HA]})\) \(pH=pK_a+log(\frac{[A^{-}]}{[HA]})\) For pOH: \(A^{-} \overset{K_b}{\rightleftharpoons} HA+OH^{-}\) \(K_b= \frac{[HA,OH^{-}]}{[A^{-}]}\) \(log(K_b)=log([OH^{-}])+log(\frac{[HA]}{[A^{-}]})\) \(-pK_b=-pOH-log(\frac{[A^{-}]}{[HA]})\) \(pOH=pK_b-log(\frac{[A^{-}]}{[HA]})\) Helpful Equations: \(pK_a=-log(K_a) \rightarrow -pK_a=log(K_a)\) \(-log([H^{+}])=pH \rightarrow log([H^{+}])=-pH\) \(pK_b=-log(K_b) \rightarrow -pK_a=log(K_b)\) \(-log([OH^{-}])=pOH \rightarrow log([OH^{-}])=-pOH\) 11. Biological systems use overlapping buffer systems to maintain a constant pH because no single buffer system can effectively maintain a constant pH value over the entire physiological range of approximately \(pH\) 5.0 to 7.4. 12. \(CO_2/HCO_3^−\) is an effective buffer when the \(pK_a\) is more than 1 unit below the \(pH\) of blood because the kidneys help control acid-base balance by excreting hydrogen ions and generating bicarbonate that helps maintain blood plasma \(pH\) within the normal range. The bicarbonate ion can combine with a proton to form carbonic acid and the more protons are absorbed from the solution the more the \(pH\) increases, resulting in hemoglobin absorbing oxygen. Carbonic acid, which can be formed from carbon dioxide and water, can dissociate into a proton and bicarbonate ion which increases the hydrogen ion concentration and lowers the blood \(pH\), resulting in hemoglobin proteins releasing oxygen. The carbonic acid, which can be formed from bicarbonate, is converted to carbon dioxide and water via a very fast enzymatic reaction. Carbon dioxide, being volatile, can be rapidly expelled from the body at varying rates by respiration. 13. \(H_{2}CO_{3} \rightleftharpoons HCO_{3}^{-}+H^{+}\) \(pK_{a1}=6.35\) left \(HCO_{3}^{-} \rightleftharpoons CO_{3}^{2-}+H^{+}\) \(pK_{a2}=10.33\) right 1.Benzenesulfonic acid (\(pKa = 0.70\)) is synthesized by treating benzene with concentrated sulfuric acid. Calculate the following: 2. Phenol has a \(pK_a\) of 9.99. Calculate the following: 3. Salicylic acid is used in the synthesis of acetylsalicylic acid, or aspirin. One gram dissolves in 460 mL of water to create a saturated solution with a \(pH\) of 2.40. 4. An intermediate used in the synthesis of perfumes is valeric acid, also called pentanoic acid. The \(pK_a\) of pentanoic acid is 4.84 at 25°C. 1. a. \(pH=-log([H^{+}])=-log(0.2388)=0.62\) \(K_a=\frac{[H^{+},A^{-}]}{[HA]} \rightarrow K_a=\frac{[H^{+}]^{2}}{[HA]} \rightarrow [H^{+}]= \sqrt{K_a\times[HA]}=\sqrt{10^{-0.70} \times 0.286\,M}=0.2388\) b. \(pH=-log([H^{+}])=-log(0.14125)=0.850\) \(K_a=\frac{[H^{+},A^{-}]}{[HA]} \rightarrow K_a=\frac{[H^{+}]^{2}}{[HA]} \rightarrow [H^{+}]= \sqrt{K_a\times[HA]}=\sqrt{10^{-0.70} \times 0.100\,M}=0.14125\) 2. a. \(pH=-log(4.46 \times 10^{-6})=5.35\) \(K_a=\frac{[C_6H_5O^{-},H_3O^{+}]}{[C_6H_5OH]}=\frac{x^2}{0.195} \rightarrow 10^{-9.99}=\frac{x^2}{0.195} \rightarrow x=4.46 \times 10^{-6}\) \(Assumption\,Valid: \frac{4.46 \times 10^{-6}}{0.195\,M} \times 100 \% <5 \%\) \(C_6H_5OH\) \(H_2O\) \(C_6H_5O^{-}\) \(H_3O^{+}\) I 0.195 - 0 0 C -x - +x +x E 0.195-x - x x b. \(Percent\,increase=\frac{3.20 \times 10^{-6}}{4.46 \times 10^{-6}}x100 \% =71.6 \% \) \(K_a=\frac{[C_6H_5O^{-},H_3O^{+}]}{C_6H_5OH}=\frac{x^2}{0.100} \rightarrow 10^{-9.99}=\frac{x^2}{0.100} \rightarrow x=3.20 \times 10^{-6}\) 3. \(K_a=\frac{[C_9H_8O_4,H_3O^{+}]}{[C_7H_6O_3]}=\frac{x^{2}}{0.015739} \rightarrow \frac{(3.98 \times 10^{-3})^{2}}{0.015739}= 1.0 \times 10^{-3}\) \(C_{7}H_{6}O_{3}:1\,g \times \frac{1\,mol}{138.121\,g\,mol} \times \frac{1}{460\,mL \times \frac {1\,L}{1000\,mL}}=0.015739\,M\) \(pH=-log([H^{+}]) \rightarrow [H^{+}]=10^{-pH}=10^{-2.40}=3.98 \times 10^{-3}\) \(pH=pK_a+log(\frac{[C_{7}H_{5}NaO_{3}]}{[C_{7}H_{6}O_{3}]})=-log(1.01 \times 10^{-3})+log(\frac{0.238}{0.015739})=4.17\) \(pH=pK_a+log(\frac{mol\,C_{7}H_{5}NaO_{3}-mol\,HCl}{mol\,C_9H_8O_4+mol\,HCl})=-log(1.01 \times 10^{-3})+log(\frac{0.0357-0.001}{0.00236085+0.001})=4.01\) \(HCl:.01\,L \times 0.100\,M=0.001\,mol\) \(C_{7}H_{5}NaO_{3}: 0.15\,L \times 0.238\,M=0.0357\,mol\) \(C_{7}H_{6}O_{3}:0.15\,L \times 0.015739\,M=0.00236085\,mol\) \(pH=pK_a+log(\frac{mol\,C_{7}H_{5}NaO_{3}+mol\,NaOH}{mol\,C_9H_8O_4-mol\,NaOH})=-log(1.01 \times 10^{-3})+log(\frac{0.0357+0.001}{0.00236085-0.001})=4.43\) \(NaOH:.01\,L \times 0.100\,M=0.001\,mol\) \(C_{7}H_{5}NaO_{3}: 0.15\,L \times 0.238\,M=0.0357\,mol\) \(C_{7}H_{6}O_{3}:0.15\,L \times 0.015739\,M=0.00236085\,mol\) 4. a. \(pH=-log([H^{+}])=-log(3.7 \times 10^{-6})=5.43\) \(K_a=\frac{[H^{+},A^{-}]}{[HA]} \rightarrow K_a=\frac{[H^{+}]^{2}}{[HA]} \rightarrow [H^{+}]=\sqrt{K_a\times[HA]}=\sqrt{10^{-4.84} \times 0.259\,M}=3.7 \times 10^{-6}\) b. \(pH=pK_a+log(\frac{B}{A})=4.84+log(\frac{0.210}{0.259})=4.75\) c. \(pH=pK_a+log(\frac{B}{A})=4.84+log(\frac{mol\,C_{5}H_{9}NaO_{2}-mol\,HCl}{mol\,C_{5}H_{10}O_{2}+mol\,HCl})=4.84+log(\frac{0.01575-8 \times 10^{-4}}{0.019425+8\times 10^{-4}})=4.71\) \(HCl:8.00\,mL \times \frac{1\,L}{1,000\,mL} \times 0.100\,M=8 \times 10^{-4}\,mol\) \(C_{5}H_{9}NaO_{2}:75.0\,mL \times \frac{1\,L}{1,000\,mL} \times 0.210\,M=0.01575\,mol\) \(C_{5}H_{10}O_{2}: 75.0 \,mL \times \frac{1\,L}{1,000\,mL} \times 0.259\,M=0.019425\,mol\) d. \(pH=pK_a+log(\frac{B}{A})=4.84+log(\frac{mol\,C_{5}H_{9}NaO_{2}+mol\,NaOH}{mol\,C_{5}H_{10}O_{2}-mol\,NaOH})=4.84+log(\frac{0.01575+8 \times 10^{-4}}{0.019425-8\times 10^{-4}})=6.07\) \(NaOH:8.00\,mL \times \frac{1\,L}{1,000\,mL} \times 0.100\,M=8 \times 10^{-4}\,mol\) \(C_{5}H_{9}NaO_{2}:75.0\,mL \times \frac{1\,L}{1,000\,mL} \times 0.210\,M=0.01575\,mol\) \(C_{5}H_{10}O_{2}: 75.0 \,mL \times \frac{1\,L}{1,000\,mL} \times 0.259\,M=0.019425\,mol\) 1. Why is the portion of the titration curve that lies below the equivalence point of a solution of a weak acid displaced upward relative to the titration curve of a strong acid? How are the slopes of the curves different at the equivalence point? Why? 2. Predict whether each solution will be neutral, basic, or acidic at the equivalence point of each titration. 3. The pKa values of phenol red, bromophenol blue, and phenolphthalein are 7.4, 4.1, and 9.5, respectively. Which indicator is best suited for each acid–base titration? 4. For the titration of any strong acid with any strong base, the \(pH\) at the equivalence point is 7.0. Why is this not usually the case in titrations of weak acids or weak bases? 5. Why are the titration curves for a strong acid with a strong base and a weak acid with a strong base identical in shape above the equivalence points but not below? 6. Describe what is occurring on a molecular level during the titration of a weak acid, such as acetic acid, with a strong base, such as \(NaOH\), at the following points along the titration curve. Which of these points corresponds to \(pH=pK_{a}\)? 7. On a molecular level, describe what is happening during the titration of a weak base, such as ammonia, with a strong acid, such as \(HCl\), at the following points along the titration curve. Which of these points corresponds to \(pOH=pK_{b}\)? 8. For the titration of a weak acid with a strong base, use the \(K_{a}\) expression to show that \(pH=pK_{a}\) at the midpoint of the titration. 9. Chemical indicators can be used to monitor \(pH\) rapidly and inexpensively. Nevertheless, electronic methods are generally preferred. Why? 10. Why does adding ammonium chloride to a solution of ammonia in water decrease the pH of the solution? 11. Given the equilibrium system \(CH_3CO_2H\,(aq) \rightleftharpoons CH_3CO_2^{-}\,(aq) + H^{+}\,(aq)\), explain what happens to the position of the equilibrium and the \(pH\) in each case. 12. Given the equilibrium system \(CH_3NH_2\,(aq) + H_2O\,(l) \rightleftharpoons CH_{3}NH_3^{+}\,(aq) + OH^{-}\,(aq)\), explain what happens to the position of the equilibrium and the \(pH\) in each case. 1. The portion of the titration curve that lies below the equivalence point of a solution of a weak acid is displaced upward relative to the titration curve of a strong acid because the starting point of a weak acid is at a higher \(pH\) compared to that of a strong acid. The slope for a strong acid is greater compared to that of a weak acid because when a weak acid is neutralized, the solution that remains is basic because the acid’s conjugate base remains in solution. 2. a. The solution will be neutral at the equivalence point because sodium hydroxide is a strong base and hydrochloric acid is a strong acid. b. The solution will be acidic at the equivalence point because ethylamine is a weak base and nitric acid is a strong acid. c. The solution will be basic at the equivalence point because aniline hydrochloride is a weak acid and potassium hydroxide is a strong base. 3. a. Barium hydroxide is a strong base and hydrochloric acid is a strong acid thus phenol red with the \(pK_a=7.4\) is best suited for the reaction. b. Trimethylamine is a weak base and nitric acid is a strong acid thus bromophenol blue with the \(pK_a=4.1\) is best suited for the reaction. c. Aniline hydrochloride is a weak acid and potassium hydroxide is a strong base thus phenolphthalein with \(pK_a=9.5\) is best suited for the reaction. 4. This is not usually the case in the titration of weak acids and weak bases because weak acids and weak bases would only ionize partially, thus complete neutralization does not occur. 5. The titration curve for a strong acid with a strong base is identical in shape above the equivalence point of the titration curve with a weak acid and strong base because both cases involve the addition of strong base. Not only does the strong base completely neutralize the acid but there is an excess of a strong base that makes the solution basic. 6. a. The titration begins with a \(pH\) is higher than the titration of a strong acid. At the beginning of the titration, there is a sharp increase in \(pH\) because the anion of the weak acid becomes a common ion that reduces the ionization of the acid. b. There is a sharp increase at the beginning of the titration that changes gradually due to the solution becoming a buffer. This continues until the strong base has overcome the buffer capacity. At the midpoint of the titration, the concentration of the weak acid is equal to the concentration of its conjugate base. This point is also known as half-neutralization because half the acid has been neutralized by the strong base. c. At the equivalence point the pH is greater than 7 because the acid (\(HA\)) has been converted to its conjugate base (\(A^{-}\)) by sodium hydroxide and equilibrium moves backward toward the acid and produces hydroxide: \(A^{-}+H_2O \rightleftharpoons AH+OH^{-}\) d. When the excess titrant has been added the solution becomes basic because sodium hydroxide completely neutralized the weak acid. 7. a. The titration begins with a \(pH\) that is lower than the titration of a strong base but higher than the \(pH=7\). At the beginning of the titration, there is a sharp decrease in \(pH\) because the cation of the weak base becomes a common ion that reduces the ionization of the base. b. The sharp decrease at the beginning of the titration changes gradually due to the solution becoming a buffer. This continues until the strong acid has overcome the buffer capacity. At the midpoint of the titration, the concentration of the weak base is equal to the concentration of its conjugate acid \(pH=pK_a\). This point is also known as half-neutralization because half the base has been neutralized by the strong acid. c. At the equivalence point the \(pH\) is less than 7 because the base \(NH_{3}\) has been converted to its conjugate acid \(NH_{4}^{+}\) by hydrochloric acid and equilibrium moves forwards toward the base and produces \(NH_{4}Cl^{-}\): \(NH_{3}+HCl \rightleftharpoons NH_{4}Cl\) d. When excess titrant has been added the solution becomes acidic because hydrochloric acid completely neutralized the weak base. 8. \(HA \overset{K_a}{\rightleftharpoons} A^{-}+H^{+}\) \(K_a= \frac{[H^{+},A^{-}]}{[HA]}\) \(pH=pK_a+log(\frac{[A^{-}]}{[HA]}\) Assume \([A^{-}]=[HA]=1\) \(pH=pK_a+log(1)\) \(pH=pK_a+0\) \(pH=pK_a\) 9. Electronic methods are preferred over chemical indicators because it is a more accurate method at determining the pH. Also, chemical indicators must be selected to observe a narrow pH range while electronic methods observe a wider range. 10. Ammonia is a weak base and dissociates in water as: \(NH_3\,(aq)+H_2O\,(l) \rightleftharpoons NH_4^{+}\,(aq)+OH^{-}\,(aq)\) When \(NH_4Cl\) is added, it 100% dissociates into \(NH_4^{+}\) and \(Cl^{-}\) in water: \(NH_{4}Cl\,(aq) \rightleftharpoons NH_{4}^{+}\,(aq)+Cl^{-}\,(aq)\) Due to common ion, \(NH_4^{+}\,(aq)\) the dissociation of \(NH_{3}\) will decrease. Thus, the concentration of \(OH^{-}\,(aq)\) decreases and the concentration of \(H_{3}O^{+}\) increases. Since \(pH=-log(H_{3}O^{+})\) the \(pH\) and \(H_{3}O^{+}\) are inversely related, the concentration of \(H_{3}O^{+}\) concentration increases and \(pH\) decreases. 11. a. The position of equilibrium shift to the left and the pH decreases. b. The position of equilibrium shifts to the right and the pH increases. c. The position of equilibrium shifts to the left and the pH increases. 12. a. The position of equilibrium shifts to the left and the pH decreases. b. The position of equilibrium shifts to the left and the pH increases. c. The position of equilibrium shifts to the left and the pH increases. 1. Calculate the pH of each solution. 2. What is the pH of a solution prepared by mixing 50.0 mL of 0.225 M HCl with 100.0 mL of a 0.184 M solution of \(NaOH\)? 3. What volume of 0.50 M HCl is needed to completely neutralize 25.00 mL of 0.86 M \(NaOH\)? 4.Calculate the final pH when each pair of solutions is mixed. 5. Calculate the final pH when each pair of solutions is mixed. 6. Calcium carbonate is a major contributor to the “hardness” of water. The amount of CaCO3 in a water sample can be determined by titrating the sample with an acid, such as HCl, which produces water and CO2. Write a balanced chemical equation for this reaction. Generate a plot of solution pH versus volume of 0.100 M HCl added for the titration of a solution of 250 mg of CaCO3 in 200.0 mL of water with 0.100 M HCl; assume that the HCl solution is added in 5.00 mL increments. What volume of HCl corresponds to the equivalence point? 7. For a titration of 50.0 mL of 0.288 M \(NaOH\), you would like to prepare a 0.200 M HCl solution. The only HCl solution available to you, however, is 12.0 M. 8. While titrating 50.0 mL of a 0.582 M solution of HCl with a solution labeled “0.500 M KOH,” you overshoot the endpoint. To correct the problem, you add 10.00 mL of the HCl solution to your flask and then carefully continue the titration. The total volume of titrant needed for neutralization is 71.9 mL. 9. Complete the following table and generate a titration curve showing the pH versus volume of added base for the titration of 50.0 mL of 0.288 M HCl with 0.321 M \(NaOH\). Clearly indicate the equivalence point. 10. The following data were obtained while titrating 25.0 mL of 0.156 M \(NaOH\) with a solution labeled “0.202 M HCl.” Plot the pH versus volume of titrant added. Then determine the equivalence point from your graph and calculate the exact concentration of your HCl solution. 11. Fill in the data for the titration of 50.0 mL of 0.241 M formic acid with 0.0982 M KOH. The pKa of formic acid is 3.75. What is the pH of the solution at the equivalence point? 12. Glycine hydrochloride, which contains the fully protonated form of the amino acid glycine, has the following structure: It is a strong electrolyte that completely dissociates in water. Titration with base gives two equivalence points: the first corresponds to the deprotonation of the carboxylic acid group and the second to loss of the proton from the ammonium group. The corresponding equilibrium equations are as follows: \[ ^{+}NH_{3}-CH_{2}-CO_{2}H\left ( aq \right ) \rightleftharpoons \;\;\;\;\; pK_{a1}=2.3\] \[ ^{+}NH_{3}-CH_{2}-CO_{2}\left ( aq \right )+ H^{+} \] \[ ^{+}NH_{3}-CH_{2}-CO_{2}\left ( aq \right ) \rightleftharpoons \;\;\;\;\; pK_{a2}=9.6\] \[ NH_{2}-CH_{2}-COO\left ( aq \right )+ H^{+} \] Given 50.0 mL of solution that is 0.430 M glycine hydrochloride, how many milliliters of 0.150 M KOH are needed to fully deprotonate the carboxylic acid group? 13. What is the pH of a solution prepared by adding 38.2 mL of 0.197 M HCl to 150.0 mL of 0.242 M pyridine? The pKb of pyridine is 8.77. 14. What is the pH of a solution prepared by adding 40.3 mL of 0.289 M \(NaOH\) to 150.0 mL of 0.564 M succinic acid (\(HO_2CCH_2CH_2CO_2H\))? (For succinic acid, pKa1 = 4.21 and pKa2 = 5.64). 14. Calculate the pH of a 0.15 M solution of malonic acid (\(HO_2CCH_2CO_2H\)), whose pKa values are as follows: pKa1 = 2.85 and pKa2 = 5.70. 1. a.\(pH=-log([H_3O^{+}])=-log(1.218)=-0.086\) \([HCl]=[H^{+}]\) \(M_{1}V_{1}=M_{2}V_{2} \rightarrow (6.09\,M)(25\,mL)=(x)(125\,mL) \rightarrow x=1.218\) b. \(pH+pOH=14 \rightarrow pH=14-pOH=14-0.80=13\) \(pOH=-log([OH^{-}])=-log(0.159)=0.80\) \([NaOH]=[OH^{-}]\) \(M_{1}V_{1}=M_{2}V_{2} \rightarrow (2.55\,M)(5.0\,mL)=(x)(80\,mL) \rightarrow x=0.159\) 2. \(pH+pOH=14 \rightarrow pH=14-pOH=14-1.321=12.7\) \(HCl:50.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.225\,mol}{1\,L}=0.01125\,mol\) \(NaOH:100.0\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.184\,mol}{1\,L}=0.0184\,mol\) Note the number of moles for \(NaOH\) is greater than \(HCl\), therefore we calculate \(pOH\). \(0.0184-0.01125=0.00715\,mol\) \(\frac{0.00715\,mol}{.15\,L}=0.0477\) \(pOH=-log([OH^{-}])=-log(0.0477)=1.321\) 3. \(M_{1}V_{1}=M_{2}V_{2} \rightarrow (0.50\,M)(x\,L)=(0.86\,M)(0.25\,L) \rightarrow x\,L=4.3 \times 10^{-2}\,L\) 4. Solution a. \(pH=pK_a+log(\frac{C_{2}H_{3}NaO_{2}}{C_{2}H_{4}O_{2}})=4.76+log(\frac{0.0935}{0.0105})=5.71\) \(HCl: 100\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.0105\,mol}{1\,L}=0.0105\,mol\) \(C_2H_3NaO_2: 100\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.115\,mol}{1\,L}=0.0115\,mol\) \(C_{2}H_{3}NaO_{2}\,(s)+HCl\,(aq) \rightarrow C_{2}H_{4}O_{2}\,(aq)+NaCl\,(s)\) \(C_{2}H_{3}NaO_{2}\) \(HCl\) \(C_{2}H_{4}O_{2}\) \(NaCl\) I 0.0115 0.0105 0 - C -0.0105 -0.0105 +0.0105 - E 0.0935 0 0.0105 - b. \(pH=pK_a+log(\frac{C_{2}H_{3}NaO_{2}}{C_{2}H_{4}O_{2}})=4.76+log(\frac{0.01}{0.005})=5.05\) \(HCl: 100\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.0105\,mol}{1\,L}=0.005\,mol\) \(C_2H_3NaO_2: 100\,mL \times \frac{1\,L}{1,000\,mL} \times \frac{0.115\,mol}{1\,L}=0.015\,mol\) \(C_{2}H_{3}NaO_{2}\,(s)+HCl\,(aq) \rightarrow C_{2}H_{4}O_{2}\,(aq)+NaCl\,(s)\) \(C_{2}H_{3}NaO_{2}\) \(HCl\) \(C_{2}H_{4}O_{2}\) \(NaCl\) I 0.015 0.005 0 0 C -0.005 -0.005 +0.005 +0.005 E 0.01 0 0.005 0.005 c. \(pH=14-pOH=14-(-log(\frac{0.0118-0.0109}{0.2\,L}))=11.7\) \(CH_{3}COOH: 0.1\,L \times 0.109\,M=0.0109\,mol\) \(NaOH: 0.1\,L \times 0.118\,M=0.0118\,mol\) \(CH_{3}COOH\,(aq)+NaOH\,(aq) \rightarrow H_{2}O\,(l)+CH_{3}COO^{-}Na\,(aq)\) \(CH_{3}COOH\) \(NaOH\) \(H_{2}O\) \(CH_{3}COONa\) I 0.0109 0.0118 - 0 C -0.0109 -0.0109 - +0.0109 E 0 \(9 \times 10^{-4}\) - 0.0109 d. \(pH=4.76+log(\frac{0.0055}{0.0943})=3.53\) \(CH_{3}COOH:0.1\,L \times 0.998\,M=0.0998\,mol\) \(NaOH: 0.05\,L \times 0.110\,M=0.0055\,mol\) \(CH_{3}COOH\,(aq)+NaOH\,(aq) \rightarrow H_{2}O\,(l)+CH_{3}COONa\,(aq)\) \(CH_{3}COOH\) \(NaOH\) \(H_{2}O\) \(CH_{3}COONa\) I 0.0998 0.0055 - 0 C -0.0055 -0.0055 - +0.0055 E 0.0943 0 - 0.0055 5. Solution a. \(pH=-log(\frac{0.9728}{0.2\,L})=4.86\) \(HCl:100\,mL \times \frac{1\,L}{1,000\,L} \times0.983\,M=0.983\,mol\) \(NaF:100\,mL \times \frac{1\,L}{1,000\,L} \times 0.102\,M=0.0102\,mol\) \(HCl\,(aq)+NaF\,(s) \rightarrow NaCl\,(aq)+HF\,(aq)\) \(HCl\) \(NaF\) \(NaCl\) \(HF\) I 0.983 0.0102 - 0 C -0.0102 -0.0102 - -0.0102 E 0.9728 0 - -0.0102 b. \(pH=pK_a+log(\frac{[NaF]}{[HF]})=3.14+log(\frac{0.00515}{0.00575})=3.09\) \(HCl:50\,mL \times \frac{1\,L}{1,000\,L} \times 0.115\,M=0.00575\,mol\) \(NaF:100\,mL \times \frac{1\,L}{1,000\,L} \times 0.109\,M=0.0109\,mol\) \(HCl\,(aq)+NaF\,(s) \rightarrow NaCl\,(aq)+HF\,(aq)\) \(HCl\) \(NaF\) \(NaCl\) \(HF\) I 0.00575 0.0109 - 0 C -0.00575 -0.00575 - +0.00575 E 0 0.00515 - 0.00575 c. \(pH=14-pOH=14-(-log(0.0498-0.0106))=12.6\) \(HF:100\,mL \times \frac{1\,L}{1,000\,L} \times 0.106\,M=0.0106\,mol\) \(NaOH: 50\,mL \times \frac{1\,L}{1,000\,L} \times 0.996\,M=0.0498\,mol\) \(HF\,(aq)+NaOH\,(aq) \rightarrow NaF\,(aq)+H_{2}O\,(aq)\) \(HCl(aq)+NaOH(aq) \rightarrow NaCl\,(aq)+H_2O\) d.\(pH=pK_a+log(\frac{[C_2H_3NaO_2]}{[CH_{3}COOH]})=4.76+log(\frac{0.0107}{0.04935})=4.10\) \(C_2H_3NaO_2: 0.1\,L \times 0.107\,M=0.0107\) \(CH_3COOH: 0.05\,mL \times 0.987\,M=0.04935\) \(CH_{3}COOH\,(aq)+H_{2}O \rightarrow C_2H_3O^{-}+H_3O^{+}\) 6. \(CaCO_{3}\,(aq)+2\,HCl\,(aq) \rightarrow CaCl_{2}\,(aq)+H_2O\,(l)+CO_{2}\,(g)\) \(CO_{3}^{2-}\,(aq)+H^{+}\,(aq) \rightleftharpoons HCO_{3}^{-}\,(aq)\) \(HCO_{3}^{2-}\,(aq)+H^{+}\,(aq) \rightleftharpoons H_{2}CO_{3}\,(aq)\) \(K_{a1}=4.3 \times 10^{-7}\) \(K_{a2}=5.0 \times 10^{-11}\) At 0 mL addition of \(HCl\): \(pH=14-pOH=14-(-log(\sqrt{0.0125 \times \frac{10^{-14}}{5.0 \times 10^{-11}}})=11.20\) \(CaCO_{3}:250\,mg \times \frac{1\,g}{1,000\,mg} \times \frac{1\,mol}{100.0869\,g} \times \frac{1}{200\,mL \times \frac{1\,L}{1,000\,mL}}=0.0125\,M\) At 5 mL addition of \(HCl\): \(pH=-log(5 \times 10^{-11})+log(\frac{0.002}{0.0005})=10.9\) \(CaCO_{3}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.005\,L \times 0.1\,M=0.0005\,mol\) \(CO_{3}^{2-}\) \(H^{+}\) \(HCO_{3}^{-}\) I 0.0025 0.0005 0 C -0.0005 -0.0005 0.0005 E 0.002 0 0.0005 At 10 mL addition of \(HCl\): \(pH=-log(5 \times 10^{-11})+log(\frac{0.001}{0.0015})=10.5\) \(CaCO_{3}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.01\,L \times 0.1\,M=0.001\,mol\) \(CO_{3}^{2-}\) \(H^{+}\) \(HCO_{3}^{-}\) I 0.0025 0.001 0 C -0.001 -0.001 0.001 E 0.0015 0 0.001 At 15 mL addition of \(HCl\):\(pH=-log(5 \times 10^{-11})+log(\frac{0.001}{0.0015})=10.1\) \(CaCO_{3}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.015\,L \times 0.1\,M=0.0015\,mol\) \(CO_{3}^{2-}\) \(H^{+}\) \(HCO_{3}^{-}\) I 0.0025 0.0015 0 C -0.0015 -0.0015 0.0015 E 0.001 0 0.0015 At 20 mL addition of \(HCl\):\(pH=-log(5 \times 10^{-11})+log(\frac{5 \times 10^{-4}}{0.002})=9.7\) \(CaCO_{3}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.02\,L \times 0.1\,M=0.002\,mol\) \(CO_{3}^{2-}\) \(H^{+}\) \(HCO_{3}^{-}\) I 0.0025 0.002 0 C -0.002 -0.002 0.002 E \(5 \times 10^{-4}\) 0 0.002 At 25 mL addition of \(HCl\): first equivalence point: \(pH=\frac{-log(K_{a1})-log(K_{a2})}{2}=\frac{-log(4.3 \times 10^{-7})-log(5.0 \times 10^{-11})}{2}=8.33\) \(HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.025\,L \times 0.1\,M=0.0025\,mol\) \(HCO_{3}^{-}\) \(H^{+}\) \(H_{2}CO_{3}\) I 0.0025 0.0025 0 C -0.0025 -0.0025 0.0025 E 0 0 0.0025 At 30 mL addition of \(HCl\): \(pH=-log(4.3 \times 10^{-7})+log(\frac{0.0025}{0.0005})=7.07\) \(HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.03\,L \times 0.1\,M=0.003\) \(HCO_{3}^{-}\) \(H^{+}\) \(H_{2}CO_{3}\) I 0.0025 0.003 0 C -0.0025 -0.0025 0.0025 E 0 \(5 \times 10^{-4}\) 0.0025 At 35 mL addition of \(HCl\): \(pH=-log(4.3 \times 10^{-7})+log(\frac{0.0025}{0.001})=6.76\) \(HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.035\,L \times 0.1\,M=0.0035\) \(HCO_{3}^{-}\) \(H^{+}\) \(H_{2}CO_{3}\) I 0.0025 0.0035 0 C -0.0025 -0.0025 0.0025 E 0 0.001 0.0025 At 40 mL addition of \(HCl\): \(pH=-log(4.3 \times 10^{-7})+log(\frac{0.0025}{0.003})=6.59\) \(HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.040\,L \times 0.1\,M=0.004\) \(HCO_{3}^{-}\) \(H^{+}\) \(H_{2}CO_{3}\) I 0.0025 0.004 0 C -0.0025 -0.0025 0.0025 E 0 0.0015 0.0025 At 45 mL addition of \(HCl\): \(pH=-log(4.3 \times 10^{-7})+log(\frac{0.0025}{0.002})=6.46\) \(HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.045\,L \times 0.1\,M=0.0045\) \(HCO_{3}^{-}\) \(H^{+}\) \(H_{2}CO_{3}\) I 0.0025 0.0045 0 C -0.0025 -0.0025 0.0025 E 0 0.002 0.0025 At 50 mL addition of \(HCl\): second equivalence point: \(pH=3.86\) \(HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.05\,L \times 0.1\,M=0.005\) \(HCO_{3}^{-}\) \(H^{+}\) \(H_{2}CO_{3}\) I 0.0025 0.005 0 C -0.0025 -0.0025 0.0025 E 0 0.0025 0.0025 At 55 mL addition of \(HCl\): \(pH=-log(\frac{0.003}{0.055})=1.26\) \(HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.055\,L \times 0.1\,M=0.0055\) \(HCO_{3}^{-}\) \(H^{+}\) \(H_{2}CO_{3}\) I 0.0025 0.0055 0 C -0.0025 -0.0025 0.0025 E 0 0.003 0.0025 At 60 mL addition of \(HCl\): \(pH=-log(\frac{0.0035}{0.06})=1.23\) \(HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.06\,L \times 0.1\,M=0.006\) \(HCO_{3}^{-}\) \(H^{+}\) \(H_{2}CO_{3}\) I 0.0025 0.006 0 C -0.0025 -0.0025 0.0025 E 0 0.0035 0.0025 At 65 mL addition of \(HCl\): \(pH=-log(\frac{0.0040}{0.065})=1.21\) \(HCO_{3}^{-}:0.2\,L \times 0.0125\,M=0.0025\,mol\) \(HCl: 0.06\,L \times 0.1\,M=0.006\) \(HCO_{3}^{-}\) \(H^{+}\) \(H_{2}CO_{3}\) I 0.0025 0.0065 0 C -0.0025 -0.0025 0.0025 E 0 0.0040 0.0025 7. Solution \(M_1V_1=M_2V_2 \rightarrow V_2=\frac{M_1V_1}{V_2}=\frac{(0.2\,M)(0.5\,L)}{12.0\,M}=8.33 \times10^{-3}\,L\) Therefore, dilute 8.33 mL of 12.0 M HCl to 500.0 mL. \(M_1V_1=M_2V_2 \rightarrow V_2=\frac{M_1V_1}{V_2}=\frac{(0.288\,M)(0.05\,L)}{0.2\,M}=7.20 \times 10^{-3}\,L\) \(M_1V_1=M_2V_2 \rightarrow V_2=\frac{M_1V_1}{V_2}=\frac{(0.288\,M)(0.05\,L)}{0.187\,M}=7.70 \times 10^{-2}\) 8. a. \(M_{1}V_{1}=M_{2}V_{2} \rightarrow M_2= \frac{M_1V_1}{V_{2}}=\frac{(0.582\,M)(0.060\,L)}{0.0719\,L} =0.49\,M\) b. \(M_{1}V_{1}=M_{2}V_{2} \rightarrow V_{2}=\frac{M_{1}V_{1}}{M_{2}} =\frac{(0.582\,M)(0.050\,L)}{0.49\,M}=59.4 \times 10^{-2}\,L\) 9. \(NaOH\,(aq)+HCl\,(aq) \rightarrow NaCl\,(s)+H_{2}O\,(l)\) At 0 mL of base added: \(pH=-log(0.288)=0.54\) At 10 mL of base added: \(pH=-log(\frac{0.0144-0.00321}{0.06})=0.73\) \(HCl:0.05\,L \times 0.288\,M=0.0144\) \(NaOH:0.01\,L \times 0.321\,M=0.00321\) At 30 mL of base added: \(pH=-log(\frac{0.0144-0.00963}{0.08})=1.22\) \(HCl:0.05\,L \times 0.288\,M=0.0144\) \(NaOH:0.03\,L \times 0.321\,M=0.00963\) At 40.0 mL of base added: \(pH=-log(\frac{0.0144-0.01284}{0.09})=1.76\) \(HCl:0.05\,L \times 0.288\,M=0.0144\) \(NaOH:0.04\,L \times 0.321\,M=0.01284\) At 45.0 mL of base added: \(pH=7\) \(HCl:0.05\,L \times 0.288\,M=0.0144\) \(NaOH:0.045\,L \times 0.321\,M=0.0144\) All \(HCl\) will neutralize at the equivalence point \([H^{+}]=[OH^{-}]\) and the pH of the solution is 7. At 50.0 mL base added: \(pH=14-pOH=14-log(\frac{0.01605-0.0144}{0.1})=12.2\) \(HCl:0.05\,L \times 0.288\,M=0.0144\) \(NaOH:0.050\,L \times 0.321\,M=0.01605\) At 55 mL base added: \(pH=14-pOH=14-log(\frac{0.01605-0.0144}{0.1})=12.5\) \(HCl:0.05\,L \times 0.288\,M=0.0144\) \(NaOH:0.055\,L \times 0.321\,M=0.017655\) At 65 mL base added: \(pH=14-pOH=14-log(\frac{0.020865-0.0144}{0.07})=13.0\) \(HCl:0.05\,L \times 0.288\,M=0.0144\) \(NaOH:0.065\,L \times 0.321\,M=0.020865\) At 75 mL base added: \(pH=14-pOH=14-log(\frac{0.024075-0.0144}{0.08})=13.1\) \(HCl:0.05\,L \times 0.288\,M=0.0144\) \(NaOH:0.075\,L \times 0.321\,M=0.024075\) 10. The equivalence point is at \(pH\) 7 and this occurs at 0.0193 L. \(V_{2}=\frac{(0.156\,M)(0.025\,L)}{0.202}=0.0193\,L\) 11. pH at equivalence point: \(14-(-log\sqrt{(\frac{10^{-14}}{10^{-3.75}} \times \frac{0.01205}{0.05+0.1227}}))=8.30\) \(0.05\,L \times 0.241\,M=0.01205\,mol\) \(V_2=\frac{(0.05\,L)(0.241\,M)}{0.0982}=0.1227\,L\) At 0 mL of base added: \(pH=-log([H^{+}])=-log(2.52 \times 10^{-6})=2.18\) \([H^{+}]= \sqrt{10^{-3.75} \times 0.241\,M}=2.52 \times 10^{-6}\) At 5 mL of base added: \(pH=3.75+log(\frac{4.91 \times 10^{-4}}{0.01205-4.91 \times 10^{-4}})=2.38\) \(KOH:0.05\,L \times 0.0982\,M=4.91 \times 10^{-4}\,mol\) \(CH_{2}O_{2}: 0.05\,L \times 0.241\,M=0.01205\,mol\) At 10 mL of base added: \(pH=3.75+log(\frac{9.82 \times 10^{-4}}{0.01205-9.82 \times 10^{-4}})=2.70\) \(KOH:0.01\,L \times 0.0982\,M=9.82 \times 10^{-4}\,mol\) \(CH_{2}O_{2}: 0.05\,L \times 0.241\,M=0.01205\,mol\) At 15 mL of base added:\(pH=3.75+log(\frac{0.001473}{0.01205-0.001473}=2.89\) \(KOH: 0.015\,L \times 0.0982\,M=0.001473\) \(CH_{2}O_{2}: 0.05\,L \times 0.241\,M=0.01205\) At 20 mL of base added: \(pH=3.75+log(\frac{0.001964}{0.01205-0.001964})=3.04\) \(KOH: 0.02\,L \times 0.0982\,M=0.001964\) \(CH_{2}O_{2}: 0.05\,L \times 0.241\,M=0.01205\) At 25 mL base added: \(pH=3.75+log(\frac{0.002455}{0.01205-0.002455})=3.16\) \(KOH:0.025\,L \times 0.0982\,M=0.002455\,mol\) \(CH_{2}O_{2}: 0.05\,L \times 0.241\,M=0.01205\) 12. \(M_1V_1=M_2V_2 \rightarrow V_2=\frac{M_1V1}{M_2}=\frac{(0.430\,M)(0.05\,L)}{0.150\,M}=1.42 \times 10^{-1}\,L\) 143 mL are needed to fully deprotonate the carboxylic acid group. a. 143 additional milliliters of KOH are needed to deprotonate the ammonium group. b. At the first equivalence point: \(pH=\frac{(pKa_{1}+pKa_{2})}{2}=5.95\) c. 143 mL of titrant are needed to obtain a solution in which glycine has no electrical charge. The isoelectric point of glycine 5.95. 13. \(pH=pK_a+log(\frac{[C_5H_5N]}{[C_5H_5NH^{+}]})=(14-8.77)+log(\frac{0.02996}{0.0063434})=5.90\) \(C_5H_5N\,(aq)+HCl\,(aq) \rightarrow C_5H_5NH^{+}\,(aq)+Cl^{-}\,(aq)\) \(HCl: 32.2\,mL \times \frac{1\,L}{1,000\,mL} \times 0.197\,M=0.0063434\,mol\) \(C_5H_5N: 150\,mL \times \frac{1\,L}{1,000\,mL} \times 0.242\,M=0.0363\,mol\) \(C_5H_5N\) \(HCl\) \(C_5H_5NH^{+}\) \(Cl^{-}\) I 0.0363 0.0063434 0 - C -0.0063434 -0.0063434 +0.0063434 - E 0.02996 0 0.0063434 - 14. \(pH=\frac{(4.21+5.64)}{2}=4.93\) 15. \(pH=\frac{(2.85+5.70)}{2}=4.25\) a. \(K_{sp}=[Ag^{+},I^{-}]\) b. \(K_{sp}=[Ca^{2+},F^{-}]^{2}\) c. \(K_{sp}=[Pb^{2+},Cl^{-}]^{2}\) d. \(K_{sp}=[Ag^{+}]^{2}[CrO_{4}^{2-}]\) 2. The solubility constant expression, \(K_{sp}\) is an equilibrium constant for a solid substance dissolving in a an aqueous solution. Thus, it is a measure of solubility and species that do not dissolve are not represented in the solubility product expression. 3. The main difference between \(Q\) and \(K_sp\) is that \(Q\) describes a reaction that is not at equilibrium unlike \(K_{sp}\). 4. An ion product can be used to determine whether a solution is saturated as it is compared to \(K_{sp}\) where there are three possible conditions for an aqueous solution of an ionic solid: \(Q<K_{sp}\). The solution is unsaturated, and more of the ionic solid, if available, will dissolve. \(Q=K_{sp}\). The solution is saturated and at equilibrium. \(Q>K_{sp}\). The solution is supersaturated, and ionic solid will precipitate. a. \(K_{sp}=[Cd^{2+},IO_{3}]^2 \rightarrow 2.5 \times 10^{-8}=x(2x)^{2} \rightarrow 2.5 \times 10^{-8}=4x^{3} \rightarrow x=1.84 \times 10^{-3}\) b. \(K_{sp}=[Ag^{+},CN^{-}] \rightarrow 1.6 \times 10^{-14}=(x)(x) \rightarrow x=7.73 \times 10^{-9}\) c. \(K_{sp}=[Hg^{2+},I^{-}]^2 \rightarrow 2.9 \times 10^{-29}=(x)(2x)^{2} \rightarrow 1.94\times 10^{-10}\) 2. a. \(K_{sp}=[Li^{+}]^{3}[PO_{4}^{3-}] \rightarrow 2.37 \times 10^{-11}=(3x)^{3}(x) \rightarrow x=9.68 \times 10^{-4}\) b. \(K_{sp}=[Ca^{2+},IO_{3}^{-}]^{2} \rightarrow 6.47 \times 10^{-6}=(x)(2x)^{2} \rightarrow 1.17 \times 10^{-2}\) c. \(K_{sp}=[Y^{3+},IO_{3}]^{-}]^{3} \rightarrow 1.12 \times 10^{-10}=(x)(3x)^{3} \rightarrow x=1.42 \times 10^{-3}\) 3. \(0.750\,L \times \frac{0.0144\,mol}{1\,L} \times \frac{311.199\,g}{1\,mol}=3.37\,g\) \(K_{sp}=[Ag^{+}]^{2}[SO_{4}^{2-}] \rightarrow 1.20 \times 10^{-5}=(2x)^{2}(x) \rightarrow x=0.0144\) 4. a. \(K_{sp}=[Ag^{+},Br^{-}] \rightarrow 5.35 \times 10^{-13}=(x)(x) \rightarrow x=7.31 \times 10^{-7}\) b. \(K_{sp}=[PB^{2+},C_{2}O_{4}] \rightarrow 8.5 \times 10^{-9}=(x)(x) \rightarrow x=9.2 \times 10^{-5}\) c. \(K_{sp}=[Fe^{2+},CO_{3}^{2-}] \rightarrow 3.13 \times 10^{-11}=(x)(x) \rightarrow x=5.60 \times 10^{-6}\) d. \(K_{sp}=[Ag^{+}]^{3}[PO_{4}^{3-}] \rightarrow 8.89 \times 10^{-17}=(3x)^{3}(x) \rightarrow x=4.26 \times 10^{-5}\) e. \(K_{sp}=[Cu,CN] \rightarrow 3.47 \times 10^{-20}=(x)(x) \rightarrow x=1.86 \times 10^{-10}\) 5. a. \(K_{sp}=[Cu,CN] \rightarrow 1.72 \times 10^{-20}=(x)(x) \rightarrow x=4.15 \times 10^{-4}\) b. \(K_{sp}=[La^{3+},IO_{3}^{-}]^{3}] \rightarrow 7.50 \times 10^{-12}=(x)(3x)^{3} \rightarrow x=7.26 \times 10^{-4}\) c. \(K_{sp}=[Mg^{2+}]^{3}[PO_{4}^{3-}]^{2} \rightarrow 1.04 \times 10^{-24}=(3x)^{3}(2x)^{2} \rightarrow x=6.26 \times 10^{-6}\) d. \(K_{sp}=[Ag^{+}]^{2}[CrO_{4}^{2-}] \rightarrow 1.12 \times 10^{-12}=(2x)^{2}(x) \rightarrow 6.54 \times 10^{-5}\) e. \(K_{sp}=[Sr^{2+},SO_{4}^{2-}] \rightarrow 3.44 \times 10^{-7}=(x)(x) \rightarrow x=5.87 \times 10^{-4}\) 6. \(SiO_{2}\,(g)+H_{2}O\,(l) \rightarrow Si(OH)_{4}\,(aq)\) \(K_{sp}=[Si^{4+},OH^{-}]^{4}\) \(K_{sp}=[Mg^{2+},OH^{-}]^{2} \rightarrow 5.61 \times 10^{-12}=(x)(2x)^{2} \rightarrow x=1.12 \times 10^{-4}\) 8. \(Predicted\,mass: \frac{5.88 \times 10^{-2}\,mol}{1\,L} \times \frac{73.891\,g}{1\,mol} \times 0.5\,L=2.17\,g\) \(K_{sp}=[Li^{+}]^2[CO_{3}^{2-}] \rightarrow 8.15 \times 10^{-4}=(2x)^{2}(x) \rightarrow x=5.88 \times 10^{-2}\) The difference between the predicted value and the actual value occurs because the carbonate from the dissociation of the salt reacts with the \(H^{+}\) from the dissociation of water. This reaction causes further dissociation of the salt (Le Chatelier’s principle). 9. \(K_{sp}=[Ba^{2+},SO_{4}^{2-}]=(x)(x)=x^{2}=(1.07 \times 10^{-5}\,M)^{2}=1.15 \times 10^{-10}\) \(BaSO_{4}:(24.6-22.1\,mg) \times \frac{1\,g}{1,000\,mg} \times \frac{1\,mol}{233.38\,g} \times \frac{1}{1.0\,L}=1.07 \times 10^{-5}\,M\) 10. \(K_{sp}=[Ag^{+}]^{2}[CrO_{4}^{2-}]=(2x)^{2}(x)=4x^{3}=4(6.54 \times 10^{-5})=1.12 \times 106{-12}\) \(K_{sp}=[Pb^{2+},Cl^{-}]^{2}=(x)(2x)^{2}=4x^{3}=4(3.24 \times 10^{-2})^{3}=1.70 \times 10^{-5}\) a. \(K_{sp}=[Ag^{+},I^{-}]=x^{2}=(1.23 \times 10^{-8})^{2})=1.52 \times 10^{-16}\) \(AgI: (2.89 \times 10^{-7}\,g \times \frac{1\,mol}{234.77\,g} \times \frac{1}{0.1\,L}=1.23 \times 10^{-8}\) b. \(K_{sp}=[Sr^{2+},F^{-}]^{2}]=(x)(2x)^{2}=4x^{3}=4(9.72 \times 10^{-4})^{3}=3.66 \times 10^{-9}\) \(SrF_{2}=1.22 \times 10^{-2}\,g \times \frac{1\,mol}{125.62\,g} \times \frac{1}{0.1\,L}=9.72 \times 10^{-4}\) c. \(K_{sp}=[Pb^{2+},OH^{-}]^{2}=4x^{3}=4(6.47 \times 10^{-4})^{3}=1.08 \times 10^{-9}\) \(Pb(OH)_{2}=0.078\,g \times \frac{1\,mol}{241.21\,g} \times \frac{1}{0.1\,L}=6.47 \times 10^{-4}\) d. \(K_{sp}=[Bi^{3+},AsO_{4}^{3-}]=x^{2}=(2.05 \times 10^{-5})^{2} =4.21 \times 10^{-10}\) \(BiAsO_{4}=0.0144\,g \times \frac{1\,mol}{350.924\,g} \times \frac{1}{2\,L}=2.05 \times 10^{-5}\) 13. a. \(K_{sp}=[Ba^{2+},CO_{3}^{2-}]=x^{2}=(1.01 \times 10^{-4})^{2}=1.03 \times 10^{-8}\) \(BaCO_{3}:0.01\,g \times \frac{1\,mol}{197.34\,g} \times \frac{1}{0.5\,L}=1.01 \times 10^{-4}\) b. \(K_{sp}=[Ca^{2+},F^{-}]^{2}=4x^{3}=4(2.24 \times 10^{-4})^{3}=4.51 \times 10^{-11}\) \(CaF_{2}: 0.0035\,g \times \frac{1\,mol}{78.07\,g} \times \frac{1}{0.2\,L}=2.24 \times 10^{-4}\) c. \(K_{sp}=[Mn^{2+},OH^{-}]^2=4x^{3}=4(2.36 \times 10^{-5})^{3}=5.26 \times 10^{-14}\) \(Mn(OH)_{2}: 6.30 \times 10^{-4}\,g \times \frac{1\,mol}{88.95\,g} \times \frac{1}{0.3\,L}=2.36 \times 10^{-5}\) d. \(K_{sp}=[Ag^{+}]^{2}[S^{2-}]=4x^{3}=4(6.46 \times 10^{-18})^{3}=1.08 \times 10^{-51}\) \(Ag_{2}S: 1.60 \times 10^{-16}\,g \times \frac{1\,mol}{247.8\,g} \times \frac{1}{0.1\,L}=6.46 \times 10^{-18}\) 14. a. \(K_{sp}=[Ba^{2+},CO_{3}^{2-}]=x^{2}=(7.0 \times 10^{-5})^{2}=4.90 \times 10^{-9}\) b. \(K_{sp}=[Ca^{2+},F^{-}]^{2}=4x^{3}=4(2.18 \times 10^{-4})^{3}=4.13 \times 10^{-11}\) \(CaF_{2}: 0.0017\,g \times \frac{1\,mol}{78.07\,g} \times \frac{1}{0.1\,L}=2.18 \times 10^{-4}\) c. \(K_{sp}=[Pb^{2+},IO_{3}^{-}]^{2}=4x^{3}=4(4.13 \times 10^{-5})^{3}=2.82 \times 10^{-13}\) \(Pb(IO_{3})_{2}: 0.0023\,g \times \frac{1\,mol}{557.0053\,g} \times \frac{1}{0.1\,L}=4.13 \times 10^{-5}\) d. \(K_{sp}=[Sr^{2+},C_2O_4^{2-}]=x^{2}=(1.58 \times 10^{-7})^{2}=2.50 \times 10^{-14}\) a. \(K_{sp}=[Ag^{+}]^{2}[SO_{4}]^{2-}=4x^{3}=4(1.35 \times 10^{-2})^{3}=9.8 \times 10^{-6}\) \(Ag_{2}SO_{4}: 4.2 \times 10^{-1}\,g \times \frac{1\,mol}{311.799\,g} \times \frac{1}{0.1\,L}=1.35 \times 10^{-2}\) b. \(K_{sp}=[Sr^{2+},SO_{4}^{2-}]=x^{2}=(8.17 \times 10^{-5})^{2}=6.7 \times 10^{-9}\) \(SrSO_{4}: 1.5 \times 10^{-3}\,g \times \frac{1\,mol}{183.68\,g} \times \frac{1}{0.1\,L}=8.17 \times 10^{-5}\) c. \(K_{sp}=[Cd^{2+},C_2O_4^{2-}]=x^{2}=(2.99 \times 10^{-4})^{2}=8.9 \times 10^{-8}\) \(CdC_{2}O_{4}: 6.0 \times 10^{-3}\,g \times \frac{1\,mol}{200.43\,g} \times \frac{1}{0.1\,L}=2.99 \times 10^{-4}\) d. \(K_{sp}=[Ba^{2+},IO_{3}^{-}]^{2}=4x^{3}=4(8.13 \times 10^{-4})^{3}=2.15 \times 10^{-9}\) \(Ba(IO_{3})_{2}: 3.96 \times 10^{-2}\,g \times \frac{1\,mol}{487.15\,g} \times \frac{1}{0.1\,L}=8.13 \times 10^{-4}\) 16. \(K_{sp}=[Ca^{2+},HPO_{4}^{2-}] \rightarrow 2.7 \times 10^{-7}=x^{2} \rightarrow x=5.2 \times 10^{-4}\) \(CaHPO4·2H2O: 3.0\,L \times \frac{2.7 \times 10^{-7}\,mol}{1\,L} \times \frac{290.1299\,g}{1\,mol}=2.35 \times 10^{-4}\,g\) 17. \(K_{sp}=[Zn^{2+},CO_{3}^{2+}] \rightarrow 5.5 \times 10^{-11}=x^{2} \rightarrow x=7.4 \times 10^{-6}\) 2.1 mg 18. \(Q=[Ag^{+}]^{2}[SO_{4}^{2-}]=(2.14 \times 10^{-3})^{2} \times 2.4 \times 10^{-3}=1.1 \times 10^{-8}<1.20 \times 10^{-5}\) Therefore, no precipitate will form. \(Ag^{+}: \frac{(0.025\,L)(0.015\,M)}{0.175\,L}=2.14 \times 10^{-3}\) \(SO_{4}^{2-}: \frac{(0.15\,L)(2.8 \times 10^{-3}}{0.175\,L}=2.4 \times 10^{-3}\) 19. a. \(Q=[Ba^{2+},F^{-}]^{2}=(0.060857)(0.05643)^{2}=1.94 \times 10^{-4}>1.84 \times 10^{-7}\) Therefore, precipitate will form. \(Ba^{2+}: \frac{(0.150)(0.142)}{0.35}=0.060857\) \(F^{-}: \frac{(0.25)(0.079)}{0.35}=0.05643\) b. \(Q=[K^{+},Cl^{-}]=(0.0465)(0.0358)=1.66 \times 10^{-3}<21.7. Therefore, precipitate will not form. \(K^{+}: \frac{(0.25)(0.079)}{0.425}=0.0465\) \(Cl^{-}: \frac{(0.175)(0.087)}{0.425}=0.0358\) c. \(Q=[Mg^{2+},C_{2}O_{4}^{2-}]=(0.0617)(0.0317)=1.95 \times 10^{-3}>8.5 \times 10^{-5}\) Therefore, precipitate will form. \(Mg^{2+}:\frac{(0.3)(0.109)}{0.53}=0.0617\) \(C_{2}O_{4}: \frac{(0.23)(0.073)}{0.53}=0.0317\) 20. 520 mL a. \(K_{sp}=\sqrt{2.4 \times 10^{-6}}=1.55 \times 10^{-3}\) b. \(K_{sp}=(\frac{5.6 \times 10^{-12}}{4})^{\frac{1}{3}}=1.12 \times 10^{-4}\) c. \(K_{sp}=(\frac{1.04 \times 10^{-24}}{108})^{\frac{1}{5}}=6.26 \times 10^{-6}\) 1. It is expected that the molar solubility of \LaPO_{4}\) be less than the value from its \(K_{sp}\) because the \(K_{sp}=[La^{3+},PO_{4}^{3-}]=x^{2}\) in which the molar solubility would be the square root of of \(K_{sp}\). 2. It would be expected that there would be no difference in the calculated molar solubility and actual molar solubility of \(Ca_{3}(PO_{4})_{2}\) and \(Mg_{3}(PO_{4})_{2}\) follow the similar equation: \(K_{sp}=108x^{5}\) . It would ultimately depend on the K_{sp} values as \(x=(\frac{K_{sp}}{108})^{\frac{1}{5}} in which the larger \(K_{sp} gives the larger molar solubility. 3. \(Mg(OH)_{2}\,(s) \rightleftharpoons Mg^{2+}\,(aq)+2\,OH^{-}\,(aq)\) \(CH_{3}CH_{2}CO_{2}H\,(aq) \rightleftharpoons CH_{3}CH_{2}CO_{2}^{-}\,(aq)+H^{+}\,(aq)\) 1. Given a solution that contains a mixture of \(NaCl\), \(CuCl_{2}\), and \(ZnCl_{2}\), we can first use \(H_{2}S\) to separate the \(Zn^{2+}\) through filtration, then use \(ZnS\) to separate the \(Cu^{2+}\) through filtration, and be left with \(Na^{+}\). | 46,391 | 2,908 |
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cides that are mainly used to keep crops healthy and prevent them from being damaged by disease and infestation. They include among others herbicides, fungicides, insecticides, acaricides, plant growth regulators and repellents (see section on ). PPPs fall under the EU Regulation (EC) No 1107/2009 which determines that PPPs cannot be placed on the market or used without prior authorization. The European Food and Safety Authority (EFSA) coordinates the EU regulation on PPPs. What are the major categories of chemicals for which regulatory frameworks are present controlling the environmental risks of these chemicals? Is a chemical producer in the EU allowed to put a new chemical on the market without a registration or authorization? Is the CLP regulation based on the hazard, exposure and/or risk of chemicals? The amount of minimum information that is required under REACH for making a proper hazard and risk assessment is dependent on what? And what do you think is the rationale behind it? Name three European agencies or authorities that are coordinating important regulatory frameworks in the EU? : Theo Vermeire : Tim Bowmer You should be able to: : REACH, chemical safety assessment, human, RCR, DNEL, DMEL The REACH Regulation aims to ensure a high level of protection of human health and the environment, including the promotion of alternative methods for assessment of hazards of substances, as well as the free circulation of substances on the internal market while enhancing competitiveness and innovation. Risk assessment under REACH aims to realize such a level of protection for humans that the likelihood of adverse effects occurring is low, taking into account the nature of the potentially exposed population (including sensitive groups) and the severity of the effect(s). Industry therefore has to prepare a risk assessment (in REACH terminology: chemical safety assessment, CSA) for all relevant stages in the life cycle of the chemical, taking into account all risk management measures envisaged, and document this in the chemical safety report (CSR). Risk characterization in the context of a CSA is the estimation of the likelihood that adverse effect levels occur due to actual or predicted exposure to a chemical. The human populations considered, or protection goals, are workers, consumers and humans exposed via the environment. In risk characterization, exposure levels are compared to reference levels to yield "risk characterization ratios" (RCRs) for each protection goal. RCRs are derived for all endpoints (e.g. skin and eye irritation, sensitization, repeated dose toxicity) and time scales. It should be noted that these RCRs have to be derived for all stages in the life-cycle of a compound. Humans can be exposed through the environment directly via inhalation of indoor and ambient air, soil ingestion and dermal contact, and indirectly via food products and drinking water (Figure 1). REACH does not consider direct exposure via soil. In the REACH exposure scenario, assessment of human exposure through the environment can be divided into three steps: A fourth step may be the consideration of aggregated exposure taking into account exposure to the same substance in consumer products and at the workplace. Moreover, there may be similar substances, acting via the same mechanism of action, that may have to be considered in the exposure assessment, for instance, as a worst case, by applying the concept of dose or concentration addition. The section on explains the concept of exposure scenarios and how concentrations in environmental compartments are derived. The aim of hazard identification is to classify chemicals and to select key data for the dose-response assessment to derive a safe reference level, which in REACH terminology is called the DNEL (Derived No Effect Level) or DMEL (Derived Minimal Effect Level). For human end-points, a distinction is made between substances considered to have a threshold for toxicity and those without a threshold. For threshold substances, a No-Observed-Adverse Effect Level (NOAEL) or Lowest-Observed-Adverse-Effect Level (LOAEL) is derived, typically from toxicity studies with laboratory animals such as rats and mice. Alternatively a Benchmark Dose (BMD) can be derived by fitting a dose-response model to all observations. These toxicity values are then extrapolated to a DNEL using assessment factors to correct for uncertainty and variability. The most frequently used assessment factors are those for interspecies differences and those for intraspecies variability (see section on ). Additionally, factors can be applied to account for remaining uncertainties such as those due to a poor database. For substances considered to exert their effect by a non-threshold mode of action, especially mutagenicity and carcinogenicity, it is generally assumed, as a default assumption, that even at very low levels of exposure residual risks cannot be excluded. That said, recent progress has been made on establishing scientific, 'health-based' thresholds for some genotoxic carcinogens. For non-threshold genotoxic carcinogens it is recommended to derive a DMEL, if the available data allow. A DMEL is a cancer risk value considered to be of very low concern, e.g. a 1 in a million tumour risk after lifetime exposure to the chemical and using a conservative linear dose-response model. There is as yet no EU-wide consensus on acceptable levels of cancer risk. Safe use of substances is demonstrated when: • RCRs are below one, both at local and regional level. For threshold substances, the RCR is the ratio of the estimated exposure (concentration or dose) and the DNEL; for non-threshold substances the DMEL is used. • The likelihood and severity of an event such as an explosion occurring due to the physicochemical properties of the substance as determined in the hazard assessment is negligible. A risk characterization needs to be carried out for each exposure scenario (see Section on ) and human population. The assessment consists of a comparison of the exposure of each human population known to be or likely to be exposed with the appropriate DNELs or DMELs and an assessment of the likelihood and severity of an event occurring due to the physicochemical properties of the substance. Based on an emission estimation for processing of dibutylphthalate (DBP) as a softener in plastics, the concentrations in environmental compartments were estimated. Based on modelling as schematically presented in Figure 1, the total human dose was determined to be 93 ug.kg . PEC-air 2.4 µg.m PEC-surface water 2.8 µg.l PEC- grassland soil 0.15 mg.kg PEC-porewater agric. soil 3.2 µg.l PEC-porewater grassl. soil 1.4 µg.l PEC-groundwater 3.2 µg.l Total Human Dose 93 µg.kg .d The total dose should be compared to a DNEL for humans. DBP is not considered a genotoxic carcinogen but is toxic to reproduction and therefore the risk assessment is based on endpoints assumed to have a threshold for toxicity. The lowest NOAEL of DBP was observed in a two-generation reproduction test in rats and at the lowest dose-level in the diet (52 mg.kg .d for males and 80 mg.kg .d for females) a reduced number of live pups per litter and decreased pup weights were seen in the absence of maternal toxicity. The lowest dose level of 52 mg.kg .d was chosen as the NOAEL. The DNEL was derived by the application of an overall assessment factor of 1000, accounting for interspecies differences, human variability and uncertainties due to a non-chronic exposure period. The deterministic estimate of the RCR would be based on the deterministic exposure estimate of 0.093 mg.kg .d and the deterministic DNEL of 0.052 mg.kg .d . The deterministic RCR would then be 1.8, based on the NOAEL. Since this is higher than one, this assessment indicates a concern, requiring a refinement of the assessment or risk management measures. Uncertainty happens! It is inherent to risk assessment. Where, in your view, are the greatest sources of uncertainty in the process of risk assessment? Are there risks identified in the example for humans indirectly exposed via the environment? To what extent are these potential or realistic risks by asking yourself the following questions: • Do you have relevant toxicological and exposure data ? • Are these fixed values or not? • How relevant or adverse are the toxicological effects observed? • Were appropriate assessment factors used? What would you recommend as a strategy to reduce the identified risks sufficiently? : Joop de Knecht : Watze de Wolf : REACH, European chemicals regulation REACH establishes procedures for collecting and assessing information on the properties, hazards and risks of substances. At quantities of > 10 tonnes, the manufacturers, importers, and down-stream users must show that their substances do not adversely affect human health or the environment for the uses and operational conditions registered. The amount of standard information required to show safe use depends on the quantity of the substance that is manufactured or imported. This section explains how risks to the environment are assessed in REACH. In the absence of ecotoxicological data for soil and/or sediment-dwelling organisms, the PNEC and/or PNEC may be provisionally calculated using the EPM. This method uses the PNEC for aquatic organisms and the suspended matter/water partitioning coefficient as inputs. For substances with a log K >5 (or with a corresponding log Kp value), the PEC/PNEC ratio resulting from the EPM is increased by a factor of 10 to take into account possible uptake through the ingestion of sediment. If the PEC/PNEC is greater than 1 a sediment test must be conducted. If one, two or three long-term No Observed Effect Concentrations (NOECs) from sediment invertebrate species representing different living and feeding conditions are available, the PNEC can be derived using default AFs of 100, 50 and 10, respectively. For data rich chemicals, the PNEC can be derived using Species Sensitivity Distributions (SSD) or other higher-tier approaches. Who is responsible within the EU that industrial chemicals do not pose a risk for the environment? In which circumstances will an environmental risk be identified? Describe how the ecotoxicogical safety levels (PNECs) for the aquatic environment are derived depending on the ecotoxicological information available? under review : Gerd Maack : Ad Ragas, Julia Fabrega, Rhys Whomsley You should be able to Human pharmaceuticals, veterinary pharmaceuticals, environmental impact, tiered approach Pharmaceuticals are a crucial element of modern medicine and confer significant benefits to society. About 4,000 active pharmaceutical ingredients are being administered worldwide in prescription medicines, over-the-counter medicines, and veterinary medicines. They are designed to be efficacious and stable, as they need to pass different barriers i.e. skin, the gastrointestinal system (GIT), or even the blood-brain barrier before reaching the target cells. Each target system has a different pH and different lipophilicity and the GIT is in addition colonised with specific bacteria, specialized to digest, dissolve and disintegrate organic molecules. As a consequence of this stability, most of the pharmaceutical ingredients are stable in the environment as well and could cause effects on non-target organisms. The active ingredients comprise a variety of synthetic chemicals produced by pharmaceutical companies in both the industrialized and the developing world at a rate of 100,000 tons per year. While pharmaceuticals are stringently regulated in terms of efficacy and safety for patients, as well as for target animal safety, user and consumer safety, the potential effects on non-target organisms and environmental effects are regulated comparably weakly. The authorisation procedure requires an environmental risk assessment (ERA) to be submitted by the applicants for each new human and veterinary medicinal product. The assessment encompasses the fate and behaviour of the active ingredient in the environment and its ecotoxicity based on a catalogue of standardised test guidelines. In the case of veterinary pharmaceuticals, constraints to reduce risk and thus ensure safe usage can be stipulated in most cases. In the case of human pharmaceuticals, it is far more difficult to ensure risk reduction through restriction of the drug's use due to practical and ethical reasons. Because of their unique benefits, a restriction is not reasonable. This is reflected in the legal framework, as a potential effect on the environment is not included in the final benefit risk assessment for a marketing authorisation. Veterinary pharmaceuticals on the other hand enter the environment mainly via soil, either indirectly, if the slurry and manure from mass livestock production is spread onto agricultural land as fertiliser, or directly from pasture animals. Moreover, pasture animals might additionally excrete directly into surface water. Pharmaceuticals can also enter the environment via the detour of manure used in biogas plants. Despite the differences mentioned above, the general scheme of the environmental risk assessment of human and veterinary pharmaceuticals is similar. Both assessments start with an exposure assessment. Only if specific trigger values are reached an in-depth assessment of fate, behaviour and effects of the active ingredient is necessary. In Phase I, the PEC calculation is restricted to the aquatic compartment. The estimation should be based on the drug substance only, irrespective of its route of administration, pharmaceutical form, metabolism and excretion. The initial calculation of the PEC in surface water assumes: The following formula is used to estimate the PEC in surface water: PEC = mg/l DOSE = Maximum daily dose consumed per capita [mg inh d ] F = Fraction of market penetration (= 1% by default) WASTE = Amount of wastewater per inhabitant per day (= 200 l by default) DILUTION = Dilution Factor (= 10 by default) Three factors of this formula, i.e. F , Waste and the Dilution Factor, are default values, meaning that the PEC in Phase I entirely depends on the dose of the active ingredient. The Fpen can be refined by providing reasonably justified market penetration data, e.g. based on published epidemiological data. If the PEC value is equal to or above 0.01 μg/l (mean dose ≥ 2 mg cap d ), a Phase II environmental fate and effect analysis should be performed. Otherwise, the ERA can stop. However, in some cases, the action limit may not be applicable. For instance, medicinal substances with a log > 4.5 are potential PBT candidates and should be screened for persistence (P), bioaccumulation potential (B), and toxicity (T) independently of the PEC value. Furthermore, some substances may affect vertebrates or lower animals at concentrations lower than 0.01 μg/L. These substances should always enter Phase II and a tailored risk assessment strategy should be followed which addresses the specific mechanism of action of the substance. This is often true for e.g. hormone active substances (see section on ). The required tests in a Phase II assessment (see below) need to cover the most sensitive life stage, and the most sensitive endpoint needs to be assessed. This means for instance that for substances affecting reproduction, the organism needs to be exposed to the substance during gonad development and the reproductive output needs to be assessed. A Phase II assessment is conducted by evaluating the PEC/PNEC ratio based on a base set of data and the predicted environmental concentration from Tier A. If a potential environmental impact is indicated, further testing might be needed to refine PEC and PNEC values in Tier B. Human pharmaceuticals are used all year round without any major fluctuations and peaks. The only exemption are substances used against cold and influenza. These substances have a clear peak in the consumption in autumn and winter times. In developed countries in Europe and North America, antibiotics display a similar peak as they are prescribed to support the substances used against viral infections. The guideline reflects this exposure scenario and asks explicitly for long-term effect tests for all three trophic levels: algae, aquatic invertebrates and vertebrates (i.e., fish). In order to assess the physio chemical fate, amongst other tests the sorption behaviour and fate in a water/sediment system should be determined. If, after refinement, the possibility of environmental risks cannot be excluded, precautionary and safety measures may consist of: Labelling should generally aim at minimising the quantity discharged into the environment by appropriate mitigation measures In the EU, the Environmental Risk Assessment (ERA) is conducted for all veterinary medicinal products. The structure of an ERA for Veterinary Medicinal Products (VMPs) is quite similar to the ERA for Human Medicinal Products. It is also tier based and starts with an exposure assessment in Phase I. Here, the potential for environmental exposure is assessed based on the intended use of the product. It is assumed that products with limited environmental exposure will have negligible environmental effects and thus can stop in Phase I. Some VMPs that might otherwise stop in Phase I as a result of their low environmental exposure, may require additional hazard information to address particular concerns associated with their intrinsic properties and use. This approach is comparable to the assessment of Human Pharmaceutical Products, see above. For the exposure assessment, a decision tree was developed (Figure 3). The decision tree consists of a number of questions, and the answers of the individual questions will conclude in the extent of the environmental exposure of the product. The goal is to determine if environmental exposure is sufficiently significant to consider if data on hazard properties are needed for characterizing a risk. Products with a low environmental exposure are considered not to pose a risk to the environment and hence these products do not need further assessment. However, if the outcome of Phase I assessment is that the use of the product leads to significant environmental exposure, then additional environmental fate and effect data are required. Examples for products with a low environmental exposure are, among others are products for companion animals only and products that result in a Predicted Environmental Concentration in soil (PECsoil) of less than 100 µg/kg, based on a worst-case estimation. A Phase II assessment is necessary if either the trigger of 100 µg/kg in the terrestrial branch or the trigger of 1 µg/L in the aquatic branch is reached. It is also necessary, if the substance is a parasiticide for food producing animals. A Phase II is also required for substances that would in principle stop in Phase I, but there are indications that an environmental risk at very low concentrations is likely due to their hazardous profile (e.g., endocrine active medicinal products). This is comparable to the assessment for Human Pharmaceutical Products. For Veterinary Pharmaceutical Products also the Phase II assessment is sub-divided into several Tiers, see Figure 4. For Tier A, a base set of studies assessing the physical-chemical properties, the environmental fate, and effects of the active ingredient is necessary. For Tier A, acute effect tests are suggested, assuming a more peak like exposure scenario due to e.g. applying manure and dung on fields and meadows, in contrast to the permanent exposure of human pharmaceuticals. If for a specific trophic level, e.g. dung fauna or algae, a risk is identified (PEC/PNEC ≥1) (see Introduction to Chapter 6), long-term tests for this level have to be conducted in Tier B. For the trophic levels, without an identified risk, the assessment can stop. If the risk still applies with these long-term studies, a further refinement with field studies in Tier C can be conducted. Here a co-operation with a competent authority is strongly recommended, as these tests are tailored, reflected by the individual design of these field studies. In addition, and independent of this, risk mitigation measures can be imposed to reduce the exposure concentration (PEC). These can be, beside others, that animals must remain stabled for a certain amount of time after the treatment, to ensure that the concentration of active ingredient in excreta is low enough to avoid adverse effects on dung fauna and their predators. Alternatively, the treated animals are denied access to water as the active ingredient has harmful effects on aquatic organisms. The Environmental Risk Assessment of Human and Veterinary Medicinal Products is a straightforward, tiered-based process with the possibility to exit at several steps in the assessment procedure. Depending on the dose, the physico-chemical properties, and the anticipated use, this can be quite early in the procedure. On the other hand, for very potent substances with specific modes of action the guidelines are flexible enough to allow specific assessments covering these modes of action. The ERA guideline for human medicinal products entered into application 2006 and many data gaps exist for products approved prior to 2006. Although there is a legal requirement for an ERA dossier for all marketing authorisation applications, new applications for pharmaceuticals on the market before 2006 are only required to submit ERA data under certain circumstances (e.g. significant increase in usage). Even for some of the blockbusters, like Ibuprofen, Diclofenac, and Metformin, full information on fate, behaviour and effects on non-target organisms is currently lacking. Furthermore, systematic post-authorisation monitoring and evaluation of potential unintended ecotoxicological effects does not exist. The market authorisation for pharmaceuticals does not expire, in contrast to e.g. an authorisation of pesticides, which needs to be renewed every 10 years. For Veterinary Medicinal Products, an in-depth ERA is necessary for food producing animals only. An ERA for non-food animals can stop with question 3 in Phase I (Figure 3) as it is considered that the use of products for companion animals leads to negligible environmental concentrations, which might not be necessarily the case. Here, the guideline does not reflect the state of the art of scientific and regulatory knowledge. For example, the market authorisation, as a pesticide or biocide, has been withdrawn or strongly restricted for some potent insecticides like imidacloprid and fipronil which both are authorised for use in companion animals. Why are pharmaceuticals a problem for non-target organisms and for the environment? How do Human pharmaceuticals enter the environment? How do Veterinary pharmaceuticals enter the environment? What is the general scheme of the environmental risk assessment of human and veterinary pharmaceuticals? Why are long-term tests needed for the assessment of human pharmaceuticals, in contrast to the assessment of veterinary pharmaceuticals? under review Frank Swartjes : Kees van Gestel, Ad Ragas, Dietmar Müller-Grabherr : You should be able to : Policy on soil contamination, Water Framework Directive, screening values comparison, Thematic Soil Strategy, Common Forum As a bomb hit, soil contamination came onto the political agenda in the United States and in Europe through a number of disasters in the late 1970s and early 1980s. Starting point was the 1978 Love Canal disaster in upper New York State, USA, in which a school and a number of residences had been built on a former landfill for chemical waste disposal with thousands of tonnes of dangerous chemical wastes, and became a national media event. In Europe in 1979, the residential site of Lekkerkerk in the Netherlands became an infamous national event. Again, a residential area had been built on a former waste dump, which included chemical waste from the painting industry, and with channels and ditches that had been filled in with chemical waste-containing materials. Since these events, soil contamination-related policies emerged one after the other in different countries in the world. Crucial elements of these policies were a benchmark date for a ban on bringing pollutants in or on the soil ('prevention'), including a strict policy, e.g. duty of care, for contaminations that are caused after the benchmark date, financial liability for polluting activities, tools for assessing the quality of soil and groundwater, and management solutions (remediation technologies and facilities for disposal). Objectives in soil policies often show evolution over time and changes go along with developing new concepts and approaches for implementing policies. In general, soil policies often develop from a maximum risk control until a functional approach. The corresponding tools for implementation usually develop from a set of screening values towards a systemic use of frameworks, enabling sound environmental protection while improving the cost-benefit-balance. Consequently, soil policy implementation usually goes through different stages. In general terms, four different stages can be distinguished, i.e., related to maximum risk control, to the use of screening values, to the use of frameworks and based on a functional approach. follows the precaution principle and is a stringent way of assessing and managing contamination by trying to avoid any risk. Procedures based on allow for a distinction in polluted and non-polluted sites for which the former, the polluted sites, require some kind of intervention. The scientific underpinning of the earliest generations of screening values was limited and expert judgement played an important role. Later, more sophisticated screening values emerged, based on risk assessment. This resulted in screening values for individual contaminants within the contaminant groups metals and metalloids, other inorganic contaminants (e.g., cyanides), polycyclic aromatic hydrocarbons (PAHs), monocyclic aromatic hydrocarbons (including BETX (benzene, toluene, xylene)), persistent organic pollutants (including PCBs and dioxins), volatile organic contaminants (including trichloroethylene, tetrachloroethylene, 1,1,1-trichloroethane, and vinyl chloride), petroleum hydrocarbons and, in a few countries only, asbestos. For some contaminants such as PAHs, sum-screening values for groups were derived in several countries, based on toxicity equivalents. In a procedure based on , often the same screening values generally act as a trigger for further, more detailed site-specific investigations in one or two additional assessment steps. In the , soil and groundwater must be suited for the land use it relates to (e.g., agricultural or residential land) and the functions (e.g., drinking water abstraction, irrigation water) it performs. Some countries skip the maximum risk control and sometimes also the screening values stages and adopt a framework and/or a functional approach. In Europe, collaboration was strengthened by concerted actions such as CARACAS (concerted action on risk assessment for contaminated sites in the European Union; 1996 - 1998) and CLARINET (Contaminated Land Rehabilitation Network for Environmental Technologies; 1998 - 2001). These concerted actions were followed up by fruitful international networks that are still are active today. These are the , which is a network of contaminated land policy makers, regulators and technical advisors from Environment Authorities in European Union member states and European Free Trade Association countries, and (Network for Industrially Co-ordinated Sustainable Land Management in Europe), which is a leading forum on industrially co-ordinated sustainable land management in Europe. NICOLE is promoting co-operation between industry, academia and service providers on the development and application of sustainable technologies. In 2000, the EU Water Framework Directive (WFD; ) was adopted by the European Commission, followed by the Groundwater Directive ( ) in 2006 (European parliament and the council of the European Union, 2019b). The environmental objectives are defined by the WFD. Moreover, 'good chemical status' and the 'no deterioration clause' account for groundwater bodies. 'Prevent and limit' as an objective aims to control direct or indirect contaminant inputs to groundwater, and distinguishes for 'preventing hazardous substances' to enter groundwater as well as 'limiting other non-hazardous substances'. Moreover, the European Commission adopted a , with soil contamination being one out of the seven identified threats. A proposal for a Soil Framework Directive, launched in 2006, with the objective to protect soils across the EU, was formally withdrawn in 2014 because of a lack of support from some countries. Today, most countries in Europe and North America, Australia and New Zealand, and several countries in Asia and Middle and South America, have regulations on soil and groundwater contamination. The policies, however, differ substantially in stage, extent and format. Some policies only cover prevention, e.g., blocking or controlling the inputs of chemicals onto the soil surface and in groundwater bodies. Other policies cover prevention, risk based quality assessment and risk management procedures and include elaborated technical tools, which enable a decent and uniform approach. In particular the larger countries such as the USA, Germany and Spain, policies differ between states or provinces within the country. And even in countries with a policy on the federal level, the responsibilities for different steps in the soil contamination chain are very different for the different layers of authorities (at the national, regional and municipal level). In Figure 1 the European countries are shown that have a procedure based on frameworks (as described above), including risk-based screening values. It is difficult, if not impossible, to summarise all policies on soil and groundwater protection worldwide. Alternatively, some general aspects of these policies are given here. A fair first basic element in nearly all soil and groundwater policies, relating to prevention of contamination, is the declaration of a formal point in time after which polluting soil and groundwater is considered an illegal act. For soil and groundwater quality assessment and management, most policies follow the risk-based land management approach as the ultimate form of the functional approach described above. Central in this approach are the risks for specific targets that need to be protected up to a specified level. Different protection targets are considered. Not surprisingly, 'human health' is the primary protection target that is adopted in nearly all countries with soil and groundwater regulations. Moreover, the ecosystem is an important protection target for soil, while for groundwater the ecosystem as a protection target is under discussion. Another interesting general characteristic of mature soil and groundwater policies is the function-specific approach. The basic principle of this approach is that land must be suited for its purpose. As a consequence, the appraisal of a contaminated site in a residential area, for instance, follows a much more stringent concept than that of an industrial site. Risk assessment tools often form the technical backbone of policies. Since the late 1980s risk assessment procedures for soil and groundwater quality appraisal were developed. In the late 1980s the exposure model CalTOX was developed by the Californian Department of Toxic Substances Control in the USA, a few years later the CSOIL model in the Netherlands (Van den Berg, 1991/ 1994/ 1995). In Figure 2, the flow chart of the Dutch CSOIL exposure model is given as an example. Three elements are recognized in CSOIL, like in most exposure models: (1) contaminant distribution over the soil compartments; (2) contaminant transfer from (the different compartments of) the soil into contact media; and (3) direct and indirect exposure to humans. The major exposure pathways are exposure through soil ingestion, crop consumption and inhalation of indoor vapours (Elert et al., 2011). Today, several exposure models exist (see Figure 3 for some 'national' European exposure models). However, these exposure models may give quite different exposure estimates for the same exposure scenario (Swartjes, 2007). Moreover, procedures were developed for ecological risk assessment, including the Species Sensitivity Distributions (see section on ), based on empirical relations between concentration in soil or groundwater and the percentage of species or ecological processes that experience adverse effects (PAF: potentially Affected Fraction). For site specific risk assessment, the TRIAD approach was developed, based on three lines of evidence, i.e., chemically-based, toxicity-based and using data from ecological field surveys (see section on the ). In the framework of the HERACLES network, another attempt was made to summarizing different EU policies on polluted soil and groundwater. A strong plea was made for harmonisation of risk assessment tools (Swartjes et al., 2009). The authors also described a procedure for harmonization based on the development of a toolbox with standardized and flexible risk assessment tools. Flexible tools are meant to cover national or regional differences in cultural, climatic and geological (e.g., soil type, depth of the groundwater table) conditions. It is generally acknowledged, however, that policy decisions should be taken on the national level. In 2007, an analysis of the differences of soil and groundwater screening values and of the underlying regulatory frameworks, human health and ecological risk assessment procedures (Carlon, 2007) was launched. Although screening values are difficult to compare, since frameworks and objectives of screening values differ significantly, a general conclusion can be drawn for e.g. the screening values at the potentially unacceptable risk level (often used as , i.e. values that trigger further research or intervention when exceeded). For the 20 metals, most soil screening values (from 13 countries or regions) show between a factor of 10 and 100 difference between the lowest and highest values. For the 23 organic pollutants considered, most soil screening values (from 15 countries or regions) differ by a factor of between 100 and 1000, but for some organic pollutants these screening values differ by more than four orders of magnitude. These conclusions are merely relevant from a policy viewpoint. Technically, these conclusions are less relevant, since, the screenings values are derived from a combination of different protection targets and tools and based on different policy decisions. Differences in screening values are explained by differences in geographical and biological and socio-cultural factors in different countries and regions, different national regulatory and policy decisions and variability in scientific/ technical tools. Carlon, C. (Ed.) (2007). Derivation methods of soil screening values in Europe. A review and evaluation of national procedures towards harmonisation, JRC Scientific and Technical report EUR 22805 EN. Elert, M., Bonnard, R., Jones, C., Schoof, R.A., Swartjes, F.A. (2011). Human Exposure Pathways. Chapter 11 in: Swartjes, F.A. (Ed.), Dealing with Contaminated Sites. From theory towards practical application. Springer Publishers, Dordrecht. Swartjes, F.A. (2007). Insight into the variation in calculated human exposure to soil contaminants using seven different European models. Integrated Environmental Assessment and Management 3, 322-332. Swartjes, F.A., D'Allesandro, M., Cornelis, Ch., Wcislo, E., Müller, D., Hazebrouck, B., Jones, C., Nathanail, C.P. (2009). . The HERACLES strategy from the end of 2009 onwards. National Institute for Public Health and the Environment (RIVM), Bilthoven, The Netherlands, RIVM Report 711701091. Van den Berg, R. (1991/1994/1995). Exposure of humans to soil contamination. A quantitative and qualitative analyses towards proposals for human toxicological C‑quality standards (revised version of the 1991/ 1994 reports). National Institute for Public Health and the Environment (RIVM), Bilthoven, The Netherlands, RIVM-report no. 725201011. Swartjes, F.A. (Ed.) (2011). . Springer Publishers, Dordrecht. Rodríguez-Eugenio, N., McLaughlin, M., Pennock, D. (2018). . Rome, FAO. What is the logical first step in policy on soil and groundwater protection, related to ' prevention' ? What are the most frequently used protection targets on policies on soil and groundwater in the world? What role should screening values play in sophisticated risk assessment procedures? What is the ideal approach when developing a new soil and groundwater policy? Regarding the harmonization of risk assessment tools: why are flexible risk assessment tools necessary? in preparation | 37,241 | 2,910 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Properties_of_Aldehydes_and_Ketones/The_Carbonyl_Group |
A carbonyl group is a chemically organic functional group composed of a carbon atom double-bonded to an oxygen atom --> [ ] The simplest carbonyl groups are usually attached to another carbon compound. These structures can be found in many aromatic compounds contributing to smell and taste. Before going into anything in depth be sure to understand that the entity itself is known as the "Carbonyl group" while the members of this group are called "carbonyl compounds" --> C=O. The carbon and oxygen are usually sp hybridized and planar. The double bonds in alkenes and double bonds in carbonyl groups are VERY different in terms of reactivity. The C=C is reactive due to C=O electronegativity attributed to the oxygen and its two lone pairs of electrons. One pair of the oxygen lone pairs are located in 2s while the other pair are in 2p orbital where its axis is directed perpendicular to the direction of the pi orbitals. The Carbonyl groups properties are directly tied to its electronic structure as well as geometric positioning. For example, the electronegativity of oxygen also polarizes the pi bond allowing the single bonded substituent connected to become electron withdrawing. The double bond lengths of a carbonyl group is about 1.2 angstroms and the strength is about 176-179 kcal/mol). It is possible to correlate the length of a carbonyl bond with its polarity; the longer the bond meaing the lower the polarity. For example, the bond length in C=O is larger in acetaldehyde than in formaldehyde (this of course takes into account the inductive effect of CH in the compound). As discussed before, we understand that oxygen has two lone pairs of electrons hanging around. These electrons make the oxygen more electronegative than carbon. The carbon is then partially postive (electrophillic) and the oxygen partially negative (nucleophillic). The polarizability is denoted by a lowercase delta and a positive or negative superscript depending. For example, carbon would have d+ and oxygen delta^(-). The polarization of carbonyl groups also effects the boiling point of aldehydes and ketones to be higher than those of hydrocarbons in the same amount. The larger the carbonyl compound the less soluble it is in water. If the compound exceeds six carbons it then becomes insoluble. C=O is prone to additions and nucleophillic attack because or carbon's positive charge and oxygen's negative charge. The resonance of the carbon partial positive charge allows the negative charge on the nucleophile to attack the Carbonyl group and become a part of the structure and a positive charge (usually a proton hydrogen) attacks the oxygen. Just a reminder, the nucleophile is a good acid therefore "likes protons" so it will attack the side with a positive charge. 1. sp ;sp 2. partial positive on the carbon and partial negative on the oxygen 3. yes; yes; yes | 2,894 | 2,912 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.03%3A_The_Dissolution_Process |
What occurs at the molecular level to cause a solute to dissolve in a solvent? The answer depends in part on the solute, but there are some similarities common to all solutes. Recall the rule that . This means that substances must have similar intermolecular forces to form solutions. When a soluble solute is introduced into a solvent, the particles of solute can interact with the particles of solvent. In the case of a solid or liquid solute, the interactions between the solute particles and the solvent particles are so strong that the individual solute particles separate from each other and, surrounded by solvent molecules, enter the solution. (Gaseous solutes already have their constituent particles separated, but the concept of being surrounded by solvent particles still applies.) This process is called and is illustrated in Figure \(\Page {1}\). When the solvent is water, the word , rather than solvation, is used. In the case of molecular solutes like glucose, the solute particles are individual molecules. However, if the solute is ionic, the individual ions separate from each other and become surrounded by solvent particles. That is, the cations and anions of an ionic solute separate when the solute dissolves. This process is referred to as (Figure \(\Page {1}\)). The dissociation of soluble ionic compounds gives solutions of these compounds an interesting property: they conduct electricity. Because of this property, soluble ionic compounds are referred to as electrolytes. Many ionic compounds dissociate completely and are therefore called . Sodium chloride is an example of a strong electrolyte. Some compounds dissolve but dissociate only partially, and solutions of such solutes may conduct electricity only weakly. These solutes are called . Acetic acid (CH ), the compound in vinegar, is a weak electrolyte. Solutes that dissolve into individual neutral molecules without dissociation do not impart additional electrical conductivity to their solutions and are called nonelectrolytes. Table sugar (C H O ) is an example of a nonelectrolyte. The term is used in medicine to mean any of the important ions that are dissolved in aqueous solution in the body. Important physiological electrolytes include Na , K , Ca , Mg , and Cl The following substances all dissolve to some extent in water. Classify each as an electrolyte or a nonelectrolyte. Each substance can be classified as an ionic solute or a nonionic solute. Ionic solutes are electrolytes, and nonionic solutes are nonelectrolytes. The following substances all dissolve to some extent in water. Classify each as an electrolyte or a nonelectrolyte. a. nonelectrolyte b. electrolyte c. nonelectrolyte d. electrolyte Our body fluids are solutions of electrolytes and many other things. The combination of blood and the circulatory system is the , because it coordinates all the life functions. When the heart stops pumping in a heart attack, the life ends quickly. Getting the heart restarted as soon as one can is crucial in order to maintain life. The primary electrolytes required in the body fluid are cations (of calcium, potassium, sodium, and magnesium) and anions (of chloride, carbonates, aminoacetates, phosphates, and iodide). These are nutritionally called . Electrolyte balance is crucial to many body functions. Here's some extreme examples of what can happen with an imbalance of electrolytes: elevated potassium levels may result in cardiac arrhythmias; decreased extracellular potassium produces paralysis; excessive extracellular sodium causes fluid retention; and decreased plasma calcium and magnesium can produce muscle spasms of the extremities. When a patient is dehydrated, a carefully prepared (commercially available) electrolyte solution is required to maintain health and well being. In terms of child health, oral electrolyte is given when a child is dehydrated due to diarrhea. The use of oral electrolyte maintenance solutions, which is responsible for saving millions of lives worldwide over the last 25 years, is one of the most important medical advances in protecting the health of children in the century, explains Juilus G.K. Goepp, , assistant director of the Pediatric Emergency Department of the Children's Center at Johns Hopkins Hospital. If a parent provides an oral electrolyte maintenance solution at the very start of the illness, dehydration can be prevented. The functionality of electrolyte solutions is related to their properties, and interest in electrolyte solutions goes far beyond chemistry. Sports drinks are designed to rehydrate the body after excessive fluid depletion. Electrolytes in particular promote normal rehydration to prevent fatigue during physical exertion. Are they a good choice for achieving the recommended fluid intake? Are they performance and endurance enhancers like they claim? Who should drink them? Typically, eight ounces of a sports drink provides between fifty and eighty calories and 14 to 17 grams of carbohydrate, mostly in the form of simple sugars. Sodium and potassium are the most commonly included electrolytes in sports drinks, with the levels of these in sports drinks being highly variable. The American College of Sports Medicine says a sports drink should contain 125 milligrams of sodium per 8 ounces as it is helpful in replenishing some of the sodium lost in sweat and promotes fluid uptake in the small intestine, improving hydration. In the summer of 1965, the assistant football coach of the University of Florida Gators requested scientists affiliated with the university study why the withering heat of Florida caused so many heat-related illnesses in football players and provide a solution to increase athletic performance and recovery post-training or game. The discovery was that inadequate replenishment of fluids, carbohydrates, and electrolytes was the reason for the “wilting” of their football players. Based on their research, the scientists concocted a drink for the football players containing water, carbohydrates, and electrolytes and called it “Gatorade.” In the next football season the Gators were nine and two and won the Orange Bowl. The Gators’ success launched the sports-drink industry, which is now a multibillion-dollar industry that is still dominated by Gatorade. | 6,309 | 2,914 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/08%3A_Solids_Liquids_and_Gases/8.04%3A_Gas_Laws |
Experience has shown that several properties of a gas can be related to each other under certain conditions. The properties are pressure ( ), volume ( ), temperature ( , in kelvins), and amount of material expressed in moles ( ). What we find is that a sample of gas cannot have any random values for these properties. Instead, only certain values, dictated by some simple mathematical relationships, will occur. The first simple relationship, referred to as a gas law, is between the pressure of a gas and its volume. If the amount of gas in a sample and its temperature are kept constant, then as the pressure of a gas is increased, the volume of the gas decreases proportionately. Mathematically, this is written as \[\mathrm{P \propto \dfrac{1}{V}} \nonumber \] where the “∝” symbol means “is proportional to.” This is one form of Boyle’s law, which relates the pressure of a gas to its volume. A more useful form of Boyle’s law involves a change in conditions of a gas. For a given amount of gas at a constant temperature, if we know the initial pressure and volume of a gas sample and the pressure or volume changes, we can calculate what the new volume or pressure will be. That form of Boyle’s law is written \[P_iV_i = P_fV_f \label{Eq1} \] where the subscript \(i\) refers to initial conditions and the subscript \(f\) refers to final conditions. To use \(\ref{Eq1}\), you need to know any three of the variables so that you can algebraically calculate the fourth variable. Also, the pressure quantities must have the same units, as must the two volume quantities. If the two similar variables don’t have the same variables, one value must be converted to the other value’s unit. What happens to the volume of a gas if its pressure is increased? Assume all other conditions remain the same. If the pressure of a gas is increased, the volume decreases in response. What happens to the pressure of a gas if its volume is increased? Assume all other conditions remain the same. If the volume of a gas is increased, the pressure decreases. If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is reduced to 0.987 atm? Assume that the amount and the temperature of the gas remain constant. The key in problems like this is to be able to identify which quantities represent which variables from the relevant equation. The way the question is worded, you should be able to tell that 1.56 atm is , 7.02 L is , and 0.987 atm is . What we are looking for is the final volume— . Therefore, substituting these values into = : The expression has atmospheres on both sides of the equation, so they cancel algebraically: Now we divide both sides of the expression by 0.987 to isolate , the quantity we are seeking: \(\mathrm{\dfrac{(1.56)(7.02\: L)}{0.987}=V_f}\) Performing the multiplication and division, we get the value of , which is 11.1 L. The volume increases. This should make sense because the pressure decreases, so pressure and volume are inversely related. If a sample of gas has an initial pressure of 3.66 atm and an initial volume of 11.8 L, what is the final pressure if the volume is reduced to 5.09 L? Assume that the amount and the temperature of the gas remain constant. 8.48 atm If the units of similar quantities are not the same, one of them must be converted to the other quantity’s units for the calculation to work out properly. It does not matter which quantity is converted to a different unit; the only thing that matters is that the conversion and subsequent algebra are performed properly. The following example illustrates this process. If a sample of gas has an initial pressure of 1.56 atm and an initial volume of 7.02 L, what is the final volume if the pressure is changed to 1,775 torr? Does the answer make sense? Assume that the amount and the temperature of the gas remain constant. This example is similar to Example \(\Page {2}\), except now the final pressure is expressed in torr. For the math to work out properly, one of the pressure values must be converted to the other unit. Let us change the initial pressure to torr: \(\mathrm{1.56\: atm\times\dfrac{760\: torr}{1\: atm}=1,190\: torr}\) Now we can use Boyle’s law: Torr cancels algebraically from both sides of the equation, leaving Now we divide both sides of the equation by 1,775 to isolate on one side. Solving for the final volume, \(\mathrm{V_f=\dfrac{(1,190)(7.02\: L)}{1,775}=4.71\: L}\) Because the pressure increases, it makes sense that the volume decreases. The answer for the final volume is essentially the same if we converted the 1,775 torr to atmospheres: \(\mathrm{1,775\: torr\times\dfrac{1\: atm}{760\: torr}=2.336\: atm}\). Using Boyle’s law: (1.56 atm)(7.02 L) = (2.335 atm) × ; \(\mathrm{V_f=\dfrac{(1.56\: atm)(7.02\: L)}{2.336\: atm}=4.69\: L}\). If a sample of gas has an initial pressure of 375 torr and an initial volume of 7.02 L, what is the final pressure if the volume is changed to 4,577 mL? Does the answer make sense? Assume that amount and the temperature of the gas remain constant. 575 torr Breathing certainly is a major contribution to your health! Without breathing, we could not survive. Curiously, the act of breathing itself is little more than an application of Boyle’s law. The lungs are a series of ever-narrowing tubes that end in a myriad of tiny sacs called alveoli. It is in the alveoli that oxygen from the air transfers to the bloodstream and carbon dioxide from the bloodstream transfers to the lungs for exhalation. For air to move in and out of the lungs, the pressure inside the lungs must change, forcing the lungs to change volume—just as predicted by Boyle’s law. The pressure change is caused by the diaphragm, a muscle that covers the bottom of the lungs. When the diaphragm moves down, it expands the size of our lungs. When this happens, the air pressure inside our lungs decreases slightly. This causes new air to rush in, and we inhale. The pressure decrease is slight—only 3 torr, or about 0.4% of an atmosphere. We inhale only 0.5–1.0 L of air per normal breath. Exhaling air requires that we relax the diaphragm, which pushes against the lungs and slightly decreases the volume of the lungs. This slightly increases the pressure of the air in the lungs, and air is forced out; we exhale. Only 1–2 torr of extra pressure is needed to exhale. So with every breath, our own bodies are performing an experimental test of Boyle’s law. Another simple gas law relates the volume of a gas to its temperature. Experiments indicate that as the temperature of a gas sample is increased, its volume increases as long as the pressure and the amount of gas remain constant. The way to write this mathematically is At this point, the concept of temperature must be clarified. Although the Kelvin scale is the preferred temperature scale, the Celsius scale is also a common temperature scale used in science. The Celsius scale is based on the melting and boiling points of water and is actually the common temperature scale used by most countries around the world (except for the United States, which still uses the Fahrenheit scale). The value of a Celsius temperature is directly related to its Kelvin value by a simple expression: Thus, it is easy to convert from one temperature scale to another. The Kelvin scale is sometimes referred to as the absolute scale because the zero point on the Kelvin scale is at absolute zero, the coldest possible temperature. On the other temperature scales, absolute zero is −260°C or −459°F. The expression relating a gas volume to its temperature begs the following question: to which temperature scale is the volume of a gas related? The answer is that gas volumes are directly related to the . Therefore, the temperature of a gas sample should always be expressed in (or converted to) a Kelvin temperature. What happens to the volume of a gas if its temperature is decreased? Assume that all other conditions remain constant. If the temperature of a gas sample is decreased, the volume decreases as well. What happens to the temperature of a gas if its volume is increased? Assume that all other conditions remain constant. The temperature increases. As with Boyle’s law, the relationship between volume and temperature can be expressed in terms of initial and final values of volume and temperature, as follows: \[\mathrm{\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}} \nonumber \] where and are the initial volume and temperature, and and are the final volume and temperature. This is Charles’s law. The restriction on its use is that the pressure of the gas and the amount of gas must remain constant. (Charles’s law is sometimes referred to as Gay-Lussac’s law, after the scientist who promoted Charles’s work.) A gas sample at 20°C has an initial volume of 20.0 L. What is its volume if the temperature is changed to 60°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant. Although the temperatures are given in degrees Celsius, we must convert them to the kelvins before we can use Charles’s law. Thus, Now we can substitute these values into Charles’s law, along with the initial volume of 20.0 L: \(\mathrm{\dfrac{20.0\: L}{293\: K}=\dfrac{V_f}{333\: K}}\) Multiplying the 333 K to the other side of the equation, we see that our temperature units will cancel: \(\mathrm{\dfrac{(333\: K)(20.0\: L)}{293\: K}=V_f}\) Solving for the final volume, = 22.7 L. So, as the temperature is increased, the volume increases. This makes sense because volume is directly proportional to the absolute temperature (as long as the pressure and the amount of the remain constant). A gas sample at 35°C has an initial volume of 5.06 L. What is its volume if the temperature is changed to −35°C? Does the answer make sense? Assume that the pressure and the amount of the gas remain constant. 3.91 L Other gas laws can be constructed, but we will focus on only two more. The combined gas law brings Boyle’s and Charles’s laws together to relate pressure, volume, and temperature changes of a gas sample: \[\mathrm{\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}} \nonumber \] To apply this gas law, the amount of gas should remain constant. As with the other gas laws, the temperature must be expressed in kelvins, and the units on the similar quantities should be the same. Because of the dependence on three quantities at the same time, it is difficult to tell in advance what will happen to one property of a gas sample as two other properties change. The best way to know is to work it out mathematically. A sample of gas has = 1.50 atm, = 10.5 L, and = 300 K. What is the final volume if = 0.750 atm and = 350 K? Using the combined gas law, substitute for five of the quantities: \(\mathrm{\dfrac{(1.50\: atm)(10.5\: L)}{300\: K}=\dfrac{(0.750\: atm)(V_f)}{350\: K}}\) We algebraically rearrange this expression to isolate on one side of the equation: \(\mathrm{V_f=\dfrac{(1.50\: atm)(10.5\: L)(350\: K)}{(300\: K)(0.750\: atm)}=24.5\: L}\) Note how all the units cancel except the unit for volume. A sample of gas has = 0.768 atm, = 10.5 L, and = 300 K. What is the final pressure if = 7.85 L and = 250 K? 0.856 atm A balloon containing a sample of gas has a temperature of 22°C and a pressure of 1.09 atm in an airport in Cleveland. The balloon has a volume of 1,070 mL. The balloon is transported by plane to Denver, where the temperature is 11°C and the pressure is 655 torr. What is the new volume of the balloon? The first task is to convert all quantities to the proper and consistent units. The temperatures must be expressed in kelvins, and the pressure units are different so one of the quantities must be converted. Let us convert the atmospheres to torr: \(\mathrm{1.09\: atm\times\dfrac{760\: torr}{1\: atm}=828\: torr = P_i}\) Now we can substitute the quantities into the combined has law: \(\mathrm{\dfrac{(828\: torr)(1,070\: mL)}{295\: K}=\dfrac{(655\: torr)\times V_f}{284\: K}}\) To solve for , we multiply the 284 K in the denominator of the right side into the numerator on the left, and we divide 655 torr in the numerator of the right side into the denominator on the left: \(\mathrm{\dfrac{(828\: torr)(1,070\: mL)(284\: K)}{(295\: K)(655\: torr)}=V_f}\) Notice that torr and kelvins cancel, as they are found in both the numerator and denominator. The only unit that remains is milliliters, which is a unit of volume. So = 1,300 mL. The overall change is that the volume of the balloon has increased by 230 mL. A balloon used to lift weather instruments into the atmosphere contains gas having a volume of 1,150 L on the ground, where the pressure is 0.977 atm and the temperature is 18°C. Aloft, this gas has a pressure of 6.88 torr and a temperature of −15°C. What is the new volume of the gas? 110,038 L So far, the gas laws we have used have focused on changing one or more properties of the gas, such as its volume, pressure, or temperature. There is one gas law that relates all the independent properties of a gas under any particular condition, rather than a change in conditions. This gas law is called the ideal gas law. The formula of this law is as follows: \[ \color{red} = nRT \nonumber \] In this equation, is pressure, is volume, is amount of moles, and is temperature. is called the ideal gas law constant and is a proportionality constant that relates the values of pressure, volume, amount, and temperature of a gas sample. The variables in this equation do not have the subscripts and to indicate an initial condition and a final condition. The ideal gas law relates the four independent properties of a gas under conditions. The value of depends on what units are used to express the other quantities. If volume is expressed in liters and pressure in atmospheres, then the proper value of is as follows: \[\mathrm{R=0.08205 \: \dfrac{L\cdot atm}{mol\cdot K}} \nonumber \] This may seem like a strange unit, but that is what is required for the units to work out algebraically. What is the volume in liters of 1.45 mol of N gas at 298 K and 3.995 atm? Using the ideal gas law where = 3.995 atm, = 1.45, and = 298, \(\mathrm{(3.995\: atm)\times V=(1.45\: mol)\left(0.08205\: \dfrac{L\cdot atm}{mol\cdot K}\right)(298\: K)}\) On the right side, the moles and kelvins cancel. Also, because atmospheres appear in the numerator on both sides of the equation, they also cancel. The only remaining unit is liters, a unit of volume. So Dividing both sides of the equation by 3.995 and evaluating, we get = 8.87 L. Note that the conditions of the gas are not changing. Rather, the ideal gas law allows us to determine what the fourth property of a gas (here, volume) be if three other properties (here, amount, pressure, and temperature) are known. What is the pressure of a sample of CO gas if 0.557 mol is held in a 20.0 L container at 451 K? 1.03 atm For convenience, scientists have selected 273 K (0°C) and 1.00 atm pressure as a set of standard conditions for gases. This combination of conditions is called standard temperature and pressure (STP). Under these conditions, 1 mol of any gas has about the same volume. We can use the ideal gas law to determine the volume of 1 mol of gas at STP: \[\mathrm{(1.00\: atm)\times V=(1.00\: mol)\left(0.08205\: \dfrac{L\cdot atm}{mol\cdot K}\right)(273\: K)} \nonumber \] This volume is 22.4 L. Because this volume is independent of the identity of a gas, the idea that 1 mol of gas has a volume of 22.4 L at STP makes a convenient conversion factor: Cyclopropane (C H ) is a gas that formerly was used as an anesthetic. How many moles of gas are there in a 100.0 L sample if the gas is at STP? We can set up a simple, one-step conversion that relates moles and liters: \(\mathrm{100.0\: L\: C_3H_6\times \dfrac{1\: mol}{22.4\: L}=4.46\: mol\: C_3H_6}\) There are almost 4.5 mol of gas in 100.0 L. Note: Because of its flammability, cyclopropane is no longer used as an anesthetic gas. Freon is a trade name for a series of fluorine- and chlorine-containing gases that formerly were used in refrigeration systems. What volume does 8.75 mol of Freon have at STP? Note: Many gases known as Freon are no longer used because their presence in the atmosphere destroys the ozone layer, which protects us from ultraviolet light from the sun. 196 L Airbags (Figure \(\Page {3}\)) are a safety feature provided in most automobiles since the 1990s. The effective operation of an airbag requires that it be rapidly inflated with an appropriate amount (volume) of gas when the vehicle is involved in a collision. This requirement is satisfied in many automotive airbag systems through use of explosive chemical reactions, one common choice being the decomposition of sodium azide, NaN . When sensors in the vehicle detect a collision, an electrical current is passed through a carefully measured amount of NaN to initiate its decomposition: This reaction is very rapid, generating gaseous nitrogen that can deploy and fully inflate a typical airbag in a fraction of a second (~0.03–0.1 s). Among many engineering considerations, the amount of sodium azide used must be appropriate for generating enough nitrogen gas to fully inflate the air bag and ensure its proper function. For example, a small mass (~100 g) of NaN will generate approximately 50 L of N . The ideal gas equation of state applies to mixtures just as to pure gases. It was in fact with a gas mixture, ordinary air, that Boyle, Gay-Lussac and Charles did their early experiments. The only new concept we need in order to deal with gas mixtures is the , a concept invented by the famous English chemist John Dalton (1766-1844). Dalton reasoned that the low density and high compressibility of gases indicates that they consist mostly of empty space; from this it follows that when two or more different gases occupy the same volume, they behave entirely independently. The contribution that each component of a gaseous mixture makes to the total pressure of the gas is known as the of that gas. The definition of Dalton's Law of Partial Pressures that address this is: which is expressed algebraically as \[P_{total}=P_1+P_2+P_3 ... = \sum_i P_i \nonumber \] or, equivalently \[ P_{total} = \dfrac{RT}{V} \sum_i n_i \nonumber \] There is also a similar relationship based on , known as . It is exactly analogous to Dalton's law, in that it states that the total volume of a mixture is just the sum of the partial volumes of its components. But there are two important differences: Amagat's law holds only for ideal gases which must all be at the same temperature and pressure. Dalton's law has neither of these restrictions. Although Amagat's law seems intuitively obvious, it sometimes proves useful in chemical engineering applications. We will make no use of it in this course. Three flasks having different volumes and containing different gases at various pressures are connected by stopcocks as shown. When the stopcocks are opened, Assume that the temperature is uniform and that the volume of the connecting tubes is negligible. The trick here is to note that the total number of moles and the temperature remain unchanged, so we can make use of Boyle's law = constant. We will work out the details for CO only, denoted by subscripts For CO , \[P_aV_a = (2.13\; atm)(1.50\; L) = 3.19\; L \cdot atm \nonumber \] Adding the products for each separate container, we obtain \[\sum_i P_iV_i = 6.36\; L \cdot atm = n_T \nonumber \] We will call this sum After the stopcocks have been opened and the gases mix, the new conditions are denoted by From Boyle's law (\(\ref{Eq1}\), \[P_1V_1 = P_2V_2 = 6.36\; L \cdot atm \nonumber \] \[V_2 = \sum_i V_i = 4.50\; L \nonumber \] Solving for the final pressure we obtain (6.36 L-atm)/(4.50 L) = . For part , note that the number of moles of each gas is = . The mole fraction of any one gas is = / . For CO , this works out to (3.19/ ) / (6.36/ ) = 0.501. Because this exceeds 0.5, we know that this is the most abundant gas in the final mixture. Dalton’s law states that in a gas mixture (\(P_{total}\)) each gas will exert a pressure independent of the other gases (\(P_n\)) and each gas will behave as if it alone occupies the total volume. By extension, the partial pressure of each gas can be calculated by multiplying the total pressure (\(P_{total}\)) by the gas percentage (%). \[P_{Total} = P_1 + P_2 + P_3 + P_4 + ... + P_n \nonumber \] or \[P_n = \dfrac{\text{% of individual gas}_n}{P_{Total}} \nonumber \] A common laboratory method of collecting the gaseous product of a chemical reaction is to conduct it into an inverted tube or bottle filled with water, the opening of which is immersed in a larger container of water. This arrangement is called a , and was widely used in the early days of chemistry. As the gas enters the bottle it displaces the water and becomes trapped in the upper part. The volume of the gas can be observed by means of a calibrated scale on the bottle, but what about its pressure? The total pressure confining the gas is just that of the atmosphere transmitting its force through the water. (An exact calculation would also have to take into account the height of the water column in the inverted tube.) But liquid water itself is always in equilibrium with its vapor, so the space in the top of the tube is a mixture of two gases: the gas being collected, and gaseous H O. The partial pressure of H O is known as the vapor pressure of water and it depends on the temperature. In order to determine the quantity of gas we have collected, we must use Dalton's Law to find the partial pressure of that gas. Oxygen gas was collected over water as shown above. The atmospheric pressure was 754 torr, the temperature was 22°C, and the volume of the gas was 155 mL. The vapor pressure of water at 22°C is 19.8 torr. Use this information to estimate the number of moles of \(O_2\) produced. From Dalton's law, \[P_{O_2} = P_{total} – P_{H_2O} = 754 – 19.8 = 734 \; torr = 0.966\; atm \nonumber \] Now use the Ideal Gas Law to convert to moles \[ n =\dfrac{PV}{RT} = \dfrac{(0.966\; atm)(0.155\;L)}{(0.082\; L atm mol^{-1} K^{-1})(295\; K)}= 0.00619 \; mol \nonumber \] Henry's law is one of the gas laws formulated by William Henry in 1803. It states: "At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid." An equivalent way of stating the law is that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. To explain this law, Henry derived the equation: \[ C =k P_{gas} \nonumber \] where Henry’s Law tells us that the greater the pressure of gas above the surface of a liquid, the higher the concentration of the gas in the liquid. Also, Henry’s law tells us that gases diffuse from areas of high gas concentration to areas of low gas concentration. Our respiratory systems are designed to maintain the proper oxygen concentration in the blood when the partial pressure of O is 0.21 atm, its normal sea-level value. Below the water surface, the pressure increases by 1 atm for each 10.3 m increase in depth; thus a scuba diver at 10.3 m experiences a total of 2 atm pressure pressing on the body. In order to prevent the lungs from collapsing, the air the diver breathes should also be at about the same pressure. But at a total pressure of 2 atm, the partial pressure of \(O_2\) in ordinary air would be 0.42 atm; at a depth of 100 ft (about 30 m), the \(O_2\) pressure of 0.8 atm would be far too high for health. For this reason, the air mixture in the pressurized tanks that scuba divers wear must contain a smaller fraction of \(O_2\). This can be achieved most simply by raising the nitrogen content, but high partial pressures of N can also be dangerous, resulting in a condition known as nitrogen narcosis. The preferred diluting agent for sustained deep diving is helium, which has very little tendency to dissolve in the blood even at high pressures. Certain diseases—such as emphysema, lung cancer, and severe asthma—primarily affect the lungs. Respiratory therapists help patients with breathing-related problems. They can evaluate, help diagnose, and treat breathing disorders and even help provide emergency assistance in acute illness where breathing is compromised. Most respiratory therapists must complete at least two years of college and earn an associate’s degree, although therapists can assume more responsibility if they have a college degree. Therapists must also pass state or national certification exams. Once certified, respiratory therapists can work in hospitals, doctor’s offices, nursing homes, or patient’s homes. Therapists work with equipment such as oxygen tanks and respirators, may sometimes dispense medication to aid in breathing, perform tests, and educate patients in breathing exercises and other therapy. Because respiratory therapists work directly with patients, the ability to work well with others is a must for this career. It is an important job because it deals with one of the most crucial functions of the body. | 25,306 | 2,915 |
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