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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Solubilty/Temperature_Effects_on_the_Solubility_of_Gases |
As with all processes under constant pressure and constant temperature, dissolving a solution into solution will occur only if \(\Delta G_{total} < 0\). \[\Delta G_{sol} = \Delta H_{sol} - T\Delta S_{sol} < 0\label{eq1}\] For dissolving solids in liquids, \(\Delta S_{sol} > 0\), but for dissolving gases solutes, the entropy of solution is negative (\(\Delta S _{sol} < 0\)) since the entropy of the gas phase solute is appreciably greater than the entropy of that solute in solution. Consequently, the only way that \(\Delta G_{sol}<0\) for a dissolving a gas in solution is if the solution process is exothermic (i.e., \(\Delta H_{sol}<0\)). This occurs due to the enthalpy differences from making and breaking in the solvent and solution. There are three basic steps involved in dissolving a solute from a condensed state (or a non-ideal gas) into a solution each with a corresponding enthalpy change. Dissolution can be viewed as occurring in three steps: The enthalpy of solution \(\Delta H_{sol}\) is the sum of these three individual steps. \[ \Delta H_{sol} = \Delta H_{\text{solute-solute}} + \Delta H_{\text{solvent-solvent}} + \Delta H_{\text{solute-solvent}} \label{eq2}\] For solids solutes, \(\Delta H_{\text{solute-solute}}\) is just the lattice energy of the solute, but for gases that follow the ideal gas equation of state, the enthalpy change associated with Step 1 is zero. \[\Delta H_{\text{solute-solute}} (gas) =0 \] This is because are no intermolecular interactions exist in ideal gases (van der Waals gases will differ as expected). In polar solvents like water, \(\Delta H_{\text{solute-solvent}} > \Delta H_{\text{solvent-solvent}}\), so the dissolution of most gases is exothermic (i.e., \(\Delta H_{sol} <0\)). Hence, when a gas dissolves in a liquid solvent, thermal energy is released which warms both the system (the solution) and the surroundings. \[\ce{ solute (gas) + water (l) \rightleftharpoons solute (aq) + water (aq)} + \Delta \label{eq4}\] where \(Delta\) is thermal enegy. Consequently, the solubility of a gas is dependent on temperature (Figure \(\Page {1}\)). The solubility of gases in liquids decreases with increasing temperature. Conversely, adding heat to the solution provides thermal energy that overcomes the attractive forces between the gas and the solvent molecules, thereby decreasing the solubility of the gas; pushes the reaction in Equation \ref{eq4} to the left. The thermodynamic perspective is that at elevated temperatures, the negative entropy term in Equation \ref{eq1} will dominate the enthalpic term that is driving the dissolution process and make \(ΔG_{soln}\) less negative and hence less spontaneous. Determine the solubility of \(\ce{N2(g)}\) when combined with \(\ce{H2O}\) at 0.0345 °C the pressure of \(\ce{N2}\) is 1.00 atm, and its solubility is 21.0 ml at STP. Begin by determining the molarity (solubility) of \(\ce{N2(g)}\) at 0 °C and STP. At STP 1 mol=22L \[\begin{align*} \text{Molarity of } \ce{N2} &= 21\,ml \left( \dfrac{1L}{1000\,ml}\right) \\[4pt] &=0.021\,L \,\ce{N2} \\[4pt] &= 0.021\,L \,\ce{N2} \left(\dfrac{1\,mol}{22\,L}\right)\\[4pt] &=\dfrac{0 .000954\,mol}{1\, L} \\[4pt] &= 9.5 \times 10^{-4}\, M\, \ce{N2} \end{align*}\] Now that the molarity of \(\ce{N2}\). \(C\) has been attained, the Henry's law constant, \(k\), can be evaluated. : \[C = k_H P_{gas} \nonumber\] where \(C\) is solubility, \(k_H\) is Henry's constant, and \(P_{gas}\) is the partial pressure of the gas being considered. Rearranging the formula to solve for \(k_H\) \[\begin{align} k_H&= \dfrac{C}{P_{gas}} \\ &= 9.5 \times 10^{-4}\, M \,N_2/ 1\,atm \nonumber \end{align} \nonumber\] Now substitute k and the partial pressure of \(\ce{N2}\) into Henry's law: \[\begin{align} C&= (9.5 \times 10^{-4}\,M\, N_2)(0.0345) \nonumber \\[4pt] &= 3.29 \times 10^{-5}\, M\, N_2 \nonumber \end{align} \nonumber\] A fish kill can occur with rapid fluctuations in temperature or sustained high temperatures. Generally, cooler water has the potential to hold more oxygen, so a period of sustained high temperatures can lead to decreased dissolved oxygen in a body of water. A short period of hot weather can increase temperatures in the surface layer of water, as the warmer water tends to stay near the surface and be further heated by the air. In this case, the top warmer layer may have more oxygen than the lower, cooler layers because it has constant access to atmospheric oxygen. Dissolving gases in non-polar organic solvents is a different situation than in polar solvents like water discussed above. For non-polar solvents, both the solvent-solvent interactions (\(\Delta H_{\text{solvent-solvent}}\)) and the solvation enthalpies (\(\Delta H_{\text{solute-solvent}}\)) are considerably weaker than in polar liquids like water due to the absence of strong dipole-dipole intermolecular interactions (or hydrogen bonding). In many cases, the enthalpy needed to break solvent-solvent interactions is comparable to the enthalpy released in making solvent-gas interactions. \[\Delta H_{\text{solvent-solvent}} \approx \Delta H_{\text{solute-solvent}}\] which means the \(\Delta H_{sol} \approx 0\) via Equation \ref{eq2}. In some solvent-solute combination \[\Delta H_{\text{solvent-solvent}} > \Delta H_{\text{solute-solvent}}\] so that \(\Delta H_{sol} > 0\). In this case, we can write the solution reactions thusly \[ \Delta + \ce{ solute (gas) + solvent (solvent) \rightleftharpoons solute (sol) + solvent (sol)} \label{eq5}\] where \(Delta\) is thermal energy. In these cases, gases dissolved in organic solvents can actually be more soluble at higher temperatures. A Le Chatelier perspective, like that used above for water, can help understand why. Increasing the temperature will shift in the equilibrium to favor dissolution (i.e, shift Equation \ref{eq5} to the right). Based off of the three steps outlined above for dissolving a solute a liquid, explain why the solubility of hydrogen is smaller in acetonitrile than in toluene and that is less than in n-hexane. Acetinitrile is a polar organic solvent and has stronger intermolecular bonds than toluene, which is a weakly polar solvent and has stronger intermolecular interactions than n-hexane. A system of cyclopentane and oxygen gas are at equilibrium with an enthalpy of -1234 kJ; predict whether the solubility of oxygen gas will be greater when heat is added to the system or when a temperature decrease occurs. Cyclopentane is an organic solvent. Oxygen gas and cyclopentane in a system at equilibrium, where the entropy is negative, will be be displaced from equilibrium when any type of temperature change is inflicted on the system. Because the dissolution of the gas is endothermic, more heat increases the solubility of the gas. | 6,782 | 1,303 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Energies_and_Potentials/Enthalpy/Heat_of_Sublimation |
The molar heat (or enthalpy) of sublimation is the amount of energy that must be added to a mole of solid at constant pressure to turn it directly into a gas (without passing through the liquid phase). Sublimation requires all the forces are broken between the molecules (or other species, such as ions) in the solid as the solid is converted into a gas. The heat of sublimation is generally expressed as \(\Delta H_{sub}\) in units of Joules per mole or kilogram of substance. is the process of changing a solid into a gas without passing through the liquid phase. To sublime a substance, a certain energy must be transferred to the substance via heat (q) or work (w). The energy needed to sublime a substance is particular to the substance's identity and temperature and must be sufficient to do all of the following: Although the process of sublimation does not involve a solid evolving through the liquid phase, the fact that enthalpy is a state function allows us to construct a "thermodynamic cycle" and add the various energies associated with the solid, liquid, and gas phases together (e.g., Hess' Law). The energies involved in can be expressed by the sum of the enthalpy changes for each step: \(\Delta H_{sub} = \Delta E_{therm_{s}} + \Delta E_{bond_{solid \rightarrow liquid}} +\Delta E_{therm_{solid}}+\Delta E_{bond_{l \rightarrow \, g}}\) Recall that for state functions, only the initial and final states of the substance are important. Say for example that state A is the initial state and state B is the final state. How a substance goes from state A to state B does not matter so much as what state A and what state B are. Concerning the state function of enthalpy, the energies associated with enthalpies (whose associated states of matter are contiguous to one another) are additive. Though in sublimation a solid does not pass through the liquid phase on its way to the gas phase, it takes the same amount of energy that it would to first melt (fuse) and then vaporize. A change in thermal energy is indicated by a change in temperature (in ) of a substance at any particular state of matter. Change in thermal energy is expressed by the equation \(\Delta E_{therm}=C_p \times \Delta T\) with For more information on heat capacity and specific heat capacity, see . Bond energy is the amount of energy that a group of atoms must absorb so that it can undergo a phase change (going from a state of lower energy to a state of higher energy). It is measured \(\Delta E_{bond}=\Delta H_{substance_{\text{phase change}}}*\Delta mass_{(substance)}\) in which \(\Delta H_{substance_{phase change}}\) is the enthalpy associated with a specific substance at a specific phase change. Common types of enthalpies include the heat of fusion (melting) and the heat of vaporization. Recall that fusion is the phase change that occurs between the solid state and the liquid state, and vaporization is the phase change that occurs between the liquid state and the gas state. that if the substance has more than type of intramolecular force holding the solid together, then the substance must absorb enough energy to break the different types of intermolecular forces before the substance can sublime. Vaporization is the transfer of molecules of a substance from the to the . Sublimation is the transfer of molecules from the to the . The solid phase is at a lower energy than the liquid phase: that is why substances always release heat when freezing, hence \(\Delta E_{fus \, (s \rightarrow l)} > 0\). Hence, although both sublimation and evaporation involve changing a substance into its gaseous state, the enthalpy change associated with sublimation is always greater than that of vaporization. This is because solid have less energy than those of a liquid, meaning it is takes more energy to excite a solid to its gaseous phase than it does to excite a liquid to its gaseous phase. Another way to look this phenomena is to take a look at the different energies involved with the heat of sublimation: Already we know that ΔH *Δm and ΔH *Δm . Hence, \(\Delta E_{vap \, (l\rightarrow g) }\) is actually one component of \(\Delta H_{sub}\). Consider the sublimation of ice: \[H_2O_{(s)} \rightarrow H_2O_{(g)}\] Sublimation can be decomposed involve two steps (assuming no change of temperature, i.e., no heat capacity issues): \[H_2O_{(s)} \rightarrow H_2O_{(l)}\] \[H_2O_{(l)} \rightarrow H_2O_{(g)}\] The enthalpy change of Step 1 is the molar , \(\Delta H_{fus}\) and the enthalpy change of Step 2 is the molar , \(\Delta H_{vap}\). Combining these two equations and canceling out anything that appears on both sides of the equation (i.e., liquid water), we're back to the sublimation equation: Step 1 + Step 2 = Sublimation Therefore the heat of sublimation, \( \Delta H_{sub}\) is equal to the sum of the heats of fusion and vaporization: \[ \Delta H_{fus} + \Delta H_{vap} = \Delta H_{sub}\] Hence, unless \(\Delta H_{fus}\) is equal to or less than zero (which it NEVER is), \(\Delta H_{sub}\) must be greater than \( \Delta H_{vap}\). Energy can be observed in many different ways. As shown above, can be expressed as Another way in which can be expressed is change in , , plus change in kinetic energy, Potential energy is the energy associated with random movement, whereas kinetic energy is the energy associated with velocity (movement with direction). and + are related by the equations
for substances in the solid and liquid states. that is divided between and ΔKE for substances in the solid and liquid states. This is because the intermolecular and intramolecular forces that exist between the atoms of the substance (i.e. atomic bond, van der Waals forces, etc) have not yet been dissociated and prevent the atomic particles from moving freely about the atmosphere (with velocity). Potential energy is just a way to have energy, and it generally describes the random movement that occurs when atoms are forced to be close to one another. Likewise, kinetic energy is just another way to have energy, which describes an atom's vigorous struggle to move and to break away from the group of atoms. The thermal energy that is added to the substance is thus divided equally between the potential and the kinetic energies because all aspects of the atoms' movement must be excited equally However, once the intermolecular and intramolecular forces which restrict the atoms' movement are dissociated (when enough energy has been added), potential energy no longer exists (for monatomic gases) because the atoms of the substance are no longer forced to vibrate and be in contact with other atoms. When a group of atoms is in the gaseous state, it's atoms can devote all their energies into moving away from one another (kinetic energy). The heat of sublimation can be useful in determining the effectiveness of medicines. Medicine is often administered in pill (solid) form, and the substances which they contain can sublime over time if the pill absorbs too much energy over time. Often times you may see the phrase "avoid excessive heat on the bottles of common painkillers (e.g. Advil). This is because in high temperature conditions, the pills can absorb heat energy, and sublimation can occur. . . . | 7,276 | 1,304 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/05%3A_Stereoisomerism_of_Organic_Molecules/5.02%3A_Configurational_Isomers |
We have defined isomers in a very general way as nonidentical molecules that possess the same number and kind of atoms. However, there are several ways in which isomers can be nonidentical. Among the alkenes, 1- and 2-butene are position isomers, because in these compounds the double bond has a different position in the carbon chain: Most, but not all alkenes, have stereoisomers that are not identical because of different arrangements of the component atoms. Thus there are two stereoisomers of 2-butene that differ in the geometric arrangement of the groups attached to the double bond. In one isomer, both methyl groups are on the side of the double bond ( -2-butene) and in the other, the methyl groups are on sides of the double bond ( -2-butene): It should be clear to you that there will be no cis-trans isomers of alkenes in which one end of the double bond carries identical groups. Thus we don not expect there to be cis-trans isomers of 1-butene or 2-methylpropene, and indeed none are known: You may wish to verify this by making ball-and-stick models of these substances. Ring formation also confers rigidity on molecular structure such that rotation about the ring bonds is prevented. As a result, stereoisomerism of the cis-trans type is possible. For example, 1,2-dimethylcyclopropane exists in two forms that differ in the arrangement of the two methyl groups with respect to the ring. In the cis isomer, the methyl groups both are situated above (or below) the plane of the ring and in the trans isomer they are situated one above and one below, as shown in Figure 5-2. Interconversion of these isomers does not occur without breaking one or more chemical bonds. Stereoisomers that do not interconvert rapidly under normal conditions, and therefore are stable enough to be separated, specifically are called . Thus - and -2-butene are configurational isomers, as are - and -1,2-dimethylcyclopropane. The terms or commonly are used to describe in compounds with double bonds and rings. When referring to the of a particular isomer, we mean to specify its geometry. For instance, the isomer of 1,2-dichloroethene shown below has the trans configuration; the isomer of 1,3-dichlorocyclobutane has the cis configuration: Cis-trans isomerism is encountered very frequently. By one convention, . Thus the following compound is -4-ethyl-3-methyl-3-heptene, despite the fact that two identical groups are cis with respect to each other, because the longest continuous chain is trans as it passes through the double bond: Notice that cis-trans isomerism is not possible at a carbon-carbon triple bond, as for 2-butyne, because the bonding arrangement at the triply bonded carbons is linear: Many compounds have more than one double bond and each may have the potential for the cis or trans arrangement. For example, 2,4-hexadiene has different configurations, which are designated as trans-trans, cis-cis, and trans-cis. Because the two ends of this molecule are identically substituted, the trans-cis becomes identical with cis-trans: The most important type of stereoisomerism is that which arises when molecules possess two structures that are not identical and also are mirror images of one another. This is not a difficult or unfamiliar concept. Many things around us, such as our hands, and pairs of shoes, are not identical and also are mirror images of one another. In the same way, nonidentical molecules exist in which the only distinction between them is that one is the mirror image of the other. A common statement is that such isomers are mirror images of one another, but these images are "superimposable." A simple example of this type of stereoisomerism is 2-chlorobutane, , which can exist in two spatial configurations, \(1\) and \(2\), that correspond to reflections of each other. These isomers are specifically called . Compounds that lack - meaning that they are identical with their mirror images - are said to be (pronounced "ki-rall", rhymes with spiral). This term is derived from the Greek word \(\chi \epsilon \iota \rho =\) hand; and "handedness" or is a property of molecules such that configurational isomers are possible that are nonidentical mirror images. Compounds that possess reflection symmetry - meaning that they are identical with their mirror images - are said to be . Enantiomers are not possible for achiral compounds. An is a pair of substances whose molecules are mirror images. The pressing question at this point is how can we tell whether a substances will be chiral or achiral. The most common origin of chirality in molecules, and the one originally recognized by van't Hoff and Le Bel, is the presence of one or more atoms, usually carbon atoms, each of which forms coplanar bonds to . This is the case for 2-chlorobutane, because the second tetrahedral carbon along the chain is bonded to four different groups: hydrogen, chlorine, methyl, and ethyl. Therefore there is a pair of enantiomers, \(1\) and \(2\). Another example is 2-bromo-2-chloro-1,1,1-trifluoroethane, which is a widely used inhalation anaesthetic. The four different groups in this case are hydrogen, chlorine, bromine, and trifluoromethyl; the pair of enantiomers is shown in Structures \(3\) and \(4\): The atom that carries the four different substituents in \(1\) and \(2\), or \(3\) and \(4\), is called a or . The latter is the more general term because, as we shall see later (Section 13-5A), dissymmetry in molecules need not be centered at an atom.\(^1\) In evaluating a chemical structure for chirality, you should look for carbons carrying four attached groups. There may be more than one chiral carbon, and you should be alert to the fact that structural differences in the attached groups do not necessarily show up at the first, or even the second, position along a chain. As an example, consider the chirality of 1,1,3-trimethylcyclohexane, Carbons \(C2\), \(C4\), \(C5\), and \(C6\) are clearly achiral because each is connected to two identical groups, which for these atoms are hydrogens. The same is true for \(C1\) because it is connected to two \(CH_3\) groups. You might conclude that \(C3\) also is an achiral position because it is connected to two \(CH_2\) groups. But this would be wrong. If you look farther, you will see that the groups attached to \(C3\) actually are different and are \(H\), \(CH_3\), \(-CH_2CH_2CH_2-\), and \(-CH_2C \left( CH_3 \right)_2\). Therefore 1,1,3-trimethylcyclohexane has a chiral center at \(C3\). In contrast, the 1,1,4-isomer has no chiral centers because the groups attached to the ring at \(C4\) are identical: Several other terms that we shall use frequently in addition to chirality are , , and . A mixture of amounts of both enantiomers is a ; separation of a racemic mixture into its component enantiomers is a , and the conversion of either enantiomer into equal parts of both is called . Until recently, the phenomenon of chirality has been better known as , and configurational isomers that are enantiomers were referred to as . The reasons for this are mainly historical. It was discovered early in the nineteenth century that many compounds, whether solid, liquid, or gas, have the property of rotating the plane of polarization of polarized light and can be said to be " ." A satisfactory explanation of the origin of optical activity came much later and developed in its modern form from the classic researches of Louis Pasteur, and from the concept of the three-dimensional carbon atoms expressed independently by J. H. van't Hoff and J. A. Le Bel.\(^2\) Pasteur's contribution to stereochemistry came as a result of his studies of the shapes of crystals of tartaric acid, \(\ce{HO_2C-CHOH-CHOH-CO_2H}\), and its salts. Tartaric acid, a by-product of wine production, was known to be optically active, and Pasteur showed that it, and nineteen different salts of it, all formed crystals that were identical with their mirror images. A different substance known as "racemic acid," for which we can write the same condensed formula, \(\ce{HO_2C-CHOH-CHOH-CO_2H}\), was known to be optically , and Pasteur expected that when he crystallized this acid or its salts he would obtain crystals that would be identical with their mirror images. However, crystallization of the sodium ammonium salt of racemic acid from water at temperatures below \(28^\text{o}\) gave crystals of different shapes and these shapes were mirror images of one another. Pasteur carefully picked apart the two kinds of crystals and showed that one of them was identical with the corresponding salt of tartaric acid, except that it rotated the plane of polarization of polarized light in the opposite direction. This separation of racemic acid into two optically active forms now is called a "resolution of racemic acid." On the basis of his discoveries, Pasteur postulated that "optical isomerism" had to be related to the molecular dissymmetry of substances such that nonidentical mirror-image forms could exist. However, it remained for van' t Hoff and Le Bel to provide, almost simultaneously, a satisfactory explanation at the molecular level. In his first published work on tetrahedral carbon van't Hoff said "...it appears more and more that the present constitutional formulae are incapable of explaining certain cases of isomerism; the reason for this is perhaps the fact that we need a more definite statement about the actual positions of the atoms."\(^3\) He goes on to discuss the consequences of the tetrahedral arrangements of atoms about carbon, explicitly in connection with optical isomerism and geometric, or cis-trans, isomerism. It is not easy for the chemist of today to appreciate fully the contributions of these early chemists because we have long accepted the tetrahedral carbon as an experimentally established fact. At the time the concept was enunciated, however, even the existence of atoms and molecules was questioned openly by many scientists, and to ascribe "shapes" to what in the first place seemed like metaphysical conceptions was too much for many to accept. Optical activity is an experimentally useful property and usually is measured as the angle of rotation (\(\alpha\)) of the plane of polarization of polarized light passing through solutions of the substances under investigation (Figure 5-4). Where measurable optical activity is present, it is found that one enantiomer rotates the plane of polarization in one direction, whereas the other causes the plane to rotate but in the opposite direction. With reference to the plane of incident light, the enantiomer that rotates the plane to the right is called and is symbolized by either or (\(+\)); the enantiomer that rotates the plane to the left is , symbolized by or (\(-\)). A racemic mixture then can be designated as or (\(\pm\)), and will have no net optical rotation. It is very important to know that , , (\(+\)), or (\(-\)) do not designate configurations. Thus, although (\(+\))-2-butanol actually has configuration \(5\) and (\(-\))-2-butanol has configuration \(6\), there is no simple way to predict that a particular sign of rotation will be associated with a particular configuration. Methods used in assigning the true configurations to enantiomers will be discussed later. A very important point to keep in mind about any pair of enantiomers is that they will have identical chemical and physical properties, except for the signs of their optical rotations, with one important proviso: All of the properties to be compared must be determined using achiral reagents in a solvent made up of achiral molecules or, in short, in an . Thus the melting and boiling points (but not the optical rotations) of \(5\) and \(6\) will be identical in an achiral environment. How a chiral environment or chiral reagents influence the properties of substances such as \(5\) and \(6\) will be considered in . \(^1\)In the older literature, chiral centers often are called and you may be confused by the difference between and . Both asymmetric and dissymmetric molecules (or objects) are chiral. An asymmetric object has no symmetry at all and looks different from all angles of view. Formulas \(3\) and \(4\) represent asymmetric molecules. A dissymmetric molecule is chiral, but looks the same from more than one angle of view. A helical spring is dissymmetric - it looks the same from each end. We will encounter dissymmetric molecules later. \(^2\)The tetrahedral carbon was first proposed by E. Paterno in 1869 (see Section1-1E), but he apparently did not recognize its implications for chirality. These implications were recognized first by van't Hoff and Le Bel, with van't Hoff proceeding on the basis of bonds to carbon being directed to the corners of a regular tetrahedron. Le Bel was opposed to such a rigid formulation of the bonds to carbon. \(^3\)An interesting account and references to van't Hoff's early work can be found in "The Reception of J. H. van't Hoff's Theory of the Asymmetric Carbon" by H. A. M. Snelders, , 2 (1974). A century has passed since van't Hoff first published his theory, which he did before he obtained his doctoral degree from the University of Utrecht. van't Hoff was the first recipient of the Nobel Prize in chemistry (1901) for his later work in thermodynamics and chemical kinetics. and (1977) | 13,421 | 1,305 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Oxides/Physical_Properties_of_Period_3_Oxides |
This page explains the relationship between the physical properties of the oxides of elements (sodium to chlorine) and their structures. Argon is obviously omitted because it does not form an oxide. The oxides we'll be looking at are: Those oxides in the top row are known as the highest oxides of the various elements. These are the oxides where the Period 3 elements are in their highest oxidation states. In these oxides, all the outer electrons in the Period 3 element are being involved in the bonding - from just the one with sodium, to all seven of chlorine's outer electrons. The trend in structure is from the metallic oxides containing giant structures of ions on the left of the period via a giant covalent oxide (silicon dioxide) in the middle to molecular oxides on the right. The giant structures (the metal oxides and silicon dioxide) will have high melting and boiling points because a lot of energy is needed to break the strong bonds (ionic or covalent) operating in three dimensions. The oxides of phosphorus, sulfur and chlorine consist of individual molecules; some are small and simple and others are polymeric. The attractive forces between these molecules will be dispersion and interactions. These vary in size depending on the size, shape and polarity of the various molecules - but will always be much weaker than the ionic or covalent bonds you need to break in a giant structure. These oxides tend to be gases, liquids or low melting point solids. None of these oxides has any free or mobile electrons. That means that none of them will conduct electricity when they are solid. The ionic oxides can, however, undergo when they are molten. They can conduct electricity because of the movement of the ions towards the electrodes and the discharge of the ions when they get there. Sodium, magnesium and aluminium oxides consist of giant structures containing metal ions and oxide ions. Magnesium oxide has a structure just like sodium chloride. The other two have more complicated arrangements. There are strong attractions between the ions in each of these oxides and these attractions need a lot of heat energy to break. These oxides therefore have high melting and boiling points. None of these conducts electricity in the solid state, but electrolysis is possible if they are molten. They conduct electricity because of the movement and discharge of the ions present. The only important example of this is in the electrolysis of aluminium oxide in the manufacture of . Whether you can electrolyse molten sodium oxide depends, of course, on whether it actually melts instead of subliming or decomposing under ordinary circumstances. If it sublimes, you will not get any liquid to electrolyse! Magnesium and aluminium oxides have melting points far too high to be able to electrolyse them in a simple lab. The of the elements increases as you go across the period, and by the time you get to silicon, there is not enough electronegativity difference between the silicon and the oxygen to form an ionic bond. Silicon dioxide is a giant covalent structure. There are three different crystal forms of silicon dioxide. The easiest one to remember and draw is based on the diamond structure. Crystalline silicon has the same structure as diamond. To turn it into silicon dioxide, all you need to do is to modify the silicon structure by including some oxygen atoms. Notice that each silicon atom is bridged to its neighbours by an oxygen atom. Don't forget that this is just a tiny part of a giant structure extending in all 3 dimensions. If you want to be fussy, the Si-O-Si bond angles are wrong in this diagram. In reality the "bridge" from one silicon atom to its neighbour is not in a straight line, but via a "V" shape (similar to the shape around the oxygen atom in a water molecule). It's extremely difficult to draw that convincingly and tidily in a diagram involving this number of atoms. The simplification is perfectly acceptable. Silicon dioxide has a high melting point - varying depending on what the particular structure is (remember that the structure given is only one of three possible structures), but they are all around 1700°C. Very strong silicon-oxygen covalent bonds have to be broken throughout the structure before melting occurs. Silicon dioxide boils at 2230°C. Because you are talking about a different form of bonding, it doesn't make sense to try to compare these values directly with the metallic oxides. What you can safely say is that because the metallic oxides and silicon dioxide have giant structures, the melting and boiling points are all high. Silicon dioxide does not have any mobile electrons or ions and hence does not conduct electricity either as a solid or a liquid. Phosphorus, sulfur and chlorine all form oxides which consist of molecules. Some of these molecules are fairly simple - others are polymeric. We are just going to look at some of the simple ones. Melting and boiling points of these oxides will be much lower than those of the metal oxides or silicon dioxide. The intermolecular forces holding one molecule to its neighbors will be van der Waals dispersion forces or dipole-dipole interactions. The strength of these will vary depending on the size of the molecules. None of these oxides conducts electricity either as solids or as liquids. None of them contains ions or free electrons. Phosphorus has two common oxides, phosphorus(III) oxide, P O , and phosphorus(V) oxide, P O . Phosphorus(III) oxide is a white solid, melting at 24°C and boiling at 173°C. The structure of its molecule is best worked out starting from a P molecule which is a little tetrahedron. Pull this apart so that you can see the bonds . . . . . . and then replace the bonds by new bonds linking the phosphorus atoms via oxygen atoms. These will be in a V-shape (rather like in water), but you probably wouldn't be penalised if you drew them on a straight line between the phosphorus atoms in an exam. The phosphorus is using only three of its outer electrons (the 3 unpaired p electrons) to form bonds with the oxygens. Phosphorus(V) oxide is also a white solid, subliming (turning straight from solid to vapour) at 300°C. In this case, the phosphorus uses all five of its outer electrons in the bonding. Solid phosphorus(V) oxide exists in several different forms - some of them polymeric. We are going to concentrate on a simple molecular form, and this is also present in the vapor. This is most easily drawn starting from P4O6. The other four oxygens are attached to the four phosphorus atoms via double bonds. If you look carefully, the shape of this molecule looks very much like the way we usually draw the repeating unit in the diamond giant structure. Don't confuse the two, though! The \(P_4O_{10}\) molecule stops here. This is not a little bit of a giant structure - it's all there is. In diamond, of course, the structure just continues almost endlessly in three dimensions. Sulfur has two common oxides, sulfur dioxide (sulfur(IV) oxide), SO , and sulfur trioxide (sulfur(VI) oxide), SO . Sulfur dioxide is a colourless gas at room temperature with an easily recognised choking smell. It consists of simple SO molecules. The sulfur uses 4 of its outer electrons to form the double bonds with the oxygen, leaving the other two as a lone pair on the sulfur. The bent shape of SO is due to this lone pair. Pure sulfur trioxide is a white solid with a low melting and boiling point. It reacts very rapidly with water vapour in the air to form sulfuric acid. That means that if you make some in the lab, you tend to see it as a white sludge which fumes dramatically in moist air (forming a fog of sulfuric acid droplets). Gaseous sulfur trioxide consists of simple SO molecules in which all six of the sulfur's outer electrons are involved in the bonding. There are various forms of solid sulfur trioxide. The simplest one is a trimer, S O , where three SO molecules are joined up and arranged in a ring. There are also other polymeric forms in which the SO molecules join together in long chains. For example: The fact that the simple molecules join up in this way to make bigger structures is what makes the sulfur trioxide a solid rather than a gas. Chlorine forms several oxides. Here we are just looking at two of them : chlorine(I) oxide (Cl O) and chlorine(VII) oxide (Cl O ). Chlorine(I) oxide is a yellowish-red gas at room temperature. It consists of simple small molecules. There's nothing in the least surprising about this molecule and it's physical properties are just what you would expect for a molecule this size. In chlorine(VII) oxide, the chlorine uses all of its seven outer electrons in bonds with oxygen. This produces a much bigger molecule, and so you would expect its melting point and boiling point to be higher than chlorine(I) oxide. Chlorine(VII) oxide is a colourless oily liquid at room temperature. In the diagram, for simplicity I have drawn a standard structural formula. In fact, the shape is tetrahedral around both chlorines, and V-shaped around the central oxygen. I intended at this point to quote values for each of the oxides, hoping to show that the melting and boiling points increase as the charges on the positive ion increase from 1+ in sodium to 3+ in aluminium. You would expect that the greater the charge, the greater the attractions. Unfortunately, the oxide with the highest melting and boiling point is magnesium oxide, not aluminium oxide! So that theory bit the dust! The reason for this probably lies in the increase in electronegativity as you go from sodium to magnesium to aluminium. That would mean that the electronegativity difference between the metal and the oxygen is decreasing. The smaller difference means that the bond won't be so purely ionic. It is also likely that molten aluminium oxide contains complex ions containing both aluminium and oxygen rather than simple aluminium and oxide ions. All this means, of course, that you aren't really comparing like with like - so wouldn't necessarily expect a neat trend. The other problems I came across lie with sodium oxide. Most sources say that this sublimes (turns straight from solid to vapour) at 1275°C. However, the usually reliable gives a melting point of 1132°C followed by a decomposition temperature (before boiling) of 1950°C. Other sources talk about it decomposing (to sodium and sodium peroxide) above 400°C. I have no idea what the truth of this is - although I suspect that the Webelements melting point value is probably for a pressure above atmospheric pressure (although it doesn't say so). | 10,631 | 1,306 |
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A plasma is an ionized gas, a gas into which sufficient energy is provided to free electrons from atoms or molecules and to allow both species, ions and electrons, to coexist. Plasma is the fourth state of matter. Many places teach that there are three states of matter; solid, liquid and gas, but there are actually four. The funny thing about that is, that as far as we know, plasmas are the most common state of matter in the universe. They are even common here on earth. A plasma is a gas that has been energized to the point that some of the electrons break free from, but travel with, their nucleus. Gases can become plasmas in several ways, but all include pumping the gas with energy. A spark in a gas will create a plasma. A hot gas passing through a big spark will turn the gas stream into a plasma that can be useful. Plasma torches like that are used in industry to cut metals. The biggest chunk of plasma you will see is the sun. The sun's enormous heat rips electrons off the hydrogen and helium molecules that make up the sun. Essentially, the sun, like most stars, is a great big ball of plasma. | 1,129 | 1,307 |
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is a galvanic process by which metals deteriorate through oxidation—usually but not always to their oxides. For example, when exposed to air, iron rusts, silver tarnishes, and copper and brass acquire a bluish-green surface called a patina. Of the various metals subject to corrosion, iron is by far the most important commercially. An estimated $100 billion per year is spent in the United States alone to replace iron-containing objects destroyed by corrosion. Consequently, the development of methods for protecting metal surfaces from corrosion constitutes a very active area of industrial research. In this section, we describe some of the chemical and electrochemical processes responsible for corrosion. We also examine the chemical basis for some common methods for preventing corrosion and treating corroded metals. Corrosion is a process. Under ambient conditions, the oxidation of most metals is thermodynamically spontaneous, with the notable exception of gold and platinum. Hence it is actually somewhat surprising that any metals are useful at all in Earth’s moist, oxygen-rich atmosphere. Some metals, however, are resistant to corrosion for kinetic reasons. For example, aluminum in soft-drink cans and airplanes is protected by a thin coating of metal oxide that forms on the surface of the metal and acts as an impenetrable barrier that prevents further destruction. Aluminum cans also have a thin plastic layer to prevent reaction of the oxide with acid in the soft drink. Chromium, magnesium, and nickel also form protective oxide films. Stainless steels are remarkably resistant to corrosion because they usually contain a significant proportion of chromium, nickel, or both. In contrast to these metals, when iron corrodes, it forms a red-brown hydrated metal oxide (\(\ce{Fe2O3 \cdot xH2O}\)), commonly known as rust, that does not provide a tight protective film (Figure \(\Page {1}\)). Instead, the rust continually flakes off to expose a fresh metal surface vulnerable to reaction with oxygen and water. Because both oxygen and water are required for rust to form, an iron nail immersed in deoxygenated water will not rust—even over a period of several weeks. Similarly, a nail immersed in an organic solvent such as kerosene or mineral oil will not rust because of the absence of water even if the solvent is saturated with oxygen. In the corrosion process, iron metal acts as the anode in a galvanic cell and is oxidized to Fe ; oxygen is reduced to water at the cathode. The relevant reactions are as follows: The \(\ce{Fe^{2+}}\) ions produced in the initial reaction are then oxidized by atmospheric oxygen to produce the insoluble hydrated oxide containing \(\ce{Fe^{3+}}\), as represented in the following equation: \[\ce{4Fe^{2+}(aq) + O2(g) + (2 + 4x)H2O \rightarrow 2Fe2O3 \cdot xH2O + 4H^{+}(aq)} \label{Eq4} \] The sign and magnitude of \(E^o_{cell}\) for the corrosion process (Equation \(\ref{Eq3}\)) indicate that there is a strong driving force for the oxidation of iron by O under standard conditions (1 M H ). Under neutral conditions, the driving force is somewhat less but still appreciable (E = 1.25 V at pH 7.0). Normally, the reaction of atmospheric CO with water to form H and HCO provides a low enough pH to enhance the reaction rate, as does acid rain. Automobile manufacturers spend a great deal of time and money developing paints that adhere tightly to the car’s metal surface to prevent oxygenated water, acid, and salt from coming into contact with the underlying metal. Unfortunately, even the best paint is subject to scratching or denting, and the electrochemical nature of the corrosion process means that two scratches relatively remote from each other can operate together as anode and cathode, leading to sudden mechanical failure (Figure \(\Page {2}\)). One of the most common techniques used to prevent the corrosion of iron is applying a protective coating of another metal that is more difficult to oxidize. Faucets and some external parts of automobiles, for example, are often coated with a thin layer of chromium using an electrolytic process. With the increased use of polymeric materials in cars, however, the use of chrome-plated steel has diminished in recent years. Similarly, the “tin cans” that hold soups and other foods are actually consist of steel container that is coated with a thin layer of tin. While neither chromium nor tin metals are intrinsically resistant to corrosion, they both form protective oxide coatings that hinder access of oxygen and water to the underlying steel (iron alloy). As with a protective paint, scratching a protective metal coating will allow corrosion to occur. In this case, however, the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for \(\ce{Sn^{2+}}\) (E° = −0.14 V) and Fe (E° = −0.45 V) in show that \(\ce{Fe}\) is more easily oxidized than \(\ce{Sn}\). As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen (Figure \(\Page {3}\)). This process is seen in some older homes where copper and iron pipes have been directly connected to each other. The less easily oxidized copper acts as the cathode, causing iron to dissolve rapidly near the connection and occasionally resulting in a catastrophic plumbing failure. One way to avoid these problems is to use a more easily oxidized metal to protect iron from corrosion. In this approach, called cathodic protection, a more reactive metal such as \(\ce{Zn}\) (E° = −0.76 V for \(\ce{Zn^{2+} + 2e^{−} -> Zn}\)) becomes the anode, and iron becomes the cathode. This prevents oxidation of the iron and protects the iron object from corrosion. The reactions that occur under these conditions are as follows: \[ \underbrace{O_{2(g)} + 4e^− + 4H^+_{(aq)} \rightarrow 2H_2O_{(l)} }_{\text{reduction at cathode}}\label{Eq5} \] \[ \underbrace{Zn_{(s)} \rightarrow Zn^{2+}_{(aq)} + 2e^−}_{\text{oxidation at anode}} \label{Eq6} \] \[ \underbrace{ 2Zn_{(s)} + O_{2(g)} + 4H^+_{(aq)} \rightarrow 2Zn^{2+}_{(aq)} + 2H_2O_{(l)} }_{\text{overall}}\label{Eq7} \] The more reactive metal reacts with oxygen and will eventually dissolve, “sacrificing” itself to protect the iron object. Cathodic protection is the principle underlying galvanized steel, which is steel protected by a thin layer of zinc. Galvanized steel is used in objects ranging from nails to garbage cans. In a similar strategy, using magnesium, for example, are used to protect underground tanks or pipes (Figure \(\Page {4}\)). Replacing the sacrificial electrodes is more cost-effective than replacing the iron objects they are protecting. Suppose an old wooden sailboat, held together with iron screws, has a bronze propeller (recall that bronze is an alloy of copper containing about 7%–10% tin). identity of metals corrosion reaction, \(E^o°_{cell}\), and preventive measures Over time, the iron screws will dissolve, and the boat will fall apart. Suppose the water pipes leading into your house are made of lead, while the rest of the plumbing in your house is iron. To eliminate the possibility of lead poisoning, you call a plumber to replace the lead pipes. He quotes you a very low price if he can use up his existing supply of copper pipe to do the job. Not unless you plan to sell the house very soon because the \(\ce{Cu/Fe}\) pipe joints will lead to rapid corrosion. Any existing \(\ce{Pb/Fe}\) joints should be examined carefully for corrosion of the iron pipes due to the \(\ce{Pb–Fe}\) junction; the less active \(\ce{Pb}\) will have served as the cathode for the reduction of \(\ce{O2}\), promoting oxidation of the more active \(\ce{Fe}\) nearby. Corrosion is a galvanic process that can be prevented using cathodic protection. The deterioration of metals through oxidation is a galvanic process called corrosion. Protective coatings consist of a second metal that is more difficult to oxidize than the metal being protected. Alternatively, a more easily oxidized metal can be applied to a metal surface, thus providing cathodic protection of the surface. A thin layer of zinc protects galvanized steel. Sacrificial electrodes can also be attached to an object to protect it. | 8,357 | 1,308 |
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The is the amount of substance reacted or produced per unit time. The is an expression indicating how the depends on the concentrations of the reactants and catalysts. The in the rate law expression is called the with respect to the reactant or catalyst. This page deals specifically with first- and second-order reaction kinetics, but you should know that other orders such as zeroth-order and 3rd order may also be involved in chemical kinetics. In acidic solutions, hydrogen peroxide and iodide ion react according to the equation: \(\ce{H2O2 + 2H+ + 3I- \rightarrow 2 H2O + I3-}\). In this reaction, the reaction can be expressed as However, from the stoichiometry, you can easily see the following relationship: \(\ce{-\dfrac{d[H2O2]}{dt}}=\ce{-\dfrac{1}{2}\dfrac{d[H+]}{dt}}=\ce{-\dfrac{1}{3}\dfrac{d[I- ]}{dt}}=\ce{\dfrac{1}{2}\dfrac{d[H2O]}{dt}}=\ce{\dfrac{d[I3- ]}{dt}}\) In reality, if the concentration of \(\ce{H2O2}\) is low, the changes in concentrations of \(\ce{H+}\) and \(\ce{H2O}\) are very difficult to detect because their quantities are so large in the solution. The merit in this equation is to show you that the rates of decreasing of reactant concentrations are governed by the stoichiometry. So are the rates of increasing of product concentrations. There are many ways to express the reaction , for example \(\ce{-\dfrac{d[H2O2]}{dt}}= \ce{-\dfrac{1}{2}\dfrac{d[H+]}{dt}} = \ce{-\dfrac{1}{3}\dfrac{d[I- ]}{dt}}=\ce{\dfrac{1}{2}\dfrac{d[H2O]}{dt}}=\ce{\dfrac{d[I3- ]}{dt}}= Rate\) or \(\ce{-\dfrac{d[I- ]}{dt}}= Rate\,'\) Obviously, \(Rate = 3 Rate\,'\) But both and ' are reasonable expressions. To generalize it, let the chemical reaction be represented by \(a\, \ce A + b\, \ce B \rightarrow c\, \ce C + d\, \ce D\) then the is represented by any one of the following \(rate = -\dfrac{1}{a}\ce{\dfrac{d[A]}{dt}}= -\dfrac{1}{b}\ce{\dfrac{d[B]}{dt}}=\dfrac{1}{c}\ce{\dfrac{d[C]}{dt}}=\dfrac{1}{d}\ce{\dfrac{d[D]}{dt}}\) Unless the rate expression is specified. For simplicity, let us consider the reactions: \(\mathrm{A + other\: reactants} \xrightarrow{\:\:\large{k}\:} \mathrm{products}\) where \(\ce{A}\) is one of the reactants, and is the rate constant. For most experiments in chemical kinetics, the concentration of one reactant or product is monitored. If the concentrations of other reactants are high, they are not greatly changed. Thus, we have a pseudo decomposition reaction. For the decomposition of \(\ce{A}\) with a rate constant , \(\mathrm{A \rightarrow products}\) The concentration \(\ce{[A]}\) can be monitored. Let the order of the reaction be n, then the expression \(\mathrm{-\dfrac{d[A]}{dt}= \mathit k [A]^n}\) is called the differential rate law. The differential expressions can be integrated to give an explicit relation of \(\ce{[A]}\) with respect to time . These explicit relations are called integrated rate laws. Depending on the value of n, the integrated equations are different. If the reaction is first order with respect to \(\ce{[A]}\), integration with respect to time, , gives: \(\mathrm{[A] = [A]_o}\, e^{\large{-k\, t}} \tag{1}\) where \(\mathrm{[A]_o}\) is the concentration of \(\ce{A}\) at = 0, and \(\ce{[A]}\) is the concentration at time . For a second order reaction, the integrated rate law is: \(\mathrm{\dfrac{1}{[A]}=\dfrac{1}{[A]_o}}+ k t \tag{2}\) Derive the above equations yourself. The usual approach to calculate the rate constant makes use of the differential rate law. A series of experiments are performed with various initial concentrations, and their rates measured. The rate constants are calculated from the initial concentration and time of measurement. Often, you should construct a graph for their evaluation. However, results so obtained contain errors due to the approximation whereas values for calculated using the integrated rate laws (1) and (2) are more accurate. Today, calculations can easily be performed with the aid of calculators and computers. Thus, we emphasize that you use the integrated rate laws whenever possible. Whether you use the differential rate laws or the integrated laws, you have to evaluate the order first. Equation (1) may be rewritten as \(k =\mathrm{\dfrac{\ln [A]_o - \ln [A]}{t}}\tag{1'}\) In a real experiment, you should plot \(\ce{- \ln [A]}\) vs. and find a line to best fit your data. The slope of the line is . Similarly, equation (2) may be rewritten as: \(k =\mathrm{\dfrac{\dfrac{1}{[A]}-\dfrac{1}{[A]_o}}{t}}\tag{2'}\) Thus, plot of \(\ce{\dfrac{1}{[A]}}\) vs. should yield a straight line, and the slope is . The half-life (\(t_{\frac{1}{2}}\)) of a reaction is the time period required to reduce the reactant to half of its original value. The half life of a first order reaction is a constant, independent of the initial concentration. The rate constant and half-life have the relationship: \(t_{\frac{1}{2}} = \dfrac{\ln(2)}{k}\) \(t_{\frac{1}{2}} \times k = \ln(2)\) For 2nd order reactions, the half life depends on the initial concentration, \(\mathrm{[A]_o}\), and the rate constant : \(t_{\frac{1}{2}} = \dfrac{1}{k \mathrm{[A]_o}}\) \(t_{\frac{1}{2}} \times k \mathrm{[A]_o} = 1\) Since the concentration is reduced to half of its original value at the end of its first half-life, the second half-life is twice as long as the first half life. Thus, a plot of \(\ce{[A]}\) vs. easily reveals the order of the reaction by tracking its half life. Either plotting \(\ce{[A]}\) vs. or plotting the appropriate linear relationship will reveal the order of the reaction. A summary of the relationships for first and second order reactions is given below: \(\mathrm{\ln [A]\: vs}\: t\) \(\textrm{slope} = - k\) \(\mathrm{\dfrac{1}{[A]}- \dfrac{1}{[A]_o}}= k t\) \(\mathrm{\dfrac{1}{[A]}\: vs}\: t\) \(\textrm{slope} = k\) Further applications of rate laws will be discussed in Integrated Rate Laws. Students must analyze the problems carefully, and appropriately apply the differential or integrated rate laws to derive the desirable results. Hint: \(\ce{[X]}\)
Since: \(\mathrm{- \dfrac{d[X]}{dt} = k}\); => \(\mathrm{[X] - [X]_o = - \mathit k t}\); The plot of \(\ce{[X]}\) and has a linear relationship. Hint: true
All radioactive decays follow first order kinetics. Hint: ln(2) = 0.693
\(k \times t_{\frac{1}{2}} = \ln(2) = 0.693\) Hint: 25 years
The first 12.5 y reduced 1.0 g of T to 1/2 g. Another 12.5 y reduced 0.5 g of T to 1/4 g.
Calculate the time for 1.0 g T to reduce to 0.9 g. Calculation of half-life from experiment is also a common exercise. Hint: 13:15 hr Hint: b. heterogeneous reaction
Chain reactions may lead to explosion, but dust explosion is not a chain reaction. Chemists explained the causes of a grain elevator explosion. Hint: 2
What is the order with respect to \(\ce{[O2]}\)? | 6,850 | 1,309 |
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Transition state theory is an equilibrium formulation of chemical reaction rates that originally comes from classical gas-phase reaction kinetics. We’ll consider a two-state system of reactant R and product P separated by a barrier ≫k T: \[R \underset{k_{r}}{\stackrel{k_{f}}{\rightleftharpoons}} P \nonumber \] which we obtain by projecting the free energy of the system onto a reaction coordinate ξ (a slow coordinate) by integrating over all the other degrees of freedom. There is a time-scale separation between the fluctuations in a state and the rare exchange events. All memory of a trajectory is lost on entering a state following a transition. Our goal is to describe the rates of crossing the transition state for the forward and reverse reactions. At thermal equilibrium, the rate constants for the forward and reverse reaction, \( k_f \) and \( k_r \), are related to the equilibrium constant and the activation barriers as \[K_{e q}=\frac{[P]}{[R]}=\frac{P_{P, e q}}{P_{R, e q}}=\frac{k_{f}}{k_{r}}=\exp \left(-\frac{\left(E_{a}^{f}-E_{a}^{r}\right)}{k_{B} T}\right) \nonumber \] \(E^f_a\), \(E^r_a\) are the activation free energies for the forward and reverse reactions, which are related to the reaction free energy through \(E^f_a - E^r_a = \Delta G^0_{rxn}\). P refers to the population or probability of occupying the reactant or product state. The primary assumptions of TST is that the transition state is well represented by an activated complex \(RP^‡\) that acts as an intermediate for the reaction from R to P, that all species are in thermal equilibrium, and that the flux across the barrier is proportional to the population of the activated complex. \[R \rightleftharpoons RP^‡ \rightleftharpoons P \nonumber \] Then, the steady state population of the activated complex can be determined by an equilibrium constant that we can express in terms of the molecular partition functions. Let’s focus on the rate of the forward reaction considering only the equilibrium \[R \rightleftharpoons RP^‡ \nonumber \] We relate the population of reactants within the reactant well to the population of the activated complex through an equilibrium constant \[K^‡_{eq} = \dfrac{[RP^‡]}{[R]} \nonumber \] which we will evaluate using partition functions for the reactant and activated complex \[ K^‡_{eq} =\dfrac{q^‡/V}{q_R/V}e^{-E^f_a/k_BT} \nonumber \] Then we write the forward flux in eq. (23.1) proportional to the population of activated complex \[\begin{aligned}\langle J^‡\rangle &= v[RP^‡]\\ &= vK^‡_{eq}[R] \end{aligned} \nonumber \] Here ν is the reaction frequency, which is the inverse of the transition state lifetime \(\tau_{mol}\). \( v^{-1}\) or \(\tau_{mol}\) reflects the time it takes to cross the transition state region. To evaluate ν, we will treat motion along the reaction coordinate ξ at the barrier as a translational degree of freedom. When the reactants gain enough energy (\(E^f_a\)), they will move with a constant forward velocity \(v_f\) through a transition state region that has a width \(\ell\). (The exact definition of \(\ell\) will not matter too much). \[ \tau_{mol} = \dfrac{\ell}{v_f} \nonumber \] Then we can write the average flux of population across the transition state in the forward direction \[\begin{aligned} \langle J^‡\rangle &= K^‡_{eq} [R] \dfrac{v_f}{\ell}\\&= \dfrac{q^‡}{q_R}e^{-E^f_a/k_BT}[R] \frac{1}{\ell} \sqrt{\dfrac{k_BT}{2\pi m}} \end{aligned}\] \[\] where \(v_f\) is obtained from a one-dimensional Maxwell–Boltzmann distribution. For a multidimensional problem, we want to factor out the slow coordinate, i.e., reaction coordinate (ξ) from partition function. \[ q^‡ = q_ξ q^{'‡} \nonumber \] \(q^{'‡}\) contains all degrees of freedom except the reaction coordinate. Next, we calculate \(q_ξ\) by treating it as translational motion: \[ q_ξ (trans) = \displaystyle\int\limits_0^{\ell} dξe^{-E_{trans}/k_BT} = \sqrt{\dfrac{2\pi m k_B T}{h^2}\ell} \] Substituting (23.1.2) into (23.1.1): \[\left\langle J_{f}^{‡}\right\rangle=\frac{k_{B} T}{h} \frac{q^{\prime ‡}}{q_{R}} e^{-E_{a}^{f} / k_{B} T}[R] \nonumber \] We recognize that the factor \(v = k_BT/h\) is a frequency whose inverse gives an absolute lower bound on the crossing time of \(~10^{-13} \) seconds. If we use the speed of sound in condensed matter this time is what is needed to propagate 1–5 Å. Then we can write \[\left\langle J_{f}^{‡}\right\rangle= k_f[R] \nonumber \] where the forward rate constant is \[k_f = Ae^{-E^f_a/k_BT} \] and the pre-exponential factor is \[ A = v\dfrac{q^{'‡}}{q_R} \nonumber \] determines the time that it takes to cross the transition state in the absence of barriers (E → 0). k is also referred to as k . To make a thermodynamic connection, we can express eq. (4) in the Eyring form \[k_{f}=v e^{\Delta S^{‡} / k_{B}} e^{-\Delta E_{f}^{‡} / k_{B} T} \nonumber \] where the transition state entropy is \[ \Delta S^‡ = k\ln{\dfrac{q^{'‡}}{q_R}} \nonumber \] \( \Delta S^‡\) represents a count (actually ratio) of the reduction of accessible microstates in making the transition from the reactant well to the transition state. For biophysical phenomena, the entropic factors are important, if not dominant! Also note implicit in TST is a dynamical picture in which every trajectory that arrives with forward velocity at the TST results in a crossing. It therefore gives an upper bound on the true rate, which may include failed attempts to cross. This is often accounted for by adding a transmission coefficient κ < 1 to k : k =κk . Kramers' theory provides a physical basis for understanding κ. | 5,590 | 1,310 |
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Thermodynamics is a powerful approach toward understanding chemical reactions, but only provides part of the picture. Specifically: Chemical change is driven by the tendency of atoms and molecules to rearrange themselves in a way that results in the maximum possible dispersion of thermal energy into the world. The observable quantity that measures this spreading and sharing of energy is the of the system. As a chemical change takes place, the quantities of reactants and products change in a way that leads to a more negative free energy. When the free energy reaches its minimum possible value, there is no more and the system is said to be in . The beauty of thermodynamics is that it enables us to unfailingly predict the net direction of a reaction and the composition of the equilibrium state even without conducting the experiment; the standard free energies of the reactants and products, which can be independently measured or obtained from tables, are all we need. When the free energy reaches its minimum possible value, there is no more and the system is said to be in : It is worth noting that the concept of "time" plays no role whatsoever in thermodynamics. But kinetics is all about time. The "speed" of a reaction — how long it takes to reach equilibrium — bears no relation at all to how it is (as given by the sign and value of Δ °) or whether it is exothermic or endothermic (given by the sign of Δ °). Moreover, there is no way that reaction rates can be predicted in advance; each reaction must be studied individually. The concept of "time" plays no role whatsoever in thermodynamics. The term "mechanism" refers to, "who does what to whom". Think of a reaction mechanism as something that goes on in a "black box" that joins reactants to products. The inner workings of the black box are ordinarily hidden from researchers, are highly unpredictable, and can only be inferred by indirect means. The stoichiometric equation for the reaction says nothing about its mechanism! Consider, for example, the gas-phase formation reactions of the hydrogen halides from the elements. The thermodynamics of these reactions are all similar (they are all highly exothermic), but their (their kinetics and mechanisms) could not be more different. \[\ce{H2(g) + I2(g) -> 2 HI(g)} \nonumber \] Careful experiments, carried out over many years, are consistent with the simplest imaginable mechanism: a collision between the two reactant molecules results in a rearrangement of the bonds. \[\ce{H2(g) + Br2(g) -> 2 HBr(g)} \nonumber\] One might be tempted to suppose that this would proceed in a similar way, but experiments reveal that the mechanism of this reaction is far more complex. The reaction takes place in a succession of steps, some of which involve atomic H and Br. \[\ce{H2(g) + Cl2(g) -> 2 HCl(g)} \nonumber \] The mechanism of this reaction is different again. Although the first two reactions reach equilibrium in minutes to an hour or so at temperatures of 300 to 600 K, a mixture of hydrogen and chlorine will not react at all in the dark, but if you shine a light on the mixture, it goes off with a bang as the instantaneous reaction releases heat and expands the gas explosively. What is particularly noteworthy is that these striking differences cannot be reliably predicted from theory; they were revealed only by experimentation. | 3,387 | 1,311 |
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It is a common observation that most of the reactions when carried out in closed vessels do not go to completion, under a given set of conditions of temperature and pressure. In fact in all such cases, in the initial state, only the reactants are present but as the reaction proceeds, the concentration of reactants decreases and that of products increases. Finally a stage is reached when no further change in concentration of the reactants and products is observed. If a mixture of gaseous hydrogen and iodine vapors is made to react at 717 k in a closed vessel for about 2 - 3 hours, gaseous hydrogen iodide is produced according to the following equation: \[ H_{2\; (g)} + I_{2\; (g)} \rightarrow 2HI_{(g)}\] But along with gaseous hydrogen iodide, there will be some amount of unreacted gaseous hydrogen and gaseous iodine left. On the other hand if gaseous hydrogen iodide is kept at 717K in a closed vessel for about 2 - 3 hours it decomposes to give gaseous hydrogen and gaseous iodine. \[ 2HI_{(g)} \rightarrow H_{2\; (g)} + I_{2\; (g)} \] In this case also some amount of gaseous hydrogen iodide will be left unreacted. This means that the products of certain reactions can be converted back to the reactants. These types of reactions are called . Thus, in reversible reactions the products can react with one another under suitable conditions to give back the reactants. In other words, in reversible reactions the reaction takes place in both the forward and backward directions. The reversible reaction may be expressed as: \[ H_{2\; (g)} + I_{2\; (g)} \rightleftharpoons 2HI_{(g)}\] These reversible reactions never go to completion if performed in a closed container. For a reversible chemical reaction, an equilibrium state is attained when the rate at which a chemical reaction is proceeding in forward direction equals the rate at which the reverse reaction is proceeding. | 1,910 | 1,312 |
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Reversible and irreversible reactions are prevalent in nature and are responsible for reactions such as the breakdown of ammonia. It was believed that all chemical reactions were irreversible until 1803, when French chemist Claude Louis Berthollet introduced the concept of reversible reactions. Initially he observed that sodium carbonate and calcium chloride react to yield calcium carbonate and sodium chloride; however, after observing sodium carbonate formation around the edges of salt lakes, he realized that large amount of salts in the evaporating water reacted with calcium carbonate to form sodium carbonate, indicating that the reverse reaction was occurring. Chemical reactions are represented by chemical equations. These equations typically have a unidirectional arrow (\(\rightarrow\)) to represent irreversible reactions. Other chemical equations may have a bidirectional harpoons (\(\rightleftharpoons\)) that represent reversible reactions (not to be confused with the double arrows \(\leftrightarrow\) used to indicate ). To review the fundamentals of chemical reactions, click here: A fundamental concept of chemistry is that chemical reactions occurred when reactants reacted with each other to form products. These unidirectional reactions are known as irreversible reactions, reactions in which the reactants convert to products and where the products convert back to the reactants. These reactions are essentially like baking. The ingredients, acting as the reactants, are mixed and baked together to form a cake, which acts as the product. This cake cannot be converted back to the reactants (the eggs, flour, etc.), just as the products in an irreversible reaction cannot convert back into the reactants. An example of an irreversible reaction is combustion. Combustion involves burning an organic compound—such as a hydrocarbon—and oxygen to produce carbon dioxide and water. Because water and carbon dioxide are stable, they do not react with each other to form the reactants. Combustion reactions take the following form: \[ C_xH_y + O_2 \rightarrow CO_2 + H_2O \] In reversible reactions, the reactants and products are never fully consumed; they are each constantly reacting and being produced. A reversible reaction can take the following summarized form: \[ A + B \underset{k_{-1}} {\overset{k_1} {\rightleftharpoons}} C + D \] Reaction 1: Reaction 2: Below is an example of the summarized form of a reversible reaction and a breakdown of the reversible reaction N O ↔ 2NO Reaction 1 and Reaction 2 happen at the same time because they are in a closed system. Nitrogen : Oxygen Reaction 1 Reaction 2 Imagine a ballroom. Let reactant A be 10 girls and reactant B be 10 boys. As each girl and boy goes to the dance floor, they pair up to become a product. Once five girls and five boys are on the dance floor, one of the five pairs breaks up and moves to the sidelines, becoming reactants again. As this pair leaves the dance floor, another boy and girl on the sidelines pair up to form a product once more. This process continues over and over again, representing a reversible reaction. Unlike irreversible reactions, reversible reactions lead to equilibrium: in reversible reactions, the reaction proceeds in both directions whereas in irreversible reactions the reaction proceeds in only one direction. To learn more about this phenomenon, click here: If the reactants are formed at the same rate as the products, a dynamic equilibrium exists. For example, if a water tank is being filled with water at the same rate as water is leaving the tank (through a hypothetical hole), the amount of water remaining in the tank remains consistent. There are four binding sites on a hemoglobin protein. Hemoglobin molecules can either bind to carbon dioxide or oxygen. As blood travels through the alveoli of the lungs, hemoglobin molecules pick up oxygen-rich molecules and bind to the oxygen. As the hemoglobin travels through the rest of the body, it drops off oxygen at the capillaries for the organ system to use oxygen. After expelling the oxygen, it picks up carbon dioxide. Because this process is constantly carried out through the body, there are always hemoglobin molecules picking or expelling oxygen and other hemoglobin molecules that are picking up or expelling carbon dioxide. Therefore, the hemoglobin molecules, oxygen, and carbon dioxide are reactants while the hemoglobin molecules with oxygen or carbon dioxide bound to them are the products. In this closed system, some reactants convert into products as some products are changing into reactants, making it similar to a reversible reaction. | 4,666 | 1,313 |
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. For a mixture of components \(A\), \(B\), \(C\), etc., Amagat’s law gives the volume as \[V_{mixture}=V_A+V_B+V_C+\dots\] For real gases, Amagat’s law is usually an even better approximation than \({}^{6}\). Again, for mixtures of ideal gases, it is exact. For an ideal gas, the partial volume is \[V_A=\frac{n_ART}{P_{mixture}}\] Since \(n_{mixture}=n_A+n_B+n_C+\dots\), we have, for a mixture of ideal gases, \[ \begin{align*} V_{mixture}&=\frac{n_{mixture}RT}{P_{mixture}} \\[4pt] &=\frac{\left(n_A+n_B+n_C+\dots \right)RT}{P_{mixture}} \\[4pt] &=V_A+V_B+V_C+\dots \end{align*}\] Applied to the mixture, the ideal-gas equation yields Amagat’s law. Also, we have \(V_A=x_AV_{mixture}\). | 703 | 1,314 |
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The thermochemical quantities that you probably encounter most often are the caloric values of food. Food supplies the raw materials that your body needs to replace cells and the energy that keeps those cells functioning. About 80% of this energy is released as heat to maintain your body temperature at a sustainable level to keep you alive. The nutritional Calorie (with a capital C) that you see on food labels is equal to 1 kcal (kilocalorie). The caloric content of food is determined from its enthalpy of combustion (Δ ) per gram, as measured in a bomb calorimeter, using the general reaction
\[ food + excess \; O_{2}(g) \rightarrow CO_{2}(g) + H_{2}O(l) + N_{2}(g) \] There are two important differences, however, between the caloric values reported for foods and the Δ of the same foods burned in a calorimeter. First, the Δ described in joules (or kilojoules) are negative for all substances that can be burned. In contrast, the caloric content of a food is always expressed as a number because it is . Therefore,
\[ caloric \;content = - \Delta H_{comb} \] Second, when foods are burned in a calorimeter, any nitrogen they contain (largely from proteins, which are rich in nitrogen) is transformed to N . In the body, however, nitrogen from foods is converted to urea [(H N) C=O], rather than N before it is excreted. The Δ of urea measured by bomb calorimetry is −632.0 kJ/mol. Consequently, the enthalpy change measured by calorimetry for any nitrogen-containing food is greater than the amount of energy the body would obtain from it. The difference in the values is equal to the Δ of urea multiplied by the number of moles of urea formed when the food is broken down. This point is illustrated schematically in the following equations:
\[ food + excess \; O_{2}\left ( g \right )\xrightarrow[]{\Delta H_{1} < 0} CO_{2}\left ( g \right )+H_{2}O\left ( l \right )+\cancel{\left ( H_{2}N \right )_{2}C=O\left ( s \right )} \] \[ \cancel{\left( H_{2}N \right )_{2}C=O\left ( s \right )} + \cancel{excess} \; \dfrac{3}{2}O_{2}\left ( g \right )\xrightarrow[]{\Delta H_{2} = 632.0 \; kJ/mol} CO_{2}\left ( g \right )+2H_{2}O\left ( l \right )+ N_{2}\left ( g \right ) \] which adds up to \[ food + excess \; O_{2}\left ( g \right )\xrightarrow[]{\Delta H_{3}=\Delta H_{1}+\Delta H_{2} < 0} 2CO_{2}\left ( g \right )+3H_{2}O\left ( l \right )+ N_{2}\left ( g \right ) \] All three Δ values are negative, and, by Hess’s law, Δ = Δ + Δ . The magnitude of Δ must be less than Δ , the calorimetrically measured Δ for a food. By producing urea rather than N , therefore, humans are excreting some of the energy that was stored in their food. Because of their different chemical compositions, foods vary widely in caloric content. As we saw in Example 5, for instance, a fatty acid such as palmitic acid produces about 39 kJ/g during combustion, while a sugar such as glucose produces 15.6 kJ/g. Fatty acids and sugars are the building blocks of fats and carbohydrates, respectively, two of the major sources of energy in the diet. Nutritionists typically assign average values of 38 kJ/g (about 9 Cal/g) and 17 kJ/g (about 4 Cal/g) for fats and carbohydrates, respectively, although the actual values for specific foods vary because of differences in composition. Proteins, the third major source of calories in the diet, vary as well. Proteins are composed of amino acids, which have the following general structure: An amino acid contains an amine group (−NH ) and a carboxylic acid group (−CO H). In addition to their amine and carboxylic acid components, amino acids may contain a wide range of other functional groups: R can be hydrogen (–H); an alkyl group (e.g., –CH ); an aryl group (e.g., –CH C H ); or a substituted alkyl group that contains an amine, an alcohol, or a carboxylic acid (Figure \(\Page {1}\) ). Of the 20 naturally occurring amino acids, 10 are required in the human diet; these 10 are called because our bodies are unable to synthesize them from other compounds. Because R can be any of several different groups, each amino acid has a different value of Δ . Proteins are usually estimated to have an average Δ of 17 kJ/g (about 4 Cal/g). Calculate the amount of available energy obtained from the biological oxidation of 1.000 g of alanine (an amino acid). Remember that the nitrogen-containing product is urea, not N , so biological oxidation of alanine will yield energy than will combustion. The value of Δ for alanine is −1577 kJ/mol. amino acid and Δ per mole caloric content per gram Write balanced chemical equations for the oxidation of alanine to CO , H O, and urea; the combustion of urea; and the combustion of alanine. Multiply both sides of the equations by appropriate factors and then rearrange them to cancel urea from both sides when the equations are added. Use Hess’s law to obtain an expression for Δ for the oxidation of alanine to urea in terms of the Δ of alanine and urea. Substitute the appropriate values of Δ into the equation and solve for Δ for the oxidation of alanine to CO , H O, and urea. Calculate the amount of energy released per gram by dividing the value of Δ by the molar mass of alanine. The actual energy available biologically from alanine is less than its Δ because of the production of urea rather than N . We know the Δ values for alanine and urea, so we can use Hess’s law to calculate Δ for the oxidation of alanine to CO , H O, and urea. We begin by writing balanced chemical equations for (1) the oxidation of alanine to CO , H O, and urea; (2) the combustion of urea; and (3) the combustion of alanine. Because alanine contains only a single nitrogen atom, whereas urea and N each contain two nitrogen atoms, it is easier to balance Equations 1 and 3 if we write them for the oxidation of 2 mol of alanine: \[ \left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+6O_{2}\left ( g \right ) \rightarrow 5CO_{2}\left ( g \right )+5H_{2}O\left ( l \right )+\left ( H_{2}N \right )_{2}C=O\left ( s \right ) \notag \] \[ \left ( 2 \right ) \; \; \left( H_{2}N \right )_{2}C=O\left ( s \right ) + \dfrac{3}{2}O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( g \right )+2H_{2}O\left ( l \right )+ N_{2}\left ( g \right ) \notag \] \[ \left ( 3 \right ) \; \; \left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+\frac{15}{2}O_{2}\left ( g \right )\rightarrow 6CO_{2}\left ( g \right )+7H_{2}O\left ( l \right )+N_{2}\left ( g \right ) \notag \] Adding Equations 1 and 2 and canceling urea from both sides give the overall chemical equation directly: \[ \left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+6O_{2}\left ( g \right ) \rightarrow 5CO_{2}\left ( g \right )+5H_{2}O\left ( l \right )+\cancel{\left ( H_{2}N \right )_{2}C=O\left ( s \right )} \notag \] \[ \cancel{ \left ( 2 \right ) \; \; \left( H_{2}N \right )_{2}C=O\left ( s \right )} + \dfrac{3}{2}O_{2}\left ( g \right ) \rightarrow CO_{2}\left ( g \right )+2H_{2}O\left ( l \right )+ N_{2}\left ( g \right ) \notag \] \[ \left ( 3 \right ) \; \; \left ( 1 \right )\; \; 2C_{3}H_{7}NO_{2}\left ( s \right )+\dfrac{15}{2}O_{2}\left ( g \right )\rightarrow 6CO_{2}\left ( g \right )+7H_{2}O\left ( l \right )+N_{2}\left ( g \right ) \notag \] By Hess’s law, Δ = Δ + Δ . We know that Δ = 2Δ (alanine), Δ = Δ (urea), and Δ = 2Δ (alanine → urea). Rearranging and substituting the appropriate values gives \[ \Delta {H_{1}} = \Delta {H_{3}} - \Delta {H_{2}} \notag \] \[ =2\left ( -1577 \; kJ/mol \right )-\left ( -632.0 \; kJ/mol \right ) \notag \]= -2522 \; kJ/\left ( 2 \;mol\; analine \right ) \notag \] Thus Δ (alanine → urea) = −2522 kJ/(2 mol of alanine) = −1261 kJ/mol of alanine. Oxidation of alanine to urea rather than to nitrogen therefore results in about a 20% decrease in the amount of energy released (−1261 kJ/mol versus −1577 kJ/mol). The energy released per gram by the biological oxidation of alanine is \[ \left (\frac{-1261 \; kJ}{1 \; \cancel{mol}} \right )\left (\frac{1 \; \cancel{mol}}{89.094 \; g} \right )= -14.15 \; kJ/g \notag \] This is equal to −3.382 Cal/g. Calculate the energy released per gram from the oxidation of valine (an amino acid) to CO , H O, and urea. Report your answer to three significant figures. The value of Δ for valine is −2922 kJ/mol. −22.2 kJ/g (−5.31 Cal/g) The reported caloric content of foods does not include Δ for those components that are not digested, such as fiber. Moreover, meats and fruits are 50%−70% water, which cannot be oxidized by O to obtain energy. So water contains no calories. Some foods contain large amounts of fiber, which is primarily composed of sugars. Although fiber can be burned in a calorimeter just like glucose to give carbon dioxide, water, and heat, humans lack the enzymes needed to break fiber down into smaller molecules that can be oxidized. Hence fiber also does not contribute to the caloric content of food. We can determine the caloric content of foods in two ways. The most precise method is to dry a carefully weighed sample and carry out a combustion reaction in a bomb calorimeter. The more typical approach, however, is to analyze the food for protein, carbohydrate, fat, water, and “minerals” (everything that doesn’t burn) and then calculate the caloric content using the average values for each component that produces energy (9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals). An example of this approach is shown in Table \(\Page {1}\) for a slice of roast beef. The compositions and caloric contents of some common foods are given in Table \(\Page {2}\) Because the Calorie represents such a large amount of energy, a few of them go a long way. An average 73 kg (160 lb) person needs about 67 Cal/h (1600 Cal/day) to fuel the basic biochemical processes that keep that person alive. This energy is required to maintain body temperature, keep the heart beating, power the muscles used for breathing, carry out chemical reactions in cells, and send the nerve impulses that control those automatic functions. Physical activity increases the amount of energy required but not by as much as many of us hope (Table \(\Page {3}\) ). A moderately active individual requires about 2500−3000 Cal/day; athletes or others engaged in strenuous activity can burn 4000 Cal/day. Any excess caloric intake is stored by the body for future use, usually in the form of fat, which is the most compact way to store energy. When more energy is needed than the diet supplies, stored fuels are mobilized and oxidized. We usually exhaust the supply of stored carbohydrates before turning to fats, which accounts in part for the popularity of low-carbohydrate diets. What is the minimum number of Calories expended by a 72.6 kg person who climbs a 30-story building? (Assume each flight of stairs is 14 ft high.) How many grams of glucose are required to supply this amount of energy? (The energy released during the combustion of glucose was calculated in Example 5.) mass, height, and energy released by combustion of glucose calories expended and mass of glucose needed Convert mass and height to SI units and then substitute these values into Equation \(\Page {6}\) to calculate the change in potential energy (in kilojoules). Divide the calculated energy by 4.184 Cal/kJ to convert the potential energy change to Calories. Use the value obtained in Example 1 for the combustion of glucose to calculate the mass of glucose needed to supply this amount of energy. The energy needed to climb the stairs equals the difference between the person’s potential energy ( ) at the top of the building and at ground level. Recall that = . Because and are given in non-SI units, we must convert them to kilograms and meters, respectively Thus \[ PE = \left ( 72.6 \; kg \right )\left ( 9.81 \; m/s^{2} \right )\left ( 128 m \right ) = 8.55 × 10^{4} \left ( kg \cdot m^{2}/s^{2} \right ) = 91.2 kJ \notag \] To convert to Calories, we divide by 4.184 kJ/kcal: \[ PE = \left ( 91.2 \; \cancel{kJ} \right ) \left ( \dfrac{1 \; kcal}{4.184 \; \cancel{kJ}} \right )=21.8 \;kcal = 21.8 \; Cal \notag \] Because the combustion of glucose produces 15.6 kJ/g (Example 5), the mass of glucose needed to supply 85.5 kJ of energy is \[ PE = \left ( 91.2 \; \cancel{kJ} \right )\left ( \frac{1 \; g \; glucose}{15.6 \; \cancel{kJ}} \right )=5.85\; g \; glucose \notag \] This mass corresponds to only about a teaspoonful of sugar! Because the body is only about 30% efficient in using the energy in glucose, the actual amount of glucose required would be higher: (100%/30%) × 5.85 g = 19.5 g. Nonetheless, this calculation illustrates the difficulty many people have in trying to lose weight by exercise alone. Calculate how many times a 160 lb person would have to climb the tallest building in the United States, the 110-story Willis Tower in Chicago, to burn off 1.0 lb of stored fat. Assume that each story of the building is 14 ft high and use a calorie content of 9.0 kcal/g of fat. About 55 times The calculations in Example2 ignore various factors, such as how fast the person is climbing. Although the rate is irrelevant in calculating the change in potential energy, it is very relevant to the amount of energy actually required to ascend the stairs. The calculations also ignore the fact that the body’s conversion of chemical energy to mechanical work is significantly less than 100% efficient. According to the average energy expended for various activities listed in Table \(\Page {1}\), a person must run more than 4.5 h at 10 mph or bicycle for 6 h at 13 mph to burn off 1 lb of fat (1.0 lb × 454 g/lb × 9.0 Cal/g = 4100 Cal). But if a person rides a bicycle at 13 mph for only 1 h per day 6 days a week, that person will burn off 50 lb of fat in the course of a year (assuming, of course, the cyclist doesn’t increase his or her intake of calories to compensate for the exercise). The nutritional is equivalent to 1 kcal (4.184 kJ). The caloric content of a food is its Δ per gram. The combustion of nitrogen-containing substances produces N (g), but the biological oxidation of such substances produces urea. Hence the actual energy available from nitrogen-containing substances, such as proteins, is less than the Δ of urea multiplied by the number of moles of urea produced. The typical caloric contents for food are 9 Cal/g for fats, 4 Cal/g for carbohydrates and proteins, and 0 Cal/g for water and minerals. Can water be considered a food? Explain your answer. Describe how you would determine the caloric content of a bag of popcorn using a calorimeter. Why do some people initially feel cold after eating a meal and then begin to feel warm? In humans, one of the biochemical products of the combustion/digestion of amino acids is urea. What effect does this have on the energy available from these reactions? Speculate why conversion to urea is preferable to the generation of N . \(\Page {9}\) Determine the amount of energy available from the biological oxidation of 1.50 g of leucine (an amino acid, Δ = −3581.7 kJ/mol). Calculate the energy released (in kilojoules) from the metabolism of 1.5 oz of vodka that is 62% water and 38% ethanol by volume, assuming that the total volume is equal to the sum of the volume of the two components. The density of ethanol is 0.824 g/mL. What is this enthalpy change in nutritional Calories? While exercising, a person lifts an 80 lb barbell 7 ft off the ground. Assuming that the transformation of chemical energy to mechanical energy is only 35% efficient, how many Calories would the person use to accomplish this task? From Figure \(\Page {2}\), how many grams of glucose would be needed to provide the energy to accomplish this task? A 30 g sample of potato chips is placed in a bomb calorimeter with a heat capacity of 1.80 kJ/°C, and the bomb calorimeter is immersed in 1.5 L of water. Calculate the energy contained in the food per gram if, after combustion of the chips, the temperature of the calorimeter increases to 58.6°C from an initial temperature of 22.1°C. | 16,097 | 1,315 |
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An electrolyte solution is a solution that generally contains ions, atoms or molecules that have lost or gained electrons, and is electrically conductive. For this reason they are often called ionic solutions, however there are some cases where the electrolytes are not ions. For this discussion we will only consider solutions of ions. A basic principle of electrostatics is that opposite charges attract and like charges repel. It also takes a great deal of force to overcome this electrostatic attraction. The general form of Coulomb's law describes the force of attraction between charges: \[F=k\frac{q_1mq_2}{r^2}\] However, we must make some changes to this physics formula to be able to use it for a solution of oppositely charged ions. In Coulomb's Law, the constant \[k=\frac{1}{4\pi\varepsilon_{0}}\], where \(\varepsilon_{0}\) is the permittivity of free space, such as in a vacuum. However, since we are looking at a solution, we must consider the effect that the medium (the solvent in this case) has on the electrostatic force, which is represented by the dielectric constant \(\varepsilon\): \[F=\frac{q_{1}q_{2}}{4\pi\varepsilon_{0}\varepsilon r^{2}}\] Polar substances such as water have a relatively high dielectric constant. Ions are not stable on their own, and thus no ions can ever be studied separately. Particularly in biology, all ions in a certain cell or tissue have a counterion that balances this charge. Therefore, we cannot measure the enthalpy or entropy of a single ion as we can atoms of a pure element. So we define a reference point. The \(\Delta_{f}\overline{H}^{\circ} \) of a hydrogen ion \(H^+\) is equal to zero, as are the other thermodynamic quantities. \[\Delta_{f}\overline{H}^{\circ}[H^{+}(aq)]=0\] \[\Delta_{f}\overline{G}^{\circ}[H^{+}(aq)]=0\] \[\overline{S}^{\circ}[H^{+}(aq)]=0\] When studying the formation of ionic solutions, the most useful quantity to describe is chemical potential \(\mu\), defined as the partial molar Gibbs energy of the ith component in a substance: \[\mu_{i}=\overline{G}_{i}=\left(\frac{\partial G}{\partial n_{i}}\right)_{T,P,n_{j}}=\mu_{i}^{\circ}+RT\ln x_{i}\] where \(x_{i}\) can be any unit of concentration of the component: mole fraction, molality, or for gases, the partial pressure divided by the pressure of pure component. To express the chemical potential of an electrolyte in solution in terms of molality, let us use the example of a dissolved salt such as magnesium chloride, \(MgCl_{2}\). \[MgCl_{2}\rightleftharpoons Mg^{2+}+2Cl^{-} \label{1}\] We can now write a more general equation for a dissociated salt: \[M_{\nu+}X_{\nu-}\rightleftharpoons\nu_{+}M^{z+}+\nu_{-}X^{z-} \label{2} \] where \(\nu_{\pm}\) represents the stoichiometric coefficient of the cation or anion and \(z_\pm\) represents the charge, and M and X are the metal and halide, respectively. The total chemical potential for these anion-cation pair would be the sum of their individual potentials multiplied by their stoichiometric coefficients: \[\mu=\nu_{+}\mu_{+}+\nu_{-}\mu_{-} \label{3} \] The chemical potentials of the individual ions are: \[\mu_{+} = \mu_+^{\circ}+RT\ln m_+ \label{4} \] \[\mu_{-} = \mu_-^{\circ}+RT\ln m_- \label{5} \] And the molalities of the individual ions are related to the original molality of the salt m by their stoichiometric coefficients \[m_{+}=\nu_{+}m\] Substituting Equations \(\ref{4}\) and \(\ref{5}\) into Equation \(\ref{3}\), \[ \mu=\left( \nu_+\mu_+^{\circ}+\nu_- mu_-^{\circ}\right)+RT\ln\left(m_+^{\nu+}m_-^{\nu-}\right) \label{6} \] since the total number of moles \(\nu=\nu_{+}+\nu_{-}\), we can define the mean ionic molality as the geometric average of the molarity of the two ions: \[ m_{\pm}=(m_+^{\nu+}m_-^{\nu-})^{\frac{1}{\nu}}\] then Equation \(\ref{6}\) becomes \[\mu=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+\nu RT\ln m_{\pm} \label{7} \] We have derived this equation for a ideal solution, but ions in solution exert electrostatic forces on one another to deviate from ideal behavior, so instead of molarities we must use the activity a to represent how the ion is behaving in solution. Therefore the mean ionic activity is defined as \[a_{\pm}=(a_{+}^{\nu+}+a_{-}^{\nu-})^{\frac{1}{\nu}}\] where \[a_{\pm}=\gamma m_{\pm} \label{mean}\] and \(\gamma_{\pm}\) is the , which is dependent on the substance.
Substituting the mean ionic activity of \Equation \(\ref{mean}\) into Equation \(\ref{7}\), \[\mu=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+\nu RT\ln a_{\pm}=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+RT\ln a_{\pm}^{\nu}=(\nu_{+}\mu_{+}^{\circ}+\nu_{-}\mu_{-}^{\circ})+RT \ln a \label{11}\] when \(a=a_{\pm}^{\nu}\). Equation \(\ref{11}\) then represents the chemical potential of a nonideal electrolyte solutions. To calculate the mean ionic activity coefficient requires the use of the Debye-Hückel limiting law, part of the . Let us now write out the chemical potential in terms of molality of the salt in our first example, \(MgCl_{2}\). First from Equation \(\ref{1}\), the stoichiometric coefficients of the ions are: \[\nu_{+} = 1,\nu_{-} = 2,\nu\; = 3 \nonumber\] The mean ionic molality is \[\begin{align*} m_{\pm} &= (m_{+}^{1}m_{-}^{2})^{\frac{1}{3}} \\[4pt] &= (\nu_{+}m\times\nu_{-}m)^{\frac{1}{3}} \\[4pt] &=m(1^{1}2^{2})^{\frac{1}{3}} \\[4pt] &=1.6\, m \end{align*}\] The expression for the chemical potential of \[MgCl_{2}\] is \[\mu_{MgCl_{2}}=\mu_{MgCl_{2}}^{\circ}+3RT\ln 1.6\m m \nonumber\] | 5,484 | 1,316 |
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The Earth's mantle is composed largely of which has high Fe or Mg content, and low silicon content. This rock may be converted to by a process (logically) called , which is an process, releasing a lot of heat energy-- about 660,000,000 joules of heat per cubic meter of rock according to the NOAA (National Oceanic and Atmospheric Administration). This is enough energy to raise the temperature of the rock by 260°C (550°F) . The "Lost City", located 20 km west of the Mid-Atlantic Ridge, is a hydrothermal vent field characterized by carbonate edifices that tower 60 m above the ocean floor and extreme conditions found nowhere else in the marine environment Serpentine itself is often chrysotile, Mg Si O (OH) . The name "serpentine" derives from the Latin ("serpent rock") because the mineral is often greenish with a smooth to scaly surface . Serpentine Polished Serpentine Sample Serpentine marbles are used in architecture and jewelry, but other serpentines include asbestos (which is a lung cancer risk when mined) and another 20 varieties of hydrous magnesium/iron phyllosilicates. Serpentinites often are toxic to plants because they may contain significant levels of nickel, chromium, and cobalt. They are often mixed, and thus treated collectively as a group called "serpentinites." In the case of the of hydrothermal vents in the mid-Atlantic, serpentinization is a greater source of energy than the radioactivity of the Earth's core . Radioactivity normally accounts for about 80% of the internal heat of the Earth . But like serpentinization, which produce or consume significant amounts of heat, are an inextricable part of all geological processes. Since serpentinites are mixtures, several equations can be used to describe their reactions. While natural minerals may have indefinite compositions like olivine, (Fe,Mg) SiO , we always look at equations for specific reactions in order to associate definite energies with them. These are called . For example, serpentinization may involve: Fayalite + water → magnetite + aqueous silica + hydrogen 3 Fe SiO + 2 H O → 2 Fe O + 3 SiO + 3 H Δ = 40.1 kJ (25°C, 1 atm pressure) (1) : Forsterite + aqueous silica → serpentine (crysotile) 3 Mg SiO + SiO + 4 H O → 2 Mg Si O (OH) Δ = -179.7 kJ (25°C, 1 atm pressure) (2) : Forsterite + water → serpentine (chrysotile) + brucite 3 Mg SiO +3 H O → 2 Mg Si O (OH) + Mg(OH) Δ = -2267.2 kJ (25°C, 1 atm pressure) (3) Here the Δ (delta subscript m) tells us whether heat energy is released or absorbed when the reaction occurs and also enables us to find the actual quantity of energy involved. By convention, if Δ is , as in Equation (1), heat is by the reaction; i.e., it is . More commonly, Δ is as in Eq. (2), indicating that heat energy is rather than absorbed by the reaction, and that the reaction is . This convention as to whether Δ is positive or negative looks at the heat change in terms of the matter actually involved in the reaction rather than its surroundings. In the reaction in Eq. (2), the strength of the bonds increases as products are formed, so the products are lower in potential energy, and the lost energy is indicated by a negative value of Δ . The values are calculated, as we'll see later, from standard tabulated values found in databases developed especially for geologists It is important to notice that Δ is the energy for the reaction as written. The quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. (2) tells us that 179.7 kJ of heat energy is given off of which is consumed, or for every 3 mol of consumed. Alternatively, it tells us that 179.7 kJ is released Mg Si O (OH) produced. Seen in this way, Δ is a conversion factor enabling us to calculate the heat absorbed when a given amount of substance is consumed or produced. If is the quantity of heat absorbed and is the amount of substance involved, then \(\Delta H_{\text{m}}=\frac{q}{n}\) (4) How much heat energy is obtained when 1 kg of the serpentine chrysotile, Mg Si O (OH) , is formed according to Equation (2)? The mass of Mg Si O (OH) is easily converted to the amount of Mg Si O (OH) from which the heat energy is easily calculated by means of Eq. (4). The value of Δ is –179.7 kJ per 2 moles of Mg Si O (OH) . The road map is \(m_{\text{Mg}_{2}\text{Si}_{2}\text{O}_{5}\text{OH}_{4}}~\) \(\xrightarrow{M}~\) \(n_{\text{Mg}_{2}\text{Si}_{2}\text{O}_{5}\text{OH}_{4}}~\) \(\xrightarrow{\Delta H_{m}}~q\) so that \(q=\text{1 }\times \text{ 10}^{\text{3}}\text{ g Mg}_{2}\text{Si}_{2}\text{O}_{5}\text{(OH)}_{4}~\) \(\times~\frac{\text{1 mol Mg}_{2}\text{Si}_{2}\text{O}_{5}\text{(OH)}_{4}}{\text{277.112 g} \text{ Si}_{2}\text{O}_{5}\text{(OH)}_{4}}~\) \(\times ~\frac{-179.7 kJ}{\text{2 mol Mg}_{2}\text{Si}_{2}\text{O}_{5}\text{(OH)}_{4}} \) Note: By convention a negative value of corresponds to a release of heat energy by the matter involved in the reaction. The quantity Δ is the . In this context the symbol Δ (delta) signifies change in” while is the symbol for the quantity being changed, namely the enthalpy. We will deal with the enthalpy in some detail in Chap. 15. For the moment we can think of it as a property of matter which increases when matter absorbs energy and decreases when matter releases energy. It is important to realize that the value of Δ given in thermochemical equations like (1), (2) or (3) depends on the physical state of both the reactants and the products. Thus, if water were present as a gas instead of a liquid in the reaction in Eq. (1), the value of Δ would be different from 40.1 kJ. These reactions may occur under conditions where water may be supercritical (above 374°C) and yet a different value would be obtained. It is also necessary to specify both the temperature and pressure since the value of Δ depends very slightly on these variables. If these are not specified [as in Eq. (3)] they usually refer to 25°C and to normal atmospheric pressure. Since geochemical processes like those above normally occur at hundreds of atmospheres (hundreds of bars) pressure and elevated temperatures, geologists adjust the standard enthalpies to give values appropriate for the conditions. Although the adjustments are not difficult, computer programs exist to do the work. Another characteristic of thermochemical equations arise from the law of conservation of energy. The first is that For example , the conversion of the two forms of calcium carbonate Therefore, the forward direction (calcite to aragonite) is exothermic, releasing heat as a more stable crystal lattice forms. Logically, in the reverse direction, disrupting the stable lattice of aragonite must require energy, so the conversion of aragonite to calcite endothermic: Melting or freezing water can release or absorb significant amounts of heat: tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This corresponds to the fact that heat is absorbed from your skin when perspiration evaporates, and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off exactly the same quantity of heat. To see why this must be true, suppose that Δ [Eq. (7)] = 44 kJ mol while Δ [Eq. (8)] = –50.0 kJ. If we took 1 mol of liquid water and allowed it to evaporate, 44 kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ of heat which had been created from nowhere! This would violate the law of conservation of energy. The only way the problem can he avoided is for Δ of the reverse reaction to be equal in magnitude but opposite in sign from Δ of the forward reaction. That is, Δ forward = –Δ reverse | 7,867 | 1,317 |
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In order to meet the requirements of normal valence, it is sometimes necessary to have , that is, more than one shared pair of electrons between two atoms. A case in point is formaldehyde, CH O. In order to provide carbon with four bonds in this molecule, we must consider carbon as forming two bonds to the oxygen as well as one to each of the two hydrogens. At the same time the oxygen atom is also provided with the two bonds its normal valence requires: Note that all four of the shared electrons in the carbon-oxygen bond are included both in the octet of carbon and in the octet of oxygen. A bond involving two electron pairs is called a . Occasionally the usual valences of the atoms in a molecule do not tell us what the skeleton structure should be. For example, in carbon monoxide, CO, it is hard to see how one carbon atom (usual valence of 4) can be matched with a single oxygen atom (usual valence of 2). In a case like this, where the valences appear to be incompatible, counting valence electrons usually leads to a structure which satisfies the octet rule. Carbon has 4 valence electrons and oxygen has 6, for a total of . We want to arrange these 10 electrons in two octets, but two separate groups of 8 electrons would require 16 electrons. Only by 16 – 10, or 6, electrons (so that those 6 electrons are part of octet, and, in effect, count twice) can we satisfy the octet rule. This leads to the structure Here three pairs of electrons are shared between two atoms, and we have a . Double and triple bonds are not merely devices for helping to fit Lewis diagrams into the octet theory. They have an objective existence, and their presence in a molecule often has a profound effect on how it reacts with other molecules. Triple bonds are invariably shorter than double bonds, which in turn are shorter than single bonds. In , for instance, the carbon-oxygen distance is 114 pm, in it is 121 pm, while in both ethyl alcohol and dimethyl ether and methanol it is 142 pm. Below are 3-D Jmol images of carbon monoxide, formaldehyde, and methanol, to compare the difference in bond length with. This agrees with the wave-mechanical picture of the chemical bond as being caused by the concentration of electron density between the nuclei. The more pairs of electrons which are shared, the greater this density and the more closely the atoms are pulled together. In line with this, we would also expect multiple bonds to be stronger than single bonds. Indeed, the bond energy of is found experimentally to be 360 kJ mol , while that of is 736 kJ mol , and that of is a gigantic 1072 kJ mol . The triple bond in carbon monoxide turns out to be the strongest known covalent bond. The formation of double and triple bonds is not as widespread among the atoms of the periodic table as one might expect. At least one of the atoms involved in a multiple bond is almost always C, N, or O, and in most cases both atoms are members of this trio. Other elements complete their octets by forming rather than multiple bonds. Since hydrogen atoms are univalent, they must certainly all be bonded to carbon atoms, presumably two to each carbon. Each carbon atom thus has the situation in which two bonds must still be accounted for. By assuming that the two carbon atoms are joined by a double bond, all the valence requirements are satisfied, and we can draw a Lewis structure containing satisfactory octets: Carbon requires four bonds, and each oxygen requires two bonds, and so two double bonds will satisfy the normal valences. The structure is Looking at the Jmol image for this molecule, the double bonds have a shorter distance than those seen in formaldehyde, but the are longer than the triple bond in carbon monoxide: Silicon also has a normal valence of 4, but it is not an element which readily forms double bonds. Each silicon can form single bonds to four oxygen atoms however, Now the silicon is satisfied, but each oxygen lacks one electron and has only formed one bond. If each of the oxygens link to another silicon, they will be satisfied, but then the added silicon atoms will have unused valences: The process of adding oxygen or silicon atoms can continue indefinitely, producing a giant lattice of covalently bonded atoms. In this giant molecule each silicon is bonded to four oxygens and each oxygen to two silicons, and so there are as many oxygen atoms as silicon. The molecular formula could be written (SiO ) where is a very large number. A portion of this giant molecule is shown below. The difference in the abilities of carbon and silicon atoms to form double bonds has important consequences in the natural environment. Because double bonds form readily, carbon dioxide consists of individual molecules—there are no “empty spaces” on either the carbon or oxygen atoms where additional electrons may be shared. Hence there is little to hold one carbon dioxide molecule close to another, and at ordinary temperatures the molecules move about independently. On a macroscopic scale this means that carbon dioxide has the properties of a gas. In silicon dioxide, on the other hand, strong covalent bonds link all silicon and oxygen atoms together in a three-dimensional network. At ordinary temperatures the atoms cannot vibrate far from their allotted positions, and silicon dioxide has the macroscopic properties of a solid. As a gas, carbon dioxide is much freer than silicon dioxide to circulate through the environment. It can be removed from the atmosphere by plants in the photosynthetic process and eventually returned to the air by means of respiration. This is one of the reasons that terrestrial life is based on carbon compounds. If a supply of carbon from atmospheric carbon dioxide were not available, living organisms would be quite different in form and structure from the ones we know on earth. Science-fiction authors are fond of suggesting, because of the periodic relationship of carbon and silicon, that life on some distant planet might be based on silicon. It is rather hard to imagine, though, the mechanism by which such life forms would obtain silicon from the rocks and soil of their planet’s surface. Certainly they would face major difficulties if the combination of silicon with oxygen to form silicon dioxide were to be used as a source of energy. Imagine breathing out a solid instead of the gaseous carbon dioxide which forms when carbon combines with oxygen during respiration in terrestrial organisms! Macroscopic properties which are determined by microscopic structure and bonding are crucial in even such fundamental activities as living and breathing. | 6,670 | 1,318 |
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where [A], [B], [C], and [D] are stoichiometric concentrations of A, B, C, and D respectively. However, dilute solutions and concentrated solutions have slight differences, and a more precise method of calculating and defining the equilibrium constant is desirable. For such an approach, the of A, B, C, and D are used in place of the in the definition of . The reactivity of A ({A}) is proportional to [A], and the proportional constant in most text is a gamma, which is called the {A} = g [A]
{B} = g [B]
{C} = g [C]
etc. The application of science (engineering) often requires some refinement, and the use of activity is an refinement base on the theory of equilibrium. The reactivities based on concentrations given above work well for non-electrolytes (or molecular compounds). In dilute solutions, the activity coefficient is unity. g = 1
or
{A} = [A] In solutions of electrolytes, the interactions of charges require some special consideration. | 983 | 1,319 |
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In the introduction to chemical kinetics, we have already defined chemical reaction rates. Rates of chemical reactions depend on the nature of the reactants, the temperature, the presence of a catalyst, and concentration. This page discusses how the concentration affects the chemical reaction rates. Concentration effect is important because chemical reactions are usually carried out in solutions. The reaction rates of chemical reactions are the amounts of a reactant reacted or the amount of a product formed per unit time (moles per second). Often, the amount can be expressed in terms of concentrations. \(Rate = \dfrac{amount\: reacted\: or\: produced}{time\: interval}\: \textrm{units: }\mathrm{g/s,\: mol/s,\: or\: \%/s}\) At certain conditions, the rates are functions of concentrations. Depending on the time interval between measurements, the rates are called However, a more realistic representation for a reaction rate is the change in concentration per unit time, either the decrease of concentration per unit time of a reactant or the increase of concentration per unit time of a product. In this case, the rate is expressed in mol/(L sec). \(Rate = \dfrac{concentration\: change\: of\: a\: reactant\: or\: product}{time\: interval}\: \textrm{units: g/(M s), M/s, ppm/s etc.}\) To measure a reaction rate, we usually monitor either a product or a reactant for its change. Any physical characteristic related to the quantity or concentration of a product or reactant can be monitored. Some of the characteristics to be monitored are: The change can be plotted on a graph, and from the graph, we can get the average rate or the instantaneous rate by either graphical methods or using a computer for the data analysis. Usually, the rate of a reaction is a function of the concentrations of reactants. For example, the rate of the reaction \(\ce{2 NO + O2 \rightarrow 2 NO2}\) has the form: \(Rate = k \ce{[O2] [NO]^2}\) The rate is proportional to the concentration of \(\ce{O2}\), usually written as \(\ce{[O2]}\) and is proportional to the square of \(\ce{[NO]}\), or \(\ce{[NO]^2}\). The orders of 1 and 2 for \(\ce{[O2]}\) and \(\ce{[NO]}\) respectively have been determined by experiment, NOT from the chemical equation. The total order of this reaction is 3 (=2+1). Note the rates and order in the following example reactions: \(\ce{H2 + I2 \rightarrow 2 HI}\),
\(Rate = k \ce{[H2] [I2]}\), Total order 2. \(\ce{H2 + Br2 \rightarrow 2 HBr}\),
\(Rate = k \ce{[H2] [Br2]}^{1/2}\), Total order 1.5. In particular, note that orders are NOT determined from the stoichiometry of the reaction equation. The order with respect to (wrt) a reactant is determined experimentally by keeping the concentration of other reactants constant, but varying the concentration of one of the reactants, say \(\ce{A}\) in a general reaction \(a \,\textrm A + b \,\textrm B + c \,\textrm C = products\) If concentrations of \(\ce{B}\) and \(\ce{C}\) are kept constant, you can measure the reaction of \(\ce{A}\) at various concentrations. You can then plot the rate as a function of \(\ce{[A]}\). For a zeroth order reaction, you will get a horizontal line, because \(rate = k\) (a horizontal line) For a first order reaction, the plot is a straight line (linear), as shown above, because \(rate = k \ce{[A]}\) (a straight line) Note that \(rate = k\) when \(\mathrm{[A] = 1}\). For a second order reaction, the plot is a branch of a parabola, because \(rate = k \ce{[A]}^2\) For a reaction with an infinite order, the plot is a step function. The is small, almost zero, when \(\ce{[A]}\) is less than 1. When \(\ce{[A]}\) is greater than or equal to 1, then the reaction rate is very large. This model applies to nuclear explosion, except that \(\mathrm{[A] = 1}\) is actually of the fission material. \(rate = k \ce{[A]}^{\infty}\) Is there a chemical process like this? Well, we all know that one of the key conditions in an atomic bomb is to have a critical mass of the fission material, \(\ce{^235U}\) or \(\ce{^239Pu}\). When such a mass is put together, the reaction rate increases dramatically, leading to an explosion. Thus, this model seems to apply; however, the mechanism for the fission reaction is not described by the order of the fission material. If only \(\ce{[A]}\) is varied in experiments, and the order wrt \(\ce{[A]}\) is , then the rate has the general expression, \(rate = k \ce{[A]}^n\) In this expression, is the specific rate constant, or the when \(\mathrm{[A] = 1.00}\). Again, the order is not necessarily an integer, but its most common values are 0.5 (1/2), 1, 2, or 3. Cases in which is a negative number are rare. Mathematical models for the effect of concentration on rates are interesting. In general, the rate of a reaction of order with respect to \(\ce{A}\) can be represented by the equation: \(y = k x^n\), ( = various values including 0.5, 1, 2, 3, ...) Plots of equations for various values of illustrate the dependence of on concentration for various orders. For a chemical reaction, we often determine the order with respect to a reagent by determining the initial rate. When more than one reactants are involved, we vary the concentrations in a systematic way so that the effect of concentration of one of the reactants can be measured. For example, in a reaction involving three reactants, \(\ce{A}\), \(\ce{B}\), and \(\ce{C}\), we vary \(\ce{[A]}\) from experiment 1 to experiment 2 and find out how the varies. Similarly, we vary concentrations of \(\ce{B}\) or \(\ce{C}\) in other experiments, keeping others constant, and investigate its effect. The example below illustrates the strategy for such an approach. Derive the rate law for the reaction \(\mathrm{A + B + C \rightarrow products}\) from the following data, where rate is measured as soon as the reactants are mixed.
Assuming the orders to be , and respectively for \(\ce{A}\), \(\ce{B}\), and \(\ce{C}\), we have \(rate = k \ce{[A]}^{\Large x} \ce{[B]}^{\Large y} \ce{[C]}^{\Large z}\) From experiment 1 and 2, we have: \(\dfrac{0.800}{0.100}=\dfrac{k\: 0.2^{\Large x}\: 0.1^{\Large y}\: 0.1^{\Large z}}{k\: 0.1^{\Large x}\: 0.1^{\Large y}\: 0.1^{\Large z}}\) Thus, \(8 = 2^{\Large x}\); and \(x = \dfrac{\ln8}{\ln2} = 3\) By similar procedures, we get \(y = 2\) and \(z = 1\). Thus, the rate law is: \(rate = k \ce{[A]}^3 \ce{[B]}^2 \ce{[C]}\)
Note the following relationships: \(x = y^{\Large z}\) \(\ln x = z \ln y\) \(z = \dfrac{\ln x}{\ln y}\) The variation of reaction s as functions of order and concentrations are summarized in the form of a Table below.
Recognize the order when is on \(\ce{[A]}\). Only when = 1 does the rate depend linearly on the concentration.
Recognize the order when \(rate = k\) when \(n = 0\). Hint: The rate increases four times.
Predict increases as the concentration doubles for reactions of various orders. Hint: Rate = rate constant k.
\(Rate = k\), when \(\mathrm{[A] = [B] = 1}\) regardless of the order of the reaction. Note the difference between and . Hint: A branch of a parabola. Hint: True
What if is a negative value? | 7,161 | 1,320 |
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One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism. In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water: \[\ce{2C8H18 (l) + 25O2(g) -> 16CO2 (g) + 18H2O(g)} \label{14.6.1} \] For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an , involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction. The overall sequence of elementary reactions is the mechanism of the reaction. To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide. \[\ce{NO2(g) + (g) -> NO(g) + CO2 (g)} \label{14.6.2} \] From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of \(\ce{NO2}\) with a molecule of \(\ce{CO}\) that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows: \[rate = k[\ce{NO2}]^2 \label{14.6.3} \] The fact that the reaction is second order in \([\ce{NO2}]\) and independent of \([\ce{CO}]\) tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be \[rate = k[\ce{NO2},\ce{CO}]. \nonumber \] The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2: According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The \(\ce{NO3}\) molecule is an in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step. The sum of the elementary reactions in a reaction mechanism give the overall balanced chemical equation of the reaction. The of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as ; if there are two reactant molecules, it is ; and if there are three reactant molecules (a relatively rare situation), it is . Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!) Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity ( \(\Page {1}\)). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is \[rate = k[A]. \nonumber \] For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in e \(\Page {1}\). For a bimolecular elementary reaction of the form A + B → products, the general rate law is \[rate = k[A,B]. \nonumber \] For elementary reactions, the order of the elementary reaction is the same as its molecularity. In contrast, the rate law be determined from the balanced chemical equation for the overall reaction (unless it is a single step mechanism and is therefore also an elementary step). Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the , that must give the experimentally determined rate law for the overall reaction.This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions. Look at the rate laws for each elementary reaction in our example as well as for the overall reaction. The experimentally determined rate law for the reaction of \(NO_2\) with \(CO\) is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so \(k\) for the overall reaction must equal \(k_1\). That is, NO is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2. Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect. In an alternative mechanism for the reaction of \(\ce{NO2}\) with \(\ce{CO}\) with \(\ce{N2O4}\) appearing as an intermediate. Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = [NO ] )? elementary reactions rate law for each elementary reaction and overall rate law The rate law for step 1 is rate = [NO ] ; for step 2, it is rate = [N O ,CO]. If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = [NO ] . This is the same as the experimentally determined rate law. Hence this mechanism, with N O as an intermediate, and the one described previously, with NO as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO and N O , directly. Iodine monochloride (\(\ce{ICl}\)) reacts with \(\ce{H2}\) as follows: \[\ce{2ICl(l) + H_2(g) \rightarrow 2HCl(g) + I_2(s)} \nonumber \] The experimentally determined rate law is \(rate = k[\ce{ICl},\ce{H2}]\). Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: \(\ce{HI}\) is an intermediate.) This mechanism is consistent with the experimental rate law if the first step is the rate-determining step. Assume the reaction between \(\ce{NO}\) and \(\ce{H_2}\) occurs via a three-step process: Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction: \[\text{rate} = k[\ce{NO}]^2[\ce{H_2}]? \tag{observed} \] The overall reaction is then \[\ce{2NO(g) + 2H_2(g) -> N2(g) + 2H2O(g)} \nonumber \] Reaction Mechanism (Slow step followed by fast step): A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction. Species that are formed in one step and consumed in another are intermediates. Each elementary reaction can be described in terms of its , the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step. | 9,520 | 1,322 |
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Covalent Network Solids are giant covalent substances like diamond, graphite and silicon dioxide (silicon(IV) oxide). This page relates the structures of covalent network solids to the physical properties of the substances. You might argue that carbon has to form 4 bonds because of its 4 unpaired electrons, whereas in this diagram it only seems to be forming 3 bonds to the neighboring carbons. This diagram is something of a simplification, and shows the arrangement of atoms rather than the bonding. Graphite Graphite Graphite Jim Clark ( ) These come in three general types: carbides are formed by elements of groups 1, 2 and aluminum. The actual for of the carbon varies, for example, aluminum carbide, based on its hydrolysis product seems to contain "C " units: \[\ce{Al4C3 + 6H2O -> 2Al(OH)3 + 3CH4}\] but calcium carbide seems to contain [C C] units: \[\ce{CaC + 2H2O -> Ca(OH)2 + HCºH}\] carbides are compounds of the transition metals with metallic properties and the C in tetrahedral holes in the metal atom lattice. The best known example is the extremely hard tungsten carbide, WC, used in cutting tools. Covalent carbides include B C and SiC (carborundum - an abrasive with a diamond-like structure) | 1,241 | 1,326 |
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Raman spectroscopy often is a highly useful adjunct to infrared spectroscopy. The experimental arrangement for Raman spectra is quite simple in principle. Monochromatic light, such as from an argon-gas laser, is passed through a sample, and the light scattered at right angles to the incident beam is analyzed by an optical spectrometer. Raman spectra arise as a result of light photons being “captured” momentarily by molecules in the sample and giving up (or gaining) small increments of energy through changes in the molecular vibrational and rotational energies before being emitted as scattered light. The changes in the vibrational and rotational energies result in changes in wavelength of the incident light. These changes are detected as lines falling both above and below the wavelength of the incident light. The line positions in Raman spectra always are reported in wave numbers. Highly efficient laser Raman spectrometers are commercially available. Although changes in wavelength in Raman scattering correspond to absorption or emission of infrared radiation, infrared and Raman spectra are not always identical. Indeed, valuable information about molecular symmetry may be obtained by comparison of infrared and Raman spectra. When a bond is it does not absorb infrared radiation and, for this reason, symmetrical diatomic molecules such as \(H_2\) and \(O_2\), which are always electrically symmetrical, do not give infrared absorption spectra. However, excitation of symmetrical vibrations does occur in Raman scattering.\(^7\) In a molecule such as ethene, \(CH_2=CH_2\), the double-bond stretching vibration is symmetrical, because both ends of the molecule are the same. As a result, the double-bond stretching absorption is not observable in the infrared spectrum of ethene and is weak in all nearly symmetrically substituted ethenes. Nonetheless, this vibration appears strongly in the Raman spectrum of ethene and provides evidence for a symmetrical structure for ethene. Absorption due to the stretching vibration of the double bond in tetrachloroethene (\(1570 \: \text{cm}^{-1}\) is strong in the Raman and absent in the infrared, whereas that arising from the less symmetrical double bond of cyclohexene (\(1658 \: \text{cm}^{-1}\)) is weak in the infrared and slightly stronger in the Raman. \(^7\)This is in accord with the spectroscopic "selection rules," derived from theoretical arguments, that predict which transitions between rotational and vibrational energy levels are "allowed" and which are "forbidden." and (1977) | 2,575 | 1,327 |
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The chemical potential can be given in terms of a partition function. Internal energy can be defined as: \[U=RT^2\left(\frac{\partial \ln Q}{\partial T}\right)_{n,V} \nonumber \] And entropy can be defined as: \[S=RT\left(\frac{\partial \ln Q}{\partial T}\right)_{n,V} + R\ln Q \nonumber \] We know that Helmholtz energy is: \[A=U-TS \nonumber \] Using our two equations above, we obtain: \[A=-RT\ln Q \nonumber \] Now, let's change gears a bit to show how Helmholtz energy is related to chemical potential. The total differential for Helmholtz energy is: \[dA = \left(\frac{\partial A}{\partial T}\right)_{n,V} + \left(\frac{\partial A}{\partial V}\right)_{n,T} + \left(\frac{\partial A}{\partial n}\right)_{V,T} \nonumber \] And the fundamental equation is: \[dA = -SdT-PdV+\left(\frac{\partial A}{\partial n}\right)_{V,T}dn \nonumber \] Using the relationship between Helmholtz energy and Gibbs energy: \[G=A+PV \nonumber \] We obtain: \[\begin{split}dG &= dA+d(PV) \\ &= -SdT+VdP+\left(\frac{\partial A}{\partial n}\right)_{V,T}dn\end{split} \nonumber \] We know that the change in Gibbs energy is: \[\begin{split}dG &= -SdT+VdP+\left(\frac{\partial G}{\partial n}\right)_{P,T}dn \\ &= -SdT+VdP+\mu dn \end{split} \nonumber \] Inspecting these equations, we see that: \[\mu = \left(\frac{\partial G}{\partial n}\right)_{P,T} = \left(\frac{\partial A}{\partial n}\right)_{V,T} \nonumber \] This shows us that, as long as the natural variables for each thermodynamic potential are held constant, the partial derivatives of Gibbs energy and Helmholtz energy with respect to the number of moles, \(n\) are equal to the chemical potential. We can now plug in our expression above for Helmholtz energy in terms of the partition function: \[\mu = -RT\left(\frac{\partial \ln Q}{\partial n}\right)_{V,T} \nonumber \] We now have chemical potential written in terms of the partition function, \(Q\). | 1,912 | 1,328 |
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A nuclear weapon is commonly defined as a device, which uses a nuclear reaction for destructive means. The first nuclear weapon was successfully detonated on July 16, 1945. The nuclear weapon, code named “Trinity”, yielded an explosion which was equivalent to 20 kilotons of Trinitrotoluene (TNT). This reaction unexpectedly had a shock wave which could be felt 100 miles away. Compared to chemical reactions, the amount of energy that can be released from nuclear reactions can be up to a million times greater. Before we can fully understand the chemical complexity and appreciate the engineering elegance of a nuclear weapon, however, it is important to grasp basic nuclear chemistry concepts. This type of nuclear reaction is caused by nuclear decay of an unstable atom that has been hit by a neutron. As a result of the instability of the atom, the nucleus splits into two fission fragments also yielding free neutrons and exorbitant amounts of energy (both in the form of electromagnetic radiation and kinetic energy). A neutron carries no electric charge but the nucleus of an atom does - a positive charge. Like with magnets, like charges repel each other, so particles that carry positive electric charges like alpha particles are repelled by the nucleus of an atom and thus do not stick to the nucleus. However, a neutrally charged neutron can combine with the nucleus of an atom which then causes the nucleus to become unstable and split into 2 nuclei. According to Einstein's formula, \(E=mc^2\), energy is released from this reaction along with other neutrons that have the same effect on nearby nuclei and a chain reaction occurs which yields an incredible amount of energy. Fusion is almost completely complimentary to fission. It is the process in which a nuclear reaction where two nuclei are joined together to form a heavier nucleus. The reaction also yields free neutrons and exorbitant amounts of energy from binding energy. Our sun is a "nuclear furnace" in the sense that it is a place where nuclear fusion occurs. Although a nuclear fusion reactor would undoubtedly be better for our environment, such reactions are not yet possible because no known material can withstand the incredible high temperatures needed for such reactions to occur. \[\ce{^2D+^3T → ^4_2He +^1_0N} + 17.6\, MeV\] A nuclear weapon can either undergo a nuclear fission reaction (atomic bomb) or a nuclear fusion reaction (H bomb or thermonuclear bomb). The first nuclear weapons built underwent pure nuclear fission. Uranium-235 and Plutonium-239 were the most common fissile isotopes used. (Uranium-235 is less than 1% naturally abundant. It requires complex methods of enrichment to be a fissile isotope). There are 2 basic nuclear fission weapon designs: The first nuclear fusion weapons (also known as thermonuclear weapons) were designed to initiate a fission-based chain reaction. The fusion reaction between tritium and deuterium would result in the free neutrons necessary to bombard a fissile isotope and start a nuclear chain reaction. The first thermonuclear weapon was detonated in November of 1952. Similar to the nuclear implosion design, thermonuclear weapons use the heat and radiation from a fission reaction to make the fissile material assume a state of supercritical mass. The supercritical mass then instantaneously undergoes a fusion chain reaction yielding exponentially more energy than a fission chain reaction. 1. Given the following nuclear fusion reaction, what particle corresponds to \(\ce{^{?}_{?}X}\)? \[\ce{^2_1H + ^2_1H \rightarrow ^{?}_{?}X + \rm{energy}}\] yields He + Energy | 3,628 | 1,329 |
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Phase transitions (e.g. melting) often occur under equilibrium conditions. We have seen that both the \(H\) and the \(S\) curves undergo a discontinuity at constant temperature during melting, because there is an enthalpy of fusion to overcome. For a general phase transition at equilibrium at constant \(T\) and \(P\), we can say that: \[Δ_{trs}G = Δ_{trs}H - T_{trs}Δ_{trs}S = 0 \nonumber \] \[Δ_{trs}H = T_{trs}Δ_{trs}S \nonumber \] \[\dfrac{Δ_{trs}H}{T_{trs}}=Δ_{trs}S \nonumber \] For melting of a crystalline solid, we now see there is a sudden jump in enthalpy. The reason is that the solid has a much more ordered structure than the crystalline solid. The decrease in order implies a finite \(Δ_{trs}S\). We should stress at this point that we are talking about transitions here. The reason for this terminology is that the discontinuity is in a function like \(S\), that is a first order derivative of \(G\) (or \(A\)): \[\left(\frac{\partial\bar{G}}{\partial T}\right)_P=-\bar{S} \nonumber \] Second order derivatives (e.g. the heat capacity) will display a singularity (+∞) at the transition point. Every phase transition will have a change in entropy associated with it. The different types of phase transitions that can occur are: | 1,265 | 1,330 |
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We previously saw that knowing the magnitude of the equilibrium constant under a given set of conditions allows chemists to predict the extent of a reaction. Often, however, chemists must decide whether a system has reached equilibrium or if the composition of the mixture will continue to change with time. In this section, we describe how to quantitatively analyze the composition of a reaction mixture to make this determination. To determine whether a system has reached equilibrium, chemists use a Quantity called the reaction Quotient (\(Q\)). The expression for the reaction Quotient has precisely the same form as the equilibrium constant expression, except that \(Q\) may be derived from a set of values measured at any time during the reaction of any mixture of the reactants and the products, regardless of whether the system is at equilibrium. Therefore, for the following general reaction: \[aA+bB \rightleftharpoons cC+dD \nonumber \] the reaction quotient is defined as follows: \[Q=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \label{15.6.1} \] To understand how information is obtained using a reaction Quotient, consider the dissociation of dinitrogen tetroxide to nitrogen dioxide, \[\ce{N2O4(g) <=> 2NO2(g)} \nonumber \] for which \(K = 4.65 \times 10^{−3}\) at 298 K. We can write \(Q\) for this reaction as follows: \[Q=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]} \label{15.6.2} \] The following table lists data from three experiments in which samples of the reaction mixture were obtained and analyzed at equivalent time intervals, and the corresponding values of \(Q\) were calculated for each. Each experiment begins with different proportions of product and reactant: As these calculations demonstrate, \(Q\) can have any numerical value between 0 and infinity (undefined); that is, \(Q\) can be greater than, less than, or equal to \(K\). Comparing the magnitudes of \(Q\) and \(K\) enables us to determine whether a reaction mixture is already at equilibrium and, if it is not, predict how its composition will change with time to reach equilibrium (i.e., whether the reaction will proceed to the right or to the left as written). All you need to remember is that the composition of a system not at equilibrium will change in a way that makes \(Q\) approach \(K\): These points are illustrated graphically in Figure \(\Page {1}\). If \(Q < K\), the reaction will proceed to the right as written. If \(Q > K\), the reaction will proceed to the left as written. If \(Q = K\), then the system is at equilibrium. A Discussing Using the Reaction Quotient (Q): At elevated temperatures, methane (\(CH_4\)) reacts with water to produce hydrogen and carbon monoxide in what is known as a steam-reforming reaction: \[\ce{CH4(g) + H2O(g) <=> CO(g) + 3H2(g)} \nonumber \] \(K = 2.4 \times 10^{−4}\) at 900 K. Huge amounts of hydrogen are produced from natural gas in this way and are then used for the industrial synthesis of ammonia. If \(1.2 \times 10^{−2}\) mol of \(CH_4\), 8.0 × 10 mol of \(H_2O\), \(1.6 \times 10^{−2}\) mol of \(CO\), and \(6.0 \times 10^{−3}\) mol of \(H_2\) are placed in a 2.0 L steel reactor and heated to 900 K, will the reaction be at equilibrium or will it proceed to the right to produce \(\ce{CO}\) and \(\ce{H_2}\) or to the left to form \(\ce{CH_4}\) and \(\ce{H_2O}\)? : balanced chemical equation, \(K\), amounts of reactants and products, and volume : direction of reaction : : We must first find the initial concentrations of the substances present. For example, we have \(1.2 \times 10^{−2} mol\) of \(\ce{CH_4}\) in a 2.0 L container, so \[[\ce{CH4}]=\dfrac{1.2\times 10^{−2} \, \text{mol}}{2.0\; \text{L}}=6.0 \times 10^{−3} M \nonumber \] We can calculate the other concentrations in a similar way: We now compute \(Q\) and compare it with \(K\): \[\begin{align*} Q&=\dfrac{[\ce{CO},\ce{H_2}]^3}{[\ce{CH_4},\ce{H_2O}]} \\[4pt] &=\dfrac{(8.0 \times 10^{−3})(3.0 \times 10^{−3})^3}{(6.0\times 10^{−3})(4.0 \times 10^{−3})} \\[4pt] &=9.0 \times 10^{−6} \end{align*} \nonumber \] Because \(K = 2.4 \times 10^{−4}\), we see that \(Q < K\). Thus the ratio of the concentrations of products to the concentrations of reactants is less than the ratio for an equilibrium mixture. The reaction will therefore proceed to the right as written, forming \(\ce{H2}\) and \(\ce{CO}\) at the expense of \(\ce{H_2O}\) and \(\ce{CH4}\). In the water–gas shift reaction introduced in Example \(\Page {1}\), carbon monoxide produced by steam-reforming reaction of methane reacts with steam at elevated temperatures to produce more hydrogen: \[\ce{CO(g) + H_2O(g) <=> CO2(g) + H2(g)} \nonumber \] \(K = 0.64\) at 900 K. If 0.010 mol of both \(\ce{CO}\) and \(\ce{H_2O}\), 0.0080 mol of \(\ce{CO_2}\), and 0.012 mol of \(\ce{H_2}\) are injected into a 4.0 L reactor and heated to 900 K, will the reaction proceed to the left or to the right as written? \(Q = 0.96\). Since (Q > K), so the reaction will proceed to the left, and \(CO\) and \(H_2O\) will form. By graphing a few equilibrium concentrations for a system at a given temperature and pressure, we can readily see the range of reactant and product concentrations that correspond to equilibrium conditions, for which \(Q = K\). Such a graph allows us to predict what will happen to a reaction when conditions change so that \(Q\) no longer equals \(K\), such as when a reactant concentration or a product concentration is increased or decreased. Lead carbonate decomposes to lead oxide and carbon dioxide according to the following equation: \[\ce{PbCO3(s) <=> PbO(s) + CO2(g)} \label{15.6.3} \] Because \(\ce{PbCO_3}\) and \(\ce{PbO}\) are solids, the equilibrium constant is simply \[K = [\ce{CO_2}]. \nonumber \] At a given temperature, therefore, any system that contains solid \(\ce{PbCO_3}\) and solid \(\ce{PbO}\) will have exactly the same concentration of \(\ce{CO_2}\) at equilibrium, regardless of the ratio or the amounts of the solids present. This situation is represented in Figure \(\Page {3}\), which shows a plot of \([\ce{CO_2}]\) versus the amount of \(\ce{PbCO_3}\) added. Initially, the added \(\ce{PbCO_3}\) decomposes completely to \(\ce{CO_2}\) because the amount of \(\ce{PbCO_3}\) is not sufficient to give a \(\ce{CO_2}\) concentration equal to \(K\). Thus the left portion of the graph represents a system that is not at equilibrium because it contains only \(\ce{CO2(g)}\) and \(\ce{PbO(s)}\). In contrast, when just enough \(\ce{PbCO_3}\) has been added to give \([CO_2] = K\), the system has reached equilibrium, and adding more \(\ce{PbCO_3}\) has no effect on the \(\ce{CO_2}\) concentration: the graph is a horizontal line. Thus any \(\ce{CO_2}\) concentration that is not on the horizontal line represents a nonequilibrium state, and the system will adjust its composition to achieve equilibrium, provided enough \(\ce{PbCO_3}\) and \(\ce{PbO}\) are present. For example, the point labeled in Figure \(\Page {2}\) lies above the horizontal line, so it corresponds to a \([\ce{CO_2}]\) that is greater than the equilibrium concentration of \(\ce{CO_2}\) (i.e., \(Q > K\)). To reach equilibrium, the system must decrease \([\ce{CO_2}]\), which it can do only by reacting \(\ce{CO_2}\) with solid \(\ce{PbO}\) to form solid \(\ce{PbCO_3}\). Thus the reaction in Equation \(\ref{15.6.3}\) will proceed to the left as written, until \([\ce{CO_2}] = K\). Conversely, the point labeled in Figure \(\Page {2}\) lies below the horizontal line, so it corresponds to a \([\ce{CO_2}]\) that is less than the equilibrium concentration of \(\ce{CO_2}\) (i.e., \(Q < K\)). To reach equilibrium, the system must increase \([\ce{CO_2}]\), which it can do only by decomposing solid \(\ce{PbCO_3}\) to form \(\ce{CO_2}\) and solid \(\ce{PbO}\). The reaction in Equation \ref{15.6.3} will therefore proceed to the right as written, until \([\ce{CO_2}] = K\). In contrast, the reduction of cadmium oxide by hydrogen gives metallic cadmium and water vapor: \[\ce{CdO(s) + H2(g) <=> Cd(s) + H_2O(g)} \label{15.6.4} \] and the equilibrium constant is \[K = \dfrac{[\ce{H_2O}]}{[\ce{H_2}]}. \nonumber \] If \([\ce{H_2O}]\) is doubled at equilibrium, then \([\ce{H2}]\) must also be doubled for the system to remain at equilibrium. A plot of \([\ce{H_2O}]\) versus \([\ce{H_2}]\) at equilibrium is a straight line with a slope of \(K\) (Figure \(\Page {3}\)). Again, only those pairs of concentrations of \(\ce{H_2O}\) and \(\ce{H_2}\) that lie on the line correspond to equilibrium states. Any point representing a pair of concentrations that does not lie on the line corresponds to a nonequilibrium state. In such cases, the reaction in Equation \(\ref{15.6.4}\) will proceed in whichever direction causes the composition of the system to move toward the equilibrium line. For example, point in Figure \(\Page {3}\) lies below the line, indicating that the \([\ce{H_2O}]/[\ce{H_2}]\) ratio is less than the ratio of an equilibrium mixture (i.e., \(Q < K\)). Thus the reaction in Equation \ref{15.6.4} will proceed to the right as written, consuming \(\ce{H_2}\) and producing \(\ce{H_2O}\), which causes the concentration ratio to move up and to the left toward the equilibrium line. Conversely, point in Figure \(\Page {3}\) lies above the line, indicating that the \([\ce{H_2O}]/[\ce{H_2}]\) ratio is greater than the ratio of an equilibrium mixture (\(Q > K\)). Thus the reaction in Equation \(\ref{15.6.4}\) will proceed to the left as written, consuming \(\ce{H_2O}\) and producing \(\ce{H_2}\), which causes the concentration ratio to move down and to the right toward the equilibrium line. In another example, solid ammonium iodide dissociates to gaseous ammonia and hydrogen iodide at elevated temperatures: \[\ce{ NH4I(s) <=> NH3(g) + HI(g)} \label{15.6.5} \] For this system, \(K\) is equal to the product of the concentrations of the two products: \[K = [\ce{NH_3},\ce{HI}]. \nonumber \] If we double the concentration of \(\ce{NH3}\), the concentration of \(\ce{HI}\) must decrease by approximately a factor of 2 to maintain equilibrium, as shown in Figure \(\Page {4}\). As a result, for a given concentration of either \(\ce{HI}\) or \(\ce{NH_3}\), only a single equilibrium composition that contains equal concentrations of both \(\ce{NH_3}\) and \(\ce{HI}\) is possible, for which \[[\ce{NH_3}] = [\ce{HI}] = \sqrt{K}. \nonumber \] Any point that lies below and to the left of the equilibrium curve (such as point in Figure \(\Page {4}\)) corresponds to \(Q < K\), and the reaction in Equation \(\ref{15.6.5}\) will therefore proceed to the right as written, causing the composition of the system to move toward the equilibrium line. Conversely, any point that lies above and to the right of the equilibrium curve (such as point in Figure \(\ref{15.6.5}\)) corresponds to \(Q > K\), and the reaction in Equation \(\ref{15.6.5}\) will therefore proceed to the left as written, again causing the composition of the system to move toward the equilibrium line. By graphing equilibrium concentrations for a given system at a given temperature and pressure, we can predict the direction of reaction of that mixture when the system is not at equilibrium. The reaction Quotient (\(Q\)) is used to determine whether a system is at equilibrium and if it is not, to predict the direction of reaction. The reaction Quotient (\(Q\) or \(Q_p\)) has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, \(Q = K\). Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Points that do not lie on the line or curve represent nonequilibrium states, and the system will adjust, if it can, to achieve equilibrium. | 11,856 | 1,331 |
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We can derive from Newtonian mechanics. This derivation assumes that gas molecules behave like point masses that do not interact with one another. The pressure of the gas results from collisions of the gas molecules with the walls of the container. The contribution of one collision to the force on the wall is equal to the change in the molecule’s momentum divided by the time between collisions. The magnitude of this force depends on the molecule’s speed and the angle at which it strikes the wall. Each such collision makes a contribution to the pressure that is equal to the force divided by the area of the wall. To find the pressure from this model, it is necessary to average over all possible molecular speeds and all possible collision angles. In Chapter 4, we derive Boyle’s law in this way. We can do a simplified derivation by making a number of assumptions. We assume that all of the molecules in a sample of gas have the same speed. Let us call it \(u\). As sketched in Figure 3, we assume that the container is a cubic box whose edge length is \(d\). If we consider all of the collisions between molecules and walls, it is clear that each wall will experience \({1}/{6}\) of the collisions; or, each pair of opposing walls will experience \({\mathrm{1}}/{\mathrm{3}}\) of the collisions. Instead of averaging over all of the possible angles at which a molecule could strike a wall and all of the possible times between collisions, we assume that the molecules travel at constant speed back and forth between opposite faces of the box. Since they are point masses, they never collide with one another. If we suppose that \({\mathrm{1}}/{\mathrm{3}}\) of the molecules go back and forth between each pair of opposite walls, we can expect to accomplish the same kind of averaging in setting up our artificial model that we achieve by averaging over the real distribution of angles and speeds. In fact, this turns out to be the case; the derivation below gets the same result as the rigorous treatment we develop in Chapter 4. Since each molecule goes back and forth between opposite walls, it collides with each wall once during each round trip. At each collision, the molecule’s speed remains constant, but its direction changes by 180\({}^{o}\); that is, the molecule’s velocity changes from \(\mathop{u}\limits^{\rightharpoonup}\) to \(-\mathop{u}\limits^{\rightharpoonup}\). Letting \(\Delta t\) be the time required for a round trip, the distance traversed in a round trip is \[\begin{aligned} 2d & =\left|\mathop{u}\limits^{\rightharpoonup}\right|\Delta t \\ ~ & =u\Delta t \end{aligned}\] The magnitude of the momentum change for a molecule in one collision is \[\begin{align*} \left|\Delta (m\mathop{u}\limits^{\rightharpoonup})\right| &=\left|m{\mathop{u}\limits^{\rightharpoonup}}_{final}-m{\mathop{u}\limits^{\rightharpoonup}}_{initial}\right| \\[4pt] &=\left|m{\mathop{u}\limits^{\rightharpoonup}}_{final}-\left({-m\mathop{u}\limits^{\rightharpoonup}}_{final}\right)\right| \\[4pt] &=2mu \end{align*}\] The magnitude of the force on the wall from one collision is \[F=\frac{\left|\Delta \left(m\mathop{u}\limits^{\rightharpoonup}\right)\right|}{\Delta t}=\frac{2mu}{\left({2d}/{u}\right)}=\frac{mu^2}{d}\] and the pressure contribution from one collision on the wall, of area \(d^2\), is \[P=\frac{F}{A}=\frac{mu^2}{d\bullet d^2}=\frac{{mu}^2}{d^3}=\frac{{mu}^2}{V}\] so that we have \[PV=mu^2\] from the collision of one molecule with one wall. If the number of molecules in the box is \(N\), \(N/3\) of them make collisions with this wall, so that the total pressure on one wall attributable to all \(N\) molecules in the box is \[P=\frac{mu^2}{V}\frac{N}{3}\] or \[PV=\frac{Nmu^2}{3}\] Since the ideal gas equation can be written as \(PV=NkT\) we see that \({Nmu^2}/{3}=NkT\) so that \(mu^2=3kT\) and \[u=\sqrt{\frac{3kT}{m}}\] Thus we have found a relationship between the molecular speed and the temperature of the gas. (The actual speed of a molecule, \(v\), can have any value between zero and—for present purposes—infinity. When we average the values of \(v^2\) for many molecules, we find the average value of the squared speeds, \(\overline{v^2}\). In Chapter 4, we find that \(u^2=\overline{v^2}\). That is, the average speed we use in our derivation turns out to be a quantity called the , \(v_{rms}=u=\sqrt{\overline{v^2}}\).) This result also gives us the (average) kinetic energy of a single gas molecule: \[KE=\frac{mu^2}{2}=\frac{3kT}{2}\] From this derivation, we have a simple mechanical model that explains Boyle’s law as the logical consequence of point-mass molecules colliding with the walls of their container. By combining this result with the ideal gas equation, we find that the average speed of ideal gas molecules depends only on the temperature. From this we have the very important result that Since our non-interacting point-mass molecules have no potential energy arising from their interactions with one another, their translational kinetic energy is the whole of their energy. (Because two such molecules neither attract nor repel one another, no work is required to change the distance between them. The work associated with changing the volume of a confined sample of an ideal gas arises because of the pressure the molecules exert on the walls of the container; the pressure arises because of the molecules’ kinetic energy.) The energy of one mole of monatomic ideal gas molecules is \[KE=\left({3}/{2}\right)RT\] When we expand our concept of ideal gases to include molecules that have rotational or vibrational energy, but which neither attract nor repel one another, it remains true that the energy of a macroscopic sample depends only on temperature. However, the molar energy of such a gas is greater than \(\left({3}/{2}\right)RT\), because of the energy associated with these additional motions. We make extensive use of the conclusion that the energy of an ideal gas depends only on temperature. As it turns out, this conclusion follows rigorously from the second law of thermodynamics. In Chapter 10, we show that \[{\left(\frac{\partial E}{\partial V}\right)}_T={\left(\frac{\partial E}{\partial P}\right)}_T=0\] for a substance that obeys the ideal gas equation; at constant temperature, the energy of an ideal gas is independent of the volume and independent of the pressure. So long as pressure, volume, and temperature are the only variables needed to specify its state, the laws of thermodynamics imply that the energy of an ideal gas depends only on temperature. While the energy of an ideal gas is independent of pressure, the energy of a real gas is a function of pressure at a given temperature. At ordinary pressures and temperatures, this dependence is weak and can often be neglected. The first experimental investigation of this issue was made by James Prescott Joule, for whom the SI unit of energy is named. Beginning in 1838, Joule did a long series of careful measurements of the mechanical equivalent of heat. These measurements formed the original experimental basis for the kinetic theory of heat. Among Joule’s early experiments was an attempt to measure the heat absorbed by a gas as it expanded into an evacuated container, a process known as a . No absorption of heat was observed, which implied that the energy of the gas was unaffected by the volume change. However, it is difficult to do this experiment with meaningful accuracy. Subsequently, Joule collaborated with William Thomson (Lord Kelvin) on a somewhat different experimental approach to essentially the same question. The provides a much more sensitive measure of the effects of intermolecular forces of attraction and repulsion on the energy of a gas during its expansion. Since our definition of an ideal gas includes the stipulation that there are no intermolecular forces, the Joule-Thomson experiment is consistent with the conclusion that the energy of an ideal gas depends only on temperature. However, since intermolecular forces are not zero for any real gas, our analysis reaches this conclusion in a somewhat indirect way. The complication arises because the Joule-Thomson results are not entirely consistent with the idea that all properties of a real gas approach those of an ideal gas at a sufficiently low pressure. (The best of models can have limitations.) We discuss the Joule-Thomson experiment in . | 8,398 | 1,332 |
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The standard of formation is defined as the change in enthalpy when one mole of a substance in the standard state (1 atm of pressure and 298.15 K) is formed from its pure elements under the same conditions. The standard enthalpy of formation is a measure of the energy released or consumed when one mole of a substance is created under standard conditions from its pure elements. The symbol of the standard enthalpy of formation is ΔH . The equation for the standard enthalpy of formation (originating from Enthalpy's being a ), shown below, is commonly used: \[\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}\] This equation essentially states that the standard enthalpy of formation is equal to the minus the . Given a simple chemical equation with the variables A, B and C representing different compounds: \(A + B \leftrightharpoons C\) and the standard enthalpy of formation values: the equation for the standard enthalpy change of formation is as follows: ΔH = ΔH [C] - (ΔH [A] + ΔH [B]) ΔH = (1 )(523 kJ/ ) - ((1 )(433 kJ/ + (1 )(-256 kJ/ )\) Because there is one mole each of A, B and C, the standard enthalpy of formation of each reactant and product is multiplied by 1 mole, which eliminates the mol denominator: ΔH = The result is 346 kJ, which is the standard enthalpy change of formation for the creation of variable "C". The standard enthalpy of formation of a pure element is in its reference form its standard enthalpy formation is . naturally exists as graphite and diamond. The enthalpy difference between graphite and diamond is too large for both to have a standard enthalpy of formation of zero. To determine which form is zero, the more stable form of carbon is chosen. This is also the form with the lowest enthalpy, so graphite has a standard enthalpy of formation equal to zero. Table 1 provides sample values of standard enthalpies of formation of various compounds. All values have units of kJ/mol and physical conditions of 298.15 K and 1 atm, referred to as the "standard state." These are the conditions under which values of standard enthalpies of formation are typically given. Note that while the majority of the values of standard enthalpies of formation are exothermic, or negative, there are a few compounds such as NO(g) and N O (g) that actually require energy from its surroundings during its formation; these endothermic compounds are generally unstable. Between Br (l) and Br (g) at 298.15 K, which substance has a nonzero standard enthalpy of formation? Br (l) is the more stable form, which means it has the lower enthalpy; thus, Br (l) has ΔH = 0. Consequently, Note: that the element phosphorus is a unique case. The reference form in phosphorus is not the most stable form, red phosphorus, but the less stable form, white phosphorus. Recall that standard enthalpies of formation can be either positive or negative. The enthalpy of formation of carbon dioxide at 298.15K is ΔH = -393.5 kJ/mol CO (g). Write the chemical equation for the formation of CO . This equation must be written for one mole of CO (g). In this case, the reference forms of the constituent elements are O (g) and graphite for carbon. \[O_{2}(g) + C(graphite) \rightleftharpoons CO_{2}(g)\] The general equation for the standard enthalpy change of formation is given below: \[\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}\] Plugging in the equation for the formation of CO gives the following: ΔH = ΔH [CO (g)] - (ΔH [O (g)] + ΔH [C(graphite)] Because O (g) and C(graphite) are in their most elementally stable forms, they each have a standard enthalpy of formation equal to 0: ΔH = -393.5 kJ = ΔH [CO (g)] - ((1 mol)(0 kJ/mol) + (1 mol)(0 kJ/mol)) ΔH [CO (g)]= -393.5 kJ Using the values in the above table of standard enthalpies of formation, calculate the ΔH for the formation of NO (g). \(NO_{2(g)}\) is formed from the combination of \(NO_{(g)}\) and \(O_{2(g)}\) in the following reaction: \(2NO(g) + O_{2}(g) \leftrightharpoons 2NO_{2}(g)\) To find the ΔH , use the formula for the standard enthalpy change of formation: \[\Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}\] The relevant standard enthalpy of formation values from Table 1 are: Plugging these values into the formula above gives the following: \[ΔH_{reaction}^o= (2 \cancel{mol})(33.18\; kJ/\cancel{mol}) - \left[(2 \cancel{mol})(90.25\ kJ/\cancel{mol}) + (1 \cancel{mol})(0\; kJ/\cancel{mol})\right]\] \[ΔH_{reaction}^o = -114.1\; kJ\] Kirchhoff's Law describes the enthalpy of a reaction's variation with temperature changes. In general, enthalpy of any substance increases with temperature, which means both the products and the reactants' enthalpies increase. The overall enthalpy of the reaction will change if the increase in the enthalpy of products and reactants is different. At constant pressure, the heat capacity is equal to change in enthalpy divided by the change in temperature. \[ c_p = \dfrac{\Delta H}{\Delta T} \label{1}\] Therefore, if the heat capacities do not vary with temperature then the change in enthalpy is a function of the difference in temperature and heat capacities. The amount that the enthalpy changes by is proportional to the product of temperature change and change in heat capacities of products and reactants. A weighted sum is used to calculate the change in heat capacity to incorporate the ratio of the molecules involved since all molecules have different heat capacities at different states. \[ H_{T_f}=H_{T_i}+\int_{T_i}^{T_f} c_{p} dT \label{2}\] If the heat capacity is temperature independent over the temperature range, then Equation \ref{1} can be approximated as \[ H_{T_f}=H_{T_i}+ c_{p} (T_{f}-T_{i}) \label{3}\] with Equation \ref{3} can only be applied to small temperature changes, (<100 K) because over a larger temperature change, the is not constant. There are many biochemical applications because it allows us to predict enthalpy changes at other temperatures by using standard enthalpy data. | 6,117 | 1,333 |
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The electron-transfer reactions that occur within and between proteins typically involve prosthetic groups separated by distances that are often greater than 10 Å. When we consider these distant electron transfers, an explicit expression for the electronic factor is required. In the nonadiabatic limit, the rate constant for reaction between a donor and acceptor held at fixed distance and orientation is: \[k_{et} = \bigg[ \frac{H_{AB} \;^{2}}{\hbar}\left(\dfrac{\pi}{\lambda RT}\right)^{1/2} \bigg]^{\frac{-(\lambda + \Delta G^{o})^{2}}{4 \lambda RT}}\ldotp \tag{6.27}\] The electronic (or tunneling) matrix element H is a measure of the electronic coupling between the reactants and the products at the transition state. The magnitude of H depends upon donor-acceptor separation, orientation, and the nature of the intervening medium. Various approaches have been used to test the validity of Equation (6.27) and to extract the parameters H and \(\lambda\). Driving-force studies have proven to be a reliable approach, and such studies have been emphasized by many workers. In the nonadiabatic limit, the probability is quite low that reactants will cross over to products at the transition-state configuration. This probability depends upon the electronic hopping frequency (determined by H ) and upon the frequency of motion along the reaction coordinate. In simple models, the electronic-coupling strength is predicted to decay exponentially with increasing donor-acceptor separation (Equation 6.28): \[H_{AB} = (H_{AB}^{o})^{\frac{- \beta}{2} (\textbf{d} - \textbf{d}^{o})} \tag{6.28} \] In Equation (6.28), H ° is the electronic coupling at close contact ( °), and \(\beta\) is the rate of decay of coupling with distance ( ). Studies of the distance dependence of electron-transfer rates in donor-acceptor complexes, and of randomly oriented donors and acceptors in rigid matrices, have suggested 0.8 \(\leq \beta \leq\) 1.2 Å . Analysis of a large number of intramolecular electron-transfer rates has suggested a \(\beta\) value of 1.4 Å for protein reactions (Figure 6.24). Assigning a single protein \(\beta\) implies that the intervening medium is homogenous. At best this is a rough approximation, because the medium separating two redox sites in a protein is a heterogenous array of bonded and nonbonded interactions. Beratan and Onuchic have developed a formalism that describes the medium in terms of "unit blocks" connected together to form a tunneling pathway. A unit block may be a covalent bond, a hydrogen bond, or a through-space jump, each with a corresponding decay factor. Dominant tunneling pathways in proteins are largely composed of bonded groups (e.g., peptide bonds), with less favorable through-space interactions becoming important when a through-bond pathway is prohibitively long (Figure 6.25). The tunneling pathway model has been used successfully in an analysis of the electron-transfer rates in modified cytochromes c (Section IV.D.1). Plastocyanin cycles between the Cu and Cu oxidation states, and transfers electrons from cytochrome f to the P component of photosystem I in the chloroplasts of higher plants and algae. The low molecular weight (10.5 kDa) and availability of detailed structural information have made this protein an attractive candidate for mechanistic studies, which, when taken together, point to two distinct surface binding sites (i.e., regions on the plastocyanin molecular surface at which electron transfer with a redox partner occurs). The first of these, the solvent-exposed edge of the Cu ligand His-87 (the adjacent site A in Figure 6.26), is ~6 Å from the copper atom and rather nonpolar. The second site (the remote site R in Figure 6.26) surrounds Tyr-83, and is much farther (~15 Å) from the copper center. Negatively charged carboxylates at positions 42-45 and 59-61 make this latter site an attractive one for positively charged redox reagents. Bimolecular electron-transfer reactions are typically run under pseudo-first-order conditions (e.g., with an inorganic redox reagent present in ~15-fold excess): \[Rate = k[plastocyanin,complex] = k_{obs}[plastocyanin] \ldotp \tag{6.29}\] For some reactions [e.g., Co(phen) oxidation of plastocyanin (Cu )] the expected linear plot of k vs. [complex] is not observed. Instead, the rate is observed to saturate (Figure 6.27). A "minimal" model used to explain this behavior involves the two pathways for electron transfer shown in Equation (6.30). \(\tag{6.30}\) Surprisingly, the rate ratio k /k is 7. Calculations indicate that, despite the significant differences in distances, H for the remote site is ~15 percent of H for the adjacent site. This figure is much higher than would be expected from distance alone, suggesting that the value of the decay parameter \(\beta\) in Equation (6.28) depends strongly on the structure of the intervening medium. Chemical modification of structurally characterized metalloproteins by transition-metal redox reagents has been employed to investigate the factors that control long-range electron-transfer reactions. In these semisynthetic multisite redox systems, the distance is fixed, and tunneling pathways between the donor and acceptor sites can be examined. Sperm-whale myoglobin can be reacted with (NH ) Ru(OH ) and then oxidized to produce a variety of ruthenated products, including a His-48 derivative whose Ru \(\leftrightarrow\) Fe tunneling pathway is depicted in Figure 6.28. Electrochemical data (Table 6.5) indicate that the (NH ) Ru group does not significantly perturb the heme center, and that equilibrium (i.e., k = k + k ) should be approached when a mixed-valent intermediate is produced by flash-photolysis techniques: \[(NH_{3})_{5}Ru^{3+}-Mb(Fe^{3+}) \xrightarrow[e^{-}]{fast} (NH_{3})_{5}Ru^{2+}-Mb(Fe^{3+}) \xrightleftharpoons[k_{-1}]{k_{1}} (NH_{3})_{5}Ru^{3+}-Mb(Fe^{2+}) \tag{6.31}\] This kinetic behavior was observed, and both the forward (k ) and reverse (k ) reactions were found to be markedly temperature-dependent: k = 0.019 s (25 °C), \(\Delta\)H = 7.4 kcal/mol, k = 0.041 s ) (25°C), \(\Delta\)H = 19.5 kcal/mol. X-ray crystallographic studies indicate that the axial water ligand dissociates upon reduction of the protein. This conformational change does not control the rates, since identical results were obtained when a second flash-photolysis technique was used to generate (NH ) Ru -Mb(Fe ) in order to approach the equilibrium from the other direction. Fe (NH ) Ru Cyanogen bromide has been used to modify the six-coordinate metmyoglobin heme site, causing the coordinated water ligand to dissociate. The CNBr-modified myoglobin heme site is thus five-coordinate in oxidation states. As expected, the self-exchange rate increased from ~1 M s to ~10 M s . Recent efforts in modeling biological electron transfers using chemically modified redox proteins point the way toward the design of semisynthetic redox enzymes for catalytic applications. An intriguing example, termed flavohemoglobin, was produced by reaction of hemoglobin with a flavin reagent designed to react with Cys-93 of the \(\beta\)-chain (i.e., the hemoglobin molecule was modified by two flavin moieties). The resulting derivative, unlike native hemoglobin, accepts electrons directly from NADPH and catalyzes the -hydroxylation of aniline in the presence of O and NADPH. In physiologically relevant precursor complexes, both redox centers are frequently buried in protein matrices. Characterization of such protein-protein complexes is clearly important, and several issues figure prominently: Most of our knowledge about the structures of protein-protein complexes comes from crystallographic studies of antigen-antibody complexes and multisubunit proteins; such systems generally exhibit a high degree of thermodynamic stability. On the other hand, complexes formed as a result of bimolecular collisions generally are much less stable, and tend to resist attempts to grow x-ray-quality crystals; the high salt conditions typically used in protein crystallizations often lead to dissociation of such complexes. One of the most widely studied protein-protein complexes is that formed between mammalian cytochrome b and cytochrome c. Using the known x-ray structures of both proteins, Salemme generated a static computer graphics model of this electron-transfer complex by docking the x-ray structures of the individual proteins. Two features of this model and its revision by molecular dynamics simulations (Figure 6.29 See color plate section, page C-12.) are noteworthy: (1) several Lys residues on cytochrome c and carboxylate-containing groups on cytochrome b form "salt bridges" (i.e., intermolecular hydrogen bonds); and (2) the hemes are nearly coplanar and are ~17 Å (Fe-Fe) apart. This distance was confirmed by an energy-transfer experiment in which the fluorescence of Zn-substituted cytochrome c was quenched by cytochrome b . Spectroscopic studies have verified the suggestion that these proteins form a 1:1 complex at low ionic strength (Figure 6.30). In addition, chemical modification and spectroscopic analyses are all in agreement with the suggestion that the complex is primarily stabilized by electrostatic interactions of the (-NH ••• O C—) type. The effect of ionic strength on the reduction of cytochrome c by cytochrome b is also in accord with this picture: lowering the ionic strength increases the reaction rate, as expected for oppositely charged molecules. A common experimental strategy for studying electron transfers proteins uses a metal-substituted heme protein as one of the reactants. In particular, the substitution of zinc for iron in one of the porphyrin redox centers allows facile initiation of electron transfer through photoexcitation of the zinc porphyrin (ZnP). The excited zinc porphyrin, ZnP* in Equation (6.32), may decay back (k ~ 10 s ) to the ground state or transfer an electron to an acceptor. \(\tag{6.32}\) The ZnP cation radical produced in the k step is a powerful oxidant; back electron transfer (k ) will thus occur and regenerate the starting material. The reactions shown in Equation (6.32) have been investigated in mixedmetal [Zn, Fe] hemoglobins. A hemoglobin molecule can be viewed as two independent electron-transfer complexes, each consisting of an \(\alpha_{1}\)-\(\beta_{2}\) subunit pair (Figure 6.31), since the \(\alpha_{1}\)-\(\alpha_{2}\), (\beta_{1}\)-(\beta_{2}\), and \(\alpha_{1}\)-(\beta_{1}\) distances are prohibitively long (> 30 Å). Both [\(\alpha\)(Zn), \(\beta\)(Fe)] and [\(\alpha\)(Fe), \(\beta\)(Zn)] hybrids have been studied. The ZnP and FeP are nearly parallel, as in the cytochrome b -cytochrome c model complex. Long-range electron transfer ( ZnP* → Fe ) between the \(\alpha_{1}\) and \(\beta_{2}\) subunits has been observed (the heme-edge/heme-edge distance is ~20 Å). The driving force for the forward electron-transfer step is ~0.8 eV, and k (see Equation 6.32) is ~100 s at room temperature, but decreases to ~9 s in the low-temperature region (Figure 6.32). Below 140-160 K the vibrations that induce electron transfer "freeze out"; nuclear tunneling is usually associated with such slow, temperature-independent rates. A complete analysis of the full temperature dependence of the rate requires a quantum-mechanical treatment of \(\lambda_{i}\) rather than that employed in the Marcus theory. It is interesting to note that the heme b vinyl groups (see Figure 6.6) for a given [\(\alpha_{1}\)(Fe), \(\beta_{2}\)Zn)] hybrid point toward each other and appear to facilitate electron transfer. | 11,639 | 1,334 |
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Precipitation reactions occur when cations and anions in aqueous solution combine to form an insoluble ionic solid called a . Whether or not such a reaction occurs can be determined by using the for common ionic solids. Because not all aqueous reactions form precipitates, one must consult the solubility rules before determining the state of the products and writing a The ability to predict these reactions allows scientists to determine which ions are present in a solution, and allows industries to form chemicals by extracting components from these reactions. Precipitates are insoluble ionic solid products of a reaction, formed when certain cations and anions combine in an aqueous solution. The determining factors of the formation of a precipitate can vary. Some reactions depend on temperature, such as solutions used for buffers, whereas others are dependent only on solution concentration. The solids produced in precipitate reactions are crystalline solids, and can be suspended throughout the liquid or fall to the bottom of the solution. The remaining fluid is called supernatant liquid. The two components of the mixture (precipitate and supernate) can be separated by various methods, such as filtration, centrifuging, or decanting. The use of solubility rules require an understanding of the way that ions react. Most precipitation reactions are single replacement reactions or double replacement reactions. A double replacement reaction occurs when two ionic reactants dissociate and bond with the respective anion or cation from the other reactant. The ions replace each other based on their charges as either a cation or an anion. This can be thought of as "switching partners"; that is, the two reactants each "lose" their partner and form a bond with a different partner: A double replacement reaction is specifically classified as a when the chemical equation in question occurs in aqueous solution and one of the of the products formed is insoluble. An example of a precipitation reaction is given below: \[CdSO_{4(aq)} + K_2S_{(aq)} \rightarrow CdS_{(s)} + K_2SO_{4(aq)}\] Both reactants are aqueous and one product is solid. Because the reactants are ionic and aqueous, they dissociate and are therefore . However, there are six solubility guidelines used to predict which molecules are insoluble in water. These molecules form a solid precipitate in solution. Whether or not a reaction forms a precipitate is dictated by the solubility rules. These rules provide guidelines that tell which ions form solids and which remain in their ionic form in aqueous solution. The rules are to be followed from the top down, meaning that if something is insoluble (or soluble) due to rule 1, it has precedence over a higher-numbered rule. If the rules state that an ion is soluble, then it remains in its aqueous ion form. If an ion is insoluble based on the solubility rules, then it forms a solid with an ion from the other reactant. If all the ions in a reaction are shown to be soluble, then no precipitation reaction occurs. To understand the definition of a , recall the equation for the double replacement reaction. Because this particular reaction is a precipitation reaction, states of matter can be assigned to each variable pair: → The first step to writing a net ionic equation is to separate the soluble (aqueous) reactants and products into their respective cations and anions. Precipitates do not dissociate in water, so the solid should not be separated. The resulting equation looks like that below: → In the equation above, and ions are present on both sides of the equation. These are called because they remain unchanged throughout the reaction Since they go through the equation unchanged, they can be eliminated to show the : → The net ionic equation only shows the precipitation reaction. A net ionic equation must be balanced on both sides not only in terms of atoms of elements, but also in terms of electric charge. Precipitation reactions are usually represented solely by net ionic equations. If all products are aqueous, a net ionic equation cannot be written because all ions are canceled out as spectator ions. Therefore, Precipitation reactions are useful in determining whether a certain element is present in a solution. If a precipitate is formed when a chemical reacts with lead, for example, the presence of lead in water sources could be tested by adding the chemical and monitoring for precipitate formation. In addition, precipitation reactions can be used to extract elements, such as magnesium from seawater. Precipitation reactions even occur in the human body between antibodies and antigens; however, the environment in which this occurs is still being studied. Complete the double replacement reaction and then reduce it to the net ionic equation. \[NaOH_{(aq)} + MgCl_{2 \;(aq)} \rightarrow \] First, of this reaction using knowledge of double replacement reactions (remember the cations and anions “switch partners”). \[2NaOH_{(aq)} + MgCl_{2\;(aq)} \rightarrow 2NaCl + Mg(OH)_2\] Second, to determine if the products are soluble. Group 1 cations (\(Na^+\)) and chlorides are soluble from rules 1 and 3 respectively, so \(NaCl\) will be soluble in water. However, rule 6 states that hydroxides are insoluble, and thus \(Mg(OH)_2\) will form a precipitate. The resulting equation is the following: \[2NaOH(aq) + MgCl_{2\;(aq)} \rightarrow 2NaCl_{(aq)} + Mg(OH)_{2\;(s)}\] Third, into their ionic forms, as they would exist in an aqueous solution. Be sure to balance both the electrical charge and the number of atoms: \[2Na^+_{(aq)} + 2OH^-_{(aq)} + Mg^{2+}_{(aq)} + 2Cl^-_{(aq)} \rightarrow Mg(OH)_{2\;(s)} + 2Na^+_{(aq)} + 2Cl^-_{(aq)}\] Lastly, (the ions that occur on both sides of the equation unchanged). In this case, they are the sodium and chlorine ions. The final is: \[Mg^{2+}_{(aq)} + 2OH^-_{(aq)} \rightarrow Mg(OH)_{2(s)}\] Complete the double replacement reaction and then reduce it to the net ionic equation. \[CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow\] The predicted products of this reaction are \(CoSO_4\) and \(NaCl\). From the solubility rules, \(CoSO_4\) is soluble because rule 4 states that sulfates (\(SO_4^{2-}\)) are soluble. Similarly, we find that \(NaCl\) is soluble based on rules 1 and 3. After balancing, the resulting equation is as follows: \[CoCl_{2\;(aq)} + Na_2SO_{4\;(aq)} \rightarrow CoSO_{4\;(aq)} + 2 NaCl_{(aq)}\] Separate the species into their ionic forms, as they would exist in an aqueous solution. Balance the charge and the atoms. Cancel out all spectator ions (those that appear as ions on both sides of the equation.): → This particular example is important because all of the reactants and the products are aqueous, meaning they cancel out of the net ionic equation. There is no solid precipitate formed; therefore, no precipitation reaction occurs. Write the net ionic equation for the potentially double displacement reactions. Make sure to include the states of matter and balance the equations. 1. Regardless of physical state, the products of this reaction are \(Fe(OH)_3\) and \(NaNO_3\). The solubility rules predict that \(NaNO_3\) is soluble because all nitrates are soluble (rule 2). However, \(Fe(OH)_3\) is insoluble, because hydroxides are insoluble (rule 6) and \(Fe\) is not one of the cations which results in an exception. After dissociation, the ionic equation is as follows: \[Fe^{3+}_{(aq)} + NO^-_{3\;(aq)} + Na^+_{(aq)} + 3OH^-_{(aq)} \rightarrow Fe(OH)_{3\;(s)} + Na^+_{(aq)} + NO^-_{3\;(aq)}\] Canceling out spectator ions leaves the net ionic equation: \[Fe^{3+}_{(aq)} + OH^-_{(aq)} \rightarrow Fe(OH)_{\;3(s)}\] 2. From the double replacement reaction, the products are \(AlCl_3\) and \(BaSO_4\). \(AlCl_3\) is soluble because it contains a chloride (rule 3); however, \(BaSO_4\) is insoluble: it contains a sulfate, but the \(Ba^{2+}\) ion causes it to be insoluble because it is one of the cations that causes an exception to rule 4. The ionic equation is (after balancing): \[2Al^{3+}_{(aq)} + 6Cl^-_{(aq)} + 3Ba^{2+}_{(aq)} + 3SO^{2-}_{4\;(aq)} \rightarrow 2 Al^{3+}_{(aq)} +6Cl^-_{(aq)} + 3BaSO_{4\;(s)}\] Canceling out spectator ions leaves the following net ionic equation: \[Ba^{2+}_{(aq)} + SO^{2-}_{4\;(aq)} \rightarrow BaSO_{4\;(s)}\] 3. From the double replacement reaction, the products \(HNO_3\) and \(ZnI_2\) are formed. Looking at the solubility rules, \(HNO_3\) is soluble because it contains nitrate (rule 2), and \(ZnI_2\) is soluble because iodides are soluble (rule 3). This means that both the products are aqueous (i.e. dissociate in water), and thus no precipitation reaction occurs. 4. The products of this double replacement reaction are \(Ca_3(PO_4)_2\) and \(NaCl\). Rule 1 states that \(NaCl\) is soluble, and according to solubility rule 6, \(Ca_3(PO_4)_2\) is insoluble. The ionic equation is: \[Ca^{2+}_{(aq)}+ Cl^-_{(aq)} + Na^+_{(aq)} + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)} + Na^+_{(aq)} + Cl^-_{(aq)}\] After canceling out spectator ions, the net ionic equation is given below: \[Ca^{2+}_{(aq)} + PO^{3-}_{4\;(aq)} \rightarrow Ca_3(PO_4)_{2\;(s)}\] 5. The first product of this reaction, \(PbSO_4\), is soluble according to rule 4 because it is a sulfate. The second product, \(KNO_3\), is also soluble because it contains nitrate (rule 2). Therefore, no precipitation reaction occurs. | 9,427 | 1,335 |
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The elements in the periodic table are arranged in order of increasing atomic number. All of these elements display several other trends and we can use the periodic law and table formation to predict their chemical, physical, and atomic properties. Understanding these trends is done by analyzing the elements electron configuration; all elements prefer an octet formation and will gain or lose electrons to form that stable configuration. We can never determine the atomic radius of an atom because there is never a zero probability of finding an electron, and thus never a distinct boundary to the atom. All that we can measure is the distance between two nuclei (internuclear distance). A covalent radius is one-half the distance between the nuclei of two identical atoms. An ionic radius is one-half the distance between the nuclei of two ions in an ionic bond. The distance must be apportioned for the smaller cation and larger anion. A metallic radius is one-half the distance between the nuclei of two adjacent atoms in a crystalline structure. The noble gases are left out of the trends in atomic radii because there is great debate over the experimental values of their atomic radii. The SI units for measuring atomic radii are the nanometer (nm) and the picometer (pm). \(1 \, nm = 1 \times 10^{-9}\, m\) and \(1\, pm = 1 \times 10^{-12}\, m\). To explain this trend, the concept of screening and penetration must be understood. Penetration is commonly known as the distance that an electron is from the nucleus. Screening is defined as the concept of the inner electrons blocking the outer electrons from the nuclear charge. Within this concept we assume that there is no screening between the outer electrons and that the inner electrons shield the outer electrons from the total positive charge of the nucleus. In order to comprehend the extent of screening and penetration within an atom, scientists came up with the \(Z_{eff}\). The equation for calculating the effective nuclear charge is shown below. \[Z_{eff}= Z - S\] In the equation S represents the number of inner electrons that screen the outer electrons. Students can easily find S by using the atomic number of the noble gas that is one period above the element. For example, the S we would use for Chlorine would be 10 (the atomic number of Neon). Z is the total number of electrons in the atom. Since we know that a neutral atom has an identical number of protons and electrons, we can use the atomic number to define Z. For example, Chlorine would have a Z value of 17 (the atomic number of Chlorine). Continuing to use Chlorine as an example, the 10 inner electrons (S) would screen out the positive charge of ten protons. Therefore there would be and effective nuclear charge of 17-10 or +7. The effective nuclear charge shows that the nucleus is pulling the outer electrons with a +7 charge and therefore the outer electrons are pulled closer to the nucleus and the atomic radii is smaller. In summary, the greater the nuclear charge, the greater pull the nucleus has on the outer electrons and the smaller the atomic radii. In contrast, the smaller nuclear charge, the lesser pull the nucleus has on the outer electrons, and the larger atomic radii. Additionally, as the atomic number increases, the effective nuclear charge also increases. Figure 3 depicts the effect that the effective nuclear charge has on atomic radii. Now we are ready to the atomic radius trend in the periodic table. The atomic number increases moving left to right across a period and subsequently so does the effective nuclear charge. Therefore, moving left to right across a period the nucleus has a greater pull on the outer electrons and the atomic radii decreases. Moving down a group in the periodic table, the number of filled electron shells increases. In a group, the valence electrons keep the same effective nuclear charge, but now the orbitals are farther from the nucleus. Therefore, the nucleus has less of a pull on the outer electrons and the atomic radii are larger. We can now use these concept to explain the atomic radius differences of cations and anions. A cation is an atom that has lost one of its outer electrons. Cations have a smaller radius than the atom that they were formed from. With the loss of an electron, the positive nuclear charge out powers the negative charge that the electrons exert. Therefore, the positive nucleus pulls the electrons tighter and the radius is smaller. An anion is an atom that has gained an outer electron. Anions have a greater radius than the atom that they were formed from. The gain of an electron does not alter the nuclear charge, but the addition of an electron causes a decrease in the effective nuclear charge. Therefore, the electrons are held more loosely and the atomic radius is increased. Expelling an electron from an atom requires enough energy to overcome the magnetic pull of the positive charge of the nucleus. Therefore, ionization energy (I.E. or I) is the energy required to completely remove an electron from a gaseous atom or ion. The Ionization Energy is always positive. The energy required to remove one valence electron is the first ionization energy, the second ionization energy is the energy required to remove a second valence electron, and so on. \[\ce{Na(g) -> Na^{+}(g) + e^{-}(g)}\] \[\ce{ Na^{+}(g) -> Na^{2+}(g) + e^{-}}\] Ionization energies increase relative to high effective charge. The highest ionization energies are the noble gases because they all have high effective charge due to their octet formation and require a high amount of energy to destroy that stable configuration. The highest amount of energy required occurs with the elements in the upper right hand corner. Additionally, elements in the left corner have a low ionization energy because losing an electron allows them to have the noble gas configuration. Therefore, it requires less energy to remove one of their valence electrons These are the ionization energies for the period three elements. Notice how Na after in the second I.E, Mg in the third I.E., Al in the fourth I.E., and so on, all have a huge increase in energy compared to the proceeding one. This occurs because the proceeding configuration was in a stable octet formation; therefore it requires a much larger amount of energy to ionize. Ionization Energies increase going left to right across a period and increase going up a group. As you go up a group, the ionization energy increases, because there are less electron shielding the outer electrons from the pull of the nucleus. Therefore, it requires more energy to out power the nucleus and remove an electron. As we move across the periodic table from left to right, the ionization energy increases , due to the effective nuclear charge increasing. This is because the larger the effective nuclear charge, the stronger the nucleus is holding onto the electron and the more energy it takes to release an electron. The ionization energy is only a general rule. There are some instances when this trend does not prove to be correct. These can typically be explained by their electron configuration. For example, Magnesium has a higher ionization energy than Aluminum. Magnesium has an electron configuration of [Ne]3s2. Magnesium has a high ionization energy because it has a filled 3s orbital and it requires a higher amount of energy to take an electron from the filled orbital. Electron affinity (E.A.) is the energy change that occurs when an electron is added to a gaseous atom. Electron affinity can further be defined as the enthalpy change that results from the addition of an electron to a gaseous atom. It can be either positive or negative value. The greater the negative value, the more stable the anion is. \[\ce{X(g) + e^{-} -> X^{-} + Energy}\] \[\ce{X(g) + e^{-} + Energy -> X^{-}} \] It is more difficult to come up with trends that describe the electron affinity. Generally, the elements on the right side of the periodic table will have large negative electron affinity. The electron affinities will become less negative as you go from the top to the bottom of the periodic table. However, Nitrogen, Oxygen, and Fluorine do not follow this trend. The noble gas electron configuration will be close to zero because they will not easily gain electrons. Electronegativity is the measurement of an atom to compete for electrons in a bond. The higher the electronegativity, the greater its ability to gain electrons in a bond. Electronegativity will be important when we later determine molecules. Electronegativity is related with ionization energy and electron affinity. Electrons with low ionization energies have low electronegativities because their nuclei do not exert a strong attractive force on electrons. Elements with high ionization energies have high electronegativities due to the strong pull exerted by the positive nucleus on the negative electrons. Therefore the electronegativity increases from bottom to top and from left to right. The metallic character is used to define the chemical properties that metallic elements present. Generally, metals tend to lose electrons to form cations. Nonmetals tend to gain electrons to form anions. They also have a high oxidation potential therefore they are easily oxidized and are strong reducing agents. Metals also form basic oxides; the more basic the oxide, the higher the metallic character. As you move across the table from left to right, the metallic character decreases, because the elements easily accept electrons to fill their valance shells. Therefore, these elements take on the nonmetallic character of forming anions. As you move up the table, the metallic character decreases, due to the greater pull that the nucleus has on the outer electrons. This greater pull makes it harder for the atoms to lose electrons and form cations. Melting Points: Trends in melting points and molecular mass of binary carbon-halogen compounds and hydrogen halides are due to intermolecular forces. Melting destroys the arrangement of atoms in a solid, therefore the amount of heat necessary for melting to occur depends on the strength of attraction between the atoms. This strength of attraction increases as the number of electrons increase. Increase in electrons increases bonding. Example: Melting point of HF should be approximately -145 °C based off melting points of HCl, HBr, and HI, but the observed value is -83.6°C. Heat and electricity conductibility vary regularly across a period. Melting points may increase gradually or reach a peak within a group then reverse direction. Example: Third period elements Na, Mg, and Al are good conductors of heat and electricity while Si is only a fair conductor and the nonmetals P, S, Cl and Ar are poor conductors. Oxidation is a reaction that results in the loss of an electron. Oxidation potential follows the same trends as the ionization energy. That is because the smaller the ionization energy, the easier it is to remove an electron. (e.g) \[K_{(s)} \rightarrow K^+ + e^-\] Reduction is a reaction that results in the gaining of an electron. Reduction potentials follow the same trend as the electron affinity. That is because the larger, negative electron affinity, the easier it is to give an electron. Example of Reduction: \[F_{(s)} + e^- \rightarrow F^-\] of Elements categorizes like elements together. Dmitri Mendeleev, a Russian scientist, was the first to create a widely accepted arrangement of the elements in 1869. Mendeleev believed that when the elements are arranged in order of increasing atomic mass, certain sets of properties recur periodically. Although most modern periodic tables are arranged in eighteen groups (columns) of elements, Mendeleev's original periodic table had the elements organized into eight groups and twelve periods (rows). On the periodic table, elements that have similar properties are in the same groups (vertical). From left to right, the atomic number (z) of the elements increases from one period to the next (horizontal). The groups are numbered at the top of each column and the periods on the left next to each row. The main group elements are groups 1,2 and 13 through 18. These groups contain the most naturally abundant elements, and are the most important for life. The elements shaded in light pink in the table above are known as transition metals. The two rows of elements starting at z=58, are sometimes called inner transition metals and have that have been extracted and placed at the bottom of the table, because they would make the table too wide if kept continuous. The 14 elements following lanthanum (z=57) are called lanthanides, and the 14 following actinium (z=89) are called actinides. Elements in the periodic table can be placed into two broad categories, metals and nonmetals. Most metals are good conductors of heat and electricity, are malleable and ductile, and are moderate to high melting points. In general, nonmetals are nonconductors of heat and electricity, are nonmalleable solids, and many are gases at room temperature. Just as shown in the table above, metals and nonmetals on the periodic table are often separated by a stairstep diagonal line, and several elements near this line are often called metalloids (Si, Ge, As, Sb, Te, and At). Metalloids are elements that look like metals and in some ways behave like metals but also have some nonmetallic properties. The group to the farthest right of the table, shaded orange, is known as the noble gases. Noble gases are treated as a special group of nonmetals. The are comprised of group 1 of the periodic table and consist of Lithium, Sodium, Rubidium, Cesium, and Francium. These metals are highly reactive and form ionic compounds (when a nonmetal and a metal come together) as well as many other compounds. Alkali metals all have a charge of +1 and have the largest atom sizes than any of the other elements on each of their respective periods. are located in group 2 and consist of Beryllium, Magnesium, Calcium, Strontium, Barium, and Radium. Unlike the Alkali metals, the earth metals have a smaller atom size and are not as reactive. These metals may also form ionic and other compounds and have a charge of +2. The range from groups IIIB to XIIB on the periodic table. These metals form positively charged ions, are very hard, and have very high melting and boiling points. Transition metals are also good conductors of electricity and are malleable. Lanthanides (shown in row ** in chart above) and Actinides (shown in row * in chart above), form the block of two rows that are placed at the bottom of the periodic table for space issues. These are also considered to be transition metals. Lanthanides are form the top row of this block and are very soft metals with high boiling and melting points. Actinides form the bottom row and are radioactive. They also form compounds with most nonmetals. To find out why these elements have their own section, check out the page. As mentioned in the introduction, metalloids are located along the staircase separating the metals from the nonmetals on the periodic table. Boron, silicon, germanium, arsenic, antimony, and tellurium all have metal and nonmetal properties. For example, Silicon has a metallic luster but is brittle and is an inefficient conductor of electricity like a nonmetal. As the metalloids have a combination of both metallic and nonmetal characteristics, they are intermediate conductors of electricity or "semiconductors". are comprised of the five nonmetal elements Flourine, Chlorine, Bromine, Iodine, and Astatine. They are located on group 17 of the periodic table and have a charge of -1. The term "halogen" means "salt-former" and compounds that contain one of the halogens are salts. The physical properties of halogens vary significantly as they can exist as solids, liquids, and gases at room temperature. However in general, halogens are very reactive, especially with the alkali metals and earth metals of groups 1 and 2 with which they form ionic compounds. The consist of group 18 (sometimes reffered to as group O) of the periodic table of elements. The noble gases have very low boiling and melting points and are all gases at room temperature. They are also very nonreactive as they already have a full valence shell with 8 electrons. Therefore, the noble gases have little tendency to lose or gain electrons. The periodic table of elements is useful in determining the charges on simple monoatomic ions. For main-group elements, those categorized in groups 1, 2, and 13-18, form ions they lose the same number of electrons as the corresponding group number to which they fall under. For example, K atoms (group 1) lose one electron to become K and Mg atoms (group 2) lose two electrons to form Mg . The other main-group elements found in group 13 and higher form more than one possible ion. The elements in groups 3-12 are called transition elements, or transition metals. Similar to the main-group elements described above, the transition metals form positive ions but due to their capability of forming more than two or more ions of differing charge, a relation between the group number and the charge is non-existent. Arrange these elements according to decreasing atomic size: Na, C, Sr, Cu, Fr Fr, Sr, Cu, Na, C Arrange these elements according to increasing negative E. A.: Ba, F, Si, Ca, O Ba, Ca, Si, O, F Arrange these elements according to increasing metallic character: Li, S, Ag, Cs, Ge Li, S, Ge, Ag, Cs Which reaction do you expect to have the greater cell potential? Second equation First equation Which equation do you expect to occur? 5. A) Yes B) No *Highlight Answer:_____ to view answers. 1. An element that is an example of a metalloid is (a) S; (b) Zn; (c) Ge; (d) Re; (e) none of these Answer: 2. In the periodic table, the vertical (up and down) columns are called (a) periods; (b) transitions; (c) families/groups; (d) metalloids; (e) none of these. Answer: 3. Why are noble gases inert (nonreactive)? Answer: 4. What are compounds that contain a halogen called? Answer: 5. Lanthanides and Actinides are: (a) alkali earth metals; (b) transition metals; (c) metalloids; (d) alkali metals; (e) none of these Answer: | 18,284 | 1,336 |
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In the early 1900's, Paul Drüde came up with the "sea of electrons" metallic bonding theory by modeling metals as a mixture of atomic cores (atomic cores = positive nuclei + inner shell of electrons) and valence electrons. Metallic bonds occur among metal atoms. Whereas ionic bonds join metals to non-metals, . A sheet of aluminum foil and a copper wire are both places where you can see metallic bonding in action. Metals tend to have high melting points and boiling points suggesting strong bonds between the atoms. Even a soft metal like sodium (melting point 97.8°C) melts at a considerably higher temperature than the element (neon) which precedes it in the Periodic Table. Sodium has the electronic structure 1s 2s 2p 3s . When sodium atoms come together, the electron in the 3s atomic orbital of one sodium atom shares space with the corresponding electron on a neighboring atom to form a molecular orbital - in much the same sort of way that a covalent bond is formed. The difference, however, is that each sodium atom is being touched by eight other sodium atoms - and the sharing occurs between the central atom and the 3s orbitals on all of the eight other atoms. Each of these eight is in turn being touched by eight sodium atoms, which in turn are touched by eight atoms - and so on and so on, until you have taken in all the atoms in that lump of sodium. of the 3s orbitals on all of the atoms overlap to give a vast number of molecular orbitals that extend over the whole piece of metal. There have to be huge numbers of molecular orbitals, of course, because any orbital can only hold two electrons. The electrons can move freely within these molecular orbitals, and so each electron becomes detached from its parent atom. The electrons are said to be delocalized. The metal is held together by the strong forces of attraction between the positive nuclei and the delocalized electrons (Figure \(\Page {1}\)). This is sometimes described as "an array of positive ions in a sea of electrons". If you are going to use this view, beware! Is a metal made up of atoms or ions? It is made of atoms Each positive center in the diagram represents all the rest of the atom apart from the outer electron, but that electron has not been lost - it may no longer have an attachment to a particular atom, but it's still there in the structure. Sodium metal is therefore written as \(\ce{Na}\), not \(\ce{Na^+}\). Use the sea of electrons model to explain why Magnesium has a higher melting point (650 °C) than sodium (97.79 °C). If you work through the same argument above for sodium with magnesium, you end up with stronger bonds and hence a higher melting point. Magnesium has the outer electronic structure 3s . Both of these electrons become delocalized, so the "sea" has twice the electron density as it does in sodium. The remaining "ions" also have twice the charge (if you are going to use this particular view of the metal bond) and so there will be more attraction between "ions" and "sea". More realistically, each magnesium atom has 12 protons in the nucleus compared with sodium's 11. In both cases, the nucleus is screened from the delocalized electrons by the same number of inner electrons - the 10 electrons in the 1s 2s 2p orbitals. That means that there will be a net pull from the magnesium nucleus of 2+, but only 1+ from the sodium nucleus. So not only will there be a greater number of delocalized electrons in magnesium, but there will also be a greater attraction for them from the magnesium nuclei. Magnesium atoms also have a slightly smaller radius than sodium atoms, and so the delocalized electrons are closer to the nuclei. Each magnesium atom also has twelve near neighbors rather than sodium's eight. Both of these factors increase the strength of the bond still further. Transition metals tend to have particularly high melting points and boiling points. The reason is that they can involve the 3d electrons in the delocalization as well as the 4s. The more electrons you can involve, the stronger the attractions tend to be. Metals have several qualities that are unique, such as the ability to conduct electricity and heat, a low , and a low (so they will give up electrons easily to form cations). Their physical properties include a lustrous (shiny) appearance, and they are and ductile. Metals have a crystal structure but can be easily deformed. In this model, the valence electrons are free, delocalized, mobile, and not associated with any particular atom. This model may account for: However, these observations are only qualitative, and not quantitative, so they cannot be tested. The "Sea of Electrons" theory stands today only as an oversimplified model of how metallic bonding works. In a molten metal, the metallic bond is still present, although the ordered structure has been broken down. The metallic bond is not fully broken until the metal boils. That means that boiling point is actually a better guide to the strength of the metallic bond than melting point is. On melting, the bond is loosened, not broken. The strength of a metallic bond depends on three things: A metallic bond will be the result of more delocalized electrons, which causes the effective nuclear charge on electrons on the cation to increase, in effect making the size of the cation smaller. Metallic bonds are strong and require a great deal of energy to break, and therefore A metallic bonding theory must explain how so much bonding can occur with such few electrons (since metals are located on the left side of the periodic table and do not have many electrons in their valence shells). The theory must also account for all of a metal's unique chemical and physical properties. Previously, we argued that bonding between atoms can classified as range of possible bonding between (fully charge transfer) and (fully shared electrons). When two atoms of slightly differing electronegativities come together to form a covalent bond, one atom attracts the electrons more than the other; this is called a polar covalent bond. However, simple “ionic” and “covalent” bonding are idealized concepts and most bonds exist on a two-dimensional continuum described by the van Arkel-Ketelaar Triangle (Figure \(\Page {4}\)). Bond triangles or triangles (named after Anton Eduard van Arkel and J. A. A. Ketelaar) are triangles used for showing different compounds in varying degrees of ionic, metallic and covalent bonding. In 1941 van Arkel recognized three extreme materials and associated bonding types. Using 36 main group elements, such as metals, metalloids and non-metals, he placed ionic, metallic and covalent bonds on the corners of an equilateral triangle, as well as suggested intermediate species. The bond triangle shows that chemical bonds are not just particular bonds of a specific type. Rather, bond types are interconnected and different compounds have varying degrees of different bonding character (for example, polar covalent bonds). Using electronegativity - two compound average electronegativity on x-axis of Figure \(\Page {4}\). \[\sum \chi = \dfrac{\chi_A + \chi_B}{2} \label{sum}\] and electronegativity difference on y-axis, \[\Delta \chi = | \chi_A - \chi_B | \label{diff}\] we can rate the dominant bond between the compounds. On the right side of Figure \(\Page {4}\) (from ionic to covalent) should be compounds with varying difference in electronegativity. The compounds with equal electronegativity, such as \(\ce{Cl2}\) (chlorine) are placed in the covalent corner, while the ionic corner has compounds with large electronegativity difference, such as \(\ce{NaCl}\) (table salt). The bottom side (from metallic to covalent) contains compounds with varying degree of directionality in the bond. At one extreme is metallic bonds with delocalized bonding and at the other are covalent bonds in which the orbitals overlap in a particular direction. The left side (from ionic to metallic) is meant for delocalized bonds with varying electronegativity difference. In general: Use the tables of electronegativities ( ) and Figure \(\Page {4}\) to estimate the following values for the selected compounds: a: \(\ce{AsH}\) Using Equations \ref{sum} and \ref{diff}: \[\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \\[4pt] &=\dfrac{2.18 + 2.22}{2} \\[4pt] &= 2.2 \end{align*}\] \[\begin{align*} \Delta \chi &= \chi_A - \chi_B \\[4pt] &= 2.18 - 2.22 \\[4pt] &= 0.04 \end{align*}\] b: \(\ce{SrLi}\) Using Equations \ref{sum} and \ref{diff}: \[\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \\[4pt] &=\dfrac{0.95 + 0.98}{2} \\[4pt] &= 0.965 \end{align*}\] \[\begin{align*} \Delta \chi &= \chi_A - \chi_B \\[4pt] &= 0.98 - 0.95 \\[4pt] &= 0.025 \end{align*}\] c: \(\ce{KF}\) Using Equations \ref{sum} and \ref{diff}: \[\begin{align*} \sum \chi &= \dfrac{\chi_A + \chi_B}{2} \\[4pt] &=\dfrac{0.82 + 3.98}{2} \\[4pt] &= 2.4 \end{align*}\] \[\begin{align*} \Delta \chi &= \chi_A - \chi_B \\[4pt] &= | 0.82 - 3.98 | \\[4pt] &= 3.16 \end{align*}\] Contrast the bonding of \(\ce{NaCl}\) and silicon tetrafluoride. \(\ce{NaCl}\) is an ionic crystal structure, and an electrolyte when dissolved in water; \(\Delta \chi =1.58\), average \(\sum \chi =1.79\), while silicon tetrafluoride is covalent (molecular, non-polar gas; \(\Delta \chi =2.08\), average \(\sum \chi =2.94\). Jim Clark ( ) Ed Vitz (Kutztown University), (University of | 9,423 | 1,337 |
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From the previous sections, the partition function for a system of \(N\) indistinguishable and independent molecules is: \[ Q(N,V,\beta) = \dfrac{\sum_i{e^{-\beta E_i}}}{N!} \label{ID1} \] And the average energy of the system is: \[ \langle E \rangle = kT^2 \left(\dfrac{\partial \ln{Q}}{\partial T}\right) \label{ID2} \] We can combine these two equations to obtain: \[ \begin{split} \langle E \rangle &= kT^2 \left(\dfrac{\partial \ln{Q}}{\partial T}\right)_{N,V} \\ &= NkT^2 \left(\dfrac{\partial \ln{q}}{\partial T}\right)_V \\ &= N\sum_i{\epsilon_i \dfrac{e^{-\epsilon_i/kT}}{q(V,T)}} \end{split} \label{ID3} \] The average energy is equal to: \[ \langle E \rangle = N \langle \epsilon \rangle \label{aveE} \] where \(\langle \epsilon \rangle\) is the average energy of a single particle. If we compare Equation \(\ref{ID2}\) with Equation \(\ref{ID2}\), we can see: \[ \langle \epsilon \rangle = \sum_i{\epsilon_i \dfrac{e^{-\epsilon_i/kT}}{q(V,T)}} \nonumber \] The probability that a particle is in state \(i\), \(\pi_i\), is given by: \[ \langle \epsilon \rangle = \dfrac{e^{-\epsilon_i/kT}}{q(V,T)} = \dfrac{e^{-\epsilon_i/kT}}{\sum_i{e^{-\epsilon_i/kT}}} \nonumber \] The energy of a particle is a sum of the energy of each degree of freedom for that particle. In the case of a molecule, the energy is: \[ \epsilon = \epsilon_\text{trans} + \epsilon_\text{rot} + \epsilon_\text{vib} + \epsilon_\text{elec} \nonumber \] The molecular partition function is the product of the degree of freedom partition functions: \[ q(V,T) = q_\text{trans} q_\text{rot} q_\text{vib} q_\text{elec} \nonumber \] The partition function for each degree of freedom follows the same is related to the Boltzmann distribution. For example, the vibrational partition function is: \[ q_\text{vib} = \sum_i{e^{-\epsilon_i/kT}} \nonumber \] The average energy of each degree of freedom follows the same pattern as before. For example, the average vibrational energy is: \[ \langle \epsilon_\text{vib} \rangle = kT^2\dfrac{\partial \ln{q_\text{vib}}}{\partial t} = -\dfrac{\partial \ln{q_\text{vib}}}{\partial \beta} \nonumber \] | 2,128 | 1,338 |
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A catalytic converter is a device used to reduce the emissions from an internal combustion engine (used in most modern day automobiles and vehicles). Not enough oxygen is available to oxidize the carbon fuel in these engines completely into carbon dioxide and water; thus toxic by-products are produced. Catalytic converters are used in exhaust systems to provide a site for the oxidation and reduction of toxic by-products (like nitrogen oxides, carbon monoxide, and hydrocarbons) of fuel into less hazardous substances such as carbon dioxide, water vapor, and nitrogen gas. Catalytic converters were first widely introduced in American production cars in 1975 due to EPA regulations on toxic emissions reductions. The United States Clean Air Act required a 75% decrease in emissions in all new model vehicles after 1975, a decrease to be carried out with the use of catalytic converters. Without catalytic converters, vehicles release hydrocarbons, carbon monoxide, and nitrogen oxide. These gases are the largest source of ground level ozone, which causes smog and is harmful to plant life. Catalytic converters can also be found in generators, buses, trucks, and trains almost everything with an internal combustion engine has a form of catalytic converter attached to its exhaust system. A catalytic converter is a simple device that uses basic redox reactions to reduce the pollutants a car makes. It converts around 98% of the harmful fumes produced by a car engine into less harmful gases. It is composed of a metal housing with a ceramic honeycomb-like interior with insulating layers. This honeycomb interior has thin wall channels that are coated with a washcoat of aluminum oxide. This coating is porous and increases the surface area, allowing more reactions to take place and containing precious metals such as platinum, rhodium, and palladium. No more than 4-9 grams of these precious metals are used in a single converter. The converter uses simple oxidation and reduction reactions to convert the unwanted fumes. Recall that oxidation is the loss of electrons and that reduction is the gaining of electrons. The precious metals mentioned earlier promote the transfer of electrons and, in turn, the conversion of toxic fumes. The last section of the converter controls the fuel-injection system. This control system is aided by an oxygen sensor that monitors how much oxygen is in the exhaust stream, and in turn tells the engine computer to adjust the air-to-fuel ratio, keeping the catalytic converter running at the stoichiometric point and near 100% efficiency. A three-way catalytic converter has three simultaneous functions: There are two types of "systems" running in a catalytic converter, "lean" and "rich." When the system is running "lean," there is more oxygen than required, and the reactions therefore favor the oxidation of carbon monoxide and hydrocarbons (at the expense of the reduction of nitrogen oxides). On the contrary, when the system is running "rich," there is more fuel than needed, and the reactions favor the reduction of nitrogen oxides into elemental nitrogen and oxygen (at the expense of the two oxidation reactions). With a constant imbalance of the reactions, the system never achieves 100% efficiency. Note: converters can store "extra" oxygen in the exhaust stream for later use. This storage usually occurs when the system is running lean; the gas is released when there is not enough oxygen in the exhaust stream. The released oxygen compensates for the lack of oxygen derived from NO reduction, or when there is hard acceleration and the air-to-fuel ratio system becomes rich faster than the catalytic converter can adapt to it. In addition, the release of the stored oxygen stimulates the the oxidation processes of CO and C H . Without the redox process to filter and convert the nitrogen oxides, carbon monoxides, and hydrocarbons, the air quality (especially in large cities) becomes harmful to the human being. These compounds are of the same family as nitrogen dioxide, nitric acid, nitrous oxide, nitrates, and nitric oxide. When NO is released into the air, it reacts, stimulated by sunlight, with organic compounds in the air; the result is smog. Smog is a pollutant and has adverse effects on children's lungs. NO reacting with sulfur dioxide produces acid rain, which is highly destructive to everything it lands on. Acid rain corrodes cars, plants, buildings, national monuments and pollutes lakes and streams to an acidity unsuitable for fish. NO can also bind with ozone to create biological mutations (such as smog), and reduce the transmission of light. This is a harmful variant of a naturally occurring gas, CO . Odorless and colorless, this gas does not have many useful functions in everyday processes. Inhaling hydrocarbons from gasoline, household cleaners, propellants, kerosene and other fuels can be fatal to children. Further complications include central nervous system impairments and cardiovascular problems. The catalytic converter is a sensitive device with precious metals coating the inside. Without these metals, the redox reactions cannot occur. There are several substances and chemicals that inhibit the catalytic converter. These contaminants prevent the catalytic converter from functioning properly. However, this process could be reversed by running the engine at a high temperature to increase the hot exhaust flow through the converter, melting or liquefying some of the contaminants and removing them from the exhaust pipe. This process does not work if the metal is coated with lead, because lead has a high boiling point. If the lead poisoning is severe enough, the whole converter is rendered useless and must be replaced. Recall that thermodynamics predicts whether or not a reaction or process is spontaneous under certain conditions, but not the rate of the process. The redox reactions below occur slowly without a catalyst; even if the processes are thermodynamically favorable, they cannot occur without proper energy. This energy is the activation energy (\(E_a\) in the figure below) required to overcome the initial energy barrier preventing the reaction. A catalyst aids in the thermodynamic process by lowering the activation energy; the catalyst itself does not produce a product, but it does affect the amount and the speed at which the products are formed. Due to the precious metals in the coating of the inner ceramic structure, many catalytic converters have been targeted for theft. The converter is the most easily-accessible component because it lies on the outside and under the car. A thief could easily slide under the car, saw the connecting tubes on each end, and leave with the catalytic converter. Depending on the type and amount of precious metals inside, a catalytic converter can be easily sold for $200 apiece. Although the catalytic converter has helped reduce toxic emissions from car engines, it also has detrimental environmental effects. In the conversion of hydrocarbons and carbon monoxide, carbon dioxide is produced. Carbon dioxide is one of the most common greenhouse gases and contributes significantly to global warming. Along with carbon dioxide, the converters sometimes rearrange the nitrogen-oxygen compounds to form nitrous oxide. This is the same compound used in laughing gas and as a speed enhancer in cars. As a greenhouse gas, nitrous oxide is a 300 times more potent than carbon dioxide, and contributes proportionally to global warming. | 7,521 | 1,339 |
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When describing the shapes of molecules, we always give the geometry rather than the geometry. Molecular geometries only include pairs. For example, NH is a tetrahedral according to electron-pair geometry because it has 3 bonds and an unshared pair of electron. However, according to molecular geometry it is a trigonal pyramidal because it has only 3 bonds. How to predict molecular geometries using VSEPR model: Bond angles decrease as the number of nonbonding electron pairs increases. Bonding pairs are attracted by the two nuclei. By contrast, nonbonding electrons move under the attractive influence of only one nucleus and thus spread out more in space. Consider acetic acid molecule : in a molecule, if its centers of negative and positive charge don’t coincide, aka \[\mu =Qr \nonumber \] Therefore, the dipole moment increases as the quantity of charge that is separated increases and as the distance between the positive and negative centers increases. Dipole moments are generally reported in debyes (D), a unit equaling 3.33 X 10 C-m. : model of chemical bonding in which an electron-pair bond is formed between two atoms by the overlap of orbitals on the two atoms The same process can apply to the creation of hybrid orbitals Same process as orbitals. For example, mixing one orbital, three orbitals, and two orbitals gives six hybrid orbitals, which are directed at the vertices of an octahedron The molecular geometry needs to be known first in order to use the concept of hybridization, then we can use the following rules: In each covalent bond we have considered so far, the electron density is concentrated symmetrically about the line that connects the two nuclei ( ). This axis passes through the middle of the overlap region Usually, single bonds are \(\sigma\) bonds. A double bond has one \(\sigma\) bond and one \(\pi\) bond. A triple bond has one \(\sigma\) bond and two \(\pi\) bonds. Consider C H as an example. After hybridization, C has 3 sp hybrid orbitals and one electron in the remaining unhybridized 2p orbital. The unhybridized 2p electron is directly perpendicular to the plane that contains the three sp hybrid orbitals Because \(\pi\) bonds require that portions of a molecule be planar, they can introduce rigidity into molecules (strongly affects the properties of substances) Although \(\pi\) bonds can be formed from orbitals, we will only consider those formed by orbitals. They can only form if unhybridized orbitals are present on the bonded atoms. Only atoms having or hybridization can be involved in \(\pi\) bonding. Double and triple bonds are more common in molecules with small atoms : electrons that are spread over a number of atoms in a molecule rather than localized between a pair of atoms Consider H as an example. Whenever two atomic orbitals overlap, two molecular orbitals form. Thus, the overlap of the 1s orbitals of two hydrogen atoms to form H produces two molecular orbitals. : molecular orbital in which the electron density is concentrated in the internuclear region. The energy of a bonding molecular orbital is lower than the energy of the separate atomic orbitals from which it forms. The bonding molecular orbital results from summing the two atomic orbitals so that the atomic orbital wave functions enhance each other in the bond region. Because an electron in this molecular orbital is strongly attracted to both nuclei, the electron is more stable (lower energy) than it is in the 1s orbital of hydrogen. Because it concentrates electron density between both nuclei, the bonding molecular orbital holds the atoms together in a covalent bond. : molecular orbital in which electron density is concentrated outside the region between the two nuclei of bonded atoms. Such orbitals, designated as σ* or π*, are less stable (of higher energy) than bonding molecular orbitals \[\text{Bond order} = \dfrac{1}{2} (\text{number of bonding electrons – number of antibonding electrons})\nonumber \] Material here applies to diatomic molecules (those composed of identical atoms) In Be , each Be has 4 electrons, for a total of eight. When we place them in molecular orbitals, we fill the σ , σ* , σ , and σ* orbitals. There is an equal number of bonding and antibonding electrons, so the bond order is 0. Thus, Be doesn’t exist. | 4,355 | 1,340 |
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1. At each toss of a die, the die lands with one face on top. This face is distinguished from the other five faces by the number of dots that appear on it. Tossing a die produces data. What is the distribution? What is the random variable of this distribution? What outcomes are possible for this distribution? How would we collect a sample of ten values of the random variable of this distribution? 2. Suppose that we toss a die three times and average the results observed. How would you describe the distribution from which this average is derived? What is the random variable of this distribution? What outcomes are possible for this distribution? What would we do to collect a sample of ten values of the random variable of this distribution? 3. Suppose that we toss three dice simultaneously and average the results observed. How would you describe the distribution from which this average is derived? What is the random variable of this distribution? What outcomes are possible for this distribution? What would we do to collect a sample of ten values of the random variable of this distribution? Suppose that some third party collects a set, call it A, of ten values from this distribution and a second set, call it B, of values from the distribution in problem 2. If we are given the data in each set but are not told which label goes with which set of data, can we analyze the data to determine which set is A and which is B? 4. The manufacturing process for an electronic component produces 3 bad components in every 1000 components produced. The bad components appear randomly. What is the probability that (a) a randomly selected component is bad? (b) a randomly selected component is good? (c) 2 bad components are produced in succession? (d) 100 good components are produced in succession? 5. A product incorporates two of the components in the previous problem. What is the probability that (a) both components are good? (b) both components are bad? (c) one component is good and one component is bad? (d) at least one component is good? 6. A card is selected at random from a well-shuffled deck. A second card is then selected at random from among the remaining 51 cards. What is the probability that (a) the first card is a heart? (b) the second card is a heart? (c) neither card is a heart? (d) both cards are hearts? (e) at least one card is a heart? 7. A graduating class has 70 men and 77 women. How many combinations of homecoming king and queen are possible? 8. After the queen is selected from the graduating class of problem 7, one woman is selected to “first attendant” to the homecoming queen. Thereafter, another woman is selected to be “second attendant.” After the queen is selected, how many ways can two attendants be selected? 9. A red die and a green die are rolled. What is the probability that (a) both come up 3? (b) both come up the same? (c) they come up different? (d) the red die comes up less than the green die? (e) the red die comes up exactly two less than the green die? (f) together they show 5? 10. A television game show offers a contestant a new car as the prize for correctly guessing which of three doors the car is behind. After the contestant selects a door, the game-show host opens an incorrect door. The host then gives the contestant the option of switching from the door he originally chose to the other door that remains unopened. Should the contestant change his selection? [Hint: Consider the final set of outcomes to result from a sequence of three choices. First, the game-show producer selects a door and places the car behind this door. Diagram the possibilities. What is the probability of each? Second, the contestant selects a door. There are now nine possible outcomes. Diagram them. What is the probability of each? Third, the host opens a door. There are now twelve possible outcomes. Diagram them. What is the probability of each? Note that these twelve possibilities are not all equally probable.] 11. For a particular distribution, possible values of the random variable, \(x\), range from zero to one. The probability density function for this distribution is \({df}/{dx}=1\). (a) Show that the probability of finding \(x\) in the range \(0\le x\le 1\) is one. (b) What is the mean of this distribution? (c) What is the variance of this distribution? The standard deviation? (d) A quantity, \(g\), is a function of x: \(g\left(x\right)=x^2\). What is the expected value of \(g\)? 12. For a particular distribution, possible values of the random variable,\(\ x\), range from one to three. The probability density function for this distribution is \({df}/{dx}=cx\), where is a constant. (a) What is the value of the constant, ? (b) What is the mean of this distribution? (c) What is the variance of this distribution? The standard deviation? (d) If \(g\left(x\right)=x^2\), what is the expected value of \(g\)? 13. For a particular distribution, possible values of the random variable, \(x\), range from two to four. The probability density function for this distribution is \({df}/{dx=cx^3}\), where is a constant. (a) What is the value of the constant, c? (b) What is the mean of this distribution? (c) What is the variance of this distribution? The standard deviation? (d) If \(g\left(x\right)=x^2\), what is the expected value of \(g\)? 14. For a particular distribution, possible values of the random variable, \(x\), range from zero to four. For \(0\le x\le \le 1\), the probability density function is \({df}/{dx}={x}/{2}\). For \(1, the probability density function is \({df}/{dx}={\left(4-x\right)}/{6}\). (a) Show that the area under this probability distribution function is one. (b) What is the mean of this distribution? (c) What is the variance of this distribution? The standard deviation? (d) If \(g\left(x\right)=x^2\), what is the expected value of \(g\)? 15. The following values, \(x_i\), of the random variable, \(x\), are drawn from a distribution: \(9.63\), \(9.00\), \(11.87\), \(10.13\), \(10.83,\ 9.50\), \(10.40\), \(9.83\), and \(10.09\). (a) Arrange these values in increasing order and calculate the “rank probability,” \({i}/{\left(N+1\right)}\), associated with each of the \(x_i\) values. (b) Plot the rank probability (on the ordinate) versus the random-variable value (on the abscissa). Sketch a smooth curve through the points on this plot. (c) What function is approximated by the curve sketched in part b? (d) Plot the data points along a horizontal axis. Then create a bar graph (histogram) by erecting bars of equal area between each pair of data points. (e) What function is approximated by the tops of the bars erected in part d? 16. For a particular distribution, possible values of the random variable range from zero to four. The following values of the random variable are drawn from this distribution: \(0.1\), \(1.0\), \(1.1\), \(1.5\), \(2.1\). Sketch an approximate probability density function for this distribution. 17. The possible values for the random variable of a particular distribution lie in the range \(0\le x\le 10\). In six trials, the following values are obtained: \(1.0\), \(1.9\), \(2.3\), \(2.7\), \(3.0\), \(3.8\). (a) Sketch an approximate probability density function for this distribution. (b) What is the best estimate we can make of the mean of this distribution? (c) What is the best estimate we can make of the variance of this distribution? (d) What is the best estimate we can make of the variance of averages-of-six drawn from this distribution? (e) What is the best estimate we can make of the variance of averages-of-sixteen drawn from this distribution? 18. A computer program generates numbers from a normal distribution with a mean of zero and a standard deviation of \(10\). Also, for any integer \(N\), the program will generate and average \(N\) values from this distribution. It will repeat this operation until it has produced 100 such averages. It will then compute the estimated standard deviation of these \(100\) average values. The table below gives various values of \(N\) and the estimated standard deviation, \(s\), that was found for \(100\) averages of that \(N\). Plot these data in a way that tests the validity of the central limit theorem. 19. If \(f\left(u\right)\) is the cumulative probability distribution function for a distribution, what is the expected value of \(f\left(u\right)\)? What interpretation can you place on this result? 20. Five replications of a volumetric analysis yield concentration estimates of \(0.3000\), \(0.3008\), \(0.3012\), \(0.3014\), and\(\ 0.3020\) mol \({\mathrm{L}}^{-1}\). Calculate the rank probability of each of these results. Sketch, over the concentration range \(0.3000 mol \({\mathrm{L}}^{-1}\), an approximation of the cumulative probability distribution function for the distribution that yielded these data. 21. The Louisville Mudhens play on a square baseball field that measures \(100\) meters on a side. Casey’s hits always fall on the field. (He never hits a foul ball or hits one out of the park.) The probability density function for the distance that a Casey hit goes parallel to the first-base line is \({df_x\left(x\right)}/{dx=\left(2\times {10}^{-4}\right)x}\). (That is, we take the first-base line as our \(x\)-axis; the third-base line as our \(y\)-axis; and home plate is at the origin. \({df_x\left(x\right)}/{dx}\) is independent of the distance that the hit goes parallel to the third-base line, our \(y\)-axis.) The probability density function for the distance that a Casey hit goes parallel to the third-base line is \({df_y\left(y\right)}/{dy}=\left(3\times {10}^{-6}\right)y^2\). (\({df_y\left(y\right)}/{dy}\) is independent of the distance that the hit goes parallel to the first-base line, our \(x\)-axis.) (a) What is the probability that a Casey hit lands at a point \(\left(x,y\right)\) such that \(x^* and \(y^* ? (b) What is the two-dimensionally probability density function that describes Casey’s hits, expressed in this Cartesian coordinate system? (c) Recall that polar coordinates transform to Cartesian coordinates according to \(x=r{\mathrm{cos} \theta \ }\) and \(y=r{\mathrm{sin} \theta \ }\). What is the probability density function for Casey’s hits expressed using polar coordinates? (d) Recall that the differential element of area in polar coordinates is \(rdrd\theta\). Find the probability that a Casey hit lands within the pie-shaped area bounded by \(0 m and \(0<\theta <{\pi }/{4}\). 22. In Chapter 2, we derived the Barometric Formula, \(\eta \left(h\right)=\eta \left(0\right)\mathrm{exp}\left({-mgh}/{kT}\right)\) for molecules of mass \(m\) in an isothermal atmosphere at a height \(h\) above the surface of the earth. \(\eta \left(h\right)\) is the number of molecules per unit volume at height \(h\) ; \(\eta \left(0\right)\) is the number of molecules per unit volume at the earth’s surface, where \(h=0\). Consider a vertical cylinder of unit cross-sectional area, extending from the earth’s surface to an infinite height. Let \(f\left(h\right)\) be the fraction of the molecules in this cylinder that is at a height less than \(h\). Prove that the probability density function is \({df}/{dh=\left({mg}/{kT}\right)\mathrm{exp}\left({-mgh}/{kT}\right)}\). 23. A particular distribution has six outcomes. These outcomes and their probabilities are \(a\) ; \(b\) ; \(c\) ; \(d\) ;\(\ e\) ; and \(f\) . (a) Partitioning I assigns these outcomes to a set of three events: Event \(A\) \(=\ a\) or \(b\) or \(c\); Event \(B\) = \(d\); and Event \(C\ =\ e\) or \(f.\) What are the probabilities of Events \(A\), \(B\), and \(C\)? (b) Partitioning II assigns the outcomes to two events: Event \(D\ =\ a\) or \(b\) or \(c\); and Event \(E\ =\ d\) or \(e\) or \(f\). What are the probabilities of Events \(D\) and \(E\)? Express the probabilities of Events \(D\) and \(E\) in terms of the probabilities of Events \(A\), \(B\), and \(C\). (c) Partitioning III assigns the outcomes to three events: Event \(F\ =\ a\) or \(b\); Event \(G\ =\ c\) or \(d\); and Event \(H\ =\ e\) or \(f\). What are the probabilities of Events \(F\), \(G\), and \(H\)? Can the probabilities of Events \(F\), \(G\), and \(H\) be expressed in terms of the probabilities of Events \(A\), \(B\), and \(C\)? 24. Consider a partitioning of outcomes into events that is not exhaustive; that is, not every outcome is assigned to an event. What problem arises when we want to describe the probabilities of these events? 25. Consider a partitioning of outcomes into events that is not mutually exclusive; that is, one (or more) outcome is assigned to two (or more) events. What problem arises when we want to describe the probabilities of these events? 26. For integer values of \(\left(p\neq 1\right)\), we find \[\int{x^p{ \ln \left(x\right)\ }dx}=\left(\frac{x^{p+1}}{p+1}\right){ \ln \left(x\right)\ }-\frac{x^{p+1}}{{\left(p+1\right)}^2}\] (a) Sketch the function, \(h\left(x\right)={df\left(x\right)}/{dx}=-4x{ \ln \left(x\right)\ }\), over the interval \(0\le x\le 1\). (b) Show that we can consider \(h\left(x\right)={df\left(x\right)}/{dx}=-4x{ \ln \left(x\right)\ }\) to be a probability density function over this interval; that is, show \(f\left(1\right)-f\left(0\right)=1\). Let us name the corresponding distribution “Sam.” (c) What is the mean, \(\mu\), of Sam? (d) What is the variance, \({\sigma }^2\), of Sam? (e) What is the standard deviation, \(\sigma\), of Sam? (f) What is the variance of averages-of-four samples taken from Sam? (g) The following four values are obtained in random sampling of an unknown distribution: 0.050; 0.010; 0.020; and 0.040. Estimate the mean, \(\mu\), variance (\({\sigma }^2\)or \(s^2\)), and the standard deviation (\(\sigma\) or ) for this unknown distribution. (h) What is the probability that a single sample drawn from Sam will lie in the interval \(0\le x\le 0.10\) ? Note: The upper limit of this interval is 0.10, not 1.0 as in part (a). (i) Is it likely that the unknown distribution sampled in part g is in fact the distribution we named Sam? Why, or why not? 27. We define the mean, \(\mu\), as the expected value of the random variable: \(\mu =\int^{\infty }_{-\infty }{u\left({df}/{du}\right)}du\). Define \(\overline{u}=\sum^N_{i=1}{\left({u_i}/{N}\right)}\), where the \(u_i\) are independent values of the random variable. Show that the expected value of \(\overline{u}\) is \(\mu\). 28. A box contains a large number of plastic balls. An integer, \(W,\) in the range \(1\le W\le 20\) is printed on each ball. There are many balls printed with each integer. The integer specifies the mass of the ball in grams. Six random samples of three balls each are drawn from the box. The balls are replaced and the box is shaken between drawings. The numbers on the balls in drawings I through VI are: I: 3, 4, 9 II: 1, 6, 17 III: 2, 5, 8 IV: 2, 6, 7 V: 3, 5, 6 VI: 2, 3, 10 (a) What are the population sets represented by the samples I through VI? (b) Sketch the probability density function as estimated from sample I. (c) Sketch the probability density function as estimated from sample II (d) Using the data from samples I through VI, estimate the probability of drawing a ball of each mass in a single trial. (e) Sketch the probability density function as estimated from the probability values in part (d). (f) From the data in sample I, estimate the average mass of a ball in the box. (g) From the data in sample II, estimate the average mass of a ball in the box. (h) From the probability values calculated in part (d), estimate the average mass of a ball in the box. | 15,605 | 1,341 |
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In this chapter, we are going to take a closer look at the families of compounds that have carbon linked through a single covalent bond to an \(\mathrm{O}\), \(\mathrm{N}\), or \(\mathrm{S}\). These are known as alcohols (\(\mathrm{R-OH}\)), amines (\(\mathrm{R-NH}_{2}\), \(\mathrm{RR}^{\prime}\mathrm{-NH}\), \(\mathrm{RR}^{\prime}\mathrm{R}^{\prime \prime}\mathrm{-N}\)), thiols (\(\mathrm{R-SH}\)), ethers (\(\mathrm{R-OR}^{\prime}\)), and sulfides (\(\mathrm{R-SR}^{\prime}\)). We group these compounds together based on the predictable similarities and differences in their chemical and physical properties, specifically the fact that each of these functional groups has a relatively electronegative element (\(\mathrm{O}\), \(\mathrm{N}\) or \(\mathrm{S}\)) attached by a single bond to carbon and each has available lone electron pairs that can be donated to H or other electrophiles. The result is that alcohols, thiols, and amines (primary and secondary) all have relatively acidic hydrogens, which influences their chemical reactivities, and all show nucleophilic properties. Table \(6.0.1\) Examples of Functional groups, their names and approximate \(\mathrm{pK}_{\mathrm{a}}\)‘s 4-methylpentan-2-ol \(\sim 15-16\) alkenes, But-3-en-2-ol Propane-1-thiol -amine Propanamine or Propyl amine Ethyl methyl ether Dimethyl sulfide We will concentrate our discussion on oxygenated compounds, but we will note reactivities across the various groups to illustrate their similarities (and differences).
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The distinction between reversible and irreversible carbonyl addition reactions may be clarified by considering the stability of alcohols having the structure shown below in the shaded box. If substituent a hydrogen, an alkyl group or an aryl group, there is a good chance the compound will be unstable (not isolable), and will decompose in the manner shown. Most hydrates and hemiacetals ( = OH & OR), for example, are known to decompose spontaneously to the corresponding carbonyl compounds. Aminols ( = NHR) are intermediates in imine formation, and also revert to their carbonyl precursors if dehydration conditions are not employed. Likewise, α-haloalcohols ( = Cl, Br & I) cannot be isolated, since they immediately decompose with the loss of HY. In all these cases addition of H– to carbonyl groups is clearly reversible. If substituent is a hydrogen, an alkyl group or an aryl group, the resulting alcohol is a stable compound and does not decompose with loss of hydrogen or hydrocarbons, even on heating. It follows then, that if nucleophilic reagents corresponding to H: , R: or Ar: add to aldehydes and ketones, the alcohol products of such additions will form irreversibly. Free anions of this kind would be extremely strong bases and nucleophiles, but their extraordinary reactivity would make them difficult to prepare and use. Fortunately, metal derivatives of these alkyl, aryl and hydride moieties are available, and permit their addition to carbonyl compounds. Addition of a hydride anion to an aldehyde or ketone would produce an alkoxide anion, which on protonation should yield the corresponding alcohol. Aldehydes would give 1º-alcohols (as shown) and ketones would give 2º-alcohols. RCH=O + H: RCH O + H O RCH OH Two practical sources of hydride-like reactivity are the complex metal hydrides lithium aluminum hydride (LiAlH ) and sodium borohydride (NaBH ). These are both white (or near white) solids, which are prepared from lithium or sodium hydrides by reaction with aluminum or boron halides and esters. Lithium aluminum hydride is by far the most reactive of the two compounds, reacting violently with water, alcohols and other acidic groups with the evolution of hydrogen gas. The following table summarizes some important characteristics of these useful reagents. Reagent Preferred Solvents Functions Reduced Reaction Work-up Some examples of aldehyde and ketone reductions, using the reagents described above, are presented in the following diagram. The first three reactions illustrate that all four hydrogens of the complex metal hydrides may function as hydride anion equivalents which bond to the carbonyl carbon atom. In the LiAlH reduction, the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the borohydride reduction the hydroxylic solvent system achieves this hydrolysis automatically. The lithium, sodium, boron and aluminum end up as soluble inorganic salts. The last reaction shows how an acetal derivative may be used to prevent reduction of a carbonyl function (in this case a ketone). Remember, with the exception of epoxides, ethers are generally unreactive with strong bases or nucleophiles. The acid catalyzed hydrolysis of the aluminum salts also effects the removal of the acetal. This equation is typical in not being balanced ( it does not specify the stoichiometry of the reagent). Reduction of α,β-unsaturated ketones by metal hydride reagents sometimes leads to a saturated alcohol, especially with sodium borohydride. This product is formed by an initial of hydride to the β-carbon atom, followed by ketonization of the enol product and reduction of the resulting saturated ketone (equation 1 below). If the saturated alcohol is the desired product, catalytic hydrogenation prior to (or following) the hydride reduction may be necessary. To avoid reduction of the double bond, cerium(III) chloride is added to the reaction and it is normally carried out below 0 ºC, as shown in equation 2. Before leaving this topic it should be noted that diborane, B H , a gas that was used in ether solution to prepare , also reduces many carbonyl groups. Consequently, selective reactions with substrates having both functional groups may not be possible. In contrast to the metal hydride reagents, diborane is a relatively electrophilic reagent, as witnessed by its ability to reduce alkenes. This difference also influences the rate of reduction observed for the two aldehydes shown below. The first, 2,2-dimethylpropanal, is less electrophilic than the second, which is activated by the electron withdrawing chlorine substituents. The two most commonly used compounds of this kind are and . They are prepared from alkyl and aryl halides, as . These reagents are powerful nucleophiles and very strong bases (pK 's of saturated hydrocarbons range from 42 to 50), so they bond readily to carbonyl carbon atoms, giving alkoxide salts of lithium or magnesium. Because of their ring strain, epoxides undergo many carbonyl-like reactions, as . Reactions of this kind are among the most important synthetic methods available to chemists, because they permit simple starting compounds to be joined to form more complex structures. Examples are shown in the following diagram. A common pattern, shown in the shaded box at the top, is observed in all these reactions. The organometallic reagent is a source of a nucleophilic alkyl or aryl group (colored blue), which bonds to the electrophilic carbon of the carbonyl group (colored magenta). The product of this addition is a metal alkoxide salt, and the alcohol product is generated by weak acid hydrolysis of the salt. The first two examples show that water soluble magnesium or lithium salts are also formed in the hydrolysis, but these are seldom listed among the products, as in the last four reactions. Ketones react with organometallic reagents to give 3º-alcohols; most aldehydes react to produce 2º-alcohols; and formaldehyde and ethylene oxide react to form 1º-alcohols (examples #5 & 6). When a chiral center is formed from achiral reactants (examples #1, 3 & 4) the product is always a racemic mixture of enantiomers. Two additional examples of the addition of organometallic reagents to carbonyl compounds are informative. The first demonstrates that active metal derivatives of function in the same fashion as alkyl lithium and Grignard reagents. The second example again illustrates the use of acetal protective groups in reactions with powerful nucleophiles. Following acid-catalyzed hydrolysis of the acetal, the resulting 4-hydroxyaldehyde is in equilibrium with its cyclic hemiacetal. | 6,710 | 1,343 |
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Nuclear reactions that transform atomic nuclei alter their identity and spontaneously emit radiation via processes of radioactive decay. In 1889, Ernest Rutherford recognized and named two modes of radioactive decay, showing the occurrence of both processes in a decaying sample of natural uranium and its daughters. Rutherford named these types of radiation based on their penetrating power: heavier alpha and lighter beta radiation. Gamma rays, a third type of radiation, were discovered by P. Villard in 1900 but weren't recognized as electromagnetic radiation until 1914. Since gamma radiation is only the discharge of a high-energy photon from an over-excited nucleus, it does not change the identity of the atom from which it originates and therefore will not be discussed in depth here. Because nuclear reactions involve the breaking of very powerful intra nuclear bonds, massive amounts of energy can be released. At such high energy levels, the matter can be converted directly to energy according to Einstein's famous Mass-Energy relationship . The sum of mass and energy are conserved in nuclear decay. The of any spontaneous reaction must be negative according to thermodynamics (ΔG < 0), and is essentially equal to the energy change of nuclear reactions because ΔE is so massive. Therefore, a nuclear reaction will occur spontaneously when: \[ΔE = Δmc^2 < 0\] \(ΔE < 0\) or \(Δm < 0\) When the mass of the products of a nuclear reaction weigh less than the reactants, the difference in mass has been converted to energy. There are three types of nuclear reactions that are classified as beta decay processes. Beta decay processes have been observed in 97% of all known unstable nuclides and are thus the most common mechanism for radioactive decay by far. The first type (here referred to as ) is also called Negatron Emission because a negatively charged beta particle is emitted, whereas the second type ( ) emits a positively charged beta particle. In , an orbital electron is captured by the nucleus and absorbed in the reaction. All these modes of decay represent changes of one in the atomic number Z of the parent nucleus but no change in the mass number A. Alpha decay is different because both the atomic and mass number of the parent nucleus decrease. In this article, the term beta decay will refer to the first process described in which a true beta particle is the product of the nuclear reaction. Nuclides can be radioactive and undergo nuclear decay for many reasons. Beta decay can occur in nuclei that are rich in neutrons - that is - the nuclide contains more neutrons than stable isotopes of the same element. These "proton deficient" nuclides can sometimes be identified simply by noticing that their mass number A (the sum of neutrons and protons in the nucleus) is significantly more than twice that of the atomic number Z (number of protons in nucleus). In order to regain some stability, such a nucleus can decay by converting one of its extra neutrons into a proton, emitting an electron and an antineutrino( ). The high energy electron emitted in this reaction is called a and is represented by \( _{-1}^{0}\textrm{e}^{-} \) in nuclear equations. Lighter atoms (Z < 60) are the most likely to undergo beta decay. The decay of a neutron to a proton, a beta particle, and an antineutrino (\(\bar{\nu}\)) is \[ \ce{_{0}^{1}n^0 \rightarrow _{0}^{1}p^+ + _{-1}^{0}e^-}+ \bar{\nu} \] Some examples of are \[ \ce{_{2}^{6}He \rightarrow _{3}^{6}Li + _{-1}^{0}e^-} +\bar{\nu} \] \[ \ce{_{11}^{24}Na \rightarrow _{12}^{24}Mg + _{-1}^{0}e^-} + \bar{\nu} \] In order for beta decay to occur spontaneously according to Δm < 0, the mass of the parent (not atom) must have a mass greater than the sum of the masses of the daughter nucleus and the beta particle: [ Z] > [ (Z+1)] + [ e ]
(Parent nucleus) > (Daughter nucleus) + (electron) The mass of the antineutrino is almost zero and can therefore be neglected. The equation above can be reached easily from any beta decay reaction, however, it is not useful because mass spectrometers measure the mass of rather than just their nuclei. To make the equation useful, we must make these nuclei into neutral atoms by adding the mass of electrons to each side of the equation. The parent nucleus then becomes the neutral atom [ Z] plus the mass of one electron, while the daughter nucleus and the beta particle on the right side of the equation become the neutral atom [ (Z+1)] plus the mass of the beta particle. The extra electron on the left cancels the mass of the beta particle on the right, leaving the inequality [ Z] > [ (Z+1)]
(Parent atom) > (Daughter atom) The change in mass then equals Δm = [ (Z+1)] - [ Z] The energy released in this reaction is carried away as kinetic energy by the beta particle and antineutrino, with an insignificant of energy causing recoil in the daughter nucleus. The beta particle can carry anywhere from all to none of this energy, therefore the maximum kinetic energy of a beta particle in any instance of beta decay is . Nuclides that are imbalanced in their ratio of protons to neutrons undergo decay to correct the imbalance. Nuclei that are rich in protons relative to their number of neutrons can decay by conversion of a proton to a neutron, emitting a (\(^0_1e^+\)) and a neutrino ( . Positrons are the antiparticles of electrons, therefore a positron has the same mass as an electron but with the opposite (positive) charge. In positron emission, the atomic number Z by 1 while the mass number A remains the same. Some examples of are \[\ce{^8_5B \rightarrow ^8_4Be + ^0_1e^{+}} + \nu_e\] \[\ce{^{50}_{25}Mg \rightarrow ^{50}_{24}Cr + ^0_1e^+} + \nu_e\] Positron emission is only one of the two types of decay that tends to happen in "neutron deficient" nuclides, therefore it is very important to establish the correct mass change criterion. Positron emission occurs spontaneously when [ Z] > [ (Z-1)] + [ e ]
(Parent nucleus) > (Daughter nucleus) + (positron) In order to rewrite this inequality in terms of the masses of neutral atoms, we add the mass of electrons to both sides of the equation, giving the mass of a neutral [ Z] atom on the left and the mass of a neutral [ (Z-1)] atom, plus an extra electron, (since only Z-1 electrons are needed to make the neutral atom), and a positron on the right. Because positrons and electrons have equal mass, the inequality can be written as [ Z] > [ (Z-1)] + 2 [ e ]
(Parent atom) > (Daughter atom) + (2 electrons) The change in mass for positron emission decay is Δm = [ (Z)] - [ Z] - 2 [ e ] As with beta decay, the kinetic energy is split between the emitted particles - in this case the positron and neutrino. As mentioned before, there are two ways in which neutron-deficient / proton-rich nuclei can decay. When the mass change \(Δm < 0\) yet is insufficient to cause spontaneous positron emission, a neutron can form by an alternate process known as electron capture. An outside electron is pulled inside the nucleus and combined with a proton to make a neutron, emitting only a neutrino. \[ \ce{^1_1p + ^0_{-1}e^{-} → ^1_0n + \nu }\] Some examples of are \[\ce{^{231}_{92}U + ^0_{-1}e^{-} → ^{231}_{91}Pa + \nu }\] \[\ce{ ^{81}{36}Kr + ^0_{-1}e^- → ^{81}_{35}Br + \nu }\] Electron capture happens most often in the heavier neutron-deficient elements where the mass change is smallest and positron emission isn't always possible. For \(Δm < 0\), the following inequality applies: [ Z] + [ e ] > [ (Z-1)]
(Parent nucleus) + (electron) > (Daughter nucleus) Adding \(Z\) electrons to each side of the inequality changes it to its useful form in which the captured electron on the left cancels out the extra electron on the right [ Z] > [ (Z-1)]
(Parent atom) > (Daughter atom) The change in mass then equals Δm = [ (Z-1)] - [ Z] When the loss of mass in a nuclear reaction is greater than zero, but less than 2 [ e ], the process cannot occur by positron emission and is spontaneous for electron capture. The other three processes of nuclear decay involve the formation of a neutron or a proton inside the nucleus to correct an existing imbalance. In alpha decay, unstable, heavy nuclei (typically \(Z > 83\)) reduce their mass number \(A\) by 4 and their atomic number \(Z\) by 2 with the emission of a helium nuclei (\(\ce{^4_2He^{2+}}\)), known as an . Some examples of are \[ \ce{^{222}_{88}Ra} \rightarrow \ce{^{218}_{86}Rn + ^4_2He^{2+}}\] \[ \ce{^{233}_{92}U} \rightarrow \ce{^{229}_{90}Th + ^4_2He^{2+}}\] As with beta decay and electron capture, Δm must only be less than zero for spontaneous alpha decay to occur. Since the number of total protons on each side of the reaction does not change, equal numbers of electrons are added to each side to make neutral atoms. Therefore, the mass of the parent atom must simply be greater than the sum of the masses of its daughter atom and the helium atom. [ Z] > [ (Z-2)] + [ He ] The change in mass then equals Δm = [ (Z)] - [ (Z-2)] - [ He ] The energy released in an alpha decay reaction is mostly carried away by the lighter helium, with a small amount of energy manifesting itself in the recoil of the much heavier daughter nucleus. Alpha decay is a form of spontaneous fission, a reaction in which a massive nuclei can lower its mass and atomic number by splitting. Other heavy unstable elements undergo fission reactions in which they split into nuclei of about equal size. Proton-deficient or neutron-deficient nuclei undergo nuclear decay reactions that serve to correct unbalanced neutron/proton ratios. Proton-deficient nuclei undergo - emitting a beta particle (electron) and an antineutrino to convert a neutron to a proton - thus raising the elements atomic number Z by one. Neutron-deficient nuclei can undergo or (depending on the mass change), either of which synthesizes a neutron - emitting a positron and a neutrino or absorbing an electron and emitting a neutrino respectively - thus lowering Z by one. Nuclei with Z > 83 which are unstable and too massive will correct by , emitting an alpha particle (helium nucleus) and decreasing both mass and atomic number. Very proton-deficient or neutron-deficient nuclei can also simply eject an excess particle directly from the nucleus. These types of decay are called and . These processes are summarized in the table below. | 10,405 | 1,344 |
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This page explains what covalent bonding is. It starts with a simple picture of the single covalent bond. At a simple level a lot of importance is attached to the electronic structures of noble gases like neon or argon which have eight electrons in their outer energy levels (or two in the case of helium). These noble gas structures are thought of as being in some way a "desirable" thing for an atom to have. You may well have been left with the strong impression that when other atoms react, they try to achieve noble gas structures. As well as achieving noble gas structures by transferring electrons from one atom to another as in ionic bonding, it is also possible for atoms to reach these stable structures by sharing electrons to give covalent bonds. For example, two chlorine atoms could both achieve stable structures by sharing their single unpaired electron as in the diagram. The fact that one chlorine has been drawn with electrons marked as crosses and the other as dots is simply to show where all the electrons come from. In reality there is no difference between them. The two chlorine atoms are said to be joined by a covalent bond. The reason that the two chlorine atoms stick together is that the shared pair of electrons is attracted to the nucleus of both chlorine atoms. Hydrogen atoms only need two electrons in their outer level to reach the noble gas structure of helium. Once again, the covalent bond holds the two atoms together because the pair of electrons is attracted to both nuclei. The hydrogen has a helium structure, and the chlorine an argon structure. For example: e \(PCl_3\), th You might perhaps wonder why boron doesn't form ionic bonds with fluorine instead. Boron doesn't form ions because the total energy needed to remove three electrons to form a B ion is simply too great to be recoverable when attractions are set up between the boron and fluoride ions. In the case of phosphorus, 5 covalent bonds are possible - as in PCl . Phosphorus forms two chlorides - PCl and PCl . When phosphorus burns in chlorine both are formed - the majority product depending on how much chlorine is available. We've already looked at the structure of PCl . The diagram of PCl (like the previous diagram of PCl ) shows only the outer electrons. Notice that the phosphorus now has 5 pairs of electrons in the outer level - certainly not a noble gas structure. You would have been content to draw PCl at GCSE, but PCl would have looked very worrying. Why does phosphorus sometimes break away from a noble gas structure and form five bonds? In order to answer that question, we need to explore territory beyond the limits of most current A'level syllabuses. Don't be put off by this! It isn't particularly difficult, and is extremely useful if you are going to understand the bonding in some important organic compounds. We are starting with methane because it is the simplest case which illustrates the sort of processes involved. You will remember that the dots-and-crossed picture of methane looks like this. There is a serious mis-match between this structure and the modern electronic structure of carbon, 1s 2s 2p 2p . The modern structure shows that there are only 2 unpaired electrons to share with hydrogens, instead of the 4 which the simple view requires. You can see this more readily using the electrons-in-boxes notation. Only the 2-level electrons are shown. The 1s electrons are too deep inside the atom to be involved in bonding. The only electrons directly available for sharing are the 2p electrons. Why then isn't methane CH ? When bonds are formed, energy is released and the system becomes more stable. If carbon forms 4 bonds rather than 2, twice as much energy is released and so the resulting molecule becomes even more stable. There is only a small energy gap between the 2s and 2p orbitals, and so it pays the carbon to provide a small amount of energy to promote an electron from the 2s to the empty 2p to give 4 unpaired electrons. The extra energy released when the bonds form more than compensates for the initial input. The carbon atom is now said to be in an excited state. Now that we've got 4 unpaired electrons ready for bonding, another problem arises. In methane all the carbon-hydrogen bonds are identical, but our electrons are in two different kinds of orbitals. You aren't going to get four identical bonds unless you start from four identical orbitals. The electrons rearrange themselves again in a process called hybridization. This reorganizes the electrons into four identical hybrid orbitals called sp hybrids (because they are made from one s orbital and three p orbitals). You should read "sp " as "s p three" - not as "s p cubed". sp hybrid orbitals look a bit like half a p orbital, and they arrange themselves in space so that they are as far apart as possible. You can picture the nucleus as being at the center of a tetrahedron (a triangularly based pyramid) with the orbitals pointing to the corners. For clarity, the nucleus is drawn far larger than it really is. Remember that hydrogen's electron is in a 1s orbital - a spherically symmetric region of space surrounding the nucleus where there is some fixed chance (say 95%) of finding the electron. When a covalent bond is formed, the atomic orbitals (the orbitals in the individual atoms) merge to produce a new molecular orbital which contains the electron pair which creates the bond. Four molecular orbitals are formed, looking rather like the original sp hybrids, but with a hydrogen nucleus embedded in each lobe. Each orbital holds the 2 electrons that we've previously drawn as a dot and a cross. The principles involved - promotion of electrons if necessary, then hybridization, followed by the formation of molecular orbitals - can be applied to any covalently-bound molecule. This diagram only shows the outer (bonding) electrons. Nothing is wrong with this! (Although it doesn't account for the shape of the molecule properly.) If you were going to take a more modern look at it, the argument would go like this: Phosphorus has the electronic structure 1s 2s 2p 3s 3p 3p 3p . If we look only at the outer electrons as "electrons-in-boxes": There are 3 unpaired electrons that can be used to form bonds with 3 chlorine atoms. The four 3-level orbitals hybridise to produce 4 equivalent sp hybrids just like in carbon - except that one of these hybrid orbitals contains a lone pair of electrons. Each of the 3 chlorines then forms a covalent bond by merging the atomic orbital containing its unpaired electron with one of the phosphorus's unpaired electrons to make 3 molecular orbitals. You might wonder whether all this is worth the bother! Probably not! It is worth it with PCl , though. You will remember that the dots-and-crosses picture of PCl looks awkward because the phosphorus doesn't end up with a noble gas structure. This diagram also shows only the outer electrons. In this case, a more modern view makes things look better by abandoning any pretense of worrying about noble gas structures. If the phosphorus is going to form PCl it has first to generate 5 unpaired electrons. It does this by promoting one of the electrons in the 3s orbital to the next available higher energy orbital. Which higher energy orbital? It uses one of the 3d orbitals. You might have expected it to use the 4s orbital because this is the orbital that fills before the 3d when atoms are being built from scratch. Not so! Apart from when you are building the atoms in the first place, the 3d always counts as the lower energy orbital. This leaves the phosphorus with this arrangement of its electrons: The 3-level electrons now rearrange (hybridise) themselves to give 5 hybrid orbitals, all of equal energy. They would be called sp d hybrids because that's what they are made from. The electrons in each of these orbitals would then share space with electrons from five chlorines to make five new molecular orbitals - and hence covalent bonds. Why does phosphorus form these extra two bonds? It puts in an amount of energy to promote an electron, which is more than paid back when the new bonds form. Put simply, it is energetically profitable for the phosphorus to form the extra bonds. The advantage of thinking of it in this way is that it completely ignores the question of whether you've got a noble gas structure, and so you don't worry about it. Nitrogen is in the same Group of the Periodic Table as phosphorus, and you might expect it to form a similar range of compounds. In fact, it doesn't. For example, the compound NCl exists, but there is no such thing as NCl . Nitrogen is 1s 2s 2p 2p 2p . The reason that NCl doesn't exist is that in order to form five bonds, the nitrogen would have to promote one of its 2s electrons. The problem is that there aren't any 2d orbitals to promote an electron into - and the energy gap to the next level (the 3s) is far too great. In this case, then, the energy released when the extra bonds are made isn't enough to compensate for the energy needed to promote an electron - and so that promotion doesn't happen. Atoms will form as many bonds as possible provided it is energetically profitable. Jim Clark ( ) | 9,224 | 1,345 |
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Most metal hydroxides are insoluble; some such as \(\ce{Ca(OH)2}\), \(\ce{Mg(OH)2}\), \(\ce{Fe(OH)2}\), \(\ce{Al(OH)3}\) etc. are sparingly soluble. However, alkali metal hydroxides \(\ce{CsOH}\), \(\ce{KOH}\), and \(\ce{NaOH}\) are very soluble, making them strong bases. When dissolved, these hydroxides are completely ionized. Since the hydroxide concentration, \(\ce{[OH- ]}\), is an integrated property of the solution, the solubility of metal hydroxide depends on pH, pOH or \(\ce{[OH- ]}\). Alkali metal hydroxides \(\ce{LiOH}\), \(\ce{NaOH}\), \(\ce{KOH}\), \(\ce{CsOH}\) are soluble, and their solutions are basic. Hydroxides of alkali earth metals are much less soluble. For example, quicklime (\(\ce{CaO}\)) reacts with water to give slaked lime, which is slightly soluble. \[\begin{array}{ccccl}
\ce{CaO &+ &H2O &\rightleftharpoons &\:Ca(OH)2}\\
\textrm{quicklime} &&&&\textrm{slaked lime (slightly soluble)}
\end{array}\] Milk of magnesia is \(\ce{Mg(OH)2}\) ( = 7e-12) suspension. In an acidic solution such as stomach juice, the following reaction takes place, \[\ce{Mg(OH)2 + H+ \rightleftharpoons Mg^2+ + 2 H2O}\] Thus, it can neutralize excess acid in the stomach. Calculate the maximum concentration of \(\ce{Mg^2+}\) in a solution which contains a buffer so that pH = 3 at 298 K. As usual, we write the equilibrium equation so that we can write the concentration below the formula. If we do not know the concentration, we assume it to be a variable x. \[\begin{array}{ccccc}
\ce{Mg(OH)2 &\rightleftharpoons &Mg^2+ &+ &2 OH-}\\
&&x && 1 \times 10^{-11}
\end{array}\] \[K_{\ce{sp}} = x (1 \times 10^{-11})^2 = 7 \times 10^{-12}\] Solving for x results in \(x = 7 \times ^{10}\) This value certainly is too large, unrealistic. Calculate the pH of a saturated \(\ce{Mg(OH)2}\) solution. We assume the concentration to be x M of \(\ce{Mg(OH)}\), and note that \(\ce{[OH- ]} = 2 x\), \[\begin{array}{ccccc}
\ce{Mg(OH)2 &\rightleftharpoons &Mg^2+ &+ &2 OH-}\\
&&x &&2 x
\end{array}\] \[K_{\ce{sp}} = x (2 x)^2 = \textrm{7e-12}\] Solving for x; x = 1.2e-4 \[\begin{align}
\ce{[OH]} &= \textrm{2.4e-4}\\
\ce{pOH} &= 3.62
\end{align}\] \[\mathrm{pH = 14 - 3.62 = 10.38}\] The pH of a saturated lime (\(\ce{Ca(OH)2}\)) solution is about 10.0. Not all metal hydroxides behave the same way - that is precipitate as hydroxide solids. Metal hydroxides such as \(\ce{Fe(OH)3}\) and \(\ce{Al(OH)3}\) react with acids and bases, and they are called . In reality, \(\ce{Al(OH)3}\) should be formulated as \(\ce{Al(H2O)3(OH)3}\), and this neutral substance has a very low solubility. It reacts in the following way as \(\ce{[H+]}\) increases. \[\begin{align}
\ce{Al(H2O)3(OH)3 + H3O+ &\rightleftharpoons Al(H2O)4(OH)2+ + HOH}\\
\ce{Al(H2O)4(OH)2+ + H3O+ &\rightleftharpoons Al(H2O)5(OH)^2+ + H2O}\\
\ce{Al(H2O)5(OH)^2+ + H3O+ &\rightleftharpoons Al(H2O)6^3+ + H2O}
\end{align}\] When the pH increases, the following reactions take place: \[\begin{align}
\ce{Al(H2O)3(OH)3 + OH- &\rightleftharpoons Al(H2O)2(OH)4- + H2O}\\
\ce{Al(H2O)2(OH)4- + OH- &\rightleftharpoons Al(H2O)(OH)5^2- + H2O}\\
\ce{Al(H2O)(OH)5^2- + OH- &\rightleftharpoons Al(OH)6^3- + H2O}
\end{align}\] The charged species are soluble in water. As a result, amphoteric hydroxides dissolve in acidic and basic solutions. \(\ce{[Mg^2+]} = \dfrac{\textrm{7e-12}}{(\textrm{1e-12})^2} =\: ?\) This value is unrealistically large. The result is correct, but meaningless. | 3,474 | 1,346 |
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Gas, liquid, and solid are known as the three states of matter or material, but each of solid and liquid states may exist in one or more forms. Thus, another term is required to describe the various forms, and the term phase is used. Each distinct form is called a phase; however, the concept of phase defined as a homogeneous portion of a system extends beyond a single material, because a phase may also involve several materials. For example, a homogeneous solution of any number of substances is a one-phase system. Phase is a concept used to explain many physical and chemical changes (reactions). A phase is a distinct and homogeneous state of a system with no visible boundary separating it into parts. Water, \(\ce{H2O}\), is such a common substance that its gas (steam), liquid (water), and solid (ice) phases are widely known. An ice water mixture has two phases, as do systems containing ice-and-vapor, and water-and-vapor. To recognize the vapor system in these systems may require a keen observation, because the vapor usually blends with air, and is not detected directly. You probably also know that several solids may exist for a substance, and each of the solid forms is also called a phase. Diamond and graphite are the most quoted examples; both are solid carbon, but they have different crystal shapes, colors, and structures. They represent two different phases of carbon. Ice is another example, under 1 atm, ice has hexagonal symmetry, while cubic ice is formed under high pressure. In fact, there are at least eight different types of ice, each being a solid phase. When you mix water and alcohol, regardless of the relative amounts that you use, they are fully miscible. The resulting mixture has only one phase (a solution). However, water and oil are normally immiscible, and their boundary of separation is visible; they form a two-phase system. Sometimes you cannot "see" the boundary, and you will need scientific reasoning to realize the number of phases present in system. Well, there is so much concept packed into one term that we can not make the definition any simpler for you. However, the term is useful because it can be used to explain many phenomena. There is no substitute for it. Learn it and use it to explain physical changes. The conversion between these phases is called a . A state change of any material due to temperature or pressure change is a phase transition. A phase transition is a physical change (or reaction). The following diagram illustrates the key phase transitions: You should know the names of the process for these phase transitions. \[\mathrm{SOLID \xrightarrow{\large{sublimation\:}} GAS \xrightarrow{\large{deposition\:}} SOLID}\] \[\mathrm{SOLID \xrightarrow{\large{melting\:}} LIQUID \underset{\large{(solidfication)}}{\xrightarrow{\large{\:\:\:freezing\:\:\:\:}}} SOLID}\] \[\mathrm{GAS \xrightarrow{\large{condensation\:}} LIQUID \xrightarrow{\large{vaporization\:}} GAS}\] The concepts of phase and phase transition introduce you into fields of materials studies. For example, if you search the internet with the phrase "phase transition", you get thousands of websites; some are related to the concept we have discussed here, but some may be using "phase transition" as a catchy phrase. The concept of phase transition is also applied to the study of nuclear matter such as protons and neutrons. Hint: more than one Skill -
Define phase and enumerate the number of phases in a system.
Milk is a . Hint: d Skill -
Identify the number of phases in a system. Hint: Three phases: solid, liquid, and vapor (gas solution). Discussion -
What will be the temperature for a system describe here? Hint: Two phases: solid and vapor (gas solution). Skill -
Define phase and enumerate the number of phases in a system. Hint: d. Four phases: liquid, \(\ce{NaCl}\), ice, gas Skill -
Enumerate the number of phases in a system.
Do you know what the temperature is for such a system to be at equilibrium? It's 0 degree F (the temperature of equal weight mixture of snow and salt). Hint: A liquid and a gas solution, 2 phases. Skill -
Identify the phases in a system. Hint: Deposition Skill -
Know the names of all phase transition processes. Hint: c. increases as the temperature of the solid increases. Skill -
This is a test of common sense.
A scientific way of looking at the vapor pressure of a solid as a function of temperature is given by the Clausius-Clapeyron Equation: \(\mathrm{P_{vap} = A e^{-\mathit H_{\large{vap}}/\mathit{RT}}}\) Hint: false Skill -
This is a test of common sense.
Only when the atmospheric pressure is 1.00 atm will the boiling temperature be 100 deg. C, which is the normal boiling point of water. Hint: false! Skill -
This is a test of common sense. You know that the density of water is the highest at 4 deg. C, and its volume for a mole is the lowest at 4 deg C.
One mole of water is 18 g, and its volume is called the molar volume. Water is a strange substance! Hint: true! Skill -
Analyze the problem carefully. Most substances expand when heated. Hint: gas Discussion -
Temperature is not specified here, but you may assume a temperature at which all three phases are stable. Under the circumstance, the pressure of vapor phase is much lower than 1 atm. For the same quantity of water, ice has a larger volume than water. | 5,427 | 1,347 |
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The law of reciprocal proportions was proposed by Jeremias Ritcher in 1792. It states that, "If two different elements combine separately with the same weight of a third element, the ratio of the masses in which they do so are either the same or a simple multiple of the mass ratio in which they combine." Oxygen and sulfur react with copper to give copper oxide and copper sulfide, respectively. Sulfur and oxygen also react with each other to form SO . Therefore, in CuS: Cu:S = 63.5:32 in CuO: Cu:O = 63.5:16 S:O = 32:16 S:O = 2:1 Now in SO : S:O = 32:32 S:O = 1:1 Thus the ratio between the two ratios is the following: \[\dfrac{2}{1} : \dfrac{1}{1} - 2:1\] This is a simple multiple ratio. | 713 | 1,348 |
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Resonance is a mental exercise and method within the of bonding that describes the delocalization of electrons within molecules. It compares and contrasts two or more possible Lewis structures that can represent a particular molecule. Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms. The net sum of valid resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. A molecule that has several resonance structures is more stable than one with fewer. Some resonance structures are more favorable than others. Electrons have no fixed position in atoms, compounds and molecules (see image below) but have probabilities of being found in certain spaces (orbitals). Resonance forms illustrate areas of higher probabilities (electron densities). This is like holding your hat in either your right hand or your left. The term Resonance is applied when there are two or more possibilities available. Chemists use Lewis diagrams to depict structure and bonding of covalent entities, such as molecules and polyatomic ions, henceforth, molecules. The Lewis diagram of many a molecule, however, is not consistent with the observed properties of the molecule. The nitrate ion, according to its Lewis diagram, has two types of nitrogen-oxygen bonds, one double bond and two single bonds, suggesting that one nitrogen-oxygen bond in the nitrate ion is shorter and stronger than each of the other two. Also, the Lewis structure implies, with respect to formal charge, that there are two types of oxygen atoms in the nitrate ion, one formally neutral and each of the other two bearing a formal charge of –1. Experimentally, however, the three nitrogen-oxygen bonds in the nitrate ion have the same bond length and the same bond energy, and the three oxygen atoms are indistinguishable. The Lewis diagram fails to explain the structure and bonding of the nitrate ion satisfactorily. Two additional Lewis diagrams can be drawn for the nitrate ion. However, none of them are consistent with the observed properties of the nitrate ion and, therefore, does not correctly depict the nitrate ion. Benzene, according to its Lewis diagram, has two types of carbon-carbon bonds, three double bonds and three single bonds, suggesting that each of three carbon-carbon bonds in benzene is shorter and stronger than each of the other three. Experimentally, however, the six carbon-carbon bonds in benzene have the same bond length and the same bond energy. The Lewis diagram fails to explain the structure and bonding of benzene satisfactorily. An attribute of molecules of which the classical Lewis diagram is not consistent with the observed properties is that other valid Lewis diagrams can be generated for them. One additional Lewis diagram can be drawn for benzene. However, none of them are consistent with the observed properties of benzene and, therefore, does not correctly depict benzene. Resonance theory is an attempt to explain the structure of a species, like the nitrate ion or benzene, no Lewis diagram of which is consistent with the observed properties of the species. The major advantage of resonance theory is that, although based on rigorous mathematical analysis, resonance theory can be applied successfully invoking little or no math. Resonance theory is explained below using the nitrate ion as the example. According to resonance theory, the structure of the nitrate ion is not 1 nor 2 nor 3 but the average of all three of them, weighted by stability. Lewis diagrams 1, 2, and 3 are called resonance forms, resonance structures, or resonance contributors of the nitrate ion. The structure of the nitrate ion is said to be a resonance hybrid or, simply, hybrid of resonance forms 1, 2, and 3. Whenever it is necessary to show the structure of the nitrate ion, resonance forms 1, 2, and 3 are drawn, connected by a double-headed arrows. The three resonance forms of the nitrate ion, 1, 2, and 3, are identical, so they have the same stability and, therefore, contribute equally to the hybrid. Since the exact extent to which each resonance form of the nitrate ion contributes to the hybrid is known, the bond order of each nitrogen-oxygen bond as well as the formal charge on each oxygen atom in the hybrid can be easily determined: According to resonance theory, each bond in the nitrate ion is one and one-third of a bond, which is consistent with the observation that the three bonds in the nitrate ion have the same bond length and the same bond energy. According to resonance theory, each oxygen atom in the nitrate ion has a formal charge of –2/3, which, in conjunction with the fact that the three nitrogen-oxygen bonds are identical, is consistent with the observation that the three oxygen atoms in the nitrate ion are indistinguishable. In each resonance form of the nitrate ion, there are two \(\pi\) electrons, and they are shared only by two atoms. An electron shared only by two atoms is said to be localized. Thus, the two \(\pi\) electrons in each resonance form of the nitrate ion are localized. The nitrate ion, as represented by the hybrid, has two \(\pi\) electrons: The two \(\pi\) electrons in the nitrate ion are shared by a total of four atoms, one nitrogen atom and three oxygen atoms. An electron shared by more than two atoms is said to be delocalized. Thus, the two \(\pi\) electrons in the nitrate ion are delocalized. Delocalization of \(\pi\) electrons in the nitrate ion requires that the four atoms be on the same plane, allowing lateral overlap of the p orbitals on them. If the energy of the nitrate ion were the weighted average of the energies of its three resonance forms, just as the structure of the nitrate ion is the weighted average of the structures of its three resonance forms, it should be equal to the energy of one of the three identical resonance forms:
If the energy of the hybrid were equal to that of a resonance form, given that all chemical entities (elementary particles, atoms, molecules, etc.) naturally tend to be in the lowest possible energy state, there would be no advantage for the nitrate ion to exist as the hybrid; it could simply exist as a resonance form. Since the nitrate ion exists as the hybrid, not as a resonance form, it can be inferred that the energy of the hybrid is lower than that of any of the resonance forms. According to resonance theory then, the energy of a molecule is lower than that of the lowest-energy resonance form. Since the nitrate ion has lower energy and, therefore, is more stable than any of its resonance forms, the nitrate ion is said to be resonance stabilized. There are two misconceptions about resonance theory among beginning students, likely due to literal interpretation of the word resonance. They are described below, using the nitrate ion as the example. The nitrate ion exists as resonance form 1 for a moment and then changes either to resonance form 2 or to resonance form 3, which interconvert, or revert to 1. In a sample of nitrate ions, at a given moment, one-third of the ions exist as resonance form 1, another one-third as resonance form 2, and the remaining one-third as resonance form 3. In a sample of nitrate ions, at a given moment, all ions have the same structure, which is the hybrid. The classic analogy used to clarify these two misconceptions is the mule (Morrison, R. T.; Boyd, R.N. Organic Chemistry, fifth edition; Allyn and Bacon: Boston, 1987, pg. 373). Biologically, a mule is a hybrid of a horse and a donkey. This does not mean that a mule resembles a horse for a moment and then changes to resemble a donkey. The appearance of a mule is a combination of that of a horse and that of a donkey and does not change with time. Nor does it mean that, in a herd, some mules resemble a horse and the others a donkey. In a herd, all mules have the same appearance, which is a combination of a horse and a donkey. The weakness of this analogy is that horses and donkeys do exist, whereas resonance forms are strictly hypothetical. A better analogy, cited in Morrison and Boyd, is the rhinoceros. Upon seeing a rhinoceros, one could describe it as the hybrid of a dragon and a unicorn, two creatures that do not exist. A consideration of resonance contributors is crucial to any discussion of the amide functional group. One of the most important examples of amide groups in nature is the ‘peptide bond’ that links amino acids to form polypeptides and proteins. Critical to the structure of proteins is the fact that, although it is conventionally drawn as a single bond, the C-N bond in a peptide linkage has a significant barrier to rotation, almost as if it were a double bond. This, along with the observation that the bonding around the peptide nitrogen has trigonal planar geometry, strongly suggests that the nitrogen is sp -hybridized. An important resonance contributor has a C=N double bond and a C-O single bond, with a separation of charge between the oxygen and the nitrogen. Although B is a minor contributor due to the separation of charges, it is still very relevant in terms of peptide and protein structure – our proteins would simply not fold up properly if there was free rotation about the peptide C-N bond. by (University of Minnesota, Morris) | 9,484 | 1,351 |
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The simplest kind of bond is that between univalent atoms in diatomic molecules, such as \(\ce{H_2}\), \(\ce{F_2}\), and so on. In the gas phase the molecules are in rapid motion, colliding with one another and the walls of the container. The atoms vibrate with respect to one another, and the molecules have rotational energy as well. Despite this activity, we can assign an average equilibrium bond distance \(\left( r_e \right)\) and an average bond energy \(\left( D_e \right)\) for normal, unexcited molecules. From calculations ( ), we learn that the energy of an \(\ce{H_2}\) molecule is a function of \(r\), the distance between the hydrogens, as shown in Figure 21-1. When the distance is reduced from \(r_e\), the energy increases very rapidly because of internuclear repulsion. As the separation between the atoms increases, the energy of the system increases more slowly and finally approaches that of the entirely free atoms. The distance \(r_e\), which corresponds to the bond length at minimum energy, with atomic number downward in a column of the periodic table as the atoms get larger. It across a horizontal row of the periodic table as the electronegativity of the atoms increases and the atomic radius becomes smaller. Other things being equal, the stronger the bond is, the shorter \(r_e\) will be, because a strong bond overcomes the repulsive forces between the nuclei and thus permits them to move closer together. For bonds between two carbon atoms, \(r_e\) usually ranges between about \(1.20\) Å and \(1.55\) Å and, if Figure 21-1 (or anything similar) applies, we should not expect significant \(\ce{C-C}\) bonding at internuclear distances greater than \(2\) Å. It is important to recognize that bonding occurs only if the electrons are paired (i.e., have opposite spins). The upper dashed curve of Figure 21-1 shows how the energy changes as two hydrogen atoms will parallel spins approach one another. That there is no net bonding can be understood by the Pauli principle ( ), which tells us that two electrons cannot be in the same orbital if they are unpaired. and (1977) | 2,130 | 1,352 |
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The usual application of mass spectroscopy to organic molecules involves bombardment with a beam of medium-energy electrons (\(50\)-\(100 \: \text{eV}\) or \(1150\)-\(2300 \: \text{kcal mol}^{-1}\)) in high vacuum, and analysis of the charged particles and fragments so produced. Most mass spectrometers are set up to analyze positively charged fragments, although negative-ion mass spectrometry also is possible. The elements of a mass spectrometer are shown in Figure 9-51. The positive ions produced by electron impact are accelerated by the negatively charged accelerating plates and sweep down to the curve of the analyzer tube where they are sorted as to their mass-to-charge (\(\left( m/e \right)\) ratio by the analyzing magnet. With good resolution, only the ions of a single mass number will pass through the slit and impinge on the collector, even when the mass numbers are in the neighborhood of several thousand. The populations of the whole range of mass numbers of interest can be determined by plotting the rate of ion collection as a function of the magnetic field of the analyzing magnet. Mass spectra of 2-propanone, 2-butanone, and propanal are shown in Figure 9-52. Each peak represents ions of particular masses formed as the result of fragmentation of the molecule produced by electron impact into \(\ce{CH_3^+}\), \(\ce{CH_3CH_2^+}\), \(\ce{CH_3CO}^\oplus\), and so on. The "cracking patterns" are, of course, functions of the energy of the bombarding electrons and serve as an extraordinarily individual fingerprint of the particular molecules. For instance, 2-propanone and propanal are isomers, yet their cracking patterns are strikingly different. . Incorrect molecular weights will be obtained if the positive ion, \(M^+\), becomes fragmented before it reaches the collector, or if two fragments combine to give a fragment heavier than \(M^+\). The peak of \(M^+\) is especially weak with alcohols and branched-chain hydrocarbons, which readily undergo fragmentation by loss of water or side-chain groups. With such compounds the peak corresponding to \(M^+\) may be \(0.1\%\) or less of the highest peak in the spectrum, which is called the and usually is assigned an arbitrary intensity of 100. The pressure of the sample in the ion source of a mass spectrometer is usually about \(10^{-5} \: \text{mm}\), and, under these conditions, buildup of fragments to give significant peaks with \(m/e\) greater than \(M^+\) is rare. One exception to this is the formation of \(\left( M + 1 \right)^+\) peaks resulting from transfer of a hydrogen atom from \(M\) to \(M^+\). The relative intensities of such \(\left( M + 1 \right)^+\) peaks are usually sensitive to the sample pressure and may be identified in this way. With the molecular weight available from the \(M^+\) peak with reasonable certainty, the next step is to determine the . If the resolution of the instrument is sufficiently high, quite exact masses can be measured, which means that ions with \(m/e\) values differing by one part in 50,000 can be distinguished. At this resolution it is possible to determine the elemental composition of each ion from its exact \(m/e\) value. Many mass spectrometers in routine use are incapable of resolving ions with \(m/e\) values that differ by less than one mass unit. In this event, the determination of elemental composition can be determined by the method of . We will illustrate this with the following simple example. The highest peaks corresponding to \(M^+\) in the mass spectrum of an unknown sample have \(m/e\) equal to 64 and 66 with relative intensities of 3:1. What is the elemental composition? The 3:1 abundance ratio is uniquely characteristic of the chlorine isotopes, \(\ce{^{35}Cl}\):\(\ce{^{37}Cl} =\) 3:1. The mass peaks at 64 and 66 are therefore both molecular ions; the 64 peak is of an ion containing \(\ce{^{35}Cl}\) and the 66 peak is of an ion containing \(\ce{^{37}Cl}\). The remaining atoms in the molecule must add up to \(\left( 64 - 35 \right) = 29\), or \(\left( 66 - 37 \right) = 29\) mass units. There are several possible combinations of \(\ce{C}\), \(\ce{H}\), \(\ce{N}\), and \(\ce{O}\) that give mass 29; they are \(\ce{N_2H}\), \(\ce{CHO}\), \(\ce{CH_3N}\), and \(\ce{C_2H_5}\).\(^{15}\) Of these, the combination with \(\ce{Cl}\) that makes the most chemical sense is \(\ce{C_2H_5}\), and the formula of the molecule therefore is \(\ce{C_2H_5Cl}\), chloroethane. This example illustrates how \(m/e\) values of . The important isotopes for this purpose in addition to those of chlorine are the stable isotopes of natural abundance, \(\ce{^{13}C}\) \(\left( 1.1\% \right)\), \(\ce{^{15}N}\) \(\left( 0.37\% \right)\), \(\ce{^{17}O}\) \(\left( 0.04\% \right)\), and \(\ce{^{18}O}\) \(\left( 0.20\% \right)\). As a further example, suppose that we have isolated a hydrocarbon and have determine from its mass spectrum that \(M^+ =\) 86 mass units. In the absence of any combination reactions there will be an \(\left( M + 1 \right)^+\) ion corresponding to the same molecular ion but with \(\ce{^{13}C}\) ion in place of \(\ce{^{12}C}\). The intensity ratio \(\left( M + 1 \right)^+/M^+\) will depend on the number of carbon atoms present, because the more carbons there are the greater the probability will be that one of them is \(\ce{^{13}C}\). The greater the probability, the larger the \(\left( M + 1 \right)^+/M^+\) ratio. For \(n\) carbons, we expect \[\frac{\text{abundance of} \left( M + 1 \right)^+}{\text{abundance of} \: M^+} = n \times \% \ce{^{13}C} \: \text{abundance}/100\] If the measured \(\left( M + 1 \right)^+/M^+\) ratio is 6.6:100, then \[\begin{align} \frac{6.6}{100} &= n \times 1.1/100 \\ n &= 6 \end{align}\] The only hydrocarbon formula with \(M^+ =\) 86 and \(n=6\) is \(\ce{C_6H_{14}}\). Nitrogen \(as \(\ce{^{15}N}\)) and oxygen (as \(\ce{^{17}O}\)) also contribute to \(\left( M + 1 \right)^+\), if present, while \(\ce{^{18}O}\) and \(\ce{^{13}C}\)'s contribute to \(\left( M + 2 \right)^+\). The calculated intensities of \(\left( M + 1 \right)^+\) and \(\left( M + 2 \right)^+\) relative to \(M^+\) (as 100) are tabulated in Table 9-5 for elemental composition of ions up to \(\ce{C_{20}}\). The table applies to fragment ions as well as molecular ions, but the intensity data from fragment ions very often is complicated by overlapping peaks. The next step in the analysis of a mass spectrum is to see what clues as to structure can be obtained from the fragment ions. It would be a serious error to imagine that in mass spectra nothing is observed but simple nonspecific fragmentation of organic molecules on electron impact. Actually, even though electron impact produces highly unstable molecular ions, there is a strong tendency for breakdown to occur by reasonable chemical processes, and this may involve straightforward fragmentation or rearrangement of atoms from one part of the molecule to another. In general, fragmentation occurs at the weakest bonds, and the most abundant fragments also are the most stable ones. For instance, hydrocarbons fragment preferentially at branch points, partly because the \(\ce{C-C}\) bonds are weaker here than elsewhere along the chain, and partly because the ionic fragments are more stable. As an example, consider 2,2-dimethylbutane. There is no molecular ion evident in its mass spectrum because it cleaves so readily at the quaternary carbon to give the \(m/e\) 57 peak corresponding to the most abundant fragment ion. This ion is presumably the -butyl cation and the alternate cleavage to the less stable ethyl cation with \(m/e =\) 29 is much less significant. An excellent example of a rearrangement with fragmentation is provided by the \(M^+\) ion of ethyl butanoate, which breaks down to give ethene and the \(M^+\) ion of an isomer of ethyl ethanoate called its "enol form". An interesting and complex rearrangement occurs on electron impact with methylbenzene (toluene). An intense peak is observed having \(m/e\) for \(\ce{C_7H_7^+}\), but the ion involved appears to be a symmetrical \(\ce{C_7H_7^+}\) ion, rather than a phenylmethyl cation. The evidence for this is that the fragmentation patterns found in the mass spectrometry of the ion itself are the same, no matter which of the monodeuteriomethylbenzenes is used as starting material. This rearrangement occurs because of the high delocalization energy of the symmetrical \(ce{C_7H_7}^\oplus\) ion (usually called "tropylium cation") and because its charge is spread out more evenly over the carbons than would be the charge for the phenylmethyl cation (see Section 8-7B). \(^{15}\)Tabulations of elemental compositions of \(\ce{C}\), \(\ce{H}\), \(\ce{N}\) and \(\ce{O}\) for mass values up to 250 are listed in many texts on mass spectrometry. Consult these tables to see all possible alternatives. See also J. H. Beynon, , Elsevier Publishing Co., Amsterdam, 1960. and (1977) | 8,932 | 1,353 |
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The validity of de Broglie’s proposal was confirmed by electron diffraction experiments of G.P. Thomson in 1926 and of C. Davisson and L. H. Germer in 1927. In these experiments it was found that electrons were scattered from atoms in a crystal and that these scattered electrons produced an interference pattern. The interference pattern was just like that produced when water waves pass through two holes in a barrier to generate separate wave fronts that combine and interfere with each other. These diffraction patterns are characteristic of wave-like behavior and are exhibited by both matter (e.g., electrons and neutrons) and electromagnetic radiation. Diffraction patterns are obtained if the wavelength is comparable to the spacing between scattering centers. Diffraction occurs when waves encounter obstacles whose size is with its wavelength. Continuing with our analysis of experiments that lead to the new quantum theory, we now look at the phenomenon of electron diffraction. It is well-known that has the ability to diffract around objects in its path, leading to an interference pattern that is particular to the object. This is, in fact, how holography works (the interference pattern is created by allowing the diffracted light to interfere with the original beam so that the hologram can be viewed by shining the original beam on the image). A simple illustration of light diffraction is the (Figure 1.7.1
). The double-slit experiments are direct demonstration of wave phenomena via observed interference. These types of experiment were first performed by Thomas Young in 1801, as a demonstration of the wave behavior of light. In the basic version of this experiment, light is illuminated only a plate pierced by two parallel slits, and the light passing through the two slits is observed on a screen behind the plate (Figures 1.7.1
and 1.7.2
). The wave nature of light causes the light waves passing through the two slits to interfere, producing bright and dark bands on the screen – a result that would not be expected if light consisted of classical particles (Figure \(\Page {0c}\)). However, the light is always found to be absorbed at the screen at discrete points, as individual particles (not waves); the interference pattern appears via the varying density of these particle hits on the screen. Furthermore, versions of the experiment that include detectors at the slits find that each detected photon passes through one slit (as would a classical particle), and not through both slits (as would a wave). Interference is a wave phenomenon in which two (or more) superimpose to form a resultant wave of greater or lower amplitude. It is the primary property used to identify wave behavior. According to classical physics, electrons should behave like particles - they travel in straight lines and do not curve in flight unless acted on by an external agent, like a magnetic field. In this model, if we fire a beam of electrons through a double slit onto a detector, we should get two bands of "hits", much as you would get if you fired a machine gun at the side of a house with two windows - you would get two areas of bullet-marked wall inside, and the rest would be intact (Figure 1.7.3
(left) and Figure 1.7.2
). However, if the slits are made small enough and close enough together, we actually observe the electrons are through the slits and with each other just like waves (Figure 1.7.3
(right) and Figure 1.7.2
a,b). This means that the electrons have , just like photons, in agreement with de Broglie's hypothesis discussed previously. In this case, they must have properties like wavelength and frequency. We can deduce the properties from the behavior of the electrons as they pass through our diffraction grating. This was a pivotal result in the development of quantum mechanics. Just as the photoelectric effect demonstrated the particle nature of light, the Davisson–Germer experiment showed the wave-nature of matter, and completed the theory of wave-particle duality. For physicists this idea was important because it meant that not only could any particle exhibit wave characteristics, but that one could use wave equations to describe phenomena in matter if one used the de Broglie wavelength. An electron microscope use a beam of accelerated electrons as a source of illumination. Since the wavelength of electrons can be up to 100,000 times shorter than that of visible light photons, electron microscopes have a higher resolving power than light microscopes and can reveal the structure of smaller objects. A transmission electron microscope can achieve better than 50 pm resolution and magnifications of up to about 10,000,000x whereas most light microscopes are limited by diffraction to about 200 nm resolution and useful magnifications below 2000x (Figure 1.7.4
). An electron, indeed any particle, is neither a nor a . Describing the electron as a particle is a mathematical model that works well in some circumstances while describing it as a wave is a different mathematical model that works well in other circumstances. When you choose to do some calculation of the electron's behavior that treats it either as a particle or as a wave, you're not saying the electron a particle or a wave: you're just choosing the mathematical model that makes it easiest to do the calculation. Like all quantum particles, neutrons can also exhibit wave phenomena and if that wavelength is short enough, atoms or their nuclei can serve as diffraction obstacles. When a beam of neutrons emanating from a reactor is slowed down and selected properly by their speed, their wavelength lies near one angstrom (0.1 nanometer), the typical separation between atoms in a solid material. Such a beam can then be used to perform a diffraction experiment. Neutrons interact directly with the nucleus of the atom, and the contribution to the diffracted intensity depends on each isotope; for example, regular hydrogen and deuterium contribute differently. It is also often the case that light (low Z) atoms contribute strongly to the diffracted intensity even in the presence of large Z atoms. Neutrons have no electric charge, so they do not interact with the atomic electrons. Hence, they are very penetrating (e.g., typically 10 cm in lead). Neutron diffraction was proposed in 1934, to exploit de Broglie’s hypothesis about the wave nature of matter. Calculate the momentum and kinetic energy of a neutron whose wavelength is comparable to atomic spacing (\(1.8 \times 10^{-10}\, m\)). This is a simple use of de Broglie’s equation \[\lambda = \dfrac{h}{p} \nonumber \] where we recognize that the wavelength of the neutron must be comparable to atomic spacing (let's assumed equal for convenience, so \(\lambda = 1.8 \times 10^{-10}\, m\)). Rearranging the de Broglie wavelength relationship above to solve for momentum (\(p\)): \[\begin{align} p &= \dfrac{h}{\lambda} \nonumber \\[4pt] &= \dfrac{6.6 \times 10^{-34} J s}{1.8 \times 10^{-10} m} \nonumber \\[4pt] &= 3.7 \times 10^{-24}\, kg \,\,m\, \,s^{-1} \nonumber \end{align} \nonumber \] The relationship for kinetic energy is \[KE = \dfrac{1}{2} mv^2 = \dfrac{p^2}{2m} \nonumber \] where \(v\) is the velocity of the particle. From the reference table of physical constants, the mass of a neutron is \(1.6749273 \times 10^{−27}\, kg\), so \[\begin{align*} KE &= \dfrac{(3.7 \times 10^{-24}\, kg \,\,m\, \,s^{-1} )^2}{2 (1.6749273 \times 10^{−27}\, kg)} \\[4pt] &=4.0 \times 10^{-21} J \end{align*} \nonumber \] The neutrons released in nuclear fission are ‘fast’ neutrons, i.e. much more energetic than this. Their wavelengths be much smaller than atomic dimensions and will not be useful for neutron diffraction. We slow down these fast neutrons by introducing a "moderator", which is a material (e.g., graphite) that neutrons can penetrate, but will slow down appreciable. (Beams Professor, , ( ) | 7,931 | 1,356 |
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A classic example of an experiment that is employed in many physical chemistry laboratory courses uses a Dumas Bulb method to measure the dissociation of N O (g) as a function of temperature (Mack & France, 1934). In this experiment, a glass bulb is used to create a constant volume container in which a volatile substance can evaporate, or achieve equilibrium with other gases present. The latter is of interest in the case of the reaction \[N_2O_4(g) \rightleftharpoons 2 NO_2(g) \label{eq1} \] The reaction is endothermic, so at higher temperatures, a larger degree of dissociation is observed. The procedure is to first calibrate the internal volume of the Dumas bulb. This is done using a heavy gas (such as SF ) and comparing the mass of the bulb when evacuated to the mass of the bulb full of the calibrant gas at a particular pressure and temperature. The Dumas bulb is then charged with a pure sample of the gas to be investigated (such as N O ) and placed in a thermalized bath. It is then allowed to come to equilibrium. Once Equilibrium is established, the stopcock is opened to allow gas to escape until the internal pressure is set to the pressure of the room. The stopcock is then closed and the bulb weighed to determine the total mass of gas remaining inside. The experiment is repeated at higher and higher temperatures (so that at each subsequent measurement, the larger degree of dissociation creates more molecules of gas and an increase in pressure in the bulb (along with the higher temperature), which then leads to the expulsion of gas when the pressure is equilibrated. The degree of dissociation is then determined based on the calculated gas density at each temperature. \[ \alpha = \dfrac{\rho_1-\rho_2}{\rho_2(n-1)} \nonumber \] where \(\rho_1\) is the measured density and \(\rho_2\) is the theoretical density if no dissociation occurs (calculated from the ideal gas law for the given temperature, pressure, and molar mass of the dissociating gas) and \(n\) is the number of fragments into which the dissociating gas dissociates (ie.g., \(n = 2\) for Equation \req{eq1}). The equilibrium constant is then calculated as \[ K = \dfrac{4 \alpha^2}{1-\alpha^2} \left( \dfrac{p}{1.00 \,atm} \right) \nonumber \] Finally, a van’t Hoff plot is generated to determine the reaction enthalpy. | 2,327 | 1,357 |
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Effects of Leaving Group An S 1 reaction speeds up with a good . This is because the leaving group is involved in the rate-determining step. A good leaving group wants to leave so it breaks the C-Leaving Group bond faster. Once the bond breaks, the carbocation is formed and the faster the carbocation is formed, the faster the nucleophile can come in and the faster the reaction will be completed. A good leaving group is a weak base because weak bases can hold the charge. They're happy to leave with both electrons and in order for the leaving group to leave, it needs to be able to accept electrons. Strong bases, on the other hand, donate electrons which is why they can't be good leaving groups. As you go from left to right on the periodic table, electron donating ability decreases and thus ability to be a good leaving group increases. Halides are an example of a good leaving group whos leaving-group ability increases as you go down the column. The two reactions below is the same reaction done with two different leaving groups. One is significantly faster than the other. This is because the better leaving group leaves faster and thus the reaction can proceed faster. Other examples of good leaving groups are sulfur derivatives such as methyl sulfate ion and other sulfonate ions (See Figure below) Methyl Sulfate Ion Mesylate Ion Triflate Ion Tosylate Ion References Outside Links Problems 1. Put the following leaving groups in order of decreasing leaving group ability 2. Which solvent would an SN1 reaction occur faster in? H2O or CH3CN 3. What kind of conditions disfavor SN1 reactions? 4. What are the products of the following reaction and does it proceed via SN1 or SN2? 5. How could you change the reactants in the problem 4 to favor the other substitution reaction? 6. Indicate the expected product and list why it occurs through SN1 instead of SN2? Answers 1. 2. An SN1 reaction would occur faster in H2O because it's polar protic and would stailize the carbocation and CH3CN is polar aprotic. 3. Polar aprotic solvents, a weak leaving group and primary substrates disfavor SN1 reactions. 4. Reaction proceeds via SN1 because a tertiary carbocation was formed, the solvent is polar protic and Br- is a good leaving group. 5. You could change the solvent to something polar aprotic like CH3CN or DMSO and you could use a better base for a nucleophile such as NH2- or OH-. 6. This reaction occurs via SN1 because Cl- is a good leaving group and the solvent is polar protic. This is an example of a solvolysis reaction because the nucleophile is also the solvent. | 2,614 | 1,358 |
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An is a little more than just adiabatic. In the latter heat cannot get in or out. In an isolated system gets in or out, neither heat nor mass nor even any radiation, such as light. The isolated system is like a little universe all to itself. Let us consider a zero law process. We have two identical blocks of metal, say aluminum. They are each at thermal equilibrium, but at different temperatures. They are brought into contact with each other but isolated from the rest of the universe. so: \[dU_A= -dU_B \nonumber \] There is also no work so: \[dU_A = \delta q_A + 0 \nonumber \] Because U is a state function this makes \(q\) a state function as well, otherwise this equality does not hold. As there is only one term on the right there is only one path (along q). So we could write: \[dU_A = dq_A \nonumber \] This implies that we do not need to worry about reversible and irreversible paths as there is only one path. Since: \[dS = \dfrac{\delta q_{rev}}{T} \nonumber \] In this particular case: \[TdS = \delta q_{rev} = dU \nonumber \] Thus we get: \[ dS = \dfrac{dU_A}{T_A} + \dfrac{dU_B}{T_B} = \dfrac{dU_A}{T_A} - \dfrac{dU_A}{T_B} = dU_A \left( \dfrac{1}{T_A} - \dfrac{1}{T_B} \right) \nonumber \] Clearly as long as the two temperatures are not the same \(dS\) is not zero and entropy is not conserved. Instead it is . Over time, the temperatures will become the same (if the blocks are identical, the final temperature is the average of \(T_A\) and \(T_B\)) and the entropy will reach a maximum. For our two identical blocks of metal (with same heat capacity, \(C_V\)), we can, in fact, derive that the entropy change: \[\Delta S = C_V \ln \left[\dfrac{T_A^2 + T_B^2}{4T_AT_B}\right]. \nonumber \] This is indeed a positive quantity. In general, we can say for an isolated system: \[dS \ge 0 \nonumber \] Thus if we are dealing with a spontaneous (and isolated) process \(dS >0\) and . This gives us a . In an isolated system \(dS\) represents the produced entropy \(dS_{prod}\) and this is a good criterion for spontaneity. Of course the requirement that the system is isolated is very restrictive and makes the criterion as good as useless... What happens in a system that can exchange heat with the rest of the universe? We do have entropy changes in that case, but part of them may have nothing to do with production, because we also have to consider the heat that is exchanged. \[dS = dS_{prod} + dS_{exchange} \nonumber \] If the process is reversible (that is completely non-spontaneous) we are dealing with \(\delta q_{rev}\) so that \(dS_{exchange} = \delta q_{rev} /T\), but that is also what \(dS_{tot}\) is equal to (by definition). This leaves no room for entropy production. So we have: Isolated: \(dS = dS_{prod} + 0\) Reversible \(dS = 0 + \delta q_{rev}/T\) Notice that this demonstrates that for non-isolated systems entropy change is a good criterion for spontaneity at all... In the case the heat exchange is of the entropy is entropy production by the system: Irreversible: \(dS = dS_{prod} + \delta q_{irrev}/T\) \(dS > \delta q_{irrev}/T\) in this case. Generalizing the isolated, irreversible and reversible cases we may say: \[dS \ge \dfrac{δq}{T} \nonumber \] This is the . | 3,244 | 1,359 |
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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 | 1,360 |
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In the mid 1920's the German physicist Werner Heisenberg showed that if we try to locate an electron within a region \(Δx\); e.g. by scattering light from it, some momentum is transferred to the electron, and it is not possible to determine exactly how much momentum is transferred, even in principle. Heisenberg showed that consequently there is a relationship between the uncertainty in position \(Δx\) and the uncertainty in momentum \(Δp\). \[\Delta p \Delta x \ge \frac {\hbar}{2} \label {5-22}\] You can see from Equation \(\ref{5-22}\) that as \(Δp\) approaches 0, \(Δx\) must approach ∞, which is the case of the free particle discussed . This uncertainty principle, which also is discussed in , is a consequence of the wave property of matter. A wave has some finite extent in space and generally is not localized at a point. Consequently there usually is significant uncertainty in the position of a quantum particle in space. Activity 1 at the end of this chapter illustrates that a reduction in the spatial extent of a wavefunction to reduce the uncertainty in the position of a particle increases the uncertainty in the momentum of the particle. This illustration is based on the ideas described in the next section. Compare the minimum uncertainty in the positions of a baseball (mass = 140 gm) and an electron, each with a speed of 91.3 miles per hour, which is characteristic of a reasonable fastball, if the standard deviation in the measurement of the speed is 0.1 mile per hour. Also compare the wavelengths associated with these two particles. Identify the insights that you gain from these comparisons. | 1,637 | 1,361 |
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The electron-transfer reactions that occur within and between proteins typically involve prosthetic groups separated by distances that are often greater than 10 Å. When we consider these distant electron transfers, an explicit expression for the electronic factor is required. In the nonadiabatic limit, the rate constant for reaction between a donor and acceptor held at fixed distance and orientation is: \[k_{et} = \bigg[ \frac{H_{AB} \;^{2}}{\hbar}\left(\dfrac{\pi}{\lambda RT}\right)^{1/2} \bigg]^{\frac{-(\lambda + \Delta G^{o})^{2}}{4 \lambda RT}}\ldotp \tag{6.27}\] The electronic (or tunneling) matrix element H is a measure of the electronic coupling between the reactants and the products at the transition state. The magnitude of H depends upon donor-acceptor separation, orientation, and the nature of the intervening medium. Various approaches have been used to test the validity of Equation (6.27) and to extract the parameters H and \(\lambda\). Driving-force studies have proven to be a reliable approach, and such studies have been emphasized by many workers. In the nonadiabatic limit, the probability is quite low that reactants will cross over to products at the transition-state configuration. This probability depends upon the electronic hopping frequency (determined by H ) and upon the frequency of motion along the reaction coordinate. In simple models, the electronic-coupling strength is predicted to decay exponentially with increasing donor-acceptor separation (Equation 6.28): \[H_{AB} = (H_{AB}^{o})^{\frac{- \beta}{2} (\textbf{d} - \textbf{d}^{o})} \tag{6.28} \] In Equation (6.28), H ° is the electronic coupling at close contact ( °), and \(\beta\) is the rate of decay of coupling with distance ( ). Studies of the distance dependence of electron-transfer rates in donor-acceptor complexes, and of randomly oriented donors and acceptors in rigid matrices, have suggested 0.8 \(\leq \beta \leq\) 1.2 Å . Analysis of a large number of intramolecular electron-transfer rates has suggested a \(\beta\) value of 1.4 Å for protein reactions (Figure 6.24). Assigning a single protein \(\beta\) implies that the intervening medium is homogenous. At best this is a rough approximation, because the medium separating two redox sites in a protein is a heterogenous array of bonded and nonbonded interactions. Beratan and Onuchic have developed a formalism that describes the medium in terms of "unit blocks" connected together to form a tunneling pathway. A unit block may be a covalent bond, a hydrogen bond, or a through-space jump, each with a corresponding decay factor. Dominant tunneling pathways in proteins are largely composed of bonded groups (e.g., peptide bonds), with less favorable through-space interactions becoming important when a through-bond pathway is prohibitively long (Figure 6.25). The tunneling pathway model has been used successfully in an analysis of the electron-transfer rates in modified cytochromes c (Section IV.D.1). Plastocyanin cycles between the Cu and Cu oxidation states, and transfers electrons from cytochrome f to the P component of photosystem I in the chloroplasts of higher plants and algae. The low molecular weight (10.5 kDa) and availability of detailed structural information have made this protein an attractive candidate for mechanistic studies, which, when taken together, point to two distinct surface binding sites (i.e., regions on the plastocyanin molecular surface at which electron transfer with a redox partner occurs). The first of these, the solvent-exposed edge of the Cu ligand His-87 (the adjacent site A in Figure 6.26), is ~6 Å from the copper atom and rather nonpolar. The second site (the remote site R in Figure 6.26) surrounds Tyr-83, and is much farther (~15 Å) from the copper center. Negatively charged carboxylates at positions 42-45 and 59-61 make this latter site an attractive one for positively charged redox reagents. Bimolecular electron-transfer reactions are typically run under pseudo-first-order conditions (e.g., with an inorganic redox reagent present in ~15-fold excess): \[Rate = k[plastocyanin,complex] = k_{obs}[plastocyanin] \ldotp \tag{6.29}\] For some reactions [e.g., Co(phen) oxidation of plastocyanin (Cu )] the expected linear plot of k vs. [complex] is not observed. Instead, the rate is observed to saturate (Figure 6.27). A "minimal" model used to explain this behavior involves the two pathways for electron transfer shown in Equation (6.30). \(\tag{6.30}\) Surprisingly, the rate ratio k /k is 7. Calculations indicate that, despite the significant differences in distances, H for the remote site is ~15 percent of H for the adjacent site. This figure is much higher than would be expected from distance alone, suggesting that the value of the decay parameter \(\beta\) in Equation (6.28) depends strongly on the structure of the intervening medium. Chemical modification of structurally characterized metalloproteins by transition-metal redox reagents has been employed to investigate the factors that control long-range electron-transfer reactions. In these semisynthetic multisite redox systems, the distance is fixed, and tunneling pathways between the donor and acceptor sites can be examined. Sperm-whale myoglobin can be reacted with (NH ) Ru(OH ) and then oxidized to produce a variety of ruthenated products, including a His-48 derivative whose Ru \(\leftrightarrow\) Fe tunneling pathway is depicted in Figure 6.28. Electrochemical data (Table 6.5) indicate that the (NH ) Ru group does not significantly perturb the heme center, and that equilibrium (i.e., k = k + k ) should be approached when a mixed-valent intermediate is produced by flash-photolysis techniques: \[(NH_{3})_{5}Ru^{3+}-Mb(Fe^{3+}) \xrightarrow[e^{-}]{fast} (NH_{3})_{5}Ru^{2+}-Mb(Fe^{3+}) \xrightleftharpoons[k_{-1}]{k_{1}} (NH_{3})_{5}Ru^{3+}-Mb(Fe^{2+}) \tag{6.31}\] This kinetic behavior was observed, and both the forward (k ) and reverse (k ) reactions were found to be markedly temperature-dependent: k = 0.019 s (25 °C), \(\Delta\)H = 7.4 kcal/mol, k = 0.041 s ) (25°C), \(\Delta\)H = 19.5 kcal/mol. X-ray crystallographic studies indicate that the axial water ligand dissociates upon reduction of the protein. This conformational change does not control the rates, since identical results were obtained when a second flash-photolysis technique was used to generate (NH ) Ru -Mb(Fe ) in order to approach the equilibrium from the other direction. Fe (NH ) Ru Cyanogen bromide has been used to modify the six-coordinate metmyoglobin heme site, causing the coordinated water ligand to dissociate. The CNBr-modified myoglobin heme site is thus five-coordinate in oxidation states. As expected, the self-exchange rate increased from ~1 M s to ~10 M s . Recent efforts in modeling biological electron transfers using chemically modified redox proteins point the way toward the design of semisynthetic redox enzymes for catalytic applications. An intriguing example, termed flavohemoglobin, was produced by reaction of hemoglobin with a flavin reagent designed to react with Cys-93 of the \(\beta\)-chain (i.e., the hemoglobin molecule was modified by two flavin moieties). The resulting derivative, unlike native hemoglobin, accepts electrons directly from NADPH and catalyzes the -hydroxylation of aniline in the presence of O and NADPH. In physiologically relevant precursor complexes, both redox centers are frequently buried in protein matrices. Characterization of such protein-protein complexes is clearly important, and several issues figure prominently: Most of our knowledge about the structures of protein-protein complexes comes from crystallographic studies of antigen-antibody complexes and multisubunit proteins; such systems generally exhibit a high degree of thermodynamic stability. On the other hand, complexes formed as a result of bimolecular collisions generally are much less stable, and tend to resist attempts to grow x-ray-quality crystals; the high salt conditions typically used in protein crystallizations often lead to dissociation of such complexes. One of the most widely studied protein-protein complexes is that formed between mammalian cytochrome b and cytochrome c. Using the known x-ray structures of both proteins, Salemme generated a static computer graphics model of this electron-transfer complex by docking the x-ray structures of the individual proteins. Two features of this model and its revision by molecular dynamics simulations (Figure 6.29 See color plate section, page C-12.) are noteworthy: (1) several Lys residues on cytochrome c and carboxylate-containing groups on cytochrome b form "salt bridges" (i.e., intermolecular hydrogen bonds); and (2) the hemes are nearly coplanar and are ~17 Å (Fe-Fe) apart. This distance was confirmed by an energy-transfer experiment in which the fluorescence of Zn-substituted cytochrome c was quenched by cytochrome b . Spectroscopic studies have verified the suggestion that these proteins form a 1:1 complex at low ionic strength (Figure 6.30). In addition, chemical modification and spectroscopic analyses are all in agreement with the suggestion that the complex is primarily stabilized by electrostatic interactions of the (-NH ••• O C—) type. The effect of ionic strength on the reduction of cytochrome c by cytochrome b is also in accord with this picture: lowering the ionic strength increases the reaction rate, as expected for oppositely charged molecules. A common experimental strategy for studying electron transfers proteins uses a metal-substituted heme protein as one of the reactants. In particular, the substitution of zinc for iron in one of the porphyrin redox centers allows facile initiation of electron transfer through photoexcitation of the zinc porphyrin (ZnP). The excited zinc porphyrin, ZnP* in Equation (6.32), may decay back (k ~ 10 s ) to the ground state or transfer an electron to an acceptor. \(\tag{6.32}\) The ZnP cation radical produced in the k step is a powerful oxidant; back electron transfer (k ) will thus occur and regenerate the starting material. The reactions shown in Equation (6.32) have been investigated in mixedmetal [Zn, Fe] hemoglobins. A hemoglobin molecule can be viewed as two independent electron-transfer complexes, each consisting of an \(\alpha_{1}\)-\(\beta_{2}\) subunit pair (Figure 6.31), since the \(\alpha_{1}\)-\(\alpha_{2}\), (\beta_{1}\)-(\beta_{2}\), and \(\alpha_{1}\)-(\beta_{1}\) distances are prohibitively long (> 30 Å). Both [\(\alpha\)(Zn), \(\beta\)(Fe)] and [\(\alpha\)(Fe), \(\beta\)(Zn)] hybrids have been studied. The ZnP and FeP are nearly parallel, as in the cytochrome b -cytochrome c model complex. Long-range electron transfer ( ZnP* → Fe ) between the \(\alpha_{1}\) and \(\beta_{2}\) subunits has been observed (the heme-edge/heme-edge distance is ~20 Å). The driving force for the forward electron-transfer step is ~0.8 eV, and k (see Equation 6.32) is ~100 s at room temperature, but decreases to ~9 s in the low-temperature region (Figure 6.32). Below 140-160 K the vibrations that induce electron transfer "freeze out"; nuclear tunneling is usually associated with such slow, temperature-independent rates. A complete analysis of the full temperature dependence of the rate requires a quantum-mechanical treatment of \(\lambda_{i}\) rather than that employed in the Marcus theory. It is interesting to note that the heme b vinyl groups (see Figure 6.6) for a given [\(\alpha_{1}\)(Fe), \(\beta_{2}\)Zn)] hybrid point toward each other and appear to facilitate electron transfer. | 11,639 | 1,362 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Radioactivity/Artificially_Induced_Radioactivity |
Radioactivity is the process by which the nucleus of an unstable atom loses energy by emitting radiation, including alpha particles, beta particles, gamma rays and conversion electrons Although is observed as a natural occurring process, it can also be artificially induced typically via the bombarding atoms of a specific element by radiating particles, thus creating new atoms. Ernest Rutherford was a prominent New Zealand scientist, and a winner of the Nobel Prize in chemistry in 1908. Amongst his vast list of discoveries, Rutherford was also the first to discover artificially induced radioactivity. Through the bombardment of alpha particles against nuclei of \(\ce{^{14}N}\) with 7 protons/electrons, Rutherford produced \(\ce{^{17}O}\) (8 protons/electrons) and protons (Figure \(\Page {1}\)). Through this observation, Rutherford concluded that atoms of one specific element can be made into atoms of another element. If the resulting element is radioactive, then this process is called Rutherford was the first researcher to create protons outside of the atomic nuclei and the \(\ce{^{17}O}\) isotope of oxygen, which is nonradioactive. Similarly, other nuclei when bombarded with alpha particles will generate new elements (Figure \(\Page {2}\)) that may be radioactive and decay naturally or that may be stable and persist like \(\ce{^{17}O}\). Before this discovery of artificial induction of radioactivity, it was a common belief that atoms of matter are unchangeable and indivisible. After the very first discoveries made by Ernest Rutherford, Irene Joliot-Curie and her husband, Frederic Joliot, a new point of view was developed. The point of view that although atoms appear to be stable, they can be transformed into new atoms with different chemical properties. Today over one thousand artificially created radioactive nuclides exist, which considerably outnumber the nonradioactive ones created. Irene Joliet-Curie and her husband Frédéric both were French scientists who shared winning the Nobel Prize award in chemistry in 1935 for artificially synthesizing a radioactive isotope of phosphorus by bombarding aluminum with alpha particles. \(\ce{^{30}P}\) with 15 protons was the first radioactive nuclide obtained through this method of artificially inducing radioactivity. \[ \ce{^27_13Al + ^4_2He \rightarrow ^30_15P + ^1_0n}\] \[ \ce{^30_15P \rightarrow ^30_14Si + ^0_{-1}\beta}\] (or radioactivation) involves making a radioactive isotope by neutron capture, e.g. the addition of a neutron to a nuclide resulting in an increase in isotope number by 1 while retaining the same atomic number (Figure \(\Page {3}\)). Activation is often an inadvertent process occurring inside or near a nuclear reactor, where there are many neutrons flying around. For example, Coba in or near a nuclear reactor will capture a neutron forming the radioactive isotope Co-60. \[\ce{ ^1_0n + ^{59}Co \rightarrow ^{60}Co }\] The \(\ce{ ^{60}Co}\) isot o \(\ce{ ^{60}Ni }\) via n Figure \(\Page {4}\). Write a nuclear equation for the creation of Mn through the bombardment of Co with neutrons. A unknown particle is produced with Mn, in order to find the mass number (A) of the unknown we must subtract the mass number of the Manganese atom from the mass number of the Cobalt atom plus the neutron being thrown. In simpler terms, Now, by referring to a periodic table to find the atomic numbers of Mn and Co, and then subtracting the atomic number of Mn from Co, we will receive the atomic number of the unknown particle Thus, the unknown particle has A = 4, and Z = 2, which would make it a Helium particle, and the nuclear formula would be as follows: \[ \ce{^{50}_{27}Co + ^1_0n \rightarrow ^{56}_{25}Mn + ^{4}_{2}\alpha } \nonumber\] Write a nuclear equation for the production of \(\ce{^{147}Eu}\) by bombardment of \(\ce{^{139}La}\) with \(\ce{^{12}Ca}\). Thus, the mass number of the unknown particle is 4. Again by referring to a periodic table and finding the atomic numbers of Lanthanum, Carbon and Europium, we are able to calculate the atomic number of the unknown particle, The atomic number for the unknown particle equals to zero, therefore 4 neutrons are emitted, and the nuclear equation is written as follows: \[ \ce{^{139}_{57}La + ^{12}_6C \rightarrow ^{147}_{63}Eu + 4 ^{0}_{1}n } \nonumber\] Induced radioactivity occurs when a previously stable material has been made radioactive by exposure to specific radiation. Most radioactivity does not induce other material to become radioactive. This Induced radioactivity was discovered by Irène Curie and F. Joliot in 1934. This is also known as man-made radioactivity. The phenomenon by which even light elements are made radioactive by artificial or induced methods is called artificial radioactivity. | 4,800 | 1,363 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Radioactivity/The_Effects_of_Radiation_on_Matter |
All radioactive particles and waves, from the entire electromagnetic spectrum, to alpha, beta, and gamma particles, possess the ability to eject electrons from atoms and molecules to create ions. There are many types of , but the two most common are and ionizing radiation. Ionizing radiation refers to radioactive particles, such as alpha and beta particles, or electromagnetic waves, such as gamma or ultraviolet rays, which have sufficient energy to detach electrons off of atoms to create ions, hence the name “ionizing radiation.” Electromagnetic radiation, which sometimes can be placed as a subcategory of ionizing radiation, deals with waves or photons from the electromagnetic spectrum. Unlike ionizing radiation, electromagnetic radiation deals with electric and magnetic field oscillations such as with X-rays, radio waves, or gamma rays. Radioactive decay of atoms creates three radioactive particles, alpha, beta, and gamma. Of the three, alpha particles are known to have the most “ionizing power,” a term describing the number of ion pairs produced per centimeter through a material, followed by beta, then gamma. However, a common misconception is that the higher ionizing power a particle has, the more damaging it is to matter. Electromagnetic waves can also ionize, hence the reason electromagnetic radiation is often placed as part of ionizing radiation. The main effect radiation has on matter is its ability to ionize atoms to become ions, a phenomenon known as , which is very similar to the . Radioactive particles or electromagnetic waves with sufficient energy collide with electrons on the atom to knock electrons off the atom. The electron ejected off the atom is called the primary electron. When the primary electrons hold energy, a particle ejecting the primary electron may cause it to eject another electron, either on their own atom or on another atom. This is known as secondary ionization. However, ionization does not have to completely eject an electron off the atom. It can raise the energy of the electron instead, raising the electron energy to a higher energy state. When the electron reverts to its normal energy level, it emits energy in the form of radiation, usually in the forms of ultraviolet rays or radio waves. Radiation can be both natural and synthetic. utilizes primary and secondary ionizations in order to emit X-rays. Most X-ray emission is due to the bombardment of electrons on a metal target. If the electrons have sufficient energy, the inner shell electrons of the atom fall out, and higher-leveled electrons fill in the hole left by the previous electrons. By doing so, packets of energy are released in the forms of X-ray photons. Other forms of ionizing radiation can produce UV and gamma rays in a similar manner. This type of radiation is known as “ionizing radiation.”
All charged particles and rays have the ability to be radioactive; however, not all rays and particles have the energy per photon to ionize atoms. This is known as “non-ionizing radiation.” Non-ionizing radiation has enough energy to excite electrons to move to a higher state, releasing photons of electromagnetic radiation such as visible light, near ultraviolet, and microwaves. Radio waves, microwaves, and neutron radiation (an important application in ) all fall under non-ionizing radiation, as their respective energies are too low to ionize atoms. Prolonged exposure to radiation often has detrimental effects on living matter. This is due to radiation’s ionizing ability, which can damage the internal functioning of cells. Radiation either ionizes or excites atoms or molecules in living cells, leading to the dissociation of molecules within an organism. The most destructive effect radiation has on living matter is ionizing radiation on DNA. Damage to DNA can cause cellular death, mutagenesis (the process by which genetic information is modified by radiation or chemicals), and genetic transformation. Effects from exposure to radiation include leukemia, birth defects, and many forms of cancer. Most external radiation is absorbed by the environment; for example, most ultraviolet radiation is absorbed by the ozone layer, preventing deadly levels of ultraviolet radiation to come in contact with the surface of the earth. Sunburn is an effect of UV radiation damaging skin cells, and prolonged exposure to UV radiation can cause genetic information in skin cells to mutate, leading to skin cancer. Alpha, beta, and gamma rays also cause damage to living matter, in varying degrees. Alpha particles have a very small absorption range, and thus are usually not harmful to life, unless ingested, due to its high ionizing power. Beta particles are also damaging to DNA, and therefore are often used in radiation therapy to mutate and kill cancer cells. Gamma rays are often considered the most dangerous type of radiation to living matter. Unlike alpha and beta particles, which are charged particles, gamma rays are instead forms of energy. They have large penetrating range and can diffuse through many cells before dissipating, causing widespread damage such as radiation sickness. Because gamma rays have such high penetrating power and can damage living cells to a great extent, they are often used in irradiation, a process used to kill living organisms. There are several methods to measure radiation; hence, there are several radiation units based on different radiation factors. Radiation units can measure radioactive decay, absorbed dosage, and human absorbed doses. Bq and Ci measure radioactive decay, while Gy and Rad measures absorbed doses. Sv and Rem measure absorbed doses in Gy and Rad equivalents. Rem takes into account different radiation types and the speed of particles. Below is a chart to help organize the different units: Measured as amount of decay at the same rate as 1 gram of radium 1 Ci = 3.70∙10 Bq 1 rem = 1 rad∙Q Q = 1 for X-rays, gamma rays, and beta particles Q = 3 for slow neutrons Q = 10 for protons and fast neutrons Q = 20 for alpha particles The most commonly used unit is the "rad," which stands for "radiation absorbed dose," and the "rem," which stands for "radiation equivalent for man." One rad corresponds to the absorption of 0.01 Joules of energy per kilogram of matter. Rem is the rad multiplied by the relative biological effectiveness, which is most often expressed as the variable "Q." The factor Q is used to take into account the different effects caused by different radiation. | 6,517 | 1,364 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/12%3A_Equilibrium_Conditions_in_Multicomponent_Systems/12.06%3A_Liquid-Liquid_Equilibria |
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\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) When two different pure liquids are unable to mix in all proportions, they are said to be . When these liquids are placed in contact with one another and allowed to come to thermal, mechanical, and transfer equilibrium, the result is two coexisting liquid mixtures of different compositions. Liquids are never actually completely . To take an extreme case, liquid mercury, when equilibrated with water, has some H\(_2\)O dissolved in it, and some mercury dissolves in the water, although the amounts may be too small to measure. The Gibbs phase rule for a multicomponent system to be described in Sec. 13.1 shows that a two-component, two-phase system at equilibrium has only two independent intensive variables. Thus at a given temperature and pressure, the mole fraction compositions of both phases are fixed; the compositions depend only on the identity of the substances and the temperature and pressure. Figure 13.5 shows a phase diagram for a typical binary liquid mixture that spontaneously separates into two phases when the temperature is lowered. The thermodynamic conditions for phase separation of this kind were discussed in Sec. 11.1.6. The phase separation is usually the result of positive deviations from Raoult’s law. Typically, when phase separation occurs, one of the substances is polar and the other nonpolar. Suppose substances A and B are both liquids when pure. In discussing the solubility of liquid B in liquid A, we can treat B as either a solute or as a constituent of a liquid mixture. The difference lies in the choice of the standard state or reference state of B. We can define the solubility of B in A as the maximum amount of B that can dissolve without phase separation in a given amount of A at the given temperature and pressure. Treating B as a solute, we can express its solubility as the mole fraction of B in the phase at the point of phase separation. The addition of any more B to the system will result in two coexisting liquid phases of fixed composition, one of which will have mole fraction \(x\B\) equal to its solubility. Experimentally, the solubility of B in A can be determined from the , the point during titration of A with B at which persistent turbidity is observed. Consider a system with two coexisting liquid phases \(\pha\) and \(\phb\) containing components A and B. Let \(\pha\) be the A-rich phase and \(\phb\) be the B-rich phase. For example, A could be water and B could be benzene, a hydrophobic substance. Phase \(\pha\) would then be an aqueous phase polluted with a low concentration of dissolved benzene, and phase \(\phb\) would be wet benzene. \(x\B\aph\) would be the solubility of the benzene in water, expressed as a mole fraction. Below, relations are derived for this kind of system using both choices of standard state or reference state. Assume that the two components have low mutual solubilities, so that B has a low mole fraction in phase \(\pha\) and a mole fraction close to 1 in phase \(\phb\). It is then appropriate to treat B as a solute in phase \(\pha\) and as a constituent of a liquid mixture in phase \(\phb\). The value of \(x\B\aph\) is the solubility of liquid B in liquid A. The equilibrium when two liquid phases are present is B(\(\phb\))\(\arrows\)B(\(\pha\)), and the expression for the thermodynamic equilibrium constant, with the solute standard state based on mole fraction, is \begin{equation} K = \frac{a\xbB\aph}{a\B\bph} = \frac{\G\xbB\aph \g\xbB\aph x\B\aph}{\G\B\bph \g\B\bph x\B\bph} \tag{12.6.1} \end{equation} The solubility of B is then given by \begin{equation} x\B\aph = \frac{\G\B\bph\g\B\bph x\B\bph}{\G\xbB\aph\g\xbB\aph}K \tag{12.6.2} \end{equation} The values of the pressure factors and activity coefficients are all close to \(1\), so that the solubility of B in A is given by \(x\B\aph \approx K\). The temperature dependence of the solubility is given by \begin{equation} \frac{\dif\ln x\B\aph}{\dif T} \approx \frac{\dif\ln K}{\dif T} = \frac{\Delsub{sol,B}H\st}{RT^2} \tag{12.6.3} \end{equation} where \(\Delsub{sol,B}H\st\) is the molar enthalpy change for the transfer at pressure \(p\st\) of pure liquid solute to the solution at infinite dilution. H\(_2\)O and -butylbenzene are two liquids with very small mutual solubilities. Figure 12.8 shows that the solubility of -butylbenzene in water exhibits a minimum at about \(12\units{\(\degC\)}\). Equation 12.6.3 allows us to deduce from this behavior that \(\Delsub{sol,B}H\st\) is negative below this temperature, and positive above. The condition for transfer equilibrium of component B is \(\mu\B\aph=\mu\B\bph\). If we use a pure-liquid reference state for B in both phases, this condition becomes \begin{equation} \mu\B^* + RT\ln(\g\B\aph x\B\aph) = \mu\B^* + RT\ln(\g\B\bph x\B\bph) \tag{12.6.4} \end{equation} This results in the following relation between the compositions and activity coefficients: \begin{equation} \g\B\aph x\B\aph = \g\B\bph x\B\bph \tag{12.6.5} \end{equation} As before, we assume the two components have low mutual solubilities, so that the B-rich phase is almost pure liquid B. Then \(x\B\bph\) is only slightly less than \(1\), \(\g\B\bph\) is close to \(1\), and Eq. 12.6.5 becomes \(x\B\aph \approx 1/\g\B\aph\). Since \(x\B\aph\) is much less than \(1\), \(\g\B\aph\) must be much greater than \(1\). In environmental chemistry it is common to use a pure-liquid reference state for a nonpolar liquid solute that has very low solubility in water, so that the aqueous solution is essentially at infinite dilution. Let the nonpolar solute be component B, and let the aqueous phase that is equilibrated with liquid B be phase \(\pha\). The activity coefficient \(\g\B\aph\) is then a or . As explained above, the aqueous solubility of B in this case is given by \(x\B\aph \approx 1/\g\B\aph\), and \(\g\B\aph\) is much greater than \(1\). We can also relate the solubility of B to its Henry’s law constant \(\kHB\aph\). Suppose the two liquid phases are equilibrated not only with one another but also with a gas phase. Since B is equilibrated between phase \(\pha\) and the gas, we have \(\g\xbB\aph=\fug\B/\kHB\aph x\B\aph\) (Table 9.4). From the equilibration of B between phase \(\phb\) and the gas, we also have \(\g\B\bph=\fug\B/x\B\bph \fug\B^*\). By eliminating the fugacity \(\fug\B\) from these relations, we obtain the general relation \begin{equation} x\B\aph = \frac{\g\B\bph x\B\bph \fug\B^*}{\g\xbB\aph \kHB\aph} \tag{12.6.6} \end{equation} If we assume as before that the activity coefficients and \(x\B\bph\) are close to 1, and that the gas phase behaves ideally, the solubility of B is given by \(x\B\aph \approx p\B^*/\kHB\aph\), where \(p\B^*\) is the vapor pressure of the pure solute. Consider a two-component system of two equilibrated liquid phases, \(\pha\) and \(\phb\). If we add a small quantity of a third component, C, it will distribute itself between the two phases. It is appropriate to treat C as a solute in phases. The thermodynamic equilibrium constant for the equilibrium \(\tx{C}(\phb) \arrows \tx{C}(\pha)\), with solute standard states based on mole fraction, is \begin{equation} K = \frac{a\xbC\aph}{a\xbC\bph} = \frac{\G\xbC\aph \g\xbC\aph x\C\aph}{\G\xbC\bph \g\xbC\bph x\C\bph} \tag{12.6.7} \end{equation} We define \(K'\) as the ratio of the mole fractions of C in the two phases at equilibrium: \begin{equation} K' \defn \frac{x\C\aph}{x\C\bph} = \frac{\G\xbC\bph \g\xbC\bph}{\G\xbC\aph \g\xbC\aph}K \tag{12.6.8} \end{equation} At a fixed \(T\) and \(p\), the pressure factors and equilibrium constant are constants. If \(x\C\) is low enough in both phases for \(\g\xbC\aph\) and \(\g\xbC\bph\) to be close to unity, \(K'\) becomes a constant for the given \(T\) and \(p\). The constancy of \(K'\) over a range of dilute composition is the . Since solute molality and concentration are proportional to mole fraction in dilute solutions, the ratios \(m\C\aph/m\C\bph\) and \(c\C\aph/c\C\bph\) also approach constant values at a given \(T\) and \(p\). The ratio of concentrations is called the or . In the limit of infinite dilution of C, the two phases have the compositions that exist when only components A and B are present. As C is added and \(x\C\aph\) and \(x\C\bph\) increase beyond the region of dilute solution behavior, the ratios \(x\B\aph/x\A\aph\) and \(x\B\bph/x\A\bph\) may change. Continued addition of C may increase the mutual solubilities of A and B, resulting, when enough C has been added, in a single liquid phase containing all three components. It is easier to understand this behavior with the help of a ternary phase diagram such as Fig. 13.17. | 16,076 | 1,366 |
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Cyclopropanecarboxylic acid and 1,4-cyclohexane-di-carboxylic acid have the same empirical formula (\(\ce{C2H3O}\)), but different molecular formulas (\(\ce{C4H6O2}\) for cyclopropanecarboxylic acid; \(\ce{C8H12O4}\) for 1,4 cyclohexane-di-carboxylic acid); the former has a three-ring aliphatic cyclic chain while the later has a six-ring aliphatic cyclic chain. The molar mass of 1,4-cyclohexane-di-carboxylic acid is twice as much as cyclopropanecarboxylic acid. The structures of these two compounds are shown below: If the heat of formation and combustion of 1,4, cyclohexane-di-carboxylic acid differ from twice as those of cyclopropanecarboxylic acid, the difference provides information regarding the structural stabilities of these two compounds. Heats of combustion of cyclopropanecarboxylic acid and 1,4 cyclohexane-di-carboxylic acid were obtained using an adiabatic bomb calorimeter. The calorimeter is initially calibrated with a known amount of benzoic acid having a known amount of heat of combustion, \(Q_m = 26414 ± 3\, J/g\). The temperature rise is measured, and the calorimeter heat capacity, \(C\), is obtained from: \[ C = \dfrac{Q_{tot}}{\Delta T} \label{1}\] Both the heat of combustion of benzoic acid and the heat released by the combustion of the nichrome wire contribute to \(Q_{tot}\), and therefore the heat due to combustion of the wire must be subtracted from \(Q_{tot}\) to obtain accurate results for the compound under investigation. Thus, we have: \[ Q_{tot} = Q_m m + q_1 \label{2}\] \(q_1\) is obtained by measuring the change in length of the nichrome fuse after ignition and multiplying this value by the conversion factor. For the No.34-gauge nichrome wire used in this experiment, this value is 9.6 J/cm; hence \[q_1 = 9.6 (L_o-L) \label{3}\] where \(L_o\) and \(L\) are the initial and final lengths of the nichrome fuse, respectively, in centimeters, Once the calibration was carried out, two trials were conducted for each compound in question. In the case of the 1,4-cyclohexane-dicarboxylic acid, a pellet of approximately 1 gram mass was created and the same procedure used to calibrate the bomb was followed again to measure the temperature change associated with the combustion of the compound. For the cyclopropanecarboxylic acid trials a benzoic acid pellet of 0.7 - 0.8 gram mass was made, and liquid cyclopropanecarboxylic acid was pipette onto it until the mass of the pellet and solution was approximately 1 gram. The following data were obtained in the physical chemistry laboratory of UW-Green Bay. | 2,573 | 1,367 |
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This page gives you the facts and simple, uncluttered mechanisms for the nucleophilic substitution reactions between halogenoalkanes and water. There is only a slow reaction between a primary halogenoalkane and water even if they are heated. The halogen atom is replaced by -OH. For example, using 1-bromoethane as a typical primary halogenoalkane: An alcohol is produced together with hydrobromic acid. Be careful not to call this hydrogen bromide. Hydrogen bromide is a gas. When it is dissolved it in water (as it will be here), it's called hydrobromic acid. The mechanism involves two steps. The first is a simple nucleophilic substitution reaction: Because the mechanism involves collision between two species in this slow step of the reaction, it is known as an S 2 reaction. The nucleophilic substitution is very slow because water isn't a very good nucleophile. It lacks the full negative charge of, say, a hydroxide ion. The second step of the reaction simply tidies up the product. A water molecule removes one of the hydrogens attached to the oxygen to give an alcohol and a hydroxonium ion (also known as a hydronium ion or an oxonium ion). The hydroxonium ion and the bromide ion (from the nucleophilic substitution stage of the reaction) make up the hydrobromic acid which is formed as well as the alcohol. If the halogenoalkane is heated under reflux with water, the halogen is replaced by -OH to give an alcohol. Heating under reflux means heating with a condenser placed vertically in the flask to prevent loss of volatile substances from the mixture. The reaction happens much faster than the corresponding one involving a primary halogenoalkane. For example: This mechanism involves an initial ionisation of the halogenoalkane: followed by a very rapid attack by the water on the carbocation (carbonium ion) formed: This is again an example of nucleophilic substitution. This time the slow step of the reaction only involves one species - the halogenoalkane. It is known as an S 1 reaction. Now there is a final stage in which the product is tidied up. A water molecule removes one of the hydrogens attached to the oxygen to give an alcohol and a hydroxonium ion - exactly as happens with primary halogenoalkanes. The rate of the overall reaction is governed entirely by how fast the halogenoalkane ionises. The fact that water isn't as good a nucleophile as, say, OH doesn't make any difference. The water isn't involved in the slow step of the reaction. Secondary halogenoalkanes use an S 2 mechanism and an S 1. Make sure you understand what happens with primary and tertiary halogenoalkanes, and then adapt it for secondary ones should ever need to. | 2,692 | 1,369 |
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As noted earlier, three solutions to the problem of dioxygen transport have evolved: hemoglobin (Hb), hemocyanin (Hc), and hemerythrin (Hr). Their remarkable distribution over plant and animal kingdoms is shown in Figure 4.8. The hemoglobins and myoglobins found in plants, snails, and vertebrates all appear to share a common, very ancient ancestor. There is some evidence now for a common ancestral hemocyanin. The appearance of hemerythrin in a few annelid worms is an evolutionary curiosity. These few words and the diagram will suffice to give some hints about how respiratory proteins evolved, a subject that is outside the scope of this book. Hemoglobins are the most evolutionarily diverse family of dioxygen carriers. They are found in some plants (e.g., leghemoglobin in the nitrogen-fixing nodules of legumes), many invertebrates (including some insect larvae), crustaceans, molluscs (especially bivalves and snails), almost all annelid worms, and in vertebrates with one possible exception, the Antarctic fish . With few exceptions the monomeric and oligomeric hemoglobins all share a basically similar building block: a single heme group is embedded in a folded polypeptide with a molecular weight of about 16 kDa (see Figure 4.2), and is anchored to the protein by coordination of the iron center to an imidazole ligand from a histidine residue. Mammalian myoglobin is often taken as the archetypical myoglobin (see Table 4.1). Sperm whale, bovine, or equine myoglobin are specific examples; the muscle tissue from which they may be extracted is more available than that from . The archetypical oligomeric hemoglobin that shows cooperative binding of O is the tetrameric hemoglobin A. It is readily available from the blood of human donors.* In some invertebrate hemoglobins, especially those of annelids, aggregates may contain as many as 192 binding sites, to give a molecular weight of about 3 x 10 Dalton. These and other high-molecular-weight hemoglobins of arthropods are often referred to as erythrocruorins (Er). In a few annelid worms, the otherwise ubiquitous heme b or protoheme is replaced by chloroheme (see Figure 4.2) to give chlorocruorins (Ch), which tum green upon oxygenation ( , Greek for green). Some organisms, for example the clam , feature a dimeric hemoglobin. The only known anomalous hemoglobin is Hb , which comes from a parasitic nematode found in the guts of pigs. It has a molecular weight of about 39 kDa per heme; this value is not a multiple of the myoglobin building block. Moreover, presumably in response to the low availability of O in pigs' guts, Hb has an extraordinarily high affinity for dioxygen, in large part owing to an extremely slow rate of dioxygen release. Leghemoglobin is another carrier with a high affinity for dioxygen, in this case because of a high rate of O binding. Since O is a poison for the nitrogenase enzyme, yet the nodules also require dioxygen, diffusion of O is facilitated, but the concentration of free dioxygen in the vicinity of nitrogen-fixing sites is minimized. Kinetic and thermodynamic data for dioxygen binding and release from a variety of hemoglobins are summarized in Table 4.2. Notice that for the hemoglobin tetramer, which comprises two pairs of slightly dissimilar subunits, the \(\alpha\) and \(\beta\) chains bind O with significantly different affinities and rate constants, especially in the T state. Isolated chains behave like monomeric vertebrate hemoglobins, such as whale myoglobin, which have affinities close to those of R-state hemoglobin. The chlorocruorins have a low affinity compared to other erythrocruorins. Especially for proteins that bind O cooperatively, a range of values is specified, since affinities and rates are sensitive to pH, ionic strength, specific anions and cations (allosteric effectors), and laboratory. For example, as we noted above, the O affinity of hemoglobin A is sensitive to the concentration of 2,3-DPG and to pH (Bohr effect). Trout hemoglobin I is insensitive to these species, whereas a second component of trout blood, trout hemoglobin IV, is so sensitive to pH (Root effect) that at pH < 7 trout hemoglobin IV is only partially saturated at P(O ) = 160 Torr. Note that O affinities span five orders of magnitude. Since heme catabolism produces carbon monoxide, and since in some environments CO is readily available exogenously, selected data for CO binding are also presented. * Blood from human donors is also a source for a variety of abnormal hemoglobins, the most famous of which is HbS, the hemoglobin giving rise to sickle-cell anemia, It was Pauling and coworkers who first found that HbS differs from HbA through the substitution of valine for glutamic acid in each of two of the four subunits comprising Hb, Sickle-cell anemia was the first condition to be denoted a "molecular disease." Hemocyanins (Hc), the copper-containing dioxygen carriers, are distributed erratically in two large phyla, (for example, octopi and snails) and (for example, lobsters and scorpions). The functional form of hemocyanin consists of large assemblies of subunits. In the mollusc family the subunit has a molecular weight of about 50 kDa and contains two copper atoms. From electron-microscopic observations, hemocyanin molecules are cylindrical assemblies about 190 or 380 Å long and 350 Å in diameter comprising 10 or 20 subunits, respectively, for a molecular weight as high as 9 x 10 Dalton. In the arthropod family, the subunit has a molecular weight of about 70 kDa with two copper atoms. Molecular aggregates are composed of 6, 12, 24, or 48 subunits. Upon oxygenation the colorless protein becomes blue (hence cyanin from , Greek for blue). Spectral changes upon oxygenation, oxygen affinities, kinetics of oxygen binding (Table 4.2), anion binding, and other chemical reactions show that the active site in the phylum and that in , although both containing a pair of copper atoms, are not identical. No monomeric hemocyanins, analogous to myoglobin and myohemerythrin (next section), are known. For some hemocyanins the binding of dioxygen is highly cooperative, if calcium or magnesium ions are present, with Hill coefficients as high as n ~ 9. However, the free energy of interaction per subunit can be small in comparison with that for tetrameric hemoglobin; 0.9 to 2.5 kcal/mol compared to 3.0 kcal/mol. Allosteric effects, at least for a 24-subunit tarantula hemocyanin, can be separated into those within a dodecamer (12 subunits)—the major contributor to overall allostery—and those between dodecamers. This has been termed . In contrast to the hemoglobin family, isolated chains have affinities typical of the T-state conformation for hemocyanin. The binding of CO, which binds to only one copper atom, is at best weakly cooperative. As alluded to above, the distribution of hemocyanins is striking, Among the molluscs exclusive use of hemocyanin as the respiratory protein occurs only with the cephalopods (squid, octopi, and cuttlefish), and in the arthropods only among the decapod (ten-footed) crustaceans (lobsters, shrimp, and crabs). The bivalve molluscs (for example, oysters and scallops) all use small dimeric or octameric hemoglobins. The edible gastropod (snail) omatia uses hemocyanin, whereas the apparently closely related fresh-water snail uses a high-oligomer hemoglobin. Both use a myoglobin as the oxygen-storage protein. The structure of the active site has been extensively probed by EXAFS methods, and the x-ray crystal structure of a hexameric deoxyhemocyanin is known. Each copper atom is coordinated to three imidazole groups from histidine residues. The pinwheel arrangement of the six subunits, the domain structure of a single subunit, and the domain containing the active site are shown in Figure 4.9. The biological occurrence of hemerythrins (Hr in Figure 4.8), the third class of dioxygen carriers, is relatively rare, being restricted to the sipunculid family (nonsegmented worms), a few members of the annelid (segmented worm) family, a couple of brachiopods (shrimps), and a couple of priapulids. The oxygen-binding site contains, like hemocyanin, a pair of metal atoms, in this case, iron. Upon oxygenation the colorless protein becomes purple-red. Monomeric (myohemerythrin), trimeric, and octameric forms of hemerythrin are known; all appear to be based on a similar subunit of about 13.5 kDa. When hemerythrin is extracted from the organism, its oxygen binding is at best only weakly cooperative, with Hill coefficients in the range 1.1 to 2.1. In coelomic cells (the tissue between the inner membrane lining the digestive tract and the outer membrane of the worm—analogous to flesh in vertebrates), oxygen apparently binds with higher cooperativity (n ~ 2.5). Perchlorate ions have been observed to induce cooperativity: since CIO has no biological role, it appears that in protein purifications the biological allosteric effector is lost. No Bohr effect occurs. Dioxygen binding data are accumulated in Table 4.2. The structure of hemerythrin in a variety of derivatives (oxy, azido, met, and deoxy) is now well-characterized. With three bridging ligands, a distinctive cofacial bioctahedral stereochemistry is seen (Figure 4.10). Solubility of O2 in water: 1.86 x 10 M/Torr Solubility of CO in water: 1.36 x 10 M/Torr a) 10 mM Ca added: necessary for cooperativity b) CO binding at pH 9.6. P (O ) Torr \(\Delta\)H kcal/mol \(\Delta\)S eu k \(\mu\)M s k s P (O ) Torr \(\Delta\)H kcal/mol \(\Delta\)S eu k \(\mu\)M s k s -6.0 CO -2.7 binding 4.1 noncooperative | 9,617 | 1,370 |
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Gas chromatography is a major separation method. It consists of the injection of a gaseous or liquid sample into a gaseous mobile phase which is passed through a column of solid support particles carrying a liquid stationary phase, maintained in an oven at a suitable temperature (which is usually above ambient but need not be above the boiling points of the analytes). Separation is the result of partition between the stationary and mobile phases and the separated constituents of the sample are usually detected by a flame ionization detector. A flow chart of the typical instrumentation for gas chromatography is illustrated in figure D9.1. Figure D9.1: Samples of increasing complexity are being analysed by gas chromatography. Universal detectors, such as the flame ionization detector, are not adequate for such a task but selective detectors can provide the additional discrimination that is needed. Nitrogen- and sulfur-containing compounds commonly occur as trace-level analytes in complex samples and highly selective detectors have been developed. Among these, the nitrogen chemiluminescence detector and the sulfur chemiluminescence detector have emerged as powerful tools in gas chromatography, supercritical fluid chromatography and high performance liquid chromatography; stand-alone nitrogen/sulfur analysers can be based on the same chemiluminescence reactions. Detectors of either element are based on the same ozone-induced gas phase chemiluminescence . The chemiluminescence is preceded by high temperature pyrolysis which oxidizes the nitrogen in the sample (RN) to nitric oxide (NO): Oxidation: \[RN + O_2 → NO + CO_2 + H_2O \label{D9.1}\] and it is believed that the sulfur in the sample (\(RS\)) is converted first into sulfur dioxide (\(SO_2\)), which is then reduced in the presence of hydrogen to sulfur monoxide (\(SO\)): Oxidation: \[RS + O_2 → SO_2 + CO_2 + H_2O\label{D9.2}\] Reduction: \[SO_2 + H_2 → SO + H_2O\label{D9.3}\] Overall: \[RS + O_2 + H_2 → SO + CO_2 + H_2O\label{D9.4}\] These reactions produce the species that react with ozone, producing excited nitrogen dioxide and excited sulfur dioxide respectively (Equations \(\ref{D9.5}\) and \(\ref{D9.7}\)): Reaction with ozone: \[NO + O_3 → NO_2^* + O_2\label{D9.5}\] Chemiluminescence: \[NO_2^* → NO_2 + light (~ 1200 nm)\label{D9.6}\] Reaction with ozone: \[SO + O_3 → SO_2^* + O_2\label{D9.7}\] Chemiluminescence: \[SO_2^* → SO_2 + light (~ 360 nm)\label{D9.8}\] The nitrogen chemiluminescence reaction emits in the near infra-red (Equation \(\ref{D9.6}\)), whereas the sulfur reaction emits in the ultra-violet (Equation \(\ref{D9.8}\)). This wide spectral separation of the emission bands enables nitrogen and sulfur to be determined selectively. A few small gaseous molecules containing sulfur also enter into the chemiluminescent reaction with ozone without undergoing preliminary pyrolysis. Figure D9.2: The instrumentation for nitrogen-sulfur chemiluminescence detection is depicted in Figure D9.2. The pyrolyser converts the analytes in the gas chromatograph column effluent into the corresponding chemiluminescent species, which pass to the reaction chamber where they react with ozone supplied by a generator. The light emitted is detected by a photomultiplier. | 3,285 | 1,372 |
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The advantage of MO theory becomes more apparent when we think about \(\pi\) bonds, especially in those situations where two or more \(\pi\) bonds are able to interact with one another. Let’s first consider the \(\pi\) bond in ethene from an MO theory standpoint (in this example we will be disregarding the various sigma bonds, and thinking about the \(\pi\) bond). According to MO theory, the two atomic 2 orbitals combine to form two \(\pi\) , one a low-energy bonding orbital and one a high-energy . These are sometimes denoted in molecular diagrams with the Greek letter psi ( ) instead of (Figure \(\Page {1}\)). In the bonding orbital, the two shaded lobes of the 2p orbitals interact constructively with each other, as do the two unshaded lobes (remember, the shading choice represents mathematical (+) and (-) signs for the wavefunction). Therefore, there is increased electron density between the nuclei in the molecular orbital – this is why it is a bonding orbital. In the higher-energy antibonding * orbital, the shaded lobe of one 2 orbital interacts destructively with the unshaded lobe of the second 2 orbital, leading to a node between the two nuclei and overall repulsion. By the , the two electrons from the two atomic orbitals will be paired in the lower-energy orbital when the molecule is in the ground state (Figure \(\Page {1}\)). Now, consider the 1,3-butadiene molecule. From valence orbital theory we might expect that the C -C bond in this molecule, because it is a \(\sigma\) bond that would rotate freely. However, it is experimentally found that there are significant barriers to rotation about this bond (as well as about the C -C and C -C double bonds), and that the entire molecule is planar. It is also observed that the C -C bond, while longer than the C -C and C -C double bonds, is significantly shorter than a typical carbon-carbon single bond. Molecular orbital theory accounts for these observations with the concept of . In this picture, the four 2 orbitals are all parallel to each other (and perpendicular to the plane of the \(\sigma\) bonds), and thus there is \(\pi\)-overlap not just between C and C and C and C , but between C and C as well. The four atomic (2 ) orbitals have combined to form four \(\pi\) molecular orbitals. The lowest energy molecular orbital, , has zero nodes, and is a bonding MO. Slightly higher in energy, but still lower than the isolated orbitals, is the orbital. This orbital has one node between C and C , but is still a bonding orbital due to the two constructive interactions between C -C and C -C . The two higher-energy MO’s are denoted * and *, and are antibonding. Notice that * has two nodes and one constructive interaction, while * has three nodes and zero constructive interactions. The energy of both of these antibonding molecular orbitals is higher than that of the 2 atomic orbitals of which they are composed. By the principle, the four electrons from the isolated 2 atomic orbitals are placed in the bonding and MO’s. Because includes constructive interaction between C and C , there is a degree, in the 1,3-butadiene molecule, of -bonding interaction between these two carbons, which accounts for the shorter length and the barrier to rotation. The simple Lewis structure picture of 1,3-butadiene shows the two bonds as being isolated from one another, with each pair of electrons ‘stuck’ in its own bond. However, molecular orbital theory predicts (accurately) that the four electrons are to some extent delocalized, or ‘spread out’, over the whole system. 1,3-butadiene is the simplest example of a system of ‘ bonds. To be considered conjugated, two or more bonds must be separated by only one single bond – in other words, there cannot be an intervening sp -hybridized carbon, because this would break up the overlapping system of parallel 2 orbitals. Molecular orbital theory is especially helpful in explaining the unique properties of a class of compounds called aromatics. Benzene, a common organic solvent, is the simplest example of an aromatic compound. Although benzene is most often drawn with three double bonds and three single bonds (Figure \(\Page {4}\)), it is known that all of the carbon-carbon bonds in benzene are exactly the same length - 1.38 Å. This is shorter than a typical carbon-carbon single bond (about 1.54 Å), and slightly longer than a typical carbon-carbon double bond (about 1.34 Å). Benzene is also a molecule in which all of the ring atoms are sp -hybridized that allows the electrons to be delocalized in molecular orbitals that extend all the way around the ring, above and below the plane of the ring. For this to happen, of course, the ring must be planar – otherwise the 2 orbitals could not overlap properly. Benzene was experimentally confirmed to be flat molecule by Dame Kathleen Londsale with X-ray crystallography. As shown in Figure \(\Page {5}\), the cyclic array of six \2P_z\)-orbitals (one on each carbon) overlap to generate six molecular orbitals, three bonding and three antibonding. The plus and minus signs shown in the diagram do not represent electrostatic charge, but refer to phase signs in the equations that describe these orbitals (in the diagram the phases are also color coded). When the phases correspond, the orbitals overlap to generate a common region of like phase, with those orbitals having the greatest overlap (e.g. π ) being lowest in energy. The remaining carbon valence electrons then occupy these molecular orbitals in pairs, resulting in a fully occupied (6 electrons) set of bonding molecular orbitals. It is this completely filled set of bonding orbitals, or , that gives the benzene ring its thermodynamic and chemical stability, just as a filled valence shell octet confers stability on the inert gases. This added stability is call . and is discuss in more detail in organic chemistry courses. Resonance structures can be used to describe the bonding in molecules such as ozone (O ) and the nitrite ion (NO ). We showed that ozone can be represented by either of these Lewis electron structures: Although the VSEPR model correctly predicts that both species are bent, it gives no information about their bond orders. Experimental evidence indicates that ozone has a bond angle of 117.5°. Because this angle is close to 120°, it is likely that the central oxygen atom in ozone is trigonal planar and hybridized. If we assume that the terminal oxygen atoms are also hybridized, then we obtain the \(\sigma\)-bonded framework shown in Figure \(\Page {6}\). Two of the three lobes on the central O are used to form O–O sigma bonds, and the third has a lone pair of electrons. Each terminal oxygen atom has two lone pairs of electrons that are also in lobes. In addition, each oxygen atom has one unhybridized 2 orbital perpendicular to the molecular plane. The \(\sigma\) bonds and lone pairs account for a total of 14 electrons (five lone pairs and two \(\sigma\) bonds, each containing 2 electrons). Each oxygen atom in ozone has 6 valence electrons, so O has a total of 18 valence electrons. Subtracting 14 electrons from the total gives us 4 electrons that must occupy the three unhybridized 2 orbitals. With a molecular orbital approach to describe the \(\pi\) bonding, three 2 atomic orbitals give us three molecular orbitals, as shown in Figure \(\Page {7}\). One of the molecular orbitals is a \(\pi\) bonding molecular orbital, which is shown as a banana-shaped region of electron density above and below the molecular plane. This region has nodes perpendicular to the O plane. The molecular orbital with the highest energy has two nodes that bisect the O–O \(\sigma\) bonds; it is a \(\pi\)* antibonding orbital. The third molecular orbital contains a single node that is perpendicular to the O plane and passes through the central O atom; it is a nonbonding molecular orbital. Because electrons in nonbonding orbitals are neither bonding nor antibonding, they are ignored in calculating bond orders. We can now place the remaining four electrons in the three energy levels shown in Figure \(\Page {7}\), thereby filling the \(\pi\) bonding and the nonbonding levels. The result is a single \(\pi\) bond holding three oxygen atoms together, or \(½ \pi\) bond per O–O. We therefore predict the overall O–O bond order to be \(½ \pi\) bond plus 1 \(\sigma\) bond), just as predicted using resonance structures. The molecular orbital approach, however, shows that the \(\pi\) nonbonding orbital is localized on the terminal O atoms, which suggests that they are more electron rich than the central O atom. The reactivity of ozone is consistent with the predicted charge localization. are a crude way of describing molecular orbitals that extend over more than two atoms. Describe the bonding in the nitrite ion in terms of a combination of hybrid atomic orbitals and molecular orbitals. Lewis dot structures and the VSEPR model predict that the NO ion is bent. chemical species and molecular geometry bonding description using hybrid atomic orbitals and molecular orbitals The lone pair of electrons on nitrogen and a bent structure suggest that the bonding in NO is similar to the bonding in ozone. This conclusion is supported by the fact that nitrite also contains 18 valence electrons (5 from N and 6 from each O, plus 1 for the −1 charge). The bent structure implies that the nitrogen is hybridized. If we assume that the oxygen atoms are hybridized as well, then we can use two hybrid orbitals on each oxygen and one hybrid orbital on nitrogen to accommodate the five lone pairs of electrons. Two hybrid orbitals on nitrogen form \(\sigma\) bonds with the remaining hybrid orbital on each oxygen. The \(\sigma\) bonds and lone pairs account for 14 electrons. We are left with three unhybridized 2 orbitals, one on each atom, perpendicular to the plane of the molecule, and 4 electrons. Just as with ozone, these three 2 orbitals interact to form bonding, nonbonding, and antibonding \(\pi\) molecular orbitals. The bonding molecular orbital is spread over the nitrogen and both oxygen atoms. Placing 4 electrons in the energy-level diagram fills both the bonding and nonbonding molecular orbitals and gives a \(\pi\) bond order of 1/2 per N–O bond. The overall N–O bond order is \(1\;\frac{1}{2}\), consistent with a resonance structure. Describe the bonding in the formate ion (HCO ), in terms of a combination of hybrid atomic orbitals and molecular orbitals. Like nitrite, formate is a planar polyatomic ion with 18 valence electrons. The \(\sigma\) bonding framework can be described in terms of hybridized carbon and oxygen, which account for 14 electrons. The three unhybridized 2 orbitals (on C and both O atoms) form three \(\pi\) molecular orbitals, and the remaining 4 electrons occupy both the bonding and nonbonding \(\pi\) molecular orbitals. The overall C–O bond order is therefore \(frac{3}{2}\) Hydrocarbons in which two or more carbon–carbon double bonds are directly linked by carbon–carbon single bonds are generally more stable than expected because of resonance. Because the double bonds are close enough to interact electronically with one another, the \(\pi\) electrons are shared over all the carbon atoms, as illustrated for 1,3-butadiene in Figure \(\Page {8}\). As the number of interacting atomic orbitals increases, the number of molecular orbitals increases, the energy spacing between molecular orbitals decreases, and the systems become more stable (Figure \(\Page {9}\)). Thus as a chain of alternating double and single bonds becomes longer, the energy required to excite an electron from the highest-energy occupied (bonding) orbital to the lowest-energy unoccupied (antibonding) orbital decreases. If the chain is long enough, the amount of energy required to excite an electron corresponds to the energy of visible light. For example, vitamin A is yellow because its chain of five alternating double bonds is able to absorb violet light. Many of the colors we associate with dyes result from this same phenomenon; most dyes are organic compounds with alternating double bonds. As the number of atomic orbitals increases, the difference in energy between the resulting molecular orbital energy levels decreases, which allows light of lower energy to be absorbed. As a result, organic compounds with long chains of carbon atoms and alternating single and double bonds tend to become more deeply colored as the number of double bonds increases. As the number of interacting atomic orbitals increases, the energy separation between the resulting molecular orbitals steadily decreases. A derivative of vitamin A called is used by the human eye to detect light and has a structure with alternating C=C double bonds. When visible light strikes retinal, the energy separation between the molecular orbitals is sufficiently close that the energy absorbed corresponds to the energy required to change one double bond in the molecule from , where like groups are on the same side of the double bond, to , where they are on opposite sides, initiating a process that causes a signal to be sent to the brain. If this mechanism is defective, we lose our vision in dim light. Once again, a molecular orbital approach to bonding explains a process that cannot be explained using any of the other approaches we have described. is central feature of molecular orbital theory where rather than the lone pair of electrons contained in localize bonds (as in the valence bond theory), electrons can exist in that are spread over the entire molecule. The \(\pi\) bonding between three or four atoms requires combining three or four unhybridized orbitals on adjacent atoms to generate \(\pi\) bonding, antibonding, and nonbonding molecular orbitals extending over all of the atoms. Filling the resulting energy-level diagram with the appropriate number of electrons explains the bonding in molecules or ions that previously required the use of resonance structures in the Lewis electron-pair approach. by (University of Minnesota, Morris) | 14,169 | 1,373 |
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Specificity in Recognition What determines the ability for a protein to recognize a specific target amongst many partners? To start, let’s run a simple calculation. Take the case that a protein (transcription factor) has to recognize a string of n sequential nucleotides among a total of N bases in a dsDNA. \[\left( \dfrac{1}{4} \right)^n \] \[\left( \dfrac{1}{4} \right)^n \geq \dfrac{1}{N} \] \[ n \geq \dfrac{\ln{N}}{\ln{4}} \] For the case that you want to define a unique binding site among N = 65k base pairs: This example illustrates that simple statistical considerations and the diversity of base combinations can provide a certain level of specificity in binding, but that other considerations are important for high fidelity binding. These considerations include the energetics of binding, the presence of multiple binding motifs for a base, and base-sequence specific binding motifs. We also need to think about the strength of interaction. Let’s assume that the transcription factor has a nonspecific binding interaction with DNA that is weak, but a strong interaction for the target sequence. We quantify these through: ∆G : nonspecific binding ∆G : specific binding Next, let’s consider the degeneracy of possible binding sites: g : number of nonspecific binding sites = (N – n) or since N ≫ n: (N – n) ≈ N g : number of sites that define the specific interaction: n The probability of having a binding partner bound to a nonspecific sequence is \[\begin{aligned}
P_{\text {nonsp }} &=\frac{g_{n} e^{-\Delta G_{1} / k T}}{g_{n} e^{-\Delta G_{1} / k T}+g_{s} e^{-\Delta G_{2} / k T}} \\ &=\frac{(N-n) e^{-\Delta G_{1} / k T}}{(N-n) e^{-\Delta G_{1} / k T}+n e^{-\Delta G_{2} / k T}} \\ &=\frac{1}{1+\frac{n}{N} e^{-\Delta G / k T}} \end{aligned}\] where ∆G = ∆G – ∆G . We do not want to have a high probability of nonspecific binding, so let’s minimize P . Solving for ΔG, and recognizing P 1, \[ \Delta G \leq -k_BT\ln{\left[ \dfrac{N}{nP_{nonsp}} \right] }\] Suppose we want to have a probability of nonspecific binding to any region of DNA that is P 1%. For N = 10 and n = 10, we find \[ \Delta G \approx -16k_BT \qquad \textrm{or} \qquad -1.6 k_BT/nucleotide \] for the probability that the partner being specifically bound with P 99%. ______________________________________________________________________________________ | 2,369 | 1,374 |
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The most common method to get an idea about the pH of solution is to use an acid base indicator. An indicator is a large organic molecule that works somewhat like a " color dye". Whereas most dyes do not change color with the amount of acid or base present, there are many molecules, known as , which do respond to a change in the hydrogen ion concentration. Most of the indicators are themselves weak acids. The most common indicator is found on "litmus" paper. It is red below pH 4.5 and blue above pH 8.2. Other commercial pH papers are able to give colors for every main pH unit. Universal Indicator, which is a solution of a mixture of indicators is able to also provide a full range of colors for the pH scale. A variety of indicators change color at various pH levels. A properly selected acid-base indicator can be used to visually "indicate" the approximate pH of a sample. An indicator is usually some weak organic acid or base dye that changes colors at definite pH values. The weak acid form (HIn) will have one color and the weak acid negative ion (In ) will have a different color. The weak acid equilibrium is: HIn → H + In See the graphic below for colors and pH ranges. Phenolphthalein is an indicator of acids (colorless) and bases (pink). Sodium hydroxide is a base, and it was in the pitcher at the beginning, so when added to the phenolphthalein in beakers 2 and 4, it turned pink (top half of the graphic). Equilibrium: HIn → H + In
colorless red The equilibrium shifts right, HIn decreases, and In increases. As the pH increase between 8.2 to 10.0 the color becomes red because of the equilibrium shifts to form mostly In ions. Neutralization reaction: HC H O + NaOH → Na(C H O ) + HOH Use equilibrium principles to explain the color change for phenolphthalein at the end of the demonstration. The simplified reaction is: H + OH → HOH As OH ions are added, they are consumed by the excess of acid already in the beaker as expressed in the above equation. The hydroxide ions keep decreasing and the hydrogen ions increase, pH decreases. See lower equation: The indicator equilibrium shifts left, In ions decrease. Below pH 8.2 the indicator is colorless. As H ions are further increased and pH decreases to pH 4-5, the indicator equilibrium is effected and changes to the colorless HIn form. Equilibrium: HIn → H + In
colorless red Color changes in molecules can be caused by changes in electron confinement. More confinement makes the light absorbed more blue, and less makes it more red. How are electrons confined in phenolphthalein? There are three benzene rings in the molecule. Every atom involved in a double bond has a p orbital which can overlap side-to-side with similar atoms next to it. The overlap creates a 'pi bond' which allows the electrons in the p orbital to be found on either bonded atom. These electrons can spread like a cloud over any region of the molecule that is flat and has alternating double and single bonds. Each of the benzene rings is such a system. See the far left graphic - The carbon atom at the center (adjacent to the yellow circled red oxygen atom) doesn't have a p-orbital available for pi-bonding, and it confines the pi electrons to the rings. The molecule absorbs in the ultraviolet, and this form of phenolphthalein is colorless. In basic solution, the molecule loses one hydrogen ion. Almost instantly, the five-sided ring in the center opens and the electronic structure around the center carbon changes (yellow circled atoms) to a double bond which now does contain pi electrons. The pi electrons are no longer confined separately to the three benzene rings, but because of the change in geometry around the yellow circled atoms, the whole molecule is now flat and electrons are free to move within the entire molecule. The result of all of these changes is the change in color to pink. Chime: Many other indicators behave on the molecular level in a similar fashion (the details may be different) but the result is a change in electronic structure along with the removal of a hydrogen ion from the molecule. Plant pigments in flowers and leaves also behave in this fashion. | 4,179 | 1,375 |
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In practice, the easiest way to measure the speed of a reaction is to measure the concentration of either the reactants or the products over time. The concentration of black dots is higher in the beaker on the right than in the beaker on the left. Reactions are often monitored by some form of spectroscopy. In spectroscopy, "light" or some other frequency of electromagnetic radiation passes through a reaction sample. The light interacts with the molecules in the sample, which absorb particular frequencies of light. Less light exits the sample than the amount that entered it; the amount that exits is measured by a detector on the other side. If the concentration of the sample is different, a different amount of light from the spectrometer is absorbed. For instance, suppose the sample is more concentrated. The more molecules there are, the more light is absorbed. Because the beam of light travels through the sample in a straight line, the more concentrated the solution, the more molecules it encounters. It is simple to calibrate the instrument to determine concentration from the amount of light absorbed. In addition, the light may interact with the reactant molecules and product molecules in different ways. This means that the absorption frequency of the reactants, and not the products, can be monitored, and thus changes in reactant concentration can be measured. This also works for the products.. The rate of reaction is written as: \[\text{rate} = \frac{d[product]}{dt} \nonumber \] In other words, the rate is equal to the change in concentration of product with change in time. Concentration can be measured in several units. Frequently, the concentration of a solution is measured, and units such as grams per liter or, much more commonly, moles per liter, are used. The change in time is most often measured in seconds. The rate of the reaction can also be written as follows: \[\text{rate} = -\frac{d[reactant]}{dt} \nonumber \] The rate is the change in concentration of reactant with change in time. The negative sign indicates that the reactant is being consumed over time as it turns into product, so its concentration is decreasing. Kinetic studies are important in understanding reactions. Not only are they important in industry, but they are also used to understand biological processes, especially enzyme-catalyzed reactions. They also play a role in environmental and atmospheric chemistry, as part of an effort to understand a variety of issues ranging from the fate of prescription pharmaceuticals in wastewater to the cascade of reactions involved in the ozone cycle. , | 2,624 | 1,376 |
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The second element in the periodic table provides our first example of a quantum-mechanical problem which be solved exactly. Nevertheless, as we will show, approximation methods applied to helium can give accurate solutions in perfect agreement with experimental results. In this sense, it can be concluded that quantum mechanics is correct for atoms more complicated than hydrogen. By contrast, the Bohr theory failed miserably in attempts to apply it beyond the hydrogen atom. The helium atom has two electrons bound to a nucleus with charge Z = 2. The successive removal of the two electrons can be diagrammed as The , the minimum energy required to remove the first electron from helium, is experimentally 24.59 eV. The second ionization energy, , is 54.42 eV. The last result can be calculated exactly since He is a hydrogen-like ion. We have \[\textit{I}_2=-\textit{E}_{ 1\textit{s}}(\ce{He}^+)=\dfrac{Z^2}{2n^2}=2 \mbox{ hartrees}=54.42\mbox{ eV}\label{2}\] The energy of the three separated particles on the right side of Equation \(\ref{1}\) is, by definition, zero. Therefore the ground-state energy of helium atom is given by \(E_0=-(\textit{I}_1+\textit{I}_2)=-79.02\mbox{ eV}=-2.90372\mbox{ hartrees}\). We will attempt to reproduce this value, as close as possible, by theoretical analysis. The the repulsive interaction between the two electrons. It is this last contribution which prevents an exact solution of the Schr dinger equation and which accounts for much of the complication in the theory. In seeking an approximation to the ground state, we might first work out the solution in the absence of the 1/ -term. In the Schr dinger equation thus simplified, we can separate the variables and to reduce the equation to two independent hydrogen-like problems. The ground state wavefunction (not normalized) for this hypothetical helium atom would be \[\psi(\text{r}_1,\text{r}_2)=\psi_{1s}(r_1)\psi_{1s}(r_2)=e^{-Z(r_1+r_2)}\label{4}\] and the energy would equal \(2\times(-Z^2/2)=-4\) hartrees, compared to the experimental value of \(-2.90\) hartrees. A significantly improved result can be obtained with the functional form ( Equation \(\ref{4}\)), but with replaced by a adjustable parameter \(\alpha\), thus: \[\tilde{\psi}(r_1,r_2)=e^{-\alpha(r_1+r_2)}\label{5}\] Using this function in the variational principle [cf. Eq (4.53)], we have where \(\hat{H}\) is the full Hamiltonian as in Equation \(\ref{3}\), including the \(1/r_{12}\)-term. The expectation values of the five parts of the Hamiltonian work out to \[\left\langle-\dfrac{1}{2}\nabla^2_1\right\rangle=\left\langle-\dfrac{1}{2}\nabla^2_2\right\rangle=\dfrac{\alpha^2}{2}\] \[\left\langle-\dfrac{Z}{r_1}\right\rangle=\left\langle-\dfrac{Z}{r_2}\right\rangle=-Z\alpha, \left\langle\dfrac{1}{r_{12}}\right\rangle=\dfrac{5}{8}\alpha\label{7}\] The sum of the integrals in Equation \(\ref{7}\) gives the variational energy \[\tilde{E}(\alpha)=\alpha^2-2Z\alpha+\dfrac{5}{8}\alpha\label{8}\] This will be always be an upper bound for the true ground-state energy. We can optimize our result by finding the value of \(\alpha\) which the energy (Equation \(\ref{8}\)). We find \[\dfrac{d\tilde{E}}{d\alpha}=2\alpha-2Z+\dfrac{5}{8}=0\label{9}\] giving the optimal value \[\alpha=Z-\dfrac{5}{16}\label{10}\] This can be given a physical interpretation, noting that the parameter \(\alpha\) in the wavefunction (Equation \(\ref{5}\)) represents an nuclear charge. Each electron partially shields the other electron from the positively-charged nucleus by an amount equivalent to \(5/8\) of an electron charge. Substituting Equation \(\ref{10}\) into Equation \(\ref{8}\), we obtain the optimized approximation to the energy For helium (\(Z = 2\)), this gives \(-2.84765\) hartrees, an error of about \(2\%\) \((E_0 = -2.90372)\). Note that the inequality \(\tilde{E} > E_0\) applies in an sense. In the late 1920's, it was considered important to determine whether the helium computation could be improved, as a test of the validity of quantum mechanics for many electron systems. The table below gives the results for a selection of variational computations on helium. The third entry refers to the , developed by Hartree. Even for the best possible choice of one-electron functions \(\psi(r)\), there remains a considerable error. This is due to failure to include the variable \(r_{12}\) in the wavefunction. The effect is known as . The fourth entry, containing a simple correction for correlation, gives a considerable improvement. Hylleraas (1929) extended this approach with a variational function of the form \[\psi(r_1, r_2, r_{12})=e^{-\alpha(r_1+r_2)} \times \textrm{polynomial in} r_1, r_2, r_{12}\] and obtained the nearly exact result with 10 optimized parameters. More recently, using modern computers, results in essentially perfect agreement with experiment have been obtained. The simpler wavefunctions for helium atom in Equation \(\ref{5}\), can be interpreted as representing two electrons in hydrogen-like 1 orbitals, designated as a 1 configuration. According to Pauli's exclusion principle, which states that no two electrons in an atom can have the same set of four quantum numbers, the two 1 electrons must have spins, one spin-up or \(\alpha\), the other spin-down or \(\beta\). A product of an orbital with a spin function is called a . For example, electron 1 might occupy a spinorbital which we designate \[\phi(1)=\psi_{1s}(1)\alpha(1) \textrm{or} \psi_{1s}(1)\beta(1)\label{12}\] Spinorbitals can be designated by a single subscript, for example, \(\phi_a\) or \(\phi_b\), where the subscript stands for a of four quantum numbers. In a two electron system the occupied spinorbitals \(\phi_a\) and \(\phi_b\) must be different, meaning that at least one of their four quantum numbers must be unequal. A two-electron spinorbital function of the form \[\Psi (1, 2) = \dfrac{1}{2} \bigg( \phi_a(1)\phi_b(2) - \phi_b(1)\phi_a(2)\bigg)\label{13}\] automatically fulflls the Pauli principle since it vanishes if \(a=b\). Moreover, this function associates each electron equally with each orbital, which is consistent with the of identical particles in quantum mechanics. The factor \(1/\sqrt{2}\) normalizes the two-particle wavefunction, assuming that \(\phi_a\) and \(\phi_b\) are normalized and mutually orthogonal. The function (Equation \(\ref{13}\)) is with respect to interchange of electron labels, meaning that This antisymmetry property is an elegant way of expressing the . Equation \(\ref{13}\) \[\Psi (1, 2) = \dfrac{1}{\sqrt{2}}\begin{vmatrix}\phi_a(1) & \phi_b(1)\\\phi_a(2) & \phi_b(2)\end{vmatrix}\label{15}\] 1 Equation \(\ref{13}\) \[\Psi (1, 2) = \psi_{1s}(1)\psi_{1s}(2) \times \dfrac{1}{\sqrt{2}}\bigg(\alpha(1)\beta(2) - \beta(1)\alpha(2)\bigg)\label{16}\] For two-electron systems (but for three or more electrons), the wavefunction can be factored into an orbital function times a spin function. The two-electron spin function \[\sigma_{0,0}(1, 2) = \dfrac{1}{\sqrt{2}}\bigg(\alpha(1)\beta(2) - \beta(1)\alpha(2)\bigg)\label{17}\] represents the two electron spins in opposing directions (antiparallel) with a total spin angular momentum of zero. The two subscripts are the quantum numbers and for the total electron spin. Eqution \(\ref{16}\) is called the spin state since there is only a single orientation for a total spin quantum number of zero. It is also possible to have both spins in the state, provided the orbitals are different. There are three possible states for two parallel spins: \[\sigma_{1,1}(1, 2) = \alpha(1)\alpha(2)\] \[\sigma_{1,0}(1, 2) = \dfrac{1}{\sqrt{2}}\bigg(\alpha(1)\beta(2) + \beta(2)\alpha(2)\bigg)\] \[\sigma_{1,-1}(1, 2) = \beta(1)\beta(2)\label{18}\] These make up the spin states, which have the three possible orientations of a total angular momentum of 1. The lowest excitated state of helium is represented by the electron configuration 1 2 . The 1 2 configuration has higher energy, even though the 2 and 2 orbitals in hydrogen are degenerate, because the 2 penetrates closer to the nucleus, where the potential energy is more negative. When electrons are in different orbitals, their spins can be either parallel or antiparallel. In order that the wavefunction satisfy the antisymmetry requirement (Equation \(\ref{14}\) ), the two-electron orbital and spin functions must have behavior under exchange of electron labels. There are four possible states from the 1 2 configuration: a singlet state \[\Psi^+ (1, 2) = \dfrac{1}{\sqrt{2}}\bigg(\psi_{1s}(1)\psi_{2s}(2) + \psi_{2s}(1)\psi_{1s}(2)\bigg) \sigma_{0, 0}(1, 2)\label{19}\] and three triplet states \[\Psi^-(1, 2) = \dfrac{1}{\sqrt{2}}\bigg(\psi_{1s}(1)\psi_{2s}(2) - \psi_{2s}(1)\psi_{1s}(2)\bigg)\begin{cases} \sigma_{1,1}(1, 2)\\ \sigma_{1,0}(1, 2) \\ \sigma_{1,-1}(1, 2)\end{cases}\label{20}\] Using the Hamiltonian in Equation \(\ref{3}\), we can compute the approximate energies \[E^{\pm}=\iint\Psi^{\pm}(1,2) \hat{H} \Psi^{\pm}(1,2)d\tau_1d\tau_2\label{21}\] After evaluating some fierce-looking integrals, this reduces to the form \[E^{\pm}=I(1s)+I(2s)+J(1s, 2s) \pm K(1s, 2s)\label{22}\] in terms of the one electron integrals \[I(a)=\int \psi_a(\textrm{r})\left\{-\dfrac{1}{2}\nabla^2-\dfrac{Z}{r}\right\} \psi_a(\textrm{r})d\tau\label{23}\] the Coulomb integrals \[J(a, b)=\iint\psi_a(\textrm{r}_1)^2\dfrac{1}{r_{12}}\psi_b(\textrm{r}_2)^2d\tau_1d\tau_2\label{24}\] and the exchange integrals \[K(a, b)=\iint\psi_a(\textrm{r}_1)\psi_b(\textrm{r}_1)\dfrac{1}{r_{12}}\psi_a(\textrm{r}_2)\psi_b(\textrm{r}_2)d\tau_1d\tau_2\label{25}\] The Coulomb integral represents the repulsive potential energy for two interacting charge distributions \(\psi_a(\textbf{r}_1)^2\) and \(\psi_b(\textbf{r}_2)^2\). The exchange integral, which has no classical analog, arises because of the exchange symmetry (or antisymmetry) requirement of the wavefunction. Both and can be shown to be positive quantities. Therefore the lower sign in (22) represents the state of lower energy, making the triplet state of the configuration 1s 2s lower in energy than the singlet state. This is an almost universal generalization and contributes to Hund's rule, to be discussed in the . (Professor Emeritus of Chemistry and Physics at the , ) | 10,322 | 1,377 |
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This is a continuation of Rate and Order of Reactions. In this module, more examples are given to show applications of integrated rate laws in problem solving. The differential rate laws and integrated rate laws are summarized in the table below to give you an overall view of reactions of these types. \(\mathrm{\ln [A]\: vs}\: t\) \(\textrm{slope} = - k\) \(\mathrm{\dfrac{1}{[A]}- \dfrac{1}{[A]_o}}= k t\) \(\mathrm{\dfrac{1}{[A]}\: vs}\: t\) \(\textrm{slope} = k\) The decomposition of \(\ce{A}\) is first order, and \(\ce{[A]}\) is monitored. The following data are recorded: The integrated rate law for 1st order is \(\mathrm{A = A_o}\, e^{- \large{k\, t}}\) Using the the first two points, \(0.0905 = 0.100\, e^{(- \large k \times 1)}\)
\(\begin{align*}
- k &= \ln \left(\dfrac{0.0905}{0.100}\right)\\
&= \mathrm{\ln(0.905) = -.0998\: min^{-1}}
\end{align*}\) Using the point when = 2 \(0.100 = 0.0819\, e^{(-\large k \times 2)}\) \(\begin{align*}
- k\, 2 &= \ln \left(\dfrac{0.0819}{0.100}\right)\\
&= \ln (0.819)\\
&= -0.200\\
k &= \mathrm{0.100\: min^{-1}}
\end{align*}\) Using the point when = 4 \(0.0670 = 0.100\, e^{(- \large k \times 4)}\) \(\begin{align*}
4\, k &= \ln 1.49\\
&= 0.400\\
k &= \mathrm{0.100\: min^{-1}}
\end{align*}\) \(= 0.693\) \(t_{\frac{1}{2}} = \dfrac{0.693}{0.1} = \mathrm{6.93\: min}\) (note the calculation of units) \(\begin{align*}
\mathrm{[A]} &= 0.100\, e^{(\large{-0.100 \times5})}\\
&= 0.100 \times 0.6065\\
&= 0.0607
\end{align*}\) \(0.01 = 0.1\, e^{(\large{-0.100 \times t})}\) \(0.100 \times t = \ln (10)\) \(\mathrm{t = \dfrac{2.303}{0.100} = 23.03\: min}\) Check: \(\begin{align*}
\mathrm{[A]} &= 0.1\, e^{(\large{-0.1 \times 23.03})}\\
&= 0.010
\end{align*}\) The dimerization reaction of butadiene is second order process: \[\ce{2 C4H_{6\large{(g)}} \rightarrow C8H_{12\large{(g)}}} \nonumber \] The rate constant at some temperature is 0.100 /min with an initial concentration of butadiene ([B]) of 1 M. Calculate the concentration of butadiene at 1, 2, 5, 10, 20, 30, and 70 minutes. Many of the following values can be evaluated without using a calculator. This is an applicaiton of this equation: \[\mathrm{\dfrac{1}{[A]}- \dfrac{1}{[A]_o}}= k t \nonumber \] From the data and results in Example 2:
Consider the relationship \(-\ce{\dfrac{d[C4H6]}{dt}} = \ce{2 \dfrac{d[C8H12]}{dt}}\)
The drop in pressure is 101-95 = 6 kPa. Consider the reaction: \(\begin{alignat}{3}
\ce{2 &C4H_{6\large{(g)}} \rightarrow\: &&C8H_{12\large{(g)}}}\\
\ce{&12\: kPa &&\:\:6\: kPa}
\end{alignat}\) Since two moles of butadiene combine to give one mole octadiene, the difference in total pressure is the partial pressure of octadiene, 12 kPa of butadiene converted to 6 kPa of octadiene. What is the partial pressure of butadiene?
The reaction is: \(\ce{2 C4H_{6\large{(g)}} \rightarrow C8H_{12\large{(g)}}}\) \(\dfrac{1}{89} - \dfrac{1}{101} = k \times 10\) \(k =\mathrm{1.33\,e^{-4}}\) What is the total pressure when the reaction is completed?
Evaluate half-life according to conditions. Apply the equation, \(k \times t_{\frac{1}{2}} = \ln 2\) | 3,242 | 1,381 |
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Any electron associated with an atom has a wavefunction that describes its position around the nucleus as well as an energy. This energy consists of two components: kinetic and potential energies. The kinetic energy is a consequence of the electron having mass and moving at a certain speed. The potential energy is a result of the electrostatic, or Coulombic, force that attracts oppositely charged particles (i.e. the electron and the proton nucleus). The hydrogen atom energy levels can be obtained through solving the Schrödinger Equation. Using the power series method, the solution of the hydrogen atom Schrödinger Equation leads to quantized energy levels \[ E_n=-\dfrac{m_ee^4}{8\epsilon{_0}^2h^2n^2}=-\dfrac{m_ee^4}{32\pi{^2}\epsilon{_0^2}\hbar{^2}n^2}\;\;n=1,2,..\] This energy can be written in terms of the Bohr radius, \( a_0\): \[ E_n=-\dfrac{e^2}{8\pi{^2}\epsilon{_0}a_0n^2}\;\; n=1,2,...\] The energy is negative due to the attractive nature of the Coulombic interaction. This is alternatively visualized as an atom whose electron has been moved infinitely far away. The potential energy of the electron is defined as zero as there is no interaction at infinite distance. As the electron approaches the nucleus, the amount of potential energy decreases as the electron gets nearer to the nucleus. One interesting feature of the energy levels are that they are not spaced evenly, but rather the spacing diminishes exponentially as \( E_n\propto 1/{n^2} \). The limit of these energies as approaches infinity is the dissociation energy, the point at which the electron escapes the Coulombic pull of the nucleus. Another important aspect to note is that the energy is solely dependent on the principal quantum number , hence the 3s, 3p and 3d orbitals all have the same energy. This is not the case for multi-electron atoms, where the s, p, or d orbitals in a given level have different energies. McQuarrie, Donald A. . 2nd ed. United States Of America: University Science Books, 2008. 321-24. 1. How much energy is required to excite the hydrogen electron from its ground state to the first excited state? 2. According to quantum mechanics, what is the dissociation energy of an H atom? | 2,224 | 1,382 |
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This page explains how to interpret the phase diagrams for simple pure substances - including a look at the special cases of the phase diagrams of water and carbon dioxide. At its simplest, a phase can be just another term for solid, liquid or gas. If you have some ice floating in water, you have a solid phase present and a liquid phase. If there is air above the mixture, then that is another phase. But the term can be used more generally than this. For example, oil floating on water also consists of two phases - in this case, two liquid phases. If the oil and water are contained in a bucket, then the solid bucket is yet another phase. In fact, there might be more than one solid phase if the handle is attached separately to the bucket rather than molded as a part of the bucket. You can recognize the presence of the different phases because there is an obvious boundary between them - a boundary between the solid ice and the liquid water, for example, or the boundary between the two liquids. A lets you work out exactly what phases are present at any given temperature and pressure. In the cases we'll be looking at on this page, the phases will simply be the solid, liquid or vapor (gas) states of a pure substance. This is the phase diagram for a typical pure substance. These diagrams (including this one) are nearly always drawn highly distorted in order to see what is going on more easily. There are usually two major distortions. We'll discuss these when they become relevant. If you look at the diagram, you will see that there are three lines, three areas marked "solid", "liquid" and "vapor", and two special points marked "C" and "T". These are easy! Suppose you have a pure substance at three different sets of conditions of temperature and pressure corresponding to 1, 2 and 3 in the next diagram. Under the set of conditions at 1 in the diagram, the substance would be a solid because it falls into that area of the phase diagram. At 2, it would be a liquid; and at 3, it would be a vapor (a gas). Suppose you had a solid and increased the temperature while keeping the pressure constant - as shown in the next diagram. As the temperature increases to the point where it crosses the line, the solid will turn to liquid. In other words, it melts. If you repeated this at a higher fixed pressure, the melting temperature would be higher because the line between the solid and liquid areas slopes slightly forward. So what actually is this line separating the solid and liquid areas of the diagram? It simply shows the effect of pressure on melting point. Anywhere on this line, there is an equilibrium between solid and liquid. You can apply Le Chatelier's Principle to this equilibrium just as if it was a chemical equilibrium. If you increase the pressure, the equilibrium will move in such a way as to counter the change you have just made. If it converted from liquid to solid, the pressure would tend to decrease again because the solid takes up slightly less space for most substances. That means that increasing the pressure on the equilibrium mixture of solid and liquid at its original melting point will convert the mixture back into the solid again. In other words, it will no longer melt at this temperature. To make it melt at this higher pressure, you will have to increase the temperature a bit. Raising the pressure raises the melting point of most solids. That's why the melting point line slopes forward for most substances. You can also play around with this by looking at what happens if you decrease the pressure on a solid at constant temperature. In the same sort of way, you can do this either by changing the temperature or the pressure. The liquid will change to a vapor - it boils - when it crosses the boundary line between the two areas. If it is temperature that you are varying, you can easily read off the boiling temperature from the phase diagram. In the diagram above, it is the temperature where the red arrow crosses the boundary line. So, again, what is the significance of this line separating the two areas? Anywhere along this line, there will be an equilibrium between the liquid and the vapor. The line is most easily seen as the effect of pressure on the boiling point of the liquid. As the pressure increases, so the boiling point increases. You will have noticed that this liquid-vapor equilibrium curve has a top limit (labeled as in the phase diagram), which is known as the . The temperature and pressure corresponding to this are known as the critical temperature and critical pressure. If you increase the pressure on a gas (vapor) at a temperature lower than the critical temperature, you will eventually cross the liquid-vapor equilibrium line and the vapor will condense to give a liquid. This works fine as long as the gas is below the critical temperature. What, though, if your temperature was above the critical temperature? There wouldn't be any line to cross! That is because, above the critical temperature, it is impossible to condense a gas into a liquid just by increasing the pressure. All you get is a highly compressed gas. The particles have too much energy for the intermolecular attractions to hold them together as a liquid. The critical temperature obviously varies from substance to substance and depends on the strength of the attractions between the particles. The stronger the intermolecular attractions, the higher the critical temperature. There's just one more line to look at on the phase diagram. This is the line in the bottom left-hand corner between the solid and vapor areas. That line represents solid-vapor equilibrium. If the conditions of temperature and pressure fell exactly on that line, there would be solid and vapor in equilibrium with each other - the solid would be subliming. (Sublimation is the change directly from solid to vapor or vice versa without going through the liquid phase.) Once again, you can cross that line by either increasing the temperature of the solid, or decreasing the pressure. The diagram shows the effect of increasing the temperature of a solid at a (probably very low) constant pressure. The pressure obviously has to be low enough that a liquid can't form - in other words, it has to happen below the point labelled as . You could read the sublimation temperature off the diagram. It will be the temperature at which the line is crossed. Point on the diagram is called the triple point. If you think about the three lines which meet at that point, they represent conditions of: Where all three lines meet, you must have a unique combination of temperature and pressure where all three phases are in equilibrium together. That's why it is called a triple point. If you controlled the conditions of temperature and pressure in order to land on this point, you would see an equilibrium which involved the solid melting and subliming, and the liquid in contact with it boiling to produce a vapor - and all the reverse changes happening as well. If you held the temperature and pressure at those values, and kept the system closed so that nothing escaped, that's how it would stay. The normal melting and boiling points are those when the pressure is 1 atmosphere. These can be found from the phase diagram by drawing a line across at 1 atmosphere pressure. Jim Clark ( ) | 7,337 | 1,383 |
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The most common sources of the hydride Nucleophile are lithium aluminum hydride (LiAlH ) and sodium borohydride (NaBH ). Note! The hydride anion is not present during this reaction; rather, these reagents serve as a source of hydride due to the presence of a polar metal-hydrogen bond. Because aluminum is less electronegative than boron, the Al-H bond in LiAlH is more polar, thereby, making LiAlH a stronger reducing agent. Addition of a hydride anion (H: ) to an aldehyde or ketone gives an alkoxide anion, which on protonation yields the corresponding alcohol. Aldehydes produce 1º-alcohols and ketones produce 2º-alcohols. In metal hydrides reductions the resulting alkoxide salts are insoluble and need to be hydrolyzed (with care) before the alcohol product can be isolated. In the sodium borohydride reduction the methanol solvent system achieves this hydrolysis automatically. In the lithium aluminum hydride reduction water is usually added in a second step. The lithium, sodium, boron and aluminum end up as soluble inorganic salts at the end of either reaction. Note! LiAlH and NaBH are both capable of reducing aldehydes and ketones to the corresponding alcohol. This mechanism is for a LiAlH reduction. The mechanism for a NaBH reduction is the same except methanol is the proton source used in the second step. 1) Nucleopilic attack by the hydride anion 2) The alkoxide is protonated Two practical sources of hydride-like reactivity are the complex metal hydrides lithium aluminum hydride (LiAlH ) and sodium borohydride (NaBH ). These are both white (or near white) solids, which are prepared from lithium or sodium hydrides by reaction with aluminum or boron halides and esters. Lithium aluminum hydride is by far the most reactive of the two compounds, reacting violently with water, alcohols and other acidic groups with the evolution of hydrogen gas. The following table summarizes some important characteristics of these useful reagents. 1) Please draw the products of the following reactions: 2) Please draw the structure of the molecule which must be reacted to produce the product. 3) Deuterium oxide (D O) is a form of water where the hydrogens have been replaced by deuteriums. For the following LiAlH reduction the water typically used has been replaced by deuterium oxide. Please draw the product of the reaction and place the deuterium in the proper location. Hint! Look at the mechanism of the reaction. 1) 2) 3) | 2,469 | 1,384 |
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An extraordinary variety of reactions of organic compounds are known to occur under the influence of visible and ultraviolet light. Some of these, such as the photochemical halogenation of alkanes and photosynthesis in green plants, already have been discussed (see and ). It is not our purpose here to review organic photochemistry in detail - rather, we shall mention a few types of important photochemical reactions and show how these can be explained by the principles discussed in the preceding section. Compounds have very different chemical behavior in their excited states compared to their ground states. Not only is the energy much higher, but the molecular geometry and electronic configurations are different. Intuitively, we expect that excited states of molecules, in which two electrons occupy orbitals, would have substantial diradical character. This is the case, especially for triplet states, as we shall see. We have mentioned how chlorine molecules dissociate to chlorine atoms on absorption of near-ultraviolet light and thereby cause radical-chain chlorination of saturated hydrocarbons ( ). Photochemical chlorination is an example of a photochemical reaction that can have a high - that is, many molecules of chlorination product can be generated per quantum of light absorbed. The quantum yield of a reaction is said to be unity when \(1 \: \text{mol}\) of reactant is converted to product(s) per einstein\(^1\) of light absorbed. The symbol for quantum yield is usually \(\Phi\). 2-Propanone (acetone) vapor undergoes a photodissociation reaction with \(313\)-\(\text{nm}\) light with \(\Phi\) somewhat less than unity. Absorption of light by 2-propanone results in the formation of an excited state that has sufficient energy to undergo cleavage of a \(\ce{C-C}\) bond (the weakest bond in the molecule) and form a methyl radical and an ethanoyl radical. This is a photochemical reaction: \(\tag{28-1}\) The subsequent steps are dark reactions. At temperatures much above room temperature, the ethanoyl radical breaks down to give another methyl radical and carbon monoxide: \(\tag{28-2}\) If this reaction goes to completion, the principal reaction products are ethane and carbon monoxide: \[2 \ce{CH_3} \cdot \rightarrow \ce{CH_3-CH_3} \tag{28-3}\] If the ethanoyl radical does not decompose completely, then some 2,3-butanedione also is formed. This reaction is quite important at room temperature or below: \(\tag{28-4}\) Lesser amounts of methane and ketene also are formed as the result of disproportionation reactions involving hydrogen-atom transfers of the types we have encountered previously in radical reactions (see ): \(\tag{28-5}\) The product-forming reactions, Equations 28-2 through 28-5, all depend on the primary photochemical event, Equation 28-1, which breaks the \(\ce{C-C}\) bond to the carbonyl group. This cleavage has been termed a after the eminent photochemist, R. G. W. Norrish:\(^2\) Another photochemical reaction is important for ketones that have at least one \(\gamma\) hydrogen on a chain connected to the carbonyl group, as in In this pathway ( ), cleavage occurs at the \(\ce{C}_\alpha \ce{-C}_\beta\) bond to give, as the major product, a ketone of shorter chain length and an alkene. Thus for 2-pentanone: This reaction occurs in an interesting way. Whatever the nature of then \(n \rightarrow \pi^*\) excited state, \(S_1\) or \(T_1\), the primary photochemical reaction is the abstraction of a hydrogen atom from the \(\gamma\) carbon by the carbonyl oxygen to give the diradical, \(1\): The subsequent dark reactions readily are understood as typical of diradicals. Cleavage of \(1\) at \(\ce{C}_\alpha \ce{-C}_\beta\) gives ethene and an enol, which rearranges to the ketone. Alternatively, \(1\) can cyclize to a cyclobutanol: A variety of photodissociation reactions have been found to take place with ketones, but the products almost always can be explained as the result of Norrish type I and/or II cleavage. Examples are: Diaryl ketones do not undergo photodissociation in the same way as alkyl ketones, probably because cleavage to phenyl and other aryl radicals is unfavorable (Table 4-6). Nevertheless, aromatic ketones are photochemically reactive in the presence of compounds that can donate a hydrogen atom, with the result that the carbonyl group is reduced. Indeed, one of the classic photochemical reactions of organic chemistry is the formation of 1,1,2,2-tetraphenyl-1,2-ethanediol (\(3\), benzopinacol) by the action of light on a solution of diphenylmethanone (\(2\), benzophenone) in isopropyl alcohol. The yield is quantitative. The light is absorbed by \(2\) and the resulting activated ketone, \(2^*\), removes a hydrogen from isopropyl alcohol: Benzopinacol results from dimerization of the radicals, \(4\): Since the quantum yields of 2-propanone and benzopinacol both are nearly unity when the light intensity is not high, it is clear that two of the radicals, \(4\), must be formed for each molecule of \(2\) that becomes activated by light. This is possible if the 2-hydroxy-2-propyl radical formed by Equation 28-7 reacts with \(2\) to give a second diphenylhydroxymethyl radical: This reaction is energetically favorable because of the greater possibility for delocalization of the odd electron in \(4\) than in the 2-hydroxy-2-propyl radical. Photochemical formation of \(3\) also can be achieved from diphenylmethanone, \(2\), and diphenylmethanol, \(5\): The mechanism is similar to that for isopropyl alcohol as the reducing agent: This reduction is believed to involve the triplet state of \(2\) by the following argument: Formation of \(3\) is reasonably efficient even when the concentration of the alcohol, \(5\) is low; therefore, whatever the excited state of the ketone, \(2^*\), that accepts a hydrogen atom from \(5\), it must be a fairly long-lived one. Because solutions of \(2\) show no visible fluorescence, they must be converted rapidly to another state of longer life than the singlet \(\left( S_1 \right)\). The long-lived state is then most reasonably a triplet state. In fact, if naphthalene is added to the reaction mixture, formation of benzopinacol, \(3\) is drastically inhibited because the benzophenone triplet transfers energy to naphthalene more rapidly than it reacts with the alcohol, \(5\) (see ). An important problem in many syntheses is to produce the desired isomer of a cis-trans pair of alkenes. The problem would not arise if it were possible to isomerize the undesired isomer to the desired isomer. In many cases such isomerizations can be carried out photochemically. A typical example is afforded by and 1,2-diphenylethene (stilbene): Here the trans form is easily available by a variety of reactions and is more stable than the cis isomer because it is less sterically hindered. However, it is possible to produce a mixture containing mostly cis isomer by irradiating a solution of the trans isomer in the presence of a suitable photosensitizer. This process in no way contravenes the laws of thermodynamics because the input of radiant energy permits the equilibrium point to be shifted from what it would be normally. Isomerization appears to occur by the following sequence: The sensitizer, usually a ketone such as benzophenone or 1-(2-naphthyl)ethanone, is raised by an \(n \rightarrow \pi^*\) transition from the singlet ground state \(\left( S_0 \right)\) to an excited state \(\left( S_1 \right)\) by absorption of light. Intersystem crossing then occurs rapidly to give the triplet state \(\left( T_1 \right)\) of the sensitizer: The next step is excitation of the alkene by energy transfer from the triplet state of the sensitizer. Remember, the net electron spin is conserved during energy transfer, which means that the alkene will be excited to the triplet state: \[^3\text{Sens}^* \left( \uparrow \uparrow \right) + ^1\text{Alkene} \left( \uparrow \downarrow \right) \rightarrow ^1\text{Sens} \left( \uparrow \downarrow \right) + ^3\text{Alkene}^* \left( \uparrow \uparrow \right)\] The triplet state of the alkene is most stable when the \(p\) orbitals, which make up the normal \(\pi\) system of the double bond, are not parallel to one another (Figure 6-17). Therefore, if the energy-transfer process leads initially to a planar triplet, this is converted rapidly to the more stable nonplanar form. The excitation of either the cis or the trans isomer of the alkene appears to lead to a common triplet state, as shown in Figure 28-4. The final step in the isomerization is decay of the alkene triplet to the ground state. This happens either by emission of light (phosphorescence) or by having the triplet energy converted to thermal energy without emission of light. Either way, the cis or trans isomer could be formed and, as can be seen from Figure 28-4, the ratio of isomers produced depends on the relative rates of decay of the alkene triplet to the ground-state isomers, \(k_c/k_t\). This ratio turns out to favor formation of the isomer. Therefore, provided both isomers can be photosensitized efficiently, sensitized irradiation of either one will lead ultimately to a mixture of both, in which the thermochemically less stable isomer predominates. The sensitizer must have a triplet energy in excess of the triplet energy of the alkene for energy transfer to occur, and the or equilibrium point is independent of the nature of the sensitizer when the latter transfers energy efficiently to cis and trans isomers. In the practical use of the sensitized photochemical equilibrium of cis and trans isomers, it is normally necessary to carry out pilot experiments to determine what sensitizers are useful. Another example of how photochemical isomerization can be used is provided by the equilibration of the \(E\) and \(Z\) form of 1-bromo-2-phenyl-1-propene: The \(E\) isomer is formed to the extent of \(95\%\) in the dehydrohalogenation of 1,2-dibromo-2-phenylpropane: Photoisomerization of the elimination product with 1-(2-naphthyl)ethanone as sensitizer produce a mixture containing \(85\%\) of the \(Z\) isomer. One may well ask why the isomerization of alkenes discussed in the preceding section requires a sensitizer. Why cannot the same result be achieved by direct irradiation? One reason is that a \(\pi \rightarrow \pi^*\) singlet excited state \(\left( S_1 \right)\) produced by direct irradiation of an alkene or arene crosses over to the triplet state \(\left( T_1 \right)\) inefficiently (compared to \(n \rightarrow \pi^*\) excitation of ketones). Also, the \(S_1\) state leads to reactions beside isomerization which, in the case of 1,2-diphenylethene and other conjugated hydrocarbons, produce cyclic products. For example, -1,2-diphenylethene irradiated in the presence of oxygen gives phenanthrene by the sequence of Equation 28-8. The primary photoreaction is cyclization to a dihydrophenanthrene intermediate, \(6\), which, in the presence of oxygen, is converted to phenanthrene: The cyclization step of Equation 28-8 is a photochemical counterpart of the electrocyclic reactions discussed in . Many similar photochemical reactions of conjugated dienes and trienes are known, and they are of great interest because, like their thermal relatives, they often are stereospecific but tend to exhibit stereochemistry opposite to what is observed for formally similar thermal reactions. For example, These reactions are \(4n\)-electron concerted processes controlled by the symmetry of the reacting orbitals. The thermal reaction is most favorable with a Mobius transition state (achieved by conrotation), whereas the photochemical reaction is most favorable with a Huckel transition state (achieved by disrotation). Conjugated dienes also undergo photochemical cycloaddition reactions. Related thermal cycloadditions of alkadienes have been discussed in , , and , but the thermal and photochemical reactions frequently give different cyclic products. Butadiene provides an excellent example of the differences: In the thermal reaction the [4 + 2] or Diels-Alder adduct is the major product, whereas in the photochemical reaction [2 + 2] cycloadditions dominate. Because the photochemical additions are sensitized by a ketone, \(\ce{C_6H_5COCH_3}\), these cycloadditions occur through the triplet state of 1,3-butadiene and, as a result, it is not surprising that these cycloadditions are stepwise, nonstereospecific, and involve diradical intermediates. Direct irradiation of 1,3-butadiene with \(254\)-\(\text{nm}\) light produces cyclobutene and small amounts of bicyclo[1.1.0]butane along with dimers. In contrast to conjugated dienes, simple alkenes such as 2-butene do not react easily by photosensitized cycloaddition. But they will form [2 + 2] cycloadducts on direct irradiation. These additions occur by way of a singlet excited state and are stereospecific: A related reaction, which has no precedent in thermal chemistry, is the cycloaddition of an alkene and an aldehyde or ketone to form an oxacyclobutane: In this kind of addition the ground-state alkene \(\left( S_0 \right)\) reacts with an excited state (usually \(T_1\)) of the carbonyl compound by way of a diradical intermediate: The ground state of molecular oxygen is unusual because it is a triplet state. Two electrons of parallel spin occupy separate \(\pi\) orbitals of equal energy (degenerate), as shown schematically in Figure 28-5,\(^3\) The next two higher electronic states both are singlet states and lie respectively \(24\) and \(37 \: \text{kcal mol}^{-1}\) above the ground state. From this we can understand why ordinary oxygen has the properties of a diradical and reacts rapidly with many radicals, as in the radical-chain oxidation of hydrocarbons (autoxidation; and ): Oxygen also efficiently quenches excited triplet states of other molecules \(\left( ^3A^* \right)\) and, in accepting triplet energy, is itself promoted to an excited state. Notice that the total spin orientation is conserved: Singlet oxygen is highly reactive toward many organic molecules and will form oxygenated addition or substitution products. As one example, conjugated dienes react with singlet oxygen to give peroxides by [4 + 2] cycloaddition. Because only singlet states are involved, this addition is quite analogous to thermal Diels-Alder reactions ( ): If the alkene or alkadiene has at least one hydrogen on the carbon adjacent to the double bond, reaction with singlet oxygen may give hydroperoxides. The mechanism of this reaction is related to [4 + 2] cycloadditions and is presumed to occur through a Huckel pericyclic transition state (see ): Many reactions of this type can be achieved by allowing the hydrocarbon to react with oxygen in the presence of a sensitizing dye that strongly absorbs visible light. The dyes most commonly used for this purpose include fluorescein (and its chlorinated analogs, eosin and rose bengal), methylene blue, and porphyrin pigments (such as chlorophyll). The overall process of photosensitized oxygenation of a substrate \(\left( \ce{A} \right)\) proceeds by the following steps: \[^1\text{Sens} \overset{h \nu}{\longrightarrow} \: ^1\text{Sens}^* \longrightarrow ^3\text{Sens}^*\] \[^3\text{Sens}^* + ^3\ce{O_2} \longrightarrow ^1\text{Sens} + ^1\ce{O_2}*\] \[^1\ce{O_2}* + \ce{A} \longrightarrow \ce{AO_2}\] Singlet oxygen can be produced chemically as well as by photochemical sensitization. There are several chemical methods available, one of the best known being the reaction of sodium hypochlorite with peroxide: \[\ce{NaOCl} + \ce{H_2O_2} \rightarrow \ce{NaCl} + \ce{H_2O} + ^1\ce{O_2} \tag{28-9}\] An alternative method of formation, which can be used in organic solvents at low temperatures, involves the thermal decomposition of triethyl phosphite ozonide (Equation 28-10): Regardless of whether singlet oxygen is formed chemically or photochemically, it gives similar products in reactions with alkenes. Photosensitized reactions of oxygen are largely damaging to living organisms. Indeed, singlet oxygen reacts destructively with amino acids, proteins, and nucleic acids. How does an organism protect itself against the damaging effects of oxygen? There are no simple answers, but green plants provide a clue. Chlorophyll is an excellent sensitizing dye for singlet oxygen; yet green plants evidently are not harmed because of it. A reason may be that singlet oxygen is quenched very efficiently by other plant pigments, especially the carotenoid pigments such as \(\beta\)-carotene ( ). That this is the case is indicated by the fact that mutant plants unable to synthesize carotene are killed rapidly by oxygen and light. That direct irradiation with ultraviolet light is damaging to single-cell organisms is well known. It also is known that the nucleic acids, DNA and RNA, are the important targets of photochemical damage, and this knowledge has stimulated much research in the field of photobiology in the hope of unravelling the chemistry involved. An interesting and significant outcome is the finding that the pyrimidine bases of nucleic acids (uracil, thymine, and cytosine) are photoreactive and undergo on irradiation with ultraviolet light. Thymine, for example, gives a dimer of structure \(7\): Comparable experiments with the nucleic acids have confirmed that cycloaddition of their pyrimidine bases also occurs with ultraviolet light and effectively cross-links the chains, a process obviously quite inimical to the functioning of the DNA ( ). A remarkable and not well understood aspect of photobiology is the repair and defense mechanism both plants and animals possess to minimize the damaging effects of radiation. On the positive side, there are photochemical reactions that are essential for human health. One of these is the formation of vitamin D (the antirachitic vitamin) by irradiation of ergosterol. This photochemical reaction is an electrocyclic ring opening of the cyclohexadiene ring of ergosterol of the type described in . The product, previtamin D , subsequently rearranges thermally to vitamin D : \(^1\)The einstein unit is defined in . \(^2\)Recipient with G. Porter of the Nobel Prize in chemistry in 1967 for work on photochemical reactions. \(^3\)For a more detailed account of the electronic configuration of molecular oxygen, see M. Orchin and H. H. Jaffe, , Houghton Mifflin Co., Boston, 1967; or H. B. Gray, , W. A. Benjamin, Inc., Menlo Park, Calif., 1973. and (1977) | 18,493 | 1,385 |
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Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY- ; ). | 817 | 1,386 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Radiative_Decay/Fluorescence |
Fluorescence, a type of luminescence, occurs in gas, liquid or solid chemical systems. Fluorescence is brought about by absorption of photons in the singlet ground state promoted to a singlet excited state. The spin of the electron is still paired with the ground state electron, unlike phosphorescence. As the excited molecule returns to ground state, it involves the emission of a photon of lower energy, which corresponds to a longer wavelength, than the absorbed photon. The energy loss is due to vibrational relaxation while in the excited state. Fluorescent bands center at wavelengths longer than the resonance line. This shift toward longer wavelengths is called a Stokes shift. Excited states are short-lived with a lifetime at about 10 seconds. Molecular structure and chemical environment affect whether or not a substance luminesces. When luminescence does occur, molecular structure and chemical environment determine the intensity of emission. Generally molecules that fluoresce are conjugated systems. Fluorescence occurs when an atom or molecules relaxes through vibrational relaxation to its ground state after being electrically excited. The specific frequencies of excitation and emission are dependent on the molecule or atom. \[ S_0 + h\nu_{ex} = S_1 \] where Figure 1 is a Jablonski energy diagram representing fluorescence. The purple arrow represents the absorption of light. The green arrow represents vibrational relaxation from singlet excited state, S to S . This process is a non-radiative relaxation in which the excitation energy is dispersed as vibrations or heat to the solvent, and no photon is emitted. The yellow arrow represents fluorescence to the singlet ground state, S . The fluorescence quantum yield ((\Phi\)) gives the efficiency of the fluorescence process. It is the ratio of photons emitted to photons absorbed. \[ \Phi = \dfrac{\text{ # emitted photons }}{\text{ # absorbed photons }} \label{Eq0} \] If every photon absorbed results in a photon emitted. The maximum fluorescence quantum yield is 1.0, and compounds with quantum yields of 0.10 are still considered fluorescent. Another way to define the fluorescence quantum yield is by the excited state decay rates: \[ \Phi = \dfrac{k_f}{\sum_i k_i} \label{Eq1}\] where \(k_f\) is the rate of spontaneous emission of radiation and the denominator is the sum of all rates of excited state decay for each deactivation process (ie phosphorescence, intersystem crossing, internal conversion…). The fluorescence lifetime is the average time the molecule remains in its excited state before emitting a photon. Fluorescence typically follows : \[ [S_1] = [S_1]_o e^{- t/\tau} \label{Eq2}\] where Various radiative and non-radiative processes can de-populate the excited state so the total decay rate is the sum over all rates: \[ \tau_{tot}= \tau_{rad} + \tau_{nrad} \label{Eq3}\] where \( \tau_{tot} \) is the total decay rate, \( \tau_{rad} \) the radiative decay rate and \(\tau_{nrad} \) the non-radiative decay rate. If the rate of spontaneous emission or any of the other rates are fast, the lifetime is short. The average lifetime of fluorescent compounds that emit photons with energies from the UV to near infrared are within the range of 0.5 to 20 nanoseconds. The fluorescence intensity, \(I_F\) is proportional to the amount of light absorbed and the fluorescence quantum yield, \(\Phi\) \[I_f=kI_o \phi [1-(10^{-εbc})] \label{Eq4}\] where If dilute solutions are used so that less than 2% of the excitation energy is absorbed, then an approximation can be made so that \[10^{x} \approx 1+x + ...\] so Equation \(\ref{Eq4}\) can be simplified to \[I_f=kI_o\Phi [εbc] \label{Eq4a}\] This relationship shows that fluorescence intensity is to concentration. Fluorescence rarely results from absorption of UV-radiation of wavelengths shorter than 250 nm because this type of radiation is sufficiently energetic to cause deactivation of the excited state by predissociation or dissociation. Most organic molecules have at least some bonds that can be ruptured by energies of this strength. Consequently, fluorescence due to \(\\sigma^* \rightarrow \sigma\) transitions is rarely observed. Instead such emission is confined to the less energetic \(\pi^* \rightarrow \pi\) and \(\pi^* \rightarrow n\) processes. Fluorescence commonly occurs from a transition from the lowest vibrational level of the first excited electronic state to the one of the vibrational levels of the electronic ground state. Quantum yield (\(\Phi\)) is greater for \(\pi^* \rightarrow \pi\) transition because these excited states show short average lifetimes (larger \(k_f\)) and because deactivation processes that compete with fluorescence is not as likely to happen. Molar absorptivity of → * transitions is 100-1000 fold greater. The average lifetime is 10 to 10 seconds for ?, ?* states and 10 to 10 seconds for n, * states. Figure 2 is a schematic of a typical filter fluorometer that uses a source beam for fluorescence excitation and a pair of photomultiplier tubes as transducers. The source beam is split near the source into a reference beam and a sample beam. The reference beam is attenuated by the aperture disk so that its intensity is roughly the same as the fluorescence intensity. Both beams pass through the primary filter, with the reference beam being reflected to the reference photomultiplier tube. The sample beam is focused on the sample by a pair of lenses and causes fluorescence emission. The emitted radiation passes through a second filter and then is focused on the sample photomultiplier tube. The electrical outputs from the two transducers are then processed by an analog to digital converter to compute the ratio of the sample to reference intensities, which can then be used for qualitative and quantitative analysis. To obtain an emission spectrum, the excitation monochromator is fixed and the emission monochromator varies. To obtain an excitation spectrum, the excitation monochromator varies while the emission monochromator is fixed. Fluorescence spectroscopy can be used to measure the concentration of a compound because the fluorescence intensity is linearly proportional to the concentration of the fluorescent molecule. Fluorescent molecules can also be used as tags. For example, fluorescence in situ hybridization (FISH) is a method of determining what genes are present in an organism's genome. Single stranded DNA encoding a gene of interest is covalently bonded to a fluorescent molecule and washed over the organism's chromosome, binding to its complementary sequence. The presence and placement of the gene in the organism then fluoresces when shined with ultraviolet light. Green fluorescence protein (GFP) is used in molecular biology to monitor the activity of proteins. The gene encoding GFP can be inserted next to a gene encoding a protein that will be studied. When the genes are expressed, the protein will be attached to GFP and can be identified in the cell by its fluorescence. | 7,043 | 1,388 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Advanced_Thermodynamics/5._Thermodynamic_Processes |
Although thermodynamics strictly speaking refers only to equilibria, by introducing the concept of work flow and heat flow, as discussed in , we can discuss processes by which a system is moved from one state to another. The concepts of heat and work are only meaningful because certain highly averaged variables are stable as a function of time. Energy changes related to such variables, like volume, are considered work. Microscopic variables, like the position of a single particle, are unstable and unpredictable as a function of time. If we treat the system classically, we would say such a variable has a Lyapunov coefficient \(L > 0\). That is, its orbit diverges from prediction as \(\Delta X = \Delta X_0 \exp[-Lt]\), where \(\Delta X_0\) is the initial measurement error in the variable, and \(\Delta X\) is the error at time \(t\). Here’s an example from meteorology: the Lyapunov coefficient for seasonal temperatures is 0; they have predictable averages (colder in the January, warmer in July in the northern hemisphere). Small weather patterns (e.g. motion of a cloud front) have Lyapunov coefficient of about (2 days) . This is not a question of better measurements, but a fundamental limitation: the error grows exponentially, so a multiplicative improvement in measurement buys you only a linear improvement in time. The situation is no better in quantum mechanics, where a particle initially started out in a position eigenstate \(\delta(x-x_0)\) evolves to ever greater position uncertainty as a matter of . Thus both classical and quantum motions are inherently unpredictable, for different reasons; the corresponding energy flow is heat flow. But when one averages over enough degrees of freedom, the averaged variables may be well behaved; that energy flow is work flow. Types of processes: A quasistatic process lies on the surface \(S(U,x_i)\) Note: this cannot be achieved in reality, but approximated by taking small steps whose endpoints lie on the fundamental surface. Note: according to postulate 2, upon change of constraints, any process must satisfy \(S_{final}>S_{initial}\). The reverse of such a process would violate postulate 2 (\(S_{final}<S_{initial}\)), and real processes are therefore irreversible. A reversible process is the idealized quasistatic limit where \(S_{final} = Si_{nitial}\). : Irreversible, quasistatic and reversible processes Thermodynamics only makes statements about equilibrium states, when the fundamental equation is satisfied. However, by using quasistatic and reversible processes as idealized limits, we can derive inequalities satisfied by real processes. As seen earlier \(dU = đQ + TdS\) for small (quasistatic) heat transfers in absence of work. The best we can do in the presence of work is therefore that all of \[ dU = Tds - pdV + \sum_i \mu_idn_i + \Gamma \,dA + H\,dM + ...\] goes into work for the first term, which corresponds to the infinitesimal heat transfer. \(–PdV\) would be simple bulk volume work (e.g. expansion of a gas), \( H dM\) would be chemical work (e.g. electrochemical if \(n\) refers to the mole number of an ion), \(\Gamma dA\) would be surface tension work (e.g. blowing a soap bubble), \(H dM\) would be magnetic work, etc. The heat transfer cannot be reduced below \(TdS\). However, part or all of the energy flow \(đW\) can of course be converted to heat \(đQ_W\). The entropy will rise further by \(T\,dS_W=đQ_W\), and correspondingly less work is done: \[ dU=TdS + đQ_W + đW_{\text{left over}}\] If all possible work is converted to heat, \(đW_{\text{left over}}=0\) and heat flow is maximized. The following theorem tells us how much work at most we can extract from a system. | 3,702 | 1,389 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/The_Scientific_Method/Science_vs._Pseudo-science%3A_Limitations_of_the_Scientific_Method |
Pseudo-science, basically "fake"-science," consists of scientific claims which are made to appear factual when they are actually false. Many people question whether Pseudo-science should even contain the word "science" as Pseudo-science isn't really even an imitation of science; it pretty much disregards the scientific method all together. Also known as alternative or fringe-science, Pseudo-science relies on invalid arguments called sophisms, a word Webster dictionary defines as "an argument apparently correct in form but actually invalid; especially : such an argument used to deceive". Pseudo-science usually lacks supporting evidence and does not abide by the scientific method. That is, pseudo-theories fail to use carefully cultivated and controlled experiments to test a hypothesis. A scientific hypothesis must include observable, empirical and testable data, and must allow other experts to test the hypothesis. Pseudo-science does not accomplish these goals. Several examples of Pseudo-Science include phrenology, astrology, homeopathy, reflexology and iridology. In order to distinguish a pseudoscience, one must look at the definition of science, and the aspects that make science what it is. Science is a process based on observations, conjectures, and assessments to provide better understanding of the natural phenomena of the world. Science generally always follows a formal system of inquiry which consists of observations, explanations, experiments, and lastly, hypothesis and predictions. Scientific theories are always challenged by experts and revised to fit new theories. Pseudo-science, however, is mostly based on beliefs and it greatly opposes contradictions. Their hypothesis are never revised to fit new data or information. Scientist continually disprove ideas to achieve a better understanding of the physical world, whereas pseudo-scienctists focus on proving theories to make their claims seem plausible. For example, science text books come out with new editions every couple of years to correct typos, update information, add new illustrations, etc. However, it has been observed that pseudo-science textbooks only come out with one edition, and is never updated or revised even if their theory has been proven to be false. Pseudo-science beliefs often tend to be greatly exaggerated and very vague. Complicated technical language is often used to sound impressive but it is usually meaningless. For example, a phrase like "energy vibrations" is used to sound remarkable but a phrase like this is insignificant and doesn't really explain anything. Furthermore, Pseudo-science often consists of outrageous, yet unprovable claims. Thus, pseudo-scientists tend to focus on confirming their ideas, rather than finding evidence that refutes them. The following dialogue contains the thought-processes behind Pseudo-Science. The dialogue above features many key characteristics of Pseudo-Science. The speaker makes his or her point valid though the two facts alone that her friend had a personal experience and that science has no proof to prove the theory wrong. Finally, the speaker insults anyone who would challenge the theory. In science, challenges to a theory are accepted as everyone has the same common goal of improving the understanding of the natural word. Below is a table that lays out the key characteristics of Science and Pseudo-Science Phrenology, also known as craniology, was a "science" popular during the early 1800s that was centered around the idea that the brain was an organ of the mind. During this time, most people believed that the brain was divided into distinct sections that all controlled different parts of a person's personality or intelligence. The basis of phrenology revolves around the concept that the brain mirrors a muscle and those parts of the brain which are "exercised" the most, will be proportionally larger than those parts of the brain that aren't often used. Thus, the scientists pictured the brain as a bumpy surface, with the make-up of the surface differing for every person depending on their personality and intelligence. By the mid 19th century, automated phrenology machines existed, which was basically a set of spring loaded probes that were placed on the head to measure the topography of one's skull. The machine then gave an automated reading about a person's characteristics based on this. Let's consider some of the key characteristics of pseudo-science from our chart, and see how they apply to phrenology. Reflexology is a way of treatment that involves physically applying pressure to the feet or hands with the belief that each are divided up into different zones that are "connected" to other parts of the body. Thus, reflexologists assert that they can make physical changes throughout the body simply by rubbing ones hands or feet. Like we did with phrenology, lets go through some of the main characteristics of Pseudo-Science and see how they apply to reflexology. An important distinction should be made between Pseudo-science and other types of defective science. Take for example, the "discovery" of N-rays. While attempting to polarize X-rays, physicist René Prosper Blondlot claimed to have discovered a new type of radiation he called N-rays. After Blondlot shared with others his exciting discovery, many other scientists confirmed his beliefs by saying they too had saw the N-rays. Though he claimed N-rays contained impossible properties, Blondlot asserted when he put a hot wire in an iron tube, he was able to detect the N-rays when he used a thread of calcium sulfite that glowed slightly when the rays were sent through a prism of aluminum. Blondlot claimed that all substances except some treated metals and green wood emit N-rays. However, Nature magazine was skeptical of Blondlot and sent physicist Robert Wood to investigate. Before Blondlot was about to show Wood the rays, Wood removed the aluminum prism from the machine without telling Blondlot. Without the prism, the rays would be impossible to detect. However, Blondlot claimed to still see the N-rays, demonstrating how the N-rays did not exist; Blondlot just wanted them to exist. This is an example of Pathological science, a phenomenon which occurs when scientists practice wishful data interpretation and come up with results they want to see. This case of Pathological science and Pseudo-science differ. For one, Blondlot asked for a confirmation by other experts, something Pseudo-science usually lacks. More importantly, in pathological science, a scientist starts by following the scientific method; Blondlot was indeed doing an experiment when he made his discovery and proceeded to experiment when he found the substances that did not emit the rays. However, Pseudo-science usually includes a complete disregard of the scientific method, while Pathological scientists includes following the scientific method but seeing the results you wish to see. Another type of invalid science, called hoax science occurred in 1999 when a team at the Lawrence Berkeley National Laboratory claimed to have discovered elements 116 and 118 when they bombarded Lead with Krypton particles. However, by 2002 it had been discovered that physicist Victor Ninov had intentionally fudged the data to get the ideal results. Thus, the concept of hoax science, which occurs when the data is intentionally falsified, differs both from pathological and pseudo science. In pathological science, scientists wishfully interpret the data and legitimately think they see what they want to see. However, in Hoax science, scientists know they don't see what they want to see, but just say they did. Finally, in Pseudo-Science, scientists don't consider the scientific method at all as they don't use valid experiments to back up their data in the first place. There have been incidents where what was once considered pseudo-science became a respectable theory. In 1911, German astronomer and meteorologist Alfred Wegener first began developing the idea of Continental Drift. The observation that the coastlines of African and South American seemed to fit together was not a new observation: scientists just couldn't believe that the continents could have drifted so far to cross the 5,000 mile Atlantic Ocean. At the time, it was a common theory that a land bridge had existed between Africa and Brazil. However, one day in the library Wegener read a study about a certain species that could not have crossed the ocean, yet had fossils appeared on both sides of the supposed land bridge. This piece of evidence lead Wegener to believe that our world had once been one piece, and had since drifted apart. However, Wegener's theory encountered much hostility and disbelief. In this time, it was the norm for scientists to stay within the scopes of their fields, meaning biologists did not study physics, chemists did not study oceanology and of course, meteorologists/astronomers like Wegener did not study geology. Thus, Wegener's theory faced much criticism just due to the fact that he was not a geologist. Also, Wegener could not explain why the continents moved, just that they did. This lack of reasoning lead to more skepticism about the theory and all these factors combined lead to the viewing of continental drift as Pseudo-Science. However, today much evidence exists that shows that Continental Drift is a perfectly acceptable scientific theory. Today, the modern ideas of plate tectonics can help explain Continental Drift, as the Plate Tectonic Theory presents the idea that the earth's surface is made up of several large plates that often move up to a few inches every year. Also, the development of paleomagnetism, which allows us to determine the earth's magnetic poles at the time a rock formed, suggests that the earth's magnetic poles have changed many times in the last 175 million years and that at one time South America and Africa were connected. Due to the need to have completely controlled experiments to test a hypothesis, science can not prove everything. For example, ideas about God and other supernatural beings can never be confirmed or denied, as no experiment exists that could test their presence. Supporters of Intelligent Design attempt to convey their beliefs as scientific, but nonetheless the scientific method can never prove this. Science is meant to give us a better understanding of the mysteries of the the natural world, by refuting previous hypotheses, and the existence of supernatural beings lies outside of science all together. Another limitation of the scientific method is when it comes making judgements about whether certain scientific phenomenons are "good" or "bad". For example, the scientific method cannot alone say that global warming is bad or harmful to the world, as it can only study the objective causes and consequences. Furthermore, science cannot answer questions about morality, as scientific results lay out of the scope of cultural, religious and social influences. Determine if each statement is true or false (see answers at bottom of the page) | 11,091 | 1,390 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/09._The_Hydrogen_Atom/Chapter_7.__Hydrogen_Atom |
When gaseous hydrogen in a glass tube is excited by a \(5000\)-volt electrical discharge, four lines are observed in the visible part of the emission spectrum: red at \(656.3\) nm, blue-green at \(486.1\) nm, blue violet at \(434.1\) nm and violet at \(410.2\) nm: Other series of lines have been observed in the ultraviolet and infrared regions. Rydberg (1890) found that all the lines of the atomic hydrogen spectrum could be fitted to a single formula \[ \dfrac{1}{\lambda} = \mathcal{R} \left( \dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}} \right), \quad n_1 = 1, \: 2, \: 3..., \: n_2 > n_1 \label{1}\] where \(\mathcal{R}\), known as the Rydberg constant, has the value \(109,677\) cm for hydrogen. The reciprocal of wavelength, in units of cm , is in general use by spectroscopists. This unit is also designated , since it represents the number of wavelengths per cm. The Balmer series of spectral lines in the visible region, shown in Figure \(\Page {1}\), correspond to the values \(n_1 = 2, \: n_2 = 3, \: 4, \: 5\) and \(6\). The lines with \(n_1 = 1\) in the ultraviolet make up the Lyman series. The line with \(n_2 = 2\), designated the Lyman alpha, has the longest wavelength (lowest wavenumber) in this series, with \(1/ \lambda = 82.258\) cm or \(\lambda = 121.57\) nm. Other atomic species have line spectra, which can be used as a "fingerprint" to identify the element. However, no atom other than hydrogen has a simple relation analogous to Equation \(\ref{1}\) for its spectral frequencies. Bohr in 1913 proposed that all atomic spectral lines arise from transitions between discrete energy levels, giving a photon such that \[ \Delta E = h \nu = \dfrac{hc}{\lambda} \label{2}\] This is called the . We now understand that the atomic transition energy \(\Delta E\) is equal to the energy of a photon, as proposed earlier by Planck and Einstein. The nuclear model proposed by Rutherford in 1911 pictures the atom as a heavy, positively-charged nucleus, around which much lighter, negatively-charged electrons circulate, much like planets in the Solar system. This model is however completely untenable from the standpoint of classical electromagnetic theory, for an accelerating electron (circular motion represents an acceleration) should radiate away its energy. In fact, a hydrogen atom should exist for no longer than \(5 \times 10^{-11}\) sec, time enough for the electron's death spiral into the nucleus. This is one of the worst quantitative predictions in the history of physics. It has been called the Hindenberg disaster on an atomic level. (Recall that the Hindenberg, a hydrogen-filled dirigible, crashed and burned in a famous disaster in 1937.) Bohr sought to avoid an atomic catastrophe by proposing that certain orbits of the electron around the nucleus could be exempted from classical electrodynamics and remain stable. The Bohr model was quantitatively successful for the hydrogen atom, as we shall now show. We recall that the attraction between two opposite charges, such as the electron and proton, is given by Coulomb's law \[F = \begin{cases} -\dfrac{e^{2}}{r^{2}} \quad \mathsf{(gaussian \: units)} \\ -\dfrac{e^{2}}{4 \pi \epsilon_0 r^{2}} \quad \mathsf{(SI \: units)} \end{cases} \label{3}\] We prefer to use the Gaussian system in applications to atomic phenomena. Since the Coulomb attraction is a central force (dependent only on r), the potential energy is related by \[F = -\dfrac{dV(r)}{dr} \label{4}\] We find therefore, for the mutual potential energy of a proton and electron, \[V(r) = -\dfrac{e^2}{r} \label{5}\] Bohr considered an electron in a circular orbit of radius \(r\) around the proton. To remain in this orbit, the electron must be experiencing a centripetal acceleration \[a = -\dfrac{v^{2}}{r} \label{6}\] where \(v\) is the speed of the electron. Using Equations \(\ref{4}\) and \(\ref{6}\) in Newton's second law, we find \[\dfrac{e^{2}}{r^{2}} = \dfrac{mv^{2}}{r} \label{7}\] where \(m\) is the mass of the electron. For simplicity, we assume that the proton mass is infinite (actually \(m_p \approx 1836 m_e\) f the hydrogen atom is the sum of the kinetic and potential energies: \[E = T + V = \dfrac{1}{2} mv^{2} - \dfrac{e^{2}}{r} \label{8}\] Using Equation \(\ref{7}\), we see that \[T = -\dfrac{1}{2} V\ \qquad \mathsf{and} \qquad E = \dfrac{1}{2} V = -T \label{9}\] This is the form of the virial theorem for a force law varying as \(r^{-2}\). Note that the energy of a bound atom is , since it is lower than the energy of the separated electron and proton, which is taken to be zero. For further progress, we need some restriction on the possible values of \(r\) or \(v\). This is where we can introduce the quantization of angular momentum \(\mathbf{L} = \mathbf{r} \times \mathbf{p}\). Since \(\mathbf{p}\) is perpendicular to \(\mathbf{r}\), we can write simply \[L = rp = mvr \label{10}\] Using Equation \(\ref{9}\), we find also that \[r = \dfrac{L^{2}}{me^{2}} \label{11}\] We introduce angular momentum quantization, writing \[L = n\hbar, \qquad n = 1, \: 2... \label{12}\] excluding \(n = 0\), since the electron would then not be in a circular orbit. The allowed orbital radii are then given by \[r_n = n^{2} a_0 \label{13}\] where \[a_0 \equiv \dfrac{\hbar^{2}}{me^{2}} = 5.29 \times 10^{-11} \: \mathsf{m} = 0.529Å \label{14}\] which is known as the . The corresponding energy is \[E_n = -\dfrac{e^{2}}{2a_0n^{2}} = -\dfrac{me^{4}}{2\hbar^{2}n^{2}}, \qquad n = 1, \: 2... \label{15}\] Rydberg's formula (Equation \(\ref{1}\)) can now be deduced from the Bohr model. We have \[ \dfrac{hc}{\lambda} = E_{n_2} - E_{n_1} = \dfrac{2\pi^{2}me^{4}}{h^{2}} \left( \dfrac{1}{n_1^{2}} - \dfrac{1}{n_2^{2}} \right) \label{16}\] and the Rydbeg constant can be identified as \[\mathcal{R} = \dfrac{2\pi^{2}me^{4}}{h^{3}c} \approx 109,737 \: \mathsf{cm}^{-1} \label{17}\] The slight discrepency with the experimental value for hydrogen \( (109,677) \) is due to the finite proton mass. This will be corrected later. The Bohr model can be readily extended to hydrogenlike ions, systems in which a single electron orbits a nucleus of arbitrary atomic number \(Z\). Thus \(Z = 1\) for hydrogen, \(Z = 2\) for \(\mathsf{He}^{+}\), \(Z = 3\) for \(\mathsf{Li}^{++}\), and so on. The Coulomb potential \(\ref{5}\) generalizes to \[V(r) = -\dfrac{Ze^{2}}{r}, \label{18}\] the radius of the orbit (Equation \(\ref{13}\)) becomes \[r_n = \dfrac{n^{2}a_0}{Z} \label{19}\] and the energy Equation \(\ref{15}\) becomes \[E_n = -\dfrac{Z^{2}e^{2}}{2a_0n^{2}} \label{20}\] De Broglie's proposal that electrons can have wavelike properties was actually inspired by the Bohr atomic model. Since \[L = rp = n\hbar = \dfrac{nh}{2\pi} \label{21}\] we find \[2\pi r = \dfrac{nh}{p} = n\lambda \label{22}\] Therefore, each allowed orbit traces out an integral number of de Broglie wavelengths. Wilson (1915) and Sommerfeld (1916) generalized Bohr's formula for the allowed orbits to \[\oint p \, dr = nh, \qquad n =1, \: 2... \label{23}\] The Sommerfeld-Wilson quantum conditions Equation \(\ref{23}\) reduce to Bohr's results for circular orbits, but allow, in addition, elliptical orbits along which the momentum \(p\) is variable. According to Kepler's first law of planetary motion, the orbits of planets are ellipses with the Sun at one focus. Figure \(\Page {2}\) shows the generalization of the Bohr theory for hydrogen, including the elliptical orbits. The lowest energy state \(n = 1\) is still a circular orbit. But \(n = 2\) allows an elliptical orbit in addition to the circular one; \(n = 3\) has three possible orbits, and so on. The energy still depends on \(n\) alone, so that the elliptical orbits represent degenerate states. Atomic spectroscopy shows in fact that energy levels with \(n > 1\) consist of multiple states, as implied by the splitting of atomic lines by an electric field (Stark effect) or a magnetic field (Zeeman effect). Some of these generalized orbits are drawn schematically in Figure \(\Page {2}\). The Bohr model was an important first step in the historical development of quantum mechanics. It introduced the quantization of atomic energy levels and gave quantitative agreement with the atomic hydrogen spectrum. With the Sommerfeld-Wilson generalization, it accounted as well for the degeneracy of hydrogen energy levels. Although the Bohr model was able to sidestep the atomic "Hindenberg disaster," it cannot avoid what we might call the "Heisenberg disaster." By this we mean that the assumption of well-defined electronic orbits around a nucleus is completely contrary to the basic premises of quantum mechanics. Another flaw in the Bohr picture is that the angular momenta are all too large by one unit, for example, the ground state actually has zero orbital angular momentum (rather than \(\hbar\)). The assumption of well-defined electronic orbits around a nucleus in the Bohr atom is completely contrary to the basic premises of quantum mechanics. In contrast to the particle in a box and the harmonic oscillator, the hydrogen atom is a real physical system that can be treated exactly by quantum mechanics. In addition to their inherent significance, these solutions suggest prototypes for atomic orbitals used in approximate treatments of complex atoms and molecules. For an electron in the field of a nucleus of charge \(+Ze\), the Schrӧdinger equation can be written \[\left\{ -\dfrac{\hbar^{2}}{2m} \nabla^{2} - \dfrac{Ze^{2}}{r} \right\} \psi(r) = E\psi(r) \label{24}\] It is convenient to introduce in which length is measured in bohrs: \[a_0 = \dfrac{\hbar^{2}}{me^{2}} = 5.29 \times 10^{-11} \: \mathsf{m} \equiv 1 \: \mathsf{bohr} \] and energy in hartrees: \[\dfrac{e^2}{a_0} = 4.358 \times 10^{-18} \: \mathsf{J} = 27.211 \: \mathsf{eV} \equiv 1 \: \mathsf{hartree} \] Electron volts \((\mathsf{eV})\) are a convenient unit for atomic energies. One \(\mathsf{eV}\) is defined as the energy an electron gains when accelerated across a potential difference of \(1 \: \mathsf{volt}\). The ground state of the hydrogen atom has an energy of \(-1/2 \: \mathsf{hartree}\) or \(-13.6 \: \mathsf{eV}\). Conversion to atomic units is equivalent to setting \[\hbar = e = m = 1\] in all formulas containing these constants. Rewriting the Sc inger equation in atomic units, we have \[\left\{ -\dfrac{1}{2} \nabla^{2} - \dfrac{Z}{r} \right\} \psi(r) = E\psi(r) \label{25}\] Since the potential energy is spherically symmetrical (a function of \(r\) alone), it is obviously advantageous to treat this problem in spherical polar coordinates \(r, \: \theta, \: \phi\). Expressing the Laplacian operator in these coordinates [cf. Eq (6-20)], \[ -\dfrac{1}{2} \left\{ \dfrac{1}{r^{2}} \dfrac{\partial}{\partial r} r^{2} \dfrac{\partial}{\partial r} + \dfrac{1}{r^{2}\sin\theta} \dfrac{\partial}{\partial \theta} \sin\theta \dfrac{\partial}{\partial \theta} + \dfrac{1}{r^{2}\sin^{2}\theta} \dfrac{\partial^{2}}{\partial\phi^{2}} \right\} \\ \times \psi(r, \: \theta, \: \phi) - \dfrac{Z}{r} \psi(r, \: \theta, \: \phi) = E\psi(r, \: \theta, \: \phi) \label{26}\] Equation \(\ref{26}\) shows that the second and third terms in the Laplacian represent the angular momentum operator \(\hat{L}^{2}\). Clearly, Equation \(\ref{26}\) will have separable solutions of the form \[\psi(r, \: \theta, \: \phi) = R(r)Y_{\ell m}(\theta, \: \phi) \label{27}\] Substituting Equation \(\ref{27}\) into Equation \(\ref{26}\) and using the angular momentum eigenvalue Equation Equation \(\ref{6-34}\), we obtain an ordinary differential equation for the radial function \(R(r)\): \[\left\{ -\dfrac{1}{2r^{2}} \dfrac{d}{dr} r^{2} \dfrac{d}{dr} + \dfrac{\ell(\ell + 1)}{2r^{2}} - \dfrac{Z}{r} \right\} R(r) = ER(r) \label{28}\] Note that in the domain of the variable \(r\), the angular momentum contribution \(\ell (\ell + 1) / 2r^{2}\) acts as an effective addition to the potential energy. It can be identified with , which pulls the electron outward, in opposition to the Coulomb attraction. Carrying out the successive differentiations in Equation \(\ref{29}\) and simplifying, we obtain \[\dfrac{1}{2}R''(r) + \dfrac{1}{r}R'(r) + \left[\dfrac{Z}{r} - \dfrac{\ell(\ell + 1)}{2r^{2}} + E\right]R(r) = 0 \label{29}\] another second-order linear differential equation with non-constant coefficients. It is again useful to explore the asymptotic solutions to Equation \(\ref{29}\), as \(r \rightarrow \infty\). In the asymptotic approximation, \[R''(r) - 2r\lvert E \rvert R(r) \approx 0 \label{30}\] having noted that the energy \(E\) is negative for bound states. Solutions to Equation \(\ref{30}\) are \[R(r) \approx \mathsf{const} \, e^{\pm\sqrt{2\lvert E \rvert}r} \label{31}\] We reject the positive exponential on physical grounds, since \(R(r) \rightarrow \infty\) as \(r \rightarrow \infty\), in violation of the requirement that the wavefunction must be finite everywhere. Choosing the negative exponential and setting \(E = -Z^{2}/2\) the ground state energy in the Bohr theory (in atomic units), we obtain \[R(r) \approx \mathsf{const} \, e^{-Zr} \label{32}\] It turns out, very fortunately, that this asymptotic approximation is also an solution of the Schrӧdinger equation (Equation \(\ref{29 }\)) with \(\ell = 0\), just what happened for the harmonic-oscillator problem in Chap. 5. The solutions to Equation \(\ref{29}\), designated \(R_{n\ell}(r)\), are labeled by \(n\), known as the , as well as by the angular momentum \(\ell\), which is a parameter in the radial equation. The solution in Equation \(\ref{32}\) corresponds to \(R_{10}(r)\). This should be normalized according to the condition \[\int_{0}^{\infty} [R_{10}(r)]^{2} \, r^{2} \, dr = 1 \label{33}\] A useful definite integral is \[\int_{0}^{\infty} r^{n} \, e^{-\alpha r} \, dr = \dfrac{n!}{\alpha^{n + 1}} \label{34}\] The normalized radial function is thereby given by \[R_{10}(r) = 2Z^{3/2} e^{-Zr} \label{35}\] Since this function is nodeless, we identify it with the ground state of the hydrogenlike atom. Multipyling Equation \(\ref{35}\) by the spherical harmonic \(Y_{00} = 1/ \sqrt{4\pi} \), we obtain the total wavefunction (Equation \(\ref{27}\)) \[\psi_{100}(r, \theta, \phi) = \left( \dfrac{Z^{3}}{\pi} \right)^{1/2} e^{-Zr} \label{36}\] This is conventionally designated as the 1 function \(\psi_{1s}(r)\). Integrals in spherical-polar coordinates over a spherically-symmetrical integrand (like the 1 orbital can be significantly simplified. We can do the reduction \[\int_{0}^{\infty} \int_{0}^{\pi} \int_{0}^{2\pi} f(r) \, r^{2} \, \sin\theta \, dr \, d\theta \, d\phi = \int_{0}^{\infty} f(r) \, 4\pi r^{2} \, dr \label{37}\] since integration over \(\theta\) and \(\phi\) gives \(4\pi\), the total solid angle of a sphere. The normalization of the 1 wavefunction can thus be written as \[\int_{0}^{\infty} [\psi_{1s}(r)]^{2} \, 4\pi r^{2} \, dr = 1 \label{38}\] There are a number of different ways of representing hydrogen-atom wavefunctions graphically. We will illustrate some of these for the 1 ground state. In atomic units, \[\psi_{1s}(r) = \dfrac{1}{\sqrt{\pi}}e^{-r} \label{39}\] is a decreasing exponential function of a single variable \(r\), and is simply plotted in Figure 3. Figure \(\Page {3}\) gives a somewhat more pictorial representation, a three-dimensional contour plot of \(\psi_{1s}(r)\) as a function of \(x\) and \(y\) in the \(x\), \(y\)-plane. According to Born's interpretation of the wavefunction, the probability per unit volume of finding the electron at the point \((r, \: \theta, \: \phi) \) is equal to the square of the normalized wavefunction \[\rho_{1s}(r) = [\psi_{1s}(r)]^{2} = \dfrac{1}{\pi}e^{-2r} \label{40}\] This is represented in Figure 5 by a scatter plot describing a possible sequence of observations of the electron position. Although results of individual measurements are not predictable, a statistical pattern does emerge after a sufficiently large number of measurements. The probability density is normalized such that \[\int_{0}^{\infty} \rho_{1s}(r) \, 4\pi r^{2} \, dr = 1 \label{41}\] In some ways \(\rho (r)\) oes provide the best description of the electron distribution, since the region around \(r = 0\), where the wavefunction has its largest values, is a relatively small fraction of the volume accessible to the electron. Larger radii \(r\) represent larger physical regions since, in spherical polar coordinates, a value of \(r\) is associated with a shell of volume \(4\pi r^{2} \, dr\) \[D_{1s}(r) = 4\pi r^{2} [\psi_{1s}(r)]^{2} \label{42}\] which represents the probability density within the entire shell of radius \(r\), normalized such that \[\int_{0}^{\infty} D_{1s}(r) \, dr = 1 \label{43}\] The functio \(\rho_{1s}(r) \) and \(D_{1s}(r) \) are both shown in Figure \(\Page {6}\). Remarkably, the 1 RDF has its maximum at \(r = a_0\), equal to the radius of the first Bohr orbit The general solution for \(R_{n\ell}(r)\) has a rather complicated form which we give without proof: \[R_{n\ell}(r) = N_{n\ell} \, \rho^{\ell} \, L_{n + \ell}^{2\ell + 1} \, (\rho) e^{-\rho /2} \qquad \rho \equiv \dfrac{2Zr}{n} \label{44}\] Here \(L_{\beta}^{\alpha}\) is an associated Laguerre p \(N_{n\ell}\) \(\ell\) \(\ell\ = 0\) \(\ell\ = 1\) \(\ell\ = 2\) \(\ell\ = 3\) \(\ell\ = 4\) inger equation
in spherical polar coordinates can now be written in full \[\psi_{n\ell m}(r, \: \theta, \: \phi) = R_{n\ell}(r)Y_{\ell m}(\theta, \: \phi) \\ n = 1, \: 2... \qquad \ell = 0, \: 1... \: n - 1 \qquad m = 0, \: \pm 1, \: \pm 2... \: \pm \ell \label{45}\] where \(Y_{\ell m}\) are the spherical harmonics tabulated in Chap. 6. Table 1 below enumerates all the hydrogenic functions we will actually need. These are called , in anticipation of their later applications to the structure of atoms and molecules. The energy levels for a hydrogenic system are given by \[E_n = -\dfrac{Z^{2}}{2n^{2}} \: \mathsf{hartrees} \label{46}\] and depends on the principal quantum number alone. Considering all the allowed values of \(\ell\) and \(m\), the level \(E_n\) has a degeneracy of \(n^{2}\). Figure 7 shows an energy level diagram for hydrogen \((Z = 1) \). For \(E \geq 0\), the energy is a continuum, since the electron is in fact a free particle. The continuum represents states of an electron and proton in interaction, but not bound into a stable atom. Figure \(\Page {7}\) also shows some of the transitions which make up the Lyman series in the ultraviolet and the Balmer series in the visible region. The \(ns\) orbitals are all spherically symmetrical, being associated with a constant angular factor, the spherical harmonic \(Y_{00} = 1/ \sqrt{4\pi} \). They have \(n - 1\) radial nodes spherical shells on which the wavefunction equals zero. The 1 ground state is nodeless and the number of nodes increases with energy, in a pattern now familiar from our study of the particle-in-a-box and harmonic oscillator. The 2 orbital, with its radial node at \(r = 2\) bohr, is also shown in Figure \(\Page {3}\). The lowest-energy solutions deviating from spherical symmetry are the 2 orbitals. Using Equations \(\ref{44}\), \(\ref{45}\) and the \(\ell = 1\) spherical harmonics, we find three degenerate eigenfunctions: \[\psi_{210}(r, \: \theta, \: \phi) = \dfrac{1}{4\sqrt{2\pi}}re^{-r/2} \cos\theta \label{47}\] and \[\psi_{21 \pm 1}(r, \: \theta, \: \phi) = \mp \dfrac{1}{4\sqrt{2\pi}}re^{-r/2} \sin\theta e^{\pm i \phi} \label{48}\] The function \(\psi_{210}\) is real and contains the factor \(r \cos\theta \), which is equal to the cartesian variable \(z\). In chemical applications, this is designated as a 2 orbital: \[\psi_{2p_z} = \dfrac{1}{4\sqrt{2\pi}}ze^{-r/2} \label{49}\] A contour plot is shown in Figure \(\Page {8}\). Note that this function is cylindrically-symmetrical about the \(z\)-axis with a node in the \(x\), \(y\)-plane. The \(\psi_{21 \pm 1}\) are complex functions and not as easy to represent graphically. Their angular dependence is that of the spherical harmonics \(Y_{1 \pm 1}\), shown in Figure 6-4. As noted in Chap. 4, any linear combination of degenerate eigenfunctions is an equally-valid alternative eigenfunction. Making use of the Euler formulas for sine and cosine \[\cos\phi = \dfrac{e^{i\phi} + e^{-i\phi}}{2} \qquad \mathsf{and} \qquad \sin\phi = \dfrac{e^{i\phi} - e^{-i\phi}}{2} \label{50}\] and noting that the combinations \(\sin\theta\cos\phi\) and \(\sin\theta\sin\phi\) correspond to the cartesian variables \(x\) and \(y\), respectively, we can define the alternative 2 orbitals \[\psi_{2p_x} = \dfrac{1}{\sqrt{2}}(\psi_{21-1} - \psi_{211}) = \dfrac{1}{4\sqrt{2\pi}} xe^{-r/2} \label{51}\] and \[\psi_{2p_y} = -\dfrac{i}{\sqrt{2}}(\psi_{21-1} + \psi_{211}) = \dfrac{1}{4\sqrt{2\pi}} ye^{-r/2} \label{52}\] Clearly, these have the same shape as the 2 -orbital, but are oriented along the \(x\)- and \(y\)-axes, respectively. The threefold degeneracy of the -orbitals is very clearly shown by the geometric equivalence the functions 2 , 2 and 2 , which is not obvious for the spherical harmonics. The functions listed in Table 1 are, in fact, the real forms for all atomic orbitals, which are more useful in chemical applications. All higher -orbitals have analogous functional forms \(x \, f(r)\), \(y \, f(r)\) and \(z \, f(r)\) and are likewise 3-fold degenerate. The orbital \(\psi_{320}\) is, like \(\psi_{210}\), a real function. It is known in chemistry as the \(d_{z^{2}}\)-orbital and can be expressed as a cartesian factor times a function of \(r\): \[\psi_{3d_{z^{2}}} = \psi_{320} = (3z^{2} - r^{2}) f(r) \label{53}\] A contour plot is shown in Figure \(\Page {9}\). This function is also cylindrically symmetric about the \(z\)-axis with angular nodes the conical surfaces with \(3z^{2} - r^{2} = 0\). The remaining four 3 orbitals are complex functions containing the spherical harmonics \(Y_{2 \pm 1} \) and \(Y_{2 \pm 2}\) pictured in Figure 6-4. We can again construct real functions from linear combinations, the result being four geometrically equivalent "four-leaf clover" functions with two perpendicular planar nodes. These orbitals are designated \(d_{x^{2} - y^{2}}, \: d_{xy}, \: d_{zx}\) and \(d_{yz}\). Two of them are shown in Figure 9. The \(d_{z^{2}}\) orbital has a different shape. However, it can be expressed in terms of two non-standard -orbitals, \(d_{z^{2} - x^{2}}\) and \(d_{y^{2} - z^{2}}\). The latter functions, along with \(d_{x^{2} - y^{2}}\) add to zero and thus constitute a linearly set. combinations of these three functions can be chosen as independent eigenfunctions. The atomic orbitals listed in Table 1 are illustrated in Figure \(\Page {20}\). Blue and red indicate, respectively, positive and negative regions of the wavefunctions (the radial nodes of the 2 and 3 orbitals are obscured). These pictures are intended as stylized representations of atomic orbitals and should be interpreted as quantitatively accurate. The electron charge distribution in an orbital \(\psi_{n\ell m}(\mathbf{r})\) is given by \[\rho(\mathbf{r}) = \lvert \psi_{n\ell m}(\mathbf{r}) \rvert ^{2} \label{54} \] which for the -orbitals is a function of \(r\) alone. The radial distribution function can be defined, even for orbitals containing angular dependence, by \[D_{n\ell}(r) = r^{2} [R_{n\ell}(r)]^{2} \label{55}\] This represents the electron density in a shell of radius \(r\), including all values of the angular variables \(\theta\), \(\phi\). Figure \(\Page {11}\) shows plots of the RDF for the first few hydrogen orbitals. (Professor Emeritus of Chemistry and Physics at the , ) | 23,571 | 1,391 |
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Isotope effects such as KIEs are invaluable tools in both physical and biological sciences and are used to aid in the understanding of reaction kinetics, mechanisms, and solvent effects. Research was first introduced on this topic over 50 years ago and has grown into an enormous field. The scientists behind much of the understanding and development of kinetic isotope effects were Jacob Bigeleisen and Maria Goeppert Mayer who published the first paper on isotope effects [J. Chem. Phys., 15, 261 (1947)]. Kinetic isotope effects specifically explore the change in rate of a reaction due to isotopic substitution. An element is identified by its symbol, mass number, and atomic number. The atomic number is the number of protons in the nucleus while the mass number is the total number of protons and neutrons in the nucleus. Isotopes are two atoms of the same element that have the same number of protons but different numbers of neutrons. Isotopes are specified by the mass number. As an example consider the two isotopes of chlorine, you can see that their mass numbers vary, with Cl being the most abundant isotope, while their atomic numbers remain the same at 17. \[ ^{35}Cl \;\text{and}\; ^{37}Cl\] The most common isotope used in light atom isotope effects is hydrogen (\(^{1}H\)) commonly replaced by its isotope deuterium (\(^{2}H\)). Note: Hydrogen also has a third isotope, tritium (\(^{2}H\)). Isotopes commonly used in heavy atom isotope effects include carbon (\(^{12}C\), \(^{13}C\), nitrogen (\(^{14}N\), \(^{15}N\)), oxygen, sulfur, and bromine. Not all elements exhibit reasonably stable isotopes (i.e. Fluorine, \(^{19}F\)), but those that due serve as powerful tools in isotope effects. Understanding potential energy surfaces is important in order to be able to understand why and how isotope effects occur as they do. The harmonic oscillator approximation is used to explain the vibrations of a diatomic molecule. The energies resulting from the quantum mechanic solution for the help to define the internuclear potential energy of a diatomic molecule and are \[ E_n = \left(n + \dfrac{1}{2}\right)h \nu \label{1}\] where The Morse potential is an analytic expression that is used as an approximation to the intermolecular potential energy curves: \[ V(l) = D_e{\left(1-e^{-\beta(l-l_o)}\right)}^2 \label{2}\] where The \(D_e\), \( \beta \), and \(l_o\) variables can be looked up in a textbook or CRC handbook. Below is an example of a Morse potential curve with the zero point vibrational energies of two isotopic molecules (for example R-H and R-D where R is a group/atom that is much heavier than H or D). The y-axis is potential energy and the x axis is internuclear distance. In this figure respond to the zero point energies of deuterium and hydrogen. The zero point energy is the lowest possible energy of a system and equates to the ground state energy. Zero point energy is dependent upon the reduced mass of the molecule as will be shown in the next section. The heavier the molecule or atom, the lower the frequency of vibration and the smaller the zero point energy. Lighter molecules or atoms have a greater frequency of vibration and a higher zero point energy. We see this is the figure below where deuterium is heavier than hydrogen and therefore has the lower zero point energy. This results in different bond dissociation energies for R-D and R-H. The bond dissociation energy for R-D (E ) is greater than the bond dissociation energy of R-H (E ). This difference in energy due to isotopic replacement results in differing rates of reaction, the effect that is measured in kinetic isotope effects. The reaction rate for the conversion of R-D is slower than the reaction rate for the conversion of R-H. p> It is important to note that isotope replacement does not change the electronic structure of the molecule or the potential energy surfaces of the reactions the molecule may undergo. Only the rate of the reaction is affected. The energy of the vibrational levels of a vibration (i.e., a bond) in a molecule is given by \[ E_n = \left(n + \dfrac{1}{2}\right)h \nu \label{3} \] where we assume that the molecule is in its ground state and we can compare zero-point vibrational energies, \[ E_o = \left(\dfrac{1}{2}\right)hv \label{4}\] Using the harmonic oscillator approximation the fundamental vibrational frequency is \[ \nu = \dfrac{1}{2 \pi} \sqrt{ \dfrac {k}{\mu} } \label{5}\] where \[ \mu = \dfrac{m_1m_2}{m_1+m_2} \label{6}\] The is used to determine reaction rates and activation energies and since we are interested in the change in rate of reactions with different isotopes, this equation is very important, \[ k = Ae^{-\frac{E_a}{kT}} \label{7} \] where The Arrhenius equation can be used to compare the rates of a reaction with R-H and R-D, \[ k_H = A_He^{-\frac{E_a^H}{kT}} \label{8}\] \[ k_D = A_De^{-\frac{E_a^D}{kT}} \label{9} \] where k and k are the rates of reaction associated with R-H and the isotope substituted R-D. We will then assume the Arrhenius constants are equal (\(A_H=A_D\)). The ratio of the rates of reaction gives an approximation for the isotope effect resulting in: \[ \dfrac{k_H}{k_D} = e^{-\frac{E_a^H - E_a^D}{kT}} \label{10}\] By using the relationship that for both R-H and R-D \[ E_o = \left(\dfrac{1}{2}\right)h\nu \label{11} \] a substitution can be made resulting in \[\dfrac{k_H}{k_D} = e^{\frac{h(\nu_H - \nu_D)}{2kT}} \label{12}\] The vibrational frequency (Equation 5) can then be substituted for R-H and R-D and the value of the expected isotope effect can be calculated. \[\dfrac{k_H}{k_D} = e^{\dfrac {h \left( \dfrac{k_{RH}}{\mu_{RH}} - \dfrac{k_{RD}}{\mu_{RD}} \right)}{4\pi kT}} \label{13}\] The same general procedure can be followed for any isotope substitution. In summary, the greater the mass the more energy is needed to break bonds. A heavier isotope forms a stronger bond. The resulting molecule has less of a tendency to dissociate. The increase in energy needed to break the bond results in a slower reaction rate and the observed isotope effect. Kinetic Isotope Effects (KIEs) are used to determine reaction mechanisms by determining rate limiting steps and transition states and are commonly measured using NMR to detect isotope location or GC/MS to detect mass changes. In a KIE experiment an atom is replaced by its isotope and the change in rate of the reaction is observed. A very common isotope substitution is when hydrogen is replaced by deuterium. This is known as a deuterium effect and is expressed by the ratio k /k (as explained above). Normal KIEs for the deuterium effect are around 1 to 7 or 8. Large effects are seen because the percentage mass change between hydrogen and deuterium is great. Heavy atom isotope effects involve the substitution of carbon, oxygen, nitrogen, sulfur, and bromine, with effects that are much smaller and are usually between 1.02 and 1.10. The difference in KIE magnitude is directly related to the percentage change in mass. Large effects are seen when hydrogen is replaced with deuterium because the percentage mass change is very large (mass is being doubled) while smaller percent mass changes are present when an atom like sulfur is replaced with its isotope (increased by two mass units). Primary kinetic isotope effects are rate changes due to isotopic substitution at a site of bond breaking in the rate determining step of a reaction. Consider the : kinetic studies have been performed that show the rate of this reaction is independent of the concentration of bromine. To determine the rate determining step and mechanism of this reaction the substitution of a deuterium for a hydrogen can be made. When hydrogen was replaced with deuterium in this reaction a \(k_H \over k_D \) of 7 was found. Therefore the rate determining step is the tautomerization of acetone and involves the breaking of a C-H bond. Since the breaking of a C-H bond is involved, a substantial isotope effect is expected. A rule of thumb for heavy atom isotope effects is that the maximum isotopic rate ratio is proportional to the square root of the inverse ratio of isotopic masses. Secondary kinetic isotope effects are rate changes due to isotopic substitutions at a site other than the bond breaking site in the rate determining step of the reaction. These come in three forms: \(\alpha\), \(\beta\), and \(\gamma\) effects. \(\beta\) secondary isotope effects occur when the isotope is substituted at a position next to the bond being broken. \[\ce{(CH3)2CHBr + H2O ->[k_H] (CH3)2CHOH}\] \[\ce{(CD3)2CHBr + H2O ->[k_D] (CD3)2CHOH}\] This is thought to be due to hyperconjugation in the transition state. Hyperconjugation involves a transfer of electron density from a sigma bond to an empty p orbital (for more on hyperconjugation see outside links). Reactions may be affected by the type of solvent used (for example H O to D O or ROH to ROD). There are three main ways solvents effect reactions: | 8,974 | 1,392 |
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Although we now have powerful spectroscopic methods available to determine the sizes of the oxide rings formed by the simple monosaccharides, the way in which this was done chemically for glucose highlights the difference in reactivity between ether and alcohol functions. The acid-catalyzed methylation of glucose with methanol to give two distinct glucosides, methyl \(\alpha\)-\(D\)-glucoside and methyl \(\beta\)-\(D\)-glucoside, corresponds to displacement of the hemiacetal hydroxyl by methoxyl to form an acetal (see left side of Figure 20-4). The remaining four hydroxyl groups can be methylated in basic solution by dimethyl sulfate or by methyl iodide and silver oxide in \(\ce{N}\),\(\ce{N}\)-dimethylmethanamide, \(\ce{HCON(CH_3)_2}\), solution. Hydrolysis of either of these pentamethyl glucose derivatives with aqueous acid affects only the acetal linkage and leads to a tetramethylated glucose, \(20\), as shown in Figure 20-4. The pyranose ring structure of \(D\)-glucose originally was established by Hirst, in 1926, by converting \(D\)-glucose to a tetra-\(\ce{O}\)-methyl-\(D\)-glucose and showing that this substance actually was 2,3,4,6-tetra-\(\ce{O}\)-methyl-\(D\)-glucose, \(20\). The key feature of \(20\) is the fact that all but the two carbons involved hemiacetal formation are protected from oxidation by being substituted with \(\ce{O}\)-methyl groups in place of hydroxy groups. The largest fragment isolated from oxidation of Hirst's tetra-\(\ce{O}\)-methyl-\(D\)-glucose was a trimethoxypentanedioic acid, \(21\), and because the two carboxyl carbons must have been the ones originally involved in ring formation, the oxide ring must be between \(\ce{C_1}\) and \(\ce{C_5}\): Reagents that specifically oxidize vicinal glycols [e.g., \(\ce{NaIO_4}\), \(\ce{Pb(O_2CCH_3)_4}\), and \(\ce{NaBiO_3}\); ] are quite helpful in determining the cyclic structures of sugars. With periodate, the numbers of moles of oxidant consumed and the moles of methanoic acid and methanal produced are different for each type of ring structure. The cleavage reactions that normally are observed follow: As we stated previously, glucose forms some, but not all, of the common carbonyl derivatives. The amount of free aldehyde present in solution is so small that it is not surprising that no hydrogen sulfite derivative forms. With amines, the product is not a Schiff's base but a glucosylamine of cyclic structure analogous to the hemiacetal structure of glucose, Equation 20-3. The Schiff's base is likely to be an intermediate that rapidly cyclizes to the glucosylamine: The reaction of glucose with an excess of phenylhydrazine (phenyldiazane) is particularly noteworthy because phenylhydrazine molecules are incorporated into one of glucose. Subsequent to the expected phenylhydrazone formation, and in a manner that is not entirely clear, the \(\ce{-CHOH}-\) group adjacent to the original aldehyde function is oxidized to a carbonyl group, which then consumes more phenylhydrazine to form a crystalline derivative called an , or specifically : The sugar osazones usually are crystalline and are useful for characterization and identification of sugars. Fischer employed them in his work that established the configuration of the sugars. The kind of information that can be obtained is illustrated by the following example: Because the phenylosazone arises from glucose, mannose, and fructose, the configurations of \(\ce{C_3}\), \(\ce{C_4}\), and \(\ce{C_5}\) must be the for all three sugars. and (1977) | 3,550 | 1,393 |
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Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY- ; ). | 817 | 1,394 |
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List all the intermolecular interactions that take place in each of the follow kings of molecules: \(CCl_3F\), \(CCl_2F_2\), \(CClF_3\), and \(CF_4\). Arrange the follow species in order of decreasing melting points: CsBr, KI, KCL, MgF2. Which has the highest boiling point I Br and Cl . Explain why? The atomic weigh of Iodine = 127, Bromine = 80, and Chlorine = 35.5. The weigh is proportion to the London dispersion force, and the higher molecular weigh, the larger the force. Thus, I has a highest boiling point. 1-Propanol C H OH and methoxyethane CH O C H have the same molecular weigh. Which has the higher boiling point? The 1-Propanol can form London Force, Dipole- Dipole, and H- bonding due to the H bonded to O atom of OH group, whereas the methoxyethane can not form the H-bonding. Therefore, the 1-Propanol has higher intermolecular attractive force and thus a higher boiling point. Why do the lightest compounds such as NH3, H2O, and HF have the highest boiling points? What kind of attractive interaction exists between atoms and between nonpolar molecules? Why nature gas CH is a good choice to storage tank in winter? It's about boiling point. The methane has the boiling point at -161 °C, making it to be a good choice for winter season. Explain why methane (CH_ is used as the primary heating gas in Alaska during wintertime instead of the more commonly used butant or propane gases use in the lower 48 states. Methane (\(CH_4\)) remains gas because its boiling point is about -160°C. Other gases, such as propane or butane, would liquefy under freezing condition. Therefore, methane is more likely to be used during wintertime at Alaska Define types of intermolecular forces and give example for each. There are 3 types of intermolecular force: London Dispersion, Dipole-Dipole (Example: Two \(NaCl\)) and Ion-Dipole (Example: \(Mg^+\) and \(HCl\)) How does the intermolecular determine the boiling point? The weakest intermolecular, the lowest boiling point. What is their dipole-dipole interaction of wo HCl molecules are co-linear head-to tail. Given: The dipole moment of HF is 1.86 D. The dipole moment of HCl is 1.05 D. The distance between the two is 1.78 H -F - - - - H -Cl \[V=-\dfrac{2\mu _{A}\mu _{B}}{4\pi \varepsilon _{0}r^{3}}\] \[V=-\dfrac{2(1.05)(1.86)}{4\pi (8.854187817\cdot 10^{-12})(1.78)^{3}}\] Calculate an ion-ion interaction energy between \(K^+\) and \(Cl^-\) at a distance of 600 pm. The ion-ion interaction energy is given by Coulomb's law. \[V = \dfrac{q_1q_2}{4 \pi \epsilon_o r}\] \[V = \dfrac{- (1.602 \times 10^{-19}\;\cancel{C})(1.602 \times 10^{-19} \cancel{C})}{4 \pi (8.853 \times 10^{-12} \cancel{C^2} \cdot N^{−1} \cdot m \cancel{^{−2}})(6 \times 10^{-10}\; \cancel{m})} = -3.84 \times 10^{-19} \; J\] Calculate an ion-dipole interaction energy between \(K^+\) and \(HCl\) at a distance of 600 pm. \(HCl\) has a dipole moment of \(1.08\;D\). \[\mu = 1.08 \cancel{D} \times \dfrac{3.3356 \times 10^{30} \; C \cdot m}{1\;\cancel{D}} = 3.6 \times 10^{-30}\; C \cdot m\] \[V = \dfrac{-q\;\mu}{4 \pi \epsilon_o r^2} = \dfrac{- (1.602 \times 10^{-19}\;\cancel{C})(3.6 \times 10^{-30} \cancel{C} \cdot \cancel{m})}{4 \pi (8.853 \times 10^{-12} \cancel{C^2} \cdot N^{−1} \cdot m \cancel{^{−2}})(6 \times 10^{-10}\; \cancel{m})^2} = -1.44 \times 10^{-20} \; J\] As expected this is appreciably smaller in energy than covalent bonds (e..g, \(HCl\) has a bond enthalpy of \(7.0 \times 10^{-19}\;J\)). Rank the interactions from weakest to strongest: A low concentration electrolytic solution behaves non-ideally while a high concentration of the same solution behaves ideally. Explain this phenomenon in terms of forces, noting that Coulomb forces depend on 1/r while van der Waals forces depend on 1/r . Which of these forces are low concentration electrolytic solutions likely to follow? High concentration? The interatomic distances in a low concentration electrolytic solution are greater than those in a high concentration solution. They follow van der Waals forces and thus behave less ideally. High concentration electrolytic solutions follow Coulomb forces. Calculate the ion-dipole interaction between H O and Li . You are given the dipole moment of H O is 1.82 D. The distance between these two is 2 Å. \[V=-\frac{q\mu }{4\pi \varepsilon _{0}r^{2}}\] \[=\frac{1.82D\cdot(\frac{3.3356\cdot 10^{-30}Cm)}{1D}}{4\pi (8.85\cdot 10^{-12})(2\cdot 10^{-10}m)^{^{2}}} =1.36\; kJ/mol\] Calculate the potential energy of interaction between a Cl ion situated 120 pm away from an \(H_2O\) molecule with a dipole moment of 1.85 D. \[\mu = 1.85 \cancel{D} \times \dfrac{3.3356 \times 10^{30} \; C \cdot m}{1\;\cancel{D}} = 6.18 \times 10^{-30}\; C \cdot m\] \[r = 1.2 \times 10^{-10}\; m\] \[V = \dfrac{q\mu}{4\pi \varepsilon _{o}r^{2}} = \dfrac{(-1.602\times10^{-19}\;C)(6.18 \times 10^{-30}\; C \cdot m)}{4\pi (8.851 \times 10^{-12}\; C^{-2}N^{-1}m^{-2})(1.2 \times 10^{-10} \; m) ^2}\] Do you expect a greater dipole-dipole interaction between two molecules that are antiparallel or between two molecules that are co-linear head-to-tail? You expect a stronger interaction when the two are co-linear head-to tail. This can be seen by looking at the formula or in the images of the two. Express the equilibrium distance r in term ð and show V = - € a. r= σ/2 = 3.40 A /2 = 1.70 A b. Volume of 2 mole of Ar 4/3 πr^3 ((6.022 x 10^23)/(2 mol))= 4/3 π (1.70 x 10^(-10) m)^3 ((6.022 x 10^23)/(2 mol)) = 6.19 x 10-6 L mol-1 V/n=RT/P= ((0.08206 L atm K^(-1) mol^(-1) (298.2 K))/1atm = 24.5 L mol-1 The fraction of this volume occupied by 2 mole of Ar (1.239 x 10^-2 L mol )/ 24.47 L mol = 2.5 x10 a) What is the original of polarity in a molecule? b) Is CO polar? Explain. Of the following compound, which one(s) is/are soluble? What makes a compound soluble in water? Explain using examples. Explain why does water have a high specific heat. The energy of a hydrogen bond for each base pair in DNA is 15 kJ/mol. Two complimentary strands has 50 base pairs each. What is the ratio of the 2 different strands to hydrogen double helix in a solution given a temperature of 300 K. First calculate the ratio of the two different strands for just one pair. \[ e^{\Delta E/RT}=exp[(15 \times 10^3\; J/mol)/(8.314\; J/K*mol)(300\; K) = 2.4 \times 10^{-3}\] Since there are 50 base pairs, we need to multiply by 50 to account for all the base pairs. exp[100X(15X10 J/mol)/(8.314 J/K*mol)(300K) = 0 Consider two pure liquids. One has strong intermolecular interactions, and the other has relatively weak intermolecular interactions. For the following properties, indicate which of the liquids you would expect to have a higher value (answer with "strong" or "weak"). Fun fact: if the DNA in a single human cell were stretched out (but still in its familiar double helix conformation), it would be approximately 2 meters long. The distance, along the helix, between nucleotides is 3.4 Å. (a) $$ 2\ m/cell \times \dfrac{bp}{3.4\ Å} \times \dfrac{10^{10}\ Å}{m} \times \dfrac{cell}{2\ haploid\ genomes} = 3 \times 10^{9} \dfrac{bp}{haploid\ genome} \] Three billion basepairs. (b) 75 trillion of the human cells in your body have genomic DNA. $$ 75 \times 10^{20}\ cells \times \dfrac{haploid\ genomes}{cell} \times \dfrac{3 \times 10^9\ bp}{haploid\ genome} \times \dfrac{mol}{6.022 \times 10^{23}} \times 650 \dfrac{g}{mol\ bp} = 200\ g \] That's about half a pound. (c) $$ \dfrac{2\ m}{cell} \times 75 \times 10^{12}\ cells \times \dfrac{km}{1000\ m} = 2 \times 10^{11}\ km \] Yes, you have way more DNA than you need to stretch it from Earth to Pluto. | 7,616 | 1,396 |
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Nuclear reactions, like chemical reactions, are accompanied by changes in energy. The energy changes in nuclear reactions, however, are enormous compared with those of even the most energetic chemical reactions. In fact, the energy changes in a typical nuclear reaction are so large that they result in a measurable change of mass. In this section, we describe the relationship between mass and energy in nuclear reactions and show how the seemingly small changes in mass that accompany nuclear reactions result in the release of enormous amounts of energy. The relationship between mass (m) and energy (E) is expressed in the following equation: \[E = mc^2 \label{20.27}\] where c is the speed of light (2.998 × 10 m/s), and E and m are expressed in units of joules and kilograms, respectively. Albert Einstein first derived this relationship in 1905 as part of his special theory of relativity: the mass of a particle is directly proportional to its energy. Thus according to , every mass has an associated energy, and similarly, any reaction that involves a change in energy must be accompanied by a change in mass. This implies that all exothermic reactions should be accompanied by a decrease in mass, and all endothermic reactions should be accompanied by an increase in mass. Given the law of conservation of mass, how can this be true? The solution to this apparent contradiction is that chemical reactions are indeed accompanied by changes in mass, but these changes are simply too small to be detected. As you may recall, all particles exhibit wavelike behavior, but the wavelength is inversely proportional to the mass of the particle (actually, to its momentum, the product of its mass and velocity). Consequently, wavelike behavior is detectable only for particles with very small masses, such as electrons. For example, the chemical equation for the combustion of graphite to produce carbon dioxide is as follows: \[\textrm{C(graphite)} + \frac{1}{2}\textrm O_2(\textrm g)\rightarrow \mathrm{CO_2}(\textrm g)\hspace{5mm}\Delta H^\circ=-393.5\textrm{ kJ/mol} \label{20.28}\] Combustion reactions are typically carried out at constant pressure, and under these conditions, the heat released or absorbed is equal to ΔH. When a reaction is carried out at constant volume, the heat released or absorbed is equal to ΔE. For most chemical reactions, however, ΔE ≈ ΔH. If we rewrite Einstein’s equation as we can rearrange the equation to obtain the following relationship between the change in mass and the change in energy: \[\Delta m=\dfrac{\Delta E}{c^2} \label{20.30}\] Because 1 J = 1 (kg·m )/s , the change in mass is as follows: This is a mass change of about 3.6 × 10 g/g carbon that is burned, or about 100-millionths of the mass of an electron per atom of carbon. In practice, this mass change is much too small to be measured experimentally and is negligible. In contrast, for a typical nuclear reaction, such as the radioactive decay of C to N and an electron (a β particle), there is a much larger change in mass: We can use the experimentally measured masses of subatomic particles and common isotopes to calculate the change in mass directly. The reaction involves the conversion of a neutral C atom to a positively charged N ion (with six, not seven, electrons) and a negatively charged β particle (an electron), so the mass of the products is identical to the mass of a neutral N atom. The total change in mass during the reaction is therefore the difference between the mass of a neutral N atom (14.003074 amu) and the mass of a C atom (14.003242 amu): \[\begin{align} \Delta m &= {\textrm{mass}_{\textrm{products}}- \textrm{mass}_{\textrm{reactants}}}
\\&=14.003074\textrm{ amu} - 14.003242\textrm{ amu} = - 0.000168\textrm{ amu}\end{align} \label{20.33}\] The difference in mass, which has been released as energy, corresponds to almost one-third of an electron. The change in mass for the decay of 1 mol of C is −0.000168 g = −1.68 × 10 g = −1.68 × 10 kg. Although a mass change of this magnitude may seem small, it is about 1000 times larger than the mass change for the combustion of graphite. The energy change is as follows: The energy released in this nuclear reaction is more than 100,000 times greater than that of a typical chemical reaction, even though the decay of C is a relatively low-energy nuclear reaction. Because the energy changes in nuclear reactions are so large, they are often expressed in kiloelectronvolts (1 keV = 10 eV), megaelectronvolts (1 MeV = 10 eV), and even gigaelectronvolts (1 GeV = 10 eV) per atom or particle. The change in energy that accompanies a nuclear reaction can be calculated from the change in mass using the relationship 1 amu = 931 MeV. The energy released by the decay of one atom of C is thus Calculate the changes in mass (in atomic mass units) and energy (in joules per mole and electronvolts per atom) that accompany the radioactive decay of U to Th and an α particle. The α particle absorbs two electrons from the surrounding matter to form a helium atom. nuclear decay reaction changes in mass and energy Use the mass values in to calculate the change in mass for the decay reaction in atomic mass units. Use to calculate the change in energy in joules per mole. Use the relationship between atomic mass units and megaelectronvolts to calculate the change in energy in electronvolts per atom. Using particle and isotope masses from we can calculate the change in mass as follows: Thus the change in mass for 1 mol of U is −0.004584 g or −4.584 × 10 kg. The change in energy in joules per mole is as follows: ΔE = (Δm)c = (−4.584 × 10 kg)(2.998 × 10 m/s) = −4.120 × 10 J/mol The change in energy in electronvolts per atom is as follows: Calculate the changes in mass (in atomic mass units) and energy (in kilojoules per mole and kiloelectronvolts per atom) that accompany the radioactive decay of tritium ( H) to He and a β particle. Δm = −2.0 × 10 amu; ΔE = −1.9 × 10 kJ/mol = −19 keV/atom We have seen that energy changes in both chemical and nuclear reactions are accompanied by changes in mass. Einstein’s equation, which allows us to interconvert mass and energy, has another interesting consequence: The mass of an atom is always less than the sum of the masses of its component particles. The only exception to this rule is hydrogen-1 ( H), whose measured mass of 1.007825 amu is identical to the sum of the masses of a proton and an electron. In contrast, the experimentally measured mass of an atom of deuterium ( H) is 2.014102 amu, although its calculated mass is 2.016490 amu: The difference between the sum of the masses of the components and the measured atomic mass is called the of the nucleus. Just as a molecule is more stable than its isolated atoms, a nucleus is more stable (lower in energy) than its isolated components. Consequently, when isolated nucleons assemble into a stable nucleus, energy is released. According to , this release of energy must be accompanied by a decrease in the mass of the nucleus. The amount of energy released when a nucleus forms from its component nucleons is the ( ). In the case of deuterium, the mass defect is 0.002388 amu, which corresponds to a nuclear binding energy of 2.22 MeV for the deuterium nucleus. Because the magnitude of the mass defect is proportional to the nuclear binding energy, both values indicate the stability of the nucleus. Just as a molecule is more stable (lower in energy) than its isolated atoms, a nucleus is more stable than its isolated components. Not all nuclei are equally stable. Chemists describe the relative stability of different nuclei by comparing the binding energy per nucleon, which is obtained by dividing the nuclear binding energy by the mass number (A) of the nucleus. As shown in , the binding energy per nucleon increases rapidly with increasing atomic number until about Z = 26, where it levels off to about 8–9 MeV per nucleon and then decreases slowly. The initial increase in binding energy is not a smooth curve but exhibits sharp peaks corresponding to the light nuclei that have equal numbers of protons and neutrons (e.g., He, C, and O). As mentioned earlier, these are particularly stable combinations. Because the maximum binding energy per nucleon is reached at Fe, all other nuclei are thermodynamically unstable with regard to the formation of Fe. Consequently, heavier nuclei (toward the right in ) should spontaneously undergo reactions such as alpha decay, which result in a decrease in atomic number. Conversely, lighter elements (on the left in ) should spontaneously undergo reactions that result in an increase in atomic number. This is indeed the observed pattern. Heavier nuclei spontaneously undergo nuclear reactions that their atomic number. Lighter nuclei spontaneously undergo nuclear reactions that their atomic number. Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for Fe. The experimental mass of the nuclide is given in . nuclide and mass nuclear binding energy and binding energy per nucleon Sum the masses of the protons, electrons, and neutrons or, alternatively, use the mass of the appropriate number of H atoms (because its mass is the same as the mass of one electron and one proton). Calculate the mass defect by subtracting the experimental mass from the calculated mass. Determine the nuclear binding energy by multiplying the mass defect by the change in energy in electronvolts per atom. Divide this value by the number of nucleons to obtain the binding energy per nucleon. A An iron-56 atom has 26 protons, 26 electrons, and 30 neutrons. We could add the masses of these three sets of particles; however, noting that 26 protons and 26 electrons are equivalent to 26 H atoms, we can calculate the sum of the masses more quickly as follows: We subtract to find the mass defect: The nuclear binding energy is thus 0.528462 amu × 931 MeV/amu = 492 MeV. The binding energy per nucleon is 492 MeV/56 nucleons = 8.79 MeV/nucleon. Calculate the total nuclear binding energy (in megaelectronvolts) and the binding energy per nucleon for U. 1800 MeV/ U; 7.57 MeV/nucleon N is the splitting of a heavy nucleus into two lighter ones. Fission was discovered in 1938 by the German scientists Otto Hahn, Lise Meitner, and Fritz Strassmann, who bombarded a sample of uranium with neutrons in an attempt to produce new elements with Z > 92. They observed that lighter elements such as barium (Z = 56) were formed during the reaction, and they realized that such products had to originate from the neutron-induced fission of uranium-235: \[_{92}^{235}\textrm U+\,_0^1\textrm n \rightarrow \,_{56}^{141}\textrm{Ba}+\,_{36}^{92}\textrm{Kr}+3_0^1\textrm n \label{20.37}\] This hypothesis was confirmed by detecting the krypton-92 fission product. The nucleus usually divides asymmetrically rather than into two equal parts, and the fission of a given nuclide does not give the same products every time. In a typical nuclear fission reaction, more than one neutron is released by each dividing nucleus. When these neutrons collide with and induce fission in other neighboring nuclei, a self-sustaining series of nuclear fission reactions known as a can result ( ). For example, the fission of U releases two to three neutrons per fission event. If absorbed by other U nuclei, those neutrons induce additional fission events, and the rate of the fission reaction increases geometrically. Each series of events is called a generation. Experimentally, it is found that some minimum mass of a fissile isotope is required to sustain a nuclear chain reaction; if the mass is too low, too many neutrons are able to escape without being captured and inducing a fission reaction. The minimum mass capable of supporting sustained fission is called the . This amount depends on the purity of the material and the shape of the mass, which corresponds to the amount of surface area available from which neutrons can escape, and on the identity of the isotope. If the mass of the fissile isotope is greater than the critical mass, then under the right conditions, the resulting supercritical mass can release energy explosively. The enormous energy released from nuclear chain reactions is responsible for the massive destruction caused by the detonation of nuclear weapons such as fission bombs, but it also forms the basis of the nuclear power industry. , in which two light nuclei combine to produce a heavier, more stable nucleus, is the opposite of nuclear fission. As in the nuclear transmutation reactions discussed in , the positive charge on both nuclei results in a large electrostatic energy barrier to fusion. This barrier can be overcome if one or both particles have sufficient kinetic energy to overcome the electrostatic repulsions, allowing the two nuclei to approach close enough for a fusion reaction to occur. The principle is similar to adding heat to increase the rate of a chemical reaction. As shown in the plot of nuclear binding energy per nucleon versus atomic number in , fusion reactions are most exothermic for the lightest element. For example, in a typical fusion reaction, two deuterium atoms combine to produce helium-3, a process known as deuterium–deuterium fusion (D–D fusion): In another reaction, a deuterium atom and a tritium atom fuse to produce helium-4 ( \(\Page {4}\)), a process known as deuterium–tritium fusion (D–T fusion): Initiating these reactions, however, requires a temperature comparable to that in the interior of the sun (approximately 1.5 × 10 K). Currently, the only method available on Earth to achieve such a temperature is the detonation of a fission bomb. For example, the so-called hydrogen bomb (or H bomb) is actually a deuterium–tritium bomb (a D–T bomb), which uses a nuclear fission reaction to create the very high temperatures needed to initiate fusion of solid lithium deuteride ( LiD), which releases neutrons that then react with Li, producing tritium. The deuterium-tritium reaction releases energy explosively. Example 9 and its corresponding exercise demonstrate the enormous amounts of energy produced by nuclear fission and fusion reactions. In fact, fusion reactions are the power sources for all stars, including our sun. Calculate the amount of energy (in electronvolts per atom and kilojoules per mole) released when the neutron-induced fission of U produces Cs, Rb, and two neutrons: balanced nuclear reaction energy released in electronvolts per atom and kilojoules per mole Following the method used in Example \(\Page {1}\), calculate the change in mass that accompanies the reaction. Convert this value to the change in energy in electronvolts per atom. Calculate the change in mass per mole of U. Then use to calculate the change in energy in kilojoules per mole. The change in mass that accompanies the reaction is as follows: The change in energy in electronvolts per atom is as follows: The change in mass per mole of \(_{92}^{235}\textrm{U}\) is −0.188386 g = −1.88386 × 10 kg, so the change in energy in kilojoules per mole is as follows: Calculate the amount of energy (in electronvolts per atom and kilojoules per mole) released when deuterium and tritium fuse to give helium-4 and a neutron: ΔE = −17.6 MeV/atom = −1.697 × 10 kJ/mol Unlike a chemical reaction, a nuclear reaction results in a significant change in mass and an associated change of energy, as described by Einstein’s equation. kiloelectronvolts megaelectronvolts electronvolts | 15,604 | 1,397 |
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Aromatics, or arenes, are derivatives of benzene or other compounds with aromatic ring systems. That is, they are cyclic, planar, fully conjugated and have an odd number of π-electron pairs. Like alkenes, aromatics have π-electrons that are loosely held and are easily attracted to electrophiles. However, aromatics don't undergo the typical reactions of alkenes. For example, bromine will not add across the double bond of benzene. Instead, a bromine atom replaces one of the hydrogen atoms on the benzene. This reaction is greatly accelerated in the presence of Lewis acids, such as ferric chloride. A similar reaction happens with chlorine. If treated with chlorine gas and a metal catalyst, a chlorine atom from chlorine gas can replace a hydrogen atom on benzene. However, the same thing doesn't work as smoothly with the other halogens, iodine and fluorine. The reactions of chlorine and bromine with benzene and other aromatics can be catalysed by a variety of Lewis acidic metal catalysts. So can the reactions of alkyl halides and acyl halides, which we don't normally think of as electrophiles for alkene addition. There are some limitations on what kind of groups can be added in this way. The carbon attached to the halide should be tetrahedral. Typically, it is much easier to add secondary or tertiary alkyls than primary ones. That is, the carbon attached to the halogen had best be attached to two or three other carbons as well. Methyls are very, very difficult to add in this way. There is an exception. The carbon attached to the halogen need not be tetrahedral, provided it is a carbonyl carbon. That reaction is called an acylation. In these cases, it is the alkyl or acyl, rather than the halogen, that replaces a hydrogen atom on the benzene. Remember, benzene is most likely acting as a nucleophile in this reaction, even though it is following a different pathway than an alkene would. It is reacting with the most electrophilic part of the alkyl halide or acyl halide. Aromatics have a limited repertoire of electrophiles with which they commonly undergo reaction. In addition to these Lewis acid-catalysed reactions, there are also reactions strong acidic media, such as a mixture of nitric and sulfuric acid. Another acidic medium, referred to as "fuming sulfuric acid," is really a mixture of sulfuric acid and sulfur trioxide. Just as with the acid-catalysed reactions, the nitro group and the sulfonate group just replace a hydrogen atom on the benzene ring. The overall reaction involves bond formation between a benzene carbon and the electrophile, and bond cleavage between the same carbon and a proton. Fill in the missing reagents in the following reactions. , | 2,709 | 1,400 |
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As noted earlier, three solutions to the problem of dioxygen transport have evolved: hemoglobin (Hb), hemocyanin (Hc), and hemerythrin (Hr). Their remarkable distribution over plant and animal kingdoms is shown in Figure 4.8. The hemoglobins and myoglobins found in plants, snails, and vertebrates all appear to share a common, very ancient ancestor. There is some evidence now for a common ancestral hemocyanin. The appearance of hemerythrin in a few annelid worms is an evolutionary curiosity. These few words and the diagram will suffice to give some hints about how respiratory proteins evolved, a subject that is outside the scope of this book. Hemoglobins are the most evolutionarily diverse family of dioxygen carriers. They are found in some plants (e.g., leghemoglobin in the nitrogen-fixing nodules of legumes), many invertebrates (including some insect larvae), crustaceans, molluscs (especially bivalves and snails), almost all annelid worms, and in vertebrates with one possible exception, the Antarctic fish . With few exceptions the monomeric and oligomeric hemoglobins all share a basically similar building block: a single heme group is embedded in a folded polypeptide with a molecular weight of about 16 kDa (see Figure 4.2), and is anchored to the protein by coordination of the iron center to an imidazole ligand from a histidine residue. Mammalian myoglobin is often taken as the archetypical myoglobin (see Table 4.1). Sperm whale, bovine, or equine myoglobin are specific examples; the muscle tissue from which they may be extracted is more available than that from . The archetypical oligomeric hemoglobin that shows cooperative binding of O is the tetrameric hemoglobin A. It is readily available from the blood of human donors.* In some invertebrate hemoglobins, especially those of annelids, aggregates may contain as many as 192 binding sites, to give a molecular weight of about 3 x 10 Dalton. These and other high-molecular-weight hemoglobins of arthropods are often referred to as erythrocruorins (Er). In a few annelid worms, the otherwise ubiquitous heme b or protoheme is replaced by chloroheme (see Figure 4.2) to give chlorocruorins (Ch), which tum green upon oxygenation ( , Greek for green). Some organisms, for example the clam , feature a dimeric hemoglobin. The only known anomalous hemoglobin is Hb , which comes from a parasitic nematode found in the guts of pigs. It has a molecular weight of about 39 kDa per heme; this value is not a multiple of the myoglobin building block. Moreover, presumably in response to the low availability of O in pigs' guts, Hb has an extraordinarily high affinity for dioxygen, in large part owing to an extremely slow rate of dioxygen release. Leghemoglobin is another carrier with a high affinity for dioxygen, in this case because of a high rate of O binding. Since O is a poison for the nitrogenase enzyme, yet the nodules also require dioxygen, diffusion of O is facilitated, but the concentration of free dioxygen in the vicinity of nitrogen-fixing sites is minimized. Kinetic and thermodynamic data for dioxygen binding and release from a variety of hemoglobins are summarized in Table 4.2. Notice that for the hemoglobin tetramer, which comprises two pairs of slightly dissimilar subunits, the \(\alpha\) and \(\beta\) chains bind O with significantly different affinities and rate constants, especially in the T state. Isolated chains behave like monomeric vertebrate hemoglobins, such as whale myoglobin, which have affinities close to those of R-state hemoglobin. The chlorocruorins have a low affinity compared to other erythrocruorins. Especially for proteins that bind O cooperatively, a range of values is specified, since affinities and rates are sensitive to pH, ionic strength, specific anions and cations (allosteric effectors), and laboratory. For example, as we noted above, the O affinity of hemoglobin A is sensitive to the concentration of 2,3-DPG and to pH (Bohr effect). Trout hemoglobin I is insensitive to these species, whereas a second component of trout blood, trout hemoglobin IV, is so sensitive to pH (Root effect) that at pH < 7 trout hemoglobin IV is only partially saturated at P(O ) = 160 Torr. Note that O affinities span five orders of magnitude. Since heme catabolism produces carbon monoxide, and since in some environments CO is readily available exogenously, selected data for CO binding are also presented. * Blood from human donors is also a source for a variety of abnormal hemoglobins, the most famous of which is HbS, the hemoglobin giving rise to sickle-cell anemia, It was Pauling and coworkers who first found that HbS differs from HbA through the substitution of valine for glutamic acid in each of two of the four subunits comprising Hb, Sickle-cell anemia was the first condition to be denoted a "molecular disease." Hemocyanins (Hc), the copper-containing dioxygen carriers, are distributed erratically in two large phyla, (for example, octopi and snails) and (for example, lobsters and scorpions). The functional form of hemocyanin consists of large assemblies of subunits. In the mollusc family the subunit has a molecular weight of about 50 kDa and contains two copper atoms. From electron-microscopic observations, hemocyanin molecules are cylindrical assemblies about 190 or 380 Å long and 350 Å in diameter comprising 10 or 20 subunits, respectively, for a molecular weight as high as 9 x 10 Dalton. In the arthropod family, the subunit has a molecular weight of about 70 kDa with two copper atoms. Molecular aggregates are composed of 6, 12, 24, or 48 subunits. Upon oxygenation the colorless protein becomes blue (hence cyanin from , Greek for blue). Spectral changes upon oxygenation, oxygen affinities, kinetics of oxygen binding (Table 4.2), anion binding, and other chemical reactions show that the active site in the phylum and that in , although both containing a pair of copper atoms, are not identical. No monomeric hemocyanins, analogous to myoglobin and myohemerythrin (next section), are known. For some hemocyanins the binding of dioxygen is highly cooperative, if calcium or magnesium ions are present, with Hill coefficients as high as n ~ 9. However, the free energy of interaction per subunit can be small in comparison with that for tetrameric hemoglobin; 0.9 to 2.5 kcal/mol compared to 3.0 kcal/mol. Allosteric effects, at least for a 24-subunit tarantula hemocyanin, can be separated into those within a dodecamer (12 subunits)—the major contributor to overall allostery—and those between dodecamers. This has been termed . In contrast to the hemoglobin family, isolated chains have affinities typical of the T-state conformation for hemocyanin. The binding of CO, which binds to only one copper atom, is at best weakly cooperative. As alluded to above, the distribution of hemocyanins is striking, Among the molluscs exclusive use of hemocyanin as the respiratory protein occurs only with the cephalopods (squid, octopi, and cuttlefish), and in the arthropods only among the decapod (ten-footed) crustaceans (lobsters, shrimp, and crabs). The bivalve molluscs (for example, oysters and scallops) all use small dimeric or octameric hemoglobins. The edible gastropod (snail) omatia uses hemocyanin, whereas the apparently closely related fresh-water snail uses a high-oligomer hemoglobin. Both use a myoglobin as the oxygen-storage protein. The structure of the active site has been extensively probed by EXAFS methods, and the x-ray crystal structure of a hexameric deoxyhemocyanin is known. Each copper atom is coordinated to three imidazole groups from histidine residues. The pinwheel arrangement of the six subunits, the domain structure of a single subunit, and the domain containing the active site are shown in Figure 4.9. The biological occurrence of hemerythrins (Hr in Figure 4.8), the third class of dioxygen carriers, is relatively rare, being restricted to the sipunculid family (nonsegmented worms), a few members of the annelid (segmented worm) family, a couple of brachiopods (shrimps), and a couple of priapulids. The oxygen-binding site contains, like hemocyanin, a pair of metal atoms, in this case, iron. Upon oxygenation the colorless protein becomes purple-red. Monomeric (myohemerythrin), trimeric, and octameric forms of hemerythrin are known; all appear to be based on a similar subunit of about 13.5 kDa. When hemerythrin is extracted from the organism, its oxygen binding is at best only weakly cooperative, with Hill coefficients in the range 1.1 to 2.1. In coelomic cells (the tissue between the inner membrane lining the digestive tract and the outer membrane of the worm—analogous to flesh in vertebrates), oxygen apparently binds with higher cooperativity (n ~ 2.5). Perchlorate ions have been observed to induce cooperativity: since CIO has no biological role, it appears that in protein purifications the biological allosteric effector is lost. No Bohr effect occurs. Dioxygen binding data are accumulated in Table 4.2. The structure of hemerythrin in a variety of derivatives (oxy, azido, met, and deoxy) is now well-characterized. With three bridging ligands, a distinctive cofacial bioctahedral stereochemistry is seen (Figure 4.10). Solubility of O2 in water: 1.86 x 10 M/Torr Solubility of CO in water: 1.36 x 10 M/Torr a) 10 mM Ca added: necessary for cooperativity b) CO binding at pH 9.6. P (O ) Torr \(\Delta\)H kcal/mol \(\Delta\)S eu k \(\mu\)M s k s P (O ) Torr \(\Delta\)H kcal/mol \(\Delta\)S eu k \(\mu\)M s k s -6.0 CO -2.7 binding 4.1 noncooperative | 9,617 | 1,401 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Material_Properties/Biological_Data |
A living organism has a material structure to provide an environment for complicated chemistry of living. Chemical and physical reactions provide energy to maintain living functions and to renew structural material. Thus, consideration of biological properties is a natural extension of physical and chemical properties. To a large extend, biological functions of any materials are related to their chemical and physical properties. However, reactions in biological systems are catalyzed by enzymes. Furthermore, products of one reaction may be reactants for another in a complicate scheme of reactions to maintain live. Malfunction of a reaction causes trouble, leading to disease or death. Thus, biological properties deserve special consideration. Generally speaking, food, medicine, and toxin can be given to a person by means of feeding, absorption, inhalation and injection. Food must be absorbed in the digestive track and transported from to the targeted organs, tissues and cells to provide energy for life. Injected substances may already be in organs or tissues, but they may not be in the targeted cells. In any cases, materials in question must pass through membranes, either by enzyme aided ( ) transport or by diffusion ( ). Transportation of a substance in a biological system is a very complicated process. Absorption and transport of a substance often involve its solubility in the medium. Substances prefer to dissolve in water type fluid are said to be , whereas those prefer to dissolve in oily fluids are . Absorption and transport of substances depend on their solubility in water and lipid media. The ultimate effect on cells, tissues, and organs must take place at the molecular level. However, effects at the molecular level are often not observable, and the symptoms of these effects may appear to be unrelated to the material in question. Solid substances such as fiber, gold and charcoal, not absorbed and used in any biological function pass out as feces. Absorbed soluble substances, but not utilized by animals are excreted with urine. Bullets and shell fragments anchored in bones of war veterans were not removed, because doctors considered them biologically inert. Stainless steel parts replace joints and bones today, because they are inert. Both inorganic and organic materials are involved in structures of living organism. Lignin, cellulose, muscles, skins, and cell walls are mostly organic, whereas bones, teeth, and shells involve mostly inorganic substances. These substances may serve only as structural materials in biological systems, and if so they can be replaced by biological inert substances. However some subsystems of structural organs are responsible for vital biological functions. For example, bone marrow is responsible for blood regeneration. Thus, replacement of biological structure material involves many disciplines. Plasma, membrane, tissue, protein, lipid, enzyme, the digestive system, and the central nervous systems are some examples of biological materials, for which properties for consideration include growth and decay, turn over time, biological half life, retention time, composition and its change, and active ingredient. These are manifestation of physical and chemical properties of biological materials. However, biological properties allow us to identify and solve the biological problems. Biological materials had been studied by biologists, chemists, and engineers from the macroscopic, molecular, and functional view points. In contrast, replacement or implant materials that imitate living tissues or organs are called biomaterial, which can be divided into two categories: soft tissue and hard tissue replacement biomaterial. The former includes sutures, surgical tapes, adhesives, skin implants etc. The latter include metals (steel, aluminum, titanium, cobalt-based alloys, and titanium-based alloys); ceramics (made up of Al O , TiO , SiO , Fe O etc.); carbon (graphite and glassy carbon); and polymers. The chemistry of living is complex, and properties of biological materials towards biomaterials are of great interest. The general reaction of biological materials towards foreign biomaterials is expel (or rejection). Living tissues form a thin layer around the inert biomaterial, but materials that irritate the tissues causes inflammation. Most pure metals evoke severe tissue reaction due to their redox reactions. However, aluminum and titanium are metals of choice, because the formation of a thin oxide layer on their surface made them inert. Similarly, ceramics are compatible to body fluid because they are made of the metal oxides. The nature of the surface also affects the biological properties, rough ones enable tight attachment of tissues. Biological activities of materials can be divided according to biological functions. Substances that provide nutrition, energy, and structural need are called food, whereas those that disrupt the normal functions are called toxins. Substances used to correct the abnormal biological functions are called medicines. In recent years, a lot of research had gone into finding quantitative structure-activity relationships (QSAR) of various substances aimed at improving drug design. As they provide an indication of some biological activity, we list some categories here: Agents affecting the central nervous system includes analgesics, anesthetics, antidepressants, convulsants, anti-convulsants, neuroleptics, and psychotomimetics. There are also steroids and hormones which interacts with genes in the nuclei of cells causing complicated developments. | 5,618 | 1,403 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.07%3A_Measuring_the_Enthalpy_Change/15.7.01%3A_Lecture_Demonstrations |
Calorimeter Constant To a styrofoam cup calorimeter containing 250 mL of water at 24.0 C is added a known amount of heat as 250 mL of water from a second styrofoam cup at 32.0 C. The final temperature, measured by a computer-interfaced thermistor, is 27.7 C. The computer-generated plot of T vs. time should be projected . What is the calorimeter constant? ΔT = 27.7 C – 24.0 C = 3.7 C ΔT = 27.5 C – 24.0 C = 3.7 C ΔT = 27.7 – 32.0 = -4.3 C q = m x S.H. x ΔT = 250 g x 4.18 J/g C x -4.3 C = -4494 J q = m x S.H. x ΔT = 250 x 4.18 x 3.7 C = 3867 J q + q + q = 0 q -4494 + 3867 = 0 q = 627 J. C = Q /T = 627 J / 3.7 C = 169 J/ C A styrofoam cup calorimeter contains 250 mL of water at 25.0 C. Solid NH NO (5 g ) is added, and the temperature falls to 23.8 C. The computer-generated plot of T vs. time should be projected . The calorimeter is found to absorb 169 J to change its temperature 1 C, so it is said to have a calorimeter constant of 169 J/ C. What is the enthalpy change for the dissolution reaction? q = m x S.H. x ΔT = 250 g x 4.18 J/g C x -1.2 C = -1254 J q = C x ΔT = 169 x -1.2 C = -203 J q = -1254 + -203 = -1457 J q = +1457 J. ΔH = q (kJ) / n (mol) n = m / M = 5 g NH NO / 80 g/mol = 0.057 mol ΔH = q/n = 1.457 kJ / 0.057 mol = +25.6 kJ/mol The reaction is spontaneous even though it is endothermic, because of the large positive entropy change resulting from water association with the separate ions in solution. Ed Vitz (Kutztown University), (University of | 1,534 | 1,404 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Nuclear_Chemistry/Applications_of_Nuclear_Chemistry/Applications |
There are many applications of nuclear technology. These are discussed in separate topics as follows: The following will be separated later in other webpages. Neutron Activation Analysis The practice of nuclear medicine involves injecting a liquid radioactive pharmaceutical (radiopharmaceutical) into a patient. Signals from the radiopharmaceutical are then detected and processed into a useful image by sophisticated instruments in order to diagnose or treat a disease state. A radiopharmaceutical is formed by combining radioactive atoms radioisotopes) with chemical or biological material formulated to collect temporarily in the part of the body to be imaged... A nuclear pharmaceutical is a physiologically active carrier to which a radioisotope is attached. It is possible to manufacture chemical or biological carriers which migrate to a particular part of the human body. Calcium, for example, is a bone 'seeker', and iodine concentrates in the thyroid gland. The radioisotope attached to these compounds emits radiation so that the relevant organ and its functioning can be 'observed'. Radiation is easy to detect. Even radiation which is many times weaker than natural background radiation can be measured. The location in the patient's body which emits the radiation can thus be very accurately pinpointed. So we see that radioactive preparations can be very useful for diagnostic examination if we choose them in such a way that they emit sufficient radiation to be easily detectable in the body, but only for a long enough time to enable completion of the examination. Nuclear pharmaceuticals for diagnosis must therefore have a rather short half-life, preferably no longer than a few hours. Useful radioisotopes for diagnostic purposes are technetium-99, gallium-67, indium-111, iodine-123, iodine-131, thallium-201, krypton-81m. Tissue dies rather quickly after receiving a large dose of radiation. This aspect of radiation can be utilized for treating tumours. The goal of therapy in nuclear medicine is to use radiation to destroy diseased or cancerous tissue while sparing adjacent healthy tissue. For certain types of cancer, this is achieved by using an external radioactive beam directed at the cancerous tumor. It is also possible to insert a small radioactive source through body openings, via the bloodstream or by means of surgery into a tumour and leave it there for a period lasting from days to weeks until sufficient dose has been given off. With the exception of radioactive iodine, which is used to treat cancer of the thyroid (see picture), few radioactive therapeutic agents are injected or swallowed. A much higher dose of radioactivity is administered in a therapeutic situation than in a diagnostic one; thus the therapeutic radiopharmaceutical must have a high affinity for the diseased tissue relative to the healthy tissue. Nuclear pharmaceuticals which are used for therapy must have a rather longer half-life. Useful radioisotopes for therapeutical purposes are iodine-131 (in NaI or in metaiodobenzylguanidine, MIBG), phosphorus-32, iridium-192, gold-198. Radioactive sources which are placed in the body near the tumour for local irradiation of the tumour are also called nuclear pharmaceuticals. Iridium-192 sources are normally used for this purpose. | 3,308 | 1,405 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/12%3A_Equilibrium_Conditions_in_Multicomponent_Systems/12.05%3A_Solid-Liquid_Equilibria |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) A (freezing point as a function of liquid composition) and a (composition of a solution in equilibrium with a pure solid as a function of temperature) are different ways of describing the same physical situation. Thus, strange as it may sound, the composition \(x\A\) of an aqueous solution at the freezing point is the mole fraction solubility of ice in the solution. Section 12.2.1 described the use of freezing-point measurements to determine the solvent chemical potential in a solution of arbitrary composition relative to the chemical potential of the pure solvent. The way in which freezing point varies with solution composition in the limit of infinite dilution was derived in Sec. 12.4.1. Now let us consider the freezing behavior over the entire composition range of an liquid mixture. Let \(T'\subs{f}\) be the freezing point of a liquid mixture of composition \(x'\A\) and \(x'\B=1-x'\A\), and let \(T''\subs{f}\) be the melting point of the solid compound of composition \(x''\A=a/(a+b)\) and \(x''\B=b/(a+b)\). Figure 12.7 shows an example of a molten metal mixture that solidifies to an alloy of fixed composition. The freezing-point curve of this system is closely approximated by Eq. 12.5.23. Consider an equilibrium between a crystalline salt (or other kind of ionic solid) and a solution containing the solvated ions: \[ \tx{M\(_{\nu_+}\)X\(_{\nu_-}\)(s)} \arrows \nu_+\tx{M\(^{z_+}\)(aq)} + \nu_-\tx{X\(^{z_-}\)(aq)} \] Here \(\nu_+\) and \(\nu_-\) are the numbers of cations and anions in the formula unit of the salt, and \(z_+\) and \(z_-\) are the charge numbers of these ions. The solution in equilibrium with the solid salt is a saturated solution. The thermodynamic equilibrium constant for this kind of equilibrium is called a , \(K\subs{s}\). We can readily derive a relation between \(K\subs{s}\) and the molalities of the ions in the saturated solution by treating the dissolved salt as a single solute substance, B. We write the equilibrium in the form B\(^*\)(s)\(\arrows\)B(sln), and write the expression for the solubility product as a proper quotient of activities: \begin{equation} K\subs{s} = \frac{a\mbB}{a\B^*} \tag{12.5.24} \end{equation} From Eq. 10.3.16, we have \(a\mbB= \G\mbB \g_{\pm}^\nu(m_+/m\st)^{\nu_+}(m_-/m\st)^{\nu_-}\). This expression is valid whether or not the ions M\(^{z_+}\) and X\(^{z_-}\) are present in solution in the same ratio as in the solid salt. When we replace \(a\mbB\) with this expression, and replace \(a\B^*\) with \(\G\B^*\) (Table 9.5), we obtain \begin{equation} K\subs{s} = \left(\frac{\G\mbB}{\G\B^*}\right) \g_{\pm}^\nu \left( \frac{m_+}{m\st} \right)^{\nu_+} \left( \frac{m_-}{m\st} \right)^{\nu_-} \tag{12.5.25} \end{equation} where \(\nu=\nu_+ + \nu_-\) is the total number of ions per formula unit. \(\g_{\pm}\) is the mean ionic activity coefficient of the dissolved salt in the saturated solution, and the molalities \(m_+\) and \(m_-\) refer to the ions M\(^{z_+}\) and X\(^{z_-}\) in this solution. The first factor on the right side of Eq. 12.5.25, the proper quotient of pressure factors for the reaction B\(^*\)(s)\(\ra\)B(sln), will be denoted \(\G\subs{r}\) (the subscript “r” stands for reaction). The value of \(\G\subs{r}\) is exactly \(1\) if the system is at the standard pressure, and is otherwise approximately \(1\) unless the pressure is very high. If the aqueous solution is produced by allowing the salt to dissolve in pure water, or in a solution of a second solute containing no ions in common with the salt, then the ion molalities in the saturated solution are \(m_+=\nu_+m\B\) and \(m_-=\nu_-m\B\) where \(m\B\) is the solubility of the salt expressed as a molality. Under these conditions, Eq. 12.5.25 becomes \begin{gather} \s{ K\subs{s} = \G\subs{r} \g_{\pm}^{\nu} \left(\nu_+^{\nu_+}\nu_-^{\nu_-}\right) \left( \frac{m\B}{m\st} \right)^{\nu} } \tag{12.5.26} \cond{(no common ion)} \end{gather} We could also have obtained this equation by using the expression of Eq. 10.3.10 for \(a\mbB\). If the ionic strength of the saturated salt solution is sufficiently low (i.e., the solubility is sufficiently low), it may be practical to evaluate the solubility product with Eq. 12.5.26 and an estimate of \(\g_{\pm}\) from the Debye–Hückel limiting law (see Prob. 12.19). The most accurate method of measuring a solubility product, however, is through the standard cell potential of an appropriate galvanic cell (Sec. 14.3.3). Since \(K\subs{s}\) is a thermodynamic equilibrium constant that depends only on \(T\), and \(\G\subs{r}\) depends only on \(T\) and \(p\), Eq. 12.5.26 shows that any change in the solution composition at constant \(T\) and \(p\) that decreases \(\g_{\pm}\) must increase the solubility. For example, the solubility of a sparingly-soluble salt increases when a second salt, lacking a common ion, is dissolved in the solution; this is a . Equation 12.5.25 is a general equation that applies even if the solution saturated with one salt contains a second salt with a common ion. For instance, consider the sparingly-soluble salt M\(_{\nu_+}\)X\(_{\nu_-}\) in transfer equilibrium with a solution containing the more soluble salt M\(_{\nu'_+}\)Y\(_{\nu'_-}\) at molality \(m\C\). The common ion in this example is the cation M\(^{z_+}\). The expression for the solubility product is now \begin{gather} \s{ K\subs{s} = \G\subs{r} \g_{\pm}^{\nu} (\nu_+m\B+\nu'_+m\C)^{\nu_+}(\nu_-m\B)^{\nu_-}/(m\st)^{\nu} } \tag{12.5.27} \cond{(common cation)} \end{gather} where \(m\B\) again is the solubility of the sparingly-soluble salt, and \(m\C\) is the molality of the second salt. \(K\subs{s}\) and \(\G\subs{r}\) are constant if \(T\) and \(p\) do not change, so any increase in \(m\C\) at constant \(T\) and \(p\) must cause a decrease in the solubility \(m\B\). This is called the . From the measured solubility of a salt in pure solvent, or in an electrolyte solution with a common cation, and a known value of \(K\subs{s}\), we can evaluate the mean ionic activity coefficient \(\g_{\pm}\) through Eq. 12.5.26 or 12.5.27. This procedure has the disadvantage of being limited to the value of \(m\B\) existing in the saturated solution. We find the temperature dependence of \(K\subs{s}\) by applying Eq. 12.1.12: \begin{equation} \frac{\dif\ln K\subs{s}}{\dif T} = \frac{\Delsub{sol,B}H\st}{RT^2} \tag{12.5.28} \end{equation} At the standard pressure, \(\Delsub{sol,B}H\st\) is the same as the molar enthalpy of solution at infinite dilution, \(\Delsub{sol,B}H^{\infty}\). | 13,900 | 1,407 |
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We can rank the strengths of acids by the extent to which they ionize in aqueous solution. The reaction of an acid with water is given by the general expression: \[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq)\] Water is the base that reacts with the acid \(\ce{HA}\), \(\ce{A^{−}}\) is the conjugate base of the acid \(\ce{HA}\), and the hydronium ion is the conjugate acid of water. A strong acid yields 100% (or very nearly so) of \(\ce{H3O+}\) and \(\ce{A^{−}}\) when the acid ionizes in water; Figure \(\Page {1}\) lists several strong acids. A weak acid gives small amounts of \(\ce{H3O+}\) and \(\ce{A^{−}}\). The relative strengths of acids may be determined by measuring their equilibrium constants in aqueous solutions. In solutions of the same concentration, stronger acids ionize to a greater extent, and so yield higher concentrations of hydronium ions than do weaker acids. The equilibrium constant for an acid is called the . For the reaction of an acid \(\ce{HA}\): \[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq)\] we write the equation for the ionization constant as: \[K_\ce{a}=\ce{\dfrac{[H3O+,A- ]}{[HA]}}\] where the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is also the solvent. If the solution is assumed to be dilute, the activity of the water is approximated by the activity of pure water, which is defined as having a value of 1. The larger the \(K_a\) of an acid, the larger the concentration of \(\ce{H3O+}\) and \(\ce{A^{−}}\) relative to the concentration of the nonionized acid, \(\ce{HA}\). Thus a stronger acid has a larger ionization constant than does a weaker acid. The ionization constants increase as the strengths of the acids increase. The following data on acid-ionization constants indicate the order of acid strength: \(\ce{CH3CO2H} < \ce{HNO2} < \ce{HSO4-}\) \[ \begin{aligned} \ce{CH3CO2H}(aq) + \ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{CH3CO2-}(aq) \quad &K_\ce{a}=1.8×10^{−5} \\[4pt] \ce{HNO2}(aq)+\ce{H2O}(l) &⇌\ce{H3O+}(aq)+\ce{NO2-}(aq) &K_\ce{a}=4.6×10^{-4} \\[4pt] \ce{HSO4-}(aq)+\ce{H2O}(aq) &⇌\ce{H3O+}(aq)+\ce{SO4^2-}(aq) & K_\ce{a}=1.2×10^{−2} \end{aligned}\] Another measure of the strength of an acid is its . The of a weak acid is the ratio of the concentration of the ionized acid to the initial acid concentration, times 100: \[\% \:\ce{ionization}=\ce{\dfrac{[H3O+]_{eq}}{[HA]_0}}×100\% \label{PercentIon} \] Because the ratio includes the initial concentration, the percent ionization for a solution of a given weak acid varies depending on the original concentration of the acid, and actually decreases with increasing acid concentration. Calculate the percent ionization of a 0.125- solution of nitrous acid (a weak acid), with a pH of 2.09. The percent ionization for an acid is: \[\ce{\dfrac{[H3O+]_{eq}}{[HNO2]_0}}×100 \nonumber\] The chemical equation for the dissociation of the nitrous acid is: \[\ce{HNO2}(aq)+\ce{H2O}(l)⇌\ce{NO2-}(aq)+\ce{H3O+}(aq). \nonumber\] Since \(10^{−pH} = \ce{[H3O+]}\), we find that \(10^{−2.09} = 8.1 \times 10^{−3}\, M\), so that percent ionization (Equation \ref{PercentIon}) is: \[\dfrac{8.1×10^{−3}}{0.125}×100=6.5\% \nonumber \] Remember, the logarithm 2.09 indicates a hydronium ion concentration with only two significant figures. Calculate the percent ionization of a 0.10 solution of acetic acid with a pH of 2.89. 1.3% ionized We can rank the strengths of bases by their tendency to form hydroxide ions in aqueous solution. The reaction of a Brønsted-Lowry base with water is given by: \[\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq)\] Water is the acid that reacts with the base, \(\ce{HB^{+}}\) is the conjugate acid of the base \(\ce{B}\), and the hydroxide ion is the conjugate base of water. A strong base yields 100% (or very nearly so) of OH and HB when it reacts with water; Figure \(\Page {1}\) lists several strong bases. A weak base yields a small proportion of hydroxide ions. Soluble ionic hydroxides such as NaOH are considered strong bases because they dissociate completely when dissolved in water. As we did with acids, we can measure the relative strengths of bases by measuring their in aqueous solutions. In solutions of the same concentration, stronger bases ionize to a greater extent, and so yield higher hydroxide ion concentrations than do weaker bases. A stronger base has a larger ionization constant than does a weaker base. For the reaction of a base, \(\ce{B}\): \[\ce{B}(aq)+\ce{H2O}(l)⇌\ce{HB+}(aq)+\ce{OH-}(aq),\] we write the equation for the ionization constant as: \[K_\ce{b}=\ce{\dfrac{[HB+,OH- ]}{[B]}}\] where the concentrations are those at equilibrium. Again, we do not include [H O] in the equation because water is the solvent. The chemical reactions and ionization constants of the three bases shown are: \[ \begin{aligned} \ce{NO2-}(aq)+\ce{H2O}(l) &⇌\ce{HNO2}(aq)+\ce{OH-}(aq) \quad &K_\ce{b}=2.17×10^{−11} \\[4pt] \ce{CH3CO2-}(aq)+\ce{H2O}(l) &⇌\ce{CH3CO2H}(aq)+\ce{OH-}(aq) &K_\ce{b}=5.6×10^{−10} \\[4pt] \ce{NH3}(aq)+\ce{H2O}(l) &⇌\ce{NH4+}(aq)+\ce{OH-}(aq) &K_\ce{b}=1.8×10^{−5} \end{aligned}\] A table of ionization constants of weak bases appears in Table E2. As with acids, percent ionization can be measured for basic solutions, but will vary depending on the base ionization constant and the initial concentration of the solution. Consider the ionization reactions for a conjugate acid-base pair, \(\ce{HA − A^{−}}\): \[\ce{HA}(aq)+\ce{H2O}(l)⇌\ce{H3O+}(aq)+\ce{A-}(aq)\] with \(K_\ce{a}=\ce{\dfrac{[H3O+,A- ]}{[HA]}}\). \[\ce{A-}(aq)+\ce{H2O}(l)⇌\ce{OH-}(aq)+\ce{HA}(aq)\] with \(K_\ce{b}=\ce{\dfrac{[HA,OH]}{[A- ]}}\). Adding these two chemical equations yields the equation for the autoionization for water: \[\begin{align*} \cancel{\ce{HA}(aq)}+\ce{H2O}(l)+\cancel{\ce{A-}(aq)}+\ce{H2O}(l) &⇌ \ce{H3O+}(aq)+\cancel{\ce{A-}(aq)}+\ce{OH-}(aq)+\cancel{\ce{HA}(aq)} \\[4pt] \ce{2H2O}(l) &⇌\ce{H3O+}(aq)+\ce{OH-}(aq) \end{align*}\] As shown in the previous chapter on equilibrium, the \(K\) expression for a chemical equation derived from adding two or more other equations is the mathematical product of the other equations’ \(K\) expressions. Multiplying the mass-action expressions together and cancelling common terms, we see that: \[K_\ce{a}×K_\ce{b}=\ce{\dfrac{[H3O+,A- ]}{[HA]}×\dfrac{[HA,OH- ]}{[A- ]}}=\ce{[H3O+,OH- ]}=K_\ce{w}\] For example, the acid ionization constant of acetic acid (CH COOH) is 1.8 × 10 , and the base ionization constant of its conjugate base, acetate ion (\(\ce{CH3COO-}\)), is 5.6 × 10 . The product of these two constants is indeed equal to \(K_w\): \[K_\ce{a}×K_\ce{b}=(1.8×10^{−5})×(5.6×10^{−10})=1.0×10^{−14}=K_\ce{w}\] The extent to which an acid, \(\ce{HA}\), donates protons to water molecules depends on the strength of the conjugate base, \(\ce{A^{−}}\), of the acid. If \(\ce{A^{−}}\) is a strong base, any protons that are donated to water molecules are recaptured by \(\ce{A^{−}}\). Thus there is relatively little \(\ce{A^{−}}\) and \(\ce{H3O+}\) in solution, and the acid, \(\ce{HA}\), is weak. If \(\ce{A^{−}}\) is a weak base, water binds the protons more strongly, and the solution contains primarily \(\ce{A^{−}}\) and \(\ce{H3O^{+}}\)—the acid is strong. Strong acids form very weak conjugate bases, and weak acids form stronger conjugate bases (Figure \(\Page {2}\)). Figure \(\Page {3}\) lists a series of acids and bases in order of the decreasing strengths of the acids and the corresponding increasing strengths of the bases. The acid and base in a given row are conjugate to each other. The first six acids in Figure \(\Page {3}\) are the most common strong acids. These acids are completely dissociated in aqueous solution. The conjugate bases of these acids are weaker bases than water. When one of these acids dissolves in water, their protons are completely transferred to water, the stronger base. Those acids that lie between the hydronium ion and water in Figure \(\Page {3}\) form conjugate bases that can compete with water for possession of a proton. Both hydronium ions and nonionized acid molecules are present in equilibrium in a solution of one of these acids. Compounds that are weaker acids than water (those found below water in the column of acids) in Figure \(\Page {3}\) exhibit no observable acidic behavior when dissolved in water. Their conjugate bases are stronger than the hydroxide ion, and if any conjugate base were formed, it would react with water to re-form the acid. The extent to which a base forms hydroxide ion in aqueous solution depends on the strength of the base relative to that of the hydroxide ion, as shown in the last column in Figure \(\Page {3}\). A strong base, such as one of those lying below hydroxide ion, accepts protons from water to yield 100% of the conjugate acid and hydroxide ion. Those bases lying between water and hydroxide ion accept protons from water, but a mixture of the hydroxide ion and the base results. Bases that are weaker than water (those that lie above water in the column of bases) show no observable basic behavior in aqueous solution. Use the \(K_b\) for the nitrite ion, \(\ce{NO2-}\), to calculate the \(K_a\) for its conjugate acid. for \(\ce{NO2-}\) is given in this section as 2.17 × 10 . The conjugate acid of \(\ce{NO2-}\) is HNO ; for HNO can be calculated using the relationship: \[K_\ce{a}×K_\ce{b}=1.0×10^{−14}=K_\ce{w} \nonumber \] Solving for , we get: \[\begin{align*} K_\ce{a} &=\dfrac{K_\ce{w}}{K_\ce{b}} \\[4pt] &=\dfrac{1.0×10^{−14}}{2.17×10^{−11}} \\[4pt] &=4.6×10^{−4} \end{align*}\] This answer can be verified by finding the for HNO in Table E1 We can determine the relative acid strengths of \(\ce{NH4+}\) and \(\ce{HCN}\) by comparing their ionization constants. The ionization constant of \(\ce{HCN}\) is given in Table E1 as 4.9 × 10 . The ionization constant of \(\ce{NH4+}\) is not listed, but the ionization constant of its conjugate base, \(\ce{NH3}\), is listed as 1.8 × 10 . Determine the ionization constant of \(\ce{NH4+}\), and decide which is the stronger acid, \(\ce{HCN}\) or \(\ce{NH4+}\). \(\ce{NH4+}\) is the slightly stronger acid ( for \(\ce{NH4+}\) = 5.6 × 10 ). The strengths of Brønsted-Lowry acids and bases in aqueous solutions can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids form stronger conjugate bases. Thus strong acids are completely ionized in aqueous solution because their conjugate bases are weaker bases than water. Weak acids are only partially ionized because their conjugate bases are strong enough to compete successfully with water for possession of protons. Strong bases react with water to quantitatively form hydroxide ions. Weak bases give only small amounts of hydroxide ion. ). | 10,819 | 1,409 |
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is defined as the upper limit of solute that can be dissolved in a given amount of solvent at equilibrium. In such an equilibrium, can be used to explain most of the main factors that affect solubility. â dictates that the effect of a stress upon a system in chemical equilibrium can be predicted in that the system tends to shift in such a way as to alleviate that stress. The relation between the solute and solvent is very important in determining . Strong solute-solvent attractions equate to greater solubility while weak solute-solvent attractions equate to lesser . In turn, polar solutes tend to dissolve best in polar solvents while non-polar solutes tend to dissolve best in non-polar solvents. In the case of a polar solute and non-polar solvent (or vice versa), it tends to be insoluble or only soluble to a miniscule degree. A general rule to remember is, "Like dissolves like." The is a term that describes the decrease in of an ionic compound when a salt that contains an ion that already exists in the chemical equilibrium is added to the mixture. This effect best be explained by . Imagine if the slightly soluble ionic compound calcium sulfate, CaSO , is added to water. The net ionic equation for the resulting chemical equilibrium is the following: \[ CaSO_{4(s)} \rightleftharpoons Ca^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \] Calcium sulfate is slightly soluble; at equilibrium, most of the calcium and sulfate exists in the solid form of calcium sulfate. Suppose the soluble ionic compound copper sulfate (CuSO ) were added to the solution. Copper sulfate is soluble; therefore, its only important effect on the net ionic equation is the addition of more sulfate (SO ) ions. \[ CuSO_{4(s)} \rightleftharpoons Cu^{2+}_{(aq)} + SO^{2-}_{4 (aq)} \] The sulfate ions dissociated from copper sulfate are already present (common to) in the mixture from the slight dissociation of calcium sulfate. Thus, this addition of sulfate ions places stress on the previously established equilibrium. Le Chatelier's principle dictates that the additional stress on this product side of the equilibrium results in the shift of equilibrium towards the reactants side in order to alleviate this new stress. Because of the shift toward the reactant side, the solubility of the slightly soluble calcium sulfate is reduced even further. Temperature changes affect the solubility of solids, liquids and gases differently. However, those effects are finitely determined only for solids and gases. The effects of temperature on the solubility of solids differ depending on whether the reaction is endothermic or exothermic. Using Le Chatelier's principle, the effects of temperature in both scenarios can be determined. In the case of liquids, there is no defined trends for the effects of temperature on the solubility of liquids. In understanding the effects of temperature on the solubility of gases, it is first important to remember that temperature is a measure of the average kinetic energy. As temperature increases, kinetic energy increases. The greater kinetic energy results in greater molecular motion of the gas particles. As a result, the gas particles dissolved in the liquid are more likely to escape to the gas phase and the existing gas particles are less likely to be dissolved. The converse is true as well. The trend is thus as follows: increased temperatures mean lesser solubility and decreased temperatures mean higher solubility. Le Chatelier's principle allows better conceptualization of these trends. First, note that the process of dissolving gas in liquid is usually . As such, temperatures result in stress on the product side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts towards the reactant side in order to alleviate this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase increases, resulting in solubility. Conversely, temperatures result in stress on the reactant side (because heat is on the product side). In turn, Le Chatelier's principle predicts that the system shifts toward the product side in order to compensate for this new stress. Consequently, the equilibrium concentration of the gas particles in gaseous phase would decrease, resulting in solubility. The effects of pressure are only significant in affecting the solubility of gases in liquids. \[ p = k_h \; c \] where: This formula indicates that (at a constant temperature) when the partial pressure decreases, the concentration of gas in the liquid decreases as well, and consequently the solubility also decreases. Conversely, when the partial pressure increases in such a situation, the concentration of gas in the liquid will increase as well; the solubility also increases. Extending the implications from , the usefulness of Le Chatelier's principle is enhanced in predicting the effects of pressure on the solubility of gases. Consider a system consisting of a gas that is partially dissolved in liquid. An increase in pressure would result in greater partial pressure (because the gas is being further compressed). This increased partial pressure means that more gas particles will enter the liquid (there is therefore less gas above the liquid, so the partial pressure decreases) in order to alleviate the stress created by the increase in pressure, resulting in greater solubility. The converse case in such a system is also true, as a decrease in pressure equates to more gas particles escaping the liquid to compensate. Consider the following exothermic reaction that is in equilibrium \[ CO_2 (g) + H_2O (l) \rightleftharpoons H_2CO_3 (aq) \] What will happen to the solubility of the carbon dioxide if: Bob is in the business of purifying silver compounds to extract the actual silver. He is extremely frugal. One day, he finds a barrel containing a saturated solution of silver chloride. Bob has a bottle of water, a jar of table salt (NaCl(s)), and a bottle of vinegar (CH COOH). Which of the three should Bob add to the solution to maximize the amount of solid silver chloride (minimizing the solubility of the silver chloride)? Bob should add table salt to the solution. According to the common-ion effect, the additional Cl ions would reduce the solubility of the silver chloride, which maximizes the amount of solid silver chloride. Allison has always wanted to start her own carbonated drink company. Recently, she opened a factory to produce her drinks. She wants her drink to "out-fizz" all the competitors. That is, she wants to maximize the solubility of the gas in her drink. What conditions (high/low temperature, high/low pressure) would best allow her to achieve this goal? She would be able to maximize the solubility of the gas, (\(CO_2\) in this case, in her drink (maximize fizz) when the pressure is high and temperature is low. Butters is trying to increase the solubility of a solid in some water. He begins to frantically stir the mixture. Should he continue stirring? Why or why not? He stop stop stirring. Stirring only affects how fast the system will reach equilibrium and does not affect the solubility of the solid at all. With respect to Henry's law, why is it a poor ideal to open a can of soda in a low pressure environment? The fizziness of soda originates from dissolved \(CO_2\), partially in the form of carbonic acid. The concentration of \(CO_2\) dissolved in the soda depends on the amount of ambient pressure pressing down on the liquid. Hence, the soda can will be under pressure to maintain the desired \(CO_2\) concentration. When the can is opened to a lower pressure environment (e.g., the ambient atmosphere), the soda will quickly "outgas" (\(CO_2\) will come out of solution) at a rate depending on the surrounding atmospheric pressure. If a can of soda were opened under a lower pressure environment, this outgassing will be faster and hence more explosive (and dangerous) than under a high pressure environment. | 8,011 | 1,410 |
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Band Theory was developed with some help from the knowledge gained during the quantum revolution in science. In 1928, Felix Bloch had the idea to take the quantum theory and apply it to solids. In 1927, Walter Heitler and Fritz London discovered bands- very closely spaced orbitals with not much difference in energy. In this image, orbitals are represented by the black horizontal lines, and they are being filled with an increasing number of electrons as their amount increases. Eventually, as more orbitals are added, the space in between them decreases to hardly anything, and as a result, a band is formed where the orbitals have been filled. Different metals will produce different combinations of filled and half filled bands. Sodium's bands are shown with the rectangles. Filled bands are colored in blue. As you can see, bands may overlap each other (the bands are shown askew to be able to tell the difference between different bands). The lowest unoccupied band is called the conduction band, and the highest occupied band is called the valence band. Bands will follow a trend as you go across a period: The probability of finding an electron in the conduction band is shown by the equation: \[ P= \dfrac{1}{e^{ \Delta E/RT}+1} \] The ∆E in the equation stands for the change in energy or energy gap. t stands for the temperature, and R is a bonding constant. That equation and this table below show how the bigger difference in energy is, or gap, between the valence band and the conduction band, the less likely electrons are to be found in the conduction band. This is because they cannot be excited enough to make the jump up to the conduction band. Metals are conductors. There is no band gap between their valence and conduction bands, since they overlap. There is a continuous availability of electrons in these closely spaced orbitals. In insulators, the band gap between the valence band the the conduction band is so large that electrons cannot make the energy jump from the valence band to the conduction band. Semiconductors have a small energy gap between the valence band and the conduction band. Electrons can make the jump up to the conduction band, but not with the same ease as they do in conductors. There are two different kinds of semiconductors: and . An intrinsic semiconductor is a semiconductor in its pure state. For every electron that jumps into the conduction band, the missing electron will generate a hole that can move freely in the valence band. The number of holes will equal the number of electrons that have jumped. In extrinsic semiconductors, the band gap is controlled by purposefully adding small impurities to the material. This process is called . Doping, or adding impurities to the lattice can change the electrical conductivity of the lattice and therefore vary the efficiency of the semiconductor. In extrinsic semiconductors, the number of holes will not equal the number of electrons jumped. There are two different kinds of extrinsic semiconductors, p-type (positive charge doped) and n-type (negative charge doped). Jim Clark ( ) | 3,106 | 1,411 |
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: The viscosity of three simple liquids is given over an extended temperature range. Each data set is analyzed in order to determine the activation energy (Ea) for viscous flow of the given liquid. : The tasks associated with this assignment can be carried out with an introductory level knowledge of chemistry. This exercise should be carried out within a software environment that is capable of data manipulation and which can generate a best-fit line for an x-y data set. You will also be graphing the data along with the fitted function. Viscosity (\(η\)) is a fluid property which indicates how resistant that fluid is to flow. Highly viscous liquids, like motor oil or molasses, take much longer to flow from their container than a relatively low viscosity liquid, like benzene or diethyl ether. To quantify viscosity, we will imagine our bulk fluid as consisting of a number of very thin layers. In order for the fluid to flow, a force will be required to slide these layers relative to one another. The amount of force (\(f\)) required is assumed to be directly proportional to the area (\(A\)) of the layers in contact with one another and the velocity difference (\(v\)) between the layers. Furthermore, the force is inversely proportional to the distance (d) between the layers. Viscosity (η) can then be introduced as a constant of proportionally, yielding a force equation of the form \[ f = \eta \dfrac{A v}{d} \label{1}\] Dimensional analysis of Equation \ref{1} give SI Units for viscosity of \(kg \,m^{-1}\, s^{-1}\). However, the unit that is typically employed in practice is called the 'poise' (P), where \[1\, P = 1\, gram \,cm^{-1}\, s^{-1}.\] Liquid viscosities are usually reported in centipoise, (\(cP\)), and gas viscosity are reported in micropoise, \(μP\).
What factors determine whether a given fluid has a high (or low) viscosity? Certainly the strength of intermolecular attractions has an influence; nitro-benzene has a much higher viscosity than regular benzene because the former is capable of dipole-dipole attractions which are considerably stronger than the dispersion forces of attraction present in bulk benzene. Other factors can contribute, such as the size and shape of molecules. For example, long chain molecules like polymers are capable of becoming entangled with each other which causes friction between the hypothetical layers of the fluid which translates into a large viscosity.
A number of experimental methods are available for measuring viscosity. Many are based upon measuring the amount of time (\(t\)) it takes for a given amount of fluid to flow through a thin glass tube or to drain from a vessel that has a small opening in the bottom. An equivalent measurement is carried on a fluid of known viscosity. The unknown viscosity is then calculated using the expression \[\dfrac{\eta_1}{\eta_2} = \dfrac{\rho_1t_1}{\rho_2 t_2} \label{2}\] where \(ρ\) represents the density of each fluid (which is usually measured separately). Viscosity varies with temperature, generally becoming smaller as temperature is elevated. This trend occurs because the increased kinetic motion at higher temperatures promotes the breaking of intermolecular bonds between adjacent layers. A considerable amount of research has been carried out in an attempt to understand the exact nature of the temperature variation of viscosity. One relatively simple model assumes that the viscosity obeys an ‘Arrhenius-like’ equation of the form \[\eta = A \exp \left(\dfrac{E_a}{RT} \right) \label{3}\] where \(A\) and \(E_a\) are constants for a given fluid. \(A\) is called the pre-exponential factor and \(E_a\) can be interpreted as the activation energy for viscous flow. Note that this expression is nearly identical to the Arrhenius equation that describes the temperature variation of the rate constant (\(k\)) of a chemical reaction, except equation (3) does not have a negative sign in the exponential which causes the viscosity to get smaller with increasing temperature.
Equation \ref{3} can be written in the logarithmic form \[\ln \eta = \ln A + \left( \dfrac{E_a}{R} \right) \dfrac{1}{T} \label{4}\] If a fluid obeys Equation \ref{4}, then a plot of viscosity versus reciprocal absolute temperature should be linear and the slope can be used to determine the activation energy for viscous flow.
In this assignment, the viscosities of three liquids are given over a certain temperature range. Each data set is analyzed in order to investigate whether the liquid obeys the simple Arrhenius model and to determine the activation energies for viscous flow for these liquids. The following table presents viscosity data for water, ethanol, and diethyl ether over specific temperature ranges. The data were obtained from the CRC Handbook of Chemistry and Physics. | 4,825 | 1,412 |
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In the first three chapters of this text, we have focused our efforts on learning about the structure of organic compounds. Now that we know what organic molecules look like, we can start to address the question of chemists are able to elucidate organic structures. The individual atoms and functional groups in organic compounds are far too small to be directly observed or photographed, even with the best electron microscope. How, then, are chemists able to draw with confidence the bonding arrangements in organic molecules, even simple ones such as acetone or ethanol? The answer lies, for the most part, in a field of chemistry called . Spectroscopy is the study of how electromagnetic radiation, across a spectrum of different wavelengths, interacts with molecules - and how these interactions can be quantified, analyzed, and ultimately interpreted to gain information about molecular structure. After first reviewing some basic information about the properties of light and introducing the basic ideas behind spectroscopy, we will move to a discussion of infrared (IR) spectroscopy, a technique which is used in organic chemistry to detect the presence or absence of common functional groups. Next, we will look at ultraviolet-visible (UV-vis) spectroscopy, in which light of a shorter wavelength is employed to provide information about organic molecules containing conjugated p-bonding systems. In the final section of this chapter, we will change tack slightly and consider another analytical technique called mass spectrometry (MS). Here, we learn about the structure of a molecule by, in a sense, taking a hammer to it and smashing it into small pieces, then measuring the mass of each piece. Although this metaphorical description makes mass spectrometry sound somewhat crude, it is in fact an extremely powerful and sensitive technique, one which has in recent years become central to the study of life at the molecular level. Looking ahead, Chapter 5 will be devoted to the study of nuclear magnetic resonance (NMR) spectroscopy, where we use ultra-strong magnets and radio frequency radiation to learn about the electronic environment of individual atoms in a molecule. For most organic chemists, NMR is the single most powerful analytical tool available in terms of the wealth of detailed information it can provide about the structure of a molecule. It is the closest thing we have to a ‘molecular camera’. In summary, the analytical techniques we will be studying in this chapter and the next primarily attempt to address the following questions about an organic molecule:
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The Heat of Reaction (also known and Enthalpy of Reaction) is the change in the of a chemical reaction that occurs at a constant pressure. It is a thermodynamic unit of measurement useful for calculating the amount of energy per mole either released or produced in a reaction. Since enthalpy is derived from pressure, volume, and internal energy, all of which are state functions, enthalpy is also a . \(ΔH\), or the change in enthalpy arose as a unit of measurement meant to calculate the change in energy of a system when it became too difficult to find the ΔU, or change in the internal energy of a system, by simultaneously measure the amount of and exchanged. Given a , the change in enthalpy can be measured as \[ΔH=q\] See section on for a more detailed explanation. The notation ΔHº or ΔHº then arises to explain the precise temperature and pressure of the heat of reaction ΔH. The standard enthalpy of reaction is symbolized by ΔHº or ΔHº and can take on both positive and negative values. The units for ΔHº are kiloJoules per mole, or kj/mol. : The standard state of a solid or liquid is the pure substance at a pressure of 1 bar ( 10 Pa) and at a relevant temperature. The ΔHº is the standard heat of reaction or standard enthalpy of a reaction, and like ΔH also measures the enthalpy of a reaction. However, ΔHº takes place under "standard" conditions, meaning that the reaction takes place at 25º C and 1 atm. The benefit of a measuring ΔH under standard conditions lies in the ability to relate one value of ΔHº to another, since they occur under the same conditions. Enthalpy can be measured experimentally through the use of a calorimeter. A calorimeter is an isolated system which has a constant pressure, so ΔH=q=cp x m x (ΔT) To calculate the standard enthalpy of reaction the standard enthalpy of formation must be utilized. Another, more detailed, form of the standard enthalpy of reaction includes the use of the standard enthalpy of formation ΔH : \[ ΔH^\ominus = \sum \Delta v_p \Delta H^\ominus_f\;(products) - \sum \Delta v_r \Delta H^\ominus_f\; (reactants)\] with Since enthalpy is a state function, the heat of reaction depends only on the final and initial states, not on the path that the reaction takes. For example, the reaction \( A \rightarrow B\) goes through intermediate steps (i.e. \(C \rightarrow D\)), but A and B remain intact. Therefore, one can measure the enthalpy of reaction as the sum of the ΔH of the three reactions by applying . Since the ΔHº represents the total energy exchange in the reaction this value can be either positive or negative. Calculate the enthalpy change for the combustion of acetylene (\(\ce{C2H2}\)) 1) The first step is to make sure that the equation is balanced and correct. Remember, the combustion of a hydrocarbon requires oxygen and results in the production of carbon dioxide and water. \[\ce{2C2H2(g) + 5O2(g) -> 4CO2(g) + 2H2O(g)}\] 2) Next, locate a table of Standard Enthalpies of Formation to look up the values for the components of the reaction (Table 7.2, Petrucci Text) 3) First find the enthalpies of the products: ΔHº CO = -393.5 kJ/mole Multiply this value by the stoichiometric coefficient, which in this case is equal to 4 mole. v ΔH CO = 4 mol (-393.5 kJ/mole) = -1574 kJ ΔH H O = -241.8 kJ/mole The stoichiometric coefficient of this compound is equal to 2 mole. So, v ΔH H O = 2 mol ( -241.8 kJ/mole) = -483.6 kJ Sum of products (Σ v ΔHº (products)) = (-1574 kJ) + (-483.6 kJ) = -2057.6 kJ ΔHº C H = +227 kJ/mole Multiply this value by the stoichiometric coefficient, which in this case is equal to 2 mole. v ΔHº C H = 2 mol (+227 kJ/mole) = +454 kJ ΔHº O = 0.00 kJ/mole The stoichiometric coefficient of this compound is equal to 5 mole. So, v ΔHº O = 5 mol ( 0.00 kJ/mole) = 0.00 kJ Sum of reactants (Δ v ΔHº (reactants)) = (+454 kJ) + (0.00 kJ) = +454 kJ ΔHº = Δ v ΔHº (products) - ? v ΔHº (reactants) = -2057.6 kJ - +454 kJ = -2511.6 kJ | 3,985 | 1,414 |
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This page looks briefly at some of the basic processes in the petrochemical industry (cracking, isomerisation and reforming) as examples of important catalytic reactions. Cracking is the name given to breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high pressures and temperatures without a catalyst, or lower temperatures and pressures in the presence of a catalyst. The source of the large hydrocarbon molecules is often the naphtha fraction or the gas oil fraction from the fractional distillation of crude oil (petroleum). These fractions are obtained from the distillation process as liquids, but are re-vaporised before cracking. The hydrocarbons are mixed with a very fine catalyst powder. These days the catalysts are zeolites (complex aluminosilicates) - these are more efficient than the older mixtures of aluminium oxide and silicon dioxide. The whole mixture is blown rather like a liquid through a reaction chamber at a temperature of about 500°C. Because the mixture behaves like a liquid, this is known as fluid catalytic cracking (or fluidised catalytic cracking). Although the mixture of gas and fine solid behaves as a liquid, this is nevertheless an example of heterogeneous catalysis - the catalyst is in a different phase from the reactants. The catalyst is recovered afterwards, and the cracked mixture is separated by cooling and further fractional distillation. There isn't any single unique reaction happening in the cracker. The hydrocarbon molecules are broken up in a fairly random way to produce mixtures of smaller hydrocarbons, some of which have carbon-carbon double bonds. One possible reaction involving the hydrocarbon C H might be: Or, showing more clearly what happens to the various atoms and bonds: This is only one way in which this particular molecule might break up. The ethene and propene are important materials for making plastics or producing other organic chemicals. The octane is one of the molecules found in petrol (gasoline). Hydrocarbons used in petrol (gasoline) are given an octane rating which relates to how effectively they perform in the engine. A hydrocarbon with a high octane rating burns more smoothly than one with a low octane rating. Molecules with "straight chains" have a tendency to pre-ignition. When the petrol / air mixture is compressed they tend to explode, and then explode a second time when the spark is passed through them. This double explosion produces knocking in the engine. Octane ratings are based on a scale on which heptane is given a rating of 0, and 2,2,4-trimethylpentane (an isomer of octane) a rating of 100. In order to raise the octane rating of the molecules found in petrol (gasoline) and so make the petrol burn better in modern engines, the oil industry rearranges straight chain molecules into their isomers with branched chains. One process uses a platinum catalyst on a zeolite base at a temperature of about 250°C and a pressure of 13 - 30 atmospheres. It is used particularly to change straight chains containing 5 or 6 carbon atoms into their branched isomers. For example: Reforming is another process used to improve the octane rating of hydrocarbons to be used in petrol, but is also a useful source of aromatic compounds for the chemical industry. Aromatic compounds are ones based on a benzene ring. Reforming uses a platinum catalyst suspended on aluminium oxide together with various promoters to make the catalyst more efficient. The original molecules are passed as vapours over the solid catalyst at a temperature of about 500°C. Isomerisation reactions occur (as above) but, in addition, chain molecules get converted into rings with the loss of hydrogen. Hexane, for example, gets converted into benzene, and heptane into methylbenzene. | 3,851 | 1,415 |
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It has been demonstrated that water adds rapidly to the carbonyl function of aldehydes and ketones. In most cases the resulting (a geminal-diol) is unstable relative to the reactants and cannot be isolated. Exceptions to this rule exist, one being formaldehyde (a gas in its pure monomeric state). Here the weaker pi-component of the carbonyl double bond, relative to other aldehydes or ketones, and the small size of the hydrogen substituents favor addition. Thus, a solution of formaldehyde in water (formalin) is almost exclusively the hydrate, or polymers of the hydrate. Similar reversible additions of alcohols to aldehydes and ketones take place. The equally unstable addition products are called . R C=O + –(R )C–O– (a hemiacetal) Acetals are geminal-diether derivatives of aldehydes or ketones, formed by reaction with two equivalents of an alcohol and elimination of water. Ketone derivatives of this kind were once called ketals, but modern usage has dropped that term. The following equation shows the overall stoichiometric change in acetal formation, but a dashed arrow is used because this conversion does not occur on simple mixing of the reactants. R C=O + 2 R C( ) + O (an acetal) In order to achieve effective acetal formation two additional features must be implemented. First, an acid catalyst must be used; and second, the water produced with the acetal must be removed from the reaction. The latter is important, since acetal formation is reversible. Indeed, once pure acetals are obtained they may be hydrolyzed back to their starting components by treatment with aqueous acid. The mechanism shown here applies to both acetal formation and acetal hydrolysis by the . Some examples of acetal formation are presented in the following diagram. As noted, p-toluenesulfonic acid (pK = -2) is often the catalyst for such reactions. Two equivalents of the alcohol reactant are needed, but these may be provided by one equivalent of a diol (example #2). Intramolecular involvement of a gamma or delta hydroxyl group (as in examples #3 and 4) may occur, and is often more facile than the intermolecular reaction. Thiols (sulfur analogs of alcohols) give thioacetals (example #5). In this case the carbonyl functions are relatively hindered, but by using excess ethanedithiol as the solvent and the Lewis acid BF as catalyst a good yield of the bis-thioacetal is obtained. Thioacetals are generally more difficult to hydrolyze than are acetals. The importance of acetals as carbonyl derivatives lies chiefly in their stability and lack of reactivity in neutral to strongly basic environments. As long as they are not treated by acids, especially aqueous acid, acetals exhibit all the lack of reactivity associated with in general.
Among the most useful and characteristic reactions of aldehydes and ketones is their reactivity toward strongly nucleophilic (and basic) metallo-hydride, alkyl and aryl reagents (to be discussed shortly). If the carbonyl functional group is converted to an acetal these powerful reagents have no effect; thus, acetals are excellent protective groups, when these irreversible addition reactions must be prevented. The reaction of aldehydes and ketones with ammonia or 1º-amines forms , also known as , (compounds having a C=N function). This reaction plays an important role in the synthesis of 2º-amines. Water is eliminated in the reaction, which is acid-catalyzed and reversible in the same sense as acetal formation. R C=O + –(R )C–O– R C= + O Imines are sometimes difficult to isolate and purify due to their sensitivity to hydrolysis. Consequently, other reagents of the type Y–NH have been studied, and found to give stable products (R C=N–Y) useful in characterizing the aldehydes and ketones from which they are prepared. Some of these reagents are listed in the following table, together with the structures and names of their carbonyl reaction products. An interesting aspect of these carbonyl derivatives is that stereoisomers are possible when the R-groups of the carbonyl reactant are different. Thus, benzaldehyde forms two stereoisomeric oximes, a low-melting isomer, having the hydroxyl group cis to the aldehyde hydrogen (called ), and a higher melting isomer in which the hydroxyl group and hydrogen are trans (the isomer). At room temperature or below the configuration of the double-bonded nitrogen atom is apparently fixed in one trigonal shape, unlike the of the sp hybridized amines. With the exception of unsubstituted hydrazones, these derivatives are easily prepared and are often crystalline solids - even when the parent aldehyde or ketone is a liquid. Since melting points can be determined more quickly and precisely than boiling points, derivatives such as these are useful for comparison and identification of carbonyl compounds. If the aromatic ring of phenylhydrazine is substituted with nitro groups at the 2- & 4-positions, the resulting reagent and the hydrazone derivatives it gives are strongly colored, making them easy to identify. It should be noted that although semicarbazide has two amino groups (–NH ) only one of them is a reactive amine. The other is amide-like and is . The previous reactions have all involved reagents of the type: , i.e. reactions with a 1º-amino group. Most aldehydes and ketones also react with 2º-amines to give products known as enamines. Two examples of these reactions are presented in the following diagram. It should be noted that, like acetal formation, these are acid-catalyzed reversible reactions in which water is lost. Consequently, enamines are easily converted back to their carbonyl precursors by acid-catalyzed hydrolysis. The last example of reversible addition is that of hydrogen cyanide (HC≡N), which adds to aldehydes and many ketone to give products called . RCH=O + RCH(O ) (a cyanohydrin) Since hydrogen cyanide itself is an acid (pK = 9.25), the addition is not acid-catalyzed. In fact, for best results cyanide anion, C≡N must be present, which means that catalytic base must be added. Cyanohydrin formation is weakly exothermic, and is favored for aldehydes, and unhindered cyclic and methyl ketones. Two examples of such reactions are shown below. The cyanohydrin from benzaldehyde is named mandelonitrile. The reversibility of cyanohydrin formation is put to use by the millipede in a remarkable defense mechanism. This arthropod releases mandelonitrile from an inner storage gland into an outer chamber, where it is enzymatically broken down into benzaldehyde and hydrogen cyanide before being sprayed at an enemy. | 6,623 | 1,416 |
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Now that we understand the difference between sigma and \(\pi\) electrons, we remember that the \(\pi\) bond is made up of loosely held electrons that form a diffuse cloud which can be easily distorted. This can be illustrated by comparing two types of double bonds, one polar and one nonpolar. The C=C double bond on the left below is nonpolar. Therefore the \(\pi\) electrons occupy a relatively symmetric molecular orbital that’s evenly distributed (shared) over the two carbon atoms. The C=O double bond, on the other hand, is polar due to the higher electronegativity of oxygen. The \(\pi\) cloud is distorted in a way that results in higher electron density around oxygen compared to carbon. Both atoms still share electrons, but the electrons spend more time around oxygen. The drawing on the right tries to illustrate that concept. Using simple Lewis formulas, or even line-angle formulas, we can also draw some representations of the two cases above, as follows. The dynamic nature of \(\pi\) electrons can be further illustrated with the use of arrows, as indicated below for the polar C=O bond: The CURVED ARROW FORMALISM is a convention used to represent the movement of electrons in molecules and reactions according to certain rules. We’ll study those rules in some detail. For now, we keep a few things in mind: We notice that the two structures shown above as a result of towards the oxygen are RESONANCE STRUCTURES. That is to say, they are both valid Lewis representations of the same species. The actual species is therefore a hybrid of the two structures. We conclude that: Just like \(\pi\) electrons have a certain degree of mobility due to the diffuse nature of \(\pi\) molecular orbitals, unshared electron pairs can also be moved with relative ease because they are not engaged in bonding. No bonds have to be broken to move those electrons. As a result, we keep in mind the following principle: Going back to the two resonance structures shown before, we can use the curved arrow formalism either to arrive from structure I to structure II, or vice versa. In case A, the arrow originates with \(\pi\) electrons, which move towards the more electronegative oxygen. In case B, the arrow originates with one of the unshared electron pairs, which moves towards the positive charge on carbon. We further notice that \(\pi\) electrons from one structure can become unshared electrons in another, and vice versa. We’ll look at additional guidelines for how to use mobile electrons later. Finally, in addition to the above, we notice that the oxygen atom, for example, is \(sp^2\) hybridized (trigonal planar) in structure I, but \(sp^3\) hybridized (tetrahedral) in structure II. So, which one is it? Again, what we are talking about is the real species. The real species is a hybrid that contains contributions from both resonance structures. In this particular case, the best we can do for now is issue a qualitative statement: since structure I is the major contributor to the hybrid, we can say that the oxygen atom in the actual species is mostly trigonal planar because it has greater \(sp^2\) character, but it still has some tetrahedral character due to the minor contribution from structure II. We’ll explore and expand on this concept in a variety of contexts throughout the course. What about sigma electrons, that is to say those forming part of single bonds? These bonds represent the “glue” that holds the atoms together and are a lot more difficult to disrupt. As a result, they are not as mobile as \(\pi\) electrons or unshared electrons, and are therefore rarely moved. There are however some exceptions, notably with highly polar bonds, such as in the case of HCl illustrated below. We will not encounter such situations very frequently. This representation better conveys the idea that the H–Cl bond is highly polar. We now go back to an old friend of ours, \(CH_3CNO\), which we introduced when we first talked about resonance structures. We use this compound to further illustrate how mobile electrons are “pushed” to arrive from one resonance structure to another. The movement of electrons that takes place to arrive at structure II from structure I starts with the triple bond between carbon and nitrogen. We’ll move one of the two \(\pi\) bonds that form part of the triple bond towards the positive charge on nitrogen, as shown: When we do this, we pay close attention to the new status of the affected atoms and make any necessary adjustments to the charges, bonds, and unshared electrons to preserve the validity of the resulting formulas. In this case, for example, the carbon that forms part of the triple bond in structure I has to acquire a positive charge in structure II because it’s lost one electron. The nitrogen, on the other hand, is now neutral because it gained one electron and it’s forming three bonds instead of four. We can also arrive from structure I to structure III by pushing electrons in the following manner. The arrows have been numbered in this example to indicate which movement starts first, but that’s not part of the conventions used in the curved arrow formalism. As we move a pair of unshared electrons from oxygen towards the nitrogen atom as shown in step 1, we are forced to displace electrons from nitrogen towards carbon as shown in step 2. Otherwise we would end up with a nitrogen with 5 bonds, which is impossible, even if only momentarily. Again, notice that in step 1 the arrow originates with an unshared electron pair from oxygen and moves towards the positive charge on nitrogen. A new \(\pi\) bond forms between nitrogen and oxygen. At the same time, the \(\pi\) electrons being displaced towards carbon in step 2 become a pair of unshared electrons in structure III. Finally, the hybridization state of some atoms also changes. For example the carbon atom in structure I is sp hybridized, but in structure III it is \(sp^3\) hybridized. You may want to play around some more and see if you can arrive from structure II to structure III, etc. However, be warned that sometimes it is trickier than it may seem at first sight. The following example illustrates how a lone pair of electrons from carbon can be moved to make a new \(\pi\) bond to an adjacent carbon, and how the \(\pi\) electrons between carbon and oxygen can be moved to become a pair of unshared electrons on oxygen. None of the previous rules has been violated in any of these examples. Now let’s look at some examples of HOW NOT TO MOVE ELECTRONS. Using the same example, but moving electrons in a different way, illustrates how such movement would result in invalid Lewis formulas, and therefore is unacceptable. Not only are we moving electrons in the wrong direction (away from a more electronegative atom), but the resulting structure violates several conventions. First, the central carbon has five bonds and therefore violates the octet rule. Second, the overall charge of the second structure is different from the first. To avoid having a carbon with five bonds we would have to destroy one of the C–C single bonds, destroying the molecular skeleton in the process. In the example below electrons are being moved towards an area of high electron density (a negative charge), rather than towards a positive charge. In addition, the octet rule is violated for carbon in the resulting structure, where it shares more than eight electrons. Additional examples further illustrate the rules we’ve been talking about. (a) (lone pairs) located on a given atom can only move to an adjacent position to make to the next atom. (b) Unless there is a positive charge on the next atom (carbon above), other electrons will have to be displaced to preserve the octet rule. In resonance structures these are almost always \(\pi\) electrons, and almost never sigma electrons. As the electrons from the nitrogen lone pair move towards the neighboring carbon to make a new \(\pi\) bond, the \(\pi\) electrons making up the C=O bond must be displaced towards the oxygen to avoid ending up with five bonds to the central carbon. c) As can be seen above, can move towards one of the two atoms they share to form a new lone pair. In the example above, the \(\pi\) electrons from the C=O bond moved towards the oxygen to form a new lone pair. Another example is: (d) \(\pi\) electrons can also move to an adjacent position to make new \(\pi\) bond. Once again, the octet rule must be observed: One of the most common examples of this feature is observed when writing resonance forms for benzene and similar rings. The presence of alternating \(\pi\) and \(\sigma\) bonds in a molecule such as benzene is known as a conjugated system, or conjugated \(\pi\) bonds. Conjugated systems can extend across the entire molecule, as in benzene, or they can comprise only part of a molecule. A conjugated system always starts and ends with a \(\pi\) bond (i.e. an \(sp^2\) or an \(sp\)-hybridized atom), or sometimes with a charge. The atoms that form part of a conjugated system in the examples below are shown in blue, and the ones that do not are shown in red. Most of the times it is \(sp^3\) hybridized atoms that break a conjugated system. Practically every time there are \(\pi\) bonds in a molecule, especially if they form part of a conjugated system, there is a possibility for having resonance structures, that is, several valid Lewis formulas for the same compound. What resonance forms show is that there is , and sometimes . All the examples we have seen so far show that electrons move around and are not static, that is, they are delocalized. because it spreads energy over a larger area rather than keeping it confined to a small area. Since electrons are charges, the presence of delocalized electrons brings extra stability to a system compared to a similar system where electrons are localized. The stabilizing effect of charge and electron delocalization is known as . Since conjugation brings up electron delocalization, it follows that (i.e. the lower its potential energy). If there are positive or negative charges, they also spread out as a result of resonance. The more resonance forms one can write for a given system, the more stable it is. That is, the greater its resonance energy. Examine the following examples and write as many resonance structures as you can for each to further explore these points: Let’s look for a moment at the three structures in the last row above. In the first structure, delocalization of the positive charge and the \(\pi\) bonds occurs over the entire ring. This becomes apparent when we look at all the possible resonance structures as shown below. In the second structure, delocalization is only possible over three carbon atoms. This is demonstrated by writing all the possible resonance forms below, which now number only two. Finally, the third structure has no delocalization of charge or electrons because no resonance forms are possible. Therefore, it is the least stable of the three. This brings us to the last topic. How do we recognize when delocalization is possible? Let’s look at some delocalization setups, that is to say, structural features that result in delocalization of electrons. There are specific structural features that bring up electron or charge delocalization. . Other common arrangements are: (a) The positive charge can be on one of the atoms that make up the \(\pi\) bond, or on an adjacent atom. The orbital view of delocalization can get somewhat complicated. For now we’re going to keep it at a basic level. We start by noting that \(sp^2\) carbons actually come in several varieties. Two of the most important and common are neutral \(sp^2\) carbons and positively charged \(sp^2\) carbons. Substances containing neutral \(sp^2\) carbons are regular alkenes. Species containing positively charged \(sp^2\) carbons are called . The central carbon in a carbocation has trigonal planar geometry, and the unhybridized p orbital is empty. The following representations convey these concepts. A combination of orbital and Lewis or 3-D formulas is a popular means of representing certain features that we may want to highlight. For example, if we’re not interested in the sp2 orbitals and we just want to focus on what the p orbitals are doing we can use the following notation. Let’s now focus on two simple systems where we know delocalization of \(\pi\) electrons exists. One is a system containing two pi bonds in conjugation, and the other has a pi bond next to a positively charged carbon. We can represent these systems as follows. If we focus on the orbital pictures, we can immediately see the potential for electron delocalization. The two \(\pi\) molecular orbitals shown in red on the left below are close enough to overlap. Overlapping is a good thing because it delocalizes the electrons and spreads them over a larger area, bringing added stability to the system. It is however time-consuming to draw orbitals all the time. The following representations are used to represent the delocalized system. A similar process applied to the carbocation leads to a similar picture. The resonance representation conveys the idea of delocalization of charge and electrons rather well. Finally, the following representations are sometimes used, but again, the simpler they are, the less accurately they represent the delocalization picture. There will be plenty of opportunity to observe more complex situations as the course progresses. | 13,503 | 1,418 |
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The solubility of gases depends on the pressure: an increase in pressure increases solubility, whereas a decrease in pressure decreases solubility. This statement is formalized in , which states that This can be expressed in the equation: \[C = k \times P_{gas} \] where \(C\) is the the solubility of a gas in solvent, \(k\)is the proportionality constant, and \(P_{gas}\) is the partial pressure of the gas above the solution. The aqueous solubility at 20 degrees Celsius of Ar at 1.00 atm is equivalent to 33.7 mL Ar(g), measured at STP, per liter of water. What is the molarity of Ar in water that is saturated with the air at 1.00 atm and 20 degrees Celsius? Air contains 0.934% Ar by volume. Assume that the volume if the water does not change when it becomes saturated with air. STP molar volume: (22.414 L = 22,414 mL) First Determine Molarity \[\begin{align*} k_{Ar} &= \dfrac{C}{P_{Ar}} \\[4pt] &= \dfrac{\left(\dfrac{33.7 mL\ Ar}{1 L}\right)\left(\dfrac{1 mol\ Ar}{22,414 mL}\right)}{1 atm} \\[4pt] &= 0.00150 M\ atm^{-1} \end{align*} \] Solve for Concentration \[\begin{align*} C &= k_{Ar}P_{Ar} \\[4pt] &= 0.0015 M\ atm^{-1} \times 0.00934 atm \\[4pt] &= 1.40 \times 10^{-5} M\ Ar \end{align*} \] Common examples of pressure effects on gas solubility can be demonstrated with carbonated beverages, such as a bottle of soda (above). Once the pressure within the unopened bottle is released, \(\ce{CO2(g)}\) is released from the solution as bubbles or fizzing. In order for deep sea divers to breathe underwater, they must inhale highly compressed air in deep water, resulting in more nitrogen dissolving in their blood, tissues, and other joints. If the diver returns to the surface too rapidly, the nitrogen gas diffuses out of the blood too quickly and causes pain and possibly death. This condition is known as "the bends." To prevent the bends, one can return to the surface slowly so that the gases will diffuse slowly and adjust to the partial decrease in pressure or breathe a mixture of compressed helium and oxygen gas, because helium is only one-fifth as soluble in blood than nitrogen. Think of a human body under water as a soda bottle under pressure. Imagine dropping the bottle and trying to open it. In order to prevent the soda from fizzing out, you open the cap slowly to let the pressure decrease. The atmosphere is approximately 78% nitrogen and 21% oxygen, but the body primarily uses the oxygen. Under water, however, the high pressure of water surrounding our bodies causes nitrogen to form in our blood and tissue. And like the bottle of soda, if the body moves or ascends to the water surface the water too quickly, the nitrogen is released too quickly and creates bubbles in the blood. | 2,743 | 1,419 |
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John Dalton (1803) stated, "'When two elements combine with each other to form two or more compounds, the ratios of the masses of one element that combines with the fixed mass of the other are simple whole numbers'. Ratio of the masses of oxygen that combines with a fixed mass of carbon (12 parts): 16:32 or 1:2 Hydrogen and oxygen are known to form 2 compounds. The hydrogen content in one is 5.93%, and that of the other is 11.2%. Show that this data illustrates the law of multiple proportions. In the first compound: hydrogen = 5.93% Oxygen = (100 -5.93) = 94.07% In the second compound: hydrogen = 11.2% Oxygen = (100 -11.2) = 88.88% Ratio of the masses of oxygen that combine with fixed mass of hydrogen: 15.86:7.9 or 2:1. This is consistent with the law of multiple proportions. | 804 | 1,420 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Coordination_Chemistry/Complex_Ion_Chemistry/Reactions_of_the_Hexaaqua_Ions_with_Hydroxide_Ions |
This page describes and explains the reactions between complex ions of the type [M(H O) ] and hydroxide ions from, for example, sodium hydroxide solution. It assumes that you know why these ions are acidic, and are happy about the equilibria involved. Although there are only minor differences, for simplicity we will look at 2+ ions and 3+ ions separately. These have the form [M(H O) ] . Their acidity is shown in the reaction of the hexaaqua ions with water molecules from the solution: \[ \ce{[M(H_2O)_6]^{2+} (aq) + H2O (l) \rightleftharpoons [M(H_2O)_5(OH)]^{+} (aq) + H3O^+ (aq)}\] They are acting as acids by donating hydrogen ions to water molecules in the solution. Because of the confusing presence of water from two different sources (the ligands and the solution), it is easier to simplify this: \[ \ce{[M(H_2O)_6]^{2+} (aq) \rightleftharpoons [M(H_2O)_5(OH)]^{+} (aq) + H^+ (aq)}\] What happens if you add hydroxide ions to this equilibrium? There are two possible reactions. Notice that this isn't a ligand exchange reaction. The hydroxide ion has removed a hydrogen ion from one of the ligand water molecules. The reaction has also become essentially irreversible. Whichever of the above reactions happens, you end up with [M(H O) (OH)] ions in solution. These are also acidic, and can lose hydrogen ions from another of the water ligands. Taking the easier version of the equilibrium: \[\ce{ [M(H_2O)_5(OH)]^{+} (aq) \rightleftharpoons [M(H_2O)_4(OH)_2](s) + H^{+} (aq) }\] Adding hydroxide ions again tips the equilibrium to the right - either by reacting with the hydrogen ions, or by reacting directly with the complex on the left-hand side. When this occurs, the new complex formed no longer has a charge - we describe it as a "neutral complex". In all the cases we are looking at, this neutral complex is insoluble in water - and so a precipitate is formed. This precipitate is often written without including the remaining water ligands. In other words we write it as M(OH) . A precipitate of the metal hydroxide has been formed. There is no logical reason why the removal of hydrogen ions from the complex should stop at this point. Further hydrogen ions can be removed by hydroxide ions to produce anionic complexes - complexes carrying negative charges. Whether this actually happens in the test tube to any extent varies from metal to metal. In fact, if you do this using sodium hydroxide solution of the usual concentrations, most of the 2+ ions that you will meet at this level don't go beyond the precipitate. The only one you are likely to come across is the zinc case - and that has a complication. The final ion is [Zn(OH) ] - a tetrahedral ion which has lost the remaining 2 water ligands. The argument here is exactly as before - the only difference is the number of hydrogen ions which have to be removed from the original hexaaqua complex to produce the neutral complex. Going beyond the neutral complex is also rather more common with 3+ than with 2+ ions, and may go as far as having a hydrogen ion removed from each of the six water molecules. This is summarized in the same sort of flow scheme as before: In each case the formula of the precipitate will be given as if it were the simple neutral complex. In fact, these "hydroxide" precipitates sometimes rearrange by losing water from combinations of the attached OH groups. This produces oxides closely associated with the lost water. These changes are beyond the scope of this site. Iron is very easily oxidized under alkaline conditions. Oxygen in the air oxidizes the iron(II) hydroxide precipitate to iron(III) hydroxide especially around the top of the tube. The darkening of the precipitate comes from the same effect. I have shown the original solution as very pale pink (the palest I can produce!), but in fact it is virtually colourless. The pale brown precipitate is oxidized to darker brown manganese(III) oxide in contact with oxygen from the air. You start and finish with colorless solutions, producing a white precipitate on the way. This looks exactly the same in the test tube as the corresponding zinc reaction above - but beware the different formulae of the precipitate and the final solution. If adding hydroxide ions removes hydrogen ions from the hexaaqua complex one at a time, it doesn't seem unreasonable that you could put them back again by adding an acid. That's just what happens! We'll look in detail at what happens in the chromium(III) case, but exactly the same principle applies to all the other examples we've looked at - whether for 2+ or 3+ ions. As long as you understand what is happening, you can work out the details if you need to These are the ions formed at the end of the sequence in which you add hydroxide ions to a solution containing hexaaquachromium(III) ions. Their formula is [Cr(OH) ] . A reminder of the color changes when you add sodium hydroxide solution to a solution containing hexaaquachromium(III) ions: If you add an acid (dilute sulphuric acid, for example), the hydrogen ions get put back on one at a time. You already know the colors of the significant stages (the beginning, the end, and the neutral complex). It isn't a separate bit of learning! You can apply this to any case. If you know the colors as you remove hydrogen ions, you automatically know them as you put the hydrogen ions back on again. It also doesn't matter where you start from either - whether, for example, you add acid to an ionic complex like [Cr(OH) ] , or a neutral one like [Fe(H O) (OH) ]. You will know that the [Fe(H O) (OH) ] is a dirty green precipitate. When you add the hydrogen ions back to it, it will revert to the very pale green solution of the [Fe(H O) ] ion. None of this is a new bit of learning - you just have to re-arrange what you already know! Jim Clark ( ) | 5,836 | 1,421 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/21%3A_Resonance_and_Molecular_Orbital_Methods/21.09%3A_Bond_Lengths_and_Double-Bond_Character |
Bond lengths frequently are cited as evidence for, or against, electron delocalization, although some caution should be exercised in this respect. For instance, if the hybrid structure of benzene is considered to be represented by the two possible Kekule structures, then each carbon-carbon bond should be halfway between a single bond and a double bond. In other words, each should possess \(50\%\) . We then may expect the carbon-carbon bond lengths for benzene to be the average of single- and double-bond lengths. However, the average of the \(\ce{C-C}\) bond in ethane \(\left( 1.534 \: \text{Å} \right)\) and in ethene \(\left( 1.337 \: \text{Å} \right)\) is \(1.436 \: \text{Å}\), which does not agree well with the measured \(\ce{C-C}\) bond distance for benzene of \(1.397 \: \text{Å}\). The discrepancy lies largely in the assumption inherent in this crude calculation that, in the absence of resonance, all \(\ce{C-C}\) single bonds are equal to \(1.534 \: \text{Å}\). Clearly, this is not a valid assumption because, as we have seen, bond energies depend upon environment, and because the energy of a bond depends upon its length (see Figure 21-1), bond lengths also must vary with environment. This can be seen from the data in Table 21-3, which gives the carbon-carbon bond lengths for several compounds. The single bonds shorten as the other bonds to carbon become progressively unsaturated, that is, as the hybridization of carbon changes from \(sp^3\) to \(sp\). Admittedly, some of this shortening may be ascribed to resonance, but not all. If we take \(1.48 \: \text{Å}\) as a reasonable \(\ce{C-C}\) bond distance between two \(sp^2\)-hybridized carbons and \(1.34 \: \text{Å}\) for \(\ce{C=C}\) bonds (see Table 2-1), the average is \(1.41 \: \text{Å}\), which is not much different from the \(1.40 \: \text{Å}\) for the carbon-carbon bonds in benzene. and (1977) | 1,906 | 1,422 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Heterogeneous_Equilibria/Metal_Coordination_Complexes |
Metals are Lewis acids because of their positive charge. When dissolved in water, they react with water to form hydrated compounds such as \(\ce{Na(H2O)6+}\) and \(\ce{Cu(H2O)6^2+}\). These are called metal complexes, or coordination compounds. These coordination reactions are reactions. The neutral molecules such as \(\ce{H2O}\) and \(\ce{NH3}\), and anions such as \(\ce{CN-}\), \(\ce{CH3COO-}\) are called ligands. The coordination reaction can be represented by \[\ce{CuSO4 + 6 H2O \rightleftharpoons} \ce{Cu(H2O)6^2+ + SO4^2-}\] Usually, copper sulfate solids are hydrated with 5 water molecules per CuSO_4 complex ( \(\mathrm{ {\color{Periwinkle} CuSO_4\cdot(H_2O)_5}}\)), and its color is light blue. When heated, it loses water of crystallization, and becomes \(\ce{CuSO4}\), which is colorless. Most people know that when ammonia is added to a \(\mathrm{ {\color{Periwinkle} Cu(H_2O)_6^{2+}}}\) , it turns deep blue. This is due to the formation of complexes: \[ \color{Periwinkle} {Cu(H_2O)_6^{2+} } + 4 NH_3 \: \rightleftharpoons {\color{Blue} Cu(H_2O)_2(NH_3)_4^{2+} + 4 H_2O}\] Actually, the above reaction takes place in steps. The \(\ce{H2O}\) molecules are displaced one at a time as the ammonia concentration, \(\ce{[NH3]}\), increases. As more \(\ce{NH3}\) is bonded to \(\ce{Cu^2+}\), the blue color deepens. Ammonia forms complexes with many metals. It forms a very strong complex with \(\ce{Ag+}\) such that \(\ce{AgCl}\) solid will dissolve in ammonia solution. \[\ce{AgCl_{\large{(s)}} + 2 NH3 \rightleftharpoons Ag(NH3)2+ + 2 Cl-}\] The silver ammonia complex is colorless, however. Other commonly encountered ligands are \(\ce{CN-}\), \(\ce{SCN-}\), \(\ce{Cl-}\), ethylene diamine (\(\ce{NH2CH3CH3NH2}\)), and acetate (\(\ce{CH3COO-}\)). For example, Formation of complexes is also a thermal dynamic phenomenon. For the equilibrium, \(\ce{Ag+ + 2 NH3 \rightleftharpoons Ag(NH3)2+}\), the formation constant is very large, \(K_{\ce f} = \ce{\dfrac{[Ag(NH3)2+]}{[Ag+] [NH3]^2}} = \textrm{1.6e7 M}^{-2}\) because \(\ce{[Ag+]}\) is very small in such an equilibrium. The formation constants of some other complexes are given in a table form on the right. Although only three metal ions are involved, the complexes are formed by five ligands, and the overall formation constants range from 1.6e7 to 7.7e33. The EDTA is a complicated organic molecule, \(\mathrm{( ^- OOCCH_2)_2}\ce{N-CH2-CH2-N(CH2COO- )2}\) with six sites (\(\ce{4\: O}\) and \(\ce{2\: N}\)) embracing the zinc ion in the zinc complex. Calculate \(\ce{[Ag+]}\) in a solution containing 0.10 M \(\ce{AgNO3}\) and 1.0 M \(\ce{NH3}\).
For simplicity of formulation, let \(\mathrm{M = [Ag^+]}\). The equilibrium equation and concentration are: \(\begin{array}{cccccl}\ce{Ag+ &+ &2 NH3 &\rightleftharpoons &Ag(NH3)2+} &\hspace{10px}K_{\ce f} = \textrm{1.6e7 M}^{-2}\\
x &&1.00-0.20+x &&0.10-x &
\end{array}\) \(\dfrac{0.10-x}{x (0.80+x)^2} = \textrm{1.6e7 M}^{-2}\) \(x = \textrm{9.7e-9 M}\)
If we assume \(x \ce M = \ce{[Ag(NH3)2+]}\), the calculation will be very difficult. Try it and find out why. \(\ce{Cl-}\) \(\ce{AgCl}\) Ans. Since 9.7e-9*0.10 = 9.7e-10 > , a precipitate will form. What is the solubility of \(\ce{AgCl}\) in a solution which contains 1.0 M \(\ce{NH3}\)? For \(\ce{Ag(NH3)2+}\), = 1.6e7, and for \(\ce{AgCl}\), = 1.8e-10.
First consider the equilibria: \(\ce{AgCl \rightleftharpoons Ag+ + Cl-} \hspace{15px} K_{\ce{sp}} = \textrm{1.8e-10 M}^2\)
\(\ce{Ag+ + 2 NH3 \rightleftharpoons Ag(NH3)2+} \hspace{15px} K_{\ce f} = \textrm{1.6e7 M}^{-2}\) Adding the two equations together results in the equilibrium equation below. \(\begin{array}{cccccccl}\ce{AgCl &+ &2 NH3 &\rightleftharpoons &Ag(NH3)2+ &+ &Cl-} &\:\:\: K = K_{\ce{sp}} K_{\ce f} = \textrm{2.9e-3}\\
&&1.0-x &&x &&x &\:\:\: \Leftarrow \textrm{equilibrium concentrations}
\end{array}\) where x is the molar solubility of \(\ce{AgCl}\). \(\dfrac{x^2}{(1.0-x)^2} = \textrm{2.9e-3}\) \(\dfrac{x}{(1.0-x)} = 0.054\) \(x = \textrm{0.051 M}\)
Answer the following questions and review the discussion in \(\ce{AgBr}\) \(\ce{AgBr}\) \(\ce{NH3}\) Ans. 2.8e-3 M As indicated earlier, the formation of a complex takes place in steps. The formation constants in these steps are called stepwise formation constants. For the reaction, \[\ce{Ag+ + NH3 \rightleftharpoons Ag(NH3)+}\] \[K_{\ce{\large f_{\Large 1}}} = \ce{\dfrac{[Ag(NH3)+]}{[Ag+] [NH3]}} = \textrm{2.2e3 M}\] And for the reaction, \[\ce{Ag(NH3)+ + NH3 \rightleftharpoons Ag(NH3)2+}\] \[K_{\ce{\large f_{\Large 2}}} = \ce{\dfrac{[Ag(NH3)2+]}{[Ag(NH3)+] [NH3]}} = \textrm{7.2e3 M}\] And obviously, for the overall reaction, \[\ce{Ag+ + 2 NH3 \rightleftharpoons Ag(NH3)2+}\] \[K_{\ce{\large f}} = \ce{\dfrac{[Ag(NH3)2+]}{[Ag+] [NH3]+}} = K_{\ce{\large f_{\Large 1}}} \times K_{\ce{\large f_{\Large 2}}} = \textrm{1.6e7 M}^2\] A generalized formula is \[K_{\ce{\large f}} = K_{\ce{\large f_{\Large 1}}} \times K_{\ce{\large f_{\Large 2}}} \times K_{\ce{\large f_{\Large 3}}} \times \: ...\] The reverse reaction of the complex formation is called dissociation, and the equilibrium constant is called dissociation constant . \p\ce{Ag(NH3)2+ \rightleftharpoons Ag+ + 2 NH3}, \hspace{10px} K_{\ce d}.\] Obviously, we have \pK_{\ce d} = \dfrac{1}{K_{\ce f}}\] Similar to stepwise formation constants, we can also apply the concept to give a stepwise dissociation constant. Calculate \(\ce{[Ag+]}\) when equal volume of 0.10 M \(\ce{AgNO3}\) and 0.10 M \(\ce{Na2S2O3}\) solutions are mixed, giving = 1.7e13 M for \(\ce{Ag(S2O3)2^3-}\).
When the solutions are mixed, \(\ce{[Ag(S2O3)2^3- ]} = \textrm{0.05 M}\). Let \(x \ce M = \ce{[Ag+]}\) at equilibrium. The dissociation equilibrium equation rather than the formation equilibrium equation is more convenient in this case. \(\begin{array}{ccccc}
\ce{Ag(S2O3)2^3- &\rightleftharpoons &Ag+ &+ &2 S2O3^3-}\\
0.05-x &&x &&2 x
\end{array}\) \(\dfrac{x (2 x)^2}{0.05-x} = \dfrac{1}{\textrm{1.7e13}} = \textrm{5.9e-14 M}^2\) Since is very small, 0.05- ~ 0.050, and \(\ce{[Ag+]} = x = \left(\dfrac{\textrm{5.9e-14}}{4} \right)^{1/3} = \textrm{2.5e-5 M}\) The approximation is justified.
If you use the formation equilibrium equation and let \(\ce{[Ag(S2O3)2^3- ]} = x\), the equation is very difficult to solve. \(\ce{CH4}\), \(\ce{H2O}\), \(\ce{NH3}\), \(\ce{Al(OH)4-}\), \(\ce{CCl4}\), \(\ce{CO3^2-}\), \(\ce{NH4+}\) \(\begin{array}{cccccl}
\ce{Ag+ &+ &2 NH3 &\rightleftharpoons &Ag(NH3)2+}, &K_{\ce f} = \textrm{1.6e7}\\
x &&1.0 &&0.01 &\leftarrow \textrm{equilibrium concentration}
\end{array}\) \(\dfrac{0.01}{x 1.0^2} = \textrm{1.6e7}\); \(x =\: ?\) Apply the concept of equilibrium in complex formation.
Derive the dissociation constant from formation constant, and show that \(K_{\ce d} = \dfrac{1}{K_{\ce f}}\).
The solubilities for \(\ce{AgCl}\), \(\ce{AgBr}\), and \(\ce{AgI}\) in 1.0 M \(\ce{NH3}\) solution are 0.051, 0.0028, and 0.000036 M respectively. | 6,992 | 1,423 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Essential_Organic_Chemistry_(Bruice)/16%3A_The_Organic_Chemistry_of_Amino_Acids_Peptides_and_Proteins/16.10____The_Secondary_Structure_of_Proteins |
Secondary structure refers to the shape of a folding protein due exclusively to hydrogen bonding between its backbone amide and carbonyl groups. Secondary structure does not include bonding between the R-groups of amino acids, hydrophobic interactions, or other interactions associated with tertiary structure. The two most commonly encountered secondary structures of a polypeptide chain are α-helices and beta-pleated sheets. These structures are the first major steps in the folding of a polypeptide chain, and they establish important topological motifs that dictate subsequent tertiary structure and the ultimate function of the protein.
Thumbnail: Structure of human hemoglobin. The proteins α and β subunits are in red and blue, and the iron-containing heme groups in green. (CC BY- ; ). | 817 | 1,424 |
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