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https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Chromatography/Chromatography |
Chromatography is a method by which a mixture is separated by distributing its components between two phases. The stationary phase remains fixed in place while the mobile phase carries the components of the mixture through the medium being used. The stationary phase acts as a constraint on many of the components in a mixture, slowing them down to move slower than the mobile phase. The movement of the components in the mobile phase is controlled by the significance of their interactions with the mobile and/or stationary phases. Because of the differences in factors such as the solubility of certain components in the mobile phase and the strength of their affinities for the stationary phase, some components will move faster than others, thus facilitating the separation of the components within that mixture. The distribution of a solute between the mobile and stationary phases in is described by \(\kappa\), the partition coefficient, defined by: \[\kappa = \dfrac{C_s }{C_m}\] where \(C_s\) is the concentration of solute in the stationary phase and \(C_m\) is the concentration of the solute in the mobile phase. The mobile phase serves to carry the sample molecules through the chromatographic column. During the sample molecules transportation through the column, each analyte is retained according to that compound's characteristic affinity for the stationary phase. The time that passes between the sample injection and peak maximum is called the retention time. The area underneath each peak is proportional to the amount of co responding analyte in solution. The retention time, \(t_R\), is given in seconds by: \[t_R = t_S + t_M\] where \(t_S\) is the time the analyte spends in the stationary phase and \(t_M\) is the time spent in the mobile phase. \(t_M\) is often referred to as the dead, or void time, as all components spend \(t_M\) in the mobile phase. Column efficiency is affected by the amount of band broadening that occurs as the sample passes through the column. Rate theory describes the shapes of the peaks in quantitative terms and is based upon the infinite number of paths that the sample may take in order to elute out of the column. Some molecules will travel through the column quickly due to their accidentally inclusion in the mobile phase while other molecules will severely lag behind because of their accidental inclusion in the stationary phase. The result of these effects is a typical Gaussian shaped chromatographic band with a spread of velocities around the mean value. Furthermore, the width of the peak increases as it move down the column because of the increased opportunity for spreading. Two additional, undesirable chromatographic features are fronting and tailing. With fronting, the front of the peak is drawn out and the tali is steepened. The opposite is true for tailing. Both effects can be cause by distribution constant that varies with the concentration. These non-ideal effects are unwanted because of they lead to poor separations. The two terms used to measure column efficiency are plate height, \(H\), and plate count, \(N\). These two terms related by the following equation where \(L\) is the length of the column: \[ N = L / H\] Greater column efficiency is characterized by a large plate count \(N\) and a small plate height \(H\). Both \(H\) and \(N\) can be determined experimentally using the following two equations: \[H = L W^2 / 16 (t_R)^2\] \[N = 16 (t_R / W)^2\] where L is the length of the column packing, W is the width of the magnitude of the base of the triangle and \(t_R\) is the retention time of the analyte. Using the theory of band broadening, the efficiency of chromatographic columns can be approximated by the equation: \[H = A + \dfrac{B}{u} + C_Su + C_Mu \] where \(H\) is the plate height in centimeters and u is the linear velocity of the mobile phase in centimeters per second. The term \(A\) describes the multiple path effect, or eddy diffusion, \(B\) describes the longitudinal diffusion coefficient and \(C_Su\) and \(C_Mu\) are the mass-transfer coefficients for the stationary and mobile phases, respectively. In chromatography, it is important that the components in solution are adequately separated so that the separate components can be collected in their purest form. This becomes easier to do as the separation between the bands for each component have a greater separation between them. Also, it is ideal to have the bands of the individual components as narrow as possible. This is to say that it is best to have each component occupying as little space as possible within the column: From this figure it can be seen that a better separation between narrow bands of components is ideal for easier collection of the individual samples. Band broadening is an especially important factor for this type of chromatography when separating colored compounds. When the bands of the components are narrow, most of the particles of that component are in close proximity with one another, which makes it easier to see the color of the bands. As the particles diffuse away from one another and broaden the component's band, the color of the band fades and can become more difficult to see, which may also make it harder to collect pure samples of the mixture's components. The extent of band broadening in chromatography is determined by the Van Deemter equation . This equation relates the efficiency of the chromatography procedure to three different factors. The Van Deemter equation is shown below: \[H = Au^{1/3} + \dfrac{B}{u} + Cu\] Where H is the height equivalent of a theoretical plate (HETP) and u is the velocity (flow rate) of the mobile phase. The lower the resulting value of \(H\) is, the greater the efficiency of the procedure. So, ideally, a scientist will want to minimize all three terms in order to minimize \(H\). The other three terms refer to factors that come into play while the chromatography is performed. The A factor is determined by a phenomenon called Eddy Diffusion . This is also called the multi-path term, as molecular particles of a certain compound have a multitude of options when it comes to finding a pathway through a packed column. The following figure helps in visualizing Eddy diffusion: Because there is an almost infinite number of different paths that a particle can travel by through a column, some paths will be longer than others. The particles that find the shortest path through the column will be eluted more quickly than those that travel a longer way. In the figure, particle \(B\) will be eluted before particle \(C\), and both will be eluted before particle A. Since it is improbable for all particles of one compound to find the shortest path, there will be fractions of the component that will behave like particles \(A\), \(B\), and \(C\). This leads to the broadening of the band. There is little a scientist can do to minimize the Eddy Diffusion factor, as it is influenced by the nature of column being used and by the particles' movement through that column. The A term is loosely affected by the flow rate of the mobile phase, and sometimes the affect of the flow rate is negligible. It is for this reason that sometimes the Van Deemter equation is written as such: \[H = A + \dfrac{B}{u} + Cu\] B/u is called the longitudinal diffusion term, and is caused by the components' natural migration from a place of high concentration (the center of the band) to a place of lower concentration (either side of the band) within the column. Diffusion occurs because molecules in a place of high concentration will tend to spread out to areas of lower concentration to achieve equilibrium. Given enough time, diffusion will result in equilibrium of the diffusing fluid via random molecular motion. The figure below helps to visualize this phenomenon: At time zero in the figure above, the particles of a compound are generally localized in a narrow band within the separating column. If the mobile phase flow rate is too small or if the system is left at rest, the particles begin to separate from one another. This causes a spread in the concentration distribution of that compound within the column, thus bringing about band broadening for the band of that particular compound. As the time that the system is left still approaches infinity, the compound reaches complete concentration equilibrium throughout the entire column. At this point, there is no definitive band for that component, as a single concentration of that compound is present throughout the entire column. Longitudinal diffusion is a chief cause of band broadening in Gas Chromatography, as the diffusion rates of gaseous species are much higher than those of liquids. It is for this reason that longitudinal diffusion is less of an issue in liquid chromatography. The magnitude of the term \(B/u\) can be minimized by increasing the flow rate of the mobile phase. Increasing the velocity of the mobile phase does not allow the components in the column to reach equilibrium, and so will hamper longitudinal diffusion. The flow rate of the mobile phase should not be increased in excess, however, as the term \(Cu\) is maximized when u is increased. Cu is referred to as the mass transfer term. Mass transfer refers to when particles are so strongly adhered to the stationary phase that the mobile phase passes over them without carrying them along. This results is particles of a component being left behind. Since it is likely that more than a single particle of any given compound will undergo this occurrence, band broadening results. This results in a phenomenon called tailing, in which a fraction a component lags behind a more concentrated frontal band. Non-equilibrium effects can be caused by two phenomena: laminar flow and turbulent flow. Laminar flow occurs in tubular capillaries, and so is most prominent in . Turbulent flow occurs as a result of particles becoming overwhelmed by the stationary phase and is more common in column chromatography. This occurrence can be visualized by observing the figure below: In the above figure, particles of the adsorbent solid become occupied by particles of the sample. If too many particles of the adsorbent are occupied, particle A will have nothing hindering it from flowing through the column. So, the particles of a single compound separate from one another. Also, as the mobile phase moves through the column, particles of the sample leave the stationary phase and migrate with the mobile phase. However, if the flow rate of the mobile phase is too high, many of the sample particles are unable to leave the stationary phase and so get left behind. These occurrences result in band broadening, as the individual particles of a single compound become less closely packed. The high flow rate of the mobile phase makes it more difficult for the components within the column to reach equilibrium between the stationary and mobile phase. It is for this reason that the Cu term is also called the non-equilibrium factor. Minimization of this factor can be achieved by decreasing the flow rate of the mobile phase. Decreasing the flow rate of the mobile phase gives sample components more time to leave the stationary phase and move with the mobile phase, thus reaching equilibrium. By observing the Van Deemter equation, it can be deduced that an ideal mobile phase flow rate must be determined to yield the best (lowest) value of H. Decreasing the flow rate too much will result in an increase of the longitudinal diffusion factor B/u, while exceedingly increasing the flow rate will increase the significance of the mass transfer term Cu. So, H can be minimized to a finite limit depending on the various parameters involved in the chromatography being performed. | 11,787 | 3,423 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Enzymes/Enzymatic_Kinetics/Catalytic_Efficiency_of_Enzymes |
Enzymes exist in all biological systems in abundant numbers, but not all of their functions are fully understood. Enzymes are important for a variety of reasons, most significantly because they are involved in many vital biochemical reactions. Increasing the reaction rate of a chemical reaction allows the reaction to become more efficient, and hence more products are generated at a faster rate. These products then become involved in some other biological pathway that initiates certain functions of the human body. This is known as the catalytic efficiency of enzymes, which, by increasing the rates, results in a more efficient chemical reaction within a biological system. An enzyme's active sites are usually composed of amino acid residues; depending on which amino acid residues are present, the specificity of the substrate can vary greatly. Depending on the pH level, the physical properties (mainly the electric charge) of an enzyme can change. A change in the electric charge can alter the interaction between the active site amino acid residues and the incoming substrate. With that said, the substrate can bind to the active site via hydrogen bonding or van der Waals forces. Once the substrate binds to the active site it forms an enzyme-substrate complex that is then involved in further chemical reactions. In order for an enzyme to be active and be energetically favorable to allow a chemical reaction to proceed forward, a substrate must bind to an enzyme's "active site". An active site can be thought of as a lock and the substrate as a key; this is known as the lock and key model. A key (substrate) must be inserted and turned (chemical reaction), then the lock (enzyme) opens (production of products). Note that an enzyme might have more than one active site. Another theory on the active site-substrate relationship is the induced fit theory, which is quite opposite of the lock and key theory (where the active site is seemingly inflexible). In the induced fit theory, the active site of the enzyme is very flexible, and only changes its conformation when the substrate binds to it. Enzymes work as a catalyst by lowering the Gibbs free energy of activation of the enzyme-substrate complex. Below are two figures showing a basic enzymatic reaction with and without a catalyst: The efficiency of the enzyme can be determined as follows: consider a simple enzymatic reaction: German biochemist Leonor Michaelis and Canadian biochemist Maud Menten derived an equation describing this system, later known as the "Michaelis-Menten Equation", shown below: \[ v_0 = \dfrac{V_{max}[S]}{K_M + [S]} \tag{1}\] This equation gives the rate of the reaction at a given substrate concentration, assuming a known V , which is the maximum rate the reaction can proceed at, and K , the Michaelis constant. However, in a practical application of the Michaelis-Menten, V is often measured, and V is observed as a saturation or plateau in a data plot. Because the substrate concentration is known, K is usually the calculated value of interest. For \(K_M\), assume \(V_0= \dfrac{V_{max}}{2}\): \[\dfrac{V_{max}}{2} = \dfrac{V_{max}[S]}{K_M + [S]} \tag{2}\] \[(K_M + [S]) \dfrac{V_{max}}{2} = V_{max}[S] \tag{3}\] \[K_M + [S] = \dfrac{V_{max}[S]}{\dfrac{V_{max}}{2}} \tag{4} \] \[K_M + [S] = 2[S] \tag{5}\] \[ K_M = [S] \tag{6}\] The Michaelis constant can be thought of as the rate at which the substrate becomes unbound from the enzyme, which can either occur in the events of substrate-enzyme complex becoming the product, or the substrate becomes unbound to the enzyme. K can be shown as an equation. \[ K_M = \dfrac{k_{-1} + k_2}{k_1} \tag{7} \] Whereas k is the rate constant at which the substrate becomes unbound to the enzyme, resulting in the dissociation of the enzyme-substrate complex, k is the rate constant where the substrate-enzyme complex disappears and turns into product, and K is the rate constant for the formation of the the substrate-enzyme complex formation. Therefore, K can be viewed as the rate of substrate-enzyme complex disappearance divided by the rate of substrate-enzyme complex formation, which is the level at which half of the substrate is bound to the enzyme. K is a useful indicator for the presence of an inhibitor because we can look for changes in K and compare to our control (biological systems that we know have zero inhibitor presence). K is a dependent variable, and its value can change due to many reasons, including the pH level of the system, temperature, or any other condition that might affect a chemical reaction. A small K indicates that the substrate has a high affinity for the enzyme. The Michaelis-Menten equation is most useful in measuring enzyme efficiency if v is plotted against [S], as follows: V is the maximum rate at which the reaction can run, regardless of [S], meaning that even if you add more substrate, the reaction cannot go any faster. That is because at V all of the active sites on the enzyme are occupied. After all the explanations on various forms of enzyme kinetic equations, we arrive at our conclusion of catalytic efficiency. Referring back to Fig 3, we have: \[ V_o = k_2 \left(\dfrac{[E]_o[S]}{\dfrac{k_{-1} + k_2}{k_1} + [S]}\right) \tag{8}\] Notice \(k_2\) describes an irreversible reaction as opposed to an equilibrium expression, when compared to k-1 and k1. k2 here is also known as kcat, the catalytic efficiency of enzyme. From the previous discussion, v0 is the measured reaction rate, which is the product formation over time, so it can be concluded that an equation would look like the following: \[ v_0 = \dfrac{d[P]}{dt} = k_2[E]_0 \tag{9}\] Where [E] is the total enzyme concentration. It is also known that V is observed when all of the enzyme-substrate complex disappear and turn into products, so we can make the following assumption: \[ V_{max} = k_2[E]_0 \tag{10} \] and after rearrangement, we have this equation: \[ k_{cat} = k_2 = \dfrac{V_{max} }{[E]_0} \tag{11}\] That is the equation for calculating catalytic efficiency, to be used after we obtain data from experiments and after using the Michaelis-Menten equation. With a larger k , the enzyme is efficient because less enzyme is needed. | 6,249 | 3,424 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Hydroboration-Oxidation_of_Alkenes |
Hydroboration-Oxidation is a two step pathway used to produce alcohols. The reaction proceeds in an Anti-Markovnikov manner, where the hydrogen (from BH or BHR ) attaches to the more substituted carbon and the boron attaches to the least substituted carbon in the bouble bond. Furthermore, the borane acts as a lewisAnti-Markovnikov acid by accepting two electrons in its empty p orbital from an alkene that is electron rich. This process allows boron to have an electron octet. A very interesting characteristic of this process is that it does not require any activation by a catalyst. The Anti-MarkovnikovHydroboration mechanism has the elements of both hydrogenation and electrophilic addition and it is a stereospecific ( meaning that the hydroboration takes place on the same face of the double bond, this leads stereochemistry. Hydroboration-oxidation of alkenes has been a very valuable laboratory method for the stereoselectivity and regioselectivity of alkenes. An Additional feature of this reaction is that it occurs without rearrangement. First off it is very imporatnt to understand little bit about the structure and the properties of the borane molecule. Borane exists naturally as a very toxic gas and it exists as dimer of the general formula B H (diborane). Additionally, the dimer B H ignites spontaneously in air. Borane is commercially available in ether and tetrahydrofuran (THF), in these solutions the borane can exist as a lewis acid-base complex, which allows boron to have an electron octet. \[ 2BH_3 \rightarrow B_2H_6\] . EpoxidationEpoxidation If you need additional visuals to aid you in understanding the mechanism, click on the outside links provided here that will take you to other pages and media that are very helpful as well. If you need clarification or a reminder on the nomenclature of alkenes refer to the link below on naming the | 1,899 | 3,426 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Molecular_Geometry/Shapes_of_Molecules_and_Ions |
This page explains how to work out the shapes of molecules and ions containing only single bonds. The examples on this page are all simple in the sense that they only contain two sorts of atoms joined by single bonds - for example, ammonia only contains a nitrogen atom joined to three hydrogen atoms by single bonds. If you are given a more complicated example, look carefully at the arrangement of the atoms before you start to make sure that there are only single bonds present. For example, if you had a molecule such as COCl , you would need to work out its structure, based on the fact that you know that carbon forms 4 covalent bonds, oxygen 2, and chlorine (normally) 1. If you did that, you would find that the carbon is joined to the oxygen by a double bond, and to the two chlorines by single bonds. That means that you couldn't use the techniques on this page, because this page only considers single bonds. The shape of a molecule or ion is governed by the arrangement of the electron pairs around the central atom. All you need to do is to work out how many electron pairs there are at the bonding level, and then arrange them to produce the minimum amount of repulsion between them. You have to include both bonding pairs and lone pairs. First you need to work out how many electrons there are around the central atom: Now work out how many bonding pairs and lone pairs of electrons there are: Divide by 2 to find the total number of electron pairs around the central atom. Work out how many of these are bonding pairs, and how many are lone pairs. You know how many bonding pairs there are because you know how many other atoms are joined to the central atom (assuming that only single bonds are formed). For example, if you have 4 pairs of electrons but only 3 bonds, there must be 1 lone pair as well as the 3 bonding pairs. Finally, you have to use this information to work out the shape: Arrange these electron pairs in space to minimize repulsions. How this is done will become clear in the examples which follow. The only simple case of this is beryllium chloride, BeCl . The electronegativity difference between beryllium and chlorine is not enough to allow the formation of ions. Beryllium has 2 outer electrons because it is in . It forms bonds to two chlorines, each of which adds another electron to the outer level of the beryllium. There is no ionic charge to worry about, so there are 4 electrons altogether - 2 pairs. It is forming 2 bonds so there are no lone pairs. The two bonding pairs arrange themselves at 180° to each other, because that's as far apart as they can get. The molecule is described as being The simple cases of this would be BF or BCl . Boron is in group 3, so starts off with 3 electrons. It is forming 3 bonds, adding another 3 electrons. There is no charge, so the total is 6 electrons - in 3 pairs. Because it is forming 3 bonds there can be no lone pairs. The 3 pairs arrange themselves as far apart as possible. They all lie in one plane at 120° to each other. The arrangement is called trigonal planar. In the diagram, the other electrons on the fluorines have been left out because they are irrelevant. There are lots of examples of this. The simplest is methane, CH . Carbon is in group 4, and so has 4 outer electrons. It is forming 4 bonds to hydrogens, adding another 4 electrons - 8 altogether, in 4 pairs. Because it is forming 4 bonds, these must all be bonding pairs. Four electron pairs arrange themselves in space in what is called a tetrahedral arrangement. A tetrahedron is a regular triangularly-based pyramid. The carbon atom would be at the centre and the hydrogens at the four corners. All the bond angles are 109.5°. It is important that you understand the use of various sorts of line to show the 3-dimensional arrangement of the bonds. In diagrams of this sort, an ordinary line represents a bond in the plane of the screen or paper. A dotted line shows a bond going away from you into the screen or paper. A wedge shows a bond coming out towards you. Nitrogen is in group 5 and so has 5 outer electrons. Each of the 3 hydrogens is adding another electron to the nitrogen's outer level, making a total of 8 electrons in 4 pairs. Because the nitrogen is only forming 3 bonds, one of the pairs must be a lone pair. The electron pairs arrange themselves in a tetrahedral fashion as in methane. In this case, an additional factor comes into play. Lone pairs are in orbitals that are shorter and rounder than the orbitals that the bonding pairs occupy. Because of this, there is more repulsion between a lone pair and a bonding pair than there is between two bonding pairs. That forces the bonding pairs together slightly - reducing the bond angle from 109.5° to 107°. Be very careful when you describe the shape of ammonia. Although the electron pair arrangement is tetrahedral, when you describe the shape, you only take notice of the atoms. Ammonia is pyramidal - like a pyramid with the three hydrogens at the base and the nitrogen at the top. Following the same logic as before, you will find that the oxygen has four pairs of electrons, two of which are lone pairs. These will again take up a tetrahedral arrangement. This time the bond angle closes slightly more to 104°, because of the repulsion of the two lone pairs. The shape is not described as tetrahedral, because we only "see" the oxygen and the hydrogens - not the lone pairs. Water is described as The nitrogen has 5 outer electrons, plus another 4 from the four hydrogens - making a total of 9. But take care! This is a positive ion. It has a 1+ charge because it has lost 1 electron. That leaves a total of 8 electrons in the outer level of the nitrogen. There are therefore 4 pairs, all of which are bonding because of the four hydrogens. The ammonium ion has exactly the same shape as methane, because it has exactly the same electronic arrangement. NH is tetrahedral. Methane and the ammonium ion are said to species (atoms, molecules or ions) are isoelectronic if they have exactly the same number and arrangement of electrons (including the distinction between bonding pairs and lone pairs). Oxygen is in group 6 - so has 6 outer electrons. Add 1 for each hydrogen, giving 9. Take one off for the +1 ion, leaving 8. This gives 4 pairs, 3 of which are bond pairs. The hydroxonium ion is isoelectronic with ammonia, and has an identical shape - pyramidal. Phosphorus (in group 5) contributes 5 electrons, and the five fluorines 5 more, giving 10 electrons in 5 pairs around the central atom. Since the phosphorus is forming five bonds, there can't be any lone pairs. (The argument for phosphorus(V) chloride, PCl , would be identical.) The 5 electron pairs take up a shape described as a trigonal bipyramid - three of the fluorines are in a plane at 120° to each other; the other two are at right angles to this plane. The trigonal bipyramid therefore has two different bond angles - 120° and 90°. Chlorine is in group 7 and so has 7 outer electrons. The three fluorines contribute one electron each, making a total of 10 - in 5 pairs. The chlorine is forming three bonds - leaving you with 3 bonding pairs and 2 lone pairs, which will arrange themselves into a trigonal bipyramid. There are actually three different ways in which you could arrange 3 bonding pairs and 2 lone pairs into a trigonal bipyramid. The right arrangement will be the one with the minimum amount of repulsion - and you can't decide that without first drawing all the possibilities. These are the only possible arrangements. Anything else you might think of is simply one of these rotated in space. We need to work out which of these arrangements has the minimum amount of repulsion between the various electron pairs. A new rule applies in cases like this: If you have more than four electron pairs arranged around the central atom, you can ignore repulsions at angles of greater than 90°. One of these structures has a fairly obvious large amount of repulsion. In this diagram, two lone pairs are at 90° to each other, whereas in the other two cases they are at more than 90°, and so their repulsions can be ignored. ClF certainly won't take up this shape because of the strong lone pair-lone pair repulsion. To choose between the other two, you need to count up each sort of repulsion. In the next structure, each lone pair is at 90° to 3 bond pairs, and so each lone pair is responsible for 3 lone pair-bond pair repulsions. Because of the two lone pairs there are therefore 6 lone pair-bond pair repulsions. And that's all. The bond pairs are at an angle of 120° to each other, and their repulsions can be ignored. Now consider the final structure. Each lone pair is at 90° to 2 bond pairs - the ones above and below the plane. That makes a total of 4 lone pair-bond pair repulsions - compared with 6 of these relatively strong repulsions in the last structure. The other fluorine (the one in the plane) is 120° away, and feels negligible repulsion from the lone pairs. The bond to the fluorine in the plane is at 90° to the bonds above and below the plane, so there are a total of 2 bond pair-bond pair repulsions. The structure with the minimum amount of repulsion is therefore this last one, because bond pair-bond pair repulsion is less than lone pair-bond pair repulsion. ClF is described as T-shaped. 6 electrons in the outer level of the sulphur, plus 1 each from the six fluorines, makes a total of 12 - in 6 pairs. Because the sulfur is forming 6 bonds, these are all bond pairs. They arrange themselves entirely at 90°, in a shape described as octahedral. Xenon forms a range of compounds, mainly with fluorine or oxygen, and this is a typical one. Xenon has 8 outer electrons, plus 1 from each fluorine - making 12 altogether, in 6 pairs. There will be 4 bonding pairs (because of the four fluorines) and 2 lone pairs. There are two possible structures, but in one of them the lone pairs would be at 90°. Instead, they go opposite each other. XeF is described as square planar. Chlorine is in group 7 and so has 7 outer electrons. Plus the 4 from the four fluorines. Plus one because it has a 1- charge. That gives a total of 12 electrons in 6 pairs - 4 bond pairs and 2 lone pairs. The shape will be identical with that of XeF . Jim Clark ( ) | 10,429 | 3,427 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./03.E%3A_Ionic_vs._Covalent_Bonding_(Exercises) |
" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact Please be sure you are familiar with the topics discussed in Essential Skills 2 (Section 3.7 "Essential Skills 2") before proceeding to the Application Problems. Problems marked with a ♦ involve multiple concepts. 1. Hydrogen sulfide is a noxious and toxic gas produced from decaying organic matter that contains sulfur. A lethal concentration in rats corresponds to an inhaled dose of 715 molecules per million molecules of air. How many molecules does this correspond to per mole of air? How many moles of hydrogen sulfide does this correspond to per mole of air? 2. Bromine, sometimes produced from brines (salt lakes) and ocean water, can be used for bleaching fibers and silks. How many moles of bromine atoms are found in 8.0 g of molecular bromine (Br )? 3. Paris yellow is a lead compound that is used as a pigment; it contains 16.09% chromium, 19.80% oxygen, and 64.11% lead. What is the empirical formula of Paris yellow? 4. A particular chromium compound used for dyeing and waterproofing fabrics has the elemental composition 18.36% chromium, 13.81% potassium, 45.19% oxygen, and 22.64% sulfur. What is the empirical formula of this compound? 5. Compounds with aluminum and silicon are commonly found in the clay fractions of soils derived from volcanic ash. One of these compounds is vermiculite, which is formed in reactions caused by exposure to weather. Vermiculite has the following formula:\(\mathrm{Ca_{0.7}[Si_6 \cdot 6Al_{1.4}]Al_4O_{20}(OH)_4}\). (The content of calcium, silicon, and aluminum are not shown as integers because the relative amounts of these elements vary from sample to sample.) What is the mass percent of each element in this sample of vermiculite? 6. ♦ Pheromones are chemical signals secreted by a member of one species to evoke a response in another member of the same species. One honeybee pheromone is an organic compound known as an alarm pheromone, which smells like bananas. It induces an aggressive attack by other honeybees, causing swarms of angry bees to attack the same aggressor. The composition of this alarm pheromone is 64.58% carbon, 10.84% hydrogen, and 24.58% oxygen by mass, and its molecular mass is 130.2 amu. a. Calculate the empirical formula of this pheromone. b. Determine its molecular formula. c. Assuming a honeybee secretes 1.00 × 10 g of pure pheromone, how many molecules of pheromone are secreted? 7. Amoxicillin is a prescription drug used to treat a wide variety of bacterial infections, including infections of the middle ear and the upper and lower respiratory tracts. It destroys the cell walls of bacteria, which causes them to die. The elemental composition of amoxicillin is 52.59% carbon, 5.24% hydrogen, 11.50% nitrogen, 21.89% oxygen, and 8.77% sulfur by mass. What is its empirical formula? 8. Monosodium glutamate (MSG; molar mass = 169 g/mol), is used as a flavor enhancer in food preparation. It is known to cause headaches and chest pains in some individuals, the so-called Chinese food syndrome. Its composition was found to be 35.51% carbon, 4.77% hydrogen, 8.28% nitrogen, and 13.59% sodium by mass. If the “missing” mass is oxygen, what is the empirical formula of MSG? 9. Ritalin is a mild central nervous system stimulant that is prescribed to treat attention deficit disorders and narcolepsy (an uncontrollable desire to sleep). Its chemical name is methylphenidate hydrochloride, and its empirical formula is \(\ce{C14H20ClNO2}\). If you sent a sample of this compound to a commercial laboratory for elemental analysis, what results would you expect for the mass percentages of carbon, hydrogen, and nitrogen? 10. Fructose, a sugar found in fruit, contains only carbon, oxygen, and hydrogen. It is used in ice cream to prevent a sandy texture. Complete combustion of 32.4 mg of fructose in oxygen produced 47.6 mg of \(\ce{CO2}\) and 19.4 mg of \(\ce{H2O}\). What is the empirical formula of fructose? 11. Coniine, the primary toxin in hemlock, contains only carbon, nitrogen, and hydrogen. When ingested, it causes paralysis and eventual death. Complete combustion of 28.7 mg of coniine produced 79.4 mg of \(\ce{CO2}\) and 34.4 mg of \(\ce{H2O}\). What is the empirical formula of the coniine? 12. Copper and tin alloys (bronzes) with a high arsenic content were presumably used by Bronze Age metallurgists because bronze produced from arsenic-rich ores had superior casting and working properties. The compositions of some representative bronzes of this type are as follows: If ancient metallurgists had used the mineral \(\ce{As2S3}\) as their source of arsenic, how much \(\ce{As2S3}\)would have been required to process 100 g of cuprite (\(\ce{Cu2O}\)) bronzes with these compositions? 13. ♦ The phrase mad as a hatter refers to mental disorders caused by exposure to mercury(II) nitrate in the felt hat manufacturing trade during the 18th and 19th centuries. An even greater danger to humans, however, arises from alkyl derivatives of mercury. a. Give the empirical formula of mercury(II) nitrate. b. One alkyl derivative, dimethylmercury, is a highly toxic compound that can cause mercury poisoning in humans. How many molecules are contained in a 5.0 g sample of dimethylmercury? c. What is the percentage of mercury in the sample? 14. Magnesium carbonate, aluminum hydroxide, and sodium bicarbonate are commonly used as antacids. Give the empirical formulas and determine the molar masses of these compounds. Based on their formulas, suggest another compound that might be an effective antacid. 15. ♦ Nickel(II) acetate, lead(II) phosphate, zinc nitrate, and beryllium oxide have all been reported to induce cancers in experimental animals. a. Give the empirical formulas for these compounds. b. Calculate their formula masses. c. Based on the location of cadmium in the periodic table, would you predict that cadmium chloride might also induce cancer? 16. ♦ Methane, the major component of natural gas, is found in the atmospheres of Jupiter, Saturn, Uranus, and Neptune. a. What is the structure of methane? b. Calculate the molecular mass of methane. c. Calculate the mass percentage of both elements present in methane. 17. Sodium saccharin, which is approximately 500 times sweeter than sucrose, is frequently used as a sugar substitute. What are the percentages of carbon, oxygen, and sulfur in this artificial sweetener? 18. Lactic acid, found in sour milk, dill pickles, and sauerkraut, has the functional groups of both an alcohol and a carboxylic acid. The empirical formula for this compound is \(\ce{CH2O}\), and its molar mass is 90 g/mol. If this compound were sent to a laboratory for elemental analysis, what results would you expect for carbon, hydrogen, and oxygen content? 19. The compound 2-nonenal is a cockroach repellant that is found in cucumbers, watermelon, and carrots. Determine its molecular mass. 20. You have obtained a 720 mg sample of what you believe to be pure fructose, although it is possible that the sample has been contaminated with formaldehyde. Fructose and formaldehyde both have the empirical formula CH O. Could you use the results from combustion analysis to determine whether your sample is pure? 21. ♦ The booster rockets in the space shuttles used a mixture of aluminum metal and ammonium perchlorate for fuel. Upon ignition, this mixture can react according to the chemical equation \[\mathrm{Al_{(s)} + NH_4ClO_{4(s)} \rightarrow Al_2O_{3(s)} + AlCl_{3(g)}+ NO_{(g)} + H_2O_{(g)}} \] Balance the equation and construct a table showing how to interpret this information in terms of the following: a. numbers of individual atoms, molecules, and ions b. moles of reactants and products c. grams of reactants and products d. numbers of molecules of reactants and products given 1 mol of aluminum metal 22. ♦ One of the byproducts of the manufacturing of soap is glycerol. In 1847, it was discovered that the reaction of glycerol with nitric acid produced nitroglycerin according to the following unbalanced chemical equation: Nitroglycerine is both an explosive liquid and a blood vessel dilator that is used to treat a heart condition known as angina. a. Balance the chemical equation and determine how many grams of nitroglycerine would be produced from 15.00 g of glycerol. b. If 9.00 g of nitric acid had been used in the reaction, which would be the limiting reactant? c. What is the theoretical yield in grams of nitroglycerin? d. If 9.3 g of nitroglycerin was produced from 9.0 g of nitric acid, what would be the percent yield? e. Given the data in part d, how would you rate the success of this reaction according to the criteria mentioned in this chapter? f. Derive a general expression for the theoretical yield of nitroglycerin in terms of x grams of glycerol. 23. ♦ A significant weathering reaction in geochemistry is hydration–dehydration. An example is the transformation of hematite (\(\ce{Fe2O3}\)) to ferrihydrite (\(\mathrm{Fe_{10}O_{15} \cdot 9H_2O}\)) as the relative humidity of the soil approaches 100%: \[\mathrm{Fe_2O_{3(s)} + H_2O_{(l)} \rightarrow Fe_{10}O_{15} \cdot 9H_2O_{(s)}}\] This reaction occurs during advanced stages of the weathering process. a. Balance the chemical equation. b. Is this a redox reaction? Explain your answer. c. If 1 ton of hematite rock weathered in this manner, how many kilograms of ferrihydrite would be formed? 24. ♦ Hydrazine (\(\ce{N2H4}\)) is used not only as a rocket fuel but also in the industry to remove toxic chromates from waste water according to the following chemical equation: \[\mathrm{4CrO_{4(aq)}^{2-}+ 3N_2H_{4(l)} + 4H_2O_{(l)} \rightarrow 4Cr(OH)_{3(s)} + 3N_{2(g)} + 8OH^-_{(aq)}}\] Identify the species that is oxidized and the species that is reduced. What mass of water is needed for the complete reaction of 15.0 kg of hydrazine? Write a general equation for the mass of chromium(III) hydroxide [Cr(OH) ] produced from x grams of hydrazine. 25. ♦ Corrosion is a term for the deterioration of metals through the chemical reaction with their environment. A particularly difficult problem for the archaeological chemist is the formation of CuCl, an unstable compound that is formed by the corrosion of copper and its alloys. Although copper and bronze objects can survive burial for centuries without significant deterioration, exposure to air can cause cuprous chloride to react with atmospheric oxygen to form Cu O and cupric chloride. The cupric chloride then reacts with the free metal to produce cuprous chloride. The continued reaction of oxygen and water with cuprous chloride causes “bronze disease,” which consists of spots of a pale green, powdery deposit of [CuCl ·3Cu(OH) ·H O] on the surface of the object that continues to grow. Using this series of reactions described, complete and balance the following equations, which together result in bronze disease: Equation 1: ___ + O → ___ + ___ Equation 2: ___ + Cu → ___ Equation 3: ___ + O + H O → CuCl ⋅ \(\underset {bronze \, disease}{3Cu(OH)_2.H_2} \) + CuCl a. Which species are the oxidants and the reductants in each equation? b. If 8.0% by mass of a 350.0 kg copper statue consisted of CuCl, and the statue succumbed to bronze disease, how many pounds of the powdery green hydrate would be formed? c. What factors could affect the rate of deterioration of a recently excavated bronze artifact? 26. ♦ Iron submerged in seawater will react with dissolved oxygen, but when an iron object, such as a ship, sinks into the seabed where there is little or no free oxygen, the iron remains fresh until it is brought to the surface. Even in the seabed, however, iron can react with salt water according to the following unbalanced chemical equation: \[Fe_{(s)} + NaCl_{(aq)} + H_2O_{(l)} \rightarrow FeCl_{2(s)} + NaOH_{(aq)}+H_{2(g)}\] The ferrous chloride and water then form hydrated ferrous chloride according to the following equation: \[FeCl_{2(s)} + 2H_2O_{(l)} \rightarrow FeCl_2 \cdot 2H_2O_{(s)} \] When the submerged iron object is removed from the seabed, the ferrous chloride dihydrate reacts with atmospheric moisture to form a solution that seeps outward, producing a characteristic “sweat” that may continue to emerge for many years. Oxygen from the air oxidizes the solution to ferric resulting in the formation of what is commonly referred to as rust (ferric oxide): \[FeCl_{2(aq)} + O_{2(g)} \rightarrow FeCl_{3(aq)} + Fe_2O_{3(s)} \] The rust layer will continue to grow until arrested. a. Balance each chemical equation. b. Given a 10.0 tn ship of which 2.60% is now rust, how many kilograms of iron were converted to FeCl , assuming that the ship was pure iron? c. What mass of rust in grams would result? d. What is the overall change in the oxidation state of iron for this process? e. In the first equation given, what species has been reduced? What species has been oxidized? 27. ♦ The glass industry uses lead oxide in the production of fine crystal glass, such as crystal goblets. Lead oxide can be formed by the following reaction: \[PbS_{(s)} + O_{2(g)} \rightarrow PbO_{(s)} + SO_{2(g)} \] Balance the equation and determine what has been oxidized and what has been reduced. How many grams of sulfur dioxide would be produced from 4.0 × 10 g of lead sulfide? Discuss some potential environmental hazards that stem from this reaction. 28. ♦ The Deacon process is one way to recover Cl on-site in industrial plants where the chlorination of hydrocarbons produces HCl. The reaction uses oxygen to oxidize HCl to chlorine, as shown. \[HCl_{(g)} + O_{2(g)} + H_2O_{(g)} \] The reaction is frequently carried out in the presence of NO as a catalyst. a. Balance the chemical equation. b. Which compound is the oxidant, and which is the reductant? c. If 26 kg of HCl was produced during a chlorination reaction, how many kilograms of water would result from the Deacon process? 29. In 1834, Eilhardt Mitscherlich of the University of Berlin synthesized benzene (C H ) by heating benzoic acid (C H COOH) with calcium oxide according to this balanced chemical equation: \[C_6H_5COOH_{(s)} + CaO_{(s)} \rightarrow \Delta C_6H_{6(l)} + CaCO_{3(s)}\] (Heating is indicated by the symbol Δ.) How much benzene would you expect from the reaction of 16.9 g of benzoic acid and 18.4 g of calcium oxide? Which is the limiting reactant? How many grams of benzene would you expect to obtain from this reaction, assuming a 73% yield? 30. Aspirin (C H O ) is synthesized by the reaction of salicylic acid (\(\ce{C7H6O3}\)) with acetic anhydride (\(\ce{C4H6O3}\)) according to the following equation: \[C_7H_6O_{3(s)} + C_4H_6O_{3(l)} \rightarrow C_9H_8O_{4(s)} + H_2O_{(l)}\] Balance the equation and find the limiting reactant given 10.0 g of acetic anhydride and 8.0 g of salicylic acid. How many grams of aspirin would you expect from this reaction, assuming an 83% yield? 31. ♦ Hydrofluoric acid etches glass because it dissolves silicon dioxide, as represented in the following chemical equation: \[SiO_{2(s)} + HF_{(aq)} \rightarrow SiF_{6(aq)}^{2-} + H_{(aq)}^+ + H_2O_{(l)} \] a. Balance the equation. b. How many grams of silicon dioxide will react with 5.6 g of HF? c. How many grams of HF are needed to remove 80% of the silicon dioxide from a 4.0 kg piece of glass? (Assume that the glass is pure silicon dioxide.) 32. ♦ Lead sulfide and hydrogen peroxide react to form lead sulfate and water. This reaction is used to clean oil paintings that have blackened due to the reaction of the lead-based paints with atmospheric hydrogen sulfide. a. Write the balanced chemical equation for the oxidation of lead sulfide by hydrogen peroxide. b. What mass of hydrogen peroxide would be needed to remove 3.4 g of lead sulfide? c. If the painting had originally been covered with 5.4 g of lead sulfide and you had 3.0 g of hydrogen peroxide, what percent of the lead sulfide could be removed? 33. ♦ It has been suggested that diacetylene (C H , HC≡C–C≡CH) may be the ozone of the outer planets. As the largest hydrocarbon yet identified in planetary atmospheres, diacetylene shields planetary surfaces from ultraviolet radiation and is itself reactive when exposed to light. One reaction of diacetylene is an important route for the formation of higher hydrocarbons, as shown in the following chemical equations: \[C_4H_{2(g)} + C_4H_{2(g)} \rightarrow C_8H_{3(g)} + H_{(g)} \] \[C_8H_{3(g)} + C_4H_{2(g)} \rightarrow C_{10}H_{3(g)} + C_2H_{2(g)} \] Consider the second reaction. a. Given 18.4 mol of C H and 1000 g of C H , which is the limiting reactant? b. Given 2.8 × 10 molecules of C H and 250 g of C H , which is the limiting reactant? c. Given 385 g of C H and 200 g of C H , which is in excess? How many grams of excess reactant would remain? d. Suggest why this reaction might be of interest to scientists. 34. ♦ Glucose (C H O ) can be converted to ethanol and carbon dioxide using certain enzymes. As alcohol concentrations are increased, however, catalytic activity is inhibited, and alcohol production ceases. a. Write a balanced chemical equation for the conversion of glucose to ethanol and carbon dioxide. b. Given 12.6 g of glucose, how many grams of ethanol would be produced, assuming complete conversion? c. If 4.3 g of ethanol had been produced, what would be the percent yield for this reaction? d. Is a heterogeneous catalyst or a homogeneous catalyst used in this reaction? e. You have been asked to find a way to increase the rate of this reaction given stoichiometric quantities of each reactant. How would you do this? 35. Early spacecraft developed by the National Aeronautics and Space Administration for its manned missions used capsules that had a pure oxygen atmosphere. This practice was stopped when a spark from an electrical short in the wiring inside the capsule of the Apollo 1 spacecraft ignited its contents. The resulting explosion and fire killed the three astronauts on board within minutes. What chemical steps could have been taken to prevent this disaster? | 18,122 | 3,429 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Significant_Digits/Propagation_of_Error |
Propagation of Error (or Propagation of Uncertainty) is defined as the effects on a function by a variable's uncertainty. Every measurement has an air of uncertainty about it, and not all uncertainties are equal. Therefore, the ability to properly combine uncertainties from different measurements is crucial. Uncertainty in measurement comes about in a variety of ways: instrument variability, different observers, sample differences, time of day, etc. Typically, error is given by the (\(\sigma_x\)) of a measurement. Anytime a calculation requires more than one variable to solve, propagation of error is necessary to properly determine the uncertainty. For example, lets say we are using a UV-Vis to determine the molar absorptivity of a molecule via : Since at least two of the variables have an uncertainty based on the equipment used, a propagation of error formula must be applied to measure a more exact uncertainty of the molar absorptivity. This example will be continued below, after the derivation. Suppose a certain experiment requires multiple instruments to carry out. These instruments each have different variability in their measurements. The results of each instrument are given as: (For simplification purposes, only the variables and will be used throughout this derivation). The end result desired is \(x\), so that \(x\) is dependent on and . It can be written that \(x\) is a function of these variables: \[x=f(a,b,c) \label{1}\] Because each measurement has an uncertainty about its mean, it can be written that the uncertainty of \(dx_i\) of the th measurement of \(x\) depends on the uncertainty of the th measurements of and \[dx_i=f(da_i,db_i,dc_i)\label{2}\] The total deviation of \(x\) is then derived from the partial derivative of with respect to each of the variables: \[dx=\left(\dfrac{\delta{x}}{\delta{a}}\right)_{b,c}da, \; \left(\dfrac{\delta{x}}{\delta{b}}\right)_{a,c}db, \; \left(\dfrac{\delta{x}}{\delta{c}}\right)_{a,b}dc \label{3}\] A relationship between the standard deviations of and is formed in two steps: In the first step, two unique terms appear on the right hand side of the equation: and . \[\left(\dfrac{\delta{x}}{\delta{a}}\right)^2(da)^2,\; \left(\dfrac{\delta{x}}{\delta{b}}\right)^2(db)^2, \; \left(\dfrac{\delta{x}}{\delta{c}}\right)^2(dc)^2\label{4}\] \[\left(\dfrac{\delta{x}}{da}\right)\left(\dfrac{\delta{x}}{db}\right)da\;db,\;\left(\dfrac{\delta{x}}{da}\right)\left(\dfrac{\delta{x}}{dc}\right)da\;dc,\;\left(\dfrac{\delta{x}}{db}\right)\left(\dfrac{\delta{x}}{dc}\right)db\;dc\label{5}\] Square terms, due to the nature of squaring, are always positive, and therefore never cancel each other out. By contrast, cross terms may cancel each other out, due to the possibility that each term may be positive or negative. If \(da\), \(db\), and \(dc\) represent random and independent uncertainties, about half of the cross terms will be negative and half positive (this is primarily due to the fact that the variables represent uncertainty about a mean). In effect, the sum of the cross terms should approach zero, especially as \(N\) increases. However, if the variables are correlated rather than independent, the cross term may not cancel out. Assuming the cross terms do cancel out, then the second step - summing from \(i = 1\) to \(i = N\) - would be: \[\sum{(dx_i)^2}=\left(\dfrac{\delta{x}}{\delta{a}}\right)^2\sum(da_i)^2 + \left(\dfrac{\delta{x}}{\delta{b}}\right)^2\sum(db_i)^2\label{6}\] Dividing both sides by \(N - 1\): \[\dfrac{\sum{(dx_i)^2}}{N-1}=\left(\dfrac{\delta{x}}{\delta{a}}\right)^2\dfrac{\sum(da_i)^2}{N-1} + \left(\dfrac{\delta{x}}{\delta{b}}\right)^2\dfrac{\sum(db_i)^2}{N-1}\label{7}\] The previous step created a situation where Equation \ref{7} could mimic the standard deviation equation. This is desired, because it creates a statistical relationship between the variable \(x\), and the other variables \(a\), \(b\), \(c\), etc... as follows: The standard deviation equation can be rewritten as the variance (\(\sigma_x^2\)) of \(x\): \[\dfrac{\sum{(dx_i)^2}}{N-1}=\dfrac{\sum{(x_i-\bar{x})^2}}{N-1}=\sigma^2_x\label{8}\] Rewriting Equation \ref{7} using the statistical relationship created yields the : \[\sigma^2_x=\left(\dfrac{\delta{x}}{\delta{a}}\right)^2\sigma^2_a+\left(\dfrac{\delta{x}}{\delta{b}}\right)^2\sigma^2_b+\left(\dfrac{\delta{x}}{\delta{c}}\right)^2\sigma^2_c\label{9}\] Thus, the end result is achieved. Equation \ref{9} shows a direct statistical relationship between multiple variables and their standard deviations. In the next section, derivations for common calculations are given, with an example of how the derivation was obtained. In the following calculations \(a\), \(b\), and \(c\) are measured variables from an experiment and \(\sigma_a\), \(\sigma_b\), and \(\sigma_c\) are the standard deviations of those variables. If \(x = a + b - c\) then \[\sigma_x= \sqrt{ {\sigma_a}^2+{\sigma_b}^2+{\sigma_c}^2} \label{10}\] If \(x = \dfrac{ a \times b}{c}\) then \[ \dfrac{\sigma_x}{x}=\sqrt{\left(\dfrac{\sigma_a}{a}\right)^2+\left(\dfrac{\sigma_b}{b}\right)^2+\left(\dfrac{\sigma_c}{c}\right)^2}\label{11} \] If \(x = a^y\) then \[\dfrac{\sigma_x}{x}=y \left(\dfrac{\sigma_a}{a}\right) \label{12}\] If \(x = \log(a)\) then \[\sigma_x=0.434 \left(\dfrac{\sigma_a}{a}\right) \label{13}\] If \(x = \text{antilog}(a)\) then \[\dfrac{\sigma_x}{x}=2.303({\sigma_a}) \label{14}\] Addition, subtraction, and logarithmic equations leads to an standard deviation, while multiplication, division, exponential, and anti-logarithmic equations lead to standard deviations. The Exact Formula for Propagation of Error in Equation \(\ref{9}\) can be used to derive the arithmetic examples noted above. Starting with a simple equation: \[x = a \times \dfrac{b}{c} \label{15}\] where \(x\) is the desired results with a given standard deviation, and \(a\), \(b\), and \(c\) are experimental variables, each with a difference standard deviation. Taking the partial derivative of each experimental variable, \(a\), \(b\), and \(c\): \[\left(\dfrac{\delta{x}}{\delta{a}}\right)=\dfrac{b}{c} \label{16a}\] \[\left(\dfrac{\delta{x}}{\delta{b}}\right)=\dfrac{a}{c} \label{16b}\] and \[\left(\dfrac{\delta{x}}{\delta{c}}\right)=-\dfrac{ab}{c^2}\label{16c}\] Plugging these partial derivatives into Equation \(\ref{9}\) gives: \[\sigma^2_x=\left(\dfrac{b}{c}\right)^2\sigma^2_a+\left(\dfrac{a}{c}\right)^2\sigma^2_b+\left(-\dfrac{ab}{c^2}\right)^2\sigma^2_c\label{17}\] Dividing Equation \(\ref{17}\) by Equation \(\ref{15}\) squared yields: \[\dfrac{\sigma^2_x}{x^2}=\dfrac{\left(\dfrac{b}{c}\right)^2\sigma^2_a}{\left(\dfrac{ab}{c}\right)^2}+\dfrac{\left(\dfrac{a}{c}\right)^2\sigma^2_b}{\left(\dfrac{ab}{c}\right)^2}+\dfrac{\left(-\dfrac{ab}{c^2}\right)^2\sigma^2_c}{\left(\dfrac{ab}{c}\right)^2}\label{18}\] Canceling out terms and square-rooting both sides yields Equation \ref{11}: \[\dfrac{\sigma_x}{x}={\sqrt{\left(\dfrac{\sigma_a}{a}\right)^2+\left(\dfrac{\sigma_b}{b}\right)^2+\left(\dfrac{\sigma_c}{c}\right)^2}} \nonumber\] Continuing the example from the introduction (where we are calculating the of a molecule), suppose we have a concentration of 13.7 (±0.3) moles/L, a path length of 1.0 (±0.1) cm, and an absorption of 0.172807 (±0.000008). The equation for molar absorptivity is dictated by Beer's law: \[ε = \dfrac{A}{lc}. \nonumber\] Since deals with multiplication/division, we'll use Equation \ref{11}: \[\begin{align*} \dfrac{\sigma_{\epsilon}}{\epsilon} &={\sqrt{\left(\dfrac{0.000008}{0.172807}\right)^2+\left(\dfrac{0.1}{1.0}\right)^2+\left(\dfrac{0.3}{13.7}\right)^2}} \\[4pt] &=0.10237 \end{align*}\] As stated in the note , Equation \ref{11} yields a relative standard deviation, or a percentage of the variable. Using Beer's Law, = 0.012614 L moles cm Therefore, the \(\sigma_{\epsilon}\) for this example would be 10.237% of , which is 0.001291. Accounting for significant figures, the final answer would be: = 0.013 L moles cm If you are given an equation that relates two different variables and given the relative uncertainties of one of the variables, it is possible to determine the relative uncertainty of the other variable by using calculus. In problems, the uncertainty is usually given as a percent. Let's say we measure the radius of a very small object. The problem might state that there is a 5% uncertainty when measuring this radius. To actually use this percentage to calculate unknown uncertainties of other variables, we must first define what uncertainty is. Uncertainty, in calculus, is defined as: \[\left(\dfrac{dx}{x}\right) = \left(\dfrac{∆x}{x}\right) = \text{uncertainty} \nonumber\] Let's look at the example of the radius of an object again. If we know the uncertainty of the radius to be 5%, the uncertainty is defined as \[\left(\dfrac{dx}{x}\right)=\left(\dfrac{∆x}{x}\right)= 5\% = 0.05.\nonumber\] Now we are ready to use calculus to obtain an unknown uncertainty of another variable. Let's say we measure the radius of an artery and find that the uncertainty is 5%. What is the uncertainty of the measurement of the volume of blood pass through the artery? Let's say the equation relating radius and volume is: \[V(r) = c(r^2) \nonumber\] where \(c\) is a constant, \(r\) is the radius and \(V(r)\) is the volume. The first step to finding the uncertainty of the volume is to understand our given information. Since we are given the radius has a 5% uncertainty, we know that (∆r/r) = 0.05. We are looking for (∆V/V). Now that we have done this, the next step is to take the derivative of this equation to obtain: \[\dfrac{dV}{dr} = \dfrac{∆V}{∆r}= 2cr \nonumber\] We can now multiply both sides of the equation to obtain: \[∆V = 2cr(∆r) \nonumber\] Since we are looking for (∆V/V), we divide both sides by V to get: \[\dfrac{∆V}{V} = \dfrac{2cr(∆r)}{V} \nonumber\] We are given the equation of the volume to be \(V = c(r)^2\), so we can plug this back into our previous equation for \(V\) to get: \[\dfrac{∆V}{V} = \dfrac{2cr(∆r)}{c(r)^2} \nonumber \] Now we can cancel variables that are in both the numerator and denominator to get: \[\dfrac{∆V}{V} = \dfrac{2∆r}{r} = 2 \left(\dfrac{∆r}{r}\right) \nonumber \] We have now narrowed down the equation so that ∆r/r is left. We know the value of uncertainty for ∆r/r to be 5%, or 0.05. Plugging this value in for ∆r/r we get: \[\dfrac{∆V}{V} = 2 (0.05) = 0.1 = 10\% \nonumber\] The uncertainty of the volume is 10%. This method can be used in chemistry as well, not just the biological example shown above. | 10,586 | 3,431 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/04%3A_Evaluating_Analytical_Data/4.05%3A_Statistical_Analysis_of_Data |
A confidence interval is a useful way to report the result of an analysis because it sets limits on the expected result. In the absence of determinate error, a confidence interval based on a sample’s mean indicates the range of values in which we expect to find the population’s mean. When we report a 95% confidence interval for the mass of a penny as 3.117 g ± 0.047 g, for example, we are stating that there is only a 5% probability that the penny’s expected mass is less than 3.070 g or more than 3.164 g. Because a confidence interval is a statement of probability, it allows us to consider comparative questions, such as these: “Are the results for a newly developed method to determine cholesterol in blood significantly different from those obtained using a standard method?” or “Is there a significant variation in the composition of rainwater collected at different sites downwind from a coal-burning utility plant?” In this section we introduce a general approach to the statistical analysis of data. Specific statistical tests are presented in . The reliability of significance testing recently has received much attention—see Nuzzo, R. “Scientific Method: Statistical Errors,” , , , 150–152 for a general discussion of the issues—so it is appropriate to begin this section by noting the need to ensure that our data and our research question are compatible so that we do not read more into a statistical analysis than our data allows; see Leek, J. T.; Peng, R. D. “What is the Question? , , , 1314-1315 for a use-ul discussion of six common research questions. In the context of analytical chemistry, significance testing often accompanies an exploratory data analysis (Is there a reason to suspect that there is a difference between these two analytical methods when applied to a common sample?) or an inferential data analysis (Is there a reason to suspect that there is a relationship between these two independent measurements?). A statistically significant result for these types of analytical research questions generally leads to the design of additional experiments better suited to making predictions or to explaining an underlying causal relationship. A significance test is the first step toward building a greater understanding of an analytical problem, not the final answer to that problem. Let’s consider the following problem. To determine if a medication is effective in lowering blood glucose concentrations, we collect two sets of blood samples from a patient. We collect one set of samples immediately before we administer the medication, and collect the second set of samples several hours later. After analyzing the samples, we report their respective means and variances. How do we decide if the medication was successful in lowering the patient’s concentration of blood glucose? One way to answer this question is to construct a normal distribution curve for each sample, and to compare the two curves to each other. Three possible outcomes are shown in Figure 4.5.1
. In Figure 4.5.1
a, there is a complete separation of the two normal distribution curves, which suggests the two samples are significantly different from each other. In Figure 4.5.1
b, the normal distribution curves for the two samples almost completely overlap, which suggests that the difference between the samples is insignificant. Figure 4.5.1
c, however, presents us with a dilemma. Although the means for the two samples seem different, the overlap of their normal distribution curves suggests that a significant number of possible outcomes could belong to either distribution. In this case the best we can do is to make a statement about the probability that the samples are significantly different from each other. The process by which we determine the probability that there is a significant difference between two samples is called significance testing or hypothesis testing. Before we discuss specific examples we will first establish a general approach to conducting and interpreting a significance test. The purpose of a is to determine whether the difference between two or more results is sufficiently large that it cannot be explained by indeterminate errors. The first step in constructing a significance test is to state the problem as a yes or no question, such as “Is this medication effective at lowering a patient’s blood glucose levels?” A null hypothesis and an alternative hypothesis define the two possible answers to our yes or no question. The , , is that indeterminate errors are sufficient to explain any differences between our results. The , , is that the differences in our results are too great to be explained by random error and that they must be determinate in nature. We test the null hypothesis, which we either retain or reject. If we reject the null hypothesis, then we must accept the alternative hypothesis and conclude that the difference is significant. Failing to reject a null hypothesis is not the same as accepting it. We retain a null hypothesis because we have insufficient evidence to prove it incorrect. It is impossible to prove that a null hypothesis is true. This is an important point and one that is easy to forget. To appreciate this point let’s return to our sample of 100 pennies in . After looking at the data we might propose the following null and alternative hypotheses. : The mass of a circulating U.S. penny is between 2.900 g and 3.200 g : The mass of a circulating U.S. penny may be less than 2.900 g or more than 3.200 g To test the null hypothesis we find a penny and determine its mass. If the penny’s mass is 2.512 g then we can reject the null hypothesis and accept the alternative hypothesis. Suppose that the penny’s mass is 3.162 g. Although this result increases our confidence in the null hypothesis, it does not prove that the null hypothesis is correct because the next penny we sample might weigh less than 2.900 g or more than 3.200 g. After we state the null and the alternative hypotheses, the second step is to choose a confidence level for the analysis. The confidence level defines the probability that we will reject the null hypothesis when it is, in fact, true. We can express this as our confidence that we are correct in rejecting the null hypothesis (e.g. 95%), or as the probability that we are incorrect in rejecting the null hypothesis. For the latter, the confidence level is given as \(\alpha\), where \[\alpha = 1 - \frac {\text{confidence interval (%)}} {100} \label{4.1}\] For a 95% confidence level, \(\alpha\) is 0.05. In this textbook we use \(\alpha\) to represent the probability that we incorrectly reject the null hypothesis. In other textbooks this probability is given as (often read as “p- value”). Although the symbols differ, the meaning is the same. The third step is to calculate an appropriate test statistic and to compare it to a critical value. The test statistic’s critical value defines a breakpoint between values that lead us to reject or to retain the null hypothesis, which is the fourth, and final, step of a significance test. How we calculate the test statistic depends on what we are comparing, a topic we cover in . The last step is to either retain the null hypothesis, or to reject it and accept the alternative hypothesis. The four steps for a statistical analysis of data using a significance test: Suppose we want to evaluate the accuracy of a new analytical method. We might use the method to analyze a Standard Reference Material that contains a known concentration of analyte, \(\mu\). We analyze the standard several times, obtaining a mean value, \(\overline{X}\), for the analyte’s concentration. Our null hypothesis is that there is no difference between \(\overline{X}\) and \(\mu\) \[H_0 \text{: } \overline{X} = \mu \nonumber\] If we conduct the significance test at \(\alpha = 0.05\), then we retain the null hypothesis if a 95% confidence interval around \(\overline{X}\) contains \(\mu\). If the alternative hypothesis is \[H_\text{A} \text{: } \overline{X} \neq \mu \nonumber\] then we reject the null hypothesis and accept the alternative hypothesis if \(\mu\) lies in the shaded areas at either end of the sample’s probability distribution curve (Figure 4.5.2
a). Each of the shaded areas accounts for 2.5% of the area under the probability distribution curve, for a total of 5%. This is a because we reject the null hypothesis for values of \(\mu\) at either extreme of the sample’s probability distribution curve. We also can write the alternative hypothesis in two additional ways \[H_\text{A} \text{: } \overline{X} > \mu \nonumber\] \[H_\text{A} \text{: } \overline{X} < \mu \nonumber\] rejecting the null hypothesis if n falls within the shaded areas shown in Figure 4.5.2
b or Figure 4.5.2
c, respectively. In each case the shaded area represents 5% of the area under the probability distribution curve. These are examples of a . For a fixed confidence level, a two-tailed significance test is the more conservative test because rejecting the null hypothesis requires a larger difference between the parameters we are comparing. In most situations we have no particular reason to expect that one parameter must be larger (or must be smaller) than the other parameter. This is the case, for example, when we evaluate the accuracy of a new analytical method. A two-tailed significance test, therefore, usually is the appropriate choice. We reserve a one-tailed significance test for a situation where we specifically are interested in whether one parameter is larger (or smaller) than the other parameter. For example, a one-tailed significance test is appropriate if we are evaluating a medication’s ability to lower blood glucose levels. In this case we are interested only in whether the glucose levels after we administer the medication are less than the glucose levels before we initiated treatment. If a patient’s blood glucose level is greater after we administer the medication, then we know the answer—the medication did not work—and do not need to conduct a statistical analysis. Because a significance test relies on probability, its interpretation is subject to error. In a significance test, a defines the probability of rejecting a null hypothesis that is true. When we conduct a significance test at \(\alpha = 0.05\), there is a 5% probability that we will incorrectly reject the null hypothesis. This is known as a , and its risk is always equivalent to \(\alpha\). A type 1 error in a two-tailed or a one-tailed significance tests corresponds to the shaded areas under the probability distribution curves in . A second type of error occurs when we retain a null hypothesis even though it is false. This is as a , and the probability of its occurrence is \(\beta\). Unfortunately, in most cases we cannot calculate or estimate the value for \(\beta\). The probability of a type 2 error, however, is inversely proportional to the probability of a type 1 error. Minimizing a type 1 error by decreasing \(\alpha\) increases the likelihood of a type 2 error. When we choose a value for \(\alpha\) we must compromise between these two types of error. Most of the examples in this text use a 95% confidence level (\(\alpha = 0.05\)) because this usually is a reasonable compromise between type 1 and type 2 errors for analytical work. It is not unusual, however, to use a more stringent (e.g. \(\alpha = 0.01\)) or a more lenient (e.g. \(\alpha = 0.10\)) confidence level when the situation calls for it. | 11,511 | 3,432 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Proteins/Protein_Structure/Secondary_Structure%3A_-Helices |
An α-helix is a right-handed coil of amino-acid residues on a polypeptide chain, typically ranging between 4 and 40 residues. This coil is held together by hydrogen bonds between the oxygen of C=O on top coil and the hydrogen of N-H on the bottom coil. Such a hydrogen bond is formed exactly every 4 amino acid residues, and every complete turn of the helix is only 3.6 amino acid residues. This regular pattern gives the α-helix very definite features with regards to the thickness of the coil and the length of each complete turn along the helix axis. The structural integrity of an α-helix is in part dependent on correct steric configuration. Amino acids whose R-groups are too large (tryptophan, tyrosine) or too small (glycine) destabilize α-helices. Proline also destabilizes α-helices because of its irregular geometry; its R-group bonds back to the nitrogen of the amide group, which causes steric hindrance. In addition, the lack of a hydrogen on Proline's nitrogen prevents it from participating in hydrogen bonding. Another factor affecting α-helix stability is the total dipole moment of the entire helix due to individual dipoles of the C=O groups involved in hydrogen bonding. Stable α-helices typically end with a charged amino acid to neutralize the dipole moment. | 1,299 | 3,434 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.11%3A_Heats_of_Vaporization_and_Condensation |
Natural resources for electric power generation have traditionally been waterfalls, oil, coal, or nuclear power. Research is being carried out to look for other renewable sources to run the generators. Geothermal sites (such as geysers) are being considered because of the steam they produce. Capabilities can be estimated by knowing how much steam is released in a given time at a particular site. Energy is absorbed in the process of converting a liquid at its boiling point into a gas. As with the melting point of a solid, the temperature of a boiling liquid remains constant and the input of energy goes into changing the state. The of a substance is the heat absorbed by one mole of that substance as it is converted from a liquid to a gas. As a gas condenses to a liquid, heat is released. The of a substance is the heat released by one mole of that substance as it is converted from a gas to a liquid. Since vaporization and condensation of a given substance are the exact opposite processes, the numerical value of the molar heat of vaporization is the same as the numerical value of the molar heat of condensation, but opposite in sign. In other words, \(\Delta H_\text{vap} = -\Delta H_\text{cond}\). When \(1 \: \text{mol}\) of water at \(100^\text{o} \text{C}\) and \(1 \: \text{atm}\) pressure is converted to \(1 \: \text{mol}\) of water vapor at \(100^\text{o} \text{C}\), \(40.7 \: \text{kJ}\) of heat is absorbed from the surroundings. When \(1 \: \text{mol}\) of water vapor at \(100^\text{o} \text{C}\) condenses to liquid water at \(100^\text{o} \text{C}\), \(40.7 \: \text{kJ}\) of heat is released into the surroundings. \[\begin{array}{ll} \ce{H_2O} \left( l \right) \rightarrow \ce{H_2O} \left( g \right) & \Delta H_\text{vap} = 40.7 \: \text{kJ/mol} \\ \ce{H_2O} \left( g \right) \rightarrow \ce{H_2O} \left( l \right) & \Delta H_\text{cond} =-40.7 \: \text{kJ/mol} \end{array}\nonumber \] Other substances have different values for their molar heats of fusion and vaporization; these substances are summarized in the table below. Notice that for all substances, the heat of vaporization is substantially higher than the heat of fusion. Much more energy is required to change the state from a liquid to a gas than from a solid to a liquid. This is because of the large separation of the particles in the gas state. The values of the heats of fusion and vaporization are related to the strength of the intermolecular forces. All of the substances in the table above, with the exception of oxygen, are capable of hydrogen bonding. Consequently, the heats of fusion and vaporization of oxygen are far lower than the others. What mass of methanol vapor condenses to a liquid as \(20.0 \: \text{kJ}\) of heat is released? First the \(\text{kJ}\) of heat released in the condensation is multiplied by the conversion factor \(\left( \frac{1 \: \text{mol}}{-35.3 \: \text{kJ}} \right)\) to find the moles of methanol that condensed. Then, moles are converted to grams. \[-20.0 \: \text{kJ} \times \frac{1 \: \text{mol} \: \ce{CH_3OH}}{-35.3 \: \text{kJ}} \times \frac{32.05 \: \text{g} \: \ce{CH_3OH}}{1 \: \text{mol} \: \ce{CH_3OH}} = 18.2 \: \text{g} \: \ce{CH_3OH}\nonumber \] Condensation is an exothermic process, so the enthalpy change is negative. Slightly more than one-half mole of methanol is condensed. | 3,356 | 3,436 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Reactions_of_Alkenes_with_Halogens |
In each case, we will look at ethene as typical of all of the alkenes. There are no complications as far as the basic facts are concerned as the alkenes get bigger. Ethene reacts explosively with fluorine to give carbon and hydrogen fluoride gas. This isn't a useful reaction. \[\ce{CH2=CH2 + 2F2 -> 2C + 4HF}\] In each case you get an addition reaction. For example, bromine adds to give 1,2-dibromoethane. The reaction with bromine happens at room temperature. If you have a gaseous alkene like ethene, you can bubble it through either pure liquid bromine or a solution of bromine in an organic solvent like tetrachloromethane. The reddish-brown bromine is decolourized as it reacts with the alkene. A liquid alkene (like cyclohexene) can be shaken with liquid bromine or its solution in tetrachloromethane. Chlorine reacts faster than bromine, but the chemistry is similar. Iodine reacts much, much more slowly, but again the chemistry is similar. You are much more likely to meet the bromine case than either of these. Alkenes decolorize bromine water. If you shake an alkene with bromine water (or bubble a gaseous alkene through bromine water), the solution becomes colorless. This is complicated by the fact that the major product is not 1,2-dibromoethane. The water also gets involved in the reaction, and most of the product is 2-bromoethanol. However, there will still be some 1,2-dibromoethane formed, so at this sort of level you can probably get away with quoting the simpler equation: Jim Clark ( ) | 1,524 | 3,437 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_OpenStax/13.E%3A_Fundamental_Equilibrium_Concepts_(Exercises) |
What does it mean to describe a reaction as “reversible”? The reaction can proceed in both the forward and reverse directions. When writing an equation, how is a reversible reaction distinguished from a nonreversible reaction? If a reaction is reversible, when can it be said to have reached equilibrium? When a system has reached equilibrium, no further changes in the reactant and product concentrations occur; the reactions continue to occur, but at equivalent rates. Is a system at equilibrium if the rate constants of the forward and reverse reactions are equal? If the concentrations of products and reactants are equal, is the system at equilibrium? The concept of equilibrium does not imply equal concentrations, though it is possible. Explain why there may be an infinite number of values for the reaction quotient of a reaction at a given temperature but there can be only one value for the equilibrium constant at that temperature. Explain why an equilibrium between Br ( ) and Br ( ) would not be established if the container were not a closed vessel shown below: Equilibrium cannot be established between the liquid and the gas phase if the top is removed from the bottle because the system is not closed; one of the components of the equilibrium, the Br vapor, would escape from the bottle until all liquid disappeared. Thus, more liquid would evaporate than can condense back from the gas phase to the liquid phase. If you observe the following reaction at equilibrium, is it possible to tell whether the reaction started with pure NO or with pure N O ? 2NO2 rightleftharpoons N2O4 Among the solubility rules previously discussed is the statement: All chlorides are soluble except Hg Cl , AgCl, PbCl , and CuCl. (a) = [Ag ,Cl ] < 1. AgCl is insoluble; thus, the concentrations of ions are much less than 1 ; (b) \(K_c=\ce{\dfrac{1}{[Pb^2+,Cl- ]^2}}\) > 1 because PbCl is insoluble and formation of the solid will reduce the concentration of ions to a low level (<1 ). Among the solubility rules previously discussed is the statement: Carbonates, phosphates, borates, and arsenates—except those of the ammonium ion and the alkali metals—are insoluble. Benzene is one of the compounds used as octane enhancers in unleaded gasoline. It is manufactured by the catalytic conversion of acetylene to benzene: \(\ce{3C2H2}(g)⟶\ce{C6H6}(g)\). Which value of would make this reaction most useful commercially? ≈ 0.01, ≈ 1, or ≈ 10. Explain your answer. Since \(K_c=\ce{\dfrac{[C6H6]}{[C2H2]^3}}\), a value of ≈ 10 means that C H predominates over C H . In such a case, the reaction would be commercially feasible if the rate to equilibrium is suitable. Show that the complete chemical equation, the total ionic equation, and the net ionic equation for the reaction represented by the equation \(\ce{KI}(aq)+\ce{I2}(aq) \rightleftharpoons \ce{KI3}(aq)\) give the same expression for the reaction quotient. KI is composed of the ions K and I . For a titration to be effective, the reaction must be rapid and the yield of the reaction must essentially be 100%. Is > 1, < 1, or ≈ 1 for a titration reaction? > 1 For a precipitation reaction to be useful in a gravimetric analysis, the product of the reaction must be insoluble. Is > 1, < 1, or ≈ 1 for a useful precipitation reaction? Write the mathematical expression for the reaction quotient, , for each of the following reactions: (a) \(Q_c=\ce{\dfrac{[CH3Cl,HCl]}{[CH4,Cl2]}}\); (b) \(Q_c=\ce{\dfrac{[NO]^2}{[N2,O2]}}\); (c) \(Q_c=\ce{\dfrac{[SO3]^2}{[SO2]^2[O2]}}\); (d) \(Q_c\) = [SO ]; (e) \(Q_c=\ce{\dfrac{1}{[P4,O2]^5}}\); (f) \(Q_c=\ce{\dfrac{[Br]^2}{[Br2]}}\); (g) \(Q_c=\ce{\dfrac{[CO2]}{[CH4,O2]^2}}\); (h) \(Q_c\) = [H O] Write the mathematical expression for the reaction quotient, , for each of the following reactions: The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium. (a) \(Q_c\) 25 proceeds left; (b) 0.22 proceeds right; (c) \(Q_c\) undefined proceeds left; (d) 1.00 proceeds right; (e) 0 proceeds right; (f) \(Q_c\) 4 proceeds left The initial concentrations or pressures of reactants and products are given for each of the following systems. Calculate the reaction quotient and determine the direction in which each system will proceed to reach equilibrium. The following reaction has = 4.50 × 10 at 720 K. rightleftharpoons If a reaction vessel is filled with each gas to the partial pressures listed, in which direction will it shift to reach equilibrium? (NH ) = 93 atm, (N ) = 48 atm, and (H ) = 52 The system will shift toward the reactants to reach equilibrium. Determine if the following system is at equilibrium. If not, in which direction will the system need to shift to reach equilibrium? rightleftharpoons [SO Cl ] = 0.12 , [Cl ] = 0.16 and [SO ] = 0.050 . for the reaction is 0.078. Which of the systems described in give homogeneous equilibria? Which give heterogeneous equilibria? (a) homogenous; (b) homogenous; (c) homogenous; (d) heterogeneous; (e) heterogeneous; (f) homogenous; (g) heterogeneous; (h) heterogeneous Which of the systems described in give homogeneous equilibria? Which give heterogeneous equilibria? For which of the reactions in does (calculated using concentrations) equal (calculated using pressures)? This situation occurs in (a) and (b). For which of the reactions in does (calculated using concentrations) equal (calculated using pressures)? Convert the values of to values of or the values of to values of . (a) = 1.6 × 10 ; (b) = 50.2; (c) = 5.31 × 10 ; (d) = 4.60 × 10 Convert the values of to values of or the values of to values of . What is the value of the equilibrium constant expression for the change \(\ce{H2O}(l) \rightleftharpoons \ce{H2O}(g)\) at 30 °C? Write the expression of the reaction quotient for the ionization of HOCN in water. Write the reaction quotient expression for the ionization of NH in water. \[Q_c=\ce{\dfrac{[NH4+,OH- ]}{[HN3]}}\] What is the approximate value of the equilibrium constant for the change \(\ce{C2H5OC2H5}(l) \rightleftharpoons \ce{C2H5OC2H5}(g)\) at 25 °C. (Vapor pressure was described in the previous chapter on liquids and solids; refer back to this chapter to find the relevant information needed to solve this problem.) The following equation represents a reversible decomposition: \(\ce{CaCO3}(s)\rightleftharpoons\ce{CaO}(s)+\ce{CO2}(g)\) Under what conditions will decomposition in a closed container proceed to completion so that no CaCO remains? The amount of CaCO must be so small that \(P_{\ce{CO2}}\) is less than when the CaCO has completely decomposed. In other words, the starting amount of CaCO cannot completely generate the full \(P_{\ce{CO2}}\) required for equilibrium. Explain how to recognize the conditions under which changes in pressure would affect systems at equilibrium. What property of a reaction can we use to predict the effect of a change in temperature on the value of an equilibrium constant? The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants' side. What would happen to the color of the solution in part (b) of if a small amount of NaOH were added and Fe(OH)3 precipitated? Explain your answer. The following reaction occurs when a burner on a gas stove is lit: Is an equilibrium among CH , O , CO , and H O established under these conditions? Explain your answer. No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere. A necessary step in the manufacture of sulfuric acid is the formation of sulfur trioxide, SO3, from sulfur dioxide, SO2, and oxygen, O2, shown here. At high temperatures, the rate of formation of \(\ce{SO3 }\)is higher, but the equilibrium amount (concentration or partial pressure) of SO3 is lower than it would be at lower temperatures. Suggest four ways in which the concentration of hydrazine, N H , could be increased in an equilibrium described by the following equation: rightleftharpoons Add N ; add H ; decrease the container volume; heat the mixture. Suggest four ways in which the concentration of PH could be increased in an equilibrium described by the following equation: \[\ce{P4}(g)+\ce{6H2}(g)\rightleftharpoons\ce{4PH3}(g) \hspace{20px} ΔH=\mathrm{110.5\:kJ}\] How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each? (a) Δ increase = shift right, Δ increase = shift left; (b) Δ increase = shift right, Δ increase = no effect; (c) Δ increase = shift left, Δ increase = shift left; (d) Δ increase = shift left, Δ increase = shift right. How will an increase in temperature affect each of the following equilibria? How will a decrease in the volume of the reaction vessel affect each? Water gas is a 1:1 mixture of carbon monoxide and hydrogen gas and is called water gas because it is formed from steam and hot carbon in the following reaction: \[\ce{H2O}(g)+\ce{C}(s)\rightleftharpoons\ce{H2}(g)+\ce{CO}(g).\] Methanol, a liquid fuel that could possibly replace gasoline, can be prepared from water gas and hydrogen at high temperature and pressure in the presence of a suitable catalyst. Nitrogen and oxygen react at high temperatures. Water gas, a mixture of H and CO, is an important industrial fuel produced by the reaction of steam with red hot coke, essentially pure carbon. (a) \(K_c=\ce{\dfrac{[CO,H2]}{[H2O]}}\); (b) [H O] no change, [CO] no change, [H ] no change; (c) [H O] decreases, [CO] decreases, [H ] decreases; (d) [H O] increases, [CO] increases, [H ] decreases; (f) [H O] decreases, [CO] increases, [H ] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change. Pure iron metal can be produced by the reduction of iron(III) oxide with hydrogen gas. Ammonia is a weak base that reacts with water according to this equation: Will any of the following increase the percent of ammonia that is converted to the ammonium ion in water and why? Only (b) Acetic acid is a weak acid that reacts with water according to this equation: rightleftharpoons Will any of the following increase the percent of acetic acid that reacts and produces \(\ce{CH3CO2-}\) ion? Suggest two ways in which the equilibrium concentration of Ag+ can be reduced in a solution of Na , Cl , Ag , and \(\ce{NO3-}\), in contact with solid AgCl. Add NaCl or some other salt that produces Cl− to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s). How can the pressure of water vapor be increased in the following equilibrium? Additional solid silver sulfate, a slightly soluble solid, is added to a solution of silver ion and sulfate ion at equilibrium with solid silver sulfate. rightleftharpoons Which of the following will occur? (a) The amino acid alanine has two isomers, α-alanine and β-alanine. When equal masses of these two compounds are dissolved in equal amounts of a solvent, the solution of α-alanine freezes at the lowest temperature. Which form, α-alanine or β-alanine, has the larger equilibrium constant for ionization \(\ce{(HX \rightleftharpoons H+ + X- )}\)? A reaction is represented by this equation: \(\ce{A}(aq)+\ce{2B}(aq)⇌\ce{2C}(aq) \hspace{20px} K_c=1×10^3\) \(K_c=\ce{\dfrac{[C]^2}{[A,B]^2}}\). [A] = 0.1 , [B] = 0.1 , [C] = 1 ; and [A] = 0.01, [B] = 0.250, [C] = 0.791. A reaction is represented by this equation: \(\ce{2W}(aq)⇌\ce{X}(aq)+\ce{2Y}(aq) \hspace{20px} K_c=5×10^{−4}\) What is the value of the equilibrium constant at 500 °C for the formation of NH according to the following equation? \[\ce{N2}(g)+\ce{3H2}(g)⇌\ce{2NH3}(g)\] An equilibrium mixture of NH ( ), H ( ), and N ( ) at 500 °C was found to contain 1.35 H , 1.15 N , and 4.12 × 10 NH . = 6.00 × 10 Hydrogen is prepared commercially by the reaction of methane and water vapor at elevated temperatures. \[\ce{CH4}(g)+\ce{H2O}(g)⇌\ce{3H2}(g)+\ce{CO}(g)\] What is the equilibrium constant for the reaction if a mixture at equilibrium contains gases with the following concentrations: CH , 0.126 ; H O, 0.242 ; CO, 0.126 ; H 1.15 , at a temperature of 760 °C? A 0.72-mol sample of PCl is put into a 1.00-L vessel and heated. At equilibrium, the vessel contains 0.40 mol of PCl ( ) and 0.40 mol of Cl ( ). Calculate the value of the equilibrium constant for the decomposition of PCl to PCl and Cl at this temperature. = 0.50 At 1 atm and 25 °C, NO with an initial concentration of 1.00 is 3.3 × 10 % decomposed into NO and O . Calculate the value of the equilibrium constant for the reaction. \[\ce{2NO2}(g)⇌\ce{2NO}(g)+\ce{O2}(g)\] Calculate the value of the equilibrium constant for the reaction \(\ce{2NO}(g)+\ce{Cl2}(g)⇌\ce{2NOCl}(g)\) from these equilibrium pressures: NO, 0.050 atm; Cl , 0.30 atm; NOCl, 1.2 atm. The equilibrium equation is × When heated, iodine vapor dissociates according to this equation: \[\ce{I2}(g)⇌\ce{2I}(g)\] At 1274 K, a sample exhibits a partial pressure of I of 0.1122 atm and a partial pressure due to I atoms of 0.1378 atm. Determine the value of the equilibrium constant, , for the decomposition at 1274 K. A sample of ammonium chloride was heated in a closed container. \[\ce{NH4Cl}(s)⇌\ce{NH3}(g)+\ce{HCl}(g)\] At equilibrium, the pressure of NH ( ) was found to be 1.75 atm. What is the value of the equilibrium constant for the decomposition at this temperature? = 3.06 At a temperature of 60 °C, the vapor pressure of water is 0.196 atm. What is the value of the equilibrium constant for the transformation at 60 °C? \[\ce{H2O}(l)⇌\ce{H2O}(g)\] Complete the changes in concentrations (or pressure, if requested) for each of the following reactions. (a) \(\begin{alignat}{3}
&\ce{2SO3}(g)⇌\:&&\ce{2SO2}(g)+\:&&\ce{O2}(g)\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&+x\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&0.125\:M
\end{alignat}\) (b) \(\begin{alignat}{3}
&\ce{4NH3}(g)+\:&&\ce{3O2}(g)⇌\:&&\ce{2N2}(g)+\:&&\ce{6H2O}(g)\\
&\underline{\hspace{40px}} &&3x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\\
&\underline{\hspace{40px}} &&0.24\:M &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\) (c) Change in pressure: \(\begin{alignat}{3}
&\ce{2CH4}(g)⇌\:&&\ce{C2H2}(g)+\:&&\ce{3H2}(g)\\
&\underline{\hspace{40px}} &&x &&\underline{\hspace{40px}}\\
&\underline{\hspace{40px}} &&\textrm{25 torr} &&\underline{\hspace{40px}}
\end{alignat}\) (d) Change in pressure: \(\begin{alignat}{3}
&\ce{CH4}(g)+\:&&\ce{H2O}(g)⇌\:&&\ce{CO}(g)+\:&&\ce{3H2}(g)\\
&\underline{\hspace{40px}} &&x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\\
&\underline{\hspace{40px}} &&\textrm{5 atm} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\) (e) \(\begin{alignat}{3}
&\ce{NH4Cl}(s)⇌\:&&\ce{NH3}(g)+\:&&\ce{HCl}(g)\\
& &&x &&\underline{\hspace{40px}}\\
& &&1.03×10^{−4}\:M &&\underline{\hspace{40px}}
\end{alignat}\) (f) change in pressure: \(\begin{alignat}{3}
&\ce{Ni}(s)+\:&&\ce{4CO}(g)⇌\:&&\ce{Ni(CO)4}(g)\\
& &&4x &&\underline{\hspace{40px}}\\
& &&\textrm{0.40 atm} &&\underline{\hspace{40px}}
\end{alignat}\) Complete the changes in concentrations (or pressure, if requested) for each of the following reactions. (a) \(\begin{alignat}{3}
&\ce{2H2}(g)+\:&&\ce{O2}(g)⇌\:&&\ce{2H2O}(g)\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&+2x\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&1.50\:M
\end{alignat}\) (b) \(\begin{alignat}{3}
&\ce{CS2}(g)+\:&&\ce{4H2}(g)⇌\:&&\ce{CH4}(g)+\:&&\ce{2H2S}(g)\\
&x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\\
&0.020\:M &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\) (c) Change in pressure: \(\begin{alignat}{3}
&\ce{H2}(g)+\:&&\ce{Cl2}(g)⇌\:&&\ce{2HCl}(g)\\
&x &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}\\
&\textrm{1.50 atm} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}}
\end{alignat}\) (d) Change in pressure: \(\begin{alignat}{3}
&\ce{2NH3}(g)+\:&&\ce{2O2}(g)⇌\:&&\ce{N2O}(g)+\:&&\ce{3H2O}(g)\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&x\\
&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\underline{\hspace{40px}} &&\textrm{60.6 torr}
\end{alignat}\) (e) \(\begin{alignat}{3}
&\ce{NH4HS}(s)⇌\:&&\ce{NH3}(g)+\:&&\ce{H2S}(g)\\
& &&x &&\underline{\hspace{40px}}\\
& &&9.8×10^{−6}\:M &&\underline{\hspace{40px}}
\end{alignat}\) (f) Change in pressure: \(\begin{alignat}{3}
&\ce{Fe}(s)+\:&&\ce{5CO}(g)⇌\:&&\ce{Fe(CO)4}(g)\\
& &&\underline{\hspace{40px}} &&x\\
& &&\underline{\hspace{40px}} &&\textrm{0.012 atm}
\end{alignat}\) Why are there no changes specified for Ni in , part (f)? What property of Ni does change? Activities of pure crystalline solids equal 1 and are constant; however, the mass of Ni does change. Why are there no changes specified for NH HS in , part (e)? What property of NH HS does change? Analysis of the gases in a sealed reaction vessel containing NH , N , and H at equilibrium at 400 °C established the concentration of N to be 1.2 and the concentration of H to be 0.24 . \[\ce{N2}(g)+\ce{3H2}(g)⇌\ce{2NH3}(g) \hspace{20px} K_c=\textrm{0.50 at 400 °C}\] Calculate the equilibrium molar concentration of NH . [NH ] = 9.1 × 10 Calculate the number of moles of HI that are at equilibrium with 1.25 mol of H and 1.25 mol of I in a 5.00−L flask at 448 °C. \(\ce{H2 + I2 ⇌ 2HI} \hspace{20px} K_c=\textrm{50.2 at 448 °C}\) What is the pressure of BrCl in an equilibrium mixture of Cl , Br , and BrCl if the pressure of Cl in the mixture is 0.115 atm and the pressure of Br in the mixture is 0.450 atm? \[\ce{Cl2}(g)+\ce{Br2}(g)⇌\ce{2BrCl}(g) \hspace{20px} K_P=4.7×10^{−2}\] = 4.9 × 10 atm What is the pressure of CO in a mixture at equilibrium that contains 0.50 atm H , 2.0 atm of H O, and 1.0 atm of CO at 990 °C? \[\ce{H2}(g)+\ce{CO2}(g)⇌\ce{H2O}(g)+\ce{CO}(g) \hspace{20px} K_P=\textrm{1.6 at 990 °C}\] Cobalt metal can be prepared by reducing cobalt(II) oxide with carbon monoxide. \(\ce{CoO}(s)+\ce{CO}(g)⇌\ce{Co}(s)+\ce{CO2}(g) \hspace{20px} K_c=4.90×10^2\textrm{ at 550 °C}\) What concentration of CO remains in an equilibrium mixture with [CO ] = 0.100 ? [CO] = 2.0 × 10 Carbon reacts with water vapor at elevated temperatures. \(\ce{C}(s)+\ce{H2O}(g)⇌\ce{CO}(g)+\ce{H2}(g) \hspace{20px} K_c=\textrm{0.2 at 1000 °C}\) Sodium sulfate 10−hydrate, \(\ce{Na2SO4 \cdot 10H2O}\), dehydrates according to the equation \[\ce{Na2SO4⋅10H2O}(s)⇌\ce{Na2SO4}(s)+\ce{10H2O}(g) \hspace{20px} \] with \(K_p=4.08×10^{−25}\) at 25°C. What is the pressure of water vapor at equilibrium with a mixture of \(\ce{Na2SO4·10H2O}\) and \(\ce{NaSO4}\)? \(P_{\ce{H2O}}=3.64×10^{−3}\:\ce{atm}\) Calcium chloride 6−hydrate, CaCl ·6H O, dehydrates according to the equation \(\ce{CaCl2⋅6H2O}(s)⇌\ce{CaCl2}(s)+\ce{6H2O}(g) \hspace{20px} K_P=5.09×10^{−44}\textrm{ at 25 °C}\) What is the pressure of water vapor at equilibrium with a mixture of CaCl ·6H O and CaCl ? A student solved the following problem and found the equilibrium concentrations to be [SO ] = 0.590 , [O ] = 0.0450 , and [SO ] = 0.260 . How could this student check the work without reworking the problem? The problem was: For the following reaction at 600 °C: \(\ce{2SO2}(g)+\ce{O2}(g)⇌\ce{2SO3}(g) \hspace{20px} K_c=4.32\) What are the equilibrium concentrations of all species in a mixture that was prepared with [SO ] = 0.500 , [SO ] = 0 , and [O ] = 0.350 ? Calculate based on the calculated concentrations and see if it is equal to . Because does equal 4.32, the system must be at equilibrium. A student solved the following problem and found [N O ] = 0.16 at equilibrium. How could this student recognize that the answer was wrong without reworking the problem? The problem was: What is the equilibrium concentration of N O in a mixture formed from a sample of NO with a concentration of 0.10 ? \[\ce{2NO2}(g)⇌\ce{N2O4}(g) \hspace{20px} K_c=160\] Assume that the change in concentration of N O is small enough to be neglected in the following problem. (a) Calculate the equilibrium concentration of both species in 1.00 L of a solution prepared from 0.129 mol of N O with chloroform as the solvent. \(\ce{N2O4}(g)⇌\ce{2NO2}(g) \hspace{20px} K_c=1.07×10^{−5}\) in chloroform (b) Show that the change is small enough to be neglected. (a) (b) Percent error \(=\dfrac{5.87×10^{−4}}{0.129}×100\%=0.455\%\). The change in concentration of N O is far less than the 5% maximum allowed. Assume that the change in concentration of COCl is small enough to be neglected in the following problem. Assume that the change in pressure of H S is small enough to be neglected in the following problem. (a) Calculate the equilibrium pressures of all species in an equilibrium mixture that results from the decomposition of H S with an initial pressure of 0.824 atm. \(\ce{2H2S}(g)⇌\ce{2H2}(g)+\ce{S2}(g) \hspace{20px} K_P=2.2×10^{−6}\) (b) Show that the change is small enough to be neglected. (a) (b) The 2 is dropped from the equilibrium calculation because 0.014 is negligible when subtracted from 0.824. The percent error associated with ignoring 2 is \(\dfrac{0.014}{0.824}×100\%=1.7\%\), which is less than allowed by the “5% test.” The error is, indeed, negligible. What are all concentrations after a mixture that contains [H O] = 1.00 and [Cl O] = 1.00 comes to equilibrium at 25 °C? \[\ce{H2O}(g)+\ce{Cl2O}(g)⇌\ce{2HOCl}(g) \hspace{20px} K_c=0.0900\] What are the concentrations of PCl , PCl , and Cl in an equilibrium mixture produced by the decomposition of a sample of pure PCl with [PCl ] = 2.00 ? \[\ce{PCl5}(g)⇌\ce{PCl3}(g)+\ce{Cl2}(g) \hspace{20px} K_c=0.0211\] [PCl ] = 1.80 ; [PC ] = 0.195 ; [PCl ] = 0.195 . Calculate the pressures of all species at equilibrium in a mixture of NOCl, NO, and Cl produced when a sample of NOCl with a pressure of 10.0 atm comes to equilibrium according to this reaction: Calculate the equilibrium concentrations of NO, O , and NO in a mixture at 250 °C that results from the reaction of 0.20 NO and 0.10 O . (Hint: is large; assume the reaction goes to completion then comes back to equilibrium.) \[\ce{2NO}(g)+\ce{O2}(g)⇌\ce{2NO2}(g) \hspace{20px} K_c=2.3×10^5\textrm{ at 250 °C}\] Calculate the equilibrium concentrations that result when 0.25 O and 1.0 HCl react and come to equilibrium. \[\ce{4HCl}(g)+\ce{O2}(g)⇌\ce{2Cl2}(g)+\ce{2H2O}(g) \hspace{20px} K_c=3.1×10^{13}\] One of the important reactions in the formation of smog is represented by the equation \[\ce{O3}(g)+\ce{NO}(g)⇌\ce{NO2}(g)+\ce{O2}(g) \hspace{20px} K_P=6.0×10^{34}\] What is the pressure of O remaining after a mixture of O with a pressure of 1.2 × 10 atm and NO with a pressure of 1.2 × 10 atm comes to equilibrium? (Hint: is large; assume the reaction goes to completion then comes back to equilibrium.) \(P_{\ce{O3}}=4.9×10^{−26}\:\ce{atm}\) Calculate the pressures of NO, Cl , and NOCl in an equilibrium mixture produced by the reaction of a starting mixture with 4.0 atm NO and 2.0 atm Cl . (Hint: is small; assume the reverse reaction goes to completion then comes back to equilibrium.) \(\ce{2NO}(g)+\ce{Cl2}(g)⇌\ce{2NOCl}(g) \hspace{20px} K_P=2.5×10^3\) Calculate the number of grams of HI that are at equilibrium with 1.25 mol of H and 63.5 g of iodine at 448 °C. \(\ce{H2 + I2 ⇌ 2HI} \hspace{20px} K_c=\textrm{50.2 at 448 °C}\) 507 g Butane exists as two isomers, −butane and isobutane. = 2.5 at 25 °C What is the pressure of isobutane in a container of the two isomers at equilibrium with a total pressure of 1.22 atm? What is the minimum mass of CaCO required to establish equilibrium at a certain temperature in a 6.50-L container if the equilibrium constant ( ) is 0.050 for the decomposition reaction of CaCO at that temperature? \(\ce{CaCO3}(s)⇌\ce{CaO}(s)+\ce{CO2}(g)\) 330 g The equilibrium constant ( ) for this reaction is 1.60 at 990 °C: \[\ce{H2}(g)+\ce{CO2}(g)⇌\ce{H2O}(g)+\ce{CO}(g)\] Calculate the number of moles of each component in the final equilibrium mixture obtained from adding 1.00 mol of H , 2.00 mol of CO , 0.750 mol of H O, and 1.00 mol of CO to a 5.00-L container at 990 °C. At 25 °C and at 1 atm, the partial pressures in an equilibrium mixture of N O and NO are \(P_{\ce{N2O4}}=0.70\:\ce{atm}\) and \(P_{\ce{NO2}}=0.30\:\ce{atm}\). (a) Both gases must increase in pressure. (b) \(P_{\ce{N2O4}}=\textrm{8.0 atm and }P_{\ce{NO2}}=1.0\:\ce{atm}\) In a 3.0-L vessel, the following equilibrium partial pressures are measured: N , 190 torr; H , 317 torr; NH , 1.00 × 10 torr. \[\ce{N2}(g)+\ce{3H2}(g)⇌\ce{2NH3}(g)\] The equilibrium constant ( ) for this reaction is 5.0 at a given temperature. \[\ce{CO}(g)+\ce{H2O}(g) <=> \ce{CO2}(g)+\ce{H2}(g)\)] (a) 0.33 mol. (b) [CO] = 0.50 Added H forms some water to compensate for the removal of water vapor and as a result of a shift to the left after H is added. Antimony pentachloride decomposes according to this equation: \(\ce{SbCl5}(g)⇌\ce{SbCl3}(g)+\ce{Cl2}(g)\) An equilibrium mixture in a 5.00-L flask at 448 °C contains 3.85 g of SbCl , 9.14 g of SbCl , and 2.84 g of Cl . How many grams of each will be found if the mixture is transferred into a 2.00-L flask at the same temperature? Consider the reaction between H and O at 1000 K 2H2 2H2O hspace 20px dfrac H2O If 0.500 atm of H and 0.500 atm of O are allowed to come to equilibrium at this temperature, what are the partial pressures of the components? \(P_{\ce{H2}}=8.64×10^{−11}\:\ce{atm}\) atm H2O atm An equilibrium is established according to the following equation \[\ce{Hg2^2+}(aq)+\ce{NO3−}(aq)+\ce{3H+}(aq)⇌\ce{2Hg^2+}(aq)+\ce{HNO2}(aq)+\ce{H2O}(l) \hspace{20px} K_c=4.6\] What will happen in a solution that is 0.20 each in \(\ce{Hg2^2+}\), \(\ce{NO3−}\), H , Hg , and HNO ? Consider the equilibrium \[\ce{4NO2}(g)+\ce{6H2O}(g)⇌\ce{4NH3}(g)+\ce{7O2}(g)\] (a) \(K_c=\ce{\dfrac{[NH3]^4[O2]^7}{[NO2]^4[H2O]^6}}\). (b) [NH ] must increase for to reach . (c) That decrease in pressure would decrease [NO ]. (d) \(P_{\ce{O2}}=49\:\ce{torr}\) The binding of oxygen by hemoglobin (Hb), giving oxyhemoglobin (HbO ), is partially regulated by the concentration of H O and dissolved CO in the blood. Although the equilibrium is complicated, it can be summarized as \(\ce{HbO2}(aq)+\ce{H3O+}(aq)+\ce{CO2}(g)⇌\ce{CO2−Hb−H+}+\ce{O2}(g)+\ce{H2O}(l)\) The hydrolysis of the sugar sucrose to the sugars glucose and fructose follows a first-order rate equation for the disappearance of sucrose. \(\ce{C12H22O11}(aq)+\ce{H2O}(l)⟶\ce{C6H12O6}(aq)+\ce{C6H12O6}(aq)\) Rate = [C H O ] In neutral solution, = 2.1 × 10 /s at 27 °C. (As indicated by the rate constant, this is a very slow reaction. In the human body, the rate of this reaction is sped up by a type of catalyst called an enzyme.) (Note: That is not a mistake in the equation—the products of the reaction, glucose and fructose, have the same molecular formulas, C H O , but differ in the arrangement of the atoms in their molecules). The equilibrium constant for the reaction is 1.36 × 10 at 27 °C. What are the concentrations of glucose, fructose, and sucrose after a 0.150 aqueous solution of sucrose has reached equilibrium? Remember that the activity of a solvent (the effective concentration) is 1. [fructose] = 0.15 The density of trifluoroacetic acid vapor was determined at 118.1 °C and 468.5 torr, and found to be 2.784 g/L. Calculate for the association of the acid. Liquid N O is dark blue at low temperatures, but the color fades and becomes greenish at higher temperatures as the compound decomposes to NO and NO . At 25 °C, a value of = 1.91 has been established for this decomposition. If 0.236 moles of N O are placed in a 1.52-L vessel at 25 °C, calculate the equilibrium partial pressures of N O ( ), NO ( ), and NO( ). \(P_{\ce{N2O3}}=\textrm{1.90 atm and }P_{\ce{NO}}=P_{\ce{NO2}}=\textrm{1.90 atm}\) A 1.00-L vessel at 400 °C contains the following equilibrium concentrations: N , 1.00 ; H , 0.50 ; and NH , 0.25 . How many moles of hydrogen must be removed from the vessel to increase the concentration of nitrogen to 1.1 ? A 0.010 solution of the weak acid HA has an osmotic pressure (see chapter on solutions and colloids) of 0.293 atm at 25 °C. A 0.010 solution of the weak acid HB has an osmotic pressure of 0.345 atm under the same conditions. (a) Which acid has the larger equilibrium constant for ionization HA \([\ce{HA}(aq)⇌\ce{A-}(aq)+\ce{H+}(aq)]\) or HB \([\ce{HB}(aq)⇌\ce{H+}(aq)+\ce{B-}(aq)]\)? (b) What are the equilibrium constants for the ionization of these acids? (Hint: Remember that each solution contains three dissolved species: the weak acid (HA or HB), the conjugate base (A or B ), and the hydrogen ion (H ). Remember that osmotic pressure (like all colligative properties) is related to the total number of solute particles. Specifically for osmotic pressure, those concentrations are described by molarities.) (a) HB ionizes to a greater degree and has the larger . (b) (HA) = 5 × 10 (HB) = 3 × 10 | 29,862 | 3,438 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.08%3A_Amorphous_Materials-_Glasses |
Some liquids become extremely viscous as the temperature falls toward their freezing points, often because they consist of macromolecules. An example is quartz, SiO , seen . Fused silica is an example of an or . It is highly rigid at room temperature, but it does not have the long-range microscopic regularity of a solid crystal lattice. Consequently it cannot be made to cleave along a plane. Instead, like ordinary window glass, it shatters into irregular fragments when struck sharply. (Window glass is primarily silica, but oxides of sodium and calcium are added to lower the melting point.) Since the microscopic structure of a glass is random, like that of a liquid, scientific purists describe glasses as highly viscous liquids, not as solids. To the left is an example of , which has many due to its high purity, high melting temperature, and high radiation resistance. | 896 | 3,439 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.19%3A_Organic_Nitrogen_Compounds |
There is a tremendous variety of organic compounds which can be derived from carbon, hydrogen, and oxygen which is evident from the numerous previous sections discussing these compounds. If we include nitrogen as a possible constituent of these molecular structures, many more possibilities arise. Most of the nitrogen-containing compounds are less important commercially, however, and we will only discuss a few of them here. may be derived from ammonia by replacing one, two, or all three hydrogens with alkyl groups. Some examples are
The terms primary (one), secondary (two), and tertiary (three) refer to the number of hydrogens that have been replaced. Both primary and secondary amines are capable of hydrogen bonding with themselves, but tertiary amines have no hydrogens on the electronegative nitrogen atom. Amines usually have unpleasant odors, smelling “fishy“. The three methylamines listed above can all be isolated from herring brine. Amines, as well as ammonia, are produced by decomposition of nitrogen-containing compounds when a living organism dies. The methylamines are obtained commercially by condensation of methanol with ammonia over an aluminum oxide catalyst: Dimethylamine is the most important, being used in the preparation of herbicides, in rubber vulcanization, and to synthesize dimethylformamide, an important solvent. are another important nitrogen containing organic compound. The key feature of an amine is a nitrogen atom bonded to a carbonyl carbon atom. Like , amides are formed in a condensation reaction. While esters are formed from the condensation reaction of an alcohol and a carboxylic acid, amides are formed from the condensation of an amine and a carboxylic acid: This general reaction is usually unfavorable, because the hydroxyl group acts as a bad leaving group. Organic chemists have devised methods to work around this by using certain chemicals to activate the carboxylic acid and allow for the addition of the amine. As amides are formed by condensation reactions, many important involve amide linkages. Nylon, for instance, is formed from the amide condensation of hexamethylenediamine and adipic acid. A second set of condensation polymers formed from amide linkages are the and found in your body and in all organisms. These polymers are formed from another organic nitrogen compound, the These molecules contain both an amine group and a carboxyl group. Examples of such are glycine and lysine: Amino acids are the constituents from which proteins are made. Some, like glycine, can be synthesized in the human body, but others cannot. Lysine is an example of an —one which must be present in the human diet because it cannot be synthesized within the body. As mentioned, the condensation of amino acids into peptides forms amide linkages. For this reason, scientists sometimes refer to the of a protein or peptide. A protein has a long series of amide bonds, as can be seen in the following figure showing the synthesis of a tri-peptide from three amino acids: Amino acids and proteins further discussed in the sections on enzymes and in a set of sections devoted to and their chemistry in living systems. The intermolecular forces and boiling points of nitrogen-containing organic compounds may be explained according to the same principles used for oxygen-containing substances. Rationalize the following boiling points: (a) 0°C for CH CH CH CH ; (b) 11°C for CH CH OCH ; (c) 97°C for CH CH CH OH; and (d) 170°C for NH CH CH OH. All four molecules have very similar geometries and the same number of electrons (26 valence electrons plus 8 core electrons), and so their London forces should be about the same. Compound (a) is an alkane and is nonpolar. By contrast compound (b) is an ether and should be slightly polar. This slight polarity results in a slightly higher boiling point. Compound (c) is isomeric with compound (b) but is an alcohol. There is hydrogen bonding between molecules of (c), and its boiling point is much higher. Molecule (d) has both an amino group and a hydroxyl group, each of which can participate in hydrogen bonding. Consequently it has the highest boiling point of all. | 4,206 | 3,441 |
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The adjective describes the very aptly, because all of them are derived from the remains of plants or animals which lived on earth millions of years ago. Coal, for example, began as plant matter in prehistoric swamps, where it was able to decompose in the absence of air. Present-day peat bogs are examples of this first stage in coal formation, and in countries such as Ireland dried peat is an important fuel. Over long periods of time, at high temperatures and pressures under the earth’s surface, peat can be transformed into lignite, a brown, soft form of coal. Continued action of geological forces converts lignite into bituminous, or soft coal, and eventually into anthracite, or hard coal. When burned, these latter two types of coal release considerably more heat per unit mass than do lignite or peat. A crucial point to realize about fossil fuels is that the energy we release by burning them came originally from the sun. The plants from which the fuels were derived grew as a result of , the combination of carbon dioxide and water under the influence of sunlight to form organic compounds whose empirical formula is approximately \[\ce{CO_{2}(g) + H_{2}O(l) \rightarrow [CH_{2}O](s) + O_{2}(g)} \label{1} \] with \(\Delta H \approx 469 \text{kJ mol}^{-1}\) Since a number of different substances are formed by photosynthesis, the empirical formula [CH O] and the Δ are only approximate. Photosynthesis is endothermic, and the necessary energy is supplied by the absorption of solar radiant energy. This energy can be released by carrying out the reverse of Equation \ref{1}, a process which is exothermic. When we burn paper, wood, or dried leaves, the heat given off is really a stored form of sunlight. Plants and animals obtain the energy they need to grow or move about from the of substances produced by photosynthesis. This oxidation process is called . After millions of years of geological change, the fossil fuels are significantly different in chemical structure from newly photosynthesized plant or animal material. The changes which occur can be approximated by the equation for formation of methane (natural gas): \[\ce{2[CH_{2}O] \rightarrow CH_{4}(g) + CO_{2}(g)} \nonumber \] with \(\Delta H \approx –47 \text{kJ mol}^{-1}\). This reaction is only slightly exothermic, and so very little of the energy captured from sunlight is lost. However, about half the carbon and all the oxygen are lost as carbon dioxide gas, and so a fossil fuel like methane can release more heat per carbon atom (and per gram) than can wood or other organic materials. This is why anthracite and bituminous coals are better fuels than the peat from which they are formed. The enthalpy changes which occur during photosynthesis, respiration, and formation and combustion of fossil fuels are summarized in Figure \(\Page {1}\). | 2,854 | 3,442 |
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Ionic bonding involves transfer of an electron from one atom (which becomes a positively charged ) to another (which becomes a negatively charged ). The two ions attract strongly to form a . Since ionic bonding requires that the atoms involved have unequal attraction for their valence electrons, an ionic compound must involve atoms of two quite different elements. Attraction for electrons depends on the distance of the electrons from the nucleus (which in turn depends on the amount of shielding by inner electrons). Ionic compounds generally form between metals toward the left and bottom of the periodic table, and nonmetals toward the right and top of the periodic table. The simplest example of a binary ionic compound is provided by the combination of elements number 1 (H) and number 3 (Li) in lithium hydride, LiH. On a microscopic level the formula LiH contains four electrons. In separate Li and H atoms these electrons are arranged as shown in part of the following figure. The H atom has the electron configuration 1 , and Li is 1 2 . When the two atoms are brought close enough together, however, the striking rearrangement of the electron clouds shown part takes place. Here the color coding shows clearly that the electron density which was associated with the 2 orbital in the individual Li atom has been transferred to a 1 orbital surrounding the H atom. As a result, two new microscopic species are formed. The extra electron transforms the H atom into a or , written H and called the . The two electrons left on the Li atom are not enough to balance the charge of +3 on the Li nucleus, and so removal of an electron produces a or , written Li and called the . The electron-transfer process can be summarized in of Lewis diagrams as follows: The opposite charges of Li and H attract each other strongly, and the ions form an (see image below) in which the two nuclei are separated by a distance of 160 pm (1.60 Å). Multiple (as seen in the image above) connect to form a , pictured below. All ionic solids form a crystal lattice and the shape of the lattice determines the properties and look of the solid. | 2,171 | 3,443 |
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Although DNA contains the necessary instructions for synthesizing all the proteins necessary for the functioning of a cell, it does not take part directly in the synthesis itself. Sections of the DNA chain are first copied into a type of RNA called , mRNA. This process is known as . In transcription, RNA polymerase opens up DNA and uses the base pair sequence of one of the DNA strands to synthesis a molecule of mRNA which is complementary to the The other strand is referred to as the , as it has the same sequence as the mRNA molecule formed. The synthesis of mRNA is governed by promoter sequences in the DNA, as well as binding of protein factors that can either promote or repress transcription. Such control is necessary for a cell produce the correct proteins at the correct time. The mRNA molecules differ from DNA in three ways: RNA is generally found in a single strand form, instead of the double helix structure of DNA. RNA has a hydroxyl group ( ) at the 2' carbon, wheras DNA simply has a hydrogen. The base uracil (U) replaces thymine (T). All three differences are displayed in Fig \(\Page {1}\). Another aspect of mRNA molecules is that they are also considerably smaller than DNA, containing the blueprints for only a few proteins at most. As the name implies, mRNA molecules are used to transport their coded instructions from the nucleus of the cell, where the DNA is situated, to the , where the process of protein synthesis actually takes place. When an mRNA molecule reaches a ribosome, a process called takes place in which the base sequence on the mRNA molecule is used to create a protein using the codon code. In translation, each codon on the mRNA base pairs with an base sequence on a RNA molecule called , tRNA. Each tRNA molecule is bound to the amino acid. Thus, the codon UUA will pair with the anticodon of the tRNA bound to leucine. The synthesis of the protein itself is performed by the ribosome, which is a protein-RNA complex, but unlike other enzymes, the catalytic activity is provided by the RNA portion, not the protein. The ribosome is thus sometimes referred to as a ribozyme, to distinguish it from the usual concept of any enzyme with protein based catalytic activity. Figure \(\Page {2}\) presents this process as a cartoon. In the cartoon, the brown structure represents the ribosome, the blue letters represent mRNA and its base sequence, the colored hooks represent tRNAs, and amino acids are represented by their three letter abbreviations. mRNA binds into the ribosome, which finds a start codon at which to begin translating. This begins with Methionine, which is bound to the tRNA which recognizes the AUG codon. The next tRNA enters in at the A site. The ribosome catalyzes the condensation reaction between the carboxyl end of methionine and the amino end of glutamate. The dipeptide is now bound to the tRNA in the A site. The tRNA for methionine shifts to the exit site and leaves the complex, while the dipeptide and bound tRNA move to the P site. The next activated tRNA, meaning the tRNA has an amino acid bound to it, enters the A site, and the process repeats until a stop codon is reached. | 3,188 | 3,445 |
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Nitration and sulfonation of are two examples of electrophilic aromatic substitution. The nitronium ion (NO ) and sulfur trioxide (SO ) are the electrophiles and individually react with benzene to give nitrobenzene and benzenesulfonic acid respectively. The source of the nitronium ion is through the protonation of nitric acid by sulfuric acid, which causes the loss of a water molecule and formation of a nitronium ion. The first step in the nitration of benzene is to activate HNO with sulfuric acid to produce a stronger electrophile, the nitronium ion. Because the nitronium ion is a good electrophile, it is attacked by benzene to produce Nitrobenzene. (Resonance forms of the intermediate can be seen in the generalized electrophilic aromatic substitution) Sulfonation is a reversible reaction that produces benzenesulfonic acid by adding sulfur trioxide and fuming sulfuric acid. The reaction is reversed by adding hot aqueous acid to benzenesulfonic acid to produce benzene. To produce benzenesulfonic acid from benzene, fuming sulfuric acid and sulfur trioxide are added. Fuming sulfuric acid, also refered to as , is a concentrated solution of dissolved sulfur trioxide in sulfuric acid. The sulfur in sulfur trioxide is electrophilic because the oxygens pull electrons away from it because oxygen is very electronegative. The benzene attacks the sulfur (and subsequent proton transfers occur) to produce benzenesulfonic acid. Sulfonation of benzene is a reversible reaction. Sulfur trioxide readily reacts with water to produce sulfuric acid and heat. Therefore, by adding heat to benzenesulfonic acid in diluted aqueous sulfuric acid the reaction is reversed. Nitration is used to add nitrogen to a benzene ring, which can be used further in substitution reactions. The nitro group acts as a ring . Having nitrogen present in a ring is very useful because it can be used as a directing group as well as a masked amino group. The products of aromatic nitrations are very important intermediates in industrial chemistry. Because sulfonation is a reversible reaction, it can also be used in further substitution reactions in the form of a directing blocking group because it can be easily removed. The sulfonic group blocks the carbon from being attacked by other substituents and after the reaction is completed it can be removed by reverse sulfonation. Benzenesulfonic acids are also used in the synthesis of detergents, dyes, and sulfa drugs. Bezenesulfonyl Chloride is a precursor to sulfonamides, which are used in chemotherapy. 1. What is/are the required reagent(s)for the following reaction: 2. What is the product of the following reaction: 3. Why is it important that the nitration of benzene by nitric acid occurs in sulfuric acid? 4. Write a detailed mechanism for the sulfonation of benzene, including all resonance forms. 5. Draw an energy diagram for the nitration of benzene. Draw the intermediates, starting materials, and products. Label the transition states. (For questions 1 and 2 see Electrophilic Aromatic Substitution for hints) For other problems involving Electrophilic Aromatic Substitution and similar reactions see: 1. SO and H SO (fuming) 2. 3. Sulfuric acid is needed in order for a good electrophile to form. Sulfuric acid protonates nitric acid to form the nitronium ion (water molecule is lost). The nitronium ion is a very good electrophile and is open to attack by benzene. Without sulfuric acid the reaction would not occur. 4. 5. | 3,501 | 3,447 |
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In beginning our study of the reactions of gases, we will assume a knowledge of the physical properties of gases as described by the and an understanding of these properties as given by the postulates and conclusions of the . We assume that we have developed a dynamic model of phase equilibrium in terms of competing rates. We will also assume an understanding of the bonding, structure, and properties of individual molecules. In performing stoichiometric calculations, we assume that we can calculate the amount of product of a reaction from the amount of the reactants we start with. For example, if we burn methane gas, \(\ce{CH_4} \left( g \right)\), in excess oxygen, the reaction \[\ce{CH_4} \left( g \right) + 2 \ce{O_2} \left( g \right) \rightarrow \ce{CO_2} \left( g \right) + 2 \ce{H_2O} \left( g \right)\] occurs, and the number of moles of \(\ce{CO_2} \left( g \right)\) produced is assumed to equal the number of moles of \(\ce{CH_4} \left( g \right)\) we start with. From our study of phase transitions we have learned the concept of equilibrium. We observed that, in the transition from one phase to another for a substance, under certain conditions both phases are found to coexist, and we refer to this as phase equilibrium. It should not surprise us that these same concepts of equilibrium apply to chemical reactions as well. In the reaction above, therefore, we should examine whether the reaction actually produces exactly one mole of \(\ce{CO_2}\) for every mole of \(\ce{CH_4}\) we start with or whether we wind up with an equilibrium mixture containing both \(\ce{CO_2}\) and \(\ce{CH_4}\). We will find that different reactions provide us with varying answers. In many cases, virtually all reactants are consumed, producing the stoichiometric amount of product. However, in many other cases, substantial amounts of reactant are still present when the reaction achieves equilibrium, and in other cases, almost no product is produced at equilibrium. Our goal will be to understand, describe, and predict the reaction equilibrium. An important corollary to this goal is to attempt to control the equilibrium. We will find that varying the conditions under which the reaction occurs can vary the amounts of reactants and products present at equilibrium. We will develop a general principle for predicting how the reaction conditions affect the amount of product produced at equilibrium. We begin by analyzing a significant industrial chemical process, the synthesis of ammonia gas, \(\ce{NH_3}|), from nitrogen and hydrogen: \[\ce{N_2} \left( g \right) + 3 \ce{H_2} \left( g \right) \rightarrow 2 \ce{NH_3} \left( g \right)\] If we start with 1 mole of \(\ce{N_2}\) and 3 moles of \(\ce{H_2}\), the balanced equation predicts that we will produce 2 moles of \(\ce{NH_3}\). In fact, if we carry out this reaction starting with these quantities of nitrogen and hydrogen at \(298 \: \text{K}\) in a \(100.0 \: \text{L}\) reaction vessel, we observe that the number of moles of \(\ce{NH_3}\) produced is \(1.91 \: \text{mol}\). This "yield" is less than predicted by the balanced equation, but the difference is not due to a limiting reagent factor. Recall that, in stoichiometry, the limiting reagent is the one that is present in less than the ratio of moles given by the balanced equation. In this case, neither \(\ce{N_2}\) nor \(\ce{H_2}\) is limiting because they are present initially in a 1:3 ratio, exactly matching the stoichiometry. Note also that this seeming deficit in the yield is not due to any experimental error or imperfection, nor is it due to poor measurements or preparation. Rather, the observation that, at \(298 \: \text{K}\), \(1.91 \: \text{mol}\) rather than \(2 \: \text{mol}\) are produced is completely reproducible: every measurement of this reaction at this temperature in this volume starting with 1 mole of \(\ce{N_2}\) and 3 moles of \(\ce{H_2}\) gives this result. We conclude that the reaction achieves in which all three gases are present in the gas mixture. We can determine the amounts of each gas at equilibrium from the stoichiometry of the reaction. When \(n_{NH_3} = 1.91 \: \text{mol}\) are created, the number of moles of \(\ce{N_2}\) remaining at equilibrium is \(n_{N_2} = 0.045 \: \text{mol}\) and \(n_{H_2} = 0.135 \: \text{mol}\). It is important to note that we can vary the relative amount of \(\ce{NH_3}\) produced by varying the temperature of the reaction, the volume of the vessel in which the reaction occurs, or the relative starting amounts of \(\ce{N_2}\) and \(\ce{H_2}\). We shall study and analyze this observation in detail in later sections. For now, though, we demonstrate that the concept of reaction equilibrium is general to all reactions. Consider the reaction \[\ce{H_2} \left( g \right) + \ce{I_2} \left( g \right) \rightarrow 2 \ce{HI} \left( g \right)\] If we begin with \(1.00 \: \text{mol}\) of \(\ce{H_2}\) and \(1.00 \: \text{mol}\) of \(\ce{I_2}\) at \(500 \: \text{K}\) in a reaction vessel of fixed volume, we observe that, at equilibrium, \(n_{HI} = 1.72 \: \text{mol}\), leaving in the equilibrium mixture \(n_{H_2} = 0.14 \: \text{mol}\) and \(n_{I_2} = 0.14 \: \text{mol}\). Similarly, consider the decomposition reaction \[\ce{N_2O_4} \left( g \right) \rightarrow 2 \ce{NO_2} \left( g \right)\] At \(298 \: \text{K}\) in a \(100.0 \: \text{L}\) reaction flask, \(1.00 \: \text{mol}\) of \(\ce{N_2O_4}\) partially decomposes to produce, at equilibrium, \(n_{NO_2} = 0.64 \: \text{mol}\) and \(n_{N_2O_4} = 0.68 \: \text{mol}\). Some chemical reaction achieve an equilibrium that appears to be very nearly complete reaction. For example, \[\ce{H_2} \left( g \right) + \ce{Cl_2} \left( g \right) \rightarrow 2 \ce{HCl} \left( g \right)\] If we begin with 1.00 mole of \(\ce{H_2}\) and 1.00 mole of \(\ce{Cl_2}\) at \(298 \: \text{K}\) in a reaction vessel of fixed volume, we observe that, at equilibrium, \(n_{HCl}\) is almost exactly \(2.00 \: \text{mol}\), leaving virtually no \(\ce{H_2}\) or \(\ce{Cl_2}\). This does not mean that the reaction has not come to equilibrium. It means instead that, at equilibrium, there are essentially no reactants remaining. In each of these cases, the amounts of reactants and products present at equilibrium vary as the conditions are varied but are completely reproducible for fixed conditions. Before making further observations that will lead to a quantitative description of the reaction equilibrium, we consider a qualitative description of equilibrium. We begin with a dynamic equilibrium description. We know from our studies of phase transitions that equilibrium occurs when the rate of the forward process (e.g. evaporation) is matched by the rate of the reverse process (e.g. condensation). Since we have now observed that gas reactions also come to equilibrium, we postulate that at equilibrium the forward reaction rate is equal to the reverse reaction rate. For example, in the decomposition reaction of \(\ce{N_2O_4}\), the rate of decomposition of \(\ce{N_2O_4}\) molecules at equilibrium must be exactly matched by the rate of recombination (or ) of \(\ce{NO_2}\) molecules. To show that the forward and reverse reactions continue to happen at equilibrium, we start with the \(\ce{NO_2}\) and \(\ce{N_2O_4}\) mixture at equilibrium and we vary the volume of the flask containing the mixture. We observe that, if we increase the volume and the reaction is allowed to come to equilibrium, the amount of \(\ce{NO_2}\) at equilibrium is larger at the expense of a smaller amount of \(\ce{N_2O_4}\). We can certainly conclude that the amounts of the gases at equilibrium depend on the reaction conditions. However, if the forward and reverse reactions stop once the equilibrium amounts of material are achieved, the molecules would not "know" that the volume of the container had increased. Since the reaction equilibrium can and does respond to a change in volume, it must be that the change in volume affect the rates of both the forward and reverse processes. This means that both reactions must be occurring at equilibrium, and that their rates must exactly match at equilibrium. This reasoning reveals that the amounts of reactant and product present at equilibrium are determined by the rates of the forward and reverse reactions. If the rate of the forward reaction (e.g. decomposition of \(\ce{N_2O_4}\)) is faster than the rate of the reverse reaction, then at equilibrium we have more product than reactant. If that difference in rates is very large, at equilibrium there will be much more product than reactant. Of course, the converse of these conclusions is also true. It must also be the case that the rates of these processes depends on, amongst other factors, the volume of the reaction flask, since the amounts of each gas present at equilibrium change when the volume is changed. It was noted above that the equilibrium partial pressures of the gases in a reaction vary depending upon a variety of conditions. These include changes in the initial numbers of moles of reactants and products, changes in the volume of the reaction flask, and changes in the temperature. We now study these variations quantitatively. Consider first the decomposition reaction of \(\ce{N_2O_4}\). Following on our previous study of this reaction, we inject an initial amount of \(\ce{N_2O_4} \left( g \right)\) into a \(100 \: \text{L}\) reaction flask at \(298 \: \text{K}\). Now, however, we vary the initial number of moles of \(\ce{N_2O_4} \left( g \right)\) in the flask and measure the equilibrium pressures of both the reactant and product gases. The results of a number of such studies are given in Table 14.1. We might have expected that the amount of \(\ce{NO_2}\) produced at equilibrium would increase in direct proportion to increases in the amount of \(\ce{N_2O_4}\) we begin with. Table 14.1 shows that this is not the case. Note that when we increase the initial amount of \(\ce{N_2O_4}\) by a factor of 10 from 0.5 moles to 5.0 moles, the pressure of \(\ce{NO_2}\) at equilibrium increases by a factor of less than 4. The relationship between the pressures at equilibrium and the initial amount of \(\ce{N_2O_4}\) is perhaps more easily seen in a graph of the data in Table 14.1, as shown in Figure 14.1. There are some interesting features here. Note that, when the initial amount of \(\ce{N_2O_4}\) is less than \(1 \: \text{mol}\), the equilibrium pressure of \(\ce{NO_2}\) is greater than that of \(\ce{N_2O_4}\). These relative pressures reverse as the initial amount increases, as the \(\ce{N_2O_4}\) equilibrium pressure of \(\ce{NO_2}\) does not increase proportionally with the initial amount of \(\ce{N_2O_4}\). In fact, the increase is slower than proportionality, suggesting perhaps a square root relationship between the pressure of \(\ce{NO_2}\) and the initial amount of \(\ce{N_2O_4}\). We test this in Figure 14.2 by plotting \(P_{NO_2}\) at equilibrium versus the square root of the initial number of moles of \(\ce{N_2O_4}\). Figure 14.2 makes it clear that this is not a simple proportional relationship, but it is closer. Note in Figure 14.1 that the equilibrium pressure \(P_{N_2O_4}\) increases close to proportionally with the initial amount of \(\ce{N_2O_4}\). This suggests plotting \(P_{NO_2}\) versus the square root of \(P_{N_2O_4}\). This is done in Figure 14.3, where we discover that there is a very simple proportional relationship between the variables plotted in this way. We have thus observed that \[P_{NO_2} = c \sqrt{2P_{N_2O_4}}\] where \(c\) is the slope of the graph. The equation can be rewritten in a standard form \[K_p = \frac{P^2_{NO_2}}{P_{N_2O_4}}\] To test the accuracy of this equation and to find the value of \(K_p\), we return to Table 14.1 and add another column in which we calculate the value of \(K_p\) for each of the data points. Table 14.2 makes it clear that the "constant" in the equation truly is independent of both the initial conditions and the equilibrium partial pressure of either one of the reactant or product. We thus refer to the constant \(K_p\) in the equation as the . It is very interesting to note the functional form of the equilibrium constant. The product \(\ce{NO_2}\) pressure appears in the numerator, and the exponent 2 on the pressure is the stoichiometric coefficient on \(\ce{NO_2}\) in the balanced chemical equation. The reactant \(\ce{N_2O_4}\) pressure appears in the denominator, and the exponent 1 on the pressure is the stoichiometric coefficient on \(\ce{N_2O_4}\) in the chemical equation. We now investigate whether other reactions have equilibrium constants and whether the form of this equilibrium constant is a happy coincidence or a general observation. We return to the reaction for the synthesis of ammonia. In a previous section, we considered only the equilibrium produced when 1 mole of \(\ce{N_2}\) is reacted with 3 moles of \(\ce{H_2}\). We now consider a range of possible initial values of these amounts, with the resultant equilibrium partial pressures given in Table 14.3. In addition, anticipating the possibility of an equilibrium constant, we have calculated the ratio of partial pressure given by: \[K_p = \frac{P^2_{NH_3}}{P_{N_2} P^3_{N_2}}\] In Table 14.3, the equilibrium partial pressures of the gases are in a very wide variety, including whether the final pressures are greater for reactants or products. However, from the data in Table 14.3, it is clear that, despite these variations, \(K_p\) is essentially a constant for all of the initial conditions examined and is thus the for this reaction. Studies of many chemical reactions of gases result in the same observations. Each reaction equilibrium can be described by an equilibrium constant in which the partial pressures of the products, each raised to their corresponding stoichiometric coefficient, are multiplied together in the numerator, and the partial pressures of the reactants, each raised to their corresponding stoichiometric coefficient, are multiplied together in the denominator. For historical reasons, this general observation is sometimes referred to as the . We have previously observed that phase equilibrium, and in particular vapor pressure, depend on the temperature, but we have not yet studied the variation of reaction equilibrium with temperature. We focus our initial study on the reaction of hydrogen gas and iodine gas and we measure the equilibrium partial pressures at a variety of temperatures. From these measurements, we can compile the data showing the temperature dependence of the equilibrium constant \(K_p\) for this reaction in Table 14.4. Note that the equilibrium constant increases dramatically with temperature. As a result, at equilibrium, the pressure of \(\ce{HI}\) must also increase dramatically as the temperature is increased. These data do not seem to have a simple relationship between \(K_p\) and temperature. We must appeal to arguments based on thermodynamics, from which it is possible to show that the equilibrium constant should vary with temperature according to the following equation: \[\text{ln} \left( K_p \right) = -\frac{\Delta H^0}{RT} + \frac{\Delta S^0}{R}\] If \(\Delta H^0\) and \(\Delta S^0\) do not depend strongly on the temperature, then this equation would predict a simple straight line relationship between \(\text{ln} \left( K_p \right)\) and \(\frac{1}{T}\). In addition, the slope of this line should be \(-\frac{\Delta H^0}{R}\). We test this possibility with the graph in Figure 14.4. In fact, we do observe a straight line through the data. In this case, the line has a negative slope. Note carefully that this means that \(K_p\) is with temperature. The negative slope means that \(-\frac{\Delta H^0}{R}\) must be negative, and indeed for this reaction in this temperature range, \(\Delta H^0 = 15.6 \frac{\text{kJ}}{\text{mol}}\). This value matches well with the slope of the line in Figure 14.4. Given the validity of the equation in describing the temperature dependence of the equilibrium constant, we can also predict that an exothermic reaction with \(\Delta H^0 < 0\) should have a positive slope in the graph of \(\text{ln} \left( K_p \right)\) versus \(\frac{1}{T}\), and thus the equilibrium constant should with increasing temperature. A good example of an exothermic reaction is the synthesis of ammonia for which \(\Delta H^0 = -99.2 \: \frac{\text{kJ}}{\text{mol}}\). Equilibrium constant data are given in Table 14.5. Note that, as predicted, the equilibrium constant for this exothermic reaction decreases rapidly with increasing temperature. The data from Table 14.5 is shown in Figure 14.5, clearly showing the contrast between the endothermic reaction and the exothermic reaction. The slope of the graph is positive for the exothermic reaction and negative for the endothermic reaction. From the equation, this is a general result for all reactions. One of our goals at the outset was to determine whether it is possible to control the equilibrium which occurs during a gas reaction. We might want to force a reaction to produce as much of the products as possible. In the alternative, if there are unwanted by-products of a reaction, we might want conditions which minimize the product. We have observed that the amount of product varies with the quantities of initial materials and with changes in the temperature. Our goal is a systematic understanding of these variations. A look back at Tables 14.1 and 14.2 shows that the equilibrium pressure of the product of the reaction increases with increasing the initial quantity of reaction. This seems quite intuitive. Less intuitive is the variation of the equilibrium pressure of the product of this reaction with variation in the volume of the container, as shown in Table 14.3. Note that the pressure of \(\ce{NH_3}\) decreases by more than a factor of ten when the volume is increased by a factor of ten. This means that, at equilibrium, there are fewer moles of \(\ce{NH_3}\) produced when the reaction occurs in a larger volume. To understand this effect, we rewrite the equilibrium constant in the equation to explicitly show the volume of the container. This is done by applying , so that each partial pressure is given by the Ideal Gas Law: \[\begin{array}{rcl} K_p & = & \frac{n^2_{NH_3} \left( \frac{RT}{V} \right)^2}{n_{N_2} \frac{RT}{V} n^3_{H_2} \left( \frac{RT}{V} \right)^3} \\ & = & \frac{n^2_{NH_3}}{n_{N_2} n^3_{H_2} \left( \frac{RT}{V} \right)^3} \end{array}\] Therefore, \[K_p \left( \frac{RT}{V} \right)^2 = \frac{n^2_{NH_3}}{n_{N_2} n^2_{H_2}}\] This form of the equation makes it clear that, when the volume increases, the left side of the equation decreases. This means that the right side of the equation must decrease also, and in turn, \(n_{NH_3}\) must decrease while \(n_{N_2}\) and \(n_{H_2}\) must increase. The equilibrium is thus shifted from products to reactants when the volume increases for the synthesis of ammonia. The effect of changing the volume must be considered for each specific reaction, because the effect depends on the stoichiometry of the reaction. One way to determine the consequence of a change in volume is to rewrite the equilibrium constant as we have done in the equation above. Finally, we consider changes in temperature. We note that \(K_p\) increases with \(T\) for endothermic reactions and decreases with \(T\) for exothermic reactions. As such, the products are increasingly favored with increasing temperature when the reaction is endothermic, and the reactants are increasingly favored with increasing temperature when the reaction is exothermic. On reflection, we note that when the reaction is exothermic, the reverse reaction is endothermic. Putting these statements together, we can say that the reaction equilibrium always shifts in the direction of the endothermic reaction when the temperature is increased. All of these observations can be collected into a single unifying concept known as . This statement is best understood by reflection on the types of "stresses" we have considered in this section. When a reactant is added to a system at equilibrium, the reaction responds by consuming some of that added reactant as it establishes a new equilibrium. This offsets some of the stress of the increase in reactant. When the temperature is raised for a reaction at equilibrium, this adds thermal energy. The system shifts the equilibrium in the endothermic direction, thus absorbing some of the added thermal energy, countering the stress. The most challenging of the three types of stress considered in this section is the change in volume. By increasing the volume containing a gas phase reaction at equilibrium, we reduce the partial pressures of all gases present and thus reduce the total pressure. Recall that the response of the synthesis of ammonia to the volume increase was to create more of the reactants at the expense of the products. One consequence of this shift is that more gas molecules are created, and this increases the total pressure in the reaction flask. Thus, the reaction responds to the stress of the volume increase by partially offsetting the pressure decrease with an increase in the number of moles of gas at equilibrium. Le Chatelier's principle is a useful mnemonic for predicting how we might increase or decrease the amount of product at equilibrium by changing the conditions of the reaction. From this principle, we can predict whether the reaction should occur at high temperature or low temperature, and whether it should occur at high pressure or low pressure. In the data given for equilibrium of the reaction of hydrogen gas and iodine gas, there is no volume given. Show that changing the volume for the reaction does not change the number of moles of reactants and products present at equilibrium, i.e. changing the volume does not shift the equilibrium. For the decomposition of \(\ce{N_2O_4}\) the number of moles of \(\ce{NO_2}\) at equilibrium increases if we increase the volume in which the reaction is contained. Explain why this must be true in terms of dynamic equilibrium and give a reason why the rates of the forward and reverse reactions might be affected differently by changes in the volume. We could balance the synthesis of ammonia equation by writing \[2 \ce{N_2} \left( g \right) + 6 \ce{H_2} \left( g \right) \rightarrow 4 \ce{NH_3} \left( g \right)\] Write the form of the equation constant for the reaction balanced as in the equation above. What is the value of the equilibrium constant? (Refer to Table 14.5.) Of course, the pressures at equilibrium do not depend on how the equation is balanced. Explain why this is true, even though the equilibrium constant can be written differently and have a different value. Show that the equilibrium constant \(K_p\) for the synthesis of ammonia can be written in terms of the concentrations or particle densities, e.g. \(\left[ \ce{N_2} \right] = \frac{n_{N_2}}{V}\), instead of the partial pressures. In this form, we call the equilibrium constant \(K_c\). Find the relationship between \(K_p\) and \(K_c\), and calculate the value of \(K_c\). For each of these reactions, predict whether increases in temperature will shift the reaction equilibrium more towards products or more towards reactants. \(2 \ce{CO} \left( g \right) + \ce{O_2} \left( g \right) \rightarrow 2 \ce{CO_2} \left( g \right)\) \(\(\ce{O_3} \left( g \right) + \ce{NO} \left( g \right) \rightarrow \ce{NO_2} \left( g \right) + \ce{O_2} \left( g \right)\) \(2 \ce{O_3} \left( g \right) \rightarrow 3 \ce{O_2} \left( g \right)\) Plot the data in Table 14.4 on a graph showing \(K_p\) on the y-axis and \(T\) on the x-axis. The shape of this graph is reminiscent of the graph of another physical property as a function of increasing temperature. Identify that property, and suggest a reason why the shapes of the graphs might be similar. Using Le Chatelier's principle, predict whether the specified "stress" will produce an increase or a decrease in the amount of product observed at equilibrium for the reaction: \[2 \ce{H_2} \left( g \right) + \ce{CO} \left( g \right) \rightarrow \ce{CH_3OH} \left( g \right)\] \[\Delta H^0 = -91 \: \frac{\text{kJ}}{\text{mol}}\] Volume of container is increased. Helium is added to container. Temperature of container is raised. Hydrogen is added to container. \(\ce{CH_3OH}\) is extracted from container as it is formed. ; Chemistry) | 24,471 | 3,449 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Addition_by_Electrophilic_Reagents |
A carbon-carbon triple bond may be located at any unbranched site within a carbon chain or at the end of a chain, in which case it is called . Because of its linear configuration (the bond angle of a sp-hybridized carbon is 180º), a ten-membered carbon ring is the smallest that can accommodate this function without excessive strain. Since the most common chemical transformation of a carbon-carbon double bond is an addition reaction, we might expect the same to be true for carbon-carbon triple bonds. Indeed, most of the also take place with alkynes, and with similar regio- and stereoselectivity. When the addition reactions of electrophilic reagents, such as strong Brønsted acids and halogens, to alkynes are studied we find a curious paradox. The reactions are even more exothermic than the additions to alkenes, and yet the rate of addition to alkynes is slower by a factor of 100 to 1000 than addition to equivalently substituted alkenes. The reaction of one equivalent of bromine with 1-penten-4-yne, for example, gave 4,5-dibromo-1-pentyne as the chief product. HC≡C-CH -CH=CH + → HC≡C-CH -CH CH Although these electrophilic additions to alkynes are sluggish, they do take place and generally display Markovnikov Rule regioselectivity and anti-stereoselectivity. One problem, of course, is that the products of these additions are themselves substituted alkenes and can therefore undergo further addition. Because of their high electronegativity, halogen substituents on a double bond act to reduce its nucleophilicity, and thereby decrease the rate of electrophilic addition reactions. Consequently, there is a delicate balance as to whether the product of an initial addition to an alkyne will suffer further addition to a saturated product. Although the initial alkene products can often be isolated and identified, they are commonly present in mixtures of products and may not be obtained in high yield. The following reactions illustrate many of these features. In the last example, 1,2-diodoethene does not suffer further addition inasmuch as vicinal-diiodoalkanes are relatively unstable. As a rule, electrophilic addition reactions to alkenes and alkynes proceed by initial formation of a , in which the electrophile accepts electrons from and becomes weakly bonded to the multiple bond. Such complexes are formed reversibly and may then reorganize to a reactive intermediate in a slower, rate-determining step. Reactions with alkynes are more sensitive to solvent changes and catalytic influences than are equivalent alkenes. For examples and a discussion of mechanisms click here. Why are the reactions of alkynes with electrophilic reagents more sluggish than the corresponding reactions of alkenes? After all, addition reactions to alkynes are generally more exothermic than additions to alkenes, and there would seem to be a higher π-electron density about the triple bond ( two π-bonds versus one ). Two factors are significant in explaining this apparent paradox. First, although there are more π-electrons associated with the triple bond, the sp-hybridized carbons exert a strong attraction for these π-electrons, which are consequently bound more tightly to the functional group than are the π-electrons of a double bond. This is seen in the ionization potentials of ethylene and acetylene. As defined by the preceding equations, an is the minimum energy required to remove an electron from a molecule of a compound. Since pi-electrons are less tightly held than sigma-electrons, we expect the ionization potentials of ethylene and acetylene to be lower than that of ethane, as is the case. Gas-phase proton affinities show the same order, with ethylene being more basic than acetylene, and ethane being less basic than either. Since the initial interaction between an electrophile and an alkene or alkyne is the formation of a pi-complex, in which the electrophile accepts electrons from and becomes weakly bonded to the multiple bond, the relatively slower reactions of alkynes becomes understandable. A second factor is presumed to be the stability of the carbocation intermediate generated by sigma-bonding of a proton or other electrophile to one of the triple bond carbon atoms. This intermediate has its positive charge localized on an unsaturated carbon, and such are less stable than their saturated analogs. Indeed, we can modify our earlier ordering of carbocation stability to include these vinyl cations in the manner shown below. It is possible that vinyl cations stabilized by conjugation with an aryl substituent are intermediates in HX addition to alkynes of the type Ar-C≡C-R, but such intermediates are not formed in all alkyne addition reactions. Application of the Hammond postulate indicates that the activation energy for the generation of a vinyl cation intermediate would be higher than that for a lower energy intermediate. This is illustrated for alkenes versus alkynes by the following energy diagrams. Despite these differences, electrophilic additions to alkynes have emerged as exceptionally useful synthetic transforms. For example, addition of HCl, acetic acid and hydrocyanic acid to acetylene give respectively the useful monomers vinyl chloride, vinyl acetate and acrylonitrile, as shown in the following equations. Note that in these and many other similar reactions transition metals, such as copper and mercury salts, are effective catalysts. Complexes formed by alkenes and alkynes with transition metals are different from the simple pi-complexes noted above. Here a synergic process involving donation of electrons from a filled π-orbital of the organic ligand into an empty d-orbital of the metal, together with back-donation of electrons from another d-orbital of the metal into the empty π*-antibonding orbital of the ligand. | 5,822 | 3,450 |
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There are many examples of reactions that involve catalysis. One that is of current importance to the chemistry of the environment is the catalytic decomposition of ozone (Fahey, 2006). The overall reaction \[ O_3 + O^{\cdot} \xrightarrow{} 2 O_2 \nonumber \] can be catalyze by atomic chlorine by the following mechanism. \[ O_3 + Cl \xrightarrow{k_1} ClO + O_2 \nonumber \] \[ ClO + O \xrightarrow{k_1} Cl + O_2 \nonumber \] The rate of change of the intermediate (\(ClO\)) concentration is given by \[ \dfrac{[ClO]}{dt} = k_1 [ O_3,Cl] - k_2 [ClO,O] \nonumber \] Applying the steady state approximation to this relationship and solving for \([ClO]\) produces \[ClO] =\dfrac{[O_3,Cl]}{k_2[O]} \label{clo} \] The rate of production of \(O_2\) (which is two times the rate of the reaction) is given by \[\dfrac{d[O_2]}{dt} = k_2[O_3,Cl] + k_2[ClO,O] \nonumber \] Substituting the expression for \([ClO]\) (Equation \ref{clo}) into the above expression yields \[\dfrac{d[O_2]}{dt} = k_2[O_3,Cl] + k_2 \left(\dfrac{[ O_3,Cl]}{k_2[O]} \right) [O] \nonumber \] \[ = k_1 [O_3[Cl] + k_1[O_3,Cl] \nonumber \] \[ = 2k_1[O_3,Cl] \nonumber \] And so the rate of the reaction is predicted to be first order in \([O_3]\), first order in the catalyst \([Cl]\), and second order overall. \[\text{rate} = k[O_3,Cl] \nonumber \] If the concentration of the catalyst is constant, the reaction kinetics will reduce to first order. \[\text{rate} = k[O_3] \nonumber \] This catalytic cycle can be represented in the following diagram: On the left, atomic oxygen picks up an oxygen atom from \(ClO\) to form \(O_2\) and generate a \(Cl\) atom, which can then react with \(O_3\) to form \(ClO\) and an \(O_2\) molecule. The closed loop in the middle is characteristic of the catalytic cycle involving \(Cl\) and \(ClO\). Further, since \(Cl\) acts as a catalyst, it can decompose many \(O_3\) molecules without being degraded through side reactions. The introduction of chlorine atoms into the upper atmosphere is a major environmental problem, leading to the annual thinning and eventual opening of the ozone layer over Antarctica. The source of chlorine is from the decomposition of chlorofluorocarbons which are sued as refrigerants and propellants due to their incredible stability near the Earth’s surface. However, in the upper atmosphere, these compounds are subjected to ultraviolet radiation emitted by the sun and decompose to form the radicals responsible for the catalytic decomposition of ozone. The world community addressed this issue by drafting the , which focused on the emission of ozone-destroying compounds. The result of this action has brought about evidence of the Antarctic ozone hole healing (K, 2015). This is one very good example science-guided political, industrial, and economic policies leading to positive changes for our environment. | 2,859 | 3,451 |
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The oxygenase enzymes catalyze reactions of dioxygen with organic substrates in which oxygen atoms from dioxygen are incorporated into the final oxidized product. These enzymes can be divided into dioxygenases, which direct both atoms of oxygen into the product (Reaction 5.53), and monooxygenases, where one atom of oxygen from dioxygen is found in the product and the other has been reduced to water (Reaction 5.54): \[Dioxygenase:\; substrate + \;^{*}O_{2} \rightarrow substrate(^{*}O)_{2} \tag{5.53}\] \[Monoxygenase:\; substrate + \;^{*}O_{2} + 2 H^{+} + 2e^{-} \rightarrow substrate(^{*}O) + H_{2}\;^{*}O \tag{5.54}\] Dioxygenase enzymes are known that contain heme iron, nonheme iron, copper, or manganese. The substrates whose oxygenations are catalyzed by these enzymes are very diverse, as are the metal-binding sites; so probably several, possibly unrelated, mechanisms operate in these different systems, For many of these enzymes, there is not yet much detailed mechanistic information. However, some of the intradiol catechol dioxygenases isolated from bacterial sources have been studied in great detail, and both structural and mechanistic information is available. These are the systems that will be described here. The role of these nonheme iron-containing enzymes is to catalyze the degradation of catechol derivatives to give muconic acids (Reaction 5.55, for example). The enzymes are induced when the only carbon sources available to the bacteria are aromatic molecules. The two best-characterized members of this class are catechol 1,2-dioxygenase (CTD) and protocatechuate 3,4-dioxygenase (PCD). \(\tag{5.55}\) Even before the x-ray crystal structure of PCD was obtained, a picture of the active site had been constructed by detailed spectroscopic work using a variety of methods. The success of the spectroscopic analyses of these enzymes is a particularly good example of the importance and usefulness of such methods in the characterization of metalloproteins. The two enzymes referred to in Reaction (5.55) have different molecular weights and subunit compositions, but apparently contain very similar active-site structures and function by very similar mechanisms. In both, the resting state of the enzyme contains one Fe ion bound at the active site. EPR spectra show a resonance at g = 4.3, characteristic of high-spin Fe in a so-called rhombic (low symmetry) environment, and the Mössbauer parameters are also characteristic of high-spin ferric. Reactions with substrate analogues (see below) cause spectral shifts of the iron chromophore, suggesting strongly that the substrate binds directly to the iron center in the course of the enzymatic reaction. It is straightforward to rule out the presence of heme in these enzymes, because the heme chromophore has characteristic electronic-absorption bands in the visible and ultraviolet regions with high extinction coefficients, which are not observed for these proteins. Likewise, the spectral features characteristic of other known cofactors or iron-sulfur centers are not observed. Instead, the dominant feature in the visible absorption spectrum is a band with a maximum near 460 nm and a molar extinction coefficient of 3000 to 4000 M cm per iron (see Figure 5.5). This type of electronic absorption spectrum is characteristic of a class of proteins, sometimes referred to as iron-tyrosinate proteins, that contain tyrosine ligands bound to iron(III) in their active sites, and which consequently show the characteristic visible absorption spectrum due to phenolate-to-iron(III) charge-transfer transitions. This assignment can be definitively proven by examination of the resonance Raman spectrum, which shows enhancement of the characteristic tyrosine vibrational modes (typically ~1170, 1270, 1500, and 1600 cm ) when the sample is irradiated in the charge-transfer band described above. Ferric complexes of phenolate ligands may be seen to give almost identical resonance Raman spectra (see Figure 5.6). These bands have been assigned as a C—H bending vibration and a C—O and two C—C stretching vibrations of the phenolate ligand. In addition, NMR studies of the relaxation rates of the proton spins of water indicate that water interacts with the paramagnetic Fe center in the enzyme. This conclusion is supported by the broadening of the Fe EPR signal in the presence of H O, due to interaction with the I = \(\frac{5}{2}\) nuclear spin of O. Thus numerous spectroscopic studies of the catechol dioxygenases led to the prediction that the high-spin ferric ion was bound to tyrosine ligands and water. In addition, EXAFS data, as well as the resemblance of the spectral properties to another, better characterized iron-tyrosinate protein, i.e., transferrin (see Chapter 1), suggested that histidines would also be found as ligands to iron in these proteins. Preliminary x-ray crystallographic results on protocatechuate 3,4-dioxygenase completely support the earlier predictions based on spectroscopic studies. The Fe center is bound to two histidine and two tyrosine ligands and a water, the five ligands being arranged in a trigonal bipyramidal arrangement, with a tyrosine and a histidine located in axial positions, and with the equatorial water or hydroxide ligand facing toward a cavity assumed to be the substrate-binding cavity. The cavity also contains the positively charged guanidinium group of an arginine side chain, in the correct position to interact with the negatively charged carboxylate group on the protocatechuate substrate (see Figure 5.7). As mentioned above, substrates and inhibitors that are substrate analogues bind to these enzymes and cause distinct changes in the spectral properties, suggesting strongly that they interact directly with the Fe center. Nevertheless, the spectra remain characteristic of the Fe oxidation state, indicating that the ferric center has not been reduced. Catecholates are excellent ligands for Fe (see, for example, the catecholate siderophores, Chapter 1) and it might therefore be assumed that the catechol substrate would bind to iron using both oxygen atoms (see 5.56). \(\tag{5.56}\) However, the observation that phenolic inhibitors p-X-C H -OH bind strongly to the enzymes suggested the possibility that the substrate binds to the iron center through only one oxygen atom (see 5.57). These results contradict an early hypothesis that the mode of substrate binding, i.e., monodentate versus bidentate, might be a crucial factor in activating the substrate for reaction with dioxygen. Spectroscopic observations of the enzymes during reactions with substrates and substrate analogues have enabled investigators to observe several intermediates along the catalytic pathway. Such studies have led to the conclusion that the iron center remains high-spin Fe throughout the entire course of the reaction. This conclusion immediately presents a problem in understanding the nature of the interaction of dioxygen with the enzyme, since dioxygen does not in general interact with highly oxidized metal ions such as Fe . The solution seems to be that this reaction represents an example of rather than activation. Studies of the oxidation of ferric catecholate coordination complexes have been useful in exploring mechanistic possibilities for these enzymes. A series of ferric complexes of 3,5-di-t-butyl-catechol with different ligands L have been found to react with O to give oxidation of the catechol ligand (Reaction 5.58) All these studies of the enzymes and their model complexes have led to the mechanism summarized in Figure 5.9. In this proposed mechanism, the catechol substrate coordinates to the ferric center in either a monodentate or a bidentate fashion, presumably displacing the water or hydroxide ligand. The resulting catechol complex then reacts with dioxygen to give a peroxy derivative of the substrate, which remains coordinated to Fe . The subsequent rearrangement of this peroxy species to give an anhydride intermediate is analogous to well-characterized reactions that occur when catechols are reacted with alkaline hydrogen peroxide. The observation that both atoms of oxygen derived from O are incorporated into the product requires that the ferric oxide or hydroxide complex formed in the step that produces the anhydride does not exchange with external water prior to reacting with the anhydride to open it up to the product diacid. It is interesting to consider how the intradiol dioxygenase enzymes overcome the kinetic barriers to oxidations by dioxygen, and why this particular mechanism is unlikely to be applicable to the monooxygenase enzymes. The first point is that the ferric catechol intermediate is paramagnetic, with resonance forms that put unpaired electron density onto the carbon that reacts with dioxygen. The spin restriction is therefore not a problem. In addition, the catechol ligand is a very good reducing agent, much more so than the typical substrates of the monooxygenase enzymes (see next section). It is possible, therefore, that the reaction of dioxygen with the ferric catechol complex results in a concerted two-electron transfer to give a peroxy intermediate, thus bypassing the relatively unfavorable one-electron reduction of O . | 9,272 | 3,452 |
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The explosive decomposition of nitrogen triiodide shown in the video is an exciting (and forceful) example of chemistry in action. For many of us, this explosion is the work of who work magic at the lab bench. In reality, chemistry is much broader. The transformations of matter can be as fast and explosive as the video shows, or involve the slow decay of living organisms into fossil fuels. The compounds around us in everyday life have many properties that would have been conceived as only a few centuries ago. Could you imagine describing to previous generations or ? Could you imagine what they would say seeing the decomposition of nitrogen triiodide? The science of chemistry is concerned with the composition, properties, and structure of matter and with the ways in which substances can change from one form to another. Since anything that has mass and occupies space can be classified as matter, this means that chemistry is involved with almost everything in the universe. But this definition is too broad to be useful. Chemistry isn't the only science that deals with the composition and transformations of matter. Some matter is composed of cells, which transform by meiosis and other processes that biologists study. Matter is also composed of subatomic particles called leptons, which transform by processes like annihilation studied by physicists. Chemists are unique because they understand or explain everything, from our bodies to our universe, in terms of the properties of just over 100 kinds of atoms found in all matter and the amazing variety of molecules and other atomic-scale structures that are created by forming and breaking bonds between atoms. In the video explosion, bonds break in nitrogen triiodide (NI ) molecules and the atoms recombine to form nitrogen (N ) and iodine (I ). . Chemistry explains or understands any subject in terms of the properties of atoms and molecules. Chemistry, in other words, is not just something that happens in laboratories. It is a unique perspective, or way of understanding, all that is around us, and even inside us. Chemistry is going on in places as diverse as the smallest bacterium, a field of ripening wheat, a modem manufacturing plant, the biospheres of planets such as Earth, the vast reaches of interstellar space, and even your eyes and brain as you read these words. ChemPRIME recognizes that the chemical perspective can add to our understanding of anything that catches our interest. Our goal will be to add another dimension--understanding in terms of the properties of atoms and molecules and how they interact--to many subjects, thereby making it clear how the study of chemistry is of importance to a wide variety of people. Biologists, for example, have examined smaller and smaller organisms, cells, and cell components, until, in the study of viruses and genes, they joined forces with chemists who were interested in larger and larger molecules. The result was a new inter-disciplinary field called molecular biology, and a reinforcement of the idea that living organisms are complicated, highly organized chemical systems. Chemists interact in similar ways with scientists in areas such as chemical physics, geochemistry, pharmacology, toxicology, ecology, meteorology, oceanography, and many others. Current practice in these fields is such that a person lacking basic chemical knowledge is at a severe disadvantage, because the perspective of the molecular level has become so important. Chemistry also underlies a great deal of modern technology. The manufacture of such basic commodities as steel, aluminum, glass, plastics, paper, textiles, and gasoline all involve chemists and chemistry. Without the abundant cheap supply of these and other substances that chemistry has helped to produce, our lives would be much less comfortable. However, as we are now beginning to discover, callous and indiscriminate use of technology can produce disadvantages as well as benefits. Automobiles, power plants, and industrial processes spew into the air harmful substances that are not always easy or cheap to eliminate. Rivers and lakes are also more easily contaminated than we once thought - substances once believed harmless now have proven to be the opposite. Issues involving the effects of technology on the environment affect everyone - not just scientists. Decisions about them are political, at least in part, and require some chemical knowledge on the part of voters as well as their elected representatives. At the very least, a citizen needs to be able to distinguish valid and invalid arguments put forward by scientific “experts” regarding such issues. (In some cases such “expertise” may be mainly a willingness to speak out rather than a command of the scientific and political issues.) It is to be hoped also that more persons will follow the example of Russell Peterson, formerly a researcher in a large chemical company, who served as Governor of Delaware and Chairman of the President’s Council on Environmental Quality. Only by a combination of scientists willing to leave their laboratories and citizens willing to master some of the basics of science can intelligent political decisions be made in a democratic society. Indeed, such a combination may be a necessity if democracy is not to degenerate into an oligarchy ruled by those who control the experts. Given the universality of chemistry, its central role among the sciences, and its importance in modern life, how is it possible to learn much about it in a short time? If everything has a chemical aspect, because atoms and molecules can aid in understanding everything, is the field of chemistry so broad and all-encompassing that one cannot master enough to make its study worth-while? We think the answer to this second question is a resounding no! This entire book has been designed to help you learn a good deal of chemistry in a short time. If it is successful, the first question will have been answered as well. An important and valuable technique of science and scientists is that of subdividing large, seemingly unsolvable problems into smaller, simpler parts. If the latter are chosen carefully, each can be mastered. Individual small advances can then be combined to yield an important, more complicated result. Thus chemists have not acted on the assumption that because all the world may be understood in terms of chemicals and chemistry, they should try to study it all at once. Rather, many chemists do much of their work under controlled, laboratory conditions, advancing in small steps toward more general, useful, and exciting results. Since people’s process of studying and understanding chemistry is far from complete, we can narrow our area of interest considerably by redefining chemistry as those things that chemists can explain on the molecular level, or are attempting to explain in terms of atoms and molecules and their properties. This more restricted view constitutes the main theme of this book. We hope that when you have finished with it, you will have a solid background in the facts, laws, and theories of chemistry, as well as those modes of behavior and thinking that chemists have found useful in solving the problems they have faced. | 7,259 | 3,453 |
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In all calculations of this sort we need to concentrate on the Since this quantity is unknown, we label it algebraically, calling it mol, where indicates a pure number. In the case under consideration, we would label the amount of cis isomer produced as the system moves to equilibrium as mol. Once this step has been taken, the amount of each of the other products and reactants transformed by the reaction can be deduced from the equation and the appropriate stoichiometric factors. In our current case the equation is We can now construct a table showing the initial amounts, the amounts transformed by reaction, the amounts present at equilibrium, and finally the equilibrium concentrations of each product and reagent: Once the final concentration of each species has been obtained in this way, an algebraic equation can be set up linking the equilibrium concentrations to the value of the equilibrium constant: \begin{align}K_{c}=\text{0}\text{.500}=\frac{\text{ }\!\![\!\!\text{ }trans\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ }cis\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }}=\frac{{\text{((1}-x\text{)}}/{\text{10) mol/L}}\;}{{\text{((1 + }x\text{)}}/{\text{10) mol/L}}\;} \\ \text{0}\text{.500}=\frac{\text{1}-x}{\text{1 + }x} \\ \end{align} At this stage we are left with an algebraic equation to solve for . Inevitably this equation involves . If any units remain, a mistake must have been made. From by cross-multiplying, we have \(\text{1 – x} = \text{0.500 + 0.500x}\) which rearranges to \(\text{0.500} = \text{1.500x}\) so that \(x=\dfrac{\text{0}\text{.500}}{\text{1}\text{.500}x}=\text{0}\text{.333}\) We thus conclude that 0.333 mol trans isomer is converted to the cis form when the original mixture is allowed to equilibrate. The final equilibrium concentrations are as follows: \[\text{ }\!\![\!\!\text{ }cis\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }=\frac{\text{1 + }x}{\text{10}}\text{ mol/L}=\frac{\text{1}\text{.333}}{\text{10}}\text{ mol/L} \nonumber \] \[[cis\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}]=\text{0}\text{.1333 mol/L}\, \nonumber \] While \[\text{ }\!\![\!\!\text{ }trans\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }=\frac{\text{1}-x}{\text{10}}\text{ mol/L}=\text{0}\text{.0667 mol/L} \nonumber \] We can easily cross check that the ratio of these two concentrations is actually equal to the equilibrium constant, that is, to 0.5. In calculating the extent of a chemical reaction from an equilibrium constant, it is often useful to realize that if the equilibrium constant is very small, the reaction proceeds to only a limited extent, while if it is very large, the reaction goes almost to completion. This point is easiest to see in the case of an equilibrium between two isomers of the type \[\text{A} \rightleftharpoons \text{B} \nonumber \] If for this reaction is very small, say 10 , then the ratio [B]/[A] = 10 . There will thus be a million times more molecules of the A isomer than of the B isomer in the equilibrium mixture. For most purposes we can regard the equilibrium mixture as being pure A. Conversely if has a very large value like 10 , the very opposite is true. In an equilibrium mixture governed by this second constant, there would be a million times more B isomer than A isomer, and for most purposes the equilibrium mixture could be regarded as pure B. The realization that an equilibrium mixture can contain only small concentrations of some of the reactants or products is often very useful in solving equilibrium problems, as the 3 example below demonstrates. When colorless hydrogen iodide gas is heated, a beautiful purple color appears, indicating that some iodine gas has been produced and that the compound has decomposed partially into its elements according to the equation \[\text{2HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \nonumber \] At 745 K (471.8°C), for this reaction has the value 0.0200. Calculate the concentration of I produced when 1.00 mol HI is heated to this temperature in a flask of volume 10.0 dm . Also calculate the fraction of the HI which has dissociated. Let us denote the amount of I produced by the reaction as mol. The equation then tells us that the amount of H produced will also be mol, while the amount of HI consumed by the decomposition will be 2 mol. The initial amount of HI, 1 mol, will thus be reduced to(1 – 2 ) mol at equilibrium. Dividing the above amounts by the volume 10 dm³, we easily obtain the equilibrium concentrations We can now write an expression for the equilibrium constant \[\text{K}_{c}=\frac{\text{ }\!\![\!\!\text{ H}_{\text{2}}\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ I}_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ HI }\!\!]\!\!\text{ }^{\text{2}}}=\frac{{\text{(}x}/{\text{10) mol dm}^{-\text{3}}\text{ }\times \text{ }{\text{(}x}/{\text{10) mol dm}^{-\text{3}}\text{ }}\;}\;}{{\text{((1}-\text{2}x\text{)}}/{\text{10) mol dm}^{-\text{3}}\text{ }\times \text{ }{\text{((1}-\text{2}x\text{)}}/{\text{10) mol dm}^{-\text{3}}}\;}\;} \nonumber \] or \[\text{0}\text{.0200}=\frac{x^{\text{2}}}{\text{(1 }-\text{ 2}x\text{)}^{\text{2}}} \nonumber \] which is the required algebraic expression, free of units. This equation is easily solved by taking the square root of both sides. \(\sqrt{\text{2 }\times \text{ 10}^{-\text{2}}}=\sqrt{\text{2}}\text{ }\times \text{ 10}^{-\text{1}}=\text{0}\text{.1414}=\frac{x^{\text{2}}}{\text{1 }-\text{ 2}x}\) Thus \(\text{0.1414 – 0.2828x} = \text{x}\) or \(\text{1.2828x} = \text{0.1414}\) so that \(x=\frac{\text{0}\text{.1414}}{\text{1}\text{.2828}}=\text{0}\text{.110}\) Thus \(\text{ }\!\![\!\!\text{ I}_{\text{2}}\text{ }\!\!]\!\!\text{ }=\frac{x}{\text{10}}\text{mol dm}^{-\text{3}}=\text{1}\text{.10 }\times \text{ 10}^{-\text{2}}\text{ mol dm}^{-\text{3}}\) Since 2 mol HI dissociated and 1 mol HI was originally present, we concluded that the fraction of HI which dissociated is \[\frac{\text{2}x\text{ mol}}{\text{1 mol}}=\text{0}\text{.220} \nonumber \] It is wise at this point to check the answer. We found = 0.110. If this is the correct value, we should also find that \[\frac{x^{\text{2}}}{\text{(1}-\text{2}x\text{)}^{\text{2}}}=\text{0}\text{.0200} \nonumber \] The value obtained using a calculator is 0.019 888 2 (which rounds to 0.0199 to three significant figures). The difference is due to errors introduced by rounding off during the calculation. A mixture of 1.00 mol HI pas and 1.00 mol H gas is heated in a 10.0-dm³ flask to 745 K. Calculate the concentration of I produced in the equilibrium mixture and also the fraction of HI which dissociates. Again we let mol represent the amount of I produced. Our table then becomes Substituting the final concentrations in an expression for the equilibrium constant, we then have \[K_{c}=\frac{\text{ }\!\![\!\!\text{ H}_{\text{2}}\text{ }\text{ }]\text{ }\!\!\!\!\text{ }\text{ }\!\!,\!\!\text{ I }_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ HI }\!\!]\!\!\text{ }^{\text{2}}}=\frac{{\text{(1 + }x}/{\text{10) mol dm}^{-\text{3}}\text{ }\times \text{ }{\text{(}x}/{\text{10) mol dm}^{-\text{3}}\text{ }}\;}\;}{{\text{((1}-\text{2}x\text{)}}/{\text{10) mol dm}^{-\text{3}}\text{ }\times \text{ }{\text{((1}-\text{2}x\text{)}}/{\text{10) mol dm}^{-\text{3}}}\;}\;} \nonumber \] or \[\text{0}\text{.0200}=\frac{\text{(1 + }x\text{)}x}{\text{(1 }-\text{ 2}x\text{)}^{\text{2}}} \nonumber \] Because of the added H , it is no longer possible to take a square root as in the previous example. Instead we need to multiply out and rearrange in order to obtain a quadratic equation of the form + + = 0. Accordingly we have \(\text{0.0200(1 – 2x)}^{2} = \text{(1 + x)x}\) or \(\text{0.0200(1 – 4x + 4x}^{2}) = \text{x} + \text{x}^{2}\) multiplying both sides by 50, we obtain \(\text{1 – 4x} + \text{4x}^{2} = \text{50x} + \text{50x}^{2}\) which on rearrangement has the required form \[\text{46x}^{2} + \text{54x – 1} = \text{0} \nonumber \] where We can now use the conventional quadratic formula \(x=\frac{-\text{b }\!\!\pm\!\!\text{ }\sqrt{\text{b}^{\text{2}}-\text{4}ac}}{\text{2}a}=\frac{-\text{54 }\!\!\pm\!\!\text{ }\sqrt{\text{54}^{\text{2}}\text{ + 4 }\times \text{ 46 }\times \text{ 1}}}{\text{2 }\times \text{ 46}}\) \(=\frac{-\text{54 }\!\!\pm\!\!\text{ 55}\text{.678}}{\text{92}}=\text{0}\text{.0182 or }-\text{1}\text{.192}\) The negative root, = – 1.192, implies that 1.192 mol I was consumed. Since no I was present to begin with, this is impossible. We conclude that the positive root, namely = 0.0182, is the correct one. Thus \[\text{ }\!\![\!\!\text{ I}_{\text{2}}\text{ }\!\!]\!\!\text{ }=\frac{x}{\text{10}}\text{mol dm}^{-\text{3}}=\text{1}\text{.82 }\times \text{ 10}^{-\text{3}}\text{ mol dm}^{-\text{3}} \nonumber \] Again the fraction dissociated is given by 2 mol/1 mol and is thus equal to 0.0364. To check this solution, we can substitute = 0.0182 in the expression \[\frac{\text{(1 + }x\text{)}x}{\text{(1}-\text{2}x\text{)}^{\text{2}}} \nonumber \] We then obtain the value 0.019 96, which rounds to the correct value of 0.0200. The inclusion of one of the products (H gas) in the mixture reduces the extent to which the hydrogen iodide dissociates quite appreciably. We will explore this phenomenon more extensively in the next section. The equilibrium constant for the dissociation of dinitrogen tetroxide according to the equation \[\text{N}_2\text{O}_4(g) \rightleftharpoons \text{2NO}_2(g) \nonumber \] changes from a very small to a very large value as the temperature is increased, as shown in the table. Calculate the fraction of N O dissociated at Let the amount of N O dissociated at the temperature under consideration be mol. From the equation, 2 mol NO will be produced. In tabular form we then have We thus have \[\text{K}_{c}=\frac{\text{ }\!\![\!\!\text{ NO}_{\text{2}}\text{ }\!\!]\!\!\text{ }^{\text{2}}}{\text{ }\!\![\!\!\text{ N}_{\text{2}}\text{O}_{\text{4}}\text{ }\!\!]\!\!\text{ }}=\frac{{\text{(2}x}/{\text{4) mol dm}^{-\text{3}}\text{ }\times \text{ }{\text{(2}x}/{\text{4) mol dm}^{-\text{3}}\text{ }}\;}\;}{{\text{((1}-x\text{)}}/{\text{4) mol dm}^{-\text{3}}}\;} \nonumber \] \[=\frac{\text{4}x^{\text{2}}}{\text{16}}\text{ }\times \text{ }\frac{\text{4}}{\text{1}-x}\text{ mol dm}^{-\text{3}}=\frac{x^{\text{2}}}{\text{1}-x}\text{ mol dm}^{-\text{3}} \nonumber \] At 200 K, = 1.09 × 10 mol dm , so that \(\frac{x^{\text{2}}}{\text{1}-x}\text{ mol dm}^{-\text{3}}=\text{1}\text{.09 }\times \text{ 10}^{-\text{7}}\text{ mol dm}^{-\text{3}}\) or \(\frac{x^{\text{2}}}{\text{1}-x}=\text{1}\text{.09 }\times \text{ 10}^{-\text{7}}\) Since is so small, we guess that very little N O is dissociated at this temperature. This means that is very small, and it is probably valid to make the approximation 1 – ≈ 1 (The symbol ≈ means approximately equal.) With this approximation our equation becomes \(\text{x}^{2} = \text{1.09 × 10}^{-7} mol\) or \[x =\sqrt{\text{1}\text{.09 }\times \text{ 10}^{-\text{8}}}\text{ mol}=\sqrt{\text{1}\text{.09}}\text{ }\times \text{ 10}^{-\text{4}}=\text{3}\text{.30 }\times \text{ 10}^{-\text{4}} \nonumber \] Our guess about was thus correct. It is a small number, especially in comparison with 1. The approximation 1 – ≈ 1 is valid to three decimal-place accuracy since 1 – = 0.9997. Since mol has dissociated, the fraction of N O dissociated is given by mol/1 mol = . The fraction dissociated is thus 3.30 × 10 . At 400 K, = 1.505 mol dm , so that \(\frac{x^{\text{2}}}{\text{1}-x}\text{ mol dm}^{-\text{3}}=\text{1}\text{.505 mol dm}^{-\text{3}}\) or \(\frac{x^{\text{2}}}{\text{1}-x}=\text{1}\text{.505}\) Since is not small, we can no longer use the approximation 1 – ≈ 1 Indeed, such an assumption leads to a ridiculous conclusion, since if 1 – x ≈ 1, then = 1.505, or = 1.227. This result is impossible since it tells us that more N O has dissociated (1.227 mol) than was originally present (1 mol). Accordingly we return to the orthodox method for solving quadratic equations. Multiplying out our equation, we obtain \(\text{x}^{2} = \text{1.505 – 1.505x}\) or \(\text{x}^{2} + \text{1.505x – 1.505} = \text{0}\) so that \[x=\frac{-\text{1}\text{.505 }\!\!\pm\!\!\text{ }\sqrt{\text{(1}\text{.505)}^{\text{2}}\text{ + 4(1}\text{.505)}}}{\text{2}}=\frac{-\text{1}\text{.505 }\!\!\pm\!\!\text{ 2}\text{.878}}{\text{2}} \nonumber \] Since a negative result has no physical meaning, we conclude that 0.6865 is the correct answer. (As a cross check we can feed this result back into our original equation.) The fraction of N O dissociated at this temperature is accordingly 0.6865. At 600 K, = 1.675 × 10 mol dm , so that \(\frac{x^{\text{2}}}{\text{1}-x}=\text{1}\text{.675 }\times \text{ 10}^{\text{3}}\) Since is fairly large, we guess that almost all the N O has dissociated at this temperature and that is accordingly close to 1. A valid approximation for these circumstances is then ≈1. With this approximation our equation becomes \(\frac{1}{\text{1}-x}=\text{1}\text{.675 }\times \text{ 10}^{\text{3}}\) or \(\text{1-x}=\frac{1}{\text{1}\text{.675 }\times \text{ 10}^{\text{3}}}=\text{5}\text{.97 }\times \text{ 10}^{-\text{4}}\) Thus \[\text{x} = \text{1 – 5.97 × 10}^{-4} = \text{0.999 40} \nonumber \] and our approximation is a good one. Since the fraction dissociated is , we conclude that 0.9994 of the original N O has dissociated. We did not use the approximation 1 to say that 1 – ≈ 0.This is not a productive way to solve the problem because we would end up dividing by zero. Common sense tells us that it will not work. Similarly, in part a of this example we used the fact that was very small to say that 1 – ≈ 1. Saying that ≈ 0 would not make sense because it would lead to the algebraic equation \(\frac{0}{\text{1}-x}=\text{1}\text{.09 }\times \text{ 10}^{-\text{7}}\) that is, that 0 = 1.09 × 10 – (1.09 × 10 ) . This would lead to the solution = 1 which is rather far from the original assumption. | 14,228 | 3,454 |
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Many people have a qualitative idea of what is meant by . Anyone who has made instant coffee or lemonade knows that too much powder gives a strongly flavored, highly concentrated drink, whereas too little results in a dilute solution that may be hard to distinguish from water. In chemistry, the concentration of a solution is the quantity of a solute that is contained in a particular quantity of solvent or solution. Knowing the concentration of solutes is important in controlling the stoichiometry of reactants for solution reactions. Chemists use many different methods to define concentrations, some of which are described in this section. The most common unit of concentration is , which is also the most useful for calculations involving the stoichiometry of reactions in solution. The molarity (M) . It is, equivalently, the number of millimoles of solute present in exactly 1 mL of solution: \( molarity = \dfrac{moles\: of\: solute}{liters\: of\: solution} = \dfrac{mmoles\: of\: solute} {milliliters\: of\: solution} \label{4.3.1}\) The units of molarity are therefore moles per liter of solution (mol/L), abbreviated as \(M\). An aqueous solution that contains 1 mol (342 g) of sucrose in enough water to give a final volume of 1.00 L has a sucrose concentration of 1.00 mol/L or 1.00 M. In chemical notation, square brackets around the name or formula of the solute represent the molar concentration of a solute. Therefore, \[[\rm{sucrose}] = 1.00\: M\] is read as “the concentration of sucrose is 1.00 molar.” The relationships between volume, molarity, and moles may be expressed as either \[ V_L M_{mol/L} = \cancel{L} \left( \dfrac{mol}{\cancel{L}} \right) = moles \label{4.3.2}\] or \[ V_{mL} M_{mmol/mL} = \cancel{mL} \left( \dfrac{mmol} {\cancel{mL}} \right) = mmoles \label{4.3.3}\] Figure 4.3.1 illustrates the use of Equation 4.3.2 and Equation 4.3.3. Calculate the number of moles of sodium hydroxide (NaOH) in 2.50 L of 0.100 M NaOH. identity of solute and volume and molarity of solution amount of solute in moles Use either Equation 4.3.2 or Equation 4.3.3, depending on the units given in the problem. Because we are given the volume of the solution in liters and are asked for the number of moles of substance, Equation 4.3.2 is more useful: \( moles\: NaOH = V_L M_{mol/L} = (2 .50\: \cancel{L} ) \left( \dfrac{0.100\: mol } {\cancel{L}} \right) = 0 .250\: mol\: NaOH \) Calculate the number of millimoles of alanine, a biologically important molecule, in 27.2 mL of 1.53 M alanine. 41.6 mmol Concentrations are often reported on a mass-to-mass (m/m) basis or on a mass-to-volume (m/v) basis, particularly in clinical laboratories and engineering applications. A concentration expressed on an m/m basis is equal to the number of grams of solute per gram of solution; a concentration on an m/v basis is the number of grams of solute per milliliter of solution. Each measurement can be expressed as a percentage by multiplying the ratio by 100; the result is reported as percent m/m or percent m/v. The concentrations of very dilute solutions are often expressed in ( ), which is grams of solute per 10 g of solution, or in ( ), which is grams of solute per 10 g of solution. For aqueous solutions at 20°C, 1 ppm corresponds to 1 μg per milliliter, and 1 ppb corresponds to 1 ng per milliliter (Table 4.3.1). Calculations Involving Molarity (M): To prepare a solution that contains a specified concentration of a substance, it is necessary to dissolve the desired number of moles of solute in enough solvent to give the desired final volume of solution. Figure 4.3.1 illustrates this procedure for a solution of cobalt(II) chloride dihydrate in ethanol. Note that the volume of the is not specified. Because the solute occupies space in the solution, the volume of the solvent needed is almost always than the desired volume of solution. For example, if the desired volume were 1.00 L, it would be incorrect to add 1.00 L of water to 342 g of sucrose because that would produce more than 1.00 L of solution. As shown in Figure 4.3.2, for some substances this effect can be significant, especially for concentrated solutions. The solution in Figure 4.3.1 contains 10.0 g of cobalt(II) chloride dihydrate, CoCl •2H O, in enough ethanol to make exactly 500 mL of solution. What is the molar concentration of CoCl •2H O? mass of solute and volume of solution concentration (M) To find the number of moles of CoCl •2H O, divide the mass of the compound by its molar mass. Calculate the molarity of the solution by dividing the number of moles of solute by the volume of the solution in liters. The molar mass of CoCl •2H O is 165.87 g/mol. Therefore, \( moles\: CoCl_2 \cdot 2H_2O = \left( \dfrac{10.0 \: \cancel{g}} {165 .87\: \cancel{g} /mol} \right) = 0 .0603\: mol \) The volume of the solution in liters is \( volume = 500\: \cancel{mL} \left( \dfrac{1\: L} {1000\: \cancel{mL}} \right) = 0 .500\: L \) Molarity is the number of moles of solute per liter of solution, so the molarity of the solution is \( molarity = \dfrac{0.0603\: mol} {0.500\: L} = 0.121\: M = CoCl_2 \cdot H_2O \) The solution shown in Figure 4.3.2 contains 90.0 g of (NH ) Cr O in enough water to give a final volume of exactly 250 mL. What is the molar concentration of ammonium dichromate? \((NH_4)_2Cr_2O_7 = 1.43\: M\) To prepare a particular volume of a solution that contains a specified concentration of a solute, we first need to calculate the number of moles of solute in the desired volume of solution using the relationship shown in Equation 4.3.2. We then convert the number of moles of solute to the corresponding mass of solute needed. This procedure is illustrated in Example 4.3.3. The so-called D5W solution used for the intravenous replacement of body fluids contains 0.310 M glucose. (D5W is an approximately 5% solution of dextrose [the medical name for glucose] in water.) Calculate the mass of glucose necessary to prepare a 500 mL pouch of D5W. Glucose has a molar mass of 180.16 g/mol. molarity, volume, and molar mass of solute mass of solute We must first calculate the number of moles of glucose contained in 500 mL of a 0.310 M solution: \( V_L M_{mol/L} = moles \) \( 500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .155\: mol\: glucose \) We then convert the number of moles of glucose to the required mass of glucose: \( mass \: of \: glucose = 0.155 \: \cancel{mol\: glucose} \left( \dfrac{180.16 \: g\: glucose} {1\: \cancel{mol\: glucose}} \right) = 27.9 \: g \: glucose \) Another solution commonly used for intravenous injections is normal saline, a 0.16 M solution of sodium chloride in water. Calculate the mass of sodium chloride needed to prepare 250 mL of normal saline solution. 2.3 g NaCl A solution of a desired concentration can also be prepared by diluting a small volume of a more concentrated solution with additional solvent. A stock solution a commercially prepared solution of known concentration, is often used for this purpose. Diluting a stock solution is preferred because the alternative method, weighing out tiny amounts of solute, is difficult to carry out with a high degree of accuracy. Dilution is also used to prepare solutions from substances that are sold as concentrated aqueous solutions, such as strong acids. The procedure for preparing a solution of known concentration from a stock solution is shown in . It requires calculating the number of moles of solute desired in the final volume of the more dilute solution and then calculating the volume of the stock solution that contains this amount of solute. Remember that diluting a given quantity of stock solution with solvent does change the number of moles of solute present. The relationship between the volume and concentration of the stock solution and the volume and concentration of the desired diluted solution is therefore \((V_s)(M_s) = moles\: of\: solute = (V_d)(M_d)\label{4.3.4}\) where the subscripts and indicate the stock and dilute solutions, respectively. Example 5.5.4 demonstrates the calculations involved in diluting a concentrated stock solution. What volume of a 3.00 M glucose stock solution is necessary to prepare 2500 mL of the D5W solution in Example 4.3.3? volume and molarity of dilute solution volume of stock solution The D5W solution in Example 4.3.3 was 0.310 M glucose. We begin by using Equation 4.3.4 to calculate the number of moles of glucose contained in 2500 mL of the solution: \( moles\: glucose = 2500\: \cancel{mL} \left( \dfrac{1\: \cancel{L}} {1000\: \cancel{mL}} \right) \left( \dfrac{0 .310\: mol\: glucose} {1\: \cancel{L}} \right) = 0 .775\: mol\: glucose \) We must now determine the volume of the 3.00 M stock solution that contains this amount of glucose: \( volume\: of\: stock\: soln = 0 .775\: \cancel{mol\: glucose} \left( \dfrac{1\: L} {3 .00\: \cancel{mol\: glucose}} \right) = 0 .258\: L\: or\: 258\: mL \) In determining the volume of stock solution that was needed, we had to divide the desired number of moles of glucose by the concentration of the stock solution to obtain the appropriate units. Also, the number of moles of solute in 258 mL of the stock solution is the same as the number of moles in 2500 mL of the more dilute solution; . The answer we obtained makes sense: diluting the stock solution about tenfold increases its volume by about a factor of 10 (258 mL → 2500 mL). Consequently, the concentration of the solute must decrease by about a factor of 10, as it does (3.00 M → 0.310 M). We could also have solved this problem in a single step by solving Equation 4.3.4 for and substituting the appropriate values: \( V_s = \dfrac{( V_d )(M_d )}{M_s} = \dfrac{(2 .500\: L)(0 .310\: \cancel{M} )} {3 .00\: \cancel{M}} = 0 .258\: L \) As we have noted, there is often more than one correct way to solve a problem. What volume of a 5.0 M NaCl stock solution is necessary to prepare 500 mL of normal saline solution (0.16 M NaCl)? 16 mL Quantitative calculations involving reactions in solution are carried out with , however, of solutions of known concentration are used to determine the number of moles of reactants. Whether dealing with volumes of solutions of reactants or masses of reactants, the coefficients in the balanced chemical equation give the number of moles of each reactant needed and the number of moles of each product that can be produced. The flowchart for stoichiometric calculations illustrated in Figure 4.3.4 shows that in the balanced chemical equation for the reaction and the masses of solid reactants and products the volumes of solutions of reactants and products can be used to determine the amounts of other species. The balanced chemical equation for a reaction and either the masses of solid reactants and products or the volumes of solutions of reactants and products can be used in stoichiometric calculations. Gold is extracted from its ores by treatment with an aqueous cyanide solution, which causes a reaction that forms the soluble [Au(CN) ] ion. Gold is then recovered by reduction with metallic zinc according to the following equation: \[ Zn(s) + 2[Au(CN)_2]^-(aq) \rightarrow [Zn(CN)_4]^{2-}(aq) + 2Au(s) \] What mass of gold can be recovered from 400.0 L of a 3.30 × 10 M solution of [Au(CN) ] ? chemical equation and molarity and volume of reactant mass of product The equation is balanced as written; proceed to the stoichiometric calculation. Figure 4.3.2 is adapted for this particular problem as follows: As indicated in the strategy, start by calculating the number of moles of [Au(CN) ] present in the solution from the volume and concentration of the [Au(CN) ] solution: \( \begin{align} moles\: [Au(CN)_2 ]^-
& = V_L M_{mol/L} \\
& = 400 .0\: \cancel{L} \left( \dfrac{3 .30 \times 10^{4-}\: mol\: [Au(CN)_2 ]^-} {1\: \cancel{L}} \right) = 0 .132\: mol\: [Au(CN)_2 ]^- \end{align} \) Because the coefficients of gold and the [Au(CN) ] ion are the same in the balanced chemical equation, assuming that Zn(s) is present in excess, the number of moles of gold produced is the same as the number of moles of [Au(CN) ] (i.e., 0.132 mol of Au). The problem asks for the mass of gold that can be obtained, so the number of moles of gold must be converted to the corresponding mass using the molar mass of gold: \( \begin{align} mass\: of\: Au &= (moles\: Au)(molar\: mass\: Au) \\
&= 0 .132\: \cancel{mol\: Au} \left( \dfrac{196 .97\: g\: Au} {1\: \cancel{mol\: Au}} \right) = 26 .0\: g\: Au \end{align}\) At a 2011 market price of over $1400 per troy ounce (31.10 g), this amount of gold is worth $1170. \( 26 .0\: \cancel{g\: Au} \times \dfrac{1\: \cancel{troy\: oz}} {31 .10\: \cancel{g}} \times \dfrac{\$1400} {1\: \cancel{troy\: oz\: Au}} = \$1170 \) What mass of solid lanthanum(III) oxalate nonahydrate [La (C O ) •9H O] can be obtained from 650 mL of a 0.0170 M aqueous solution of LaCl by adding a stoichiometric amount of sodium oxalate? 3.89 g In Example 4.3.2, the concentration of a solution containing 90.00 g of ammonium dichromate in a final volume of 250 mL were calculated to be 1.43 M. Let’s consider in more detail exactly what that means. Ammonium dichromate is an ionic compound that contains two NH ions and one Cr O ion per formula unit. Like other ionic compounds, it is a strong electrolyte that dissociates in aqueous solution to give hydrated NH and Cr O ions: \[ (NH_4 )_2 Cr_2 O_7 (s) \xrightarrow {H_2 O(l)} 2NH_4^+ (aq) + Cr_2 O_7^{2-} (aq)\label{4.3.5} \] Thus 1 mol of ammonium dichromate formula units dissolves in water to produce 1 mol of Cr O anions and 2 mol of NH cations (see Figure 4.3.4). When carrying out a chemical reaction using a solution of a salt such as ammonium dichromate, it is important to know the concentration of each ion present in the solution. If a solution contains 1.43 M (NH ) Cr O , then the concentration of Cr O must also be 1.43 M because there is one Cr O ion per formula unit. However, there are two NH ions per formula unit, so the concentration of NH ions is 2 × 1.43 M = 2.86 M. Because each formula unit of (NH ) Cr O produces ions when dissolved in water (2NH + 1Cr O ), the concentration of ions in the solution is 3 × 1.43 M = 4.29 M. What are the concentrations of all species derived from the solutes in these aqueous solutions? molarity concentrations Classify each compound as either a strong electrolyte or a nonelectrolyte. If the compound is a nonelectrolyte, its concentration is the same as the molarity of the solution. If the compound is a strong electrolyte, determine the number of each ion contained in one formula unit. Find the concentration of each species by multiplying the number of each ion by the molarity of the solution. Because each formula unit of NaOH produces one Na ion and one OH ion, the concentration of each ion is the same as the concentration of NaOH: [Na ] = 0.21 M and [OH ] = 0.21 M. The only solute species in solution is therefore (CH ) CHOH molecules, so [(CH ) CHOH] = 3.7 M. \( In(NO _3 ) _3 (s) \xrightarrow {H_ 2 O(l)} In ^{3+} (aq) + 3NO _3^- (aq) \) One formula unit of In(NO ) produces one In ion and three NO ions, so a 0.032 M In(NO ) solution contains 0.032 M In and 3 × 0.032 M = 0.096 M NO —that is, [In ] = 0.032 M and [NO ] = 0.096 M. What are the concentrations of all species derived from the solutes in these aqueous solutions? Concentration of Ions in Solution from a Soluble Salt: The of a substance is the quantity of solute present in a given quantity of solution. Concentrations are usually expressed in terms of , defined as the number of moles of solute in 1 L of solution. Solutions of known concentration can be prepared either by dissolving a known mass of solute in a solvent and diluting to a desired final volume or by diluting the appropriate volume of a more concentrated solution (a ) to the desired final volume. ( ) | 16,010 | 3,456 |
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Carboxylic acid derivatives are a group of functional groups whose chemistry is closely related. The main difference is the presence of an electronegative substituent that can act as a leaving group during nucleophile substitution reactions. Although there are many types of carboxylic acid derivatives known we will be focusing on just four: Acid halides, Acid anhydrides, Esters, and Amides. 1) Nucleophilic attack on the carbonyl 2) Leaving group is removed Although and ketones also contain a carbonyl their chemistry is distinctly different because they do not contain a suitable leaving group. Once the tetrahedral intermediate is formed aldehydes and ketones cannot reform the carbonyl. Because of this aldehydes and ketones typically undergo nucleophilic additions and not substitutions. The relative reactivity of carboxylic acid derivatives toward nucleophile substitutions is related to the electronegative leaving group’s ability to activate the carbonyl. The more electronegative leaving groups withdrawn electron density from the carbonyl, thereby, increasing its electrophilicity. ) | 1,118 | 3,457 |
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List common applications of radioactive isotopes Radioactive isotopes have the same chemical properties as stable isotopes of the same element, but they emit radiation, which can be detected. If we replace one (or more) atom(s) with radioisotope(s) in a compound, we can track them by monitoring their radioactive emissions. This type of compound is called a (or ). Radioisotopes are used to follow the paths of biochemical reactions or to determine how a substance is distributed within an organism. Radioactive tracers are also used in many medical applications, including both diagnosis and treatment. They are used to measure engine wear, analyze the geological formation around oil wells, and much more. Radioisotopes have revolutionized medical practice, where they are used extensively. Over 10 million nuclear medicine procedures and more than 100 million nuclear medicine tests are performed annually in the United States. Four typical examples of radioactive tracers used in medicine are technetium-99 \(\ce{(^{99}_{43}Tc)}\), thallium-201 \(\ce{(^{201}_{81}Tl)}\), iodine-131 \(\ce{(^{131}_{53}I)}\), and sodium-24 \(\ce{(^{24}_{11}Na)}\). Damaged tissues in the heart, liver, and lungs absorb certain compounds of technetium-99 preferentially. After it is injected, the location of the technetium compound, and hence the damaged tissue, can be determined by detecting the γ rays emitted by the Tc-99 isotope. Thallium-201 (Figure \(\Page {1}\)) becomes concentrated in healthy heart tissue, so the two isotopes, Tc-99 and Tl-201, are used together to study heart tissue. Iodine-131 concentrates in the thyroid gland, the liver, and some parts of the brain. It can therefore be used to monitor goiter and treat thyroid conditions, such as Grave’s disease, as well as liver and brain tumors. Salt solutions containing compounds of sodium-24 are injected into the bloodstream to help locate obstructions to the flow of blood. Radioisotopes used in medicine typically have short half-lives—for example, the ubiquitous Tc-99m has a half-life of 6.01 hours. This makes Tc-99m essentially impossible to store and prohibitively expensive to transport, so it is made on-site instead. Hospitals and other medical facilities use Mo-99 (which is primarily extracted from U-235 fission products) to generate Tc-99. Mo-99 undergoes β decay with a half-life of 66 hours, and the Tc-99 is then chemically extracted (Figure \(\Page {2}\)). The parent nuclide Mo-99 is part of a molybdate ion, \(\ce{MoO4^2-}\); when it decays, it forms the pertechnetate ion, \(\ce{TcO4-}\). These two water-soluble ions are separated by column chromatography, with the higher charge molybdate ion adsorbing onto the alumina in the column, and the lower charge pertechnetate ion passing through the column in the solution. A few micrograms of Mo-99 can produce enough Tc-99 to perform as many as 10,000 tests. Radioisotopes can also be used, typically in higher doses than as a tracer, as treatment. is the use of high-energy radiation to damage the of cancer cells, which kills them or keeps them from dividing (Figure \(\Page {3}\)). A cancer patient may receive delivered by a machine outside the body, or from a radioactive substance that has been introduced into the body. Note that is similar to internal radiation therapy in that the cancer treatment is injected into the body, but differs in that chemotherapy uses chemical rather than radioactive substances to kill the cancer cells. Cobalt-60 is a synthetic radioisotope produced by the neutron activation of Co-59, which then undergoes β decay to form Ni-60, along with the emission of γ radiation. The overall process is: \[\ce{^{59}_{27}Co + ^1_0n⟶ ^{60}_{27}Co⟶ ^{60}_{28}Ni + ^0_{−1}β + 2^0_0γ} \nonumber \] The overall decay scheme for this is shown graphically in Figure \(\Page {4}\). Radioisotopes are used in diverse ways to study the mechanisms of chemical reactions in plants and animals. These include labeling fertilizers in studies of nutrient uptake by plants and crop growth, investigations of digestive and milk-producing processes in cows, and studies on the growth and metabolism of animals and plants. For example, the radioisotope C-14 was used to elucidate the details of how photosynthesis occurs. The overall reaction is: \[\ce{6CO2}(g)+\ce{6H2O}(l)⟶\ce{C6H12O6}(s)+\ce{6O2}(g), \nonumber \] but the process is much more complex, proceeding through a series of steps in which various organic compounds are produced. In studies of the pathway of this reaction, plants were exposed to CO containing a high concentration of \(\ce{^{14}_6C}\). At regular intervals, the plants were analyzed to determine which organic compounds contained carbon-14 and how much of each compound was present. From the time sequence in which the compounds appeared and the amount of each present at given time intervals, scientists learned more about the pathway of the reaction. Commercial applications of radioactive materials are equally diverse (Figure \(\Page {5}\)). They include determining the thickness of films and thin metal sheets by exploiting the penetration power of various types of radiation. Flaws in metals used for structural purposes can be detected using high-energy gamma rays from cobalt-60 in a fashion similar to the way X-rays are used to examine the human body. In one form of pest control, flies are controlled by sterilizing male flies with γ radiation so that females breeding with them do not produce offspring. Many foods are preserved by radiation that kills microorganisms that cause the foods to spoil. Americium-241, an α emitter with a half-life of 458 years, is used in tiny amounts in ionization-type smoke detectors (Figure \(\Page {6}\)). The α emissions from Am-241 ionize the air between two electrode plates in the ionizing chamber. A battery supplies a potential that causes movement of the ions, thus creating a small electric current. When smoke enters the chamber, the movement of the ions is impeded, reducing the conductivity of the air. This causes a marked drop in the current, triggering an alarm. Compounds known as radioactive tracers can be used to follow reactions, track the distribution of a substance, diagnose and treat medical conditions, and much more. Other radioactive substances are helpful for controlling pests, visualizing structures, providing fire warnings, and for many other applications. Hundreds of millions of nuclear medicine tests and procedures, using a wide variety of radioisotopes with relatively short half-lives, are performed every year in the US. Most of these radioisotopes have relatively short half-lives; some are short enough that the radioisotope must be made on-site at medical facilities. Radiation therapy uses high-energy radiation to kill cancer cells by damaging their DNA. The radiation used for this treatment may be delivered externally or internally. | 6,925 | 3,458 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/24%3A_Complex_Ions_and_Coordination_Compounds/24.02%3A_Ligands |
A metal ion in solution does not exist in isolation, but in combination with ligands (such as solvent molecules or simple ions) or chelating groups, giving rise to complex ions or coordination compounds. These complexes contain a central atom or ion, often a transition metal, and a cluster of ions or neutral molecules surrounding it. Ligands are ions or neutral molecules that bond to a central metal atom or ion. Ligands act as Lewis bases (electron pair donors), and the central atom acts as a Lewis acid (electron pair acceptor). Ligands have at least one donor atom with an electron pair used to form covalent bonds with the central atom. The term ligand come from the latin word (which meaning to bind) was first used by Alfred Stock in 1916 in relation to silicon chemistry. Ligands can be anions, cations, or neutral molecules. Ligands can be further characterized as monodentate, bidentate, tridentate etc. where the concept of teeth is introduced, hence the idea of bite angle etc. A monodentate ligand has only one donor atom used to bond to the central metal atom or ion. The term "monodentate" can be translated as "one tooth," referring to the ligand binding to the center through only one atom. Some examples of monodentate ligands are: chloride ions (referred to as chloro when it is a ligand), water (referred to as aqua when it is a ligand), hydroxide ions (referred to as hydroxo when it is a ligand), and ammonia (referred to as ammine when it is a ligand). Bidentate ligands have two donor atoms which allow them to bind to a central metal atom or ion at two points. Common examples of bidentate ligands are ethylenediamine (en), and the oxalate ion (ox). Shown below is a diagram of ethylenediamine: the nitrogen (blue) atoms on the edges each have two free electrons that can be used to bond to a central metal atom or ion. Polydentate ligands Unlike polydentate ligands, ambidentate ligands can attach to the central atom in two places. A good example of this is thiocyanate, \(SCN^−\), which can attach at either the sulfur atom or the nitrogen atom. Draw metal complexes using the ligands below and metal ions of your choice. Chelation is a process in which a polydentate ligand bonds to a metal ion, forming a ring. The complex produced by this process is called a chelate, and the polydentate ligand is referred to as a chelating agent. The term chelate was first applied in 1920 by Sir Gilbert T. Morgan and H.D.K. Drew, who stated: "The adjective chelate, derived from the great claw or chela (chely- Greek) of the lobster or other crustaceans, is suggested for the caliperlike groups which function as two associating units and fasten to the central atom so as to produce heterocyclic rings." As the name implies, chelating ligands have high affinity for metal ions relative to ligands with only one binding group (which are called monodentate = "single tooth") ligands. Both Ethylenediamine (Figure \(\Page {2}\)) and Ethylenediaminetetraaceticacid acid (Figure \(\Page {3}\)) are examples of chelating agents, but many others are commonly found in the inorganic laboratory. The chelate effect is the enhanced affinity of chelating ligands for a metal ion compared to the affinity of a collection of similar nonchelating (monodentate) ligands for the same metal. The follows the same principle as the chelate effect, but the effect is further enhanced by the cyclic conformation of the ligand. Macrocyclic ligands are not only multi-dentate, but because they are covalently constrained to their cyclic form, they allow less conformational freedom. The ligand is said to be "pre-organized" for binding, and there is little entropy penalty for wrapping it around the metal ion. For example heme b is a tetradentate cyclic ligand that strongly complexes transition metal ions, including Fe in (Figure \(\Page {4}\)). Some other common chelating and cyclic ligands are shown below: Adapted from the Wikibook constructed by students at . | 3,992 | 3,459 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/11%3A_Gases/11.09%3A_Mean_Free_Path_Diffusion_and_Effusion_of_Gases |
As you have learned, the molecules of a gas are stationary but in constant and random motion. If someone opens a bottle of perfume in the next room, for example, you are likely to be aware of it soon. Your sense of smell relies on molecules of the aromatic substance coming into contact with specialized olfactory cells in your nasal passages, which contain specific receptors (protein molecules) that recognize the substance. How do the molecules responsible for the aroma get from the perfume bottle to your nose? You might think that they are blown by drafts, but, in fact, molecules can move from one place to another even in a draft-free environment. Diffusion is the gradual mixing of gases due to the motion of their component particles even in the absence of mechanical agitation such as stirring. The result is a gas mixture with uniform composition. Diffusion is also a property of the particles in liquids and liquid solutions and, to a lesser extent, of solids and solid solutions. The related process, effusion, is the escape of gaseous molecules through a small (usually microscopic) hole, such as a hole in a balloon, into an evacuated space. The phenomenon of effusion had been known for thousands of years, but it was not until the early 19th century that quantitative experiments related the rate of effusion to molecular properties. The rate of effusion of a gaseous substance is inversely proportional to the square root of its molar mass. This relationship , after the Scottish chemist Thomas Graham (1805–1869). The ratio of the effusion rates of two gases is the square root of the inverse ratio of their molar masses Heavy molecules effuse through a porous material more slowly than light molecules, as illustrated schematically in Figure 10.8.1 for ethylene oxide and helium. Helium ( = 4.00 g/mol) effuses much more rapidly than ethylene oxide ( = 44.0 g/mol). Because helium is less dense than air, helium-filled balloons “float” at the end of a tethering string. Unfortunately, rubber balloons filled with helium soon lose their buoyancy along with much of their volume. In contrast, rubber balloons filled with air tend to retain their shape and volume for a much longer time. Because helium has a molar mass of 4.00 g/mol, whereas air has an average molar mass of about 29 g/mol, pure helium effuses through the microscopic pores in the rubber balloon \(\sqrt{\dfrac{29}{4.00}}=2.7\) times faster than air. For this reason, high-quality helium-filled balloons are usually made of Mylar, a dense, strong, opaque material with a high molecular mass that forms films that have many fewer pores than rubber. Hence, mylar balloons can retain their helium for days. At a given temperature, heavier molecules move more slowly than lighter molecules. During World War II, scientists working on the first atomic bomb were faced with the challenge of finding a way to obtain large amounts of U. Naturally occurring uranium is only 0.720% U, whereas most of the rest (99.275%) is U, which is not fissionable (i.e., it will not break apart to release nuclear energy) and also actually poisons the fission process. Because both isotopes of uranium have the same reactivity, they cannot be separated chemically. Instead, a process of gaseous effusion was developed using the volatile compound UF (boiling point = 56°C). isotopic content of naturally occurring uranium and atomic masses of U and U ratio of rates of effusion and number of effusion steps needed to obtain 99.0% pure UF The first step is to calculate the molar mass of UF containing U and U. Luckily for the success of the separation method, fluorine consists of a single isotope of atomic mass 18.998. The molar mass of UF is The molar mass of UF is The difference is only 3.01 g/mol (less than 1%). The ratio of the effusion rates can be calculated from Graham’s law. \[\rm\dfrac{\text{rate }^{235}UF_6}{\text{rate }^{238}UF_6}=\sqrt{\dfrac{352.04\;g/mol}{349.03\;g/mol}}=1.0043\] Graham’s law is an empirical relationship that states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses. The relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy. We can write the expression for the average kinetic energy of two gases with different molar masses: \[KE=\dfrac{1}{2}\dfrac{M_{\rm A}}{N_A}v_{\rm rms,A}^2=\dfrac{1}{2}\dfrac{M_{\rm B}}{N_A}v_{\rm rms,B}^2\tag{10.8.1}\] Multiplying both sides by 2 and rearranging give \[\dfrac{v_{\rm rms, B}^2}{v_{\rm rms,A}^2}=\dfrac{M_{\rm A}}{M_{\rm B}}\tag{10.8.2}\] Taking the square root of both sides gives \[\dfrac{v_{\rm rms, B}}{v_{\rm rms,A}}=\sqrt{\dfrac{M_{\rm A}}{M_{\rm B}}}\tag{10.8.2}\] Thus the rate at which a molecule, or a mole of molecules, diffuses or effuses is directly related to the speed at which it moves. Equation 10.8.3 shows that Graham’s law is a direct consequence of the fact that gaseous molecules at the same temperature have the same average kinetic energy. Typically, gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour). The effect of molar mass on these speeds is dramatic, as illustrated in Figure 10.8.3 for some common gases. Because all gases have the same average kinetic energy, according to the Boltzmann distribution, molecules with lower masses, such as hydrogen and helium, have a wider distribution of speeds. Figure 10.8.3 The Wide Variation in Molecular Speeds Observed at 298 K for Gases with Different Molar Masses The lightest gases have a wider distribution of speeds and the highest average speeds. Molecules with lower masses have a wider distribution of speeds and a higher average speed. Gas molecules do not diffuse nearly as rapidly as their very high speeds might suggest. If molecules actually moved through a room at hundreds of miles per hour, we would detect odors faster than we hear sound. Instead, it can take several minutes for us to detect an aroma because molecules are traveling in a medium with other gas molecules. Because gas molecules collide as often as 10 times per second, changing direction and speed with each collision, they do not diffuse across a room in a straight line, as illustrated schematically in Figure 10.8.4. The average distance traveled by a molecule between collisions is the mean free path. The denser the gas, the shorter the mean free path; conversely, as density decreases, the mean free path becomes longer because collisions occur less frequently. At 1 atm pressure and 25°C, for example, an oxygen or nitrogen molecule in the atmosphere travels only about 6.0 × 10 m (60 nm) between collisions. In the upper atmosphere at about 100 km altitude, where gas density is much lower, the mean free path is about 10 cm; in space between galaxies, it can be as long as 1 × 10 m (about 6 million miles). The denser the gas, the shorter the mean free path. is the gradual mixing of gases to form a sample of uniform composition even in the absence of mechanical agitation. In contrast, is the escape of a gas from a container through a tiny opening into an evacuated space. The rate of effusion of a gas is inversely proportional to the square root of its molar mass ( ), a relationship that closely approximates the rate of diffusion. As a result, light gases tend to diffuse and effuse much more rapidly than heavier gases. The of a molecule is the average distance it travels between collision | 7,539 | 3,460 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Defects/Frenkel_Defect |
The (also known as the Frenkel pair/disorder) is a defect in the lattice crystal where an atom or ion occupies a normally vacant site other than its own. As a result the atom or ion leaves its own lattice site vacant. The Frenkel Defect explains a defect in the molecule where an atom or ion (normally the cation) leaves its own lattice site vacant and instead occupies a normally vacant site (Figure \(\Page {1}\)). The cation leaves its own lattice site open and places itself between the area of all the other cations and anions. This defect is only possible if the cations are smaller in size when compared to the anions. The number of Frenkel Defects can be calculated using the equation: Frenkel The crystal lattices are relatively open and the coordination number is low. | 798 | 3,461 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.02%3A_Ionization_of_Water |
In the section on , we saw that water can act as a very weak acid and a very weak base, donating protons to itself to a limited extent: \[\text{2H}_{2}\text{O}\text{ }({l}) \rightleftharpoons \text{H}_{3}\text{O}^{+} ({aq}) + \text{OH}^{-} ({aq}) \nonumber \] The equilibrium constant \(K\) for this reaction can be written as follows: \[K_{a}=\dfrac{a_{H_3O^+}·a_{OH^-}}{a_{H_2O}^2} \approx \frac{[H_{3}O^{+},HO^{-}]}{(1)^{2}}=[H_{3}O^{+},HO^{-}] \label{16.3.4} \] where \(a\) is the activity of a species. Because water is the solvent, and the solution is assumed to be dilute, the activity of the water is approximated by the activity of pure water, which is defined as having a value of 1. The activity of each solute is approximated by the molarity of the solute. It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that K = K [H O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation K = K (\(\textit{a}_{H_2O}\)). Because \(\textit{a}_{H_2O}\) = 1 for a dilute solution, K = K (1), or K = K . In this reaction, one water molecule acts as an acid and one water molecule acts as a base. Thus, this reaction actually can be designated as the \(K_a\) of water and as the \(K_b\) of water. It is most common, however, to designate this reaction and the associated law of mass action as the \(K_w\) of water: \[K_{w}=[H_{3}O^{+},HO^{-}] \label{16.3.5} \] Measurements of the electrical conductivity of carefully purified water indicate that at 25°C [H O ] = [OH ] = 1.00 × 10 mol/L, so that \[\begin{align}K_{w}={1.00}\times{10}^{-7}\text{ mol L}^{-1}\times{1.00}\times{10}^{-7}\text{ mol L}^{-1} \nonumber \\\text{ }\\\text{ }={1.00}\times{10}^{-14}\text{ mol}^2\text{L}^{-2} \nonumber \end{align} \nonumber \] (Since the equilibrium law is not obeyed exactly, even in dilute solutions, results of most equilibrium calculations are rounded to three significant figures. Hence the value of = 1.00 × 10 mol /L is sufficiently accurate for all such calculations.) The equilibrium constant applies not only to pure water but to any aqueous solution at 25°C. Thus, for example, if we add 1.00 mol of the strong acid HNO to H O to make a total volume of 1 L, essentially all the HNO molecules donate their protons to H O: \[\text{HNO}_{3} + \text{H}_{2}\text{O} \rightarrow \text{NO}_{3}^{-} + \text{H}_{3}\text{O}^{+} \nonumber \] and a solution in which [H O ] = 1.00 mol/L is obtained. Although this solution is very acidic, there are still hydroxide ions present. We can calculate their concentration by rearranging Eq. \(\ref{3}\): \[\begin{align}\text{ }[\text{OH}^{-}]=\dfrac{K_{w}}{[\text{ H}_{3}\text{O}^{+}]}=\dfrac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ L}^{-2}}{\text{1.00 mol L}^{-1}}\\\text{ }\\\text{ }=\text{1.00 }\times \text{ 10}^{-14}\text{ mol L}^{-1}\end{align} \nonumber \] The addition of the HNO to H O not only increases the hydronium-ion concentration but also reduces the hydroxide-ion concentration from an initially minute 10 mol/L to an even more minute 10 mol/L. Calculate the hydronium-ion concentration in a solution of 0.306 Ba(OH) . Since 1 mol Ba(OH) produces 2 mol OH in solution, we have \[[OH^-] = 2 \times 0.306 \dfrac{mol}{L} = 0.612 \dfrac{mol}{L} \nonumber \] Then \[\begin{align}\text{ }[\text{ H}_{3}\text{O}^{+}]=\dfrac{K_{w}}{[\text{OH}^{-}]}=\dfrac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ L}^{-2}}{\text{0.612 mol L}^{-1}} \nonumber \\\text{ }\\\text{ }=\text{1.63 }\times \text{ 10}^{-14}\text{ mol L}^{-1} \nonumber \end{align} \nonumber \] Note that since strong acids like HNO are completely converted to H O in aqueous solution, it is a simple matter to determine [H O ], and from it, [OH ]. Similarly, when a strong base dissolves in H O it is entirely converted to OH , so that [OH ], and from it [H O ] are easily obtained. | 4,097 | 3,464 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/06%3A_Electron_Transfer/6.04%3A_Electron-Transfer_Theory |
The simplest electron transfer occurs in an outer-sphere reaction. The changes in oxidation states of the donor and acceptor centers result in a change in their equilibrium nuclear configurations. This process involves geometric changes, the magnitudes of which vary from system to system. In addition, changes in the interactions of the donor and acceptor with the surrounding solvent molecules will occur. The Franck-Condon principle governs the coupling of the electron transfer to these changes in nuclear geometry: during an electronic transition, the electronic motion is so rapid that the nuclei (including metal ligands and solvent molecules) do not have time to move. Hence, electron transfer occurs at a fixed nuclear configuration. In a self-exchange reaction, the energies of the donor and acceptor orbitals (hence, the bond lengths and bond angles of the donor and acceptor) must be the same before efficient electron transfer can take place. The incorporation of the Franck-Condon restriction leads to the partitioning of an electron-transfer reaction into reactant (precursor complex) and product (successor complex) configurations. The steps in Equations \ref{6.13} to \ref{6.15} go from reactants to products: \(K\) is the equilibrium constant for the formation of the precursor complex [A , B ], and \(k_{et}\) is the forward electron transfer rate to produce the successor complex [A , B ]. \[A_{ox} + B_{red} \xrightleftharpoons{K} [A_{ox},\; B_{red}] \label{6.13}\] \[[A_{ox},\; B_{red}] \xrightarrow{k_{et}} [A_{red},\; B_{ox}] \label{6.14}\] \[[A_{red},\; B_{ox}] \xrightarrow{fast} A_{red} + B_{ox} \label{6.15}\] pioneered the use of potential energy diagrams as an aid in describing electron-transfer processes. For the sake of simplicity, the donor and acceptor are assumed to behave like collections of harmonic oscillators. Instead of two separate potential energy surfaces being used for the reactants, they are combined into a single surface that describes the potential energy of the precursor complex as a function of its nuclear configuration (i.e., the sum of the translational, rotational, and vibrational degrees of freedom of the reactant molecules and the molecules in the surrounding solvent-3N coordinates, where N is the number of nuclei present). Similarly, a single potential energy (3N-dimensional) surface is used to describe the potential energy of the successor complex as a function of its nuclear configuration. It has become conventional to simplify such potential energy diagrams by using one-dimensional slices through the reactant and product surfaces in order to visualize the progress of a reaction, as illustrated in Figure 6.21. The intersection of the reactant and product surfaces (point S) represents the transition state (or "activated complex"), and is characterized by a loss of one degree of freedom relative to the reactants or products. The actual electron-transfer event occurs when the reactants reach the transition-state geometry. For bimolecular reactions, the reactants must diffuse through the solvent, collide, and form a precursor complex prior to electron transfer. Hence, disentangling the effects of precursor complex formation from the observed reaction rate can pose a serious challenge to the experimentalist; unless this is gone, the factors that determine the kinetic activation barrier for the electron-transfer step cannot be identified with certainty. The surfaces depicted in Figure 6.21 presume that the electrons remain localized on the donor and acceptor; as long as this situation prevails, no electron transfer is possible. Thus some degree of electronic interaction, or coupling, is required if the redox system is to pass from the precursor to the successor complex. This coupling removes the degeneracy of the reactant and product states at the intersection of their respective zero-order surfaces (points S in Figure 6.21) and leads to a splitting in the region of the intersection of the reactant and product surfaces (Figure 6.22). If the degree of electronic interaction is sufficiently small, first-order perturbation theory can be used to obtain the energies of the new first-order surfaces, which do not cross. The splitting at the intersection is equal to 2H , where H is the electronic-coupling matrix element. The magnitude of \(H_{AB}\) determines the behavior of the reactants once the intersection region is reached. Two cases can be distinguished. First, \(H_{AB}\) is very small; for these so-called , there is a high probability that the reactants will "jump" to the upper first-order potential energy surface, leading to very little product formation. If the electronic interaction is sufficiently large, as it is for "adiabatic" reactions, the reactants will remain on the lower first-order potential energy surface upon passage through the transition-state region. The term adiabatic (Greek: , not able to go through) is used in both thermodynamics and quantum mechanics, and the uses are analogous. In the former, it indicates that there is no heat flow in or out of the system. In the latter, it indicates that a change occurs such that the system makes no transition to other states. Hence, for an adiabatic reaction, the system remains on the same (i.e., lower) first-order electronic surface for the entire reaction. The probability of electron transfer occurring when the reactants reach the transition state is unity. The degree of adiabaticity of the reaction is given by a transmission coefficient, \(\kappa\), whose value ranges from zero to one. For systems whose H is sufficiently large (>k T, where k is the Boltzmann constant), \(\kappa\) = 1. This situation occurs when the reacting centers are close together, the orbital symmetries are favorable, and no substantial changes in geometry are involved. The transmission coefficient is generally very small (\(\kappa\) < 1) for electron-transfer reactions of metalloproteins, owing to the long distances involved. | 6,014 | 3,465 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Analytical_Chemistry_2.1_(Harvey)/11%3A_Electrochemical_Methods/11.07%3A_Chapter_Summary_and_Key_Terms |
In this chapter we introduced three electrochemical methods of analysis: potentiometry, coulometry, and voltammetry. In potentiometry we measure the potential at an indicator electrode without allowing any significant current to pass through the electrochemical cell, and use the Nernst equation to calculate the analyte’s activity after accounting for junction potentials. There are two broad classes of potentiometric electrodes: metallic electrodes and membrane electrodes. The potential of a metallic electrode is the result of a redox reaction at the electrode’s surface. An electrode of the first kind responds to the concentration of its cation in solution; thus, the potential of a Ag wire is determined by the activity of Ag in solution. If another species is in equilibrium with the metal ion, the electrode’s potential also responds to the concentration of that species. For example, the potential of a Ag wire in a solution of Cl responds to the concentration of Cl because the relative concentrations of Ag and Cl are fixed by the solubility product for AgCl. We call this an electrode of the second kind. The potential of a membrane electrode is determined by a difference in the composition of the solution on each side of the membrane. Electrodes that use a glass membrane respond to ions that bind to negatively charged sites on the membrane’s surface. A pH electrode is one example of a glass membrane electrode. Other kinds of membrane electrodes include those that use insoluble crystalline solids or liquid ion-exchangers incorporated into a hydrophobic membrane. The F ion-selective electrode, which uses a single crystal of LaF as the ion-selective membrane, is an example of a solid-state electrode. The Ca ion-selective electrode, in which the chelating ligand di-( -decyl)phosphate is immobilized in a PVC membrane, is an example of a liquid-based ion-selective electrode. Potentiometric electrodes are designed to respond to molecules by using a chemical reaction that produces an ion whose concentration is determined using a traditional ion-selective electrode. A gas-sensing electrode, for example, includes a gas permeable membrane that isolates the ion-selective electrode from the gas. When a gas-phase analyte diffuses across the membrane it alters the composition of the inner solution, which is monitored with an ion-selective electrode. An enzyme electrodes operate in the same way. Coulometric methods are based on Faraday’s law that the total charge or current passed during an electrolysis is proportional to the amount of reactants and products participating in the redox reaction. If the electrolysis is 100% efficient—which means that only the analyte is oxidized or reduced—then we can use the total charge or total current to determine the amount of analyte in a sample. In controlled-potential coulometry we apply a constant potential and measure the resulting current as a function of time. In controlled-current coulometry the current is held constant and we measure the time required to completely oxidize or reduce the analyte. In voltammetry we measure the current in an electrochemical cell as a function of the applied potential. There are several different voltammetric methods that differ in terms of the choice of working electrode, how we apply the potential, and whether we include convection (stirring) as a means for transporting of material to the working electrode. Polarography is a voltammetric technique that uses a mercury electrode and an unstirred solution. Normal polarography uses a dropping mercury electrode, or a static mercury drop electrode, and a linear potential scan. Other forms of polarography include normal pulse polarography, differential pulse polarography, staircase polarography, and square-wave polarography, all of which use a series of potential pulses. In hydrodynamic voltammetry the solution is stirred using either a magnetic stir bar or by rotating the electrode. Because the solution is stirred a dropping mercury electrode is not used; instead we use a solid electrode. Both linear potential scans and potential pulses can be applied. In stripping voltammetry the analyte is deposited on the electrode, usually as the result of an oxidation or reduction reaction. The potential is then scanned, either linearly or using potential pulses, in a direction that removes the analyte by a reduction or oxidation reaction. Amperometry is a voltammetric method in which we apply a constant potential to the electrode and measure the resulting current. Amperometry is most often used in the construction of chemical sensors for the quantitative analysis of single analytes. One important example is the Clark O electrode, which responds to the concentration of dissolved O in solutions such as blood and water. amalgam anodic current cathode controlled-current coulometry coulometric titrations current efficiency diffusion layer electrochemically irreversible electrode of the second kind enzyme electrodes galvanostat hanging mercury drop electrode ionophore limiting current mediator migration overpotential potentiometer redox electrode salt bridge silver/silver chloride electrode static mercury drop electrode voltammetry amperometry asymmetry potential cathodic current controlled-potential coulometry coulometry cyclic voltammetry dropping mercury electrode electrochemically reversible electrochemistry faradaic current gas-sensing electrode hydrodynamic voltammetry ion selective electrode liquid-based ion-selective electrode membrane potential nonfaradaic current peak current potentiostat reference electrode saturated calomel electrode solid-state ion-selective electrodes stripping voltammetry voltammogram anode
auxiliary electrode charging current convection counter electrode diffusion electrical double layer electrode of the first kind electrogravimetry Faraday’s law glass electrode indicator electrode junction potential mass transport mercury film electrode Ohm’s law polarography pulse polarography residual current selectivity coefficient standard hydrogen electrode total ionic strength adjustment buffer working electrode | 6,176 | 3,466 |
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The word microscopy comes from the Greek words for small and to view. On April 13, 1625, Giovanni Faber coined the term microscope. A microscope is an instrument that enables us to view small objects that are otherwise invisible to our naked eye. One way that microscopes allow us to see smaller objects is through the process of magnification, i.e. enlarging the image of the object. When a microscope enlarges an image of a 1 mm object to 10 mm, this is a 10 x magnification. Lens: The lens is the part of a microscope that bends a beam of light and focuses this on the object or sample. The resolution of a microscope is the smallest distance between two objects that results in two images that are distinguishable from each other. For example, the resolution of our eyes ranges from 0.1 to 0.2 mm. This means that our eyes can distinguish between two objects that are separated by 0.1 to 0.2 mm. Early Light Microscopes Light has both a particle and a wave property. A beam of light can be polarized by lining up its vibrations with each other. Thus, the polarizing microscope polarizes light in order to magnify images. This microscope also determines properties of materials that transmit light, whether they are crystalline or non crystalline. The optical features of transparent material were recognized when William Henry Fox Talbot added two Nicol prisms (prisms that can polarize light) to a microscope. However, it was Henry Clifton Sorby (1826-1908) who used polarized light microscopy to study thinned sections of transparent rocks. He showed that through their optical properties, these thinned sections of minerals could be analyzed. The polarizing microscope can be divided into three major component sets: The quality of magnification depends on the objective lens and the smaller the diameter of the outermost lens, the higher the magnification. In 1740, Dr. Johann N. Lieberkuhn authenticated an instrument for illuminating opaque materials that had a cup shaped mirror encircling the objective lens of a microscope. This mirror is called a reflector. A reflector has a concave reflecting surface and a lens in its center. This evenly illuminates the specimen when the specimen is fixed up to the light and the light rays reflected from it and to the specimen. Henry Clifton Sorby used a small reflector and attached this over the objective lens of his microscope. When he used this to study steel, he was able to see residues and distinguish these from the hard components of the steel. From then on, several scientists that study minerals also used reflected light microscopes and this technology improved throughout time. Professor Michael Isaacson of Cornell University invented this type of a microscope. This microscope also uses light but not lenses. In order to focus the light on a sample, Isaacson passed light through a very tiny hole. The hole and the sample are so closed together that the light beam does not spread out. This type of a microscope enabled Isaacson's team to resolve up to 40 nm when they used yellow-green light. In this type of a microscope, the resolution is not really limited by the wavelength of light but the amount of the sample since it is very small. Because they only have resolutions in the micrometer range by using visible light, the light microscopes cannot be used to see in the nanometer range. In order to see in the nanometer range, we would need something that has higher energy than visible light. A physicist named came up with an equation that shows the shorter the wavelength of a wave, the higher the energy it has. From the wave-particle duality, we know that matter, like light, can have both wave and particle properties. This means that we can also use matter, like electrons, instead of light. Electrons have shorter wavelengths than light and thus have higher energy and better resolution. Electron microscopes use electrons to focus on a sample. In 1926-1927, Busch demonstrated that an appropriately shaped magnetic field could be used as a lens. This discovery made it possible to use magnetic fields to focus the electron beam for electron microscopes. After Busch’s discovery and development of electron microscopes, companies in different parts of the world developed and produced a prototype of an electron microscope called Transmission Electron Microscopes (TEM). In TEM, the beam of electrons goes through the sample and their interactions are seen on the side of the sample where the beam exits. Then, the image is gathered on a screen. TEMs consist of three major parts: TEM has a typical resolution of approximately 2 nm. However, the sample has to be thin enough to transmit electrons so it cannot be used to look at living cells. In 1942, Zworykin, Hillier, and R.L. Snyder developed another type of an electron microscope called Scanning Electron Microscope (SEM). SEM is another example of an electron microscope and is arguably the most widely used electron beam instrument. In SEM, the electron beam excites the sample and its radiation is detected and photographed. SEM is a mapping device—a beam of electrons scanning across the surface of the sample creates the overall image. SEM also consists of major parts: SEM’s resolution is about 20 nm and its magnification is about 200,000x. SEM cannot be used to study living cells as well since the sample for this process must be very dry. Scanning probe microscopes are also capable of magnifying or creating images of samples in the nanometer range. Some of them can even give details up to the atomic level. Briab Josephson shared when he explained . This phenomenon eventually led to the development of Scanning Tunneling Microscopes by Heinrich Rohrer and Gerd Binnig around 1979. Rohrer and Binnig received the Nobel Prize in physics in 1986. The STM uses an electron conductor needle, composed of either platinum-rhodium or tungsten, as a probe to scan across the surface of a solid that conducts electricity as well. The tip of the needle is usually very fine; it may even be a single atom that is 0.2 nm wide. Electrons tunnel across the space between the tip of the needle and the specimen surface when the tip and the surface are very close to each other. The tunneling current is very sensitive to the distance of the tip from the surface. As a result, the needle moves up and down depending on the surface of the solid—a piezoelectric cylinder monitors this movement. The three-dimensional image of the surface is then projected on a computer screen. The STM has a resolution of about 0.1 nm. However, the fact that the needle-tip and the sample must be electrical conductors limits the amount of materials that can be studied using this technology. In 1986, Binnig, Berger, and Calvin Quate invented the first derivative of the STM—the Atomic Force Microscope. The AFM is another type of a scanning microscope that scans the surface of the sample. It is different from the STM because it does not measure the current between the tip of the needle and the sample. The AFM has a stylus with a sharp tip that is attached on the end of a long a cantilever. As the stylus scans the sample, the force of the surface pushes or pulls it. The cantilever deflects as a result and a laser beam is used to measure this deflection. This deflection is then turned into a three dimensional topographic image by a computer. With AFM, a much higher resolution is attained with less sample damage. The AFM can be used on non-conducting samples as well as on liquid samples because there is no current applied on the sample. Thus the AFM can be used to study biological molecules such as cells and proteins. | 7,670 | 3,469 |
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In we add a reagent, called the , to a solution that contains another reagent, called the , and allow them to react. The type of reaction provides us with a simple way to divide titrimetry into four categories: acid–base titrations, in which an acidic or basic titrant reacts with a titrand that is a base or an acid; complexometric titrations , which are based on metal–ligand complexation; redox titrations, in which the titrant is an oxidizing or reducing agent; and precipitation titrations, in which the titrand and titrant form a precipitate. We will deliberately avoid the term analyte at this point in our introduction to titrimetry. Although in most titrations the analyte is the titrand, there are circumstances where the analyte is the titrant. Later, when we discuss specific titrimetric methods, we will use the term analyte where appropriate. Despite their difference in chemistry, all titrations share several common features. Before we consider individual titrimetric methods in greater detail, let’s take a moment to consider some of these similarities. As you work through this chapter, this overview will help you focus on the similarities between different titrimetric methods. You will find it easier to understand a new analytical method when you can see its relationship to other similar methods. If a titration is to give an accurate result we must combine the titrand and the titrant in stoichiometrically equivalent amounts. We call this stoichiometric mixture the . Unlike precipitation gravimetry, where we add the precipitant in excess, an accurate titration requires that we know the exact volume of titrant at the equivalence point, . The product of the titrant’s equivalence point volume and its molarity, , is equal to the moles of titrant that react with the titrand. \[\text { moles titrant }=M_{T} \times V_{e q} \nonumber\] If we know the stoichiometry of the titration reaction, then we can calculate the moles of titrand. Unfortunately, for most titration reactions there is no obvious sign when we reach the equivalence point. Instead, we stop adding the titrant at an of our choosing. Often this end point is a change in the color of a substance, called an , that we add to the titrand’s solution. The difference between the end point’s volume and the equivalence point’s volume is a determinate . If the end point and the equivalence point volumes coincide closely, then this error is insignificant and is safely ignored. Clearly, selecting an appropriate end point is of critical importance. Instead of measuring the titrant’s volume, we may choose to measure its mass. Although generally we can measure mass more precisely than we can measure volume, the simplicity of a volumetric titration makes it the more popular choice. Almost any chemical reaction can serve as a titrimetric method provided that it meets the following four conditions. The first condition is that we must know the stoichiometry between the titrant and the titrand. If this is not the case, then we cannot convert the moles of titrant used to reach the end point to the moles of titrand in our sample. Second, the titration reaction effectively must proceed to completion; that is, the stoichiometric mixing of the titrant and the titrand must result in their complete reaction. Third, the titration reaction must occur rapidly. If we add the titrant faster than it can react with the titrand, then the end point and the equivalence point will differ significantly. Finally, we must have a suitable method for accurately determining the end point. These are significant limitations and, for this reason, there are several common titration strategies. Depending on how we are detecting the endpoint, we may stop the titration too early or too late. If the end point is a function of the titrant’s concentration, then adding the titrant too quickly leads to an early end point. On the other hand, if the end point is a function of the titrand's concentration, then the end point exceeds the equivalence point. A simple example of a titration is an analysis for Ag using thiocyanate, SCN , as a titrant. \[\mathrm{Ag}^{+}(a q)+\mathrm{SCN}^{-}(a q)\rightleftharpoons\mathrm{Ag}(\mathrm{SCN})(s) \nonumber\] This reaction occurs quickly and with a known stoichiometry, which satisfies two of our requirements. To indicate the titration’s end point, we add a small amount of Fe to the analyte’s solution before we begin the titration. When the reaction between Ag and SCN is complete, formation of the red-colored Fe(SCN) complex signals the end point. This is an example of a since the titrant reacts directly with the analyte. This is an example of a precipitation titration. You will find more information about precipitation titrations in this chapter. If the titration’s reaction is too slow, if a suitable indicator is not available, or if there is no useful direct titration reaction, then an indirect analysis may be possible. Suppose you wish to determine the concentration of formaldehyde, H CO, in an aqueous solution. The oxidation of H CO by \(\text{I}_3^-\) \[\mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{I}_{3}^-(a q)+3 \mathrm{OH}^{-}(a q)\rightleftharpoons\mathrm{HCO}_{2}^{-}(a q)+3 \mathrm{I}^{-}(a q)+2 \mathrm{H}_{2} \mathrm{O}(1) \nonumber\] is a useful reaction, but it is too slow for a titration. If we add a known excess of \(\text{I}_3^-\) and allow its reaction with H CO to go to completion, we can titrate the unreacted \(\text{I}_3^-\) with thiosulfate, \(\text{S}_2\text{O}_3^{2-}\). \[\mathrm{I}_{3}^{-}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q)\rightleftharpoons\mathrm{S}_{4} \mathrm{O}_{6}^{2-}(a q)+3 \mathrm{I}^{-}(a q) \nonumber\] The difference between the initial amount of \(\text{I}_3^-\) and the amount in excess gives us the amount of \(\text{I}_3^-\) that reacts with the formaldehyde. This is an example of a . This is an example of a redox titration. You will find more information about redox titrations in this chapter. Calcium ions play an important role in many environmental systems. A direct analysis for Ca might take advantage of its reaction with the ligand ethylenediaminetetraacetic acid (EDTA), which we represent here as Y . \[\mathrm{Ca}^{2+}(a q)+\mathrm{Y}^{4-}(a q)\rightleftharpoons\mathrm{CaY}^{2-}(a q) \nonumber\] Unfortunately, for most samples this titration does not have a useful indicator. Instead, we react the Ca with an excess of MgY \[\mathrm{Ca}^{2+}(a q)+\mathrm{MgY}^{2-}(a q)\rightleftharpoons\mathrm{Ca} \mathrm{Y}^{2-}(a q)+\mathrm{Mg}^{2+}(a q) \nonumber\] releasing an amount of Mg equivalent to the amount of Ca in the sample. Because the titration of Mg with EDTA \[\mathrm{Mg}^{2+}(a q)+\mathrm{Y}^{4-}(a q)\rightleftharpoons\mathrm{MgY}^{2-}(a q) \nonumber\] has a suitable end point, we can complete the analysis. The amount of EDTA used in the titration provides an indirect measure of the amount of Ca in the original sample. Because the species we are titrating was displaced by the analyte, we call this a . MgY is the Mg –EDTA metal–ligand complex. You can prepare a solution of MgY by combining equimolar solutions of Mg and EDTA. This is an example of a complexation titration. You will find more information about complexation titrations in this chapter. If a suitable reaction with the analyte does not exist it may be possible to generate a species that we can titrate. For example, we can determine the sulfur content of coal by using a combustion reaction to convert sulfur to sulfur dioxide \[\mathrm{S}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{2}(g) \nonumber\] and then convert the SO to sulfuric acid, H SO , by bubbling it through an aqueous solution of hydrogen peroxide, H O . \[\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{SO}_{4}(a q) \nonumber\] Titrating H SO with NaOH \[\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaOH}(a q)\rightleftharpoons2 \mathrm{H}_{2} \mathrm{O}(l )+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \nonumber\] provides an indirect determination of sulfur. This is an example of an acid–base titration. You will find more information about acid–base titrations in this chapter. To find a titration’s end point, we need to monitor some property of the reaction that has a well-defined value at the equivalence point. For example, the equivalence point for a titration of HCl with NaOH occurs at a pH of 7.0. A simple method for finding the equivalence point is to monitor the titration mixture’s pH using a pH electrode, stopping the titration when we reach a pH of 7.0. Alternatively, we can add an indicator to the titrand’s solution that changes color at a pH of 7.0. Why a pH of 7.0 is the equivalence point for this titration is a topic we will cover later in the section on acid–base titrations. Suppose the only available indicator changes color at a pH of 6.8. Is the difference between this end point and the equivalence point small enough that we safely can ignore the titration error? To answer this question we need to know how the pH changes during the titration. A provides a visual picture of how a property of the titration reaction changes as we add the titrant to the titrand. The titration curve in Figure 9.1.1
, for example, was obtained by suspending a pH electrode in a solution of 0.100 M HCl (the titrand) and monitoring the pH while adding 0.100 M NaOH (the titrant). A close examination of this titration curve should convince you that an end point pH of 6.8 produces a negligible titration error. Selecting a pH of 11.6 as the end point, however, produces an unacceptably large titration error. For the titration curve in Figure 9.1.1
, the volume of titrant to reach a pH of 6.8 is 24.99995 mL, a titration error of \(-2.00 \times 10^{-4}\)% relative to the equivalence point of 25.00 mL. Typically, we can read the volume only to the nearest ±0.01 mL, which means this uncertainty is too small to affect our results. The volume of titrant to reach a pH of 11.6 is 27.07 mL, or a titration error of +8.28%. This is a significant error. The shape of the titration curve in Figure 9.1.1
is not unique to an acid–base titration. Any titration curve that follows the change in concentration of a species in the titration reaction (plotted logarithmically) as a function of the titrant’s volume has the same general sigmoidal shape. Several additional examples are shown in Figure 9.1.2
. The titrand’s or the titrant’s concentration is not the only property we can use to record a titration curve. Other parameters, such as the temperature or absorbance of the titrand’s solution, may provide a useful end point signal. Many acid–base titration reactions, for example, are exothermic. As the titrant and the titrand react, the temperature of the titrand’s solution increases. Once we reach the equivalence point, further additions of titrant do not produce as exothermic a response. Figure 9.1.3
shows a typical where the intersection of the two linear segments indicates the equivalence point. The only essential equipment for an acid–base titration is a means for delivering the titrant to the titrand’s solution. The most common method for delivering titrant is a (Figure 9.1.4
), which is a long, narrow tube with graduated markings and equipped with a stopcock for dispensing the titrant. The buret’s small internal diameter provides a better defined meniscus, making it easier to read precisely the titrant’s volume. Burets are available in a variety of sizes and tolerances (Table 9.1.1
), with the choice of buret determined by the needs of the analysis. You can improve a buret’s accuracy by calibrating it over several intermediate ranges of volumes using the method described in for calibrating pipets. Calibrating a buret corrects for variations in the buret’s internal diameter. ±0.01 ±0.01 An automated titration uses a pump to deliver the titrant at a constant flow rate (Figure 9.1.5
). Automated titrations offer the additional advantage of using a microcomputer for data storage and analysis. | 12,089 | 3,470 |
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The earliest examples of metal–ligand are Liebig’s determinations, in the 1850s, of cyanide and chloride using, respectively, Ag and Hg as the titrant. Practical analytical applications of complexation titrimetry were slow to develop because many metals and ligands form a series of metal–ligand complexes. Liebig’s titration of CN with Ag was successful because they form a single, stable complex of \(\text{Ag(CN)}_2^-\), which results in a single, easily identified end point. Other metal–ligand complexes, such as \(\text{CdI}_4^{2-}\), are not analytically useful because they form a series of metal–ligand complexes (CdI , CdI , \(\text{CdI}_3^-\) and \(\text{CdI}_4^{2-}\)) that produce a sequence of poorly defined end points. Recall that an acid–base titration curve for a diprotic weak acid has a single end point if its two values are not sufficiently different. See for an example. In 1945, Schwarzenbach introduced aminocarboxylic acids as multidentate ligands. The most widely used of these new ligands—ethylenediaminetetraacetic acid, or EDTA—forms a strong 1:1 complex with many metal ions. The availability of a ligand that gives a single, easily identified end point made complexation titrimetry a practical analytical method. Ethylenediaminetetraacetic acid, or EDTA, is an aminocarboxylic acid. EDTA, the structure of which is shown in Figure 9.3.1
a in its fully deprotonated form, is a Lewis acid with six binding sites—the four negatively charged carboxylate groups and the two tertiary amino groups—that can donate up to six pairs of electrons to a metal ion. The resulting metal–ligand complex, in which EDTA forms a cage-like structure around the metal ion (Figure 9.3.1
b), is very stable. The actual number of coordination sites depends on the size of the metal ion, however, all metal–EDTA complexes have a 1:1 stoichiometry. To illustrate the formation of a metal–EDTA complex, let’s consider the reaction between Cd and EDTA \[\mathrm{Cd}^{2+}(a q)+\mathrm{Y}^{4-}(a q)\rightleftharpoons\mathrm{Cd} \mathrm{Y}^{2-}(a q) \label{9.1}\] where Y is a shorthand notation for the fully deprotonated form of EDTA shown in Figure 9.3.1
a. Because the reaction’s formation constant \[K_{f}=\frac{\left[\mathrm{CdY}^{2-}\right]}{\left[\mathrm{Cd}^{2+}\right]\left[\mathrm{Y}^{4-}\right]}=2.9 \times 10^{16} \label{9.2}\] is large, its equilibrium position lies far to the right. Formation constants for other metal–EDTA complexes are found in . In addition to its properties as a ligand, EDTA is also a weak acid. The fully protonated form of EDTA, H Y , is a hexaprotic weak acid with successive p values of \[\mathrm{p} K_\text{a1}=0.0 \quad \mathrm{p} K_\text{a2}=1.5 \quad \mathrm{p} K_\text{a3}=2.0 \nonumber\] \[\mathrm{p} K_\text{a4}=2.66 \quad \mathrm{p} K_\text{a5}=6.16 \quad \mathrm{p} K_\text{a6}=10.24 \nonumber\] The first four values are for the carboxylic acid protons and the last two values are for the ammonium protons. Figure 9.3.2
shows a ladder diagram for EDTA. The specific form of EDTA in reaction \ref{9.1} is the predominate species only when the pH is more basic than 10.24. The formation constant for CdY in Equation \ref{9.2} assumes that EDTA is present as Y . Because EDTA has many forms, when we prepare a solution of EDTA we know it total concentration, , not the concentration of a specific form, such as Y . To use Equation \ref{9.2}, we need to rewrite it in terms of . At any pH a mass balance on EDTA requires that its total concentration equal the combined concentrations of each of its forms. \[C_{\mathrm{EDTA}}=\left[\mathrm{H}_{6} \mathrm{Y}^{2+}\right]+\left[\mathrm{H}_{5} \mathrm{Y}^{+}\right]+\left[\mathrm{H}_{4} \mathrm{Y}\right]+\left[\mathrm{H}_{3} \mathrm{Y}^-\right]+\left[\mathrm{H}_{2} \mathrm{Y}^{2-}\right]+\left[\mathrm{HY}^{3-}\right]+\left[\mathrm{Y}^{4-}\right] \nonumber\] To correct the formation constant for EDTA’s acid–base properties we need to calculate the fraction, \(\alpha_{\text{Y}^{4-}}\), of EDTA that is present as Y . \[\alpha_{\text{Y}^{4-}}=\frac{\left[\text{Y}^{4-}\right]}{C_\text{EDTA}} \label{9.3}\] Table 9.3.1
provides values of \(\alpha_{\text{Y}^{4-}}\) for selected pH levels. Solving Equation \ref{9.3} for [Y ] and substituting into Equation \ref{9.2} for the CdY formation constant \[K_{\mathrm{f}}=\frac{\left[\mathrm{CdY}^{2-}\right]}{\left[\mathrm{Cd}^{2+}\right] (\alpha_{\mathrm{Y}^{4-}}) C_{\mathrm{EDTA}}} \nonumber\] and rearranging gives \[K_{f}^{\prime}=K_{f} \times \alpha_{\text{Y}^{4-}}=\frac{\left[\mathrm{CdY}^{2-}\right]}{\left[\mathrm{Cd}^{2+}\right] C_{\mathrm{EDTA}}} \label{9.4}\] where \(K_f^{\prime}\) is a pH-dependent . As shown in Table 9.3.2
, the conditional formation constant for CdY becomes smaller and the complex becomes less stable at more acidic pHs. To maintain a constant pH during a complexation titration we usually add a buffering agent. If one of the buffer’s components is a ligand that binds with Cd , then EDTA must compete with the ligand for Cd . For example, an \(\text{NH}_4^+ / \text{NH}_3\) buffer includes NH , which forms several stable Cd –NH complexes. Because EDTA forms a stronger complex with Cd than does NH , it displaces NH ; however, the stability of the Cd –EDTA complex decreases. We can account for the effect of an , such as NH , in the same way we accounted for the effect of pH. Before adding EDTA, the mass balance on Cd , , is \[C_{\mathrm{Cd}} = \left[\mathrm{Cd}^{2+}\right] + \left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)^{2+}\right] + \left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{2}^{2+}\right] + \left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{3}^{2+}\right] + \left[\mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\right] \nonumber\] and the fraction of uncomplexed Cd , \(\alpha_{Cd^{2+}}\), is \[\alpha_{\mathrm{Cd}^{2+}}=\frac{\left[\mathrm{Cd}^{2+}\right]}{C_{\mathrm{Cd}}} \label{9.5}\] The value of \(\alpha_{\mathrm{Cd}^{2+}}\) depends on the concentration of NH . Contrast this with \(\alpha_{\text{Y}^{4-}}\), which depends on pH. Solving Equation \ref{9.5} for [Cd ] and substituting into Equation \ref{9.4} gives \[K_{f}^{\prime}=K_{f} \times \alpha_{Y^{4-}} = \frac {[\text{CdY}^{2-}]} {\alpha_{\text{Cd}^{2+}} C_\text{Cd} C_\text{EDTA}} \nonumber\] Because the concentration of NH in a buffer essentially is constant, we can rewrite this equation \[K_{f}^{\prime \prime}=K_{f} \times \alpha_{\mathrm{Y}^{4-}} \times \alpha_{\mathrm{Cd}^{2+}}=\frac{\left[\mathrm{CdY}^{2-}\right]}{C_{\mathrm{Cd}} C_{\mathrm{EDTA}}} \label{9.6}\] to give a conditional formation constant, \(K_f^{\prime \prime}\), that accounts for both pH and the auxiliary complexing agent’s concentration. Table 9.3.3
provides values of \(\alpha_{\text{M}^{2+}}\) for several metal ion when NH is the complexing agent. Now that we know something about EDTA’s chemical properties, we are ready to evaluate its usefulness as a titrant. To do so we need to know the shape of a complexometric titration curve. In we learned that an acid–base titration curve shows how the titrand’s pH changes as we add titrant. The analogous result for a complexation titration shows the change in pM, where M is the metal ion’s concentration, as a function of the volume of EDTA. In this section we will learn how to calculate a titration curve using the equilibrium calculations from . We also will learn how to sketch a good approximation of any complexation titration curve using a limited number of simple calculations. pM = –log[M ] Let’s calculate the titration curve for 50.0 mL of \(5.00 \times 10^{-3}\) M Cd using a titrant of 0.0100 M EDTA. Furthermore, let’s assume the titrand is buffered to a pH of 10 using a buffer that is 0.0100 M in NH . Because the pH is 10, some of the EDTA is present in forms other than Y . In addition, EDTA will compete with NH for the Cd . To evaluate the titration curve, therefore, we first need to calculate the conditional formation constant for CdY . From and we find that \(\alpha_{\text{Y}^{4-}}\) is 0.367 at a pH of 10, and that \(\alpha_{\text{Cd}^{2+}}\) is 0.0881 when the concentration of NH is 0.0100 M. Using these values, the conditional formation constant is \[K_{f}^{\prime \prime}=K_{f} \times \alpha_{\text{Y}^{4-}} \times \alpha_{\text{Cd}^{2+}}=\left(2.9 \times 10^{16}\right)(0.367)(0.0881)=9.4 \times 10^{14} \nonumber\] Because \(K_f^{\prime \prime}\) is so large, we can treat the titration reaction \[\mathrm{Cd}^{2+}(a q)+\mathrm{Y}^{4-}(a q) \longrightarrow \mathrm{CdY}^{2-}(a q) \nonumber\] as if it proceeds to completion. The next task is to determine the volume of EDTA needed to reach the equivalence point. At the equivalence point we know that the moles of EDTA added must equal the moles of Cd in our sample; thus \[\operatorname{mol} \mathrm{EDTA}=M_{\mathrm{EDTA}} \times V_{\mathrm{EDTA}}=M_{\mathrm{Cd}} \times V_{\mathrm{Cd}}=\mathrm{mol} \ \mathrm{Cd}^{2+} \nonumber\] Substituting in known values, we find that it requires \[V_{eq}=V_{\mathrm{EDTA}}=\frac{M_{\mathrm{Cd}} V_{\mathrm{Cd}}}{M_{\mathrm{EDTA}}}=\frac{\left(5.00 \times 10^{-3} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})}{0.0100 \ \mathrm{M}}=25.0 \ \mathrm{mL} \nonumber\] of EDTA to reach the equivalence point. Before the equivalence point, Cd is present in excess and pCd is determined by the concentration of unreacted Cd . Because not all unreacted Cd is free—some is complexed with NH —we must account for the presence of NH . For example, after adding 5.0 mL of EDTA, the total concentration of Cd is \[C_{\mathrm{Cd}} = \frac {(\text{mol Cd}^{2+})_\text{initial} - (\text{mol EDTA})_\text{added}} {\text{total volume}} = \frac {M_\text{Cd}V_\text{Cd} - M_\text{EDTA}V_\text{EDTA}} {V_\text{Cd} + V_\text{EDTA}} \nonumber\] \[C_{\mathrm{Cd}}=\frac{\left(5.00 \times 10^{-3} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})-(0.0100 \ \mathrm{M})(5.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+5.0 \ \mathrm{mL}} \nonumber\] \[C_{\mathrm{Cd}}=3.64 \times 10^{-3} \ \mathrm{M} \nonumber\] To calculate the concentration of free Cd we use Equation \ref{9.5} \[\left[\mathrm{Cd}^{2+}\right]=\alpha_{\mathrm{Cd}^{2+}} \times C_{\mathrm{Cd}}=(0.0881)\left(3.64 \times 10^{-3} \ \mathrm{M}\right)=3.21 \times 10^{-4} \ \mathrm{M} \nonumber\] which gives a pCd of \[\mathrm{pCd}=-\log \left[\mathrm{Cd}^{2+}\right]=-\log \left(3.21 \times 10^{-4}\right)=3.49 \nonumber\] At the equivalence point all Cd initially in the titrand is now present as CdY . The concentration of Cd , therefore, is determined by the dissociation of the CdY complex. First, we calculate the concentration of CdY . \[\left[\mathrm{CdY}^{2-}\right]=\frac{\left(\mathrm{mol} \ \mathrm{Cd}^{2+}\right)_{\mathrm{initial}}}{\text { total volume }} = \frac {M_\text{Cd}V_\text{Cd}} {V_\text{Cd} + V_\text{EDTA}} \nonumber\] \[\left[\mathrm{CdY}^{2-}\right]=\frac{\left(5.00 \times 10^{-3} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+25.0 \ \mathrm{mL}}=3.33 \times 10^{-3} \ \mathrm{M} \nonumber\] Next, we solve for the concentration of Cd in equilibrium with CdY . \[K_{\mathrm{f}}^{\prime \prime}=\frac{\left[\mathrm{CdY}^{2-}\right]}{C_{\mathrm{Cd}} C_{\mathrm{EDTA}}}=\frac{3.33 \times 10^{-3}-x}{(x)(x)}=9.5 \times 10^{14} \nonumber\] \[x=C_{\mathrm{Cd}}=1.87 \times 10^{-9} \ \mathrm{M} \nonumber\] In calculating that [CdY ] at the equivalence point is \(3.33 \times 10^{-3}\) M, we assumed the reaction between Cd and EDTA went to completion. Here we let the system relax back to equilibrium, increasing and from 0 to , and decreasing the concentration of CdY by . Once again, to find the concentration of uncomplexed Cd we must account for the presence of NH ; thus \[\left[\mathrm{Cd}^{2+}\right]=\alpha_{\mathrm{Cd}^{2+}} \times C_{\mathrm{Cd}}=(0.0881)\left(1.87 \times 10^{-9} \ \mathrm{M}\right)=1.64 \times 10^{-10} \ \mathrm{M} \nonumber\] and pCd is 9.78 at the equivalence point. After the equivalence point, EDTA is in excess and the concentration of Cd is determined by the dissociation of the CdY complex. First, we calculate the concentrations of CdY and of unreacted EDTA. For example, after adding 30.0 mL of EDTA the concentration of CdY is \[\left[\mathrm{CdY}^{2-}\right]=\frac{\left(\mathrm{mol} \mathrm{Cd}^{2+}\right)_{\mathrm{initial}}}{\text { total volume }} = \frac{M_{\mathrm{Cd}} V_{\mathrm{Cd}}}{V_{\mathrm{Cd}}+V_{\mathrm{EDTA}}} \nonumber\] \[\left[\mathrm{CdY}^{2-}\right]=\frac{\left(5.00 \times 10^{-3} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+30.0 \ \mathrm{mL}}=3.12 \times 10^{-3} \ \mathrm{M} \nonumber\] and the concentration of EDTA is \[C_{\mathrm{EDTA}} = \frac {(\text{mol EDTA})_\text{added} - (\text{mol Cd}^{2+})_\text{initial}} {\text{total volume}} = \frac{M_{\mathrm{EDTA}} V_{\mathrm{EDTA}}-M_{\mathrm{Cd}} V_{\mathrm{Cd}}}{V_{\mathrm{Cd}}+V_{\mathrm{EDTA}}} \nonumber\] \[C_{\text{EDTA}} = \frac {(0.0100 \text{ M})(30.0 \text{ mL}) - (5.00 \times 10^{-3} \text{ M})(50.0 \text{ mL})} {50.0 \text{ mL} + 30.0 \text{ mL}} \nonumber\] \[C_{\mathrm{EDTA}}=6.25 \times 10^{-4} \ \mathrm{M} \nonumber\] Substituting into Equation \ref{9.6} and solving for [Cd ] gives \[\frac{\left[\mathrm{CdY}^{2-}\right]}{C_{\mathrm{Cd}} C_{\mathrm{EDTA}}} = \frac{3.12 \times 10^{-3} \ \mathrm{M}}{C_{\mathrm{Cd}}\left(6.25 \times 10^{-4} \ \mathrm{M}\right)} = 9.5 \times 10^{14} \nonumber\] \[C_{\text{Cd}} = 5.27 \times 10^{-15} \text{ M} \nonumber\] \[ \left[ \text{Cd}^{2+} \right] = \alpha_{\text{Cd}^{2+}} \times C_{\text{Cd}} = (0.0881)(5.27 \times 10^{-15} \text{ M}) = 4.64 \times 10^{-16} \text{ M} \nonumber\] a pCd of 15.33. Table 9.3.4
and Figure 9.3.3
show additional results for this titration. After the equilibrium point we know the equilibrium concentrations of CdY and of EDTA in all its forms, . We can solve for using \(K_f^{\prime \prime}\) and then calculate [Cd ] using \(\alpha_{\text{Cd}^{2+}}\). Because we used the same conditional formation constant, \(K_f^{\prime \prime}\), for other calculations in this section, this is the approach used here as well. There is a second method for calculating [Cd ] after the equivalence point. Because the calculation uses only [CdY ] and , we can use \(K_f^{\prime}\) instead of \(K_f^{\prime \prime}\); thus \[\frac{\left[\mathrm{CdY}^{2-}\right]}{\left[\mathrm{Cd}^{2+}\right] C_{\mathrm{EDTA}}}=\alpha_{\mathrm{Y}^{4-}} \times K_{\mathrm{f}} \nonumber\] \[\frac{3.13 \times 10^{-3} \ \mathrm{M}}{\left[\mathrm{Cd}^{2+}\right]\left(6.25 \times 10^{-4}\right)}=(0.367)\left(2.9 \times 10^{16}\right) \nonumber\] Solving gives [Cd ] = \(4.71 \times 10^{-16}\) M and a pCd of 15.33. We will use this approach when we learn how to sketch a complexometric titration curve. Calculate titration curves for the titration of 50.0 mL of \(5.00 \times 10^{-3}\) M Cd with 0.0100 M EDTA (a) at a pH of 10 and (b) at a pH of 7. Neither titration includes an auxiliary complexing agent. Compare your results with Figure 9.3.3
and comment on the effect of pH on the titration of Cd with EDTA. Let’s begin with the calculations at a pH of 10 where some of the EDTA is present in forms other than Y . To evaluate the titration curve, therefore, we need the conditional formation constant for CdY , which, from is \(K_f^{\prime} = 1.1 \times 10^{16}\). Note that the conditional formation constant is larger in the absence of an auxiliary complexing agent. The titration’s equivalence point requires \[V_{e q}=V_{\mathrm{EDTA}}=\frac{M_{\mathrm{Cd}} V_{\mathrm{Cd}}}{M_{\mathrm{EDTA}}}=\frac{\left(5.00 \times 10^{-3} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})}{(0.0100 \ \mathrm{M})}=25.0 \ \mathrm{mL} \nonumber\] of EDTA. Before the equivalence point, Cd is present in excess and pCd is determined by the concentration of unreacted Cd . For example, after adding 5.00 mL of EDTA, the total concentration of Cd is \[\left[\mathrm{Cd}^{2+}\right]=\frac{\left(5.00 \times 10^{-3} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})-(0.0100 \ \mathrm{M})(5.00 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+5.00 \ \mathrm{mL}} \nonumber\] which gives [Cd ] as \(3.64 \times 10^{-3}\) and pCd as 2.43. At the equivalence point all Cd initially in the titrand is now present as CdY . The concentration of Cd , therefore, is determined by the dissociation of the CdY complex. First, we calculate the concentration of CdY . \[\left[\mathrm{CdY}^{2-}\right]=\frac{\left(5.00 \times 10^{-3} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+25.00 \ \mathrm{mL}}=3.33 \times 10^{-3} \ \mathrm{M} \nonumber\] Next, we solve for the concentration of Cd in equilibrium with CdY . \[K_{f}^{\prime}=\frac{\left[\mathrm{CdY}^{2-}\right]}{\left[\mathrm{Cd}^{2+}\right] C_{\mathrm{EDTA}}}=\frac{3.33 \times 10^{-3}-x}{(x)(x)}=1.1 \times 10^{16} \nonumber\] Solving gives [Cd ] as \(5.50 \times 10^{-10}\) M or a pCd of 9.26 at the equivalence point. After the equivalence point, EDTA is in excess and the concentration of Cd is determined by the dissociation of the CdY complex. First, we calculate the concentrations of CdY and of unreacted EDTA. For example, after adding 30.0 mL of EDTA \[\left[\mathrm{CdY}^{2-}\right]=\frac{\left(5.00 \times 10^{-3} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+30.00 \ \mathrm{mL}}=3.12 \times 10^{-3} \ \mathrm{M} \nonumber\] \[C_{\mathrm{EDTA}}=\frac{(0.0100 \ \mathrm{M})(30.00 \ \mathrm{mL})-\left(5.00 \times 10^{-3} \ \mathrm{M}\right)(50.0 \ \mathrm{mL})}{50.0 \ \mathrm{mL}+30.00 \ \mathrm{mL}} \nonumber\] \[C_{\mathrm{EDTA}}=6.25 \times 10^{-4} \ \mathrm{M} \nonumber\] Substituting into the equation for the conditional formation constant \[K_{f}^{\prime}=\frac{\left[\mathrm{CdY}^{2-}\right]}{\left[\mathrm{Cd}^{2+}\right] C_{\mathrm{EDTA}}}=\frac{3.12 \times 10^{-3} \ \mathrm{M}}{(\mathrm{x})\left(6.25 \times 10^{-4} \ \mathrm{M}\right)}=1.1 \times 10^{16} \nonumber\] and solving for [Cd ] gives \(4.54 \times 10^{-16}\) M or a pCd of 15.34. The calculations at a pH of 7 are identical, except the conditional formation constant for CdY is \(1.5 \times 10^{13}\) instead of \(1.1 \times 10^{16}\). The following table summarizes results for these two titrations as well as the results from for the titration of Cd at a pH of 10 in the presence of 0.0100 M NH as an auxiliary complexing agent. Volume of EDTA (mL) pCd at pH 10 pCd at pH 10 w/ 0.0100 M NH pCd at pH 7 2.30 3.36 2.30 5.00 2.43 3.49 2.43 2.60 3.66 2.60 2.81 3.87 2.81 3.15 4.20 3.15 3.56 4.62 3.56 9.26 9.77 7.83 12.08 Examining these results allows us to draw several conclusions. First, in the absence of an auxiliary complexing agent the titration curve before the equivalence point is independent of pH (compare columns 2 and 4). Second, for any pH, the titration curve after the equivalence point is the same regardless of whether an auxiliary complexing agent is present (compare columns 2 and 3). Third, the largest change in pH through the equivalence point occurs at higher pHs and in the absence of an auxiliary complexing agent. For example, from 23.0 mL to 27.0 mL of EDTA the change in pCd is 11.38 at a pH of 10, 10.33 at a pH of 10 in the presence of 0.0100 M NH , and 8.52 at a pH of 7. To evaluate the relationship between a titration’s equivalence point and its end point, we need to construct only a reasonable approximation of the exact titration curve. In this section we demonstrate a simple method for sketching a complexation titration curve. Our goal is to sketch the titration curve quickly, using as few calculations as possible. Let’s use the titration of 50.0 mL of \(5.00 \times 10^{-3}\) M Cd with 0.0100 M EDTA in the presence of 0.0100 M NH to illustrate our approach. This is the same example we used in developing the calculations for a complexation titration curve. You can review the results of that calculation in and . We begin by calculating the titration’s equivalence point volume, which, as we determined earlier, is 25.0 mL. Next, we draw our axes, placing pCd on the -axis and the titrant’s volume on the -axis. To indicate the equivalence point’s volume, we draw a vertical line that intersects the -axis at 25.0 mL of EDTA. Figure 9.3.4
a shows the result of the first step in our sketch. Before the equivalence point, Cd is present in excess and pCd is determined by the concentration of unreacted Cd . Because not all unreacted Cd is free—some is complexed with NH —we must account for the presence of NH . The calculations are straightforward, as we saw earlier. Figure 9.3.4
b shows the pCd after adding 5.00 mL and 10.0 mL of EDTA. The third step in sketching our titration curve is to add two points after the equivalence point. Here the concentration of Cd is controlled by the dissociation of the Cd –EDTA complex. Beginning with the conditional formation constant \[K_{f}^{\prime}=\frac{\left[\mathrm{CdY}^{2-}\right]}{\left[\mathrm{Cd}^{2+}\right] C_{\mathrm{EDTA}}} = \alpha_{\text{Y}^{4-}} \times K_{f}=(0.367)\left(2.9 \times 10^{16}\right)=1.1 \times 10^{16} \nonumber\] we take the log of each side and rearrange, arriving at \[\begin{array}{c}{\log K_{f}^{\prime}=-\log \left[\mathrm{Cd}^{2+}\right]+\log \frac{\left[\mathrm{CdY}^{2-}\right]}{C_{\mathrm{EDTA}}}} \\ {\mathrm{pCd}=\log K_{f}^{\prime}+\log \frac{C_{\mathrm{EDTA}}}{\left[\mathrm{CdY}^{2-}]\right.}}\end{array} \nonumber\] Recall that we can use either of our two possible conditional formation constants, \(K_f^{\prime}\) or \(K_f^{\prime \prime}\), to determine the composition of the system at equilibrium. Note that after the equivalence point, the titrand is a metal–ligand complexation buffer, with pCd determined by and [CdY ]. The buffer is at its lower limit of \(\text{pCd} = \log{K_f^{\prime}} - 1\) when \[\frac{C_{\mathrm{EDTA}}}{\left[\mathrm{CdY}^{2-}\right]} = \frac {(\text{mole EDTA})_\text{added} - (\text{mol Cd}^{2+})_\text{initial}} {(\text{mol Cd}^{2+})_\text{initial}} = \frac {1} {10} \nonumber\] Making appropriate substitutions and solving, we find that \[\frac{M_{\mathrm{EDTA}} V_{\mathrm{EDTA}}-M_{\mathrm{Cd}} V_{\mathrm{Cd}}}{M_{\mathrm{Cd}} V_{\mathrm{Cd}}}=\frac{1}{10} \nonumber\] \[M_{\mathrm{EDTA}} V_{\mathrm{EDTA}}-M_{\mathrm{Cd}} V_{\mathrm{Cd}}=0.1 \times M_{\mathrm{Cd}} V_{\mathrm{Cd}} \nonumber\] \[V_{\mathrm{EDTA}}=\frac{1.1 \times M_{\mathrm{Cd}} V_{\mathrm{Cd}}}{M_{\mathrm{EDTA}}}=1.1 \times V_{e q} \nonumber\] Thus, when the titration reaches 110% of the equivalence point volume, pCd is \(\log{K_f^{\prime}} - 1\). A similar calculation should convince you that pCd is \(\log{K_f^{\prime}} + 1\) when the volume of EDTA is \(2 \times V_\text{eq}\). Figure 9.3.4
c shows the third step in our sketch. First, we add a ladder diagram for the CdY complex, including its buffer range, using its \(\log{K_f^{\prime}}\) value of 16.04. Next, we add two points, one for pCd at 110% of (a pCd of 15.04 at 27.5 mL) and one for pCd at 200% of (a pCd of 16.04 at 50.0 mL). Next, we draw a straight line through each pair of points, extending each line through the vertical line that indicates the equivalence point’s volume (Figure 9.3.4
d). Finally, we complete our sketch by drawing a smooth curve that connects the three straight-line segments (Figure 9.3.4
e). A comparison of our sketch to the exact titration curve (Figure 9.3.4
f) shows that they are in close agreement. Our treatment here is general and applies to any complexation titration using EDTA as a titrant. Sketch titration curves for the titration of 50.0 mL of \(5.00 \times 10^{-3}\) M Cd with 0.0100 M EDTA (a) at a pH of 10 and (b) at a pH of 7. Compare your sketches to the calculated titration curves from . The figure below shows a sketch of the titration curves. The two black points before the equivalence point ( = 5 mL, pCd= 2.43 and = 15 mL, pCd= 2.81) are the same for both pHs and taken from the results of Exercise 9.3.1
. The two black points after the equivalence point for a pH of 7 ( = 27.5 mL, pCd= 12.2 and = 50 mL, pCd= 13.2) are plotted using the \(\log{K_f^{\prime}}\) of 13.2 for CdY . The two points after the equivalence point for a pH of 10 ( = 27.5 mL, pCd= 15.0 and = 50 mL, pCd= 16.0) are plotted using the \(\log{K_f^{\prime}}\) of 16.0 for CdY . The points in are the calculations from for a pH of 10, and the points in are the calculations from for a pH of 7. The equivalence point of a complexation titration occurs when we react stoichiometrically equivalent amounts of the titrand and titrant. As is the case for an acid–base titration, we estimate the equivalence point for a complexation titration using an experimental end point. A variety of methods are available for locating the end point, including indicators and sensors that respond to a change in the solution conditions. Most indicators for complexation titrations are organic dyes—known as —that form stable complexes with metal ions. The indicator, In , is added to the titrand’s solution where it forms a stable complex with the metal ion, MIn . As we add EDTA it reacts first with free metal ions, and then displaces the indicator from MIn . \[\text{MIn}^{n-}(aq) + \text{Y}^{4-}(aq) \rightarrow \text{MY}^{2-}(aq) + \text{In}^{m-}(aq) \nonumber\] If MIn and In have different colors, then the change in color signals the end point. The accuracy of an indicator’s end point depends on the strength of the metal–indicator complex relative to the strength of the metal–EDTA complex. If the metal–indicator complex is too strong, the change in color occurs after the equivalence point. If the metal–indicator complex is too weak, however, the end point occurs before we reach the equivalence point. Most metallochromic indicators also are weak acids. One consequence of this is that the conditional formation constant for the metal–indicator complex depends on the titrand’s pH. This provides some control over an indicator’s titration error because we can adjust the strength of a metal–indicator complex by adjusted the pH at which we carry out the titration. Unfortunately, because the indicator is a weak acid, the color of the uncomplexed indicator also may change with pH. Figure 9.3.5
, for example, shows the color of the indicator calmagite as a function of pH and pMg, where H In , HIn , and In are different forms of the uncomplexed indicator, and MgIn is the Mg –calmagite complex. Because the color of calmagite’s metal–indicator complex is red, its use as a metallochromic indicator has a practical pH range of approximately 8.5–11 where the uncomplexed indicator, HIn , has a blue color. Table 9.3.5
provides examples of metallochromic indicators and the metal ions and pH conditions for which they are useful. Even if a suitable indicator does not exist, it often is possible to complete an EDTA titration by introducing a small amount of a secondary metal–EDTA complex if the secondary metal ion forms a stronger complex with the indicator and a weaker complex with EDTA than the analyte. For example, calmagite has a poor end point when titrating Ca with EDTA. Adding a small amount of Mg –EDTA to the titrand gives a sharper end point. Because Ca forms a stronger complex with EDTA, it displaces Mg , which then forms the red-colored Mg –calmagite complex. At the titration’s end point, EDTA displaces Mg from the Mg –calmagite complex, signaling the end point by the presence of the uncomplexed indicator’s blue form. all metal ions carry a +2 charge expect for iron, which is +3 metal ions in font have poor end points An important limitation when using a metallochromic indicator is that we must be able to see the indicator’s change in color at the end point. This may be difficult if the solution is already colored. For example, when titrating Cu with EDTA, ammonia is used to adjust the titrand’s pH. The intensely colored \(\text{Cu(NH}_3)_2^{4+}\) complex obscures the indicator’s color, making an accurate determination of the end point difficult. Other absorbing species present within the sample matrix may also interfere. This often is a problem when analyzing clinical samples, such as blood, or environmental samples, such as natural waters. If at least one species in a complexation titration absorbs electromagnet- ic radiation, then we can identify the end point by monitoring the titrand’s absorbance at a carefully selected wavelength. For example, we can identify the end point for a titration of Cu with EDTA in the presence of NH by monitoring the titrand’s absorbance at a wavelength of 745 nm, where the \(\text{Cu(NH}_3)_2^{4+}\) complex absorbs strongly. At the beginning of the titration the absorbance is at a maximum. As we add EDTA, however, the reaction \[\text{Cu(NH}_3)_4^{2+}(aq) + \text{Y}^{4-} \rightleftharpoons \text{CuY}^{2-}(aq) + 4\text{NH}_3(aq) \nonumber\] decreases the concentration of \(\text{Cu(NH}_3)_2^{4+}\) and decreases the absorbance until we reach the equivalence point. After the equivalence point the absorbance essentially remains unchanged. The resulting spectrophotometric titration curve is shown in Figure 9.3.6
a. Note that the titration curve’s -axis is not the measured absorbance, , but a corrected absorbance, \[A_\text{corr} = A_\text{meas} \times \frac {V_\text{EDTA} + V_\text{Cu}} {V_\text{Cu}} \nonumber\] where and are, respectively, the volumes of EDTA and Cu. Correcting the absorbance for the titrand’s dilution ensures that the spectrophotometric titration curve consists of linear segments that we can extrapolate to find the end point. Other common spectrophotometric titration curves are shown in Figures 9.3.6
b-f. The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical complexation titrimetric method. Although each method is unique, the following description of the determination of the hardness of water provides an instructive example of a typical procedure. The description here is based on Method 2340C as published in , 20th Ed., American Public Health Association: Washington, D. C., 1998. The operational definition of water hardness is the total concentration of cations in a sample that can form an insoluble complex with soap. Although most divalent and trivalent metal ions contribute to hardness, the two most important metal ions are Ca and Mg . Hardness is determined by titrating with EDTA at a buffered pH of 10. Calmagite is used as an indicator. Hardness is reported as mg CaCO /L. Select a volume of sample that requires less than 15 mL of titrant to keep the analysis time under 5 minutes and, if necessary, dilute the sample to 50 mL with distilled water. Adjust the sample’s pH by adding 1–2 mL of a pH 10 buffer that contains a small amount of Mg –EDTA. Add 1–2 drops of indicator and titrate with a standard solution of EDTA until the red-to-blue end point is reached (Figure 9.3.7
). 1. Why is the sample buffered to a pH of 10? What problems might you expect at a higher pH or a lower pH? Of the two primary cations that contribute to hardness, Mg forms the weaker complex with EDTA and is the last cation to react with the titrant. Calmagite is a useful indicator because it gives a distinct end point when titrating Mg (see ). Because of calmagite’s acid–base properties, the range of pMg values over which the indicator changes color depends on the titrand’s pH ( ). Figure 9.3.8
shows the titration curve for a 50-mL solution of 10 M Mg with 10 M EDTA at pHs of 9, 10, and 11. Superimposed on each titration curve is the range of conditions for which the average analyst will observe the end point. At a pH of 9 an early end point is possible, which results in a negative determinate error. A late end point and a positive determinate error are possible if the pH is 11. 2. Why is a small amount of the Mg –EDTA complex added to the buffer? The titration’s end point is signaled by the indicator calmagite. The indicator’s end point with Mg is distinct, but its change in color when titrating Ca does not provide a good end point (see ). If the sample does not contain any Mg as a source of hardness, then the titration’s end point is poorly defined, which leads to an inaccurate and imprecise result. Adding a small amount of Mg –EDTA to the buffer ensures that the titrand includes at least some Mg . Because Ca forms a stronger complex with EDTA, it displaces Mg from the Mg –EDTA complex, freeing the Mg to bind with the indicator. This displacement is stoichiometric, so the total concentration of hardness cations remains unchanged. The displacement by EDTA of Mg from the Mg –indicator complex signals the titration’s end point. 3. Why does the procedure specify that the titration take no longer than 5 minutes? A time limitation suggests there is a kinetically-controlled interference, possibly arising from a competing chemical reaction. In this case the interference is the possible precipitation of CaCO at a pH of 10. Although many quantitative applications of complexation titrimetry have been replaced by other analytical methods, a few important applications continue to find relevance. In the section we review the general application of complexation titrimetry with an emphasis on applications from the analysis of water and wastewater. First, however, we discuss the selection and standardization of complexation titrants. EDTA is a versatile titrant that can be used to analyze virtually all metal ions. Although EDTA is the usual titrant when the titrand is a metal ion, it cannot be used to titrate anions, for which Ag or Hg are suitable titrants. Solutions of EDTA are prepared from its soluble disodium salt, Na H Y•2H O, and standardized by titrating against a solution made from the primary standard CaCO . Solutions of Ag and Hg are prepared using AgNO and Hg(NO ) , both of which are secondary standards. Standardization is accomplished by titrating against a solution prepared from primary standard grade NaCl. Complexation titrimetry continues to be listed as a standard method for the determination of hardness, Ca , CN , and Cl in waters and wastewaters. The evaluation of hardness was described earlier in Representative Method 9.3.1. The determination of Ca is complicated by the presence of Mg , which also reacts with EDTA. To prevent an interference the pH is adjusted to 12–13, which precipitates Mg as Mg(OH) . Titrating with EDTA using murexide or Eriochrome Blue Black R as the indicator gives the concentration of Ca . Cyanide is determined at concentrations greater than 1 mg/L by making the sample alkaline with NaOH and titrating with a standard solution of AgNO to form the soluble \(\text{Ag(CN)}_2^-\) complex. The end point is determined using -dimethylaminobenzalrhodamine as an indicator, with the solution turning from a yellow to a salmon color in the presence of excess Ag . Chloride is determined by titrating with Hg(NO ) , forming HgCl . The sample is acidified to a pH of 2.3–3.8 and diphenylcarbazone, which forms a colored complex with excess Hg , serves as the indicator. The pH indicator xylene cyanol FF is added to ensure that the pH is within the desired range. The initial solution is a greenish blue, and the titration is carried out to a purple end point. The quantitative relationship between the titrand and the titrant is determined by the titration reaction’s stoichiometry. For a titration using EDTA, the stoichiometry is always 1:1. The concentration of a solution of EDTA is determined by standardizing against a solution of Ca prepared using a primary standard of CaCO . A 0.4071-g sample of CaCO is transferred to a 500-mL volumetric flask, dissolved using a minimum of 6 M HCl, and diluted to volume. After transferring a 50.00-mL portion of this solution to a 250-mL Erlenmeyer flask, the pH is adjusted by adding 5 mL of a pH 10 NH –NH Cl buffer that contains a small amount of Mg –EDTA. After adding calmagite as an indicator, the solution is titrated with the EDTA, requiring 42.63 mL to reach the end point. Report the molar concentration of EDTA in the titrant. The primary standard of Ca has a concentration of \[\frac {0.4071 \text{ g CaCO}_3}{0.5000 \text{ L}} \times \frac {1 \text{ mol Ca}^{2+}}{100.09 \text{ g CaCO}_3} = 8.135 \times 10^{-3} \text{ M Ca}^{2+} \nonumber\] The moles of Ca in the titrand is \[8.135 \times 10^{-3} \text{ M Ca}^{2+} \times 0.05000 \text{ L} = 4.068 \times 10^{-4} \text{ mol Ca}^{2+} \nonumber\] which means that \(4.068 \times 10^{-4}\) moles of EDTA are used in the titration. The molarity of EDTA in the titrant is \[\frac {4.068 \times 10^{-4} \text{ mol Ca}^{2+}}{0.04263 \text{ L}} = 9.543 \times 10^{-3} \text{ M EDTA} \nonumber\] A 100.0-mL sample is analyzed for hardness using the procedure outlined in , requiring 23.63 mL of 0.0109 M EDTA. Report the sample’s hardness as mg CaCO /L. In an analysis for hardness we treat the sample as if Ca is the only metal ion that reacts with EDTA. The grams of Ca in the sample, therefore, are \[(0.0109 \text{ M EDTA})(0.02363 \text{ L}) \times \frac {1 \text{ mol Ca}^{2+}}{\text{mol EDTA}} = 2.58 \times 10^{-4} \text{ mol Ca}^{2+} \nonumber\] \[2.58 \times 10^{-4} \text{ mol Ca}^{2+} \times \frac {1 \text{ mol CaCO}_3}{\text{mol Ca}^{2+}} \times \frac {100.09 \text{ g CaCO}_3}{\text{mol CaCO}_3} = 0.0258 \text{ g CaCO}_3 \nonumber\] and the sample’s hardness is \[\frac {0.0258 \text{ g CaCO}_3}{0.1000 \text{ L}} \times \frac {1000 \text{ mg}}{\text{g}} = 258 \text{ g CaCO}_3\text{/L} \nonumber\] As shown in the following example, we can extended this calculation to complexation reactions that use other titrants. The concentration of Cl in a 100.0-mL sample of water from a freshwater aquifer is tested for the encroachment of sea water by titrating with 0.0516 M Hg(NO ) . The sample is acidified and titrated to the diphenylcarbazone end point, requiring 6.18 mL of the titrant. Report the concentration of Cl , in mg/L, in the aquifer. The reaction between Cl and Hg produces a metal–ligand complex of HgCl . Each mole of Hg + reacts with 2 moles of Cl ; thus \[\frac {0.0516 \text{ mol Hg(NO}_3)_2}{\text{L}} \times 0.00618 \text{ L} \times \frac {2 \text{ mol Cl}^-}{\text{mol Hg(NO}_3)_2} \times \frac {35.453 \text{ g Cl}^-}{\text{mol Cl}^-} = 0.0226 \text{ g Cl}^- \nonumber\] are in the sample. The concentration of Cl in the sample is \[\frac {0.0226 \text{ g Cl}^-}{0.1000 \text{ L}} \times \frac {1000 \text{ mg}}{\text{g}} = 226 \text{ mg/L} \nonumber\] A 0.4482-g sample of impure NaCN is titrated with 0.1018 M AgNO , requiring 39.68 mL to reach the end point. Report the purity of the sample as %w/w NaCN. The titration of CN with Ag produces the metal-ligand complex \(\text{Ag(CN)}_2^-\); thus, each mole of AgNO reacts with two moles of NaCN. The grams of NaCN in the sample is \[(0.1018 \text{ M AgNO}_3)(0.03968 \text{ L}) \times \frac {2 \text{ mol NaCN}}{\text{mol AgNO}_3} \times \frac {49.01 \text{ g NaCN}}{\text{mol NaCN}} = 0.3959 \text{ g NaCN} \nonumber\] and the purity of the sample is \[\frac {0.3959 \text{ g NaCN}}{0.4482 \text{ g sample}} \times 100 = 88.33 \text{% w/w NaCN} \nonumber\] Finally, complex titrations involving multiple analytes or back titrations are possible. An alloy of chromel that contains Ni, Fe, and Cr is analyzed by a complexation titration using EDTA as the titrant. A 0.7176-g sample of the alloy is dissolved in HNO and diluted to 250 mL in a volumetric flask. A 50.00-mL aliquot of the sample, treated with pyrophosphate to mask the Fe and Cr, requires 26.14 mL of 0.05831 M EDTA to reach the murexide end point. A second 50.00-mL aliquot is treated with hexamethylenetetramine to mask the Cr. Titrating with 0.05831 M EDTA requires 35.43 mL to reach the murexide end point. Finally, a third 50.00-mL aliquot is treated with 50.00 mL of 0.05831 M EDTA, and back titrated to the murexide end point with 6.21 mL of 0.06316 M Cu . Report the weight percents of Ni, Fe, and Cr in the alloy. The stoichiometry between EDTA and each metal ion is 1:1. For each of the three titrations, therefore, we can write an equation that relates the moles of EDTA to the moles of metal ions that are titrated. titration 1: mol Ni = mol EDTA
titration 2: mol Ni +mol Fe = mol EDTA
titration 3: mol Ni + mol Fe + mol Cr + mol Cu = mol EDTA We use the first titration to determine the moles of Ni in our 50.00-mL portion of the dissolved alloy. The titration uses \[\frac {0.05831 \text{ mol EDTA}}{\text{L}} \times 0.02614 \text{ L} = 1.524 \times 10^{-3} \text{ mol EDTA} \nonumber\] which means the sample contains \(1.524 \times 10^{-3}\) mol Ni. Having determined the moles of EDTA that react with Ni, we use the second titration to determine the amount of Fe in the sample. The second titration uses \[\frac {0.05831 \text{ mol EDTA}}{\text{L}} \times 0.03543 \text{ L} = 2.066 \times 10^{-3} \text{ mol EDTA} \nonumber\] of which \(1.524 \times 10^{-3}\) mol are used to titrate Ni. This leaves \(5.42 \times 10^{-4}\) mol of EDTA to react with Fe; thus, the sample contains \(5.42 \times 10^{-4}\) mol of Fe. Finally, we can use the third titration to determine the amount of Cr in the alloy. The third titration uses \[\frac {0.05831 \text{ mol EDTA}}{\text{L}} \times 0.0500 \text{ L} = 3.926 \times 10^{-3} \text{ mol EDTA} \nonumber\] of which \(1.524 \times 10^{-3}\) mol are used to titrate Ni and \(5.42 \times 10^{-4}\) mol are used to titrate Fe. This leaves \(8.50 \times 10^{-4}\) mol of EDTA to react with Cu and Cr. The amount of EDTA that reacts with Cu is \[\frac {0.06316 \text{ mol Cu}^{2+}}{\text{L}} \times 0.006211 \text{ L} \times \frac {1 \text{ mol EDTA}}{\text{mol Cu}^{2+}} = 3.92 \times 10^{-4} \text{ mol EDTA} \nonumber\] leaving \(4.58 \times 10^{-4}\) mol of EDTA to react with Cr. The sample, therefore, contains \(4.58 \times 10^{-4}\) mol of Cr. Having determined the moles of Ni, Fe, and Cr in a 50.00-mL portion of the dissolved alloy, we can calculate the %w/w of each analyte in the alloy. \[\frac {1.524 \times 10^{-3} \text{ mol Ni}}{50.00 \text{ mL}} \times \frac {58.69 \text{ g Ni}}{\text{mol Ni}} = 0.4472 \text{ g Ni} \nonumber\] \[\frac {0.4472 \text{ g Ni}}{0.7176 \text{ g sample}} \times 100 = 62.32 \text{% w/w Ni} \nonumber\] \[\frac {5.42 \times 10^{-4} \text{ mol Fe}}{50.00 \text{ mL}} \times \frac {55.845 \text{ g Fe}}{\text{mol Fe}} = 0.151 \text{ g Fe} \nonumber\] \[\frac {0.151 \text{ g Fe}}{0.7176 \text{ g sample}} \times 100 = 21.0 \text{% w/w Fe} \nonumber\] \[\frac {4.58 \times 10^{-4} \text{ mol Cr}}{50.00 \text{ mL}} \times \frac {51.996 \text{ g Cr}}{\text{mol Cr}} = 0.119 \text{ g Cr} \nonumber\] \[\frac {0.119 \text{ g Cr}}{0.7176 \text{ g sample}} \times 100 = 16.6 \text{% w/w Cr} \nonumber\] An indirect complexation titration with EDTA can be used to determine the concentration of sulfate, \(\text{SO}_4^{2-}\), in a sample. A 0.1557-g sample is dissolved in water and any sulfate present is precipitated as BaSO by adding Ba(NO ) . After filtering and rinsing the precipitate, it is dissolved in 25.00 mL of 0.02011 M EDTA. The excess EDTA is titrated with 0.01113 M Mg , requiring 4.23 mL to reach the end point. Calculate the %w/w Na SO in the sample. The total moles of EDTA used in this analysis is \[(0.02011 \text{ M EDTA})(0.02500 \text{ L}) = 5.028 \times 10^{-4} \text{ mol EDTA} \nonumber\] Of this, \[(0.01113 \text{ M Mg}^{2+})(0.00423 \text{ L}) \times \frac {1 \text{ mol EDTA}}{\text{mol Mg}^{2+}} = 4.708 \times 10^{-5} \text{ mol EDTA} \nonumber\] are consumed in the back titration with Mg , which means that \[5.028 \times 10^{-4} \text{ mol EDTA} - 4.708 \times 10^{-5} \text{ mol EDTA} = 4.557 \times 10^{-4} \text{ mol EDTA} \nonumber\] react with the BaSO . Each mole of BaSO reacts with one mole of EDTA; thus \[4.557 \times 10^{-4} \text{ mol EDTA} \times \frac {1 \text{ mol BaSO}_4}{\text{mol EDTA}} \times \frac {1 \text{ mol Na}_2\text{SO}_4}{\text{mol BaSO}_4} \times \frac {142.04 \text{ g Na}_2\text{SO}_4}{\text{mol Na}_2\text{SO}_4} = 0.06473 \text{ g Na}_2\text{SO}_4 \nonumber\] \[\frac{0.06473 \text{ g Na}_2\text{SO}_4}{0.1557 \text{ g sample}} \times 100 = 41.57 \text{% w/w Na}_2\text{SO}_4 \nonumber\] The scale of operations, accuracy, precision, sensitivity, time, and cost of a complexation titration are similar to those described earlier for acid–base titrations. Complexation titrations, however, are more selective. Although EDTA forms strong complexes with most metal ion, by carefully controlling the titrand’s pH we can analyze samples that contain two or more analytes. The reason we can use pH to provide selectivity is shown in Figure 9.3.9
a. A titration of Ca at a pH of 9 has a distinct break in the titration curve because the conditional formation constant for CaY of \(2.6 \times 10^9\) is large enough to ensure that the reaction of Ca and EDTA goes to completion. At a pH of 3, however, the conditional formation constant of 1.23 is so small that very little Ca reacts with the EDTA. Suppose we need to analyze a mixture of Ni and Ca . Both analytes react with EDTA, but their conditional formation constants differ significantly. If we adjust the pH to 3 we can titrate Ni with EDTA without titrating Ca (Figure 9.3.9
b). When the titration is complete, we adjust the titrand’s pH to 9 and titrate the Ca with EDTA. A spectrophotometric titration is a particularly useful approach for analyzing a mixture of analytes. For example, as shown in Figure 9.3.10
, we can determine the concentration of a two metal ions if there is a difference between the absorbance of the two metal-ligand complexes. | 45,596 | 3,471 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Significant_Digits/Significant_Figures |
Significant figures are used to keep track of the quality (variability) of measurements. This includes propagating that information during calculations using the measurements. The purpose of this page is to help you organize the information about significant figures -- to help you set priorities. Sometimes students are overwhelmed by too many rules, and lack guidance about how to sort through them. What is the purpose? Which rules are most important? The following points as being most important: I will de-emphasize the following: Let's break that into two parts. One is about the information per se, and the other is about priorities, about the approach to thinking about Significant Digits. The information here should agree, for the most part. However, what may be different is the order of presenting things, with a different perspective in the approach -- the steps -- to learning Significant Digits. We will all end up in the same place. If you were completely happy with how the Significant Digits topic is presented in your own course, you probably wouldn't be reading this page. Think of it as another approach -- to the same thing. Sometimes, looking at things differently can help. Trying two approaches can be better than trying only one. There is no claim that one approach is "right" or even "better". If there is a discrepancy between any information here and your own course, please let me know -- or check with your own instructor. Some details are a matter of preference. In the lab. When you take a measurement, you record not only the value of the measurement, but also some information about its quality. Using Significant Digits is one simple way to record the quality of the information. A simple and useful statement is that the significant figures (Significant Digits) are the digits that are certain in the measurement plus one uncertain digit. Significant Digits is a set of arbitrary rules. Almost everything about Significant Digits follows from how you make the measurements, and then from understanding how numbers work when you do calculations. Unfortunately, there are "special cases" that can come up with Significant Digits. If all the rules are presented together, it is easy to get lost in the rules. Better -- and what we will do here -- is to emphasize the logic of using Significant Digits. This involves a few basic ideas, which can be stated as rules. We will leave special cases for a while, so they do not confuse the big picture. The number of high priority rules about Significant Digits is small. The best way to start with Significant Digits is in the lab, taking measurements. An alternative is to use an activity that simulates taking measurements -- of various accuracy. We will do that here, using drawings of measurement scales. A bad way to start with Significant Digits is to learn a list of rules. When you take a measurement, you write down the correct number of digits. You write down the significant digits. That is, the way you write a number conveys some information about how accurate it is. It is up to you to determine how many digits are worth writing down. It is important that you do so, since what you write conveys not only the measurement but something about its quality. For many common lab instruments, the proper procedure is to estimate one digit beyond those shown directly by the measurement scale. If that one estimated digit seems meaningful, then it is indeed a significant digit. The scale shown here is a "typical" measurement scale. The specific scale is from a 10 mL graduated cylinder -- shown horizontally here for convenience. The arrow marks the position of a measurement.
Our goal is to read the scale at the position of the arrow. Let's go through this in detail. How meaningful is a drawing of a measurement scale, such as the one in the example above? It illustrates one particular issue very well: how to read a scale per se, figure out what the marks and labels mean, and how to estimate the final digit. Real measuring instruments, such as graduated cylinders, have those issues. Depending on the situation, there may be other issues that affect the ease of reading. In the drawing above, the goal is to read a well-defined arrow. With a real graduated cylinder, you may need to deal with a meniscus (curved surface) and parallax. Those issues are beyond our topic here. In estimating that last digit, be sure to write down the zero if your best estimate is indeed zero. For example, if the last digit reflects hundredths of a mL, you might estimate in one case that there are 6 hundredths; thus you would write 6 as the last digit (e.g., 8.16 mL -- 3 Significant Digits). But you might (in another case) estimate that there are 0 hundredths; it is important that you write that zero (e.g., 8.10 mL -- 3 Significant Digits). That final zero says you looked for hundredths and found none. If you wrote only 8.1 mL (2 Significant Digits), it would imply that you did not look for hundredths. The arrow below appears to be "right on" the "4.7" line. (Let's assume that. The point here is to deal with the case where you think the arrow is "on" the line.) Thus we estimate that the hundredths place is 0. The proper reading, then, is 4.70 mL (3 Significant Digits). That final zero means that we looked for hundredths, and found none. If we wrote 4.7 mL (2 Significant Digits), it would imply that we didn't look for hundredths. The scale shown in Example 2 is the same scale as in Example 1. In Example 1 our proper reading had 3 Significant Digits. That is also true in Example 2. That final 0 in Example 2 is an estimate; it is entirely equivalent to the final 8 estimated in Example 1. There are a couple of ways to approach this: Both approaches will work. They reflect the same principles. Often, simply looking at the number will be sufficient. However, when you are not sure, it helps to go back to basics: think about the underlying measurement. We will illustrate this in the next section, on zeroes -- the situation most likely to cause confusion. We tend to spend more time on this issue than it really is worth. Only one tenth of all digits are zeroes, yet the bulk of a list of Significant Digits rules may be about how to treat the zeroes. Many zeroes are clear enough, but indeed it can take a bit of thought to decide whether some zeroes are or are not significant. If you understand where Significant Digits come from, then whether a zero is significant should be clear -- at least most of the time. If you are learning Significant Digits by memorizing rules, then you are doing it the hard way -- not understanding the meaning. If, for whatever reason, you are struggling with Significant Digits, the problem of the zeroes is a low priority problem. Here is what I usually suggest to students. Don't worry too much about the rules for zeroes, especially when you are just starting. As you go on, ask about specific cases where you are not sure about the zeroes. That way, you will gradually learn how to deal with the zeroes, but not get bogged down with what can seem to be a bunch of picky rules. The key point in deciding whether a zero is significant is to decide if it is part of the measurement, or simply a digit that is there to "fill space". The next section will help with much of the "zeroes problem". When a number is written in standard scientific (exponential) notation format, there should be no problem with zeroes. In this format, with one digit before the decimal point and only Significant Digits after the decimal point, all digits shown are significant. How many Significant Digits are in the measurement 0.00023456 m? In scientific notation that is 2.3456x10 m. 5 Significant Digits. Scientific notation makes clear that all the zeroes to the left are not significant. The first zero is just decorative and could be omitted; the others are place-holders, so you can show that the 2 is the fourth decimal place. The "rule" that covers this case may be stated: zeroes on the left end of a number are not significant -- regardless of where the decimal point is. Hopefully, the example, showing how this plays out in scientific notation, makes this rule clearer. How many Significant Digits are in the measurement 0.00023450 m? In scientific notation that is 2.3450x10 m. 5 Significant Digits. That final zero is part of the measurement. If it weren't, why would it be there? The "rule" that covers this case may be stated: zeroes on the right end of a number are significant -- they are to the right of the decimal point. This rule may seem confusing in words, but showing the case in scientific notation should make it clearer. How many Significant Digits are in the measurement 234000 m? Why did I choose to not consider the zeroes significant? Maybe they are significant. Or maybe one of them is significant. The problem is that there is no way to tell from the number 234000 whether those zeroes are significant or are merely place holders, telling us (for example) that the 4 is in the thousands place. So why choose to make them not significant? First, that is the conservative position. I don't know whether they are significant, and to claim that they are is an unwarranted claim of quality. Second, 3 Significant Digits is reasonable -- a common way to measure distances; 6 Significant Digits is not likely. What if the person making the measurement knows that the measurement is good to 4 Significant Digits, with the first zero being significant? Then, somehow, they need to say so. One good way is to put the measurement in proper scientific notation in the first place: 2.340x10 m, 4 Significant Digits. It depends on the type of calculation. Each math operation has its own rules for handling Significant Digits. More precisely, there is one rule each for: Those three rules are distinct; you must be careful to use the right rule for the right operation. But there is good news: The multiplication rule is by far the most important in basic chemistry -- and it is perhaps also the simplest. So, as a matter of priority, emphasize the multiplication rule. When you have mastered it, you can go on and learn the addition rule. It is useful, though much less important. Whether you need the rule for logs will depend on your course; some courses manage to avoid this rule completely. In summary ... there are three rules, but there is a clear set of priorities with them. Emphasize the multiplication rule. It is the most important rule, and the easiest one. If you multiply two numbers with the same number of Significant Digits, then the answer should have that same number of Significant Digits. If you multiply together two numbers that each have 4 Significant Digits, then the answer should have 4 Significant Digits. Multiply 12.3 cm by 2.34 cm. Doing the arithmetic on the calculator gives 28.782. In this case, each number has 3 Significant Digits. Thus we report the result to 3 Significant Digits. Proper rounding of 28.782 to 3 Significant Digits gives 28.8. With the units, the final answer is 28.8 cm . If you multiply together two numbers with different numbers of Significant Digits, then the answer should have the same number of Significant Digits as the "weaker" number. Hm, that is a lot of words. An example should help. Multiply a number with 3 Significant Digits and a number with 4 Significant Digits. Keep 3 Significant Digits in the answer. Multiply 24 cm by 268 cm. Doing the arithmetic on the calculator gives 6432. One measurement has 2 Significant Digits and one has 3 Significant Digits. The 2 Significant Digits number is "weaker": it has less information; it has only two digits of information in it. That is, the 2 Significant Digits number limits the calculation. Thus we report the result to 2 Significant Digits. Proper rounding of 6432 to 2 Significant Digits gives 6400. That is clearer in scientific notation, as 6.4x10 . With the units, the final answer is 6.4x10 cm . [Recall section , especially Example 5.] The following two examples serve as reminders that it is important to understand the context of the particular problem. In Example 7, we reported the product of 24 & 268 to 2 Significant Digits. But in Example 8, which follows, we report the product of those same two numbers to 3 Significant Digits. Both are correct -- because the contexts are different. Example 9 reminds us of another issue in carefully recording measurements. You have an object that is 268 cm long. What would be the total length of 24 such objects? The calculator gives 6432, as in Example 7. Now we look at the Significant Digits; we must carefully think about what each number means. "268 cm" is an ordinary measurement; it has 3 Significant Digits. But the "24" is a count, and is taken as exact (with no uncertainty). That is, the "24" does not limit the calculation, and we report 3 Significant Digits. With the units, the final answer is 6.43x103 cm. You measure the sides of a rectangle. The sides are 28.2 cm and 25 cm. What is the area? But before you calculate the area... There is probably something wrong with the statement of this question. What? What's wrong? Well, we have an object, approximately square. Someone has measured two sides. One would think they used the same measuring instrument -- the same ruler. But the two reported measurements are inconsistent. One is reported to the nearest cm, and one is reported to the nearest tenth. That is suspicious. Why were they not reported the same way? The purpose of this example is to remind you of the importance of reading the measuring instrument carefully and consistently, and recording the final zero if indeed that is your estimate. There is no need to carry out the calculation in this case. For students who are just starting chemistry, the addition rule for Significant Digits is not as important as the multiplication rule. The intent of that statement is to help you set priorities. Learn one thing at a time -- especially if you are finding the topic difficult. The multiplication rule is more important; learn it first and get comfortable with it. Most instructors will want you to learn the addition rule. I am not suggesting otherwise. Again, the emphasis here is to guide you to learn one thing at a time. Here is an example of a basic chem situation that would seem to involve the addition rule, yet where using that rule is not really needed. Consider calculating the molar mass (formula weight) of a compound, say KOH. Using the atomic masses shown on the periodic table, the molar mass of KOH is 39.10 + 16.00 + 1.008 = 56.108 (in g/mol). One answer might be to use the Significant Digits rule for addition and note that the result is only good to the hundredths place. Therefore, we round it to 56.11 g/mol. However, that may be unnecessary -- and even undesirable. The reason for calculating a molar mass is to use it in a real calculation. In real cases, it is usually fine to calculate molar mass by using the atomic masses shown on your periodic table. No rounding, at least now. When you use the molar mass for a calculation, you round the final result. At this step, you should -- in principle -- consider the quality of the molar mass number. However, in practice, it is likely to not matter. It is most likely -- especially in beginning chemistry -- that the Significant Digits of the final result will be limited by other parts of the calculation, not the molar mass. Therefore, I encourage beginning students to use the procedure above... Use all the digits of the atomic weights shown on their periodic table. Just add them up, and use the molar mass you get. Don't round the molar mass. Round the final result for the overall calculation, assuming that the molar mass Significant Digits is not a concern. This is usually fine, and lets you worry about the addition rule a bit later. Now, it is easy enough for the textbook to make up problems where the above method would not be satisfactory. My point is that such cases are uncommon in real problems, especially in introductory chemistry. In fact, a simple example of a question is "Calculate the molar mass of ... [some chemical]." How many Significant Digits do you report? Well, you'll need to use the addition rule for Significant Digits. But that is an artificial question; in the real world one almost always wants to know a molar mass in the context of a specific calculation involving some measurement, and it is quite likely that the measurement will limit the quality of the result. The logarithm of 74 is 1.87. (We will use base 10 logs here, but the Significant Digits rule is the same in any case.) 74 has 2 Significant Digits, and the log shown, 1.87, has 2 Significant Digits. Why? Because the 1 in the log (the part before the decimal point -- the "characteristic") relates to the exponent, and is an "exact" number. Whoa! What exponent? Well, it will help to put the number in standard scientific notation. 74 is 7.4x10 . Now consider the log of each part: the log of 10 is 1, an exact number; the log of 7.4 is 0.87 -- with a proper 2 Significant Digits. Add those together, and you get log 74 = 1.87 -- with 2 Significant Digits. Log of 740,000? That is log of 7.4x10 . 5.87. In scientific notation only the exponent is different from the previous number; therefore in the logarithm, only the leading integer is different. This log rule is often skipped in an intro chem course for a couple of reasons. First, logs may come up only once, with pH. Second, students in an intro chem course often are weak with using exponents -- and may not have learned about logs at all. So, sometimes one just suggests that pH be reported to two decimal places -- a usable if rough approximation. The short answer is "no". It is common now that most calculations are done on a calculator. Just do all the steps with the calculator, letting the machine keep track of the intermediate results. There is no need to even write down intermediates, much less round them. Why avoid rounding at each step? Each time you round, you are throwing away some information. If you do it over and over, it gets worse and worse; you accumulate rounding errors -- and that is not so good. Imagine that we want to calculate 1.00 * (1.127) . For our purposes here, the numbers are measurements, and we are to give the answer with proper Significant Digits. Proper Significant Digits in this case is 3 (because 1.00 is 3 Significant Digits). We might consider two ways to do this: Well, those two calculations give answers that are quite different! How can we judge them? Here is one approach... The original number 1.127, by convention, means 1.127 +/- 0.001. That is, this measurement might be 1.126 to 1.128. If we do the calculation with 1.126, we get 3.28. If we do the calculation with 1.128, we get 3.34. Thus it seems that the result should be in the range of those two numbers, 3.28-3.34. In fact, method 1 (calculate with the original number and round only at the end) gives 3.31 -- which is in the middle of that range. However, method 2 (round first), gives 3.39 -- which is outside the range, by quite a bit. The reason should be clear enough in this example: we have rounded "up" ten times, and thus biased the result upwards. This is an example of how rounding errors can accumulate. It is better to round only at the end. At the start of this example we said that the proper number of Significant Digits in this case was 3. As we went on, we found that the range of possible answers was 3.28-3.34, or 3.31 +/- 0.03. Obviously, this means that stating the answer as 3.31, to 3 Significant Digits with an implication of +/- 0.01, is not so good. This illustrates a limitation of Significant Digits; it is not so good when there are many error terms to keep track of (10, in this case). The main point of this example was to show the effect of compounding rounding errors -- hence the desirability of not rounding off at intermediate stages. (For more about such limitations of Significant Digits, see the section below: .) The discussion of Significant Digits when adding up atomic weights to calculate a molecular weight, in the section , is consistent with this point. The question of how to round when the -- or at least appears to be a 5 -- is discussed below in the Special cases section on . How many Significant Digits do conversion factors have? Well, it depends. Conversion factors within the metric system, i.e., involving only metric prefixes, are exact. Similarly, conversion factors between large and small units within the American system (e.g., 12 inches per foot, are exact). Conversion factors between metric and American systems are typically not exact, and it is your responsibility to try to make sure you use a conversion factor that has enough Significant Digits for your case. It is generally not good to allow a conversion factor to limit the quality of a calculation. The conversion factor between centimeters and inches, 2.54 cm = 1 inch, is -- because it has been defined to be exact. If you convert 14.626 cm to inches, at 2.54 cm/inch, you can properly report the result as 5.7583 inches -- 5 Significant Digits, like the original measurement -- because the conversion factor is exact. Many conversion factors we use in chemistry relate one property to another. Examples are density (mass per volume, g/mL) and molar mass (mass per mole, g/mol). These conversion factors are based on measurements, and their Significant Digits must be considered. It is your responsibility to think about the Significant Digits of a conversion factor. The best approach is usually to think about where the number came from. Is it a definition? a measurement? Using Significant Digits can be a good simple way to introduce students to the idea of measurement errors. It allows us to begin to relate the measurement scale to measurement quality, and does not require much math to implement. However, Significant Digits are only an approximation to true error analysis, and it is important to avoid getting bogged down in trying to make Significant Digits work well when they really don't. One type of difficulty with Significant Digits can be seen with reading a scale to the nearest "tenth". (The scale shown with illustrates this case.) In this case, 1.1 and 9.1 are both proper measurements. If we assume for simplicity that each measurement is good to +/- 0.1, the uncertainty in the first measurement is about 10% and the uncertainty in the second measurement is about 1%. Clearly, simply saying that both numbers are good to two Significant Digits is only a rough indication of the quality of the measurement. Further, Significant Digits does not convey the magnitude of the reading uncertainty for any specific scale. The common statement, which I used in the previous paragraph, is that readings are assumed to be good to 1 in the last place shown. But on some scales, it would be much more realistic to suggest that the uncertainty is 2 or even 5 in the last place shown. A similar problem can occur when the errors from many numbers are accumulated in one calculation. illustrated this. Another limitation of Significant Digits is that it deals with only one source of error, that inherent in reading the scale. Real experimental errors have many contributions, including operator error and sometimes even hidden systematic errors. One cannot do better than what the scale reading allows, but the total uncertainty may well be more than what the Significant Digits of the measurements would suggest. I have found that, even in introductory courses, some of the students will realize some of these limitations. When they point them out to me, I am happy to compliment them on their understanding. I then explain that Significant Digits is a simple and approximate way to start looking at measurement errors, and assure them that more sophisticated -- but more labor-intensive -- ways are available. Some modern measuring instruments have a digital scale. Electronic balances are particularly common. How do you know how many Significant Digits to write down from a digital scale? Good question. Most such instruments will display the proper number of digits. However, you should watch the instrument and see if that seems reasonable. Remember that we usually estimate one digit beyond what is certain. With a digital scale, this is reflected in some fluctuation of the last digit. So if you see the last digit fluctuating by 1 or 2, that is fine. Write down that last digit; you should try to write down a value that is about in the middle of the range the scale shows. If the fluctuation is more than 2 or so in the last digit, it may mean that the instrument is not working properly. For example, if the balance display is fluctuating much, it may mean that the balance is being influenced by air currents -- or by someone bumping the bench. Regardless of the reason, a large fluctuation may mean that a displayed digit is not really significant. These measuring instruments have only one calibration line. You adjust the liquid level to the calibration line -- as close as you can; you then have the volume that is shown on the device. A 10 mL volumetric pipet measures 10 mL; that is the only thing it can do. So, how many Significant Digits do we report in such a measurement? Obviously the usual procedures for determining Significant Digits are not applicable. One key determinant of the quality of a measurement with a volumetric pipet is the -- the accuracy of the device as guaranteed by the manufacturer. The tolerance may be shown on the instrument; if not, it can be obtained from the catalog or other reference source. There is no necessary relationship between the tolerance and measurement error. However, it turns out that these instruments have been designed so that the tolerance is close to the typical measurement error. Thus, as an approximation, but a useful one, one can treat the stated tolerance as the measurement error. As a rule of thumb, high quality ("Class A") volumetric glassware will give 4 Significant Digits measurements. (In contrast, ordinary glassware will give about 3 Significant Digits at best.) Of course, this assumes that the instrument is being used by trained personnel. In serious work, one would take care to measure actual experimental errors. There are two points to be made here. The first is to make sure that the final 5 really is a final 5. And then, if it is, what to do. This might seem to be simple enough, but with common calculators it is easy to be misled. Calculators know nothing about Significant Digits; how many digits they display depends on various things, including how you set them. It is easy for a calculator to mislead you about a final 5. For example, imagine that the true result of a calculation is 8.347, but that the calculator is set to display two decimal places (two digits beyond the decimal point). It will show 8.35. If you want 2 Significant Digits, you would be tempted to round to 8.4. However, that is clearly incorrect, if you look at the complete result 8.347, which should round to 8.3 for 2 Significant Digits. How do you avoid this problem? If you see a final 5 that you want to round off, increase the number of digits displayed before making your decision. There are two schools of thought on this. What should you do? Well, this is really a rather arcane point, not worth much attention. If your instructor prefers a particular way, do it. It really is not a big deal, one way or the other. If you are looking to decide your own preferred approach, I'd suggest you read a bit about what various people suggest, and why. If you just want my opinion, well, I suggest "rounding even". | 27,760 | 3,478 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Key_Elements_of_Green_Chemistry_(Lucia)/03%3A_Hazards/3.01%3A_Introduction_to_Hazards_of_Chemistry |
What we as a population must realize is that no substance in and of itself is a poison or a remedy. It is the dosage that defines the activity of the substance. For example, let’s look at the concept of “hormesis” as a precursor to the concept of poison/remedy by dosage. We all have heard at one time or another the phrase, “What doesn't kill you makes you stronger”. This phrase contains truth and contains at its essence the theory of hormesis: when organisms are exposed to low levels of stressors or toxins, they become more resistant to larger levels of stressors or toxins. This theory has been met with skepticism. Recently, however, biologists have put together a molecular explanation of its function and it has finally been accepted as a fundamental principle of biomedicine. For example, exposing mice to low levels of gamma ray radiation before irradiating them with high levels actually decreases their likelihood of cancer. Similarly, when dioxin is given to rats we find the same situation. However, the biochemical mechanisms are not well understood. It is believed that a low dose of a toxin can trigger repair mechanisms that are efficient enough to not only neutralize the toxin, but repair other defects not caused by the toxin. Thus, hormesis is nature’s way of dealing with harmful agents; in fact, antibodies are a natural consequence of hormesis. However, the toxins/poisons that were once absolute, are NO LONGER. For example, thalidomide was found to be a very dangerous chemical for embryo development, but has recently found great promise in a number of ailments according to the Mayo Clinic including HIV, skin lesions, and multiple myeloma (please see: ). Such a fact is outstanding considering the horrific aftermath (a few photographs shown below(Figure \(\Page {1}\)) of its use in the middle part of last century: In fact, the list of former “pure” toxins is extremely interesting: snake venom, bacteria (botulin), fungi (penicillin), leeches (Hirudin), maggots (gangrene), etc. The toxic aspect notwithstanding, we all live in the wake of a world and society that is rife with potential hazards. A hazard is “threat” to life, health, property, or environment. These can come in many forms, but they are classified according to their modalities or nature of operation. The modalities of hazards are the following: Within these modalities of hazards, hazards can be further refined as to their types. The following types of hazards classify the threats to life: | 2,509 | 3,479 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Molecular_Geometry/Molecular_Geometry_Overview |
The specific three dimensional arrangement of atoms in molecules is referred to as . We also define molecular geometry as the positions of the atomic nuclei in a molecule. There are various instrumental techniques such as and other experimental techniques which can be used to tell us where the atoms are located in a molecule. Using advanced techniques, very complicated structures for proteins, enzymes, DNA, and RNA have been determined. Molecular geometry is associated with the chemistry of vision, smell and odors, taste, drug reactions and enzyme controlled reactions to name a few. It is common practice to represent bonding patterns by "generic" formulas such as \(AX_4\), \(AX_2E_2\), etc., in which "X" stands for bonding pairs and "E" denotes lone pairs. This convention is known as the "AXE Method." Molecular geometry is associated with the specific orientation of bonding atoms. A careful analysis of electron distributions in orbitals will usually result in correct molecular geometry determinations. In addition, the simple writing of Lewis diagrams can also provide important clues for the determination of molecular geometry. on a picture to link to a page with the GIF file and a short discussion of the molecule. The valence shell is the outermost electron-occupied shell of an atom that holds the electrons involved in bonding. In a covalent bond, a pair of electrons is shared between two atoms. In a polyatomic molecule, several atoms are bonded to a central atom using two or more electron pairs. The repulsion between negatively charged electron pairs in bonds or as lone pairs causes them to spread apart as much as possible. The idea of "electron pair repulsion can be demonstrated by tying several inflated balloons together at their necks. Each balloon represents an electron pair. The balloons will try to minimize the crowding and will spread as far apart as possible. According to VSEPR theory, molecular geometry can be predicted by starting with the electron pair geometry about the central atom and adding atoms to some or all of the electron pairs. This model produces good agreement with experimental determinations for simple molecules. With this model in mind, the molecular geometry can be determined in a systematic way. Molecules can then be divided into two groups:
| 2,335 | 3,480 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Exercises%3A_General_Chemistry/Exercises%3A_Averill_et_al./01.E%3A_Introduction_to_Chemistry_(Exerciese) |
" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9 "Essential Skills 1") before proceeding to the Application Problems. Problems marked with a ♦ involve multiple concepts. 1. In 1953, James Watson and Francis Crick spent three days analyzing data to develop a model that was consistent with the known facts about the structure of DNA, the chemical substance that is the basis for life. They were awarded the Nobel Prize in Physiology or Medicine for their work. Based on this information, would you classify their proposed model for the structure of DNA as an experiment, a law, a hypothesis, or a theory? Explain your reasoning. 2. In each scenario, state the observation and the hypothesis: a. A recently discovered Neanderthal throat bone has been found to be similar in dimensions and appearance to that of modern humans; therefore, some scientists believe that Neanderthals could talk. b. Because DNA profiles from samples of human tissue are widely used in criminal trials, DNA sequences from plant residue on clothing can be used to place a person at the scene of a crime. 3. Small quantities of gold from far underground are carried to the surface by groundwater, where the gold can be taken up by certain plants and stored in their leaves. By identifying the kinds of plants that grow around existing gold deposits, one should be able to use this information to discover potential new gold deposits. a. State the observation. b. State the hypothesis. c. Devise an experiment to test the hypothesis. 4. Large amounts of nitrogen are used by the electronics industry to provide a gas blanket over a component during production. This ensures that undesired reactions with oxygen will not occur. Classify each statement as an extensive property or an intensive property of nitrogen. a. Nitrogen is a colorless gas. b. A volume of 22.4 L of nitrogen gas weighs 28 g at 0°C. c. Liquid nitrogen boils at 77.4 K. d. Nitrogen gas has a density of 1.25 g/L at 0°C. 5. Oxygen is the third most abundant element in the universe and makes up about two-thirds of the human body. Classify each statement as an extensive property or an intensive property of oxygen. a. Liquid oxygen boils at 90.2 K. b. Liquid oxygen is pale blue. c. A volume of 22.4 L of oxygen gas weighs 32 g at 0°C. d. Oxygen has a density of 1.43 g/L at 0°C. 6. One of the first high-temperature superconductors was found to contain elements in the ratio 1Y:2Ba:3Cu:6.8O. A material that contains elements in the ratio 1Y:2Ba:3Cu:6O, however, was not a high-temperature superconductor. Do these materials obey the law of multiple proportions? Is the ratio of elements in each compound consistent with Dalton’s law of indivisible atoms? 7. ♦ There has been increased evidence that human activities are causing changes in Earth’s atmospheric chemistry. Recent research efforts have focused on atmospheric ozone (O ) concentrations. The amount of ozone in the atmosphere is influenced by concentrations of gases that contain only nitrogen and oxygen, among others. The following table gives the masses of nitrogen that combine with 1.00 g of oxygen to form three of these compounds. a. Determine the ratios of the masses of nitrogen that combine with 1.00 g of oxygen in these compounds. Is this data consistent with the law of multiple proportions? b. Predict the mass of nitrogen that would combine with 1.00 g of oxygen to form another possible compound in the series. 8. Indium has an average atomic mass of 114.818 amu. One of its two isotopes has an atomic mass of 114.903 amu with a percent abundance of 95.70. What is the mass of the other isotope? 9. Earth’s core is largely composed of iron, an element that is also a major component of black sands on beaches. Iron has four stable isotopes. Use the data to calculate the average atomic mass of iron. 10. ♦ Because ores are deposited during different geologic periods, lead ores from different mining regions of the world can contain different ratios of isotopes. Archaeologists use these differences to determine the origin of lead artifacts. For example, the following table lists the percent abundances of three lead isotopes from one artifact recovered from Rio Tinto in Spain. a. If the only other lead isotope in the artifact is Pb, what is its percent abundance? b. What is the average atomic mass of lead if the only other isotope in the artifact is Pb? c. An artifact from Laurion, Greece, was found to have a Pb: Pb ratio of 0.8307. From the data given, can you determine whether the lead in the artifact from Rio Tinto came from the same source as the lead in the artifact from Laurion, Greece? 11. The macrominerals (sodium, magnesium, potassium, calcium, chlorine, and phosphorus) are widely distributed in biological substances, although their distributions are far from uniform. Classify these elements by both their periods and their groups and then state whether each is a metal, a nonmetal, or a semimetal. If a metal, is the element a transition metal? 12. The composition of fingernails is sensitive to exposure to certain elements, including sodium, magnesium, aluminum, chlorine, potassium, calcium, selenium, vanadium, chromium, manganese, iron, cobalt, copper, zinc, scandium, arsenic, and antimony. Classify these elements by both their periods and their groups and then determine whether each is a metal, a nonmetal, or a semimetal. Of the metals, which are transition metals? Based on your classifications, predict other elements that could prove to be detectable in fingernails. 13. Mercury levels in hair have been used to identify individuals who have been exposed to toxic levels of mercury. Is mercury an essential element? a trace element? 14. Trace elements are usually present at levels of less than 50 mg/kg of body weight. Classify the essential trace elements by their groups and periods in the periodic table. Based on your classifications, would you predict that arsenic, cadmium, and lead are potential essential trace elements? | 6,190 | 3,482 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Ozonolysis_of_Alkenes_and_Alkynes |
Ozonolysis is a method of oxidatively cleaving or using (\(O_3\)), a reactive allotrope of oxygen. The process allows for carbon-carbon double or triple bonds to be replaced by double bonds with oxygen. This reaction is often used to identify the structure of unknown alkenes by breaking them down into smaller, more easily identifiable pieces. Ozonolysis also occurs naturally and would break down repeated units used in rubber and other polymers. On an industrial scale, azelaic acid and pelargonic acids are produced from ozonolysis. The gaseous ozone is first passed through the desired alkene solution in either methanol or dichloromethane. The first intermediate product is an ozonide molecule which is then further reduced to carbonyl products. This results in the breaking of the Carbon-Carbon double bond and is replaced by a Carbon-Oxygen double bond instead. The first step in the mechanism of ozonolysis is the initial electrophilic addition of ozone to the Carbon-Carbon double bond, which then form the molozonide intermediate. Due to the unstable molozonide molecule, it continues further with the reaction and breaks apart to form a carbonyl and a carbonyl oxide molecule. The carbonyl and the carbonyl oxide rearranges itself and reforms to create the stable ozonide intermediate. A reductive workup could then be performed to convert convert the ozonide molecule into the desired carbonyl products. | 1,437 | 3,483 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity_of_Alpha_Hydrogens/Malonic_Ester_Synthesis |
Malonic ester is a reagent specifically used in a reaction which converts alkyl halides to carboxylic acids called the Malonic Ester Synthesis. Malonic ester synthesis is a synthetic procedure used to convert a compound that has the general structural formula 1 into a carboxylic acid that has the general structural formula 2. However, the generation of from ethyl acetate quantitatively in high yield is not an easy task because the reaction requires a very strong base, such as LDA, and must be carried out at very low temperature under strictly anhydrous conditions. Due to the fact that Malonic ester’s α hydrogens are adjacent to two carbonyls, they can be deprotonated by sodium ethoxide (NaOEt) to form Sodio Malonic Ester. Because Sodio Malonic Ester is an enolate, it can then be alkylated with alkyl halides. After alkylation the product can be converted to a dicarboxylic acid through saponification and subsequently one of the carboxylic acids can be removed through a decarboxylation step. 1) Saponification 2) Decarboxylation 3) Tautomerization All of the steps together form the Malonic ester synthesis. \[RX \rightarrow RCH_2CO_2H\] Example | 1,179 | 3,484 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 3,486 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Redox_Chemistry/The_Fall_of_the_Electron |
In oxidation-reduction ("redox") reactions, electrons are transferred from a donor ( ) to an acceptor ( ). But how one predict whether, or in which direction, such a reaction will actually go? Presented below is a very simple way of understanding how different redox reactions are related. When discussing acid-base reactions, the picture can be constructed of minimizing free energy for a acid/base base in which protons "Fall" from higher-energy sources (acids) to lower-energy sinks (bases). Similarly, electrons-transfer reactions spontaneously proceed in the direction in which electrons fall (in free energy) from sources (reducing agents) to sinks (oxidizing agents). For example: Will the reaction the following reaction go in the forward or reverse direction as written? \[ \ce{ Zn + Cu^{2+} -> Zn^{2+} + Cu}\] Assume equal effective concentrations of the two ions to avoid favoring one or the other direction. We make can use of an , the relevant portion of which is shown here: Because electrons have a higher energy on Zn than they do on Cu, copper ions will serve as an to Zn, and the reaction will go to the right: the Zn gets oxidized to Zn , and the Cu is reduced to metallic copper. This is just minimizing free energy of the entire system as a guiding principle in all reactions viewed from a thermodynamics perspective. Since free energy is related to potential of a reaction via recall the relation \( \Delta{G} = -nFE \), the free energy (or Gibbs energy) on the y-axes can be substituted with the potential. In this diagram, electron donors (otherwise known as reducing agents or reductants) are shown on the left, and their conjugate oxidants (acceptors) are on the right. The vertical location of each redox couple represents the free energy of an electron in the reduced form of the couple, relative to the free energy of the electron when attached to the hydrogen ion (and thus in H ). An oxidant can be regarded as a substance possessing vacant electron levels; the "stronger" the oxidizing agent, the lower the energy of the vacancy (the sink). If a reductant is added to a solution containing several oxidants, it will supply electrons to the various empty levels below it, filling them from the lowest up. Note however, that electron transfer reactions can be very slow, so kinetic factors may alter the order in which these steps actually take place. . Locate the couples involving these two elements within the vertical section labeled "water stability range" (light blue background). The metals above the H /H couple are known as the because they can all donate electrons to\(H^+\), reducing it to \(H_2\) and leaving the metal cation. In other words, \(H^+\) can serve as an electron sink to these metals, which are therefore attacked by acid. But since some \(H^+\) is always present in water, all of these metals can react with water. Generally, the higher they are, the more readily they react. With zinc and below, reaction with water is so slow at room temperature as to be negligible, but these metals will be attacked by acidic solutions, in which the concentration of \(H^+\) ions is much greater. Those metals that are below hydrogen in this table are not attached by \(H^+\) and are referred to as the . (Gold, Au, just below chlorine, is the noblest of all.) The species on the right side below the H O/O couple can all serve as electron sinks to water and will oxidize it to O . However, this reaction can be extremely slow; only F , the strongest of all the oxidizing agents (at the very bottom of the table) reacts quickly. It turns out, then, that only those redox pairs situated within the water stability region are thermodynamically stable in aqueous solution; all others will tend to decompose the water. Three scales of free energy are shown in the figure on the right. This diagram provides an overview of the major redox couples that provide the energy that drives the life process. Most organisms derive their metabolic energy from respiration, a process in which electrons from foodstuffs (nominally glucose), fall to lower-free energy acceptors on the right. In eucaryotic organisms this electron sink is dioxygen. Aerobic is the most efficient of all because the electron falls so far (as it does so, part of the energy is captured by a series of intermediates and used for the synthesis of ATP). To make the glucose, animals rely on plants, which utilize the energy of sunlight to force electrons from \(O_2\) back up to the top left of the diagram. This, of course, is , which is just respiration driven in reverse. is a fairly recent development in the history of life. There still exist a host of primitive organisms (all bacteria) that inhabit anoxic environments and must employ other electron sinks that reside on the scale, and thus yield amounts of energy. Among the more familiar of these sinks: Not all organisms start with glucose; H , just below it, can serve as an electron source and was likely an important one during the earliest stages of life, as were most of the sources below it. If you already know some electrochemistry, you probably know how to use the to carry out quantitative calculations. Nevertheless, this is still a very helpful picture when you have to deal with multiple redox systems which occur very commonly in environmental chemistry, analytical chemistry, and biochemistry. | 5,401 | 3,487 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/11%3A_Chemical_Kinetics_I/11.03%3A_Rate_Laws |
A rate law is any mathematical relationship that relates the concentration of a reactant or product in a chemical reaction to time. Rate laws can be expressed in either derivative (or ratio, for finite time intervals) or integrated form. One of the more common general forms a rate law for the reaction \[A + B \rightarrow products \nonumber \] may take is \[\text{rate}=k[A]^\alpha[B]^\beta \nonumber \] where \(k\), \(\alpha\), and \(\beta\) are experimentally determined values. However, a rate law can take many different forms, some of which can be quite intricate and complex. The powers \(\alpha\) and \(\beta\) need not be integers. For example \[\text{rate}=k[A]^\alpha [B]^{1/2} \label{ex2} \] is a rate law that is observed for some reactions. Sometimes, the concentrations of products must be included. \[\text{rate}=\dfrac{k[A]^{1/2}[B]}{[P]} \nonumber \] In some cases, the concentration for a catalyst or enzyme is important. For example, many enzyme mitigated reactions in biological systems follow the Michaelis-Menten rate law, which is of the form \[\text{rate}=\dfrac{V_{max}[S]}{K_m + [S]} \nonumber \] where \(V_{max}\) and \(K_M\) are factors that are determined experimentally, and \([S]\) is the concentration of the substrate in the reaction. For those cases where the rate law can be expressed in the form \[\text{rate}=k[A]^\alpha[B]^\beta[C]^\gamma \nonumber \] where \(A\), \(B\), and \(C\) are reactants (or products or catalysts, etc.) involved in the reaction, the reaction is said to be of \(\alpha\) order in \(A\), \(\beta\) order in \(B\), and \(\gamma\) order in \(C\). The reaction is said to be \(\alpha +\beta + \gamma\) order overall. Some examples are shown in the following table: Reaction orders can also be fractional such as for Equation \ref{ex2} which is 1 order in \(A\), and half order in \(B\). The order can also be negative such as which is 1 order in A, and -1 order in B. In this case, an build-up of the concentration of B will retard (slow) the reaction. In all cases, the order of the reaction with respect to a specific reactant or product (or catalyst, or whatever) must be determined experimentally. As a general rule, the stoichiometry cannot be used to predict the form of the rate law. However, the rate law can be used to gain some insight into the possible pathways by which the reaction can proceed. That is the topic of Chapter 12. For now we will focus on three useful methods that are commonly used in chemistry to determine the rate law for a reaction from experimental data. Perhaps the simplest of the methods to be used are the empirical methods, which rely on the qualitative interpretation of a graphical representation of the concentration vs time profile. In these methods, some function of concentration is plotted as a function of time, and the result is examined for a linear relationship. For the following examples, consider a reaction of the form \[A + B \rightarrow products \nonumber \] in which A is one of the reactants. In order to employ these empirical methods, one must generate the forms of the integrated rate laws. | 3,125 | 3,489 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Reactivity_of_Alkanes/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 3,491 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Properties_of_Alkanes/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 3,492 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/06%3A_Chemical_Bonding_-_Electron_Pairs_and_Octets/6.19%3A_Atomic_Sizes |
The sizes of atoms and ions are important in determining the properties of both covalent and ionic compounds. You should already have some appreciation of the factors which govern atomic sizes from the color-coded dot-density diagrams of Hydrogen, Helium, and Lithium and of Beryllium, Boron and Carbon. From the leftmost diagram below, one can see that Hydrogen seems to have a larger atomic radius than Helium, but not larger than Lithium (although it's "radius" is quite spread out). On the right side, one can see that the atomic radius steadily decreases as you go down the line. Note the relative locations of these elements on the periodic table and predict what the trend might be. By far the largest atom illustrated in these color plates is Li. Because Li has an electron in the = 2 shell, it is larger than H or He whose 1 electron clouds are much closer to the nucleus. Li is also larger than Be, B, or C. In the latter atoms, the 2 and 2 electron clouds are attracted by a greater nuclear charge and hence are held closer to the center of the atom than the 2 cloud in Li. Thus two important rules may be applied to the prediction of atomic sizes. As one moves from top to bottom of the periodic table, the principal quantum number increases and electrons occupy orbitals whose electron clouds are successively farther from the nucleus. The atomic radii increase. As one moves from left to right across a horizontal period, then value of the outermost electron clouds remains the same, but the nuclear charge increases steadily. The increased nuclear attraction contracts the electron cloud, and hence the atomic size decreases. It is difficult to measure the size of an atom very exactly. As the dot-density diagrams show, an atom is not like a billiard ball which has a definite radius. Instead of stopping suddenly, an electron cloud gradually fades out so that one cannot point to a definite radius at which it ends. One way out of this difficulty is to find out how closely atoms are packed together in a crystal lattice. Figure \(\Page {2}\) illustrates part of a crystal of solid Cl at a very low temperature. The distance AA′ has the value of 369 pm. Since this represents the distance between adjacent atoms in Cl molecules, we can take it as the distance at which different Cl atoms just “touch.” Half this distance, 184 pm, is called the of Cl. The van der Waals radius gives an approximate idea of how closely atoms in molecules can approach each other. Commonly accepted values of the van der Waals radii for the representative elements are shown in the Figure \(\Page {3}\). Note how these radii and the periodic table. Also given are values for the of each atom. Returning to the figure of Cl ( \(\Page {2}\), we see that the distance AB between two Cl atoms in the molecule (i.e., the Cl—Cl ) has a value of 202 pm. The covalent radius is one-half of this bond length, or 101 pm. Covalent radii are approximately additive and enable us to predict rough values for the internuclear distances in a variety of molecules. For example, if we add the covalent radius of C (77 pm) to that of O (66 pm), we obtain an estimate for the length of the C―O bond, namely, 143 pm. This is in exact agreement with the measured value in ethyl alcohol and dimethyl ether . | 3,328 | 3,493 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Synthesis_of_Alkenes/Alkenes_by_Dehydration_of_Alcohols |
Alkenes are usually prepared from either alcohols or haloalkanes (alkyl halides). Alkenes are obtained by the dehydration of alcohols. The dehydration of alcohols can be affected by two common methods. By passing the vapors of an alcohol over alumina (\(Al_2O_3\)) at 623 K (350°C). The order of the ease of dehydration of alcohols is: tertiary > secondary > primary. Secondary and tertiary alcohols are best dehydrated by dilute sulfuric acid. By heating an alcohol with concentrated sulfuric acid at 453 K (180°C). Other dehydrating agents like phosphoric acid and anhydrous zinc chloride may also be used. Cyclohexanol on dehydration gives cyclohexene. cyclohexanol cyclohexene The loss of water from an alcohol to give an alkene does not occur in just one step; a series of steps are involved in the mechanism of dehydration of alcohols. In the dehydration reaction given above, the following steps are involved. The mechanism of dehydration of ethyl alcohol is described below. | 1,002 | 3,494 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/04%3A_Transport/14%3A_Hydrodynamics/14.02%3A_Stokes_Law |
How is a fluid’s macroscopic resistance to flow related to microscopic friction originating in random forces between the fluid’s molecules? In discussing the Langevin equation, we noted that the friction coefficient \(\zeta\) was the proportionality constant between the drag force experienced by an object and its velocity through the fluid: \(f_d=-\zeta v\). Since this drag force is equal and opposite to the stress exerted on an object as it moves through a fluid, there is a relationship of the drag force to the fluid viscosity. Specifically, we can show that Einstein’s friction coefficient ζ is related to the dynamic viscosity of the fluid \(\eta\), as well as other factors describing the size and shape of the object (but not its mass). This connection is possible as a result of George Stokes’ description of the fluid velocity field around a sphere moving through a viscous fluid at a constant velocity. He considered asphere of radius R moving through a fluid with laminar flow: that in which the fluid exhibits smooth parallel velocity profiles without lateral mixing. Under those conditions, and no-slip boundary conditions, one finds that the drag force on a sphere is \( f_d = 6\pi \eta R_hv \) and viscous force per unit area is entirely uniform across the surface of the sphere. This gives us Stokes’ Law \[ \zeta = 6\pi \eta R_h \] Here Rh is referred to as the hydrodynamic radius of the sphere, the radius at which one can apply the no-slip boundary condition, but which on a molecular scale may include water that is strongly bound to the molecule. Combining eq. (1) with the Einstein formula for diffusion coefficient, \(D=k_BT/\zeta\) gives the Stokes–Einstein relationship for the translation diffusion constant of a sphere \[D_{trans} = \dfrac{k_BT}{6\pi \eta R_h} \] One can obtain a similar a Stokes–Einstein relationship for orientational diffusion of a sphere in a viscous fluid. Relating the orientational diffusion constant and the drag force that arises from resistance to shear, one obtains \[ D_{rot} = \dfrac{k_BT}{6V_h\eta } \nonumber \] ________________________________________ | 2,133 | 3,495 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Volumetric_Chemical_Analysis_(Shiundu) |
The objective of this course is to introduce the student to the fundamental concepts of analytical chemistry with particular emphasis on volumetric chemical analysis. The module is designed to familiarize the learner with the principles that underpin chemical reactivity of different types of chemical reactions. The theories, concepts of volumetric analysis and measurements of data as they apply to analytical chemistry are examined. Special emphasis is placed on the application of basic principles of chemical equilibria to acid-base reactions, precipitation reactions, oxidation-reduction (electron –transfer) reactions, and complex ion reactions. We will look in more detail at the quantitative aspects of acid-base titrations.
| 748 | 3,499 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.8%3A_Standard_Enthalpies_of_Formation |
One way to report the heat absorbed or released by chemical reactions would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, starting from the elemental forms of each atom at 25 C and 1 atm pressure. (\(ΔH_f\)) is the enthalpy change for the formation of 1 mol of a compound from its component elements, such as the formation of carbon dioxide from carbon and oxygen. The formation of any chemical can be as a reaction from the corresponding elements: \[ \text{elements} \rightarrow \text{compound} \nonumber\] which in terms of the the Enthalpy of formation becomes \[\Delta H_{rxn} = \Delta H_{f} \label{7.8.1} \] For example, consider the combustion of carbon: \[ \ce{ C(s) + O2 (g) -> CO2 (g)} \nonumber\] then \[ \Delta H_{rxn} = \Delta H_{f}\left [CO_{2}\left ( g \right ) \right ] \nonumber \] The sign convention for Δ is the same as for any enthalpy change: \(ΔH_f < 0\) if heat is released when elements combine to form a compound and \(ΔH_f > 0\) if heat is absorbed. The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system. The magnitude of Δ for a reaction depends on the physical states of the reactants and the products (gas, liquid, solid, or solution), the pressure of any gases present, and the temperature at which the reaction is carried out. To avoid confusion caused by differences in reaction conditions and ensure uniformity of data, the scientific community has selected a specific set of conditions under which enthalpy changes are measured. These standard conditions serve as a reference point for measuring differences in enthalpy, much as sea level is the reference point for measuring the height of a mountain or for reporting the altitude of an airplane. The standard conditions for which most thermochemical data are tabulated are a of 1 atmosphere (atm) for all gases and a of 1 M for all species in solution (1 mol/L). In addition, each pure substance must be in its standard state is usually its most stable form at a pressure of 1 atm at a specified temperature. We assume a temperature of 25°C (298 K) for all enthalpy changes given in this text, unless otherwise indicated. Enthalpies of formation measured under these conditions are called (\(ΔH^o_f\)) The standard enthalpy of formation of any element in its standard state is zero by definition. For example, although oxygen can exist as ozone (O ), atomic oxygen (O), and molecular oxygen (O ), O is the most stable form at 1 atm pressure and 25°C. Similarly, hydrogen is H (g), not atomic hydrogen (H). Graphite and diamond are both forms of elemental carbon, but because graphite is more stable at 1 atm pressure and 25°C, the standard state of carbon is graphite (Figure \(\Page {1}\)). Therefore, \(\ce{O2(g)}\), \(\ce{H2(g)}\), and graphite have \(ΔH^o_f\) values of zero. The standard enthalpy of formation of glucose from the elements at 25°C is the enthalpy change for the following reaction: \[ 6C\left (s, graphite \right ) + 6H_{2}\left (g \right ) + 3O_{2}\left (g \right ) \rightarrow C_{6}H_{12}O_{6}\left (s \right )\; \; \; \Delta H_{f}^{o} = - 1273.3 \; kJ \label{7.8.2} \] It is not possible to measure the value of \(ΔH^oo_f \) for glucose, −1273.3 kJ/mol, by simply mixing appropriate amounts of graphite, \(\ce{O2}\), and \(\ce{H2}\) and measuring the heat evolved as glucose is formed since the reaction shown in Equation \(\ref{7.8.2}\) does not occur at a measurable rate under any known conditions. Glucose is not unique; most compounds cannot be prepared by the chemical equations that define their standard enthalpies of formation. Instead, values of \(ΔH^oo_f \) are obtained using and standard enthalpy changes that have been measured for other reactions, such as combustion reactions. Values of \(ΔH^o_f\) for an extensive list of compounds are given in . Note that \(ΔH^o_f\) values are always reported in kilojoules per mole of the substance of interest. Also notice in that the standard enthalpy of formation of O (g) is zero because it is the most stable form of oxygen in its standard state. For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. compound formula and phase. balanced chemical equation for its formation from elements in standard states Use to identify the standard state for each element. Write a chemical equation that describes the formation of the compound from the elements in their standard states and then balance it so that 1 mol of product is made. To calculate the standard enthalpy of formation of a compound, we must start with the elements in their standard states. The standard state of an element can be identified in : by a \(ΔH^o_f\) Hydrogen chloride contains one atom of hydrogen and one atom of chlorine. Because the standard states of elemental hydrogen and elemental chlorine are \(\ce{H2(g)}\) and \(\ce{Cl2(g)}\), respectively, the unbalanced chemical equation is \[\ce{H2(g) + Cl2(g) \rightarrow HCl(g)} \nonumber\] Fractional coefficients are required in this case because values are reported for of the product, \(\ce{HCl}\). Multiplying both \(\ce{H2(g)}\) and \(\ce{Cl2(g)}\) by 1/2 balances the equation: \[ \ce{1/2 H_{2} (g) + 1/2 Cl_{2} (g) \rightarrow HCl (g)} \nonumber\] The standard states of the elements in this compound are \(\ce{Mg(s)}\), \(\ce{C(s, graphite)}\), and \(\ce{O2(g)}\). The unbalanced chemical equation is thus \[\ce{Mg(s) + C (s, graphite) + O2 (g) \rightarrow MgCO3 (s)} \nonumber\] This equation can be balanced by inspection to give \[ \ce{Mg (s) + C (s, graphite ) + 3/2 O2 (g)\rightarrow MgCO3 (s)} \nonumber\] Palmitic acid, the major fat in meat and dairy products, contains hydrogen, carbon, and oxygen, so the unbalanced chemical equation for its formation from the elements in their standard states is as follows: \[\ce{C(s, graphite) + H2(g) + O2(g) \rightarrow CH3(CH2)14CO2H(s)} \nonumber\] There are 16 carbon atoms and 32 hydrogen atoms in 1 mol of palmitic acid, so the balanced chemical equation is \[\ce{16C (s, graphite) + 16 H2(g) + O2(g) -> CH3(CH2)14CO2H(s) } \nonumber\] For the formation of each compound, write a balanced chemical equation corresponding to the standard enthalpy of formation of each compound. \[ \ce{ Na (s) + 1/2 Cl2 (g) \rightarrow NaCl (s)} \nonumber \] \[ \ce{H_{2} (g) + 1/8 S8 (s) + 2O2 ( g) \rightarrow H2 SO4( l) } \nonumber\] \[\ce{2C(s) + O2(g) + 2H2(g) -> CH3CO2H(l)} \nonumber \] Definition of Heat of Formation Reactions: Tabulated values of standard enthalpies of formation can be used to calculate enthalpy changes for reaction involving substances whose \(\Delta{H_f^o}\) values are known. The standard enthalpy of reaction \(\Delta{H_{rxn}^o}\) is the enthalpy change that occurs when a reaction is carried out with all reactants and products in their standard states. Consider the general reaction \[ aA + bB \rightarrow cC + dD \label{7.8.3}\] where \(A\), \(B\), \(C\), and \(D\) are chemical substances and \(a\), \(b\), \(c\), and \(d\) are their stoichiometric coefficients. The magnitude of \(ΔH^ο\) is the sum of the standard enthalpies of formation of the products, each multiplied by its appropriate coefficient, minus the sum of the standard enthalpies of formation of the reactants, also multiplied by their coefficients: \[ \Delta H_{rxn}^{o} = \underbrace{ \left [c\Delta H_{f}^{o}\left ( C \right ) + d\Delta H_{f}^{o}\left ( D \right ) \right ] }_{\text{products} } - \underbrace{ \left [a\Delta H_{f}^{o}\left ( A \right ) + b\Delta H_{f}^{o}\left ( B \right ) \right ]}_{\text{reactants }} \label{7.8.4} \] More generally, we can write \[ \Delta H_{rxn}^{o} = \sum m\Delta H_{f}^{o}\left ( products \right ) - \sum n\Delta H_{f}^{o}\left ( reactants \right ) \label{7.8.5} \] where the symbol \(\sum\) means “sum of” and \(m\) and \(n\) are the stoichiometric coefficients of each of the products and the reactants, respectively. “Products minus reactants” summations such as Equation \(\ref{7.8.5}\) arise from the fact that enthalpy is a state function. Because many other thermochemical quantities are also state functions, “products minus reactants” summations are very common in chemistry; we will encounter many others in subsequent chapters. "Products minus reactants" summations are typical of state functions. To demonstrate the use of tabulated values, we will use them to calculate \(ΔH_{rxn}\) for the combustion of glucose, the reaction that provides energy for your brain: \[ \ce{ C6H12O6 (s) + 6O2 (g) \rightarrow 6CO2 (g) + 6H2O (l)} \label{7.8.6} \] Using Equation \(\ref{7.8.5}\), we write \[ \Delta H_{f}^{o} =\left \{ 6\Delta H_{f}^{o}\left [ CO_{2}\left ( g \right ) \right ] + 6\Delta H_{f}^{o}\left [ H_{2}O\left ( g \right ) \right ] \right \} - \left \{ \Delta H_{f}^{o}\left [ C_{6}H_{12}O_{6}\left ( s \right ) \right ] + 6\Delta H_{f}^{o}\left [ O_{2}\left ( g \right ) \right ] \right \} \label{7.8.7} \] From , the relevant values are , , and . Because O (g) is a pure element in its standard state, Inserting these values into Equation \(\ref{7.8.7}\) and changing the subscript to indicate that this is a combustion reaction, we obtain \[ \begin{align} \Delta H_{comb}^{o} &= \left [ 6\left ( -393.5 \; kJ/mol \right ) + 6 \left ( -285.8 \; kJ/mol \right ) \right ] - \left [-1273.3 + 6\left ( 0 \; kJ\;mol \right ) \right ] \label{7.8.8} \\[4pt] &= -2802.5 \; kJ/mol \end{align} \] As illustrated in Figure \(\Page {2}\), we can use Equation \(\ref{7.8.8}\) to calculate \(ΔH^ο_f\) for glucose because enthalpy is a state function. The figure shows two pathways from reactants (middle left) to products (bottom). The more direct pathway is the downward green arrow labeled \(ΔH^ο_{comb}\). The alternative hypothetical pathway consists of that convert the reactants to the elements in their standard states (upward purple arrow at left) and then convert the elements into the desired products (downward purple arrows at right). The reactions that convert the reactants to the elements are the reverse of the equations that define the \(ΔH^ο_f\) values of the reactants. Consequently, the enthalpy changes are \[ \begin{align} \Delta H_{1}^{o} &= \Delta H_{f}^{o} \left [ glucose \left ( s \right ) \right ] \nonumber \\[4pt] &= -1 \; \cancel{mol \; glucose}\left ( \dfrac{1273.3 \; kJ}{1 \; \cancel{mol \; glucose}} \right ) \nonumber \\[4pt] &= +1273.3 \; kJ \nonumber \\[4pt] \Delta H_{2}^{o} &= 6 \Delta H_{f}^{o} \left [ O_{2} \left ( g \right ) \right ] \nonumber \\[4pt] & =6 \; \cancel{mol \; O_{2}}\left ( \dfrac{0 \; kJ}{1 \; \cancel{mol \; O_{2}}} \right ) \nonumber \\[4pt] &= 0 \; kJ \end{align} \label{7.8.9} \] Recall that when we reverse a reaction, we must also reverse the of the accompanying enthalpy change (Equation \ref{7.8.4} since the products are now reactants and vice versa. The overall enthalpy change for conversion of the reactants (1 mol of glucose and 6 mol of O ) to the elements is therefore +1273.3 kJ. The reactions that convert the elements to final products (downward purple arrows in Figure \(\Page {2}\)) are identical to those used to define the values of the products. Consequently, the enthalpy changes (from \[ \begin{matrix} \Delta H_{3}^{o} = \Delta H_{f}^{o} \left [ CO_{2} \left ( g \right ) \right ] = 6 \; \cancel{mol \; CO_{2}}\left ( \dfrac{393.5 \; kJ}{1 \; \cancel{mol \; CO_{2}}} \right ) = -2361.0 \; kJ \\ \Delta H_{4}^{o} = 6 \Delta H_{f}^{o} \left [ H_{2}O \left ( l \right ) \right ] = 6 \; \cancel{mol \; H_{2}O}\left ( \dfrac{-285.8 \; kJ}{1 \; \cancel{mol \; H_{2}O}} \right ) = -1714.8 \; kJ \end{matrix} \] The overall enthalpy change for the conversion of the elements to products (6 mol of carbon dioxide and 6 mol of liquid water) is therefore −4075.8 kJ. Because enthalpy is a state function, the difference in enthalpy between an initial state and a final state can be computed using pathway that connects the two. Thus the enthalpy change for the combustion of glucose to carbon dioxide and water is the sum of the enthalpy changes for the conversion of glucose and oxygen to the elements (+1273.3 kJ) and for the conversion of the elements to carbon dioxide and water (−4075.8 kJ): \[ \Delta H_{comb}^{o} = +1273.3 \; kJ +\left ( -4075.8 \; kJ \right ) = -2802.5 \; kJ \label{7.8.10} \] This is the same result we obtained using the “products minus reactants” rule (Equation \(\ref{7.8.5}\)) and values. The two results must be the same because Equation \(\ref{7.8.10}\) is just a more compact way of describing the thermochemical cycle shown in Figure \(\Page {1}\). Long-chain fatty acids such as palmitic acid (\(\ce{CH3(CH2)14CO2H}\)) are one of the two major sources of energy in our diet (\(ΔH^o_f\) =−891.5 kJ/mol). Use the data in to calculate for the combustion of palmitic acid. Based on the energy released in combustion , which is the better fuel — glucose or palmitic acid? compound and \(ΔH^ο_{f}\) values \(ΔH^ο_{comb}\) per mole and per gram To determine the energy released by the combustion of palmitic acid, we need to calculate its \(ΔH^ο_f\). As always, the first requirement is a balanced chemical equation: \[C_{16}H_{32}O_{2(s)} + 23O_{2(g)} \rightarrow 16CO_{2(g)} + 16H_2O_{(l)} \nonumber \] Using Equation \(\ref{7.8.5}\) (“products minus reactants”) with values from (and omitting the physical states of the reactants and products to save space) gives \[ \begin{align*} \Delta H_{comb}^{o} &= \sum m \Delta H^o_f\left( {products} \right) - \sum n \Delta H^o_f \left( {reactants} \right) \\[4pt] &= \left [ 16\left ( -393.5 \; kJ/mol \; CO_{2} \right ) + 16\left ( -285.8 \; kJ/mol \; H_{2}O \; \right ) \right ] \\[4pt] & - \left [ -891.5 \; kJ/mol \; C_{16}H_{32}O_{2} + 23\left ( 0 \; kJ/mol \; O_{2} \; \right ) \right ] \\[4pt] &= -9977.3 \; kJ/mol \nonumber \end{align*} \] This is the energy released by the combustion of 1 mol of palmitic acid. The energy released by the combustion of 1 g of palmitic acid is \( \Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{9977.3 \; kJ}{\cancel{1 \; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{256.42 \; g} \right )= -38.910 \; kJ/g \nonumber \) As calculated in Equation \(\ref{7.8.8}\), \( of glucose is −2802.5 kJ/mol. The energy released by the combustion of 1 g of glucose is therefore \( \Delta H_{comb}^{o} \; per \; gram =\left ( \dfrac{-2802.5 \; kJ}{\cancel{1\; mol}} \right ) \left ( \dfrac{\cancel{1 \; mol}}{180.16\; g} \right ) = -15.556 \; kJ/g \nonumber \) The combustion of fats such as palmitic acid releases more than twice as much energy per gram as the combustion of sugars such as glucose. This is one reason many people try to minimize the fat content in their diets to lose weight. Use to calculate \(ΔH^o_{rxn}\) for the , which is used industrially on an enormous scale to obtain H (g): \[ \ce{ CO ( g ) + H2O (g ) -> CO2 (g) + H2 ( g )} \nonumber\] −41.2 kJ/mol We can also measure the enthalpy change for another reaction, such as a combustion reaction, and then use it to calculate a compound’s \(ΔH^ο_f\) which we cannot obtain otherwise. This procedure is illustrated in Example \(\Page {3}\). Beginning in 1923, [\(\ce{(C2H5)4Pb}\)] was used as an antiknock additive in gasoline in the United States. Its use was completely phased out in 1986 because of the health risks associated with chronic lead exposure. Tetraethyllead is a highly poisonous, colorless liquid that burns in air to give an orange flame with a green halo. The combustion products are \(\ce{CO2(g)}\), \(\ce{H2O(l)}\), and red \(\ce{PbO(s)}\). What is the standard enthalpy of formation of tetraethyllead, given that \(ΔH^ο_f\) is −19.29 kJ/g for the combustion of tetraethyllead and \(ΔH^ο_f\) of red PbO(s) is −219.0 kJ/mol? reactant, products, and \(ΔH^ο_{comb}\) values \(ΔH^ο_f\) of the reactants The balanced chemical equation for the combustion reaction is as follows: \[\ce{2(C2H5)4Pb(l) + 27O2(g) → 2PbO(s) + 16CO2(g) + 20H2O(l)} \nonumber\] Using Equation \(\ref{7.8.5}\) gives \[ \Delta H_{comb}^{o} = \left [ 2 \Delta H_{f}^{o}\left ( PbO \right ) + 16 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 20 \Delta H_{f}^{o}\left ( H_{2}O \right )\right ] - \left [2 \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) + 27 \Delta H_{f}^{o}\left ( O_{2} \right ) \right ] \nonumber \] Solving for \(ΔH^o_f [\ce{(C2H5)4Pb}]\) give \[ \Delta H_{f}^{o}\left ( \left ( C_{2}H_{5} \right ) _{4} Pb \right ) = \Delta H_{f}^{o}\left ( PbO \right ) + 8 \Delta H_{f}^{o}\left ( CO_{2} \right ) + 10 \Delta H_{f}^{o}\left ( H_{2}O \right ) - \dfrac{27}{2} \Delta H_{f}^{o}\left ( O_{2} \right ) - \dfrac{\Delta H_{comb}^{o}}{2} \nonumber \] The values of all terms other than \(ΔH^o_f [\ce{(C2H5)4Pb}]\) are given in . The magnitude of \(ΔH^o_{comb}\) is given in the problem in kilojoules per of tetraethyl lead. We must therefore multiply this value by the molar mass of tetraethyl lead (323.44 g/mol) to get \(ΔH^o_{comb}\) for 1 mol of tetraethyl lead: \[\begin{align*} \Delta H_{comb}^{o} &= \left ( \dfrac{-19.29 \; kJ}{\cancel{g}} \right )\left ( \dfrac{323.44 \; \cancel{g}}{mol} \right ) \\[4pt] &= -6329 \; kJ/mol \end{align*} \] Because the balanced chemical equation contains 2 mol of tetraethyllead, \(ΔH^o_{rxn}\) is \[\begin{align*} \Delta H_{rxn}^{o} &= 2 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb} \left ( \dfrac{-6329 \; kJ}{1 \; \cancel{mol \; \left ( C_{2}H{5}\right )_4 Pb }} \right ) \\[4pt] &= -12,480 \; kJ \end{align*}\] Inserting the appropriate values into the equation for \(ΔH^o_f [\ce{(C2H5)4Pb}]\) \[ \begin{align*} \Delta H_{f}^{o} \left [ \left (C_{2}H_{4} \right )_{4}Pb \right ] & = \left [1 \; mol \;PbO \;\times 219.0 \;kJ/mol \right ]+\left [8 \; mol \;CO_{2} \times \left (-393.5 \; kJ/mol \right )\right ] +\left [10 \; mol \; H_{2}O \times \left ( -285.8 \; kJ/mol \right )\right ] + \left [-27/2 \; mol \; O_{2}) \times 0 \; kJ/mol \; O_{2}\right ] \left [12,480.2 \; kJ/mol \; \left ( C_{2}H_{5} \right )_{4}Pb \right ]\\[4pt]
&= -219.0 \; kJ -3148 \; kJ - 2858 kJ - 0 kJ + 6240 \; kJ = 15 kJ/mol \end{align*}\] Ammonium sulfate, \(\ce{(NH4)2SO4}\), is used as a fire retardant and wood preservative; it is prepared industrially by the highly exothermic reaction of gaseous ammonia with sulfuric acid: \[ \ce{2NH3(g) + H2SO4(aq) \rightarrow (NH4)2SO4(s)} \nonumber \] The value of \(ΔH^o_{rxn}\) is -179.4 \(\ce{H2SO4}\). kilojoules −1181 kJ/mol Calculating DH° using DHf°: The (\(ΔH_{f}\)) is the enthalpy change that accompanies the formation of a compound from its elements. (\(ΔH^o_{f}\)) are determined under : a pressure of 1 atm for gases and a concentration of 1 M for species in solution, with all pure substances present in their (their most stable forms at 1 atm pressure and the temperature of the measurement). The standard heat of formation of any element in its most stable form is defined to be zero. The (\(ΔH^o_{rxn}\)) can be calculated from the sum of the of the products (each multiplied by its stoichiometric coefficient) minus the sum of the standard enthalpies of formation of the reactants (each multiplied by its stoichiometric coefficient)—the “products minus reactants” rule. The (\(ΔH_{soln}\)) is the heat released or absorbed when a specified amount of a solute dissolves in a certain quantity of solvent at constant pressure. ( ) | 19,742 | 3,500 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkynes/Reactivity_of_Alkynes/Ozonolysis_of_Alkenes_and_Alkynes |
Ozonolysis is a method of oxidatively cleaving or using (\(O_3\)), a reactive allotrope of oxygen. The process allows for carbon-carbon double or triple bonds to be replaced by double bonds with oxygen. This reaction is often used to identify the structure of unknown alkenes by breaking them down into smaller, more easily identifiable pieces. Ozonolysis also occurs naturally and would break down repeated units used in rubber and other polymers. On an industrial scale, azelaic acid and pelargonic acids are produced from ozonolysis. The gaseous ozone is first passed through the desired alkene solution in either methanol or dichloromethane. The first intermediate product is an ozonide molecule which is then further reduced to carbonyl products. This results in the breaking of the Carbon-Carbon double bond and is replaced by a Carbon-Oxygen double bond instead. The first step in the mechanism of ozonolysis is the initial electrophilic addition of ozone to the Carbon-Carbon double bond, which then form the molozonide intermediate. Due to the unstable molozonide molecule, it continues further with the reaction and breaks apart to form a carbonyl and a carbonyl oxide molecule. The carbonyl and the carbonyl oxide rearranges itself and reforms to create the stable ozonide intermediate. A reductive workup could then be performed to convert convert the ozonide molecule into the desired carbonyl products. | 1,437 | 3,501 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.13%3A_Conjugate_Acid-Base_Pairs |
Through examples found in the sections on acids and bases proton-transfer processes are broken into two hypothetical steps: (1) donation of a proton by an acid, and (2) acceptance of a proton by a base. (Water served as the base in the acid example and as the acid in the base example [ ]). The hypothetical steps are useful because they make it easy to see what species is left after an acid donated a proton and what species is formed when a base accepted a proton. We shall use hypothetical steps or in this section, but you should bear in mind that free protons never actually exist in aqueous solution. Suppose we first consider a , the ammonium ion. When it donates a proton to any other species, we can write the half-equation: \[ \text{NH}_4^+ \rightarrow \text{H}^+ +\text{NH}_3 \nonumber \] The submicroscopic representations below show the donation of the proton of ammonium. The removal of this proton results in NH , which is easily seen at the submicroscopic level. But NH is one of the compounds we know as a . In other words, when it donates a proton, the weak acid NH is transformed into a weak base NH . Another example, this time starting with a weak base, is provided by fluoride ion: \[\text{F}^{-} + \text{H}^{+} \rightarrow \text{HF} \nonumber \] The submicroscopic representation above shows how the addition of a proton to fluoride converts a weak base (F in green) into a weak acid (HF). The situation just described for NH and NH or for F and HF applies to acids and bases. Whenever an acid donates a proton, the acid changes into a base, and whenever a base accepts a proton, an acid is formed. An acid and a base which differ only by the presence or absence of a proton are called a . Thus NH is called the conjugate base of NH , and NH is the conjugate acid of NH . Similarly, HF is the conjugate acid of F , and F the conjugate base of HF. What is the conjugate acid or the conjugate base of (a) HCl; (b) CH NH ; (c) OH ; (d) HCO . The use of conjugate acid-base pairs allows us to make a very simple statement about relative strengths of acids and bases. , and, conversely, Table \(\Page {1}\) gives a list of some of the more important conjugate acid-base pairs in order of increasing strength of the base. This table enables us to see how readily a given acid will react with a given base. The reactions with most tendency to occur are between the strong acids in the top left-hand comer of the table and the strong bases in the bottom right-hand comer. If a line is drawn from acid to base for such a reaction, it will have a slope. By contrast, reactions with little or no tendency to occur (between the weak acids at the bottom left and the weak bases at the top right) correspond to a line from acid to base with an slope. When the slope of the line is not far from horizontal, the conjugate pairs are not very different in strength, and the reaction goes only part way to completion. Thus, for example, if the acid HF is compared with the base CH COO , we expect the reaction to go part way to completion since the line is barely downhill. A like HCl donates its proton so readily that there is essentially no tendency for the conjugate base Cl to reaccept a proton. Consequently, Cl is a very weak base. A strong base like the H ion accepts a proton and holds it so firmly that there is no tendency for the conjugate acid H to donate a proton. Hence, H is a very weak acid. Write a balanced equation to describe the reaction which occurs when a solution of potassium hydrogen sulfate, KHSO , is mixed with a solution of sodium bicarbonate, NaHCO . The Na ions and K ions have no acid-base properties and function purely as spectator ions. Therefore any reaction which occurs must be between the hydrogen sulfate ion, HSO and the hydrogen carbonate ion, HCO . Both HSO and HCO are amphiprotic, and either could act as an acid or as a base. The reaction between them is thus either \[\text{HCO}_3^- + \text{HSO}_4^- \rightarrow \text{CO}_3^{2-} + \text{H}_2\text{SO}_4 \nonumber\] or \[\text{HSO}_4^- + \text{HCO}_3^- \rightarrow \text{SO}_4^{2-} + \text{H}_2\text{CO}_3 \nonumber\] Table \(\Page {1}\) tells us immediately that the second reaction is the correct one. A line drawn from HSO as an acid to HCO as a base is downhill. The first reaction cannot possibly occur to any extent since HCO is a very weak acid and HSO is a base whose strength is negligible What reactions will occur when an excess of acetic acid is added to a solution of potassium phosphate, K PO ? The line joining CH COOH to PO in Table \(\Page {1}\) is downhill, and so the reaction \[\text{CH}_3\text{COOH} + \text{PO}_4^{3-} \rightarrow \text{CH}_3\text{COO}^- \text{HPO}_4^{2-} \nonumber\] should occur. There is a further possibility because HPO is itself a base and might accept a second proton. The line from CH COOH to HPO is also downhill, but just barely, and so the reaction \[\text{CH}_3\text{COOH} + \text{HPO}_4^{2-} \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{PO}_4^- \nonumber\] can occur, but it does not go to completion. Hence double arrows are used. Although H PO is a base and might be protonated to yield phosphoric acid, H PO , a line drawn from CH COOH to H PO is , and so this does not happen. | 5,298 | 3,502 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Defects/Solid_Defects |
Few, if any, crystals are in that all unit cells consist of the ideal arrangement of atoms or molecules and all cells line up in a three dimensional space with no distortion. Some cells may have one or more atoms less whereas others may have one or more atoms than the ideal unit cell. The imperfection of crystals are called . Crystal defects are results of thermodynamic equilibrium contributed also by the increase in entropy term of the Gibb's free energy: \[\Delta G = \Delta H - T \Delta S\] Only at the unattainable absolute zero K will a crystal be perfect, in other words, no crystals are absolutely perfect. However, the degree of imperfection vary from compound to compound. On the other hand, some solid-like structure called also exist in a liquid. For example, the density of water is the highest at 277 K. The flickering clusters increase as temperature drops below 277 K, and the water density decreases as a result. The missing and lacking of atoms or ions in an ideal or imaginary crystal structure or lattice and the misalignment of unit cells in real crystals are called crystal defects or solid defects. Crystal defects occur as points, along lines, or in the form of a surface, and they are called respectively. Point defects can be divided into Frenkel defects and , and these often occur in ionic crystals. The former are due to misplacement of ions and vacancies. Charges are balanced in the whole crystal despite the presence of interstitial or extra ions and vacancies. On the other hand, when only vacancies of cation and anions are present with no interstitial or misplaced ions, the defects are called Schottky defects. Point defects are common in crystals with large anions such as AgBr, AgI, RbAgI . Due to the defects, the ions have some freedom to move about in crystals, making them relatively good conductors. These are called ionic conductors, unlike metals in which electrons are responsible for electric conductivity. Recently, ionic conductors have attracted a lot of attention because the fuel cell and battery technologies require conducting solids to separate the electrodes. Line defects are mostly due to misalignment of ions or presence of vacancies along a line. When lines of ions are missing in an otherwise perfect array of ions, an appeared. Edge dislocation is responsible for the ductility and malleability. In fact the hammering and stretching of materials often involve the movement of edge dislocation. Movements of dislocations give rise to their plastic behavior. Line dislocations usually do not end inside the crystal, and they either form loops or end at the surface of a single crystal. A dislocation is characterized by its Burgers vector: If you imagine going around the dislocation line, and exactly going back as many atoms in each direction as you have gone forward, you will not come back to the same atom where you have started. The Burgers vector points from start atom to the end atom of your journey (This "journey" is called Burgers circuit in dislocation theory). In this electron microscope image of the surface of a crystal, you see point defects and a Burger journey around an edge dislocation. The dislocation line is in the crystal, and the image shows its ending at the surface. A Burger vector is approximately perpendicular to the dislocation line, and the missing line of atoms is somewhere within the block of the Buerger journey. If the misalignment shifts a block of ions gradually downwards or upwards causing the formation of a screw like deformation, a is formed. The diagram here shows the idealized screw dislocation. Line defects weakens the structure along a one-dimensional space, and the defects type and density affects the mechanical properties of the solids. Thus, formation and study of dislocations are particularly important for structural materials such as metals. This link gives some impressive images of dislocations. Chemical etching often reveal pits which are visible under small magnifications. The Table of X-ray Crystallographic Data of Minerals (The CRC Handbook of Chemistry and Physics) list the following for bunsenite (NiO): Crystal system: cubic, structure type: rock salt, a = 4.177*10-8 cm. In the table of Physical Constant of Inorganic Compounds, the density of bunsenite (NiO) is 6.67g / cm3. From these values, evaluate the cell volume (volume of the unit cell), sum of Ni and O radii (rNi + rO), 2 Molar volumes, (X-ray) density, and the Schottky defect vacancy rate.
Actually, most of the required values have been listed in the table, but their evaluations illustrate the methods. These values are evaluated below: Cell volume = a3
= (4.177*10-8)3
= 72.88*10-24 cm3 \[ r_{Ni}+r_O = \dfrac{a}{ 2}\]
\[ = 2.088 \times 10^{-8}\] X-ray density = 4*(58.69+16.00) / (6.023*10 *72.88*10 )
= 6.806 g / cm , Compared to the observed density = 6.67 g / cm . The molar volumes (58.69+16.00) / density are thus 74.69 / 6.806 = 10.97 cm ; and 74.69 / 6.67 = 11.20 cm The vacancy rate = 6.806 - 6.67 / 6.806
= 0.02 (or 2%) These methods are hints to assignments. From the given conditions, we cannot calculate individual radii of Ni and O, but their sum is calculable. Plane defects occur along a 2-dimensional surface. The of a crystal is an obvious imperfection, because these surface atoms are different from those deep in the crystals. When a solid is used as a catalyst, the catalytic activity depends very much on the surface area per unit mass of the sample. For these powdery material, methods have been developed for the determination of unit areas per unit mass. Another surface defects are along the . A grain is a single crystal. If many seeds are formed when a sample starts to crystallize, each seed grow until they meet at the boundaries. Properties along these boundaries are different from the grains. A third plane defects are the . For example, in the close packing arrangement, the adjancent layers always have the AB relationship. In a ccp (fcc) close packing sequence, ...ABCABC..., one of the layer may suddenly be out of sequence, and become ..ABABCABC.... Similarly, in the hcp sequence, there is a possibility that one of the layer accidentally startes in the C location and resulting in the formation of a grain boundary. You already know that to obtain a perfectly pure substance is almost impossible. Purification is a costly process. In general, analytical reagent-grade chemicals are of high purity, and yet few of them are better than 99.9% pure. This means that a foreign atom or molecule is present for every 1000 host atoms or molecules in the crystal. Perhaps the most demanding of purity is in the electronic industry. Silicon crystals of 99.999 (called 5 nines) or better are required for IC chips productions. These crystal are with nitrogen group elements of P and As or boron group elements B, Al etc to form n- aand p-type semiconductors. In these crystals, the impurity atom substitute atoms of the host crystals. Presence minute foreign atoms with one electron more or less than the valence four silicon and germanium host atoms is the key of making n- and p-type semiconductors. Having many semiconductors connected in a single chip makes the integrated circuit a very efficient information processor. The electronic properties change dramatically due to these impurities. This is further described in Inorganic Chemistry by Swaddle. In other bulk materials, the presence of impurity usually leads to a lowering of melting point. For example, Hall and Heroult tried to electrolyze natural aluminum compounds. They discovered that using a 5% mixture of Al O (melting point 273 K) in cryolite Na AlF (melting point 1273 K) reduced the melting point to 1223 K, and that enabled the production of aluminum in bulk. Recent modifications lowered melting temperatures below 933 K. Some types of glass are made by mixing silica (SiO ), alumina (Al O ), calcium oxide (CaO), and sodium oxide (Na O). They are softer, but due to lower melting points, they are cheaper to produce. Color centers are imperfections in crystals that cause color (defects that cause color by absorption of light). Due to defects, metal oxides may also act as semiconductors, because there are many different types of electron traps. Electrons in defect region only absorb light at certain range of wavelength. The color seen are due to lights not absorbed. For example, a diamond with C vacancies (missing carbon atoms) absorbs light, and these centers give green color as shown here. Replacement of Al for Si in quartz give rise to the color of smoky quartz. A high temperature phase of ZnO , (x < 1), has electrons in place of the O vacancies. These electrons are color centers, often referred to as -centers (from the German word meaning color). Similarly, heating of ZnS to 773 K causes a loss of sulfur, and these material fluoresces strongly in ultraviolet light. Some non-stoichiometric solids are engineered to be n-type or p-type semiconductors. Nickel oxide NiO gain oxygen on heating in air, resulting in having Ni sites acting as electron trap, a p-type semiconductor. On the other hand, ZnO lose oxygen on heating, and the excess Zn metal atoms in the sample are ready to give electrons. The solid is an n-type semiconductor.
Correlate properties of a material to its structure.
Explain Frendel and Schottky defects.
Correlate properties of a material to its structure.
Specify the desirable properties of a material. | 9,547 | 3,503 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/05%3A_The_Electronic_Structure_of_Atoms/5.16%3A_Electron_Configurations |
In order to examine the results of for each element in the periodic table, we must recall the general rules that are necessary in order to predict for all atoms of the elements. To review, these rules are as follows: The order in which the subshells are filled merits some discussion. As can be seen in Figure \(\Page {1}\) within a given shell the energies of the subshells increase in the order < < < . When we discussed the boron atom, we saw that a orbital is higher in energy than the orbital in the same shell because the orbital is more effectively screened from the nucleus. Similar reasoning explains why orbitals are higher in energy than orbitals but lower than orbitals. Not only are the energies of a given shell spread out in this way, but there is sometimes an overlap in energy between shells. As can be seen from Figure \(\Page {1}\) the subshell of highest energy in the third shell, namely, 3 , is above the subshell of lowest energy in the fourth shell, namely, 4 . Similar overlaps occur among subshells of the fourth, fifth, sixth, and seventh shells. These cause exceptions to the expected order of filling subshells. The 6 orbital, for example, starts to fill before the 4 . Although the order in which the subshells fill seems hopelessly complex at first sight, there is a very simple device available for remembering it. This is shown in Figure \(\Page {1}\) . The rows in this table consist of all possible subshells within each shell. For example, the second row from the bottom contains 2 and 2 , the two subshells in the second shell. Insertion of diagonal lines in the manner shown gives the right order for filling the subshells. Predict the electron configuration for each of the following atoms: (a) \({}_{\text{15}}^{\text{31}}\text{P}\); (b) \({}_{\text{27}}^{\text{59}}\text{Co}\). In each case we follow the rules just stated. For \({}_{\text{15}}^{\text{31}}\text{P}\) there would be 15 protons and 16 neutrons in the nucleus and 15 extra nuclear electrons. Using Figure 1 to predict the order in which orbitals are filled, we have 2 2 electrons, leaving 11 more to add 2 , 2 , 2 , (or 2 ) 6 electrons, leaving 5 more to add 3 2 electrons, leaving 3 more to add 3 , 3 , 3 3 electrons [Ar] 9 more to add We now come to an energy overlap between the third and fourth shells. Because the 3 orbitals are so well shielded from the nucleus, they are in energy than the 4 orbitals. Accordingly the next orbitals to be filled are the 4 orbitals: Once the 4 orbital is filled, the 3d orbitals are next in line to be filled. The 7 remaining electrons are insufficient to fill this subshell so that we have the final result Electron configurations of the atoms may be determined experimentally. Table 1 in lists the results that have been obtained. There are some exceptions to the four rules enunciated above, but they are usually relatively minor. An obvious example of such an exception is the structure of chromium. It is found to be [Ar]3 4 , whereas our rules would have predicted [Ar]3 4 . Chromium adopts this structure because it allows the electrons to avoid each other more effectively. A complete discussion of this and other exceptions is beyond the scope of an introductory text. | 3,274 | 3,504 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.05%3A_Thermodynamics_of_Mixing_and_Dilution |
The free energy of a pure liquid or solid at 1 atm pressure is just its molar free energy of formation Δ multiplied by the number of moles present. For gases and substances in solution, we have to take into account the (which, in the case of gases, is normally expressed in terms of the pressure). We know that the lower the concentration, the greater the entropy, and thus the smaller the free energy. The following excerpt from this lesson serves as the starting point for the rest of the present lesson. As a substance becomes more dispersed in space, the thermal energy it carries is also spread over a larger volume, leading to an increase in its entropy. Because entropy, like energy, is an extensive property, a dilute solution of a given substance may well possess a smaller entropy than the same volume of a more concentrated solution, but the entropy of solute (the entropy) will of course always increase as the solution becomes more dilute. For gaseous substances, the volume and pressure are respectively direct and inverse measures of concentration. For an ideal gas that expands at a constant temperature (meaning that it absorbs heat from the surroundings to compensate for the work it does during the expansion), the increase in entropy is given by \[\Delta S = R\ln \left( \dfrac{V_2}{V_1} \right) \label{2-4}\] (If the gas is allowed to cool during the expansion, the relation becomes more complicated and will best be discussed in a more advanced course.) Because the pressure of a gas is inversely proportional to its volume, we can easily alter the above relation to express the entropy change associated with a change in the pressure of a perfect gas: \[\Delta S = R\ln \left( \dfrac{P_1}{P_2} \right) \label{2-5}\] Expressing the entropy change directly in concentrations \(c\), we have the similar relation \[\Delta S = R\ln \left( \dfrac{C_1}{C_2} \right) \label{2-6}\] Although these equations strictly apply only to perfect gases and cannot be used at all for liquids and solids, it turns out that in a dilute solution, the solute can often be treated as a gas dispersed in the volume of the solution, so the last equation can actually give a fairly accurate value for the entropy of dilution of a solution. We will see later that this has important consequences in determining the equilibrium concentrations in a homogeneous reaction mixture. The pressure of a perfect gas does not affect its enthalpy, but it does affect the entropy (box at left), and thus, through the Δ term, the free energy. When the pressure of such a gas changes from \(P_1\) to \(P_2\), the Gibbs energy change is \[ \Delta G = \Delta H - T \Delta S = 0 - RT \ln \left( \dfrac{P_1}{P_2} \right) \label{4.8}\] How can we evaluate the free energy of a specific sample of a gas at some arbitrary pressure? First, recall that the standard molar free energy ° that you would look up in a table refers to a pressure of 1 atm. The free energy per mole of our sample is just the sum of this value and any change in free energy that would occur if the pressure were changed from 1 atm to the pressure of interest \[ G = G^o + RT \ln \left( \dfrac{P_1}{1\; atm} \right) \label{4.9}\] which we normally write in abbreviated form \[G = G^o + RT \ln P \label{4-10}\] The term is not commonly used in traditional thermodynamics because it is essentially synonymous with the free energy, but it is worth knowing because it helps us appreciate the physical significance of free energy in certain contexts. The higher the pressure of a gas, the greater will be the tendency of its molecules to leave the confines of the container; we will call this the . The above equation tells us that the pressure of a gas is a directly observable measure of its free energy ( , not °!). Combining these two ideas, we can say that the . All substances, given the opportunity to form a homogeneous mixture with other substances, will tend to become more dilute. This can be rationalized simply from elementary statistics; there are more equally probable ways of arranging one hundred black marbles and one hundred white marbles, than two hundred marbles of a single color. For massive objects like marbles this has nothing to do with entropy, of course. However, when we are dealing with huge numbers of molecules capable of storing, exchanging and spreading thermal energy, mixing and expansion are definitely entropy-driven processes. It can be argued, in fact, that ; after all, when we mix two gases, each is expanding into the space formerly occupied exclusively by the other. Suppose, for example, that we have a gas initially confined to one half of a box, and we then remove the barrier so that it can expand into the full volume of the container. We know that the entropy of the gas will increase as the thermal energy of its molecules spreads into the enlarged space; the actual increase, according to Equations \(\ref{2-4}\) and \(\ref{2-5}\) above, is \(R \ln 2\). And from Equation \(\ref{4-10}\), the change in \(G\) will be \(–RT \ln 2\). Now let us repeat the experiment, but starting this time with "red" molecules in one half of the container and "blue" ones in the right half. Because we now have two gases expanding into twice their initial volumes, the changes in and will be twice as great: \[ΔS = 2 R \ln 2\] \[ΔG = –2 RT \ln 2\] However, notice that although each gas underwent an expansion, . What is true for gaseous molecules can, in principle, apply also to solute molecules dissolved in a solvent. An important qualification here is that the solution must be an one, meaning that the strength of interactions between all type of molecules (solutes A and B, and the solvent) must be the same. Remember that the enthalpy associated with the expansion of a perfect gas is by definition zero. In contrast, the \(ΔH_{mixing}\) of two liquids or of dissolving a solute in a solvent have finite values which may limit the miscibility of liquids or the solubility of a solute. Given this proviso, we can define the Gibbs energy of dilution or mixing by substituting this equation into the definition of Δ : \[ \Delta G_{dilution} = \Delta H_{dilution} - RT\ln \left( \dfrac{C_1}{C_2} \right) \label{4-11}\] If the substance in question forms an with the other components, then \(ΔH_{diution}\) is zero, and we can write \[ \Delta G_{dilution} = RT\ln \left( \dfrac{C_2}{C_1} \right) \label{4-12}\] These relations tell us that the dilution of a substance from an initial concentration \(C_1\) to a more dilute concentration \(C_2\) is accompanied by a , and thus will occur spontaneously. By the same token, the spontaneous “un-dilution” of a solution will not occur (we do not expect the tea to diffuse back into the tea bag!) However, un-dilution can be forced to occur if some means can be found to supply to the system an amount of energy (in the form of work) equal to \(\Delta G_{dilution} \). An important practical example of this is the metabolic work performed by the kidneys in concentrating substances from the blood for excretion in the urine. To find the Gibbs energy of a solute at some arbitrary concentration, we proceed in very much the same way as we did for a gas: we take the sum of the standard free energy plus any change in the free energy that would accompany a change in concentration from the standard state to the actual state of the solution. From Equation \(\ref{4-12}\) it is easy to derive an expression analogous to Equation \(\ref{4-10}\): \[G = G^o + RT \ln C \label{4-13}\] which gives the free energy of a solute at some arbitrary concentration \(C\) in terms of its value \(G^o\) in its standard state. Although Equation \(\ref{4-13}\) has the same simple form as Equation \(\ref{4-10}\), its practical application is fraught with difficulties, the major one being that it does not usually give values of \(G\) that are consistent with experiment, especially for solutes that are ionic or are slightly soluble. This is because most solutions (especially those containing dissolved ions) are far from ; intermolecular interactions between solute molecules and between solute and solvent bring back the enthalpy term that we left out in deriving Equation \(\ref{4-12}\). In addition, the structural organization of the solution becomes concentration dependent, so that the entropy depends on concentration in a more complicated way than is implied by the concentration analog of Equation \(\ref{4-12}\). We characterize the tendency for a chemical reaction A → B to occur at constant temperature and pressure by the value of its standard Gibbs energy change Δ °. If this quantity is negative, we know that the reaction will take place spontaneously. However, have you ever wondered why it is that substance A is not completely transformed into B if the latter is thermodynamically more stable? The answer is that if the reaction takes place in a single phase (gas or liquid), something else is going on: A and B are mixing together, and this process creates its own Gibbs energy change \(ΔG_{mixing}\). For a simple binary mixture of A and B (without any reaction), the changes in \(S\) and \(G\) can be represented by these simple plots: We will not try to prove it here, but it turns out that no matter how much lower the Gibbs energy of the products compared to that of the reactants, the free energy of the system can always be reduced even more if some of the reactants remain in the solution to contribute a Δ term. This is the reason that a plot of G as a function of the composition of such a system has a minimum at some point short of complete conversion. refers to the transport of a substance across a concentration gradient. The direction is always toward the region of lower concentration. You should now see that from a thermodynamic standpoint, these processes are identical in that they both represent the spontaneous "escape" of molecules from a region of higher concentration (lower entropy, higher Gibbs energy) to a region of lower concentration. Instead of complicating G° by trying to correct for all of these effects, chemists have chosen to retain its simple form by making a single small change in the form of \(\ref{4-13}\): \[G = G^o + RT \ln a \label{4-14}\] This equation is guaranteed to work, because , the of the solute, is its concentration. The relation between the activity and the concentration is given by \[a = \gamma c \label{4-15}\] where \(\gamma\) is the . As the solution becomes more dilute, the activity coefficient approaches unity: \[ \lim _{c \rightarrow 0} \gamma =1 \label{4.16}\] The price we pay for this simplicity is that the relation between the concentration and the activity at higher concentrations can be quite complicated, and must be determined experimentally for every different solution. The question of what standard state we choose for the solute (that is, at what concentration is ° defined, and in what units is it expressed?) is one that you will wish you had never asked. We might be tempted to use a concentration of 1 molar, but a solution this concentrated would be subject to all kinds of intermolecular interaction effects, and would not make a very practical standard state. These effects could be eliminated by going to the opposite extreme of an “infinitely dilute” solution, but by Equation \(\ref{4-12}\) this would imply a free energy of minus infinity for the solute, which would be awkward. Chemists have therefore agreed to define the standard state of a solute as one in which the concentration is 1 molar, but all solute-solute interactions are magically switched off, so that \(\gamma\) is effectively unity. Since this is impossible, no solution corresponding to this standard state can actually exist, but this turns out to be only a small drawback, and seems to be the best compromise between convenience, utility, and reality. | 11,894 | 3,505 |
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Some may be balanced using , but most are rather difficult to handle. Therefore it is useful to have some rules, albeit somewhat arbitrary ones, to help find appropriate coefficients. These rules depend on whether the reaction occurs in acidic or basic solution. In either situation we must make sure that the number of electrons accepted by the oxidizing agent exactly equals the number of electrons donated by the reducing agent. We shall apply the rules to the equation Reduction: \(\ce{IO^{3–} -> I2}\) Reduction: \(\ce{2IO3^{–} -> I2}\) Reduction: \(\ce{2IO3^{–} -> I2 + 6H2O}\) Reduction: \(\ce{12H^+ + 2IO3^{–} -> I2 + 6H2O}\) (The total charge on the left side was 0, but on the right it was – 2 + 4 = +2. Therefore 2 were needed on the right.) Reduction: \(\ce{10e^{–} + 12H^+ + 2IO3^{–} -> I2 + 6H2O}\) (The total charge on the left was 12 - 2 = +10, but on the right it was 0. Therefore 10 were needed on the left.) Oxidation: The oxidation number of electrons increases from +4 to +6, corresponding to a loss of 2 . Reduction: The oxidation number of I falls from +5 to 0, corresponding to a gain of 5 for each I. Since there are 2 I atoms, 10 must be added. Reduction: \(\cancel{10e^–} + \ce{12H^+ + 2IO3^{–} -> I2 + 6H2O}\) Note that when the half-equations were summed, the number of electrons was the same on both sides, and so no free electrons (which could not exist in aqueous solution) appear in the final result. It also would be more accurate to write H O instead of H for the hydronium ion. This can be done by adding 8H O to both sides of the equation: (On the right, the 8H O molecules are protonated to 8H O . It is also a good idea at this point to check that all atoms, as well as the electrical charges, balance. Potassium permanganate KMnO , can be used to oxidize alcohols to carboxylic acids. An example is \(\overset{\text{+7 }-\text{2}}{\mathop{\text{2MnO}_{\text{4}}^{-}}}\,\text{ + }\overset{-\text{2 +1 }-\text{2 +1}}{\mathop{\text{CH}_{\text{3}}\text{OH}}}\,\text{ + }\to \text{ }\overset{\text{+4 }-\text{2}}{\mathop{\text{MnO}_{\text{2}}}}\,\text{ + }\overset{\text{+1 }-\text{2}}{\mathop{\text{H}_{\text{2}}\text{O}}}\,\text{ + }\overset{\text{+1 +2 }-\text{2 }-\text{2 }}{\mathop{\text{HCOO}^{-}}}\,\text{ }\overset{\text{+2}}{\mathop{\text{2Mn}^{\text{2+}}}}\,\text{ + }\overset{-\text{2 +1}}{\mathop{\text{OH}^{-}}}\,\text{ (2)}\) Reduction: \(\ce{MnO4^{–} -> MnO2}\) Reduction: \(\ce{MnO4^{–} -> MnO2 + 2OH^–}\) Oxidation: \(\ce{CH3OH + 5OH^{–} -> HCOO^{–} + 4H2O}\) Reduction: \(\ce{MnO4^{–} + 2H2O -> MnO2 + 4OH^{–}}\) Oxidation: \(\ce{CH3OH + 5OH^{–} -> HCOO^{–} + 4H2O + 4e^{–}}\) Reduction: \(\ce{MnO4^{–} + 2H2O + 3e^{–} -> MnO2 + 4OH^{–}}\) Oxidation: C goes from -2 to +2, corresponding to a loss of 4 . Reduction: Mn goes from +7 to +4, corresponding to a gain of 3 . Reduction: \(\ce{4MnO4^{–} + 8H2O + 12e^{–} -> 4MnO2 + 16OH^{–}}\) The steps for balancing a redox reaction in an acidic or basic solution are summarized below for reference. | 3,100 | 3,506 |
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We have stated that the change in energy (\(ΔU\)) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is called , (or just . Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows: \[ \ce{Cu(s) + 4HNO3(aq) \rightarrow Cu(NO3)2(aq) + 2H_2O(l) + 2NO2(g)} \nonumber \] If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure \(\Page {1}\)). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., ). We find the amount of \(PV\) work done by multiplying the external pressure \(P\) by the change in volume caused by movement of the piston (\(ΔV\)). At a constant external pressure (here, atmospheric pressure), \[w = −PΔV \label{5.4.2} \] The negative sign associated with \(PV\) work done indicates that the system loses energy when the volume increases. If the volume increases at constant pressure (\(ΔV > 0\)), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (\(ΔV < 0\)), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy. The internal energy \(U\) of a system is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called (\(H\)) (from the Greek , meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy \(U\) plus the product of its pressure \(P\) and volume \(V\): \[H =U + \label{5.4.3} \] Because internal energy, pressure, and volume are all state functions, enthalpy is also a . So we can define a change in enthalpy (\(\Delta H\)) accordingly \[ΔH = H_{final} − H_{initial} \nonumber \] If a chemical change occurs at constant pressure (i.e., for a given \(P\), \(ΔP = 0\)), the change in enthalpy (\(ΔH\)) is \[ \begin{align} ΔH &= Δ(U + PV) \\[4pt] &= ΔU + ΔPV \\[4pt] &= ΔU + PΔV \label{5.4.4} \end{align} \] Substituting \(q + w\) for \(ΔU\) (First Law of Thermodynamics) and \(−w\) for \(PΔV\) (Equation \(\ref{5.4.2}\)) into Equation \(\ref{5.4.4}\), we obtain \[ \begin{align} ΔH &= ΔU + PΔV \\[4pt] &= q_p + \cancel{w} −\cancel{w} \\[4pt] &= q_p \label{5.4.5} \end{align} \] The subscript \(p\) is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation \(\ref{5.4.5}\) we see that at constant pressure the change in enthalpy, \(ΔH\) of the system, is equal to the heat gained or lost. \[ \begin{align} ΔH &= H_{final} − H_{initial} \\[4pt] &= q_p \label{5.4.6} \end{align} \] Just as with \(ΔU\), because enthalpy is a state function, the magnitude of \(ΔH\) depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does occur at constant pressure. To find \(ΔH\) , measure \(q_p\). When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction (\(ΔH_{rxn}\)), the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so \(ΔH_{rxn}\) is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so \(ΔH_{rxn}\) is positive. Thus: In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table: If Δ is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, (Figure \(\Page {2}a\)). Conversely, if Δ is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, (Figure \(\Page {2b}\)). Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion. Bond breaking requires an input of energy; bond making releases energy.y. \( \begin{matrix}
heat+ H_{2}O(s) \rightarrow H_{2}O(l) & \Delta H > 0
\end{matrix} \label{5.4.7} \) \( \begin{matrix}
H_{2}O(l) \rightarrow H_{2}O(s) + heat & \Delta H < 0
\end{matrix} \label{5.4.8} \) In both cases, the of the enthalpy change is the same; only the is different. \[ \ce{2Al(s) + Fe2O3(s) -> 2Fe (s) + Al2O3 (s) } + 815.5 \; kJ \label{5.4.9} \] Thus Δ = −851.5 kJ/mol of Fe O . We can also describe Δ for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for Δ , in kilojoules rather than kilojoules per mole, is written after the reaction, as in Equation \(\ref{5.4.10}\), it is the value of Δ corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation: \[ \ce{ 2Al(s) + Fe2O3(s) -> 2Fe(s) + Al2O3 (s)} \quad \Delta H_{rxn}= - 851.5 \; kJ \label{5.4.10} \] If 4 mol of Al and 2 mol of \(\ce{Fe2O3}\) react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows: \[ - \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \label{5.4.6a} \] The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example \(\Page {1}\). Certain parts of the world, such as southern California and Saudi Arabia, are short of freshwater for drinking. One possible solution to the problem is to tow icebergs from Antarctica and then melt them as needed. If \(ΔH\) is 6.01 kJ/mol for the reaction at 0°C and constant pressure: \[\ce{H2O(s) → H_2O(l)} \nonumber \] How much energy would be required to melt a moderately large iceberg with a mass of 1.00 million metric tons (1.00 × 10 metric tons)? (A metric ton is 1000 kg.) energy per mole of ice and mass of iceberg energy required to melt iceberg Because enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given Δ for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—so we need to calculate the number of moles of ice in the iceberg and multiply that number by Δ (+6.01 kJ/mol): \[ \begin{align*} moles \; H_{2}O & = 1.00\times 10^{6} \; \cancel{\text{metric ton }} \ce{H2O} \left ( \dfrac{1000 \; \cancel{kg}}{1 \; \cancel{\text{metric ton}}} \right ) \left ( \dfrac{1000 \; \cancel{g}}{1 \; \cancel{kg}} \right ) \left ( \dfrac{1 \; mol \; H_{2}O}{18.015 \; \cancel{g \; H_{2}O}} \right ) \\[4pt] & = 5.55\times 10^{10} \; mol \,\ce{H2O} \end{align*} \nonumber \] The energy needed to melt the iceberg is thus \[ \left ( \dfrac{6.01 \; kJ}{\cancel{mol \; H_{2}O}} \right )\left ( 5.55 \times 10^{10} \; \cancel{mol \; H_{2}O} \right )= 3.34 \times 10^{11} \; kJ \nonumber \] Because so much energy is needed to melt the iceberg, this plan would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of different energy sources equivalent to the amount of energy needed to melt the iceberg are shown below. Possible sources of the approximately \(3.34 \times 10^{11}\, kJ\) needed to melt a \(1.00 \times 10^6\) metric ton iceberg Alternatively, we can rely on ambient temperatures to slowly melt the iceberg. The main issue with this idea is the cost of dragging the iceberg to the desired place. If 17.3 g of powdered aluminum are allowed to react with excess \(\ce{Fe2O3}\), how much heat is produced? 273 kJ One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, since enthalpy is a state function, all we have to know is the initial and final states of the reaction. This allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following: The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system. For a chemical reaction, the (\(ΔH_{rxn}\)) is the difference in enthalpy between products and reactants; the units of \(ΔH_{rxn}\) are kilojoules per mole. Reversing a chemical reaction reverses the sign of \(ΔH_{rxn}\). ( ) | 9,155 | 3,507 |
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The experiences you have had in the chemical laboratory have probably already taught you that there is an uphill character to some reactions and a downhill character to others. A simple example is the combination of mercury with bromine which we considered in detail in the : \[\ce{Hg(l) +Br2(l)→HgBr2(s)} \nonumber \] This reaction proceeds from Hg( ) and Br ( ) to HgBr ( ), much as a ball rolls downhill. On the other hand, decomposition of HgBr to the elements is an uphill process, and we must “push” this reaction to force it to occur. One way to do this is to dissolve HgBr in water and pass an electrical current through the solution. Mercury will appear at one electrode and bromine at the other. We want to understand why some chemical reactions are downhill while others are uphill. In particular we want to know what kinds of atomic and molecular processes proceed downhill, and which processes require a “push” to go uphill. We find that these questions of uphill and downhill nature of a reaction can be answered using concepts such as spontaneity, entropy, and free energy. Chemical reactions which proceed downhill, are said to be . With spontaneity, we are considering a process will occur, but can say nothing about the at which it will occur. Spontaneous process may also require energy to overcome an activation barrier. This means that some processes may be spontaneous, but do not occur at a noticeable rate. An everyday example is burning a log of wood where the burning reaction is clearly "downhill" and thus defined spontaneous, but without a flame to supply heat to overcome the activation barrier, the reaction will not occur. A spontaneous process corresponds to a , as measured by the of the reactants and products. is defined as the number of alternative microscopic arrangements which correspond to the same macroscopic state. Because most macroscopic samples of matter contain 10 particles or more, very large values of are encountered. Therefore it is convenient to use the entropy , which is proportional to the logarithm of , as a measure of spontaneity. According to the second law of thermodynamics, when a spontaneous process occurs, there must be an increase in total entropy. The entropy of a perfectly ordered crystal at absolute zero is zero according to the third law of thermodynamics. . For a given amount of substance, and , the greater the entropy, while stronger forces between atoms, molecules, or ions result in lower entropy. In general the of a substance. To determine whether a chemical reaction is spontaneous we must calculate the change in total entropy of the chemical system its . This can be done directly using Δ ° and –ΔH°/ or indirectly using the change in Δ °. For a spontaneous reaction occurring at constant temperature and pressure, Δ ° must be negative. Since Δ ° = Δ ° – Δ °, a negative enthalpy change is the most important factor governing spontaneity of a reaction at low temperatures. At high values of , – Δ ° becomes large and an increase in entropy of the system is essential for a spontaneous process. The change in Gibbs free energy at standard pressure can be calculated from tables of Δ °. Once obtained it can be used to determine two other useful values, the either the amount of (or the amount of energy needed to drive a process in a non-spontaneous direction), and the . For a reaction in the gas phase ° is numerically the same as , provided the latter is expressed in atmospheres. Knowledge of Δ ° thus allows calculation of the equilibrium partial pressures of reactants and products. | 3,623 | 3,508 |
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With some familiarity with the properties of single electrons, such as the single electron around the hydrogen nucleus above, we can discuss atoms containing more than one electron. The images found here depict electron wave density by number of dots. Thus, more dots indicates more electron density 'cloud' in that region. In these diagrams, each electron wave is displayed with a different color and you can toggle the view of each electron wave on and off using the buttons below the dot density diagrams. The first element in the periodic table with more than one electron is helium, which has two electrons. Dot-density diagrams for both these electrons are shown below. One electron is color coded in blue, and the other in green. Note that both electrons occupy the same orbital, namely, a 1 orbital. It turns out that 2 is the maximum number of electrons any orbital can hold. This restriction is connected with a property of the electrons not yet discussed, namely, their . Electrons can not only move about from place to place, but they can also rotate or spin about themselves. Two orientations (clockwise and counterclockwise, referred to as spin up or spin down) are possible for this spin. According to the , if two electrons occupy the same orbital, they have opposite spins. Two such electrons are said to be and are often represented by arrows pointing in different directions, i.e., by the symbol . Two electrons spinning in the same direction are said to have their spins and are indicated by . The Pauli principle implies that if two electrons have parallel spins, they occupy different orbitals. An obvious feature of the helium atom shown below is that it is somewhat than the hydrogen atom drawn to the same scale above. This contraction is caused by the increase in the charge on the nucleus from +1 in the hydrogen atom to +2 in the helium atom. This pulls both the green and the blue electron clouds in more tightly. This effect is offset, but to a lesser extent, by the mutual repulsion of the two electron clouds. In the dot density image below, the three electrons of the lithium atom are color-coded blue, green, and red. As in the previous atom, two electrons (blue and green) occupy the 1 orbital. The Pauli principle prevents more than two electrons from occupying this orbital, and so the third (red) electron must occupy the next higher orbital in energy, namely, the 2s orbital. A convenient shorthand form for indicating this is 1 2 The superscripts and indicate that there are two electrons in the 1 orbital and one electron in the 2 orbital, respectively. As in the case of helium, the increase in nuclear charge to +3 produces a corresponding reduction in the size of the lithium 1 orbital. In sharp contrast to this compact inner orbital is the very large and very diffuse cloud of the outer 2 electron. There are two reasons why this 2 cloud is so large. The first reason is that the principal quantum number has increased from 1 to 2. As shown in , the 2 electron cloud is bigger than 1 even in the hydrogen atom with a nuclear charge of only +1. A second reason is that the two 1 electrons are usually closer to the nucleus than the 2 electron. These two 1 electrons have the effect of or the outer electron from the full attractive force of the +3 charge on the nucleus. When the 2 electron is some distance from the nucleus, it “sees” not only the +3 charge on the nucleus but also the two negative charges close by. The overall effect is almost as though two of the three positive charges on the nucleus are canceled, leaving a net charge of + 1 to hold the outer electron to the atom. This situation can also be described by saying that the is close to +1. It should be clear from Plate 4 that when a lithium atom interacts with another atom, the 2 electron is far more likely to be involved than either of the two 1 electrons. In Lewis’ terminology, it is a and occupies a . The pair of 1 electrons are a complete shell and form the kernel of the lithium atom. There is thus a close correspondence between the wave-mechanical picture and Lewis’ earlier, less mathematical ideas. It is also worth noting that the wave model of lithium gives a atom―a great advance over the elongated orbits which were needed to describe the alkali-metal atoms in the Bohr theory (see ). | 4,379 | 3,509 |
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Guystav Kirchoff and Robert Bunsen first used atomic absorption—along with atomic emission—in 1859 and 1860 as a means for identify atoms in flames and hot gases. Although atomic emission continued to develop as an analytical technique, progress in atomic absorption languished for almost a century. Modern atomic absorption spectroscopy has its beginnings in 1955 as a result of the independent work of A. C. Walsh and C. T. J. Alkemade [(a) Walsh, A. , , 933A–941A; (b) Koirtyohann, S. R. , , 1024A–1031A; (c) Slavin, W. , , 1033A–1038A]. Commercial instruments were in place by the early 1960s, and the importance of atomic absorption as an analytical technique soon was evident. Atomic absorption spectrophotometers use the same single-beam or double-beam optics described earlier for molecular absorption spectrophotometers (see and ). There is, however, an important additional need in atomic absorption spectroscopy: we first must covert the analyte into free atoms. In most cases the analyte is in solution form. If the sample is a solid, then we must bring the analyte into solution before the analysis. When analyzing a lake sediment for Cu, Zn, and Fe, for example, we bring the analytes into solution as Cu , Zn , and Fe by extracting them with a suitable reagent. For this reason, only the introduction of solution samples is considered in this chapter. What reagent we choose to use to bring an analyte into solution depends on our research goals. If we need to know the total amount of metal in the sediment, then we might try a microwave digestion using a mixture of concentrated acids, such as HNO , HCl, and HF. This destroys the sediment’s matrix and brings everything into solution. On the other hand, if our interest is biologically available metals, we might extract the sample under milder conditions using, for example, a dilute solution of HCl or CH COOH at room temperature. The process of converting an analyte to a free gaseous atom is called . Converting an aqueous analyte into a free atom requires that we strip away the solvent, volatilize the analyte, and, if necessary, dissociate the analyte into free atoms. Desolvating an aqueous solution of CuCl , for example, leaves us with solid particulates of CuCl . Converting the particulate CuCl to gas phases atoms of Cu and Cl requires thermal energy. \[\mathrm{CuCl}_{2}(a q) \rightarrow \mathrm{CuCl}_{2}(s) \rightarrow \mathrm{Cu}(g)+2 \mathrm{Cl}(g) \nonumber\] There are two common atomization methods: flame atomization and electrothermal atomization, although a few elements are atomized using other methods. Figure 10.4.1
shows a typical flame atomization assembly with close-up views of several key components. In the unit shown here, the aqueous sample is drawn into the assembly by passing a high-pressure stream of compressed air past the end of a capillary tube immersed in the sample. When the sample exits the nebulizer it strikes a glass impact bead, which converts it into a fine aerosol mist within the spray chamber. The aerosol mist is swept through the spray chamber by the combustion gases—compressed air and acetylene in this case—to the burner head where the flame’s thermal energy desolvates the aerosol mist to a dry aerosol of small, solid particulates. The flame’s thermal energy then volatilizes the particles, producing a vapor that consists of molecular species, ionic species, and free atoms. The slot burner in Figure 10.4.1
a provides a long optical pathlength and a stable flame. Because absorbance is directly proportional to pathlength, a long pathlength provides greater sensitivity. A stable flame minimizes uncertainty due to fluctuations in the flame. The burner is mounted on an adjustable stage that allows the entire assembly to move horizontally and vertically. Horizontal adjustments ensure the flame is aligned with the instrument’s optical path. Vertical adjustments change the height within the flame from which absorbance is monitored. This is important because two competing processes affect the concentration of free atoms in the flame. The more time an analyte spends in the flame the greater the atomization efficiency; thus, the production of free atoms increases with height. On the other hand, a longer residence time allows more opportunity for the free atoms to combine with oxygen to form a molecular oxide. As seen in Figure 10.4.2
, for a metal this is easy to oxidize, such as Cr, the concentration of free atoms is greatest just above the burner head. For a metal, such as Ag, which is difficult to oxidize, the concentration of free atoms increases steadily with height. The flame’s temperature, which affects the efficiency of atomization, depends on the fuel–oxidant mixture, several examples of which are listed in Table 10.4.1
. Of these, the air–acetylene and the nitrous oxide–acetylene flames are the most popular. Normally the fuel and oxidant are mixed in an approximately stoichiometric ratio; however, a fuel-rich mixture may be necessary for easily oxidized analytes. Figure 10.4.3
shows a cross-section through the flame, looking down the source radiation’s optical path. The primary combustion zone usually is rich in gas combustion products that emit radiation, limiting is useful- ness for atomic absorption. The interzonal region generally is rich in free atoms and provides the best location for measuring atomic absorption. The hottest part of the flame typically is 2–3 cm above the primary combustion zone. As atoms approach the flame’s secondary combustion zone, the decrease in temperature allows for formation of stable molecular species. The most common means for introducing a sample into a flame atomizer is a continuous aspiration in which the sample flows through the burner while we monitor absorbance. Continuous aspiration is sample intensive, typically requiring from 2–5 mL of sample. Flame microsampling allows us to introduce a discrete sample of fixed volume, and is useful if we have a limited amount of sample or when the sample’s matrix is incompatible with the flame atomizer. For example, continuously aspirating a sample that has a high concentration of dissolved solids—sea water, for example, comes to mind—may build-up a solid de- posit on the burner head that obstructs the flame and that lowers the absorbance. Flame microsampling is accomplished using a micropipet to place 50–250 μL of sample in a Teflon funnel connected to the nebulizer, or by dipping the nebulizer tubing into the sample for a short time. Dip sampling usually is accomplished with an automatic sampler. The signal for flame microsampling is a transitory peak whose height or area is proportional to the amount of analyte that is injected. The principal advantage of flame atomization is the reproducibility with which the sample is introduced into the spectrophotometer; a significant disadvantage is that the efficiency of atomization is quite poor. There are two reasons for poor atomization efficiency. First, the majority of the aerosol droplets produced during nebulization are too large to be carried to the flame by the combustion gases. Consequently, as much as 95% of the sample never reaches the flame, which is the reason for the waste line shown at the bottom of the spray chamber in . A second reason for poor atomization efficiency is that the large volume of combustion gases significantly dilutes the sample. Together, these contributions to the efficiency of atomization reduce sensitivity because the analyte’s concentration in the flame may be a factor of \(2.5 \times 10^{-6}\) less than that in solution [Ingle, J. D.; Crouch, S. R. , Prentice-Hall: Englewood Cliffs, NJ, 1988; p. 275]. A significant improvement in sensitivity is achieved by using the resistive heating of a graphite tube in place of a flame. A typical electrothermal atomizer, also known as a , consists of a cylindrical graphite tube approximately 1–3 cm in length and 3–8 mm in diameter. As shown in Figure 10.4.4
, the graphite tube is housed in an sealed assembly that has an optically transparent window at each end. A continuous stream of inert gas is passed through the furnace, which protects the graphite tube from oxidation and removes the gaseous products produced during atomization. A power supply is used to pass a current through the graphite tube, resulting in resistive heating. Samples of between 5–50 μL are injected into the graphite tube through a small hole at the top of the tube. Atomization is achieved in three stages. In the first stage the sample is dried to a solid residue using a current that raises the temperature of the graphite tube to about 110 C. In the second stage, which is called ashing, the temperature is increased to between 350–1200 C. At these temperatures organic material in the sample is converted to CO and H O, and volatile inorganic materials are vaporized. These gases are removed by the inert gas flow. In the final stage the sample is atomized by rapidly increasing the temperature to between 2000–3000 C. The result is a transient absorbance peak whose height or area is proportional to the absolute amount of analyte injected into the graphite tube. Together, the three stages take approximately 45–90 s, with most of this time used for drying and ashing the sample. Electrothermal atomization provides a significant improvement in sensitivity by trapping the gaseous analyte in the small volume within the graphite tube. The analyte’s concentration in the resulting vapor phase is as much as \(1000 \times\) greater than in a flame atomization [Parsons, M. L.; Major, S.; Forster, A. R. , , 411–418]. This improvement in sensitivity—and the resulting improvement in detection limits—is offset by a significant decrease in precision. Atomization efficiency is influenced strongly by the sample’s contact with the graphite tube, which is difficult to control reproducibly. A few elements are atomized by using a chemical reaction to produce a volatile product. Elements such as As, Se, Sb, Bi, Ge, Sn, Te, and Pb, for example, form volatile hydrides when they react with NaBH in the presence of acid. An inert gas carries the volatile hydride to either a flame or to a heated quartz observation tube situated in the optical path. Mercury is determined by the cold-vapor method in which it is reduced to elemental mercury with SnCl . The volatile Hg is carried by an inert gas to an unheated observation tube situated in the instrument’s optical path. Atomic absorption is used widely for the analysis of trace metals in a variety of sample matrices. Using Zn as an example, there are standard atomic absorption methods for its determination in samples as diverse as water and wastewater, air, blood, urine, muscle tissue, hair, milk, breakfast cereals, shampoos, alloys, industrial plating baths, gasoline, oil, sediments, and rocks. Developing a quantitative atomic absorption method requires several considerations, including choosing a method of atomization, selecting the wavelength and slit width, preparing the sample for analysis, minimizing spectral and chemical interferences, and selecting a method of standardization. Each of these topics is considered in this section. The most important factor in choosing a method of atomization is the analyte’s concentration. Because of its greater sensitivity, it takes less analyte to achieve a given absorbance when using electrothermal atomization. Table 10.4.2
, which compares the amount of analyte needed to achieve an absorbance of 0.20 when using flame atomization and electrothermal atomization, is useful when selecting an atomization method. For example, flame atomization is the method of choice if our samples contain 1–10 mg Zn /L, but electrothermal atomization is the best choice for samples that contain 1–10 μg Zn /L. Varian Cookbook, SpectraAA Software Version 4.00 Pro. As: 10 mg/L by hydride vaporization; Hg: 11.5 mg/L by cold-vapor; and Sn:18 mg/L by hydride vaporization The source for atomic absorption is a hollow cathode lamp that consists of a cathode and anode enclosed within a glass tube filled with a low pressure of an inert gas, such as Ne or Ar (Figure 10.4.5
). Applying a potential across the electrodes ionizes the filler gas. The positively charged gas ions collide with the negatively charged cathode, sputtering atoms from the cathode’s surface. Some of the sputtered atoms are in the excited state and emit radiation characteristic of the metal(s) from which the cathode is manufactured. By fashioning the cathode from the metallic analyte, a hollow cathode lamp provides emission lines that correspond to the analyte’s absorption spectrum. Because atomic absorption lines are narrow, we need to use a line source instead of a continuum source (compare, for example, with ). The effective bandwidth when using a continuum source is roughly \(1000 \times\) larger than an atomic absorption line; thus, ≈ , % ≈ 100, and ≈ 0. Because a hollow cathode lamp is a line source, and have different values giving a % < 100 and > 0. Each element in a hollow cathode lamp provides several atomic emission lines that we can use for atomic absorption. Usually the wavelength that provides the best sensitivity is the one we choose to use, although a less sensitive wavelength may be more appropriate for a sample that has higher concentration of analyte. For the Cr hollow cathode lamp in Table 10.4.3
, the best sensitivity is obtained using a wavelength of 357.9 nm. Another consideration is the emission line's intensity. If several emission lines meet our requirements for sensitivity, we may wish to use the emission line with the largest relative because there is less uncertainty in measuring and . When analyzing a sample that is ≈10 mg Cr/L, for example, the first three wavelengths in Table 10.4.3
provide an appropriate sensitivity; the wavelengths of 425.4 nm and 429.0 nm, however, have a greater and will provide less uncertainty in the measured absorbance. The emission spectrum for a hollow cathode lamp includes, in addition to the analyte's emission lines, additional emission lines from impurities present in the metallic cathode and from the filler gas. These additional lines are a potential source of stray radiation that could result in an instrumental deviation from Beer’s law. The monochromator’s slit width is set as wide as possible to improve the throughput of radiation and narrow enough to eliminate these sources of stray radiation. Flame and electrothermal atomization require that the analyte is in solution. Solid samples are brought into solution by dissolving in an appropriate solvent. If the sample is not soluble it is digested, either on a hot-plate or by microwave, using HNO , H SO , or HClO . Alternatively, we can extract the analyte using a Soxhlet extractor. Liquid samples are analyzed directly or the analytes extracted if the matrix is in- compatible with the method of atomization. A serum sample, for instance, is difficult to aspirate when using flame atomization and may produce an unacceptably high background absorbance when using electrothermal atomization. A liquid–liquid extraction using an organic solvent and a chelating agent frequently is used to concentrate analytes. Dilute solutions of Cd , Co , Cu , Fe , Pb , Ni , and Zn , for example, are concentrated by extracting with a solution of ammonium pyrrolidine dithiocarbamate in methyl isobutyl ketone. A spectral interference occurs when an analyte’s absorption line overlaps with an interferent’s absorption line or band. Because they are so narrow, the overlap of two atomic absorption lines seldom is a problem. On the other hand, a molecule’s broad absorption band or the scattering of source radiation is a potentially serious spectral interference. An important consideration when using a flame as an atomization source is its effect on the measured absorbance. Among the products of combustion are molecular species that exhibit broad absorption bands and particulates that scatter radiation from the source. If we fail to compensate for these spectral interferences, then the intensity of transmitted radiation is smaller than expected. The result is an apparent increase in the sample’s absorbance. Fortunately, absorption and scattering of radiation by the flame are corrected by analyzing a blank. Spectral interferences also occur when components of the sample’s matrix other than the analyte react to form molecular species, such as oxides and hydroxides. The resulting absorption and scattering constitutes the sample’s background and may present a significant problem, particularly at wavelengths below 300 nm where the scattering of radiation becomes more important. If we know the composition of the sample’s matrix, then we can prepare our samples using an identical matrix. In this case the background absorption is the same for both the samples and the standards. Alternatively, if the background is due to a known matrix component, then we can add that component in excess to all samples and standards so that the contribution of the naturally occurring interferent is insignificant. Finally, many interferences due to the sample’s matrix are eliminated by increasing the atomization temperature. For example, switching to a higher temperature flame helps prevents the formation of interfering oxides and hydroxides. If the identity of the matrix interference is unknown, or if it is not possible to adjust the flame or furnace conditions to eliminate the interference, then we must find another method to compensate for the background interference. Several methods have been developed to compensate for matrix interferences, and most atomic absorption spectrophotometers include one or more of these methods. One of the most common methods for is to use a continuum source, such as a D lamp. Because a D lamp is a continuum source, absorbance of its radiation by the analyte’s narrow absorption line is negligible. Only the background, therefore, absorbs radiation from the D lamp. Both the analyte and the background, on the other hand, absorb the hollow cathode’s radiation. Subtracting the absorbance for the D lamp from that for the hollow cathode lamp gives a corrected absorbance that compensates for the background interference. Although this method of background correction is effective, it does assume that the background absorbance is constant over the range of wavelengths passed by the monochromator. If this is not true, then subtracting the two absorbances underestimates or overestimates the background. Other methods of background correction have been developed, including Zeeman effect background correction and Smith–Hieftje background correction, both of which are included in some commercially available atomic absorption spectrophotometers. Consult the chapter’s additional resources for additional information. The quantitative analysis of some elements is complicated by chemical interferences that occur during atomization. The most common chemical interferences are the formation of nonvolatile compounds that contain the analyte and ionization of the analyte. One example of the formation of a nonvolatile compound is the effect of \(\text{PO}_4^{3-}\) or Al on the flame atomic absorption analysis of Ca . In one study, for example, adding 100 ppm Al to a solution of 5 ppm Ca decreased calcium ion’s absorbance from 0.50 to 0.14, while adding 500 ppm \(\text{PO}_4^{3-}\) to a similar solution of Ca decreased the absorbance from 0.50 to 0.38. These interferences are attributed to the formation of nonvolatile particles of Ca (PO ) and an Al–Ca–O oxide [Hosking, J. W.; Snell, N. B.; Sturman, B. T. , , 128–130]. When using flame atomization, we can minimize the formation of non-volatile compounds by increasing the flame’s temperature by changing the fuel-to-oxidant ratio or by switching to a different combination of fuel and oxidant. Another approach is to add a releasing agent or a protecting agent to the sample. A is a species that reacts preferentially with the interferent, releasing the analyte during atomization. For example, Sr and La serve as releasing agents for the analysis of Ca in the presence of \(\text{PO}_4^{3-}\) or Al . Adding 2000 ppm SrCl to the Ca / \(\text{PO}_4^{3-}\) and to the Ca /Al mixtures described in the previous paragraph increased the absorbance to 0.48. A reacts with the analyte to form a stable volatile complex. Adding 1% w/w EDTA to the Ca / \(\text{PO}_4^{3-}\) solution described in the previous paragraph increased the absorbance to 0.52. An ionization interference occurs when thermal energy from the flame or the electrothermal atomizer is sufficient to ionize the analyte \[\mathrm{M}(s)\rightleftharpoons \ \mathrm{M}^{+}(a q)+e^{-} \label{10.1}\] where M is the analyte. Because the absorption spectra for M and M are different, the position of the equilibrium in reaction \ref{10.1} affects the absorbance at wavelengths where M absorbs. To limit ionization we add a high concentration of an , which is a species that ionizes more easily than the analyte. If the ionization suppressor's concentration is sufficient, then the increased concentration of electrons in the flame pushes reaction \ref{10.1} to the left, preventing the analyte’s ionization. Potassium and cesium frequently are used as an ionization suppressor because of their low ionization energy. Because Beer’s law also applies to atomic absorption, we might expect atomic absorption calibration curves to be linear. In practice, however, most atomic absorption calibration curves are nonlinear or linear over a limited range of concentrations. Nonlinearity in atomic absorption is a consequence of instrumental limitations, including stray radiation from the hollow cathode lamp and the variation in molar absorptivity across the absorption line. Accurate quantitative work, therefore, requires a suitable means for computing the calibration curve from a set of standards. When possible, a quantitative analysis is best conducted using external standards. Unfortunately, matrix interferences are a frequent problem, particularly when using electrothermal atomization. For this reason the method of standard additions often is used. One limitation to this method of standardization, however, is the requirement of a linear relationship between absorbance and concentration. Most instruments include several different algorithms for computing the calibration curve. The instrument in my lab, for example, includes five algorithms. Three of the algorithms fit absorbance data using linear, quadratic, or cubic polynomial functions of the analyte’s concentration. It also includes two algorithms that fit the concentrations of the standards to quadratic functions of the absorbance. The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of Cu and Zn in biological tissues provides an instructive example of a typical procedure. The description here is based on Bhattacharya, S. K.; Goodwin, T. G.; Crawford, A. J. , , 1567–1593, and Crawford, A. J.; Bhattacharya, S. K. Varian Instruments at Work, Number AA–46, April 1985. . Copper and zinc are isolated from tissue samples by digesting the sample with HNO after first removing any fatty tissue. The concentration of copper and zinc in the supernatant are determined by atomic absorption using an air-acetylene flame. . Tissue samples are obtained by a muscle needle biopsy and dried for 24–30 h at 105 C to remove all traces of moisture. The fatty tissue in a dried sample is removed by extracting overnight with anhydrous ether. After removing the ether, the sample is dried to obtain the fat-free dry tissue weight (FFDT). The sample is digested at 68 C for 20–24 h using 3 mL of 0.75 M HNO . After centrifuging at 2500 rpm for 10 minutes, the supernatant is transferred to a 5-mL volumetric flask. The digestion is repeated two more times, for 2–4 hours each, using 0.9-mL aliquots of 0.75 M HNO . These supernatants are added to the 5-mL volumetric flask, which is diluted to volume with 0.75 M HNO . The concentrations of Cu and Zn in the diluted supernatant are determined by flame atomic absorption spectroscopy using an air-acetylene flame and external standards. Copper is analyzed at a wavelength of 324.8 nm with a slit width of 0.5 nm, and zinc is analyzed at 213.9 nm with a slit width of 1.0 nm. Background correction using a D lamp is necessary for zinc. Results are reported as μg of Cu or Zn per gram of FFDT. . 1. Describe the appropriate matrix for the external standards and for the blank? The matrix for the standards and the blank should match the matrix of the samples; thus, an appropriate matrix is 0.75 M HNO . Any interferences from other components of the sample matrix are minimized by background correction. 2. Why is a background correction necessary for the analysis of Zn, but not for the analysis of Cu? Background correction compensates for background absorption and scattering due to interferents in the sample. Such interferences are most severe when using a wavelength less than 300 nm. This is the case for Zn, but not for Cu. 3. A Cu hollow cathode lamp has several emission lines, the properties of which are shown in the following table. Explain why this method uses the line at 324.8 nm. With 1.5 mg Cu/L giving an absorbance of 0.20, the emission line at 324.8 nm has the best sensitivity. In addition, it is the most intense emission line, which decreases the uncertainty in the measured absorbance. To evaluate the method described in Representative Method 10.4.1, a series of external standard is prepared and analyzed, providing the results shown here [Crawford, A. J.; Bhattacharya, S. K. “Microanalysis of Copper and Zinc in Biopsy-Sized Tissue Specimens by Atomic Absorption Spectroscopy Using a Stoichiometric Air-Acetylene Flame,” Varian Instruments at Work, Number AA–46, April 1985]. A bovine liver standard reference material is used to evaluate the method’s accuracy. After drying and extracting the sample, a 11.23-mg FFDT tissue sample gives an absorbance of 0.023. Report the amount of copper in the sample as μg Cu/g FFDT. Linear regression of absorbance versus the concentration of Cu in the standards gives the calibration curve shown below and the following calibration equation. \[A=-0.0002+0.0661 \times \frac{\mu \mathrm{g} \ \mathrm{Cu}}{\mathrm{mL}} \nonumber\] Substituting the sample’s absorbance into the calibration equation gives the concentration of copper as 0.351 μg/mL. The concentration of copper in the tissue sample, therefore, is \[\frac { \frac{0.351 \mu \mathrm{g} \ \mathrm{Cu}}{\mathrm{mL}} \times 5.000 \ \mathrm{mL}} {0.01123 \text{ g sample}}=156 \ \mu \mathrm{g} \ \mathrm{Cu} / \mathrm{g} \ \mathrm{FDT} \nonumber\] Atomic absorption spectroscopy is ideally suited for the analysis of trace and ultratrace analytes, particularly when using electrothermal atomization. For minor and major analytes, sample are diluted before the analysis. Most analyses use a macro or a meso sample. The small volume requirement for electrothermal atomization or for flame microsampling, however, makes practical the analysis of micro and ultramicro samples. If spectral and chemical interferences are minimized, an accuracy of 0.5–5% is routinely attainable. When the calibration curve is nonlinear, accuracy is improved by using a pair of standards whose absorbances closely bracket the sample’s absorbance and assuming that the change in absorbance is linear over this limited concentration range. Determinate errors for electrothermal atomization often are greater than those obtained with flame atomization due to more serious matrix interferences. For an absorbance greater than 0.1–0.2, the relative standard deviation for atomic absorption is 0.3–1% for flame atomization and 1–5% for electrothermal atomization. The principle limitation is the uncertainty in the concentration of free analyte atoms that result from variations in the rate of aspiration, nebulization, and atomization for a flame atomizer, and the consistency of injecting samples for electrothermal atomization. The sensitivity of a flame atomic absorption analysis is influenced by the flame’s composition and by the position in the flame from which we monitor the absorbance. Normally the sensitivity of an analysis is optimized by aspirating a standard solution of analyte and adjusting the fuel-to-oxidant ratio, the nebulizer flow rate, and the height of the burner, to give the greatest absorbance. With electrothermal atomization, sensitivity is influenced by the drying and ashing stages that precede atomization. The temperature and time at each stage is optimized for each type of sample. Sensitivity also is influenced by the sample’s matrix. We already noted, for example, that sensitivity is decreased by a chemical interference. An increase in sensitivity may be realized by adding a low molecular weight alcohol, ester, or ketone to the solution, or by using an organic solvent. Due to the narrow width of absorption lines, atomic absorption provides excellent selectivity. Atomic absorption is used for the analysis of over 60 elements at concentrations at or below the level of μg/L. The analysis time when using flame atomization is short, with sample throughputs of 250–350 determinations per hour when using a fully automated system. Electrothermal atomization requires substantially more time per analysis, with maximum sample throughputs of 20–30 determinations per hour. The cost of a new instrument ranges from between $10,000– $50,000 for flame atomization, and from $18,000–$70,000 for electrothermal atomization. The more expensive instruments in each price range include double-beam optics, automatic samplers, and can be programmed for multielemental analysis by allowing the wavelength and hollow cathode lamp to be changed automatically. | 30,102 | 3,510 |
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1. Because of the risk of lead poisoning, the exposure of children to lead-based paint is a significant public health concern. The first step in the quantitative analysis of lead in dried paint chips is to dissolve the sample. Corl evaluated several dissolution techniques [Corl, W. E. , , 40–43]. Samples of paint were collected and then pulverized using a Pyrex mortar and pestle. Replicate portions of the powdered paint were taken for analysis. The following table shows results for a paint sample and for a standard reference material. Both samples and standards were digested with HNO on a hot plate. Replicate % w/w Pb in Sample % w/w Pb in Standard 1 5.09 11.48 2 6.29 11.62 3 6.64 11.47 4 4.63 11.86 (a) Determine the overall variance, the variance due to the method and the variance due to sampling. (b) What percentage of the overall variance is due to sampling? (c) How might you decrease the variance due to sampling? 2. To analyze a shipment of 100 barrels of an organic solvent, you plan to collect a single sample from each of 10 barrels selected at random. From which barrels should you collect samples if the first barrel is given by the twelfth entry in the random number table in , with subsequent barrels given by every third entry? Assume that entries in the random number table are arranged by rows. 3. The concentration of dissolved O in a lake shows a daily cycle from the effect of photosynthesis, and a yearly cycle due to seasonal changes in temperature. Suggest an appropriate systematic sampling plan to monitor the daily change in dissolved O . Suggest an appropriate systematic sampling plan for monitoring the yearly change in dissolved O . 4. The data in the following table were collected during a preliminary study of the pH of an industrial wastewater stream. 0.5 1.0 Prepare a figure showing how the pH changes as a function of time and suggest an appropriate sampling frequency for a long-term monitoring program. 5. You have been asked to monitor the daily fluctuations in atmospheric ozone in the downtown area of a city to determine if there is relationship between daily traffic patterns and ozone levels. (a) Which of the following sampling plans will you use and why: random, systematic, judgmental, systematic–judgmental, or stratified? (b) Do you plan to collect and analyze a series of grab samples, or will you form a single composite sample? (c) Will your answers to these questions change if your goal is to determine if the average daily ozone level exceeds a threshold value? If yes, then what is your new sampling strategy? 6. The distinction between a homogeneous population and a heterogeneous population is important when we develop a sampling plan. (a) Define homogeneous and heterogeneous. (b) If you collect and analyze a single sample, can you determine if the population is homogeneous or is heterogeneous? 7. Beginning with , derive . Assume that the particles are spherical with a radius of and a density of . 8. The sampling constant for the radioisotope Na in homogenized human liver is approximately 35 g [Kratochvil, B.; Taylor, J. K. , , 924A–938A]. (a) What is the expected relative standard deviation for sampling if we analyze 1.0-g samples? (b) How many 1.0-g samples must we analyze to obtain a maximum sampling error of ±5% at the 95% confidence level? 9. Engels and Ingamells reported the following results for the % w/w K O in a mixture of amphibolite and orthoclase [Engels, J. C.; Ingamells, C. O. , , 1007–1017]. 0.247 0.300 0.247 0.275 0.258 0.311 0.258 0.330 Each of the 12 samples had a nominal mass of 0.1 g. Using this data, calculate the approximate value for , and then, using this value for , determine the nominal mass of sample needed to achieve a percent relative standard deviation of 2%. 10. The following data was reported for the determination of KH PO in a mixture of KH PO and NaCl [Guy, R. D.; Ramaley, L.; Wentzell, P. D. , , 1028–1033]. 0.2515 0.847 0.2465 0.598 0.2770 0.431 0.2460 0.842 0.2485 0.964 0.2590 1.178 0.5084 1.009 0.4954 0.947 0.5286 0.618 0.5232 0.744 0.4965 0.572 0.4995 0.709 1.027 0.987 0.991 0.998 0.997 (a) Prepare a graph of % w/w KH PO vs. the actual sample mass. Is this graph consistent with your understanding of the factors that affect sampling variance. (b) For each nominal mass, calculate the percent relative standard deviation, , based on the data. The value of for this analysis is estimated as 350. Use this value of to determine the theoretical percent relative standard deviation, , due to sampling. Considering these calculations, what is your conclusion about the importance of indeterminate sampling errors for this analysis? (c) For each nominal mass, convert to an absolute standard deviation. Plot points on your graph that correspond to ±1 absolute standard deviations about the overall average % w/w KH PO for all samples. Draw smooth curves through these two sets of points. Does the sample appear homogeneous on the scale at which it is sampled? 11.In this problem you will collect and analyze data to simulate the sampling process. Obtain a pack of M&M’s (or other similar candy). Collect a sample of five candies and count the number that are red (or any other color of your choice). Report the result of your analysis as % red. Return the candies to the bag, mix thoroughly, and repeat the analysis for a total of 20 determinations. Calculate the mean and the standard deviation for your data. Remove all candies from the bag and determine the true % red for the population. Sampling in this exercise should follow binomial statistics. Calculate the expected mean value and the expected standard deviation, and compare to your experimental results. 12. Determine the error (\(\alpha = 0.05\)) for the following situations. In each case assume that the variance for a single determination is 0.0025 and that the variance for collecting a single sample is 0.050. (a) Nine samples are collected, each analyzed once. (b) One sample is collected and analyzed nine times. (c) Five samples are collected, each analyzed twice. 13. Which of the sampling schemes in problem 12 is best if you wish to limit the overall error to less than ±0.30 and the cost to collect a single sample is $1 and the cost to analyze a single sample is $10? Which is the best sampling scheme if the cost to collect a single sample is $7 and the cost to analyze a single sample is $3? 14. Maw, Witry, and Emond evaluated a microwave digestion method for Hg against the standard open-vessel digestion method [Maw, R.; Witry, L.; Emond, T. , , 39–41]. The standard method requires a 2-hr digestion and is operator-intensive. The microwave digestion is complete in approximately 0.5 hr and requires little monitoring by the operator. Samples of baghouse dust from air-pollution-control equipment were collected from a hazardous waste incinerator and digested in triplicate before determining the concentration of Hg in ppm. Results are summarized in the following two tables. Does the microwave digestion method yields acceptable results when compared to the standard digestion method? 15. Simpson, Apte, and Batley investigated methods for preserving water samples collected from anoxic (O -poor) environments that have high concentrations of dissolved sulfide [Simpson, S. L.: Apte, S. C.; Batley, G. E. , , 4202–4205]. They found that preserving water samples with HNO (a common method for preserving aerobic samples) gave significant negative determinate errors when analyzing for Cu . Preserving samples by first adding H O and then adding HNO eliminated the determinate error. Explain their observations. 16. In a particular analysis the selectivity coefficient, , is 0.816. When a standard sample with an analyte-to-interferent ratio of 5:1 is carried through the analysis, the error when determining the analyte is +6.3%. (a) Determine the apparent recovery for the analyte if =0. (b) Determine the apparent recovery for the interferent if = 0. 17. The amount of Co in an ore is determined using a procedure for which Fe in an interferent. To evaluate the procedure’s accuracy, a standard sample of ore known to have a Co/Fe ratio of 10.2 is analyzed. When pure samples of Co and Fe are taken through the procedure the following calibration relationships are obtained \[S_{\mathrm{Co}}=0.786 \times m_{\mathrm{Co}} \text { and } S_{\mathrm{Fe}}=0.699 \times m_{\mathrm{Fe}} \nonumber\] where is the signal and is the mass of Co or Fe. When 278.3 mg of Co are taken through the separation step, 275.9 mg are recovered. Only 3.6 mg of Fe are recovered when a 184.9 mg sample of Fe is carried through the separation step. Calculate (a) the recoveries for Co and Fe; (b) the separation factor; (c) the selectivity ratio; (d) the error if no attempt is made to separate the Co and Fe; (e) the error if the separation step is carried out; and (f ) the maximum possible recovery for Fe if the recovery for Co is 1.00 and the maximum allowed error is 0.05%. 18. The amount of calcium in a sample of urine is determined by a method for which magnesium is an interferent. The selectivity coefficient, , for the method is 0.843. When a sample with a Mg/Ca ratio of 0.50 is carried through the procedure, an error of \(-3.7 \%\) is obtained. The error is +5.5% when using a sample with a Mg/Ca ratio of 2.0. (a) Determine the recoveries for Ca and Mg. (b) What is the expected error for a urine sample in which the Mg/Ca ratio is 10.0? 19. Using the formation constants in , show that F is an effective masking agent for preventing a reaction between Al and EDTA. Assume that the only significant forms of fluoride and EDTA are F and Y . 20. Cyanide is frequently used as a masking agent for metal ions. Its effectiveness as a masking agent is better in more basic solutions. Explain the reason for this dependence on pH. 21. Explain how we can separate an aqueous sample that contains Cu , Sn , Pb , and Zn into its component parts by adjusting the pH of the solution. 22. A solute, , has a distribution ratio between water and ether of 7.5. Calculate the extraction efficiency if we extract a 50.0-mL aqueous sample of using 50.0 mL of ether as (a) a single portion of 50.0 mL; (b) two portions, each of 25.0 mL; (c) four portions, each of 12.5 mL; and (d) five portions, each of 10.0 mL. Assume the solute is not involved in any secondary equilibria. 23. What volume of ether is needed to extract 99.9% of the solute in problem 23 when using (a) 1 extraction; (b) 2 extractions; (c) four extrac- tions; and (d) five extractions. 24. What is the minimum distribution ratio if 99% of the solute in a 50.0-mL sample is extracted using a single 50.0-mL portion of an organic solvent? Repeat for the case where two 25.0-mL portions of the organic solvent are used. 25. A weak acid, HA, with a of \(1.0 \times 10^{-5}\) has a partition coefficient, , of \(1.2 \times 10^3\) between water and an organic solvent. What restriction on the sample’s pH is necessary to ensure that 99.9% of the weak acid in a 50.0-mL sample is extracted using a single 50.0-mL portion of the organic solvent? 26. For problem 25, how many extractions are needed if the sample’s pH cannot be decreased below 7.0? 27. A weak base, B, with a of \(1.0 \times 10^{-3}\) has a partition coefficient, , of \(5.0 \times 10^2\) between water and an organic solvent. What restriction on the sample’s pH is necessary to ensure that 99.9% of the weak base in a 50.0-mL sample is extracted when using two 25.0-mL portions of the organic solvent? 28. A sample contains a weak acid analyte, HA, and a weak acid interferent, HB. The acid dissociation constants and the partition coefficients for the weak acids are = \(1.0 \times 10^{-3}\), = \(1.0 \times 10^{-7}\), = = \(5.0 \times 10^2\). (a) Calculate the extraction efficiency for HA and HB when a 50.0-mL sample, buffered to a pH of 7.0, is extracted using 50.0 mL of the organic solvent. (b) Which phase is enriched in the analyte? (c) What are the recoveries for the analyte and the interferent in this phase? (d) What is the separation factor? (e) A quantitative analysis is conducted on the phase enriched in analyte. What is the expected relative error if the selectivity coefficient, , is 0.500 and the initial ratio of HB/HA is 10.0? 29. The relevant equilibria for the extraction of I from an aqueous solution of KI into an organic phase are shown below. (a) Is the extraction efficiency for I better at higher or at a lower concentrations of I ? (b) Derive an expression for the distribution ratio for this extraction. 30. The relevant equilibria for the extraction of the metal-ligand complex ML from an aqueous solution into an organic phase are shown below. (a) Derive an expression for the distribution ratio for this extraction. (b) Calculate the extraction efficiency when a 50.0-mL aqueous sample that is 0.15 mM in M and 0.12 M in L is extracted using 25.0 mL of the organic phase. Assume that is 10.3 and that \(\beta_2\) is 560. 31. Derive for the extraction scheme outlined in . 32. The following information is available for the extraction of Cu by CCl and dithizone: = \(7 \times 10^4\); \(\beta_2 = 5 \times 10^{22}\); = \(3 \times 10^{-5}\); = \(1.1 \times 10^4\); and = 2. What is the extraction efficiency if a 100.0-mL sample of an aqueous solution that is \(1.0 \times 10^{-7}\) M Cu and 1 M in HCl is extracted using 10.0 mL of CCl containing \(4.0 \times 10^{-4}\) M dithizone (HL)? 33. Cupferron is a ligand whose strong affinity for metal ions makes it useful as a chelating agent in liquid–liquid extractions. The following table provides pH-dependent distribution ratios for the extraction of Hg , Pb , and Zn from an aqueous solution to an organic solvent. (a) Suppose you have a 50.0-mL sample of an aqueous solution that contains Hg , Pb , and Zn . Describe how you can separate these metal ions. (b) Under the conditions for your extraction of Hg , what percent of the Hg remains in the aqueous phase after three 50.0-mL extractions with the organic solvent? (c) Under the conditions for your extraction of Pb , what is the minimum volume of organic solvent needed to extract 99.5% of the Pb in a single extraction? (d) Under the conditions for your extraction of Zn , how many extractions are needed to remove 99.5% of the Zn if each extraction uses 25.0 mL of organic solvent? | 14,433 | 3,511 |
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1. The following data were obtained for four compounds separated on a 20-m capillary column. (a) Calculate the number of theoretical plates for each compound and the average number of theoretical plates for the column, in mm. (b) Calculate the average height of a theoretical plate. (c) Explain why it is possible for each compound to have a different number of theoretical plates. 2. Using the data from Problem 1, calculate the resolution and the selectivity factors for each pair of adjacent compounds. For resolution, use both equation 12.2.1 and equation 12.3.3, and compare your results. Discuss how you might improve the resolution between compounds B and C. The retention time for an nonretained solute is 1.19 min. 3. Use the chromatogram in Figure 12.8.1
, obtained using a 2-m column, to determine values for , , \(t_r^{\prime}\), , , and . 4. Use the partial chromatogram in Figure 12.8.2
to determine the resolution between the two solute bands. 5. The chromatogram in Problem 4 was obtained on a 2-m column with a column dead time of 50 s. Suppose you want to increase the resolution between the two components to 1.5. Without changing the height of a theoretical plate, what length column do you need? What height of a theoretical plate do you need to achieve a resolution of 1.5 without increasing the column’s length? 6. Complete the following table. 7. Moody studied the efficiency of a GC separation of 2-butanone on a dinonyl phthalate packed column [Moody, H. W. , , 218–219]. Evaluating plate height as a function of flow rate gave a van Deemter equation for which is 1.65 mm, is 25.8 mm•mL min , and is 0.0236 mm•min mL . (a) Prepare a graph of versus for flow rates between 5 –120 mL/min. (b) For what range of flow rates does each term in the Van Deemter equation have the greatest effect? (c) What is the optimum flow rate and the corresponding height of a theoretical plate? (d) For open-tubular columns the term no longer is needed. If the and terms remain unchanged, what is the optimum flow rate and the corresponding height of a theoretical plate? (e) Compared to the packed column, how many more theoretical plates are in the open-tubular column? 8. Hsieh and Jorgenson prepared 12–33 μm inner diameter HPLC columns packed with 5.44-μm spherical stationary phase particles [Hsieh, S.; Jorgenson, J. W. , , 1212–1217]. To evaluate these columns they measured reduced plate height, , as a function of reduced flow rate, , \[b=\frac{H}{d_{p}} \quad v=\frac{u d_{p}}{D_{m}} \nonumber\] where is the particle diameter and is the solute’s diffusion coefficient in the mobile phase. The data were analyzed using van Deemter plots. The following table contains a portion of their results for norepinephrine. (a) Construct separate van Deemter plots using the data in the first row and in the last row for reduced flow rates in the range 0.7–15. Determine the optimum flow rate and plate height for each case given = 5.44 μm and = \(6.23 \times 10^{-6}\) cm s . (b) The term in the van Deemter equation is strongly correlated with the column’s inner diameter, with smaller diameter columns providing smaller values of . Offer an explanation for this observation. When comparing columns, chromatographers often use dimensionless, reduced parameters. By including particle size and the solute’s diffusion coefficient, the reduced plate height and reduced flow rate correct for differences between the packing material, the solute, and the mobile phase. 9. A mixture of -heptane, tetrahydrofuran, 2-butanone, and -propanol elutes in this order when using a polar stationary phase such as Carbowax. The elution order is exactly the opposite when using a nonpolar stationary phase such as polydimethyl siloxane. Explain the order of elution in each case. 10. The analysis of trihalomethanes in drinking water is described in . A single standard that contains all four trihalomethanes gives the following results. Analysis of water collected from a drinking fountain gives areas of \(1.56 \times 10^4\), \(5.13 \times10^4\), \(1.49 \times 10^4\), and \(1.76 \times 10^4\) for, respectively, CHCl , CHCl Br, CHClBr , and CHBr . All peak areas were corrected for variations in injection volumes using an internal standard of 1,2-dibromopentane. Determine the concentration of each of the trihalomethanes in the sample of water. 11. Zhou and colleagues determined the %w/w H O in methanol by capillary column GC using a nonpolar stationary phase and a thermal conductivity detector [Zhou, X.; Hines, P. A.; White, K. C.; Borer, M. W. , , 390–394]. A series of calibration standards gave the following results. (a) What is the %w/w H O in a sample that has a peak height of 8.63? (b) The %w/w H O in a freeze-dried antibiotic is determined in the following manner. A 0.175-g sample is placed in a vial along with 4.489 g of methanol. Water in the vial extracts into the methanol. Analysis of the sample gave a peak height of 13.66. What is the %w/w H O in the antibiotic? 12. Loconto and co-workers describe a method for determining trace levels of water in soil [Loconto, P. R.; Pan, Y. L.; Voice, T. C. , , 128–132]. The method takes advantage of the reaction of water with calcium carbide, CaC , to produce acetylene gas, C H . By carrying out the reaction in a sealed vial, the amount of acetylene produced is determined by sampling the headspace. In a typical analysis a sample of soil is placed in a sealed vial with CaC . Analysis of the headspace gives a blank corrected signal of \(2.70 \times 10^5\). A second sample is prepared in the same manner except that a standard addition of 5.0 mg H O/g soil is added, giving a blank-corrected signal of \(1.06 \times 10^6\). Determine the milligrams H O/g soil in the soil sample. 13. Van Atta and Van Atta used gas chromatography to determine the %v/v methyl salicylate in rubbing alcohol [Van Atta, R. E.; Van Atta, R. L. , , 230–231]. A set of standard additions was prepared by transferring 20.00 mL of rubbing alcohol to separate 25-mL volumetric flasks and pipeting 0.00 mL, 0.20 mL, and 0.50 mL of methyl salicylate to the flasks. All three flasks were diluted to volume using isopropanol. Analysis of the three samples gave peak heights for methyl salicylate of 57.00 mm, 88.5 mm, and 132.5 mm, respectively. Determine the %v/v methyl salicylate in the rubbing alcohol. 14. The amount of camphor in an analgesic ointment is determined by GC using the method of internal standards [Pant, S. K.; Gupta, P. N.; Thomas, K. M.; Maitin, B. K.; Jain, C. L. , , 322–325]. A standard sample is prepared by placing 45.2 mg of camphor and 2.00 mL of a 6.00 mg/mL internal standard solution of terpene hydrate in a 25-mL volumetric flask and diluting to volume with CCl . When an approximately 2-μL sample of the standard is injected, the FID signals for the two components are measured (in arbitrary units) as 67.3 for camphor and 19.8 for terpene hydrate. A 53.6-mg sample of an analgesic ointment is prepared for analysis by placing it in a 50-mL Erlenmeyer flask along with 10 mL of CCl . After heating to 50 C in a water bath, the sample is cooled to below room temperature and filtered. The residue is washed with two 5-mL portions of CCl and the combined filtrates are collected in a 25-mL volumetric flask. After adding 2.00 mL of the internal standard solution, the contents of the flask are diluted to volume with CCl . Analysis of an approximately 2-μL sample gives FID signals of 13.5 for the terpene hydrate and 24.9 for the camphor. Report the %w/w camphor in the analgesic ointment. 15. The concentration of pesticide residues on agricultural products, such as oranges, is determined by GC-MS [Feigel, C. , Number 52]. Pesticide residues are extracted from the sample using methylene chloride and concentrated by evaporating the methylene chloride to a smaller volume. Calibration is accomplished using anthracene-d as an internal standard. In a study to determine the parts per billion heptachlor epoxide on oranges, a 50.0-g sample of orange rinds is chopped and extracted with 50.00 mL of methylene chloride. After removing any insoluble material by filtration, the methylene chloride is reduced in volume, spiked with a known amount of the internal standard and diluted to 10 mL in a volumetric flask. Analysis of the sample gives a peak–area ratio ( / ) of 0.108. A series of calibration standards, each containing the same amount of anthracene-d as the sample, gives the following results. Report the nanograms per gram of heptachlor epoxide residue on the oranges. 16. The adjusted retention times for octane, toluene, and nonane on a particular GC column are 15.98 min, 17.73 min, and 20.42 min, respectively. What is the retention index for each compound? 17. The following data were collected for a series of normal alkanes using a stationary phase of Carbowax 20M. What is the retention index for a compound whose adjusted retention time is 9.36 min? 18. The following data were reported for the gas chromatographic analysis of -xylene and methylisobutylketone (MIBK) on a capillary column [Marriott, P. J.; Carpenter, P. D. , , 96–99]. Explain the difference in the retention times, the peak areas, and the peak widths when switching from a split injection to a splitless injection. 19. Otto and Wegscheider report the following retention factors for the reversed-phase separation of 2-aminobenzoic acid on a C column when using 10% v/v methanol as a mobile phase [Otto, M.; Wegscheider, W. , , 11–22]. Explain the effect of pH on the retention factor for 2-aminobenzene. 20. Haddad and associates report the following retention factors for the reversed-phase separation of salicylamide and caffeine [Haddad, P.; Hutchins, S.; Tuffy, M. , , 166-168]. (a) Explain the trends in the retention factors for these compounds. (b) What is the advantage of using a mobile phase with a smaller %v/v methanol? Are there any disadvantages? 21. Suppose you need to separate a mixture of benzoic acid, aspartame, and caffeine in a diet soda. The following information is available. (a) Explain the change in each compound’s retention time. (b) Prepare a single graph that shows retention time versus pH for each compound. Using your plot, identify a pH level that will yield an acceptable separation. 22. The composition of a multivitamin tablet is determined using an HPLC with a diode array UV/Vis detector. A 5-μL standard sample that contains 170 ppm vitamin C, 130 ppm niacin, 120 ppm niacinamide, 150 ppm pyridoxine, 60 ppm thiamine, 15 ppm folic acid, and 10 ppm riboflavin is injected into the HPLC, giving signals (in arbitrary units) of, respectively, 0.22, 1.35, 0.90, 1.37, 0.82, 0.36, and 0.29. The multivitamin tablet is prepared for analysis by grinding into a powder and transferring to a 125-mL Erlenmeyer flask that contains 10 mL of 1% v/v NH in dimethyl sulfoxide. After sonicating in an ultrasonic bath for 2 min, 90 mL of 2% acetic acid is added and the mixture is stirred for 1 min and sonicated at 40 C for 5 min. The extract is then filtered through a 0.45-μm membrane filter. Injection of a 5-μL sample into the HPLC gives signals of 0.87 for vitamin C, 0.00 for niacin, 1.40 for niacinamide, 0.22 for pyridoxine, 0.19 for thiamine, 0.11 for folic acid, and 0.44 for riboflavin. Report the milligrams of each vitamin present in the tablet. 23. The amount of caffeine in an analgesic tablet was determined by HPLC using a normal calibration curve. Standard solutions of caffeine were prepared and analyzed using a 10-μL fixed-volume injection loop. Results for the standards are summarized in the following table. The sample is prepared by placing a single analgesic tablet in a small beaker and adding 10 mL of methanol. After allowing the sample to dissolve, the contents of the beaker, including the insoluble binder, are quantitatively transferred to a 25-mL volumetric flask and diluted to volume with methanol. The sample is then filtered, and a 1.00-mL aliquot transferred to a 10-mL volumetric flask and diluted to volume with methanol. When analyzed by HPLC, the signal for caffeine is found to be 21 469. Report the milligrams of caffeine in the analgesic tablet. 24. Kagel and Farwell report a reversed-phase HPLC method for determining the concentration of acetylsalicylic acid (ASA) and caffeine (CAF) in analgesic tablets using salicylic acid (SA) as an internal standard [Kagel, R. A.; Farwell, S. O. , , 163–166]. A series of standards was prepared by adding known amounts of ace- tylsalicylic acid and caffeine to 250-mL Erlenmeyer flasks and adding 100 mL of methanol. A 10.00-mL aliquot of a standard solution of salicylic acid was then added to each. The following results were obtained for a typical set of standard solutions. A sample of an analgesic tablet was placed in a 250-mL Erlenmeyer flask and dissolved in 100 mL of methanol. After adding a 10.00-mL portion of the internal standard, the solution was filtered. Analysis of the sample gave a peak height ratio of 23.2 for ASA and of 17.9 for CAF. (a) Determine the milligrams of ASA and CAF in the tablet. (b) Why is it necessary to filter the sample? (c) The directions indicate that approximately 100 mL of methanol is used to dissolve the standards and samples. Why is it not necessary to measure this volume more precisely? (d) In the presence of moisture, ASA decomposes to SA and acetic acid. What complication might this present for this analysis? How might you evaluate whether this is a problem? 25. Bohman and colleagues described a reversed-phase HPLC method for the quantitative analysis of vitamin A in food using the method of standard additions Bohman, O.; Engdahl, K. A.; Johnsson, H. , , 251–252]. In a typical example, a 10.067-g sample of cereal is placed in a 250-mL Erlenmeyer flask along with 1 g of sodium ascorbate, 40 mL of ethanol, and 10 mL of 50% w/v KOH. After refluxing for 30 min, 60 mL of ethanol is added and the solution cooled to room temperature. Vitamin A is extracted using three 100-mL portions of hexane. The combined portions of hexane are evaporated and the residue containing vitamin A transferred to a 5-mL volumetric flask and diluted to volume with methanol. A standard addition is prepared in a similar manner using a 10.093-g sample of the cereal and spiking with 0.0200 mg of vitamin A. Injecting the sample and standard addition into the HPLC gives peak areas of, respectively, \(6.77 \times10^3\) and \(1.32 \times 10^4\). Report the vitamin A content of the sample in milligrams/100 g cereal. 26. Ohta and Tanaka reported on an ion-exchange chromatographic method for the simultaneous analysis of several inorganic anions and the cations Mg and Ca in water [Ohta, K.; Tanaka, K. , , 189–195]. The mobile phase includes the ligand 1,2,4-benzenetricarboxylate, which absorbs strongly at 270 nm. Indirect detection of the analytes is possible because its absorbance decreases when complexed with an anion. (a) The procedure also calls for adding the ligand EDTA to the mobile phase. What role does the EDTA play in this analysis? (b) A standard solution of 1.0 mM NaHCO , 0.20 mM NaNO , 0.20 mM MgSO , 0.10 mM CaCl , and 0.10 mM Ca(NO ) gives the following peak areas (arbitrary units). Analysis of a river water sample (pH of 7.49) gives the following results. Determine the concentration of each ion in the sample. (c) The detection of \(\text{HCO}_3^-\) actually gives the total concentration of carbonate in solution ([\(\text{CO}_3^{2-}\)]+[\(\text{HCO}_3^-\)]+[H CO ]). Given that the pH of the water is 7.49, what is the actual concentration of \(\text{HCO}_3^-\)? (d) An independent analysis gives the following additional concentrations for ions in the sample: [Na ] = 0.60 mM; [\(\text{NH}_4^+\)] = 0.014 mM; and [K ] = 0.046 mM. A solution’s ion balance is defined as the ratio of the total cation charge to the total anion charge. Determine the charge balance for this sample of water and comment on whether the result is reasonable. 27. The concentrations of Cl , \(\text{NO}_2^-\), and \(\text{SO}_4^{2-}\) are determined by ion chromatography. A 50-μL standard sample of 10.0 ppm Cl , 2.00 ppm \(\text{NO}_2^-\), and 5.00 ppm \(\text{SO}_4^{2-}\) gave signals (in arbitrary units) of 59.3, 16.1, and 6.08 respectively. A sample of effluent from a wastewater treatment plant is diluted tenfold and a 50-μL portion gives signals of 44.2 for Cl , 2.73 for \(\text{NO}_2^-\), and 5.04 for \(\text{SO}_4^{2-}\). Report the parts per million for each anion in the effluent sample. 28. A series of polyvinylpyridine standards of different molecular weight was analyzed by size-exclusion chromatography, yielding the following results. When a preparation of polyvinylpyridine of unknown formula weight is analyzed, the retention volume is 8.45 mL. Report the average formula weight for the preparation. 29. Diet soft drinks contain appreciable quantities of aspartame, benzoic acid, and caffeine. What is the expected order of elution for these compounds in a capillary zone electrophoresis separation using a pH 9.4 buffer given that aspartame has p values of 2.964 and 7.37, benzoic acid has a p of 4.2, and the p for caffeine is less than 0. Figure 12.8.3
provides the structures of these compounds. 30. Janusa and coworkers describe the determination of chloride by CZE [Janusa, M. A.; Andermann, L. J.; Kliebert, N. M.; Nannie, M. H. , , 1463–1465]. Analysis of a series of external standards gives the following calibration curve. \[\text { area }=-883+5590 \times \mathrm{ppm} \text{ Cl}^{-} \nonumber\] A standard sample of 57.22% w/w Cl is analyzed by placing 0.1011-g portions in separate 100-mL volumetric flasks and diluting to volume. Three unknowns are prepared by pipeting 0.250 mL, 0.500 mL, an 0.750 mL of the bulk unknown in separate 50-mL volumetric flasks and diluting to volume. Analysis of the three unknowns gives areas of 15 310, 31 546, and 47 582, respectively. Evaluate the accuracy of this analysis. 31. The analysis of \(\text{NO}_3^-\) in aquarium water is carried out by CZE using \(\text{IO}_4^-\) as an internal standard. A standard solution of 15.0 ppm \(\text{NO}_3^-\) and 10.0 ppm \(\text{IO}_4^-\) gives peak heights (arbitrary units) of 95.0 and 100.1, respectively. A sample of water from an aquarium is diluted 1:100 and sufficient internal standard added to make its concentration 10.0 ppm in \(\text{IO}_4^-\). Analysis gives signals of 29.2 and 105.8 for \(\text{NO}_3^-\) and \(\text{IO}_4^-\), respectively. Report the ppm \(\text{NO}_3^-\) in the sample of aquarium water. 32. Suggest conditions to separate a mixture of 2-aminobenzoic acid (p = 2.08, p = 4.96), benzylamine (p = 9.35), and 4-methylphenol (p = 10.26) by capillary zone electrophoresis. Figure \(Page {4}\) provides the structures of these compounds. 33. McKillop and associates examined the electrophoretic separation of some alkylpyridines by CZE [McKillop, A. G.; Smith, R. M.; Rowe, R. C.; Wren, S. A. C. , , 497–503]. Separations were carried out using either 50-μm or 75-μm inner diameter capillaries, with a total length of 57 cm and a length of 50 cm from the point of injection to the detector. The run buffer was a pH 2.5 lithium phosphate buffer. Separations were achieved using an applied voltage of 15 kV. The electroosmotic mobility, μ as measured using a neutral marker, was found to be \(6.398 \times 10^{-5}\) cm V s . The diffusion coefficient for alkylpyridines is \(1.0 \times 10^{-5}\) cm s . (a) Calculate the electrophoretic mobility for 2-ethylpyridine given that its elution time is 8.20 min. (b) How many theoretical plates are there for 2-ethylpyridine? (c) The electrophoretic mobilities for 3-ethylpyridine and 4-ethylpyridine are \(3.366 \times 10^{-4}\) cm V s and \(3.397 \times 10^{-4} \text{ cm}^2 \text{ V}^{-1} \text{ s}^{-1}\), respectively. What is the expected resolution between these two alkylpyridines? (d) Explain the trends in electrophoretic mobility shown in the following table. (e) Explain the trends in electrophoretic mobility shown in the following table. (f) The p for pyridine is 5.229. At a pH of 2.5 the electrophoretic mobility of pyridine is \(4.176 \times 10^{-4}\) cm V s . What is the expected electrophoretic mobility if the run buffer’s pH is 7.5? | 20,351 | 3,512 |
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1. What are the three components of the scientific method? Is it necessary for an individual to conduct experiments to follow the scientific method? 2. Identify each statement as a theory or a law and explain your reasoning. a. The ratio of elements in a pure substance is constant. b. An object appears black because it absorbs all the visible light that strikes it. c. Energy is neither created nor destroyed. d. Metals conduct electricity because their electrons are not tightly bound to a particular nucleus and are therefore free to migrate. 3. Identify each statement as a theory or a law and explain your reasoning. a. A pure chemical substance contains the same proportion of elements by mass. b. The universe is expanding. c. Oppositely charged particles attract each other. d. Life exists on other planets. 4. Classify each statement as a qualitative observation or a quantitative observation. a. Mercury and bromine are the only elements that are liquids at room temperature. b. An element is both malleable and ductile. c. The density of iron is 7.87 g/cm3. d. Lead absorbs sound very effectively. e. A meteorite contains 20% nickel by mass. 5. Classify each statement as a quantitative observation or a qualitative observation. a. Nickel deficiency in rats is associated with retarded growth. b. Boron is a good conductor of electricity at high temperatures. c. There are 1.4–2.3 g of zinc in an average 70 kg adult. d. Certain osmium compounds found in air in concentrations as low as 10.7 µg/m3 can cause lung cancer. 3. a. law b. theory c. law d. theory 5. a. qualitative b. qualitative c. quantitative d. quantitative | 1,684 | 3,513 |
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Because energy takes many forms, only some of which can be seen or felt, it is defined by its effect on matter. For example, microwave ovens produce energy to cook food, but we cannot see that energy. In contrast, we can see the energy produced by a light bulb when we switch on a lamp. In this section, we describe the forms of energy and discuss the relationship between energy, heat, and work. The forms of energy include thermal energy, radiant energy, electrical energy, nuclear energy, and chemical energy (Figure \(\Page {1}\)). Thermal energy results from atomic and molecular motion; the faster the motion, the greater the thermal energy. The temperature of an object is a measure of its thermal energy content. Radiant energy is the energy carried by light, microwaves, and radio waves. Objects left in bright sunshine or exposed to microwaves become warm because much of the radiant energy they absorb is converted to thermal energy. Electrical energy results from the flow of electrically charged particles. When the ground and a cloud develop a separation of charge, for example, the resulting flow of electrons from one to the other produces lightning, a natural form of electrical energy. Nuclear energy is stored in the nucleus of an atom, and chemical energy is stored within a chemical compound because of a particular arrangement of atoms. Electrical energy, nuclear energy, and chemical energy are different forms of potential energy (PE), which is energy stored in an object because of the relative positions or orientations of its components. A brick lying on the windowsill of a 10th-floor office has a great deal of potential energy, but until its position changes by falling, the energy is contained. In contrast, kinetic energy (KE) is energy due to the motion of an object. When the brick falls, its potential energy is transformed to kinetic energy, which is then transferred to the object on the ground that it strikes. The electrostatic attraction between oppositely charged particles is a form of potential energy, which is converted to kinetic energy when the charged particles move toward each other. Energy can be converted from one form to another (Figure \(\Page {2}\)) or, as we saw with the brick, transferred from one object to another. For example, when you climb a ladder to a high diving board, your body uses chemical energy produced by the combustion of organic molecules. As you climb, the chemical energy is converted to to overcome the force of gravity. When you stand on the end of the diving board, your potential energy is greater than it was before you climbed the ladder: the greater the distance from the water, the greater the potential energy. When you then dive into the water, your potential energy is converted to kinetic energy as you fall, and when you hit the surface, some of that energy is transferred to the water, causing it to splash into the air. Chemical energy can also be converted to radiant energy; one common example is the light emitted by fireflies, which is produced from a chemical reaction. Although energy can be converted from one form to another, . This is known as the : One definition of energy is the capacity to do work. The easiest form of work to visualize is mechanical work (Figure \(\Page {3}\)), which is the energy required to move an object a distance d when opposed by a force F, such as gravity: \[w=F\,d \label{5.1.1} \] Because the force (F) that opposes the action is equal to the mass (m) of the object times its acceleration (a), we can also write Equation \(\ref{5.1.1}\) as follows: Recall from that weight is a force caused by the gravitational attraction between two masses, such as you and Earth. Consider the mechanical work required for you to travel from the first floor of a building to the second. Whether you take an elevator or an escalator, trudge upstairs, or leap up the stairs two at a time, energy is expended to overcome the force of gravity. The amount of work done ( ) and thus the energy required depends on three things: In contrast, heat ( ) is thermal energy that can be transferred from an object at one temperature to an object at another temperature. The net transfer of thermal energy stops when the two objects reach the same temperature. Energy is an property of matter—for example, the amount of in an object is proportional to both its mass and its temperature. A water heater that holds 150 L of water at 50°C contains much more thermal energy than does a 1 L pan of water at 50°C. Similarly, a bomb contains much more chemical energy than does a firecracker. We now present a more detailed description of kinetic and potential energy. The kinetic energy of an object is related to its mass \(m\) and velocity \(v\): \[KE=\dfrac{1}{2}mv^2 \label{5.1.4} \] For example, the kinetic energy of a 1360 kg (approximately 3000 lb) automobile traveling at a velocity of 26.8 m/s (approximately 60 mi/h) is \[KE=\dfrac{1}{2}(1360 kg)(26.8 ms)^2= 4.88 \times 10^5 g \cdot m^2 \label{5.1.5} \] Because all forms of energy can be interconverted, energy in any form can be expressed using the same units as kinetic energy. The SI unit of energy, the joule (J), is defined as 1 kilogram·meter /second (kg·m /s ). Because a joule is such a small quantity of energy, chemists usually express energy in kilojoules (1 kJ = 10 J). For example, the kinetic energy of the 1360 kg car traveling at 26.8 m/s is 4.88 × 10 J or 4.88 × 10 kJ. It is important to remember that , whether thermal, radiant, chemical, or any other form. Because heat and work result in changes in energy, their units must also be the same. To demonstrate, let’s calculate the potential energy of the same 1360 kg automobile if it were parked on the top level of a parking garage 36.6 m (120 ft) high. Its potential energy is equivalent to the amount of work required to raise the vehicle from street level to the top level of the parking garage, which is = . According to Equation \(\ref{5.1.2}\), the force ( ) exerted by gravity on any object is equal to its mass ( , in this case, 1360 kg) times the acceleration ( ) due to gravity ( , 9.81 m/s at Earth’s surface). The distance ( ) is the height ( ) above street level (in this case, 36.6 m). Thus the potential energy of the car is as follows: \[ PE= F\;d = m\,a\;d = m\,g\,h \label{5.1.6a} \] \[PE=(1360, Kg)\left(\dfrac{9.81\, m}{s^2}\right)(36.6\;m) = 4.88 \times 10^5\; \frac{Kg \cdot m}{s^2} \label{5.1.6b} \] \[=4.88 \times 10^5 J = 488\; kJ \label{5.1.6c} \] The units of potential energy are the same as the units of kinetic energy. Notice that in this case the potential energy of the stationary automobile at the top of a 36.6 m high parking garage is the same as its kinetic energy at 60 mi/h. If the vehicle fell from the roof of the parking garage, its potential energy would be converted to kinetic energy, and it is reasonable to infer that the vehicle would be traveling at 60 mi/h just before it hit the ground, neglecting air resistance. After the car hit the ground, its potential and kinetic energy would both be zero. The units of energy are the same for all forms of energy. Energy can also be expressed in the non-SI units of calories (cal), where 1 cal was originally defined as the amount of energy needed to raise the temperature of exactly 1 g of water from 14.5°C to 15.5°C. The name is derived from the Latin , meaning “heat.” Although energy may be expressed as either calories or joules, calories were defined in terms of heat, whereas joules were defined in terms of motion. Because calories and joules are both units of energy, however, the calorie is now defined in terms of the joule: \[1 \;cal = 4.184 \;J \;\text{exactly} \label{5.1.7a} \] \[1 \;J = 0.2390\; cal \label{5.1.7b} \] In this text, we will use the SI units—joules (J) and kilojoules (kJ)—exclusively, except when we deal with nutritional information. mass and velocity or height kinetic and potential energy Use Equation 5.1.4 to calculate the kinetic energy and Equation 5.1.6 to calculate the potential energy, as appropriate. The kinetic energy of the baseball is therefore \[
KE= 1492 \;\cancel{g} \left(\dfrac{1\; kg}{1000 \;\cancel{g}} \right) \left(\dfrac{44.7 \;m}{s} \right)^2= 1.49 \times 10^2 \dfrac{kg⋅m^2}{s^2}= 1.49 \times10^2\; J \nonumber \] 3.10 × 10 J 65 J All forms of energy can be interconverted. Three things can change the energy of an object: the transfer of heat, work performed on or by an object, or some combination of heat and work. is a branch of chemistry that qualitatively and quantitatively describes the energy changes that occur during chemical reactions. is the capacity to do work. is the amount of energy required to move an object a given distance when opposed by a force. is due to the random motions of atoms, molecules, or ions in a substance. The of an object is a measure of the amount of thermal energy it contains. is the transfer of thermal energy from a hotter object to a cooler one. Energy can take many forms; most are different varieties of , energy caused by the relative position or orientation of an object. is the energy an object possesses due to its motion. The most common units of energy are the , defined as 1 (kg·m )/s , and the , defined as the amount of energy needed to raise the temperature of 1 g of water by 1°C (1 cal = 4.184 J). | 9,402 | 3,514 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Metabolism/Important_High_Energy_Molecules_in_Metabolism |
The complicated processes of metabolism wouldn't be possible without the help of certain high-energy molecules. The main purpose of these molecules is to transfer either inorganic phosphate groups (Pi) or hydride (H ) ions. The inorganic phosphate groups are used to make high energy bonds with many of the intermediates of metabolism. These bonds can then be broken to yield energy, thus driving the metabolic processes of life. Hydride ions can be transferred from one intermediate to another resulting in a net oxidation or reduction of the intermediate. Oxidation corresponds to a loss of hydride and reduction to the gaining of hydride. Certain reduced forms of high energy molecules such as NADH and [FADH ] can donate their electrons to the electron carriers of the (ETC) which results in the production of ATP (only under aerobic conditions). ATP (Adenosine Triphosphate) contains high energy bonds located between each phosphate group. These bonds are known as phosphoric anhydride bonds. There are three reasons these bonds are high energy: ADP (Adenosine Diphosphate) also contains high energy bonds located between each phosphate group. It has the same structure as ATP, with one less phosphate group. The same three reasons that ATP bonds are high energy apply to ADP's bonds. NAD (Nicotinamide adenine dinucleotide (oxidized form)) is the major electron acceptor for catabolic reactions. It is strong enough to oxidize alcohol groups to carbonyl groups, while other electron acceptors (like [FAD]) are only able to oxidize saturated carbon chains from to . It is an important molecule in many metabolic processes like beta-oxidation, , and TCA cycle. With out NAD the aforementioned processes would be unable to occur. NADH (reduced form) is an NAD that has accepted electrons in the form of hydride ions. NADH is also one of the molecules responsible for donating electrons to the ETC to drive oxidative phosphorolation and also pyruvate during fermentation processes. NADP (Nicotinamide adenine dinucleotide phosphate (oxidized form)) is the major electron donator for anabolic reactions. Nicotinamide adenine dinucleotide phosphate (reduced form) | 2,193 | 3,515 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Quantifying_Nature/Units_of_Measure/SI_Units_-_A_Summary |
List all the basic quantities and their units you know of and search for those that you do not know yet. Understanding and proper expression of quantities are basic skills for any modern educated person. You have to master all quantities described here. Quantities form the basis for science and engineering and any moment of our lives. Unless you have expressed the quantities in numbers and units, you have not expressed anything. . Missing units and improper use of units are serious omissions and errors. Years ago, physicists used either the mks (meter-kilogram-second) system of units or the cgs (centimeter-gram-second) systems for , and . In addition to these three basic quantities are four others: the and the . Chemical quantities are mostly based on the last one. Thus, these are seven basic quantities, and each has an unit. The international system of units (Systeme International d'Units) was adopted by the General Conference on Weights and Measure in 1960, and the SI units are widely used today. All SI units are based on these basic units. Close your eyes, and see if you can name the 7 fundamental quantities in science and their (SI) Units. Science is based on only 7 basic quantities; for each, we have to define a standard unit. Think why these are the basic quantities. Are these related to any other quantities? Can they be derived from other quantities? There are other quantities aside from the seven basic quantities mentioned above. However, all other quantities are related to the basic quantities. Thus, their units can be derived from the seven SI units above. For this reason, other units are called The table below lists some examples: Derived units can be expressed in terms of basic quantities. From the specific derived unit, you can reason its relationship with the basic quantities. For some specific common quantities, the SI units have special symbols. As you use these often, you will feel at home with them. To remember it is very hard. However, you will encounter them during your study of these quantities. They are collected here to point out to you that these are special SI symbols. The following units are still in common use for chemistry. There are some other commonly used units too, but their meanings are clear by the time you use them. The following units are used in special technologies or disciplines. Since most people are not familiar with them, they are explained in more detail here. The unit erg is for energy, 1 J = 10,000,000 erg. Newton (N), he defined force One N is the gravitational pull of 98 g mass Pascal (Pa), who studied effect of pressure on fluid 1 atm = 101325 Pa = 101.3 kPa Joule (J) is an energy unit 1 J = 1 N m = 10e7 ergs Kelvin (K) 1C is the same as 273.15 K mole (mol), derived from Latin, meaning mass one mole has 6.023e23 atoms or molecules m, kg, s, A, K, cd, mol m, k, s, current, temperature, luminous, mole M stands for mol/L, a concentration unit 1 C/s, for an electric current 10 C/s V (J/s = watt) watt is the unit for power Becquerel (B), he discovered radioactivity 1 Ci = 3.7e10 B | 3,107 | 3,520 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Instrumentation_and_Analysis/Mass_Spectrometry/MALDI-TOF |
Proteins and peptides have been characterized by high pressure liquid chromatography (HPLC) or SDS PAGE by generating peptide maps. These peptide maps have been used as fingerprints of protein or as a tool to know the purity of a known protein in a known sample. Mass spectrometry gives a peptide map when proteins are digested with amino end specific, carboxy end specific, or amino acid specific digestive enzymes. This peptide map can be used to search a sequence database to find a good match from the existing database. This is because the more accurately the peptide masses are known, the less chance there is of bad matches. | 649 | 3,521 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkanes/Synthesis_of_Alkanes/00%3A_Front_Matter/02%3A_InfoPage |
Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being integrated. and are supported by the Department of Education Open Textbook Pilot 1525057, and 1413739. Unless otherwise noted, LibreTexts content is licensed by . Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not ). and are supported by the Department of Education Open Textbook Pilot Program, and Merlot. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. . . | 1,070 | 3,522 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Basics_of_General_Organic_and_Biological_Chemistry_(Ball_et_al.)/09%3A_Solutions/9.01%3A_Solutions |
A solution is another name for a homogeneous mixture. A as a material composed of two or more substances. In a solution, the combination is so intimate that the different substances cannot be differentiated by sight, even with a microscope. Compare, for example, a mixture of salt and pepper and another mixture consisting of salt and water. In the first mixture, we can readily see individual grains of salt and the flecks of pepper. A mixture of salt and pepper is not a solution. However, in the second mixture, no matter how carefully we look, we cannot see two different substances. Salt dissolved in water is a solution. The major component of a solution, called the , is typically the same phase as the solution itself. Each minor component of a solution (and there may be more than one) is called the . In most of the solutions we will describe in this textbook, there will be no ambiguity about whether a component is the solvent or the solute. For example, in a solution of salt in water, the solute is salt, and solvent is water. Solutions come in all phases, and the solvent and the solute do not have to be in the same phase to form a solution (such as salt and water). For example, air is a gaseous solution of about 80% nitrogen and about 20% oxygen, with some other gases present in much smaller amounts. An alloy is a solid solution consisting of a metal (like iron) with some other metals or nonmetals dissolved in it. Steel, an alloy of iron and carbon and small amounts of other metals, is an example of a solid solution. Table \(\Page {1}\) lists some common types of solutions, with examples of each. What causes a solution to form? The simple answer is that the solvent and the solute must have similar intermolecular interactions. When this is the case, the individual particles of solvent and solute can easily mix so intimately that each particle of solute is surrounded by particles of solute, forming a solution. However, if two substances have very different intermolecular interactions, large amounts of energy are required to force their individual particles to mix intimately, so a solution does not form. Thus two alkanes like -heptane, C H , and -hexane, C H , are completely miscible in all proportions. The C H and C H molecules are so similar (recall Section 4.6) that there are only negligible differences in intermolecular forces. For a similar reason, methanol, CH OH, is completely miscible with water. In this case both molecules are polar and can form hydrogen bonds among themselves, and so there are strong intermolecular attractions within each liquid. However, CH OH dipoles can align with H O dipoles, and CH OH molecules can hydrogen bond to H O molecules, and so the attractions among unlike molecules in the solution are similar to those among like molecules in each pure liquid. This process leads to a simple rule of thumb: . Solvents that are very polar will dissolve solutes that are very polar or even ionic. Solvents that are nonpolar will dissolve nonpolar solutes. Thus water, being polar, is a good solvent for ionic compounds and polar solutes like ethanol (C H OH). However, water does not dissolve nonpolar solutes, such as many oils and greases (Figure \(\Page {1}\)). We use the word soluble to describe a solute that dissolves in a particular solvent, and the word insoluble for a solute that does not dissolve in a solvent. Thus, we say that sodium chloride is soluble in water but insoluble in hexane (C H ). If the solute and the solvent are both liquids and soluble in any proportion, we use the word miscible, and the word immiscible if they are not. Water is considered a polar solvent. Which substances should dissolve in water? Because water is polar, substances that are polar or ionic will dissolve in it. Toluene (C H CH ) is widely used in industry as a nonpolar solvent. Which substances should dissolve in toluene? Octane only. Predict which of the following compounds will be most soluble in water: Since ethanol contains an group, it can hydrogen bond to water. Although the same is true of hexanol, the OH group is found only at one end of a fairly large molecule. The rest of the molecule can be expected to behave much as though it were a nonpolar alkane. This substance should thus be much less soluble than the first. Experimentally we find that ethanol is completely miscible with water, while only 0.6 g hexanol dissolves in 100 g water. Would I be more soluble in CCl or H O? I is nonpolar. Of the two solvents, CCl is nonpolar and H O is polar, so I would be expected to be more soluble in CCl . | 4,614 | 3,523 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/10%3A_Electrochemistry/10.01%3A_Electricity |
Electricity has been known for some time. Ancient Egyptians, for example, referred to electric fish in the Nile River as early as 2750 BC (Moller & Kramer, 1991). In 1600, William Gilbert studied what would later be seen to be electrostatic attraction, by creating static charges rubbing amber (Stewart, 2001). And Benjamin Franklin’s famous experiment (although it is actually uncertain if he performed the experiment) of attaching a metal key to a kite string occurred in 1752, and showed that lightening is an electrical phenomenon (Uman, 1987). One of the biggest breakthroughs in the study of electricity as a chemical phenomenon was made by Alessandro Volta, who in 1799 showed that electricity could be generated by stacking copper and zinc disks submerged in sulfuric acid (Routledge, 1881). The reactions that Volta produced in his included both and processes that could be considered as . The half-reactions can be classified as (the loss of electrons) which happens at the and (the gain of electrons) which occurs at the . Those half reactions were \[\underbrace{Zn \rightarrow Zn^{2+} + 2 e^-}_{\text{aanode}} \nonumber \] \[\underbrace{2 H^+ + 2 e^- \rightarrow H_2}_{\text{cathode}} \nonumber \] The propensity of zinc to oxidize coupled with that of hydrogen to reduce creates a potential energy difference between the electrodes at which these processes occur. And like any potential energy difference, it can create a force which can be used to do work. In this case, the work is that of pushing electrons through a circuit. The work of such a process can be calculated by integrating \[ dw_e - -E \,dQ \nonumber \] where \(E\) is the potential energy difference, and \(dQ\) is an infinitesimal amount of charge carried through the circuit. The infinitesimal amount of charge carried through the circuit can be expressed as \[dQ = e\,dN \nonumber \] where \(e\) is the charge carried on one electron (\(1.6 \times 10^{-19} C\)) and \(dN\) is the infinitesimal change in the number of electrons. Thus, if the potential energy difference is constant \[w_e = -e\,E \int_o^{N} dN = -N\,e\,E \nonumber \] But since the number of electrons carried through a circuit is an enormous number, it would be far more convenient to express this in terms of the number of moles of electrons carried through the circuit. Noting that the number of moles (\(n\)) is given by \[n=\dfrac{N}{N_A} \nonumber \] and that the charge carried by one mole of electrons is given by \[ F = N_A e = 96484\,C \nonumber \] where \(F\) is and has the magnitude of one Faraday (or the total charge carried by one mole of electrons.) The Faraday is named after Michael Faraday (1791-1867) (Doc, 2014), a British physicist who is credited with inventing the electric motor, among other accomplishments. Putting the pieces together, the total electrical work accomplished by pushing n moles of electrons through a circuit with a potential difference \(E\), is \[w_e = -nFE \nonumber \] | 2,993 | 3,524 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Medicinal_Chemistry/Antihistamines_and_Local_Anesthetics |
Histamine is 2-(4-imidazolyl)ethylamine and is a hydrophilic molecule comprised of an imadazole ring and an amino group connected by two methylene groups. It arises in vivo by decarboxylation of the amino acid histadine. Histamine is concentrated in mast cells, cells whose function is essentially to release histamine and immunoglobins when tissue damage occurs. They are especially numerous in parts of the body that are injured often, such as the fingers and toes, or which enjoy frequent contact with the environment, such as the mucosa of the lips, nose, etc. Histamine is also a neurotransmitter in the CNS and a typical problem with some antihistamines is drowsiness. The effort has been to produce compounds that do not enter the brain very well. There are many drugs with histamine-like properties, and most contain the following fragment: Histamine contracts many smooth muscles, such as those of the bronchi and gut, but powerfully relaxes others, including those of fine blood vessels. It is also a very potent stimulus to gastric acid production. Effects attributable to these actions dominate the overall response to the drug; however, there are several others, of which edema formation and stimulation of sensory nerve endings are perhaps the most familiar. Some of these effects as bronchoconstriction and contraction of the gut, are mediated by one type of histamine receptor, the H1 receptors, which are readily blocked by pyrilamine and other such classical antihistamines, now more properly described as histamine H1-receptor blocking drugs of simply H1 blockers. Other effects, most notably gastric secretion, are completely refractory to such antagonists, involve activation of H2 receptors, and are susceptible to inhibition by the more recently developed histamine H2-receptor blocking drugs. Still others, such as the hypotension resulting from vascular dilation, are mediated by receptors of both H1 and H2 types, since they are annulled only by a combination of H1 and H2 blockers. The two classes of histamine receptors also reveal themselves by differential responses to various histamine-like agonists. Thus, 2-methylhistamine preferentially elicits responses mediated by H1 receptors, whereas 4-methylhistamine has corresponingly preferential effect mediated through H2 receptors. These conpounds are representatives of two classes of histtamin-like dtugs, the H1-agonists and H2-agonists. All of the available antagonists are reversable, competative inhibitors of the actions of histamine. The structure of almost all of the "classic" antihistamines have a tertiary amino group linked by two- or three-atom chain to two aromatic substituens and confrom to the general formula shown below, where Ar is aryl and X is a nitrogen or a carbon aton or a C-O- ether linkage. Two common examples of H1 antagonists are shown below. H1-blocking drugs have an established and valued place in the symptomatic treatment of various immediate hypersensitivity reactions, in which their usefulness is attributable to their antagonism of endogenously released histamine, one of several autoacids that elicit allergic response. In addition, the central properties of some of the series are of considerable therapeutic value in suppressing motion sickness. In treating diseases of allergy, the effect of antihistamines is purely palliative and confined to the suppression in varying degree of symptoms attributable to the pharmacological activity of histamine released by the antigen-antibody reaction. The drugs do not diminish the intensity of this reaction, which is the root cause of the various hypersensitivity diseases. This limitation must be clearly recognized. In bronchial asthma, histamine blockers are singularly ineffectual. The have no role in the therapy of severe attacks in which chief reliance must be placed on epinephrine, isoproterenol, and theophylline. Equally, in the treat5ment of systemic anaphylaxis, in which autoacids other than histamine are again important, the mainstay of therapy is once more epinephrine, with histamine antagonists having only a subordinate and adjuvant role. Other allergies of the respiratory tract are more amenable to therapy with H1 blockers. The best results are obtained in seasonal rhinitis (hay fever) and conjunctivitis, in which these drugs relieve the sneezing, rhinorrhea, and itching of the eyes, nose and throat. The H2 blockers are reversible, competitive antagonists of the actions of histamine on H2 receptors. They are highly selective in their action and are highly selective in their action and are virtually without effect on H1 receptors. The most prominent of the effects of histamine that are mediated by H2 receptors is stimulation of gastric acid secretion, and it is the ability of the H2 blockers to inhibit this effect that explains much of their importance. Despite the widespread distribution of H2 receptors in the body, H2 blockers interfere remarkably little with physiological function other than gastric secretion, implying that extragastric H2 receptors are of minor physiological importance H2 blockers are used in treatment of peptic ulcer disease PUD; a disease in which ulceration occurs in the lower esophagus, stomach, duodenum, or jejunum. The most prominent symptom is gnawing pain that is relieved by food and alkali, but worsened by alcohol and condiments. The proximate cause of PUD is gastric acid hypersecretion. The synthesis of H2 antagonists was acheived by stepwise modifications of the histamine molecule, which resulted, some 200 compounds later, in the first highly effective drug with potent H2-blocking activity, burimamide. This, like later compounds, retained the imidazole ring of histamine byt possessed a much bulkier side chain. Cimetidine, the first H2 blocker to be introduced for general clinical use, won rapid acceptance for the treatment of ulcers and other gastric hypersecretory conditions and soon became one of the most widely prescribed of all drugs. this success led to the synthesis of numerous congeners. Some of the more popular drugs are shown below. The synthesis of rantadine is not difficult. The figure below shows the three steps of its synthesis from common starting chemicals. The parietal cells secrete acid by means of a membrane pump, identified as an H+, K+-ATPase, that exchanges hydrogen ions for potassium ions. By analogy with the familiar Na+, K+-ATPase, whose function can be inhibited by digitalis, this proton pump can likewise by inhibited by a newly discovered family. Omeprazole is the prototypical "acid pump" inhibitor which was allowed for clinical use in 1989. Its effects on gastric acid reduction are profound, showing a greater decrease of daily acid secretion than is obtained with four cimetidine given four times a day. It has been demonstrated in vitro that under acid conditions as high as 0.5M, HCl, omeprazole cyclized reversibly to a spiro-dyhydroimidazole intermediate (see figure below), which opens to sulfenic acid (not isolated). Cyclization, by the loss of H2O leads to a cyclic sulfenamide that was isolated and identified. Treatment of the sulfenamide with mercaptoehanol (HSCH2CH2OH) opened the ring to produce the predicted disylfide adduct shown. Since these conditions simulate the gastric environment and H+, K+ATPase was known to have an essential -SH group, it has been proposed that the sulfenamide produced from omeprazole is the chemical species that forms a covalent drug-enzyme complex with H+, K-ATPase in the acid compartment of the parietal cell, thereby blocking \(\ce{H^{+}}\) release. Cromolyn sodium, the disodium salt of 1,3-bis(2-carboxychromone-5-yloxy)-2-hydroxypropane, has the following structure: Cromolyn does not relax bronchial or other smooth muscle. Nor does it inhibit significantly responses to these muscles to any of a variety of pharmacological spasmogens. It does, however, inhibit the release of histamine and other autocoids (including leukotrienes) from human lung during allergic responses mediated by IgE antibodies and thereby the stimulus for bronchospasm. Inhibition of the liberation of leukotrienes is particularly important in allergic bronchial asthma, where these products appear to be the principal cause of bronchoconstriction. Cromolyn acts on the pulmonary for the immediate hypersensitivity reaction. Cromolyn does not inhibit the binding of IgE to mast cells nor the interaction between cell-bound IgE and specific antigen; rather, it suppresses the secretory response to this reaction. Cromolyn sodium is insufflated into the lings by a special device as a micronized powder. The drug is strictly prophylactic; it will not abort an asthmatic attack in progress. es of , a shrub grow The structures some of the typical anesthetics are shown below. These structures contain hydrophilic and hydrophobic domains that are separated by an intermediate alkyl chain. Linkage of these two domains is of either the ester or amide type. the ester link is important because this bond is readily hydrolyzed during metabolic degradation and inactivation in the body. Procaine, for example, can be divided into three main portions: the aromatic acid (para-aminobenzoic), the alcohol (ethanol), and the tertiary amino group (diethylamino). Changes in any part of the molecule alter the anesthetic potency and the toxicity of the compound. Increasing the length of the alcohol group leads to a greater anesthetic potency. It also leads to an increase in toxicity. Local anesthetics prevent the generation and the conduction of the nerve impulse. Their site of action is the cell membrane. Local anesthetic and other classes of agents (e.g., alcohols and barbiturates) block conduction by decreasing or preventing the large transient increase in the permeability of the membrane to sodium ions that is produced by a slight depolarization of the membrane. As anesthetic action progressively develops in a nerve, the threshold for electrical excitability gradually increases and the safety factor for conduction decreases; when this action is sufficiently well developed, block of conduction is produced. The local anesthetics also reduce the permeability of resting nerve to potassium as well as to sodium ions. Since changes in permeability to potassium require higher concentration of local anesthetic, blockade of conduction is not accompanied by any large or consistent change in the resting potential. All the commonly used local anesthetics contain a tertiary or secondary nitrogen atom and, therefore, can exist either as the uncharged tertiary of secondary amine or as the positively charged substituted ammonium cation, depending on the dissociation content of the compound and the pH of the solutions. The pKa of a typical local anesthetic lies between 8.0 and 9.0, so that only 5 to 20% will be protonated at the pH of the tissues. this fraction, although small, is important because the drug usually has to diffuse through connective tissue and other cellular membranes to reach its site of action, and it is generally agreed that it can do so only in the form of the uncharged amine. Once the anesthetic has reached the nerve, the form of the molecule active in nerve fibers is the cation which combines with some receptor in the membreane to prevent the generationof an action potential. | 11,354 | 3,525 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.06%3A_Thermochemistry |
When a chemical reaction occurs, there is usually a change in temperature of the chemicals themselves and of the beaker or flask in which the reaction is carried out. If the temperature increases, the reaction is —energy is given off as heat when the container and its contents cool back to room temperature. ( is energy transferred from one place to another solely because of a difference in temperature.) An reaction produces a decrease in temperature. In this case heat is absorbed from the surroundings to return the reaction products to room temperature. , a word derived from the Greek , “heat,” is the measurement and study of energy transferred as heat when chemical reactions take place. It is extremely important in a technological world where a great deal of work is accomplished by transforming and harnessing heat given off during combustion of coal, oil, and natural gas. If your workstation is authorized to view JCE Software, you will see a video below which shows an example of an endothermic reaction. Ammonium thiocyanate is mixed with barium hydroxide, and the reaction takes in enough heat to freeze water. Find others videos by searching YouTube for "endothermic reaction". | 1,213 | 3,526 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/17%3A_Thermochemistry/17.03%3A_Exothermic_and_Endothermic_Processes |
A campfire is an example of basic thermochemistry. The reaction is initiated by the application of heat from a match. The reaction converting wood to carbon dioxide and water (among other things) continues, releasing heat energy in the process. This heat energy can then be used to cook food, roast marshmallows, or simply to keep warm when it's cold outside. When physical or chemical changes occur, they are generally accompanied by a transfer of energy. The states that in any physical or chemical process, energy is neither created nor destroyed. In other words, the entire energy in the universe is conserved. In order to better understand the energy changes taking place during a reaction, we need to define two parts of the universe: and . The is the specific portion of matter in a given space that is being studied during an experiment or an observation. The is everything in the universe that is not part of the system. In practical terms for a laboratory chemist, the system is the particular chemicals being reacted, while the surroundings is the immediate vicinity within the room. During most processes, energy is exchanged between the system and the surroundings. If the system loses a certain amount of energy, that same amount of energy is gained by the surroundings. If the system gains a certain amount of energy, that energy is supplied by the surroundings. A chemical reaction or physical change is if heat is absorbed by the system from the surroundings. In the course of an endothermic process, the system gains heat from the surroundings, and so the temperature of the surroundings decreases. The quantity of heat for a process is represented by the letter \(q\). The sign of \(q\) for an endothermic process is positive because the system is gaining heat. A chemical reaction or physical change is if heat is released by the system into the surroundings. Because the surroundings are gaining heat from the system, the temperature of the surroundings increases. The sign of \(q\) for an exothermic process is negative because the system is losing heat. Heat flow is measured in one of two common units: the calorie and the joule. The joule \(\left( \text{J} \right)\) is the unit of energy. The calorie is familiar because it is commonly used when referring to the amount of energy contained within food. A is the quantity of heat required to raise the temperature of 1 gram of water by \(1^\text{o} \text{C}\). For example, raising the temperature of \(100 \: \text{g}\) of water from \(20^\text{o} \text{C}\) to \(22^\text{o} \text{C}\) would require \(100 \times 2 = 200 \: \text{cal}\). Calories contained within food are actually kilocalories \(\left( \text{kcal} \right)\). In other words, if a certain snack contains 85 food calories, it actually contains \(85 \: \text{kcal}\) or \(85,000 \: \text{cal}\). In order to make the distinction, the dietary calorie is written with a capital C. \[1 \: \text{kilocalorie} = 1 \: \text{Calorie} = 1000 \: \text{calories}\nonumber \] To say that the snack "contains" 85 Calories means that \(85 \: \text{kcal}\) of energy are released when that snack is processed by the human body. Heat changes in chemical reactions are typically measured in joules rather than calories. The conversion between a joule and a calorie is shown below. \[1 \: \text{J} = 0.2390 \: \text{cal or} \: 1 \: \text{cal} = 4.184 \: \text{J}\nonumber \] We can calculate the amount of heat released in kilojoules when a 400 Calorie hamburger is digested: \[400. \: \text{Cal} = 400. \: \text{kcal} \times \frac{4.184 \: \text{kJ}}{1 \: \text{kcal}} = 1.67 \times 10^3 \: \text{kJ}\nonumber \] | 3,675 | 3,527 |
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When two substances whose molecules are very similar form a liquid solution, the vapor pressure of the mixture is very simply related to the vapor pressures of the pure substances. Suppose, for example, we mix 1 mol benzene with 1 mol toluene as shown in the figure below. The mole fraction of benzene, , and the mole fraction of toluene, , are both equal to 0.5. At 79.6°C the measured vapor pressure of this mixture is 516 mmHg, slightly less than 517 mmHg, the average of the vapor pressures of pure benzene (744 mmHg) and of pure toluene (290 mmHg) at the same temperature. It is easy to explain this behavior if we assume that because benzene and toluene molecules are so nearly alike, they behave the same way in solution as they do in the pure liquids. Since there are only half as many benzene molecules in the mixture as in pure benzene, the rate at which benzene molecules escape from the surface of the solution will be half the rate at which they would escape from the pure liquid. In consequence the partial vapor pressure of benzene above the mixture will be one-half the vapor pressure of pure benzene. By a similar argument the partial vapor pressure of the toluene above the solution is also one-half that of pure toluene. Accordingly, we can write \[p_b =\frac{1}{2} P_b^*\] and \[P_t=\frac{1}{2} P_t^*\] where and are the partial pressures of benzene and toluene vapors, respectively, and and are the vapor pressures of the pure liquids. The total vapor pressure of the solution is \[P=p_{\text{b}}\text{ + }p_{\text{t}}=\frac{\text{1}}{\text{2}}P_{\text{b}}^{*}\text{ + }\frac{\text{1}}{\text{2}}P_{\text{t}}^{*}=\frac{P_{\text{b}}^{*}\text{ + }P_{\text{t}}^{*}}{\text{2}}\] The vapor pressure of the mixture is equal to the mean of the vapor pressures of the two pure liquids. We can generalize the above argument to apply to a liquid solution of any composition involving any two substances A and B whose molecules are very similar. The partial vapor pressure of A above the liquid mixture, , will then be the vapor pressure of pure A, , multiplied by the fraction of the molecules in the liquid which are of type A, that is, the mole fraction of A, . In equation form \[p_A=x_AP_A^* \label{3}\] Similarly for component B \[p_B=x_BP_B^* \label{4}\] Adding these two partial pressures, we obtain the total vapor pressure \[P=p_A + p_B = x_AP_A^* + x_BP_B^* \label{5}\] Liquid solutions which conform to Eqs. \(\ref{3}\) and \(\ref{5}\) are said to obey and to be or . In addition to its use in predicting the vapor pressure of a solution, Raoult’s law may be applied to the solubility of a gas in a liquid. Dividing both sides of Equation \(\ref{3}\) by gives \[x_{\text{A}}=\frac{\text{1}}{P_{\text{A}}^{*}}\text{ }\times \text{ }p_{\text{A}}=k_{\text{A}}\times \text{ }p_{\text{A}}\label{6}\] Since the vapor pressure of any substance has a specific value at a given temperature, Equation \(\ref{6}\) tells us that the mole fraction of a gaseous solute is proportional to the partial pressure of that gas above the solution. For an ideal solution the proportionality constant is the reciprocal of the vapor pressure of the pure solute at the temperature in question. Since vapor pressure increases as temperature increases, , which is 1/ , must decrease. Thus we expect the solubility of a gas in a liquid to increase as the partial pressure of gas above the solution increases, but to decrease as temperature increases. Equation \(\ref{6}\) is known as . It also applies to gaseous solutes which do not form ideal solutions, but in such cases the Henry’s-law constant does not equal the reciprocal of the vapor pressure. The video below shows the effect of varied pressure on the amount of CO dissolved in soda. The amount of dissolved CO is monitored by a pH indicator. The more dissolved CO , the lower the pH (the more red the solution). Watch the video to find out how the solubility of CO is related to the pressure, paying particular attention to the color of the solution. In actual fact very few liquid mixtures obey Raoult’s law exactly. Even for molecules as similar as benzene and toluene, we noted a deviation of 517 mmHg – 516 mmHg, or 1 mmHg at 79.6°C. Much larger deviations occur if the molecules are not very similar. These deviations are of two kinds. As can be seen from Figure \(\Page {2}\) , a plot of the vapor pressure against the mole fraction of one component yields a straight line for an ideal solution. For non-ideal mixtures the actual vapor pressure can be larger than the ideal value (positive deviation from Raoult’s law) or smaller (negative deviation). Negative deviations correspond to cases where attractions between unlike molecules are greater than those between like molecules. Because of this extra intermolecular attraction, molecules have more difficulty escaping the solution and the vapor pressure is lower. The opposite is true of a mixture of benzene and methanol. When C H molecules are randomly distributed among CH OH molecules, the latter cannot hydrogen bond effectively. Molecules can escape more readily from the solution, and the vapor pressure is higher than Raoult’s law would predict. Ed Vitz (Kutztown University), (University of | 5,272 | 3,529 |
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An important consequence of the kinetic molecular theory is what it predicts in terms of and effects. Effusion is defined as a loss of material across a boundary. A common example of effusion is the loss of gas inside of a balloon over time. The rate at which gases will effuse from a balloon is affected by a number of factors. But one of the most important is the frequency with which molecules collide with the interior surface of the balloon. Since this is a function of the average molecular speed, it has an inverse dependence on the square root of the molecular weight. \[\text{Rate of effusion} \propto \dfrac{1}{\sqrt{MW}} \nonumber \] This can be used to compare the relative rates of effusion for gases of different molar masses. A Knudsen cell is a chamber in which a thermalized sample of gas is kept, but allowed to effuse through a small orifice in the wall. The gas sample can be modeled using the Kinetic Molecular Theory model as a collection of particles traveling throughout the cell, colliding with one another and also with the wall. If a small orifice is present, any molecules that would collide with that portion of the wall will be lost through the orifice. This makes a convenient arrangement to measure the vapor pressure of the material inside the cell, as the total mass lost by effusion through the orifice will be proportional to the vapor pressure of the substance. The vapor pressure can be related to the mass lost by the expression \[ p = \dfrac{g}{A \Delta t} \sqrt{\dfrac{2 \pi RT}{MW}} \nonumber \] where \(g\) is the mass lost, \(A\) is the area of the orifice, \(\Delta t\) is the time the effusion is allowed to proceed, \(T\) is the temperature and \(MW\) is the molar mass of the compound in the vapor phase. The pressure is then given by \(p\). A schematic of what a Knudsen cell might look like is given below. | 1,870 | 3,531 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Supplemental_Modules_(Biological_Chemistry)/Carbohydrates/Carbohydrates_Fundamentals/Haworth_Formula |
Traditionally, in carbohydrate chemistry, the furanose rings and the pyranose rings in carbohydrate molecules are shown in the planar conformation, placed on the plane perpendicular to the plane of the paper. This representation of rings is known as the eg: cyclic forms of D-glucose To generate the Haworth formulas of the cyclic forms of a monosaccharide, use the following procedure, explained using the pyranoses of D-glucose. : Draw the Fischer projection of the acyclic form of D-glucose. (See D,L convention) : Number the carbon chain in 1 starting at the top. : To generate the pyranose ring, the oxygen atom on C-5 in 1 needs to be attached to C-1 by a single bond. In 1, C-1 is behind the plane of the paper and the hydroxy group on C-5 is in front. For the pyranose ring to be planar, both C-1 and the hydroxy group on C-5 have to be either behind or in front of the plane of the paper. C-5 is a chiral center. In order to bring the hydroxy group on C-5 to the site occupied by the \(CH_2OH\) group without changing the absolute configuration at C-5, rotate the three ligands H, OH, and CH2OH on C-5 in 1 clockwise without moving the fourth ligand. (See Fischer projection) 1 and 2 both represent D-glucose, but, in 2, unlike in 1, C-1 and the hydroxy group on C-5 are on the same side of the plane of the paper. : Ignore that 2 is a Fischer projection and rotate it clockwise by 90º. : Redraw the atom chain along the horizontal axis as follows. : Add the ligands on C-2 through C-5 in 4. The ligands pointing up in 3 are pointing up in 4; those pointing down in 3 are pointing down in 4. : Remove the hydrogen atom and the oxygen atom on C-1 and the hydrogen atom in the hydroxy group on C-5 in 5 and connect the two atoms by a single bond. : Add the two remaining bonds to C-1 in 6. Attach a hydrogen atom to the bond pointing up and a hydroxy group to the bond pointing down on C-1 in 7. Interchange the hydrogen atom and the hydroxy group on C-1 in 8. 8 and 9 are the Haworth formulas of the pyranoses of D-glucose. If, in the acyclic form of a monosaccharide, the hydroxy group that reacts with the carbonyl carbon is not on a chiral carbon (eg: D-fructose→pyranoses), skip step 3. | 2,226 | 3,532 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.03%3A_Collision_and_activation-_the_Arrhenius_law |
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the green-highlighted terms in the context of this topic. Why are some reactions so much faster than others, and why are reaction rates independent of the thermodynamic tendency of the reaction to take place? These are the central questions we address in this unit. In doing so, we open the door to the important topic of : what happens at the microscopic level when chemical reactions take place? We can thank Prof. for unlocking this door! To keep things as simple as possible, we will restrict ourselves to reactions that take place in the The same principles will apply to reactions in liquids and solids, but with added complications that we will discuss in a later unit. The of a chemical reaction is the sequence of actual events that take place as reactant molecules are converted into products. Each of these events constitutes an that can be represented as a coming-together of discrete particles ("collison") or as the breaking-up of a molecule ("dissociation") into simpler units. The molecular entity that emerges from each step may be a final product of the reaction, or it might be an — a species that is created in one elementary step and destroyed in a subsequent step, and therefore does not appear in the net reaction equation. A reaction mechanism must ultimately be understood as a "blow-by-blow" description of the molecular-level events whose sequence leads from reactants to products. These elementary steps (also called ) are almost always very simple ones involving one, two, or [rarely] three chemical species which are classified, respectively, as This fundamental rule must guide any analysis of an ordinary chemical reaction mechanism. This explains why termolecular processes are so uncommon. The kinetic theory of gases tells us that for every 1000 binary collisions, there will be only one event in which three molecules simultaneously come together. Four-way collisions are so improbable that this process has never been demonstrated in an elementary reaction. Consider a simple bimolecular step \[\ce{A + B -> products} \nonumber\] Clearly, if two molecules A and B are to react, they must approach closely enough to disrupt some of their existing bonds and to permit the creation of any new ones that are needed in the products. We call such an encounter a . The frequency of collisions between A and B in a gas will be proportional to the concentration of each; if we double [A], the frequency of A-B collisions will double, and doubling [B] will have the same effect. So if all collisions lead to products, than the rate of a bimolecular process will be first-order in A and B, or second-order overall: \[\text{rate} = k[\ce{A},\ce{B}] \nonumber\] However, not all collisions are equal. In a gas at room temperature and normal atmospheric pressure, there will be about 10 collisions in each cubic centimetre every second. If every collision between two reactant molecules yielded products, all reactions would be complete in a fraction of a second. When two billiard balls collide, they simply bounce off of each other. This is also the most likely outcome if the reaction between A and B requires a significant disruption or rearrangement of the bonds between their atoms. In order to effectively initiate a reaction, (kinetic energy) to bring about this bond disruption. More about this further on. And there is often one additional requirement. In many reactions, especially those involving more complex molecules, the reacting species must be oriented in a manner that is appropriate for the particular process. For example, in the gas-phase reaction of dinitrogen oxide with nitric oxide, the oxygen end of N O must hit the nitrogen end of NO; reversing the orientation of either molecule prevents the reaction. Owing to the extensive randomization of molecular motions in a gas or liquid, there are always enough correctly-oriented molecules for some of the molecules to react. But of course, the more critical this orientational requirement is, the fewer collisions will be effective. Energetic collisions between molecules cause interatomic bonds to stretch and bend farther, temporarily weakening them so that they become more susceptible to cleavage. Distortion of the bonds can expose their associated electron clouds to interactions with other reactants that might lead to the formation of new bonds. Chemical bonds have some of the properties of mechanical springs, whose potential energy depends on the extent to which they are stretched or compressed. Each atom-to-atom bond can be described by a potential energy diagram that shows how its energy changes with its length. When the bond absorbs energy (either from heating or through a collision), it is elevated to a higher quantized vibrational state (indicated by the horizontal lines) that weakens the bond as its length oscillates between the extended limits corresponding to the curve. A particular collision will typically excite a number of bonds in this way. Within about 10 second this excitation gets distributed among the other bonds in the molecule in rather complex and unpredictable ways that can concentrate the added energy at a particularly vulnerable point. The affected bond can stretch and bend farther, making it more susceptible to cleavage. Even if the bond does not break by pure stretching, it can become distorted or twisted so as to expose nearby electron clouds to interactions with other reactants that might encourage a reaction. Consider, for example, the isomerization of cyclopropane to propene which takes place at fairly high temperatures in the gas phase. We can imagine the collision-to-product sequence in the following [grossly oversimplified] way: Note that The cyclopropane isomerization described above is typical of many decomposition reactions that are found to follow first-order kinetics, implying that the process is unimolecular. Until about 1921, chemists did not understand the role of collisions in unimolecular processes. It turns out that the mechanisms of such reactions are really rather complicated, and that at very low pressures they do follow second-order kinetics. Such reactions are more properly described as . The chemical reactions associated with most food spoilage are catalyzed by enzymes produced by the bacteria which mediate these processes. It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Everyone knows that milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator, butter goes rancid more quickly in the summer than in the winter, and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be noticeably more lethargic on cold days. It is not hard to understand why this should be. Thermal energy relates direction to motion at the molecular level. As the temperature rises, molecules move faster and collide more vigorously, greatly increasing the likelyhood of bond cleavages and rearrangements as described above. Most reactions involving neutral molecules cannot take place at all until they have acquired the energy needed to stretch, bend, or otherwise distort one or more bonds. This critical energy is known as the of the reaction. of the kind shown below plot the total energy input to a reaction system as it proceeds from reactants to products. In examining such diagrams, take special note of the following: Activation energy diagrams can describe both exothermic and endothermic reactions: ... and the activation energies of the forward reaction can be large, small, or zero (independently, of course, of the value of Δ ): Processes with zero activation energy most commonly involve the combination of oppositely-charged ions or the pairing up of electrons in free radicals, as in the dimerization of nitric oxide (which is an odd-electron molecule). In this plot for the dissociation of bromine, the is just the enthalpy of atomization Br → 2 Br· and the reaction coordinate corresponds roughly to the stretching of the vibrationally-excited bond. The "activated complex", if it is considered to exist, is just the last, longest "stretch". The reverse reaction, being the recombination of two radicals, occurs immediately on contact. In most cases, the activation energy is supplied by thermal energy, either through intermolecular collisions or (in the case of thermal dissocation) by thermal excitation of a bond-stretching vibration to a sufficiently high quantum level. As products are formed, the activation energy is returned in the form of vibrational energy which is quickly degraded to heat. It's worth noting, however, that other sources of activation energy are sometimes applicable: A is usually defined as a substance that speeds up a reaction without being consumed by it. More specifically, a catalyst provides an alternative, lower activation energy pathway between reactants and products. As such, they are vitally important to chemical technology; approximately 95% of industrial chemical processes involve catalysts of various kind. In addition, most biochemical processes that occur in living organisms are mediated by , which are catalysts made of proteins. It is important to understand that a catalyst affects only the of a reaction; it does alter the thermodynamic tendency for the reaction to occur. Thus there is a single value of Δ for the two pathways depicted in the plot on the right In the vast majority of cases, we depend on thermal actvation, so the major factor we need to consider is what fraction of the molecules possess enough kinetic energy to react at a given temperature. According to kinetic molecular theory, a opulation of molecules at a given temperature is distributed over a variety of kinetic energies that is described by the Maxwell-Boltzman distribution law. The two distribution plots shown here are for a lower temperature and a higher temperature . The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of that are shown. It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) are more rapid at higher temperatures. By 1890 it was common knowledge that higher temperatures speed up reactions, often doubling the rate for a 10-degree rise, but the reasons for this were not clear. Finally, in 1899, the Swedish chemist Svante Arrhenius (1859-1927) combined the concepts of activation energy and the Boltzmann disribution law into one of the most important relationships in physical chemistry: Take a moment to focus on the meaning of this equation, neglecting the factor for the time being. First, note that this is another form of the exponential decay law we discussed in the previous section of this series. What is "decaying" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent . And what is the significance of this quantity? If you recall that is the , it will be apparent that the exponent is just the ratio of the activation energy to the average kinetic energy. The larger this ratio, the smaller the rate (hence the negative sign.) This means that high temperature and low activation energy favor larger rate constants, and thus speed up the reaction. And because these terms occur in an exponent, their effects on the rate are quite substantial. The two plots below show the effects of the activation energy (denoted here by ) on the rate constant. Even a modest activation energy of 50 kJ/mol reduces the rate by a factor of 10 . The logarithmic scale in the right-hand plot leads to nice straight lines, as described under the next heading below. Looking at the role of temperature, we see a similar effect. (If the -axis were in "kilodegrees" the slopes would be more comparable in magnitude with those of the kilojoule plot at the above right.) The Arrhenius equation \[k=A \mathrm{e}^{-E_{a} / R T}\] can be written in a non-exponential form which is often more convenient to use and to interpret graphically. Taking the logarithms of both sides and separating the exponential and pre-exponential terms yields \[\ln k=\ln \left(A \mathrm{e}^{-E_{\mathrm{a}} / R T}\right)=\ln A+\ln \left(\mathrm{e}^{-E_{\mathrm{a}} / R T}\right)\] \[\ln k=\ln A-\dfrac{E_{a}}{R T}\] which is the equation of a straight line whose slope is \(–E_a /R\). This affords a simple way of determining the activation energy from values of \(k\) observed at different temperatures; we just plot \(\ln k\) as a function of \(1/T\). Thus for the isomerization of cyclopropane to propene the following data were obtained (calculated values shaded in pink): From the calculated slope, we have – ( / ) = –3.27 × 10 K =– (8.314 J mol K ) (–3.27 × 10 K) = 273 kJ mol This activation energy is rather high, which is not surprising because a carbon-carbon bond must be broken in order to open the cyclopropane ring. (C–C bond energies are typically around 350 kJ/mol.) This is why the reaction must be carried out at high temperature. (... if you are willing to live a bit dangerously!) Since the ln -vs.-1/ plot yields a straight line, it is often convenient to estimate the activation energy from experiments at only two temperatures. To see how this is done, consider that \[\ln k_{2}-\ln k_{1}=\left(\ln A-\frac{E_{a}}{R T_{2}}\right)-\left(\ln A-\frac{E_{a}}{R T_{1}}\right)=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) \nonumber\] (... in which we have made the ln- term disappear by subtracting the expressions for the two ln- terms.) Solving the expression on the right for the activation energy yields \[E_{a}=\dfrac{R \ln \dfrac{k_{2}}{k_{1}}}{\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}}} \nonumber\] A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a ten-C° rise in the temperature approximately doubles the rate. (This is obviously not generally true, especially when a strong covalent bond must be broken.) But for a reaction that does show this behavior, what would the activation energy be? We will center our ten-degree interval at 300 K. Substituting into the above expression yields = (8.314)(0.693) / (.00339 - 0.00328) = (5.76 J K ) / (0.00011 K ) = 52400 J = It takes about 3.0 minutes to cook a hard-boiled egg in Los Angeles, but at the higher altitude of Denver, where water boils at 92°C, the cooking time is 4.5 minutes. Use this information to estimate the activation energy for the coagulation of egg albumin protein. The ratio of the rate constants at the elevations of LA and Denver is 4.5/3.0 = 1.5, and the respective temperatures are and . With the subscripts 2 and 1 referring to LA and Denver respectively, we have = (8.314)(ln 1.5) / (1/365 – 1/273) = (8.314)(.405) / (0.00274 – 0.00366) = (3.37 J K ) / (0.000923 K ) = 3650 J = : This rather low value seems reasonable because protein denaturation involves the disruption of relatively weak hydrogen bonds; no covalent bonds are broken It is now time to focus in on the pre-exponential term in the Arrhenius equation. We have been neglecting it because it is not directly involved in relating temperature and activation energy, which is the main practical use of the equation. But since multiplies the exponential term, its value clearly contributes to the value of the rate constant and thus of the rate. and of the temperature. If this fraction were unity, the Arrhenius law would reduce to \[k = A\] In other words, is the fraction of molecules that would react if either the activation energy were zero, or if the kinetic energy of all molecules exceeded — admittedly, an uncommon scenario. So what would limit the rate constant if there were no activation energy requirements? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known as the or . In some reactions, the relative orientation of the molecules at the point of collision is important, so we can also define a geometrical or (commonly denoted by ρ (Greek lower case ). In general, we can express as the product of these two factors: \[A = Zρ\] Values of \(ρ\) are generally very difficult to assess; they are sometime estimated by comparing the observed rate constant with the one in which is assumed to be the same as . The more complicated the structures of the reactants, the more likely that the value of the rate constant will depend on the trajectories at which the reactants approach each other. We showed one example of this near the top of the page, but for another, consider the addition of a hydrogen halide such as HCl to the double bond of an alkene, converting it to a chloroalkane. This kind of is well-known to all students of organic chemistry. Experiments have shown that the reaction only takes place when the HCl molecule approaches the alkene with its hydrogen-end, and in a direction that is approximately perpendicular to the double bond, as shown at below. The reason for this becomes apparent when we recall that HCl is highly polar owing to the high electronegativity of chlorine, so that the hydrogen end of the molecule is slightly positive. The double bond of ethene consists of two clouds of negative charge corresponding to the σ ( ) and π ( ) molecular orbitals. The latter, which extends above and below the plane of the C H molecule, interacts with and attracts the HCl molecule. If, instead, the HCl approaches with its chlorine end leading as in , electrostatic repulsion between the like charges causes the two molecules to bounce away from each other before any reaction can take place. The same thing happens in ; the electronegativity difference between carbon and hydrogen is too small to make the C–H bond sufficiently polar to attract the incoming chlorine atom. The lesson you should take from this example is that once you start combining a variety of chemical principles, you gradually develop what might be called "chemical intuition" which you can apply to a wide variety of problems. This is far more important than memorizing specific examples. Now that you know what it takes to get a reaction started, you are ready for the next lesson that describes their actual . | 18,849 | 3,533 |
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The size of an ion is governed not only by its electronic structure but also by its . This relationship is evident in the following figure comparing ionic radii. Ions in the first row of this figure, H , Li , and Be , all have the same 1 electronic structure as the helium (He) atom, but differ in size due to the different number of protons each has in their nucleus. Species which have the same electronic structure but different charges are said to be . For any electronic series, such as H , He, Li , Be , in which the nuclear charge increases by 1 each time, we find a progressive decrease in size due to the increasingly strong attraction of the nucleus for the electron cloud. Each row in the figure corresponds to an isoelectronic series involving a different noble-gas electron configuration. As we move from the more negative to the more positive ions in each row, there is a steady decrease in size. If we move any of the columns, ionic sizes increase due to the increasing principal quantum number of the outermost electrons. The sizes of singly charged cations, for example, increase in the following order: Li < Na < K < Rb < Cs . A further point of interest is the size of an ion relative to the atom from which it was formed. Recall the electron dot density diagram showing the formation of LiH. when the Li atom lost its electron and became a Li ion, its size decreased dramatically. Comparing atomic radii to the ionic radii above reveals that this is also true for the other alkali metals. For example, check out the image below, which compares the atomic radii of the uncharged element to its ion. On the left are cations, which have lost an electron to become an ion and show decreases in size. On the right side of the figure are the anions, which form when an element gains an electron. The added electron, as can be observed from the pattern above, increases the atomic radii. Since the added electron goes into a subshell that already has occupants, rather than starting on a new subshell, there is often very little change in size. This is clearly seen in the electron dot density diagram mentioned above, where the formation of an H ion from an H atom produces no perceptible increase in size. Comparing the figure of atomic radii to ionic radii above, we also find that the van der Waals radii of nonmetals are only slightly smaller than the radii of their anions. However, some ions still experience fairly significant size changes, as can be seen in the figure above (anions are on the right side). The sizes of the ions involved have a considerable influence on both the chemical and physical properties of ionic compounds. There is a strong correlation, for example, between ionic size and the melting point of an ionic compound. Among the halides of sodium the melting point decreases in the order of NaF (995°C) > NaCl (808°C) > NaBr (750°C) > NaI (662°C). The larger the anion, the farther it is from the sodium ion, and the weaker the coulombic force of attraction between them. Hence the lower the melting point. When a very small cation combines with a very large anion, the resulting compound is less likely to exhibit the characteristic macroscopic properties of an ionic substance. | 3,247 | 3,534 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.21%3A_The_Separation_of_Mixtures |
Now let us imagine that you want to collect all the red marbles again. If you simply shake the can, it is unlikely that you will ever divide the marbles into two layers, each with only one kind of marble. Similarly, if two miscible liquids are combined, a chemist cannot simply un-mix the liquids into pure components. Continuing the analogy, what if a few green marbles and blue marbles are placed into the can? Given enough red and white marbles, it may be difficult to determine that the green marbles and blue marbles are actually there. Similarly, when chemists have a multi-component solution which may contain traces of important chemical species, they are faced with the challenge of detecting whether these chemicals are present in solution. To deal with these difficulties, chemists employ different methods to separate solutions into their components. Two essential techniques are and . | 912 | 3,535 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_General_Chemistry_(Petrucci_et_al.)/07%3A_Thermochemistry/7.5%3A_The_First_Law_of_Thermodynamics |
To study the flow of energy during a chemical reaction, we need to distinguish between a system, the small, well-defined part of the universe in which we are interested (such as a chemical reaction), and its surroundings, the rest of the universe, including the container in which the reaction is carried out (Figure \(\Page {1}\)). In the discussion that follows, the mixture of chemical substances that undergoes a reaction is always the system, and the flow of heat can be from the system to the surroundings or vice versa. Three kinds of systems are important in chemistry. An open system can exchange both matter and energy with its surroundings. A pot of boiling water is an open system because a burner supplies energy in the form of heat, and matter in the form of water vapor is lost as the water boils. A closed system can exchange energy but not matter with its surroundings. The sealed pouch of a ready-made dinner that is dropped into a pot of boiling water is a closed system because thermal energy is transferred to the system from the boiling water but no matter is exchanged (unless the pouch leaks, in which case it is no longer a closed system). An isolated system exchanges neither energy nor matter with the surroundings. Energy is always exchanged between a system and its surroundings, although this process may take place very slowly. A truly isolated system does not actually exist. An insulated thermos containing hot coffee approximates an isolated system, but eventually the coffee cools as heat is transferred to the surroundings. In all cases, the amount of heat lost by a system is equal to the amount of heat gained by its surroundings and vice versa. That is, , which must be true if . The state of a system is a complete description of a system at a given time, including its temperature and pressure, the amount of matter it contains, its chemical composition, and the physical state of the matter. A state function is a property of a system whose magnitude depends on only the present state of the system, not its previous history. Temperature, pressure, volume, and potential energy are all state functions. The temperature of an oven, for example, is independent of however many steps it may have taken for it to reach that temperature. Similarly, the pressure in a tire is independent of how often air is pumped into the tire for it to reach that pressure, as is the final volume of air in the tire. Heat and work, on the other hand, are not state functions because they are . For example, a car sitting on the top level of a parking garage has the same potential energy whether it was lifted by a crane, set there by a helicopter, driven up, or pushed up by a group of students (Figure \(\Page {2}\)). The amount of work expended to get it there, however, can differ greatly depending on the path chosen. If the students decided to carry the car to the top of the ramp, they would perform a great deal more work than if they simply pushed the car up the ramp (unless, of course, they neglected to release the parking brake, in which case the work expended would increase substantially!). The potential energy of the car is the same, however, no matter which path they choose. The reaction of powdered aluminum with iron(III) oxide, known as the thermite reaction, generates an enormous amount of heat—enough, in fact, to melt steel (see chapter opening image). The balanced chemical equation for the reaction is as follows: \[ 2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s) \tag{7.5.1}\] We can also write this chemical equation as \[2Al(s) + Fe_2O_3(s) \rightarrow 2Fe(s) + Al_2O_3(s) + \text{heat} \tag{7.5.2}\] to indicate that heat is one of the products. Chemical equations in which heat is shown as either a reactant or a product are called . In this reaction, the system consists of aluminum, iron, and oxygen atoms; everything else, including the container, makes up the surroundings. During the reaction, so much heat is produced that the iron liquefies. Eventually, the system cools; the iron solidifies as heat is transferred to the surroundings.A process in which heat ( ) is transferred a system its surroundings is described as exothermic. By convention, < 0\) for an exothermic reaction. When you hold an ice cube in your hand, heat from the surroundings (including your hand) is transferred to the system (the ice), causing the ice to melt and your hand to become cold. We can describe this process by the following thermochemical equation: \[ heat + H_2O_{(s)} \rightarrow H_2O_{(l)} \tag{7.5.3}\] When heat is transferred a system its surroundings, the process is endothermic. By convention, \(q > 0\) for an endothermic reaction. Technically, it is poor form to have a \(heat\) term in the chemical reaction like in Equations 7.5.2 and 7.5.3 since is it not a true species in the reaction. However, this is a convenient approach to represent exothermic and endothermic behavior and is commonly used by chemists. The relationship between the energy change of a system and that of its surroundings is given by the first law of thermodynamics, which states that the energy of the universe is constant. We can express this law mathematically as follows: \[U_{univ}=ΔU_{sys}+ΔU_{surr}=0 \tag{7.5.4a}\] where the subscripts , , and refer to the universe, the system, and the surroundings, respectively. Thus the change in energy of a system is identical in magnitude but opposite in sign to the change in energy of its surroundings. The tendency of all systems, chemical or otherwise, is to move toward the state with the lowest possible energy. An important factor that determines the outcome of a chemical reaction is the tendency of all systems, chemical or otherwise, to move toward the lowest possible overall energy state. As a brick dropped from a rooftop falls, its potential energy is converted to kinetic energy; when it reaches ground level, it has achieved a state of lower potential energy. Anyone nearby will notice that energy is transferred to the surroundings as the noise of the impact reverberates and the dust rises when the brick hits the ground. Similarly, if a spark ignites a mixture of isooctane and oxygen in an internal combustion engine, carbon dioxide and water form spontaneously, while potential energy (in the form of the relative positions of atoms in the molecules) is released to the surroundings as heat and work. The internal energy content of the \(CO_2/H_2O\) product mixture is less than that of the isooctane \(O_2\) reactant mixture. The two cases differ, however, in the form in which the energy is released to the surroundings. In the case of the falling brick, the energy is transferred as work done on whatever happens to be in the path of the brick; in the case of burning isooctane, the energy can be released as solely heat (if the reaction is carried out in an open container) or as a mixture of heat and work (if the reaction is carried out in the cylinder of an internal combustion engine). Because heat and work are the only two ways in which energy can be transferred between a system and its surroundings, any change in the internal energy of the system is the sum of the heat transferred (q) and the work done (w): \[ΔU_{sys} = q + w \tag{7.5.5}\] Although \(q\) and \(w\) are not state functions on their own, their sum (\(ΔU_{sys}\)) is independent of the path taken and is therefore a state function. A major task for the designers of any machine that converts energy to work is to maximize the amount of work obtained and minimize the amount of energy released to the environment as heat. An example is the combustion of coal to produce electricity. Although the maximum amount of energy available from the process is fixed by the energy content of the reactants and the products, the fraction of that energy that can be used to perform useful work is not fixed. Because we focus almost exclusively on the changes in the energy of a system, we will not use “sys” as a subscript unless we need to distinguish explicitly between a system and its surroundings. Although \(q\) and \(w\) are not state functions, their sum (\(ΔU_{sys}\)) is independent of the path taken and therefore is a state function. A sample of an ideal gas in the cylinder of an engine is compressed from 400 mL to 50.0 mL during the compression stroke against a constant pressure of 8.00 atm. At the same time, 140 J of energy is transferred from the gas to the surroundings as heat. What is the total change in the internal energy (ΔU) of the gas in joules? : initial volume, final volume, external pressure, and quantity of energy transferred as heat Asked for: total change in internal energy A From Equation 7.5.5, we know that ΔU = q + w. We are given the magnitude of q (140 J) and need only determine its sign. Because energy is transferred from the system (the gas) to the surroundings, q is negative by convention. B Because the gas is being compressed, we know that work is being done on the system, so \(w\) must be positive. From Equation 7.5.5, Thus ΔU = q + w = −140 J + 284 J = 144 J In this case, although work is done on the gas, increasing its internal energy, heat flows from the system to the surroundings, decreasing its internal energy by 144 J. The work done and the heat transferred can have opposite signs. A sample of an ideal gas is allowed to expand from an initial volume of 0.200 L to a final volume of 3.50 L against a constant external pressure of 0.995 atm. At the same time, 117 J of heat is transferred from the surroundings to the gas. What is the total change in the internal energy (ΔU) of the gas in joules? : −216 J By convention, both heat flow and work have a negative sign when energy is transferred from a system to its surroundings and vice versa. The first law of thermodynamics states that the energy of the universe is constant. The change in the internal energy of a system is the sum of the heat transferred and the work done. At constant pressure, heat flow (q) and internal energy (U) are related to the system’s enthalpy (H). The heat flow is equal to the change in the internal energy of the system plus the PV work done. When the volume of a system is constant, changes in its internal energy can be calculated by substituting the ideal gas law into the equation for ΔU. | 10,348 | 3,536 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Arenes/Synthesis_of_Arenes/Electrophilic_Aromatic_Substitution |
This section is on the general mechanism of how an electrophilic atom becomes a part of a benzene ring through the substitution of a hydrogen. Common reactions that proceed by electrophilic aromatic substitution include the nitration and sulfonation of benzene, hydration of benzene, friedel-crafts acylation and friedel-crafts alkylation. Benzene is an aromatic compound that is greatly stabilized by its resonance forms. Stable compounds are much harder to react with, therefore a strong electrophile will be needed to attack the \(pi\) electrons in the benzene ring. Electrophiles used in alkene reactions will typically not be strong enough on their own, therefore a catalyst is required to help generate the electrophile. Although the general mechanism listed here starts with a non-substituted benzene ring, it should make sense that this same reaction could still occur even if there was already a constituent present on the ring, creating polysubstituted benzene rings. Below in a potential energy diagram showing the reaction course of benzene with an electrophile. Notice that the INTERMEDIATE IS NOT . Aromatic cyclic compounds are much more stable than cyclic alkenes, which is why the reaction will continue after the substitution until the aromaticity of the ring as been regained. Step 1 has a high activation energy barrier because a non-aromatic intermediate is formed. Aromatic molecules are lower in energy than their non-aromatic counterparts, so this step is endothermic (product has higher energy than reactant). This step is slow and not thermodynamically favored because the product has higher energy than the reactant. From this intermediate there is a smaller energy of activation- this is because the reaction wants to proceed forward and regain the aromaticity lost in the first step, so this occurs quickly. The aromatic product has lower energy, so this step is exothermic. If you still have trouble interpreting this diagram, check out this site on to review how to read potential energy diagrams. | 2,051 | 3,537 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/20%3A_Molecules_in_Living_Systems/20.15%3A_Secondary_Protein_Structure |
One might expect a long-chain protein molecule to be rather floppy, adopting a variety of molecular shapes and changing rapidly from one conformation to another. In practice this seldom happens. Instead the protein chain stays more or less in the same conformation all the time. It is held in this shape by the cooperative effect of a large number of hydrogen bonds between different segments of the chain. A particularly important conformation of the polypeptide chain is the spiral structure shown in Figure \(\Page {1}\). This is called an . Many fibrous proteins like hair, skin, and nails consist almost entirely of α helices. In globular proteins too, although the overall structure is more complex, short lengths of the chain often have this configuration. In an a helix the polypeptide chain is twisted into a right-hand spiral—the chain turns around clockwise as one moves along it. The spiral is held together by hydrogen bonds from the amido ( ) group of one peptide bond to the carbonyl group of a peptide bond three residues farther along the chain. Two factors contribute toward making this a particularly stable structure. One is the involvement of all the and groups in the chain in the hydrogen bonding. Spirals with slightly more or slightly less twist do not permit this. The second factor is the way in which the side chains project outward from an α helix. Bulky side chains therefore do not interfere with the hydrogen bonding, enabling a fairly rigid cylinder to be formed. A second regular arrangement of the polypeptide chain is the , the β-keratin structure found in silk and shown in Figure \(\Page {2}\). As in the α helix, this structure allows all the amido and carbonyl groups to participate in hydrogen bonds. This hydrogen bonding structure can be accomplished in two manners, either a parallel or antiparallel β sheet, which are compared in Figure \(\Page {3}\). Unlike the α helix, though, the side chains are squeezed rather close together in a pleated-sheet arrangement. In consequence very bulky side chains make the structure unstable. This explains why silk is composed almost entirely of glycine, alanine, and serine, the three amino acids with the smallest side chains. Most other proteins contain a much more haphazard collection of amino acid residues. | 2,327 | 3,539 |
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As an example of the application of Le Chatelier's principle, consider the effect of tripling the pressure on an equilibrium mixture of NO and N O : \[\text{N}_2\text{O}_4(g) \rightleftharpoons \text{2NO}_2(g) \label{1} \] This could be done using the piston and cylinder shown in Figure \(\Page {1}\), in which case tripling the pressure would be expected to reduce the volume of the mixture to one-third its former value. Under the new conditions, however, Le Chatelier’s principle tells us that a new equilibrium will be achieved which counteracts the alteration of pressure. That is, the concentrations of N O and NO should change in such a way as to lessen the pressure increase. This can happen if some of the NO reacts to form N O because two molecules of NO are consumed for every one molecule of N O produced. This reduction in the number of gas molecules will reduce the pressure at the new volume. Thus Le Chatelier’s principle predicts that the reverse of Equation \(\ref{1}\) will occur, producing more N O and using up some NO . We say that N O NO . This agrees with the experimental data on this equilibrium, already given in . Note that the value of the equilibrium constant remains the same, even though the equilibrium shifts. Notice that the effect just described occurs because the gas volume decreased and the concentrations of NO and N O both increased. If we had increased the total pressure on the equilibrium system by pumping in an inert gas such as N ( ), the volume would have remained the same, as would the partial pressures and concentrations of NO and N O In such a case no shift in the equilibrium would be expected. Note also the words in the statement of Le Chatelier’s principle. If the equilibrium reaction had not involved a change in the number of molecules in the gas phase, no shift in the concentrations could have made any difference in the pressure. Thus for the reaction \[\text{2HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \nonumber \] changing the pressure by changing the size of the container would have no effect. You can check this for yourself by redoing using several different volumes and the same initial amount of HI. In general, whenever a gaseous equilibrium involves a change in the number of molecules (Δn ≠ 0), increasing the pressure by reducing the volume will shift the equilibrium in the direction of fewer molecules. This applies even if pure liquids or solids are involved in the reaction. An example is the reaction \[\text{C}(s) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + \text{H}_2(g) \nonumber \] in which superheated steam is passed over carbon obtained from coal to produce carbon monoxide and hydrogen. Since the volume of solid carbon is negligible compared with the volumes of the gases, we need consider only the latter. Hence Δn = 1 (two gas molecules on the right for every one on the left), and an increase in pressure should favor the reverse reaction. This reaction is an important industrial process, and for the reason we have just outlined, is carried out at a low pressure. The video below demonstrates what this process looks like, using the reaction of N O and NO once again. Here we see visual confirmation of the fact that an increase in the volume of a container causes a shift in the equilibrium based on the number of moles of gas, according to Le Chatelier's principle. | 3,426 | 3,540 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/07%3A_Alkenes-_Structure_and_Reactivity/7.11%3A_The_Hammond_Postulate |
After completing this section, you should be able to use the Hammond postulate to explain the formation of the most stable carbocation during the addition of a protic acid, HX, to an alkene. Make certain that you can define, and use in context, the key term below. So far in this chapter the following points have been made about the electrophilic addition of HX to a double bond. It appears that the stability of the carbocation reactive intermediate has a direct effect on the products of a reaction. However, it is the activation energy required to reach the transition state of the reaction's rate determine step which determines which determines which product is produced. This implies that there is a relationship between the transition state and the carbocation reactive intermediate in the mechanism of electrophilic addition. Chemists are often very interested in the structures of the transition states in a reaction's mechanism. In particular, the transition state for a mechanism's rate determining step directly determines the energy of activation barrier and thereby the rate for the overall reaction. Understanding the structure of a transition state allows chemists to consider structural features which might stabilize or destabilize the transition state causing a corresponding change in the rate of reaction. However, transition state structures cannot be directly observed because they are highly unstable activated complexes which instantly convert to a more stable species. In order to gain some insight into the structure of particular transition state, chemists often invoke the , which states that (reactant, intermediate or product). For an exergonic reaction, the transition state is closer in energy to the reactants. Therefore, the structure of the transition state can assumed to resemble the reactants more than the products. Shown below is a hypothetical exergonic reaction between reactant compounds A and B to form the product AB. The Hammond postulate would theorize that the distance between A and B in the transition state would be relatively large thus resembling the reactants where A and B are two isolated species. For an endergonic reaction, the transition state is closer in energy to the product. Therefore, the structure of the transition state can assumed to resemble the products more than the reactants. In the hypothetical endergonic reaction shown below, reactant compounds C and D react to form the product CD. The Hammond postulate would predict that the distance between C and D in the transition state would be relatively small thus resembling the products where C and D are bonded together as a single product CD. By applying the Hammond postulate and other ideas cultivated in this chapter the reason why electrophilic additions tend to follow Markovnikov's rule. When the energy diagram of an electrophilic addition was discussed in , it was noted the first step of the mechanism was the rate determining step. The first step of the mechanism also is endergonic and results in the formation of a carbocation intermediate. The Hammond postulate suggests that the transition state structure for the first step of the mechanism resembles that of the carbocation intermediate because they are the closest in energy. A transitions state, seen below, is typically drawn as a theoretical structure part way between the reactants and the product. For this transition state the pi bonds and the H-Br bond are in the process of being broken and are represented with a dashed line. The C-H bond is in the process of being formed so it also represented with a dashed line. The bromine is shown with a partial negative charge (sigma-) because it is becoming a bromide ion (Br ) which has a full negative charge. Most importantly, the carbon is in the process of becoming a carbocation so it is shown to have a partial positive change (sigma+). Because the Hammond postulate predicts this transitions state closely resembles the carbocation intermediate, the partial positive charge can said to closely resemble the full positive charge of the carbocation. Consequently, any structural feature that stabilized the carbocation intermediate will also stabilize the transition state. The partial positive charge of the transition state is stabilized by adjacent alkyl groups thorough inductive effects and hyperconjugation much like the carbocation intermediate. Adding more alkyl substituents to the partially positive charged carbon stabilizes the transition state, causing it to become lower in energy. This in turn, decreases the energy of activation and increases the rate of the reaction. In short, during an electrophilic addition, the double bond carbon with the most alkyl substituents will for a carbocation intermediate and therefore its C-X bond faster than the double bond carbon with fewer alkyl substituents. These effects cause electrophillic additions to follow Markovnikov's rule and place the halogen (X) group on the more substituted carbon of asymmetrically alkyl substituted double bond. 1) Exergonic and the transition state (second step) represents the reactant (cation). As shown to go from intermediate cation to final product the step is exergonic. | 5,263 | 3,541 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.12%3A_Measuring_the_Entropy |
This method for determining the entropy centers around a very simple relationship between , the heat energy absorbed by a body, the temperature at which this absorption takes place, and Δ , the resultant increase in entropy: It is easy to see how Eq. \(\ref{1}\) can be used to measure the entropy. We start with our substance as close to the absolute zero of temperature as is technically feasible and heat it in many stages, measuring the heat absorbed at each stage, until we arrive at the desired temperature, say 298 K. The initial value of the entropy is zero, and we can calculate the entropy increase for each stage by means of Eq. \(\ref{1}\)and so the sum of all these increases is the entropy value for 298 K. In the case of simple gases, values of entropy measured in this way agree very well with those calculated from knowledge of molecular structure. Equation \(\ref{1}\) was discovered long before the statistical nature of entropy was realized. Scientists and engineers began to appreciate the importance of the quantity / very early in the nineteenth century because of its connection with the efficiency of steam engines. These arguments were developed by both Lord Kelvin in England and Rudolf Clausius (1822 to 1888) in Germany. It was Clausius who first formulated the second law in terms of the entropy , but Clausius had only a vague idea that entropy was in any way connected with molecules or probability. The statistical nature of entropy was first suggested by Boltzmann in 1877 and then developed into an elegant system in 1902 by Josiah Willard Gibbs (1839 to 1903), one of the real giants among American scientists. An important feature of Eq. \(\ref{1}\) is the relationship between the entropy increase and the temperature. A given quantity of heat energy produces a very large change of entropy when absorbed at a very temperature but only a small change when absorbed at a temperature. Calculate the increase in entropy which a substance undergoes when it absorbs 1 kJ of heat energy at the following temperatures: (a) 3 K; (b) 300 K; (c) 3000 K. An amusing analogy to this behavior can be drawn from everyday life. If a 10-year-old boy is allowed to play in his bedroom for half an hour, the increase in disorder is scarcely noticeable because the room is already disordered (i.e., at a higher “temperature”). By contrast, if the boy is let loose for half an hour in a neat and tidy living room (i.e., at a lower “temperature”), the effect is much more dramatic. | 2,521 | 3,542 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introductory_Chemistry_(CK-12)/14%3A_The_Behavior_of_Gases/14.02%3A_Factors_Affecting_Gas_Pressure |
The pressure of the air in a basketball has to be adjusted so that the ball bounces to the correct height. Before a game, officials check the ball by dropping it from shoulder height and seeing how far back up it bounces. What would an official do if the ball did not bounce up high enough, or if it bounced too high? The pressure inside a container is dependent on the amount of gas inside the container. If a basketball does not bounce high enough, an official could remedy the situation by using a hand pump and adding more air to the ball. Conversely, if it bounces too high, the official could let some air out of the ball. Recall from the kinetic-molecular theory that gas particles move randomly and in straight lines until they elastically collide with either other gas particles, or with one of the walls of the container. It is these collisions with the walls of the container that define the pressure of the gas. Four variables are used to describe the condition of a gas: pressure \(\left( P \right)\), volume \(\left( V \right)\), temperature \(\left( T \right)\), and the amount of the gas as measured by the number of moles (\left( n \right)\). We will examine separately how the volume, temperature, and amount of gas each affect the pressure of an enclosed gas sample. The figure below shows what happens when air is added to a . A rigid container is one that is incapable of expanding or contracting. A steel canister is an example of a rigid container. The canister on the left contains a gas at a certain pressure. The attached air pump is then used to double the amount of gas in the canister. Since the canister cannot expand, the increased number of air molecules will strike the inside walls of the canister twice as frequently as they did before. The result is that the pressure inside the canister doubles. As you might imagine, if more and more air is continually added to a rigid container, it may eventually burst. Reducing the number of molecules in a rigid container has the opposite effect, and the pressure decreases. Pressure is also affected by the volume of the container. If the volume of a container is decreased, the gas molecules have less space in which to move around. As a result, they will strike the walls of the container more often, and the pressure increases. The figure below shows a cylinder of gas whose volume is controlled by an adjustable piston. On the left, the piston is pulled mostly out and the gauge reads a certain pressure. On the right, the piston has been pushed so that the volume of the enclosed portion of the container where the gas is located has been cut in half. The pressure of the gas doubles. Increasing the volume of the container would have the opposite effect, and the pressure of the gas would decrease. It would be very inadvisable to place a can of soup over a campfire without venting the can. As the can heats up, it may explode. The kinetic-molecular theory explains why. The air inside the rigid can of soup is given more kinetic energy by the heat coming from the campfire. The kinetic energy causes the air molecules to move faster; they impact the container walls more frequently and with more force. The increase in pressure inside may eventually exceed the strength of the can and it will explode. An additional factor is that the soup may begin boiling, which will then aid even more gas and more pressure to the inside of the can. Shown in the figure below is a cylinder of gas (left) that is at room temperature \(\left( 300 \: \text{K} \right)\). On the right, the cylinder has been heated until the Kelvin temperature has doubled to \(600 \: \text{K}\). The kinetic energy of the gas molecules increases, so collisions with the walls of the container are now more forceful than they were before. As a result, the pressure of the gas doubles. Decreasing the temperature would have the opposite effect, and the pressure of an enclosed gas would decrease. | 3,968 | 3,543 |
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As shown below, there are two 1 electrons and two 2 electrons in the Be atom. Its electron configuration is thus \[1s^{2}2s^{2} \ce{or [He]}2s^{2} \nonumber \] The symbol [He] denotes the inner shell of two 1 electrons which have the same configuration as the noble gas He. The beryllium atom is noticeably smaller than the lithium atom. This is because of the increase in nuclear charge from +3 to +4. Since the two outer 2 electrons (red and orange) do not often come between each other and the nucleus, they do not screen each other from the nucleus very well. Only the two inner electrons are effective in this respect. The effective nuclear charge holding a 2 electron to the nucleus is thus nearly +2, about twice the value for lithium, and the 2 electron clouds are drawn closer to the center of the atom. The next element after beryllium is boron. Since the 2 orbital is completely filled, a new type of orbital must be used for the fifth electron. There are three 2 orbitals available, and any of them might be used. Plate 5 shows the fifth electron (color-coded purple) occupying the 2p orbital. Note carefully the differences between the 2p and 2 electron density distributions in the boron atom. Although on the average both electron clouds extend about the same distance from the nucleus, the 2p electron wave has a node passing through the center of the atom. Thus the 2p electron cloud has a much smaller probability density very close to the nucleus than does a 2 cloud. This means that the 2p electron cloud is more effectively screened by the 1 electrons from the nuclear charge. The atom exerts a slightly smaller overall pull on the 2 electron than it does on the 2 electron. The presence of the inner electrons thus has the effect of making the 2 orbital somewhat in energy than the 2 orbital. This difference in energy between 2 and 2 electrons in the boron atom is an example of a more general behavior. In any atom with sufficient electrons we always find that a orbital is somewhat higher in energy than an orbital with the same value of . In the lithium atom, for example, the third electron occupies a 2 rather than a 2 orbital because this gives it a somewhat lower energy. Further on in the periodic table we will find a similar difference between 3 and 3 orbitals and between 4 and 4 orbitals. We shall examine the electron configuration of one more atom, carbon, with the aid of the color-coded diagrams. In this case six electrons must be distributed among the orbitals—four will be paired in the 1 and 2 orbitals, leaving two -type electron clouds. These are shown color-coded purple and cyan in Plate 5 as 2p and 2p , although the choice of , or directions is arbitrary. The choice of orbitals is not arbitrary, however. It can be shown experimentally that both electrons in the carbon atom have the same spin. Hence they cannot occupy the same orbital. This illustrates another general rule regarding electron configurations. . This is known as . Thus the electron configuration of carbon is \[\ce{[He]}2s^{2} 2p_{x}^{1} 2p_{y}^{1} \nonumber \] This might also be written (using arrows to indicate the orientations of electron spins):
The notation \(\ce{[He]}2s^{2}2p^{2}\) may also be found. In such a case it is assumed that the reader knows that the two 2 electrons are not spin paired. It is worth noting that the arrangement of electrons in different 2 orbitals, necessitated by Hund’s rule, produces a configuration of . If both 2 electrons could occupy the same orbital, say the 2p orbital, they would often be close to each other, and their mutual repulsion would correspond to a higher potential energy. If each is forced to occupy an orbital of different orientation, though, the electrons keep out of each other’s way much more effectively. Their mutual repulsion and hence their potential energy is less. In talking about polyelectronic atoms, the terms shell and subshell are often used. When the two electrons have the same principal quantum number, they are said to belong to the same . In the carbon atom, for example, the two 2 electrons and the 2p and the 2p electrons all belong to the second shell, while the two 1 electrons belong to the first shell. Shells defined in this manner can be further divided into according to whether the electrons being discussed occupy , or orbitals. We can thus divide the second shell into 2 and 2 subshells. The third shell can similarly be divided into 3 , 3 , and 3 subshells, and so on. | 4,564 | 3,544 |
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In a galvanic cell, current is produced when electrons flow externally through the circuit from the anode to the cathode because of a difference in potential energy between the two electrodes in the electrochemical cell. In the Zn/Cu system, the valence electrons in zinc have a substantially higher potential energy than the valence electrons in copper because of shielding of the s electrons of zinc by the electrons in filled d orbitals. Hence electrons flow spontaneously from zinc to copper(II) ions, forming zinc(II) ions and metallic copper. Just like water flowing spontaneously downhill, which can be made to do work by forcing a waterwheel, the flow of electrons from a higher potential energy to a lower one can also be harnessed to perform work. Because the potential energy of valence electrons differs greatly from one substance to another, the voltage of a galvanic cell depends partly on the identity of the reacting substances. If we construct a galvanic cell similar to the one in part (a) in Figure 19.3 but instead of copper use a strip of cobalt metal and 1 M Co in the cathode compartment, the measured voltage is not 1.10 V but 0.51 V. Thus we can conclude that the difference in potential energy between the valence electrons of cobalt and zinc is less than the difference between the valence electrons of copper and zinc by 0.59 V. The measured potential of a cell also depends strongly on the concentrations of the reacting species and the temperature of the system. To develop a scale of relative potentials that will allow us to predict the direction of an electrochemical reaction and the magnitude of the driving force for the reaction, the potentials for oxidations and reductions of different substances must be measured under comparable conditions. To do this, chemists use the (E° ), defined as the potential of a cell measured under standard conditions—that is, with all species in their standard states (1 M for solutions,Concentrated solutions of salts (about 1 M) generally do not exhibit ideal behavior, and the actual standard state corresponds to an activity of 1 rather than a concentration of 1 M. Corrections for nonideal behavior are important for precise quantitative work but not for the more qualitative approach that we are taking here. 1 atm for gases, pure solids or pure liquids for other substances) and at a fixed temperature, usually 25°C. Measured redox potentials depend on the potential energy of valence electrons, the concentrations of the species in the reaction, and the temperature of the system. It is physically impossible to measure the potential of a single electrode: only the difference between the potentials of two electrodes can be measured. (This is analogous to measuring absolute enthalpies or free energies. Recall that only differences in enthalpy and free energy can be measured.) We can, however, compare the standard cell potentials for two different galvanic cells that have one kind of electrode in common. This allows us to measure the potential difference between two dissimilar electrodes. For example, the measured standard cell potential (E°) for the Zn/Cu system is 1.10 V, whereas E° for the corresponding Zn/Co system is 0.51 V. This implies that the potential difference between the Co and Cu electrodes is 1.10 V − 0.51 V = 0.59 V. In fact, that is exactly the potential measured under standard conditions if a cell is constructed with the following cell diagram: \[Co_{(s)} ∣ Co^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M) ∣ Cu (s)\;\;\; E°=0.59\; V \label{19.9}\] This cell diagram corresponds to the oxidation of a cobalt anode and the reduction of Cu in solution at the copper cathode. All tabulated values of standard electrode potentials by convention are listed for a reaction written as a reduction, not as an oxidation, to be able to compare standard potentials for different substances ( ). The standard cell potential (E° ) is therefore the difference between the tabulated reduction potentials of the two half-reactions, not their sum: \[E°_{cell} = E°_{cathode} − E°_{anode} \label{19.10}\] In contrast, recall that half-reactions are written to show the reduction and oxidation reactions that actually occur in the cell, so the overall cell reaction is written as the sum of the two half-reactions. According to Equation \(\ref{19.10}\), when we know the standard potential for any single half-reaction, we can obtain the value of the standard potential of many other half-reactions by measuring the standard potential of the corresponding cell. The overall cell reaction is the sum of the two half-reactions, but the cell potential is the difference between the reduction potentials: \[E°_{cell} = E°_{cathode} − E°_{anode}\] Although it is impossible to measure the potential of any electrode directly, we can choose a reference electrode whose potential is defined as 0 V under standard conditions. The is universally used for this purpose and is assigned a standard potential of 0 V. It consists of a strip of platinum wire in contact with an aqueous solution containing 1 M H . The [H ] in solution is in equilibrium with H gas at a pressure of 1 atm at the Pt-solution interface (Figure \(\Page {2}\)). Protons are reduced or hydrogen molecules are oxidized at the Pt surface according to the following equation: \[2H^+_{(aq)}+2e^− \rightleftharpoons H_{2(g)} \label{19.11}\] One especially attractive feature of the SHE is that the Pt metal electrode is not consumed during the reaction. Figure \(\Page {3}\) shows a galvanic cell that consists of a SHE in one beaker and a Zn strip in another beaker containing a solution of Zn ions. When the circuit is closed, the voltmeter indicates a potential of 0.76 V. The zinc electrode begins to dissolve to form Zn , and H ions are reduced to H in the other compartment. Thus the hydrogen electrode is the cathode, and the zinc electrode is the anode. The diagram for this galvanic cell is as follows: \[Zn_{(s)}∣Zn^{2+}_{(aq)}∥H^+(aq, 1 M)∣H_2(g, 1 atm)∣Pt_{(s)} \label{19.12}\] The half-reactions that actually occur in the cell and their corresponding electrode potentials are as follows: \[E°_{cell}=E°_{cathode}−E°_{anode}=0.76\; V\] Although the reaction at the anode is an oxidation, by convention its tabulated E° value is reported as a reduction potential. The potential of a half-reaction measured against the SHE under standard conditions is called the for that half-reaction.In this example, the standard reduction potential for Zn (aq) + 2e → Zn(s) is −0.76 V, which means that the standard electrode potential for the reaction that occurs at the anode, the oxidation of Zn to Zn , often called the Zn/Zn redox couple, or the Zn/Zn couple, is −(−0.76 V) = 0.76 V. We must therefore subtract E° from E° to obtain E° : 0 − (−0.76 V) = 0.76 V. Because electrical potential is the energy needed to move a charged particle in an electric field, standard electrode potentials for half-reactions are intensive properties and do not depend on the amount of substance involved. Consequently, E° values are independent of the stoichiometric coefficients for the half-reaction, and, most important, the coefficients used to produce a balanced overall reaction do not affect the value of the cell potential. E° values do depend on the stoichiometric coefficients for a half-reaction, because it is an property. To measure the potential of the Cu/Cu couple, we can construct a galvanic cell analogous to the one shown in Figure \(\Page {3}\) but containing a Cu/Cu couple in the sample compartment instead of Zn/Zn . When we close the circuit this time, the measured potential for the cell is negative (−0.34 V) rather than positive. The negative value of E° indicates that the direction of spontaneous electron flow is the opposite of that for the Zn/Zn couple. Hence the reactions that occur spontaneously, indicated by a positive E° , are the reduction of Cu to Cu at the copper electrode. The copper electrode gains mass as the reaction proceeds, and H is oxidized to H at the platinum electrode. In this cell, the copper strip is the cathode, and the hydrogen electrode is the anode. The cell diagram therefore is written with the SHE on the left and the Cu /Cu couple on the right: \[Pt_{(s)}∣H_2(g, 1 atm)∣H^+(aq, 1\; M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{19.16}\] The half-cell reactions and potentials of the spontaneous reaction are as follows: \[E°_{cell} = E°_{cathode}− E°_{anode} = 0.34\; V\] Thus the standard electrode potential for the Cu /Cu couple is 0.34 V. Electrode Potentials and ECell: Previously, we described a method for balancing redox reactions using oxidation numbers. Oxidation numbers were assigned to each atom in a redox reaction to identify any changes in the oxidation states. Here we present an alternative approach to balancing redox reactions, the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. This method more closely reflects the events that take place in an electrochemical cell, where the two half-reactions may be physically separated from each other. We can illustrate how to balance a redox reaction using half-reactions with the reaction that occurs when Drano, a commercial solid drain cleaner, is poured into a clogged drain. Drano contains a mixture of sodium hydroxide and powdered aluminum, which in solution reacts to produce hydrogen gas: \[Al_{(s)} + OH^−_{(aq)} \rightarrow Al(OH)^−_{4(aq)} + H_{2(g)} \label{19.20}\] In this reaction, \(Al_{(s)}\) is oxidized to Al , and H in water is reduced to H gas, which bubbles through the solution, agitating it and breaking up the clogs. The overall redox reaction is composed of a reduction half-reaction and an oxidation half-reaction. From the standard electrode potentials listed , we find the corresponding half-reactions that describe the reduction of H ions in water to H and the oxidation of Al to Al in basic solution: The half-reactions chosen must exactly reflect the reaction conditions, such as the basic conditions shown here. Moreover, the physical states of the reactants and the products must be identical to those given in the overall reaction, whether gaseous, liquid, solid, or in solution. In Equation \(\ref{19.21}\), two H ions gain one electron each in the reduction; in Equation \(\ref{19.22}\), the aluminum atom loses three electrons in the oxidation. The charges are balanced by multiplying the reduction half-reaction (Equation \(\ref{19.21}\)) by 3 and the oxidation half-reaction (Equation \(\ref{19.22}\)) by 2 to give the same number of electrons in both half-reactions: \[6H_2O_{(l)} + 2Al_{(s)} + 8OH^−_{(aq)} \rightarrow 2Al(OH)^−{4(aq)} + 3H_{2(g)} + 6OH^−_{(aq)} \label{19.25}\] Simplifying by canceling substances that appear on both sides of the equation, \[6H_2O_{(l)} + 2Al_{(s)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{19.26}\] We have a −2 charge on the left side of the equation and a −2 charge on the right side. Thus the charges are balanced, but we must also check that atoms are balanced: \[2Al + 8O + 14H = 2Al + 8O + 14H \label{19.27}\] The atoms also balance, so Equation \(\ref{19.26}\) is a balanced chemical equation for the redox reaction depicted in Equation \(\ref{19.20}\). The half-reaction method requires that half-reactions exactly reflect reaction conditions, and the physical states of the reactants and the products must be identical to those in the overall reaction. We can also balance a redox reaction by first balancing the atoms in each half-reaction and then balancing the charges. With this alternative method, we do not need to use the half-reactions listed in but instead focus on the atoms whose oxidation states change, as illustrated in the following steps: Write the reduction half-reaction and the oxidation half-reaction. For the reaction shown in Equation \(\ref{19.20}\), hydrogen is reduced from H in OH to H , and aluminum is oxidized from Al° to Al : Elements other than O and H in the previous two equations are balanced as written, so we proceed with balancing the O atoms. We can do this by adding water to the appropriate side of each half-reaction: Balance the charges in each half-reaction by adding electrons. Two electrons are gained in the reduction of H ions to H , and three electrons are lost during the oxidation of Al° to Al : In this case, we multiply Equation \(\ref{19.34}\) (the reductive half-reaction) by 3 and Equation \(\ref{19.35}\) (the oxidative half-reaction) by 2 to obtain the same number of electrons in both half-reactions: Adding and, in this case, canceling 8H , 3H O, and 6e , \[2Al_{(s)} + 5H_2O_{(l)} + 3OH^−_{(aq)} + H^+_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{19.38}\] We have three OH and one H on the left side. Neutralizing the H gives us a total of 5H O + H O = 6H O and leaves 2OH on the left side: \[2Al_{(s)} + 6H_2O_{(l)} + 2OH^−_{(aq)} \rightarrow 2Al(OH)^−_{4(aq)} + 3H_{2(g)} \label{19.39}\] Check to make sure that all atoms and charges are balanced. Equation \(\ref{19.39}\) is identical to Equation \(\ref{19.26}\), obtained using the first method, so the charges and numbers of atoms on each side of the equation balance. In acidic solution, the redox reaction of dichromate ion (\(Cr_2O_7^{2−}\)) and iodide (\(I^−\)) can be monitored visually. The yellow dichromate solution reacts with the colorless iodide solution to produce a solution that is deep amber due to the presence of a green \(Cr^{3+}_{(aq)}\) complex and brown I (aq) ions (Figure \(\Page {4}\)): \[Cr_2O^{2−}_{7(aq)} + I^−_{(aq)} \rightarrow Cr^{3+}_{(aq)} + I_{2(aq)}\] Balance this equation using half-reactions. redox reaction and balanced chemical equation using half-reactions Follow the steps to balance the redox reaction using the half-reaction method. From the standard electrode potentials listed in we find the half-reactions corresponding to the overall reaction: Balancing the number of electrons by multiplying the oxidation reaction by 3, Adding the two half-reactions and canceling electrons, \[Cr_2O^{2−}_{7(aq)} + 14H^+_{(aq)} + 6I^−_{(aq)} \rightarrow 2Cr^{3+}_{(aq)} + 7H_2O_{(l)} + 3I_{2(aq)}\] We must now check to make sure the charges and atoms on each side of the equation balance: The charges and atoms balance, so our equation is balanced. We can also use the alternative procedure, which does not require the half-reactions listed in . Chromium is reduced from \(Cr^{6+}\) in \(Cr_2O_7^{2−}\) to \(Cr^{3+}\), and \(I^−\) ions are oxidized to \(I_2\). Dividing the reaction into two half-reactions, Balancing the atoms other than oxygen and hydrogen, We now balance the O atoms by adding H O—in this case, to the right side of the reduction half-reaction. Because the oxidation half-reaction does not contain oxygen, it can be ignored in this step. Next we balance the H atoms by adding H to the left side of the reduction half-reaction. Again, we can ignore the oxidation half-reaction. We must now add electrons to balance the charges. The reduction half-reaction (2Cr to 2Cr ) has a +12 charge on the left and a +6 charge on the right, so six electrons are needed to balance the charge. The oxidation half-reaction (2I to I ) has a −2 charge on the left side and a 0 charge on the right, so it needs two electrons to balance the charge: To have the same number of electrons in both half-reactions, we must multiply the oxidation half-reaction by 3: Adding the two half-reactions and canceling substances that appear in both reactions, This is the same equation we obtained using the first method. Thus the charges and atoms on each side of the equation balance. Copper is found as the mineral covellite (\(CuS\)). The first step in extracting the copper is to dissolve the mineral in nitric acid (\(HNO_3\)), which oxidizes sulfide to sulfate and reduces nitric acid to \(NO\): \[CuS_{(s)} + HNO_{3(aq)} \rightarrow NO_{(g)} + CuSO_{4(aq)}\] Balance this equation using the half-reaction method. \[3CuS_{(s)} + 8HNO{3(aq)} \rightarrow 8NO_{(g)} + 3CuSO_{4(aq)} + 4H_2O_{(l)}\] Balancing a Redox Reaction in Acidic Conditions: The standard cell potential for a redox reaction (E° ) is a measure of the tendency of reactants in their standard states to form products in their standard states; consequently, it is a measure of the driving force for the reaction, which earlier we called voltage. We can use the two standard electrode potentials we found earlier to calculate the standard potential for the Zn/Cu cell represented by the following cell diagram: \[ Zn{(s)}∣Zn^{2+}(aq, 1 M)∥Cu^{2+}(aq, 1 M)∣Cu_{(s)} \label{19.40}\] We know the values of E° for the reduction of Zn and E° for the reduction of Cu , so we can calculate E° : \[E°_{cell} = E°_{cathode} − E°_{anode} = 1.10\; V\] This is the same value that is observed experimentally. If the value of E° is positive, the reaction will occur spontaneously as written. If the value of E° is negative, then the reaction is not spontaneous, and it will not occur as written under standard conditions; it will, however, proceed spontaneously in the opposite direction. As we shall see, this does not mean that the reaction cannot be made to occur at all under standard conditions. With a sufficient input of electrical energy, virtually any reaction can be forced to occur. Example 4 and its corresponding exercise illustrate how we can use measured cell potentials to calculate standard potentials for redox couples. A positive E° means that the reaction will occur spontaneously as written. A negative E° means that the reaction will proceed spontaneously in the opposite direction. A galvanic cell with a measured standard cell potential of 0.27 V is constructed using two beakers connected by a salt bridge. One beaker contains a strip of gallium metal immersed in a 1 M solution of GaCl , and the other contains a piece of nickel immersed in a 1 M solution of NiCl . The half-reactions that occur when the compartments are connected are as follows: cathode: Ni (aq) + 2e → Ni(s) anode: Ga(s) → Ga (aq) + 3e If the potential for the oxidation of Ga to Ga is 0.55 V under standard conditions, what is the potential for the oxidation of Ni to Ni ? galvanic cell, half-reactions, standard cell potential, and potential for the oxidation half-reaction under standard conditions standard electrode potential of reaction occurring at the cathode We have been given the potential for the oxidation of Ga to Ga under standard conditions, but to report the standard electrode potential, we must reverse the sign. For the reduction reaction Ga (aq) + 3e → Ga(s), E° = −0.55 V. Using the value given for E° and the calculated value of E° , we can calculate the standard potential for the reduction of Ni to Ni from Equation \(\ref{19.10}\): This is the standard electrode potential for the reaction Ni (aq) + 2e → Ni(s). Because we are asked for the potential for the oxidation of Ni to Ni under standard conditions, we must reverse the sign of E° . Thus E° = −(−0.28 V) = 0.28 V for the oxidation. With three electrons consumed in the reduction and two produced in the oxidation, the overall reaction is not balanced. Recall, however, that standard potentials are independent of stoichiometry. A galvanic cell is constructed with one compartment that contains a mercury electrode immersed in a 1 M aqueous solution of mercuric acetate \(Hg(CH_3CO_2)_2\) and one compartment that contains a strip of magnesium immersed in a 1 M aqueous solution of \(MgCl_2\). When the compartments are connected, a potential of 3.22 V is measured and the following half-reactions occur: If the potential for the oxidation of Mg to Mg is 2.37 V under standard conditions, what is the standard electrode potential for the reaction that occurs at the anode? 0.85 V We can use this procedure described to measure the standard potentials for a wide variety of chemical substances, some of which are listed in . These data allow us to compare the oxidative and reductive strengths of a variety of substances. The half-reaction for the standard hydrogen electrode (SHE) lies more than halfway down the list in Table \(\Page {1}\). All reactants that lie below the SHE in the table are stronger oxidants than H , and all those that lie above the SHE are weaker. The strongest oxidant in the table is F , with a standard electrode potential of 2.87 V. This high value is consistent with the high electronegativity of fluorine and tells us that fluorine has a stronger tendency to accept electrons (it is a stronger oxidant) than any other element. \[Ce^{4+}(aq) + e^− \rightleftharpoons Ce^{3+}(aq)\] Similarly, all species in Table \(\Page {1}\) that lie above H reductants H H reductant ΔH reductant Species in Talbe Table \(\Page {1}\) (or Table ) that lie above H are stronger reducing agents (more easily oxidized) than H . Species that lie below H are stronger oxidizing agents. Because the half-reactions shown in Table \(\Page {1}\) are arranged in order of their E° values, we can use the table to quickly predict the relative strengths of various oxidants and reductants. Any species on the left side of a half-reaction will spontaneously oxidize any species on the right side of another half-reaction that lies below it in the table. Conversely, any species on the right side of a half-reaction will spontaneously reduce any species on the left side of another half-reaction that lies above it in the table. We can use these generalizations to predict the spontaneity of a wide variety of redox reactions (E° > 0), as illustrated below. The black tarnish that forms on silver objects is primarily Ag S. The half-reaction for reversing the tarnishing process is as follows: reduction half-reaction, standard electrode potential, and list of possible reductants reductants for Ag S, strongest reductant, and potential reducing agent for removing tarnish From their positions inTable \(\Page {1}\), decide which species can reduce Ag S. Determine which species is the strongest reductant. Use Table \(\Page {1}\) to identify a reductant for Ag S that is a common household product. We can solve the problem in one of two ways: (1) compare the relative positions of the four possible reductants with that of the Ag S/Ag couple in Table \(\Page {1}\) or (2) compare E° for each species with E° for the Ag S/Ag couple (−0.69 V). Use the data in Table \(\Page {1}\) to determine whether each reaction is likely to occur spontaneously under standard conditions: redox reaction and list of standard electrode potentials ( ) reaction spontaneity Adding the two half-reactions gives the overall reaction: \(\textrm{cathode:} \; \mathrm{Be^{2+}(aq)} +\mathrm{2e^-} \rightarrow \mathrm{Be(s)}\) \(\textrm{anode:} \; \mathrm{Sn(s) \rightarrow \mathrm{Sn^{2+}}(s)} +\mathrm{2e^-} \) \(\textrm{total:} \; \mathrm{Sn(s)+ \mathrm{Be^{2+}(aq)} \rightarrow \mathrm{Sn^{2+}}(aq)} + \mathrm{Be(s)}\) The standard cell potential is quite negative, so the reaction will not occur spontaneously as written. That is, metallic tin cannot reduce Be to beryllium metal under standard conditions. Instead, the reverse process, the reduction of stannous ions (Sn ) by metallic beryllium, which has a positive value of E° , will occur spontaneously. The two half-reactions and their corresponding potentials are as follows The standard potential for the reaction is positive, indicating that under standard conditions, it will occur spontaneously as written. Hydrogen peroxide will reduce MnO , and oxygen gas will evolve from the solution. Use the data in Table \(\Page {1}\) to determine whether each reaction is likely to occur spontaneously under standard conditions: Although the sign of E° tells us whether a particular redox reaction will occur spontaneously under standard conditions, it does not tell us to what extent the reaction proceeds, and it does not tell us what will happen under nonstandard conditions. To answer these questions requires a more quantitative understanding of the relationship between electrochemical cell potential and chemical thermodynamics. When using a galvanic cell to measure the concentration of a substance, we are generally interested in the potential of only one of the electrodes of the cell, the so-called , whose potential is related to the concentration of the substance being measured. To ensure that any change in the measured potential of the cell is due to only the substance being analyzed, the potential of the other electrode, the , must be constant. You are already familiar with one example of a reference electrode: the SHE. The potential of a reference electrode must be unaffected by the properties of the solution, and if possible, it should be physically isolated from the solution of interest. To measure the potential of a solution, we select a reference electrode and an appropriate indicator electrode. Whether reduction or oxidation of the substance being analyzed occurs depends on the potential of the half-reaction for the substance of interest (the sample) and the potential of the reference electrode. The potential of any reference electrode should be affected by the properties of the solution to be analyzed, and it should also be physically isolated. There are many possible choices of reference electrode other than the SHE. The SHE requires a constant flow of highly flammable hydrogen gas, which makes it inconvenient to use. Consequently, two other electrodes are commonly chosen as reference electrodes. One is the , which consists of a silver wire coated with a very thin layer of AgCl that is dipped into a chloride ion solution with a fixed concentration. The cell diagram and reduction half-reaction are as follows: \[Cl^−_{(aq)}∣AgCl_{(s)}∣Ag_{(s)} \label{19.44}\] \[AgCl_{(s)}+e^− \rightarrow Ag_{(s)} + Cl^−_{(aq)}\] If a saturated solution of KCl is used as the chloride solution, the potential of the silver–silver chloride electrode is 0.197 V versus the SHE. That is, 0.197 V must be subtracted from the measured value to obtain the standard electrode potential measured against the SHE. A second common reference electrode is the , which has the same general form as the silver–silver chloride electrode. The SCE consists of a platinum wire inserted into a moist paste of liquid mercury (Hg Cl ; called calomel in the old chemical literature) and KCl. This interior cell is surrounded by an aqueous KCl solution, which acts as a salt bridge between the interior cell and the exterior solution (part (a) in Figure \(\Page {4}\)). Although it sounds and looks complex, this cell is actually easy to prepare and maintain, and its potential is highly reproducible. The SCE cell diagram and corresponding half-reaction are as follows: \[Pt_{(s)} ∣ Hg_2Cl_{2(s)}∣KCl_{(aq, sat)} \label{19.45}\] \[Hg_2Cl_{2(s)} + 2e^− \rightarrow 2Hg_{(l)} + 2Cl^−{(aq)} \label{19.46}\] At 25°C, the potential of the SCE is 0.2415 V versus the SHE, which means that 0.2415 V must be subtracted from the potential versus an SCE to obtain the standard electrode potential. One of the most common uses of electrochemistry is to measure the H ion concentration of a solution. A is generally used for this purpose, in which an internal Ag/AgCl electrode is immersed in a 0.10 M HCl solution that is separated from the solution by a very thin glass membrane (part (b) in Figure \(\Page {5}\)). The glass membrane absorbs protons, which affects the measured potential. The extent of the adsorption on the inner side is fixed because [H ] is fixed inside the electrode, but the adsorption of protons on the outer surface depends on the pH of the solution. The potential of the glass electrode depends on [H ] as follows (recall that pH = −log[H ]: \[E_{glass} = E′ + (0.0591\; V \times \log[H^+]) = E′ − 0.0591\; V \times pH \label{19.47}\] The voltage E′ is a constant that depends on the exact construction of the electrode. Although it can be measured, in practice, a glass electrode is calibrated; that is, it is inserted into a solution of known pH, and the display on the pH meter is adjusted to the known value. Once the electrode is properly calibrated, it can be placed in a solution and used to determine an unknown pH. are used to measure the concentration of a particular species in solution; they are designed so that their potential depends on only the concentration of the desired species (part (c) in Figure \(\Page {5}\)). These electrodes usually contain an internal reference electrode that is connected by a solution of an electrolyte to a crystalline inorganic material or a membrane, which acts as the sensor. For example, one type of ion-selective electrode uses a single crystal of Eu-doped \(LaF_3\) as the inorganic material. When fluoride ions in solution diffuse to the surface of the solid, the potential of the electrode changes, resulting in a so-called fluoride electrode. Similar electrodes are used to measure the concentrations of other species in solution. Some of the species whose concentrations can be determined in aqueous solution using ion-selective electrodes and similar devices are listed in Table \(\Page {2}\). The Standard Hydrogen Electrode (SHE): The flow of electrons in an electrochemical cell depends on the identity of the reacting substances, the difference in the potential energy of their valence electrons, and their concentrations. The potential of the cell under standard conditions (1 M for solutions, 1 atm for gases, pure solids or liquids for other substances) and at a fixed temperature (25°C) is called the standard cell potential (E° ). Only the difference between the potentials of two electrodes can be measured. By convention, all tabulated values of standard electrode potentials are listed as standard reduction potentials. The overall cell potential is the reduction potential of the reductive half-reaction minus the reduction potential of the oxidative half-reaction (E° = E° − E° ). The potential of the standard hydrogen electrode (SHE) is defined as 0 V under standard conditions. The potential of a half-reaction measured against the SHE under standard conditions is called its standard electrode potential. The standard cell potential is a measure of the driving force for a given redox reaction. All E° values are independent of the stoichiometric coefficients for the half-reaction. Redox reactions can be balanced using the half-reaction method, in which the overall redox reaction is divided into an oxidation half-reaction and a reduction half-reaction, each balanced for mass and charge. The half-reactions selected from tabulated lists must exactly reflect reaction conditions. In an alternative method, the atoms in each half-reaction are balanced, and then the charges are balanced. Whenever a half-reaction is reversed, the sign of E° corresponding to that reaction must also be reversed. The oxidative and reductive strengths of a variety of substances can be compared using standard electrode potentials. Apparent anomalies can be explained by the fact that electrode potentials are measured in aqueous solution, which allows for strong intermolecular electrostatic interactions, and not in the gas phase. If E° is positive, the reaction will occur spontaneously under standard conditions. If E° is negative, then the reaction is not spontaneous under standard conditions, although it will proceed spontaneously in the opposite direction. The potential of an indicator electrode is related to the concentration of the substance being measured, whereas the potential of the reference electrode is held constant. Whether reduction or oxidation occurs depends on the potential of the sample versus the potential of the reference electrode. In addition to the SHE, other reference electrodes are the silver–silver chloride electrode; the saturated calomel electrode (SCE); the glass electrode, which is commonly used to measure pH; and ion-selective electrodes, which depend on the concentration of a single ionic species in solution. Differences in potential between the SHE and other reference electrodes must be included when calculating values for E°. | 32,275 | 3,545 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Lattice_Defects/Metal_Oxides |
Metal oxides are very common commodities, widely applied, and have many different varieties. For example, zinc oxide sintered together with other metal oxide additives have been made into nonlinear resistors, which are called Varistors for surge suppressing function. The suppressing function has been applied for switching and for protection of random voltage protections. Iron oxide and other metal oxides are used in thermite reactions, and this has been applied in many ways, including welding in spaceship repairs. Iron oxides are also the raw material for all magnets and magnetic materials used for computer disks and recording tapes. Metals are protected from further oxidation by forming a hard scale of oxides when it is being oxidized. Not all metal oxides form a scale. In general, when the oxide formed is not very dense, it is not under stress, and the oxide layer forms a scale. Usually, a mole of metal oxide should occupy more volume than a mole of the metal itself. If this is true, the oxide is not under stress, and a protective scale is formed. In general, . On the other hand, if the oxide formed occupies a smaller volume than the volume occupied by the metal itself, the oxide layer will be under tension and at some point it will crack. Thus, the oxide offers no protection for further oxidation. The molar volume is easily calculated by dividing the molar mass by the density: \[\text{molar volume} =\dfrac{\text{Molar mass}}{density}\] The densities of \(Mg\) and \(MgO\) are 1.74 and 3.58 g/mL respectively. Calculate their molar volumes. The molar volumes are given below: Molar volume of Mg = 24.31 (g/mol) / 1.74 (g/mL) = 14.0 mL / mol
Molar volume of MgO = (24.31+16.00) (g/mol) / 3.58 (g/mL) = 11.3 mL / mol Since the molar volume of the oxide is less than that of the metal, the oxide does not form a protective scale. The densities of \(Al\) and \(Al_2O_3\) are 2.702 and 3.965 g/mL respectively. Calculate their volumes per mole of Al. The volumes per mole of La are given below: Molar volume of Al = 26.98 (g/mol) / 2.702 (g/mL) = 9.985 mL / mol
Volume of Al2O3 = (26.98+24.00) (g/mol) / 3.965 (g/mL =12.86 mL / mol It has been a well known fact that aluminum oxide forms a protective scale. These data confirm the fact, and now you have an explanation for corrosion. However, we should realize that sometimes the metal oxide does not form a protective layer even if the oxide is not under tension. A major technological concern of metals is the corrosion due to oxidation. The rate of oxidation is usually expressed in terms of depth of the oxidation layer. Several models have been proposed to express the thickness of the oxide layer as a function of time . In the following discussion, is a rate constant. The linear rule. When the oxide offers absolutely no protection, the progress of oxidation is a linear relationship with time. This has been called the by Swaddle. \[dy = k dt\] or \[y = k t\] When the oxide layer gives some protection, the parabolic law apply. This law is formulated with the consideration, \[y dy = k dt\] or \[y_2 = k' t + c\] with \(k' = 2 k\) and c is another constant or \[y = k" t 1/2 + c"\] with k" and c" other constants Oxidation of copper has been shown to follow this rule. The thicker the oxide layer, the more protection the oxide offers in this case. When the oxide layer forms a protective layer, but large flakes crack and leas to faster oxidation as a result. Then, the rate is a combination of the linear rule and the parabolic law. When the oxide forms a good protective layer, the and the ) have been applied. Some suggested formulas are: \[ \dfrac{1}{y} = a \ln (k t +1)\] or \[y = b ln (k t + 1)\] The , and are just some parameters to be determined by experimental methods. The properties of metals and its oxide play roles in the rate of corrosion. There is no set rate laws, and every case must be studied carefully. So far, we have hinted a few has models for exploration and analysis of corrosion problems. The problem with iron as well as many other metals is that the oxide formed by oxidation does not firmly adhere to the surface of the metal and flakes off easily causing "pitting". Extensive pitting eventually causes structural weakness and disintegration of the metal. Most people consider the oxidation of iron results in the formation of a film of iron sesquioxide, which is a term given to red iron(III) oxide, Fe O , which is also called ferric oxide. In reality, the oxidation and corrosion of iron is a very complicated process. The corrosion takes place due to oxidation and galvanic actions. The rust formed is generally represented by \[Fe_ O_ \cdot H_ O\] where is an unspecific amount of water. Depending on the amount of water, the oxides appear in various colors. The pH of water and electrolytes present in the water affect the rate of iron corrosion, because the presence of electrolytes increases the conductivity of the solution. As mentioned above, iron can be at oxidation state II or III in the form of Fe or Fe in its oxides. As a result, iron oxides tends to be somewhat non-stoichiometric. The common iron oxides are: ferrous oxide FeO
hematite a-Fe2O3
maghemite g-Fe2O3
magnetite Fe3O4 However, when water is involved, several oxyhydroxides are possible: goethite a-FeOOH
akaganeite b-FeOOH
lepidocrocite g-FeOOH There is no need to memorize all these minerals, but you get the idea of varieties of iron oxides and oxyhydroxides. The basic structures of iron oxides and iron oxy hydroxides can be described as a close packing of oxygen (or hydroxide) with iron ions in the octahedral sites. The structures of goetite and hematite may be describe as an approximate hcp close packing of O or HO ions with some of the octahedral sites occupied by iron ions. Thus, these two types of structures are usually designated as phases. The structures of lepidocrocite and maghemite may be described as an approximate ccp (fcc) packing of O or HO ions with some of the octahedral sites occupied by iron ions. Thus, these two types of structures are usually designated as phases. Some of the iron ions can be replaced by other metal ions, forming solid solutions in a process known as in terms of structural chemistry. y = b ln (k t + 1)
Explain protection of metal from corrosion.
Explain protection of metal from corrosion. | 6,454 | 3,546 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Book3A_Bioinorganic_Chemistry_(Bertini_et_al.)/03%3A_Calcium_in_Biological_Systems/3.15%3A_Troponin_C |
The contraction of striated muscle is triggered by Ca ions. Muscle cells are highly specialized, and contain two types of filaments that may slide past each other in an energy-consuming process. One of the filaments, the thin filament, is built up by actin molecules (M ≈ 42 kDa) polymerized end-to-end in a double helix. In the grooves of this helix runs a long rod-like molecule, tropomyosin; and located on this molecule at every seventh actin, is a complex of three proteins, . The three proteins in the troponin complex are (TnI), (TnT), and (TnC). A schematic picture of the organization of the thin filament is shown in Figure 3.20. Troponin C is the Ca -binding subunit of troponin, and it is structurally highly homologous to calmodulin. Skeletal-muscle troponin C (sTnC; M ≈ 18 kDa) can bind four Ca ions, but cardiac-muscle troponin C (cTnC) has one of the four calcium sites modified, so that it binds only three Ca ions. The x-ray structures of sTnC from turkey and chicken skeletal muscle have been determined to resolutions of 2.8 and 3.0 Å, respectively. The structure of turkey sTnC is shown in Figure 3.21. The similarity between the structures of CaM (Figure 3.17) and sTnC is obvious. In sTnC we again find two domains, each with two potential Ca sites, separated by a 9-turn \(\alpha\)-helix. The crystals were grown in the presence of Ca at a low pH (pH = 5), and only two Ca ions are found in the C-terminal domain. The two Ca -binding sites in this domain have the same helix-loop-helix motif that is found in CaM, and they both conform to the archetypal EF-hand structure. The interhelix angles between helices E and F and between G and H are close to 110°. By contrast, the helices in the N-terminal domain, where no Ca ions are bound, are closer to being antiparallel, with interhelix angles of 133° (helices A and B) and 151° (helices C and D). Both sTnC and cTnC have two high-affinity Ca -binding sites (see Table 3.2) that also bind Mg ions competitively, although with a much lower affinity. These two sites are usually called " ." In sTnC there are also two (in cTnC, only one) Ca -binding sites of lower affinity (K ≈ 105 M ) that bind Mg weakly and therefore have been called " ." Since Ca binding to the latter sites is assumed to be the crucial step in the contractile event, they are often referred to as " " (see below). The existence of additional weak Mg sites (K ≈ 300 M ) on sTnC, not in direct competition with Ca , has also been inferred. Spectroscopic studies have shown that the two strong Ca - Mg sites are located in the C-terminal domain, and the weaker Ca -specific sites in the N-terminal domain of sTnC. This pattern is similar to that observed with CaM. NMR spectroscopic studies strongly suggest that binding of Ca to both sTnC and cTnC is cooperative. In sTnC the C-terminal domain binds Mg much more strongly than the N-terminal domain, by contrast to CaM, where the reverse is true. The rates of dissociation of Ca and Mg from sTnC have been measured by both stopped-flow and Ca NMR techniques. As with CaM, the actual numbers depend on the solution conditions, ionic strength, presence of Mg , etc. (see Table 3.4). On the rate of Mg dissociation from the Ca - Mg sites, quite different results have been obtained by stopped-flow studies of fluorescence-labeled sTnC (k ≈ 8 s ) and by Mg NMR (k \(\simeq\) 800-1000 s ). This apparent discrepancy seems to have been resolved by the observation that both binding and release of Mg ions to the Ca -Mg sites occur , with k < 20 s for one of the ions, and k ≥ 800 s for the other. The rates of dissociation of the Mg ions are important, since under physiological conditions the Ca -Mg sites of sTnC are likely to be predominantly occupied by Mg ions, release of which determines the rate at which Ca can enter into these sites. Spectroscopic and biochemical data indicate that upon binding Ca , sTnC and cTnC undergo significant conformation changes. Comparisons of NMR spectroscopic changes on Ca binding to intact sTnC, as well as to the two fragments produced by tryptic cleavage (essentially the N-terminal and C-terminal halves of the molecule, just as was the case with CaM), have shown that the conformation changes induced are mainly localized within the domain that is binding added ions. Thus the central \(\alpha\)-helix connecting the domains seems unable to propagate structural changes from one domain to the other. It has been suggested that the structural differences found in the x-ray structure of turkey sTnC between the C-terminal domain, which in the crystal contains two bound Ca ions, and the N-terminal domain, in which no Ca ions were found, may represent these conformational changes. This rather substantial conformational change is schematically depicted in Figure 3.22. However, preliminary structure calculations of the calcium-saturated and calcium-free forms of calbindin D indicate that much more subtle conformational changes take place upon binding Ca in calbindin D . Interestingly, H NMR spectroscopy has provided evidence for the concept that the structural change induced by Mg binding to the C-terminal domain of sTnC must be very similar to that induced by Ca ions. Another result obtained by Cd NMR studies is that the cadmium-loaded N-terminal domain of sTnC in solution undergoes a rapid interchange between two or more conformations, with an exchange rate on the order of 10 -10 s . Just as CaM exerts its biological function in complexes with other proteins, TnC participates in the three-protein troponin complex. It presently appears that TnC and TnI form a primary complex that is anchored by TnT to a binding site on tropomyosin. In the troponin complex the Ca affinity is increased by a factor of about ten over that in isolated sTnC, both at the Ca -Mg sites and at the Ca -specific sites. A similar increase in affinity is found for Mg . Given the amounts of "free" Mg inside muscle cells (1 to 3 mM), it seems likely that the Ca -Mg sites in the resting state of troponin are filled with Mg , so that a transitory release of Ca leads primarily to rapid Ca binding to the Ca -specific sites, and subsequently to conformation change and contraction. | 6,282 | 3,547 |
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The spectrophotometric methods of analysis covered in this chapter include those based on the absorption, emission, or scattering of electromagnetic radiation. When a molecule absorbs UV/Vis radiation it undergoes a change in its valence shell electron configuration. A change in vibrational energy results from the absorption of IR radiation. Experimentally we measure the fraction of radiation transmitted, , by the sample. Instrumentation for measuring absorption requires a source of electromagnetic radiation, a means for selecting a wavelength, and a detector for measuring transmittance. Beer’s law relates absorbance to both transmittance and to the concentration of the absorbing species (\(A = - \log T = \varepsilon b C\)). In atomic absorption we measure the absorption of radiation by gas phase atoms. Samples are atomized using thermal energy from either a flame or a graphite furnace. Because the width of an atom’s absorption band is so narrow, the continuum sources common for molecular absorption are not used. Instead, a hollow cathode lamp provides the necessary line source of radiation. Atomic absorption suffers from a number of spectral and chemical interferences. The absorption or scattering of radiation from the sample’s matrix are important spectral interferences that are minimized by background correction. Chemical interferences include the formation of nonvolatile forms of the analyte and ionization of the analyte. The former interference is minimized by using a releasing agent or a protecting agent, and an ionization suppressor helps minimize the latter interference. When a molecule absorbs radiation it moves from a lower energy state to a higher energy state. In returning to the lower energy state the molecule may emit radiation. This process is called photoluminescence. One form of photoluminescence is fluorescence in which the analyte emits a photon without undergoing a change in its spin state. In phosphorescence, emission occurs with a change in the analyte’s spin state. For low concentrations of analyte, both fluorescent and phosphorescent emission intensities are a linear function of the analyte’s concentration. Thermally excited atoms also emit radiation, forming the basis for atomic emission spectroscopy. Thermal excitation is achieved using either a flame or a plasma. Spectroscopic measurements also include the scattering of light by a particulate form of the analyte. In turbidimetry, the decrease in the radiation’s transmission through the sample is measured and related to the ana- lyte’s concentration through an equation similar to Beer’s law. In nephelometry we measure the intensity of scattered radiation, which varies linearly with the analyte’s concentration. absorbance amplitude background correction chromophore double-beam electromagnetic spectrum excitation spectrum fiber-optic probe fluorescence frequency interferometer ionization suppressor line source mole-ratio method nephelometry phosphorescence photoluminescence polychromatic relaxation self-absorption signal-to-noise ratio slope-ratio method spectrophotometer transducer turbidimetry wavenumber absorbance spectrum attenuated total reflectance Beer’s law continuum source effective bandwidth emission external conversion filter fluorescent quantum yield graphite furnace internal conversion Jacquinot’s advantage method of continuous variations monochromatic nominal wavelength phosphorescent quantum yield photon protecting agent releasing agent signal averaging single-beam spectral searching spectroscopy transmittance vibrational relaxation absorptivity atomization chemiluminescence dark current electromagnetic radiation emission spectrum Fellgett’s advantage filter photometer fluorimeter interferogram intersystem crossing lifetime molar absorptivity monochromator phase angle photodiode array plasma radiationless deactivation resolution signal processor singlet excited state spectrofluorometer stray radiation triplet excited state wavelength | 4,008 | 3,548 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Key_Elements_of_Green_Chemistry_(Lucia)/02%3A_Life-Cycle_Analysis/2.03%3A_Conclusions_and_Review |
There are many tools to assist life cycle assessments. There are software and data packages designed for performing LCAs. No matter the form of the software, the use of some sort of LCA software and data management system is needed in LCAs. The life cycle inventory step of an LCA often requires a large data set listing hundreds of emissions to the environment. Keeping track of these flows manually is arduous, so LCA software is designed to manage these flows and perform specific functions such as impact assessments based on the inventory as well as uncertainty analysis. There is a large list of LCA software emerging with various features. A basic overview of how data and LCA software will first be provided then a list of software packages. LCA software can be split into several components: Software packages such as SimaPro, openLCA, and Gabi are frameworks or calculators that keep track of data and performs intensive numerical calculations. With the many flows and detailed 49 data, much effort has been invested in creating efficient calculation methods to speed up analysis time. This framework, however, is not useful without inventory data. There are many premade datasets provided from sources such as Ecoinvent, Gabi and United States Department of Agriculture (USDA) that contain previous life cycle inventory results for various chemicals, materials, energy, services, and waste treatment processes. LCA software can access this previously developed data and include a chemical or other process from a dataset in their LCA without needing to perform an entire LCA on that particular material or process. This fundamental aspect of LCA, the leveraging of previous study results for new studies, is a key benefit of LCA software and can save countless hours on the LCI step. LCIA methods are procedures and conversions that are used in performing a LCIA such as global warming potential characterizations and weighting methods. There are many accepted LCIA methods that calculate LCA results using different impact categories, types of impacts, and weighting methods. Figure \(\Page {1}\) visually depicts how the different components of LCA software and data interact. The life cycle inventory step requires data from datasets (e.g., Ecoinvent) and primary data gathered by the LCA practitioner surrounding the process or product under analysis. The combination of these two types of data with the use of LCA software calculations gives an LCI. The LCI data can then be used to perform an impact assessment using the LCIA methods (e.g., TRACI). www.openLCA.org | 2,593 | 3,549 |
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" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact Problems marked with a ♦ involve multiple concepts. \[2H_2O_2(l) + N_2H_4(l) \rightarrow N_2(g) + 4H_2O(g)\] \[2B_5H_9(l) + 12O_2(g) \rightarrow 5B_2O_3(s) + 9H_2O(g)\] The combustion of sucrose, however, occurs as follows: C H O (s) + 12O (g) → 12CO (g) + 11H O(l) The earliest known tin artifacts were discovered in Egyptian tombs of the 18th dynasty (1580–1350 BC), although archaeologists are surprised that so few tin objects exist from earlier eras. It has been suggested that many early tin objects were either oxidized to a mixture of stannous and stannic oxides or transformed to powdery, gray tin. Sketch a thermodynamic cycle similar to part (b) in Figure 18.15 "Two Forms of Elemental Sulfur and a Thermodynamic Cycle Showing the Transition from One to the Other" to show the conversion of liquid tin to gray tin. Then calculate the change in entropy that accompanies the conversion of Sn(l) to α-Sn using the following data: C (white) = 26.99, C (gray) = 25.77 J/(mol·K), ΔH = 7.0 kJ/mol, ΔH → α = −2.2 kJ/mol. Based strictly on thermodynamic arguments, which would you choose to administer to a patient suffering from cadmium toxicity? Why? Assume a body temperature of 37°C. 2N H (l) + N O (l) → 4H O(g) + 3N (g) ΔH° = −249 kJ/mol, ΔS° = 218 J/(mol·K) 2CsCl(l) + CaC (s) → CaCl (l) + 2C(s) + 2Cs(g) Compare the free energy produced from this reaction at 25°C and at 1227°C, the temperature at which it is normally run, given these values: ΔH = 32.0 kJ/mol, ΔS = 8.0 J/(mol·K); ΔH = −0.6 kJ/mol, ΔS = 3.6 J/(mol·K). Regeneration is carried out at 250°C. \[8H^+ + 8e^− + N_2 + 16ATP \rightarrow H_2 + 2NH_3 + 16ADP + 16P_i\] Solution | 1,794 | 3,552 |
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Resonance structures are a set of two or more Lewis Structures that collectively describe the electronic bonding a single polyatomic species including fractional bonds and fractional charges. Resonance structure are capable of describing delocalized electrons that cannot be expressed by a single Lewis formula with an integer number of covalent bonds. Sometimes, even when are considered, the bonding in some molecules or ions cannot be described by a single Lewis structure. Resonance is a way of describing delocalized electrons within certain molecules or polyatomic ions where the bonding cannot be expressed by a single Lewis formula. A molecule or ion with such delocalized electrons is represented by several contributing structures (also called resonance structures or canonical forms). Such is the case for ozone (O ), an allotrope of oxygen with a V-shaped structure and an O–O–O angle of 117.5°. 1. We know that ozone has a V-shaped structure, so one O atom is central: 2. Each O atom has 6 valence electrons, for a total of 18 valence electrons. 3. Assigning one bonding pair of electrons to each oxygen–oxygen bond gives with 14 electrons left over. 4. If we place three lone pairs of electrons on each terminal oxygen, we obtain and have 2 electrons left over. 5. At this point, both terminal oxygen atoms have octets of electrons. We therefore place the last 2 electrons on the central atom: 6. The central oxygen has only 6 electrons. We must convert one lone pair on a terminal oxygen atom to a bonding pair of electrons—but which one? Depending on which one we choose, we obtain either Which is correct? In fact, neither is correct. Both predict one O–O single bond and one O=O double bond. As you will learn, if the bonds were of different types (one single and one double, for example), they would have different lengths. It turns out, however, that both O–O bond distances are identical, 127.2 pm, which is shorter than a typical O–O single bond (148 pm) and longer than the O=O double bond in O (120.7 pm). Equivalent Lewis dot structures, such as those of ozone, are called resonance structures. The position of the is the same in the various resonance structures of a compound, but the position of the is different. Double-headed arrows link the different resonance structures of a compound: The double-headed arrow indicates that the actual electronic structure is an of those shown, not that the molecule oscillates between the two structures. When it is possible to write more than one equivalent resonance structure for a molecule or ion, the actual structure is the of the resonance structures. Like ozone, the electronic structure of the carbonate ion cannot be described by a single Lewis electron structure. Unlike O , though, the actual structure of CO is an average of resonance structures. 1. Because carbon is the least electronegative element, we place it in the central position: 2. Carbon has 4 valence electrons, each oxygen has 6 valence electrons, and there are 2 more for the −2 charge. This gives 4 + (3 × 6) + 2 = 24 valence electrons. 3. Six electrons are used to form three bonding pairs between the oxygen atoms and the carbon: 4. We divide the remaining 18 electrons equally among the three oxygen atoms by placing three lone pairs on each and indicating the −2 charge: 5. No electrons are left for the central atom. 6. At this point, the carbon atom has only 6 valence electrons, so we must take one lone pair from an oxygen and use it to form a carbon–oxygen double bond. In this case, however, there are possible choices: As with ozone, none of these structures describes the bonding exactly. Each predicts one carbon–oxygen double bond and two carbon–oxygen single bonds, but experimentally all C–O bond lengths are identical. We can write resonance structures (in this case, three of them) for the carbonate ion: The actual structure is an average of these three resonance structures. 1. Count up the valence electrons: (1*5) + (3*6) + 1(ion) = electrons 2. Draw the bond connectivities: 3. Add octet electrons to the atoms bonded to the center atom: 4. Place any leftover electrons (24-24 = ) on the center atom: 5. Does the central atom have an octet? 6. Does the central atom have an octet? Note: We would expect that the bond lengths in the \(NO_3^-\) ion to be somewhat shorter than a single bond Benzene is a common organic solvent that was previously used in gasoline; it is no longer used for this purpose, however, because it is now known to be a carcinogen. The benzene molecule (C H ) consists of a regular hexagon of carbon atoms, each of which is also bonded to a hydrogen atom. Use resonance structures to describe the bonding in benzene. molecular formula and molecular geometry resonance structures Each hydrogen atom contributes 1 valence electron, and each carbon atom contributes 4 valence electrons, for a total of (6 × 1) + (6 × 4) = 30 valence electrons. If we place a single bonding electron pair between each pair of carbon atoms and between each carbon and a hydrogen atom, we obtain the following: Each carbon atom in this structure has only 6 electrons and has a formal charge of +1, but we have used only 24 of the 30 valence electrons. If the 6 remaining electrons are uniformly distributed pairwise on alternate carbon atoms, we obtain the following: Three carbon atoms now have an octet configuration and a formal charge of −1, while three carbon atoms have only 6 electrons and a formal charge of +1. We can convert each lone pair to a bonding electron pair, which gives each atom an octet of electrons and a formal charge of 0, by making three C=C double bonds. There are, however, two ways to do this: Each structure has alternating double and single bonds, but experimentation shows that each carbon–carbon bond in benzene is identical, with bond lengths (139.9 pm) intermediate between those typically found for a C–C single bond (154 pm) and a C=C double bond (134 pm). We can describe the bonding in benzene using the two resonance structures, but the actual electronic structure is an average of the two. The existence of multiple resonance structures for aromatic hydrocarbons like benzene is often indicated by drawing either a circle or dashed lines inside the hexagon: The sodium salt of nitrite is used to relieve muscle spasms. Draw two resonance structures for the nitrite ion (NO ). Resonance structures are particularly common in oxoanions of the -block elements, such as sulfate and phosphate, and in aromatic hydrocarbons, such as benzene and naphthalene. Resonance Structures: Some molecules have two or more chemically equivalent Lewis electron structures, called resonance structures. Resonance is a mental exercise and method within the of bonding that describes the delocalization of electrons within molecules. These structures are written with a between them, indicating that none of the Lewis structures accurately describes the bonding but that the actual structure is an average of the individual resonance structures. 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. | 7,610 | 3,553 |
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Photoluminescence is divided into two categories: fluorescence and phosphorescence. A pair of electrons that occupy the same electronic ground state have opposite spins and are in a singlet spin state (Figure 10.6.1
a). When an analyte absorbs an ultraviolet or a visible photon, one of its valence electrons moves from the ground state to an excited state with a conservation of the electron’s spin (Figure 10.6.1
b). Emission of a photon from a to the singlet ground state—or between any two energy levels with the same spin—is called . The probability of fluorescence is very high and the average lifetime of an electron in the excited state is only 10 –10 s. Fluorescence, therefore, rapidly decays once the source of excitation is removed. In some cases an electron in a singlet excited state is transformed to a (Figure 10.6.1
c) in which its spin no is longer paired with the ground state. Emission between a triplet excited state and a singlet ground state—or between any two energy levels that differ in their respective spin states–is called . Because the average lifetime for phosphorescence ranges from 10 –10 s, phosphorescence may continue for some time after we remove the excitation source. The use of molecular fluorescence for qualitative analysis and for semi-quantitative analysis dates to the early to mid 1800s, with more accurate quantitative methods appearing in the 1920s. Instrumentation for fluorescence spectroscopy using a filter or a monochromator for wavelength selection appeared in, respectively, the 1930s and 1950s. Although the discovery of phosphorescence preceded that of fluorescence by almost 200 years, qualitative and quantitative applications of molecular phosphorescence did not receive much attention until after the development of fluorescence instrumentation. As you might expect, the persistence of long-lived phosphorescence made it more noticeable. To appreciate the origin of fluorescence and phosphorescence we must consider what happens to a molecule following the absorption of a photon. Let’s assume the molecule initially occupies the lowest vibrational energy level of its electronic ground state, which is the singlet state labeled S in Figure 10.6.2
. Absorption of a photon excites the molecule to one of several vibrational energy levels in the first excited electronic state, S , or the second electronic excited state, S , both of which are singlet states. Relaxation to the ground state occurs by a number of mechanisms, some of which result in the emission of a photon and others that occur without the emission of a photon. These relaxation mechanisms are shown in Figure 10.6.2
. The most likely relaxation pathway from any excited state is the one with the shortest lifetime. When a molecule relaxes without emitting a photon we call the process . One example of radiationless deactivation is , in which a molecule in an excited vibrational energy level loses energy by moving to a lower vibrational energy level in the same electronic state. Vibrational relaxation is very rapid, with an average lifetime of <10 s. Because vibrational relaxation is so efficient, a molecule in one of its excited state’s higher vibrational energy levels quickly returns to the excited state’s lowest vibrational energy level. Another form of radiationless deactivation is an in which a molecule in the ground vibrational level of an excited state passes directly into a higher vibrational energy level of a lower energy electronic state of the same spin state. By a combination of internal conversions and vibrational relaxations, a molecule in an excited electronic state may return to the ground electronic state without emitting a photon. A related form of radiationless deactivation is an in which excess energy is transferred to the solvent or to another component of the sample’s matrix. Let’s use Figure 10.6.2
to illustrate how a molecule can relax back to its ground state without emitting a photon. Suppose our molecule is in the highest vibrational energy level of the second electronic excited state. After a series of vibrational relaxations brings the molecule to the lowest vibrational energy level of S , it undergoes an internal conversion into a higher vibrational energy level of the first excited electronic state. Vibrational relaxations bring the molecule to the lowest vibrational energy level of S . Following an internal conversion into a higher vibrational energy level of the ground state, the molecule continues to undergo vibrational relaxation until it reaches the lowest vibrational energy level of S . A final form of radiationless deactivation is an in which a molecule in the ground vibrational energy level of an excited electronic state passes into one of the higher vibrational energy levels of a lower energy electronic state with a different spin state. For example, an intersystem crossing is shown in Figure 10.6.2
between the singlet excited state S and the triplet excited state T . Fluorescence occurs when a molecule in an excited state’s lowest vibrational energy level returns to a lower energy electronic state by emitting a photon. Because molecules return to their ground state by the fastest mechanism, fluorescence is observed only if it is a more efficient means of relaxation than a combination of internal conversions and vibrational relaxations. A quantitative expression of fluorescence efficiency is the fluorescent , \(\Phi_f\), which is the fraction of excited state molecules that return to the ground state by fluorescence. The fluorescent quantum yields range from 1 when every molecule in an excited state undergoes fluorescence, to 0 when fluorescence does not occur. The intensity of fluorescence, , is proportional to the amount of radiation absorbed by the sample, – , and the fluorescent quantum yield \[I_{f}=k \Phi_{f}\left(P_{0}-P_{\mathrm{T}}\right) \label{10.1}\] where is a constant that accounts for the efficiency of collecting and detecting the fluorescent emission. From Beer’s law we know that \[\frac{P_{\mathrm{T}}}{P_{0}}=10^{-\varepsilon b C} \label{10.2}\] where is the concentration of the fluorescing species. Solving Equation \ref{10.2} for and substituting into Equation \ref{10.1} gives, after simplifying \[I_{f}=k \Phi_{f} P_{0}\left(1-10^{-\varepsilon b C}\right) \label{10.3}\] When \(\varepsilon bC\) < 0.01, which often is the case when the analyte's concentration is small, Equation \ref{10.3} simplifies to \[I_{f}=2.303 k \Phi_{f} \varepsilon b C P_{0}=k^{\prime} P_{0} \label{10.4}\] where ′ is a collection of constants. The intensity of fluorescence, therefore, increases with an increase in the quantum efficiency, the source’s incident power, and the molar absorptivity and the concentration of the fluorescing species. Fluorescence generally is observed when the molecule’s lowest energy absorption is a \(\pi \rightarrow \pi^*\) transition, although some \(n \rightarrow \pi^*\) transitions show weak fluorescence. Many unsubstituted, nonheterocyclic aromatic compounds have a favorable fluorescence quantum yield, although substitutions on the aromatic ring can effect \(\Phi_f\) significantly. For example, the presence of an electron-withdrawing group, such as –NO , decreases \(\Phi_f\), while adding an electron-donating group, such as –OH, increases \(\Phi_f\). Fluorrescence also increases for aromatic ring systems and for aromatic molecules with rigid planar structures. Figure 10.6.3
shows the fluorescence of quinine under a UV lamp. A molecule’s fluorescent quantum yield also is influenced by external variables, such as temperature and solvent. Increasing the temperature generally decreases \(\Phi_f\) because more frequent collisions between the molecule and the solvent increases external conversion. A decrease in the solvent’s viscosity decreases \(\Phi_f\) for similar reasons. For an analyte with acidic or basic functional groups, a change in pH may change the analyte’s structure and its fluorescent properties. As shown in Figure 10.6.2
, fluorescence may return the molecule to any of several vibrational energy levels in the ground electronic state. Fluorescence, therefore, occurs over a range of wavelengths. Because the change in energy for fluorescent emission generally is less than that for absorption, a molecule’s fluorescence spectrum is shifted to higher wavelengths than its absorption spectrum. A molecule in a triplet electronic excited state’s lowest vibrational energy level normally relaxes to the ground state by an intersystem crossing to a singlet state or by an external conversion. Phosphorescence occurs when the molecule relaxes by emitting a photon. As shown in Figure 10.6.2
, phosphorescence occurs over a range of wavelengths, all of which are at lower energies than the molecule’s absorption band. The intensity of phosphorescence, \(I_p\), is given by an equation similar to Equation \ref{10.4} for fluorescence \[\begin{align} I_{P} &=2.303 k \Phi_{P} \varepsilon b C P_{0} \\[4pt] &=k^{\prime} P_{0} \label{10.5}\end{align}\] where \(\Phi_p\) is the . Phosphorescence is most favorable for molecules with \(n \rightarrow \pi^*\) transitions, which have a higher probability for an intersystem crossing than \(\pi \rightarrow \pi^*\) transitions. For example, phosphorescence is observed with aromatic molecules that contain carbonyl groups or heteroatoms. Aromatic compounds that contain halide atoms also have a higher efficiency for phosphorescence. In general, an increase in phosphorescence corresponds to a decrease in fluorescence. Because the average lifetime for phosphorescence can be quite long, ranging from 10 –10 s, the phosphorescent quantum yield usually is quite small. An improvement in \(\Phi_p\) is realized by decreasing the efficiency of external conversion. This is accomplished in several ways, including lowering the temperature, using a more viscous solvent, depositing the sample on a solid substrate, or trapping the molecule in solution. Figure 10.6.4
shows an example of phosphorescence. Photoluminescence spectra are recorded by measuring the intensity of emitted radiation as a function of either the excitation wavelength or the emission wavelength. An is obtained by monitoring emission at a fixed wavelength while varying the excitation wavelength. When corrected for variations in the source’s intensity and the detector’s response, a sample’s excitation spectrum is nearly identical to its absorbance spectrum. The excitation spectrum provides a convenient means for selecting the best excitation wavelength for a quantitative or qualitative analysis. In an a fixed wavelength is used to excite the sample and the intensity of emitted radiation is monitored as function of wavelength. Although a molecule has a single excitation spectrum, it has two emission spectra, one for fluorescence and one for phosphorescence. Figure 10.6.5
shows the UV absorption spectrum and the UV fluorescence emission spectrum for quinine. The basic instrumentation for monitoring fluorescence and phosphorescence—a source of radiation, a means of selecting a narrow band of radiation, and a detector—are the same as those for absorption spectroscopy. The unique demands of fluorescence and phosphorescence, however, require some modifications to the instrument designs seen earlier in (filter photometer), (single-beam spectrophotometer), (double-beam spectrophotometer), and (diode array spectrometer). The most important difference is that the detector cannot be placed directly across from the source. Figure 10.6.6
shows why this is the case. If we place the detector along the source’s axis it receives both the transmitted source radiation, , and the fluorescent, , or phosphorescent, , radiation. Instead, we rotate the director and place it at 90 to the source. Figure 10.6.7
shows the basic design of an instrument for measuring fluorescence, which includes two wavelength selectors, one for selecting the source's excitation wavelength and one for selecting the analyte's emission wavelength. In a the excitation and emission wavelengths are selected using absorption or interference filters. The excitation source for a fluorimeter usually is a low-pressure Hg vapor lamp that provides intense emission lines distributed throughout the ultraviolet and visible region. When a monochromator is used to select the excitation and the emission wavelengths, the instrument is called a . With a monochromator the excitation source usually is a high-pressure Xe arc lamp, which has a continuous emission spectrum. Either instrumental design is appropriate for quantitative work, although only a spectrofluorometer can record an excitation or emission spectrum. A Hg vapor lamp has emission lines at 254, 312, 365, 405, 436, 546, 577, 691, and 773 nm. The sample cells for molecular fluorescence are similar to those for molecular absorption (see ). Remote sensing using a fiber optic probe (see ) is possible using with either a fluorimeter or spectrofluorometer. An analyte that is fluorescent is monitored directly. For an analyte that is not fluorescent, a suitable fluorescent probe molecule is incorporated into the tip of the fiber optic probe. The analyte’s reaction with the probe molecule leads to an increase or decrease in fluorescence. An instrument for molecular phosphorescence must discriminate between phosphorescence and fluorescence. Because the lifetime for fluorescence is shorter than that for phosphorescence, discrimination is achieved by incorporating a delay between exciting the sample and measuring the phosphorescent emission. Figure 10.6.8
shows how two out-of-phase choppers allow us to block fluorescent emission from reaching the detector when the sample is being excited and to prevent the source radiation from causing fluorescence when we are measuring the phosphorescent emission. Because phosphorescence is such a slow process, we must prevent the excited state from relaxing by external conversion. One way this is accomplished is by dissolving the sample in a suitable organic solvent, usually a mixture of ethanol, isopentane, and diethylether. The resulting solution is frozen at liquid-N temperatures to form an optically clear solid. The solid matrix minimizes external conversion due to collisions between the analyte and the solvent. External conversion also is minimized by immobilizing the sample on a solid substrate, making possible room temperature measurements. One approach is to place a drop of a solution that contains the analyte on a small disc of filter paper. After drying the sample under a heat lamp, the sample is placed in the spectrofluorometer for analysis. Other solid substrates include silica gel, alumina, sodium acetate, and sucrose. This approach is particularly useful for the analysis of thin layer chromatography plates. Molecular fluorescence and, to a lesser extent, phosphorescence are used for the direct or indirect quantitative analysis of analytes in a variety of matrices. A direct quantitative analysis is possible when the analyte’s fluorescent or phosphorescent quantum yield is favorable. If the analyte is not fluorescent or phosphorescent, or if the quantum yield is unfavorable, then an indirect analysis may be feasible. One approach is to react the analyte with a reagent to form a product that is fluorescent or phosphorescent. Another approach is to measure a decrease in fluorescence or phosphores- cence when the analyte is added to a solution that contains a fluorescent or phosphorescent probe molecule. A decrease in emission is observed when the reaction between the analyte and the probe molecule enhances radiationless deactivation or results in a nonemitting product. The application of fluorescence and phosphorescence to inorganic and organic analytes are considered in this section. Except for a few metal ions, most notably \(\text{UO}_2^+\), most inorganic ions are not sufficiently fluorescent for a direct analysis. Many metal ions are determined indirectly by reacting with an organic ligand to form a fluorescent or, less commonly, a phosphorescent metal–ligand complex. One example is the reaction of Al with the sodium salt of 2, 4, 3′-trihydroxyazobenzene-5′-sulfonic acid—also known as alizarin garnet R—which forms a fluorescent metal–ligand complex (Figure 10.6.9
). The analysis is carried out using an excitation wavelength of 470 nm, with fluorescence monitored at 500 nm. Table 10.6.1
provides additional examples of chelating reagents that form fluorescent metal–ligand complexes with metal ions. A few inorganic nonmetals are determined by their ability to decrease, or quench, the fluorescence of another species. One example is the analysis for F based on its ability to quench the fluorescence of the Al –alizarin garnet R complex. 8-hydroxyquinoline flavonal 2-( -hydroxyphenyl) benzoxazole As noted earlier, organic compounds that contain aromatic rings generally are fluorescent and aromatic heterocycles often are phosphorescent. Table 10.6.2
provides examples of several important biochemical, pharmaceutical, and environmental compounds that are analyzed quantitatively by fluorimetry or phosphorimetry. If an organic analyte is not naturally fluorescent or phosphorescent, it may be possible to incorporate it into a chemical reaction that produces a fluorescent or phosphorescent product. For example, the enzyme creatine phosphokinase is determined by using it to catalyze the formation of creatine from phosphocreatine. Reacting the creatine with ninhydrin produces a fluorescent product of unknown structure. phenylalanine (F) tyrosine (F) tryptophan (F, P) vitamins vitamin A (F) vitamin B2 (F) vitamin B6 (F) vitamin B12 (F) vitamin E (F) folic acid (F) catecholamines dopamine (F) norepinephrine (F) pharmaceuticals and drugs quinine (F) salicylic acid (F, P) morphine (F) barbiturates (F) LSD (F) codeine (P) caffeine (P) sulfanilamide (P) environmental pollutants pyrene (F) benzo[a]pyrene (F) organothiophosphorous pesticides (F) carbamate insecticides (F) DDT (P) From Equation \ref{10.4} and Equation \ref{10.5} we know that the intensity of fluorescence or phosphorescence is a linear function of the analyte’s concentration provided that the sample’s absorbance of source radiation (\(A = \varepsilon bC\)) is less than approximately 0.01. Calibration curves often are linear over four to six orders of magnitude for fluorescence and over two to four orders of magnitude for phosphorescence. For higher concentrations of analyte the calibration curve becomes nonlinear because the assumptions that led to Equation \ref{10.4} and Equation \ref{10.5} no longer apply. Nonlinearity may be observed for smaller concentrations of analyte fluorescent or phosphorescent contaminants are present. As discussed earlier, quantum efficiency is sensitive to temperature and sample matrix, both of which must be con- trolled when using external standards. In addition, emission intensity depends on the molar absorptivity of the photoluminescent species, which is sensitive to the sample matrix. The best way to appreciate the theoretical and the practical details discussed in this section is to carefully examine a typical analytical method. Although each method is unique, the following description of the determination of quinine in urine provides an instructive example of a typical procedure. The description here is based on Mule, S. J.; Hushin, P. L. , , 708–711, and O’Reilly, J. E.; , , 610–612. Figure 10.6.3
shows the fluorescence of the quinine in tonic water. Quinine is an alkaloid used to treat malaria. It is a strongly fluorescent compound in dilute solutions of H SO (\(\Phi_f = 0.55\)). Quinine’s excitation spectrum has absorption bands at 250 nm and 350 nm and its emission spectrum has a single emission band at 450 nm. Quinine is excreted rapidly from the body in urine and is determined by measuring its fluorescence following its extraction from the urine sample. Transfer a 2.00-mL sample of urine to a 15-mL test tube and use 3.7 M NaOH to adjust its pH to between 9 and 10. Add 4 mL of a 3:1 (v/v) mixture of chloroform and isopropanol and shake the contents of the test tube for one minute. Allow the organic and the aqueous (urine) layers to separate and transfer the organic phase to a clean test tube. Add 2.00 mL of 0.05 M H SO to the organic phase and shake the contents for one minute. Allow the organic and the aqueous layers to separate and transfer the aqueous phase to the sample cell. Measure the fluorescent emission at 450 nm using an excitation wavelength of 350 nm. Determine the concentration of quinine in the urine sample using a set of external standards in 0.05 M H SO , prepared from a 100.0 ppm solution of quinine in 0.05 M H SO . Use distilled water as a blank. 1. Chloride ion quenches the intensity of quinine’s fluorescent emission. For example, in the presence of 100 ppm NaCl (61 ppm Cl ) quinine’s emission intensity is only 83% of its emission intensity in the absence of chloride. The presence of 1000 ppm NaCl (610 ppm Cl ) further reduces quinine’s fluorescent emission to less than 30% of its emission intensity in the absence of chloride. The concentration of chloride in urine typically ranges from 4600–6700 ppm Cl . Explain how this procedure prevents an interference from chloride. The procedure uses two extractions. In the first of these extractions, quinine is separated from urine by extracting it into a mixture of chloroform and isopropanol, leaving the chloride ion behind in the original sample. 2. Samples of urine may contain small amounts of other fluorescent compounds, which will interfere with the analysis if they are carried through the two extractions. Explain how you can modify the procedure to take this into account? One approach is to prepare a blank that uses a sample of urine known to be free of quinine. Subtracting the blank’s fluorescent signal from the measured fluorescence from urine samples corrects for the interfering compounds. The fluorescent emission for quinine at 450 nm can be induced using an excitation frequency of either 250 nm or 350 nm. The fluorescent quantum efficiency is the same for either excitation wavelength. Quinine’s absorption spectrum shows that \(\varepsilon_{250}\) is greater than \(\varepsilon_{350}\). Given that quinine has a stronger absorbance at 250 nm, explain why its fluorescent emission intensity is greater when using 350 nm as the excitation wavelength. From Equation \ref{10.4} we know that is a function of the following terms: , \(\Phi_f\), , \(\varepsilon\), , and . We know that \(\Phi_f\), , and are the same for both excitation wavelengths and that \(\varepsilon\) is larger for a wavelength of 250 nm; we can, therefore, ignore these terms. The greater emission intensity when using an excitation wavelength of 350 nm must be due to a larger value for or . In fact, at 350 nm for a high-pressure Xe arc lamp is about 170% of that at 250 nm. In addition, the sensitivity of a typical photomultiplier detector (which contributes to the value of ) at 350 nm is about 140% of that at 250 nm. To evaluate the method described in Representative Method 10.6.1, a series of external standard are prepared and analyzed, providing the results shown in the following table. All fluorescent intensities are corrected using a blank prepared from a quinine-free sample of urine. The fluorescent intensities are normalized by setting for the highest concentration standard to 100. After ingesting 10.0 mg of quinine, a volunteer provides a urine sample 24-h later. Analysis of the urine sample gives a relative emission intensity of 28.16. Report the concentration of quinine in the sample in mg/L and the percent recovery for the ingested quinine. Linear regression of the relative emission intensity versus the concentration of quinine in the standards gives the calibration curve shown below and the following calibration equation. \[I_{f}=0.122+9.978 \times \frac{\mathrm{g} \text { quinine }}{\mathrm{mL}} \nonumber\] Substituting the sample’s relative emission intensity into the calibration equation gives the concentration of quinine as 2.81 μg/mL. Because the volume of urine taken, 2.00 mL, is the same as the volume of 0.05 M H SO used to extract the quinine, the concentration of quinine in the urine also is 2.81 μg/mL. The recovery of the ingested quinine is \[\frac{\frac{2.81 \ \mu \mathrm{g} \text { quinine }}{\mathrm{mL} \text { urine }} \times 2.00 \ \mathrm{mL} \text { urine } \times \frac{1 \mathrm{mg}}{1000 \ \mu \mathrm{g}}} {10.0 \ \mathrm{mg} \text { quinine ingested }} \times 100=0.0562 \% \nonumber\] It can take 10–11 days for the body to completely excrete quinine so it is not surprising that such a small amount of quinine is recovered from this sample of urine. Photoluminescence spectroscopy is used for the routine analysis of trace and ultratrace analytes in macro and meso samples. Detection limits for fluorescence spectroscopy are influenced by the analyte’s quantum yield. For an analyte with \(\Phi_f > 0.5\), a picomolar detection limit is possible when using a high quality spectrofluorometer. For example, the detection limit for quinine sulfate, for which \(\Phi\) is 0.55, generally is between 1 part per billion and 1 part per trillion. Detection limits for phosphorescence are somewhat higher, with typical values in the nanomolar range for low-temperature phosphorimetry and in the micromolar range for room-temperature phosphorimetry using a solid substrate. The accuracy of a fluorescence method generally is between 1–5% when spectral and chemical interferences are insignificant. Accuracy is limited by the same types of problems that affect other optical spectroscopic methods. In addition, accuracy is affected by interferences that affect the fluorescent quantum yield. The accuracy of phosphorescence is somewhat greater than that for fluorescence. The relative standard deviation for fluorescence usually is between 0.5–2% when the analyte’s concentration is well above its detection limit. Precision usually is limited by the stability of the excitation source. The precision for phosphorescence often is limited by reproducibility in preparing samples for analysis, with relative standard deviations of 5–10% being common. From Equation \ref{10.4} and Equation \ref{10.5} we know that the sensitivity of a fluorescent or a phosphorescent method is affected by a number of parameters. We already have considered the importance of quantum yield and the effect of temperature and solution composition on \(\Phi_f\) and \(\Phi_p\). Besides quantum yield, sensitivity is improved by using an excitation source that has a greater emission intensity, , at the desired wavelength, and by selecting an excitation wavelength for which the analyte has a greater molar absorptivity, \(\varepsilon\). Another approach for improving sensitivity is to increase the volume from which emission is monitored. Figure 10.6.10
shows how rotating a monochromator’s slits from their usual vertical orientation to a horizontal orientation increases the sampling volume. The result can in- crease the emission from the sample by \(5-30 \times\). The selectivity of fluorescence and phosphorescence is superior to that of absorption spectrophotometry for two reasons: first, not every compound that absorbs radiation is fluorescent or phosphorescent; and, second, selectivity between an analyte and an interferent is possible if there is a difference in either their excitation or their emission spectra. The total emission intensity is a linear sum of that from each fluorescent or phosphorescent species. The analysis of a sample that contains analytes, therefore, is accomplished by measuring the total emission intensity at wavelengths. As with other optical spectroscopic methods, fluorescent and phosphorescent methods provide a rapid means for analyzing samples and are capable of automation. Fluorimeters are relatively inexpensive, ranging from several hundred to several thousand dollars, and often are satisfactory for quantitative work. Spectrofluorometers are more expensive, with models often exceeding $50,000. | 28,271 | 3,554 |
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Chemistry is a quantitative science. Amounts of substances and energies must always be expressed in numbers and units (in order to make some sense of what you are talking about). You should also develop a sensation about quantities every time you encounter them; you should be familiar with the name, prefix, and symbol used for various quantities. However, due to the many different units we use, expression of quantities is rather complicated. We will deal with the number part of quantities on this page, using SI Units. By now, you probably realized that every time the number increases by a factor of a thousand, we give a new name, a new prefix, and a new symbol in its expression. After you are familiar with the words associated with these numbers, you should be able to communicate numbers with ease. Consider the following number: 123,456,789,101,234,567 In words, this 18-digit number takes up a few lines: One hundred twenty three quindrillions, four hundred fifty six trillions, seven hundred eighty nine billions, one hundred and one millions, two hundred and thirty four thousands, five hundred and sixty seven. If a quantity makes use of this number, the quantity has been measured precisely. Most quantities do not have a precise measurement to warrant so many significant figures. The above number may often be expressed as 123e15 or read as one hundred twenty three quindrillions. *The unit ampere, A, is equal to Coulombs per second, (A = C/s). | 1,485 | 3,555 |
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The most definitive set of organic nomenclature rules currently in use were evolved through several international conferences and are known as the (IUPAC rules). We first shall describe this system for naming the hydrocarbons known as - the so-called saturated hydrocarbons that have no double or triple bonds, or rings, and conform to the general formula \(C_nH_{2n+2}\). The alkanes are classified as "continuous chain" (that is, " ") if all the carbon atoms in the chain are linked to no more than two other carbons; or " chain" if one or more carbon atoms are linked to more than two other carbons: The first four continuous-chain hydrocarbons have nonsystematic names: The higher members, beginning with pentane, are named systematically with a numerical prefix (pent-, hex-, hept-, etc., to denote the number of carbon atoms) and with the ending to classify the compound as a paraffin hydrocarbon, as in Table 3-1. To specify a continuous-chain hydrocarbon, the prefix (for normal) sometimes is used. However, in the absence of any qualifying prefix, the hydrocarbon is considered to be "normal" or unbranched and we shall not use this prefix henceforth. You should memorize the names up to \(C_{10}H_{22}\). The possibility of having branched-chain hydrocarbons that are structural isomers of the continuous-chain hydrocarbons begins with butane (\(n = 4\)). The IUPAC rules for the systematic naming of these hydrocarbons follow. 1. The continuous chain of carbon atoms is taken as the parent hydrocarbon and is the framework on which the various substituent groups are attached. Thus the hydrocarbon \(1\) is a substituted pentane rather than a substituted butane because the longest chain has five carbons: 2. The substituent groups attached to the main chain are named by replacing the ending of the alkane by . We then have the (or , the simplest examples being the methyl (\(CH_3-\)) and ethyl (\(CH_3CH_3-\)) groups. 3. The parent hydrocarbon then is numbered starting from the end of the chain, and the substituent groups are assigned numbers corresponding to their positions on the chain. The direction of numbering is chosen to give the lowest numbers to the side-chain substituents.\(^2\) Thus hydrocarbon \(1\) is 2,3-dimethylpentane rather than 3,4-dimethylpentane. The prefix di- signifies that there are identical substituents: The prefixes used to designate the number of substituents follow up to ten. Mono- is not used to designate a single substituent in systematic nomenclature, but may be used in conversation for emphasis. 4. Where there are two identical substituents at one position as in \(2\), numbers are supplied for each, and the prefix, di-, tri-, and so on, is included to signify the number of groups of the same kind: 5. Branched-chain substituent groups are given appropriate names by a simple extension of the system used for branched-chain hydrocarbons. The longest chain of the substituent is numbered . Parentheses are used to separate the numbering of the substituent and of the main hydrocarbon chain: Additional examples of alkyl substituents and their names are listed in Table 3-2. These are further classified according to whether they are primary, secondary, or tertiary. An alkyl group is described as if the carbon at the point of attachment is bonded to only other carbon, as if bonded to other carbons, and if bonded to other carbons. Thus if \(R\) is any hydrocarbon radical, the different kinds of alkyl groups are Confusion can arise here because we often refer to specific carbons rather than whole alkyl groups as , , and so on. In this context, a carbon is if it is bonded to other carbon, if bonded to , if bonded to , and if bonded to . Thus, either carbon of ethane is primary, the \(C2\) carbon in propane, \(CH_3CH_2CH_3\), is secondary, and the \(C2\) carbon of 2,2-dimethylpropane, \((CH_3)_4C\), is quaternary. The situation with regard to naming alkyl substituents has been muddied considerably by the fact that the IUPAC rules allow use of trivial names for a few alkyl groups. Thus -butyl sometimes is used in place of 1-methylpropyl, and -butyl in place of 1,1-dimethylethyl These and other examples are included in parentheses in Table 3-2. Further odd-ball but less official customs are the prefix , which is reserved for substituents with two methyl groups at the end of an otherwise straight chain (e.g., isopropyl), and the prefix , which is used to denote three methyl groups at the end of a chain (e.g., neopentyl, which is more properly called 2,2-dimethylpropyl). Also in common use are the names and for the hydrocarbons 2-methyl-propane and 2,2-dimethylpropane, respectively There is no ambiguity involved in the use of and prefixes here, but the practice of using the name "isooctane" for 2,2,4-trimethylpentane is erroneous. Fortunately, use of these special names is declining. 6. When there are two or more different substituents present, the question arises as to what order they should be cited in naming the compound. The system adopted by IUPAC and long practiced by cites them in alphabetical order without regard for whether there is a multiplying prefix such as di- or tri-. Examples are given below. When a hydrocarbon is substituted with other than alkyl groups a new problem arises, which can be illustrated by \(CH_3CH_2Cl\) This substance can be called either chloroethane or ethyl chloride, and both names are used in conversation and in print almost interchangeably. In the IUPAC system, halogens, nitro groups, and a few other monovalent groups are considered to be and are named as haloalkanes, nitroalkanes, and so on. The alphabetical order of precedence is preferred for substituents of different types when two or more are attached to a hydrocarbon chain because this makes indexing and using indexes more straightforward: The rest of this chapter is devoted to names of compounds we will not discuss for several chapters ahead and you may wish to stop at this point and return later as necessary. However, you should test your knowledge of alkane nomenclature by doing Exercises 3-1, 3-2, 3-3, 3-9, 3-10, and 3-11. \(^2\)Confusion is possible when the numbering from each end is similar. The rule is that when the series of substituent locants are compared term by term, the "lowest" series has the lowest number at the point of difference. The compound is 2,3,5-trimethylhexane, 2,4,5-trimethylhexane. and (1977) | 6,509 | 3,556 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/04%3A_Alkanes/4.07%3A_Nitration_of_Alkanes |
Another reaction of commercial importance is the nitration of alkanes to give nitroparaffins. Such reactions usually are carried out in the vapor phase at elevated temperatures using nitric acid (\(HNO_3\)) or nitrogen tetroxide (\(N_2O_4\)) as the nitrating agent: All available evidence points to a radical mechanism for nitration, but many aspects of the reaction are not fully understood. Mixtures are obtained; nitration of propane gives not only 1- and 2-nitropropanes but nitroethane and nitromethane: In commercial practice, the yield and product distribution in nitration of alkanes is controlled as far as possible by the judicious addition of catalysts (e.g., oxygen and halogens), which are believed to raise the concentration of alkyl radicals. The products are separated from the mixtures by . and (1977) | 838 | 3,558 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electrochemistry/Electrodes/Standard_Hydrogen_Electrode |
An electrode by definition is a point where current enters and leaves the electrolyte. When the current leaves the electrodes it is known as the cathode and when the current enters it is known as the anode. Electrodes are vital components of electrochemical cells. They transport produced electrons from one half-cell to another, which produce an electrical charge. This charge is based off a standard electrode system (SHE) with a reference potential of 0 volts and serves as a medium for any cell potential calculation. An electrode is a metal whose surface serves as the location where oxidation-reduction equilibrium is established between the metal and what is in the solution. The electrode can either be an anode or a cathode. An anode receives current or electrons from the electrolyte mixture, thus becoming oxidized. When the atoms or molecules get close enough to the surface of the electrode, the solution in which the electrode is placed into, donates electrons. This causes the atoms/molecules to become positive ions. An electrode has to be a good electrical conductor so it is usually a metal. Now what this metal is made out of is dependent on whether or not it is involved in the reaction. Some reactions require an inert electrode that does not participate. An example of this would be platinum in the SHE reaction(described later). While other reactions utilize solid forms of the reactants, making them the electrodes. An example of this type of cell would be: Here you can see that a solid form of the reactant, copper, is used. The copper, as well as the silver, are participating as reactants and electrodes. Some commonly used inert electrodes: graphite(carbon), platinum, gold, and rhodium. Some commonly used reactive (or involved) electrodes: copper, zinc, lead, and silver. A Standard Hydrogen Electrode (SHE) is an electrode that scientists use for reference on all half-cell potential reactions. The value of the standard electrode potential is zero, which forms the basis one needs to calculate cell potentials using different electrodes or different concentrations. It is important to have this common reference electrode just as it is important for the International Bureau of Weights and Measures to keep a sealed piece of metal that is used to reference the S.I. Kilogram. SHE is composed of a 1.0 M H (aq) solution containing a square piece of platinized platinum (connected to a platinum wire where electrons can be exchanged) inside a tube. During the reaction, hydrogen gas is then passed through the tube and into the solution causing the reaction: 2H (aq) + 2e <==> H (g). Platinum is used because it is inert and does not react much with hydrogen. First an initial discharge allows electrons to fill into the highest occupied energy level of Pt. As this is done, some of the H+ ions form H O ions with the water molecules in the solution. These hydrogen and hydronium ions then get close enough to the Pt electrode (on the platinized surface of this electrode) to where a hydrogen is attracted to the electrons in the metal and forms a hydrogen atom. Then these combine with other hydrogen atoms to create H2(g). This hydrogen gas is released from the system. In order to keep the reaction going, the electrode requires a constant flow of H (g). The Pt wire is connected to a similar electrode in which the opposite process is occurring, thus producing a charge that is referenced at 0 volts. Other standard electrodes are usually preferred because the SHE can be a difficult electrode to set up. The difficulty arises in the preparation of the platinized surface and in controlling the concentration of the reactants. For this reason the SHE is referred to as a hypothetical electrode. The three electrode system is made up of the working electrode, reference electrode, and the auxiliary electrode. The three electrode system is important in voltammetry. All three of these electrodes serve a unique roll in the three electrode system. A reference electrode refers to an electrode that has an established electrode potential. In an electrochemical cell, the reference electrode can be used as a half cell. When the reference electrode acts as a half cell, the other half cell's electrode potential can be discovered. An auxiliary electrode is an electrode makes sure that current does not pass through the reference cell. It makes sure the current is equal to that of the working electrode's current. The working electrode is the electrode that transports electrons to and from the substances that are present. Some examples of reference cells include: This reference electrode consists of a mercury and mercury-chloride molecules. This electrode can be relatively easier to make and maintain compared to the SHE. It is composed of a solid paste of Hg Cl and liquid elemental mercury attached to a rod that is immersed in a saturated KCl solution. It is necessary to have the solution saturated because this allows for the activity to be fixed by the potassium chloride and the voltage to be lower and closer to the SHE. This saturated solution allows for the exchange of chlorine ions to take place. All this is usually placed inside a tube that has a porous salt bridge to allow the electrons to flow back through and complete the circuit. \[\dfrac{1}{2} Hg_2Cl_{2(s)}+e- \rightleftharpoons Hg_{(l)}+Cl^-_{(aq)}\] : An electrode of this sort precipitates a salt in the solution that participates in the electrode reaction. This electrode consists, of solid silver and its precipitated salt AgCl. This a widely used reference electrode because it is inexpensive and not as toxic as the Calomel electrode that contains mercury. A Silver-Silver Chloride electrode is made by taking a wire of solid silver and coding it in AgCl. Then it is placed in a tube of KCl and AgCl solution. This allows ions to be formed (and the opposite) as electrons flow in and out of the electrode system. \[AgCl_{(s)}+e^- \rightleftharpoons Ag^+_{(aq)}+Cl^-_{(aq)}\] 1. Which electrode oxidizes the solution in the half-cell? Anode or Cathode? 2. Why is the Standard Hydrogen Electrode important to calculating cell potentials? 3. Identify the which side is the cathode and which side is the anode. Ag(s) | Ag+(aq)(.5M) || Ag (aq) (.05M) | Ag(s) 4. Why is it important to use an inert electrode in situations like the SHE? 5. What is the standard half cell potential for the SHE? Answers (highlight to see): | 6,455 | 3,559 |
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