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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(Fleming)/04%3A_Putting_the_First_Law_to_Work/4.06%3A_Useful_Definitions_and_Relationships |
In this chapter (and in the previous chapter), several useful definitions have been stated. The following “measurable quantities” have been defined: The following relation has been derived: \[ \dfrac{ \alpha}{\kappa_T} = \left( \dfrac{\partial p}{\partial T} \right)_V \nonumber \] And the following relationships were given without proof (yet!): \[\left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial p}{\partial T} \right)_V - p \nonumber \] and \[\left( \dfrac{\partial H}{\partial p} \right)_T = - T \left( \dfrac{\partial V}{\partial T} \right)_p - p \nonumber \] Together, these relationships and definitions make a powerful set of tools that can be used to derive a number of very useful expressions. Derive an expression for \(\left( \dfrac{\partial H}{\partial V} \right)_T\) in terms of measurable quantities. Begin by using the total differential of \(H(p, T)\): \[ dH = \left( \dfrac{\partial H}{\partial p} \right)_T dp + \left( \dfrac{\partial H}{\partial T} \right)_p dT \nonumber \] Divide by \(dV\) and constrain to constant \(T\) (to generate the partial of interest on the left): \[\left.\dfrac{dH}{dV} \right\rvert_{T}= \left( \dfrac{\partial H}{\partial p} \right)_T \left.\dfrac{dp}{dV} \right\rvert_{T} + \cancelto{0}{\left( \dfrac{\partial H}{\partial T} \right)_p \left.\dfrac{dT}{dV} \right\rvert_{T}} \nonumber \] The last term on the right will vanish (since \(dT = 0\) for constant \(T\)). After converting to partial derivatives \[ \left(\dfrac{\partial H}{\partial V} \right)_{T} = \left( \dfrac{\partial H}{\partial p} \right)_T \left(\dfrac{\partial p}{\partial V} \right)_{T} \label{eq5} \] This result is simply a demonstration of the ! But now we are getting somewhere. We can now substitute for \(\left(\dfrac{\partial H}{\partial V} \right)_{T}\) using our “toolbox of useful relationships”: \[ \left(\dfrac{\partial H}{\partial V} \right)_{T} = \left[ -T \left(\dfrac{\partial V}{\partial T} \right)_{p} +V \right] \left(\dfrac{\partial p}{\partial V} \right)_{T} \nonumber \] Using the , this expression becomes \[ \left(\dfrac{\partial H}{\partial V} \right)_{T} = -T \left(\dfrac{\partial V}{\partial T} \right)_{p}\left(\dfrac{\partial p}{\partial V} \right)_{T} + V \left(\dfrac{\partial p}{\partial V} \right)_{T} \label{eq7} \] Using the (Transformation Type II), the middle term of Equation \ref{eq7} can be simplified \[ \left(\dfrac{\partial H}{\partial V} \right)_{T} = T \left(\dfrac{\partial p}{\partial T} \right)_{V} + V \left(\dfrac{\partial p}{\partial V} \right)_{T} \nonumber \] And now all of the partial derivatives on the right can be expressed in terms of \(\alpha\) and \(\kappa_T\) (along with \(T\) and \(V\), which are also “measurable properties”. \[ \left(\dfrac{\partial H}{\partial V} \right)_{T} = T \dfrac{\alpha}{\kappa_T} + V \dfrac{1}{-V \kappa_T} \nonumber \] or \[ \left(\dfrac{\partial H}{\partial V} \right)_{T} = \dfrac{1}{\kappa_T} ( T \alpha -1) \nonumber \] Calculate \(\Delta H\) for the isothermal compression of ethanol which will decrease the molar volume by \(0.010\, L/mol\) at 300 K. (For ethanol, \(\alpha = 1.1 \times 10^{-3 }K^{-1}\) and \(\kappa_T = 7.9 \times 10^{-5} atm^{-1}\)). Integrating the total differential of \(H\) at constant temperature results in \[ \Delta H = \left(\dfrac{\partial H}{\partial V} \right)_{T} \Delta V \nonumber \] From Example \(\Page {1}\), we know that \[ \Delta H = \left [ \dfrac{1}{ 7.9 \times 10^{-5} atm^{-1}} \left( (300 \,K) (1.1 \times 10^{-3 }K^{-1}) -1 \right) \right] ( - 0.010\, L/mol ) \nonumber \] \[ \Delta H = \left( 84.81 \, \dfrac{\cancel{atm\,L}}{mol}\right) \underbrace{\left(\dfrac{8.314\,J}{0.8206\, \cancel{atm\,L}}\right)}_{\text{conversion factor}} = 9590 \, J/mol \nonumber \] | 3,787 | 1,927 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Real_(Non-Ideal)_Systems/Salting_Out |
Salting out is a purification method that utilizes the reduced solubility of certain molecules in a solution of very high . Salting out is typically, but not limited to, the precipitation of large biomolecules such as proteins. In contrast to , salting out occurs in aqueous solutions of high ionic strength that reduce the molecule's solubility causing certain proteins to precipitate. Ideally, the type of salt being used and the concentration of the salt can be varied to selectively precipitate a the molecule. In reality, salting out is an effective means for initial molecule purification, but lacks the ability for precise isolation of a specific protein. The conformation of large biomolecules is typically controlled by hydrophobic and hydrophilic interactions with the cellular environment. These interactions largely govern the molecule's final conformation by folding in such a way that most hydrophobic functional groups are shielded from the polar cellular environment. To achieve this conformation the molecule folds in such a way that all of the hydrophobic parts of a molecule are aggregated together and the hydrophilic groups are left to interact with the water. In the case of proteins it is the charged amino acids that allow selective salting out to occur. Charged and polar amino acids such as glutamate, lysine, and tyrosine require water molecules to surround them to remain dissolved. In an aqueous environment with a high ionic strength, the water molecules surround the charges of the ions and proteins. At a certain ionic strength, the water molecules are no longer able to support the charges of both the ions and the proteins. The result is the precipitation of the least soluble solute, such as proteins and large organic molecules. Salting out can be a powerful tool to separate classes of proteins that vary in size, charge, and surface area among other characteristics. One method of controlling the precipitation is the utilize the different effects of various salts and their respective concentrations. A salt's ability to induce selective precipitation is dependent on many interactions with the water and solutes. Research by Franz Hofmeister in the early 20th century organized various anions and cations by their ability to salt out. The ordering of cations and anions is called the Hoffmeister Series (1). The cations are arranged as follows \[\ce{NH4^{+}> K^{+}> Na^{+} >Li^{+} >Mg^{2+} >Ca^{2+}} \nonumber\] where ammonium has the highest ability to precipitate other proteinaceous solutes. Likewise, the order for anions is \[\ce{F^{-} ≥ SO4^{2-}> H2PO4^{-}> H3CCOO^{-}> Cl^{-}> NO3^{-}> Br^{-}> ClO3^{-}> I^{-}>ClO^{-}} \nonumber\] Between cations and anions in solution the concentration of the anion typically has the greatest effect on protein precipitation. One of the most commonly used salts is ammonium sulfate, which is typically used because the ions produced in an aqueous solution are very high on the series, and their interaction with the protein itself is relatively low. Other ions such as iodide are very good at precipitating proteins, but are not used due to their propensity to denature or modify the protein. Salting out relies on changes in solubility based on ionic strength. The ionic strength of a solution, I, is defined as \[I =\dfrac{1}{2} \sum_i \ m_i {z_i}^{2} \label{1}\] where Total ionic strength of multiple ions is the sum of the ionic strengths of all of the ions. Using the limiting law given by \[ \log \gamma_\pm = -\dfrac{1.824 \times 10^6} { \left( \epsilon T \right)^{3/2}} | z_+ z_- | \sqrt I \label{2}\] where which can be adapted for for an aqueous solution at 298 K, \[ \log \gamma_\pm = - 0.509 | z_+ z_- | \sqrt I \label{3}\] the solubility, \(S\), of a particular solute can be defined as \[ \log \dfrac{S}{S_o}\ = - 0.509 | z_+ z_- | \sqrt I -K' I \label{4}\] where A common salt used in protein precipitation is ammonium sulfate, calculate the ionic strength of a 4g being added to 1230 ml 0.1M NaCl. 0.191M Which of the following polypeptides would likely precipitate first at pH of 4: AAVKI or DDEKVK Although Ionic strength matters, one cannot forget that normal solubility rules still hold and the polypeptide AAVKI would likely precipitate first given its almost complete non-polar nature. Since protein precipitation is dependent on which salt is used, which of the following salts would precipitate protein at the lowest concentration of the salt solution? c Protein Y was just discovered by scientists at a national lab. The scientists managed to purify the protein with some precipitate in their flask. However, their yields were very low, to solve their problem to extracted the rest of the protein from the solution. They added 58 g of NaCl to 1 L of their protein solution in order to salt out the protein Y. After addition of the NaCl, they noticed that the solution no longer had some of the previously precipitated protein. What is the reason for the disappearance of the precipitate? The ionic strength of the solution after addition of 58g of NaCl was about 1. In this case the ionic strength was in the region where salting in occurs. Hence the disappearance of the precipitate. | 5,218 | 1,928 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Crystal_Lattices/Thermodynamics_of_Lattices/Lattice_Energy%3A_The_Born-Haber_cycle |
Ionic solids tend to be very stable compounds. The enthalpies of formation of the ionic molecules cannot alone account for this stability. These compounds have an additional stability due to the lattice energy of the solid structure. However, lattice energy cannot be directly measured. The Born-Haber cycle allows us to understand and determine the lattice energies of ionic solids. is a type of potential energy that may be defined in two ways. In one definition, the lattice energy is the energy required to break apart an ionic solid and convert its component atoms into gaseous ions. This definition causes the value for the lattice energy to always be positive, since this will always be an endothermic reaction. The other definition says that lattice energy is the reverse process, meaning it is the energy released when gaseous ions bind to form an ionic solid. As implied in the definition, this process will always be exothermic, and thus the value for lattice energy will be negative. Its values are usually expressed with the units kJ/mol. Lattice Energy is used to explain the stability of ionic solids. Some might expect such an ordered structure to be less stable because the entropy of the system would be low. However, the crystalline structure allows each ion to interact with multiple oppositely charge ions, which causes a highly favorable change in the enthalpy of the system. A lot of energy is released as the oppositely charged ions interact. It is this that causes ionic solids to have such high melting and boiling points. Some require such high temperatures that they decompose before they can reach a melting and/or boiling point. There are several important concept to understand before the Born-Haber Cycle can be applied to determine the lattice energy of an ionic solid; ionization energy, electron affinity, dissociation energy, sublimation energy, heat of formation, and Hess's Law. The values used in the Born-Haber Cycle are all predetermined changes in enthalpy for the processes described in the section above. Hess' Law allows us to add or subtract these values, which allows us to determine the lattice energy. Determine the energy of the metal and nonmetal in their elemental forms. (Elements in their natural state have an energy level of zero.) Subtract from this the heat of formation of the ionic solid that would be formed from combining these elements in the appropriate ration. This is the energy of the ionic solid, and will be used at the end of the process to determine the lattice energy. The Born-Haber Cycle requires that the elements involved in the reaction are in their gaseous forms. Add the changes in enthalpy to turn one of the elements into its gaseous state, and then do the same for the other element. Metals exist in nature as single atoms and thus no dissociation energy needs to be added for this element. However, many nonmetals will exist as polyatomic species. For example, Cl exists as Cl in its elemental state. The energy required to change Cl into 2Cl atoms must be added to the value obtained in Step 2. Both the metal and nonmetal now need to be changed into their ionic forms, as they would exist in the ionic solid. To do this, the ionization energy of the metal will be added to the value from Step 3. Next, the of the nonmetal will be subtracted from the previous value. It is subtracted because it is a release of energy associated with the addition of an electron. *This is a common error due to confusion caused by the definition of electron affinity, so be careful when doing this calculation. Now the metal and nonmetal will be combined to form the ionic solid. This will cause a release of energy, which is called the lattice energy. The value for the lattice energy is the difference between the value from Step 1 and the value from Step 4. -------------------------------------------------------------------------------------------------------------------------------------------- The diagram below is another representation of the Born-Haber Cycle. The Born-Haber Cycle can be reduced to a single equation: Heat of formation= Heat of atomization+ Dissociation energy+ (sum of Ionization energies)+ (sum of Electron affinities)+ Lattice energy *Note: In this general equation, the electron affinity is added. However, when plugging in a value, determine whether energy is released (exothermic reaction) or absorbed (endothermic reaction) for each electron affinity. If energy is released, put a negative sign in front of the value; if energy is absorbed, the value should be positive. Rearrangement to solve for lattice energy gives the equation: Lattice energy= Heat of formation- Heat of atomization- Dissociation energy- (sum of Ionization energies)- (sum of Electron Affinities) | 4,794 | 1,929 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Book%3A_Thermodynamics_and_Chemical_Equilibrium_(Ellgen)/02%3A_Gas_Laws/2.08%3A_Real_Gases_Versus_Ideal_Gases |
Now, we need to expand on the qualifications with which we begin this chapter. We imagine that the results of a large number of experiments are available for our analysis. Our characterization of these results has been that all gases obey the same equations—Boyle’s law, Charles’ law, and the ideal gas equation—and do so exactly. This is an oversimplification. In fact they are always approximations. They are approximately true for all gases under all “reasonable” conditions, but they are not exactly true for any real gas under any condition. It is useful to introduce the idea of hypothetical gases that obey the classical gas equations exactly. In the previous section, we call the combination of Boyle’s law and Charles’ law the ideal gas equation. We call the hypothetical substances that obey this equation . Sometimes we refer to the classical gas laws collectively as the . At very high gas densities, the classical gas laws can be very poor approximations. As we have noted, they are better approximations the lower the density of the gas. In fact, experiments show that the pressure—volume—temperature behavior of any real gasreal gas becomes arbitrarily close to that predicted by the ideal gas equation in the limit as the pressure goes to zero. This is an important observation that we use extensively. At any given pressure and temperature, the ideal gas laws are better approximations for a compound that has a lower boiling point than they are for a compound with a higher boiling point. Another way of saying this is that they are better approximations for molecules that are weakly attracted to one another than they are for molecules that are strongly attracted to one another. Forces between molecules cause them to both attract and repel one another. The net effect depends on the distance between them. If we assume that there are no intermolecular forcesintermolecular forces acting between gas molecules, we can develop exact theories for the behavior of macroscopic amounts of the gas. In particular, we can show that such substances obey the ideal gas equation. (We shall see that a complete absence of repulsive forces implies that the molecules behave as point masses.) Evidently, the difference between the behavior of a real gas and the behavior it would exhibit if it were an ideal gas is just a measure of the effects of intermolecular forces. The ideal gas equation is not the only equation that gives a useful representation for the interrelation of gas pressure–volume–temperature data. There are many such . They are all approximations, but each can be a particularly useful approximation in particular circumstances. We discuss equation and the later in this chapter. Nevertheless, we use the ideal gas equation extensively. We will see that much of chemical thermodynamics is based on the behavior of ideal gases. Since there are no ideal gases, this may seem odd, at best. If there are no ideal gases, why do we waste time talking about them? After all, we don’t want to slog through tedious, long-winded, pointless digressions. We want to understand how real stuff behaves! Unfortunately, this is more difficult. The charm of ideal gases is that we can understand their behavior; the ideal gas equation expresses this understanding in a mathematical model. Real gases are another story. We can reasonably say that we can best understand the behavior of a real gas by understanding how and why it is different from the behavior of a (hypothetical) ideal gas that has the same molecular structure. | 3,557 | 1,930 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/01%3A_Introduction_to_Organic_Chemistry/1.03%3A_What_Preparation_Should_You_Have |
We have tried to give you a taste of the beginnings of organic chemistry and a few of the important principles that brought order out of the confusion that existed as to the nature of organic compounds. Before moving on to other matters, it may be well to give you some ideas of what kind of preparation will be helpful to you in learning about organic chemistry from this textbook. The most important thing you can bring is a strong desire to master the subject. We hope you already have some knowledge of general chemistry and that you already will have had experience with simple inorganic compounds. That you will know, for example, that elemental bromine is \(Br_2\) and a noxious, dark red-brown, corrosive liquid; that sulfuric acid is \(H_2SO_4\), a syrupy colorless liquid that reacts with water with the evolution of considerable heat and is a strong acid; that sodium hydroxide is \(NaOH\), a colorless solid that dissolves in water to give a strongly alkaline solution. It is important to know the characteristics of acids and bases, how to write simple, balanced chemical reactions, such as \(2H_2 + O_2 \rightarrow 2H_2O\), and \(2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O\), what the concept of a mole of a chemical substance is, and to be somewhat familiar with the periodic table of the elements as well as with the metric system, at least insofar as grams, liters, and degrees centigrade are concerned. Among other things, you also should understand the basic ideas of the differences between salts and covalent compounds, as well as between gases, liquids, and solids; what a solution is; the laws of conservation of mass and energy; the elements of how to derive the Lewis electron structures of simple molecules such as \(H : \underset{\cdot \cdot}{\ddot{O}} : H =\) water; that \(PV = nRT\); and how to calculate molecular formulas from percentage compositions and molecular weights. We shall use no mathematics more advanced than simple algebra but we do expect that you can use logarithms and are able to carry through the following conversions forward and backward: \(\text{log}_{10} \: 510,000 = \: \text{log}_{10} \left( 5.1 \times 10^5 \right) = 5.708\) The above is an incomplete list, given to illustrate the level of preparation we are presuming in this text. If you find very much of this list partly or wholly unfamiliar, you don't have to give up, but have a good general chemistry textbook available for study and reference - and use it! Some useful general chemistry books are listed at the end of the chapter. A four-place table of logarithms will be necessary; a set of ball-and-stick models and a chemical handbook will be very helpful, as would be a small electronic calculator or slide rule to carry out the simple arithmetic required for many of the exercises. In the next section, we review some general chemistry regarding saltlike and covalent compounds that will be of special relevance to our later discussions. and (1977) | 2,991 | 1,931 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Addition_Reactions/Nucleophilic_Addition_Reactions/The_Reduction_of_Aldehydes_and_Ketones |
This page gives you the facts and mechanisms for the reduction of carbonyl compounds (specifically aldehydes and ketones) using sodium tetrahydridoborate (sodium borohydride) as the reducing agent. Sodium tetrahydridoborate (previously known as sodium borohydride) has the formula NaBH , and contains the BH ion. That ion acts as the reducing agent. There are several quite different ways of carrying out this reaction. Two possible variants (there are several others!) are: In each case, reduction essentially involves the addition of a hydrogen atom to each end of the carbon-oxygen double bond to form an alcohol. Reduction of aldehydes and ketones lead to two different sorts of alcohol. For example, with ethanal you get ethanol: Notice that this is a simplified equation - perfectly acceptable to examiners. The H in square brackets means "hydrogen from a reducing agent". In general terms, reduction of an aldehyde leads to a primary alcohol. A primary alcohol is one which only has one alkyl group attached to the carbon with the -OH group on it. They all contain the grouping -CH OH. For example, with propanone you get propan-2-ol: Reduction of a ketone leads to a secondary alcohol. A secondary alcohol is one which has two alkyl groups attached to the carbon with the -OH group on it. They all contain the grouping -CHOH. The BH ion is essentially a source of hydride ions, H . The simplification used is to write H instead of BH . Doing this not only makes the initial attack easier to write, but avoids you getting involved with some quite complicated boron compounds that are formed as intermediates. The reduction is an example of nucleophilic addition. The carbon-oxygen double bond is highly polar, and the slightly positive carbon atom is attacked by the hydride ion acting as a nucleophile. A hydride ion is a hydrogen atom with an extra electron - hence the lone pair. In the first stage, there is a nucleophilic attack by the hydride ion on the slightly positive carbon atom. The lone pair of electrons on the hydride ion forms a bond with the carbon, and the electrons in one of the carbon-oxygen bonds are repelled entirely onto the oxygen, giving it a negative charge. What happens now depends on whether you add an acid or water to complete the reaction. When the acid is added, the negative ion formed picks up a hydrogen ion to give an alcohol. This time, the negative ion takes a hydrogen ion from a water molecule. As before, the reaction starts with a nucleophilic attack by the hydride ion on the slightly positive carbon atom. Again, what happens next depends on whether you add an acid or water to complete the reaction. The negative ion reacts with a hydrogen ion from the acid added in the second stage of the reaction. This time, the negative ion takes a hydrogen ion from a water molecule. Jim Clark ( ) | 2,856 | 1,932 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.03%3A_The_Atomic_Theory |
The development of the atomic theory owes much to the work of two men: Antoine Lavoisier, who did not himself think of matter in terms of atoms but whose work laid organization groundwork for thinking about elements, and John Dalton, to whom the atomic theory is attributed. Much of Lavoisier’s work as a chemist was devoted to the study of combustion. He became convinced that when a substance is burned in air, it combines with some component of the air. Eventually he realized that this component was the which had been discovered by Joseph Priestly (1733 to 1804) a few years earlier. Lavoisier renamed this substance . In an important series of experiments he showed that when mercury is heated in oxygen at a moderate temperature, a red substance, , is obtained. (A is the ash left when a substance burns in air.) At a higher temperature this calx decomposes into mercury and oxygen. Lavoisier’s careful experiments also revealed that the combined masses of mercury and oxygen were exactly equal to the mass of calx of mercury. That is, there was no change in mass upon formation or decomposition of the calx. Lavoisier hypothesized that this should be true of all chemical changes, and further experiments showed that he was right. This principle is now called the . As Lavoisier continued his experiments with oxygen, he noticed something else. Although oxygen combined with many other substances, it never behaved as though it were itself a combination of other substances. Lavoisier was able to decompose the red calx into mercury and oxygen, but he could find no way to break down oxygen into two or more new substances. Because of this he suggested that oxygen must be an —an ultimately simple substance which could not be decomposed by chemical changes. Lavoisier did not originate the idea that certain substances (elements) were fundamental and all others could be derived from them. This had first been proposed in Greece during the fifth century B.C. by Empedocles, who speculated that all matter consisted of combinations of earth, air, fire, and water. These ideas were further developed and taught by Aristotle and remained influential for 2000 years. Lavoisier did produce the first table of the elements which contained a large number of substances that modern chemists would agree should be classifies as elements. He published it with the knowledge that further research might succeed decomposing some of the substances listed, thus showing them not to be elements. One of his objectives was to prod his contemporaries into just that kind of research. Sure enough the “earth substances” listed at the bottom were eventually shown to be combinations of certain metals with oxygen. It is also interesting to note that not even Lavoisier could entirely escape from Aristotle’s influence. The second element in his list is Aristotle’s “fire,” which Lavoisier called “caloric,” and which we now call “heat.” Both heat and light, the first two items in the table, are now regarded as forms of energy rather than of matter. Although his table of elements was incomplete, and even incorrect in some instances, Lavoisier’s work represented a major step forward. By classifying certain substances as elements, he stimulated much additional chemical research and brought order and structure to the subject where none had existed before. His contemporaries accepted his ideas very readily, and he became known as the father of chemistry. John Dalton (1766 to 1844) was a generation younger than Lavoisier and different from him in almost every respect. Dalton came from a working class family and only attended elementary school. Apart from this, he was entirely self-taught. Even after he became famous, he never aspired beyond a modest bachelor’s existence in which he supported himself by teaching mathematics to private pupils. Dalton made many contributions to science, and he seems not to have realized that his atomic theory was the most important of them. In his “New System of Chemical Philosophy” published in 1808, only the last seven pages out of a total of 168 are devoted to it! The postulates of the atomic theory are given below. The first is no advance on the ancient Greek philosopher Democritus who had theorized almost 2000 years earlier that matter consists of very small particles. The second postulate, however, shows the mark of an original genius; here Dalton links the idea of to the idea of . Lavoisier’s criterion for an element had been essentially a macroscopic, experimental one. If a substance could not be decomposed chemically, then it was probably an element. By contrast, Dalton defines an element in theoretical, sub-microscopic terms. . Different elements have different atoms. There are just as many different kinds of elements as there are different kinds of atoms. Now look back a moment to the physical states of mercury, where sub-microscopic pictures of solid, liquid, and gaseous mercury were given. Applying Dalton’s second postulate to this figure, you can immediately conclude that mercury is an element, because only one kind of atom appears. Although mercury atoms are drawn as spheres in the figure, it would be more common today to represent them using chemical symbols. The chemical symbol for an element (or an atom of that element) is a one- or two-letter abbreviation of its name. Usually, but not always, the first two letters are used. To complicate matters further, chemical symbols are sometimes derived from a language other than English. For example the symbol for Hg for mercury comes from the first and seventh letters of the element’s Latin name, The chemical symbols for all the currently known elements are listed above in the table, which also includes atomic weights. These symbols are the basic vocabulary of chemistry because the atoms they represent make up all matter. You will see symbols for the more important elements over and over again, and the sooner you know what element they stand for, the easier it will be for you to learn chemistry. These more important element have been indicated in the above table by colored shading around their names. Dalton’s fourth postulate states that atoms may combine to form molecules. An example of this is provided by bromine, the only element other than mercury which is a liquid at ordinary room temperature (20°C). Macroscopically, bromine consists of dark-colored crystals below –7.2°C and a reddish brown gas above 58.8°C. The liquid is dark red-brown and has a pungent odor similar to the chlorine used in swimming pools. It can cause severe burns on human skin and should not be handled without the protection of rubber gloves. Figure \(\Page {1}\) Sub-microscopic view of the diatomic molecules of the element bromine (a) in the gaseous state (above 58°C); (b) in liquid form (between -7.2 and 58.8°C); and (c) in solid form (below -7.2°C). The sub-microscopic view of bromine in the following figure is in agreement with its designation as an element—only one kind of atom is present. Except at very high temperatures, though, bromine atoms always double up. Whether in solid, liquid, or gas, they go around in pairs. Such a tightly held combination of two or more atoms is called a . The composition of a molecule is indicated by a . A subscript to the right of the symbol for each element tells how many atoms of that element are in the molecule. For example, the atomic weights table gives the chemical Br for bromine, but each molecule contains two bromine atoms, and so the chemical is Br . According to Dalton’s fourth postulate, atoms combine in the ratio of small whole numbers, and so the subscripts in a formula should be small whole numbers. | 7,722 | 1,933 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/01%3A_Introduction_to_Organic_Chemistry/1.05%3A_The_Breadth_of_Organic_Chemistry |
Organic chemistry originally was defined as the chemistry of those substances formed by living matter and, for quite a while, there was a firm belief that it would never be possible to prepare organic compounds in the laboratory outside of a living system. However, after the discovery by Wohler, in 1828, that a supposedly typical organic compound, urea, could be prepared by heating an inorganic salt, ammonium cyanate, this definition gradually lost significance and organic chemistry now is broadly defined as the chemistry of carbon-containing compounds. Nonetheless, the designation "organic" is still very pertinent because the chemistry of organic compounds is also the chemistry of living organisms. Each of us and every other living organism is comprised of, and endlessly manufactures, organic compounds. Further, all organisms consume organic compounds as raw materials, except for those plants that use photosynthesis or related processes to synthesize their own from carbon dioxide. To understand every important aspect of this chemistry, be it the details of photosynthesis, digestion, reproduction, muscle action, memory or even the thought process itself, is a primary goal of science and it should be recognized that only through application of organic chemistry will this goal be achieved. Modern civilization consumes vast quantities of organic compounds. Coal, petroleum, and natural gas are primary sources of carbon compounds for use in production of energy and as starting materials for the preparation of plastics, synthetic fibers, dyes, agricultural chemicals, pesticides, fertilizers, detergents, rubbers and other elastomers, paints and other surface coatings, medicines and drugs, perfumes and flavors, antioxidants and other preservatives, as well as asphalts, lubricants, and solvents that are derived from petroleum. Much has been done and you soon may infer from the breadth of the material that we will cover that most everything worth doing already has been done. However, many unsolved scientific problems remain and others have not even been thought of but, in addition, there are many technical and social problems to which answers are badly needed. Some of these include problems of pollution of the environment, energy sources, overpopulation and food production, insect control, medicine, drug action, and improved utilization of natural resources. and (1977) | 2,420 | 1,934 |
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The calculated energy of the electron-pair bond of the hydrogen molecule as a function of \(\ce{H-H}\) intermolecular distance \(r\) by the (exact), MO, and VB procedures is shown in Figure 21-11. The results show that neither the MO nor the VB calculations come close to the calculation in reproducing the experimental dissociation energy, \(D_e\), or the variation of the energy with the intermolecular distance. The VB method gives a little better energy value at the minimum and the MO method gives poor results at larger values of \(r\). We can say that, as calculated by the MO method, the molecule does not "dissociate properly". Within the calculations of the MO method, the molecule does not "dissociate properly". Why do these calculations yield results so far from the curve? There are two reasons. First, atomic orbitals are used that are appropriate for atoms, but are hardly expected to be the best orbitals for the electrons when two or more atoms are in close proximity. It is convenient to use atomic orbitals in simple calculations because they are mathematically simple, but more complicated orbitals are known to give better results. Second, neither treatment properly takes into account electron-electron repulsions. For two electrons, a term of the form \(\frac{e^2}{r^2_{12}}\) (in which \(e\) is the electronic charge and \(r_{12}\) is the distance between the electrons) is required to describe the repulsion between electrons. The exact calculations avoid both difficulties but are so complex mathematically as to be devoid of any capability for providing qualitative understanding. The VB method gives a slightly lower energy than the MO method at the minimum, because in the simple MO method, when we calculate the energy resulting from two electrons going into the lowest molecular orbital, we put no restraints on their being close together. As a result there is a \(50\%\) probability for electrons being simultaneously in either half of the molecular orbital. In contrast, the simple VB method combines configurations \(1\) and \(2\), each having just electron per atomic orbital, and no account is taken of the possibility of either atomic orbital containing more than one electron. This is equivalent to neglecting the pairing schemes \(\ce{H}^\ominus \ce{H}^\oplus \leftrightarrow \ce{H}^\oplus \ce{H}^\ominus\). Neither the VB nor the MO approximation is the best possible; the simple MO method tends to take too little account of interelectronic repulsion, whereas the VB method tends to take too much account of it. However, as can be seen in Figure 21-11, taking too much account of electron repulsion is the better approximation. Why does an electron-pair bond calculated by the MO method not dissociate properly? We have seen that half of the time both electrons in the low-energy molecular orbital are in the vicinity of just of the nuclei. But as the nuclei move , this corresponds to a far greater energy than having only one electron in the vicinity of each nucleus, as the VB method suggests. There is no unequivocal answer to the question as to which is the better method. Calculations by the VB method are likely to be more reliable than those by the MO method, but in practice are much more difficult to carry out. For many-electron molecules the MO procedure is simpler to visualize because we combine atomic orbitals into molecular orbitals and then populate the lower-energy orbitals with electrons. In the VB method, atomic orbitals are occupied, but the electrons of different atoms are paired to form bonds, a process that requires explicit consideration of many-electron wave functions. To put it another way, it is easier to visualize a system of molecular orbitals containing \(N\) electrons than it is to visualize a hybrid wave function of \(N\) electrons. How can the MO and VB methods be improved? The answer depends on what one wants - more accurate calculations or better qualitative understanding. To improve VB calculations we need orbitals that allow the electrons to spread out over more than one atom. The suit this purpose and give an energy curve only slightly above the exact curve of Figure 21-11. In the GVB treatment the orbitals delocalize less as \(r\) increases. When atomic orbitals are derived for each carbon of the \(\pi\)-electron system of benzene by the GVB method, they are somewhat more spread out than simple carbon \(p\) orbitals (Figure 21-12). Use of these orbitals in VB calculations gives excellent results with just the two pairing schemes of benzene, \(9\) and \(10\). Improvement of the MO method involves better orbitals, better account of interelectronic repulsion, and introduction of mixing of different electron configurations in the molecular orbitals (" "). Improved MO calculations give much more accurate energies at the minimum of a plot such as Figure 21-11, but the bonds still do not dissociate properly, for the same reason as with the simple MO method. We cannot say that either the VB or the MO method is more ; only that one approximation may be more useful than the other in attempting to solve a particular problem. The fact is, the more each is refined, the more they appear to merge into a common procedure; but, unfortunately, in the refinement process the mathematics become so complex that qualitative understanding of what is being done tends to disappear altogether. We cannot say that either the VB or the MO method is more ; only that one approximation may be more useful than the other in attempting to solve a particular problem. and (1977) | 5,604 | 1,935 |
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The development of the atomic theory owes much to the work of two men: Antoine Lavoisier, who did not himself think of matter in terms of atoms but whose work laid organization groundwork for thinking about elements, and John Dalton, to whom the atomic theory is attributed. Much of Lavoisier’s work as a chemist was devoted to the study of combustion. He became convinced that when a substance is burned in air, it combines with some component of the air. Eventually he realized that this component was the which had been discovered by Joseph Priestly (1733 to 1804) a few years earlier. Lavoisier renamed this substance . In an important series of experiments he showed that when mercury is heated in oxygen at a moderate temperature, a red substance, , is obtained. (A is the ash left when a substance burns in air.) At a higher temperature this calx decomposes into mercury and oxygen. Lavoisier’s careful experiments also revealed that the combined masses of mercury and oxygen were exactly equal to the mass of calx of mercury. That is, there was no change in mass upon formation or decomposition of the calx. Lavoisier hypothesized that this should be true of all chemical changes, and further experiments showed that he was right. This principle is now called the . As Lavoisier continued his experiments with oxygen, he noticed something else. Although oxygen combined with many other substances, it never behaved as though it were itself a combination of other substances. Lavoisier was able to decompose the red calx into mercury and oxygen, but he could find no way to break down oxygen into two or more new substances. Because of this he suggested that oxygen must be an —an ultimately simple substance which could not be decomposed by chemical changes. Lavoisier did not originate the idea that certain substances (elements) were fundamental and all others could be derived from them. This had first been proposed in Greece during the fifth century B.C. by Empedocles, who speculated that all matter consisted of combinations of earth, air, fire, and water. These ideas were further developed and taught by Aristotle and remained influential for 2000 years. Lavoisier did produce the first table of the elements which contained a large number of substances that modern chemists would agree should be classifies as elements. He published it with the knowledge that further research might succeed decomposing some of the substances listed, thus showing them not to be elements. One of his objectives was to prod his contemporaries into just that kind of research. Sure enough the “earth substances” listed at the bottom were eventually shown to be combinations of certain metals with oxygen. It is also interesting to note that not even Lavoisier could entirely escape from Aristotle’s influence. The second element in his list is Aristotle’s “fire,” which Lavoisier called “caloric,” and which we now call “heat.” Both heat and light, the first two items in the table, are now regarded as forms of energy rather than of matter. Although his table of elements was incomplete, and even incorrect in some instances, Lavoisier’s work represented a major step forward. By classifying certain substances as elements, he stimulated much additional chemical research and brought order and structure to the subject where none had existed before. His contemporaries accepted his ideas very readily, and he became known as the father of chemistry. John Dalton (1766 to 1844) was a generation younger than Lavoisier and different from him in almost every respect. Dalton came from a working class family and only attended elementary school. Apart from this, he was entirely self-taught. Even after he became famous, he never aspired beyond a modest bachelor’s existence in which he supported himself by teaching mathematics to private pupils. Dalton made many contributions to science, and he seems not to have realized that his atomic theory was the most important of them. In his “New System of Chemical Philosophy” published in 1808, only the last seven pages out of a total of 168 are devoted to it! The postulates of the atomic theory are given below. The first is no advance on the ancient Greek philosopher Democritus who had theorized almost 2000 years earlier that matter consists of very small particles. The second postulate, however, shows the mark of an original genius; here Dalton links the idea of to the idea of . Lavoisier’s criterion for an element had been essentially a macroscopic, experimental one. If a substance could not be decomposed chemically, then it was probably an element. By contrast, Dalton defines an element in theoretical, sub-microscopic terms. . Different elements have different atoms. There are just as many different kinds of elements as there are different kinds of atoms. Now look back a moment to the physical states of mercury, where sub-microscopic pictures of solid, liquid, and gaseous mercury were given. Applying Dalton’s second postulate to this figure, you can immediately conclude that mercury is an element, because only one kind of atom appears. Although mercury atoms are drawn as spheres in the figure, it would be more common today to represent them using chemical symbols. The chemical symbol for an element (or an atom of that element) is a one- or two-letter abbreviation of its name. Usually, but not always, the first two letters are used. To complicate matters further, chemical symbols are sometimes derived from a language other than English. For example the symbol for Hg for mercury comes from the first and seventh letters of the element’s Latin name, The chemical symbols for all the currently known elements are listed above in the table, which also includes atomic weights. These symbols are the basic vocabulary of chemistry because the atoms they represent make up all matter. You will see symbols for the more important elements over and over again, and the sooner you know what element they stand for, the easier it will be for you to learn chemistry. These more important element have been indicated in the above table by colored shading around their names. Dalton’s fourth postulate states that atoms may combine to form molecules. An example of this is provided by bromine, the only element other than mercury which is a liquid at ordinary room temperature (20°C). Macroscopically, bromine consists of dark-colored crystals below –7.2°C and a reddish brown gas above 58.8°C. The liquid is dark red-brown and has a pungent odor similar to the chlorine used in swimming pools. It can cause severe burns on human skin and should not be handled without the protection of rubber gloves. Figure \(\Page {1}\) Sub-microscopic view of the diatomic molecules of the element bromine (a) in the gaseous state (above 58°C); (b) in liquid form (between -7.2 and 58.8°C); and (c) in solid form (below -7.2°C). The sub-microscopic view of bromine in the following figure is in agreement with its designation as an element—only one kind of atom is present. Except at very high temperatures, though, bromine atoms always double up. Whether in solid, liquid, or gas, they go around in pairs. Such a tightly held combination of two or more atoms is called a . The composition of a molecule is indicated by a . A subscript to the right of the symbol for each element tells how many atoms of that element are in the molecule. For example, the atomic weights table gives the chemical Br for bromine, but each molecule contains two bromine atoms, and so the chemical is Br . According to Dalton’s fourth postulate, atoms combine in the ratio of small whole numbers, and so the subscripts in a formula should be small whole numbers. | 7,722 | 1,936 |
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In the attempt to measure the heat equivalent of mechanical work (as Joule did in his famous experiment) a student uses an apparatus similar to that shown below: The 1.50 kg weight is lifted 30.0 cm against the force due to gravity (9.8 N). If the specific heat of water is 4.184 J/(g °C), what is the expected temperature increase of the 1.5 kg of water in the canister? 1.00 mol of an ideal gas, initially occupying 12.2 L at 298 K, expands isothermally against a constant external pressure of 1.00 atm until the pressure of the gas is equal to the external pressure. Calculate \(\Delta p\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the expansion. Consider 1.00 mol of an ideal gas expanding isothermally at 298 K from an initial volume of 12.2 L to a final volume of 22.4 L. Calculate \(\Delta p\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the expansion. Consider 1.00 mol of an ideal gas (C = 3/2 R) Occupying 22.4 L that undergoes an isochoric (constant volume) temperature increase from 298 K to 342 K. Calculate \(\Delta p\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the change. Consider 1.00 mol of an ideal gas (C = 5/2 R) initially at 1.00 atm that undergoes an isobaric expansion from 12.2 L to 22.4 L. Calculate \(\Delta T\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the change. Consider 1.00 mol of an ideal gas (C = 3/2 R) initially at 12.2 L that undergoes an adiabatic expansion to 22.4 L. Calculate \(\Delta T\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the change. Derive an expression for the work of an isothermal, reversible expansion of a gas that follows the equation of state (in which \(a\) is a parameter of the gas) \[ pV = nRT -\dfrac{an^2}{V}\] from \(V_1\) to \(V_2\). Use the following data [Huff, Squitieri, and Snyder, J. Am. Chem. Soc., , 3380 (1948)] to calculate the standard enthalpy of formation of tungsten carbide, \(WC(s)\). The standard molar enthalpy of combustion (\(\Delta H_c\)) of propane gas is given by \[C_3H_8(g) + 5 O_2(g) \rightarrow 3 CO_2(g) + 4 H_2O(l)\] with \(\Delta H_c = -2220 \,kJ/mol\) The standard molar enthalpy of vaporization (\(\Delta H_{vap}\)) for liquid propane \[C_3H_8(l) \rightarrow C_3H_8(g)\] with \(\Delta H_{vap} = 15\, kJ/mol\) The enthalpy of combustion (\(\Delta H_c\)) of aluminum borohydride, \(Al(BH_4)_3(l)\), was measured to be -4138.4 kJ/mol [Rulon and Mason, , , 5491 (1951)]. The combustion reaction for this compound is given by \[ Al(BH_4)_3(l) + 6 O_2(g) \rightarrow ½ Al_2O_3(s) + 3/2 B_2O_3(s) + 6 H_2O(l)\] Given the following additional data, calculate the enthalpy of formation of \(Al(BH_4)_3(g)\). The standard enthalpy of formation (\(\Delta H_f^o\)) for water vapor is -241.82 kJ/mol at 25 °C. Use the data in the following table to calculate the value at 100 °C. \(\Delta C_p = (1.00 + 2.00 \times 10^{-3} T)\, J/K\) and \(\Delta H_{298} = -5.00\, kJ\) for a dimerization reaction \[2 A \rightarrow A_2\] Find the temperature at which \(\ H = 0\). From the following data, determine the lattice energy of \(BaBr_2\). \[Ca(s) \rightarrow Ca(g)\] with \(\Delta H_{sub} = 129\, kJ/mol\) \[Br_2(l) \rightarrow Br_2(g)\] with \(\Delta H_{vap} = 31\, kJ/mol\) \[Br_2(g) \rightarrow 2 Br(g)\] with \(D(Br-Br) = 193 \, kJ/mol\) \[Ca(g) \rightarrow Ca^+(g) + e^-\] with \(1^{st} \, IP(K) = 589.8 \, kJ/mol\) \[Ca^+(g) \rightarrow Ca^{2+}(g) + e^-\] with \(2^{nd} IP(K) = 1145.4 \,kJ/mol\) \[Br(g) + e^- \rightarrow Br-(g) \] with \(1^{st} EA(Br) = 194 \, kJ/mol\) \[Ca(s) + Br_2^-(l) \rightarrow CaBr_2(s)\] with \(\Delta H_f = -675 \, kJ/mol\) Using average bond energies ( ) estimate the reaction enthalpy for the reaction \[C_2H_4 + HBr \rightarrow C_2H_5Br\] | 3,716 | 1,937 |
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What is the entropy changed when 10.2 moles of diatomic gas is cooled from a temperature of 25 °C to 10 °C at constant pressure? Determine the value of while reversibly heating 5 moles of an ideal gas from 25 °C to 73 °C at constant volume. At constant volume, If 1.50 moles of an ideal gas were compressed isothermally from 6.6 L to 2.5 L: Imagine you are to carry out the isothermal and reversible gas expansion with an ideal gas. You have 0.5 moles of the gas at 280K. You wish to expand this gas from its initial volume of 2 liters to a new volume of 8 liters. What would your values for heat, work, change in entropy, and change in internal energy have to be? What if you wanted to instead carry his experiment out as isothermal and irreversible with 2.0 atm of external pressure? Consider 0.50 moles of an ideal gas at 30 C. It expands from a pressure of 4.5 atm to a pressure of 2.0 atm at the same temperature. (\dfrac{q}{T}\)=(\dfrac{-W}{T}\)=(\dfrac{-1}{T}*(-nRT\ln\dfrac{V2}{V1})\)=(\nR\ln\dfrac{P1}{P}\) : delta S of system is 3.37 J/K Delta S(surrouding)=(\dfrac{-1}{T}*P2*V2*(1-\dfrac{V1}{V2})\)= (\-nR(1-\dfrac{P2}{P1})\) delta S of surrounding = -1/T * P2*V2*(1 - V1/V2) = -nR(1 - P2/P1) =-0.5 * 8.3145 *(1-2/4.5)=2.31 J/K delta of the universe is : 3.37 J/K +2.31J/K=5.68 J/K Calculate ΔS for the heating of 2.00 moles of nitrogen from 25 C to 200 C. The heat capacity of oxygen is: C = (3.268 + 0.00325T) J K mol \[\Delta S = \int n\dfrac{\bar{C_{P}}}{T}dT\] \[\Delta S = \int_{(25+273)K}^{(200+273)K} (2 mol)\dfrac{(3.268 + 0.00325T)J \ mol^{-1} \ K^{-1}}{T}dT\] \[\Delta S = \int_{298K}^{473K} (\dfrac{6.536}{T}+ 0.0065)J \ mol^{-1}dT\] \[\Delta S = {(6.536}\ln(T)+ 0.0065T)_{298}^{473}J \ mol^{-1}\] \[\Delta S = 4.16 J \ mol^{-1}\] The value of (CO ) is less than that of (CO ) at 298 K. (Note: Refer to table below). Would this be the case at a temperature of 0 K as well? Since the gaseous form of a compound always has a greater enthalpy than the more ordered liquid or aqueous form, this would still be the case at a temperature of 0 K as well. Is the standard entropy of CO (g) higher than the standard entropy of CO (s) at 298 K? What about at 0 K? You have an ideal gas that expands 3.56 L with the initial volume at 22 L. All you know is that you have 0.45 moles and that the temperature is 22 ºC. What is the change in Gibbs free energy for this system? =167.3 0.70 moles of an ideal gas expands adiabatically from 1.0 atm to 2.5 atm at a temperature of 30 C. Calculate the values of q, w, U, S, and G. \(q=0\) \(w=-nRT\ln\dfrac{P_1}{P_2}=-(.70mol)(8.314\dfrac{J}{mol\cdot K})(303K)(\ln\dfrac{1.0atm}{2.5atm})(\dfrac{1kJ}{1000kJ}=1.62\dfrac{kJ}{mol}\) \(\bigtriangleup U=w=1.62\dfrac{kJ}{mol}\) \(\bigtriangleup S_{univ}=\bigtriangleup S_{sys}=nR\ln\dfrac{P_1}{P_2}=(.70mol)(8.314\dfrac{J}{mol\cdot K})\ln\dfrac{1.0atm}{2.5atm}=-5.33\dfrac{J}{mol}\) \(dH=VdP\rightarrow \dfrac{dH}{dP}= nRT\ln\dfrac{P_2}{P_1}=(.70mol)(8.314\dfrac{J}{mol\cdot K}(303K)(\ln\dfrac{2.5atm}{1.0atm}(\dfrac{1kJ}{1000J}\) \(=1.62\dfrac{kJ}{mol}\) \(\bigtriangleup G= \bigtriangleup H- T\bigtriangleup S= 1.62 \times 10^3\dfrac{J}{mol}-(303K)(-5.33\dfrac{J}{mol/cdot K}=3234\dfrac{J}{mol}\) \(=3.23\dfrac{kJ}{mol}\) Use the values listed listed below to calculate the value of G° for the following reaction: pyruvate(aq) 2 lactate ion(aq) + 2H (aq) (Δ G°[pyruvate(aq)] = -472.4 kJ mol , Δ G°[lactate ion] = -516.7 kJ mol ). Is pyruvate or the lactate ion favored under standard conditions? Explain. Use the values listed listed below to calculate the value of G° for the following reaction: pyruvate(aq) → 2 lactate ion(aq) + 2H (aq) (Δ G°[pyruvate(aq)] = -472.4 kJ mol , Δ G°[lactate ion] = -516.7 kJ mol ) A scientist measures studies the thermodynamics of a protein of interest. He interested in finding the Free Energy of unfolding in the quaternary structure of protein at a concentration of 5µM. He finds that the Gibbs free energy of the reaction -0.84 Kcal/mol. Why does the scientist want to know the value of (\Delta G\) Explain how the reaction proceeds. The value of (\Delta G\) of the reaction will let the scientist know how likely the reaction of the unfolding of the protein is to occur. If (\Delta G\) is negative the reaction will go forward, towards the unfolded state. In this case (\Delta G\) is negative so the reaction goes forward to the unfolding of the protein. For the reaction \[H_{2 (g)} + Cl_{2 (g)} \rightarrow 2HCl_{(g)}\] Using the formula Consider the conversion of graphite carbon into diamond \[C_{(graphite)}\rightarrow C_{(diamond)}\] (a) $$\Delta rH^{o}=\Delta _f\bar{H}^{o}[C_{(diamond)}]-\Delta _f\bar{H}^{o}[C_{(graphite)}]\] $$=1.90Kjmol^{-1}-0kjmol^{-1}\] $$=1.90kj\,mol^{-1}\] $$\Delta rS^{o}=\Delta S^{o}[C_{(diamond)}]-\Delta S^{o}[C_{(graphite)}]\] $$=2.4J\,K^{-1}mol^{-1}-5.7JK^{-1}mol^{-1}\] $$=-3.3J\,K^{-1}mol^{-1}\] At \(0^oC\), $$\Delta rG^{o}=\Delta rH^o-(273K)\Delta rS^{o}\] $$=(1.90kj\,mol^{-1})-(273K)(-3.3\times10^{-3}kj\,K^{-1}mol^{-1})\] $$=2.8834kj\,mol^{-1}\] (b) Step 1: $$G_2=G_1+V(P_2-P_1)\] Step 2: (Hint: Apply molar quantities) $$\bar{G}_2=\bar{G}_1+\bar{V}\Delta P\] Step 3: (Hint: Apply the equation to diamond and graphite) $$\bar{G}_2[Graphite]=\bar{G}_1[graphite]+\bar{V}[graphite]\Delta P\] $$\bar{G}_2[diamond]=\bar{G}_1[diamond]+\bar{V}[diamond]\Delta P\] Step 4: (Hint: Combine the equations) $$\Delta rG_2=\Delta rG_1+[\bar{V}[diamond]-\bar{V}[graphite]]\Delta P\] Step 5: (Hint: At \(25^oC,P_1=1bar\; so\; \Delta rG_1=\Delta rG^o=2.883kj\,mol^{-1}\)) $$\Delta rG_2=2.883kj\,mol^{-1}+(-2.1cm^3mol^{-1})\Delta P(\dfrac{1L}{1000cm^{3}})(\dfrac{1atm}{1.01 bar})(\dfrac{101.3J}{1L\,atm})(\dfrac{1kj}{1000J})\] $$\Delta rG_2=2.883kj\,mol^{-1}-2.1\times10^{-4}\Delta P\,kj\,bar^{-1}mol^{-1}\] Step 6: (Hint: To make the process spontaneous) $$\Delta rG_2=2.883kj\,mol^{-1}-2.1\times10^{-4}\Delta P\,kj\,bar^{-1}mol^{-1}\lt0\] $$2.883kj\,mol^{-1}\lt2.1\times10^{-4}\Delta P\,kj\,bar^{-1}mol^{-1}\] $$\Delta P\lt\dfrac{2.883kj\,mol^{-1}}{2.1\times10^{-4}}=1.4\times10^{4} bar\] Predict the signs of \(\Delta H,\Delta S\) and \(\Delta G\) of the system for the following processes at 1atm: \(\Delta H\) is positive for all because melting is endothermic. \(\Delta S\) is positive for all because going from solid to liquid is becoming more disordered. \(\Delta G\) for (a) is positive because reaction is not spontaneous when melting at below the melting point, \(\Delta G\) for (b) is 0 because reaction is in equilibrium at the melting point, \(\Delta G\) for (c) is negative because reaction is spontaneous when melting at above the melting point. Crystals of AgCl(s) form spontaneously when aqueous solutions of silver(I)nitrate and sodium chloride are combined. What can you deduce about the signs of the changes in enthalpy and entropy? The formation reaction to form magnesium oxide is as follow: \[ \ce{Mg(s) + 1/2 O_2 (g) \rightarrow MgO(s) }\] Verify the equation ΔG° = Δ \[\Delta_{r}\bar{G}^{\circ} = \sum \Delta_f\bar{G}^{\circ}[products] - \sum \Delta_f\bar{G}^{\circ}[reactants] = -569.5 \dfrac{kJ}{mol} -0-0 = {\color{red}-569.5 \dfrac{kJ}{mol}}\] \[\Delta_{r}\bar{H}^{\circ} = \sum \Delta_f\bar{H}^{\circ}[products] - \sum \Delta_f\bar{H}^{\circ}[reactants] = -601.8 \dfrac{kJ}{mol} -0-0 = -601.8 \dfrac{kJ}{mol} \] \[\Delta_{r}\bar{S}^{\circ} = \sum \Delta_{f}\bar{S}^{\circ} [products] - \sum \Delta_{f}\bar{S}^{\circ} [reactants] = (26.78 - 32.68 - \dfrac{205.0}{2})\dfrac{J}{mol \cdot K} = -108.4 \] \[\Delta_{r}\bar{H}^{\circ} - T\Delta_{r}\bar{S}^{\circ} = -601.8 \dfrac{kJ}{mol} - (298K)(-0.1084 \dfrac{kJ}{mol \cdot K}) = {\color{red}-569.5 \dfrac{kJ}{mol}}\] At , a particular reaction is spontaneous. What would be the smallest possible value of for the reaction if Use the following equation \[ \Delta G = \Delta H - T \Delta S\] Since the process is spontaneous: Note: Don't forget to flip the inequality when dividing by a negative number. \[ \begin{align} \Delta G &< 0 \\ \Delta H - T \Delta S &< 0 \\ (10 \ kJ) - (322 \ K) \Delta S &< 0 \\ - (322 \ K) \Delta S &< 10 \ kJ \\ \Delta S &>0.031 \end{align}\] Thus the minimum value is the closest number above kJ K . for more information on Gibb's free energy. Calculate the minimum change in entropy of reaction for a spontaneous process at 340K with an enthalpy change of reaction equal to 800 J. Your answer must be in KJ/K and have 2 significant figures. The entropy change for a reaction is found to be 47.50 J/K. The enthalpy change for the reaction is 21.25 KJ. At what temperature is this reaction spontaneous? \[\Delta _{r}G=\Delta _{r}H-T\Delta _{r}S\] Because we want to find the temperature at equilibrium we set the Gibbs Free Energy equal to 0. Thus: \[\Delta _{r}H=T\Delta _{r}S\] \[T=\frac{\Delta _{r}H}{\Delta _{r}S}=\frac{21.25\times 10^{3}J}{47.50\frac{J}{K}}=447.4K\] The standard Gibbs energy change, ∆ G , for the reaction is +108 kJ. What can you deduce about the state of reaction when the reactants are mixed together? When the two gases are mixed, there is no interaction between them. So, the products would not form. The Gibbs' free energy is an indicator of a reaction's favorability, but does not indicate that that reaction will proceed quickly. If the rate constant for the reaction is low, or the concentration of one or more of the reagents is low, the reaction might not proceed at a pace that the scientist would observe over a few hours or even longer. If water changes from a liquid to a gas at 50 C and 0.123 bar, what must the pressure be to convert water from a liquid to a gas at 75 C? The molar enthalpy of vaporization for water (Δ H) is 42.3 kJ mol \[\ln \left( \dfrac{P_{2}}{P_{1}} \right) = \dfrac{\Delta \bar{H}_{vap}}{R} \left( \dfrac{1}{T_{1}}-\dfrac{1}{T_{2}} \right) \] \[T_1= \left( 50 + 273 \right) K = 323\,K\] \[T_2 = \left( 75 + 273 \right) K= 348\,K\] \[\ln \left( \dfrac{P_{2}}{0.123\,bar} \right) = \dfrac{42.3\times 10^3\,J mol^{-1} }{8.314 J \ mol^{-1} \ K^{-1}} \left( \dfrac{1}{323\,K}-\dfrac{1}{348\,K} \right) \] \[\ln \left( \dfrac{P_{2}}{0.123\,bar} \right) = 1.132\] \[\dfrac{P_{2}}{0.123\,bar} = 3.10\] \[P_{2}= 0.381 \,bar\] What is the change in pressure when ice melts ast room temperature if the density of ice is 0.919 g/ml and the Δ H = 6.01 KJ/mol? Calculate the freezing point depression of ice if a pressure of 450 atm is exerted on it by a car that weighs 1500-kg. Note: The molar volumes are as follows: molar volume of liquid water is 18.03 mL mol and the molar volume of ice is 19.65 mL mol . Use the slope of the S-L curve from the phase diagram: This means that the freezing point depression determined for when 450 atm of pressure is applied to ice by a car is 3.30 K. Here is a phase diagram for carbon. Look at the phase diagram of water. Which phase is more dense, liquid or solid? How can you tell by just looking at the graph? The link below contains a phase diagram of water: Through looking at the phase diagram, we can see that the liquid phase is more dense. You can see this because the line separating the two phases is negative. Explain what happens to Δ H as the temperature increases significantly. You go on an adventure and find yourself stranded on the top of a mountain where the air pressure is 0.68 atm. You find water but you want to make sure is safe to drink so you want to boil the water. At what temperature would you expect the water to start boiling? Hint: The molar heat of vaporization of H O is 41.0 kJ mol The equations are as follows Determine the conditions needed for each equation. Calculate the entropy change when nitrogen gas is heated from 25°C to 75°C while simultaneously allowed to expand from 0.562 L to 1.245 L. Assume ideal gas behavior. 1. Find how many moles of gas we have: \(PV=nRT\) \[n=\dfrac{PV}{RT}=\dfrac{1 atm*0.562L}{(0.08206L*atm/K*mol)(298.15K)}=0.02297mol \] 2. entropy change during expansion: \[Delta S=nR\ln{\dfrac{V_2}{V_1}}=(0.02297mol)(8.314J/K*mol)\ln{\dfrac{1.245L}{0.562L}}=0.1519J/K\] 3. entropy change during heating at constant volume: \[\begin{align*}\Delta S&=n\bar{C}_v\ln{\dfrac{T_2}{T_1}}=n(\dfrac{5}{2}R)\ln{\dfrac{T_2}{T_1}}\\
&=(0.02297mol)(\dfrac{5}{2})(8.314J/K*mol)\ln{\dfrac{348.15K}{283.15K}}=0.0987J/K\end{align*}\] 4. Combine 2&3: \[\Delta S_{tot}=\Delta S_{expansion}+\Delta S_{heating}=0.1519J/K+0.0987J/K=0.2506J/K\] A \(4 \times 10^{-2}\) moles sample of Xenon (Xe) is heated from 300 K to 400 K in a 1 L container at 1.5 atm. What will be the change in entropy if Xe is able to expand to twice the volume of the container? Photosynthesis depends upon the absorption of visible light. However, the infrared radiation can not be used for photosynthesis. Explain why. Whereas, for the visible light, e Photosynthesis is carried out by using the visible light. An amount of ideal diatomic gas is expanded from 4.8 atm to 1.7 atm while experiencing the change of temperature from 35°C to 128ºC. The change in entropy of the gas expansion process is 15.0 J/K. Determine the ΔU, ΔH, and ΔS of the entire process. Assume the temperature is not enough to unleash the rotational and vibrational energy of the gas. Hint: Break the problem into 2 processes (isothermic and isobaric) Step 1: Consider the isothermal process of the ideal gas (expanding from 4.8 atm to 1.7atm at constant temperature of 35°C ΔU = 0, ΔH = 0, ΔS = 15.0 J/K \[\Delta S = nRln\dfrac{V_2}{V_1} = nRln\dfrac{P_1}{P_2}\] \[n = \dfrac{\Delta S}{Rln\dfrac{P_1}{P_2}} = \dfrac{15.0 \dfrac{J}{K}}{( 8.314 \dfrac{J}{mol \cdot K})ln\dfrac{4.8 atm}{1.7 atm}} = 1.738 moles\] Step 2: Consider the change in temperature at constant pressure of 1.7 atm \[\Delta H = C_p \Delta T = \dfrac{7}{5}nR\Delta T = \dfrac{7}{5} (1.738 moles)(8.314\dfrac{J}{K\cdot mol})(128-35)K = 1.88 \times 10^{3} J\] \[\Delta U = \Delta H - nR\Delta T = (1.88 \times 10^{3} J ) -(1.738 moles)(8.314 \dfrac{J}{K \cdot mol})(128-35)K = 537 J\] \[\Delta S = C_p ln\dfrac{T_2}{T_1} = \dfrac{7}{5}nRln\dfrac{T_2}{T_1} = \dfrac{7}{5}(1.738 moles)(8.314 \dfrac{J}{K \cdot mol})ln\dfrac{(128+273)K}{(35+273)K} = 5.34 \dfrac{J}{K}\] For the entire process \[ \Delta H = 0 J + 1.88 \times 10^3 J = {\color{red}1.88 \times 10^3 J} \] \[ \Delta U = 0 J + 537 J = {\color{red}537 J} \] \[ \Delta S = 15.0 \dfrac{J}{K} + 5.34 \dfrac{J}{K} = {\color{red}20.34 \dfrac{J}{K}} \] Calculate the ΔU, ΔH, and ΔS of 3.00 moles of an ideal monatomic gas that are compressed from 7.00 L to 1.00 L while being heated from 25 C to 100 C. Break the process down into two steps: isothermal compression from 7 L to 1 L at 298 K and heating at a constant pressure from 298 K to 373 K. Because the compressions is isothermal, ΔU=0 & ΔH=0 for this step \[\Delta S = nR\ln(\dfrac{V_{2}}{V_{1}})=(3mol)(8.314J\ mol^{-1}\ K^{-1})\ln(\dfrac{1}{7})\] \[\Delta S = -48.5 J \ K^{-1}\] \[\Delta H = C_{P}\Delta T = \dfrac{5}{2}nR\Delta T=\dfrac{5}{2}(3 mol)(8.314 J\ mol^{-1}\ K^{-1})(373-298)K\] \[\Delta H = 4677 J\] \[\Delta U = \Delta H - nR\Delta T = 4677 J - (3mol)(8.314 J\ mol^{-1}\ K^{-1})(373-298)K\] \[\Delta U = 2806 J\] \[\Delta S = C_{P}\ln(\dfrac{T_{2}}{T_{1}}) = \dfrac{5}{2}nR\ln(\dfrac{T_{2}}{T_{1}})\] \[\Delta S = \dfrac{5}{2}(3mol)(8.314 J\ mol^{-1}\ K^{-1})\ln(\dfrac{373}{298}) = 14.0 J \ K^{-1}\] \[\Delta H = 0J + 4677 J= 4.68kJ\] \[\Delta U = 0J+ 2806 J=2.81kJ\] \[\Delta S = -48.5 J \ K^{-1}+14.0 J \ K^{-1} = -34.5 J \ K^{-1}\] Briefly provide examples and descriptions for each of the three laws of thermodynamics. The are often generally stated to be: With this knowledge, which of the three laws of thermodynamics apply to the examples below? There may be more than one. a.) A graduate student in UC davis claims to have created the world's first perpetual motion engine. This machine produces no heat and can be used, (he claims), to power the campus for the next decade! This is not possible. b.) If T becomes 0, entropy becomes undefined. Hence, it is not possible to reach absolute zero. c.) In an adiabatic gas expansion where: \[\Delta U=Q+w\] No thermal energy or matter is exchanged. Hence: \[\Delta U=w\] d.) In an imperfect crystal of CO, carbon monoxide, there are multiple energy states at 0K. Thus, it experiences residual entropy and, quite unusually, follows: \[\lim _{T\to0K}S>0\] a.) first and second laws; The existence of the perpetual motion machine violates the first and second laws of thermodynamics. Because \[\Delta U=Q+w\] and heat is a form of entropy, a perpetual motion machine does not increase the entropy of the universe. By definition, a machine which does work must do less work than it has internal energy. b.) second and third law; In order for absolute zero to be reached, the system would have to be removed from the universe. Because \[\Delta S=n\bar{C}_{p}ln\left ( \frac{T_{2}}{T_{1}} \right )\] and entropy must always increase, a system within the universe would always be at a temperature lower than its surroundings and hence could experience heat transfer which would mitigate the absolute zero condition. c.) first law; Energy which is "lost" from the system due to lack of heat transfer is compensated by the system doing more work. d.) third law; In a perfect crystal, where there is only one energy state, there is no residual entropy at 0K In a reversible adiabatic expansion, what do you expect to happen to the volume? What about the temperature? We expect both volume and temperature to increase. Aniline can hydrogen bond, whereas benzene is nonpolar. Explain why, contrary to our expectation, benzene melts at a higher temperature than aniline. Why is the boiling point of aniline higher than that of benzene? Molar Gibbs energies of solid, liquid and vapor depend on temperature and pressure. Briefly explain how temperature affects molar Gibbs energy. Explain how pressure changes affect molar Gibbs energy. Why is water an exemption? When ice melts to liquid water, entropy is increasing and the the surrounding becomes cooler since ice absorbs heat. Explain the relationship between those two facts. For any spontaneous endothermic process, what must be true about the entropy of the process? For the following situation take a rubber band and stretch it quickly. It will feel warm. Next stretch it out for a few seconds and then release it. Consequently, it will feel cool. Why do you think this happens? Explain thermodynamically. \[\Delta G =\Delta H-T\Delta S\] Stretching the rubber band creates a nonspontaneous process \((\Delta G\gt0)\) and \(-(\Delta G\lt 0)\). The warming effect means that \(\Delta H\lt0)\). This makes \(\Delta S\) negative. T must be positive and \(\Delta S\) must be negative. This tells us that the rubber band under tension is more disordered in its natural state. If theres no tension and the rubber band snaps back to its natural state, \(\Delta G\) is negative and \(-\Delta G\) is positive. Since it cools the \(\Delta H\gt0\) so \(T\Delta S\) is positive. Entropy increases when the rubber band goes from stretched state to natural state. Cold packs utilizes the chemical reaction of water and ammonium nitrate (or similar substances) to turn cold. Predict the signs of \(\Delta H, \Delta S\) and \(\Delta G\) for the reaction when the membrane that separates the two substances in a cold pack breaks. The reaction is spontaneous so \(\Delta G\) is negative; the cold pack feels cold because it's taking in heat so \(\Delta H\) is positive; \(\Delta S\) is positive because disorder increases as ammonium nitrate breaks down to ammonium ion and nitrate ion. At what temperature will toluene have a vapor pressure of 517 torr. The normal boiling point of toluene is 110.6°C with the molar enthalpy of vaporization being 35.2 kJ/mol. (Hint: Use equation). \[ ln\dfrac{P_2}{P_1} = \dfrac{\Delta _vap \bar H}{R} (\dfrac{1}{T_2} - \dfrac{1}{T_1}) \] \[ ln\dfrac{517 torr}{760 torr} = \dfrac{35.2 \times 10^3 \dfrac{J}{mol}}{8.314 \dfrac{J}{K \cdot mol}} (\dfrac{1}{T_2} - \dfrac{1}{(110.6+273)K}) \] \[ \Rightarrow {\color{red}T_2 = 397.5 K = 124^{\circ}C} \] You have discovered a way to create a new organic compound of the formula \(C_xH_yN_zO_α\). What measurements would you need to take to find out values for the following:properties: \[ \triangle_{f}\overline{H}^{\circ} = \ ? \ \ \triangle _{f}\overline{S}^{\circ} = \ ? \ \ \triangle _{f}\overline{G}^{\circ} = \ ?\] For enthalpy, you need to take measurements of the change of enthalpy of one mole of the unknown compound. Thus you would need to take measurements for the molar heat capacity and how much temperature changes in for formation of one mole. See and . For entropy of formation you would need to divide enthalpy by the temperature at which the formation takes place because the change in entropy can equate to enthalpy divided by temperature. See . For Gibb's free energy of formation, you simply need the enthalpy and entropy of formation along with the temperature where one mole of the substance is formed. See . \[ \Delta{G} = \Delta H - T \Delta S \] You are given hexane, C H , and a constant pressure calorimeter. What steps would you take to determine standard , for this substance? An equation for the complete combustion of hexane can be written as follows: \[C_{6}H_{14(l)}+O_{2(g)}\rightarrow CO_{2(l)}+H_{2}O_{(l)}\] The calorimeter would then be used to find the heat released by this reaction by the following rule: \[\Delta _{r}H=-q_{P, calorimeter}=-\left ( C_{calorimeter} \Delta T\right )\] after which the for hexane could be calculated using the known heat of formation for H O, O , and CO . Under the 3rd law of dynamics, the standard entropy would then be calculated by \[\Delta \bar{S}^{\circ}=-\frac{q_{rxn}}{T}\] where T=298K From this, standard entropy for the formation of hexane should be obtained from the known standard entropies of H O, O , and CO through the following reaction for the formation of hexane from elemental reactants: \[6C_{(graphite)}+7H_{2(g)}\rightarrow C_{6}H_{14(l)}\] The standard , for this substance is then determined through the relationship: \[\Delta _{f}\bar{G}^{\circ}=\Delta _{f}\bar{H}^{\circ}-T\Delta _{r}S^{\circ}\] \[\dfrac{PV}{/RT} = n\] | 22,236 | 1,938 |
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The pressure and temperature dependence of \(G\) is also easy to describe. The best starting place is the definition of \(G\). \[G = U + pV -TS \label{eq1} \] \[dG = dU + pdV – pdV + Vdp – TdS – SdT \nonumber \] The differential can be simplified by substituting the combined first and second law statement for \(dU\) (consider a reversible process and \(pV\) work only). \[dG = \cancel{TdS} \bcancel{– pdV} + \bcancel{pdV} + Vdp – \cancel{TdS} – SdT \nonumber \] Canceling the \(TdS\) and \(pdV\) terms leaves \[dG = V\,dp – S\,dT \label{Total1} \] This suggests that the natural variables of \(G\) are \(p\) and \(T\). So the total differential \(dG\) can also be expressed \[ dG = \left( \dfrac{\partial G}{\partial p} \right)_T dp + \left( \dfrac{\partial G}{\partial T} \right)_p dT \label{Total2} \] And by inspection of Equations \ref{Total1} and \ref{Total2}, it is clear that \[\left( \dfrac{\partial G}{\partial p} \right)_T = V \nonumber \] and \[ \left( \dfrac{\partial G}{\partial T} \right)_p = -S \nonumber \] It is also clear that the Maxwell relation on \(G\) is given by \[\left( \dfrac{\partial V}{\partial T} \right)_p = \left( \dfrac{\partial S}{\partial p} \right)_T \nonumber \] which is an extraordinarily useful relationship, since one of the terms is expressible entirely in terms of measurable quantities! \[\left( \dfrac{\partial V}{\partial T} \right)_p = V\alpha \nonumber \] The pressure dependence of \(G\) is given by the pressure derivative at constant temperature \[\left( \dfrac{\partial G}{\partial p} \right)_T = V \label{Max2} \] which is simply the molar volume. For a fairly incompressible substance (such as a liquid or a solid) the molar volume will be essentially constant over a modest pressure range. The density of gold is 19.32 g/cm . Calculate \(\Delta G\) for a 1.00 g sample of gold when the pressure on it is increased from 1.00 atm to 2.00 atm. The change in the Gibbs function due to an isothermal change in pressure can be expressed as \[ \Delta G =\int_{p_1}^{p_2} \left( \dfrac{\partial G}{\partial p} \right)_T dp \nonumber \] And since substituting Equation \ref{Max2}, results in \[ \Delta G =\int_{p_1}^{p_2} V dp \nonumber \] Assuming that the molar volume is independent or pressure over the stated pressure range, \(\Delta G\) becomes \[\Delta G = V(p_2-p_1) \nonumber \] So, the molar change in the Gibbs function can be calculated by substituting the relevant values. \[ \begin{align} \Delta G & = \left( \dfrac{197.0\, g}{mol} \times \dfrac{1\,}{19.32\,g} \times \dfrac{1\,L}{1000\,cm^3} \right) (2.00 \,atm -1.00 \,atm) \underbrace{ \left(\dfrac{8.315 \,J}{0.08206\, atm\,L}\right)}_{\text{conversion unit}}\\ &= 1.033\,J \end{align} \nonumber \] | 2,744 | 1,939 |
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Suspensions and colloids are two common types of mixtures whose properties are in many ways intermediate between those of true solutions and heterogeneous mixtures. A suspension is a heterogeneous mixture of particles with diameters of about 1 µm (1000 nm) that are distributed throughout a second phase. Common suspensions include paint, blood, and hot chocolate, which are solid particles in a liquid, and aerosol sprays, which are liquid particles in a gas. If the suspension is allowed to stand, the two phases will separate, which is why paints must be thoroughly stirred or shaken before use. A colloid is also a heterogeneous mixture, but the particles of a colloid are typically smaller than those of a suspension, generally in the range of 2 to about 500 nm in diameter. Colloids include fog and clouds (liquid particles in a gas), milk (solid particles in a liquid), and butter (solid particles in a solid). Other colloids are used industrially as catalysts. Unlike in a suspension, the particles in a colloid do not separate into two phases on standing. The only combination of substances that cannot produce a suspension or a colloid is a mixture of two gases because their particles are so small that they always form true solutions. The properties of suspensions, colloids, and solutions are summarized in Table 13.9. Properties of Liquid Solutions, Colloids, and Suspensions | 1,402 | 1,940 |
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The behavior of coordination compounds cannot be adequately explained by the same theories used for main group element chemistry. The observed geometries of coordination complexes are not consistent with hybridized orbitals on the central metal overlapping with ligand orbitals, as would be predicted by valence bond theory. The observed colors indicate that the orbitals often occur at different energy levels rather than all being degenerate, that is, of equal energy, as are the three orbitals. To explain the stabilities, structures, colors, and magnetic properties of transition metal complexes, a different bonding model has been developed. Just as valence bond theory explains many aspects of bonding in main group chemistry, crystal field theory is useful in understanding and predicting the behavior of transition metal complexes. To explain the observed behavior of transition metal complexes (such as how colors arise), a model involving electrostatic interactions between the electrons from the ligands and the electrons in the unhybridized orbitals of the central metal atom has been developed. This electrostatic model is (CFT). It allows us to understand, interpret, and predict the colors, magnetic behavior, and some structures of coordination compounds of transition metals. CFT focuses on the nonbonding electrons on the central metal ion in coordination complexes not on the metal-ligand bonds. Like valence bond theory, CFT tells only part of the story of the behavior of complexes. However, it tells the part that valence bond theory does not. In its pure form, CFT ignores any covalent bonding between ligands and metal ions. Both the ligand and the metal are treated as infinitesimally small point charges. All electrons are negative, so the electrons donated from the ligands will repel the electrons of the central metal. Let us consider the behavior of the electrons in the unhybridized orbitals in an octahedral complex. The five orbitals consist of lobe-shaped regions and are arranged in space, as shown in Figure \(\Page {1}\). In an octahedral complex, the six ligands coordinate along the axes. In an uncomplexed metal ion in the gas phase, the electrons are distributed among the five orbitals in accord with Hund's rule because the orbitals all have the same energy. However, when ligands coordinate to a metal ion, the energies of the orbitals are no longer the same. In octahedral complexes, the lobes in two of the five orbitals, the \(d_{z^2}\) and \(d_{x^2−y^2}\) orbitals, point toward the ligands (Figure \(\Page {1}\)). These two orbitals are called the (the symbol actually refers to the symmetry of the orbitals, but we will use it as a convenient name for these two orbitals in an octahedral complex). The other three orbitals, the , , and orbitals, have lobes that point between the ligands and are called the (again, the symbol really refers to the symmetry of the orbitals). As six ligands approach the metal ion along the axes of the octahedron, their point charges repel the electrons in the orbitals of the metal ion. However, the repulsions between the electrons in the orbitals (the \(d_{z^2}\) and \(d_{x^2−y^2}\) orbitals) and the ligands are greater than the repulsions between the electrons in the orbitals (the , , and orbitals) and the ligands. This is because the lobes of the orbitals point directly at the ligands, whereas the lobes of the orbitals point between them. Thus, electrons in the orbitals of the metal ion in an octahedral complex have higher potential energies than those of electrons in the orbitals. The difference in energy may be represented as shown in Figure \(\Page {2}\). The difference in energy between the and the orbitals is called the and is symbolized by , where oct stands for octahedral. The magnitude of Δ depends on many factors, including the nature of the six ligands located around the central metal ion, the charge on the metal, and whether the metal is using 3 , 4 , or 5 orbitals. Different ligands produce different crystal field splittings. The increasing crystal field splitting produced by ligands is expressed in the , a short version of which is given here: \[\large \underset{\textrm{a few ligands of the spectrochemical series, in order of increasing field strength of the ligand}}{\xrightarrow{\ce{I- <Br- <Cl- <F- <H2O<C2O4^2- <NH3<\mathit{en}<NO2- <CN-}}} \nonumber \] In this series, ligands on the left cause small crystal field splittings and are , whereas those on the right cause larger splittings and are . Thus, the Δ value for an octahedral complex with iodide ligands (I ) is much smaller than the Δ value for the same metal with cyanide ligands (CN ). Electrons in the orbitals follow the aufbau (“filling up”) principle, which says that the orbitals will be filled to give the lowest total energy, just as in main group chemistry. When two electrons occupy the same orbital, the like charges repel each other. The energy needed to pair up two electrons in a single orbital is called the . Electrons will always singly occupy each orbital in a degenerate set before pairing. P is similar in magnitude to Δ . When electrons fill the orbitals, the relative magnitudes of Δ and P determine which orbitals will be occupied. In [Fe(CN) ] , the strong field of six cyanide ligands produces a large Δ . Under these conditions, the electrons require less energy to pair than they require to be excited to the orbitals (Δ > P). The six 3 electrons of the Fe ion pair in the three orbitals (Figure \(\Page {3}\)). Complexes in which the electrons are paired because of the large crystal field splitting are called because the number of unpaired electrons (spins) is minimized. In [Fe(H O) ] , on the other hand, the weak field of the water molecules produces only a small crystal field splitting (Δ < P). Because it requires less energy for the electrons to occupy the orbitals than to pair together, there will be an electron in each of the five 3 orbitals before pairing occurs. For the six electrons on the iron(II) center in [Fe(H O) ] , there will be one pair of electrons and four unpaired electrons (Figure \(\Page {3}\)). Complexes such as the [Fe(H O) ] ion, in which the electrons are unpaired because the crystal field splitting is not large enough to cause them to pair, are called because the number of unpaired electrons (spins) is maximized. A similar line of reasoning shows why the [Fe(CN) ] ion is a low-spin complex with only one unpaired electron, whereas both the [Fe(H O) ] and [FeF ] ions are high-spin complexes with five unpaired electrons. Predict the number of unpaired electrons. The complexes are octahedral. The size of the crystal field splitting only influences the arrangement of electrons when there is a choice between pairing electrons and filling the higher-energy orbitals. For which -electron configurations will there be a difference between high- and low-spin configurations in octahedral complexes? , , , and CFT is applicable to molecules in geometries other than octahedral. In octahedral complexes, remember that the lobes of the set point directly at the ligands. For tetrahedral complexes, the orbitals remain in place, but now we have only four ligands located between the axes (Figure \(\Page {4}\)). None of the orbitals points directly at the tetrahedral ligands. However, the set (along the Cartesian axes) overlaps with the ligands less than does the set. By analogy with the octahedral case, predict the energy diagram for the orbitals in a tetrahedral crystal field. To avoid confusion, the octahedral set becomes a tetrahedral set, and the octahedral set becomes a set. Since CFT is based on electrostatic repulsion, the orbitals closer to the ligands will be destabilized and raised in energy relative to the other set of orbitals. The splitting is less than for octahedral complexes because the overlap is less, so Δ is usually small \(\left(Δ_\ce{tet}=\dfrac{4}{9}Δ_\ce{oct}\right)\): Explain how many unpaired electrons a tetrahedral ion will have. 4; because Δ is small, all tetrahedral complexes are high spin and the electrons go into the orbitals before pairing The other common geometry is square planar. It is possible to consider a square planar geometry as an octahedral structure with a pair of ligands removed. The removed ligands are assumed to be on the -axis. This changes the distribution of the orbitals, as orbitals on or near the -axis become more stable, and those on or near the or -axes become less stable. This results in the octahedral and the sets splitting and gives a more complicated pattern with no simple Δ . The basic pattern is: Experimental evidence of magnetic measurements supports the theory of high- and low-spin complexes. Remember that molecules such as O that contain unpaired electrons are paramagnetic. Paramagnetic substances are attracted to magnetic fields. Many transition metal complexes have unpaired electrons and hence are paramagnetic. Molecules such as N and ions such as Na and [Fe(CN) ] that contain no unpaired electrons are diamagnetic. Diamagnetic substances have a slight tendency to be repelled by magnetic fields. When an electron in an atom or ion is unpaired, the magnetic moment due to its spin makes the entire atom or ion paramagnetic. The size of the magnetic moment of a system containing unpaired electrons is related directly to the number of such electrons: the greater the number of unpaired electrons, the larger the magnetic moment. Therefore, the observed magnetic moment is used to determine the number of unpaired electrons present. The measured magnetic moment of low-spin [Fe(CN) ] confirms that iron is diamagnetic, whereas high-spin [Fe(H O) ] has four unpaired electrons with a magnetic moment that confirms this arrangement. When atoms or molecules absorb light at the proper frequency, their electrons are excited to higher-energy orbitals. For many main group atoms and molecules, the absorbed photons are in the ultraviolet range of the electromagnetic spectrum, which cannot be detected by the human eye. For coordination compounds, the energy difference between the orbitals often allows photons in the visible range to be absorbed. The human eye perceives a mixture of all the colors, in the proportions present in sunlight, as white light. Complementary colors, those located across from each other on a color wheel, are also used in color vision. The eye perceives a mixture of two complementary colors, in the proper proportions, as white light. Likewise, when a color is missing from white light, the eye sees its complement. For example, when red photons are absorbed from white light, the eyes see the color green. When violet photons are removed from white light, the eyes see lemon yellow. The blue color of the [Cu(NH ) ] ion results because this ion absorbs orange and red light, leaving the complementary colors of blue and green (Figure \(\Page {5}\)). The octahedral complex [Ti(H O) ] has a single electron. To excite this electron from the ground state orbital to the orbital, this complex absorbs light from 450 to 600 nm. The maximum absorbance corresponds to Δ and occurs at 499 nm. Calculate the value of Δ in Joules and predict what color the solution will appear. Using Planck's equation (refer to the section on electromagnetic energy), we calculate: \[v=\dfrac{c}{λ}\mathrm{\:so\:\dfrac{3.00×10^8\: m/s}{\dfrac{499\: nm×1\: m}{10^9\:nm}}=6.01×10^{14}\:Hz} \nonumber \] \[E=hnu\mathrm{\:so\:6.63×10^{−34}\:\textrm{J⋅s}×6.01×10^{14}\:Hz=3.99×10^{−19}\:Joules/ion} \nonumber \] Because the complex absorbs 600 nm (orange) through 450 (blue), the indigo, violet, and red wavelengths will be transmitted, and the complex will appear purple. A complex that appears green, absorbs photons of what wavelengths? red, 620–800 nm Small changes in the relative energies of the orbitals that electrons are transitioning between can lead to drastic shifts in the color of light absorbed. Therefore, the colors of coordination compounds depend on many factors. As shown in Figure \(\Page {6}\), different aqueous metal ions can have different colors. In addition, different oxidation states of one metal can produce different colors, as shown for the vanadium complexes in the link below. The specific ligands coordinated to the metal center also influence the color of coordination complexes. For example, the iron(II) complex [Fe(H O) ]SO appears blue-green because the high-spin complex absorbs photons in the red wavelengths (Figure \(\Page {7}\)). In contrast, the low-spin iron(II) complex K [Fe(CN) ] appears pale yellow because it absorbs higher-energy violet photons. In general, strong-field ligands cause a large split in the energies of orbitals of the central metal atom (large Δ ). Transition metal coordination compounds with these ligands are yellow, orange, or red because they absorb higher-energy violet or blue light. On the other hand, coordination compounds of transition metals with weak-field ligands are often blue-green, blue, or indigo because they absorb lower-energy yellow, orange, or red light. A coordination compound of the Cu ion has a configuration, and all the orbitals are filled. To excite an electron to a higher level, such as the 4 orbital, photons of very high energy are necessary. This energy corresponds to very short wavelengths in the ultraviolet region of the spectrum. No visible light is absorbed, so the eye sees no change, and the compound appears white or colorless. A solution containing [Cu(CN) ] , for example, is colorless. On the other hand, octahedral Cu complexes have a vacancy in the orbitals, and electrons can be excited to this level. The wavelength (energy) of the light absorbed corresponds to the visible part of the spectrum, and Cu complexes are almost always colored—blue, blue-green violet, or yellow (Figure \(\Page {8}\)). Although CFT successfully describes many properties of coordination complexes, molecular orbital explanations (beyond the introductory scope provided here) are required to understand fully the behavior of coordination complexes. Crystal field theory treats interactions between the electrons on the metal and the ligands as a simple electrostatic effect. The presence of the ligands near the metal ion changes the energies of the metal orbitals relative to their energies in the free ion. Both the color and the magnetic properties of a complex can be attributed to this crystal field splitting. The magnitude of the splitting (Δ ) depends on the nature of the ligands bonded to the metal. Strong-field ligands produce large splitting and favor low-spin complexes, in which the orbitals are completely filled before any electrons occupy the orbitals. Weak-field ligands favor formation of high-spin complexes. The and the orbitals are singly occupied before any are doubly occupied. | 15,027 | 1,941 |
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Mercury (10 cc) and 10 cc of water in test tubes at room temperature (with thermistors interfaced to computer, if possible). Allow both tubes to equilibrate to room temperature (~17 C), then immerse them simultaneously in a water bath at 34 C. Will 136 g of mercury heat up faster or slower than 10 g of water? Measure T every 1 second with interfaced computer. Plot. Paradoxically, mercury increases in T much faster. The amount of heat absorbed by each is roughly equivalent, since the test tubes and volumes are identical. Small differences result from the change in rate of heat transfer across the glass with ΔT. Figure \(\Page {1}\) Mercury and water in test tubes Choose T at any time, and calculate ΔT for Hg and water. Assume q absorbed is same for both, and calculate Cp for Hg from q = m x C x ΔT for water. A small mass of water at 0 C is added to a measured mass of liquid nitrogen, and the amount that evaporates is compared to the mass that evaporates when a larger mass of water at 95 C is added to liquid nitrogen. This demonstration requires knowledge of both specific heat and heat capacity. | 1,125 | 1,942 |
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As we shall see, the two heaviest members of each group usually exhibit substantial similarities in chemical behavior and are quite different from the lightest member. As shown in Table \(\Page {1}\), the observed trends in the properties of the group 3 elements are similar to those of groups 1 and 2. Due to their ns (n − 1)d valence electron configurations, the chemistry of all four elements is dominated by the +3 oxidation state formed by losing all three valence electrons. As expected based on periodic trends, these elements are highly electropositive metals and powerful reductants, with La (and Ac) being the most reactive. In keeping with their highly electropositive character, the group 3 metals react with water to produce the metal hydroxide and hydrogen gas: \[2M_{(s)} + 6H_2O_{(l)} \rightarrow 2M(OH)_{3(s)} + 3H_{2(g)}\label{Eq1}\] The chemistry of the group 3 metals is almost exclusively that of the M ion; the elements are powerful reductants. Moreover, all dissolve readily in aqueous acid to produce hydrogen gas and a solution of the hydrated metal ion: M (aq). The group 3 metals react with nonmetals to form compounds that are primarily ionic in character. For example, reacting group 3 metals with the halogens produces the corresponding trihalides: MX . The trifluorides are insoluble in water because of their high lattice energies, but the other trihalides are very soluble in water and behave like typical ionic metal halides. All group 3 elements react with air to form an oxide coating, and all burn in oxygen to form the so-called sesquioxides (M O ), which react with H O or CO to form the corresponding hydroxides or carbonates, respectively. Commercial uses of the group 3 metals are limited, but “mischmetal,” a mixture of lanthanides containing about 40% La, is used as an additive to improve the properties of steel and make flints for cigarette lighters. Because the elements of group 4 have a high affinity for oxygen, all three metals occur naturally as oxide ores that contain the metal in the +4 oxidation state resulting from losing all four ns (n − 1)d valence electrons. They are isolated by initial conversion to the tetrachlorides, as shown for Ti: \[2FeTiO_{3(s)} + 6C_{(s)} + 7Cl_{2(g)} \rightarrow 2TiCl_{4(g)} + 2FeCl_{3(g)} + 6CO_{(g)}\label{Eq2}\] followed by reduction of the tetrachlorides with an active metal such as Mg. The chemistry of the group 4 metals is dominated by the +4 oxidation state. Only Ti has an extensive chemistry in lower oxidation states. In contrast to the elements of group 3, the group 4 elements have important applications. Titanium (melting point = 1668°C) is often used as a replacement for aluminum (melting point = 660°C) in applications that require high temperatures or corrosion resistance. For example, friction with the air heats the skin of supersonic aircraft operating above Mach 2.2 to temperatures near the melting point of aluminum; consequently, titanium is used instead of aluminum in many aerospace applications. The corrosion resistance of titanium is increasingly exploited in architectural applications, as shown in the chapter-opening photo. Metallic zirconium is used in UO -containing fuel rods in nuclear reactors, while hafnium is used in the control rods that modulate the output of high-power nuclear reactors, such as those in nuclear submarines. Consistent with the periodic trends shown in Figure 23.2, the group 4 metals become denser, higher melting, and more electropositive down the column (Table \(\Page {1}\)). Unexpectedly, however, the atomic radius of Hf is slightly smaller than that of Zr due to the lanthanide contraction. Because of their ns (n − 1)d valence electron configurations, the +4 oxidation state is by far the most important for all three metals. Only titanium exhibits a significant chemistry in the +2 and +3 oxidation states, although compounds of Ti are usually powerful reductants. In fact, the Ti (aq) ion is such a strong reductant that it rapidly reduces water to form hydrogen gas. Reaction of the group 4 metals with excess halogen forms the corresponding tetrahalides (MX ), although titanium, the lightest element in the group, also forms dihalides and trihalides (X is not F). The covalent character of the titanium halides increases as the oxidation state of the metal increases because of increasing polarization of the anions by the cation as its charge-to-radius ratio increases. Thus TiCl is an ionic salt, whereas TiCl is a volatile liquid that contains tetrahedral molecules. All three metals react with excess oxygen or the heavier chalcogens (Y) to form the corresponding dioxides (MO ) and dichalcogenides (MY ). Industrially, TiO , which is used as a white pigment in paints, is prepared by reacting TiCl with oxygen at high temperatures: \[TiCl_{4(g)} + O_{2(g)} \rightarrow TiO_{2(s)} + 2Cl_{2(g)}\label{Eq3}\] The group 4 dichalcogenides have unusual layered structures with no M–Y bonds holding adjacent sheets together, which makes them similar in some ways to graphite (Figure \(\Page {1}\)). The group 4 metals also react with hydrogen, nitrogen, carbon, and boron to form hydrides (such as TiH ), nitrides (such as TiN), carbides (such as TiC), and borides (such as TiB ), all of which are hard, high-melting solids. Many of these binary compounds are nonstoichiometric and exhibit metallic conductivity. Like the group 4 elements, all group 5 metals are normally found in nature as oxide ores that contain the metals in their highest oxidation state (+5). Because of the lanthanide contraction, the chemistry of Nb and Ta is so similar that these elements are usually found in the same ores. Three-fourths of the vanadium produced annually is used in the production of steel alloys for springs and high-speed cutting tools. Adding a small amount of vanadium to steel results in the formation of small grains of V C , which greatly increase the strength and resilience of the metal, especially at high temperatures. The other major use of vanadium is as V O , an important catalyst for the industrial conversion of SO to SO in the contact process for the production of sulfuric acid. In contrast, Nb and Ta have only limited applications, and they are therefore produced in relatively small amounts. Although niobium is used as an additive in certain stainless steels, its primary application is in superconducting wires such as Nb Zr and Nb Ge, which are used in superconducting magnets for the magnetic resonance imaging of soft tissues. Because tantalum is highly resistant to corrosion, it is used as a liner for chemical reactors, in missile parts, and as a biologically compatible material in screws and pins for repairing fractured bones. The chemistry of the two heaviest group 5 metals (Nb and Ta) is dominated by the +5 oxidation state. The chemistry of the lightest element (V) is dominated by lower oxidation states, especially +4. As indicated in Table \(\Page {1}\), the trends in properties of the group 5 metals are similar to those of group 4. Only vanadium, the lightest element, has any tendency to form compounds in oxidation states lower than +5. For example, vanadium is the only element in the group that forms stable halides in the lowest oxidation state (+2). All three metals react with excess oxygen, however, to produce the corresponding oxides in the +5 oxidation state (M O ), in which polarization of the oxide ions by the high-oxidation-state metal is so extensive that the compounds are primarily covalent in character. Vanadium–oxygen species provide a classic example of the effect of increasing metal oxidation state on the protonation state of a coordinated water molecule: vanadium(II) in water exists as the violet hydrated ion [V(H O) ] ; the blue-green [V(H O) ] ion is acidic, dissociating to form small amounts of the [V(H O) (OH)] ion and a proton; and in water, vanadium(IV) forms the blue vanadyl ion [(H O) VO] , which contains a formal V=O bond (Figure \(\Page {2}\)). Consistent with its covalent character, V O is acidic, dissolving in base to give the vanadate ion ([VO ] ), whereas both Nb O and Ta O are comparatively inert. Oxides of these metals in lower oxidation states tend to be nonstoichiometric. Because vanadium ions with different oxidation states have different numbers of d electrons, aqueous solutions of the ions have different colors: in acid V(V) forms the pale yellow [VO ] ion; V(IV) is the blue vanadyl ion [VO] ; and V(III) and V(II) exist as the hydrated V (blue-green) and V (violet) ions, respectively. Although group 5 metals react with the heavier chalcogens to form a complex set of binary chalcogenides, the most important are the dichalcogenides (MY ), whose layered structures are similar to those of the group 4 dichalcogenides. The elements of group 5 also form binary nitrides, carbides, borides, and hydrides, whose stoichiometries and properties are similar to those of the corresponding group 4 compounds. One such compound, tantalum carbide (TiC), has the highest melting point of any compound known (3738°C); it is used for the cutting edges of high-speed machine tools. As an illustration of the trend toward increasing polarizability as we go from left to right across the d block, in group 6 we first encounter a metal (Mo) that occurs naturally as a sulfide ore rather than as an oxide. Molybdenite (MoS ) is a soft black mineral that can be used for writing, like PbS and graphite. Because of this similarity, people long assumed that these substances were all the same. In fact, the name molybdenum is derived from the Greek molybdos, meaning “lead.” More than 90% of the molybdenum produced annually is used to make steels for cutting tools, which retain their sharp edge even when red hot. In addition, molybdenum is the only second- or third-row transition element that is essential for humans. The major chromium ore is chromite (FeCr O ), which is oxidized to the soluble [CrO ] ion under basic conditions and reduced successively to Cr O and Cr with carbon and aluminum, respectively. Pure chromium can be obtained by dissolving Cr O in sulfuric acid followed by electrolytic reduction; a similar process is used for electroplating metal objects to give them a bright, shiny, protective surface layer. Pure tungsten is obtained by first converting tungsten ores to WO , which is then reduced with hydrogen to give the metal. The metals become increasing polarizable across the d block. Consistent with periodic trends, the group 6 metals are slightly less electropositive than those of the three preceding groups, and the two heaviest metals are essentially the same size because of the lanthanide contraction (Table \(\Page {2}\)). All three elements have a total of six valence electrons, resulting in a maximum oxidation state of +6. Due to extensive polarization of the anions, compounds in the +6 oxidation state are highly covalent. As in groups 4 and 5, the lightest element exhibits variable oxidation states, ranging from Cr , which is a powerful reductant, to CrO , a red solid that is a powerful oxidant. For Mo and W, the highest oxidation state (+6) is by far the most important, although compounds in the +4 and +5 oxidation states are known. The chemistry of the two heaviest group 6 metals (Mo and W) is dominated by the +6 oxidation state. The chemistry of the lightest element (Cr) is dominated by lower oxidation states. As observed in previous groups, the group 6 halides become more covalent as the oxidation state of the metal increases: their volatility increases, and their melting points decrease. Recall that as the electronegativity of the halogens decreases from F to I, they are less able to stabilize high oxidation states; consequently, the maximum oxidation state of the corresponding metal halides decreases. Thus all three metals form hexafluorides, but CrF is unstable at temperatures above −100°C, whereas MoF and WF are stable. Consistent with the trend toward increased stability of the highest oxidation state for the second- and third-row elements, the other halogens can oxidize chromium to only the trihalides, CrX (X is Cl, Br, or I), while molybdenum forms MoCl , MoBr , and MoI , and tungsten gives WCl , WBr , and WI . Both Mo and W react with oxygen to form the covalent trioxides (MoO and WO ), but Cr reacts to form only the so-called sesquioxide (Cr O ). Chromium will form CrO , which is a highly toxic compound that can react explosively with organic materials. All the trioxides are acidic, dissolving in base to form the corresponding oxoanions ([MO ] ). Consistent with periodic trends, the sesquioxide of the lightest element in the group (Cr O ) is amphoteric. The aqueous chemistry of molybdate and tungstate is complex, and at low pH they form a series of polymeric anions called isopolymetallates, such as the [Mo O ] ion, whose structure is as follows: An isopolymolybdate cluster. The [Mo O ] ion, shown here in both side and top views, is typical of the oxygen-bridged clusters formed by Mo(VI) and W(VI) in aqueous solution. Reacting molybdenum or tungsten with heavier chalcogens gives binary chalcogenide phases, most of which are nonstoichiometric and electrically conducting. One of the most stable is MoS ; it has a layered structure similar to that of TiS (Figure \(\Page {1}\)), in which the layers are held together by only weak van der Waals forces, which allows them to slide past one another rather easily. Consequently, both MoS and WS are used as lubricants in a variety of applications, including automobile engines. Because tungsten itself has an extraordinarily high melting point (3380°C), lubricants described as containing “liquid tungsten” actually contain a suspension of very small WS particles. As in groups 4 and 5, the elements of group 6 form binary nitrides, carbides, and borides whose stoichiometries and properties are similar to those of the preceding groups. Tungsten carbide (WC), one of the hardest compounds known, is used to make the tips of drill bits. Continuing across the periodic table, we encounter the group 7 elements (Table \(\Page {2}\)). One group 7 metal (Mn) is usually combined with iron in an alloy called ferromanganese, which has been used since 1856 to improve the mechanical properties of steel by scavenging sulfur and oxygen impurities to form MnS and MnO. Technetium is named after the Greek technikos, meaning “artificial,” because all its isotopes are radioactive. One isotope, Tc (m for metastable), has become an important biomedical tool for imaging internal organs. Because of its scarcity, Re is one of the most expensive elements, and its applications are limited. It is, however, used in a bimetallic Pt/Re catalyst for refining high-octane gasoline. All three group 7 elements have seven valence electrons and can form compounds in the +7 oxidation state. Once again, the lightest element exhibits multiple oxidation states. Compounds of Mn in oxidation states ranging from −3 to +7 are known, with the most common being +2 and +4 (Figure \(\Page {3}\)). In contrast, compounds of Tc and Re in the +2 oxidation state are quite rare. Because the electronegativity of Mn is anomalously low, elemental manganese is unusually reactive. In contrast, the chemistry of Tc is similar to that of Re because of their similar size and electronegativity, again a result of the lanthanide contraction. Due to the stability of the half-filled 3d electron configuration, the aqueous Mn ion, with a 3d valence electron configuration, is a potent oxidant that is able to oxidize water. It is difficult to generalize about other oxidation states for Tc and Re because their stability depends dramatically on the nature of the compound. Like vanadium, compounds of manganese in different oxidation states have different numbers of d electrons, which leads to compounds with different colors: the Mn (aq) ion is pale pink; Mn(OH) , which contains Mn(III), is a dark brown solid; MnO , which contains Mn(IV), is a black solid; and aqueous solutions of Mn(VI) and Mn(VII) contain the green manganate ion [MnO ] and the purple permanganate ion [MnO ] , respectively. Consistent with higher oxidation states being more stable for the heavier transition metals, reacting Mn with F gives only MnF , a high-melting, red-purple solid, whereas Re reacts with F to give ReF , a volatile, low-melting, yellow solid. Again, reaction with the less oxidizing, heavier halogens produces halides in lower oxidation states. Thus reaction with Cl , a weaker oxidant than F , gives MnCl and ReCl . Reaction of Mn with oxygen forms only Mn O , a mixed-valent compound that contains two Mn(II) and one Mn(III) per formula unit and is similar in both stoichiometry and structure to magnetite (Fe O ). In contrast, Tc and Re form high-valent oxides, the so-called heptoxides (M O ), consistent with the increased stability of higher oxidation states for the second and third rows of transition metals. Under forced conditions, manganese will form Mn O , an unstable, explosive, green liquid. Also consistent with this trend, the permanganate ion [MnO ] is a potent oxidant, whereas [TcO ] and [ReO ] are much more stable. Both Tc and Re form disulfides and diselenides with layered structures analogous to that of MoS , as well as more complex heptasulfides (M S ). As is typical of the transition metals, the group 7 metals form binary nitrides, carbides, and borides that are generally stable at high temperatures and exhibit metallic properties. The chemistry of the group 7 metals (Mn, Tc, and Re) is dominated by lower oxidation states. Compounds in the maximum possible oxidation state (+7) are readily reduced. In many older versions of the periodic table, groups 8, 9, and 10 were combined in a single group (group VIII) because the elements of these three groups exhibit many horizontal similarities in their chemistry, in addition to the similarities within each column. In part, these horizontal similarities are due to the fact that the ionization potentials of the elements, which increase slowly but steadily across the d block, have now become so large that the oxidation state corresponding to the formal loss of all valence electrons is encountered only rarely (group 8) or not at all (groups 9 and 10). As a result, the chemistry of all three groups is dominated by intermediate oxidation states, especially +2 and +3 for the first-row metals (Fe, Co, and Ni). The heavier elements of these three groups are called precious metals because they are rather rare in nature and mostly chemically inert. The chemistry of groups 8, 9, and 10 is dominated by intermediate oxidation states such as +2 and +3. The chemistry of group 8 is dominated by iron, whose high abundance in Earth’s crust is due to the extremely high stability of its nucleus. Ruthenium and osmium, on the other hand, are extremely rare elements, with terrestrial abundances of only about 0.1 ppb and 5 ppb, respectively, and they were not discovered until the 19th century. Because of the high melting point of iron (1538°C), early humans could not use it for tools or weapons. The advanced techniques needed to work iron were first developed by the Hittite civilization in Asia Minor sometime before 2000 BC, and they remained a closely guarded secret that gave the Hittites military supremacy for almost a millennium. With the collapse of the Hittite civilization around 1200 BC, the technology became widely distributed, however, leading to the Iron Age. Cobalt is one of the least abundant of the first-row transition metals. Its oxide ores, however, have been used in glass and pottery for thousands of years to produce the brilliant color known as “cobalt blue,” and its compounds are consumed in large quantities in the paint and ceramics industries. The heavier elements of group 9 are also rare, with terrestrial abundances of less than 1 ppb; they are generally found in combination with the heavier elements of groups 8 and 10 in Ni–Cu–S ores. Nickel silicates are easily processed; consequently, nickel has been known and used since antiquity. In fact, a 75:25 Cu:Ni alloy was used for more than 2000 years to mint “silver” coins, and the modern US nickel uses the same alloy. In contrast to nickel, palladium and platinum are rare (their terrestrial abundance is about 10–15 ppb), but they are at least an order of magnitude more abundant than the heavier elements of groups 8 and 9. Platinum and palladium are used in jewelry, the former as the pure element and the latter as the Pd/Au alloy known as white gold. Some properties of the elements in groups 8–10 are summarized in Table \(\Page {3}\). As in earlier groups, similarities in size and electronegativity between the two heaviest members of each group result in similarities in chemistry. We are now at the point in the d block where there is no longer a clear correlation between the valence electron configuration and the preferred oxidation state. For example, all the elements of group 8 have eight valence electrons, but only Ru and Os have any tendency to form compounds in the +8 oxidation state, and those compounds are powerful oxidants. The predominant oxidation states for all three group 8 metals are +2 and +3. Although the elements of group 9 possess a total of nine valence electrons, the +9 oxidation state is unknown for these elements, and the most common oxidation states in the group are +3 and +1. Finally, the elements of group 10 all have 10 valence electrons, but all three elements are normally found in the +2 oxidation state formed by losing the ns2 valence electrons. In addition, Pd and Pt form numerous compounds and complexes in the +4 oxidation state. We stated that higher oxidation states become less stable as we go across the d-block elements and more stable as we go down a group. Thus Fe and Co form trifluorides, but Ni forms only the difluoride NiF . In contrast to Fe, Ru and Os form a series of fluorides up to RuF and OsF . The hexafluorides of Rh and Ir are extraordinarily powerful oxidants, and Pt is the only element in group 10 that forms a hexafluoride. Similar trends are observed among the oxides. For example, Fe forms only FeO, Fe O , and the mixed-valent Fe O (magnetite), all of which are nonstoichiometric. In contrast, Ru and Os form the dioxides (MO ) and the highly toxic, volatile, yellow tetroxides, which contain formal M=O bonds. As expected for compounds of metals in such high oxidation states, the latter are potent oxidants. The tendency of the metals to form the higher oxides decreases rapidly as we go farther across the d block. Higher oxidation states become less stable across the d-block, but more stable down a group. Reactivity with the heavier chalcogens is rather complex. Thus the oxidation state of Fe, Ru, Os, Co, and Ni in their disulfides is +2 because of the presence of the disulfide ion (S ), but the disulfides of Rh, Ir, Pd, and Pt contain the metal in the +4 oxidation state together with sulfide ions (S ). This combination of highly charged cations and easily polarized anions results in substances that are not simple ionic compounds and have significant covalent character. The groups 8–10 metals form a range of binary nitrides, carbides, and borides. By far the most important of these is cementite (Fe C), which is used to strengthen steel. At high temperatures, Fe C is soluble in iron, but slow cooling causes the phases to separate and form particles of cementite, which gives a metal that retains much of its strength but is significantly less brittle than pure iron. Palladium is unusual in that it forms a binary hydride with the approximate composition PdH . Because the H atoms in the metal lattice are highly mobile, thin sheets of Pd are highly permeable to H but essentially impermeable to all other gases, including He. Consequently, diffusion of H through Pd is an effective method for separating hydrogen from other gases. The coinage metals—copper, silver, and gold—occur naturally (like the gold nugget shown here); consequently, these were probably the first metals used by ancient humans. For example, decorative gold artifacts dating from the late Stone Age are known, and some gold Egyptian coins are more than 5000 yr old. Copper is almost as ancient, with objects dating to about 5000 BC. Bronze, an alloy of copper and tin that is harder than either of its constituent metals, was used before 3000 BC, giving rise to the Bronze Age. Deposits of silver are much less common than deposits of gold or copper, yet by 3000 BC, methods had been developed for recovering silver from its ores, which allowed silver coins to be widely used in ancient times. Deposits of gold and copper are widespread and numerous, and for many centuries it was relatively easy to obtain large amounts of the pure elements. For example, a single gold nugget discovered in Australia in 1869 weighed more than 150 lb. Because the demand for these elements has outstripped their availability, methods have been developed to recover them economically from even very low-grade ores (as low as 1% Cu content for copper) by operating on a vast scale, as shown in the photo of an open-pit copper mine. Copper is used primarily to manufacture electric wires, but large quantities are also used to produce bronze, brass, and alloys for coins. Much of the silver made today is obtained as a by-product of the manufacture of other metals, especially Cu, Pb, and Zn. In addition to its use in jewelry and silverware, silver is used in Ag/Zn and Ag/Cd button batteries. Gold is typically found either as tiny particles of the pure metal or as gold telluride (AuTe ). It is used as a currency reserve, in jewelry, in the electronics industry for corrosion-free contacts, and, in very thin layers, as a reflective window coating that minimizes heat transfer. Some properties of the coinage metals are listed in Table \(\Page {4}\). The electronegativity of gold (χ = 2.40) is close to that of the nonmetals sulfur and iodine, which suggests that the chemistry of gold should be somewhat unusual for a metal. The coinage metals have the highest electrical and thermal conductivities of all the metals, and they are also the most ductile and malleable. With an ns (n − 1)d valence electron configuration, the chemistry of these three elements is dominated by the +1 oxidation state due to losing the single ns electron. Higher oxidation states are also known, however: +2 is common for Cu and, to a lesser extent, Ag, and +3 for Au because of the relatively low values of the second and (for Au) third ionization energies. All three elements have significant electron affinities due to the half-filled ns orbital in the neutral atoms. As a result, gold reacts with powerful reductants like Cs and solutions of the alkali metals in liquid ammonia to produce the gold anion Au with a 6s 5d valence electron configuration. All group 11 elements are relatively unreactive, and their reactivity decreases from Cu to Au. Hence they are noble metals that are particularly well suited for use in coins and jewelry. Copper reacts with O at high temperatures to produce Cu O and with sulfur to form Cu S. Neither silver nor gold reacts directly with oxygen, although oxides of these elements can be prepared by other routes. Silver reacts with sulfur compounds to form the black Ag S coating known as tarnish. Gold is the only metal that does not react with sulfur; it also does not react with nitrogen, carbon, or boron. All the coinage metals do, however, react with oxidizing acids. Thus both Cu and Ag dissolve in HNO and in hot concentrated H SO , while Au dissolves in the 3:1 HCl:HNO mixture known as aqua regia. Furthermore, all three metals dissolve in basic cyanide solutions in the presence of oxygen to form very stable [M(CN) ] ions, a reaction that is used to separate gold from its ores. Although the most important oxidation state for group 11 is +1, the elements are relatively unreactive, with reactivity decreasing from Cu to Au. All the monohalides except CuF and AuF are known (including AgF). Once again, iodine is unable to stabilize the higher oxidation states (Au and Cu ). Thus all the copper(II) halides except the iodide are known, but the only dihalide of silver is AgF . In contrast, all the gold trihalides (AuX ) are known, again except the triiodide. No binary nitrides, borides, or carbides are known for the group 11 elements. We next encounter the group 12 elements. Because none of the elements in group 12 has a partially filled (n − 1)d subshell, they are not, strictly speaking, transition metals. Nonetheless, much of their chemistry is similar to that of the elements that immediately precede them in the d block. The group 12 metals are similar in abundance to those of group 11, and they are almost always found in combination with sulfur. Because zinc and cadmium are chemically similar, virtually all zinc ores contain significant amounts of cadmium. All three metals are commercially important, although the use of Cd is restricted because of its toxicity. Zinc is used for corrosion protection, in batteries, to make brass, and, in the form of ZnO, in the production of rubber and paints. Cadmium is used as the cathode in rechargeable NiCad batteries. Large amounts of mercury are used in the production of chlorine and NaOH by the chloralkali process, while smaller amounts are consumed in mercury-vapor streetlights and mercury batteries. As shown in Table \(\Page {4}\), the group 12 metals are significantly more electropositive than the elements of group 11, and they therefore have less noble character. They also have much lower melting and boiling points than the preceding transition metals. In contrast to trends in the preceding groups, Zn and Cd are similar to each other, but very different from the heaviest element (Hg). In particular, Zn and Cd are rather active metals, whereas mercury is not. Because mercury, the only metal that is a liquid at room temperature, can dissolve many metals by forming amalgams, medieval alchemists especially valued it when trying to transmute base metals to gold and silver. All three elements in group 12 have ns (n − 1)d valence electron configurations; consequently, the +2 oxidation state, corresponding to losing the two ns electrons, dominates their chemistry. In addition, mercury forms a series of compounds in the +1 oxidation state that contain the diatomic mercurous ion Hg . The most important oxidation state for group 12 is +2; the metals are significantly more electropositive than the group 11 elements, so they are less noble. All the possible group 12 dihalides (MX ) are known, and they range from ionic (the fluorides) to highly covalent (such as HgCl ). The highly covalent character of many mercuric and mercurous halides is surprising given the large size of the cations, and this has been attributed to the existence of an easily distorted 5d subshell. Zinc and cadmium react with oxygen to form amphoteric MO, whereas mercury forms HgO only within a narrow temperature range (350–400°C). Whereas zinc and cadmium dissolve in mineral acids such as HCl with the evolution of hydrogen, mercury dissolves only in oxidizing acids such as HNO and H SO . All three metals react with sulfur and the other chalcogens to form the binary chalcogenides; mercury also has an extraordinarily high affinity for sulfur. For each reaction, explain why the indicated products form. : balanced chemical equation Asked for: why the indicated products form Refer to the periodic trends in this section. : Predict the products of each reactions and then balance each chemical equation. The group 3 transition metals are highly electropositive metals and powerful reductants. They react with nonmetals to form compounds that are largely ionic and with oxygen to form sesquioxides (M O ). The group 4 metals also have a high affinity for oxygen. In their reactions with halogens, the covalent character of the halides increases as the oxidation state of the metal increases because the high charge-to-radius ratio causes extensive polarization of the anions. The dichalcogenides have layered structures similar to graphite, and the hydrides, nitrides, carbides, and borides are all hard, high-melting-point solids with metallic conductivity. The group 5 metals also have a high affinity for oxygen. Consistent with periodic trends, only the lightest (vanadium) has any tendency to form compounds in oxidation states lower than +5. The oxides are sufficiently polarized to make them covalent in character. These elements also form layered chalcogenides, as well as nitrides, carbides, borides, and hydrides that are similar to those of the group 4 elements. As the metals become more polarizable across the row, their affinity for oxygen decreases. The group 6 metals are less electropositive and have a maximum oxidation state of +6, making their compounds in high oxidation states largely covalent in character. As the oxidizing strength of the halogen decreases, the maximum oxidation state of the metal also decreases. All three trioxides are acidic, but Cr O is amphoteric. The chalcogenides of the group 6 metals are generally nonstoichiometric and electrically conducting, and these elements also form nitrides, carbides, and borides that are similar to those in the preceding groups. The metals of group 7 have a maximum oxidation state of +7, but the lightest element, manganese, exhibits an extensive chemistry in lower oxidation states. As with the group 6 metals, reaction with less oxidizing halogens produces metals in lower oxidation states, and disulfides and diselenides of Tc and Re have layered structures. The group 7 metals also form nitrides, carbides, and borides that are stable at high temperatures and have metallic properties. In groups 8, 9, and 10, the ionization potentials of the elements are so high that the oxidation state corresponding to the formal loss of all valence electrons is encountered rarely (group 8) or not at all (groups 9 and 10). Compounds of group 8 metals in their highest oxidation state are powerful oxidants. The reaction of metals in groups 8, 9, and 10 with the chalcogens is complex, and these elements form a range of binary nitrides, carbides, and borides. The coinage metals (group 11) have the highest electrical and thermal conductivities and are the most ductile and malleable of the metals. Although they are relatively unreactive, they form halides but not nitrides, borides, or carbides. The group 12 elements, whose chemistry is dominated by the +2 oxidation state, are almost always found in nature combined with sulfur. Mercury is the only metal that is a liquid at room temperature, and it dissolves many metals to form amalgams. The group 12 halides range from ionic to covalent. These elements form chalcogenides and have a high affinity for soft ligands. | 34,993 | 1,943 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/DeVoes_Thermodynamics_and_Chemistry/11%3A_Reactions_and_Other_Chemical_Processes/11.10%3A_Chapter_11_Problems |
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\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) An underlined problem number or problem-part letter indicates that the numerical answer appears in Appendix I. Use values of \(\Delsub{f}H\st\) and \(\Delsub{f}G\st\) in Appendix H to evaluate the standard molar reaction enthalpy and the thermodynamic equilibrium constant at \(298.15\K\) for the oxidation of nitrogen to form aqueous nitric acid: \[ \ce{1/2N2}\tx{(g)} + \ce{5/4O2}\tx{(g)} + \ce{1/2H2O}\tx{(l)} \arrow \ce{H+}\tx{(aq)} + \ce{NO3-}\tx{(aq)} \] In 1982, the International Union of Pure and Applied Chemistry recommended that the value of the standard pressure \(p\st\) be changed from \(1\units{atm}\) to \(1\br\). This change affects the values of some standard molar quantities of a substance calculated from experimental data. (a) Find the changes in \(H\m\st\), \(S\m\st\), and \(G\m\st\) for a gaseous substance when the standard pressure is changed isothermally from \(1.01325\br\) (\(1\units{atm}\)) to exactly \(1\br\). (Such a small pressure change has an entirely negligible effect on these quantities for a substance in a condensed phase.) What are the values of the corrections that need to be made to the standard molar enthalpy of formation, the standard molar entropy of formation, and the standard molar Gibbs energy of formation of N\(_2\)O\(_4\)(g) at \(298.15\K\) when the standard pressure is changed from \(1.01325\br\) to \(1\br\)? From data for mercury listed in Appendix H, calculate the saturation vapor pressure of liquid mercury at both \(298.15\K\) and \(273.15\K\). You may need to make some reasonable approximations. Given the following experimental values at \(T = 298.15\K\), \(p=1\br\): \begin{alignat*}{2} & \tx{H\(^+\)(aq)} + \tx{OH\(^-\)(aq)} \arrow \tx{H\(_2\)O(l)} & & \Delsub{r}H\st = -55.82\units{kJ mol\(^{-1}\)} \cr & \tx{Na(s)} + \tx{H\(_2\)O(l)} \arrow \tx{Na\(^+\)(aq}) + \tx{OH\(^-\)(aq)} + \textstyle \frac{1}{2}\tx{H\(_2\)(g)} & \qquad & \Delsub{r}H\st = -184.52\units{kJ mol\(^{-1}\)} \cr & \tx{NaOH(s)} \arrow \tx{NaOH(aq)} & & \Delsub{sol}H^{\infty} = -44.75\units{kJ mol\(^{-1}\)} \cr & \tx{NaOH in 5 H\(_2\)O} \arrow \tx{NaOH in \(\infty\) H\(_2\)O} & & \Del H\m\dil = -4.93\units{kJ mol\(^{-1}\)} \cr & \tx{NaOH(s)} & & \Delsub{f}H\st = -425.61\units{kJ mol\(^{-1}\)} \end{alignat*} Using only these values, calculate: \(\Delsub{f}H\st\) for Na\(^+\)(aq), NaOH(aq), and OH\(^-\)(aq); \(\Delsub{f}H\) for NaOH in 5 H\(_2\)O; \(\Del H\m\sol\) for the dissolution of \(1\mol\) NaOH(s) in \(5\mol\) H\(_2\)O. States 1 and 2 referred to in this problem are the initial and final states of the isothermal bomb process. The temperature is the reference temperature of \(298.15\K\). Parts (a)–(c) consist of simple calculations of some quantities needed in later parts of the problem. Begin by using the masses of C\(_6\)H\(_{14}\) and H\(_2\)O placed in the bomb vessel, and their molar masses, to calculate the amounts (moles) of C\(_6\)H\(_{14}\) and H\(_2\)O present initially in the bomb vessel. Then use the stoichiometry of the combustion reaction to find the amount of O\(_2\) consumed and the amounts of H\(_2\)O and CO\(_2\) present in state 2. (There is not enough information at this stage to allow you to find the amount of O\(_2\) present, just the change.) Also find the final mass of H\(_2\)O. Assume that oxygen is present in excess and the combustion reaction goes to completion. From the molar masses and the densities of liquid C\(_6\)H\(_{14}\) and H\(_2\)O, calculate their molar volumes. From the amounts present initially in the bomb vessel and the internal volume, find the volumes of liquid C\(_6\)H\(_{14}\), liquid H\(_2\)O, and gas in state 1 and the volumes of liquid H\(_2\)O and gas in state 2. For this calculation, you can neglect the small change in the volume of liquid H\(_2\)O due to its vaporization. When the bomb vessel is charged with oxygen and before the inlet valve is closed, the pressure at \(298.15\K\) measured on an external gauge is found to be \(p_1 = 30.00\br\). To a good approximation, the gas phase of state 1 has the equation of state of pure O\(_2\) (since the vapor pressure of water is only \(0.1\units{\(\%\)}\) of \(30.00\br\)). Assume that this equation of state is given by \(V\m=RT/p+B\subs{BB}\) (Eq. 2.2.8), where \(B\subs{BB}\) is the second virial coefficient of O\(_2\) listed in Table 11.3. Solve for the amount of O\(_2\) in the gas phase of state 1. The gas phase of state 2 is a mixture of O\(_2\) and CO\(_2\), again with a negligible partial pressure of H\(_2\)O. Assume that only small fractions of the total amounts of O\(_2\) and CO\(_2\) dissolve in the liquid water, and find the amount of O\(_2\) in the gas phase of state 2 and the mole fractions of O\(_2\) and CO\(_2\) in this phase. You now have the information needed to find the pressure in state 2, which cannot be measured directly. For the mixture of O\(_2\) and CO\(_2\) in the gas phase of state 2, use Eq. 9.3.23 to calculate the second virial coefficient. Then solve the equation of state of Eq. 9.3.21 for the pressure. Also calculate the partial pressures of the O\(_2\) and CO\(_2\) in the gas mixture. Although the amounts of H\(_2\)O in the gas phases of states 1 and 2 are small, you need to know their values in order to take the energy of vaporization into account. In this part, you calculate the fugacities of the H\(_2\)O in the initial and final gas phases, in part (g) you use gas equations of state to evaluate the fugacity coefficients of the H\(_2\)O (as well as of the O\(_2\) and CO\(_2\)), and then in part (h) you find the amounts of H\(_2\)O in the initial and final gas phases. The pressure at which the pure liquid and gas phases of H\(_2\)O are in equilibrium at \(298.15\K\) (the saturation vapor pressure of water) is \(0.03169\br\). Use Eq. 7.8.18 to estimate the fugacity of H\(_2\)O(g) in equilibrium with pure liquid water at this temperature and pressure. The effect of pressure on fugacity in a one-component liquid–gas system is discussed in Sec. 12.8.1; use Eq. 12.8.3 to find the fugacity of H\(_2\)O in gas phases equilibrated with liquid water at the pressures of states 1 and 2 of the isothermal bomb process. (The mole fraction of O\(_2\) dissolved in the liquid water is so small that you can ignore its effect on the chemical potential of the water.) Calculate the fugacity coefficients of H\(_2\)O and O\(_2\) in the gas phase of state 1 and of H\(_2\)O, O\(_2\), and CO\(_2\) in the gas phase of state 2. For state 1, in which the gas phase is practically-pure O\(_2\), you can use Eq. 7.8.18 to calculate \(\phi\subs{O\(_2\)}\). The other calculations require Eq. 9.3.29, with the value of \(B_i'\) found from the formulas of Eq. 9.3.26 or Eqs. 9.3.27 and 9.3.28 (\(y\A\) is so small that you can set it equal to zero in these formulas). Use the fugacity coefficient and partial pressure of O\(_2\) to evaluate its fugacity in states 1 and 2; likewise, find the fugacity of CO\(_2\) in state 2. [You calculated the fugacity of the H\(_2\)O in part (f).] From the values of the fugacity and fugacity coefficient of a constituent of a gas mixture, you can calculate the partial pressure with Eq. 9.3.17, then the mole fraction with \(y_i=p_i/p\), and finally the amount with \(n_i=y_i n\). Use this method to find the amounts of H\(_2\)O in the gas phases of states 1 and 2, and also calculate the amounts of H\(_2\)O in the liquid phases of both states. Next, consider the O\(_2\) dissolved in the water of state 1 and the O\(_2\) and CO\(_2\) dissolved in the water of state 2. Treat the solutions of these gases as ideal dilute with the molality of solute \(i\) given by \(m_i=\fug_i/k_{m,i}\) (Eq. 9.4.21). The values of the Henry’s law constants of these gases listed in Table 11.3 are for the standard pressure of \(1\br\). Use Eq. 12.8.35 to find the appropriate values of \(k_{m,i}\) at the pressures of states 1 and 2, and use these values to calculate the amounts of the dissolved gases in both states. At this point in the calculations, you know the values of all properties needed to describe the initial and final states of the isothermal bomb process. You are now able to evaluate the various Washburn corrections. These corrections are the internal energy changes, at the reference temperature of \(298.15\K\), of processes that connect the standard states of substances with either state 1 or state 2 of the isothermal bomb process. First, consider the gaseous H\(_2\)O. The Washburn corrections should be based on a pure-liquid standard state for the H\(_2\)O. Section 7.9 shows that the molar internal energy of a pure gas under ideal-gas conditions (low pressure) is the same as the molar internal energy of the gas in its standard state at the same temperature. Thus, the molar internal energy change when a substance in its pure-liquid standard state changes isothermally to an ideal gas is equal to the standard molar internal energy of vaporization, \(\Delsub{vap}U\st\). Using the value of \(\Delsub{vap}U\st\) for H\(_2\)O given in Table 11.3, calculate \(\Del U\) for the vaporization of liquid H\(_2\)O at pressure \(p\st\) to ideal gas in the amount present in the gas phase of state 1. Also calculate \(\Del U\) for the condensation of ideal gaseous H\(_2\)O in the amount present in the gas phase of state 2 to liquid at pressure \(p\st\). Next, consider the dissolved O\(_2\) and CO\(_2\), for which gas standard states are used. Assume that the solutions are sufficiently dilute to have infinite-dilution behavior; then the partial molar internal energy of either solute in the solution at the standard pressure \(p\st=1\br\) is equal to the standard partial molar internal energy based on a solute standard state (Sec. 9.7.1). Values of \(\Delsub{sol}U\st\) are listed in Table 11.3. Find \(\Del U\) for the dissolution of O\(_2\) from its gas standard state to ideal-dilute solution at pressure \(p\st\) in the amount present in the aqueous phase of state 1. Find \(\Del U\) for the desolution (transfer from solution to gas phase) of O\(_2\) and of CO\(_2\) from ideal-dilute solution at pressure \(p\st\), in the amounts present in the aqueous phase of state 2, to their gas standard states. Calculate the internal energy changes when the liquid phases of state 1 ( -hexane and aqueous solution) are compressed from \(p\st\) to \(p_1\) and the aqueous solution of state 2 is decompressed from \(p_2\) to \(p\st\). Use an approximate expression from Table 7.4, and treat the cubic expansion coefficient of the aqueous solutions as being the same as that of pure water. The final Washburn corrections are internal energy changes of the gas phases of states 1 and 2. H\(_2\)O has such low mole fractions in these phases that you can ignore H\(_2\)O in these calculations; that is, treat the gas phase of state 1 as pure O\(_2\) and the gas phase of state 2 as a binary mixture of O\(_2\) and CO\(_2\). One of the internal energy changes is for the compression of gaseous O\(_2\), starting at a pressure low enough for ideal-gas behavior (\(U\m=U\m\st\)) and ending at pressure \(p_1\) to form the gas phase present in state 1. Use the approximate expression for \(U\m-U\m\st\gas\) in Table 7.5 to calculate \(\Del U = U(p_1) - nU\m\st\gas\); a value of \(\dif B/\dif T\) for pure O\(_2\) is listed in Table 11.3. The other internal energy change is for a process in which the gas phase of state 2 at pressure \(p_2\) is expanded until the pressure is low enough for the gas to behave ideally, and the mixture is then separated into ideal-gas phases of pure O\(_2\) and CO\(_2\). The molar internal energies of the separated low-pressure O\(_2\) and CO\(_2\) gases are the same as the standard molar internal energies of these gases. The internal energy of unmixing ideal gases is zero (Eq. 11.1.11). The dependence of the internal energy of the gas mixture is given, to a good approximation, by \(U = \sum_i U_i\st\gas - npT\dif B/\dif T\), where \(B\) is the second virial coefficient of the gas mixture; this expression is the analogy for a gas mixture of the approximate expression for \(U\m-U\m\st\gas\) in Table 7.5. Calculate the value of \(\dif B/\dif T\) for the mixture of O\(_2\) and CO\(_2\) in state 2 (you need Eq. 9.3.23 and the values of \(\dif B_{ij}/\dif T\) in Table 11.3) and evaluate \(\Del U = \sum_i n_i U_i\st\gas -U(p_2)\) for the gas expansion. Add the internal energy changes you calculated in parts (j)–(m) to find the total internal energy change of the Washburn corrections. Note that most of the corrections occur in pairs of opposite sign and almost completely cancel one another. Which contributions are the greatest in magnitude? The internal energy change of the isothermal bomb process in the bomb vessel, corrected to the reference temperature of \(298.15\K\), is found to be \(\Del U(\tx{IBP},T\subs{ref}) = -32.504\units{kJ}\). Assume there are no side reactions or auxiliary reactions. From Eqs. 11.5.9 and 11.5.10, calculate the standard molar internal energy of combustion of -hexane at \(298.15\K\). From Eq. 11.5.13, calculate the standard molar enthalpy of combustion of -hexane at \(298.15\K\). By combining the results of Prob. 11.7(p) with the values of standard molar enthalpies of formation from Appendix H, calculate the standard molar enthalpy of formation of liquid -hexane at \(298.15\K\). Consider the combustion of methane: \[ \ce{CH4}\tx{(g)} + \ce{2O2}\tx{(g)} \arrow \ce{CO2}\tx{(g)} + \ce{2H2O}\tx{(g)} \] Suppose the reaction occurs in a flowing gas mixture of methane and air. Assume that the pressure is constant at \(1\br\), the reactant mixture is at a temperature of \(298.15\K\) and has stoichiometric proportions of methane and oxygen, and the reaction goes to completion with no dissociation. For the quantity of gaseous product mixture containing \(1\mol\) CO\(_2\), \(2\mol\) H\(_2\)O, and the nitrogen and other substances remaining from the air, you may use the approximate formula \(C_p(\tx{P})=a+bT\), where the coefficients have the values \(a=297.0\units{J K\(^{-1}\)}\) and \(b=8.520\timesten{-2}\units{J K\(^{-2}\)}\). Solve Eq. 11.6.1 for \(T_2\) to estimate the flame temperature to the nearest kelvin. The standard molar Gibbs energy of formation of crystalline mercury(II) oxide at \(600.00\K\) has the value \(\Delsub{f}G\st=-26.386\units{kJ mol\(^{-1}\)}\). Estimate the partial pressure of O\(_2\) in equilibrium with HgO at this temperature: \(\ce{2HgO}\tx{(s)} \arrows \ce{2Hg}\tx{(l)} + \ce{O2}\tx{(g)}\). The combustion of hydrogen is a reaction that is known to “go to completion.” Use data in Appendix H to evaluate the thermodynamic equilibrium constant at \(298.15\K\) for the reaction \[ \ce{H2}\tx{(g)}+\ce{1/2O2}\tx{(g)} \arrow \ce{H2O}\tx{(l)} \] Assume that the reaction is at equilibrium at \(298.15\K\) in a system in which the partial pressure of O\(_2\) is \(1.0\br\). Assume ideal-gas behavior and find the equilibrium partial pressure of H\(_2\) and the number of H\(_2\) in \(1.0\units{m\(^3\)}\) of the gas phase. In the preceding part, you calculated a very small value (a fraction) for the number of H\(_2\) molecules in \(1.0\units{m\(^3\)}\). Statistically, this fraction can be interpreted as the fraction of a given length of time during which one molecule is present in the system. Take the age of the universe as \(1.0\timesten{10}\) years and find the total length of time in seconds, during the age of the universe, that a H\(_2\) molecule is present in the equilibrium system. (This hypothetical value is a dramatic demonstration of the statement that the limiting reactant is essentially entirely exhausted during a reaction with a large value of \(K\).) Let G represent carbon in the form of and D represent the crystal form. At \(298.15\K\), the thermodynamic equilibrium constant for G\(\rightleftharpoons\)D, based on a standard pressure \(p\st=1\br\), has the value \(K=0.31\). The molar volumes of the two crystal forms at this temperature are \(V\m(\tx{G})=5.3\timesten{-6}\units{m\(^3\) mol\(^{-1}\)}\) and \(V\m(\tx{D})=3.4\timesten{-6}\units{m\(^3\) mol\(^{-1}\)}\). (a) Write an expression for the reaction quotient \(Q\subs{rxn}\) as a function of pressure. Use the approximate expression of the pressure factor given in Table 9.6. Use the value of \(K\) to estimate the pressure at which the D and G crystal forms are in equilibrium with one another at \(298.15\K\). (This is the lowest pressure at which graphite could in principle be converted to diamond at this temperature.) Consider the dissociation reaction \(\ce{N2O4}\tx{(g)} \arrow \ce{2NO2}\tx{(g)}\) taking place at a constant temperature of \(298.15\K\) and a constant pressure of \(0.0500\br\). Initially (at \(\xi=0\)) the system contains \(1.000\mol\) of N\(_2\)O\(_4\) and no NO\(_2\). Other needed data are found in Appendix H. Assume ideal-gas behavior. (a) For values of the advancement \(\xi\) ranging from 0 to \(1\mol\), at an interval of \(0.1\mol\) or less, calculate \([ G(\xi)-G(0) ]\) to the nearest \(0.01\units{kJ}\). A computer spreadsheet would be a convenient way to make the calculations. (b) Plot your values of \(G(\xi)-G(0)\) as a function of \(\xi\), and draw a smooth curve through the points. On your curve, indicate the estimated position of \(\xi\eq\). Calculate the activities of N\(_2\)O\(_4\) and NO\(_2\) for this value of \(\xi\), use them to estimate the thermodynamic equilibrium constant \(K\), and compare your result with the value of \(K\) calculated from Eq. 11.8.11. | 24,928 | 1,944 |
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A significant portion of enzymes function such that their properties can be studied using the equation. However, a particular class of enzymes exhibit kinetic properties that cannot be studied using the . The rate equation of these unique enzymes is characterized by an “S-shaped” sigmoidal curve, which is different from the majority of enzymes whose rate equation exhibits hyberbolic curves. Allosteric regulation is the regulation of an enzyme or other protein by binding an effector molecule at the protein's allosteric site (that is, a site other than the protein's active site). Effectors that enhance the protein's activity are referred to as allosteric activators, whereas those that decrease the protein's activity are called allosteric inhibitors. The term allostery refers to the fact that the regulatory site of an allosteric protein is from its active site. Allosteric regulations are a natural example of control loops, such as feedback from downstream products or feedforward from upstream substrates. Long-range allostery is especially important in cell signaling. Cooperativity is a phenomenon displayed by enzymes or receptors that have multiple binding sites where the affinity of the binding sites for a ligand is increased, positive cooperativity, or decreased, negative cooperativity, upon the binding of a ligand to a binding site. We also see cooperativity in large chain molecules made of many identical (or nearly identical) subunits (such as DNA, proteins, and phospholipids), when such molecules undergo phase transitions such as melting, unfolding or unwinding. This is referred to as subunit cooperativity (discussed below). An example of positive cooperativity is the binding of oxygen to hemoglobin. One oxygen molecule can bind to the ferrous iron of a heme molecule in each of the four chains of a hemoglobin molecule. Deoxy-hemoglobin has a relatively low affinity for oxygen, but when one molecule binds to a single heme, the oxygen affinity increases, allowing the second molecule to bind more easily, and the third and fourth even more easily. The oxygen affinity of 3-oxy-hemoglobin is ~300 times greater than that of deoxy-hemoglobin. This behavior leads the affinity curve of hemoglobin to be sigmoidal, rather than hyperbolic as with the monomeric myoglobin. By the same process, the ability for hemoglobin to lose oxygen increases as fewer oxygen molecules are bound. Negative allosteric modulation (also known as allosteric inhibition) occurs when the binding of one ligand the affinity for substrate at other active sites. For example, when 2,3-BPG binds to an allosteric site on hemoglobin, the affinity for oxygen of all subunits decreases. This is when a regulator is absent from the binding site. Another instance in which negative allosteric modulation can be seen is between ATP and the enzyme Phosphofructokinase within the negative feedback loop that regulates glycolysis. Phosphofructokinase (generally referred to as PFK) is an enzyme that catalyses the third step of : the phosphorylation of Fructose-6-phosphate into Fructose 1,6-bisphosphate. PFK can be allosterically inhibited by high levels of ATP within the cell. When ATP levels are high, ATP will bind to an allosteric site on phosphofructokinase, causing a change in the enzyme's three-dimensional shape. This change causes its affinity for substrate (fructose-6-phosphate and ATP) at the active site to decrease, and the enzyme is deemed inactive. This causes glycolysis to cease when ATP levels are high, thus conserving the body's glucose and maintaining balanced levels of cellular ATP. In this way, ATP serves as a negative allosteric modulator for PFK, despite the fact that it is also a substrate of the enzyme. Sigmoidal kinetic profiles are the result of enzymes that demonstrate positive cooperative binding. cooperativity refers to the observation that binding of the substrate or ligand at one binding site affects the affinity of other sites for their substrates. For enzymatic reactions with multiple substrate binding sites, this increased affinity for the substrate causes a rapid and coordinated increase in the velocity of the reaction at higher \([S]\) until \(V_{max}\) is achieved. Plotting the \(V_0\) vs. \([S]\) for a cooperative enzyme, we observe the characteristic sigmoidal shape with low enzyme activity at low substrate concentration and a rapid and immediate increase in enzyme activity to \(V_{max}\) as \([S]\) increases. The phenomenon of cooperativity was initially observed in the oxygen-hemoglobin interaction that functions in carrying oxygen in blood. Positive cooperativity implies allosteric binding – binding of the ligand at one site increases the enzyme’s affinity for another ligand at a site different from the other site. Enzymes that demonstrate cooperativity are defined as allosteric. There are several types of allosteric interactions: homotropic (positive) and heterotropic (negative).
Positive and negative allosteric interactions (as illustrated through the phenomenon of cooperativity) refer to the enzyme's binding affinity for other ligands at other sites, as a result of ligand binding at the initial binding site. When the ligands interacting are all the same compounds, the effect of the allosteric interaction is considered homotropic. When the ligands interacting are different, the effect of the allosteric interaction is considered heterotropic. It is also very important to remember that allosteric interactions tend to be driven by ATP hydrolysis. The degree of cooperativity is determined by Hill equation (Equation \(\ref{Eq1}\)) for non-Michaelis-Menten kinetics. The Hill equation accounts for allosteric binding at sites other than the active site. \(n\) is the "Hill coefficient." \[ \theta = \dfrac{[L]^n}{K_d+[L]^n} = \dfrac{[L]^n}{K_a^n+[L]^n} \label{Eq1}\] where Taking the logarithm of both sides of the equation leads to an alternative formulation of the Hill Equation. \[ \log \left( \dfrac{\theta}{1-\theta} \right) = n\log [L] - \log K_d \label{Eq2}\] Currently, there are 2 models for illustrating cooperativity: the concerted model and the sequential model. Most allosteric effects can be explained by the concerted MWC model put forth by Monod, Wyman, and Changeux, or by the sequential model described by Koshland, Nemethy, and Filmer. Both postulate that enzyme subunits exist in one of two conformations, tensed (T) or relaxed (R), and that relaxed subunits bind substrate more readily than those in the tense state. The two models differ most in their assumptions about subunit interaction and the preexistence of both states. The concerted model of allostery, also referred to as the symmetry model or MWC model, postulates that enzyme subunits are connected in such a way that a conformational change in one subunit is necessarily conferred to all other subunits. Thus, all subunits must exist in the same conformation. The model further holds that, in the absence of any ligand (substrate or otherwise), the equilibrium favours one of the conformational states, T or R. The equilibrium can be shifted to the R or T state through the binding of one ligand (the allosteric effector or ligand) to a site that is different from the active site (the allosteric site). The sequential model of allosteric regulation holds that subunits are not connected in such a way that a conformational change in one induces a similar change in the others. Thus, all enzyme subunits do not necessitate the same conformation. Moreover, the sequential model dictates that molecules of substrate bind via an induced fit protocol. In general, when a subunit randomly collides with a molecule of substrate, the active site, in essence, forms a glove around its substrate. While such an induced fit converts a subunit from the tensed state to relaxed state, it does not propagate the conformational change to adjacent subunits. Instead, substrate-binding at one subunit only slightly alters the structure of other subunits so that their binding sites are more receptive to substrate. To summarize: Allostery is a direct and efficient means for regulation of biological macromolecule function, produced by the binding of a ligand at an allosteric site topographically distinct from the orthosteric site. Due to the often high receptor selectivity and lower target-based toxicity, allosteric regulation is also expected to play an increasing role in drug discovery and bioengineering. The AlloSteric Database (ASD, provides a central resource for the display, search and analysis of the structure, function and related annotation for allosteric molecules. Currently, ASD contains allosteric proteins from more than 100 species and modulators in three categories (activators, inhibitors, and regulators). Each protein is annotated with detailed description of allostery, biological process and related diseases, and each modulator with binding affinity, physicochemical properties and therapeutic area. Integrating the information of allosteric proteins in ASD should allow the prediction of allostery for unknown proteins, to be followed with experimental validation. In addition, modulators curated in ASD can be used to investigate potential allosteric targets for a query compound, and can help chemists to implement structure modifications for novel allosteric drug design. Allosteric enzymes are an exception to the . Because they have more than two subunits and active sites, they do not obey the Michaelis-Menten kinetics, but instead have sigmoidal kinetics. Since allosteric enzymes are cooperative, a sigmoidal plot of \(V_0\) versus \([S]\) results: There are distinct properties of Allosteric Enzymes that makes it different compared to other enzymes. There are two primary models for illustrating cooperativity. (also called the model) illustrates cooperativity by assuming that proteins have two or more subunits, and that each part of the protein molecule is able to exist in either the relaxed (R) state or the tense (T) state - the tense state of a protein molecule is favored when it doesn't have any substrates bound. All aspects, including binding and dissociation constants are the same for each ligand at the respective binding sites. aims to demonstrate cooperativity by assuming that the enzyme/protein molecule affinity is relative and changes as substrates bind. Unlike the concerted model, the sequential model accounts for different species of the protein molecule. | 10,505 | 1,945 |
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This page looks at the the different types of catalyst (heterogeneous and homogeneous) with examples of each kind, and explanations of how they work. You will also find a description of one example of autocatalysis - a reaction which is catalysed by one of its products. Catalysts can be divided into two main types - heterogeneous and homogeneous. In a heterogeneous reaction, the catalyst is in a different phase from the reactants. In a homogeneous reaction, the catalyst is in the same phase as the reactants. What is a phase? If you look at a mixture and can see a boundary between two of the components, those substances are in different phases. A mixture containing a solid and a liquid consists of two phases. A mixture of various chemicals in a single solution consists of only one phase, because you can't see any boundary between them. You might wonder why phase differs from the term physical state (solid, liquid or gas). It includes solids, liquids and gases, but is actually a bit more general. It can also apply to two liquids (oil and water, for example) which don't dissolve in each other. You could see the boundary between the two liquids. If you want to be fussy about things, the diagrams actually show more phases than are labelled. Each, for example, also has the glass beaker as a solid phase. All probably have a gas above the liquid - that's another phase. We don't count these extra phases because they aren't a part of the reaction. This involves the use of a catalyst in a different phase from the reactants. Typical examples involve a solid catalyst with the reactants as either liquids or gases. Most examples of heterogeneous catalysis go through the same stages: One or more of the reactants are adsorbed on to the surface of the catalyst at active sites. There is some sort of interaction between the surface of the catalyst and the reactant molecules which makes them more reactive. The reaction happens. The product molecules are desorbed. Adsorption is where something sticks to a surface. It isn't the same as absorption where one substance is taken up within the structure of another. Be careful! An active site is a part of the surface which is particularly good at adsorbing things and helping them to react. This might involve an actual reaction with the surface, or some weakening of the bonds in the attached molecules. At this stage, both of the reactant molecules might be attached to the surface, or one might be attached and hit by the other one moving freely in the gas or liquid. Desorption simply means that the product molecules break away. This leaves the active site available for a new set of molecules to attach to and react. A good catalyst needs to adsorb the reactant molecules strongly enough for them to react, but not so strongly that the product molecules stick more or less permanently to the surface. Silver, for example, isn't a good catalyst because it doesn't form strong enough attachments with reactant molecules. Tungsten, on the other hand, isn't a good catalyst because it adsorbs too strongly. Metals like platinum and nickel make good catalysts because they adsorb strongly enough to hold and activate the reactants, but not so strongly that the products can't break away. The simplest example of this is the reaction between ethene and hydrogen in the presence of a nickel catalyst. In practice, this is a pointless reaction, because you are converting the extremely useful ethene into the relatively useless ethane. However, the same reaction will happen with any compound containing a carbon-carbon double bond. One important industrial use is in the hydrogenation of vegetable oils to make margarine, which also involves reacting a carbon-carbon double bond in the vegetable oil with hydrogen in the presence of a nickel catalyst. Ethene molecules are adsorbed on the surface of the nickel. The double bond between the carbon atoms breaks and the electrons are used to bond it to the nickel surface. Hydrogen molecules are also adsorbed on to the surface of the nickel. When this happens, the hydrogen molecules are broken into atoms. These can move around on the surface of the nickel. If a hydrogen atom diffuses close to one of the bonded carbons, the bond between the carbon and the nickel is replaced by one between the carbon and hydrogen. That end of the original ethene now breaks free of the surface, and eventually the same thing will happen at the other end. As before, one of the hydrogen atoms forms a bond with the carbon, and that end also breaks free. There is now space on the surface of the nickel for new reactant molecules to go through the whole process again. Catalytic converters change poisonous molecules like carbon monoxide and various nitrogen oxides in car exhausts into more harmless molecules like carbon dioxide and nitrogen. They use expensive metals like platinum, palladium and rhodium as the heterogeneous catalyst. The metals are deposited as thin layers onto a ceramic honeycomb. This maximises the surface area and keeps the amount of metal used to a minimum. Taking the reaction between carbon monoxide and nitrogen monoxide as typical: Catalytic converters can be affected by catalyst poisoning. This happens when something which isn't a part of the reaction gets very strongly adsorbed onto the surface of the catalyst, preventing the normal reactants from reaching it. Lead is a familiar catalyst poison for catalytic converters. It coats the honeycomb of expensive metals and stops it working. In the past, lead compounds were added to petrol (gasoline) to make it burn more smoothly in the engine. But you can't use a catalytic converter if you are using leaded fuel. So catalytic converters have not only helped remove poisonous gases like carbon monoxide and nitrogen oxides, but have also forced the removal of poisonous lead compounds from petrol. During the Contact Process for manufacturing sulphuric acid, sulphur dioxide has to be converted into sulphur trioxide. This is done by passing sulphur dioxide and oxygen over a solid vanadium(V) oxide catalyst. This example is slightly different from the previous ones because the gases actually react with the surface of the catalyst, temporarily changing it. It is a good example of the ability of transition metals and their compounds to act as catalysts because of their ability to change their oxidation state. The sulphur dioxide is oxidised to sulphur trioxide by the vanadium(V) oxide. In the process, the vanadium(V) oxide is reduced to vanadium(IV) oxide. The vanadium(IV) oxide is then re-oxidised by the oxygen. This is a good example of the way that a catalyst can be changed during the course of a reaction. At the end of the reaction, though, it will be chemically the same as it started. This has the catalyst in the same phase as the reactants. Typically everything will be present as a gas or contained in a single liquid phase. The examples contain one of each of these . . . This is a solution reaction that you may well only meet in the context of catalysis, but it is a lovely example! Persulphate ions (peroxodisulphate ions), S O , are very powerful oxidising agents. Iodide ions are very easily oxidised to iodine. And yet the reaction between them in solution in water is very slow. If you look at the equation, it is easy to see why that is: The reaction needs a collision between two negative ions. Repulsion is going to get seriously in the way of that! The catalysed reaction avoids that problem completely. The catalyst can be either iron(II) or iron(III) ions which are added to the same solution. This is another good example of the use of transition metal compounds as catalysts because of their ability to change oxidation state. For the sake of argument, we'll take the catalyst to be iron(II) ions. As you will see shortly, it doesn't actually matter whether you use iron(II) or iron(III) ions. The persulphate ions oxidise the iron(II) ions to iron(III) ions. In the process the persulphate ions are reduced to sulfate ions. The iron(III) ions are strong enough oxidising agents to oxidise iodide ions to iodine. In the process, they are reduced back to iron(II) ions again. Both of these individual stages in the overall reaction involve collision between positive and negative ions. This will be much more likely to be successful than collision between two negative ions in the uncatalysed reaction. What happens if you use iron(III) ions as the catalyst instead of iron(II) ions? The reactions simply happen in a different order. This is a good example of homogeneous catalysis where everything is present as a gas. Ozone, O , is constantly being formed and broken up again in the high atmosphere by the action of ultraviolet light. Ordinary oxygen molecules absorb ultraviolet light and break into individual oxygen atoms. These have unpaired electrons, and are known as free radicals. They are very reactive. The oxygen radicals can then combine with ordinary oxygen molecules to make ozone. Ozone can also be split up again into ordinary oxygen and an oxygen radical by absorbing ultraviolet light. This formation and breaking up of ozone is going on all the time. Taken together, these reactions stop a lot of harmful ultraviolet radiation penetrating the atmosphere to reach the surface of the Earth. The catalytic reaction we are interested in destroys the ozone and so stops it absorbing UV in this way. Chlorofluorocarbons (CFCs) like CF Cl , for example, were used extensively in aerosols and as refrigerants. Their slow breakdown in the atmosphere produces chlorine atoms - chlorine free radicals. These catalyse the destruction of the ozone. This happens in two stages. In the first, the ozone is broken up and a new free radical is produced. The chlorine radical catalyst is regenerated by a second reaction. This can happen in two ways depending on whether the ClO radical hits an ozone molecule or an oxygen radical. If it hits an oxygen radical (produced from one of the reactions we've looked at previously): Or if it hits an ozone molecule: Because the chlorine radical keeps on being regenerated, each one can destroy thousands of ozone molecules. In autocatalysis, the reaction is catalysed by one of its products. One of the simplest examples of this is in the oxidation of a solution of ethanedioic acid (oxalic acid) by an acidified solution of potassium manganate(VII) (potassium permanganate). The reaction is very slow at room temperature. It is used as a titration to find the concentration of potassium manganate(VII) solution and is usually carried out at a temperature of about 60°C. Even so, it is quite slow to start with. The reaction is catalysed by manganese(II) ions. There obviously aren't any of those present before the reaction starts, and so it starts off extremely slowly at room temperature. However, if you look at the equation, you will find manganese(II) ions amongst the products. More and more catalyst is produced as the reaction proceeds and so the reaction speeds up. You can measure this effect by plotting the concentration of one of the reactants as time goes on. You get a graph quite unlike the normal rate curve for a reaction. Concentrations are high at the beginning and so the reaction is fast - shown by a rapid fall in the reactant concentration. As things get used up, the reaction slows down and eventually stops as one or more of the reactants are completely used up. You can see the slow (uncatalysed) reaction at the beginning. As catalyst begins to be formed in the mixture, the reaction speeds up - getting faster and faster as more and more catalyst is formed. Eventually, of course, the rate falls again as things get used up. Don't assume that a rate curve which looks like this necessarily shows an example of autocatalysis. There are other effects which might produce a similar graph. For example, if the reaction involved a solid reacting with a liquid, there might be some sort of surface coating on the solid which the liquid has to penetrate before the expected reaction can happen. A more common possibility is that you have a strongly exothermic reaction and aren't controlling the temperature properly. The heat evolved during the reaction speeds the reaction up. | 12,580 | 1,946 |
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is a common form of air pollution found mainly in urban areas and large population centers. The term refers to any type of atmospheric pollution—regardless of source, composition, or concentration—that creates a significant reduction in atmospheric visibility. encompasses a broad category of air pollutants created through a multitude of processes that relate specifically to the atmospheric conditions of the formation region. In the early 1900s, London was plagued by a particular type of smog that resulted from a combination of dense fog and soot from coal combustion. In modern times, the Los Angeles Basin is often associated with dense photochemical smog, produced through a combination of vehicle exhaust and sunlight. These are two of many examples of pollution classified as smog, but they are in no way chemically related. Smog refers to a diverse category of air pollutants with varying chemical composition; however, all types of smog form a visible haze that reduces atmospheric visibility. The term was first coined in 1905 in a paper by Dr. Henry Antoine Des Voeux to describe the combination of smoke and fog that had been plaguing London during that time. London has since enacted strict air pollution regulations which have drastically reduced incidents of smog in that region; however, London-type smog is still a major problem in areas of the world that burn large quantities of coal for heat. In the United States, smog is most typically associated with the Los Angeles Basin of Southern California and its photochemical smog. In Los Angeles, a combination of orographic features, ample sunlight, and a dense population combine to form some of the worst air quality in the United States. when volatile organic compounds (VOCs) are present in the atmosphere, the equation changes entirely. Highly reactive VOCs oxidize nitrogen oxide into nitrogen dioxide without break The following substances are identified in photochemical smog: 1. Nitrogen Dioxide (\(NO_2\)) from vehicle exhaust, is photolyzed by ultraviolet (UV) radiation (\(h\nu\)) from the sun and decomposes into Nitrogen Oxide (\(NO\) and an oxygen radical: \[NO_2 + h\nu \rightarrow NO + O^. \label{1} \] 2. The oxygen radical then reacts with an atmospheric oxygen molecule to create ozone, O3: \[O^. + O_2 \rightarrow O_3 \label{2} \] 3. Under normal conditions, O reacts with NO, to produce NO2 and an oxygen molecule: \[O_3 + NO \rightarrow O_2 + NO_2 \label{3} \] This is a continual cycle that leads only to a temporary increase in net ozone production. To create photochemical smog on the scale observed in Los Angeles, the process must include Volatile organic compounds (VOC's). 4. VOC's react with hydroxide in the atmosphere to create water and a reactive VOC molecule: \[RH + OH^. \rightarrow R^. + H_2O \label{4} \] 5. The reactive VOC can then bind with an oxygen molecule to create an oxidized VOC: \[R^. + O_2 \rightarrow RO_2 \label{5} \] 6. The oxidized VOC can now bond with the nitrogen oxide produced in the earlier set of equations to form nitrogen dioxide and a reactive VOC molecule: \[RO_2+ NO \rightarrow RO-. + NO_2 \label{6} \] In the second set of equations, it is apparent that nitrogen oxide, produced in equation 1, is oxidized in equation 6 without the destruction of any ozone. This means that in the presence of VOCs, equation 3 is essentially eliminated, leading to a large and rapid build up in the photochemical smog in the lower atmosphere. Figure 1, courtesy of the EPA, depicts concentrations and constituents of photochemical smog throughout the course of an average work day. In the morning, NO and VOC concentrations are high, as people fill their cars with gas and drive to work. By midmorning , VOC's begin to oxidize NO into NO , thus reducing their respective concentrations. At midday, NO concentrations peak just before solar radiation becomes intense enough to photolyze the NO bond, releasing an oxygen atom that quickly gets converted into O . By late afternoon, peak concentrations of photochemical smog are present. Every new vehicle sold in the United States must include a catalytic converter to reduce photochemical emissions. Catalytic converters force CO and incompletely combusted hydrocarbons to react with a metal catalyst, typically platinum, to produce CO and H O. Additionally, catalytic converters reduce nitrogen oxides from exhaust gases into O and N , eliminating the cycle of ozone formation. Many scientists have suggested that pumping gas at night could reduce photochemical ozone formation by limiting the amount of exposure VOCs have with sunlight. London-type smog is mainly a product of burning large amounts of high sulfur coal. Clean air laws passed in 1956 have greatly reduced smog formation in the United Kingdom; however, in other parts of the world London-type smog is still very prevalent. The main constituent of London-type smog is soot; however, these smogs also contain large quantities of fly ash, sulfur dioxide, sodium chloride and calcium sulfate particles. If concentrations are high enough, sulfur dioxide can react with atmospheric hydroxide to produce sulfuric acid, which will precipitate as acid rain. \[SO_2 + OH^. \rightarrow HOSO_2 \label{1} \] \[HOSO_2 + O_2 \rightarrow HO_2 + SO_3 \label{2} \] \[SO_3 + H_2O \rightarrow H_2SO_4 \label{3} \] Because ozone is highly reactive, it has the ability to oxidize and destroy lung tissue. Short term exposures to elevated levels of ozone (above .75 ppm) have been linked to a host of respiratory irritations including coughing, wheezing, substernal soreness, pharyngitis, and dyspnea. Prolonged exposure to the molecule has been proven to cause a permanent reduction in lung function, as well as elevate the risk of developing asthma. Sulfur dioxide is a common component of London smog. Epidemiological studies have linked short term sulfur dioxide exposure to respiratory irritations including coughing, wheezing, and pharyngitis. 2. List three factors that make the Los Angeles Basin an ideal place for photochemical smog formation. 3. How can pumping your gas at night reduce photochemical ozone formation? 4. True or False: Photochemical smog is created from a combination of soot, fly ash, and sulfur dioxide. 5. Why is it not a good idea to breath ozone? | 6,330 | 1,948 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/15%3A_Chemical_Equilibrium/15.03%3A_Solving_Equilibrium_Problems |
There are two fundamental kinds of equilibrium problems: (1) those in which we are given the concentrations of the reactants and the products at equilibrium (or, more often, information that allows us to calculate these concentrations), and we are asked to calculate the equilibrium constant for the reaction; and (2) those in which we are given the equilibrium constant and the initial concentrations of reactants, and we are asked to calculate the concentration of one or more substances at equilibrium. In this section, we describe methods for solving both kinds of problems. We saw in the exercise in Example 6 in that the equilibrium constant for the decomposition of CaCO (s) to CaO(s) and CO (g) is = [CO ]. At 800°C, the concentration of CO in equilibrium with solid CaCO and CaO is 2.5 × 10 M. Thus at 800°C is 2.5 × 10 . (Remember that equilibrium constants are unitless.) A more complex example of this type of problem is the conversion of -butane, an additive used to increase the volatility of gasoline, to isobutane (2-methylpropane). This reaction can be written as follows:
\( n-butane \left ( g \right ) \rightleftharpoons isobutane \left ( g \right ) \tag{15.3.1} \) and the equilibrium constant = [isobutane]/[ -butane]. At equilibrium, a mixture of -butane and isobutane at room temperature was found to contain 0.041 M isobutane and 0.016 M -butane. Substituting these concentrations into the equilibrium constant expression,
\( K= \dfrac{isobutane}{n-butane}=\dfrac{0.041\;\cancel{M}}{0.016 \; \cancel{M}} = 2.6 \tag{15.3.2} \) Thus the equilibrium constant for the reaction as written is 2.6. The reaction between gaseous sulfur dioxide and oxygen is a key step in the industrial synthesis of sulfuric acid: \( 2SO_{2}\left ( g \right ) + O_{2}\left ( g \right ) \rightleftharpoons 2SO_{3}\left ( g \right ) \) A mixture of SO and O was maintained at 800 K until the system reached equilibrium. The equilibrium mixture contained 5.0 × 10 M SO , 3.5 × 10 M O , and 3.0 × 10 M SO . Calculate and at this temperature. balanced equilibrium equation and composition of equilibrium mixture equilibrium constant Write the equilibrium constant expression for the reaction. Then substitute the appropriate equilibrium concentrations into this equation to obtain . Substituting the appropriate equilibrium concentrations into the equilibrium constant expression, \( K=\dfrac{\left [ SO_{3} \right ]^{2}}{\left [ SO_{2} \right ]^{2}\left [ O_{2} \right ]}=\dfrac{\left ( 5.0\times 10^{-2} \right )^{2}}{\left ( 3.0\times 10^{-3} \right )^{2}\left ( 3.5\times 10^{-3} \right )}=7.9\times 10^{4} \) To solve for , we use , where Δ = 2 − 3 = −1: \( K_{p}= K\left ( RT \right )^{\Delta n} \)
\( =7.9\times 10^{4}\left [ \left (0.082606\; L\cdot atm/mol\cdot \cancel{K} \right ) \left ( 800 \; \cancel{K} \right )\right ] \)
\( =1.2\times 10^{3}\) Exercise Hydrogen gas and iodine react to form hydrogen iodide via the reaction \( H_{2}\left ( g \right ) + I_{2}\left ( g \right ) \rightleftharpoons 2HI\left ( g \right ) \) A mixture of H and I was maintained at 740 K until the system reached equilibrium. The equilibrium mixture contained 1.37 × 10 M HI, 6.47 × 10 M H , and 5.94 × 10 M I . Calculate and for this reaction. = 48.8; = 48.8 Chemists are not often given the concentrations of all the substances, and they are not likely to measure the equilibrium concentrations of all the relevant substances for a particular system. In such cases, we can obtain the equilibrium concentrations from the initial concentrations of the reactants and the balanced chemical equation for the reaction, as long as the equilibrium concentration of one of the substances is known. Example 9 shows one way to do this. A 1.00 mol sample of NOCl was placed in a 2.00 L reactor and heated to 227°C until the system reached equilibrium. The contents of the reactor were then analyzed and found to contain 0.056 mol of Cl . Calculate at this temperature. The equation for the decomposition of NOCl to NO and Cl is as follows: \( 2NOCl \left ( g \right ) \rightleftharpoons 2NO\left ( g \right ) + Cl_{2}\left ( g \right ) \) balanced equilibrium equation, amount of reactant, volume, and amount of one product at equilibrium Write the equilibrium constant expression for the reaction. Construct a table showing the initial concentrations, the changes in concentrations, and the final concentrations (as initial concentrations plus changes in concentrations). Calculate all possible initial concentrations from the data given and insert them in the table. Use the coefficients in the balanced chemical equation to obtain the changes in concentration of all other substances in the reaction. Insert those concentration changes in the table. Obtain the final concentrations by summing the columns. Calculate the equilibrium constant for the reaction. The first step in any such problem is to balance the chemical equation for the reaction (if it is not already balanced) and use it to derive the equilibrium constant expression. In this case, the equation is already balanced, and the equilibrium constant expression is as follows: \( K=\dfrac{\left [ NO_{2} \right ]^{2}\left [ Cl_{2} \right ]}{\left [ NOCl \right ]^{2}} \) To obtain the concentrations of NOCl, NO, and Cl at equilibrium, we construct a table showing what is known and what needs to be calculated. We begin by writing the balanced chemical equation at the top of the table, followed by three lines corresponding to the initial concentrations, the changes in concentrations required to get from the initial to the final state, and the final concentrations. Initially, the system contains 1.00 mol of NOCl in a 2.00 L container. Thus [NOCl] = 1.00 mol/2.00 L = 0.500 M. The initial concentrations of NO and Cl are 0 M because initially no products are present. Moreover, we are told that at equilibrium the system contains 0.056 mol of Cl in a 2.00 L container, so [Cl ] = 0.056 mol/2.00 L = 0.028 M. We insert these values into the following table: We use the stoichiometric relationships given in the balanced chemical equation to find the change in the concentration of Cl , the substance for which initial and final concentrations are known: According to the coefficients in the balanced chemical equation, 2 mol of NO are produced for every 1 mol of Cl , so the change in the NO concentration is as follows: \( \Delta \left [ NO \right ] = \left ( \dfrac{0.028 \; \cancel{mol\;Cl_{2}}}{L} \right )\left ( \dfrac{2\; mol\; NO}{1\;\cancel{mol\;Cl_{2}}} \right )=0.056\; M \) Similarly, 2 mol of NOCl are consumed for every 1 mol of Cl produced, so the change in the NOCl concentration is as follows: \( \Delta \left [ NOCl \right ] = \left ( \dfrac{0.028 \; \cancel{mol\;Cl_{2}}}{L} \right )\left ( \dfrac{-2\; mol\; NO}{1\;\cancel{mol\;Cl_{2}}} \right )=-0.056\; M \) We insert these values into our table: We sum the numbers in the [NOCl] and [NO] columns to obtain the final concentrations of NO and NOCl: We can now complete the table: We can now calculate the equilibrium constant for the reaction: \( K=\dfrac{\left [ NO_{2} \right ]^{2}\left [ Cl_{2} \right ]}{\left [ NOCl \right ]^{2}}=\dfrac{\left ( 0.056 \right )^{2}\left ( 0.028 \right )}{0.444}^{2}=4.5\times 10^{-4} \) Exercise The German chemist Fritz Haber (1868–1934; Nobel Prize in Chemistry 1918) was able to synthesize ammonia (NH ) by reacting 0.1248 M H and 0.0416 M N at about 500°C. At equilibrium, the mixture contained 0.00272 M NH . What is for the reaction N + 3 H ⇌ 2NH at this temperature? What is ? = 0.105; = 2.61 × 10 A metal catalyst bed, where ammonia was produced, is in the large cylinder at the left. The Haber-Bosch process used for the industrial production of ammonia uses essentially the same process and components but on a much larger scale. Unfortunately, Haber’s process enabled Germany to prolong World War I when German supplies of nitrogen compounds, which were used for explosives, had been exhausted in 1914. To describe how to calculate equilibrium concentrations from an equilibrium constant, we first consider a system that contains only a single product and a single reactant, the conversion of -butane to isobutane (Equation 15.26), for which = 2.6 at 25°C. If we begin with a 1.00 M sample of -butane, we can determine the concentration of -butane and isobutane at equilibrium by constructing a table showing what is known and what needs to be calculated, just as we did in Example 9. The initial concentrations of the reactant and product are both known: [ -butane] = 1.00 M and [isobutane] = 0 M. We need to calculate the equilibrium concentrations of both -butane and isobutane. Because it is generally difficult to calculate final concentrations directly, we focus on the in the concentrations of the substances between the initial and the final (equilibrium) conditions. If, for example, we define the change in the concentration of isobutane (Δ[isobutane]) as + , then the change in the concentration of -butane is Δ[ -butane] = − . This is because the balanced chemical equation for the reaction tells us that 1 mol of -butane is consumed for every 1 mol of isobutane produced. We can then express the final concentrations in terms of the initial concentrations and the changes they have undergone. Substituting the expressions for the final concentrations of -butane and isobutane from the table into the equilibrium equation, \( K=\dfrac{\left [ isobutane \right]}{\left [ n-butane \right ]}=\dfrac{x}{1.00-x}=2.6 \) Rearranging and solving for , \( x = 2.6\left ( 1.00-x \right )=2.6-2.6x \)
\( x + 2.6x =2.6 \)
\( x = 0.72 \) We obtain the final concentrations by substituting this value into the expressions for the final concentrations of -butane and isobutane listed in the table: We can check the results by substituting them back into the equilibrium constant expression to see whether they give the same that we used in the calculation: \( K=\dfrac{\left [ isobutane \right]}{\left [ n-butane \right ]}=\dfrac{0.72 \; \cancel{M}}{0.28 \; \cancel{M}}=2.6 \) This is the same we were given, so we can be confident of our results. Example 10 illustrates a common type of equilibrium problem that you are likely to encounter. The is important in several chemical processes, such as the production of H for fuel cells. This reaction can be written as follows: \( H_{2}\left ( g \right ) + CO_{2}\left ( g \right ) \rightleftharpoons H_{2}O\left ( g \right ) + CO\left ( g \right )\) = 0.106 at 700 K. If a mixture of gases that initially contains 0.0150 M H and 0.0150 M CO is allowed to equilibrate at 700 K, what are the final concentrations of all substances present? balanced equilibrium equation, , and initial concentrations final concentrations Construct a table showing what is known and what needs to be calculated. Define as the change in the concentration of one substance. Then use the reaction stoichiometry to express the changes in the concentrations of the other substances in terms of . From the values in the table, calculate the final concentrations. Write the equilibrium equation for the reaction. Substitute appropriate values from the table to obtain . Calculate the final concentrations of all species present. Check your answers by substituting these values into the equilibrium constant expression to obtain . The initial concentrations of the reactants are [H ] = [CO ] = 0.0150 M. Just as before, we will focus on the in the concentrations of the various substances between the initial and final states. If we define the change in the concentration of H O as , then Δ[H O] = + . We can use the stoichiometry of the reaction to express the changes in the concentrations of the other substances in terms of . For example, 1 mol of CO is produced for every 1 mol of H O, so the change in the CO concentration can be expressed as Δ[CO] = + . Similarly, for every 1 mol of H O produced, 1 mol each of H and CO are consumed, so the change in the concentration of the reactants is Δ[H ] = Δ[CO ] = − . We enter the values in the following table and calculate the final concentrations. We can now use the equilibrium equation and the given to solve for : \( K=\dfrac{\left [ H_{2}O] \right ] \left [ CO \right ]}{\left [ H_{2} \right ]\left [ CO_{2} \right ]}=\dfrac{\left (x \right )\left ( x \right ) }{\left ( 0.0150-x \right )\left ( 0.0150-x \right )}=\dfrac{x^{2}}{\left ( 0.0150-x \right )^{2}}=0.160 \notag \) We could solve this equation with the quadratic formula, but it is far easier to solve for by recognizing that the left side of the equation is a perfect square; that is, \[\dfrac{x^2}{(0.0150−x)^2}=\left(\dfrac{x}{0.0150−x}\right)^2=0.106 \notag \] (The quadratic formula is presented in Essential Skills 7 in .) Taking the square root of the middle and right terms, \[\dfrac{x^2}{(0.0150−x)^2} =(0.106)^{1/2}=0.326 \notag \] \[x =(0.326)(0.0150)−0.326x \notag \] \[1.326x=0.00489 \notag \] \[x =0.00369=3.69 \times 10^{−3} \notag \] The final concentrations of all species in the reaction mixture are as follows: We can check our work by inserting the calculated values back into the equilibrium constant expression: \[K=\dfrac{[H_2O,CO]}{[H_2,CO_2]}=\dfrac{(0.00369)^2}{(0.0113)^2}=0.107 \notag \] To two significant figures, this is the same as the value given in the problem, so our answer is confirmed. Exercise Hydrogen gas reacts with iodine vapor to give hydrogen iodide according to the following chemical equation: \[H_{2(g)}+I_{2(g)} \rightleftharpoons 2HI_{(g)} \notag \] = 54 at 425°C. If 0.172 M H and I are injected into a reactor and maintained at 425°C until the system equilibrates, what is the final concentration of each substance in the reaction mixture? [HI] = 0.270 M; [H ] = [I ] = 0.037 M In Example 10, the initial concentrations of the reactants were the same, which gave us an equation that was a perfect square and simplified our calculations. Often, however, the initial concentrations of the reactants are not the same, and/or one or more of the products may be present when the reaction starts. Under these conditions, there is usually no way to simplify the problem, and we must determine the equilibrium concentrations with other means. Such a case is described in Example 11. In the water–gas shift reaction shown in Example 10, a sample containing 0.632 M CO and 0.570 M H is allowed to equilibrate at 700 K. At this temperature, = 0.106. What is the composition of the reaction mixture at equilibrium? balanced equilibrium equation, concentrations of reactants, and composition of reaction mixture at equilibrium Write the equilibrium equation. Construct a table showing the initial concentrations of all substances in the mixture. Complete the table showing the changes in the concentrations ( ) and the final concentrations. Write the equilibrium constant expression for the reaction. Substitute the known value and the final concentrations to solve for . Calculate the final concentration of each substance in the reaction mixture. Check your answers by substituting these values into the equilibrium constant expression to obtain . [CO ] = 0.632 M and [H ] = 0.570 M. Again, is defined as the change in the concentration of H O: Δ[H O] = + . Because 1 mol of CO is produced for every 1 mol of H O, the change in the concentration of CO is the same as the change in the concentration of H O, so Δ[CO] = + . Similarly, because 1 mol each of H and CO are consumed for every 1 mol of H O produced, Δ[H ] = Δ[CO ] = − . The final concentrations are the sums of the initial concentrations and the changes in concentrations at equilibrium. We can now use the equilibrium equation and the known value to solve for : \[K=\dfrac{[H_2O,CO]}{[H_2,CO_2]}=\dfrac{x^2}{(0.570−x)(0.632−x)}=0.106 \notag \] In contrast to Example 10, however, there is no obvious way to simplify this expression. Thus we must expand the expression and multiply both sides by the denominator: \[x^2 = 0.106(0.360 − 1.20x + x^2) \notag \] Collecting terms on one side of the equation, \[0.894x^2 + 0.127x − 0.0382 = 0 \notag \] This equation can be solved using the quadratic formula: \[ x = \dfrac{-b \pm \sqrt{b^2-4ac}}{2a} = \dfrac{−0.127 \pm \sqrt{(0.127)^2−4(0.894)(−0.0382)}}{2(0.894)} \notag \] \[x =0.148 \text{ and } −0.290 \notag \] Only the answer with the positive value has any physical significance, so Δ[H O] = Δ[CO] = +0.148 M, and Δ[H ] = Δ[CO ] = −0.148 M. The final concentrations of all species in the reaction mixture are as follows: We can check our work by substituting these values into the equilibrium constant expression: \[K=\dfrac{[H_2O,CO]}{[H_2,CO_2]}=\dfrac{(0.148)^2}{(0.422)(0.484)}=0.107 \notag \] Because is essentially the same as the value given in the problem, our calculations are confirmed. Exercise The exercise in Example 8 showed the reaction of hydrogen and iodine vapor to form hydrogen iodide, for which = 54 at 425°C. If a sample containing 0.200 M H and 0.0450 M I is allowed to equilibrate at 425°C, what is the final concentration of each substance in the reaction mixture? [HI] = 0.0882 M; [H ] = 0.156 M; [I ] = 9.2 × 10 M In many situations it is not necessary to solve a quadratic (or higher-order) equation. Most of these cases involve reactions for which the equilibrium constant is either very small ( ≤ 10 ) or very large ( ≥ 10 ), which means that the change in the concentration (defined as ) is essentially negligible compared with the initial concentration of a substance. Knowing this simplifies the calculations dramatically, as illustrated in Example 12. Atmospheric nitrogen and oxygen react to form nitric oxide: \[N_{2(g)}+O_{2(g)} \rightleftharpoons 2NO_{(g)} \notag \] = 2.0 × 10 at 25°C. What is the partial pressure of NO in equilibrium with N and O in the atmosphere (at 1 atm, P{N } = 0.78 atm and P{O } = 0.21 atm balanced equilibrium equation and values of , P{O } and P{N } partial pressure of NO Construct a table and enter the initial partial pressures, the changes in the partial pressures that occur during the course of the reaction, and the final partial pressures of all substances. Write the equilibrium equation for the reaction. Then substitute values from the table to solve for the change in concentration ( ). Calculate the partial pressure of NO. Check your answer by substituting values into the equilibrium equation and solving for . Because we are given and partial pressures are reported in atmospheres, we will use partial pressures. The initial partial pressure of O is 0.21 atm and that of N is 0.78 atm. If we define the change in the partial pressure of NO as 2 , then the change in the partial pressure of O and of N is − because 1 mol each of N and of O is consumed for every 2 mol of NO produced. Each substance has a final partial pressure equal to the sum of the initial pressure and the change in that pressure at equilibrium. Substituting these values into the equation for the equilibrium constant, \[K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(2x)^2}{(0.78−x)(0.21−x)}=2.0 \times 10^{−31} \notag \] In principle, we could multiply out the terms in the denominator, rearrange, and solve the resulting quadratic equation. In practice, it is far easier to recognize that an equilibrium constant of this magnitude means that the extent of the reaction will be very small; therefore, the value will be negligible compared with the initial concentrations. If this assumption is correct, then to two significant figures, (0.78 − ) = 0.78 and (0.21 − ) = 0.21. Substituting these expressions into our original equation, \[\dfrac{(2x)^2}{(0.78)(0.21)} = 2.0 \times 10^{−31} \notag \] \[\dfrac{4x^2}{0.16} =2.0 \times10^{−31} \notag \] \[x^2=\dfrac{0.33 \times 10^{−31}}{4} \notag \] \[x^=9.1 \times 10^{−17} \notag \] Substituting this value of into our expressions for the final partial pressures of the substances, From these calculations, we see that our initial assumption regarding was correct: given two significant figures, 2.0 × 10 is certainly negligible compared with 0.78 and 0.21. When can we make such an assumption? As a general rule, if is less than about 5% of the total, or 10 > > 10 , then the assumption is justified. Otherwise, we must use the quadratic formula or some other approach. The results we have obtained agree with the general observation that toxic NO, an ingredient of smog, does not form from atmospheric concentrations of N and O to a substantial degree at 25°C. We can verify our results by substituting them into the original equilibrium equation: \[K_p=\dfrac{(P_{NO})^2}{(P_{N_2})(P_{O_2})}=\dfrac{(1.8 \times 10^{−16})^2}{(0.78)(0.21)}=2.0 times 10^{−31} \notag \] The final agrees with the value given at the beginning of this example. Exercise Under certain conditions, oxygen will react to form ozone, as shown in the following equation: \[H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \notag \] = 2.5 × 10 at 25°C. What ozone partial pressure is in equilibrium with oxygen in the atmosphere P(O ) ? 4.8 × 10 atm Another type of problem that can be simplified by assuming that changes in concentration are negligible is one in which the equilibrium constant is very large ( ≥ 10 ). A large equilibrium constant implies that the reactants are converted almost entirely to products, so we can assume that the reaction proceeds 100% to completion. When we solve this type of problem, we view the system as equilibrating from the side of the reaction rather than the reactants side. This approach is illustrated in Example 13. The chemical equation for the reaction of hydrogen with ethylene (C H ) to give ethane (C H ) is as follows: \[H_{2(g)}+C_2H_{4(g)} \overset{Ni}{\rightleftharpoons} C_2H_{6(g)} \notag \] = 9.6 × 10 at 25°C. If a mixture of 0.200 M H and 0.155 M C H is maintained at 25°C in the presence of a powdered nickel catalyst, what is the equilibrium concentration of each substance in the mixture? balanced chemical equation, , and initial concentrations of reactants equilibrium concentrations Construct a table showing initial concentrations, concentrations that would be present if the reaction were to go to completion, changes in concentrations, and final concentrations. Write the equilibrium constant expression for the reaction. Then substitute values from the table into the expression to solve for (the change in concentration). Calculate the equilibrium concentrations. Check your answers by substituting these values into the equilibrium equation. From the magnitude of the equilibrium constant, we see that the reaction goes essentially to completion. Because the initial concentration of ethylene (0.155 M) is less than the concentration of hydrogen (0.200 M), ethylene is the limiting reactant; that is, no more than 0.155 M ethane can be formed from 0.155 M ethylene. If the reaction were to go to completion, the concentration of ethane would be 0.155 M and the concentration of ethylene would be 0 M. Because the concentration of hydrogen is greater than what is needed for complete reaction, the concentration of unreacted hydrogen in the reaction mixture would be 0.200 M − 0.155 M = 0.045 M. The equilibrium constant for the forward reaction is very large, so the equilibrium constant for the reverse reaction must be very small. The problem then is identical to that in Example 12. If we define − as the change in the ethane concentration for the reverse reaction, then the change in the ethylene and hydrogen concentrations is + . The final equilibrium concentrations are the sums of the concentrations for the forward and reverse reactions. Substituting values into the equilibrium constant expression, \[K=\dfrac{[C_2H_6]}{[H_2,C_2H_4]}=\dfrac{0.155−x}{(0.045+x)x}=9.6 \times 10^{18} \notag \] Once again, the magnitude of the equilibrium constant tells us that the equilibrium will lie far to the right as written, so the reverse reaction is negligible. Thus is likely to be very small compared with either 0.155 M or 0.045 M, and the equation can be simplified [(0.045 + ) = 0.045 and (0.155 − ) = 0.155] as follows: \[K=\dfrac{0.155}{0.045x} = 9.6 \times 10^{18} \notag \] \[x=3.6 \times 10^{−19} \notag \] The small value indicates that our assumption concerning the reverse reaction is correct, and we can therefore calculate the final concentrations by evaluating the expressions from the last line of the table: We can verify our calculations by substituting the final concentrations into the equilibrium constant expression: \[K=\dfrac{[C_2H_6]}{[H_2,C_2H_4]}=\dfrac{0.155}{(0.045)(3.6 \times 10^{−19})}=9.6 \times 10^{18} \notag \] This value agrees with our initial value at the beginning of the example. Exercise Hydrogen reacts with chlorine gas to form hydrogen chloride: \[H_{2(g)}+Cl_{2(g)} \rightleftharpoons 2HCl_{(g)} \notag \] = 4.0 × 10 at 47°C. If a mixture of 0.257 M H and 0.392 M Cl is allowed to equilibrate at 47°C, what is the equilibrium composition of the mixture? When an equilibrium constant is calculated from equilibrium concentrations, molar concentrations or partial pressures are substituted into the equilibrium constant expression for the reaction. Equilibrium constants can be used to calculate the equilibrium concentrations of reactants and products by using the quantities or concentrations of the reactants, the stoichiometry of the balanced chemical equation for the reaction, and a tabular format to obtain the final concentrations of all species at equilibrium. Describe how to determine the magnitude of the equilibrium constant for a reaction when not all concentrations of the substances are known. Calculations involving systems with very small or very large equilibrium constants can be dramatically simplified by making certain assumptions about the concentrations of products and reactants. What are these assumptions when is (a) very large and (b) very small? Illustrate this technique using the system for which you are to calculate the concentration of the product at equilibrium starting with only A and B. Under what circumstances should simplifying assumptions not be used? In the equilibrium reaction A + B what happens to if the concentrations of the reactants are doubled? tripled? Can the same be said about the equilibrium reaction The following table shows the reported values of the equilibrium P{O } at three temperatures for the reaction for which Δ ° = 31 kJ/mol. Are these data consistent with what you would expect to occur? Why or why not? Given the equilibrium system N O what happens to if the initial pressure of N O is doubled? If is 1.7 × 10 at 2300°C, and the system initially contains 100% N O at a pressure of 2.6 × 10 atm, what is the equilibrium pressure of each component? At 430°C, 4.20 mol of HI in a 9.60 L reaction vessel reaches equilibrium according to the following equation: H (g) + I (g) At equilibrium, [H ] = 0.047 M and [HI] = 0.345 M. What are and for this reaction? Methanol, a liquid used as an automobile fuel additive, is commercially produced from carbon monoxide and hydrogen at 300°C according to the following reaction: and = 1.3 × 10 . If 56.0 g of CO is mixed with excess hydrogen in a 250 mL flask at this temperature, and the hydrogen pressure is continuously maintained at 100 atm, what would be the maximum percent yield of methanol? What pressure of hydrogen would be required to obtain a minimum yield of methanol of 95% under these conditions? Starting with pure A, if the total equilibrium pressure is 0.969 atm for the reaction what is ? The decomposition of ammonium carbamate to NH and CO at 40°C is written as If the partial pressure of NH at equilibrium is 0.242 atm, what is the equilibrium partial pressure of CO ? What is the total gas pressure of the system? What is ? At 375 K, for the reaction is 2.4, with pressures expressed in atmospheres. At 303 K, is 2.9 × 10 . For the gas-phase reaction show that = ( ) assuming ideal gas behavior. For the gas-phase reaction show that the total pressure is related to the equilibrium pressure by the following equation: \[P_T=\sqrt{K_pP_{I_2}} + P_{I_2} \notag \] Experimental data on the system are given in the following table. Graph [Br ] versus moles of Br (l) present; then write the equilibrium constant expression and determine . Data accumulated for the reaction n- at equilibrium are shown in the following table. What is the equilibrium constant for this conversion? If 1 mol of -butane is allowed to equilibrate under the same reaction conditions, what is the final number of moles of -butane and isobutane? Solid ammonium carbamate (NH CO NH ) dissociates completely to ammonia and carbon dioxide when it vaporizes: \[ NH_4CO_2NH_{2(s)} \rightleftharpoons 2NH_{3(g)}+CO_{2(g)} \notag \] At 25°C, the total pressure of the gases in equilibrium with the solid is 0.116 atm. What is the equilibrium partial pressure of each gas? What is ? If the concentration of CO is doubled and then equilibrates to its initial equilibrium partial pressure + atm, what change in the NH concentration is necessary for the system to restore equilibrium? The equilibrium constant for the reaction is = 2.2 × 10 at 100°C. If the initial concentration of COCl is 3.05 × 10 M, what is the partial pressure of each gas at equilibrium at 100°C? What assumption can be made to simplify your calculations? Aqueous dilution of IO results in the following reaction: \[IO^−_{4(aq)}+2H_2O_{(l)} \rightleftharpoons H_4IO^−_{6(aq)} \notag \] and = 3.5 × 10 . If you begin with 50 mL of a 0.896 M solution of IO that is diluted to 250 mL with water, how many moles of H IO are formed at equilibrium? Iodine and bromine react to form IBr, which then sublimes. At 184.4°C, the overall reaction proceeds according to the following equation: \[I_{2(g)}+Br_{2(g)} \rightleftharpoons 2IBr_{(g)} \notag \] = 1.2 × 10 . If you begin the reaction with 7.4 g of I vapor and 6.3 g of Br vapor in a 1.00 L container, what is the concentration of IBr(g) at equilibrium? What is the partial pressure of each gas at equilibrium? What is the total pressure of the system? For the reaction \[C_{(s)} + 12N_{2(g)}+\frac{5}{2}H_{2(g)} \rightleftharpoons CH3NH2(g) \notag \] = 1.8 × 10 . If you begin the reaction with 1.0 mol of N , 2.0 mol of H , and sufficient C(s) in a 2.00 L container, what are the concentrations of N and CH NH at equilibrium? What happens to if the concentration of H is doubled? | 30,708 | 1,949 |
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Skills to Develop Here is one of the nonsensical ways of remembering the Group I and Group II elements. For the moment, please read your text on these elements. This page is still under construction, but some questions have been made up to test your skills. Hint: 92% of the mass of the sun is hydrogen. Hint: More than 90% of the Earth's crust is silica. Silica is \(\ce{SiO2}\). Quartz and sand are typical silica, and silicon oxide is an important ingredient of many other minerals. Hint: Using a nickel catalyst to convert light hydrocarbons to \(\ce{H}\) and \(\ce{CO}\). Hint: The production of ammonia using the Haber process. Hint: potassium hydride Hint: -1 Hint: +1 Hint: Melting points increase in the order: \(\ce{Cs}\) < \(\ce{Rb}\) < \(\ce{K}\) < \(\ce{Na}\) < \(\ce{Li}\) Hint: First ionization potential decreases in the order: \(\ce{Be}\) > \(\ce{Mg}\) > \(\ce{Ca}\) > \(\ce{Sr}\) > \(\ce{Ba}\) Hint: \(\ce{Ba}\) has the highest density. Check their values up. Hint: Atomic radii of the noble gases increase with atomic number: \(\ce{He}\) < \(\ce{Ne}\) < \(\ce{Ar}\) < \(\ce{Kr}\) < \(\ce{Xe}\) Hint: \(\ce{Li}\) is the only metal that reacts with air to form \(\ce{Li3N}\). \(\ce{Li3N}\) is called lithium nitride. Nitrogen has an oxidation state of -3. Hint: By electrolysis of concentrated aqueous sodium chloride (brine). Hint: For the production of glass; things may change in the future. Hint: Barium ions form a precipitate with sulfate ion. \(\ce{Ba^2+ + SO4^2- \rightarrow BaSO4\: (insoluble)}\) Hint: By electrolysis of molten salt \(\ce{NaCl}\). Describe the chemical reaction in this process. Hint: Beryllium, \(\ce{Be}\), is stable towards it, but \(\ce{Mg}\) reacts with hot water. Hint: Calcium hydroxide and ammonia. Write the equation for the described process. Hint: Most calcium on Earth is calcium carbonate. Hint: Calcium carbonate (the ingredient of sea and egg shells) Hint: Quick lime is \(\ce{CaO}\), calcium oxide Hint: Check them out from your periodic table. Hint: By fractional distillation of air. Hint: Xenon fluoride has been made. Xenon fluoride has the formula: \(\ce{XeF4}\). | 2,154 | 1,950 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Reactions/Reactivity/Oxidative_Addition_and_Reductive_Elimination/OA4._Concerted_Oxidative_Addition |
Concerted oxidative addition is a more general reaction than polar addition, in the sense that it is not restricted to compounds that can undergo aliphatic nucleophilic substitution. It could also be thought of as non-polar oxidative addition, because it does not involve charged intermediates as seen in the polar mechanism. Aryl halides, for example, do not undergo nucleophilic substitution, but they do undergo concerted oxidative addition. Instead of proceeding step by step, the addition of both fragments is synchronized. They add to the metal at the same time. Figure OA4.1. At first, it's difficult to understand this mechanism in terms of nucleophiles and electrophiles. The reaction is generally explained in terms of molecular orbital interactions, however, that can be thought of as nucleophile-electrophile interactions. There are interactions involved in a concerted or non-polar oxidative addition. Figure OA4.2. Molecular orbital interactions in a non-polar oxidative addition. Figure OA4.3. A curved arrow representation of non-polar oxidative addition. Provide a mechanism, with curved arrows, for the following reaction. Propose a reason for the fact that one of the following dimethyl palladium compounds undergoes reductive elimination, but the other one does not. Frequently, oxidative addition and reductive elimination are combined with other reactions into catalytic cycles. These cycles form the basis of important processes used to make valuable materials. Propose catalytic cycles for the following reactions. You don't need to draw curved arrows; just provide the intermediate formed after each reaction step. , | 1,654 | 1,951 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.E%3A_Matter_and_Measurement_(Exercises) |
. In addition to these individual basis; please contact Please be sure you are familiar with the topics discussed in Essential Skills 1 ( ) before proceeding to the Conceptual Problems. Please be sure you are familiar with the topics discussed in Essential Skills 1 (Section 1.9) before proceeding to the Numerical Problems. 1. Unlike weight, mass does not depend on location. The mass of the person is therefore the same on Earth and Mars: 176 lb ÷ 2.2 lb/kg = 80 kg. 3. a. Cu: 1.12 cm b. Ca: 6.49 cm c. Ti: 2.22 cm d. Ir: 0.4376 cm Volume decreases: Ca > Ti > Cu > Ir 5. 629 g 9. 1.74 g/cm 1. Milk turns sour. This is a ________________ 2. HCl being a strong acid is a __________, Wood sawed in two is ___________ 3. CuSO is dissolved in water 4. Aluminum Phosphate has a density of 2.566 g/cm3 5. Which of the following are examples of matter? 6. The formation of gas bubbles is a sign of what type of change? 7. True or False: Bread rising is a physical property. 8. True or False: Dicing potatoes is a physical change. 9. Is sunlight matter? 10. The mass of lead is a _____________property. 1)chemical change 2) chemical property, physical change 3) physical change 4) physical property 5) All of the above 6) chemical 7) False 8) True 9) No 10) physical property | 1,302 | 1,952 |
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When it comes to buffer solution one of the most common equation is the . An important point that must be made about this equation is it's useful only if stoichiometric or initial concentration can be substituted into the equation for equilibrium concentrations. Where the Henderson-Hasselbalch approximation comes from \[HA + H_2O \rightleftharpoons H_3O^+ + A^- \label{1} \] where, We know that \(K_a\) is equal to the products over the reactants and, by definition, H O is essentially a pure liquid that we consider to be equal to one. \[K_a = [H_3O^+,A^-] \label{2} \] Take the \(-\log\) of both sides: \[-\log \; K_a = -\log([H_3O^+,A^-]) \label{3} \] \[-\log \; K_a = -\log[H_3O^+] \; -\log[A^-] \label{4} \] Using the following two relationships: \[-\log[K_a] = pK_a \label{5} \] \[-\log[H_3O^+] = pH \label{6} \] We can simplify the above equation: \[pK_a = pH - \log[A^-] \label{7} \] If we add \(\log[A^-]\) to both sides, we get the \[pH = pK_a + \log[A^-] \label{8} \] This approximation is only when: Suppose we needed to make a buffer solution with a pH of 2.11. In the first case, we would try and find a weak acid with a pK value of 2.11. However, at the same time the molarities of the acid and the its salt must be equal to one another. This will cause the two molarities to cancel; leaving the \(\log [A^-]\) equal to \(\log(1)\) which is zero. \[pH = pK_a + \log[A^-] = 2.11 + \log(1) = 2.11 \nonumber \] This is a very unlikely scenario, however, and you won't often find yourself with Case #1 What mass of \(NaC_7H_5O_2\) must be dissolved in 0.200 L of 0.30 M HC H O to produce a solution with pH = 4.78? (Assume solution volume is constant at 0.200L) \(HC_7H_5O_2 + H_20 \rightleftharpoons H_3O^+ + C_7H_5O_2\) \(K_a =6.3 \times 10^{-5}\) \(K_a = \dfrac{[H_3O^+,C_7H_5O_2]}{[HC_7H_5O_2]} = 6.3 \times 10^{-5}\) \([H_3O^+] = 10^{-pH} = 10^{-4.78} = 16.6 \times 10^{-6}\;M\;[HC_7H_5O_2] = 0.30\;M\;[C_7H_5O_2] =\) \[[C_7H_5O_2^-] = K_a \times \dfrac{[HC_7H_5O_2]}{[H_3O^+]} \nonumber \] \[1.14 \; M = 6.3 \times 10^{-5} \times \dfrac{0.30}{16.6 \times 10^{-6}} \nonumber \] Mass = 0.200 L x 1.14 mol C H O / 1L x 1mol NaC H O / 1 mol C H O x 144 g NaC H O / 1 mol NaC H O = 32.832 g NaC H O | 2,243 | 1,954 |
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In writing the conventional Lewis structures for molecules, we assume that a covalent chemical bond between two atoms involves sharing a pair of elections, one from each atom. Figure 6-5 shows how atomic orbitals can be considered to be used in bond formation. Here, we postulate that a bond is formed by the pulling together of two atomic nuclei by attractive forces exerted by the nuclei for the two paired electrons in overlapping atomic orbitals. Because atomic orbitals can hold a maximum of electrons, it is reasonable to ask why it is that two rather than one, three, or four electrons normally are involved in a bond. The answer is that two overlapping atomic orbitals can be considered to combine to give one low-energy and one high-energy (see the top part of Figure 6-6(a)).\(^2\) Orbitals that overlap as shown in Figure 6-6(a) are said to overlap in the sigma manner,\(^3\) and the bonding orbital is called a (\(\sigma\)); the antibonding orbital is called a (read "sigma star"). Two paired electrons suffice to fill the \(\sigma\) orbital. Any additional electrons must go into the high-energy \(\sigma^*\) orbital and contribute not to bonding but to repulsion between the atoms. The hydrogen molecule-ion, \(H_2^\oplus\), can be regarded as having one electron in a \(\sigma\) orbital. It has been studied in the vapor state by spectroscopic means and found to have a dissociation energy to \(H^\oplus\) and \(H \cdot\) of \(61 \: \text{kcal mol}^{-1}\) compared to the \(104.2 \: \text{kcal mol}^{-1}\) bond energy for \(H_2\). Several possible combinations of two hydrogen orbitals and from one to four electrons are shown in Figure 6-6(b). \(^2\)More about the difference between bonding and antibonding orbitals is given in . For now we will say that the property of orbitals that leads to bonding or antibonding is a property analogous to . An combination of two orbitals is bonding, and an combination is antibonding. \(^3\)The designation sigma (\(\sigma\)) denotes that orbital overlap and electron density are greatest along the internuclear axis. and (1977) | 2,123 | 1,955 |
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A mixture of metal ions in a solution can be separated by precipitation with anions such as \(\ce{Cl-}\), \(\ce{Br-}\), \(\ce{SO4^2-}\), \(\ce{CO3^2-}\), \(\ce{S^2-}\), \(\ce{Cr2O4^2-}\), \(\ce{PO4^2-}\), \(\ce{OH-}\) etc. When a metal ion or a group of metal ions form insoluble salts with a particular anion, they can be separated from others by precipitation. We can also separate the anions by precipitating them with appropriate metal ions. There are no definite dividing lines between , and , but concentrations of their saturated solutions are small, medium, and large. Solubility products are usually listed for insoluble and sparingly soluble salts, but they are not given for soluble salts. Solubility products for soluble salts are very large. What type of salts are usually soluble, sparingly soluble and insoluble? The following are some general guidelines, but these are not precise laws. These are handy rules for us to have if we deal with salts often. On the other hand, solubility is an important physical property of a substance, and these properties are listed in handbooks. Formation of crystals from a saturated solution is a phenomenon, and it can be applied to separate various chemicals or ions in a solution. When solubilities of two metal salts are very different, they can be separated by precipitation. The values for various salts are valuable information, and some data are given in the Handbook of this website. On the Handbook Menu, clicking the item Salts will give the 's of some salts. In the first two examples, we show how barium and strontium can be separated as chromate. The for strontium chromate is 3.6E-5 and the for barium chromate is 1.2E-10. What concentration of potassium chromate will precipitate the maximum amount of either the barium or the strontium chromate from an equimolar 0.30 M solution of barium and strontium ions without precipitating the other?
Since the for barium chromate is smaller, we know that \(\ce{BaCrO4}\) will form a precipitate first as \(\ce{[CrO4^2- ]}\) increases so that for \(\ce{BaCrO4}\) also increases from zero to of \(\ce{BaCrO4}\), at which point, \(\ce{BaCrO4}\) precipitates. As \(\ce{[CrO4^2- ]}\) increases, \(\ce{[Ba^2+]}\) decreases. Further increase of \(\ce{[CrO4^2- ]}\) till for \(\ce{SrCrO4}\) increases to of \(\ce{SrCrO4}\); it then precipitates. Let us write the equilibrium equations and data down to help us think. Further, let be the concentration of chromate to precipitate \(\ce{Sr^2+}\), and be that to precipitate \(\ce{Ba^2+}\). According to the definition of we have, \[\begin{array}{cccccl}
\ce{SrCrO4 &\rightarrow &Sr^2+ &+ &CrO4^2-}, &K_{\ce{sp}} = 3.6\times 10^{-5}\\
&&0.30 &&x &
\end{array}\] \[x = \dfrac{\textrm{3.6e-5}}{0.30} = 1.2 \times 10^{-4} M\] \(\begin{array}{cccccl}
\ce{BaCrO4 &\rightarrow &Ba^2+ &+ &CrO4^2-}, &K_{\ce{sp}} = 1.2 \times 10^{-10}\\
&&0.30 &&y &
\end{array}\) \(y = \dfrac{\textrm{1.2e-10}}{0.30} = 4.0 \times 10^{-10} \;M\) The 's for the two salts indicate \(\ce{BaCrO4}\) to be much less soluble, and it will precipitate before any \(\ce{SrCrO4}\) precipitates. If chromate concentration is maintained a little less than 1.2e-4 M, \(\ce{Sr^2+}\) ions will remain in the solution. In reality, to control the increase of \(\ce{[CrO4^2- ]}\) is very difficult. The for strontium chromate is \(3.6\times 10^{-5}\) and the for barium chromate is \(1.2\times 10^{-10}\). Potassium chromate is added a small amount at a time to first precipitate \(\ce{BaCrO4}\). Calculate \(\ce{[Ba^2+]}\) when the first trace of \(\ce{SrCrO4}\) precipitate starts to form in a solution that contains 0.30 M each of \(\ce{Ba^2+}\) and \(\ce{Sr^2+}\) ions.
From the solution given in , \(\ce{[CrO4^2- ]} = 3.6\times 10^{-4}\; M\) when \(\ce{SrCrO4}\) starts to form. At this concentration, the \(\ce{[Ba^2+]}\) is estimated as follows. \(\ce{[Ba^2+]\,} \textrm{3.6\times 10^{-4} = 1.2\times 10^{-10}}\) The of \(\ce{BaCrO4}\). Thus, \(\ce{[Ba^2+]} = \textrm{3.33e-7 M}\) Very small indeed, compared to 0.30. In the fresh precipitate of \(\ce{SrCrO4}\), the mole ratio of \(\ce{SrCrO4}\) to \(\ce{BaCrO4}\) is 0.30 / 3.33e-7 = 9.0e5. In other words, the amount of \(\ce{Ba^2+}\) ion in the solid is only 1e-6 (1 ppm) of all metal ions, providing that all the solid was removed when \(\ce{[CrO4^2- ]} = \textrm{3.6e-4 M}\). The calculation shown here indicates that the separation of \(\ce{Sr}\) and \(\ce{Ba}\) is pretty good. In practice, an impurity level of 1 ppm is a very small value. What reagent should you use to separate silver and lead ions that are present in a solution? What data or information will be required for this task?
The 's for salts of silver and lead are required. We list the 's for chlorides and sulfates in a table here. These value are found in the Handbook Menu of our website as Salts . Because the 's \(\ce{AgCl}\) and \(\ce{PbCl2}\) are very different, chloride, \(\ce{Cl-}\), apppears a good choice of negative ions for their separation. The literature also indicates that \(\ce{PbCl2}\) is rather soluble in warm water, and by heating the solution to 350 K (80 C), you can keep \(\ce{Pb^2+}\) ions in solution and precipitate \(\ce{AgCl}\) as a solid. The solubility of \(\ce{AgCl}\) is very small even at high temperatures. Find more detailed information about the solubility of lead chloride as a function of temperature. Can sulfate be used to separate silver and lead ions? Which one will form a precipitate first as the sulfate ion concentration increases? What is the \(\ce{[Pb^2+]}\) when \(\ce{Ag2SO4}\) begins to precipitate in a solution that contains 0.10 M \(\ce{Ag+}\)? | 5,709 | 1,956 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/09%3A_Chemical_Kinetics/9.04%3A_More_Complex_Reactions |
A major goal in chemical kinetics is to determine the sequence of elementary reactions, or the reaction mechanism, that comprise complex reactions. For example, Sherwood Rowland and Mario Molina won the Nobel Prize in Chemistry in 1995 for proposing the elementary reactions involving chlorine radicals that contribute to the overall reaction of \(O_3 \rightarrow O_2\) in the troposphere. In the following sections, we will derive rate laws for complex reaction mechanisms, including reversible, parallel and consecutive reactions. Consider the reaction in which chemical species \(\text{A}\;\) undergoes one of two irreversible first order reactions to form either species \(\text{B}\;\) or species \(\text{C}\;\): \[\begin{align} \text{A} &\overset{k_1}{\rightarrow} \text{B} \\ \text{A} &\overset{k_2}{\rightarrow} \text{C} \end{align}\] The overall reaction rate for the consumption of \(\text{A}\) can be written as: \[\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] - k_2 \left[ \text{A} \right] = - \left( k_1 + k_2 \right) \left[ \text{A} \right] \label{21.1}\] Integrating \(\left[ \text{A} \right]\) with respect to \(t\), we obtain the following equation: \[\left[ \text{A} \right] = \left[ \text{A} \right]_0 e^{-\left( k_1 + k_2 \right) t} \label{21.2}\] Plugging this expression into the equation for \(\dfrac{d \left[ \text{B} \right]}{dt}\), we obtain: \[\dfrac{d \left[ \text{B} \right]}{dt} = k_1 \left[ \text{A} \right] = k_1 \left[ \text{A} \right]_0 e^{- \left( k_1 + k_2 \right) t} \label{21.3}\] Integrating \(\left[ \text{B} \right]\) with respect to \(t\), we obtain: \[\left[ \text{B} \right] = -\dfrac{k_1 \left[ \text{A} \right]_0}{k_1 + k_2} \left( e^{-\left( k_1 + k_2 \right) t} \right) + c_1 \label{21.4}\] At \(t = 0\), \(\left[ \text{B} \right] = 0\). Therefore, \[c_1 = \dfrac{k_1 \left[ \text{A} \right]_0}{k_1 + k_2} \label{21.5}\] \[\left[ \text{B} \right] = \dfrac{k_1 \left[ \text{A} \right]_0}{k_1 + k_2} \left( 1 - e^{-\left( k_1 + k_2 \right) t} \right) \label{21.6}\] Likewise, \[\left[ \text{C} \right] = \dfrac{k_2 \left[ \text{A} \right]_0}{k_1 + k_2} \left( 1 - e^{-\left( k_1 + k_2 \right) t} \right) \label{21.7}\] The ratio of \(\left[ \text{B} \right]\) to \(\left[ \text{C} \right]\) is simply: \[\dfrac{\left[ \text{B} \right]}{\left[ \text{C} \right]} = \dfrac{k_1}{k_2} \label{21.8}\] An important parallel reaction in industry occurs in the production of ethylene oxide, a reagent in many chemical processes and also a major component in explosives. Ethylene oxide is formed through the partial oxidation of ethylene: \[2 \: C_2 H_4 + O_2 \overset{k_1}{\longrightarrow} 2 \: C_2 H_4 O\] However, ethylene can also undergo a combustion reaction: \[C_2 H_4 + 3 \: O_2 \overset{k_2}{\longrightarrow} 2 \: CO_2 + 2 \: H_2 O\] To select for the first reaction, the oxidation of ethylene takes place in the presence of a silver catalyst, which significantly increases \(k_1\) compared to \(k_2\). Figure \(\Page {1}\) displays the concentration profiles for species \(\text{A}\), \(\text{B}\), and \(\text{C}\) in a parallel reaction in which \(k_1 > k_2\). Consider the following series of first-order irreversible reactions, where species \(\text{A}\) reacts to form an intermediate species, \(\text{I}\), which then reacts to form the product, \(\text{P}\): \[\text{A} \overset{k_1}{\longrightarrow} \text{I} \overset{k_2}{\longrightarrow} \text{P}\] We can write the reaction rates of species \(\text{A}\), \(\text{I}\) and \(\text{P}\) as follows: \[\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] \label{21.9}\] \[\dfrac{d \left[ \text{I} \right]}{dt} = k_1 \left[ \text{A} \right] - k_2 \left[ \text{I} \right] \label{21.10}\] \[\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{I} \right] \label{21.11}\] As before, integrating \(\left[ \text{A} \right]\) with respect to \(t\) leads to: \[\left[ \text{A} \right] = \left[ \text{A} \right]_0 e^{-k_1 t} \label{21.12}\] The concentration of species \(\text{I}\) can be written as \[\left[ \text{I} \right] = \dfrac{k_1 \left[ \text{A} \right]_0}{k_2 - k_1} \left( e^{-k_1 t} - e^{-k_2 t} \right) \label{21.13}\] Then, solving for \(\left[ \text{P} \right]\), we find that: \[\left[ \text{P} \right] = \left[ \text{A} \right]_0 \left[ 1 + \dfrac{1}{k_1 - k_2} \left( k_2 e^{-k_1 t} - k_1 e^{-k_2 t} \right) \right] \label{21.14}\] Figure \(\Page {2}\) displays the concentration profiles for species \(\text{A}\), \(\text{I}\), and \(\text{P}\) in a consecutive reaction in which \(k_1 = k_2\). As can be seen from the figure, the concentration of species \(\text{I}\) reaches a maximum at some time, \(t_\text{max}\). Oftentimes, species \(\text{I}\) is the desired product. Returning to the oxidation of ethylene into ethylene oxide, it is important to note another reaction in which ethylene oxide can decompose into carbon dioxide and water through the following reaction \[C_2 H_4 O + \dfrac{5}{2} \: O_2 \overset{k_3}{\longrightarrow} 2 \: CO_2 + 2 \: H_2 O\] Thus, to maximize the concentration of ethylene oxide, the oxidation of ethylene is only allowed proceed to partial completion before the reaction is stopped. Finally, in the limiting case when \(k_2 \gg k_1\), we can write the concentration of \(\text{P}\) as \[\left[ \text{P} \right] \approx \left[ \text{A} \right]_0 \left\{ 1 + \dfrac{1}{-k_2} k_2 e^{-k_1 t} \right\} = \left[ \text{A} \right]_0 \left( 1 - e^{-k_1 t} \right) \label{21.15}\] Thus, when \(k_2 \gg k_1\), the reaction can be approximated as \(\text{A} \rightarrow \text{P}\) and the apparent rate law follows \(1^{st}\) order kinetics. Consider the reactions \[\text{A} \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{I} \overset{k_2}{\rightarrow} \text{P}\] We can write the reaction rates as: \[\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] + k_{-1} \left[ \text{I} \right] \label{21.16}\] \[\dfrac{d \left[ \text{I} \right]}{dt} = k_1 \left[ \text{A} \right] - k_{-1} \left[ \text{I} \right] - k_2 \left[ \text{I} \right] \label{21.17}\] \[\dfrac{ d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{I} \right] \label{21.18}\] The exact solutions of these is straightforward, in principle, but rather involved, so we will just state the exact solutions, which are \[\begin{align} \left[ \text{A} \right] \left( t \right) &= \dfrac{ \left[ \text{A} \right]_0}{2 \lambda} \left[ \left( \lambda - k_1 + K \right) e^{-\left( k_1 + K - \lambda \right) t/2} + \left( \lambda + k_1 - K \right) e^{-\left( k_1 + K + \lambda \right) t/2} \right] \\ \left[ \text{I} \right] \left( t \right) &= \dfrac{ k_1 \left[ \text{A} \right]_0}{\lambda} \left[ e^{- \left( k_1 + K - \lambda \right) t/2} - e^{-\left( k_1 + K + \lambda \right) t/2} \right] \\ \left[ \text{P} \right] \left( t \right) &= 2 k_1 k_2 \left[ \text{A} \right]_0 \left[ \dfrac{2}{\left( k_1 + K \right)^2 - \lambda^2} - \dfrac{1}{\lambda} \left( \dfrac{ e^{-\left( k_1 + K - \lambda \right) t/2}}{k_1 + K - \lambda} - \dfrac{e^{-\left( k_1 + K + \lambda \right) t/2}}{k_1 + K + \lambda} \right) \right] \end{align} \label{21.19}\] where \[\begin{align} K &= k_2 + k_{-1} \\ \lambda &= \sqrt{\left( k_1 - K \right)^2 - 4k_1 k_{-1}} \end{align} \label{21.20}\] Consider the following consecutive reaction in which the first step is reversible: \[\text{A} \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{I} \overset{k_2}{\rightarrow} \text{P}\] We can write the reaction rates as: \[\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] + k_{-1} \left[ \text{I} \right] \label{21.21}\] \[\dfrac{ d \left[ \text{I} \right]}{dt} = k_1 \left[ \text{A} \right] - k_{-1} \left[ \text{I} \right] - k_2 \left[ \text{I} \right] \label{21.22}\] \[\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{I} \right] \label{21.23}\] These equations can be solved explicitly in terms of \(\left[ \text{A} \right]\), \(\left[ \text{I} \right]\), and \(\left[ \text{P} \right]\), but the math becomes very complicated quickly. If, however, \(k_2 + k_{-1} \gg k_1\) (in other words, the rate of consumption of \(\text{I}\) is much faster than the rate of production of \(\text{I}\)), we can make the approximation that the concentration of the intermediate species, \(\text{I}\), is small and constant with time: \[\dfrac{d \left[ \text{I} \right]}{dt} \approx 0 \label{21.24}\] Equation 21.22 can now be written as \[\dfrac{d \left[ \text{I} \right]}{dt} = k_1 \left[ \text{A} \right] - k_{-1} \left[ \text{I} \right]_{ss} - k_2 \left[ \text{I} \right]_{ss} \approx 0 \label{21.25}\] where \(\left[ \text{I} \right]_{ss}\) is a constant represents the steady state concentration of intermediate species, \(\left[ \text{I} \right]\). Solving for \(\left[ \text{I} \right]_{ss}\), \[\left[ \text{I} \right]_{ss} = \dfrac{k_1}{k_{-1} + k_2} \left[ \text{A} \right] \label{21.26}\] We can then write the rate equation for species \(\text{A}\) as \[\dfrac{d \left[ \text{A} \right]}{dt} = -k_1 \left[ \text{A} \right] + k_{-1} \left[ \text{I} \right]_{ss} = -k_1 \left[ \text{A} \right] + k_{-1} \dfrac{k_1}{k_{-1} + k_2} \left[ \text{A} \right] = -\dfrac{k_1 k_2}{k_{-1} + k_2} \left[ \text{A} \right] \label{21.27}\] Integrating, \[\left[ \text{A} \right] = \left[ \text{A} \right]_0 e^{-\dfrac{k_1 k_2}{k_{-1} + k_2} t} \label{21.28}\] Equation 21.28 is the same equation we would obtain for apparent \(1^{st}\) order kinetics of the following reaction: \[\text{A} \overset{k'}{\longrightarrow} \text{P}\] where \[k' = \dfrac{k_1 k_2}{k_{-1} + k_2} \label{21.29}\] Figure \(\Page {3}\) displays the concentration profiles for species, \(\text{A}\), \(\text{I}\), and \(\text{P}\) with the condition that \(k_2 + k_{-1} \gg k_1\). These types of reaction kinetics appear when the intermediate species, \(\text{I}\), is highly reactive. Consider the isomerization of methylisonitrile gas, \(CH_3 NC\), to acetonitrile gas, \(CH_3 CN\): \[CH_3 NC \overset{k}{\longrightarrow} CH_3 CN\] If the isomerization is a unimolecular elementary reaction, we should expect to see \(1^{st}\) order rate kinetics. Experimentally, however, \(1^{st}\) order rate kinetics are only observed at high pressures. At low pressures, the reaction kinetics follow a \(2^{nd}\) order rate law: \[\dfrac{d \left[ CH_3 NC \right]}{dt} = -k \left[ CH_3 NC \right]^2 \label{21.30}\] To explain this observation, J.A. Christiansen and F.A. Lindemann proposed that gas molecules first need to be energized via intermolecular collisions before undergoing an isomerization reaction. The reaction mechanism can be expressed as the following two elementary reactions \[\begin{align} \text{A} + \text{M} &\overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{A}^* + \text{M} \\ \text{A}^* &\overset{k_2}{\rightarrow} \text{B} \end{align}\] where \(\text{M}\) can be a reactant molecule, a product molecule or another inert molecule present in the reactor. Assuming that the concentration of \(\text{A}^*\) is small, or \(k_1 \ll k_2 + k_{-1}\), we can use a steady-state approximation to solve for the concentration profile of species \(\text{B}\) with time: \[\dfrac{d \left[ \text{A}^* \right]}{dt} = k_1 \left[ \text{A} \right] \left[ \text{M} \right] - k_{-1} \left[ \text{A}^* \right]_{ss} \left[ \text{M} \right] - k_2 \left[ \text{A}^* \right]_{ss} \approx 0 \label{21.31}\] Solving for \(\left[ \text{A}^* \right]\), \[\left[ \text{A}^* \right] = \dfrac{k_1 \left[ \text{M} \right] \left[ \text{A} \right]}{k_2 + k_{-1} \left[ \text{M} \right]} \label{21.32}\] The reaction rates of species \(\text{A}\) and \(\text{B}\) can be written as \[-\dfrac{d \left[ \text{A} \right]}{dt} = \dfrac{d \left[ \text{B} \right]}{dt} = k_2 \left[ \text{A}^* \right] = \dfrac{k_1 k_2 \left[ \text{M} \right] \left[ \text{A} \right]}{k_2 + k_{-1} \left[ \text{M} \right]} = k_\text{obs} \left[ \text{A} \right] \label{21.33}\] where \[k_\text{obs} = \dfrac{k_1 k_2 \left[ \text{M} \right]}{k_2 + k_{-1} \left[ \text{M} \right]} \label{21.34}\] At high pressures, we can expect collisions to occur frequently, such that \(k_{-1} \left[ \text{M} \right] \gg k_2\). Equation 21.33 then becomes \[-\dfrac{d \left[ \text{A} \right]}{dt} = \dfrac{k_1 k_2}{k_{-1}} \left[ \text{A} \right] \label{21.35}\] which follows \(1^{st}\) order rate kinetics. At low pressures, we can expect collisions to occurs infrequently, such that \(k_{-1} \left[ \text{M} \right] \ll k_2\). In this scenario, equation 21.33 becomes \[-\dfrac{d \left[ \text{A} \right]}{dt} = k_1 \left[ \text{A} \right] \left[ \text{M} \right] \label{21.36}\] which follows second order rate kinetics, consistent with experimental observations. Consider again the following consecutive reaction in which the first step is reversible: \[\text{A} \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \text{I} \overset{k_2}{\rightarrow} \text{P}\] \[K_\text{eq} = \dfrac{k_1}{k_{-1}} \approx \dfrac{ \left[ \text{I} \right]}{\left[ \text{A} \right]} \label{21.37}\] \[\left[ \text{I} \right] = K_\text{eq} \left[ \text{A} \right] \label{21.38}\] \[\begin{align} K &\approx k_{-1} \\ \lambda &\approx \sqrt{ \left( k_1 - k_{-1} \right)^2 + 4k_1 k_{-1}} = \sqrt{k_1^2 + 2k_1 k_{-1} + k_{-1}^2} = k_1 + k_{-1} \\ \lambda - k_1 + K &\approx k_1 + k_{-1} + k_1 - k_{-1} = 2k_1 \\ \lambda + k_1 - K &\approx k_1 + k_{-1} + k_1 - k_{-1} = 2k_1 \\ k_1 + K - \lambda &\approx k_1 + k_{-1} - k_1 - k_{-1} = 0 \\ k_1 + K + \lambda &\approx k_1 + k_{-1} + k_1 + k_{-1} = 2 \left( k_1 + k_{-1} \right) \end{align} \label{21.39}\] \[\begin{align} \left[ \text{A} \right] \left( t \right) &= \dfrac{ \left[ \text{A} \right]_0}{2 \left( k_1 + k_{-1} \right)} \left[ 2k_{-1} + 2k_1 e^{-\left( k_1 + k_{-1} \right) t} \right] \\ \left[ \text{I} \right] \left( t \right) &= \dfrac{k_1 \left[ \text{A} \right]_0}{\left( k_1 + k_{-1} \right)} \left[ 1 - e^{-\left( k_1 + k_{-1} \right) t} \right] \end{align} \label{21.40}\] \[\dfrac{ \left[ \text{I} \right]}{\left[ \text{A} \right]} \equiv K_\text{eq} \rightarrow \dfrac{k_1}{k_{-1}} \label{21.41}\] \[\dfrac{d \left[ \text{P} \right]}{dt} = k_2 \left[ \text{I} \right] = k_2 K_\text{eq} \left[ \text{A} \right] = \dfrac{k_1 k_2}{k_{-1}} \left[ \text{A} \right] \label{21.42}\] \[\text{A} \overset{k'}{\longrightarrow} \text{P}\] \[k' = \dfrac{k_1 k_2}{k_{-1}} \label{21.43}\] | 14,322 | 1,957 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Basic_Principles_of_Organic_Chemistry_(Roberts_and_Caserio)/01%3A_Introduction_to_Organic_Chemistry/1.06%3A_Some_Philosophical_Observations |
As you proceed with your study of organic chemistry, you may well feel confused as to what it is you are actually dealing with. On the one hand, there will be exhortations to remember how organic chemistry pervades our everyday life. And yet, on the other hand, you also will be exhorted to think about organic compounds in terms of abstract structural formulas representing molecules when there is absolutely no way at all to deal with molecules as single entities. Especially if you are not studying organic compounds in the laboratory concurrently, you may come to confuse the abstraction of formulas and ball-and-stick models of the molecules with the reality of organic compounds, and this would be most undesirable. At each stage of the way, you should try to make, or at least visualize, a juncture between a structural formula and an actual substance in a bottle. This will not be easy - it takes time to reach the level of experience that a practicing organic chemist has so that he can tell you with some certainty that the structural formula \(21\) represents in actuality, a limpid, colorless liquid with a pleasant odor, slightly soluble in water, boiling somewhere about \(100^\text{o}\). A useful method for developing this sort of feeling for the relationship between structures and actual compounds is to check your perception of particular substances with their properties as given in a chemical handbook. One, perhaps comforting, thought for you at this time is that differences between the chemical behaviors of relatively similar organic compounds usually are ascribed to just three important and different kinds of effects - two of which have root in common experience. One, called , is a manifestation of experience that two solid objects cannot occupy the same space at once. Another is the , which boils down to a familiar catechism that like electrical charges repel each other and unlike charges attract each other. The remaining important effect, the one that has no basis in common experience, derives from quantum mechanics. The explains why benzene is unusually stable, how and why many reactions occur in special ways and, probably most important of all, the ways that organic compounds interact with electromagnetic radiation of all kinds - from radio waves to x rays. We shall try to give as clear explanations as possible of the quantum mechanical effects, but some of it will just have to be accepted as fact that we cannot ourselves experience directly nor understand intuitively. For example, when a grindstone rotates, so far as our experience goes, it can have an infinitely variable rate of rotation and, consequently, infinitely variable rotation (angular) momentum. However, molecules in the gas phase have only rotation rates and corresponding rotational momentum values. No measurement technique can detect in-between values of these quantities. Molecules are "quantized rotators." About all you can do is try to accept this fact, and if you try long enough, you may be able to substitute familiarity for understanding and be happy with that. All of us have some concepts we use continually (even perhaps unconsciously) about energy and work. Thermodynamics makes these concepts quantitative and provides very useful information about what might be called the potential for any process to occur, be it production of electricity from a battery, water running uphill, photosynthesis, or formation of nitrogen oxides in combustion of gasoline. In the past, most organic chemists seldom tried to apply thermodynamics to the reactions in which they were interested. Much of this was due to the paucity of thermodynamic data for more than a few organic compounds, but some was because organic chemists often liked to think of themselves as artistic types with little use for quantitative data on their reactions (which may have meant that they didn't really know about thermodynamics and were afraid to ask). Times have changed. Extensive thermochemical data are now available, the procedures are well understood, and the results both useful and interesting. We shall make considerable use of thermodynamics in our exposition of organic chemistry. We believe it will greatly improve your understanding of why some reactions go and others do not. Finally, you should recognize that you almost surely will have some problems with the following chapters in making decisions as to how much time and emphasis you should put on the various concepts, principles, facts, and so on, that we will present for you. As best we can, we try to help you by pointing out that this idea, fact, and so on, is "especially important," or words to that effect. Also, we have tried to underscore important information by indicating the breadth of its application to other scientific disciplines as well as to technology. In addition, we have caused considerable material to be set in smaller type and indented. Such material includes extensions of basic ideas and departments of fuller explanation. In many places, the exposition is more complete than it needs to be for you at the particular location in the book. However, you will have need for the extra material later and it will be easier to locate and easier to refresh your memory on what came before, if it is one place. We will try to indicate clearly what you should learn immediately and what you will want to come back for later. The problem is, no matter what we think is important, you or your professor will have your own judgments about relevance. And because it is quite impossible to write an individual text for your particular interests and needs, we have tried to accommodate a range of interests and needs through providing a rather rich buffet of knowledge about modern organic chemistry. Hopefully, all you will need is here, but there is surely much more, too. So, to avoid intellectual indigestion, we suggest you not try to learn everything as it comes, but rather try hardest to understand the basic ideas and concepts to which we give the greatest emphasis. As you proceed further, the really important facts, nomenclature, and so on (the kind of material that basically requires memorization), will emerge as that which, in your own course of study, you will find you use over and over again. In hope that you may wish either to learn more about particular topics or perhaps gain better understanding through exposure to a different perspective on how they can be presented, we have provided supplementary reading lists at the end of each chapter. Our text contains many exercises. You will encounter some in the middle of the chapters arranged to be closely allied to the subject at hand. Others will be in the form of supplementary exercises at the end of the chapters. Many of the exercises will be drill; many others will extend and enlarge upon the text. The more difficult problems are marked with a star (\(^\mathbf{*}\)). and (1977) | 6,963 | 1,958 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid_Base_Reactions/Neutralization |
A neutralization reaction is when an and a react to form water and a salt and involves the combination of H ions and OH ions to generate water. The neutralization of a strong acid and strong base has a pH equal to 7. The neutralization of a strong acid and weak base will have a pH of less than 7, and conversely, the resulting pH when a strong base neutralizes a weak acid will be greater than 7. When a solution is neutralized, it means that salts are formed from equal weights of acid and base. The amount of acid needed is the amount that would give one mole of protons (H ) and the amount of base needed is the amount that would give one mole of (OH ). Because salts are formed from neutralization reactions with equivalent concentrations of weights of acids and bases: parts of acid will always neutralize parts of base. Consider the reaction between \(\ce{HCl}\) and \(\ce{NaOH}\) in water: \[\underset{acid}{HCl(aq)} + \underset{base}{NaOH_{(aq)}} \leftrightharpoons \underset{salt}{NaCl_{(aq)}} + \underset{water}{H_2O_{(l)}}\] This can be written in terms of the ions (and canceled accordingly) \[\ce{H^{+}(aq)} + \cancel{\ce{Cl^{-}(aq)}} + \cancel{\ce{Na^{+}(aq)}} + \ce{OH^{-} (aq)} → \cancel{\ce{Na^{+}(aq)}} + \cancel{\ce{Cl^{-}_(aq)}} + \ce{H_2O(l)}\] When the spectator ions are removed, the net ionic equation shows the \(H^+\) and \(OH^-\) ions forming water in a strong acid, strong base reaction: \(H^+_{(aq)} + OH^-_{(aq)} \leftrightharpoons H_2O_{(l)} \) When a strong acid and a strong base fully neutralize, the pH is neutral. Neutral pH means that the pH is equal to 7.00 at 25 ºC. At this point of neutralization, there are equal amounts of \(OH^-\) and \(H_3O^+\). There is no excess \(NaOH\). The solution is \(NaCl\) at the equivalence point. When a strong acid completely neutralizes a strong base, the pH of the salt solution will always be 7. A weak acid, weak base reaction can be shown by the net ionic equation example: \(H^+ _{(aq)} + NH_{3(aq)} \leftrightharpoons NH^+_{4 (aq)} \) The equivalence point of a neutralization reaction is when both the acid and the base in the reaction have been completely consumed and neither of them are in excess. When a strong acid neutralizes a weak base, the resulting solution's pH will be less than 7. When a strong base neutralizes a weak acid, the resulting solution's pH will be greater than 7. One of the most common and widely used ways to complete a neutralization reaction is through . In a titration, an acid or a base is in a flask or a beaker. We will show two examples of a titration. The first will be the titration of an acid by a base. The second will be the titration of a base by an acid. Suppose 13.00 mL of a weak acid, with a molarity of 0.1 M, is titrated with 0.1 M NaOH. How would we draw this titration curve? First, we need to find out where our titration curve begins. To do this, we find the initial pH of the weak acid in the beaker before any NaOH is added. This is the point where our titration curve will start. To find the initial pH, we first need the concentration of H O . Set up an to find the concentration of H3O+: \[Ka=(7)(10^{-3})\] \[K_a=(7)(10^{-3})=\dfrac{(x^2)M}{(0.1-x)M}\] \[x=[H_3O^+]=0.023\;M\] Solve for pH: \[pH=-\log_{10}[H_3O^+]=-\log_{10}(0.023)=1.64\] To accurately draw our titration curve, we need to calculate a data point between the starting point and the equivalence point. To do this, we solve for the pH when neutralization is 50% complete. Solve for the moles of OH- that is added to the beaker. We can to do by first finding the volume of OH- added to the acid at half-neutralization. 50% of 13 mL= 6.5mL Use the volume and molarity to solve for moles (6.5 mL)(0.1M)= 0.65 mmol OH Now, Solve for the moles of acid to be neutralized (10 mL)(0.1M)= 1 mmol HX Set up an ICE table to determine the equilibrium concentrations of HX and X: To calculate the pH at 50% neutralization, use the . pH=pKa+log[mmol Base/mmol Acid] pH=pKa+ log[0.65mmol/0.65mmol] pH=pKa+log(1) \[pH=pKa\] Therefore, when the weak acid is 50% neutralized, pH=pKa Solve for the pH at the equivalence point. The concentration of the weak acid is half of its original concentration when neutralization is complete 0.1M/2=.05M HX Set up an ICE table to determine the concentration of OH-: Kb=(x^2)M/(0.05-x)M Since Kw=(Ka)(Kb), we can substitute Kw/Ka in place of Kb to get Kw/Ka=(x^2)/(.05) \[x=[OH^-]=(2.67)(10^{-7})\] \[pOH=-\log_{10}((2.67)(10^{-7}))=6.57\] \[pH=14-6.57=7.43\] Solve for the pH after a bit more NaOH is added past the equivalence point. This will give us an accurate idea of where the pH levels off at the endpoint. The equivalence point is when 13 mL of NaOH is added to the weak acid. Let's find the pH after 14 mL is added. Solve for the moles of OH- \[ (14 mL)(0.1M)=1.4\; mmol OH^-\] Solve for the moles of acid \[(10\; mL)(0.1\;M)= 1\;mmol \;HX\] Set up an ICE table to determine the \(OH^-\) concentration: \[[OH-]=\frac{0.4\;mmol}{10\;mL+14\;mL}=0.17\;M\] \[pOH=-log_{10}(0.17)=1.8\] \[pH=14-1.8=12.2\] We have now gathered sufficient information to construct our titration curve. In this case, we will say that a base solution is in an Erlenmeyer flask. To neutralize this base solution, you would add an acid solution from a buret into the flask. At the beginning of the titration, before adding any acid, it is necessary to add an indicator, so that there will be a color change to signal when the equivalence point has been reached. We can use the equivalence point to find molarity and vice versa. For example, if we know that it takes 10.5 mL of an unknown solution to neutralize 15 mL of 0.0853 M NaOH solution, we can find the molarity of the unknown solution using the following formula: \[M_1V_1 = M_2V_2\] where is the molarity of the first solution, is the volume in liters of the first solution, is the molarity of the second solution, and is the volume in liters of the second solution. When we plug in the values given to us into the problem, we get an equation that looks like the following: \[(0.0835)(0.015) = M_2(0.0105)\] After solving for M , we see that the molarity of the unknown solution is 0.119 M. From this problem, we see that in order to neutralize 15 mL of 0.0835 M NaOH solution, 10.5 mL of the .119 M unknown solution is needed. 1. Will the salt formed from the following reaction have a pH greater than, less than, or equal to seven? \(CH3COOH_{(aq)} + NaOH_{(s)} \leftrightharpoons Na^+ + CH3COO^- + H2O_{(l)}\) 2. How many mL of .0955 M Ba(OH) solution are required to titrate 45.00 mL of .0452 M HNO ? 3. Will the pH of the salt solution formed by the following chemical reaction be greater than, less than, or equal to seven? \(NaOH + H_2SO_4 \leftrightharpoons H_2O + NaSO_4\) 4. We know that it takes 31.00 mL of an unknown solution to neutralize 25.00 mL of .135 M KOH solution. What is the molarity of the unknown solution? 1. After looking at the net ionic equation, \[CH_3CO_2H_{(aq)} + OH^- \leftrightharpoons CH_3COO^- + H_2O_{(l)}\] we see that a weak acid, \(CH_3CO_2H\), is being neutralized by a strong base, \(OH^-\). By looking at the chart above, we can see that when a strong base neutralizes a weak acid, the . 2. By plugging the numbers given in the problem in the the equation: \[M_1V_1= M_2V_2\] we can solve for \(V_2\). \[V_2= \dfrac{M_1V_1}{M_2} = \dfrac{(0.0452)(0.045)}{0.0955} = 21.2\; mL\] Therefore it takes \(Ba(OH)_2\) to titrate 45.00 mL \(HNO_3\). 3. We know that NaOH is a strong base and H SO is a strong acid. Therefore, we know the . 4. By plugging the numbers given in the problem into the equation: \[M_1V_2 = M_2V_2\] we can solve for M . (0.135)(0.025) = M (0.031) M = 0.108 M. Therefore, the molarity of the unknown solution is . | 7,788 | 1,959 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_for_the_Biosciences_(Chang)/04%3A_The_Second_Law_of_Thermodynamics/4.02%3A_Entropy |
Entropy is a that is often erroneously referred to as the 'state of disorder' of a system. Qualitatively, entropy is simply a measure how much the energy of atoms and molecules become more spread out in a process and can be defined in terms of statistical probabilities of a system or in terms of the other thermodynamic quantities. Entropy is also the subject of the and laws of thermodynamics, which describe the changes in entropy of the universe with respect to the system and surroundings, and the entropy of substances, respectively. Entropy is a thermodynamic quantity that is generally used to describe the course of a process, that is, whether it is a spontaneous process and has a probability of occurring in a defined direction, or a non-spontaneous process and will not proceed in the defined direction, but in the reverse direction. To define entropy in a statistical manner, it helps to consider a simple system such as in Figure \(\Page {1}\). Two atoms of hydrogen gas are contained in a volume of \( V_1 \). Since all the hydrogen atoms are contained within this volume, the probability of finding any one hydrogen atom in \( V_1 \) is 1. However, if we consider half the volume of this box and call it \( V_2 \),the probability of finding any one atom in this new volume is \( \frac{1}{2}\), since it could either be in \( V_2 \) or outside. If we consider the two atoms, finding both in \( V_2 \), using the multiplication rule of probabilities, is \[ \frac{1}{2} \times \frac{1}{2} =\frac{1}{4}.\] For finding four atoms in \( V_2 \) would be \[ \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}= \frac{1}{16}.\] Therefore, the probability of finding N number of atoms in this volume is \( \frac{1}{2}^N \). Notice that the probability decreases as we increase the number of atoms. If we started with volume \( V_2 \) and expanded the box to volume \( V_1 \), the atoms would eventually distribute themselves evenly because this is the most probable state. In this way, we can define our direction of spontaneous change from the lowest to the highest state of probability. Therefore, entropy \(S\) can be expressed as \[ S=k_B \ln{\Omega} \label{1}\] where \(\Omega\) is the probability and \( k_B \) is a proportionality constant. This makes sense because entropy is an extensive property and relies on the number of molecules, when \(\Omega\) increases to \( W^2 \), S should increase to 2S. Doubling the number of molecules doubles the entropy. So far, we have been considering one system for which to calculate the entropy. If we have a process, however, we wish to calculate the change in entropy of that process from an initial state to a final state. If our initial state 1 is \( S_1=K_B \ln{\Omega}_1 \) and the final state 2 is \( S_2=K_B\ln{\Omega}_2 \), \[ \Delta S=S_2-S_1=k_B \ln \dfrac{\Omega_2}{\Omega_1} \label{2}\] using the rule for subtracting logarithms. However, we wish to define \(\Omega\) in terms of a measurable quantity. Considering the system of expanding a volume of gas molecules from above, we know that the probability is proportional to the volume raised to the number of atoms (or molecules), \( \alpha V^{N}\). Therefore, \[ \Delta S=S_2-S_1=k_B \ln \left(\dfrac{V_2}{V_1} \right)^N=Nk_B \ln \dfrac{V_2}{V_1} \label{3} \] We can define this in terms of moles of gas and not molecules by setting the \( k_{B} \) or Boltzmann constant equal to \( \frac{R}{N_A} \), where R is the gas constant and \( N_A \) is Avogadro's number. So for a expansion of an ideal gas and holding the temperature constant, \[ \Delta S=\dfrac{N}{N_A} R \ln \left(\dfrac{V_2}{V_1} \right)^N=nR\ln \dfrac{V_2}{V_1} \label{4}\] because \( \frac{N}{N_A}=n \), the number of moles. This is only defined for constant temperature because entropy can change with temperature. Furthermore, since S is a state function, we do not need to specify whether this process is reversible or irreversible. Using the statistical definition of entropy is very helpful to visualize how processes occur. However, calculating probabilities like \(\Omega\) can be very difficult. Fortunately, entropy can also be derived from thermodynamic quantities that are easier to measure. Recalling the concept of from the first law of thermodynamics, the q) absorbed by an ideal gas in a reversible, isothermal expansion is \[ q_{rev}=nRT\ln \dfrac{V_2}{V_1} \; . \label{5}\] If we divide by T, we can obtain the same equation we derived above for \( \Delta S\): \[ \Delta S=\dfrac{q_{rev}}{T}=nR\ln \dfrac{V_2}{V_1} \;. \label{6}\] We must restrict this to a reversible process because entropy is a state function, however the heat absorbed is path dependent. An irreversible expansion would result in less heat being absorbed, but the entropy change would stay the same. Then, we are left with \[ \Delta S> \dfrac{q_{irrev}}{T} \] for an irreversible process because \[ \Delta S=\Delta S_{rev}=\Delta S_{irrev} .\] This apparent discrepancy in the entropy change between an irreversible and a reversible process becomes clear when considering the changes in entropy of the surrounding and system, as described in the second law of thermodynamics. It is evident from our experience that ice melts, iron rusts, and gases mix together. However, the entropic quantity we have defined is very useful in defining whether a given reaction will occur. Remember that the rate of a reaction is independent of spontaneity. A reaction can be spontaneous but the rate so slow that we effectively will not see that reaction happen, such as diamond converting to graphite, which is a spontaneous process. Energy of all types -- in chemistry, most frequently the kinetic energy of molecules (but also including the phase change/potential energy of molecules in fusion and vaporization, as well as radiation) changes from being localized to becoming more in space if that energy is not constrained from doing so. The simplest example stereotypical is the expansion illustrated in Figure 1. The initial motional/kinetic energy (and potential energy) of the molecules in the first bulb is unchanged in such an isothermal process, but it becomes in the final larger volume. Further, this concept of energy dispersal equally applies to heating a system: a spreading of molecular energy from the volume of greater-motional energy (“warmer”) molecules in the surroundings to include the volume of a system that initially had “cooler” molecules. It is not obvious, but true, that this distribution of energy in greater space is implicit in the Gibbs free energy equation and thus in chemical reactions. “Entropy change is the measure of , whether isothermal gas expansion, gas or liquid mixing, reversible heating and phase change, or chemical reactions.” There are for entropy change. Thus, “information probability” is only one of the two requisites for entropy change. Some current approaches regarding “information entropy” are either misleading or truly fallacious, if they do not include explicit statements about the essential inclusion of molecular kinetic energy in their treatment of chemical reactions. In the early 19th century, steam engines came to play an increasingly important role in industry and transportation. However, a systematic set of theories of the conversion of thermal energy to motive power by steam engines had not yet been developed. Nicolas Léonard Sadi Carnot (1796-1832), a French military engineer, published in 1824. The book proposed a generalized theory of heat engines, as well as an idealized model of a thermodynamic system for a heat engine that is now known as the Carnot cycle. Carnot developed the foundation of the second law of thermodynamics, and is often described as the "Father of thermodynamics." The Carnot cycle consists of the following four processes: The P-V diagram of the Carnot cycle is shown in Figure \(\Page {2}\). In isothermal processes I and III, ∆U=0 because ∆T=0. In adiabatic processes II and IV, q=0. Work, heat, ∆U, and ∆H of each process in the Carnot cycle are summarized in Table \(\Page {1}\). The T-S diagram of the Carnot cycle is shown in Figure \(\Page {3}\). In isothermal processes I and III, ∆T=0. In adiabatic processes II and IV, ∆S=0 because dq=0. ∆T and ∆S of each process in the Carnot cycle are shown in Table \(\Page {2}\). The Carnot cycle is the most efficient engine possible based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures. The efficiency of the carnot engine is defined as the ratio of the energy output to the energy input. \[\begin{align*} \text{efficiency} &=\dfrac{\text{net work done by heat engine}}{\text{heat absorbed by heat engine}} =\dfrac{-w_{sys}}{q_{high}} \\[4pt] &=\dfrac{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)+nRT_{low}\ln \left(\dfrac{V_{4}}{V_{3}}\right)}{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)} \end{align*}\] Since processes II (2-3) and IV (4-1) are adiabatic, \[\left(\dfrac{T_{2}}{T_{3}}\right)^{C_{V}/R}=\dfrac{V_{3}}{V_{2}}\] and \[\left(\dfrac{T_{1}}{T_{4}}\right)^{C_{V}/R}=\dfrac{V_{4}}{V_{1}}\] And since = and = , \[\dfrac{V_{3}}{V_{4}}=\dfrac{V_{2}}{V_{1}}\] Therefore, \[\text{efficiency}=\dfrac{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)-nRT_{low}\ln\left(\dfrac{V_{2}}{V_{1}}\right)}{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)}\] \[\boxed{\text{efficiency}=\dfrac{T_{high}-T_{low}}{T_{high}}}\] The Carnot cycle has the greatest efficiency possible of an engine (although other cycles have the same efficiency) based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures. | 9,849 | 1,960 |
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Calculate the work done by the reaction \[\ce{Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)}\] when 1 mole of hydrogen gas is collected at 273 K and 1.0 atm. (Neglect volume changes other than the change in gas volume, and assume that the has behaves ideally.) If the internal energy of 3 moles of Cl is 3.96 KJ, what is the temperature of the system? The internal energy of 3 moles of neon gas (Ne) Since we can assume that conditions are ideal, we know that the temperature is 25°C or 298 K. This assumption allows us to ignore the small change in volume of the gas. Some ice cubes are placed inside a sealed plastic bag on a hot day, what is the sign of \(w\), \(q\), \(ΔH\), \(ΔU\)? What does this mean about the system? A sports bottle containing Gatorade is held in a man's hand while running on a treadmill. Suppose the Gatorade is the system in question. Answer the subsequent questions. Consider an ice cube melting. 2.5 moles of CO expands against an external pressure of 1.5 atm. It was initially at 9.0 atm and 30 C. The final volume is 35.0 L. a) \(T=\dfrac{PV}{Rn}= \dfrac{(1.5atm)(35L)}{(.08206\dfrac{L\cdot atm}{mol\cdot K})(2.5 mol)}= 255K\) b) \(w=-(nRT)ln\dfrac{P_1}{P_2}=-(2.5mol)(8.314\dfrac{J}{mol\cdot K})(255K)ln\dfrac{9.0atm}{1.5atm}=-9.5\dfrac{kJ}{mol}\) \(q_p=nCpT= n\dfrac{3}{2}RT= (2.5mol)\dfrac{3}{2}(8.314\dfrac{J}{mol\cdot K})(255K)= 7.95\dfrac{kJ}{mol}\) \(\Delta U=q+w= 7.95\dfrac{kJ}{mol} + -9.5\dfrac{kJ}{mol}= -1.55\dfrac{kJ}{mol}\)
Determine the at 343.15 K for . (Note: The heat of vaporization for water is known to be 40.79 kJ mol at 273.15 K, the of was determined to be 85.6 J K mol and the of was determined to be 54.35 J K mol ). Use the reaction below and give your final answer in kJ mol . ,Is \((q= -w\) true for a cyclic process? Yes, \(q= -w\) is a true statement. It is true because \(\Delta U\) for any cyclic process is equal to zero. A 10.0 g sheet of gold with a temperature of 18.0°C is laid flat on a sheet of iron that weights 20.0 g and has a temperature of 55.6°C. Given that the specific heats of Au and Fe are 0.129 J g °C and 0.444 J g °C , respectively, what is the final temperature of the combined metals? (Hint: The heat gained by the gold must be equal to the heat lost by the iron.) You are cooking 100 grams of chicken soup at 300K. You know that it will take 500 J to raise the temperature of your soup to the desired 350K. Assuming constant pressure, calculate the molar heat capacity of this chicken soup. Why does the molar heat of vaporization for water decrease as temperature increases? The value of ∆H for heating 1 mole of hexane is 28.9kJ/mole at 342 K and 25.6 kJ/mole at 298 K. Explain why those values are different. What is ΔH if of 3.00 moles of O (g) are cooled from 175 C to 35.0 C and \[\bar{C_{p}} = (7.50 \times 10^{-3}T - 2.30 \times 10^{-6}T^{2} J mol^{-1} K^{-1}\] \[dH = n\bar{C_{p}}dT\] \[\Delta H = (3 mol) \int_{448K}^{308K}(7.50 \times 10^{-3}T - 2.30 \times 10^{-6}T^{2}) J mol^{-1} K^{-1}dT\] \[\Delta H = (3 mol) (\dfrac{7.50 \times 10^{-3}}{2}T^{2} - \dfrac{2.30 \times 10^{-6}}{3}T^{3} J mol^{-1} K^{-1} | _{448K}^{308K})\] \[\Delta H = (3 mol) [(3.75 \times 10^{-3}(308K)^{2}-7.67 \times 10^{-7}(308K)^{3})-(3.75 \times 10^{-3}(448K)^{2}-7.67 \times 10^{-7}(448K)^{3})]J mol^{-1} K^{-1}\] \[\Delta H = (3 mol) [(333.3)-(683.7)]J mol^{-1}\] \[\Delta H = -1050 J \] For a constant-pressure process, the heat capacity per mole is given by the expression: C per mole = (16.0 + 4.35 x 10 T + 1.15 T ) J K mol Find the molar heat capacity of oxygen gas from 38°C to 115 C \[\begin{align} \Delta H &=\int_{38}^{115}(16+4.35 \times 10^{-3}T+1.15T^{2})dT \\ &= (16*115+(4.35 \times 10^{-3})/2 * 115^2 + 1.15/3 *115^3) - (16*38 +(4.35 \times 10^{-3})/2 *38^2 +1.15/3 * 38^3) \\ &= 5.63 J/mol \end{align}\] What is the heat capacity ratio (\(γ\)) if of an ideal gas if its \(C_V = 42\, J\, K^{-1} mol^{-1}\)? since \(\overline{C_{P}} = \overline{C_{V}} + R\) \[\begin{align} \gamma = \dfrac{\overline{C_{P}}}{\overline{C_{V}}} &= \dfrac{\overline{C_{V}}+R}{\overline{C_{V}}} \\ \gamma &= \dfrac{(42+8.314) J K^{-1} mol^{-1}}{42J K^{-1} mol^{-1}} \\ \gamma &= 1.2 \end{align}\] Explain (in terms of thermo chemistry) why Emperor Penguins huddle together in the Antarctic. Through laboratory experiment, a group of students found out that molar mass of a solid element can be calculated in term of the specific heat capacity : Molar mass x the specific heat of metals= 25 (Eq.1) They were also obtaining the values of the three metals, Aluminum (0.900 J/g C), Zinc (0.381 J/g C), and Arsenic (0.329 J/g C). Determine which metal violate the above equation (Eq.1) Aluminum (Al) can not be applied in this equation because the specific heat capacity of Al is approximately equal 1.0 J g ( C ) What is the molar heat capacity of lead given its specific heat is 0.128J/(gºC) at 25ºC? The first thing we do is find the molar mass of lead which is 207.2 g/mol. We multiply our molar mass by the specific heat to get the molar heat capacity. What is the total work done on 2.00 moles of an ideal monatomic gas at 3.00 atm and 4.00 L when it adiabatically expands to 7.00 L? The equation for work is \[w = n \bar{C_{V}} (T_{2} - T_{1})\] For a monatomic gas, \[\bar{C_{V}} = \dfrac{3}{2}R \ and \ \bar{C_{P}} = \dfrac{5}{2}R\] \[\gamma = \dfrac{\bar{C_{P}}}{\bar{C_{V}}} = 1.67\] Solve for T : \[PV=nRT\] \[T_{1}=\dfrac{PV}{nR}=\dfrac{(3 atm)(4L)}{(2 moles)(0.08206 \dfrac{L \ atm}{mol \ K})}\] \[T_{1} = 73.12 K\] Solve for T : \[P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}\] \[P_{2}=\dfrac{(3atm)(4L)^{1.67}}{(7L)^{1.67}}=1.178atm\] \[T_{2}=\dfrac{PV}{nR}=\dfrac{(1.178 atm)(7L)}{(2 moles)(0.08206 \dfrac{L \ atm}{mol \ K})}\] \[T_{2}=50.24K\] Solve for work: \[w = n \bar{C_{V}} (T_{2} - T_{1})\] \[w = (2 mol)(\dfrac{3}{2})(8.314 \dfrac{J}{mol \ K})(50.24K - 73.12 K)\] \[w = -571 J\] You notice your system changes volume from 10 L to 30 L with a constant pressure at 770 torr. Then, there is a decrease in pressure to 740 torr with the volume staying constant at 30 L. Calculate the total amount of work done. We split up the solution into two parts. Work done during the volume change and the work done during the pressure change. First part: Second part: Convert pressure to atm: =107.4 J Then add up the two works w = -2046.3+107.4=-1938.9 J Calculate the values of q, w, ΔU, and ΔH for the reversible adiabatic expansion of 1 mole of a diatomic ideal gas from 10.00 m to 35.0 m with an initial temperature of 298 K. Calculate the values of q, w, ΔU, and ΔH for the reversible adiabatic expansion of 1 mole of a diatomic ideal gas from 10.00 m to 40.0 m with an initial temperature of 298K. A quantity of 0.27 mole of argon is confined in a container at 3.0 atm and 298 K and then allowed to expand adiabatically under two different conditions: (a) reversibly to 0.80 atm and (b) against a constant pressure of 0.80 atm. Calculate the final temperature in each case. In an adiabatic process 0.5 moles of He in a container are allowed to expand against a constant pressure of 1.00 atm and 298 K. The final temperature after the gas has expanded is 205 K. Calculate the initial pressure inside the container. In an adiabatic process \(C_{V}=(T_{2}-T_{1})=-P_{ex}(V_{2}-V_{1})\) we get \(\dfrac{3}{2}(T_{2}-T_{1})=-P_{ex}\left ( \dfrac{T_{2}}{P_{2}}-\dfrac{T_{1}}{P_{1}} \right )\) we rearrange and get \(\dfrac{3}{2}(T_{2}-T_{1})=\dfrac{P_{ex}T_{1}}{P_{1}}-T_2\) we solve for \(P_{1}\) and get \(P_{1}=\dfrac{P_{2}}{\dfrac{T_{2}}{T_{1}}(\dfrac{5}{2})-\dfrac{3}{2}}\) \(P_{1}=\dfrac{1.00atm}{\dfrac{205K}{298K}(\dfrac{5}{2})-\dfrac{3}{2}}\) \(P_{1}=4.5 atm\) A quantity of 0.27 mole of argon is confined in a container at 3.0 atm and 298 K and then allowed to expand adiabatically under two different conditions: Calculate the final temperature in each case. Two moles of ideal diatomic gas \(H_2\) expand from 2.00 atm to 1.30 atm while being cooled from 550 K to 250 K. Compute the Change of Enthalpy (∆H) and Change of Internal Energy (∆U) for 2 different processes: a. reversible and b. irreversible. Explain and calculate the difference a and b? 0.5 mole of Helium at 298 K are allowed to expand reversibly from 5 atm to 1 atm. Find q, w and ΔU if 808 kJ/mol of heat is released by burning x (gram) amount of Iron in a constant volume bomb calorimeter. The calorimeter containing 100 g water inside has a heat capacity of 2100 JoC . Evaluate the amount of Iron needed to increase the water by 5.20C? In your constant-volume bomb calorimeter you have a 0.3677 g sample of Potassium. The water surrounding the sample increase from 18.08 ºC to 20.667 ºC. The calorimeter has a heat capacity of 1515 J/(gºC). What is the change in internal energy? Since the bomb calorimeter has a constant volume \(\Delta V\), there is no (PV) work so the first law of thermodynamics \[\Delta U =q +w\] can be simplified to \[ \Delta U = q\] Then the relationship between heat and temperature change \[q = c_v \Delta T\] can be changed \[\Delta U = m \,c_v\, \Delta T\] which for this case is \[\Delta U= ( 0.3677\;g ) (1,515\; J/(gºC)) (20.667 \;ºC -18.08 \; ºC)\] \[\Delta U =1411.1\] The enthalpy of combustion for benzoic acid (\(C_6H_5COOH\)) is commonly used as the standard \(\Delta H_{combustion} = -3226.7\,kJ\,mol^{-1}\), for calibrating bomb calorimeters at a constant-volume process. (a) \[q_{calorimeter}=-q_{benzoic\,acid} \] \[q_{calorimeter}(3.678^{o}C)=-(\dfrac{1.567g}{122.1g\,mol{-1}})(-3226.7kj\,mol^{-1})\] \[c_{calorimeter}=\dfrac{41.41kj}{3.678^{o}C}=11.26kj^{o}C^{-1}\] (b) (Hint: Calorimeter absorbs heat transfered from the combustion of maltose) \[q_{calorimeter}=-q{maltose}\] Step 1: Calculate \(\Delta rU\) \[(11.26kj\,^{o}C^{-1})(26.58^{o}C-16.42^{o}C)=-(\dfrac{.6792g}{46.06844gmol^{-1}})(\Delta rU^{o})\] \[\Delta\,rU^{o}=-\dfrac{114.4016kj}{.014743282mol}=-7.759\times10^{3} jk\, mol^{-1}\] Step 2: Calculation of enthalpy of combustion \[C_2H5OH_{(s)}+3O_{2\,(g)}\rightarrow2CO_{2(g)}+3H_2O_{(l)}\] \(\Delta n=-1\) Therefore, \[\Delta rH^{o}=\Delta rU^{o}+RT\Delta n=(-7.759\times10^{3}kj\,mol^-1)+(8.3146\times10^{-3}kj\,mol^{-1}K^{-1})(298.15K)(-1)=-7.761\times10^{3}kj\,mol^{-1}\] Step 3: Calculate the molar enthalpy of formation. \[\Delta rH=2\Delta _f\bar{H}^{o}[CO_{2(g)}]+3\Delta _f\bar{H}^{o}[H_2O_{(l)}]-\Delta _f\bar{H}^{o}[C_2H_5OH_{(s)}]-3\Delta _f\bar{H}^{o}[O_{2(g)}]\] \[=2(-393.5kj\,mol^{-1})+3(-285.8kj\,mol^{-1})-3(0kj\,mol^{-1})-(-7.761\times10^{3}kj\,mol^{-1})\] \[=6.116\times10^{3}kj\,mol^{-1}\] A quantity of 2.50 x 10 mL of 0.468 M HBr is mixed with 2.50 x 10 mL of 0.425 M KOH in a constant-pressure calorimeter that has a heat capacity of 437 J °C . The initial temperature if the HBr and KOH solutions is the same at 21.35°C. For the process \[H^+(aq) + OH^-(aq) \rightleftharpoons H_2O(l)\] the heat of neutralization is -56.2 kJ mol . What is the final temperature of the mixed solution? In a constant-pressure calorimeter with a heat capacity of 395\(J\,^{o}C^{-1}\), you mix \(1.5\times10^{3}\,mL\) of HBr at 0.788M and \(1.5\times10^{3}mL\) \(Ca(OH)_2\) at .394M. For this process to occur the heat of neutralization is \(-56.2kj\,mol^{-1}\). The initial temperature of both solutions is \(23.08^{o}C\). \[H^+_{(aq)}+OH^-_{(aq)}\rightarrow H_2O_{(l)}\] Find the final temperature of the mixed solution. Step 1: (Hint: Determine the number of moles of the reactants) \[n_{H^{+}}=(1.5L)(0.788\,mol\,L^{-1})=1.182\,mol\] \[n_{OH^{-}}=2(1.5L)(0.394\,mol\,L^{-1})=1.182\,mol\] Step 2: (Hint: Calculate the the thermal energy released by the reaction) \[q_P=(-52.2\,kJ\,mol^{-1})(1.182\,mol)=-66.42\, kJ\] Step 3: (Hint: Assume the the density of the solutions are the same as water and specific heat is same as water too.) \[s_{solution}=4.184\,J\, g^{-1}^{o}C^{-1}\] \[m_{solution}=m_{HBr}+m_{Ca(OH)_2}\] \[=(1.5 \times 10^{3}mL)(1.00gmL^{-1})+(1.5 \times 10^{3}mL)(1.00gmL^{-1})=3.00 \times 10^3g\] Step 4: \[66.4284\, kJ=395J^{o}C^{-1}+(3.00 \times 10^{3}g)(4.184\, J\, g^{-1}\,^{o}C^{-1})(\dfrac{1kJ}{1000\,J})\Delta T\] \[66.4284=(12.947^{o}C^{-1})\Delta T\] \[\Delta T=5.13^{o}C\] Step 5: \[\Delta T=T_f-T_i\] \[T_f=28.21^{o}C\] A quantity of \(2.50 \times 10^2\; mL\) of \(0.468;\; M \;HBr\) is mixed with \(2.50 \times 10^2\; mL\) of \(0.425\; M\; KOH\) in a constant-pressure calorimeter that has a heat capacity of 437 J °C . The initial temperatures of the \(HBr\) and \(KOH\) solutions are the same at 21.35 °C. The heat of neutralization for the below reaction is \(-56.2\; kJ/mol \[H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)}\] What is the final temperature of the mixed solution? 1.52 g of toluene \((C_7H_8)\) is combusted in a constant volume bomb calorimeter at 25°C. Calculate the \(\Delta _r\bar{U}\) and \(\Delta _r\bar{H}\) for this reaction given that 39.183 kJ of heat was released. First we need to construct the balanced reaction that occurs \[C_7H_8(l)+9O_2(g)\rightarrow7CO_2(g)+4H_2O(l)\] \(\Delta _r\bar{U}=q_r=-q_{cal}=\dfrac{-39.183kJ}{\dfrac{1.52g}{92.14(g/mol)}}=-2375\; kJ/mol\) \(\begin{align*}
\Delta _r\bar{H}&=\Delta _r\bar{U}+\Delta (PV)=\Delta _r\bar{U}+\Delta n(RT)\\&=-2375 \; kJ/mol + (-2)(8.314 \times 10^{-3}\; kJ/K*mol)(298.15\; K)=-2380 \; kJ/mol \end{align*} \) The of ΔU a of the combustion of naphthalene (C H ) are -480 kj mol . The heat evolved from the combustion in a constant-volume calorimeter is 7.5 KJ. Calculate the mass of naphthalene in grams in the calorimeter. For the reaction \[C_2H_5OH_{(l)} + O_{2 (g)} \rightarrow CO_{2 (g)} + H_2O_{(l)}\] Consider the following reaction: \[\begin{align*}C_2H_6(g)+\dfrac{7}{2}O_2(g)\rightarrow &2CO_2(g)+3H_2O(l)\\
&\Delta_rH^\circ=-1560kJ/mol\end{align*}\] What is the value of \(\Delta_rH^\circ\) if On what basis is the 0 value for standard enthalpy of formation assigned to a substance at a specific temperature? Base on that, identify which substances have Δ H° = 0 and which do not at 1 atm and 25°C (give explanation for those that do not). H (g), O (g), Al(s), Br (g), NaCl(s), Cl 0 value for standard enthalpy of formation is assigned to elemental substances that are in their natural/standard physical state at that specific condition of temperature and pressure. Which of the following substances has a standard enthalpy of formation, at 298 K, of 0 K? O (g), Br (g), Hg(s), CH (g) Why is the standard enthalpy of formation of O zero at a temperature of 298 K? The standard enthalpy of formation of O is zero at 298 K since O is the most stable allotropic form of oxygen at that particular temperature. a) The ionization of a weak acetic acid is written as: \[ CH_{3}COOH(aq) \rightleftharpoons CH_{3}COO^{-} + H^+ \] Suppose the change of reaction enthalpy for this reaction is +1.0 kJ/mol and the H° of CH COO is -486.01 kJ/mol. What would be the molar enthalpy of formation of aqueous acetic acid? b) The acetic acid is reacted with a strong base NaOH. Calculate the enthalpy of neutralization for this reaction of weak acid strong base. Δ H°[OH (aq)] = -229.6 kJ/mol and Δ H°[H O(l)] = -285.8 kJ/mol Hint: Look at \[\Delta _r \bar{H}^{\circ} = \sum{[\Delta_{f} \bar{H}^{\circ}}]reactants - \sum{[\Delta_{f} \bar{H}^{\circ}}]products\] \[\Delta _r \bar{H}^{\circ} = \Delta_{f} \bar{H}^{\circ}[CH_{3}COO^-] + \Delta_{f} \bar{H}^{\circ}[H^+] - \Delta_{f} \bar{H}^{\circ}[CH_{3}COOH]\] \[+1.0 \dfrac{kJ}{mol} = -486.01 \dfrac{kJ}{mol} + 0 \dfrac{kJ}{mol} - \Delta _f \bar{H}^{\circ}[CH_{3}COOH]\] \[ {\color{red} \Delta _f \bar{H}^{\circ}[CH_{3}COOH] = -487.01 \dfrac{kJ}{mol}}\] b) Note that acetic acid is a weak acid, therefore only dissociates partially or does not dissociate at all. Therefore the complete ionic reaction of reaction is: \[CH_{3}COOH(aq) + Na^+ (aq) + OH^- (aq) = CH_3COO^- (aq) + Na^+ (aq) + H^2O(l)\] Since Na cancels out, the net equation is \[CH_{3}COOH(aq) + OH^- (aq) = CH_3COO^- (aq) + H^2O(l)\] \[\Delta _r \bar{H}^{\circ} = \Delta_{f} \bar{H}^{\circ}[CH_{3}COO^-] + \Delta_{f} \bar{H}^{\circ}[H_2O] - \Delta_{f} \bar{H}^{\circ}[CH_{3}COOH] - \Delta_{f} \bar{H}^{\circ}[OH^-] \] \[\Delta _r \bar{H}^{\circ} = (-486.01 -285.5 + 486.01+229.6) \dfrac{kJ}{mol}\] \[ {\color{red}\Delta _r \bar{H}^{\circ} = -55.9 \dfrac{kJ}{mol}}\] Calculate the enthalpy of formation for Br ions in the following reactions: \[\Delta _rH^\circ =\Delta _f\overline{H}^\circ (H^+)+\Delta _f\overline{H}^\circ (Br^-)-\Delta _f\overline{H}^\circ (HBr)\] \[ \Delta _f\overline{H}^\circ (Br^-)=\Delta _rH^\circ -\Delta _f\overline{H}^\circ (H^+)+\Delta _f\overline{H}^\circ (HBr)\] \[\Delta _f\overline{H}^\circ (Br^-)=36.4\dfrac{kJ}{mol} -0+(-.105.7\dfrac{kJ}{mol})\] \[\Delta _f\overline{H}^\circ (Br^-)=69.3\dfrac{kJ}{mol}\] When determining standard enthalpies of formation of ions in aqueous solutions, set the value of \(H^+\) ion to 0 for \(\Delta _{f}\bar{H}^o[H^{+}_{(aq)}]=0.\) (a) $$\Delta rH^{o}=-98.4kj\,mol^{-1}=\Delta _f\bar{H}^{o}[H^+_{(aq)}]+\Delta _f\bar{H}^{o}[Br^-_{(aq)}]-\Delta _f\bar{H}^{o}[HBr_{(g)}]\] $$\Delta _f\bar{H}^{o}[Br^-_{(aq)}]=-98.4kj\,mol^{-1}-\Delta _f\bar{H}^{o}[H^+_{(aq)}]+\Delta _f\bar{H}^{o}[HBr_{(g)}]\] $$=-98.4kj\,mol^{-1}-0kj\,mol^{-1}+(-36.29kj\,mol^{-1})\] $$=-134.69kj\,mol^{-1}\] (b) The neutralization for 1 mole of \(H_2O\) is \(\Delta rH^{o}=-53.4kj\,mol^{-1}\) $$\Delta rH^{o}=\Delta_f\bar{H}^{o}[H_2O_{(l)}]-\Delta_f\bar{H}^{o}[H^+_{(aq)}]-\Delta_f\bar{H}^{o}[OH^-_{(aq)}]\] $$\Delta_f\bar{H}^{o}[OH^-_{(aq)}]=\Delta_f\bar{H}^{o}[H_2O_{(l)}]-\Delta_f\bar{H}^{o}[H^+_{(aq)}]+53.4kj\,mol^{-1}\] $$=(-285.8kj\,mol^{-1})-0kj\,mol^{-1}+53.4kj\,mol^{-1}\] $$=-232.4kj\,mol^{-1}\] For the following chemical reaction, calculate the . \[ \begin{align} H_{3}PO_{4} (aq) & \rightarrow H_{2}PO_{4}^{2-}(aq) + H^{+} (aq) \\ \triangle _{f}H^{\circ}_{H_{3}PO_{4}(aq)} & = -1277 \ \ kJ \ mol^{-1} \\ \triangle _{f}H^{\circ}_{H^{+}(aq)} & = \ \ 0 \ \ kJ \ mol^{-1} \\ \triangle _{f}H^{\circ}_{H_{2}PO_{4}^{2-}(aq)} & = -1302.5 \ \ kJ \ mol^{-1} \end{align}\] Use the following equation: \[ \Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}\] Solve \[ \begin{align} \Delta H_{reaction}^o & = \Delta H_{f}^o(H_2PO_4^{2-})+\Delta H_{f}^o(H^{+}) - \Delta H_{f}^o(H_3PO_4)\\ & = -1302.5 \ kJ \ mol^{-1} - (-1277 \ kJ \ mol^{-1}) \\ & \underline {= -25.5 \ kJ \ mol^{-1}} \end{align}\] for more information about standard enthalpies of reaction. The reaction for the dissociation of hydrobromic acid in water is as follows: \[HBr_{(g)}\rightleftharpoons H^{+}_{(aq)}+Br_{(aq)}^{-}\, \, \, \, \, \, \, \, \, \, \, \, \Delta _{r}H^{\circ}=-85.2\frac{KJ}{mol}\] \[\Delta _{f}H^{\circ}(HBr)=-36.4\frac{KJ}{mol}\] a.) Given the enthalpy of formation for HBr, what is the enthalpy of formation for the Bromine ion? Assume that the enthalpy of formation for the Hydrogen ion is 0. b.) What is the enthalpy of reaction for the neutralization of HBr with KOH? Assume that the potassium and bromine ions remain in solution following the . \[\Delta _{f}H^{\circ}(OH_{(aq)}^{-})=-36.4\frac{KJ}{mol} \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \Delta _{f}H^{\circ}(H_{2}O_{(l)})=-285.8\frac{KJ}{mol} \] a.) For any reaction, \[\Delta _{r}H^{\circ}=\sum \nu \Delta _{f}\bar{H}^{\circ}(products)-\sum \nu \Delta _{f}\bar{H}^{\circ}(reactants)\] For this reaction, \[\Delta _{r}H^{\circ}= \Delta _{f}\bar{H}^{\circ}(Br^{-}_{(aq)})+\Delta _{f}\bar{H}^{\circ}(H_{(aq)}^{+})-\Delta _{f}\bar{H}^{\circ}(HBr_{(aq)})\] \[\Delta _{f}\bar{H}^{\circ}(Br^{-}_{(aq)})=\Delta _{r}H^{\circ}+\Delta _{f}\bar{H}^{\circ}(HBr_{(aq)})\] \[=\left (-85.2\frac{KJ}{mol} \right )+\left (-36.4 \frac{KJ}{mol} \right )=-121.6\frac{KJ}{mol}\] b.) For any neutraliztion reaction in solution, \[OH_{(aq)}^{-}+H_{(aq)}^{+}\rightleftharpoons H_{2}O_{(l)}\] \[\Delta _{r}H^{\circ}= \Delta _{f}\bar{H}^{\circ}(H_{2}O_{(l)})-\Delta _{f}\bar{H}^{\circ}(OH_{(aq)}^{-})\] \[=-285.8\frac{KJ}{mol}-\left ( -36.4 \frac{KJ}{mol}\right )=-249.4\frac{KJ}{mol}\] How much heat is given off when one explodes 24.65 grams of ? The equation for the combustion of is given below: \[ \begin{align} & 4 C_3H_5N_3O_9(l) \rightarrow 12 CO2(g) + 10 H_2O(g) + 6 N_2(g) + O_2(g) \\ & \triangle _{r}H^{\circ}_{C_3H_5N_3O_9(l)}= -1.529 \ MJ \ mol^{-1} \end{align}\] The molar mass of nitroglycerin is 227.0865 g/mol. To solve this question simply multiple the enthalpy of reaction by the amount moles of nitroglycerin. \[ \begin{align} 24.65 \ g \div 227.0865 \ g \ mol^{-1} \ &= 0.109 \ moles \ of \ nitroglycerin \\ 1.529 \ MJ \ mol^{-1} \cdot 0.109 \ mol &= \underline{ 0.166 MJ \ of \ heat } \end{align}\] via CO + H O → H CO ? The change in enthalpy associated with this reaction is +360.19 kJ/mol. ΔH The following reaction was carried out at standard temperature and pressure: \[2C_{(graphite)}+3H_{2(g)}\rightarrow C_{2}H_{6(g)}\, \, \, \, \, \, \, \, \, \, \Delta _{r}H^{\circ}=-84.7\frac{KJ}{mol}\] If -148.16 KJ of heat was released during this reaction, how many grams of Ethane were formed? For any reaction: \[q=n\Delta _{f}H^{\circ}=\frac{m}{\mathfrak{M}}\Delta _{f}H^{\circ}\] This can be rearranged as \[m=\frac{q\mathfrak{M}}{\Delta _{f}H^{\circ}}\] The molar mass of ethane is \[\mathfrak{M}=2(12.01 \frac{g}{mol})+6(1.008\frac{g}{mol})=30.07 \frac{g}{mol}\] The mass of ethane produced, then, is \[m=\frac{q\mathfrak{M}}{\Delta _{f}H^{\circ}}=\frac{(-148.16 KJ)(30.07\frac{g}{mol})}{-84.7\frac{KJ}{mol}}=52.6g\] When 2.3 g of calcium carbonate is decomposed under constant pressure, 178.3 kJ of heat were released. What is the \(\Delta_rH^\circ \) for this reaction? \[CaCO_3(s)\rightarrow CaO(s)+CO_2(g)\] \[\Delta _rH^\circ = q_p=\dfrac{178.3kJ}{\dfrac{2.3g}{100.09g/mol}}=7759.15kJ/mol\] The isomerization process from 1-butene to 2-butene has a reaction enthalpy of -7.1 kJ/mol. 1-butene 2-butene If the hydrogenation of 1 butene has a reaction enthalpy of -126.8 kJ/mol. How much energy would the hydrogenation of 2-butene takes? Keep in mind that both processes yield the same product. Hint: Look at example problems of 1-butene + 2H → butane 1-butene + 2H → butane From the reaction below, calculate the enthalpy of formation of methane gas. \[ \begin{align} & CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + 2 H_2O(l) \\ & \triangle _{r}H^{\circ} = -891 kJ \ mol^{-1} \\ & \triangle _{f}H^{\circ}_{CO_2(g)} = -393.5 kJ \ mol^{-1} \\ & \triangle _{f}H^{\circ}_{H_2O (l)} = -285.8 kJ \ mol^{-1} \end{align}\] Use the following equation: \[ \Delta H_{reaction}^o = \sum {\Delta H_{f}^o(products)} - \sum {\Delta H_{f}^o(Reactants)}\] Plug in appropriate values and solve. Change in enthalpy of oxygen is 0 because it is the reference form of a pure element. \[ \Delta H_{r}^o = \Delta H_{f}^o(CO_2) + (2 \cdot \Delta H_{f}^o(H_2O)) - (2 \cdot \Delta H_{f}^o(O_2)) - \Delta H_{f}^o(CH_4)\] \[ \begin{align} -891 \ kJ \ mol^{-1}&=(-393.5 \ kJ \ mol^{-1}) + (-285.8 \ kJ \ mol^{-1}) - (\Delta H_{f}^o(CH_4)) \\ \Delta H_{f}^o(CH_4)&= \underline{ 211.7\ kJ \ mol^{-1} } \end{align}\] for more information on standard enthalpies. For the following combustion reaction: \[C_{3}H_{7}OH_{(l)}+3O_{2(g)}\rightarrow 4H_{2}O_{(l)}+3CO_{2(g)}\, \, \, \, \, \, \, \, \, \, \,\Delta _{r}H^{\circ}=-2005.8\frac{KJ}{mol}\] Calculate the enthalpy of formation for 2-propanol, given the following enthalpies of formation \[\Delta _{f}\bar{H}^{\circ}(H_{2}O)=-285.8 \frac{KJ}{mol}\, \, \, \, \, \, \, \, \, \, \, \, \, \, \Delta _{f}\bar{H}^{\circ}(CO_{2})=-393.5 \frac{KJ}{mol}\] In general: \[\Delta _{r}H^{\circ}=\sum \nu \Delta _{f}\bar{H}^{\circ}(products)-\sum \nu \Delta _{f}\bar{H}^{\circ}(reactants)\] In general, \[\Delta _{r}H^{\circ}=3 \Delta _{f}\bar{H}^{\circ}(CO_{2})+4\Delta _{f}\bar{H}^{\circ}(H_{2}O)-\Delta _{f}\bar{H}^{\circ}(C_{3}H_{7}OH)\] \[\Delta _{f}\bar{H}^{\circ}(C_{3}H_{7}OH)=3 \Delta _{f}\bar{H}^{\circ}(CO_{2})+4\Delta _{f}\bar{H}^{\circ}(H_{2}O)-\Delta _{r}H^{\circ}\] \[=3\left (-393.5\frac{KJ}{mol} \right )+4\left (-285.8\frac{KJ}{mol} \right )-\left (-2005.8\frac{KJ}{mol} \right )=-317.9 \frac{KJ}{mol}\] At constant pressure, the change in enthalpy with respect to temperature is given by Cp. By determining the change in temperature and multiplying it by Cp we can find the change in enthalpy. ΔH = 31.3 J/K*mol * 440K = 13.772 kJ/mol The combustion of graphite is C(gr) + O (g) → CO (g) Calculate the change in the enthalpy of combustion from 298 K to 398 K. The molar C values are (in J K mol ): C(gr):8.52, O (g):29.4, and CO (g):37.1 What values are needed to calculate the change in standard enthalpy of a reaction? Consider the following reactions: C(graphite) + O → CO (s) ∆ H = -110.5 kJ/mole CO (g) → Co (s) + O ∆ H = 283.0 kJ/mole Find the enthalpy of formation of carbon dioxide (CO reaction is the sum of \(\Delta H_1\) and \(\Delta H_2\) \[2C + O_2 \rightarrow 2CO \,\,, \Delta H_1 = -110.5 kJ/mol \[2CO + O2 \rightarrow 2CO_2 \Delta H_2 = -283 kJ/mol thus the \(\Delta H\) of reaction is (-110.5)+(-283.0) = -393.5 kJ/mol What is Δ H of the following reaction? Δ H for O(g) is 249.4 kJ mol and Δ H for O (g) is 142.7 kJ mol \[2O_{2(g)} \rightarrow O_{(g)} + O_{3(g) }\] \[\Delta H^{\circ}_{rxn} = \sum \Delta H^{\circ}_{prod} - \sum \Delta H^{\circ}_{reactants}\] \[\Delta H^{\circ}_{rxn} = [\Delta H^{\circ}(O(g)) +\Delta H^{\circ}(O_{3}(g))]-2\Delta H^{\circ}(O_{2}(g))\] \[\Delta H^{\circ}_{rxn} = (249.4 + 142.7) \dfrac{kJ}{mol} \] \[\Delta H^{\circ}_{rxn} = 392.1 \dfrac{kJ}{mol} \] Methane combusts, what is the enthalpy of combustion for this reaction? CH (g) Write the balanced formula for this reaction to identify the coefficients \[CH_{4(g)} +2 O_{2(g)} \rightarrow 2H_2O_{(l)} + CO_{2(g)}\] \[\Delta H= \sum H_{products} - sum H_{reactants}\] \[= 2x H(H_2O) + H(CO_2) - H(CH_4)\] We do not include \(O_2\) because it is a standard state \[=[(2 x -285.8 + -393.5) - (-74.85)] kJ/mol\] \[=-890.25\; kJ/mol\] Predict whether the values of q, w, ΔU, and ΔH are positive, zero, or negative for each of the following processes: (a) freezing of ice at 1 atm and 273 K, (b) melting of solid butanol at 1 atm and the normal melting point, (c) reversible isothermal compression of an ideal gas, and (d) reversible adiabatic compression of an ideal gas. Predict whether the values of q, w, ΔU, and ΔH are positive, zero, or negative for each of the following processes: (a) freezing of ice at 1 atm and 273 K, (b) melting of solid butanol at 1 atm and the normal melting point, (c) reversible isothermal compression of an ideal gas, and (d) reversible adiabatic compression of an ideal gas. The first law of thermodynamics states that energy is conserved. It cannot be destroyed nor created. Using Einstein’s equation E= mc explain how large amounts of energy are necessary to fuse atoms together. If energy is conserved where did all this energy go? When large amounts of energy are used to fuse atoms together (eg: fusion) we can observe that the mass of the atoms is very small compared to the amount of energy used. However, using Einstein's equation E= mc we see that the mass is multiplied times the speed of light squared which will result in a very large number. The energy was not destroyed but rather stored in the chemical bonds of the atoms. For the nuclear process we cannot apply the assumption that \(∆_fH = 0\) for most stable elements because there are different components in the reactant and product sides. Explain this difference? You eat 2 pounds of fries for lunch and the fuel value is approximately \(2.9kcal\,g^{-1}\). If you don't store any of the energy, calculate the amount of water (evaporated perspiration) needed to keep your body temperature constant. Step1: (Hint: Calculate the fuel value for 2 pound of fries) \[(2\;pounds\;of\;fries)(\dfrac{454g\,fries}{1\,pound\,fries})(\dfrac{2.9kcal\,fries}{g\,fries})(\dfrac{4.184kJ}{1kcal})=1.1017\times10^4kj\] Step 2: (Hint: Use \(\Delta _{vap}H\) to calculate grams of \(H_2O\)) \[H_2O_{(l)}\rightarrow H_2O{(g)}\] \[\Delta _{vap}H=44.01kjmol^{-1} at\;298K\] \[(\dfrac{1.1017\times10^4jk}{44.01kj\,mol^{-1}})(\dfrac{18.02H_2O}{1mol\,H_2O})=4.51\times10^3g\,H_2O\] What is the final temperature of the following reaction: \[C_3H_{8(g)}+5O_{2(g)} \rightarrow 3CO_{2(g)} +4H_2O_{(l)}\] Hint: first find \(\Delta_rH^\circ\), then look up the standard molar heat capacities of the products. \(\begin{align*}\Delta _rH^\circ &=\sum n\Delta _fH^\circ_{product}-\sum n\Delta _fH^\circ_{reactant}\\&=3(-393.5kJ/mol)+4(-241.8kJ/mol)-(-104.6kJ/mol)-5(0kJ/mol)\\&=-2043.1kJ/mol\end{align*}\) \(\Delta_rH^\circ=q_{p,rxn}=-q_{p,absorbed by product}= (n\bar{C}_{p,CO_2}+n\bar{C}_{p,H_2O})\Delta T\\
\Delta T=T_f-T_i=\dfrac{-q_{p,abs by prod}}{n\bar{C}_{p,CO_2}+n\bar{C}_{p,H_2O}}\) \(\begin{align*}T_f&=T_i+\dfrac{-q_{p,abs by prod}}{n\bar{C}_{p,CO_2}+n\bar{C}_{p,H_2O}}\\
&=298.15K+\dfrac{2043.1kJ/mol}{3(37.1*10^{-3}kJ/K*mol)+4(33.6*10^{-3}kJ/K*mol)}=8613.6K\end{align*}\) The reaction below is a oxidation of 2-propanol to make acetone a) Calculate the change of enthalpy of reaction by using bond enthalpies b) Compare the result with the reaction of the combustion 2-propanol that has an enthalpy of -1825.5 kJ/mol. What does this say about the two reactions. (Hint: Look at ). \[\Delta _{r} H^{\circ} = \sum BE(reactants) - \sum BE(products) =[(2898 + 694+351+460) - (2484+694+724+460)] \frac{kJ}{mol} = {\color{red}+41 \frac{kJ}{mol}} \] The molar enthalpy of sublimation for an unknown organic substance is found to be 58.05 KJ/mol, and the molar enthalpy of vaporization is 37.12 KJ/mol. Using these values, provide an estimate for the molar enthalpy of fusion for this substance. In general, \[\Delta _{sub}\bar{H}^{\circ}=\Delta _{vap}\bar{H}^{\circ}+\Delta _{fus}\bar{H}^{\circ}\] \[\Delta _{fus}\bar{H}^{\circ}=\Delta _{sub}\bar{H}^{\circ}-\Delta _{vap}\bar{H}^{\circ}=58.05\frac{KJ}{mol}-37.12\frac{KJ}{mol}=20.93\frac{KJ}{mol}\] Calculate the standard enthalpy of sublimation of ethanol. and You can estimate the standard enthalpy of sublimation by adding up the standard enthalpies of fusion and vaporization since sublimation consists of two simultaneous phase changes. See for more information on it. H (ethanol) = 4.9+38.56 = → The dissociation of water is H O(l) -> H (aq) + OH (aq), which when added to the neutralization reaction above, gives the overall reaction \[HNO_3 (aq) \rightarrow H^+(aq) + NO_3^-(aq)\] which is the dissociation reaction for nitric acid. The enthalpy change of the first reaction is given by the difference in the total enthalpies of formation of the products and reactants, i.e. Δ H=(-206.6 kJ/mol -285.8 kJ/mol) - (-229.6 kJ/mol -207.6 kJ/mol) = -55.2 kJ/mol. When added to the enthalpy change for the dissociation of water to obtain the enthalpy change for the dissociation of nitric acid, we have: Δ H=-55.2 kJ/mol + 55.8 kJ/mol = 0.6 kJ/mol | 30,911 | 1,964 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Reactivity_of_Alkenes/Addition_Reactions_of_Alkenes/Addition_of_Strong_Br%C3%B8nsted_Acids |
Strong Brønsted acids such as HCl, HBr, HI & H SO , rapidly add to the C=C functional group of alkenes to give products in which new covalent bonds are formed to hydrogen and to the conjugate base of the acid. Using the above equation as a guide, write the addition products expected on reacting each of these reagents with cyclohexene. Weak Brønsted acids such as water (pK = 15.7) and acetic acid (pK = 4.75) do not normally add to alkenes. However, the addition of a strong acid serves to catalyze the addition of water, and in this way alcohols may be prepared from alkenes. For example, if sulfuric acid is dissolved in water it is completely ionized to the hydronium ion, H O , and this strongly acidic (pK = -1.74) species effects hydration of ethene and other alkenes. CH =CH + CH –CH + The importance of choosing an appropriate solvent for these addition reactions should now be clear. If the addition of HCl, HBr or HI is desired, water and alcohols should not be used. These strong acids will ionize in such solvents to give ROH and the nucleophilic oxygen of the solvent will compete with the halide anions in the final step, giving alcohol and ether products. By using inert solvents such as hexane, benzene and methylene chloride, these competing solvent additions are avoided. Because these additions proceed by way of polar or ionic intermediates, the rate of reaction is greater in polar solvents, such as nitromethane and acetonitrile, than in non-polar solvents, such as cyclohexane and carbon tetrachloride. Only one product is possible from the addition of these strong acids to symmetrical alkenes such as ethene and cyclohexene. However, if the double bond carbon atoms are not structurally equivalent, as in molecules of 1-butene, 2-methyl-2-butene and 1-methylcyclohexene, the reagent conceivably may add in two different ways. This is shown for 2-methyl-2-butene in the following equation. When addition reactions to such unsymmetrical alkenes are carried out, we find that one of the two possible constitutionally isomeric products is formed preferentially. Selectivity of this sort is termed . In the above example, 2-chloro-2-methylbutane is nearly the exclusive product. Similarly, 1-butene forms 2-bromobutane as the predominant product on treatment with HBr. After studying many addition reactions of this kind, the Russian chemist Vladimir Markovnikov noticed a trend in the structure of the favored addition product. He formulated this trend as an empirical rule we now ca When a Brønsted acid, HX, adds to an unsymmetrically substituted double bond, the acidic hydrogen of the acid bonds to that carbon of the double bond that has the greater number of hydrogen atoms already attached to it.
In more homelier vernacular this rule may be restated as, " " Empirical rules like the Markovnikov Rule are useful aids for remembering and predicting experimental results. Indeed, empirical rules are often the first step toward practical mastery of a subject, but they seldom constitute true understanding. The Markovnikov Rule, for example, suggests there are common and important principles at work in these addition reactions, but it does not tell us what they are. The next step in achieving an understanding of this reaction must be to construct a rational mechanistic model that can be tested by experiment. All the reagents discussed here are strong Brønsted acids so, as a first step, it seems sensible to find a base with which the acid can react. Since we know that these acids do not react with alkanes, it must be the pi-electrons of the alkene double bond that serve as the base. As shown in the diagram on the right, the pi-orbital extends into the space immediately above and below the plane of the double bond, and the electrons occupying this orbital may be attracted to the proton of a Brønsted acid. The resulting acid-base equilibrium generates a carbocation intermediate (the conjugate acid of the alkene) which then combines rapidly with the anionic conjugate base of the Brønsted acid. This two-step mechanism is illustrated for the reaction of ethene with hydrogen chloride by the following equations. H C=CH + H C–CH + An energy diagram for this two-step addition mechanism is shown to the left. From this diagram we see that the slow or rate-determining step (the first step) is also the product determining step (the anion will necessarily bond to the carbocation site). Electron donating double bond substituents increase the reactivity of an alkene, as evidenced by the increased rate of hydration of 2-methylpropene (two alkyl groups) compared with 1-butene (one alkyl group). Evidently, alkyl substituents act to increase the rate of addition by lowering the activation energy, ΔE of the rate determining step, and it is here we should look for a rationalization of Markovnikov's rule. As expected, electron withdrawing substituents, such as fluorine or chlorine, reduce the reactivity of an alkene to addition by acids (vinyl chloride is less reactive than ethene). George Hammond formulated a useful principle that relates the nature of a transition state to its location on the reaction path. This states that . In strongly exothermic reactions the transition state will resemble the reactant species. In strongly endothermic conversions, such as that shown to the right, the transition state will resemble the high-energy intermediate or product, and will track the energy of this intermediate if it changes. This change in transition state energy and activation energy as the stability of the intermediate changes may be observed by clicking the higher or lower buttons to the right of the energy diagram. Three examples may be examined, and the reference curve is changed to gray in the diagrams for higher (magenta) and lower (green) energy intermediates. The carbocation intermediate formed in the first step of the addition reaction now assumes a key role, in that it directly influences the activation energy for this step. Independent research shows that the stability of carbocations varies with the nature of substituents, in a manner similar to that seen for alkyl radicals. The exceptional stability of allyl and benzyl cations is the result of charge delocalization, and the stabilizing influence of alkyl substituents, although less pronounced, has been interpreted in a similar fashion. From this information, applying the Hammond Postulate, we arrive at a plausible rationalization of Markovnikov's rule. , because the activation energy of the path to the former is the lower of the two possibilities. This is illustrated by the following equation for the addition of hydrogen chloride to propene. Note that the initial acid-base equilibrium leads to a pi-complex which immediately reorganizes to a sigma-bonded carbocation intermediate. The more stable 2º-carbocation is formed preferentially, and the conjugate base of the Brønsted acid (chloride anion in the example shown below) then rapidly bonds to this electrophilic intermediate to form the final product. The following energy diagram summarizes these features. Note that the pi-complex is not shown, since this rapidly and reversibly formed species is common to both possible reaction paths. A more extensive discussion of the factors that influence carbocation stability may be accessed by . | 7,370 | 1,965 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/08%3A_Properties_of_Organic_Compounds/8.08%3A_Properties_of_Alkanes/8.8.02%3A_Table_of_Electronegativities |
Data from Linus Pauling. “The Nature of the Chemical Bond,” 3d ed., Cornell University Press, Ithaca, N.Y., 1960. This table displays the Linus Pauling derivation of electronegativities. Fluorine, the most electronegative element, has arbitrarily been given a value of 4.0. Every other element's electronegativity has been scaled accordingly. Elements with electronegativities of 2.5 or more are all nonmetals in the top right-hand comer of the periodic table. These have been color-coded dark red. By contrast, elements with negativities of l.3 or less are all metals on the lower left of the table. These elements have been coded in dark gray. They are often referred to as the most electropositive elements, and they are the metals which invariably form binary ionic compounds. Between these two extremes we notice that most of the remaining metals (largely transition metals) have electronegativities between l.4 and l.9 (light gray), while most of the remaining nonmetals have electronegativities between 2.0and 2.4 (light red). Another feature worth noting is the very large differences in electronegativities in the top right-hand comer of the table. Fluorine, with an electronegativity of 4, is by far the most electronegative element. At 3.5 oxygen is a distant second, while chlorine and nitrogen are tied for third place at 3.0. This table is found on CoreChem:Electronegativity. | 1,402 | 1,966 |
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The idea that ethanoic acid (acetic acid) is a possible common starting material for the biosynthesis of many organic compounds was first proposed by Collie (1893) on purely structural grounds. He recognized a structural connection between a linear chain of recurring \(\ce{CH_3CO}\) units (a polyketomethylene chain, \(\ce{CH_3COCH_2COCH_2COCH_2CO}-\)) and certain cyclic natural products. In the example given below, orsellinic acid is represented as if it were derived from a chain of four \(\ce{CH_3CO}\) units by a condensation-cyclization reaction: Experimental verification of Collie's hypothesis came many years later when isotopic hydrogen and carbon (\(\ce{^2H}\), \(\ce{^3H}\), \(\ce{^{13}C}\), and \(\ce{^{14}C}\)) became available. Tracer studies showed that long-chain fatty acids are made by plants and animals from \(\ce{CH_3CO}\) units by successively linking together the carbonyl group of one to the methyl group of another (K. Bloch and F. Lynen, Nobel Prize, 1964). If ethanoic acid supplied to the organism is labeled at the carboxyl group with \(\ce{^{14}C} \: \left( \overset{*}{\ce{C}} \right)\), the fatty acid has the label at alternate carbons: \[\ce{CH_3} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{O_2H}\] However, if the carbon of the methyl group is labeled, the product comes out labeled at the other set of alternate carbons: \[\overset{*}{\ce{C}} \ce{H_3CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CH_2} \overset{*}{\ce{C}} \ce{H_2CO_2H}\] Ethanoic acid is activated for biosynthesis by combination with the thiol, coenzyme A (\textbf{CoA} \ce{SH}\), Figure 18-7) to give the thioester, ethanoyl (acetyl) coenzyme A \(\left( \ce{CH_3COS} \textbf{CoA} \right)\). You may recall that the metabolic degradation of fats also involves this coenzyme ( ) and it is tempting to assume that fatty acid biosynthesis is simply the reverse of fatty acid metabolism to \(\ce{CH_3COS} \textbf{CoA}\). However, this is not quite the case. In fact, it is a general observation in biochemistry that primary metabolites are synthesized by different routes from those by which they are metabolized (for example, compare the pathways of carbon in photosynthesis and metabolism of carbohydrates in and ). A brief description of the main events in fatty-acid biosynthesis follows, and all of these steps must be understood to be under control of appropriate enzymes and their coenzymes even though they are omitted here. The \(\ce{CH_3CO}\) group of ethanoyl coenzyme A is first transferred to a protein having a free thiol \(\left( \ce{SH} \right)\) group to make another thioester, represented here as \(\ce{CH_3COS}-\textbf{ACP}\), where stands for cyl- arrier- rotein. The growing carbon chain remains bound to this protein throughout the synthesis: \[\ce{CH_3COS} \textbf{CoA} + \ce{HS}-\textbf{ACP} \rightarrow \ce{CH_3COS}-\textbf{ACP} + \textbf{CoA} \ce{SH}\] Carboxylation of \(\ce{CH_3COS}-\textbf{ACP}\) yields a propanedioyl thioester, \(6\), which then undergoes a Claisen condensation with a second mole of \(\ce{CH_3COS}-\textbf{ACP}\) accompanied by decarboxylation to yield a 3-oxobutanoyl thioester, \(7\): Reduction of the ketone group of the thioester (by \(\ce{NADPH}\)) leads to a thiol ester of a four-carbon carboxylic acid. Repetitive condensations with thioester \(6\) followed by reduction eventually lead to fatty acids. Each repetition increases the chain length by two carbons: The preceding scheme is representative of fatty acid biosynthesis in plants, animals, and bacteria. The major difference is that plant and bacterial fatty acids usually contain more double bonds (or even triple bonds) than do animal fatty acids. Collie's hypothesis that aromatic compounds are made biologically from ethanoic acid was greatly expanded by A. J. Birch to include an extraordinary number of diverse compounds. The generic name "acetogenin" has been suggested as a convenient classification for ethanoate (acetate)-derived natural products, but the name "polyketides" also is used. Naturally occurring aromatic compounds and quinones are largely made in this way. An example is 2-hydroxy-6-methylbenzoic acid formed as a metabolite of the mold ; using \(\ce{^{14}C}\)-carboxyl-labeled ethanoic acid, the label has been shown to be at the positions indicated below: The biosynthesis of terpenes clearly follows a somewhat different course from fatty acids in that branched-chain compounds are formed. One way that this can come about is for 2-oxobutanoyl coenzyme A to undergo an aldol addition at the keto carbonyl group with the ethanoyl coenzyme A to give the 3-methyl-3-hydroxypentanedioic acid derivative, \(8\): The next step is reduction of one of the carboxyl groups of \(8\) to give mevalonic acid: This substance has been shown by tracer studies to be an efficient precursor of terpenes and steroids. Mevalonic acid has six carbon atoms, whereas the isoprene unit has only five. Therefore, if mevalonic acid is the precursor of isoprene units, it must lose one carbon atom at some stage. Synthesis of mevalonic acid labeled at the carboxyl group with \(\ce{^{14}C}\), and use of this material as a starting material for production of cholesterol, gives cholesterol. Therefore, the carboxyl carbon is the one that is lost: Formation of the "biological isoprene unit" from mevalonic acid has been shown to proceed by stepwise phosphorylation of both alcohol groups, then elimination and decarboxylation to yield 3-methyl-3-butenyl pyrophosphate, \(9\) (often called \(\Delta^3\)-isopentenyl pyrophosphate): The coupling of the five-carbon units, \(9\), to give isoprenoid compounds has been suggested to proceed by the following steps. First, isomerization of the double bond is effected by an enzyme \(\left( \ce{E} \right)\) carrying an \(\ce{SH}\) group: The ester, \(10\), then becomes connected to the double bond of a molecule of \(9\), probably in an enzyme-induced carbocation type of polymerization ( ): The product of the combination of two units of the pyrophosphate, \(9\), through this sequence is if, as shown, the proton is lost to give a trans double bond. Formation of a cis double bond would give neryl pyrophosphate ( ). Continuation of the head-to-tail addition of five-carbon units to geranyl (or neryl) pyrophosphate can proceed in the same way to farnesyl pyrophosphate and so to gutta-percha (or natural rubber). At some stage, a new process must be involved because, although many isoprenoid compounds are head-to-tail type polymers of isoprene, others, such as squalene, lycopene, and \(\beta\)- and \(\gamma\)-carotene (Table 30-1), are formed differently. Squalene, for example, has a structure formed from head-to-head reductive coupling of two farnesyl pyrophosphates: Since squalene can be produced from farnesyl pyrophosphate with \(\ce{NADPH}\) and a suitable enzyme system, the general features of the above scheme for terpene biosynthesis are well supported by experiment. In summary, the sequence from ethanoate to squalene has been traced as \[\text{ethanoyl coenzyme A} \rightarrow \text{mevalonic acid} \rightarrow \text{isopentenyl pyrophosphate} \rightarrow \text{farnesyl pyrophosphate} \rightarrow \text{squalene}\] Isotopic labeling experiments show that cholesterol is derived from ethanoate by way of squalene and lanosterol. The evidence for this is that homogenized liver tissue is able to convert labeled squalene to labeled lanosterol and thence to labeled cholesterol. The conversion of squalene to lanosterol is particularly interesting because, although squalene is divisible into isoprene units, lanosterol is not - a methyl being required at \(\ce{C_8}\) and not \(\ce{C_{13}}\): As a result, some kind of rearrangement must be required to get from squalene to lanosterol. The nature of this rearrangement becomes clearer if we write the squalene formula so as to take the shape of lanosterol: When squalene is written in this form, we see that it is beautifully constructed for cyclization to lanosterol. The key intermediate that initiates the cyclization is the 2,3-epoxide of squalene. Enzymatic cleavage of the epoxide ring is followed by cyclization and then manifold hydride \(\left( \ce{H} \colon \right)\) and methide \(\left( \ce{CH_3} \colon \right)\) shifts to give lanosterol: The evidence is strong that the biosynthesis of lanosterol actually proceeds by a route of this type. With squalene made from either methyl- or carboxyl-labeled ethanoate, all the carbons of lanosterol and cholesterol are labeled just as predicted from the mechanism. Furthermore, ingenious double-labeling experiments have shown that the methyl at \(\ce{C_{13}}\) of lanosterol is the one that was originally located at \(\ce{C_{14}}\), whereas the one at \(\ce{C_{14}}\) is the one that came from \(\ce{C_8}\). The conversion of lanosterol to cholesterol involves removal of the three methyl groups at the 4,4- and 14-positions, shift of the double bond at the \(B\)/\(C\) junction between \(\ce{C_5}\) and \(\ce{C_6}\), and reduction of the \(\ce{C_{24}}\)-\(\ce{C_{25}}\) double bond. The methyl groups are indicated by tracer experiments to be eliminated by oxidation to carbon dioxide. The biosynthetic connection between ethanoyl coenzyme A and the complex natural products briefly discussed is summarized in Figure 30-1. and (1977) | 9,655 | 1,967 |
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Alkenes are found throughout nature. They form the basis of many natural products, such as terpenes, which play a variety of roles in the lives of plants and insects. The C=C bonds of alkenes are very different from the C=O bonds that are also common in nature. The C=C bonds of alkenes are electron-rich and nucleophilic, in contrast to the electron-poor C=O bonds of carbohydrates, fatty acids and proteins. That difference plays a role in how terpenes form in nature. Alkenes, or olefins, are also a major product of the petroleum industry. Reactions of alkenes form the basis for a significant porion of our manufacturing economy. Commonly used plastics such as polyethylene, polypropylene and polystyrene are all formed through the reactions of alkenes. These materials continue to find use in our society because of their valuable properties, such as high strength, flexibility and low weight. Alkenes undergo addition reactions like carbonyls do. Often, they add a proton to one end of the double bond and another group to the other end. These reactions happen in slightly different ways, however. Alkenes are reactive because they have a high-lying pair of π-bonding electrons. These electrons are loosely held, being high in energy compared to σ-bonds. The fact that they are not located between the carbon nuclei, but are found above and below the plane of the double bond, also makes these electrons more accessible. Alkenes can donate their electrons to strong electrophiles other than protons, too. Sometimes their reactivity pattern is a little different than the simple addition across the double bond, but that straightforward pattern is what we will focus on in this chapter. The reaction of 2-methylpropene (or isobutylene) with HBr, as depicted above, is really a 2-step process. Draw this mechanism again and in each of the two steps label both the nucleophile and the electrophile (so, that's four labels). Draw a reaction progress diagram for the reaction of 2-methylpropene with hydrogen bromide. Predict the rate law for the reaction of 2-methylpropene with hydrogen bromide. Based on the reaction shown above, draw products for the following reactions. Acids other than HBr can add to alkenes. Based on the reaction with HBr, draw products for the following reactions. In the following reaction, the addition has gone a slightly different way. Draw a mechanism for this reaction. , | 2,419 | 1,968 |
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Spherical balloons are inflated and tied off, and their inlets are twisted together so that four balloons are attached. They assume tetrahedral geometry by mutual repulsion, modeling VSERPR approach to repulsion of four charge centers. If one balloon is broken, the remaining three assume a planar triangular geometry, and if another balloon is broken, the remaining two assume a linear geometry. Prepare nitrogen triiodide, See Molecules with Lone Pairs Lecture Demonstrations. The instability of NI (or better, NI *NH ) is due to the same repulsions that are the basis for the VSEPR approach (3 large I atoms and a lone pair bonded to the relatively small N atom) Ed Vitz (Kutztown University), (University of | 735 | 1,969 |
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As was previously demonstrated, the spontaneity of a process may depend upon the temperature of the system. Phase transitions, for example, will proceed spontaneously in one direction or the other depending upon the temperature of the substance in question. Likewise, some chemical reactions can also exhibit temperature dependent spontaneities. To illustrate this concept, the equation relating free energy change to the enthalpy and entropy changes for the process is considered: \[ ΔG=ΔH−TΔS \] The spontaneity of a process, as reflected in the arithmetic sign of its free energy change, is then determined by the signs of the enthalpy and entropy changes and, in some cases, the absolute temperature. Since T is the absolute (kelvin) temperature, it can only have positive values. Four possibilities therefore exist with regard to the signs of the enthalpy and entropy changes: These four scenarios are summarized in Figure \(\Page {1}\). The incomplete combustion of carbon is described by the following equation: \[\ce{2C}(s)+\ce{O2}(g)⟶\ce{2CO}(g) \nonumber\] How does the spontaneity of this process depend upon temperature? Combustion processes are exothermic (Δ < 0). This particular reaction involves an increase in entropy due to the accompanying increase in the amount of gaseous species (net gain of one mole of gas, Δ > 0). The reaction is therefore spontaneous (Δ < 0) at all temperatures. Popular chemical hand warmers generate heat by the air-oxidation of iron: How does the spontaneity of this process depend upon temperature? Δ and Δ are negative; the reaction is spontaneous at low temperatures. When considering the conclusions drawn regarding the temperature dependence of spontaneity, it is important to keep in mind what the terms “high” and “low” mean. Since these terms are adjectives, the temperatures in question are deemed high or low relative to some reference temperature. A process that is nonspontaneous at one temperature but spontaneous at another will necessarily undergo a change in “spontaneity” (as reflected by its Δ ) as temperature varies. This is clearly illustrated by a graphical presentation of the free energy change equation, in which Δ is plotted on the axis versus on the axis: \[ΔG=ΔH−TΔS\] \[y=b+mx\] Such a plot is shown in Figure \(\Page {2}\). A process whose enthalpy and entropy changes are of the same arithmetic sign will exhibit a temperature-dependent spontaneity as depicted by the two yellow lines in the plot. Each line crosses from one spontaneity domain (positive or negative Δ ) to the other at a temperature that is characteristic of the process in question. This temperature is represented by the -intercept of the line, that is, the value of for which Δ is zero: \[ΔG=0=ΔH−TΔS\] \[T=\dfrac{ΔH}{ΔS}\] And so, saying a process is spontaneous at “high” or “low” temperatures means the temperature is above or below, respectively, that temperature at which Δ for the process is zero. As noted earlier, this condition describes a system at equilibrium. As defined in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its solid and liquid phases are in equilibrium (that is, when vaporization and condensation occur at equal rates). Use the information in to estimate the boiling point of water. The process of interest is the following phase change: When this process is at equilibrium, Δ = 0, so the following is true: Using the standard thermodynamic data from , The accepted value for water’s normal boiling point is 373.2 K (100.0 °C), and so this calculation is in reasonable agreement. Note that the values for enthalpy and entropy changes data used were derived from standard data at 298 K ( ). If desired, you could obtain more accurate results by using enthalpy and entropy changes determined at (or at least closer to) the actual boiling point. Use the information in to estimate the boiling point of CS . 313 K (accepted value 319 K) The fact that ΔG° and K are related provides us with another explanation of why equilibrium constants are temperature dependent. This relationship can be expressed as follows: \[\ln K=-\dfrac{\Delta H^\circ}{RT}+\dfrac{\Delta S^\circ}{R} \label{18.40}\] Assuming ΔH° and ΔS° are for an exothermic reaction (ΔH° < 0), the magnitude of K decreases with increasing temperature, whereas for an endothermic reaction (ΔH° > 0), the magnitude of K increases with increasing temperature. The quantitative relationship expressed in Equation \(\ref{18.40}\) agrees with the qualitative predictions made by applying Le Chatelier’s principle. Because heat is produced in an exothermic reaction, adding heat (by increasing the temperature) will shift the equilibrium to the left, favoring the reactants and decreasing the magnitude of K. Conversely, because heat is consumed in an endothermic reaction, adding heat will shift the equilibrium to the right, favoring the products and increasing the magnitude of K. Equation \(\ref{18.40}\) also shows that the magnitude of ΔH° dictates how rapidly K changes as a function of temperature. In contrast, the magnitude and sign of ΔS° affect the magnitude of K but not its temperature dependence. If we know the value of K at a given temperature and the value of ΔH° for a reaction, we can estimate the value of K at any other temperature, even in the absence of information on ΔS°. Suppose, for example, that K and K are the equilibrium constants for a reaction at temperatures T and T , respectively. Applying Equation \(\ref{18.40}\) gives the following relationship at each temperature: Subtracting \(\ln K_1\) from \(\ln K_2\), \[\ln K_2-\ln K_1=\ln\dfrac{K_2}{K_1}=\dfrac{\Delta H^\circ}{R}\left(\dfrac{1}{T_1}-\dfrac{1}{T_2}\right) \label{18.41}\] Thus calculating ΔH° from tabulated enthalpies of formation and measuring the equilibrium constant at one temperature (K ) allow us to calculate the value of the equilibrium constant at any other temperature (K ), assuming that ΔH° and ΔS° are independent of temperature. The linear relation between \(\ln K \)and the standard and in Equation \(\ref{18.41}\) is known as the van’t Hoff equation. It shows that a plot of \(\ln K\) vs. \(1/T\) should be a line with slope \(-\Delta_r{H^o}/R\) and intercept \(\Delta_r{S^o}/R\). Hence, these thermodynamic enthalpy and entropy changes for a reversible reaction can be determined from plotting \(\ln K\) vs. \(1/T\) data without the aid of . Of course, the main assumption here is that \(\Delta_r{H^o}\) and \(\Delta_r{S^o}\) are only very weakly dependent on \(T\), which is usually valid over a narrow temperature range. The equilibrium constant for the formation of NH from H and N at 25°C is K = 5.4 × 10 . What is K at 500°C? : balanced chemical equation, ΔH°, initial and final T, and K at 25°C K at 500°C : Convert the initial and final temperatures to kelvins. Then substitute appropriate values into Equation \(\ref{18.41}\) to obtain K , the equilibrium constant at the final temperature. : The value of ΔH° for the reaction obtained using Hess’s law is −91.8 kJ/mol of N . If we set T = 25°C = 298.K and T = 500°C = 773 K, then from Equation \(\ref{18.41}\) we obtain the following: Thus at 500°C, the equilibrium strongly favors the reactants over the products. In the exercise in Example \(\Page {3}\), you calculated K = 2.2 × 10 for the reaction of NO with O to give NO at 25°C. Use the ΔH values in the exercise in Example 10 to calculate K for this reaction at 1000°C. : 5.6 × 10 The Van't Hoff Equation: For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature. If we assume ideal gas behavior, the ideal gas law allows us to express ΔG in terms of the partial pressures of the reactants and products, which gives us a relationship between ΔG and K , the equilibrium constant of a reaction involving gases, or K, the equilibrium constant expressed in terms of concentrations. If ΔG° < 0, then K or K > 1, and products are favored over reactants. If ΔG° > 0, then K or K < 1, and reactants are favored over products. If ΔG° = 0, then K or K = 1, and the system is at equilibrium. We can use the measured equilibrium constant K at one temperature and ΔH° to estimate the equilibrium constant for a reaction at any other temperature. ( ) | 8,412 | 1,970 |
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Δ is meaningful only for changes in which the temperature and pressure remain constant. These are the conditions under which most reactions are carried out in the laboratory; the system is usually open to the atmosphere (constant pressure) and we begin and end the process at room temperature (after any heat we have added or which is liberated by the reaction has dissipated.) The importance of the Gibbs function can hardly be over-stated: Thus if the free energy of the reactants is greater than that of the products, the entropy of the world will increase when the reaction takes place as written, and so the reaction will tend to take place spontaneously. Conversely, if the free energy of the products exceeds that of the reactants, then the reaction will not take place in the direction written, but it will tend to proceed in the reverse direction. In a spontaneous change, Gibbs energy always decreases and never increases. This of course reflects the fact that the entropy of the world behaves in the exact opposite way (owing to the negative sign in the Δ term). \[H_2O_{(l)} \rightarrow H_2O_{(s)} \label{23.5.6}\] water below its freezing point undergoes a decrease in its entropy, but the heat released into the surroundings more than compensates for this, so the entropy of the world increases, the free energy of the H O diminishes, and the process proceeds spontaneously. In a spontaneous change, Gibbs energy decreases and never increases. An important consequence of the one-way downward path of the free energy is that once it reaches its minimum possible value, all net change comes to a halt. This, of course, represents the state of chemical equilibrium. These relations are nicely summarized as follows: Recall the condition for spontaneous change \[ΔG = ΔH – TΔS < 0 \label{Master}\] it is apparent that the temperature dependence of Δ depends almost entirely on the entropy change associated with the process. (We say "almost" because the values of Δ and Δ are themselves slightly temperature dependent; both gradually increase with temperature). In particular, notice that in the above equation For any given reaction, the sign of Δ can also be positive or negative. This means that there are four possibilities for the influence that temperature can have on the spontaneity of a process. The following cases generalizes these relations for the four sign-combinations of Δ and Δ . (Note that use of the standard Δ ° and Δ ° values in the example reactions is not strictly correct here, and can yield misleading results when used generally.) Under these conditions, both the Δ and Δ terms will be negative, so Δ will be negative regardless of the temperature. An exothermic reaction whose entropy increases will be spontaneous at all temperatures. Example Reaction \[C_{(graphite)} + O_{2(g)} \rightleftharpoons CO_{2(g)}\] The positive entropy change is due mainly to the greater mass of CO molecules compared to those of O . Δ Example reaction: \[3 H_2 + N_2 \rightleftharpoons 2 NH_{3(g)}\] The decrease in moles of gas in the drives the entropy change negative, making the reaction . Thus higher , which speeds up the reaction, also reduces its extent. This is the reverse of the previous case; the entropy increase must overcome the handicap of an endothermic process so that TΔS > ΔH. Since the effect of the temperature is to "magnify" the influence of a positive ΔS, the process will be spontaneous at temperatures above T = ΔH / ΔS. (Think of melting and boiling.) Example reaction: \[N_2O_{4(g)} \rightleftharpoons 2 NO_{2(g)}\] are typically endothermic with positive entropy change, and are therefore . With both Δ Δ working against it, this kind of process will not proceed spontaneously at any temperature. Substance A always has a greater number of accessible energy states, and is therefore always the preferred form. Example reaction: \[½ N_2 + O_2 \rightleftharpoons NO_{2(g)}\] This reaction is , meaning that . But because the reverse reaction is kinetically inhibited, NO can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable. The above cases and associated plots are the important ones; do not try to memorize them, but make sure you understand and can explain or reproduce them for a given set of Δ and Δ . The other two plots on each diagram are only for the chemistry-committed. To further understand how the various components of ΔG dictate whether a process occurs spontaneously, we now look at a simple and familiar physical change: the conversion of liquid water to water vapor. If this process is carried out at 1 atm and the normal boiling point of 100.00°C (373.15 K), we can calculate ΔG from the experimentally measured value of ΔH (40.657 kJ/mol). For vaporizing 1 mol of water, \(ΔH = 40,657; J\), so the process is highly endothermic. From the definition of ΔS, we know that for 1 mol of water, Hence there is an increase in the disorder of the system. At the normal boiling point of water, \[\Delta G_{100^\circ\textrm C}=\Delta H_{100^\circ\textrm C}-T\Delta S_{100^\circ\textrm C}=\textrm{40,657 J}-[(\textrm{373.15 K})(\textrm{108.96 J/K})]=\textrm{0 J}\] The energy required for vaporization offsets the increase in disorder of the system. Thus ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions. Now suppose we were to superheat 1 mol of liquid water to 110°C. The value of ΔG for the vaporization of 1 mol of water at 110°C, assuming that ΔH and ΔS do not change significantly with temperature, becomes \[\Delta G_{110^\circ\textrm C}=\Delta H-T\Delta S=\textrm{40,657 J}-[(\textrm{383.15 K})(\textrm{108.96 J/K})]=-\textrm{1091 J}\] At 110°C, ΔG < 0, and vaporization is predicted to occur spontaneously and irreversibly. We can also calculate ΔG for the vaporization of 1 mol of water at a temperature below its normal boiling point—for example, 90°C—making the same assumptions: \[\Delta G_{90^\circ\textrm C}=\Delta H-T\Delta S=\textrm{40,657 J}-[(\textrm{363.15 K})(\textrm{108.96 J/K})]=\textrm{1088 J}\] At 90°C, ΔG > 0, and water does not spontaneously convert to water vapor. When using all the digits in the calculator display in carrying out our calculations, ΔG = 1090 J = −ΔG , as we would predict. We can also calculate the temperature at which liquid water is in equilibrium with water vapor. Inserting the values of ΔH and ΔS into the definition of ΔG (Equation \(\ref{Master}\)), setting ΔG = 0, and solving for T, Thus ΔG = 0 at T = 373.15 K and 1 atm, which indicates that liquid water and water vapor are in equilibrium; this temperature is called the normal boiling point of water. At temperatures greater than 373.15 K, ΔG is negative, and water evaporates spontaneously and irreversibly. Below 373.15 K, ΔG is positive, and water does not evaporate spontaneously. Instead, water vapor at a temperature less than 373.15 K and 1 atm will spontaneously and irreversibly condense to liquid water. Figure \(\Page {1}\) shows how the ΔH and TΔS terms vary with temperature for the vaporization of water. When the two lines cross, ΔG = 0, and ΔH = TΔS. A similar situation arises in the conversion of liquid egg white to a solid when an egg is boiled. The major component of egg white is a protein called albumin, which is held in a compact, ordered structure by a large number of hydrogen bonds. Breaking them requires an input of energy (ΔH > 0), which converts the albumin to a highly disordered structure in which the molecules aggregate as a disorganized solid (ΔS > 0). At temperatures greater than 373 K, the TΔS term dominates, and ΔG < 0, so the conversion of a raw egg to a hard-boiled egg is an irreversible and spontaneous process above 373 K. Δ is key in determining whether or not a reaction will take place in a given direction. It turns out, however, that it is almost never necessary to explicitly evaluate Δ . It is far more convenient to work with the equilibrium constant of a reaction, within which Δ is "hidden". This is just as well, because for most reactions (those that take place in solutions or gas mixtures) the value of Δ depends on the of the various reaction components in the mixture; it is not a simple sum of the "products minus reactants" type, as is the case with Δ . Because ΔH° and ΔS° determine the magnitude of ΔG° and because K is a measure of the ratio of the concentrations of products to the concentrations of reactants, we should be able to express K in terms of ΔG° and vice versa. ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. For a reversible process that does not involve external work, we can express the change in free energy in terms of volume, pressure, entropy, and temperature, thereby eliminating ΔH from the equation for ΔG. The general relationship can be shown as follow (derivation not shown): \[ \Delta G = V \Delta P − S \Delta T \label{18.29}\] If a reaction is carried out at constant temperature (ΔT = 0), then Equation \(\ref{18.29}\) simplifies to \[\Delta{G} = V\Delta{P} \label{18.30}\] Under normal conditions, the pressure dependence of free energy is not important for solids and liquids because of their small molar volumes. For reactions that involve gases, however, the effect of pressure on free energy is very important. Assuming ideal gas behavior, we can replace the \(V\) in Equation \(\ref{18.30}\) by nRT/P (where n is the number of moles of gas and R is the ideal gas constant) and express \(\Delta{G}\) in terms of the initial and final pressures (\(P_i\) and \(P_f\), respectively): \[\Delta G=\left(\dfrac{nRT}{P}\right)\Delta P=nRT\dfrac{\Delta P}{P}=nRT\ln\left(\dfrac{P_\textrm f}{P_\textrm i}\right) \label{18.31}\] If the initial state is the standard state with P = 1 atm, then the change in free energy of a substance when going from the standard state to any other state with a pressure P can be written as follows: \[G − G^° = nRT\ln{P}\] This can be rearranged as follows: \[G = G^° + nRT\ln {P} \label{18.32}\] As you will soon discover, Equation \(\ref{18.32}\) allows us to relate ΔG° and K . Any relationship that is true for \(K_p\) must also be true for \(K\) because \(K_p\) and \(K\) are simply different ways of expressing the equilibrium constant using different units. Let’s consider the following hypothetical reaction, in which all the reactants and the products are ideal gases and the lowercase letters correspond to the stoichiometric coefficients for the various species: \[aA+bB \rightleftharpoons cC+dD \label{18.33}\] Because the free-energy change for a reaction is the difference between the sum of the free energies of the products and the reactants, we can write the following expression for ΔG: \[\delta{G}=\sum_m G_{products}−\sum_n G_{reactants}=(cG_C+dG_D)−(aG_A+bG_B) \label{18.34}\] Substituting Equation \(\ref{18.32}\) for each term into Equation \(\ref{18.34}\), \[ΔG=[(cG^o_C+cRT \ln P_C)+(dG^o_D+dRT\ln P_D)]−[(aG^o_A+aRT\ln P_A)+(bG^o_B+bRT\ln P_B)]\] Combining terms gives the following relationship between ΔG and the reaction quotient Q: \[\Delta G=\Delta G^\circ+RT \ln\left(\dfrac{P^c_\textrm CP^d_\textrm D}{P^a_\textrm AP^b_\textrm B}\right)=\Delta G^\circ+RT\ln Q \label{18.35}\] where ΔG° indicates that all reactants and products are in their standard states. For gases at equilibrium (\(Q = K_p\),), and as you’ve learned in this chapter, ΔG = 0 for a system at equilibrium. Therefore, we can describe the relationship between ΔG° and K for gases as follows: \[ 0 = ΔG° + RT\ln K_p \label{18.36a}\] \[ \color{red} ΔG°= −RT\ln K_p \label{18.36b}\] If the products and reactants are in their standard states and ΔG° < 0, then K > 1, and products are favored over reactants. Conversely, if ΔG° > 0, then K < 1, and reactants are favored over products. If ΔG° = 0, then \(K_p = 1\), and neither reactants nor products are favored: the system is at equilibrium. For a spontaneous process under standard conditions, \(K_{eq}\) and \(K_p\) are greater than 1. We previosuly calculated that ΔG° = −32.7 kJ/mol of N for the reaction \[N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \nonumber\] This calculation was for the reaction under standard conditions—that is, with all gases present at a partial pressure of 1 atm and a temperature of 25°C. Calculate ΔG for the same reaction under the following nonstandard conditions: Does the reaction favor products or reactants? : balanced chemical equation, partial pressure of each species, temperature, and ΔG° : whether products or reactants are favored : : A The relationship between ΔG° and ΔG under nonstandard conditions is given in Equation \(\ref{18.35}\). Substituting the partial pressures given, we can calculate Q: \[Q=\dfrac{P^2_{\textrm{NH}_3}}{P_{\textrm N_2}P^3_{\textrm H_2}}=\dfrac{(0.021)^2}{(2.00)(7.00)^3}=6.4\times10^{-7} \nonumber\] B Substituting the values of ΔG° and Q into Equation \(\ref{18.35}\), Because ΔG < 0 and Q < 1.0, the reaction is spontaneous to the right as written, so products are favored over reactants. Calculate ΔG for the reaction of nitric oxide with oxygen to give nitrogen dioxide under these conditions: T = 50°C, P = 0.0100 atm, \(P_{\mathrm{O_2}}\) = 0.200 atm, and \(P_{\mathrm{NO_2}}\) = 1.00 × 10 atm. The value of ΔG° for this reaction is −72.5 kJ/mol of O . Are products or reactants favored? : −92.9 kJ/mol of O ; the reaction is spontaneous to the right as written, so products are favored. Calculate K for the reaction of H with N to give NH at 25°C. As calculated in Example 10, ΔG° for this reaction is −32.7 kJ/mol of N . : balanced chemical equation from Example 10, ΔG°, and temperature : K Substitute values for ΔG° and T (in kelvins) into Equation \(\ref{18.36}\) to calculate K , the equilibrium constant for the formation of ammonia. In Example 10, we used tabulated values of ΔG to calculate ΔG° for this reaction (−32.7 kJ/mol of N ). For equilibrium conditions, rearranging Equation \(\ref{18.36b}\), Inserting the value of ΔG° and the temperature (25°C = 298 K) into this equation, Thus the equilibrium constant for the formation of ammonia at room temperature is favorable. However, the rate at which the reaction occurs at room temperature is too slow to be useful. Calculate K for the reaction of NO with O to give NO at 25°C. As calculated in the exercise in Example 10, ΔG° for this reaction is −70.5 kJ/mol of O . : 2.2 × 10 Although K is defined in terms of the partial pressures of the reactants and the products, the equilibrium constant K is defined in terms of the concentrations of the reactants and the products. We described the relationship between the numerical magnitude of K and K in Chapter 15 and showed that they are related: \[K_p = K(RT)^{Δn} \label{18.37}\] where Δn is the number of moles of gaseous product minus the number of moles of gaseous reactant. For reactions that involve only solutions, liquids, and solids, Δn = 0, so K = K. For all reactions that do not involve a change in the number of moles of gas present, the relationship in Equation \(\ref{18.36b}\) can be written in a more general form: \[ΔG° = −RT \ln K \label{18.38}\] Only when a reaction results in a net production or consumption of gases is it necessary to correct Equation \(\ref{18.38}\) for the difference between K and K. Although we typically use concentrations or pressures in our equilibrium calculations, recall that equilibrium constants are generally expressed as unitless numbers because of the use of or in precise thermodynamic work. Systems that contain gases at high pressures or concentrated solutions that deviate substantially from ideal behavior require the use of fugacities or activities, respectively. Combining Equations \(\ref{18.38}\) with \(ΔG^o = ΔH^o − TΔS^o\) provides insight into how the components of ΔG° influence the magnitude of the equilibrium constant: \[ΔG° = ΔH° − TΔS° = −RT \ln K \label{18.39}\] Equation \(\ref{18.39}\) is quite powerful and connected the nature of the system under equilibrium \(K\) to the condition of the system under standard conditions \(\Delta G^o\).; that is quite powerful. Notice that \(K\) becomes larger as ΔS° becomes more positive, indicating that the magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum disorder. Moreover, K increases as ΔH° decreases. Thus the magnitude of the equilibrium constant is also directly influenced by the tendency of a system to seek the lowest energy state possible. The magnitude of the equilibrium constant is directly influenced by the tendency of a system to move toward maximum entropy and seek the lowest energy state possible. To further illustrate the relation between these two essential thermodynamic concepts, consider the observation that reactions spontaneously proceed in a direction that ultimately establishes equilibrium. As may be shown by plotting the free energy change versus the extent of the reaction (for example, as reflected in the value of ), equilibrium is established when the system’s free energy is minimized (Figure \(\Page {2}\)). If a system is present with reactants and products present in nonequilibrium amounts ( ≠ ), the reaction will proceed spontaneously in the direction necessary to establish equilibrium. Relating Grxn and Kp: We have seen that there is no way to measure absolute enthalpies, although we can measure changes in enthalpy (ΔH) during a chemical reaction. Because enthalpy is one of the components of Gibbs free energy, we are consequently unable to measure absolute free energies; we can measure only changes in free energy. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free-energy change can be calculated from the definition of free energy, if the standard enthalpy and entropy changes are known, using Equation \(\ref{Eq5}\): \[ΔG° = ΔH° − TΔS° \label{Eq5}\] If ΔS° and ΔH° for a reaction have the same sign, then the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. It is important to recognize that a positive value of ΔG° for a reaction does not mean that no products will form if the reactants in their standard states are mixed; it means only that at equilibrium the concentrations of the products will be less than the concentrations of the reactants. A positive ΔG° means that the equilibrium constant is less than 1. Calculate the standard free-energy change (ΔG°) at 25°C for the reaction \[H_{2(g)}+O_{2(g)} \rightleftharpoons H_2O_{2(l)} \nonumber\] At 25°C, the standard enthalpy change (ΔH°) is −187.78 kJ/mol, and the absolute entropies of the products and reactants are: Is the reaction spontaneous as written? : balanced chemical equation, ΔH° and S° for reactants and products spontaneity of reaction as written : A To calculate ΔG° for the reaction, we need to know ΔH°, ΔS°, and T. We are given ΔH°, and we know that T = 298.15 K. We can calculate ΔS° from the absolute molar entropy values provided using the “products minus reactants” rule: As we might expect for a reaction in which 2 mol of gas is converted to 1 mol of a much more ordered liquid, ΔS° is very negative for this reaction. B Substituting the appropriate quantities into Equation \(\ref{Eq5}\), \[\begin{align}\Delta G^\circ=\Delta H^\circ -T\Delta S^\circ &=-187.78\textrm{ kJ/mol}-(\textrm{298.15 K}) [-226.3\;\mathrm{J/(mol\cdot K)}\times\textrm{1 kJ/1000 J}] \nonumber \\ &=-187.78\textrm{ kJ/mol}+67.47\textrm{ kJ/mol}=-120.31\textrm{ kJ/mol} \nonumber \end{align}\] The negative value of ΔG° indicates that the reaction is spontaneous as written. Because ΔS° and ΔH° for this reaction have the same sign, the sign of ΔG° depends on the relative magnitudes of the ΔH° and TΔS° terms. In this particular case, the enthalpy term dominates, indicating that the strength of the bonds formed in the product more than compensates for the unfavorable ΔS° term and for the energy needed to break bonds in the reactants. Calculate the standard free-energy change (ΔG°) at 25°C for the reaction \[2H_2(g)+N_2(g) \rightleftharpoons N_2H_4(l) \nonumber \] . At 25°C, the standard enthalpy change (ΔH°) is 50.6 kJ/mol, and the absolute entropies of the products and reactants are S°(N H ) = 121.2 J/(mol•K), S°(N ) = 191.6 J/(mol•K), and S°(H ) = 130.7 J/(mol•K). Is the reaction spontaneous as written? : 149.5 kJ/mol; no Tabulated values of standard free energies of formation allow chemists to calculate the values of ΔG° for a wide variety of chemical reactions rather than having to measure them in the laboratory. The standard free energy of formation (\(ΔG^∘_f\))of a compound is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. By definition, the standard free energy of formation of an element in its standard state is zero at 298.15 K. One mole of Cl gas at 298.15 K, for example, has \(ΔG^∘_f = 0\). The standard free energy of formation of a compound can be calculated from the standard enthalpy of formation (ΔH ) and the standard entropy of formation (ΔS ) using the definition of free energy: \[Δ^o_{f} =ΔH^o_{f} −TΔS^o_{f} \label{Eq6}\] Using standard free energies of formation to calculate the standard free energy of a reaction is analogous to calculating standard enthalpy changes from standard enthalpies of formation using the familiar “products minus reactants” rule: \[ΔG^o_{rxn}=\sum mΔG^o_{f} (products)− \sum nΔ^o_{f} (reactants) \label{Eq7a}\] where m and n are the stoichiometric coefficients of each product and reactant in the balanced chemical equation. A very large negative ΔG° indicates a strong tendency for products to form spontaneously from reactants; it does not, however, necessarily indicate that the reaction will occur rapidly. To make this determination, we need to evaluate the kinetics of the reaction. Calculate ΔG° for the reaction of isooctane with oxygen gas to give carbon dioxide and water (described in Example 7). Use the following data: : balanced chemical equation and values of ΔG° for isooctane, CO , and H O : spontaneity of reaction as written : Use the “products minus reactants” rule to obtain ΔG , remembering that ΔG° for an element in its standard state is zero. From the calculated value, determine whether the reaction is spontaneous as written. The balanced chemical equation for the reaction is as follows: \[\mathrm{C_8H_{18}(l)}+\frac{25}{2}\mathrm{O_2(g)}\rightarrow\mathrm{8CO_2(g)}+\mathrm{9H_2O(l)} \nonumber\] We are given ΔG values for all the products and reactants except O (g). Because oxygen gas is an element in its standard state, ΔG (O ) is zero. Using the “products minus reactants” rule, Because ΔG° is a large negative number, there is a strong tendency for the spontaneous formation of products from reactants (though not necessarily at a rapid rate). Also notice that the magnitude of ΔG° is largely determined by the ΔG of the stable products: water and carbon dioxide. Calculate ΔG° for the reaction of benzene with hydrogen gas to give cyclohexane using the following data Is the reaction spontaneous as written? : Calculated values of ΔG° are extremely useful in predicting whether a reaction will occur spontaneously if the reactants and products are mixed under standard conditions. We should note, however, that very few reactions are actually carried out under standard conditions, and calculated values of ΔG° may not tell us whether a given reaction will occur spontaneously under nonstandard conditions. What determines whether a reaction will occur spontaneously is the free-energy change (ΔG) under the actual experimental conditions, which are usually different from ΔG°. If the ΔH and TΔS terms for a reaction have the same sign, for example, then it may be possible to reverse the sign of ΔG by changing the temperature, thereby converting a reaction that is not thermodynamically spontaneous, having K < 1, to one that is, having a K > 1, or vice versa. Because ΔH and ΔS usually do not vary greatly with temperature in the absence of a phase change, we can use tabulated values of ΔH° and ΔS° to calculate ΔG° at various temperatures, as long as no phase change occurs over the temperature range being considered. In the absence of a phase change, neither \(ΔH\) nor \(ΔS\) vary greatly with temperature. Calculate (a) ΔG° and (b) ΔG for the reaction N (g)+3H (g)⇌2NH (g), assuming that ΔH and ΔS do not change between 25°C and 300°C. Use these data: : balanced chemical equation, temperatures, S° values, and ΔH for NH : ΔG° and ΔG at 300°C : A To calculate ΔG° for the reaction using Equation \(\ref{Eq5}\), we must know the temperature as well as the values of ΔS° and ΔH°. At standard conditions, the temperature is 25°C, or 298 K. We can calculate ΔS° for the reaction from the absolute molar entropy values given for the reactants and the products using the “products minus reactants” rule: We can also calculate ΔH° for the reaction using the “products minus reactants” rule. The value of ΔH (NH ) is given, and ΔH is zero for both N and H : \[\begin{align}\Delta H^\circ_{\textrm{rxn}}&=2\Delta H^\circ_\textrm f(\mathrm{NH_3})-[\Delta H^\circ_\textrm f(\mathrm{N_2})+3\Delta H^\circ_\textrm f(\mathrm{H_2})] \nonumber \\ &=[2\times(-45.9\textrm{ kJ/mol})]-[(1\times0\textrm{ kJ/mol})+(3\times0 \textrm{ kJ/mol})] \nonumber \\ &=-91.8\textrm{ kJ(per mole of N}_2) \nonumber\end{align} \nonumber\] B Inserting the appropriate values into Equation \(\ref{Eq5}\) \[\Delta G^\circ_{\textrm{rxn}}=\Delta H^\circ-T\Delta S^\circ=(-\textrm{91.8 kJ})-(\textrm{298 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J})=-\textrm{32.7 kJ (per mole of N}_2) \nonumber\] C To calculate ΔG for this reaction at 300°C, we assume that ΔH and ΔS are independent of temperature (i.e., ΔH = H° and ΔS = ΔS°) and insert the appropriate temperature (573 K) into Equation \(\ref{Eq2}\): \[\begin{align}\Delta G_{300^\circ\textrm C}&=\Delta H_{300^\circ\textrm C}-(\textrm{573 K})(\Delta S_{300^\circ\textrm C})=\Delta H^\circ -(\textrm{573 K})\Delta S^\circ \nonumber \\ &=(-\textrm{91.8 kJ})-(\textrm{573 K})(-\textrm{198.1 J/K})(\textrm{1 kJ/1000 J})=21.7\textrm{ kJ (per mole of N}_2) \nonumber \end{align} \nonumber \] In this example, changing the temperature has a major effect on the thermodynamic spontaneity of the reaction. Under standard conditions, the reaction of nitrogen and hydrogen gas to produce ammonia is thermodynamically spontaneous, but in practice, it is too slow to be useful industrially. Increasing the temperature in an attempt to make this reaction occur more rapidly also changes the thermodynamics by causing the −TΔS° term to dominate, and the reaction is no longer spontaneous at high temperatures; that is, its K is less than one. This is a classic example of the conflict encountered in real systems between thermodynamics and kinetics, which is often unavoidable. Calculate for the following reaction \[2NO_{(g)}+O_{2\; (g)} \rightleftharpoons 2NO_{2\; (g)} \nonumber\] which is important in the formation of urban smog. Assume that \(ΔH\) and \(ΔS\) do not change between 25.0°C and 750°C and use these data: The effect of temperature on the spontaneity of a reaction, which is an important factor in the design of an experiment or an industrial process, depends on the sign and magnitude of both ΔH° and ΔS°. The temperature at which a given reaction is at equilibrium can be calculated by setting ΔG° = 0 in Equation \(\ref{Eq5}\), as illustrated in Example \(\Page {4}\). The reaction of nitrogen and hydrogen gas to produce ammonia is one in which ΔH° and ΔS° are both negative. Such reactions are predicted to be thermodynamically spontaneous at low temperatures but nonspontaneous at high temperatures. Use the data in Example \(\Page {3}\) to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous, assuming that ΔH° and ΔS° are independent of temperature. : ΔH° and ΔS° : temperature at which reaction changes from spontaneous to nonspontaneous : Set ΔG° equal to zero in Equation \(\ref{Eq5}\) and solve for T, the temperature at which the reaction becomes nonspontaneous. In Example \(\Page {3}\), we calculated that ΔH° is −91.8 kJ/mol of N and ΔS° is −198.1 J/K per mole of N , corresponding to ΔG° = −32.7 kJ/mol of N at 25°C. Thus the reaction is indeed spontaneous at low temperatures, as expected based on the signs of ΔH° and ΔS°. The temperature at which the reaction becomes nonspontaneous is found by setting ΔG° equal to zero and rearranging Equation \(\ref{Eq5}\) to solve for T: This is a case in which a chemical engineer is severely limited by thermodynamics. Any attempt to increase the rate of reaction of nitrogen with hydrogen by increasing the temperature will cause reactants to be favored over products above 463 K. ΔH° and ΔS° are both negative for the reaction of nitric oxide and oxygen to form nitrogen dioxide. Use those data to calculate the temperature at which this reaction changes from spontaneous to nonspontaneous. : 792.6 K We can predict whether a reaction will occur spontaneously by combining the entropy, enthalpy, and temperature of a system in a new state function called Gibbs free energy (G). The change in free energy (ΔG) is the difference between the heat released during a process and the heat released for the same process occurring in a reversible manner. If a system is at equilibrium, ΔG = 0. If the process is spontaneous, ΔG < 0. If the process is not spontaneous as written but is spontaneous in the reverse direction, ΔG > 0. At constant temperature and pressure, ΔG is equal to the maximum amount of work a system can perform on its surroundings while undergoing a spontaneous change. The standard free-energy change (ΔG°) is the change in free energy when one substance or a set of substances in their standard states is converted to one or more other substances, also in their standard states. The standard free energy of formation (ΔG ), is the change in free energy that occurs when 1 mol of a substance in its standard state is formed from the component elements in their standard states. Tabulated values of standard free energies of formation are used to calculate ΔG° for a reaction. | 30,559 | 1,971 |
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In this module, the relationship between Pressure, Temperature, Volume, and Amount of a gas are described and how these relationships can be combined to give a general expression that describes the behavior of a gas. Any set of relationships between a single quantity (such as V) and several other variables (\(P\), \(T\), and \(n\)) can be combined into a single expression that describes all the relationships simultaneously. The three individual expressions are as follows: \[V \propto \dfrac{1}{P} \;\; \text{@ constant n and T}\] \[V \propto T \;\; \text{@ constant n and P}\] \[V \propto n \;\; \text{@ constant T and P}\] Combining these three expressions gives \[V \propto \dfrac{nT}{P} \tag{6.3.1}\] which shows that the volume of a gas is proportional to the number of moles and the temperature and inversely proportional to the pressure. This expression can also be written as \[V= {\rm Cons.} \left( \dfrac{nT}{P} \right) \tag{6.3.2}\] By convention, the proportionality constant in Equation 6.3.1 is called the gas constant, which is represented by the letter \(R\). Inserting R into Equation 6.3.2 gives \[ V = \dfrac{Rnt}{P} = \dfrac{nRT}{P} \tag{6.3.3}\] Clearing the fractions by multiplying both sides of Equation 6.3.4 by \(P\) gives \[PV = nRT \tag{6.3.4}\] This equation is known as the . An ideal gas is defined as a hypothetical gaseous substance whose behavior is independent of attractive and repulsive forces and can be completely described by the ideal gas law. In reality, there is no such thing as an ideal gas, but an ideal gas is a useful conceptual model that allows us to understand how gases respond to changing conditions. As we shall see, under many conditions, most real gases exhibit behavior that closely approximates that of an ideal gas. The ideal gas law can therefore be used to predict the behavior of real gases under most conditions. The ideal gas law does not work well at very low temperatures or very high pressures, where deviations from ideal behavior are most commonly observed. Significant deviations from ideal gas behavior commonly occur at low temperatures and very high pressures. Before we can use the ideal gas law, however, we need to know the value of the gas constant R. Its form depends on the units used for the other quantities in the expression. If V is expressed in liters (L), P in atmospheres (atm), T in kelvins (K), and n in moles (mol), then \[R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol} \tag{6.3.5}\] Because the product PV has the units of energy, R can also have units of J/(K•mol): \[R = 8.3145 \dfrac{\rm J}{\rm K\cdot mol}\tag{6.3.6}\] Scientists have chosen a particular set of conditions to use as a reference: 0°C (273.15 K) and \(\rm1\; bar = 100 \;kPa = 10^5\;Pa\) pressure, referred to as standard temperature and pressure ( ). \[\text{STP:} \hspace{2cm} T=273.15\;{\rm K}\text{ and }P=\rm 1\;bar=10^5\;Pa\] Please note that STP was defined differently in the part. The old definition was based on a standard pressure of 1 atm. We can calculate the volume of 1.000 mol of an ideal gas under standard conditions using the variant of the ideal gas law given in Equation 6.3.4: \[V=\dfrac{nRT}{P}\tag{6.3.7}\] Thus the volume of 1 mol of an ideal gas is and , approximately equivalent to the volume of three basketballs. The molar volumes of several real gases at 0°C and 1 atm are given in Table 10.3, which shows that the deviations from ideal gas behavior are quite small. Thus the ideal gas law does a good job of approximating the behavior of real gases at 0°C and 1 atm. The relationships described in Section 10.3 as Boyle’s, Charles’s, and Avogadro’s laws are simply special cases of the ideal gas law in which two of the four parameters (P, V, T, and n) are held fixed. The ideal gas law allows us to calculate the value of the fourth variable for a gaseous sample if we know the values of any three of the four variables ( , , , and ). It also allows us to predict the of a sample of a gas (i.e., its final temperature, pressure, volume, and amount) following any changes in conditions if the parameters ( , , , and ) are specified for an Some applications are illustrated in the following examples. The approach used throughout is always to start with the same equation—the ideal gas law—and then determine which quantities are given and which need to be calculated. Let’s begin with simple cases in which we are given three of the four parameters needed for a complete physical description of a gaseous sample. The balloon that Charles used for his initial flight in 1783 was destroyed, but we can estimate that its volume was 31,150 L (1100 ft ), given the dimensions recorded at the time. If the temperature at ground level was 86°F (30°C) and the atmospheric pressure was 745 mmHg, how many moles of hydrogen gas were needed to fill the balloon? volume, temperature, and pressure amount of gas We are given values for , , and and asked to calculate . If we solve the ideal gas law (Equation 6.3.4) for , we obtain \[\rm745\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.980\;atm\] and are given in units that are not compatible with the units of the gas constant [ = 0.08206 (L•atm)/(K•mol)]. We must therefore convert the temperature to kelvins and the pressure to atmospheres: \[T=273+30=303{\rm K}\] Substituting these values into the expression we derived for , we obtain \[n=\dfrac{PV}{RT}=\rm\dfrac{0.980\;atm\times31150\;L}{0.08206\dfrac{atm\cdot L}{\rm mol\cdot K}\times 303\;K}=1.23\times10^3\;mol\] Suppose that an “empty” aerosol spray-paint can has a volume of 0.406 L and contains 0.025 mol of a propellant gas such as CO . What is the pressure of the gas at 25°C? 1.5 atm In Example \(\Page {1}\), we were given three of the four parameters needed to describe a gas under a particular set of conditions, and we were asked to calculate the fourth. We can also use the ideal gas law to calculate the effect of in any of the specified conditions on any of the other parameters, as shown in Example \(\Page {5}\). The Ideal Gas Law: When a gas is described under two different conditions, the ideal gas equation must be applied twice - to an initial condition and a final condition. This is: \[\begin{array}{cc}\text{Initial condition }(i) & \text{Final condition} (f)\\P_iV_i=n_iRT_i & P_fV_f=n_fRT_f\end{array}\] Both equations can be rearranged to give: \[R=\dfrac{P_iV_i}{n_iT_i} \hspace{1cm} R=\dfrac{P_fV_f}{n_fT_f}\] The two equations are equal to each other since each is equal to the same constant \(R\). Therefore, we have: \[\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\tag{6.3.8}\] The equation is called the . The equation is particularly useful when one or two of the gas properties are held constant between the two conditions. In such cases, the equation can be simplified by eliminating these constant gas properties. Suppose that Charles had changed his plans and carried out his initial flight not in August but on a cold day in January, when the temperature at ground level was −10°C (14°F). How large a balloon would he have needed to contain the same amount of hydrogen gas at the same pressure as in Example \(\Page {1}\)? temperature, pressure, amount, and volume in August; temperature in January volume in January To see exactly which parameters have changed and which are constant, prepare a table of the initial and final conditions: Both \(n\) and \(P\) are the same in both cases (\(n_i=n_f,P_i=P_f\)). Therefore, Equation can be simplified to: \[\dfrac{V_i}{T_i}=\dfrac{V_f}{T_f}\] This is the relationship first noted by Charles. Solving the equation for \(V_f\), we get: \[V_f=V_i\times\dfrac{T_f}{T_i}=\rm31150\;L\times\dfrac{263\;K}{303\;K}=2.70\times10^4\;L\] It is important to check your answer to be sure that it makes sense, just in case you have accidentally inverted a quantity or multiplied rather than divided. In this case, the temperature of the gas decreases. Because we know that gas volume decreases with decreasing temperature, the final volume must be less than the initial volume, so the answer makes sense. We could have calculated the new volume by plugging all the given numbers into the ideal gas law, but it is generally much easier and faster to focus on only the quantities that change. At a laboratory party, a helium-filled balloon with a volume of 2.00 L at 22°C is dropped into a large container of liquid nitrogen (T = −196°C). What is the final volume of the gas in the balloon? : 0.52 L Example \(\Page {1}\) illustrates the relationship originally observed by Charles. We could work through similar examples illustrating the inverse relationship between pressure and volume noted by Boyle ( = constant) and the relationship between volume and amount observed by Avogadro ( / = constant). We will not do so, however, because it is more important to note that the historically important gas laws are only special cases of the ideal gas law in which two quantities are varied while the other two remain fixed. The method used in Example \(\Page {1}\) can be applied in such case, as we demonstrate in Example \(\Page {2}\) (which also shows why heating a closed container of a gas, such as a butane lighter cartridge or an aerosol can, may cause an explosion). Aerosol cans are prominently labeled with a warning such as “Do not incinerate this container when empty.” Assume that you did not notice this warning and tossed the “empty” aerosol can in Exercise 5 (0.025 mol in 0.406 L, initially at 25°C and 1.5 atm internal pressure) into a fire at 750°C. What would be the pressure inside the can (if it did not explode)? initial volume, amount, temperature, and pressure; final temperature final pressure Follow the strategy outlined in Example \(\Page {5}\). Prepare a table to determine which parameters change and which are held constant: Both \(V\) and \(n\) are the same in both cases (\(V_i=V_f,n_i=n_f\)). Therefore, Equation can be simplified to: \[P_iT_i=P_fT_f\] By solving the equation for \(P_f\), we get: \[P_f=P_i\times\dfrac{T_i}{T_f}=\rm1.5\;atm\times\dfrac{1023\;K}{298\;K}=5.1\;atm\] This pressure is more than enough to rupture a thin sheet metal container and cause an explosion! Suppose that a fire extinguisher, filled with CO to a pressure of 20.0 atm at 21°C at the factory, is accidentally left in the sun in a closed automobile in Tucson, Arizona, in July. The interior temperature of the car rises to 160°F (71.1°C). What is the internal pressure in the fire extinguisher? : 23.4 atm We saw in Example \(\Page {1}\) that Charles used a balloon with a volume of 31,150 L for his initial ascent and that the balloon contained 1.23 × 10 mol of H gas initially at 30°C and 745 mmHg. Suppose that Gay-Lussac had also used this balloon for his record-breaking ascent to 23,000 ft and that the pressure and temperature at that altitude were 312 mmHg and −30°C, respectively. To what volume would the balloon have had to expand to hold the same amount of hydrogen gas at the higher altitude? initial pressure, temperature, amount, and volume; final pressure and temperature final volume Follow the strategy outlined in Example \(\Page {5}\). Begin by setting up a table of the two sets of conditions: By eliminating the constant property (\(n\)) of the gas, Equation 6.3.8 is simplified to: \[\dfrac{P_iV_i}{T_i}=\dfrac{P_fV_f}{T_f}\] By solving the equation for \(V_f\), we get: \[V_f=V_i\times\dfrac{P_i}{P_f}\dfrac{T_f}{T_i}=\rm3.115\times10^4\;L\times\dfrac{0.980\;atm}{0.411\;atm}\dfrac{243\;K}{303\;K}=5.96\times10^4\;L\] Does this answer make sense? Two opposing factors are at work in this problem: decreasing the pressure tends to the volume of the gas, while decreasing the temperature tends to the volume of the gas. Which do we expect to predominate? The pressure drops by more than a factor of two, while the absolute temperature drops by only about 20%. Because the volume of a gas sample is directly proportional to both and 1/ , the variable that changes the most will have the greatest effect on . In this case, the effect of decreasing pressure predominates, and we expect the volume of the gas to increase, as we found in our calculation. We could also have solved this problem by solving the ideal gas law for and then substituting the relevant parameters for an altitude of 23,000 ft: Except for a difference caused by rounding to the last significant figure, this is the same result we obtained previously. A steel cylinder of compressed argon with a volume of 0.400 L was filled to a pressure of 145 atm at 10°C. At 1.00 atm pressure and 25°C, how many 15.0 mL incandescent light bulbs could be filled from this cylinder? (Hint: find the number of moles of argon in each container.) 4.07 × 10 Second Type of Ideal Gas Law Problems: The ideal gas law can also be used to calculate molar masses of gases from experimentally measured gas densities. To see how this is possible, we first rearrange the ideal gas law to obtain \[\dfrac{n}{V}=\dfrac{P}{RT}\tag{6.3.9}\] The left side has the units of moles per unit volume (mol/L). The number of moles of a substance equals its mass (\(m\), in grams) divided by its molar mass (\(M\), in grams per mole): \[n=\dfrac{m}{M}\tag{6.3.10}\] Substituting this expression for \(n\) into Equation 6.3.9 gives \[\dfrac{m}{MV}=\dfrac{P}{RT}\tag{6.3.11}\] Because \(m/V\) is the density \(d\) of a substance, we can replace \(m/V\) by \(d\) and rearrange to give \[\rho=\dfrac{m}{V}=\dfrac{MP}{RT}\tag{6.3.12}\] The distance between particles in gases is large compared to the size of the particles, so their densities are much lower than the densities of liquids and solids. Consequently, gas density is usually measured in grams per liter (g/L) rather than grams per milliliter (g/mL). Calculate the density of butane at 25°C and a pressure of 750 mmHg. compound, temperature, and pressure density The molar mass of butane (C H ) is Using 0.08206 (L•atm)/(K•mol) for means that we need to convert the temperature from degrees Celsius to kelvins ( = 25 + 273 = 298 K) and the pressure from millimeters of mercury to atmospheres: \[P=\rm750\;mmHg\times\dfrac{1\;atm}{760\;mmHg}=0.987\;atm\] Substituting these values into Equation 6.3.12 gives \[\rho=\rm\dfrac{58.123\;g/mol\times0.987\;atm}{0.08206\dfrac{L\cdot atm}{K\cdot mol}\times298\;K}=2.35\;g/L\] Radon (Rn) is a radioactive gas formed by the decay of naturally occurring uranium in rocks such as granite. It tends to collect in the basements of houses and poses a significant health risk if present in indoor air. Many states now require that houses be tested for radon before they are sold. Calculate the density of radon at 1.00 atm pressure and 20°C and compare it with the density of nitrogen gas, which constitutes 80% of the atmosphere, under the same conditions to see why radon is found in basements rather than in attics. radon, 9.23 g/L; N , 1.17 g/L A common use of Equation 6.3.12 is to determine the molar mass of an unknown gas by measuring its density at a known temperature and pressure. This method is particularly useful in identifying a gas that has been produced in a reaction, and it is not difficult to carry out. A flask or glass bulb of known volume is carefully dried, evacuated, sealed, and weighed empty. It is then filled with a sample of a gas at a known temperature and pressure and reweighed. The difference in mass between the two readings is the mass of the gas. The volume of the flask is usually determined by weighing the flask when empty and when filled with a liquid of known density such as water. The use of density measurements to calculate molar masses is illustrated in Example \(\Page {6}\). The reaction of a copper penny with nitric acid results in the formation of a red-brown gaseous compound containing nitrogen and oxygen. A sample of the gas at a pressure of 727 mmHg and a temperature of 18°C weighs 0.289 g in a flask with a volume of 157.0 mL. Calculate the molar mass of the gas and suggest a reasonable chemical formula for the compound. pressure, temperature, mass, and volume molar mass and chemical formula Solving Equation 6.3.12 for the molar mass gives Density is the mass of the gas divided by its volume: \[\rho=\dfrac{m}{V}=\dfrac{0.289\rm g}{0.17\rm L}=1.84 \rm g/L\] We must convert the other quantities to the appropriate units before inserting them into the equation: \[T=18+273=291 K\] \[P=727\rm mmHg\times\dfrac{1\rm atm}{760\rm mmHg}=0.957\rm atm\] The molar mass of the unknown gas is thus \[\rho=\rm\dfrac{1.84\;g/L\times0.08206\dfrac{L\cdot atm}{K\cdot mol}\times291\;K}{0.957\;atm}=45.9 g/mol\] \[M({\rm NO})=14 + 16=30 \rm\; g/mol\] \[M({\rm N_2O})=(2)(14)+16=44 \rm\;g/mol\] \[M({\rm NO_2})=14+(2)(16)=46 \rm\;g/mol\] The most likely choice is NO which is in agreement with the data. The red-brown color of smog also results from the presence of NO gas. You are in charge of interpreting the data from an unmanned space probe that has just landed on Venus and sent back a report on its atmosphere. The data are as follows: pressure, 90 atm; temperature, 557°C; density, 58 g/L. The major constituent of the atmosphere (>95%) is carbon. Calculate the molar mass of the major gas present and identify it. : 44 g/mol; \(CO_2\) Density and the Molar Mass of Gases: The ideal gas law is derived from empirical relationships among the pressure, the volume, the temperature, and the number of moles of a gas; it can be used to calculate any of the four properties if the other three are known. : \(PV = nRT\), where \(R = 0.08206 \dfrac{\rm L\cdot atm}{\rm K\cdot mol}=8.3145 \dfrac{\rm J}{\rm K\cdot mol}\) : \(\dfrac{P_iV_i}{n_iT_i}=\dfrac{P_fV_f}{n_fT_f}\) \(\rho=\dfrac{MP}{RT}\) The empirical relationships among the volume, the temperature, the pressure, and the amount of a gas can be combined into the , = . The proportionality constant, , is called the and has the value 0.08206 (L•atm)/(K•mol), 8.3145 J/(K•mol), or 1.9872 cal/(K•mol), depending on the units used. The ideal gas law describes the behavior of an , a hypothetical substance whose behavior can be explained quantitatively by the ideal gas law and the kinetic molecular theory of gases. is 0°C and 1 atm. The volume of 1 mol of an ideal gas at STP is 22.41 L, the . All of the empirical gas relationships are special cases of the ideal gas law in which two of the four parameters are held constant. The ideal gas law allows us to calculate the value of the fourth quantity ( , , , or ) needed to describe a gaseous sample when the others are known and also predict the value of these quantities following a change in conditions if the original conditions (values of , , , and ) are known. The ideal gas law can also be used to calculate the density of a gas if its molar mass is known or, conversely, the molar mass of an unknown gas sample if its density is measured. | 18,924 | 1,972 |
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Most of the O consumed by aerobic organisms is used to produce energy in a process referred to as "oxidative phosphorylation," a series of reactions in which electron transport is coupled to the synthesis of ATP and in which the driving force for the reaction is provided by the four-electron oxidizing power of O (Reaction 5.1). (This subject is described in any standard text on biochemistry and will not be discussed in detail here.) The next to the last step in the electrontransport chain produces reduced cytochrome c, a water-soluble electron-transfer protein. Cytochrome c then transfers electrons to cytochrome c oxidase, where they are ultimately transferred to O . (Electron-transfer reactions are discussed in Chapter 6.) Cytochrome c oxidase is the terminal member of the respiratory chain in all animals and plants, aerobic yeasts, and some bacteria. This enzyme is always found associated with a membrane: the inner mitochondrial membrane in higher organisms or the cell membrane in bacteria. It is a large, complex, multisubunit enzyme whose characterization has been complicated by its size, by the fact that it is membrane-bound, and by the diversity of the four redox metal sites, i.e., two copper ions and two heme iron units, each of which is found in a different type of environment within the protein. Because of the complexity of this system and the absence of detailed structural information, spectroscopic studies of this enzyme and comparisons of spectral properties with O -binding proteins and with model iron-porphyrin and copper complexes have been invaluable in its characterization. Iron-porphyrin complexes of imidazole are a logical starting point in the search for appropriate spectroscopic models for heme centers in metalloproteins, since the histidyl imidazole side chain is the most common axial ligand bound to iron in such enzymes. Iron-porphyrin complexes with two axial imidazole ligands are known for both the ferrous and ferric oxidation states. \(\tag{5.32}\) Monoimidazole complexes of iron porphyrins are also known for both the ferrous and the ferric oxidation states. The design of these model complexes has been more challenging than for six-coordinate complexes because of the high affinity of the five-coordinate complexes for a sixth ligand. In the ferrous complex, five coordination has been achieved by use of 2-methylimidazole ligands, as described in Chapter 4. The ferrous porphyrin binds a single 2-methylimidazole ligand, and, because the Fe center is raised out of the plane of the porphyrin ring, the 2-methyl substituent suffers minimal steric interactions with the porphyrin. However, the affinity of the five-coordinate complex for another 2-methylimidazole ligand is substantially lower, because the Fe must drop down into the plane of the porphyrin to form the six-coordinate complexes, in which case the 2-methyl substitutents on both axial ligands suffer severe steric interactions with the porphyrin. Using this approach, five-coordinate monoimidazole complexes can be prepared. They are coordinatively unsaturated, and will bind a second axial ligand, such as O and CO. They have been extensively studied as models for O -binding heme proteins such as hemoglobin and myoglobin. Monoimidazole ferrous porphyrins thus designed are high-spin d6 with four unpaired electrons. They are even-spin systems and EPR spectra have not been observed. Five-coordinate monoimidazole ferric-porphyrin complexes have also been prepared in solution by starting with a ferric porphyrin complex with a very poorly coordinating anion, e.g., Fe P(SbF ). Addition of one equivalent of imidazole results in formation of the five-coordinate monoimidazole complex (Reaction 5.33). \[Fe^{III}(TPP)(SbF_{6}) + ImH \rightarrow [Fe^{III}(TPP)(ImH)]^{+} + SbF_{6}^{-} \tag{5.33}\] When imidazole is added to ferric-porphyrin complexes of other anionic ligands, e.g., CI , several equivalents of imidazole are required to displace the more strongly bound anionic ligand; consequently, only six-coordinate complexes are observed (Reaction 5.34). \[FE^{III}(TPP)Cl + 2 ImH \rightarrow [Fe^{III}(TPP)(ImH)_{2}]^{+} + Cl^{-} \tag{5.34}\] Monoimidazole ferric porphyrins are coordinatively unsaturated, readily bind a second axial ligand, and thus are appropriate models for methemoglobin or metmyoglobin. The five-coordinate complexes are high-spin d , but usually become low-spin upon binding another axial ligand to become six-coordinate. The oxidized form of cytochrome c oxidase contains two Cu and two Fe heme centers. It can be fully reduced to give a form of the enzyme containing two Cu and two Fe heme centers. The heme found in cytochrome c oxidase is different from that found in other heme proteins. It is heme a, closely related to heme b, which is found in hemoglobin, myoglobin, and cytochrome P-450, but has one of the vinyl groups replaced by a farnesyl substituent and one of the methyl groups replaced by a formyl substituent (see 5.35). Each of the four metal centers has a different coordination environment appropriate to its function. Cytochrome a and Cu appear solely to carry out an electron-transfer function without interacting directly with dioxygen. Cytochrome a and Cu appear to be part of a binuclear center that acts as the site for dioxygen binding and reduction. A schematic describing the probable nature of these four metal sites within cytochrome oxidase is given in Figure 5.3 and a description of the evidence supporting the formulation of each center then follows. Cytochrome a in both oxidation states has spectral characteristics that are entirely consistent with a low-spin ferric heme center with two axial imidazole ligands. In its oxidized form, it gives an EPR spectrum with g values similar to those obtained with model ferric-porphyrin complexes with two axial imidazole ligands (see above, Section IV.B.1). Moreover, addition of cyanide anion to the oxidized enzyme or CO to the reduced enzyme does not perturb this center, indicating that cyanide does not bind to the heme, again consistent with a six-coordinate heme. The absence of ligand binding is characteristic of six-coordinate heme sites found in electron-transfer proteins and suggests strongly that cytochrome a functions as an electron-transfer center within cytochrome c oxidase. Cu is also believed to act as an electron-transfer site. It has quite remarkable EPR spectroscopic characteristics, with g values at g = 2.18, 2.03, and 1.99, and no hyperfine splitting, resembling more an organic free radical than a typical Cu center (see Figures 5.2 and 5.4A). ENDOR studies of yeast cytochrome c oxidase containing H-cysteine or N-histidine (from yeast grown with the isotopically substituted amino acids) showed shifts relative to the unsubstituted enzyme, indicating that both of these ligands are bound to Cu . But the linear electric-field effect of Cu did not give the patterns characteristic of Cu -histidine complexes, indicating that the unpaired electron is not on the copper ion. The current hypothesis about this center is that copper is bonded in a highly covalent fashion to one, or more likely two, sulfur ligands, and that the unpaired electron density is principally on sulfur, i.e., [Cu - SR\(\leftrightarrow\)Cu - • SR]. Copper-thiolate model complexes with spectroscopic properties similar to Cu have never been synthesized, presumably because such complexes are unstable with respect to disulfide bond formation, i.e, 2 RS• \(\rightarrow\) RS-SR. In the enzyme, RS• radicals are presumably constrained in such a way that they cannot couple to form disulfide bonds. The other heme center, cytochrome a , does bind ligands such as cyanide to the Fe form and carbon monoxide to the Fe form, indicating that it is either five-coordinate or that it has a readily displaceable ligand. Reaction with CO, for example, produces spectral changes characteristic of a five-coordinate ferrous heme binding CO to give the six-coordinate carbonmonoxy product analogous to MbCO. The cytochrome a site is therefore an excellent candidate for O binding within cytochrome oxidase. The EPR spectrum of fully oxidized cytochrome c oxidase might be expected to give signals corresponding to two Cu centers and two ferric heme centers. In fact, all that is observed in the EPR spectrum of the oxidized enzyme is the typical low-spin six-coordinate ferric heme spectrum due to cytochrome G and the EPR signal attributed to Cu (see Figure 5.4A). The fact that signals attributable to cytochrome a and Cu are not observed in the EPR spectrum led to the suggestion that these two metal centers are antiferromagnetically coupled. The measured magnetic susceptibility for the isolated enzyme was found to be consistent with this hypothesis, suggesting that these two metal centers consist of an S = \(\frac{1}{2}\) Cu antiferromagnetically coupled through a bridging ligand to a high-spin S = \(\frac{5}{2}\) Fe to give an S = 2 binuclear unit. EXAFS measurements indicating a copper-iron separation of 3-4 Å as well as the strength of the magnetic coupling suggest that the metal ions are linked by a single-atom ligand bridge, but there is no general agreement as to the identity of this bridge. The cytochrome a -Cu coupling can be disrupted by reduction of the individual metal centers. In this fashion, a g = 6 ESR signal can be seen for cytochrome a or g = 2.053, 2.109, and 2.278 signals for Cu . Nitric-oxide binding to Cu also decouples the metals, allowing the g = 6 signal to be seen (see Figure 5.4B). Mössbauer spectroscopy also indicates that cytochrome a is high-spin in the oxidized as well as the reduced state. ENDOR studies suggest that Cu has three nitrogens from imidazoles bound to it with water or hydroxide as a fourth ligand. Studies using N-labeled histidine in yeast have demonstrated that histidine is a ligand to cytochrome a . All of these features have been incorporated into Figure 5.3. Before we consider the reactions of cytochrome c oxidase with dioxygen, it is instructive to review the reactions of dioxygen with iron porphyrins and copper complexes. Dioxygen reacts with ferrous-porphyrin complexes to make mononuclear dioxygen complexes (Reaction 5.36; see preceding chapter for discussion of this important reaction). Such dioxygen complexes react rapidly with another ferrous porphyrin, unless sterically prevented from doing so, to form binuclear peroxo-bridged complexes (Reaction 5.37). These peroxo complexes are stable at low temperature, but, when the temperature is raised, the O—O bond cleaves and two equivalents of an iron(IV) oxo complex are formed (Reaction 5.38). Subsequent reactions between the peroxo-bridged complex and the Fe oxo complex produce the\(\mu\)-oxo dimer (see Reactions 5.39-5.40). \[3Fe^{II}(P) + 3O_{2} \rightarrow 3Fe(P)(O_{2}) \tag{5.36}\] \[3Fe(P)(O_{2}) + 3Fe^{II}(P) \rightarrow 3(P)Fe^{III}—O—O—Fe^{III}(P) \tag{5.37}\] \[(P)Fe^{III}—O—O—Fe^{III}(P) \rightarrow 2Fe^{IV}(P)(O) \tag{5.38}\] \[2Fe^{IV}(P)(O) + 2(P)Fe^{III}—O—O—Fe^{III}(P) \rightarrow 2(P)Fe^{III}—O—Fe^{III}(P) + 2Fe(P)(O_{2}) \tag{5.39}\] \[2Fe(P)(O_{2}) \rightarrow 2 Fe^{II}(P) + 2O_{2} \tag{5.40}\] \[4Fe^{II}(P) + O_{2} \rightarrow 2(P)Fe^{III}—O—Fe^{III}(P) \tag{5.41}\] The reaction sequence (5.36) to (5.40) thus describes a four-electron reduction of O in which the final products, two oxide, O , ligands act as bridging ligands in binuclear ferric-porphyrin complexes (Reaction 5.41). Copper(l) complexes similarly react with dioxygen to form peroxo-bridged binuclear complexes. Such complexes do not readily undergo O—O bond cleavage, apparently because the copper(III) oxidation state is not as readily attainable as the Fe(IV) oxidation state in an iron-porphyrin complex. Nevertheless, stable peroxo complexes of copper(II) have been difficult to obtain, because, as soon as it is formed, the peroxo complex either is protonated to give free hydrogen peroxide or is itself reduced by more copper(l) (Reactions 5.42 to 5.46). \[2Cu^{I} + O_{2} \rightarrow Cu^{II}—O—O—Cu^{II} \tag{5.42}\] \[Cu^{II}—O—O—Cu^{II} + 2H^{+} \rightarrow 2 Cu^{II} + H_{2}O_{2} \tag{5.43}\] \[2Cu^{I} + H_{2}O_{2} + 2H^{+} \rightarrow 2 Cu^{II} + 2 H_{2}O \tag{5.44}\] or $$Cu^{II}—O—O—Cu^{II} + 2 Cu^{I} + 4 H^{+} \rightarrow 4 Cu^{II} + 2 H_{2}O \tag{5.45}\] \[4 Cu^{I} + O_{2} + 4 H^{+} \rightarrow 4 Cu^{II} + 2 H_{2}O \tag{5.46}\] Recently, however, examples of the long-sought stable binuclear copper(II) peroxo complex have been successfully synthesized and characterized, and interestingly enough, two entirely different structural types have been identified, i.e., \(\mu\)-1,2 and \(\mu\)-\(\eta^{2}\):\(\eta^{2}\) dioxygen complexes (see 5.47) \(\tag{5.47}\) A single turnover in the reaction of cytochrome c oxidase involves (1) reduction of the four metal centers by four equivalents of reduced cytochrome c, (2) binding of dioxygen to the partially or fully reduced enzyme, (3) transfer of four electrons to dioxygen, coupled with (4) protonation by four equivalents of protons to produce two equivalents of water, all without the leakage of any substantial amount of potentially harmful partially reduced dioxygen byproducts such as superoxide or hydrogen peroxide. At low temperatures, the reaction can be slowed down, so that the individual steps in the dioxygen reduction can be observed. Such experiments are carried out using the fully reduced enzyme to which CO has been bound. Binding of CO to the Fe heme center in reduced cytochrome c oxidase inhibits the enzyme and makes it unreactive to dioxygen. The CO-inhibited derivative can then be mixed with dioxygen and the mixture cooled. Photolysis of metal-CO complexes almost always leads to dissociation of CO, and CO-inhibited cytochrome c oxidase is no exception. Photolytic dissociation of CO frees the Fe heme, thereby initiating the reaction with dioxygen, which can then be followed spectroscopically. Dioxygen reacts very rapidly with the fully reduced enzyme to produce a species that appears to be the dioxygen adduct of cytochrome a (Reaction 5.48). Such a species is presumed to be similar to other mononuclear oxyheme derivatives. The dioxygen ligand in this species is then rapidly reduced to peroxide by the nearby Cu , forming what is believed to be a binuclear \(\mu\)-peroxo species (Reaction 5.49). These steps represent a two-electron reduction of dioxygen to the peroxide level, and are entirely analogous to the model reactions discussed above (Reactions 5.36 to 5.46), except that the binuclear intermediates contain one copper and one heme iron. The \(\mu\)-peroxo Fe - (O ) - Cu species is then reduced by a third electron, resulting in cleavage of the O—O bond (Reaction 5.50). One of the oxygen atoms remains with iron in the form of a ferryl complex, i. e., an Fe oxo, and the other is protonated and bound to copper in the form of a Cu aquo complex. Reduction by another electron leads to hydroxo complexes of both the Fe heme and the Cu centers (Reaction 5.51). Protonation then causes dissociation of two water molecules from the oxidized cytochrome a -Cu center (Reaction 5.52). \[(cyt\; a_{3})\overbrace{Fe^{II} \quad Cu_{B}^{I}} + O_{2} \rightarrow (cyt\; a_{3})\overbrace{Fe^{III}(O_{2}^{-}) \quad Cu_{B}^{I}} \tag{5.48}\] \[(cyt\; a_{3})\overbrace{Fe^{III}(O_{2}^{-}) \quad Cu_{B}^{I}} \rightarrow (cyt\; a_{3})\overbrace{Fe^{III}-(O_{2}^{2-}) \quad Cu_{B}^{II}} \tag{5.49}\] \[(cyt\; a_{3})\overbrace{Fe^{III}-(O_{2}^{2-})-Cu_{B}^{II}} + e^{-} + 2H^{+} \rightarrow (cyt\; a_{3})\overbrace{Fe^{IV}=O \quad H_{2}O-Cu_{B}^{II}} \tag{5.50}\] \[(cyt\; a_{3})\overbrace{Fe^{IV}=O \quad H_{2}O-Cu_{B}^{II}} +e^{-} \rightarrow (cyt\; a_{3})\overbrace{Fe^{III}-(OH^{-}) \quad (HO^{-})-Cu_{B}^{II}} \tag{5.51}\] \[(cyt\; a_{3})\overbrace{Fe^{III}-(OH^{-}) \quad (HO^{-})-Cu_{B}^{II}} + 2H^{+} \rightarrow (cyt\; a_{3})\overbrace{Fe^{III} \quad Cu_{B}^{II}} + 2 H_{2}O \tag{5.52}\] Several important questions remain to be resolved in cytochrome c oxidase research. One is the nature of the ligand bridge that links cytochrome a and Cu in the oxidized enzyme. Several hypotheses have been advanced (imidazolate, thiolate sulfur, and various oxygen ligands), but then discarded or disputed, and there is consequently no general agreement concerning its identity. However, EXAFS measurements of metal-metal separation and the strength of the magnetic coupling between the two metal centers provide evidence that a single atom bridges the two metals. Another issue, which is of great importance, is to find out how the energy released in the reduction of dioxygen is coupled to the synthesis of ATP. It is known that this occurs by coupling the electron-transfer steps to a proton-pumping process, but the molecular mechanism is unknown. Future research should provide some interesting insights into the mechanism of this still mysterious process. | 16,986 | 1,973 |
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We introduced the concept of equilibrium in Chapter 11, where you learned that a liquid and a vapor are in equilibrium when the number of molecules evaporating from the surface of the liquid per unit time is the same as the number of molecules condensing from the vapor phase. Vapor pressure is an example of a physical equilibrium because only the physical form of the substance changes. Similarly, in Chapter 13, we discussed saturated solutions, another example of a physical equilibrium, in which the rate of dissolution of a solute is the same as the rate at which it crystallizes from solution. In this chapter, we describe the methods chemists use to quantitatively describe the composition of chemical systems at equilibrium, and we discuss how factors such as temperature and pressure influence the equilibrium composition. As you study these concepts, you will also learn how urban smog forms and how reaction conditions can be altered to produce \(H_2\) rather than the combustion products \(CO_2\) and \(H_2O\) from the methane in natural gas. You will discover how to control the composition of the gases emitted in automobile exhaust and how synthetic polymers such as the polyacrylonitrile used in sweaters and carpets are produced on an industrial scale. | 1,282 | 1,974 |
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You may have noticed "E85" gasoline selling for $2.39/gallon for FlexFuel vehicles while regular gas was selling for $2.79. Is E85 use advantageous if you have a FlexFuel vehicle capable of using either E85 (85% ethanol) or regular gas? Note that many states sell regular gas that is E10 (10% ethanol) for standard (non-FlexFuel) vehicles. 85% Ethanol at a modern pump The model T was first designed to run on ethanol There are really (at least) two questions here: 1. Is it a good deal for the consumer (is energy from ethanol economical)? 2. Is it a good deal for the environment (is the favorable)? The energy balance is the ratio of the energy produced by 1 kg of the fuel (i.e. ethanol), to the energy necessary to produce it (cultivation of plants, fermentation, transportation, irrigation, etc.). The energy balance for ethanol in the US is only 1.3 to 1.6 (we get 1.3-1.6 J out for 1 J energy input), while in Brazil it's 8.3 to 10.2. The energy balance for oil is about 5: Today, about 5 barrels of oil extracted for every 1 barrel of oil is consumed in the process (a century ago, when oil was more plentiful the ratio was 50:1). The reason for the poor energy balance for fuel ethanol in the US is that we use corn. We extract the cornstarch, and then hydrolyze it at high temperatures (over 90 C and ferment the resulting sugars to give the ethanol (simultaneous saccharification and fermentation, SSF). More energy efficient processes are being developed. Starch is a polymer of glucose composed of amylose (shown below) and amylopectin, a similar polymer with more branching. The chemical reactions that occur are first, hydrolysis of starches to maltose (C H O ): The complex sugar maltose is further hydrolyzed (sometimes in the same step) to glucose (C H O ): Finally, during ethanol fermentation, glucose is decomposed into ethanol (C H OH) and carbon dioxide. Switchgrass requires similar processing, but has the potential for a higher energy balance. In Brazil, sugarcane is used in ethanol production, eliminating the high energy cost reaction (1), because sugarcane produces sucrose, a sugar similar to maltose. In all cases, a large energy expenditure is necessary to separate the ethanol from the water in which the reactions take place, and other impurities.
↑ Sugarcane ← Switchgrass But even in Brazil, rising sugar costs led to a rebound in reliance on petroleum, and E100 capable vehicles declined sharply in the late 1980s . Nonetheless, the existence of ethanol pumps at a large percentage of gas stations has helped sustain the use of other blends. Clearly, the economics and history of fuel use should be studied carefully when investing in future fuel development. So which is a better deal: filling your 10-gallon tank with E85 costing $2.69/gallon or with gasoline costing $2.99/gallon? . Energy changes which accompany chemical reactions are almost always expressed by . For example, during combustion ethanol reacts with oxygen to produce carbon dioxide, water vapor, and heat according to the thermochemical equation: The quantity Δ is the . In this context the symbol Δ (delta) signifies change in” while is the symbol for the quantity being changed, namely the enthalpy. We will deal with the enthalpy in some detail in Chap. 15. For the moment we can think of it as a property of matter which increases when matter absorbs energy and decreases when matter releases energy. Here the Δ (delta subscript m) tells us whether heat energy is released or absorbed when the reaction occurs and also enables us to find the actual quantity of energy involved. By convention, if Δ is , heat is by the reaction; i.e., it is . More commonly, Δ is as in Eq. (3), indicating that heat energy is rather than absorbed by the reaction, and that the reaction is . This convention as to whether Δ is positive or negative looks at the heat change in terms of the matter actually involved in the reaction rather than its surroundings. In the reaction in Eq. (4), stronger bonds have formed, leading to a decrease in potential energy, and it is this decrease which is indicated by a negative value of Δ . It is important to notice that the quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. (4) tells us that 1367 kJ of heat energy is given off of C H OH which is consumed. Alternatively, it tells us that 1367 kJ is released H O produced, or every 2 mol of carbon dioxide produced, or every 3 mol of oxygen consumed. Seen in this way, Δ is a conversion factor enabling us to calculate the heat absorbed when a given amount of substance is consumed or produced. If is the quantity of heat absorbed and is the amount of substance involved, then \(\Delta H_{\text{m}}=\frac{q}{n}\) (5) Equation (4) represents the standard heat of combustion, which is related to the "higher heating value" of a fuel. Because liquid water is produced in Equation (4) and the heat (1367 kJ/mol) is given for a theoretical reaction occurring at 25 C, it is only an approximation of the heat produced in an actual combustion, where the reaction takes place at a high temperature and produces water vapor. A better estimate may be the Lower Heating Value (LHV) which is adjusted by adding the heat of vaporization of water, and heat required to raise the temperature of reactants to the combustion temperature and products to 150 C (an arbitrarily chosen standard). We'll use the LHV for ethanol, -1330 kJ/mol, and abbreviate ethanol (C H OH) as EtOH to calculate the heat in 1 gallon of ethanol: \(V_{\text{EtOH}}~\xrightarrow{M}~m_{\text{EtOH}}\text{ }\xrightarrow{M}\text{ }n_{\text{EtOH}}~\xrightarrow{\Delta H_{m}}~q\) so that \(q=\text {1 gallon EtOH}
~\times~\frac{\text{3.79 L}}{\text{1 gallon}}
~\times~\frac{\text{1000 mL}}{\text{1 L}}
~\times~\frac{\text{0.789 g}}{\text{mL}}
~\times~\frac{\text{1 mol EtOH}}{\text{46.07 g EtOH}}
~\times~\frac{-\text{1330 kJ}}{\text{mol EtOH}} \) \(=-86~330\text{ kJ}=-\text{86}\text{.33 MJ (estimate)}\) We can calculate the energy value for octane, which represents gasoline fairly well, in a similar way. The thermochemical equation for the combustion of octane (C H ) is: C H + 25/2 O → 8 CO + 9 H O ( ) (6) ΔH ~ -5430 kJ/mol Again, we'll use the LHV for octane (-5064 kJ/mol ) in our calculation: \(q=\text {1 gallon octane}
~\times~\frac{\text{3.79 L}}{\text{1 gallon}}
~\times~\frac{\text{1000 mL}}{\text{1 L}}
~\times~\frac{\text{0.737 g}}{\text{mL}}
~\times~\frac{\text{1 mol octane}}{\text{114.23 g octane}}
~\times~\frac{\text{-5 064 kJ}}{\text{mol octane}} \)
\(=-123~800\text{ kJ}=-\text{123}\text{.80 MJ (estimate)}\) So 1 gallon of gasoline has (123,800 / 86,330) or 1.4 times as much heating value, but costs only $2.79 / $2.39 or 1.2 times as much. It's the better buy. If the Energy Balance may only be about 1.3 in the US, it actually is a net loss to burn ethanol, . As the Energy Balance and availability of petroleum decreases, we had better develop more energy efficient means of ethanol production. Indeed, based on EPA tests for all 2006 E85 models, the average fuel economy for E85 vehicles was 25.56% lower than unleaded gasoline. It is important to realize that the value of Δ given in thermochemical equations like (4) or (6) depends on the physical state of both the reactants and the products. Thus, if water were obtained as a gas instead of a liquid in the reaction in Eq. (4), the value of Δ would be different from -1367 kJ. It is also necessary to specify both the temperature and pressure since the value of Δ depends very slightly on these variables. If these are not specified [as in Eq. (3)] they usually refer to 25°C and to normal atmospheric pressure. Two more characteristics of thermochemical equations arise from the law of conservation of energy. The first is that For example, H O( ) → H O( ) Δ = 44 kJ (7 ) tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This corresponds to the fact that heat is absorbed from your skin when perspiration evaporates, and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off exactly the same quantity of heat. H O( ) → H O( ) Δ = –44 kJ (7 ) To see why this must be true, suppose that Δ [Eq. (7a)] = 44 kJ while Δ [Eq. (7b)] = –50.0 kJ. If we took 1 mol of liquid water and allowed it to evaporate, 44 kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ of heat which had been created from nowhere! This would violate the law of conservation of energy. The only way the problem can he avoided is for Δ of the reverse reaction to be equal in magnitude but opposite in sign from Δ of the forward reaction. That is, Δ forward = –Δ reverse Since Reaction (4) produces 3 mol of H O( ), it would produce 3 mol x 44 kJ/mol = 132 kJ less energy if the water were produced as the vapor, because the heat released in condensation to the liquid would not be included. The enthalpy change would then be -1367 + 132 kJ or -1235 kJ. This is still different from the LHV because of the heat required to change the temperature of the reactants and products from the standard temperature (25 C) to the combustion temperature. In 2007, only 3.3% of American cars were FlexFuel and only 1% of the filling stations provided FlexFuel, but the percentage has been increasing rapidly. World ethanol production for transport fuel tripled between 2000 and 2007 from 17 billion to more than 52 billion litres, with 89% produced in Brazil and the US. The price differential shows large changes (it was about 30% in 2007). The National Alcohol Program in Brazil, a world leader in using ethanol fuel, mandated decreased reliance on petroleum after the first oil crisis in 1973, and by 1979 several automakers provided cars that ran on pure ethanol (E100). After reaching more than 4 million cars and light trucks running on pure ethanol by the late 1980s,[3] the use of E100-only vehicles sharply declined after increases in sugar prices produced shortages of ethanol fuel. The emphasis has been on E85 (85% ethanol) vehicles in the US, because engines won't start reliably on E100 below about 60 F. Even E85 cannot be used below about 30 F, and in northern states E70 is delivered by E85 pumps without changing the label. | 10,483 | 1,975 |
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One of the most useful features of thermochemical equations is that they can be combined to determine Δ values for other chemical reactions which have never been observed. We might want to see what Δ would be if we could carry out a reaction that has never been done, or it might be interesting for theoretical reasons. For example, we have noted that the body would have to store up to 67.5 lb of sugar complexes for the energy equivalent of 10 lb of fat. In Example 1 below, we calculate the energy for the in which a fat is converted to sugar: C H O ( ) + 8 O ( ) → 3 C H O ( ) Δ for this reaction is the extra energy our body can get from a fat. But first, a simpler example may help to make the method clear. Instead of the oxidation of a complicated fat molecule, we'll consider the simplest possible oxidation, a sequence in which carbon itself is oxidized. Step 1 is the oxidation of 1 mol C( ) and 0.5 mol O ( ) to form 1 mol CO( ): C( ) + ½O ( ) → CO( ) Δ = –110.5 kJ = Δ (Note that since the equation refers to moles, not molecules, fractional coefficients are permissible.) In step 2 the some of the mole of CO reacts with an additional 0.5 mol O yielding 1 mol CO : CO( ) + ½O ( ) → CO ( ) Δ = –283.0 kJ = Δ The net result of this two-step process is production of 1 mol CO from the original 1 mol C and 1 mol O (0.5 mol in each step). All the CO produced in step 1 is used up in step 2. On paper this net result can be obtained by the two chemical equations as though they were algebraic equations. The CO produced is canceled by the CO consumed since it is both a reactant and a product of the overall reaction Experimentally it is found that the enthalpy change for the net reaction is the of the enthalpy changes for steps 1 and 2: Δ = –110.5 kJ + (–283.0 kJ ) = –393.5 kJ = Δ + Δ That is, the thermochemical equation C( ) + O ( ) → CO ( ) Δ = –393.5 kJ is the correct one for the overall reaction. In the general case it is always true that . This principle is known as . If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain Δ values for reactions which cannot be carried out experimentally, as the next example shows. Although fat metabolism is a complicated process (called "beta oxidation") which yields the ATP that releases energy to muscle, we could imagine a reaction that helps us understand why fats store so much energy compared to sugar. We could imagin the combustion of steric acid to the sugar, glucose, according to the equation (1) C H O ( ) + 8 O ( ) → 3 C H O ( ) Δ This would represent the "extra" energy that fats provide, over the energy that metabolism of a sugar like glucose provides. Calculate Δ for this reaction from the following thermochemical equations, (which are heats of combustion that are easily determined experimentally): (2) C H O ( ) + 26 O ( ) → 18 CO ( ) + 18 H O( ) (25°C, 1 atm pressure) Δ = –11 407 kJ (3) C H O ( ) + 6 O ( ) → 6 CO ( ) + 6 H O( ) (25°C, 1 atm pressure) Δ = –2 800 kJ We see that reaction (3) has glucose (C H O ) on the left, but the target reaction (1) has it on the right. We'll need to reverse equation (3), and then combine it with equation (2) to get the target equation (1). If we reverse (3), we change the sign on Δ : (3a) 6 CO ( ) + 6 H O( )→ C H O ( ) + 6 O ( ) (-) Δ = +2 800 kJ But we also might notice that the target equation contains no CO ( ) or H O( ), so we'll need to multiply equation (3) by 3, so that there will be an equal amount of CO ( ) or H O( ) on the left and right, and they will cancel. Multiplying equation (3a) by 3: (3b) 18 CO ( ) + 18 H O( )→ 3 C H O ( ) + 18 O ( ) (-3)Δ = +8 400 kJ When we combine this equation, and its associated Δ with Equation (2), we get the target reaction, (1): \(\text{C}_{18}\text{H}_{36}\text{O}_{2}\text{(s)} + \text{26 O}_{\text{2}}\text{(g)}\to \)\(\text{18 CO}_{\text{2}}\text{(g)}~+~\text{18 H}_{2}\text{O}~~~~~~~~~~~~~~~~\Delta H_{\text{m}}=\text{11 407 kJ }\)
\(
\underline{\text{18 CO}_{\text{2}}\text{(g) + 18 H}_{\text{2}}\text{O}(l)}\underline{\to \text{3 C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{6}\text{(s) +}\text{18 O}_{\text{2}}\text{(g)}}~~~~~~~~\Delta H_{\text{m}}=-3\text{(}-\text{2 800 kJ)}\)
\(
\text{C}_{18}\text{H}_{36}\text{O}_{2}\text{(s)} + \text{8 O}_{\text{2}}\text{(g)}\to \text{3 C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{6}\text{(g)} \) \(\Delta H_{\text{m}}=\text{(}\text{-11 407}+\text{8 400 kJ}\text{)}~=~\text{-3007 kJ}\) Δ = Δ + (-3)Δ = –11 407 + 8400 kJ Δ = –3 007 kJ So one mole (284.48 g) of stearic acid releases 3 007 kJ when it's oxidized to 3 mol of glucose. This is 10.57 kJ/g, or 2.5 Cal/g that we get from fat but not from sugar. Additionally, for every gram of stearic acid, we get the energy from 1.900 g of glucose (see the stoichiometry summary table below), which provides 4 Cal per gram. This is 1.90 g x 4 Cal/g = 7.59 Cal, so the total energy from 1 g of fat is 2.5 + 7.6 = 10.1 Cal in this case (similar to the 9 Cal/g estimate for typical fats). | 5,226 | 1,976 |
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Thus far, we have been discussing problems and answers in equilibria--perhaps the most popular type of problem being how to find the of a weak solution given a certain concentration of a molecule. However, those problems in particular usually only involve what is called a monoprotic acid. “Mono” in the word “monoprotic” indicates that there is only one ionizeable hydrogen atom in an acid when immersed in water, whereas the concept of allows for two or more ionizeable hydrogen atoms. Consider the following chemical equation as the molecule acetic acid equilibrates in the solution: \[\ce{ CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}} \nonumber \] Although acetic acid carries a four hydrogen atoms, only a single becomes ionized. Not to get into too much detail between monoprotic and polyprotic acids, but if you desire to find the pH given a concentration of a weak acid (in this case, acetic acid), you would create and complete an adjusting for how much acetic acid disassociates. However, if you wish to find the pH of a solution after a disassociates, there are extra steps that would need to be done. Let's first take a look at a unique example: What is the pH of 0.75 M sulfuric acid? In sulfuric acid (H SO ), there are two ionizable hydrogen atoms. What makes this molecule interesting is that its ionization constant for the first hydrogen (\(K_{a1}\)) ionized is significantly larger than is the second ionization constant (\(K_{a2}\)). The \(K_{a1}\) constant for sulfuric acid is conveniently dubbed “very large” while the \(K_{a2}\) constant is 1.1 x 10 . As such, the sulfuric acid will completely disassociate into HSO and H O ions (as a strong acid). \[ \ce{H2SO4 (aq) + H2O -> HSO4^{-} (aq) + H_3O^{+}} \nonumber \] Since the sulfuric acid completely disassociates in the solution, we can skip the ICE table process for sulfuric acid, and assert that the concentration \(\ce{HSO4^{-}}\) and \(\ce{H3O^{+}}\) are the same as that of H SO , that is 0.75 M. (This neglects the background concentration of \(\ce{H_3O^{+}}\) in water of \(1 \times 10^{-7}M\)). Equation: \[ \ce{HSO4^{-} (aq) + H2O <=> SO4^{2-} (aq) + H_3O^{+}} \nonumber \] ICE Table: \(K_{a2} = \dfrac{[SO_4^{2-},H_3O^+]}{[HSO_4^-]} = 1.1 \times 10^{-2} = 0.011 = \dfrac{x(0.75+x)}{0.75 - x}\) Assume \(x\) in the denominator is negligible. Therefore, \[x = 0.011 M = [SO_4^{2-}] \nonumber \] Since we know the value of x, we can use the equation from the ICE table to find the value of [HSO ]. \[ [HSO_4^-] = 0.75 \; M - x = 0.75 - 0.011 = 0.74 \; M \nonumber \] We can also find [H O ] using the equation from the ICE table. \[[H_3O+] = 0.75 \; M + x = 0.75 + 0.011 = 0.76 \; M \nonumber \] We can then find the pH from the calculated [H O ] value. \[ pH = -log[H_3O^+] = -log0.76 = 0.119 \nonumber \] Let's say our task is to find the pH given a polyprotic which gains protons in water. Thankfully, the process is essentially the same as finding the pH of a polyprotic acid except in this case we deal with the concentration of OH instead of H O . Let's take a look at how to find the pH of C H O N , a diprotic base with a concentration of 0.00162 M, and a \(K_{b1}\) of 10 and a \(K_{b2}\) of 10 . Equation: \(C_{20}H_{24}O_2N \; (aq) + H_2O \rightleftharpoons C_{20}H_{24}O_2N_2H^+ + OH^- \) \[K_{b1} = \dfrac{[C_{20}H_{24}O_2N_2H^+,OH^-]}{[C_{20}H_{24}O_2N_2]} = 10^{-6} = 0.011 = \dfrac{(x)(x)}{0.00162 - x} \nonumber \] Again, assume x in the denominator is negligible. Therefore, \(0.011 \approx \dfrac{x^2}{0.00162}\) Then, \(x \approx 4 \times 10^{-5}\) We can then find the pH. \(pOH = -log(4 \times 10^{-5}) = 4.4\) \(pH = 14 - 4.4 = 9.6\) As we determine the pH of the solution, we realize that the OH gained using the second ionization constant is so insignificant that it does not impact the final pH value. For good measure, the following is the process to determine the pH in case the second use of the ICE table would indeed make a difference. Equation: \(C_{20}H_{24}O_2N_2H^+ + H_2O \rightleftharpoons C_{20}H_{24}O_2N_2H_2^{2+} + OH^-\) ICE Table: \(K_{b2} = \dfrac{[C_{20}H_{24}O_2N_2H_2^{2^+},OH^-]}{[C_{20}H_{24}O_2N_2H^+]} = 10^{-9.8}\) \[10^{-9.8} = \dfrac{(4 \times 10^{-5} + x)(x)}{(4 \times 10^{-5} - x)} \nonumber \] \[10^{-9.8} = \dfrac{0.00004 \; x + x^2}{0.00004 - x} \nonumber \] \[10^{-9.8}(0.00004 - x) = 0.00004 x + x^2 \nonumber \] \[x^2 + (4 \times 10^{-5})x - 6.3 \times 10^{-15} = 0 \nonumber \] \[ \begin{align*} x &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\[4pt] &= \dfrac{-4 \times 10^{-5} \pm \sqrt{(4 \times 10^{-5})^2 - 4(1)(6.3 \times 10^{-15})}}{2(1)} \\[4pt] & = 0 \end{align*} \] | 4,657 | 1,977 |
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Write the electron configurations for each of the following elements: The electron configuration of an atom is the representation of the arrangement of electrons distributed among the orbital shells and sub-shells. The electron configuration of each element is unique to its position on the periodic table where the energy level is determined by the period and the number of electrons is given by the atomic number of the element. There are four different types of orbitals (s, p, d, and f) which have different shapes and each orbital can hold a maximum of 2 electrons, but the p, d and f orbitals have different sub-levels, meaning that they are able to hold more electrons. The periodic table is broken up into groups which we can use to determine orbitals and thus, write electron configurations: Group 1 & 2: S orbital Group 13 - 18: P orbital Group 3 - 12: D orbital Lanthanide & Actinides: F orbital Each orbital (s, p, d, f) has a maximum number of electrons it can hold. An easy way to remember the electron maximum of each is to look at the periodic table and count the number of periods in each collection of groups. Group 1 & 2: 2 (2 electrons total = 1 orbital x max of 2 electrons = 2 electrons) Group 13 - 18: 6 (6 electrons total = 3 orbitals x 2 electrons max = 6 electrons) Group 3 - 12: 10 (10 electrons total = 5 orbitals x 2 electrons max = 10 electrons) Lanthanide & Actinides: 14 (14 electrons total = 7 orbitals x 2 electrons max = 14 electrons) Electron fills the orbitals in a specific pattern that affects the order in which the long-hand versions are written: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f An easier and faster way to write electron configurations is to use noble gas configurations as short-cuts. We are able to do this because the electron configurations of the noble gases always have all filled orbitals. He: 1s 2s Ne: 1s 2s 2p Ar: 1s 2s 2p 3s 3p Kr: 1s 2s 2p 3s 3p 4s 3d 4p Xe: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p Rn: 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 4f 5d 6p The most common noble gas configuration used is Ar. When you want to use the noble gas configuration short-cut, you place the noble gas's symbol inside of brackets: [Ar] and then write it preceding the rest of the configuration, which is solely the orbitals the proceed after that of the noble gas. a. Sc Let's start off by identifying where Scandium sits on the periodic table: row 4, group 3. This identification is the critical basis we need to write its electron configuration. By looking at Scandium's atomic number, 21, it gives us both the number of protons and the number of electrons. At the end of writing its electron configuration, the electrons should add up to 21. At row 4, group 3 Sc, is a transition metal; meaning that its electron configuration will include the D orbital. Now, we can begin to assign the 21 electrons of Sc to orbitals. As you assign electrons to their orbitals, you move right across the periodic table. Its first 2 electrons are in the 1s orbital which is denoted as 1s where the "1" preceding the s denotes the fact that it is of row one, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 21-2=19 more electrons to assign. Its next 2 electrons are in the 2s orbital which is denoted as 2s where the "2" preceding the s indicates that it is of row two, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 19-2=17 more electrons to assign. Its next 6 electrons are in the 2p orbital which is denoted as 2p where the "2" preceding the p indicates that it is of row two, and it has an exponent of 6 because it fulfills the p orbital's maximum electron number. Now we have 17-6=11 more electrons to assign. Its next 2 electrons are in the 3s orbital which is denoted as 3s where the "3" preceding the s indicates that it is of row three, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 11-2=9 more electrons to assign. Its next 6 electrons are in the 3p orbital which is denoted as 3p where the "3" preceding the p indicates that it is of row three, and it has an exponent of 6 because it fulfills the p orbital's maximum electron number. Now we have 9-6=3 more electrons to assign. Its next 2 electrons are in the 4s orbital which is denoted as 4s where the "4" preceding the s indicates that it is of row four, and it has an exponent of 2 because it fulfills the s orbital's maximum electron number. Now we have 3-2=1 more electron to assign. Its last electron would be alone in the 3 d orbital which is denoted as 3d where the "3" preceding the d indicates that, even though it is technically of row 4, by disregarding the first row of H and He, this is the third row and it has an exponent of 1 because there is only 1 electron to be placed in the d orbital. Now we have assigned all of the electrons to the appropriate orbitals and sub-orbitals, so that the final, entire electron configuration is written as: 1s 2s 2p 3s 3p 4s 3d This is the long-hand version of its electron configuration. So for Sc, its short-hand version of its electron configuration would therefore be: [Ar] 4s 3d b. Ti Start off by identifying where Titanium sits on the periodic table: row 4, group 4, meaning it has 22 electrons total. Titanium is one element to the right of the previous problem's Sc, so we will basically use the same method except, in the end, there will be 2 electrons remaining, so therefore the final orbital will be denoted as: 3d If needed, look above to the exact steps for how to do it in detail again; the long-hand electron configuration for Titanium will be: 1s 2s 2p 3s 3p 4s 3d So for Ti, its short-hand version of its electron configuration would therefore be: [Ar] 4s 3d c. Cr Start off by identifying where Chromium sits on the periodic table: row 4, group 6, that means it has a total of 24 electrons. But first, Cr, along with Mo, Nb, Ru, Rh, Pd, Cu, Sg, Pt and Au, is a special case. You would think that since it has 24 electrons that its configuration would look like: 1s 2s 2p 3s 3p 4s 3d which is how we learned it earlier. However, this electron configuration is very unstable because of the fact that there are 4 electrons in its 3 d orbital. The most stable configurations are half-filled (d ) and full orbitals (d ), so the elements with electrons resulting in ending with the d or d are so unstable that we write its stable form instead, where an electron from the preceding s orbital will be moved to fill the d orbital, resulting in a stable orbital. If needed, look above to the exact steps for how to do the beginning of the configuration in detail again. However we have to apple the new rule to attain stability so that the long-hand electron configuration for Chromium will be: 1s 2s 2p 3s 3p So for Cr, its short-hand version of its electron configuration would therefore be: [Ar] 4s 3d d. Fe Start off by identifying where Iron sits on the periodic table: row 4, group 8, meaning it has 26 electrons total. This is 5 elements to the right of the previous problem's Sc, so we will basically use the same method except, in the end, there will be 6 electrons remaining, so therefore the final orbital will be denoted as: 3d If needed, look above to the exact steps for how to do it in detail again; the long-hand electron configuration for Iron will be: 1s 2s 2p 3s 3p 4s 3d So for Fe, its short-hand version of its electron configuration would therefore be: [Ar] 4s 3d e. Ru Start off by identifying where Ruthenium sits on the periodic table: row 5, group 8, that means it has a total of 44 electrons. But first, as stated earlier, Ru, along with Cr, Mo, Nb, Rh, Pd, Cu, Sg, Pt and Au, is a special case. You would think that since it has 44 electrons that its configuration would look like: 1s 2s 2p 3s 3p 4s 3d 4p which is how we learned it earlier. However, this electron configuration is very unstable because of the fact that, even though there are 4 paired electrons, there are also 4 electrons unpaired. This results in a very unstable configuration, so to restore stability, we have to use a configuration that has the most paired electrons, which would be to take an electron from the s orbital and place it in the d orbital to create: 5s 4d If needed, look above to the exact steps for how to do the beginning of the configuration in detail again. However we have to apple the new rule to attain stability so that the long-hand electron configuration for Ru will be: 1s 2s 2p 3s 3p 4s 3d 4p So for Cr, its short-hand version of its electron configuration would therefore be: [Kr] 5s 4d Write the electron configurations for each of the following elements and its ions: Electrons are distributed into molecular orbitals, the \(s, p, d, and f\) blocks. An orbital will have a number in front of it and a letter that corresponds to the block. The s block holds two electrons, the p block holds six, the d block holds ten, and the f block holds fourteen. So, based on the number of electrons an atom has, the molecular orbitals are filled up in a certain way. The order of the orbitals is \(1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p\). An exponent will be put after the letter for each orbital to signify how many electrons are in that orbital. Noble gas notation can also be used by putting the noble gas prior to the element you are writing the configuration for, and then proceed by writing the orbitals filled after the noble gas. Metal ions of the d-block will have the two electrons removed from the s block prior to any electrons being removed from the proceeding d-block. Solutions: 1. \(Ti\) Titanium has an atomic number of 22, meaning it has 22 electrons. The noble gas prior to Titanium is Argon. Looking at row 4 of the periodic table, Titanium still has 4 electrons to be placed in orbitals since Argon has 18 electrons that are already placed. The remaining electrons will fill the \(4s\) orbital and the remaining two electrons will go into the \(3d\) orbital. [Ar]4s 3d 2. \(Ti^{+2}\) This is an ion with a plus 2 charge, meaning 2 electrons have been removed. The electrons will be removed from the \(4s\) orbital and the 2 remaining electrons will be placed in the \(3d\) orbital. Like number 1, the prior noble gas is Argon. [Ar]3d 3. \(Ti^{+3}\) This is an ion with a plus 3 charge, meaning 3 electrons have been removed. The first 2 electrons will be removed from the \(4s\) orbital, and the third will be taken from the \(3d\) orbital, and the 1 remaining electron will be placed in the 3d orbital. Like number 1, the prior noble gas is Argon. [Ar]3d 4. \(Ti^{+4}\) This is an ion with a plus 4 charge, meaning 4 electrons have been removed. The first 2 electrons will be removed from the \(4s\) orbital and the second 2 will be removed from the \(3d\) orbital. This results in the ion having the same electron configuration as Argon. [Ar] Answers: Write the electron configurations for each of the following elements and its 3+ ions: In order to write the electron configuration, we begin by finding the element on the periodic table. Since La, Sm, and Lu are all a period below the noble gas Xenon, we can abbreviate \({1s^2}{2s^2}{2p^6}{3s^2}{3p^6}{3d^{10}}{4s^2}{4p^6}{4d^{10}}{5s^2}{5p^6}\) as [Xe] when writing the orbital configurations. We then find the remaining of the orbital configurations using the Aufbau Principle. For other elements not just those in period 6, the shorthand notation using noble gases would be the noble gas in the period above the given element. 1. La has three additional electrons. Two of them fill the 6s shell and the other single electron is placed on the 5d shell. \(La:\) [Xe] \({6{s}^2} {5{d}^1}\) 2. Sm has eight more electrons. The 6s orbital is filled as previously and the 4f orbital receives 6 electrons because pairing electrons requires lower energy on the 4f shell than on the 5d shell. \(Sm:\) [Xe] \({6{s}^2} {4{f}^6}\) 3. Lu has seventeen more electrons. Two electrons fill the 6s orbital, 14 electrons fill the 4f orbital, and extra single electron goes to the 5d orbital . \(Lu:\) [Xe] \({6s^2}{4f^{14}}{5d^1}\) To find the 3+ ion electron configuration, we remove 3 electrons from the neutral configuration, starting with the 6s orbital. 1. The ionization of La removes the three extra electrons. So it reverts back to the stable Xenon configuration. \({La^{3+}:}\) [Xe] 2. The ionization of Sm removes two electrons from the 6s shell and one from the outermost (4f) shell \({Sm^{3+}}:\) [Xe] \({4f^5}\) 3. The ionization of Lu removes its two 6s shell and one from the outermost (5d) shell, leaving only a full 4f shell \(Lu^{3+}:\) [Xe] \(4f^{14}\) La: [Xe]6 5 , La : [Xe]; Sm: [Xe]6 4 , Sm : [Xe]4 ; Lu: [Xe]6 4 5 , Lu : [Xe]4 Why are the lanthanoid elements not found in nature in their elemental forms? Lanthanides are rarely found in their elemental forms because they readily give their electrons to other more electronegative elements, forming compounds instead of staying in a pure elemental form. They have very similar chemical properties with one another, are often found deep within the earth, and difficult to extract. They are the inner transition elements and have partially filled d orbitals that can donate electrons. Because of this, they are very reactive and electropositive. Which of the following elements is most likely to be used to prepare La by the reduction of La O : Al, C, or Fe? Why? An is a list of elements in decreasing order of their reactivity. Elements on the top of the list are good because they easily give up an electron, and elements on the bottom of the series are good because they are highly electronegative would really want to accept an electron. Compare Aluminum, Carbon, and Iron on an activity series. Many activity series include carbon and hydrogen as references. An activity series can be found here The activity series goes in the order (from top to bottom): Aluminum, Carbon, and Iron. Identify which element is the best reducing agent. Elements on the top of the list are the best reducing agents, because they give up electrons the best. Aluminum is the best reducing agent of the options available. Therefore aluminum will be the best reducing agent to prepare La by the reduction of La O because it is the most reactive in the series amongst the three elements. Al is used because it is the strongest reducing agent and the only option listed that can provide sufficient driving force to convert La(III) into La. Which of the following is the strongest oxidizing agent: \(\ce{VO4^{3-}}\), \(\ce{CrO4^2-}\), or \(\ce{MnO4-}\)? Oxidizing agents oxidize other substances. In other words, they gain electrons or become reduced. These agents should be in their highest oxidation state. In order to determine, the strength of the compounds above as oxidizing agents, determine the oxidation numbers of each constituent elements. \(\\\mathrm{VO_4^{3-}}\) We know that \(\mathrm{O}\) has a -2 oxidation state and the overall charge of the ion is -3. We just need to determine Vanadate's oxidation number in this compound. \(\\\mathrm{V} + \mathrm{-2(4)} = \mathrm{-3}\) \(\\\mathrm{V} = \mathrm{+5}\) Vanadate has an oxidation number of +5, which is its highest possible oxidation state. \(\\\mathrm{CrO_4^{2-}}\) Like in the previous calculation, \(\mathrm{O}\) has a -2 oxidation state. The overall charge is -2. So calculate for chromium. \(\\\mathrm{Cr} + \mathrm{-2(4)} = \mathrm{-2}\) \(\\\mathrm{Cr} = \mathrm{+6}\) Chromium is in its highest possible oxidation state of +6 in this compound. \(\\\mathrm{MnO_4^-}\) \(\mathrm{O}\) has a -2 oxidation state and the overall charge is -1. \(\\\mathrm{Mn} + \mathrm{-2(4)} = \mathrm{-1}\) \(\\\mathrm{Mn} = \mathrm{+7}\) Manganese is also in its highest oxidation state, +7. An oxidizing agent has to be able to gain electrons which, in turn, reduces its oxidation state. Here manganese has the greatest oxidation state which allows it to experience a greater decrease in its oxidation state if needed, meaning it can gain the most electrons. So among the three compounds, \(\mathrm{MnO_4^-}\) is the strongest oxidizing agent. This method assumes the metals have similar electronegativities. Alternatively, check a redox table. \(MnO_4^-\) Which of the following elements is most likely to form an oxide with the formula MO : Zr, Nb, or Mo? Mo because Zr has an oxidation state of +4 and Nb has a oxidation state of +5 and those would not balance out the charge of 3 oxygens in the state of -2 which creates a charge of -6. Mo however has multiple oxidation states, the most common being +6 which balances out the -6 charge created by 3 oxygen ions. This is why its most likely to form an oxide with the formula MO or \(\ce{MoO3}\). Mo The following reactions all occur in a blast furnace. Which of these are redox reactions? o identify redox reaction, we have to determine if have to see if the equation is an oxidation-reduction reaction-meaning that the species are changing oxidation states during the reaction, which involves the transfer of electrons between two species. If a species is losing electrons, then that species is being oxidized. If a species is gaining electrons, then that species is being reduced. A way to remember this is using the acronyms OIL RIG. xidation s oss, and eduction s ain, referring to electrons. Both of these must occur for an equation to be a redox reaction. Let's see if these equations are redox reactions or not: a. In the reactants side \(\ce{Fe2O3}\), Fe is has an oxidation number of +3. In the product \(\ce{Fe3O4}\), Fe has an oxidation number of +2.67. Since Fe changed from +3 to +2.67, we can say that Fe had gained electrons and therefore reduced. In the reactant, CO, carbon has an oxidation number of +2, and in \(\ce{CO2}\) (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this . b. In the reactant, \(\ce{Fe3O4}\), Fe has an oxidation number of +2.67. In the product, FeO, Fe has an oxidation number of +2. Since the oxidation of Fe has changed from +2.67 to +2, electrons have been added therefore Fe has been reduced. In the reactant, CO, carbon has an oxidation number of +2, and in \(\ce{CO2}\) (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this . c. In the reactant side, in FeO, Fe has an oxidation number of +2 and in the products side Fe has an oxidation number of 0. Since the oxidation number of Fe changed from +2 to 0, electrons have been gained and therefore Fe has been reduced. In the reactant, CO, carbon has an oxidation number of +2, and in \(\ce{CO2}\) (product) carbon has an oxidation number of +4. Therefore, carbon has lost electrons and it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this . d. In the reactants C has an oxidation number of 0, and in the products side in \(\ce{CO2}\), C has an oxidation number of +4. Since the oxidation number of C has changed from 0 to +4, we can say that C has been oxidized. In the reactants, in \(\ce{O2}\) oxygen has an oxidation number of 0, and in the products CO2, oxygen has an oxidation number of -2. Since the oxidation number of oxygen has changed from 0 to -2, oxygen has been reduced. Since there is oxidation and reduction of species- we can conclude that this . e. In the reactants \(\ce{CO2 }\) has an oxidation number of +4, and in the products side in CO, C has an oxidation number of +2. Since carbon went from +4 to +2, carbon has been reduced. In the reactants, in \(\ce{CO2 }\) oxygen has an oxidation number of -4 and in the products CO carbon has an oxidation number of -2. Since oxygen went from -4 to -2, it has been oxidized. Since there is oxidation and reduction of species- we can conclude that this . f. In the reactants, \(\ce{CaCO3}\) Ca has an oxidation number of +2, and in products CaO Ca has an oxidation number of +2. Since the oxidization number doesn't change- we can conclude that this equation g. In the products CaO Ca has an oxidation number of +2, and in the products \(\ce{CaSiO3}\) Ca has an oxidation number of +2. Since the oxidization number doesn't change- we can conclude that this equation a, b, c, d, e Why is the formation of slag useful during the smelting of iron? Slag is a substance formed as a byproduct of iron ore or iron pellets melting together in a blast furnace. Slag is also the byproduct that is formed when a desired metal has been separated from its raw ore. It is important to note that slag from steel mills is created in a manner that reduces the loss of the desired iron ore. The \(\ce{CaSiO3}\)slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to \(\ce{O2}\), which would oxidize the \(\ce{Fe}\) back to \(\ce{Fe2O3}\). Since Fe has a low reduction potential of -0.440 this means it has a high oxidation potential so it would easily oxidize in the presence of O2. Creating a barrier between iron and oxygen allows the maximum product of iron to be obtained in the end of smelting. The CaSiO slag is less dense than the molten iron, so it can easily be separated. Also, the floating slag layer creates a barrier that prevents the molten iron from exposure to O , which would oxidize the Fe back to Fe O . Would you expect an aqueous manganese(VII) oxide solution to have a pH greater or less than 7.0? Justify your answer. Manganese(VII) oxide, can be written as Mn O In relation to the Lewis acid-base theory, a Lewis acid accepts lone pair electrons, and is also known as the electron pair acceptor. Based on this theory, acidity can be measured by the element's ability to accept electron pairs. By doing the math, we find that Manganese has an oxidation state of +7 (Oxygen has an oxidation state of -2, and 2x-7=-14 , in order to balance the ion's charge, Mn must be +7). Therefore Mn has high capability of accepting electrons due to its high positive charge. For most metals, as the oxidation number increases, so does its acidity, because of its increased ability to accept electrons. In relation to the Lewis acid-base theory, the Lewis acid accepts lone pair electrons; thus, it is also known as the electron pair acceptor. This may be any chemical species. Acids are substances that must be lower than 7. Therefore, oxides of manganese is most likely going to become more acidic in (aq) solutions if the oxidation number increases. Iron(II) can be oxidized to iron(III) by dichromate ion, which is reduced to chromium(III) in acid solution. A 2.5000-g sample of iron ore is dissolved and the iron converted into iron(II). Exactly 19.17 mL of 0.0100 Na Cr O is required in the titration. What percentage of the ore sample was iron? To answer this question, we must first identify the net ionic equation from the given half-reactions. We can write the oxidation and reduction half-reactions: \[ \text{ oxidation:} \text{ Fe}^\text{ 2+} \rightarrow \text{ Fe}^\text{ 3+} \] \[ \text{ reduction: }\ce{Cr2O7^2-} \rightarrow \text{ Cr}^\text{ 3+} \] We can quickly balance the oxidation half-reaction by adding the appropriate number of electrons to get \[ \text{ Fe}^\text{ 2+} \rightarrow \text{ Fe}^\text{ 3+}+\text{ e}^- \] The first step in balancing the reduction half-reaction is to balance elements in the equation other than O and H. In doing so, we get \[\ce{Cr2O7^2-} \rightarrow 2 \text{ Cr}^\text{ 3+} \] The second step would be to add enough water molecules to balance the oxygen. \[ \ce{Cr2O7^2-} \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O} \] Next, we add the correct amount of H to balance the hydrogen atoms. \[ \ce{Cr2O7^2-}+14 \text{ H}^+ \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O} \] Finally, we add enough electrons to balance charge. \[ \ce{Cr2O7^2-}+14 \text{ H}^++6 \text{ e}^- \rightarrow 2 \text{ Cr}^\text{ 3+}+ \ce{7H2O} \] The electrons involved in both half-reactions must be equal in order for us to combine the two to get the net ionic equation. This can be done by multiplying each equation by the appropriate coefficient. Scaling the oxidation half-reaction by 6, we get \[ 6 \text{ Fe}^\text{ 2+} \rightarrow 6 \text{ Fe}^\text{ 3+}+6 \text{ e}^- \] Now we can combine both half-reactions to get \[ 6 \text{ Fe}^\text{ 2+}+\ce{Cr2O7^2-}+14 \text{ H}^++6 \text{ e}^- \rightarrow 6 \text{ Fe}^\text{ 3+}+6 \text{ e}^-+2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}\] The electrons cancel out, so you get: \[ 6 \text{ Fe}^\text{ 2+}+\ce{Cr2O7^2-}+14 \text{ H}^+ \rightarrow 6 \text{ Fe}^\text{ 3+}+2 \text{ Cr}^\text{ 3+}+ \ce{7H2O}\] From this we can see that the mole ratio of Cr O to Fe is 1:6. Given that 19.17 mL (or 0.01917 L) of 0.01 M Na Cr O was needed for titration we know that \[ 0.01917\text{ L} \times 0.01 \text{ M} = 1.917 \times 10^{-4} \text{ mol} \] of Na Cr O reacted. Also, since any number of moles of Na Cr O produces the same number of moles of Cr O in solution \[ 1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Na2Cr2O7} =1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Cr2O7^2-}\] We can use the mole ratio of Cr O to Fe to determine how many moles of iron (ii) was in the solution. The number of moles of iron (ii) is the same as the number of moles of pure iron in the sample since all of the iron was converted into iron (ii). \[ 1.917 \times 10^{-4} \text{ mol} \text{ of } \ce{Cr2O7^2-}\times \frac{6\text{ mol}\text{ of } \text{ Fe}^\text{ 2+}}{1\text{ mol}\text{ of }\ce{Cr2O7^2-}} = 0.0011502\text{ mol}\text{ of } \text{ Fe}^\text{ 2+} \] \[ 0.0011502\text{ mol}\text{ of } \text{ Fe}^\text{ 2+} = 0.0011502\text{ mol}\text{ of } \text{ Fe} \] Now we can find the number of grams of iron that were present in the 2.5 g iron ore sample. \[ 0.0011502\text{ mol}\text{ of } \text{ Fe}\times\frac{55.847\text { g}}{1\text{ mol}} = 0.0642352194\text{ g}\text{ of }\text{ Fe} \] Finally, we can answer the question and find what percentage of the ore sample was iron. \[ \frac{0.0642352194\text{ g}}{2.5\text{ g}} \times 100 \approx 2.57\text{%} \] 2.57% How many cubic feet of air at a pressure of 760 torr and 0 °C is required per ton of Fe O to convert that Fe O into iron in a blast furnace? For this exercise, assume air is 19% oxygen by volume. This question uses a series of unit conversions and the \(PV=nRT\) equation. The first step is to write out the balanced chemical equation for the conversion of Fe O to pure iron. \[2\;Fe_2O_3(s)\rightarrow 4\;Fe(s)+3\;O_2(g)\] Next, we need to analyze the original question to determine the value that we need to solve for. Because the question asks for a value of cubic feet, we know we need to solve for volume. We can manipulate \(PV=nRT\) to solve for volume. \[V={nRT}/P\] Now determine the known variables and convert into units that will be easy to deal with. \[n = 2000\:lbs\; Fe_2O_3\frac{453.592\: grams\: Fe_2O_3}{1\: lb \:Fe_2O_3}\frac{1\: mole \:Fe_2O_3}{159.69\: grams \:Fe_2O_3}\frac{3 \;moles\: O_2}{2\; moles \:Fe_2O_3}\] \[n=8521\: moles\: of \:O_2\] Convert to atm for easier calculations \[R=\frac{.0821\:L\:atm}{mol\:K}\] \[T=0^{\circ}C=273\:K\] \[P=760 \:torr= 1 \:atm\] Now plug the numbers into the manipulated gas law to get to an answer for V. \[V=190991.8\: liters \:of\: O_2\] From here we convert liters to cubic feet. use the conversion \[1\;L=.0353 ft^3\] thus we have 6744.811 ft of O We then refer back to the initial question and remember that this value is only 19% of the volume of the total air. So use a simple equation to determine the total volume of air in cubic feet. \[6744.811ft^3=.19x\] x=35499 ft of air 35499 ft of air Find the potentials of the following electrochemical cell: Cd | Cd ( = 0.10) ‖ Ni ( = 0.50) | Ni Write out your two half reactions and identify which is oxidation and which is reduction using the acronym OIL RIG to remember that oxidation is loss of electrons and reduction is gain of electrons Cd(s)⟶Cd (aq)+2e (oxidation) Ni (aq)+2e ⟶Ni(s) (reduction) Write out complete balanced equation Cd(s)+ Ni (aq)⟶Cd (aq)+Ni(s) Find E E = E -E oxidation: Cd(s)⟶Cd (aq)+2e E =-0.40V reduction: Ni (aq)+2e ⟶Ni(s) E =-0.26V * E values come from standard reduction potentials table given above. Also, remember anode is where oxidation happens, and cathode is where reduction happens. E cell=-0.26-(-.40) E =0.14V Find Q Q=[products]/[reactants] (look at complete balanced equation) (remember that [x] means the concentration of x typically given in molarity and that we ignore solids or liquids) Q=[Cd ]/[Ni ] Q=0.10/0.50 Q=0.2 Calculate E using E= E -(.0592/n)logQ (n is number of moles of electrons transferred and in our case the balanced reaction transfers 2 electrons) E= 0.14-(.0592/2)log(0.2) E= 0.14-(-.207) 0.16 V A 2.5624-g sample of a pure solid alkali metal chloride is dissolved in water and treated with excess silver nitrate. The resulting precipitate, filtered and dried, weighs 3.03707 g. What was the percent by mass of chloride ion in the original compound? What is the identity of the salt? Assuming that metal chloride is XCl The balance equation for the reaction would be: \[XCl(aq)+AgNO_{3}(aq)\rightarrow XNO_{3}(aq) +AgCl(s)\] The mass of AgCl = 3.03707g To find the moles of AgCl present: Next, we can determine the moles of AgCl present in the reaction since 1) the mass of the precipitate is given to us and 2) this value can help us determine the moles of alkali metal chloride compound present. Given the mass of AgCl is 3.03707g in the problem and the molecular mass of AgCl per mole is 143.32g, we can solve for how many moles of AgCl is in the reaction: \[moles of Agcl=\tfrac{3.03707g}{143.32g/mol}=0.0211 mol\] Since the molar ratio of the compounds are 1:1 so the number of moles of XCl used = 0.0211 mol We can calculate the weight of Cl with the equation: \[0.0211 mol \times 35.5g/mol = 0.7490g\] the amount of metal present in the original compound is the weight of the compound subtracted by weight of the Cl ion: \[(2.5624- 0.7490)g= 1.8134g\] And the percentage can be calculate by \[\frac{0.7490}{2.5624}\times 100= 29.23 \%\] the molar ratio of XCl is 1:1 so then Atomic mass of metal = \(=\frac{1.8134\;g\; metal}{0.0211\; mol\; RbCl}=85.943g/mol\) So the atomic mass is 85.943 g/mol which is of Rb hence the identity of the salt is RbCl The standard reduction potential for the reaction \(\ce{[Co(H2O)6]^3+}(aq)+\ce{e-}⟶\ce{[Co(H2O)6]^2+}(aq)\) is about 1.8 V. The reduction potential for the reaction \(\ce{[Co(NH3)6]^3+}(aq)+\ce{e-}⟶\ce{[Co(NH3)6]^2+}(aq)\) is +0.1 V. Calculate the cell potentials to show whether the complex ions, [Co(H O) ] and/or [Co(NH ) ] , can be oxidized to the corresponding cobalt(III) complex by oxygen. To calculate the cell potential, we need to know the potentials for each half reaction. After doing so, we need to determine which one is being oxidized and which one is being reduced. The one that is oxidized is the anode and the one that is reduced is the cathode. To find the cell potential, you use this formula and the reduction potential values found in a reduction potential table. If E° is positive, \(\Delta\)G is negative and the reaction is spontaneous. E° = E°cathode - E°anode Because it states that \([Co(H_{2}O)_{6}]^{3+}\) will be oxidized, this means it is the anode. \(O_{2}(g) + 4 H^{+}(aq) + 4 e^{-} \rightarrow 2 H_{2}O\) +1.229 V \(O_{2}\) is being reduced, so it is the cathode. 1.229V - 1.8V= -.571 V, or -0.6 V using significant figures. This cannot happen spontaneously because E° is negative. For \([Co(NH_{3})_{6}]^{3+}\), it is again being oxidized, meaning it’s the anode. 1.229-.1= 1.129 V or 1.1 V using significant figures. This reaction is spontaneous because E° is positive. ° = −0.6 V, ° is negative so this reduction is not spontaneous. ° = +1.1 V, ° is positive so this reduction is spontaneous. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.) There is a myriad of reactions that can occur, which include: single replacement, double replacement, combustion, acid-base/neutralization, decomposition or synthesis. The first step to determine the products of a reaction is to identify the type of reaction. From then on, the next steps you take to predict the products will vary based on the reaction type. Predict the products of each of the following reactions. (Note: In addition to using the information in this chapter, also use the knowledge you have accumulated at this stage of your study, including information on the prediction of reaction products.) Predict the products of each of the following reactions. Whenever a metal reacts with an acid, the products are salt and hydrogen. Because Fe is lower on the activity series, we know that when it reacts with an acid it will result in the formation of Hydrogen gas. To simplify the equation is: \[Metal + Acid ⟶ Salt + Hydrogen\] The salt produced will depend on the metal and in this case, the metal is iron (Fe) so the resulting equation would be: \[\ce{Fe}(s)+\ce{H2SO4}(aq)⟶ \ce{FeSO4}(aq) + \ce{H2}(g)\] This equation works out as the H2 is removed from H SO , resulting in a SO ion where Fe will take on an oxidation state of Fe to form FeSO which will be the salt in this example. But since FeSO and H2SO are aqueous, the reactants and products can also be written as its ions where the overall equation can be: \[\ce{Fe}(s)+\ce{2H3O+}(aq)+\ce{SO2^{−4}}(aq)⟶\ce{Fe2+}(aq)+\ce{SO2^{−4}}(aq)+\ce{H2}(g)+\ce{2H2O}(l)\] In this case, adding a metal hydroxide (NaOH) to a solution with a transition metal ion (Fe) will form a transition metal hydroxide (XOH). As iron is bonded to three chlorine atoms in the reactants side, it has the oxidation state of +3 where three hydroxide ions (OH ) are needed to balance out the charges when they are bonded in the products. The remaining ions are Na and Cl where they bond together in a 1:1 ratio where there are 3 molecules of NaCl once the reaction is balanced. The overall reaction will be: \[\ce{FeCl3}(aq)+\ce{NaOH}(aq)⟶ \ce{Fe(OH)3}(s) + \ce{3NaCl}(aq)\] NOTE: Fe(OH) (s) is a solid as it is a rule that all all transition metal hydroxides are insoluble and a precipitate is formed. Since NaOH(aq) and NaCl(aq) are aqueous, we can write them out in their ion forms: \[\ce{FeCl3}(aq)+\ce{3Na+}(aq)+\ce{3OH^{−}}(aq)+\ce{Fe(OH)3}(s)+\ce{3Na+}(aq)+\ce{3Cl+}(aq)\] This is an example of a metal hydroxide reacting with an acid where a metal salt and water will always be formed: \[Metal Hydroxide + Acid ⟶ Metal Salt + Water\] When this rule is applied to this equation, we will get the following: \[\ce{Mn(OH)2}(s)+\ce{HBr}(aq)⟶ \ce{MnBr2}(aq)+\ce{2H2O}(l)\] But to follow through with this question, the aqueous solutions such as HBr(aq) and MnBr (aq) can be re-written as: \[\ce{Mn(OH)2}(s)+\ce{2H3O+}(aq)+\ce{2Br-}(aq)⟶\ce{Mn2+}(aq)+\ce{2Br-}(aq)+\ce{4H2O}(l)\] This is the general reaction of a metal reacting with oxygen which will always result in a metal oxide. However, the metal oxide is determined by the oxidation state of the metal so there may be several outcomes of this reaction such as: \[\ce{4Cr}(s)+\ce{3O2}(g)⟶\ce{2Cr2O3}(s)\] \[\ce{Cr}(s)+\ce{O2}(g)⟶\ce{2CrO}(s)\] \[\ce{Cr}(s)+\ce{O2}(g)⟶\ce{CrO2}(s)\] \[\ce{2Cr}(s)+\ce{3O2}(g)⟶\ce{CrO3}(s)\] However, Cr O is the main oxide of chromium so it can be assumed that this is the general product of this reaction. This follows the general reaction of a metal oxide and an acid will always result in a salt and water \[Metal Oxide + Acid ⟶ Salt + Water\] Using this general reaction, similar to the general reactions above, the reaction will result in: \[\ce{Mn2O3}(s)+\ce{HCl}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)\] However, since HCl is an aqueous solution, the overall equation can also be re-written as: \[\ce{Mn2O3}(s)+\ce{6H3O+}(aq)+\ce{6Cl-}(aq)⟶\ce{2MnCl3}(s)+\ce{9H2O}(l)\] Titanium is able to react with the halogens where there are two oxidation state that titanium can be: +3 and +4. The following reactions follow each oxidation state accordingly: \[\ce{2Ti}(s)+\ce{3F2}(g)⟶ \ce{2TiF3}(s)\] \[\ce{Ti}(s)+\ce{2F2}(g)⟶\ce{TiF4}(s)\] However, since there is the symbol "xs", this indicates that F is added in excess so the second reaction is favored more as it drives the reaction to completion. OVERALL: \[\ce{Ti}(s)+\ce{ Describe the electrolytic process for refining copper. By electrolysis, copper can be refined and purely made. The reason why copper needs to remove the impurities is because it helps increase the electrical conductivity in electrical wire. You can refine copper and remove the impurities through electrolysis. Pure copper is important in making electrical wire, because it creates better electrical conductivity when transferring electricity. In order for better electrical conductivity, the impurities needs to be removed and this can be done by firing the impure copper to remove the impurities, such as sulfur, oxygen, etc. and shaping them into electrical anodes that can be used in electrolysis. Then the copper electrodes are placed into an electrical cell (into separate beakers) where electrical current can pass through the beakers and onto the electrodes. Through this process, the copper is stripped off of the anode and deposited onto the cathode. This process helps remove the impurities and refine copper because all the copper has been deposited onto the cathode all in one electrode. This process increases the weight of the cathode due to copper being deposited onto the cathode. This is a prime example of how to tell if an electrode is a cathode or an anode, as stated in Q17.2.9 above. Predict the products of the following reactions and balance the equations. What is the gas produced when iron(II) sulfide is treated with a nonoxidizing acid? Formula for iron(II) sulfide: \[FeS\] Definition of non-oxidizing acid: A non-oxidizing acid is an acid that doesn't act the oxidizing agent. Its anion is a weaker oxidizing agent than H+, thus it can't be reduced. Examples of non-oxidizing acids: \[HCl, HI, HBr, H_3PO_4, H_2SO_4\] Step 2: Choose one of the non-oxidizing acid, in this case HCl, and write the chemical reaction: \[FeS(s)+2HCl(aq)\rightarrow FeCl_2(s)+H_2S(g)\] The gas produced when iron (II) sulfide treated with a non-oxidizing acid, HCl, is H S (dihydrogen sulfide) gas. Predict the products of each of the following reactions and then balance the chemical equations. Steam is water (\(\ce{H_{2}O}\)) We can write out the reaction as: \(\ce{Fe}\) + \(\ce{H_{2}O}\) → ? This is a single replacement reaction, so \(\ce{Fe}\) replaces \(\ce{H_{2}}\). So, one of the products is \(\ce{Fe_{3}O_{4}}\) since it is a combination of iron(II) oxide, \(\ce{FeO}\), and iron(III) oxide, \(\ce{Fe_{2}O_{3}}\). The \(\ce{Fe}\) is heated in an atmosphere of steam. \(\ce{H_{2}}\) becomes neutrally charged and becomes another product. After balancing the coefficients, the final reaction is: \(\ce{3Fe}(s)\) + \(\ce{4H_{2}O}(g)\) → \(\ce{Fe_{3}O_{4}}(s)\) + \(\ce{4H_{2}}(g)\) \(\ce{NaOH}\) added to a solution of \(\ce{Fe(NO_{3})_{3}}\) is a double replacement and precipitation reaction. We can write out the reaction as: \(\ce{NaOH}\) + \(\ce{Fe(NO_{3})_{3}}\) → ? The \(\ce{Na}\) and \(\ce{Fe}\) switch to form \(\ce{Fe(OH)_{3}}(s)\) and \(\ce{NaNO_{3}}(aq)\). \(\ce{Fe(OH)_{3}}\) is solid because it is insoluble according to solubility rules. After balancing the coefficients in the reaction, the final reaction is: \(\ce{Fe(NO_{3})_{3}}(aq)\) + \(\ce{3NaOH}(aq)\) → \(\ce{Fe(OH)_{3}}(s)\) + \(\ce{NaNO_{3}}(aq)\) For instance, the acid used to make the acidic solution is \(\ce{H_{2}SO_{4}}\), then the reaction is: \(\ce{FeSO_{4}}\) + \(\ce{KMnO_{4}}\) + \(\ce{H_{2}SO_{4}}\) → \(\ce{Fe_{2}(SO_{4})_{3}}\) + \(\ce{MnSO_{4}}\) + \(\ce{H_{2}O}\) + \(\ce{K_{2}SO_{4}}\) Next, the net ionic reaction has to be written to get rid of the spectator ions in the reaction, this is written as: \(\ce{Fe^{2+}}\) + \(\ce{MnO_{4}^{-}}\) + \(\ce{H^{+}}\) → \(\ce{Fe^{3+}}\) + \(\ce{Mn^{2+}}\) + \(\ce{H_{2}O}\) As seen in the net ionic equation above, \(\ce{Fe^{2+}}\) is oxidized to \(\ce{Fe^{3+}}\) and \(\ce{MnO_{4}^{-}}\) is reduced to \(\ce{Mn^{2+}}\). These can be written as two half reactions: \(\ce{Fe^{2+}}\) → \(\ce{Fe^{3+}}\) \(\ce{MnO_{4}^{-}}\) → \(\ce{Mn^{2+}}\) To balance the oxidation half reaction, one electron as to be added to the \(\ce{Fe^{3+}}\), this is shown as: \(\ce{Fe^{2+}}\) → \(\ce{Fe^{3+}}\) + \(\ce{e^{-}}\) The reduction half reaction also has to be balanced, but with \(\ce{H^{+}}\) ions and \(\ce{H_{2}O}\), this is shown as: \(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) → \(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\) After the charge of the \(\ce{Mn}\) atoms are balanced, the overall charge has to be balanced on both sides because on the reactants side, the charge is \(\ce{7+}\), and the charge on the products side is \(\ce{2+}\). The overall charge can be balanced by adding electrons, this is shown as: \(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) + \(\ce{5e^{-}}\) → \(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\) Now since both half reactions are balanced, the electrons in both half reactions have to be equal, and then the half reactions are added together. After this is done, the reaction looks like this: \(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) + \(\ce{5Fe^{2+}}\) + \(\ce{5e^{-}}\) → \(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\) + \(\ce{5Fe^{3+}}\) + \(\ce{5e^{-}}\) The \(\ce{5e^{-}}\) on both sides cancel out and the final balanced reaction is: \(\ce{MnO_{4}^{-}}\) + \(\ce{8H^{+}}\) + \(\ce{5Fe^{2+}}\) →\(\ce{Mn^{2+}}\) + \(\ce{4H_{2}O}\) + \(\ce{5Fe^{3+}}\) \(\ce{Fe}\) added to a dilute solution of \(\ce{H_{2}SO_{4}}\) is a single replacement reaction. The \(\ce{Fe}\) is added to a dilute solution so the \(\ce{H_{2}SO_{4}}\) is written as separate ions. We can write out the reaction as: \(\ce{Fe}(s)\) + \(\ce{2H^+}(aq)\) + \(\ce{(SO_{4})^{2-}}(aq)\) → ? The Fe replaces the \(\ce{H^+}\) ion, and becomes an \(\ce{Fe^{2+}}\) ion. \(\ce{H_{2}O}\) is also a product because the solution is dilute. Furthermore, the \(\ce{FeSO_{4}}\) also has to be separated into ions as a result of the \(\ce{Fe}\) being added to a dilute solution. After balancing all of the coefficients, the final reaction is: \(\ce{Fe}(s)\) + \(\ce{(2H_{3}O)^+}(aq)\) + \(\ce{(SO_{4})^{2-}}(aq)\) → \(\ce{Fe^{2+}}(aq)\) + \(\ce{SO_{4}^{2-}}(aq)\) + \(\ce{H_{2}}(g)\) + \(\ce{2H_{2}O}(l)\) \(\ce{H^+}\) can also be written as the the hydronium ion, \(\ce{(H_{3}O)^{+}}\). We initially can initially write out: \(\ce{4Fe(NO_{3})_{2}}\) + \(\ce{4HNO_{3}}\) + \(\ce{O_{2}}\) → ? We write the oxygen term in the reactants because it is stated that the solution is allowed to stand in air. We just have to analyze the possible products that can be formed and we can see that the hydrogen from nitric acid can combine with oxygen gas to form water and then combining everything together, we get the final reaction to be: \(\ce{4Fe(NO_{3})_{2}}(aq)\) + \(\ce{4HNO_{3}}(aq)\) + \(\ce{O_{2}}(g)\) → \(\ce{2H_{2}O}(l)\) + \(\ce{4Fe(NO_{3})_{3}}(aq)\) When \(\ce{FeCO_{3}}\) is added to \(\ce{HClO_{4}}\), a double replacement reaction occurs. The \(\ce{Fe^{2+}}\) ion switches spots with the \(\ce{H^+}\) ion to form \(\ce{Fe(ClO_{4})_{2}}\) as a product. When the \(\ce{H^+}\) ion is added to the \(\ce{(CO_{3})^{2-}}\) ion, \(\ce{H_{2}CO_{3}}\) is formed. After balancing the coefficients, the final reaction is: \(\ce{FeCO_{3}}(s)\) + \(\ce{HClO_{4}}(aq)\) → \(\ce{Fe(ClO_{4})_{2}}(aq)\) + \(\ce{H_{2}O}(l)\) + \(\ce{CO_{2}}(g)\) Air is composed of oxygen gas, which is a diatomic molecule, so it is \(\ce{O_{2}}\). Adding \(\ce{Fe}\) to \(\ce{O_{2}}\) will cause a synthesis reaction to occur forming \(\ce{Fe_{2}O_{3}}\). After balancing coefficients, the final reaction is: \(\ce{3Fe}(s)\) + \(\ce{2O_{2}}(g)\) → \(\ce{Fe_{2}O_{3}}(s)\) Balance the following equations by oxidation-reduction methods; note that elements change oxidation state. \[\ce{Co(NO3)2}(s)⟶\ce{Co2O3}(s)+\ce{NO2}(g)+\ce{O2}(g)\] Balance the following equations by oxidation-reduction methods; note that three elements change oxidation state. \[Co(NO_3){_2(s)}⟶Co_2O{_3(s)}+NO{_2(g)}+O{_2(g)}\] In this reaction, N changes oxidation states from +5 to +4 (reduced), Co changes oxidation states from +2 to +3 (oxidized), and O changes oxidation states from -2 to 0 (also oxidized). First, split this reaction into an oxidation and reduction half reaction set, and balance all of the elements that are not hydrogen or oxygen (we will deal with these later): \[Reduction: 2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2\] Now, for the oxidation reaction, we are only dealing with O on the products side. In order to balance this, we will need to add water and hydrogen to both sides: \[Oxidation: 2H_2O\rightarrow O_2+4H^+ \] Balance the amount of oxygens on each side by adding the correct number of water molecules (H O), and balance the amount of hydrogen by adding the correct number of H atoms: \[Reduction: 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O\] \[Oxidation: 2H_2O\rightarrow O_2+4H^+\] Finally, balance the charges by adding electrons to each side of the equation. For the reduction reaction, we will add 2 electrons to balance out the 2H , and to the oxidation reaction, we will add 4 electrons to balance out the 4H . Remember, the goal of this step is to make sure that the charges are balanced, so we can cancel them out in the end. \[Reduction: 2e^- + 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O\] \[Oxidation: 2H_2O\rightarrow O_2+4H^+ +4e^-\] Multiply the reduction reaction by two, in order to balance the charges so there are 4 electrons on each side of the reaction. \[Reduction: 2(2e^- + 2H^++2Co(NO_3){_2}\rightarrow Co_2O_{3}+4NO_2+H_2O)\] and combine both reactions which comes out to: \[2H_2O + 4Co(NO_3)_2 + 4H^+ \rightarrow 2CO_2O_3 + 8NO_2 + 2H_2O + O_2 + 4H^+\] Cancel out like terms: \[4Co(NO_3){_2(s)} \rightarrow 2CO_2O{_3(s)} + 8NO{_2(g)} + O{_2(g)}\] Both sides have overall charges of 0 and can be checked to see if they are balanced. \[4Co(NO_3){_2(s)} \rightarrow 2CO_2O{_3(s)} + 8NO{_2(g)} + O{_2(g)}\] Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility. Dilute sodium cyanide solution is slowly dripped into a slowly stirred silver nitrate solution. A white precipitate forms temporarily but dissolves as the addition of sodium cyanide continues. Use chemical equations to explain this observation. Silver cyanide is similar to silver chloride in its solubility. A: Step 1: look at the question and begin to write out a general product to reactant formula for this reaction. Step 2: try to reason out why a precipitate will form but only for a finite period of time before reforming in an aqueous substance. Step 3: With step 2 you should have noticed that the reaction is a multiple step reaction and using the rough formula that you derived in step 1, you should try and see what the series of steps are that lead to the overall product of liquid AgCN In this reaction we see how NaCN is added to AgNO .A precipitate forms but then disappears with the addition of even more NaCN, this must mean that its an intermediate reaction which will not appear as the final product. The silver and the cyanide temporarily bond, but the bond is too weak to hold them together so they are pulled apart again when NaCN is added because a new, more stronger and stable compound is formed: [Ag(CN) ] (aq). The actual reaction equation when it is first taking place is \[AgCl(aq)+NaCN(aq)\rightarrow AgCN(s)+NaCl(aq)\] This can be written out in the following way: as CN is added, the silver and the cyanide combine : Ag (aq)+CN (aq)→AgCN (s) As more CN is added the silver and two cyanide combine to create a more stable compound: Ag (aq)+2CN (aq)→[Ag(CN) ] (aq) AgCN(s) + CN (aq) → [Ag(CN) ] (aq) As CN is added, \[\ce{Ag+}(aq)+\ce{CN-}(aq)⟶\ce{AgCN}(s)\] As more CN is added, \[\ce{Ag+}(aq)+\ce{2CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)\] \[\ce{AgCN}(s)+\ce{CN-}(aq)⟶\ce{[Ag(CN)2]-}(aq)\] Predict which will be more stable, [CrO ] or [WO ] , and explain. According to the rules associated with Crystal Field Stabilizing Energies, stable molecules contain more electrons in the lower-energy molecular orbitals than in the high-energy molecular orbitals. In this case, both complexes have O as ligands, and both have a -2 charge. Therefore, you determine stability by comparing the metals. Chromium is in the 3d orbital, according to the periodic table. Tungsten (W) is in the 5d orbital. 3d is a lower energy level than 5d.Higher-level orbitals are more easily ionized, and make their base elemental form more stable. If the elemental form is more stable the oxidized form is less stable. Therefore, [CrO ] is more stable than [WO ] . [CrO ] is more stable because Chromium is in the 3d orbital while Tungsten is in the 4d orbital, which has a higher energy level and makes it less stable. Give the oxidation state of the metal for each of the following oxides of the first transition series. (Hint: Oxides of formula M O are examples of in which the metal ion is present in more than one oxidation state. It is possible to write these compound formulas in the equivalent format MO·M O , to permit estimation of the metal’s two oxidation states.) The first step to solving this problem is looking at the rules of Oxidizing states for various elements: chem.libretexts.org/Core/Analytical_Chemistry/Electrochemistry/Redox_Chemistry/Oxidation_State The main rules that will be used in these problems will be the oxidation state rule 6 which states that oxidation state for Oxygen is (-2) and rule 2 which is that the total sum of the oxidation state of all atoms in any given species is equal to the net charge on that species. Solving these problems requires simple algebra. The oxidation states of both elements in the compound is equal to zero, so set the unknown oxidation of the element that is not oxygen to a variable \({x}\), and the oxidation state of Oxygen equal to \({-2}\). Then multiply both oxygen states by the number of atoms of the element present. Add the values together, set the equation equal to zero and solve for \({x}\). \(\ce{FeO}={-2}+{x}={0}⟶{x}={Fe}={+2}\) \(Fe^{2+}\) \(\ce{Fe2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2x}={0}⟶{x}={Fe}={+3}\) \(Fe^{3=}\) (One Fe Atom has an oxidation state of +2 and the other 2 Fe atoms have an oxidation state of +3) 7. \(\ce{Co3O4}=\ce{CoO}·\ce{Co2O3}=\) \(\ce{CoO}={-2}+{x}={0}⟶{x}={Co}={+2}\) \(Co^{2+}\) \(\ce{Co2O3}={3{(-2)}}+{2{x}}={0}⟶{-6}+{2x}={0}⟶{x}={Co}={+3}\) \(Co^{3+}\) (One Co Atom has an oxidation state of +2 and the other 2 Co atoms have an oxidation state of +3) 8. \(\ce{NiO}={-2}+{x}={0}⟶{x}={Ni}={+2}\) \(Ni^{2+}\) 9. \(\ce{Cu2O}={-2}+{2{x}}={0}⟶{-2}+{2x}={0}⟶{x}={Cu}={+1}\) \(Cu^{1+}\) Sc ; Ti ; V ; Cr ; Mn ; Fe and Fe ; Co and Co ; Ni ; Cu Indicate the coordination number for the central metal atom in each of the following coordination compounds: First we must identify whether or not the ligand has more than one bonded atom (bidentate/polydentate). Using the table below we are able to do this. Now that we have identified the number of bonded atoms from each ligand, we can find the total number of atoms bonded to the central metal ion, giving us the coordination number. Give the coordination numbers and write the formulas for each of the following, including all isomers where appropriate: To determine coordination numbers we must count the total number of ligands bonded to the central metal and distinguish monodentate and polydentate ligands. To determine the formulas, we use the nomenclature rules and work backwards. Give the coordination number for each metal ion in the following compounds: You can determine a compound's coordination number based on how many ligands are bound to the central atom. 1) In this compound, Cobalt is the central atom, and it has 3 CO molecules attached to it. However, CO is a bidentate ligand, which means it binds to the central atom in two places rather than one. This means that the coordination number of [Co(CO ) ] is 6. A coordination number of 6 means that the structure is most likely octahedral. 2) In this compound, Copper is the central atom. 4 ammonia molecules are attached to it. This means the coordination number is 4, and the structure is likely tetrahedral. 3) For this compound, we can ignore the (SO ) because it is not bound to the central atom. The central atom is cobalt, and it has 4 ammonia molecules and 2 bromine molecules bound to it. The coordination number is 6. 4) There are two compounds here, indicated by the brackets. The central atom for both is platinum. One of them has 4 ammonia molecules attached, and the other has 4 chlorine atoms attached. Both complexes have a coordination number of 4. 5) We can ignore (NO ) for this compound. The central atom is Chromium. There are 3 ethylenediamine molecules attached to the chromium. Ethylenediamine is a bidentate ligand, so the coordination number is 6. 6) Palladium is the central atom. 2 ammonia molecules and 2 bromine atoms are bound to the palladium atom. The coordination number is 4. 7) We can ignore the K structure. Copper is the central atom, and there are 5 chlorine molecules attached to it. The coordination number is 5, so the structure is either trigonal bipyramidal or square pyramidal. 8) In this compound, zinc is the central atom. There are 2 ammonia molecules and 2 chlorine atoms attached. This means that the coordination number is 4. Sketch the structures of the following complexes. Indicate any , , and optical isomers. Cis and trans are a type of geometric isomer, meaning there is a difference in the orientation in which the ligands are attached to the central metal. In cis, two of the same ligands are adjacent to one another and in trans, two of the same ligands are directly across from one another. Optical isomers → have the ability to rotate light, optical isomers are also chiral. Only chiral complexes have optical isomers Chiral → asymmetric, structure of its mirror image is not superimposable Enantiomers: chiral optical isomers (compound can have multiple enantiomers) Tetrahedral complex with 4 distinct ligands → always chiral Solutions: a. \([Pt(H_2O)_2Br_2]\) (square planar) This complex has 2 kinds of ligands. The matching ligands can either be adjacent to each other and be cis, or they can be across from each other and be trans. b. \([Pt(NH_3)(py)(Cl)(Br)]\) (square planar, py = pyridine, \(C_5H_5N\)) This complex has 4 different ligands. There is no plane of symmetry in any of the enantiomers, making the structures chiral and therefore has optical isomers. c. \([Zn(NH_3)_3Cl]^+\) (tetrahedral) There is a plane of symmetry from \(NH_3\) through \(Zn\) to the other \(NH_3\), therefore it is not chiral. d. \([Pt(NH_3)_3Cl]^+\) (square planar) There is a plane of symmetry from \(NH_3\) through \(Pt\) to the other \(NH_3\), therefore it is not chiral. e. \([Ni(H_2O)_2Cl_2]\) The \(Cl\) ligands can either be right next to each other, or directly across from one another allowing for both cis and trans geometries. f. \([Co(C_2O_4)_2Cl_2]^-3\) (note that \(C_2O_4^-2\) is the bidentate oxalate ion, \(^−O_2CCO_2^-\) There is a plane of symmetry from \(Cl\) through Co to the other \(Cl\) in a "trans" chlorine configuration, therefore it is not chiral in a chlorine "trans" configuration. However, there is no symmetry in the chlorine "cis" configuration, indicating multiple "cis" isomers. a. [Pt(H O) Br ]: b. [Pt(NH )(py)(Cl)(Br)]: c. [Zn(NH ) Cl] : d. [Pt(NH ) Cl] : e. [Ni(H O) Cl ]: f. [Co(C O ) Cl ] : Draw diagrams for any , , and optical isomers that could exist for the following (en is ethylenediamine): We are instructed to draw all geometric isomers and optical isomers for the specified compound. Optical isomers exist when an isomer configuration is not superimposable on its mirror image. This means there are two distinct molecular shapes. Often a left and right hand are cited as an example; if you were to take your right hand and place it upon your left, you cannot make the major parts of your hand align on top of one another. The basic idea when deciding whether something is optically active is to look for a plane of symmetry--if you are able to bisect a compound in a manner that establishes symmetry, then the compound does not have an optical isomer. Cis isomers exist when there are 2 ligands of the same species placed at 90 degree angles from each other. Trans isomers exist when there are 2 ligands of the same species placed at 180 degree angles from each other. Problem 1 This compound is an octahedral molecule, so the six ligands (atoms in the complex that are not the central transition metal) are placed around the central atom at 90 degree angles. Two optical isomers exist for [Co(en) (NO )Cl] . The second isomer is drawn by taking the mirror image of the first. Problem 2 This compound is also an octahedral molecule. Two cis (optical) isomers and one trans isomer exist for [Co(en) Cl ] . The trans isomer can be drawn by placing the chlorine ligands in positions where they form a 180 degree angle with the central atom. The first cis isomer can be drawn by placing the chlorine ligands in positions where they form a 90 degree angle with the central atom. The second cis isomer can be found by mirroring the first cis isomer, like we did in problem 1. Problem 3 This compound is also an octahedral molecule. One trans isomer and one cis isomer of [Pt(NH ) Cl ] exist. The trans isomer can be drawn by placing the ammonia ligands in positions where they form a 180 degree angle with the central atom. The cis isomer can be drawn by placing the ammonia ligands in positions where they form a 90 degree angle with the central atom. Problem 4 This compound is also an octahedral molecule. Two optical isomers for [Cr(en) ] exist. The second optical isomer can be drawn by taking the mirror image of the first optical isomer. Problem 5 This compound is a square planar complex, so the ligands are placed around the central atom in a plane, at 90 angles. A trans isomer and a cis isomer exist for the complex [Pt(NH ) Cl ]. The trans isomer can be drawn by placing the ammonia ligands in positions where they form a 180 degree angle in the plane with the central atom. The cis isomer can be drawn by placing the ammonia ligands in positions where they form a 90 degree angle in the plane with the central atom. Name each of the compounds or ions given in Exercise Q19.2.3, including the oxidation state of the metal. Rules to follow for coordination complexes 1. Cations are always named before the anions. 2. Ligands are named before the metal atom or ion. 3. Ligand names are modified with an ‐o added to the root name of an anion. For neutral ligands the name of the molecule is used, with the exception of OH2, NH3, CO and NO. 4. The prefixes mono‐, di‐, tri‐, tetra‐, penta‐, and hexa‐ are used to denote the number of simple ligands. 5. The prefixes bis‐, tris‐, tetrakis‐, etc., are used for more complicated ligands or ones that already contain di‐, tri‐, etc. 6. The oxidation state of the central metal ion is designated by a Roman numeral in parentheses. 7. When more than one type of ligand is present, they are named alphabetically. Prefixes do not affect the order. 8. If the complex ion has a negative charge, the suffix –ate is added to the name of the metal. 9. In the case of complex‐ion isomerism the names cis, trans, fac, or mer may precede the formula of the complex‐ion name to indicate the spatial arrangement of the ligands. Cis means the ligands occupy adjacent coordination positions, and trans means opposite positions just as they do for organic compounds. The complexity of octahedral complexes allows for two additional geometric isomers that are peculiar to coordination complexes. Fac means facial, or that the three like ligands occupy the vertices of one face of the octahedron. Mer means meridional, or that the three like ligands occupy the vertices of a triangle one side of which includes the central metal atom or ion. Name each of the compounds or ions given in Exercise Q19.2.5. S19.2.7 Names of the above compounds. 1. [Co(en) (NO )Cl] Attain the names of the ligands and metal cation. Names can be found here. Co: Cobalt en: Ethylenediamine NO : Nitro Cl: Chloro Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here. (en) : bis(Ethylenediamine) Find the charges of the ligands. Charges can be found here. en: 0 NO : -1 Cl: -1 Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges. Co + 2(en) + (NO ) + Cl = 1 Co +2(0) + (-1) + (-1) = 1 Co = 3 For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last. 2. [Co(en) Cl ] Attain the names of the ligands and metal cation. Names can be found here. Co: Cobalt en: Ethylenediamine Cl: Chloro Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here. (en) : bis(Ethylenediamine) Cl : dichloro Find the charges of the ligands. Charges can be found here. en: 0 Cl: -1 Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges. Co + 2(en) +2(Cl) = 1 Co + 2(0) + 2(-1) = 1 Co = 3 For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last. 3. [Pt(NH ) Cl ] Attain the names of the ligands and metal cation. Names can be found here. Pt: Platinum NH : Ammine Cl: Chloro Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here. (NH ) : diammine Cl : tetrachloro Find the charges of the ligands. Charges can be found here. NH : 0 Cl: -1 Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges. Pt + 2(NH ) + 4(Cl) = 0 Pt + 2(0) + 4(-1) = 0 Pt = 4 For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last. 4. [Cr(en) ] Attain the names of the ligands and metal cation. Names can be found here. Cr: Cromium en: ethylenediamine Add the appropriate pre-fixes to each ligand depending on the number. Pre-fixes can be found here. (en) : tris(ethylenediamine) Find the charges of the ligands. Charges can be found here. en: 0 Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges. Cr + 3(en) = 3 Cr + 3(0) = 3 Cr = 3 For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last. 5. [Pt(NH ) Cl ] Attain the names of the ligands and metal cation. Names can be found here. NH : Ammine Cl: Chloro Pt: Platinum Add the appropriate pre-fixes to each ligand depending on the number. (NH ) : diammine Cl : dichloro Find the charges of the ligands. Charges can be found here. NH : 0 Cl: -1 Algebraically attain the charge of the metal cation using the overall charge of the complex ion and the individual ligand charges. Pt + 2(NH ) + 2(Cl) = 0 Pt + 2(0) + 2(-1) = 0 Pt = 2 For the name alphabetically place the ligands, pre-fixes should not be accounted, and use roman numerals for the metal cation which should be placed last. Specify whether the following complexes have isomers. Isomers are compounds that have the same number of atoms, but have different structures. Structural isomers (linkage, ionization, coordination) and stereoisomers (geometric and optical) can occur with several compounds. 1. tetrahedral \(\mathrm{[Ni(CO)_2(Cl)_2]}\) (Fig 1.) Immediately, we can cancel out the possibility of linkage, ionization, and coordination isomers. There are no other coordination complex for coordination isomerism, there is no ligand that can bond to the atom in more than one way for it to exhibit linkage isomerism, and there are no ions outside the coordination sphere for ionization isomerism. This is a tetrahedral structure which immediately rules out any geometric isomers since they require 90° and/or 180° bond angles. Tetrahedral structures have 109.5° angles. To confirm that the structure has no optical isomer, we must determine if there is a plane of symmetry. Structures that have no plane of symmetry are considered chiral and would have optical isomers. (Fig 2.) Since there is a plane of symmetry, we can conclude that there are no optical isomers. Overall, there are no isomers that exist for this compound. 2. trigonal byprimidal \(\mathrm{[Mn(CO)_4(NO)]}\) (Fig 3.) There are no ions, other coordination complex, and ambidentate ligands. Therefore, no structural isomers exist for this structure. Geometric isomers do not exist for this compound because there is only one nitrosyl ligand. (Fig 4.) In the image above, after the structure has been rotated, we can see that there is a plane of symmetry. Thus, there are no optical isomers. No isomers (the ones mentioned above) exist for this compound. 3. \(\mathrm{[Pt(en)_{2}Cl_2]Cl_2}\) (Fig 5.) Coordination isomerism cannot exist for this complex because there are no other complexes. There are no linkage isomers because there are no ambidentate ligands. Ionization isomers cannot exist in this complex either, even though there is a neutral molecule outside the coordination sphere. If we exchange \(\mathrm{Cl_2}\) with one ethyldiamine molecule, There would be 5 ligands in the coordination sphere instead of 4. This difference in the ratio of metal atom to ligands means that an ionization isomer cannot exist. (Fig 6.) The image above, shows the chloro and ethyldiamine ligands at a 90° angle with its other identical ligand. This is the isomer, while Fig. 5 shows the isomer. Fig 5. shows that there is a plane of symmetry in the isomer. Therefore, that structure does not have an optical isomer. On the other hand, the isomer does not have a plane of symmetry and therefore has an optical isomer. none; none; The two Cl ligands can be or . When they are , there will also be an optical isomer. Predict whether the carbonate ligand \(\ce{CO3^2-}\) will coordinate to a metal center as a monodentate, bidentate, or tridentate ligand. \(\ce{CO3^2−}\) can be either monodentate or bidentate, since two of its oxygen atoms have lone pairs as shown above and can form covalent bonds with a transition metal ion. In most cases carbonate is monodentate because of its trigonal planar geometry (there is 120 degrees between the oxygens so it's hard for both to bind to the same metal). However, in some cases it will bind to two different metals, making it bidentate. CO will coordinate to a metal center as a monodentate ligand. Draw the geometric, linkage, and ionization isomers for [CoCl CN,CN]. Isomers are compounds with same formula but different atom arrangement. There are two subcategories: , which are isomers that contain the same number of atoms of each kind but differ in which atoms are bonded to one another, and , isomers that have the same molecular formula and ligands, but differ in the arrangement of those ligands in 3D space. There are three subcategories under structural isomers: , which are isomers that are identical except for a ligand has exchanging places with an anion or neutral molecule that was originally outside the coordination complex; , isomers that have an interchange of some ligands from the cationic part to the anionic part; and , in two or more coordination compounds in which the donor atom of at least one of the ligands is different. There are also two main kinds of stereoisomers: , metal complexes that differ only in which ligands are adjacent to one another (cis) or directly across from one another (trans) in the coordination sphere of the metal, and , which occurs when the mirror image of an object is non-superimposable on the original object.
Some of the isomers look almost identical, but that is because the CN ligand can be attached by both (but not at the same time) the C or N. Determine the number of unpaired electrons expected for [Fe(NO ) ] and for [FeF ] in terms of crystal field theory. Step 1: Determine the oxidation state of the Fe For \([Fe(NO_{2}){_{6}}]^{3-}\) and \([FeF_{6}]^{3-}\), both \(NO_{2}\) and \(F_{6}\) have a charge of -1. Since there is 6 of them then that means the charge is -6 and in order for there to be an overall charge of -3, Fe has to have a +3 charge. Step 2: Determine type of ligand Based on the spectrochemical series we can see that \(NO_{2}^{-}\) is a stronger field ligand than F\(^{-}\), and therefore is a low spin complex because it has a high \(\Delta_{\circ}\) unlike F\(^{-}\) which is a high spin. Step 3: Draw the crystal field \([Fe(NO_{2}){_{6}}]^{3-}\) \([FeF_{6}]^{3-}\) There is 1 unpaired electron for \([Fe(NO_{2}){_{6}}]^{3-}\), and 5 for \([FeF_{6}]^{3-}\) based on the crystal field theory. [Fe(NO ) ] :1 electron [FeF ] :5 electrons Draw the crystal field diagrams for [Fe(NO ) ] and [FeF ] . State whether each complex is high spin or low spin, paramagnetic or diamagnetic, and compare Δ to P for each complex. a) \[[Fe(NO_2)_6]^{_{-4}}\] b) \[[FeF_6]^{_{-3}}\] Give the oxidation state of the metal, number of electrons, and the number of unpaired electrons predicted for [Co(NH ) ]Cl . The oxidation state of the metal can be found by identifying the charge of one of each molecule in the coordinate compound, multiplying each molecule's charge by the respective number of molecules present, and adding the products. This final sum represents the charge of the overall coordination compound. You can then solve for the oxidation state of the metal algebraically. In this case, one chloride anion Cl has a charge of -1. So three chloride anions have a total charge of -3. One ammine ligand NH has no charge so six ammine ligands have a total charge of zero. Finally, we are trying to solve for the oxidation state of a cobalt ion. Now we can write the equation that adds the total charges of each molecule or ion and is equal to the total charge of the overall coordinate compound. \[ (\text{oxidation}\text{ state}\text{ of}\text{ Co})+(-3)+0=0 \] \[ \text{oxidation}\text{ state}\text{ of}\text{ Co}=+3 \] . Now we need to identify the number of d-electrons in the Co ion. The electron configuration for cobalt that has no charge is \[ [\text{Ar}]4\text{s}^23\text{d}^7 \] However, a Co ion has 3 less electrons than its neutral counterpart and has an electron configuration of \[ [\text{Ar}]3\text{d}^6 \] For transition metals, the \( \text{s} \) electrons are lost first. So cobalt loses its two \( 4\text{s} \) electrons first and then loses a single \( 3\text{d} \) electron meaning To predict the number of unpaired electrons, we must first determine if the complex is high spin or low spin. Whether the complex is high spin or low spin is determined by the ligand in the coordinate complex. Specifically, the ligand must be identified as either a weak-field ligand or a strong-field ligand based on the spectrochemical series. Weak-field ligands induce high spin while strong-field ligands induce low spin. We can then construct the energy diagram or crystal field diagram of the designated spin that has the proper electron placings. The geometric shape of the compound must also be identified to construct the correct diagram. Finally, from this crystal field diagram we can determine the number of unpaired electrons. The number of electrons in the diagram is equal to the number of \( \text{d} \) electrons of the metal. The ligand in this case is NH , which is a strong field ligand according to the spectrochemical series. This means that the complex is low spin. Additionally, six monodentate ligands means the ligand field is octahedral. The number of electrons that will be in the diagram is 6 since the metal ion Co has 6 \( \text{d} \) electrons. Now the proper crystal field diagram can be constructed. From the crystal field diagram, we can tell that . a) 3+ b) 6 d electrons c) No unpaired electrons The solid anhydrous solid CoCl is blue in color. Because it readily absorbs water from the air, it is used as a humidity indicator to monitor if equipment (such as a cell phone) has been exposed to excessive levels of moisture. Predict what product is formed by this reaction, and how many unpaired electrons this complex will have. From our knowledge of ligands and coordination compounds (or if you need a refresher Coordination Compounds), we can assume the product of CoCl in water. H O is a common weak field ligand that forms six ligand bonds around the central Cobalt atom while the Chloride stays on the outer sphere. We can use this to determine the complex: \([Co(H_2O)_6]Cl_2\) From this formation, we can use the Crystal Field Theory (CFT)(Crystal Field Theory) to determine the number of unpaired electrons. This coordination compound has six ligand bonds attached to the central atom which means the CFT model will follow the octahedral splitting. Keep in mind that we know H O is a weak field ligand and will produce a high spin. High spin is when the electrons pairing energy (P) is greater than the octahedral splitting energy. Thus, the electrons spread out and maximize spin. In order to fill out our crystal field diagram, we need to determine the charge of cobalt. Because the H O ligand is neutral, and there are two chlorine ions, we can deduce the charge of cobalt is plus two in order to make the coordination complex neutral. From here, we can use the electron configuration of Co [Ar]4s 3d . The electrons that are taken away from the cobalt atom in order to form the plus two charge will from the 4s orbital and leave the 3d orbital untouched. Thus, there will be 7 electrons in the crystal field diagram and appear as: We can see here that there are 3 unpaired electrons. [Co(H O) ]Cl with three unpaired electrons. Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain. Is it possible for a complex of a metal in the transition series to have six unpaired electrons? Explain. It is not possible for a metal in the transition series to have six unpaired electrons. This is because transition metals have a general electron configuration of (n-1)d ns where n is the quantum number. The last electron will go into the d orbital which has 5 orbitals that can each contain 2 electrons, yielding 10 electrons total. According to Hund's Rule, electrons prefer to fill each orbital singly before they pair up. This is more energetically favorable. Since there are only 5 orbitals and due to Hund's Rule, the maximum number of unpaired electrons a transition metal can have is 5. Therefore, there cannot be a complex of a transition metal that has 6 unpaired electrons. For example, lets look at iron's electron configuration. Iron has an electron configuration of 1s 2s 2p 3s 3p 4s 3d . Now the most important orbital to look at is the d orbital which has 6 electrons in it, but there are only 4 unpaired electrons as you can see by this diagram: 3d: [↿⇂] [↿] [↿,↿,↿] Each [ ] represents an orbital within the d orbital. This diagram follows Hund's rule and shows why no transition metal can have 6 unpaired electrons. How many unpaired electrons are present in each of the following? 1. For [CoF ] , we first found the oxidation state of Co, which is 3+ since F has a 1- charge and since there is 6 F, Co's charge has to be 3+ for the overall charge to be 3-. \[\textrm{charge of Co} + \textrm{-6} = \textrm{-6}\] \[\textrm{charge of Co} = \textrm{+3}\] After finding the oxidation state, I then go to the periodic table to find its electron configuration: [Ar]3d We distribute the 6 d-orbital electrons along the complex and since it is high spin, the electrons is distributed once in each energy level before it is paired. There is only one pair and the other 4 electrons are unpaired, making the answer 4. 2. The same process is repeated. We find the charge of Mn, which is 3+, making the electron configuration: [Ar]3d \[\textrm{charge of Mn} + \textrm{-6} = \textrm{-3}\] \[\textrm{charge of Mn} = \textrm{+3}\] There is a difference between this and number 1. This is low spin so instead of distributing one electron in each level before pairing it, I must distribute one electron on the bottom and then pair them all up before I'm able to move to the top portion. So since there is 4, there is only a pair at d and the other two electrons are unpaired. Making the number of unpaired electrons 2. 3.The same process as number 2 is applied. The only difference is that the charge of Mn is now 2+ so the electron configuration: [Ar]3d . \[\textrm{charge of Mn} + \textrm{-6} = \textrm{-4}\] \[\textrm{charge of Mn} = \textrm{+2}\] Since is it low spin like number 2, I only need to add an extra electron to the next level, making that 2 pairs of electron and only 1 electron unpaired. 4. Since Cl has a -1 charge like CN, Mn's charge is also 2+ with the same electron configuration as number 3, which is 5. \[\textrm{charge of Mn} + \textrm{-6} = \textrm{-4}\] \[\textrm{charge of Mn} = \textrm{+2}\] With 5 electrons, this is high spin instead of low. So as stated in number 1, we pair distribute the electrons on all levels first. Since there are 5 electrons and 5 levels and they are al distributed, there are zero pairs, making that 5 unpaired electrons. 5. Using the same process as the problems above, Rh's charge is 3+, with the electron configuration: [Kr]4d . \[\textrm{charge of Rh} + \textrm{-6} = \textrm{-3}\] \[\textrm{charge of Mn} = \textrm{+3}\] With a low spin and 6 electrons, all electrons are paired up, making it 0 electrons that are unpaired. 4; 2; 1; 5; 0 Explain how the diphosphate ion, [O P−O−PO ] , can function as a water softener that prevents the precipitation of Fe as an insoluble iron salt. The diphosphate ion, [O P−O−PO ] can function as a water softener keeping the iron in a water soluble form because of its more negative electrochemical potential than water's. This is similar to the way plating prevents metals from reacting with oxygen to corrode. Mineral deposits are formed by ionic reactions. The Fe will form an insoluble iron salt of iron(III) oxide-hydroxide when a salt of ferric iron hydrolyzes water. However, with the addition of [O P−O−PO ] , the Fe cations are more attracted to the PO group, forming a Fe(PO ) complex. The excess minerals in this type of water is considered hard thus its name hard water. For complexes of the same metal ion with no change in oxidation number, the stability increases as the number of electrons in the orbitals increases. Which complex in each of the following pairs of complexes is more stable? The Spectrochemical Series is as follows \[I^{-}<Br^{-}<SCN^{-}\approx Cl^{-}<F^{-}<OH^{-}<ONO^{-}<ox<H_{2}O<SCN^{-}<EDTA<NH_{3}<en<NO_{2}^{-}<CN\] The strong field ligands (on the right) are low spin which fills in more electrons in the orbitals. The weak field ligands (on the left) are high spin so it can fill electrons in the orbitals and orbitals. In conclusion, more electrons are filled up from the strong field ligands because the electrons don't move up to the orbitals. a. \([Fe(CN)_{6}]^{4-}\) \(CN\) is a stronger ligand than \(H_{2}O\) so it is low spin, which fills up the orbitals. b. \([Co(NH_{3})_{6}]^{3+}\) \(NH_{3}\) is a stronger ligand than \(F\). c. \([Mn(CN)_{6}]^{4-}\) \(CN\) is a stronger ligand than \(Cl^{-}\). For more information regarding the shape of the complex and d-electron configuration, libretext provides more information on how to classify high and low spin complexes. [Fe(CN) ] ; [Co(NH ) ] ; [Mn(CN) ] Trimethylphosphine, P(CH ) , can act as a ligand by donating the lone pair of electrons on the phosphorus atom. If trimethylphosphine is added to a solution of nickel(II) chloride in acetone, a blue compound that has a molecular mass of approximately 270 g and contains 21.5% Ni, 26.0% Cl, and 52.5% P(CH ) can be isolated. This blue compound does not have any isomeric forms. What are the geometry and molecular formula of the blue compound? a) b) Tetrahedral Would you expect the complex [Co(en) ]Cl to have any unpaired electrons? Any isomers? Assign oxidation states to each element. Cl- has a -1 oxidation state. En is neutral, so 0. The entire complex is also neutral, so in order to balance the charges out, Co must be +3 because there are 3 chlorides, which gives a -3 charge. Write the electron configuration for \(Co^{3+}\). \([Ar]3d^6\). There are 6 electrons. Check where en lies on the spectrochemical series. Does it have a strong field strength? It does, so these electrons will exist at the d-level with high splitting energy because the magnitude of the pairing energy is less than the crystal field splitting energy in the octahedral field. You will the notice that there aren't any unpaired electrons when you draw the Crystal Field Theory (CFT) diagram. This complex does not have any geometric isomers because cis-trans structures cannot be formed. The mirror image is nonsuperimpoasable, which means the enantiomers are chiral molecules; if the mirror image is placed on top on the original molecule, then they will never be perfectly aligned to give the same molecule. The complex does not have any unpaired electrons. The complex does not have any geometric isomers, but the mirror image is nonsuperimposable, so it has an optical isomer. Would you expect the Mg [Cr(CN) ] to be diamagnetic or paramagnetic? Explain your reasoning. The first step to determine the magnetism of the complex is to calculate the oxidation state of the transition metal. In this case, the transition metal is Cr. Before doing so, we need to find charge of the of the complex ion [Cr(CN) ] given that the oxidation state of Mg [Cr(CN) ] [Cr(CN) ] Mg [Cr(CN) ] A19.3.11 a) Paramagnetic Would you expect salts of the gold ion, Au , to be colored? Explain. No. . A partially filled d orbital, for example, can yield various colors. After completing the noble gas configuration, we see that Au has a configuration of [Xe] 4f . Since Au has a , we are certain that any salts of the gold ion, Au will be *An example of a colored ion would be copper(II). Cu has an electron configuration of [Ar]3d . It has one unpaired electron. Copper(II) appears blue. No. Au has a complete 5 sublevel. [CuCl ] is green. [Cu(H O) ] is blue. Which absorbs higher-energy photons? Which is predicted to have a larger crystal field splitting? Although a color might appear a certain way, it actual absorbs a different color, opposite of it on the color wheel. In this case; [CuCl ] appears green but is opposite of red on the color wheel which is absorbed and is characterized by wavelengths 620-800 nanometers. [Cu(H O) ] appears blue but is opposite of orange on the color wheel which is absorbed and is characterized by wavelengths 580-620 nanometers. When determining which absorbs the higher energy photons, one must look at the complex itself. A higher energy indicates a high energy photon absorbed and a lower energy indicates a lower energy photon absorbed. How can we determine this? By looking at the complex and more specifically the ligand attached and its location in the spectrochemical series. The ligands attached are Water and Chlorine and since Water is a stronger ligand than Chlorine according to the series, it also has larger energy, indicating a higher energy. This means that the complex [Cu(H O) ] absorbs a higher energy photon because of its a stronger ligand than chlorine. Part 2 of this question also asks which complex is predicted to have a larger crystal field splitting. To determine this you also use the spectrochemical series and see which ligand is stronger. Since H O is stronger than Cl on the spectrochemical series, we can say [Cu(H O) ] has a higher crystal field splitting. a) [Cu(H O) ] [Cu(H O) ] has a higher crystal field splitting | 84,938 | 1,978 |
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Thus far, we have assumed nothing about the value of the temperature corresponding to any particular volume of our standard fluid. We could define one unit of temperature to be any particular change in the volume of our standard fluid. Historically, defined one unit (degree) of temperature to be one one-hundredth of the increase in volume of a fixed quantity of standard fluid as he warmed it from the lowest temperature he could achieve, which he elected to call 0 degrees, to the temperature of his body, which he elected to call 100 degrees. Fahrenheit’s zero of temperature was achieved by mixing salt with ice and water. This is not a very reproducible condition, so the temperature of melting ice (with no salt present), soon became the calibration standard. Fahrenheit’s experiments put the melting point of ice at 32 F. The normal temperature for a healthy person is now taken to be 98.6 F; possibly Fahrenheit had a slight fever when he was doing his calibration experiments. In any case, human temperatures vary enough so that Fahrenheit’s 100-degree point was not very practical either. The boiling point of water, which Fahrenheit’s experiments put at 212 F, became the calibration standard. Later, the centigrade scale was developed with fixed points at 0 degrees and 100 degrees at the melting point of ice and the boiling point of water, respectively. The centigrade scale is now called the after Anders Celsius, Anders, a Swedish astronomer. In 1742, Celsius proposed a scale on which the temperature interval between the boiling point and the freezing point of water was divided into 100 degrees; however, a more positive number corresponded to a colder condition. Further reflection convinces us that the Charles’ law equation can be simplified by defining a new temperature scale. When we extend the straight line in any of our volume- -temperature plots, it always intersects the zero-volume horizontal line at the same temperature. Since we cannot associate any meaning with a negative volume, we infer that the temperature at zero volume represents a natural minimum point for our temperature scale. Let the value of \(T^*\) at this intersection be \(T^*_0\). Substituting into our volume-temperature relationship, we have \[0=n\beta \left(P\right)T^*_0+n\gamma (P)\] or \[\gamma \left(P\right)=-\beta (P)T^*_0\] So that \[\begin{align} V&= n\beta \left(P\right)T^*-n\beta (P)T^*_0 \\[4pt] &=n\beta \left(P\right)[T^*-T^*_0] \\[4pt] &=n\beta \left(P\right)T \end{align}\] where we have created a new temperature scale. Temperature values on our new temperature scale, T, are related to temperature values on the old temperature scale, \(T^*\), by the equation \[T=T^*-T^*_0\] When the size of one unit of temperature is defined using the Celsius scale ( ., \(T^*\) is the temperature in degrees Celsius), this is the origin of the \({}^{2}\). Then, on the Kelvin temperature scale, \(T^*_0\) is -273.15 degrees. (That is, \(T=0\) when \(T^*_0\) = 273.15; 0 K is \(-\)273.15 degrees Celsius.) The temperature at which the volume extrapolates to zero is called the of temperature. When the size of one unit of temperature is defined using the Fahrenheit scale and the zero of temperature is set at absolute zero, the resulting temperature scale is called the , after William Rankine, a Scottish engineer who proposed it in 1859. | 3,368 | 1,981 |
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The factors that influence whether an elimination reaction proceeds through an E1 or E2 reaction are almost exactly the same as the factors that influence the S 1/S 2 pathway. Cation stability, solvents and basicity play prominent roles. By analogy with substitution reaction, in which elimination mechanism does cation stability play a strong role: E1 or E2? Draw an example of an alkyl halide that is likely to undergo an E1 elimination. By analogy with substitution reactions, what mechanism would be promoted by protic solvents: E1 or E2? Basicity refers to the strength of the base. Which mechanism is more likely to occur with strong bases: E1 or E2? , | 671 | 1,982 |
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The STM doesn’t work the way a conventional microscope does, using optics to magnify a sample. Instead a sharp (1-10 nm) probe that is electrically conductive is scanned just above the surface of an electrically conductive sample. The principle of STM is based on tunneling of electrons between this conductive sharp probe and sample. Tunneling is a phenomenon that describes how electrons flow (or tunnel) across two objects of differing electric potentials when they are brought into close proximity to each other. When a voltage is applied between probe and surface electrons will flow across the gap (probe-sample distance) generating a measurable current. The tunneling phenomena can be explained by quantum mechanics. Tunneling originates from the wavelike properties of electrons. When two conductors are close enough there is an overlapping of the electron wavefunctions. Electrons can then diffuse across the barrier between the probe (tip-terminating ideally in a single atom) and the sample when a small voltage is applied. The resulting diffusion of electrons is called tunneling. A more detailed quantum mechanics explanation can be found. An important characteristic of tunneling is that the amplitude of the current exhibits an exponential decay with the distance, d. One way to describe this relationship is by the equation: \[\mathrm{I \sim Ve^{–cd}}\nonumber\] This dependency on tunneling current and probe-sample distance allows for precise control of probe-sample separation, resulting in high . Furthermore, is only carried out by the outermost single atom of the probe. This allows for high . Commercial probes are available but often users make their own probes. A common method is to tungsten, W, wire in NaOH to create a sharp probe. A problem with W probes is that they oxidize over time. Platinum iridium (Pt-Ir) is preferred for use in air because platinum does not easily oxidize. The tiny fraction of Iridium in the alloy makes it much harder. The Pt-Ir tips are usually shaped by cutting Pt-Ir wire with a wire cutter. It should be noted that a tip does not necessarily have to be one perfect point. STM Probe A simple analogy to describe SPM is to think of a stylus of a turntable scanning across a record, Figure 4. However unlike the stylus in a turntable, the probe in SPM make direct contact with the surface. In STM a voltage is applied between the metallic probe and the sample, typically (0-3 V). When the probe is close to the surface (2-4 ) the voltage will result in a current, due to tunneling between the probe and sample. When the probe is far away from the surface, the current is zero. The tunneling current produced is low (pA-nA) but can be monitored using amplifiers. A 3D scanner with an electronic is used to raster the probe across the sample to obtain a topographical image and monitor the tunneling current. The probe is attached to a 3-D . By adjusting the voltage applied to the scanner the position of the probe can be controlled. This is due to the unique properties of piezoelectric materials that are incorporated into the scanner. Piezoelectric materials have a permanent dipole moment across unit cell (Example: PbZrTiO (PZT)). If the dipoles are oriented, the material changes length in applied electric field. Each scanner responds differently to applied voltage because of the differences in the material properties and dimensions of each piezoelectric element. Sensitivity is a measure of this response, a ratio of how far the piezo extends or contracts per applied volt. A 10 to 10 % length change per V allows < 1 Å positioning. Most SPM instruments use a piezoelectric scan tube technology which combines independent piezos to control directions in the x, y, and z. AC voltages applied to the different electrodes produce a scanning (raster) motion in x and y. This motion is controlled by a computer. There are some factors to consider with piezoelectric scanners: piezo scanners are more sensitive at the end of travel. Therefore opposite scans will behave differently and display some hysteresis. Drift of piezo displacement may occur with large changes in x, y offsets Piezoelectric materials' sensitivity to voltages decreases over time. Therefore scanners need to be calibrated on a standard basis This is due to the scanner swinging in an arc motion to measure x,y displacement. This is often compensated with the z piezo or by using flattening algorithms. There are two methods of imaging in STM: 1) Constant Current A constant tunneling current is maintained during scanning (typically 1 nA). This is done by vertically (z) moving the probe at each (x,y) data point until a “setpoint” current is reached. The vertical position of the probe at each (x,y) data point is stored by the computer to form the topographic image of the sample surface. This method is most common in STM. 2) Constant Height In this approach the probe-sample distance is fixed. A variation in tunneling current forms the image. This approaching allows for faster imaging, but only works for flat samples. To view an interactive STM model and see how the STM probe moves across a surface you can go to the following website: To see how you could possibly build your own STM go here: Although the STM itself does not need vacuum to operate (it works in air as well as under liquids), ultrahigh vacuum (UHV) is required to avoid contamination of the samples from the surrounding medium. | 5,462 | 1,983 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/07%3A_Periodic_Properties_of_the_Elements/7.08%3A_Group_Trends_for_Selected_Nonmetals |
Non-metallic character is the ability to be reduced (be an oxidizing agent), form acidic hydroxides, form covalent compounds with non-metals. These characteristics increase with a larger nuclear charge and smaller radius, with no increase in shielding. The most active non-metal would be the one farthest up and to the right -- not including the noble gases (non-reactive.) Hydrogen has a 1 electron configuration and is placed above the alkali metal group. Hydrogen is a , which occurs as a (H ) under normal conditions. \[2Na_{(s)} + H_{2(g)} \rightarrow 2NaH_{(s)} \label{7.8.1} \] As we proceed down group 16 the elements become more metallic in nature: Oxygen can be found in two molecular forms, O and O (ozone). These two forms of oxygen are called ( ) \[3O_{2(g)} \rightarrow 2O_{3(g)}\;\;\; \Delta H = 284.6\; kJ / mol \label{7.8.2} \] Oxygen has a great tendency to attract electrons from other elements (i.e. to "oxidize" them) Sulfur also exists in several allotropic forms, the most common stable allotrope is the yellow solid S (an 8 member ring of sulfur atoms). Like oxygen, from other elements, and to form (which contain the S ion). This is particular true for the active metals: \[16Na_{(s)} + S_{8(s)} \rightarrow 8Na_2S_{(s)}\label{7.8.3} \] Note: most sulfur in nature is present as a metal-sulfur compound. Sulfur chemistry is more complex than that of oxygen. "Halogen" is derived from the Greek meaning "salt formers" Colors of diatomic halogens: ( flame colors) The halogens have some of the most negative electron affinities (i.e. large exothermic process in gaining an electron from another element) \[X_2 + 2e^- \rightarrow 2X^-\label{7.8.4} \] The chemistry of the halogens is dominated by their tendency to gain electrons from other elements (forming a halide ion) In 1992, of chlorine was produced. Both chlorine and sodium can be produced by electrolysis of molten sodium chloride (table salt). The electricity is used to strip electrons from chloride ions and transfer them to sodium ions to produce chlorine gas and solid sodium metal Chlorine reacts slowly with water to produce hydrochloric acid and hypochlorous acid: \[Cl_{2(g)} + H_2O_{(l)} \rightarrow HCl_{(aq)} + HOCl_{(aq)}\label{7.8}.5 \] Hypochlorous acid is a disinfectant, thus chlorine is a useful addition to swimming pool water The halogens react with most metals to form ionic halides: \[Cl_{2(g)} + 2Na_{(s)} \rightarrow 2NaCl_{(s)}\label{7.8.6} \] Rn is highly radioactive and some of its properties are unknown They are exceptionally unreactive. It was reasoned that if any of these were reactive, they would most likely be Rn, Xe or Kr where the first ionization energies were lower. In order to react, they would have to be combined with an element which had a high tendency to remove electrons from other atoms. Such as fluorine. Compounds of noble gases to date: \(XeF_2\) \(XeF_4\) \(XeF_6\) only one compound with Kr has been made \(KrF_2\) No compounds observed with He, Ne, or Ar; they are truly inert gases. ( ) | 3,066 | 1,984 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Physical_Chemistry_(LibreTexts)/17%3A_Boltzmann_Factor_and_Partition_Functions/17.01%3A_The_Boltzmann_Factor |
\[\textbf{v}_i(t) = \dfrac{d \textbf{r}_i}{dt} = \dot{\textbf{r}}_i \label{2.12} \] \[\textbf{F}_i = \textbf{F}_i(\textbf{r}_1, \ldots, \textbf{r}_N) \label{2.13} \] \[\dfrac{3}{2} NkT = \sum_{i=1}^N \dfrac{1}{2}m_i \textbf{v}_i^2 \label{2.14} \] \[T = \dfrac{2}{3k} \left( \dfrac{1}{N} \sum_{i=1}^N \dfrac{1}{2}m_i \textbf{v}_i^2 \right) \label{2.15} \] A confined monatomic gas can be seen as a box with a whole bunch of atoms in it. Each of these particles can be in one of the states given by the last formula. If all of them have large \(n\) values there is obviously a lot of kinetic energy in the system. The lowest energy is when all atoms have 1,1,1 as quantum numbers. Boltzmann realized that this should relate to temperature. When we add energy to the system (by heating it up without changing the volume of the box), the temperature goes up. At higher temperatures we would expect higher quantum numbers, and at lower \(T\), lower ones. But how exactly are the atoms distributed over the various states? This is a good example of a problem involving a discrete probability distribution. The probability that a certain level (e.g., \(n = ( n_1,n_2,n_3)\) with energy \(E_i\)) is occupied should be a function of temperature: \(P_i(T)\). Boltzmann postulated that you could look at temperature as a form of energy. The thermal energy of a system is directly proportional to an absolute temperature. \[E_{thermal} = k T \nonumber \] The proportionality constant \(k\) (or \(k_B\)) is named after him: the Boltzmann constant. It plays a central role in all statistical thermodynamics. The Boltzmann factor is used to approximate the fraction of particles in a large system. The Boltzmann factor is given by: \[ e^{-\beta E_i} \label{17.1} \] where: As the following section demonstrates, the term \( \beta \) in Equation \(\ref{17.1}\) is expressed as: \[ \beta=\frac{1}{k T} \nonumber \] The rates of many physical processes are also determined by the Boltzmann factor. For a random particle, its thermal energy of a particle is a small multiple of the energy \(k T\). An increase in temperature results in more particles crossing the energy barrier characteristic of activation processes. If this is to occur, particles need to carry sufficient energy. This energy is needed for the particles to successfully cross the energy barrier and is usually called activation energy. The fraction of molecules that have sufficient energy to escape the original material surface is approximately proportional to the Boltzmann factor. | 2,562 | 1,985 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/24%3A_Chemistry_of_Life-_Organic_and_Biological_Chemistry/24.05%3A_Functional_Groups |
You were previously introduced to several structural units that chemists use to classify organic compounds and predict their reactivities. These functional groups, which determine the chemical reactivity of a molecule under a given set of conditions, can consist of a single atom (such as Cl) or a group of atoms (such as CO H). The major families of organic compounds are characterized by their functional groups. Figure \(\Page {1}\) summarizes five families introduced in earlier chapters, gives examples of compounds that contain each functional group, and lists the suffix or prefix used in the systematic nomenclature of compounds that contain each functional group. The first family listed in Figure \(\Page {1}\) is the hydrocarbons. These include alkanes, with the general molecular formula C H where n is an integer; alkenes, represented by C H ; alkynes, represented by C H ; and arenes. Halogen-substituted alkanes, alkenes, and arenes form a second major family of organic compounds, which include the alkyl halides and the aryl halides. Oxygen-containing organic compounds, a third family, may be divided into two main types: those that contain at least one C–O bond, which include alcohols, phenols (derivatives of benzene), and ethers, and those that contain a carbonyl group (C=O), which include aldehydes, ketones, and carboxylic acids. Carboxylic acid derivatives, the fourth family listed, are compounds in which the OH of the –CO H functional group is replaced by either an alkoxy (–OR) group, producing an ester, or by an amido (–NRR′, where R and R′ can be H and/or alkyl groups), forming an amide. Nitrogen-containing organic compounds, the fifth family, include amines; nitriles, which have a C≡N bond; and nitro compounds, which contain the –NO group. The systematic nomenclature of organic compounds indicates the positions of substituents using the lowest numbers possible to identify their locations in the carbon chain of the parent compound. If two compounds have the same systematic name, then they are the same compound. Although systematic names are preferred because they are unambiguous, many organic compounds are known by their common names rather than their systematic names. Common nomenclature uses the prefix form—for a compound that contains no carbons other than those in the functional group, and acet—for those that have one carbon atom in addition [two in the case of acetone, (CH ) C=O]. Thus methanal and ethanal, respectively, are the systematic names for formaldehyde and acetaldehyde. Recall that in the systematic nomenclature of aromatic compounds, the positions of groups attached to the aromatic ring are indicated by numbers, starting with 1 and proceeding around the ring in the direction that produces the lowest possible numbers. For example, the position of the first CH group in dimethyl benzene is indicated with a 1, but the second CH group, which can be placed in any one of three positions, produces 1,2-dimethylbenzene, 1,3-dimethylbenzene, or 1,4-dimethylbenzene (Figure \(\Page {2}\)). In common nomenclature, in contrast, the prefixes ortho-, meta-, and para- are used to describe the relative positions of groups attached to an aromatic ring. If the CH groups in dimethylbenzene, whose common name is xylene, are adjacent to each other, the compound is commonly called ortho-xylene, abbreviated o-xylene. If they are across from each other on the ring, the compound is commonly called para-xylene or p-xylene. When the arrangement is intermediate between those of ortho- and para- compounds, the name is meta-xylene or m-xylene. We begin our discussion of the structure and reactivity of organic compounds by exploring structural variations in the simple saturated hydrocarbons known as alkanes. These compounds serve as the scaffolding to which the various functional groups are most often attached. Functional groups determine the chemical reactivity of an organic molecule. Functional groups are structural units that determine the chemical reactivity of a molecule under a given set of conditions. Organic compounds are classified into several major categories based on the functional groups they contain. In the systematic names of organic compounds, numbers indicate the positions of functional groups in the basic hydrocarbon framework. Many organic compounds also have common names, which use the prefix form—for a compound that contains no carbons other than those in the functional group and acet—for those that have one additional carbon atom. | 4,548 | 1,988 |
https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Introduction_to_Organic_and_Biochemistry_(Malik)/09%3A_Food_to_energy_metabolic_pathways/9.03%3A_Oxidation_of_glucose_-the_glycolysis |
Oxidation of glucose is the 2 stage of catabolism of carbohydrates that takes place in the cytoplasm outside the mitochondria. It is a series of ten reactions, i.e., a metabolic pathway called glycolysis, as illustrated in Figure \(\Page {1}\). The ten reactions of glycolysis are the following, modified from a Public Domain resource at . 1. Glucose is converted to glucose-6-phosphate by an SN reaction mechanism between a primary alcohol acting as a nucleophile and a \(\ce{P}\) of \(\ce{ATP}\) as an electrophile, where \(\ce{ADP}\) acts as a good leaving group, as shown below. 2. Glucose-6-phosphate, which is an aldohexose, isomerizes to fructose-6-phosphate, which is a ketohexose. 3. The primary alcohol group of fructose-6-phosphate is phosphorylated by the same mechanism as in the first step producing fructose-1,6-bisphosphate. 4. The open-chain form of fructose-1,6-bisphosphate, which is in equilibrium with the cyclic form, splits into two compounds glyceraldehyde-3-phosphate and dihydroxyacetone phosphate. 5. This reaction is an isomerization reaction between glyceraldehyde-3-phosphate and its isomer dihydroxyacetone phosphate. The aldehyde or ketone in these two isomers, tautomerize to their enol forms, which either revert to the initial compound or to its isomers when the enol form changes back to an aldo or keto form. 6. The aldehyde (\(\ce{-HC=O}\)) group of glyceraldehyde-3-phosphate is oxidized in three steps: activated by the addition of \(\ce{HS-CoA}\) (step 6i); followed by oxidized of \(\ce{-OH}\) to a carbonyl (\(\ce{C=O}\) group at the expense of reduction of \(\ce{NAD^{+}}\) to \(\ce{NADH}\) (step 6ii); and finally displacement of \(\ce{HS-CoA}\) by a phosphate (\(\ce{-PO4^{2-}}\)) group (step 6iii). The overall reaction is the following. The ketone group in dihydroxyacetone phosphate does not oxidize directly, but it converts to glyceraldehyde-3-phosphate as the latter is consumed due to the equilibrium of creation#5. So, reaction#6 and reactions after this happen twice for each glucose molecule processed. 7. A phosphate group is transferred from 1,3-bisphosphoglycerate to ADP producing an ATP and a 3-phosphoglycerate. Since this step happens twice for each glucose molecule processed, it compensates for the two ATP consumed, one in step#1 and the other in step#3. 8. 3-Phosphoglycerate is isomerized to 2-phosphoglycerate. In this step, the enzyme phosphoglycerate mutase first transfers its phosphate group to the \(\ce{-OH}\) at position#2 and then receives the phosphate from position#3 of the intermediate, resulting in the isomerization of 3-Phosphoglycerate to 2-phosphoglycerate. 9. 2-Phosphoglycerate is dehydrated, i.e., \(\ce{H2O}\) is eliminated form it, producing phosphoenolpyruvate. 10. Phosphoenolpyruvate transfers its phosphate group to ADP producing pyruvate and ATP. Since the reaction happens two times for each glucose, two ATP are produced in this step. In summary, glycolysis consists of two phases. The first phase, comprising reactions#1 to 5, is the preparatory phase, which is also energy consuming phase in which a six \(\ce{C}\) glucose (\(\ce{C6H12O6}\)) is phosphorylated twice at the expense of two ATP's and converts to two glyceraldehyde-3-phosphate molecules which are three\(\ce{C}\) compounds. The second phase, comprising reactions#6 to 10, is the payoff phase in which four ATP's are produced, i.e., two more than the ATP consumed in the first phase and two \(\ce{NAD^{+}}\) are reduced to two \(\ce{NADH}\) along with the production of two pyruvates \(\ce{CH3COCOO^{-}}\) as shown in the following reaction. \(\ce{\underbrace{C6H12O6}_{Glucose} + 2NAD^{+} + 2ADP + 2Pi -> 2\underbrace{CH3-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Pyruvate} + 2NADH + 2ATP + 2H^{+} + 2H2O}\) Two \(\ce{NAD^{+}}\) are reduced to two \(\ce{NADH}\) which need to be oxidized back to \(\ce{NAD^{+}}\) for the process to continue. It happens in different ways depending on whether sufficient oxygen is present, i.e., or no oxygen is present, i.e., . The fate of pyruvate also depends on whether the condition is aerobic or anaerobic, as described below. In aerobic conditions \(\ce{NADH}\) is oxidized to \(\ce{NAD^{+}}\) at the expense of oxygen in mitochondria in the fourth stage of catabolism. Pyruvate is also transferred to mitochondria and undergoes oxidation at the expense of \(\ce{NAD^{+}}\) reduced to \(\ce{NADH}\) through a series of relations catalyzed by a complex of three enzymes and five coenzymes, collectively called the pyruvate dehydrogenase complex. The overall reaction is the decarboxylation of pyruvate, i.e., carbon dioxide (\(\ce{CO2}\)) eliminates and the acetyl (\(\ce{CH3-CO-}\)) group is transferred to \(\ce{HS-CoA}\) producing acetyl-\(\ce{CoA}\), as shown in the following reaction. \(\ce{\underbrace{CH3-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Pyruvate} + HS-CoA + NAD^{+} ->[\text{Pyruvate dehydrogenase complex}] \underbrace{CH3-\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-S-CoA}_{Acetyl\;CoA} + CO2 + NADH + H^{+}}\) Acetyl-\(\ce{CoA}\) enters into the citric acid cycle, also called the Krebs cycle, in mitochondria which is the third stage of catabolism. The processing of pyruvate under aerobic conditions is also called which links glycolysis to the citric acid cycle as explained in the video below. Overall reaction of one glucose molecule under aerobic condtions before entry into the citric acid cycle is the following. \(\ce{\underbrace{C6H12O6}_{Glucose} + 2HS-CoA + 4NAD^{+} + 2ADP + 2Pi -> 2\underbrace{CH3-\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-S-CoA}_{Acetyl\;CoA} + 2CO2 + 4NADH + 2ATP + 4H^{+} + 2H2O}\) During vigorous exercises, as shown in the figure on the right, or hard physical work, oxygen depletes in muscles creating anaerobic (oxygen-free) conditions. Under anaerobic conditions \(\ce{NADH}\) is oxidized to \(\ce{NAD^{+}}\) in the cytoplasm at the expense of the reduction of pyruvate to lactate as shown in the reaction below. \(\ce{\underbrace{CH3-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Pyruvate} + NADH + H^{+} <->[\text{Lactate dehydrogenase}] \underbrace{CH3-\!\!\!\!\!{\overset{\overset{\huge\;\;\;{OH}}|}{C}}\!\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Lactate} + NAD^{+}}\) Overall glycolysis of glucose under anaerobic conditions is the following reaction. \(\ce{\underbrace{C6H12O6}_{Glucose} + 2ADP + 2Pi -> 2\underbrace{CH3-\!\!\!\!\!{\overset{\overset{\huge\;\;\;{OH}}|}{C}}\!\!\!-\!\!{\overset{\overset{\huge\enspace\!{O}}|\!\!|\enspace}{C}}\!\!-O^{-}}_{Lactate} + 2ATP}\) The accumulation of lactate makes muscles tire and sour. the person keeps beating heavily after the exercise to pay the oxygen debt. Most of the pyruvate is transported to the liver where it is re-oxidized to pyruvate. Instead of converting pyruvate to lactate as in humans and animals, yeast has an enzyme called pyruvate decarboxylase that decarboxylates pyruvate to acetaldehyde. Then \(\ce{NADH}\) is oxidized to \(\ce{NAD^{+}}\) at the expense of the reduction of acetaldehyde to ethanol in a process called fermentation, as shown in the reaction below. Copyright; User:Cwernert at en. , vectorized by User:GKFX., CC BY- , via Wikimedia Commons. The overall reaction of glycolysis of glucose in fermentation process by yeast is the folloiwng. \(\ce{\underbrace{C6H12O6}_{Glucose} + 2ADP + 2Pi -> 2\underbrace{CH3-CH2-OH}_{Ethanol} + 2CO2 + 2ATP}\) | 7,795 | 1,990 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Microscopy/Scanning_Probe_Microscopy/04_Additional_SPM_Methods/03_Magnetic_Force_Microscopy |
Magnetic force microscopy (MFM) is a mode that maps the spatial distribution of magnetic materials on a surface, by measuring the magnetic interaction between a sample and a tip. MFM is conducting cantilever that is equipped with a tip that has a magnetic coating such as (Co-Cr). The interactions between the probe and surface can be detected via the deflection of the cantilever (contact mode). More commonly the magnetic interaction can be monitored through monitoring the oscillation of the probe (tapping mode). The changes in magnetic field shift the resonant frequency of the cantilever. This shift can be monitored as shown in Figure 1. A more detailed description of the various ways the actually shift frequency can be monitored can be found. A key aspect of this measurement is that a “background is taken” prior to the MFM measurement to ensure the MFM measurement is due to magnetic domains and not topography. First the topography of the sample is measured, then the probe is “lifted” a distance from the sample and the response of the tip to magnetic domain is measured (taking into account cantilever associated with topography). | 1,157 | 1,991 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Microscopy/Scanning_Probe_Microscopy/04_Additional_SPM_Methods/02_Chemical_Force_Microscopy |
Chemical force microscopy (CFM) is a technique, which combines the force sensitivity of the AFM with chemical discrimination. This is achieved by modifying probes with covalently linked molecules that terminate in well-defined functional groups or biological molecules. By using a suitable tip modification, chemically specific probing of the surface based on a defined tip-surface interaction can be achieved. For example, CFM experiments have been used to probe fundamental adhesion and friction forces at the solid-liquid interface and biological interactions such as biotin and streptavidin. Interesting application of CFM to evaluate surface chemistry of skeletal tissue | 689 | 1,993 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Thermodynamic_Cycles/Brayton_Cycle |
The Brayton Cycle is a thermodynamic cycle that describes how gas turbines operate. The idea behind the Brayton Cycle is to extract energy from flowing air and fuel to generate usuable work which can be used to power many vehicles by giving them thrust. The most basic steps in extracting energy is compression of flowing air, combustion, and then expansion of that air to create work and also power the compression at the same time. The usefulness of the Brayton Cycle is tremendous due to the fact it is the backbone in driving many vehicles such as jets, helicopters, and even submarines. The first gas turbine that implemented the Brayton Cycle (not knowingly however, because it was created before the Brayton Cycle was even established) was John Barber's gas turbine patented in 1791. The idea of the machine was to compress atmospheric air in one chamber and fuel in another chamber and both chambers would be connected to a combustion vessel. Once the air has mixed with the fuel and reacted, the energy from the combustion would be used to spin a turbine to do useful work. However, because back in the late 18th century there was lack of technological advances and such, the gas turbine did not have enough energy to pressurize the gases and do useful work at the same time therefore it was not used. George Brayton was an engineer that designed the first continuous ignition combustion engine which was a two-stroke engine that was sold under the name "Brayton's Ready Motors." The design employed the thermodynamic processes that is now considered "The Brayton Cycle," but is also coined The Joule Cycle. The gas turbine was patented in 1872. The design was a engine connected to a reservoir of pressurized atmospheric air and gas which would only turn on if a valve was turned. This would release the pressurized gas to a combustion vessel, which would turn pistons to create mechanical work and re-compress the gas in the reservoir. The Brayton Cycle for John barber's gas turbine is incomplete due to the fact energy is not redirected into compression of the initial gases, yet because this was one of the first prominent gas turbine engines ever created, it still holds much significance. Fuel and atmospheric gas is held in different chambers and heated to increase pressure. This is in part to the ideal gas law PV = nRT, and since the volume of the vessel stays constant, an increase in temperature increases pressure. The gas combine into a square compartment where a spark or flame ignites the mixture which then rapidly increases temperature (but not pressure because the gas quickly escapes to spin the turbine). Although this is a very crude gas turbine engine, it nontheless was a great foundations for further scientific advancements and the development of the Brayton Cycle. George Brayton's gas turbine was the first and most prominent fully operational gas turbine that implemented the Brayton Cycle. Gas is pressurized and held in reservoir A, where a valve would release it to move through tunnel B and be ignited in chamber C. This is an due to the fact any increase in pressure would just push the gas out of the engine. When work is done on the D piston, mechanical work can be employed for a variety of things, such as generation of electricity or movement, and there is also a piston in compartment E that sucks in atmospheric air from valve F. Valve G is a fitting to a fuel cell where the fuel to air ratio can be set so that a desired ratio for maximum efficiency is kept. The gas is then mechanically compressed back into reservoir A through an due to the piston E. This is one of the many modern day gas turbine engines that utilize the Brayton Cycle in order to power many vehicles or generate power. At the front of the engine is the inlet of the compression chamber so that air is sucked in by the many turbines that are constantly spinning and angled in a specific location for optimum air compression. Enough the air is compressed enough in the middle of the engine (the combustion vessel), fuel is added to the combustion chamber and an ignition is initiated, where the extremely exothermic reaction causes the gas to violently exits the engine in the expansion chamber in the back of the engine. There are turbines right in front of the expansion chamber that is connected to the turbins in the compression chamber so the whole engine is a continuous cycle as long as there is a steady stream of fuel being introduced into the combustion chamber. A quick qualitative look at how the Brayton Cycle works it by reviewing how a jet engine works. A gas turbine from a jet's wing sucks in atmospheric air from the back of its engine and is compressed in the mixing/combustion chamber. In the mixing/combustion chamber, fuel is mixed with the compressed atmospheric air, where it is ignited and left to exit in the expansion chamber. The energy that comes out of the back of the gas turbine is work used to power the compression step as well as give thrust to the jet. The Brayton Cycle can then be described quantitatively in the gas turbine engine of a jet by two diagrams, the Temperature/Entropy Diagram and the Pressure/Volume Diagram. In this diagram, we see that there are 8 processses to describe the Brayton cycle in terms of temperature, entropy, and pressure. (1) Ambiant air in the atmosphere that is currently undisturbed. (1 -> 2) Ambiant air comes into contact with the compressor of the gas turbine and the pressure and temperature rises dramatically. The rise in pressure comes from work being done the air by the compressor which packs the air into the mixer/combustion chamber, and the rise in pressure causes a rise in temperature in the gas molecules because volume of the vessel stays constant (PV = nRT). Because this is an ideal process, entropy is believed to stay the same, thus this is an (in reality though, entropy does increase due to the flow and movement of the gas molecules). (3 -> 5) The atmospheric air has been compacted into the combustion chamber where gaseous fuel is mixed with the air. Once this mixture has been ignited, we a see a steep rise in the temperature and entropy (not the pressure, because the curves represent a specific value of pressure, so this is an ) due to the combustion reaction of the fuel and air. The energy from the chemical bonds in the fuel are broken due to ignition and a highly exothermic reaction occurs which raises entropy because of the breaking down of hydrocarbon chains to water and air (more molecules) and raises the temperature due to increased ambiant energy from the exothermic reaction. (5 -> 8) At point 5, the pressurized fuel and air leave the combustion chamber to the expansion chamber, where we see a quick drop in pressure due to a larger volume and exposure to the surroundings. The energy from the combustion chamber is used two for two purposes: spinning a turbine that is connected to the compressor (which keeps the Brayton Cycle running continuously) and as thrust. These two purposes represent point 6 and ideally is an . The quick drop in pressure shows how the energy from the air in the combustion energy is used mechanically to turn a turbine that will run the compressor process because the energy it takes to compress the atmospheric energy is lesser than the energy produced from the ignition of the fuel. The energy left over from spinning the turbine is the used as thrust to do work (such as flight in a jet). The expelled air then becomes ambiant air that is of a higher energy level than the air from point 1, but will eventually lose energy to the surroundings ( ) and become the initial ambiant air. In this diagram, there are six processes that describe the pressure and volume of the gas. A common mistake is by thinking the volume relates to the vessel of the reaction, when in fact it is the volume of the gas. This graph coincides with the TS diagram but do not progress through the points at the same time (because this diagram only has 6 points). (1 -> 3) The ambiant air is sucked into the compressor where the volume of the gas quickly falls due to the compression into the combustion chamber. As compression continues, the pressure of the gas begins to quickly rise at point 2 after the volume of the combustion chamber is filled and peaks at point 3. At point 3, ignition occurs. (4 -> 6) As ignition occurs, the pressure of the gas remains constant due to the face the gas is able to escape into the expansion chamber (notice that even though gas is leaving, the compression process is still working, so any pressure lost from the gas leaving is held constant from the gas entering the combustion chamber) which results in a rise in volume of the energized gas. As the gas leaves into the atmosphere, the pressure drops and the volume of the gas expands to what it was in point 1. Work is done from the expansion of the gas which pushes out of the expansion chamber with high force. This force is then used to spin turbins and give thrust. The ideal processes shown in the above diagrams are used to study and understand the Brayton Cycle better. However, some corrections are needed when applied to real world problems. The first problem is the fact that in the compression process, it is believed to be isentropic. This is wrong, because the high speed flow of the ambiant air increases the entropy (higher energy molecules) and thus is not an isentropic process. It is actually an because no heat exchange occurs in the gas and only mechanical work is done for compression. This also relates back to the expansion process where the gas expands in the expansion chamber but has yet to leave into the atmosphere. Ideally it is isentropic, but the expansion of gas does increase entropy, therefore it is actually an adiabatic process because again, there is not heat exchange (only mechanical work done by expansion). The non-ideal processes of the Brayton Cycle points out a problem; that the work used to raise entropy is thus a leak in the amount of work that could have been used for useful mechanical energy. A set of equations is then used to calculate the efficiency of the Brayton Cycle at certain pressures and temperatures. To find the efficiency of the Brayton Cycle, we must find out how much work each process contributes to the total internal energy. We will be analyzing the PV diagram above to do this. First, the internal energy \[U = q_1 + q_2 - w = 0\] is equal to zero because the states that energy is not destroyed or created, and because in the Brayton cycle the final state function of the gas is the initial, U = 0. This means \[w = q_1 + q_2\] where \(q_1\) is the heat recieved by the combustion (so it is negative) and \(q_2\) is the heat released after expansion. If you treat the gas as a perfect gas with constant specific heats, we can find the heat addition from the combuster to be \[q_1 = c_p(T_I - T_F)\] and the heat lost to the atmosphere \[q_2 = c_p(T_F - T_I)\] Where \(T_F\) is the final temperature of the combustion or "heat lost to the atmosphere" part and the latter is the initial. (So in the PV curve, the combustion process would have \(q_1 = c_p(T_4 - T_3)\) So now we have expressed the amount of heat lost and gained in terms of temperatures, we can re-establish the equation to find eta (thermal efficiency) \[\eta = \dfrac{\textrm{Net work}}{\textrm{Heat in}} = \dfrac{c_p[(T_c -T_c) - (T_d - T_a)]}{c_p(T_[T_c = T_b]} = 1 - \frac{(T_d-T_a)}{(T_c-T_b)} = 1 - \frac{T_a(T_d/T_a-1)}{T_b(T_c/T_b-1)}.\] where c is the final temperature of the combustion process and b is the initial temperature before combustion and a is the initial temperature of the undisturbed gas and d is the temperature of the gas after it has be expelled. The corresponding numbers to letters from the PV graph is a = 2; b = 3; c = 4; d = 6. The smaller the temperature ratio is the higher the efficiency of the Braytons cycle is. So that is, the more heat input into the system and the smaller amount of heat lost to the atmosphere will significantly reduce the temperature ratio and have a higher percentage of efficiency. 1. Ideal Non-Ideal 2. 3. \(1 - (4000 - 300)/5000 = 0.26\). Which means it is running at 26% efficiency. 4. ANS for combustion phase = 5(300 - 4000) = -18500 \(q_2\) cannot be found because the initial temperature of the air was not given, therefore no solution. 5. Ideally, compression or expansion does not cause entropy (disorder in the system) to change because nothing is causing the the molecules to increase in energy was they are being pushed together or going apart, thus entropy stays constant. However, in the real world, the mechanical energy pushing the molecules together (the compression turbine) causes them to gain molecular energy and does increase entropy in the system. | 12,833 | 1,994 |
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There are a number of recurring types of structural features in organic compounds that commonly are known as . In fact, a traditional approach to the subject of organic chemistry involves the classification of compounds according to their functional groups. Thus the structural features \(C = C\), \(C \equiv C\), \(C = O\), \(OH\), \(NH_2\), and \(C \equiv N\) are the functional groups of alkenes, alkynes, carbonyl compounds, alcohols, amines, and nitriles, respectively. It will be helpful to look at the structural features of some of the major types of organic compounds even though the details of their chemistry will not be discussed until later chapters. Examples of the structures arranged in accord with their functional groups are given in Table 2-2. The examples chosen are representative of compounds containing carbon and hydrogen ( ) as well as compounds containing halogens, oxygen, nitrogen, and sulfur. We do not expect you to memorize this table. In time you will become familiar with all of the types of structures in it. In Table 2-2 we generally have used as first-choice names because these names emphasize the relationships between the compounds and ease the burden fo the beginning student in having to remember many special names. We have little hope that systematic names such as methanal, 2-propanone, and ethanoic acid soon will replace the commonly used nonsystematic names formaldehyde, acetone, and acetic acid. But there is no question that every organic chemist knows what compounds the names methanal, 2-propanone, and ethanoic acid represent, so the beginner can communicate with these names and later become familiar with and use the special names. We will have more to say on this subject in Chapter 3. One of the main reasons for classifying compounds by their functional groups is that it also classifies their chemical behavior. By this we mean that the reactions of compounds and, to some extent, their physical properties are influenced profoundly by the nature of the functional groups present. Indeed, many organic reactions involve transformations of the functional group that do not affect the rest of the molecule. For instance, alcohols, \(R-OH\), can be transformed into a number of other compounds, such as organic halides, \(R-Cl\) or \(R-Br\); ethers, \(R-O-R\); and amines, \(R-NH_2\) without changing the structure of the hydrocarbon group \(R\). Furthermore, any compound possessing a particular functional group may be expected to exhibit reactions characteristic of that group and, to some extent at least, of inorganic compounds with similar functional groups. A good example of the use of the functional-group concept is for acid-base properties. Alcohols, \(ROH\), are structurally related to water, \(HOH\), in that both possess a hydroxyl function. We may then expect the chemistry of alcohols to be similar to that of water. In fact, both are weak because the \(OH\) group has a reactive proton that it can donate to a sufficiently strongly basic substance, written as \(:B\) here: Water and alcohols both are weak because the oxygens of their \(OH\) groups have unshared electron pairs to use in bonding with a proton donated by an acid, \(HA\): We can carry the analogy further to include carboxylic acids, \(RCO_2H\), which also have a hydroxyl function. They also should possess acidic and basic properties. They do have these properties and they are, in fact, stronger acids than either water or alcohols and form salts with bases: Amines, \(RNH_2\), are structurally related to ammonia, \(NH_3\), and we therefore may predict that they will have similar properties. A property of ammonia that you probably will have encountered earlier is that it acts as a base and forms salts with acids. Amines behave likewise: It is with logic of this kind - inferring chemical behavior from structural analogies - that much of organic chemistry can be understood. There are other logical classification schemes, however, and one of these depends more on of reactions than on functional groups. The rationale of classification by reaction types is that different functional groups may show the same kinds of reactions. Thus, as we have just seen, alcohols, carboxylic acids, and amines all can accept a proton from a suitably strong acid. Fortunately, there are very few different types of organic reactions - at least as far as the overall result that they produce. The most important are acid-base, substitution, addition, elimination, and rearrangement reactions. Some examples of these are given below, and you should understand that these are descriptive of the overall chemical change and nothing is implied as to or the reaction occurs (also see ). of one atom or group of atoms for another: , usually to a double or triple bond: , which is the reverse of addition: where one structure is converted to an isomeric structure: Certain reactions commonly are described as either or reactions and most simply can be thought of as reactions that result in changes in the oxygen or hydrogen content of a molecule by direct or indirect reactions with oxygen or hydrogen, respectively. They frequently fall into one of the categories already mentioned. Reduction of ethene to ethane is clearly , as is oxidation of ethene to oxacyclopropane: Reactions that lead to substantial degradation of molecules into smaller fragments are more difficult to classify. An example is the combustion of ethane to carbon dioxide and water. All of the chemical bonds in the reactants are broken in this reaction. It seems pointless to try to classify this as anything but a complete combustion or oxidation reaction: and (1977) | 5,708 | 1,995 |
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The law of mass action only rigorously applies to elementary reactions (in that otherwise the exponents do match stoichiometric coefficients). These are reactions whose equations match which particles are colliding on the microscopic scale. Consider the reaction \[ r_1 R_1 + r_2 R_2 + \ldots + r_n R_n <=> p_1 P_1 + p_2 P_2 + \ldots + p_m P_m \label {1} \] Reaction \((1)\) is elementary if and only if it proceeds in a single step. At some time \(t = t_c\) all reactant particles \( R_1, R_2, \ldots, R_n \) collide. Quantum mechanically this collision is not too well defined, so view it to be Newtonian in essence. Not to mention: the probability \(P\) of the simultaneous collision goes to zero as \(n \to \infty \). This is why in actuality most elementary steps involve at most three individual constituents. We therefore sacrifice some generality and instead consider the easier to handle \[ R_1(g) + R_2(g) <=> P_1(g) + P_2(g) \label {1'} \] The reactant molecules must collide in order for a reaction to take place. Therefore we must know the in order to determine the reaction rate. This rate must be the product of some probability and the collision frequency. gives a way to calculate this . In doing so Maxwell-Boltzmann distributions are used to calculate the ratio of atoms that can react within a given concentration. Most of this 'stuff' gets swept up in the reaction rate constant. Collision theory postulates that particles only initiate a transformation when they have enough energy. This is called and denoted by \(E_a\). Maxwell-Boltzmann distribution \[\frac{\mathrm{d}N_v}{N} = 4\pi\left(\frac{m}{2\pi k_bT}\right)^{3/2}v^2\exp\left(-\frac{mv^2}{2k_bT}\right)\mathrm{d}v\label{2}\] allows us to estimate the fraction of particles that have the required amount of movement. Note we are technically assuming total energy = kinetic energy. Similarly relativistic effects are ignored, and no cap is placed on maximum velocity. With these simplifications, the fraction of particles is \[f = \frac{N^*}{N} = \frac{1}{k_bT}\int_{E_a}^\infty\exp\left(-\frac{E}{k_bT}\right)\mathrm{d}E\label{3}\] which evaluates to \[f = \exp\left(-\frac{E_a}{k_bT}\right)\label{3'}.\] Say \(Z^*\) is the number of collisions that matter. We will look for a function \(Z\) such that \[Z^* = Z\cdot f. \label {4}\] The Maxwell-Boltzmann distribution (Equation \(\ref{2}\)) gives for a single kind of particle (say two \(\ce{R1}\)'s) that there are \[ Z = 2N^2 \left( \sigma_{\ce{R1}} \right)^2 \sqrt{\dfrac{\pi RT}{M_{\ce{R1}}}} \label{5}\] collisions in \(1\) second per \(1\ \mathrm{cm^3}\) if \(N\) is the number of particles per square centimeter. (This is the traditional starting point unit.) Hence the rate of such a reaction would be \[v_r = \frac{2Z^*}{N_A} \cdot 10^3\ \ \ \left(\frac{\mathrm{mol}}{\mathrm{s\cdot dm^3}}\right). \label{6}\] If you like, you are now able to substitute \(f\) from Equation \(\ref{3'}\) and \(Z\) via Equation \(\ref{5}\) into Equation \(\ref{4}\). Then insert Equation \(\ref{4}\) into Equation \(\ref{6}\). After necessary manipulation, this yields Equation \(\ref{7}\). \[v_r = \underbrace{4 \times 10^{-3} N_A \left(\sigma_{\ce{R1}} \right)^2\sqrt{\frac{\pi RT}{M_{\ce{R1}}}}\exp\left(-\frac{E_a}{k_bT}\right) }_k \cdot \overbrace{\left(\frac{N}{N_A}\cdot 10^3\right)^2}^{c^2}. \label{7}\] Coming back to our actual reaction (Equation \(\ref{1'}\)) involves more relative quantities. For example, \(\sigma_{\ce{R1}}\) becomes \[\sigma_{\text{aver}} = 0.5\left(d_{\ce{R1}} + d_{\ce{R2}} \right).\] So it is a bit more difficult. But the result is analogous to Equation \(\ref{7}\). \[v_r = \underbrace{2\sqrt{2} \cdot 10^{-3} N_A\left(\sigma_\text{aver}\right)^2\sqrt{\pi RT\left(\frac{1}{M_{\ce{R1}}} + \frac{1}{M_{\ce{R2}}}\right)}\exp\left(-\frac{E_a}{k_bT}\right)}_k \cdot c_{\ce{R1}} \cdot c_{\ce{R2}}\] More compactly, if \(\ce{R1} = A\) and \(\ce{R2} = B\) \[v_r = k[A,B]\label{8}\] which is what we set out to prove. As your quote suggests, collision theory is a first theoretical explanation for the proportionality to products of concentrations. It does generally explain various exponentiation. It either. The higher (or non-integer or negative) exponents usually derive from the mechanism itself. In other words, the fact that common reactions are comes into play. For instance, the transition \[\ce{2Br- + H2O2 + 2H+ -> Br2 + H2O}\] is experimentally found to follow \[v_r = k\ce{[H2O2,H+,Br^{-}]}\label {a}.\] Mathematically, we can verify that mechanism is \[\ce{H+ + H2O2 <=>[K] H2O+-OH,} \tag{fast equilibrium} \] \[\ce{H2O+-OH + Br- ->[k_2] HOBr + H2O,}\tag{slow}\] \[\ce{HOBr + H+ + Br- ->[k_3] Br2 + H2O.}\tag{fast}\] Indeed, \[K_c = \frac{\ce{[H2O+-OH]}}{\ce{[H+]}\ce{[H2O2]}} \implies \ce{[H2O+-OH]} = K_c\ce{[H+]}\ce{[H2O2]}\label {b}.\] Applying the method of stationary concentration gives \[v\left(\ce{HOBr}\right) = 0 \implies k_2\ce{[H2O+-OH]}\ce{[Br-]} = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]}\label {c}.\] The overall rate of the reaction is characterised by the rate of formation of bromine \(\ce{Br2}\). So, \[v_r = v\left(\ce{Br2}\right) = k_3\ce{[H2O2]}\ce{[H+]}\ce{[Br-]} \overset{(c)}{=} k_2\ce{[H2O+-OH]}\ce{[Br-]} \overset{(b)}{=} k_2K_c\ce{[H+,H2O2]}\ce{[Br-]}\] or more briefly using \(k_2K_c = k\) \[v_r = k\ce{[H2O2,H+]}\ce{[Br-]}\label {d}.\] This only shows that it be a valid mechanism, not that the reaction actually follows such a pathway. Therefore, while we can use collision theory to derive the law of mass action as a first approximation, the exponents for non-elementary reactions are determined via experiment. | 5,644 | 1,996 |
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The Michaelis –Menten model of enzyme kinetics was derived for single substrate reactions. Enzymatic reactions requiring multiple substrates and yielding multiple products are more common and yielding multiple products are more common than single-substrate reaction. In these types of reactions, the all the substrates involved are bound to the enzyme before catalysis of the reaction takes place to release the products. Sequential reactions can be either ordered or random. In contrast to the Michealis-Menton kinetics where a binary Enzyme-Substrate complex is generated in the mechanism (\([ES]\), in bisubstrate enzyme reactions, a ternary complex of the enzyme and two substrates is generated: \[ A + B \overset{E}{\longrightleftharpoons} P+ Q\] Bisubstrate reactions account for ~ 60% of the known enzymatic reactions. Multi-substrate reactions follow complex rate equations that describe how the substrates bind and in what sequence. The analysis of these reactions is much simpler if the concentration of substrate \(A\) is kept constant and substrate \(B\) varied. Under these conditions, the enzyme behaves just like a single-substrate enzyme and a plot of \(v\) by \([S]\) gives apparent \(K_M\) and \(V_{max}\) constants for substrate B. If a set of these measurements is performed at different fixed concentrations of A, these data can be used to work out what the mechanism of the reaction is. For an enzyme that takes two substrates A and B and turns them into two products P and Q, there are two types of mechanism: ternary complex and ping–pong. How do you resolve the enzymes kinetics of these more complicated systems? The answer is fairly straightforward. You keep one of the substrates (B, for example) fixed, and vary the other substrate (A) and obtain a of hyperbolic plots of \(v_o\) vs \(A\) at different fixed \(B\) concentrations. This would give a series of linear \(1/v\) vs. \(1/A\) double-reciprocal plots ( ) as well. The pattern of Lineweaver-Burk plots depends on how the reactants and products interact with the enzyme. In this mechanism, both substrates must bind to the enzyme before any products are made and released.The substrates might bind to the enzyme in a random fashion (A first then B or vice-versa) or in an ordered fashion (A first followed by B). An abbreviated notation scheme developed by W.W. Cleland is shown in Figure \(\Page {1}\) for the sequential random and sequential ordered mechanisms (Figure \(\Page {1}\)). For both mechanisms, Lineweaver-Burk plots at varying A and different fixed values of B give a series of intersecting lines. Derivative curves can be solved to obtain appropriate kinetic constants. In ordered sequential reactions, all the substrates are first bound to the enzyme in a . The products, too, are released after catalysis in a defined order or sequence. An example is the lactate dehydrogenase enzyme, which is a protein that catalyzes glucose metabolism. In this ordered mechanism, the coenzyme, NADH, always binds first, with pyruvate binding afterward. During the reaction, the pyruvate is reduced to lactate while NADH is oxidized to NAD by the enzyme. Lactate is then released first, followed by the release of NAD . This is a characteristic of a ternary complex, which consists of three molecules that are bound together. Before catalysis, the substrates and coenzyme are bound to the enzyme. After catalysis, the complex consists of the enzyme and products, NAD and lactate. In random sequential reactions, the substrates and products are bound and then released in no preferred order, or "random" order (Figure \(\Page {1}\)). An example is the creatine kinase enzyme, which catalyzes creatine and ATP (the two substrates) to form phosphocreatine and ADP (teh Products) in Figure \(\Page {4}\). In this case, substrates may bind first and either products may be released first. A ternary complex is still observed for random sequential reactions. Before catalysis, the complex is generated that includes the enzyme, ATP, and creatine. After catalysis, the complex consists of the enzyme, ADP, and phosphocreatine. In this mechanism, one substrate bind first to the enzyme followed by product P release. Typically, product P is a fragment of the original substrate A.The rest of the substrate is covalently attached to the enzyme E, which is designated as E'. Now the second reactant, B, binds and reacts with the enzyme to form a covalent adduct with the A as it is covalentattached to the enzyme to form product Q. This is now released and the enzyme is restored to its initial form, E. This represents a ping-pong mechanism. An abbreviated notation scheme is shown below for the ping-pong mechanisms. For this mechanism, Lineweaver-Burk plots at varying A and different fixed values of B give a series of parallel lines. An example of this type of reaction might be low molecular weight protein tyrosine phosphatase against the small substrate p-initrophenylphosphate (A) which binds to the enzyme covalently with the expulsion of the product P, the p-nitrophenol leaving group. Water (B) then comes in and covalently attacks the enzyme, forming an adduct with the covalently bound phosphate releasing it as inorganic phosphate. In this particular example, however, you cannot vary the water concentration and it would be impossible to generate the parallel plots characteristic of ping-pong kinetics. | 5,424 | 1,999 |
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It should be clear by now that chemical kinetics is governed by the mathematics of systems of differential equations. Thus far, we have only looked at reaction systems that give rise to purely differential equations, however, in many instances the rate equations are nonlinear. When the differential equations are nonlinear, the behavior is considerably more complex. In particular, nonlinear equations can lead to oscillatory solutions and can also exhibit the phenomenon of . Chaotic systems are systems that are highly sensitive to small changes in the parameters of the equations or in the initial conditions. Basically, this means that the behavior of a chaotic system can be unpredictable, since such small changes can occur in the form of small errors in determining the parameters (rounding to the nearest tenth or hundredth) or in specifying the initial conditions, and these small changes can cause the system to evolve in time in a very different way. The iodine clock reaction is a popular chemistry experiment in which one can visualize how different rate constants in consecutive reactions affect the concentration of species during the reaction. Iodine anions (\(I^-\)) are colorless. When \(I^-\) is reacted with hydrogen peroxide and protons, triiodide is formed, which has a dark blue color. Consider the following series of irreversible reactions: \[H_2 O_2 + 3 \: I^- + 2 \: H^+ \overset{k_1}{\longrightarrow} I_3^- + 2 \: H_2O\] \[I_3^- + 2 \: S_2 O_3^{2-} \overset{k_2}{\longrightarrow} 3 \: I^- + S_4 O_6^{2-}\] The rate laws for this system are \[\begin{align} \dfrac{ d \left[ I_3^- \right]}{dt} &= k_1 \left[ I^- \right]^3 - k_2 \left[ I_3^- \right] \left[ S_2 O_3^{2-} \right]^2 \\ \dfrac{d \left[ I^- \right]}{dt} &= -k_1 \left[ I^- \right]^3 + 3k_2 \left[ I_3^- \right] \left[ S_2 O_3^{2-} \right]^2 \\ \dfrac{d \left[ S_2 O_3^{2-} \right]}{dt} &= -k_2 \left[ I_3^- \right] \left[ S_2 O_3^{2-} \right]^2 \end{align} \label{Eq2}\] In order to make the equations look a little simpler, let us introduce the variables: \[x = \left[ I^- \right], \: \: \: \: \: \: \: y = \left[ I_3^- \right], \: \: \: \: \: \: \: z = \left[ S_2 O_3^{2-} \right] \label{Eq3}\] In terms of these, the rate equations are \[\begin{align} \dfrac{dx}{dt} &= -k_1 x^3 + 3 k_2 yz^2 \\ \dfrac{dy}{dt} &= k_1 x^3 - k_2 yz^2 \\ \dfrac{dz}{dt} &= -k_2 yz^2 \end{align} \label{Eq4}\] If we solve these numerically, we find the following time dependence of the three concentrations: This is a clear example of nonlinearity. Note how the concentration of \(I_3^-\) remains close to \(0\) for a period of time and then suddenly starts to increase. In a sense, think of the “straw that broke the camel’s back”. As we pile straws on the back of the camel, the camel remains upright until that last straw, which suddenly breaks the back of the camel, and the camel suddenly falls to the ground. This is also an illustration of nonlinearity. Despite the complexity of the rate equations, we can still analyze the approximately and predict the behavior seen in Figure \(\Page {1}\). In this reaction mechanism, \(k_2 \gg k_1\). Given the rate law for \(I_3^-\), \[\dfrac{d \left[ I_3^- \right]}{dt} = k_1 \left[ I^- \right]^3 - k_2 \left[ I_3^- \right] \left[ S_2 O_3^{2-} \right]^2 \label{Eq5}\] if we use the steady-state approximation, we can set the \(\dfrac{d \left[ I_3^- \right]}{dt}\) equal to \(0\), yielding \[\left[ I_3^- \right] = \dfrac{k_1}{k_2} \dfrac{\left[ I^- \right]^3}{\left[ S_2 O_3^{2-} \right]^2} \label{Eq6}\] Since \(k_2 \gg k_1\), the concentration of \(\left[ I_3^- \right]\) is approximately \(0\) as long as there are \(S_2 O_3^{2-}\) ions present. As soon as all of the \(S_2 O_3^{2-}\) is consumed, the concentration of \(I_3^-\) can build up in the solution, changing the solution to a dark blue color. Figure \(\Page {1}\) displays the concentration profiles for \(I^-\), \(I_3^-\), and \(S_3 O_3^{2-}\). As can be seen from the figure, the concentration of \(I_3^-\) (red line) remains at approximately \(0 \: \text{mol/L}\) until all of the \(S_2 O_3^{2-}\) (blue line) has been depleted. In all of the examples we have seen thus far, the concentration of intermediate species displays a single maximum during the course of the reaction. There is another class of reactions called in which the concentration of intermediate species oscillates with time. Consider the following series of reactions \[\text{A} + \text{Y} \overset{k_1}{\longrightarrow} \text{X}\] \[\text{X} + \text{Y} \overset{k_2}{\longrightarrow} \text{P}\] \[\text{B} + \text{X} \overset{k_3}{\longrightarrow} 2 \: \text{X} + \text{Z}\] \[2 \: \text{X} \overset{k_4}{\longrightarrow} \text{Q}\] \[\text{Z} \overset{k_5}{\longrightarrow} \text{Y}\] In the above reaction mechanism, \(\text{A}\) and \(\text{B}\) are reactants; \(\text{X}\), \(\text{Y}\), and \(\text{Z}\) are intermediates; and \(\text{P}\) and \(\text{Q}\) are products. The third reaction in which \(\text{B}\) and \(\text{X}\) react to form \(\text{X}\) and \(\text{Z}\) is known as an ”autocatalytic reaction” in which at least one of the reactants is also a product. Such reactions are a key feature of oscillating reactions, as will be discussed below. Let us assume that the concentrations of \(\text{A}\) and \(\text{B}\) are large, such that we can approximate them to be constant with time. The rate equation for species \(\text{X}\) can be written as \[\dfrac{d \left[ \text{X} \right]}{dt} = k_1 \left[ \text{A} \right] \left[ \text{Y} \right] - k_2 \left[ \text{X} \right] \left[ \text{Y} \right] + k_3 \left[ \text{B} \right] \left[ \text{X} \right] - 2k_4 \left[ \text{X} \right]^2 \label{Eq7}\] Using the steady-state approximation, we can set \(d \text{X}/dt = 0\) and rewrite Equation \(\ref{Eq7}\) as \[\left( -2 k_4 \right) \left[ \text{X} \right]^2 + \left( k_2 \left[ \text{Y} \right] - k_3 \left[ \text{B} \right] \right) \left[ \text{X} \right] + k_1 \left[ \text{A} \right] \left[ \text{Y} \right] = 0 \label{Eq8}\] We can then use the quadratic formula to solve for \(\text{X}\): \[\left[ \text{X} \right] = -\dfrac{ \left( k_2 \left[ \text{Y} \right] - k_3 \left[ \text{B} \right] \right) \pm \sqrt{\left( k_2 \left[ \text{Y} \right] - k_3 \left[ \text{B} \right] \right)^2 - 4 \left( -2k_4 \right) \left( k_1 \left[ \text{A} \right] \left[ \text{Y} \right] \right)}}{2 \left( -2k_4 \right)} \label{Eq9}\] Thus, there are two solutions for the concentration of \(\text{X}\) accessible to the reaction system. To examine solutions for \(\left[ \text{X} \right]\), let us first assume that \(\left[ \text{Y} \right]\) is large. Under these conditions, the first two reactions in the reaction mechanism largely determine the concentration of \(\left[ \text{X} \right]\). We can thus approximate Equation \(\ref{Eq7}\) as \[0 \approx k_1 \left[ \text{A} \right] \left[ \text{Y} \right] - k_2 \left[ \text{X} \right] \left[ \text{Y} \right] \label{Eq10}\] Solving for \(\left[ \text{X} \right]\) yields \[\left[ \text{X} \right] \approx \dfrac{k_1 \left[ \text{A} \right]}{k_2} \label{Eq11}\] As the reaction continues, species \(\text{Y}\) is depleted and the assumption that \(\left[ \text{Y} \right]\) is large becomes invalid. Instead the \(3^{rd}\) and \(4^{th}\) steps of the reaction mechanism determine the concentration of \(\text{X}\). In this limit, we can approximate Equation \(\ref{Eq7}\) as \[0 \approx k_3 \left[ \text{B} \right] \left[ \text{X} \right] - 2 k_4 \left[ \text{X} \right]^2 \label{Eq12}\] Solving for \(\left[ \text{X} \right]\) yields \[\left[ \text{X} \right] \approx \dfrac{k_3 \left[ \text{B} \right]}{2k_4} \label{Eq13}\] In the second mechanism, the autocatalytic reaction step leads to an increase in the concentration of \(\text{X}\) and \(\text{Z}\), which in turn leads to an increase in the concentration of \(\text{Y}\). The feedback loop between the production of species \(\text{X}\) and \(\text{Y}\) leads to oscillatory behavior in the system. This reaction mechanism is known as the Belousov-Zhabotinksii reaction first discovered in the 1950s. The actual Belousov-Zhabotinskii reaction is complex, involving many individual steps, and involves the oscillation between the concentration of \(HBrO_2\) and \(Br^-\). The reaction equations are \[\begin{align} BrO_3^- + Br^- + 2 \: H^+ &\longrightarrow HBrO_2 + HOBr \\ HBrO_2 + Br^- + H^+ &\longrightarrow 2 \: HOBr \\ HOBr + Br^- + H^+ &\longrightarrow Br_2 + H_2 O \\ 2 \: HBrO_2 &\longrightarrow BrO_3^- + HOBr + H^+ \\ BrO_3^- + HBrO_2 + H^+ &\longrightarrow 2 \: BrO_2^. + H_2 O \\ BrO_2^. + Ce^{3+} + H^+ &\longrightarrow HBrO_2 + Ce^{4+} \\ BrO_2^. + Ce^{4+} + H_2 O &\longrightarrow BrO_3^- + Ce^{3+} + 2 \: H^+ \end{align} \label{Eq14}\] The essential steps in this mechanism can be reduced to the following set of reactions. Note that we leave this unbalanced and only include the species whose concentrations as functions of time we seek. \[\begin{align} BrO_3^- + Br^- &\overset{k_1}{\longrightarrow} HBrO_2 + HOBr \\ HBrO_2 + Br^- &\overset{k_2}{\longrightarrow} 2 \: HOBr \\ BrO_3^- + HBrO_2 &\overset{k_3}{\longrightarrow} 2 \: HBrO_2 + 2 \: Ce^{4+} \\ 2 \: HBrO_2 &\overset{k_4}{\longrightarrow} BrO_3^- + HOBr \\ Ce^{4+} &\overset{k_5}{\longrightarrow} fBr^- \end{align} \label{Eq15}\] Setting the variables as follows: \[x = \left[ HBrO_2 \right], \: \: \: \: \: \: \: y = \left[ Br^- \right], \: \: \: \: \: \: \: z = \left[ Ce^{4+} \right] \label{Eq16}\] We make the approximation that \(\left[ BrO_3^- \right]\) to be a constant \(a\). In this case, the rate equations become \[\begin{align} \dfrac{dx}{dt} &= k_1 ay - k_2 xy + k_3 ax - k_4 x^2 \\ \dfrac{dy}{dt} &= -k_1 ay - k_2 xy + f k_5 z \\ \dfrac{dz}{dt} &= 2k_3 ax - k_5 z \end{align} \label{Eq17}\] Solving these equations numerically, we obtain the trajectory of two of the species show in the Figure \(\Page {2}\). On the other hand, we can drive this system to become chaotic by changing the parameters a little. When this is done, we find the follow plot of the concentration of \(x\): ( ) | 10,072 | 2,000 |
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Although most of the known elements have at least one isotope whose atomic nucleus is stable indefinitely, all elements have isotopes that are unstable and disintegrate, or , at measurable rates by emitting radiation. Some elements have no stable isotopes and eventually decay to other elements. In contrast to the chemical reactions that were the main focus of earlier chapters and are due to changes in the arrangements of the valence electrons of atoms, the process of nuclear decay results in changes inside an atomic nucleus. We begin our discussion of nuclear reactions by reviewing the conventions used to describe the components of the nucleus. As you learned in , each element can be represented by the notation where , the mass number, is the sum of the number of protons and the number of neutrons, and , the atomic number, is the number of protons. The protons and neutrons that make up the nucleus of an atom are called nucleons , and an atom with a particular number of protons and neutrons is called a nuclide . Nuclides with the same number of protons but different numbers of neutrons are called . Isotopes can also be represented by an alternative notation that uses the name of the element followed by the mass number, such as carbon-12. The stable isotopes of oxygen, for example, can be represented in any of the following ways: \[ \begin{matrix}
_{Z}^{A}X: & _{8}^{16}O & _{8}^{17}O & _{8}^{18}O\\
^{A}X: & ^{16}O &^{17}O & ^{18}O\\
element -A & oxygen-16 & oxygen-17 & oxygen-18
\end{matrix} \notag \] Because the number of neutrons is equal to − , we see that the first isotope of oxygen has 8 neutrons, the second isotope 9 neutrons, and the third isotope 10 neutrons. Isotopes of all naturally occurring elements on Earth are present in nearly fixed proportions, with each proportion constituting an isotope’s . For example, in a typical terrestrial sample of oxygen, 99.76% of the O atoms is oxygen-16, 0.20% is oxygen-18, and 0.04% is oxygen-17. Any nucleus that is unstable and decays spontaneously is said to be radioactive , emitting subatomic particles and electromagnetic radiation. The emissions are collectively called and can be measured. Isotopes that emit radiation are called radioisotopes As you learned in , the rate at which radioactive decay occurs is characteristic of the isotope and is generally reported as a ( ), the amount of time required for half of the initial number of nuclei present to decay in a first-order reaction. (For more information on half-life, see .) An isotope’s half-life can range from fractions of a second to billions of years and, among other applications, can be used to measure the age of ancient objects. Example 1 and its corresponding exercise review the calculations involving radioactive decay rates and half-lives. Fort Rock Cave in Oregon is the site where archaeologists discovered several Indian sandals, the oldest ever found in Oregon. Analysis of the C content of the sagebrush used to make the sandals gave an average decay rate of 5.1 disintegrations per minute (dpm) per gram of carbon. The current C/ C ratio in living organisms is 1.3 × 10 , with a decay rate of 15 dpm/g C. How long ago was the sagebrush in the sandals cut? The half-life of C is 5730 yr. radioisotope, current C/ C ratio, initial decay rate, final decay rate, and half-life age Use to calculate / , the ratio of the number of atoms of C originally present in the sample to the number of atoms now present. Substitute the value for the half-life of C into to obtain the rate constant for the reaction. Substitute the calculated values for / and the rate constant into to obtain the elapsed time . We can use the integrated rate law for a first-order nuclear reaction ( to calculate the amount of time that has passed since the sagebrush was cut to make the sandals: \[ ln \dfrac{N}{N_{o}}=-kt \] From , we know that = . We can therefore use the initial and final activities ( = 15 and = 5.1) to calculate / : \[ \dfrac{A_{o}}{A}=\dfrac{kN_{o}}{kN}=\dfrac{N_{o}}{N}=dfrac{1.5}{5.1} \] Now we can calculate the rate constant from the half-life of the reaction (5730 yr) using : \[ t_{1/2}=\dfrac{0.693}{k} \] Rearranging this equation to solve for , \[ k =\dfrac{0.693}{t_{1/2}}=\dfrac{0.693}{5730 \; yr}=1.21\times 10^{-4} \; yr^{-1} \] Substituting the calculated values into the equation for , \[ t =\dfrac{ln\left (N/N_{o} \right )}{k}=\dfrac{ln\left (15/5.1 \right )}{1.21\times 10^{-4} \; yr^{-1}} = 8900 \; yr \] Thus the sagebrush in the sandals is about 8900 yr old. Exercise While trying to find a suitable way to protect his own burial chamber, the ancient Egyptian pharaoh Sneferu developed the pyramid, a burial structure that protected desert graves from thieves and exposure to wind. Analysis of the C content of several items in pyramids built during his reign gave an average decay rate of 8.6 dpm/g C. When were the objects in the chamber created? about 4600 yr ago, or about 2600 BC As discussed in , the nucleus of an atom occupies a tiny fraction of the volume of an atom and contains the number of protons and neutrons that is characteristic of a given isotope. Electrostatic repulsions would normally cause the positively charged protons to repel each other, but the nucleus does not fly apart because of the strong nuclear force , an extremely powerful but very short-range attractive force between nucleons ( ). All stable nuclei except the hydrogen-1 nucleus ( H) contain at least one neutron to overcome the electrostatic repulsion between protons. As the number of protons in the nucleus increases, the number of neutrons needed for a stable nucleus increases even more rapidly. Too many protons (or too few neutrons) in the nucleus result in an imbalance between forces, which leads to nuclear instability. The relationship between the number of protons and the number of neutrons in stable nuclei, arbitrarily defined as having a half-life longer than 10 times the age of Earth, is shown graphically in . The stable isotopes form a “peninsula of stability” in a “sea of instability.” Only two stable isotopes, H and He, have a neutron-to-proton ratio less than 1. Several stable isotopes of light atoms have a neutron-to-proton ratio equal to 1 (e.g., He, B, Ca. All other stable nuclei have a higher neutron-to-proton ratio, which increases steadily to about 1.5 for the heaviest nuclei. Regardless of the number of neutrons, however, all elements with > 83 are unstable and radioactive. Data source: National Nuclear Data Center, Brookhaven National Laboratory, Evaluated Nuclear Structure Data File (ENSDF), Chart of Nuclides, . As shown in , more than half of the stable nuclei (166 out of 279) have numbers of both neutrons and protons; only 6 of the 279 stable nuclei do not have odd numbers of both. Moreover, certain numbers of neutrons or protons result in especially stable nuclei; these are the so-called 2, 8, 20, 50, 82, and 126. For example, tin ( = 50) has 10 stable isotopes, but the elements on either side of tin in the periodic table, indium ( = 49) and antimony ( = 51), have only 2 stable isotopes each. Nuclei with magic numbers of protons neutrons are said to be “doubly magic” and are even more stable. Examples of elements with doubly magic nuclei are with 2 protons and 2 neutrons, and with 82 protons and 126 neutrons, which is the heaviest known stable isotope of any element. The pattern of stability suggested by the magic numbers of nucleons is reminiscent of the stability associated with the closed-shell electron configurations of the noble gases in group 18 and has led to the hypothesis that the nucleus contains shells of nucleons that are in some ways analogous to the shells occupied by electrons in an atom. As shown in , the “peninsula” of stable isotopes is surrounded by a “reef” of radioactive isotopes, which are stable enough to exist for varying lengths of time before they eventually decay to produce other nuclei. Classify each nuclide as stable or radioactive. mass number and atomic number predicted nuclear stability Use the number of protons, the neutron-to-proton ratio, and the presence of even or odd numbers of neutrons and protons to predict the stability or radioactivity of each nuclide. Exercise Classify each nuclide as stable or radioactive. In addition to the “peninsula of stability,” shows a small “island of stability” that is predicted to exist in the upper right corner. This island corresponds to the superheavy elements , with atomic numbers near the magic number 126. Because the next magic number for neutrons should be 184, it was suggested that an element with 114 protons and 184 neutrons might be stable enough to exist in nature. Although these claims were met with skepticism for many years, since 1999 a few atoms of isotopes with = 114 and = 116 have been prepared and found to be surprisingly stable. One isotope of element 114 lasts 2.7 seconds before decaying, described as an “eternity” by nuclear chemists. Moreover, there is recent evidence for the existence of a nucleus with = 292 that was found in Th. With an estimated half-life greater than 10 years, the isotope is particularly stable. Its measured mass is consistent with predictions for the mass of an isotope with = 122. Thus a number of relatively long-lived nuclei may well be accessible among the superheavy elements. Subatomic particles of the nucleus (protons and neutrons) are called . A is an atom with a particular number of protons and neutrons. An unstable nucleus that decays spontaneously is , and its emissions are collectively called . Isotopes that emit radiation are called . Each nucleon is attracted to other nucleons by the . Stable nuclei generally have even numbers of both protons and neutrons and a neutron-to-proton ratio of at least 1. Nuclei that contain of protons and neutrons are often especially stable. , with atomic numbers near 126, may even be stable enough to exist in nature. What distinguishes a nuclear reaction from a chemical reaction? Use an example of each to illustrate the differences. What do chemists mean when they say a substance is ? What characterizes an isotope? How is the mass of an isotope of an element related to the atomic mass of the element shown in the periodic table? In a typical nucleus, why does electrostatic repulsion between protons not destabilize the nucleus? How does the neutron-to-proton ratio affect the stability of an isotope? Why are all isotopes with > 83 unstable? What is the significance of a of protons or neutrons? What is the relationship between the number of stable isotopes of an element and whether the element has a magic number of protons? Do you expect Bi to have a large number of stable isotopes? Ca? Explain your answers. Potassium has three common isotopes, K, K, and K, but only potassium-40 is radioactive (a beta emitter). Suggest a reason for the instability of K. Samarium has 11 relatively stable isotopes, but only 4 are nonradioactive. One of these 4 isotopes is Sm, which has a lower neutron-to-proton ratio than lighter, radioactive isotopes of samarium. Why is Sm more stable? Isotopes with magic numbers of protons and/or neutrons tend to be especially stable. Elements with magic numbers of protons tend to have more stable isotopes than elements that do not. Potassium-40 has 19 protons and 21 neutrons. Nuclei with odd numbers of both protons and neutrons tend to be unstable. In addition, the neutron-to-proton ratio is very low for an element with this mass, which decreases nuclear stability. Write the nuclear symbol for each isotope using notation. Write the nuclear symbol for each isotope using \( _{Z}^{A}X \) notation. Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope. Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope. Which of these nuclides do you expect to be radioactive? Explain your reasoning. Which of these nuclides do you expect to be radioactive? Explain your reasoning. | 12,169 | 2,001 |
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When molecules go through a mass spectrometer, some of them arrive intact at the detector, but many of them break into pieces in a variety of different ways. To establish a charge on a molecule, an electron had to be removed; removal of that electron is effected through a collision, usually with a high-energy electron. During that collision, energy is transferred from the high-energy electron to the molecule, and that energy has to go somewhere. Part of it gets partitioned into various bond vibrations, so bonds start to vibrate quite a lot, until some of them snap completely. The molecular ion breaks apart and forms a fragment ion. Some fragment ions are very common in mass spectrometry. These ions are seen frequently for either of two reasons: Fragmentations occur through well-defined pathways or mechanisms. A mechanism is a step-by-step series of events that happens in a reaction. It is important to understand how reactions happen, but we will look at fragmentations when we study radical reactions. However, it is useful to know what factors make cations stable. There are a number of ions commonly seen in mass spectrometry that tell you a little bit about the structure. Just like with anions, there are a couple of common factors influence cation stability: However, in most cases, we will be looking at carbon with a positive charge, and there are additional factors to distinguish between them Molecular orbital calculations suggest that the cation is stabilized through interaction with neighboring C-H bonds in the alkyl groups. Specifically, a C-H sigma bonding orbital has symmetry similar to the empty p orbital on the positive carbon. The lobes on the two orbitals can overlap such that they are in phase, and that allows electrons to be donated from the C-H bond to the central, electron-deficient carbon. Formally, there is a bonding interaction and an antibonding interaction between these two orbitals. Since one of these orbitals is empty, the antibonding combination remains unoccupied. The bonding combination is populated, however, and since it is lower in energy than either the p orbital or the C-H sigma bond (all bonding combinations are lower in energy than the orbitals that combine to form them), there is a net decrease in energy. Draw as many resonance structures as you can that help explain teh stability of the following cations: a) allyl cation b) benzyl cation c) tropylium cation d) an acylium ion e) an iminium ion , | 2,493 | 2,002 |
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All hydrocarbons are attacked by oxygen at elevated temperatures and, if oxygen is in excess, complete combustion occurs to carbon dioxide and water: \[ CH_4 + 2O_2 \rightarrow CO_ 2 + 2H_2O\] The heat evolved in this process - the heat of the combustion reaction, \(\Delta H\) - is a measure of the amount of energy stored in the \(C-C\) and \(C-H\) bonds of the hydrocarbon compared to the energy stored in the products, carbon dioxide and water. It can be measured experimentally with considerable accuracy and generally is reported as \(\Delta H^\text{0}\) the amount of heat (in kilocalories)\(^2\) liberated on complete combustion of one mole of hydrocarbon when the reactants and the products are in standard states, and at the same temperature, usually \(25^\text{o}\).\(^3\) Not all chemical reactions that occur spontaneously liberate heat - some actually absorb heat. By convention, \(\Delta H^\text{0}\) is given a sign when heat is evolved ( ) and a sign when heat is absorbed ( ). The heat evolved or absorbed also is called the change. For the combustion of \(1 \: \text{mol}\) of methane at \(15^\text{o}\), we find by experiment (corrected from constant volume to constant pressure, if necessary) that the reaction is exothermic by \(212.8 \: \text{kcal}\). This statement can be expressed as follows: \[ CH_4(g) + 2O_2(g) \rightarrow CO_2 (g) + 2H_2O (l)\] with \(\Delta H^o = -212.8 \, kcal\). The symbol \(\left( g \right)\) denotes that the reactants and products are in the gaseous state except for the water, which is liquid \(\left( l \right)\). If we wish to have \(\Delta H^\text{0}\) with gaseous water \(H_2O \: \left( g \right)\) as the product we have to make a correction for the heat of vaporization of water (\(10.5 \: \text{kcal} \: \text{mol}^{-1}\) at \(25^\text{o}\)): The task of measuring the heats of all chemical reactions is a formidable one and about as practical as counting grains of sand on the beach. However, it is of practical interest to be able to estimate heats of reaction, and this can be done quite simply with the aid of bond energies. The necessary bond energies are given in Table 4-3, and it is important to notice that they apply only to \(25^\text{o}C\). Also, they do not apply, without suitable corrections, to many compounds, such as benzene, that have more than one double bond. This limitation will be discussed in Chapters 6 and 21. To calculate \(\Delta H^\text{0}\) for the combustion of one mole of methane, first we break bonds as follows, using \(98.7 \: \text{kcal} \: \text{mol}^{-1}\) for the energy of each of the \(C-H\) bonds, and then \(118.9 \: \text{kcal} \: \text{mol}^{-1}\) for the energy of the double bond in oxygen: Then we make bonds, using \(192 \: \text{kcal} \: \text{mol}^{-1}\) for each \(O=C\) bond in carbon dioxide, and \(110.6 \: \text{kcal} \: \text{mol}^{-1}\) for each of the \(H-O\) bonds in water: The net sum of these \(\Delta H^\text{0}\) values is \(394.8 + 237.8 - 384.0 - 442.4 = -193.8 \: \text{kcal}\), which is reasonably close to the value of \(-191.8 \: \text{kcal}\) for the heat of combustion of one mole of methane determined experimentally. The same type of procedure can be used to estimate \(\Delta H^\text{0}\) values for many other kinds of reactions of organic compounds in the vapor phase at \(25^\text{o}\). Moreover, if appropriate heats of vaporization are available, it is straightforward to compute \(\Delta H^\text{0}\) for vapor-phase reactions of substances which are normally liquids or solids at \(25^\text{o}\). The special problems that arise when solutions and ionic substances are involved are considered in Chapters 8 and 11. It is important to recognize that the bond energies listed in Table 4-3 for all molecules other than diatomic molecules are values. That the \(C-H\) bond energy is stated to be \(98.7 \: \text{kcal}\) does not mean that, if the hydrogens of methane were detached one by one, \(98.7 \: \text{kcal}\) would have to be put in at each step. Actually, the experimental evidence is in accord with quite different energies for the separate dissociation steps: The moral is that we should try to avoid using the bond energies in Table 4-3 as a measure of \(\Delta H^\text{0}\) for the dissociation of just one bond in a polyatomic molecule. For this we need what are called , some of which are given in Table 4-6. The values given have been selected to emphasize how structure influences bond energy. Thus, \(C-H\) bond energies in alkanes decrease in the order primary \(>\) secondary \(>\) tertiary; likewise, \(C-H\) bonds decrease in strength along the series \(C \equiv C-H \: > \: C=C-H \: > \: C-C-H\). How accurate are \(\Delta H^\text{0}\) values calculated from bond energies? Generally quite good provided nonbonded interactions between the atoms are small and the bond angles and distances are close to the normal values ( ). A few examples of calculated and experimental heats of combustion of some hydrocarbons are given in Table 4-4. Negative discrepancies represent heats of combustion than expected from the average bond energies and positive values correspond to than expected heats of combustion. Comparing isomers in Table 4-4, we see that 2-methylpropane and 2,2,3,3-tetramethylbutane give off less heat when burned than do butane and octane, and this is a rather general characteristic result of chain branching. Cyclopropane has a \(\Delta H^\text{0}\) of combustion of \(27.7 \: \text{kcal} \: \text{mol}^{-1}\) than expected from bond energies, and this clearly is associated with the abnormal \(C-C-C\) bond angles in the ring. These matters will be discussed in detail in Chapter 12. For cyclohexane, which has normal bond angles, the heat of combustion is close to the calculated value. \(^2\)In this book we use kilocalories in place of the presently recommended (SI) joules for units of energy. As of the date of writing, it is not clear just how general the use of the joule will become among chemists. To convert calories to joules (or \(\text{kcal}\) to \(\text{kJ}\)), multiply by 4.184. \(^3\)You may wonder how a reaction, such as combustion of methane, can occur at \(25^\text{o}\). The fact is that the reaction can be carried out at any desired temperature. The important thing is that the \(\Delta H^\text{0}\) value we are talking about here is the heat liberated or absorbed when you start with the reactants at \(25^\text{o}\) and finish with the products at \(25^\text{o}\). As long as \(\Delta H^\text{0}\) is defined this way, it does not matter at what temperature the reaction actually occurs. Standard states for gases are \(1 \: \text{atm}\) partial pressure. Standard states for liquids or solids usually are the pure liquid or solid at \(1 \: \text{atm}\) external pressure. and (1977) | 6,827 | 2,003 |
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Make sure you thoroughly understand the following essential ideas: The kinetics fundamentals we covered in the earlier sections of this lesson group relate to processes that take place in the gas phase. But chemists and biochemists are generally much more concerned with solutions. This lesson will take you through some of the extensions of basic kinetics that you need in order to understand the major changes that occur when reactions take place in liquid solutions. Most of the added complications of kinetics and rate processes in liquid solutions arise from the of the liquid phase. In a typical gas at atmospheric pressure, the molecules occupy only about 0.2 per cent of the volume; the other 99.8 percent is empty space. In a liquid, molecules may take up more than half the volume, and the "empty" spaces are irregular and ever-changing as the solvent molecules undergo thermal motions of their own. In a typical liquid solution, the solvent molecules massively outnumber the reactant solute molecules, which tend to find themselves momentarily (~10 sec) confined to a "hole" within the liquid. This trapping becomes especially important when the solvent is strongly hydrogen-bonded as is the case with water or alcohol. When thermal motions occasionally release a solute molecule from this trap, it will jump to a new location. The jumps are very fast (10 - 10 sec) and short (usually a few solvent-molecule diameters), and follow an entirely random pattern, very much as in . Consider a simple bimolecular process A + B → products. The reactant molecules will generally be jumping from hole to hole in the solvent matrix, only occasionally finding themselves in the same where thermal motions are likely to bring them into contact. The process can be represented as \[A + B \rightarrow \{AB\} → \text{products}\] in which the \(\{AB\}\) term represents the caged reactants including the and the activated complex. Contrast this scenario with a similar reaction taking place in the gas phase; the molecules involved in the reaction will often be the only ones present, so a significant proportion of the collisions will be \(A\)-\(B\) encounters. However, if the collision should fail to be energetically or geometrically viable, the reactant molecules fly apart and are unlikely to meet again anytime soon. In a liquid, however, the solute molecules are effectively in a constant state of collision — if not with other reactants, then with solvent molecules which can exchange kinetic energy with the reactants. So once an A-B encounter pair forms, the two reactants get multiple whacks at each other, greatly increasing the probability that they will obtain the kinetic energy needed to kick them over the activation hump before the encounter pair disintegrates. The encounter pair model introduces some new rate parameters: \[\ce{A + B <=>[k1,k_{-1}] {AB} -> products}\] The first step is an equilibrium between the reactants outside and inside the solvent cage. The rate constants \(k_1\) and \(k_2\) reflect those relating to diffusion of molecules through the solvent; their values are strongly dependent on the viscosity (and thus the temperature) of the solvent. (Note that \(k_1\) is a second-order rate constant, while \(k_2\) is first-order.) is the transport of a substance through a concentration gradient; that is, from a region of higher concentration to one of lower concentration. Think of the way the color of tea spreads out when a tea bag is immersed in hot water. Diffusion occurs because random thermal motions are statistically more likely to move molecules out of a region of higher concentration than in the reverse direction, simply because in the latter case fewer molecules are available to make the reverse trip. Eventually the concentrations become uniform and equilibrium is attained. As molecules diffuse through a liquid, they must nudge neighboring molecules out of the way. The work required to do this amounts to an activation energy, so diffusion can be thought of as a kinetic process with its own rate constant and activation energy. These parameters depend on the sizes of the solute and solvent molecules and on how strongly the latter interact with each other. This suggests two important for reactions in solution. For water at room temperature, is typically 10 -10 dm mol s and is around 10 -10 dm mol s . Given these values, > 10 s implies diffusion control, while values < 10 s are indicative of activation control. Several general kinds of reactions are consistently very "fast" and thus are commonly found to be diffusion-controlled in most solvents: Gas-phase rate constants are normally expressed in units of mol s , but rate constants of reactions in solution are conventionally given in mol/L units, or dm mol s . Conversion between them depends on a number of assumptions and is non-trivial. For example the formation of I from I atoms in hexane at 298 K has = 1.3×10 dm mol s . The most famous of these is one of the fastest reactions known:\[H^+ + OH^– → H_2O \nonumber \] for which = 1.4×10 dm mol s at 298 K. Polar solvents such as water and alcohols interact with ions and polar molecules through attractive dipole-dipole and ion-dipole interactions, leading to lower-energy solvated forms which stabilize these species. In this way, a polar solvent can alter both the thermodynamics and kinetics (rate) of a reaction. If the products of the reaction are markedly more or less polar than the reactants, solvent polarity can change the overal thermodynamics (equilibrium constant) of the reaction. Nowhere is this more apparent than when an ionic solid such as salt dissolves in water. The Na and Cl ions are bound together in the solid through strong coulombic forces; pulling the solid apart in a vacuum or in a nonpolar solvent is a highly endothermic process. In contrast, dissolution of NaCl in water is slightly exothermic and proceeds spontaneously. The water facilitates this process in two important ways. First, its high dielectric constant of 80 reduces the force between the separated ions to 1/80 of its normal value. Secondly, the water molecules form a around the ions (lower left), rendering them energetically (thermodynamically) more stable than they were in the NaCl solid. In the same way, a reaction whose mechanism involves the formation of an intermediate or activated complex having a polar or ionic character will have its activation energy, and thus its rate, subject to change as the solvent polarity is altered. As an example we will consider an important class of reactions that you will hear much about if you take a course in organic chemistry. When an aqueous solution of a strong base such as KOH is added to a solution of -butyl chloride in ethanol, the chlorine is replaced by a hydroxyl group, leaving -butyl alcohol as a product: This reaction is one of a large and important class known as that are discussed in most organic chemistry courses. In these reactions, a species that possesses a pair of non-bonding electrons (also called a or ) uses them to form a new bond with an — a compound in which a carbon atom has a partial positive charge owing to its bonds to electron-withdrawing groups. In the example here, other nucleophiles such as NH or even H O would serve as well. In order to reflect the generality of this process and to focus on the major changes that take place, we will represent this reaction as Extensive studies of this class of reactions in the 1930's revealed that it proceeds in two activation energy-controlled steps, followed by a simple dissociation into the products: In step , which is rate-determining, the chlorine leaves the alkyl chloride which becomes an intermediate known as a ("cat-ion"). These ions, in which the central carbon atom lacks a complete octet, are highly reactive, and in step the carbocation is attacked by the hydroxide ion which supplies the missing electron. The immediate product is another cation in which the positive charge is on the oxygen atom. This is unstable and rapidly dissociates ( )into the alcohol and a hydrogen ion. The reaction coordinate diagram helps us understand the effect of solvent polarity on this reaction. Polar solvent molecules interact most strongly with species in which the electric charge is concentrated in one spot. Thus the carbocation is stabilized to a greater degree than are the activated complexes in which the charge is spread out between the positive and negative ends. As the heavy green arrows indicate, a more polar solvent will stabilize the carbocation more than it will either of the activated complexes; the effect is to materially reduce the activation energy of the rate-determining step, and thus speed up the reaction. Because neither the alkyl chloride nor the alcohol is charged, the change in solvent polarity has no effect on the equilibrium constant of the reaction. This is dramatically illustrated by observing the rate of the reaction in solvents composed of ethanol and water in varying amounts: | 9,102 | 2,004 |
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We can address the question of what happens when a molecule becomes adsorbed onto a surface at two levels; specifically we can aim to identify The latter topic is covered in detail in , while this section will consider only the local adsorption geometry and adsorbate structure. Chemisorption, by definition, involves the formation of new chemical bonds between the adsorbed species and the surface atoms of the substrate - basically the same type of bonds that are present in any molecular complex. In considering what type of species may be formed on a metal surface, therefore, it is important not to abandon chemical common sense and, if in doubt, to look for inspiration at the structures of known metal-organic complexes. In the H molecule, the valence electrons are all involved in the H-H σ-bond and there are no additional electrons which may interact with the substrate atoms. Consequently, chemisorption of hydrogen on metals is almost invariably a dissociative process in which the H-H bond is broken, thereby permitting the hydrogen atoms to independently interact with the substrate (see for a description of the energetics of this process). The adsorbed species in this instance therefore are hydrogen atoms. The exact nature of the adsorbed hydrogen atom complex is generally difficult to determine experimentally, and the very small size of the hydrogen atom does mean that migration of hydrogen from the interface into sub-surface layers of the substrate can occur with relative ease on some metals (e.g. Pd, rare earth metals). The possibility of molecular H chemisorption at low temperatures cannot be entirely excluded, however, as demonstrated by the discovery of molecular hydrogen transition metal compounds, such as W(η -H )(CO) (P Pr ) , in which both atoms of the hydrogen molecule are coordinated to a single metal centre. Halogens also chemisorb in a dissociative fashion to give adsorbed halogen atoms. The reasons for this are fairly clear - in principle a halogen molecule could act as a Lewis base and bind to the surface without breakage of the X-X bond, in practice the lone pairs are strongly held by the highly electronegative halogen atom so any such interaction would be very weak and the thermodynamics lie very heavily in favour of dissociative adsorption [ i.e. D(X-X) + D(M-X ) << 2 D(M-X) ]. Clearly the kinetic barrier to dissociation must also be low or non-existent for the dissociative adsorption to occur readily. Another way of looking at the interaction of a halogen molecule with a metal surface is as follows: the significant difference in electronegativity between a typical metal and halogen is such that substantial electron transfer from the metal to halogen is favoured. If a halogen molecule is interacting with a metal surface then this transferred electron density will enter the σ* antibonding orbital of the molecule, thereby weakening the X-X bond. At the same time the build-up of negative charge on the halogen atoms enhances the strength of the metal-halogen interaction. The net result of these two effects when taken to their limit is that the halogen molecule dissociates and the halogen atoms interact with the metal with a strong ionic contribution to the bonding. Halogen atoms tend to occupy high co-ordination sites on the surface - for example, the 3-fold hollow site on (111) surfaces ( ) and the 4-fold hollow site on (100) surfaces ( ). ( ) ( ) This behavior is typical of atomic adsorbates which almost invariably endeavor to maximize their co-ordination and hence prefer to occupy the highest-available co-ordination site on the surface. As a result of the electron transfer from the metal substrate to the halogen atoms, each adsorbed atom is associated with a significant . One consequence of this is that there are repulsive (dipole-dipole) interactions between the adsorbed atoms, which are especially evident at higher surface coverages and which can lead to a substantial reduction in the enthalpy of adsorption at specific coverages (if these coverages mark a watershed, above which the atoms are forced to occupy sites which are much closer together). Another feature of the halogen adsorption chemistry of some metals is the transition from an adsorbed surface layer to surface compound formation at high gas exposures. Oxygen is an example of a molecule which usually adsorb dissociatively, but are also found to adsorb molecularly on some metals (e.g. Ag, Pt). In those cases where both types of adsorption are observed it is the dissociative process that corresponds to the higher adsorption enthalpy. As noted above, in the molecular adsorption state the interaction between the molecule and the surface is relatively weak. Molecules aligned such that the internuclear axis is parallel to the surface plane may bond to a single metal atom of the surface via both Although the interaction of the molecule with the surface is generally weak, one might expect that there might be a substantial barrier to dissociation due to the high strength (and high dissociation enthalpy) of the O=O bond. Nevertheless on most metal surfaces, dissociation of oxygen is observed to be facile which is related to the manner in which the interaction with the surface can mitigate the high intrinsic bond energy (see ) and thereby facilitate dissociation. Once formed, oxygen atoms are strongly bound to the surface and, as noted previously, will tend to occupy the highest available co-ordination site. The strength of the interaction between adsorbate and substrate is such that the adjacent metal atoms are often seen to undergo significant displacements from the equilibrium positions that they occupy on the clean metal surface. This displacement may simply lead to a distortion of the substrate surface in the immediate vicinity of the adsorbed atom (so that, for example, the adjacent metal atoms are drawn in towards the oxygen and the metal-oxygen bond distance is reduced) or to a more extended surface reconstruction (see ) Dissociative oxygen adsorption is frequently irreversible - rather than simply leading to desorption, heating of an adsorbed oxygen overlayer often results in either the gradual removal of oxygen from the surface by diffusion into the bulk of the substrate (e.g. Si(111) or Cu(111)) or to the formation of a surface oxide compound. Even at ambient temperatures, extended oxygen exposure often leads to the nucleation of a surface oxide. Depending on the reactivity of the metal concerned, further exposure at low temperatures may result either in a progressive conversion of the bulk material to oxide or the oxidation process may effectively stop after the formation of a passivating surface oxide film of a specific thickness (e.g. Al). The interaction of nitrogen with metal surfaces shows many of the same characteristics as those described above for oxygen. However, in general N is less susceptible to dissociation as a result of the lower M-N bond strength and the substantial kinetic barrier associated with breaking the N≡N triple bond. Depending upon the metal surface, carbon monoxide may adsorb either in a molecular form or in a dissociative fashion - in some cases both states coexist on particular surface planes and over specific ranges of temperature. Molecularly chemisorbed CO has been found to bond in various ways to single crystal metal surfaces - analogous to its behaviour in isolated metal carbonyl complexes. Terminal ("Linear")
(all surfaces) Bridging ( 2f site )
(all surfaces) Bridging / 3f hollow
( (111) ) Bridging / 4f hollow
(rare- (100) ?) Whilst the above structural diagrams amply demonstrate the inadequacies of a simple valence bond description of the bonding of molecules to surface, they do to an extent also illustrate one of its features and strengths - namely that a given element, in this case carbon, tends to have a specific valence. Consequently, as the number of metal atoms to which the carbon is co-ordinated increases, so there is a corresponding reduction in the C-O bond order. However, it must be emphasised that a molecule such as CO does not necessarily prefer to bind at the highest available co-ordination site. So, for example, the fact that there are 3-fold hollow sites on an (111) surface does not mean that CO will necessarily adopt this site - the preferred site may still be a terminal or 2-fold bridging site, and the site or site(s) which is(are) occupied may change with either surface coverage or temperature. The energy difference between the various adsorption sites available for molecular CO chemisorption appears therefore to be very small. A description of the nature of the bonding in a terminal CO-metal complex, in terms of a simple molecular orbital model, is given in . Ammonia has lone pairs available for bonding on the central nitrogen atom and may bond without dissociation to a single metal atom of a surface, acting as a Lewis base, to give a pseudo-tetrahedral co-ordination for the nitrogen atom. Alternatively, progressive dehydrogenation may occur to give surface NH ( = 2,1,0) species and adsorbed hydrogen atoms, i.e. NH → NH + H → NH + 2 H → N + 3 H As the number of hydrogens bonded to the nitrogen atom is reduced, the adsorbed species will tend to move into a higher co-ordination site on the surface (thereby tending to maintain the valence of nitrogen). Other Group V and Group VI hydrides (e.g. PH , H O, H S) exhibit similar adsorption characteristics to ammonia. Unsaturated hydrocarbons (alkenes, alkynes, aromatic molecules etc.) all tend to interact fairly strongly with metal atom surfaces. At low temperatures (and on less reactive metal surfaces) the adsorption may be molecular, albeit perhaps with some distortion of bond angles around the carbon atom. Ethene, for example, may bond to give both a π-complex ( ) or a di-σ adsorption complex ( ): ( ) ( ) [ Further examples: ] As the temperature is raised, or even at low temperatures on more reactive surfaces (in particular those that bind hydrogen strongly), a stepwise dehydrogenation may occur. One particularly stable surface intermediate found in the dehydrogenation of ethene is the complex, whose formation also involves H-atom transfer between the carbon atoms. The ultimate product of complete dehydrogenation, and the loss of molecular hydrogen by desorption, is usually either carbidic or graphitic surface carbon. | 10,492 | 2,005 |
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Essential Skills 1 and Essential Skills 2 described some fundamental mathematical operations used for solving problems in chemistry. This section introduces you to base-10 logarithms, a topic with which you must be familiar to do the Questions and Problems for end of Chapter 4. Essential Skills 1 introduced exponential notation, in which a base number is multiplied by itself the number of times indicated in the exponent. The number 10 , for example, is the base 10 multiplied by itself three times (10 × 10 × 10 = 1000). Now suppose that we do not know what the exponent is—that we are given only a base of 10 and the final number. If our answer is 1000, the problem can be expressed as \(10^a = 1000 \) We can determine the value of by using an operation called the , or , abbreviated as , that represents the power to which 10 is raised to give the number to the right of the equals sign. This relationship is stated as log 10 = . In this case, the logarithm is 3 because 10 = 1000: \(log \: 10^3 = 3\) \( log\: 1000 = 3 \) Now suppose you are asked to find when the final number is 659. The problem can be solved as follows (remember that any operation applied to one side of an equality must also be applied to the other side): \(10^a = 659 \) \(log\: 10^a = log\: 659 \) \(a = log\: 659 \) If you enter 659 into your calculator and press the “log” key, you get 2.819, which means that = 2.819 and 10 = 659. Conversely, if you enter the value 2.819 into your calculator and press the “10 ” key, you get 659. You can decide whether your answer is reasonable by comparing it with the results you get when = 2 and = 3: \(a = 2 \textrm : \: 10^2 = 100 \) \(a = 2.819 \textrm : \: 10^{2.819} = 659 \) \(a = 3 \textrm : \: 10^3 = 1000 \) Because the number 659 is between 100 and 1000, must be between 2 and 3, which is indeed the case. Table \(\Page {1}\) lists some base-10 logarithms, their numerical values, and their exponential forms. \(\Page {1}\) Relationships in Base-10 Logarithms Base-10 logarithms may also be expressed as log , in which the base is indicated as a subscript. We can write log 10 = in either of two ways: \(log\: 10^a = a \) \(log_{10} = (10^a) = a \) The second equation explicitly indicates that we are solving for the base-10 logarithm of 10 . The number of significant figures in a logarithmic value is the same as the number of digits the decimal point in its logarithm, so log 62.2, a number with three significant figures, is 1.794, with three significant figures after the decimal point; that is, 10 = 62.2, 62.23. Skill Builder ES1 provides practice converting a value to its exponential form and then calculating its logarithm. Express each number as a power of 10 and then find the common logarithm. Solution Convert each base-10 logarithm to its numerical value. Solution Because logarithms are exponents, the properties of exponents that you learned in Essential Skills 1 apply to logarithms as well, which are summarized in Table \(\Page {1}\). The logarithm of (4.08 × 20.67), for example, can be computed as follows: \(log(4.08 \times 20.67) = log\: 4.08 + log\: 20.67 = 0.611 + 1.3153 = 1.926 \) We can be sure that this answer is correct by checking that 10 is equal to 4.08 × 20.67, and it is. In an alternative approach, we multiply the two values before computing the logarithm: \(4.08 \times 20.67 = 84.3\) \( log\: 84.3 = 1.926 \) We could also have expressed 84.3 as a power of 10 and then calculated the logarithm: \(log\: 84.3 = log(8.43 \times 10) = log\: 8.43 + log\: 10 = 0.926 + 1 = 1.926 \) As you can see, . We can use the properties of exponentials and logarithms to show that the logarithm of the inverse of a number (1/ ) is the negative logarithm of that number (−log ): \( log\left ( \frac{1}{B} \right )=-log\left ( B \right ) \) If we use the formula for division given and recognize that log 1 = 0, then the logarithm of 1/ is \( log\left ( \frac{1}{B} \right )=log\left ( 1 \right )-log\left ( B \right )=-log\left ( B \right ) \) Convert each number to exponential form and then calculate the logarithm (assume all trailing zeros on whole numbers are not significant). Solution log[(1 × 10 )(1 × 10 )] = 2.0 + 3.0 = 5.0 Alternatively, (1 × 10 )(1 × 10 ) = 1 × 10 = 1 × 10 log(1 × 10 ) = 5.0 log[(1.00 × 10 ) ÷ (1 × 10 )] = 1 × 10 = 1 × 10 Alternatively, (1.00 × 10 ) ÷ (1 × 10 ) = 1 × 10 = 1 × 10 log(1 × 10 ) = −3.0 log[(1 × 10 )(1 × 10 )] = 3.0 + (−2.0) = 1.0 Alternatively, (1 × 10 )(1.0 × 10 ) = 1 × 10 = 1 × 10 log(1 × 10 ) = 1.0 log[(2 × 10 )(3 × 10 )] = log(2 × 10 ) + log(3 × 10 ) = (log 2 + log 10 ) + (log 3 + log 10 ) = 0.30 + 2 + 0.48 + 3 = 5.8 Alternatively, (2 × 10 )(3 × 10 ) = 6 × 10 = 6 × 10 log(6 × 10 ) = log 6 + log 10 = 0.78 + 5 = 5.8 log[(2.05 × 10) ÷ (2.6 × 10 )] = (log 2.05 + log 10) − (log 2.6 + log 10 ) = (0.3118 + 1) − [0.415 + (−2)] = 1.3118 + 1.585 = 2.90 Alternatively, (2.05 × 10) ÷ (2.6 × 10 ) = 0.788 × 10 = 0.788 × 10 log(0.79 × 10 ) = log 0.79 + log 10 = −0.102 + 3 = 2.90 Convert each number to exponential form and then calculate its logarithm (assume all trailing zeros on whole numbers are not significant). Solution | 5,188 | 2,007 |
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The most common example of work in the systems discussed in this book is the work of expansion. It is also convenient to use the work of expansion to exemplify the difference between work that is done reversibly and that which is done irreversibly. The example of expansion against a constant external pressure is an example of an irreversible pathway. It does not mean that the gas cannot be re-compressed. It does, however, mean that there is a definite direction of spontaneous change at all points along the expansion. Imagine instead a case where the expansion has no spontaneous direction of change as there is no net force push the gas to seek a larger or smaller volume. The only way this is possible is if the pressure of the expanding gas is the same as the external pressure resisting the expansion at all points along the expansion. With no net force pushing the change in one direction or the other, the change is said to be or to occur . The work of a reversible expansion of an ideal gas is fairly easy to calculate. If the gas expands reversibly, the external pressure (\(p_{ext}\)) can be replaced by a single value (\(p\)) which represents both the pressure of the gas and the external pressure. \[ dw = -pdV \nonumber \] or \[ w = - \int p dV \nonumber \] But now that the external pressure is not constant, \(p\) cannot be extracted from the integral. Fortunately, however, there is a simple relationship that tells us how \(p\) changes with changing \(V\) – ! If the gas is assumed to be an ideal gas \[ w = - \int p dV -\int \left( \dfrac{nRT}{V}\right) dV \nonumber \] And if the temperature is held constant (so that the expansion follows an pathway) the nRT term can be extracted from the integral. \[ w = -nRT \int_{V_1}^{V_2} \dfrac{dV}{V} = -nRT \ln \left( \dfrac{V_1}{V_2} \right) \label{isothermal} \] Equation \ref{isothermal} is derived for ideal gases only; a van der Waal gas would result in a different version. What is the work done by 1.00 mol an ideal gas expanding reversibly from a volume of 22.4 L to a volume of 44.8 L at a constant temperature of 273 K? Using Equation \ref{isothermal} to calculate this \[\begin{align*} w & = -(1.00 \, \cancel{mol}) \left(8.314\, \dfrac{J}{\cancel{mol}\,\cancel{ K}}\right) (273\,\cancel{K}) \ln \left( \dfrac{44.8\,L}{22.4 \,L} \right) \nonumber \\[4pt] & = -1570 \,J = 1.57 \;kJ \end{align*} \] : A reversible expansion will require more work than an irreversible expansion (such as an expansion against a constant external pressure) when the final states of the two expansions are the same! The work of expansion can be depicted graphically as the area under the p-V curve depicting the expansion. Comparing examples \(\Page {1}\) and \(3.1.2\), for which the initial and final volumes were the same, and the constant external pressure of the irreversible expansion was the same as the final pressure of the reversible expansion, such a graph looks as follows. The work is depicted as the shaded portion of the graph. It is clear to see that the reversible expansion (the work for which is shaded in both light and dark gray) exceeds that of the irreversible expansion (shaded in dark gray only) due to the changing pressure of the reversible expansion. In general, it will always be the case that the work generated by a reversible pathway connecting initial and final states will be the maximum work possible for the expansion. It should be noted (although it will be proven in a later chapter) that \(\Delta U\) for an isothermal reversible process involving only p-V work is 0 for an ideal gas. This is true because the internal energy, U, is a measure of a system’s capacity to convert energy into work. In order to do this, the system must somehow store that energy. The only mode in which an ideal gas can store this energy is in the translational kinetic energy of the molecules (otherwise, molecular collisions would not need to be elastic, which as you recall, was a postulate of the kinetic molecular theory!) And since the average kinetic energy is a function only of the temperature, it (and therefore \(U\)) can only change if there is a change in temperature. Hence, for any isothermal process for an ideal gas, \(\Delta U=0\). And, perhaps just as usefully, for an isothermal process involving an ideal gas, \(q = -w\), as any energy that is expended by doing work must be replaced with heat, lest the system temperature drop. One common pathway which processes can follow is that of constant volume. This will happen if the volume of a sample is constrained by a great enough force that it simply cannot change. It is not uncommon to encounter such conditions with gases (since they are highly compressible anyhow) and also in geological formations, where the tremendous weight of a large mountain may force any processes occurring under it to happen at constant volume. If reversible changes in which the only work that can be done is that of expansion (so-called p-V work) are considered, the following important result is obtained: \[ dU = dq + dw = dq - pdV \nonumber \] However, \(dV = 0\) since the volume is constant! As such, \(dU\) can be expressed only in terms of the heat that flows into or out of the system at constant volume \[ dU = dq_v \nonumber \] Recall that \(dq\) can be found by \[ dq = \dfrac{dq}{\partial T} dT = C\, dt \label{eq1} \] This suggests an important definition for the (\(C_V\)) which is \[C_V \equiv \left( \dfrac{\partial U}{\partial T}\right)_V \nonumber \] When Equation \ref{eq1} is integrated the \[q = \int _{T_1}^{T_2} nC_V dt \label{isochoric} \] Consider 1.00 mol of an ideal gas with \(C_V = 3/2 R\) that undergoes a temperature change from 125 K to 255 K at a constant volume of 10.0 L. Calculate \(\Delta U\), \(q\), and \(w\) for this change. Since this is a constant volume process \[w = 0 \nonumber \] Equation \ref{isochoric} is applicable for an isochoric process, \[q = \int _{T_1}^{T_2} nC_V dt \nonumber \] Assuming \(C_V\) is independent of temperature: \[\begin{align*} q & = nC_V \int _{T_1}^{T_2} dt \\[4pt] &= nC_V ( T_2-T_1) \\[4pt] & = (1.00 \, mol) \left( \dfrac{3}{2} 8.314\, \dfrac{J}{mol \, K}\right) (255\, K - 125 \,K) \\[4pt] & = 1620 \,J = 1.62\, kJ \end{align*} \] Since this a constant volume pathway, \[ \begin{align*} \Delta U & = q + \cancel{w} \\ & = 1.62 \,kJ \end{align*} \] Most laboratory-based chemistry occurs at constant pressure. Specifically, it is exposed to the constant air pressure of the laboratory, glove box, or other container in which reactions are taking place. For constant pressure changes, it is convenient to define a new thermodynamic quantity called . \[ H \equiv U+ pV \nonumber \] or \[\begin{align*} dH &\equiv dU + d(pV) \\[4pt] &= dU + pdV + Vdp \end{align*} \] For reversible changes at constant pressure (\(dp = 0\)) for which only p-V work is done \[\begin{align} dH & = dq + dw + pdV + Vdp \\[4pt] & = dq - \cancel{pdV} + \cancel{pdV} + \cancelto{0}{Vdp} \\ & = dq \label{heat} \end{align} \] And just as in the case of constant volume changes, this implies an important definition for the \[C_p \equiv \left( \dfrac{\partial H}{\partial T} \right)_p \nonumber \] Consider 1.00 mol of an ideal gas with \(C_p = 5/2 R\) that changes temperature change from 125 K to 255 K at a constant pressure of 10.0 atm. Calculate \(\Delta U\), \(\Delta H\), \(q\), and \(w\) for this change. \[q = \int_{T_1}^{T_2} nC_p dT \nonumber \] assuming \(C_p\) is independent of temperature: \[ \begin{align*} q & = nC_p \int _{T_1}^{T_2} dT \\ & = nC_p (T_2-T_1) \\ & = (1.00 \, mol) \left( \dfrac{5}{2} 8.314 \dfrac{J}{mol \, K}\right) (255\, K - 125\, K) = 2700\, J = 1.62\, kJ \end{align*} \] So via Equation \ref{heat} (specifically the integrated version of it using differences instead of differentials) \[ \Delta H = q = 1.62 \,kJ \nonumber \] \[ \begin{align*} \Delta U & = \Delta H - \Delta (pV) \\ & = \Delta H -nR\Delta T \\ & = 2700\, J - (1.00 \, mol) \left( 8.314\, \dfrac{J}{mol \, K}\right) (255\, K - 125 \,K) \\ & = 1620 \,J = 1.62\, kJ \end{align*} \] Now that \(\Delta U\) and \(q\) are determined, then work can be calculated \[\begin{align*} w & =\Delta U -q \\ & = 1.62\,kJ - 2.70\,kJ = -1.08\;kJ \end{align*} \] It makes sense that \(w\) is negative since this process is an gas expansion. Calculate \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for 1.00 mol of an ideal gas expanding reversibly and isothermally at 273 K from a volume of 22.4 L and a pressure of 1.00 atm to a volume of 44.8 L and a pressure of 0.500 atm. Since this is an isothermal expansion, Equation\ref{isothermal} is applicable \[ \begin{align*} w & = -nRT \ln \dfrac{V_2}{V_1} \\ & = (1.00 \, mol) \left( 8.314\, \dfrac{J}{mol \, K}\right) (255\, K) \ln \left(\dfrac{44.8\,L}{22.4\,L} \right) \\ & = 1572\,J = 1.57\,kJ \\[4pt] \Delta U & = q + w \\ & = q + 1.57\,KJ \\ & = 0 \\[4pt] q &= -1.57\,kJ \end{align*} \] Since this is an isothermal expansion \[\Delta H = \Delta U + \Delta (pV) = 0 + 0 \nonumber \] where \(\Delta (pV) = 0\) due to Boyle’s Law! An pathway is defined as one in which no heat is transferred (\(q = 0\)). Under these circumstances, if an ideal gas expands, it is doing work (\(w < 0\)) against the surroundings (provided the external pressure is not zero!) and as such the internal energy must drop (\(\Delta U <0 \)). And since \(\Delta U\) is negative, there must also be a decrease in the temperature (\(\Delta T < 0\)). How big will the decrease in temperature be and on what will it depend? The key to answering these questions comes in the solution to how we calculate the work done. If the adiabatic expansion is reversible and done on an ideal gas, \[dw = -pdV \nonumber \] and \[dw = nC_vdT \label{Adiabate2} \] Equating these two terms yields \[- pdV = nC_v dT \nonumber \] Using the ideal gas law for an expression for \(p\) (\(p = nRT/V\)) \[ - \dfrac{nRT}{V} dV = nC_vdT \nonumber \] And rearranging to gather the temperature terms on the right and volume terms on the left yields \[\dfrac{dV}{V} = -\dfrac{C_V}{R} \dfrac{dT}{T} \nonumber \] This expression can be integrated on the left between \(V_1\) and \(V_2\) and on the right between \(T_1\) and \(T_2\). Assuming that \(C_v/nR\) is independent of temperature over the range of integration, it can be pulled from the integrand in the term on the right. \[ \int_{V_1}^{V_2} \dfrac{dV}{V} = -\dfrac{C_V}{R} \int_{T_1}^{T_2} \dfrac{dT}{T} \nonumber \] The result is \[ \ln \left(\dfrac{V_2}{V_1} \right) = - \dfrac{C_V}{R} \ln \left( \dfrac{T_2}{T_1} \right) \nonumber \] or \[ \left(\dfrac{V_2}{V_1} \right) = \left(\dfrac{T_2}{T_1} \right)^{- \frac{C_V}{R}} \nonumber \] or \[ V_1T_1^{\frac{C_V}{R}} = V_2T_2^{\frac{C_V}{R}} \nonumber \] or \[T_1 \left(\dfrac{V_1}{V_2} \right)^{- \frac{R} {C_V}} = T_2 \label{Eq4Alternative} \] Once \(\Delta T\) is known, it is easy to calculate \(w\), \(\Delta U\) and \(\Delta H\). 1.00 mol of an ideal gas (C = 3/2 R) initially occupies 22.4 L at 273 K. The gas expands adiabatically and reversibly to a final volume of 44.8 L. Calculate \(\Delta T\), \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for the expansion. Since the pathway is adiabatic: \[q =0 \nonumber \] Using Equation \ref{Eq4Alternative} \[ \begin{align*} T_2 & = T_1 \left(\dfrac{V_1}{V_2} \right)^{- \frac{R} {C_V}} \\ & =(273\,K) \left( \dfrac{22.4\,L}{44.8\,L} \right)^{2/3} \\ & = 172\,K \end{align*} \] So \[\Delta T = 172\,K - 273\,K = -101\,K \nonumber \] For calculating work, we integrate Equation \ref{Adiabate2} to get \[ \begin{align*} w & = \Delta U = nC_v \Delta T \\ & = (1.00 \, mol) \left(\dfrac{3}{2} 8.314\, \dfrac{J}{mol \, K} \right) (-101\,K ) \\ & = 1.260 \,kJ \end{align*} \] \[ \begin{align*} \Delta H & = \Delta U + nR\Delta T \\ & = -1260\,J + (1.00 \, mol) \left(\dfrac{3}{2} 8.314\, \dfrac{J}{mol \, K} \right) (-101\,K ) \\ & = -2100\,J \end{align*} \] The following table shows recipes for calculating \(q\), \(w\), \(\Delta U\), and \(\Delta H\) for an ideal gas undergoing a reversible change along the specified pathway. | 12,067 | 2,008 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/05%3A_Energy_Changes_in_Chemical_Reactions/5.06%3A_Calorimetry |
Thermal energy itself cannot be measured easily, but the temperature change caused by the flow of thermal energy between objects or substances can be measured. Calorimetry describes a set of techniques employed to measure enthalpy changes in chemical (reactions) and physical(melting for example) processes using devices called . To have any meaning, the quantity that is actually measured in a calorimetric experiment, the change in the temperature of the device, must be related to the heat evolved or consumed in a chemical reaction. We begin this section by explaining how the flow of thermal energy affects the temperature of an object. We have seen that the temperature of an object changes when it absorbs or loses thermal energy. The of the temperature change depends on both the of thermal energy transferred ( ) and the of the object. Its heat capacity ( )J/ C. is the amount of energy needed to raise the temperature of the object exactly 1°C; the units of are joules per degree Celsius (J/°C). Note that a degree Celsius is exactly the same as a Kelvin, so the heat capacities can be expresses equally well, and perhaps a bit more correctly in SI, as joules per Kelvin, J/K The change in temperature (Δ ) is
\[ \Delta T = \frac{q}{C} \] where is the amount of heat (in joules), is the heat capacity (in joules per degree Celsius), and Δ is − (in degrees Celsius). Note that Δ is written as the final temperature minus the initial temperature. Since Kelvin and degrees Celcius are exactly the same the DIFFERENCE Δ = − is the same whether one uses Kelving (K) or Celcius (°C) for BOTH and , but make sure not to mix these two temperature units. The value of is intrinsically a positive number because it is defined as energy necessary to RAISE the temperature, but Δ and can be either positive or negative, and they both must have the sign. If Δ and are positive, then . If Δ and are negative, then . The heat capacity of an object depends on both its and its . For example, doubling the mass of an object doubles its heat capacity. Consequently, the amount of substance must be indicated when the heat capacity of the substance is reported. The molar heat capacity ( ) is t of are thus J/(mol·°C). The specific heat ( ) is t ts units are thus J/(g·°C). We can relate the quantity of a substance, the amount of heat transferred, its heat capacity, and the temperature change in two ways:
\[ q=nC_{p}\Delta T \]
\[ q=mC_{s}\Delta T \] The specific heats of some common substances are given in Table \(\Page {1}\) Note that the specific heat values of most solids are less than 1 J/(g·°C), whereas those of most liquids are about 2 J/(g·°C). Water in its solid and liquid states is an exception. The heat capacity of ice is twice as high as that of most solids; the heat capacity of liquid water, 4.184 J/(g·°C), is one of the highest known. The high specific heat of liquid water has important implications for life on Earth. A given mass of water releases more than five times as much heat for a 1°C temperature change as does the same mass of limestone or granite. Consequently, coastal regions of our planet tend to have less variable climates than regions in the center of a continent. After absorbing large amounts of thermal energy from the sun in summer, the water slowly releases the energy during the winter, thus keeping coastal areas warmer than otherwise would be expected (Figure \(\Page {1}\) ). Water’s capacity to absorb large amounts of energy without undergoing a large increase in temperature also explains why swimming pools and waterbeds are usually heated. Heat must be applied to raise the temperature of the water to a comfortable level for swimming or sleeping and to maintain that level as heat is exchanged with the surroundings. Moreover, because the human body is about 70% water by mass, a great deal of energy is required to change its temperature by even 1°C. Consequently, the mechanism for maintaining our body temperature at about 37°C does not have to be as finely tuned as would be necessary if our bodies were primarily composed of a substance with a lower specific heat. The high specific heat of liquid water has important implications for life on Earth. A given mass of water releases more than five times as much heat for a 1°C temperature change as does the same mass of limestone or granite. Consequently, coastal regions of our planet tend to have less variable climates than regions in the center of a continent. After absorbing large amounts of thermal energy from the sun in summer, the water slowly releases the energy during the winter, thus keeping coastal areas warmer than otherwise would be expected (Figure \(\Page {1}\) ). Water’s capacity to absorb large amounts of energy without undergoing a large increase in temperature also explains why swimming pools and waterbeds are usually heated. Heat must be applied to raise the temperature of the water to a comfortable level for swimming or sleeping and to maintain that level as heat is exchanged with the surroundings. Moreover, because the human body is about 70% water by mass, a great deal of energy is required to change its temperature by even 1°C. Consequently, the mechanism for maintaining our body temperature at about 37°C does not have to be as finely tuned as would be necessary if our bodies were primarily composed of a substance with a lower specific heat. A home solar energy storage unit uses 400 L of water for storing thermal energy. On a sunny day, the initial temperature of the water is 22.0°C. During the course of the day, the temperature of the water rises to 38.0°C as it circulates through the water wall. How much energy has been stored in the water? (The density of water at 22.0°C is 0.998 g/mL.) volume and density of water and initial and final temperatures amount of energy stored Use the density of water at 22.0°C to obtain the mass of water ( ) that corresponds to 400 L of water. Then compute Δ for the water. Determine the amount of heat absorbed by substituting values for , , and Δ into Equation \(\Page {1}\) The mass of water is \[ mass \; of \; H_{2}O=400 \; \cancel{L}\left ( \dfrac{1000 \; \cancel{mL}}{1 \; \cancel{L}} \right ) \left ( \dfrac{0.998 \; g}{1 \; \cancel{mL}} \right ) = 3.99\times 10^{5}g\; H_{2}O \notag \] The temperature change (Δ ) is 38.0°C − 22.0°C = +16.0°C. From Table \(\Page {1}\), the specific heat of water is 4.184 J/(g·°C). From Equation \(\Page {3}\) the heat absorbed by the water is thus \[ q=mC_{s}\Delta T=\left ( 3.99X10^{5} \; \cancel{g} \right )\left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right ) \left ( 16.0 \; \cancel{^{o}C} \right ) = 2.67 \times 10^{7}J = 2.67 \times 10^{4}kJ \notag \] Both and Δ are positive, consistent with the fact that the water has absorbed energy. Some solar energy devices used in homes circulate air over a bed of rocks that absorb thermal energy from the sun. If a house uses a solar heating system that contains 2500 kg of sandstone rocks, what amount of energy is stored if the temperature of the rocks increases from 20.0°C to 34.5°C during the day? Assume that the specific heat of sandstone is the same as that of quartz (SiO ) in Table \(\Page {1}\). 2.7 × 10 kJ (Even though the mass of sandstone is more than six times the mass of the water in Example 7, the amount of thermal energy stored is the same to two significant figures.) When two objects at different temperatures are placed in contact, heat flows from the warmer object to the cooler one until the temperature of both objects is the same. The law of conservation of energy says that the total energy cannot change during this process:
\[ q_{cold} + q_{hot} = 0 \] The equation implies that the amount of heat that flows a warmer object is the same as the amount of heat that flows a cooler object. Because the direction of heat flow is opposite for the two objects, the sign of the heat flow values must be opposite:
\[ q_{cold} = -q_{hot} \] Thus heat is conserved in any such process, consistent with the law of conservation of energy. Substituting for from Equation \(\Page {2}\) gives
\[ \left [ mC_{s} \Delta T \right ] _{hot} + \left [ mC_{s} \Delta T \right ] _{hot}=0 \] which can be rearranged to give
\[ \left [ mC_{s} \Delta T \right ] _{hot} = - \left [ mC_{s} \Delta T \right ] _{hot} \] When two objects initially at different temperatures are placed in contact, we can use Equation \(\Page {7}\) to calculate the final temperature if we know the chemical composition and mass of the objects. If a 30.0 g piece of copper pipe at 80.0°C is placed in 100.0 g of water at 27.0°C, what is the final temperature? Assume that no heat is transferred to the surroundings. mass and initial temperature of two objects final temperature Using Equation \(\Page {6}\) and writing Δ as − for both the copper and the water, substitute the appropriate values of , , and into the equation and solve for . We can adapt Equation \(\Page {7}\) to solve this problem, remembering that Δ is defined as − : \[ \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{Cu} + \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{H_{2}O} =0 \notag \] Substituting the data provided in the problem and Table \(\Page {1}\) gives \[ \left [ \left (30 \; g \right ) \left (0.385 \; J \right ) \left (T_{final} - T_{initial} \right ) \right ] _{Cu} + \left [ mC_{s} \left (T_{final} - T_{initial} \right ) \right ] _{H_{2}O} =0 \notag \] \[ T_{final}\left ( 11.6 \; J/ ^{o}C \right ) -924 \; J + T_{final}\left ( 418.4 \; J/ ^{o}C \right ) -11,300 \; J \notag \] \[ T_{final}\left ( 430 \; J/\left ( g\cdot ^{o}C \right ) \right ) = -12,224 \; J \notag \] \[ T_{final} = -28.4 \; ^{o}C \notag \] Exercise (a) If a 14.0 g chunk of gold at 20.0°C is dropped into 25.0 g of water at 80.0°C, what is the final temperature if no heat is transferred to the surroundings? A 28.0 g chunk of aluminum is dropped into 100.0 g of water with an initial temperature of 20.0°C. If the final temperature of the water is 24.0°C, what was the initial temperature of the aluminum? (Assume that no heat is transferred to the surroundings.) 80.0°C In Example 7, radiant energy from the sun was used to raise the temperature of water. A calorimetric experiment uses essentially the same procedure, except that the thermal energy change accompanying a chemical reaction is responsible for the change in temperature that takes place in a calorimeter. If the reaction releases heat ( < 0), then heat is absorbed by the calorimeter ( > 0) and its temperature increases. Conversely, if the reaction absorbs heat ( > 0), then heat is transferred from the calorimeter to the system ( < 0) and the temperature of the calorimeter decreases. In both cases, . The heat capacity of the calorimeter or of the reaction mixture may be used to calculate the amount of heat released or absorbed by the chemical reaction. The amount of heat released or absorbed per gram or mole of reactant can then be calculated from the mass of the reactants. Because Δ is defined as the heat flow at constant pressure, measurements made using a constant-pressure calorimeter give Δ values directly. This device is particularly well suited to studying reactions carried out in solution at a constant atmospheric pressure. A “student” version, called a (Figure \(\Page {2}\) ), is often encountered in general chemistry laboratories. Commercial calorimeters operate on the same principle, but they can be used with smaller volumes of solution, have better thermal insulation, and can detect a change in temperature as small as several millionths of a degree (10 °C). Because the heat released or absorbed at constant pressure is equal to Δ , the relationship between heat and Δ is
\[ \Delta H_{rxn}=q_{rxn}=-q_{calorimater}=-mC_{s} \Delta T \] The use of a constant-pressure calorimeter is illustrated in Example 3. When 5.03 g of solid potassium hydroxide are dissolved in 100.0 mL of distilled water in a coffee-cup calorimeter, the temperature of the liquid increases from 23.0°C to 34.7°C. The density of water in this temperature range averages 0.9969 g/cm . What is Δ (in kilojoules per mole)? Assume that the calorimeter absorbs a negligible amount of heat and, because of the large volume of water, the specific heat of the solution is the same as the specific heat of pure water. mass of substance, volume of solvent, and initial and final temperatures Δ Calculate the mass of the solution from its volume and density and calculate the temperature change of the solution. Find the heat flow that accompanies the dissolution reaction by substituting the appropriate values into Equation \(\Page {8}\). Use the molar mass of KOH to calculate Δ . To calculate Δ , we must first determine the amount of heat released in the calorimetry experiment. The mass of the solution is \[ \left (100.0 \; mL\; H2O \right ) \left ( 0.9969 \; g/ \cancel{mL} \right )+ 5.03 \; g \; KOH=104.72 \; g \notag \] The temperature change is (34.7°C − 23.0°C) = +11.7°C. Because the solution is not very concentrated (approximately 0.9 M), we assume that the specific heat of the solution is the same as that of water. The heat flow that accompanies dissolution is thus \[ q_{calorimater}=mC_{s} \Delta T =\left ( 104.72 \; \cancel{g} \right ) \left ( \dfrac{4.184 \; J}{\cancel{g}\cdot \cancel{^{o}C}} \right )\left ( 11.7 \; ^{o}C \right )=5130 \; J =5.13 \; lJ \notag \] The temperature of the solution increased because heat was absorbed by the solution ( > 0). Where did this heat come from? It was released by KOH dissolving in water. From Equation \(\Page {1}\), we see that This experiment tells us that dissolving 5.03 g of KOH in water is accompanied by the of 5.13 kJ of energy. Because the temperature of the solution increased, the dissolution of KOH in water must be exothermic. The last step is to use the molar mass of KOH to calculate Δ —the heat released when dissolving 1 mol of KOH: \[ \Delta H_{soln}= \left ( \dfrac{5.13 \; kJ}{5.03 \; \cancel{g}} \right )\left ( \dfrac{56.11 \; \cancel{g}}{1 \; mol} \right )=-57.2 \; kJ/mol \notag \] A coffee-cup calorimeter contains 50.0 mL of distilled water at 22.7°C. Solid ammonium bromide (3.14 g) is added and the solution is stirred, giving a final temperature of 20.3°C. Using the same assumptions as in Example 9, find Δ for NH Br (in kilojoules per mole). 16.6 kJ/mol Constant-pressure calorimeters are not very well suited for studying reactions in which one or more of the reactants is a gas, such as a combustion reaction. The enthalpy changes that accompany combustion reactions are therefore measured using a constant-volume calorimeter, such as the bomb calorimeter shown schematically in Figure 9.6.3). The reactant is placed in a steel cup inside a steel vessel with a fixed volume (the “bomb”). The bomb is then sealed, filled with excess oxygen gas, and placed inside an insulated container that holds a known amount of water. Because combustion reactions are exothermic, the temperature of the bath and the calorimeter increases during combustion. If the heat capacity of the bomb and the mass of water are known, the heat released can be calculated. Because the volume of the system (the inside of the bomb) is fixed, the combustion reaction occurs under conditions in which the volume, but not the pressure, is constant. As we noted in Chapter 9.2 the heat released by a reaction carried out at constant volume is identical to the change in (Δ ) rather than the enthalpy change (Δ ); Δ is related to Δ by an expression that depends on the change in the number of moles of gas during the reaction. The difference between the heat flow measured at constant volume and the enthalpy change is usually quite small, however (on the order of a few percent). Assuming that Δ < Δ , the relationship between the measured temperature change and Δ is given in Equation \(\Page {9}\), where is the total heat capacity of the steel bomb and the water surrounding it:
\[ \Delta H_{comb} < q_{comb} = q_{calorimater} = C_{bomb} \Delta T \] To measure the heat capacity of the calorimeter, we first burn a carefully weighed mass of a standard compound whose enthalpy of combustion is accurately known. Benzoic acid (C H CO H) is often used for this purpose because it is a crystalline solid that can be obtained in high purity. The combustion of benzoic acid in a bomb calorimeter releases 26.38 kJ of heat per gram (i.e., its Δ = −26.38 kJ/g). This value and the measured increase in temperature of the calorimeter can be used in Equation \(\Page {9}\) to determine . The use of a bomb calorimeter to measure the Δ of a substance is illustrated in Example 10. The combustion of 0.579 g of benzoic acid in a bomb calorimeter caused a 2.08°C increase in the temperature of the calorimeter. The chamber was then emptied and recharged with 1.732 g of glucose and excess oxygen. Ignition of the glucose resulted in a temperature increase of 3.64°C. What is the Δ of glucose? mass and Δ for combustion of standard and sample Δ of glucose Calculate the value of for benzoic acid by multiplying the mass of benzoic acid by its Δ . Then determine the heat capacity of the calorimeter ( ) from and Δ . Calculate the amount of heat released during the combustion of glucose by multiplying the heat capacity of the bomb by the temperature change. Determine the Δ of glucose by multiplying the amount of heat released per gram by the molar mass of glucose. The first step is to use Equation \(\Page {9}\) and the information obtained from the combustion of benzoic acid to calculate . We are given Δ , and we can calculate from the mass of benzoic acid: \[ q{comb} = \left ( 0.579 \; \cancel{g} \right )\left ( -26.38 \; kJ/\cancel{g} \right ) = - 15.3 \; kJ \notag \] From Equation \(\Page {9}\), \[ -C{bomb} = \dfrac{q_{comb}}{\Delta T} = \dfrac{-15.3 \; kJ}{2.08 \; ^{o}C} =- 7.34 \; kJ/^{o}C \notag \] According to the strategy, we can now use the heat capacity of the bomb to calculate the amount of heat released during the combustion of glucose: \[ q_{comb}=-C_{bomb}\Delta T = \left ( -7.34 \; kJ/^{o}C \right )\left ( 3.64 \; ^{o}C \right )=- 26.7 \; kJ \notag \] Because the combustion of 1.732 g of glucose released 26.7 kJ of energy, the Δ of glucose is \[ \Delta H_{comb}=\left ( \dfrac{-26.7 \; kJ}{1.732 \; \cancel{g}} \right )\left ( \dfrac{180.16 \; \cancel{g}}{mol} \right )=-2780 \; kJ/mol =2.78 \times 10^{3} \; kJ/mol \notag \] This result is in good agreement (< 1% error) with the value of Δ = −2803 kJ/mol that calculated using enthalpies of formation. When 2.123 g of benzoic acid is ignited in a bomb calorimeter, a temperature increase of 4.75°C is observed. When 1.932 g of methylhydrazine (CH NHNH ) is ignited in the same calorimeter, the temperature increase is 4.64°C. Calculate the Δ of methylhydrazine, the fuel used in the maneuvering jets of the US space shuttle. −1.30 × 10 kJ/mol Equation \(\Page {2}\) : = Δ Equation \(\Page {3}\) : = Δ Equation \(\Page {8}\) : Δ = = − = − Δ Equation \(\Page {9}\) : Δ < = − = − Δ is the set of techniques used to measure enthalpy changes during chemical processes. It uses devices called , which measure the change in temperature when a chemical reaction is carried out. The magnitude of the temperature change depends on the amount of heat released or absorbed and on the heat capacity of the system. The of an object is the amount of energy needed to raise its temperature by 1°C; its units are joules per degree Celsius. The of a substance is the amount of energy needed to raise the temperature of 1 g of the substance by 1°C, and the is the amount of energy needed to raise the temperature of 1 mol of a substance by 1°C. Liquid water has one of the highest specific heats known. Heat flow measurements can be made with either a , which gives Δ values directly, or a , which operates at constant volume and is particularly useful for measuring enthalpies of combustion. Can an object have a negative heat capacity? Why or why not? What two factors determine the heat capacity of an object? Does the specific heat also depend on these two factors? Explain your answer. Explain why regions along seacoasts have a more moderate climate than inland regions do. Although soapstone is more expensive than brick, soapstone is frequently the building material of choice for fireplaces, particularly in northern climates with harsh winters. Propose an explanation for this. Using Equation \(\Page {2}\) and Equation \(\Page {3}\), derive a mathematical relationship between and . Complete the following table for 28.0 g of each element at an initial temperature of 22.0°C. Using , how much heat is needed to raise the temperature of a 2.5 g piece of copper wire from 20°C to 80°C? How much heat is needed to increase the temperature of an equivalent mass of aluminum by the same amount? If you were using one of these metals to channel heat away from electrical components, which metal would you use? Once heated, which metal will cool faster? Give the specific heat for each metal. Gold has a molar heat capacity of 25.418 J/(mol·K), and silver has a molar heat capacity of 23.350 J/(mol·K). In an exothermic reaction, how much heat would need to be evolved to raise the temperature of 150 mL of water 7.5°C? Explain how this process illustrates the law of conservation of energy. How much heat must be evolved by a reaction to raise the temperature of 8.0 oz of water 5.0°C? What mass of lithium iodide would need to be dissolved in this volume of water to produce this temperature change? A solution is made by dissolving 3.35 g of an unknown salt in 150 mL of water, and the temperature of the water rises 3.0°C. The addition of a silver nitrate solution results in a precipitate. Assuming that the heat capacity of the solution is the same as that of pure water, use the information in Table 9.5.1 and solubility rules to identify the salt. Using the data in Table 9.8.2, calculate the change in temperature of a calorimeter with a heat capacity of 1.78 kJ/°C when 3.0 g of charcoal is burned in the calorimeter. If the calorimeter is in a 2 L bath of water at an initial temperature of 21.5°C, what will be the final temperature of the water after the combustion reaction (assuming no heat is lost to the surroundings)? A 3.00 g sample of TNT (trinitrotoluene, C H N O ) is placed in a bomb calorimeter with a heat capacity of 1.93 kJ/°C; the Δ of TNT is −3403.5 kJ/mol. If the initial temperature of the calorimeter is 19.8°C, what will be the final temperature of the calorimeter after the combustion reaction (assuming no heat is lost to the surroundings)? What is the Δ of TNT? = × (molar mass) For Cu: = 58 J; For Al: = 130 J; Even though the values of the molar heat capacities are very similar for the two metals, the specific heat of Cu is only about half as large as that of Al, due to the greater molar mass of Cu versus Al: = 0.385 and 0.897 (g·K) for Cu and Al, respectively. Thus loss of one joule of heat will cause almost twice as large a decrease in temperature of Cu versus Al. 4.7 kJ Δ = −0.56 kJ/g; based on reaction with AgNO , salt contains halide; dividing Δ values in Table 5.2 by molar mass of salts gives lithium bromide as best match, with −0.56 kJ/g. = 43.1°C; the combustion reaction is \[4C_7H_5N_3O_{6(s)} + 21O_{2(g)} \rightarrow 28CO_{2(g)} + 10H_2O_{(g)} + 6N_{2(g)} \notag \] with \[Δ_f^οH (TNT) = −65.5\; kJ/mol \notag \] | 23,666 | 2,009 |
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This page gives you the facts and a simple, uncluttered mechanism for the electrophilic substitution reaction between benzene and chlorine or bromine in the presence of a catalyst such as aluminum chloride or iron. Benzene reacts with chlorine or bromine in an electrophilic substitution reaction, but only in the presence of a catalyst. The catalyst is either aluminum chloride (or aluminum bromide if you are reacting benzene with bromine) or iron. Strictly speaking iron is not a catalyst, because it gets permanently changed during the reaction. It reacts with some of the chlorine or bromine to form iron(III) chloride, \(FeCl_3\), or iron(III) bromide, \(FeBr_3\). \[ 2Fe + 3Cl_2 \rightarrow 2FeCl_3\] \[ 2Fe + 3Br_2 \rightarrow 2FeBr_3\] These compounds act as the catalyst and behave exactly like aluminum chloride in these reactions. The reaction between benzene and chlorine in the presence of either aluminum chloride or iron gives chlorobenzene. \[C_6H_6 + Cl_2 \rightarrow C_6H_5Cl + HCl\] or: The reaction between benzene and bromine in the presence of either aluminum bromide or iron gives bromobenzene. Iron is usually used because it is cheaper and more readily available. \[C_6H_6 + Br_2 \rightarrow C_6H_5Br + HBr\] or: We are going to explore the reaction using chlorine and aluminum chloride. If you want one of the other combinations, all you have to do is to replace each \(Cl\) by \(Br\), or each \(Al\) by \(Fe\). As a chlorine molecule approaches the benzene ring, the delocalized electrons in the ring repel electrons in the chlorine-chlorine bond. It is the slightly positive end of the chlorine molecule which acts as the electrophile. The presence of the aluminum chloride helps this polarization. The hydrogen is removed by the \(AlCl_4^-\) ion which was formed in the first stage. The aluminum chloride catalyst is re-generated in this second stage. Jim Clark ( ) | 1,907 | 2,010 |
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This page explains the origin of hydrogen bonding - a relatively strong form of intermolecular attraction. Many elements form compounds with hydrogen. If you plot the boiling points of the compounds of the with hydrogen, you find that the boiling points increase as you go down the group. The increase in boiling point happens because the molecules are getting larger with more electrons, and so van der Waals become greater. If you repeat this exercise with the compounds of the elements in Groups 5, 6 and 7 with hydrogen, something odd happens. Although for the most part the trend is exactly the same as in group 4 (for exactly the same reasons), the boiling point of the compound of hydrogen with the first element in each group is abnormally high. In the cases of NH , H O and HF there must be some additional intermolecular forces of attraction, requiring significantly more heat energy to break. These relatively powerful intermolecular forces are described as hydrogen bonds. The molecules which have this extra bonding are: The solid line represents a bond in the plane of the screen or paper. Dotted bonds are going back into the screen or paper away from you, and wedge-shaped ones are coming out towards you. Notice that in each of these molecules: Consider two water molecules coming close together. The + hydrogen is so strongly attracted to the lone pair that it is almost as if you were beginning to form a co-ordinate (dative covalent) bond. It doesn't go that far, but the attraction is significantly stronger than an ordinary dipole-dipole interaction. Hydrogen bonds have about a tenth of the strength of an average covalent bond, and are being constantly broken and reformed in liquid water. If you liken the covalent bond between the oxygen and hydrogen to a stable marriage, the hydrogen bond has "just good friends" status. Notice that each water molecule can potentially form four hydrogen bonds with surrounding water molecules. There are exactly the right numbers of + hydrogens and lone pairs so that every one of them can be involved in hydrogen bonding. This is why the boiling point of water is higher than that of ammonia or hydrogen fluoride. In the case of ammonia, the amount of hydrogen bonding is limited by the fact that each nitrogen only has one lone pair. In a group of ammonia molecules, there aren't enough lone pairs to go around to satisfy all the hydrogens. In hydrogen fluoride, the problem is a shortage of hydrogens. In water, there are exactly the right number of each. Water could be considered as the "perfect" hydrogen bonded system. The diagram shows the potential hydrogen bonds formed to a chloride ion, Cl . Although the lone pairs in the chloride ion are at the 3-level and wouldn't normally be active enough to form hydrogen bonds, in this case they are made more attractive by the full negative charge on the chlorine. However complicated the negative ion, there will always be lone pairs that the hydrogen atoms from the water molecules can hydrogen bond to. An is an organic molecule containing an -O-H group. Any molecule which has a hydrogen atom attached directly to an oxygen or a nitrogen is capable of hydrogen bonding. Such molecules will always have higher boiling points than similarly sized molecules which don't have an -O-H or an -N-H group. The hydrogen bonding makes the molecules "stickier", and more heat is necessary to separate them. Ethanol, CH CH -O-H, and methoxymethane, CH -O-CH , both have the same molecular formula, C H O. They have the same number of electrons, and a similar length to the molecule. The van der Waals attractions (both dispersion forces and dipole-dipole attractions) in each will be much the same. However, ethanol has a hydrogen atom attached directly to an oxygen - and that oxygen still has exactly the same two lone pairs as in a water molecule. Hydrogen bonding can occur between ethanol molecules, although not as effectively as in water. The hydrogen bonding is limited by the fact that there is only one hydrogen in each ethanol molecule with sufficient + charge. In methoxymethane, the lone pairs on the oxygen are still there, but the hydrogens aren't sufficiently + for hydrogen bonds to form. Except in some rather unusual cases, the hydrogen atom has to be attached directly to the very electronegative element for hydrogen bonding to occur. The boiling points of ethanol and methoxymethane show the dramatic effect that the hydrogen bonding has on the stickiness of the ethanol molecules: The hydrogen bonding in the ethanol has lifted its boiling point about 100°C. It is important to realize that hydrogen bonding exists in addition to van der Waals attractions. For example, all the following molecules contain the same number of electrons, and the first two are much the same length. The higher boiling point of the butan-1-ol is due to the additional hydrogen bonding. Comparing the two alcohols (containing -OH groups), both boiling points are high because of the additional hydrogen bonding due to the hydrogen attached directly to the oxygen - but they are not the same. The boiling point of the 2-methylpropan-1-ol isn't as high as the butan-1-ol because the branching in the molecule makes the van der Waals attractions less effective than in the longer butan-1-ol. Hydrogen bonding also occurs in organic molecules containing N-H groups - in the same sort of way that it occurs in ammonia. Examples range from simple molecules like CH NH (methylamine) to large molecules like proteins and . The two strands of the famous double helix in DNA are held together by hydrogen bonds between hydrogen atoms attached to nitrogen on one strand, and lone pairs on another nitrogen or an oxygen on the other one. Jim Clark ( ) | 5,777 | 2,011 |
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In differential form, the free energy functions can be expressed as \[dA = -pdV - SdT \nonumber \] and \[dG = -Vdp - SdT \nonumber \] So by inspection, it is easy to see that \[\left( \dfrac{\partial A}{\partial T} \right)_V = -S \nonumber \] and \[\left( \dfrac{\partial G}{\partial T} \right)_p = -S \nonumber \] And so, it should be fairly straightforward to determine how each changes with changing temperature: \[\Delta A = - \int_{T_1}^{T_2} \left( \dfrac{\partial A}{\partial T} \right)_V dT = - \int_{T_1}^{T_2} S\,dT \nonumber \] and \[\Delta G = - \int_{T_1}^{T_2} \left( \dfrac{\partial G}{\partial T} \right)_p dT = - \int_{T_1}^{T_2} S\,dT \nonumber \] But the temperature dependence of the entropy needed to be known in order to evaluate the integral. A convenient work-around can be obtained starting from the definitions of the free energy functions. \[ A= U -TS \nonumber \] and \[G = H -TS \nonumber \] Dividing by \(T\) yields \[\dfrac{A}{T} = \dfrac{U}{T} -S \nonumber \] and \[\dfrac{G}{T} = \dfrac{H}{T} -S \nonumber \] Now differentiating each expression with respect to \(T\) at constant \(V\) or \(p\) respectively yields \[ \left( \dfrac{\partial \left( \frac{A}{T}\right)}{\partial T} \right)_V = - \dfrac{U}{T^2} \nonumber \] and \[ \left( \dfrac{\partial \left( \frac{G}{T}\right)}{\partial T} \right)_p = - \dfrac{H}{T^2} \nonumber \] Or differentiating with respect to \(1/T\) provides a simpler form that is mathematically equivalent: \[ \left( \dfrac{\partial \left( \frac{A}{T}\right)}{\partial \left( \frac{1}{T}\right)} \right)_V = U \nonumber \] and \[ \left( \dfrac{\partial \left( \frac{G}{T}\right)}{\partial \left( \frac{1}{T}\right)} \right)_p = H \nonumber \] Focusing on the second expression (since all of the arguments apply to the first as well), we see a system that can be integrated. Multiplying both sides by \(d(1/T)\) yields: \[ d \left( \dfrac{G}{T} \right) = H d \left( \dfrac{1}{T} \right) \nonumber \] Or for finite changes \(\Delta G\) and \(\Delta H\): \[ d \left( \dfrac{\Delta G}{T} \right) = \Delta H d \left( \dfrac{1}{T} \right) \nonumber \] and integration, assuming the interval yields \[ \int_{T_1}^{T_2} d \left( \dfrac{\Delta G}{T} \right) = \Delta H \int_{T_1}^{T_2} d \left( \dfrac{1}{T} \right) \nonumber \] \[\dfrac{\Delta G_{T_2}}{T_2} - \dfrac{\Delta G_{T_1}}{T_1} = \Delta H \left( \dfrac{1}{T_2} - \dfrac{1}{T_1} \right) \label{GH1} \] Equation \ref{HG1} is the and can be used to determine how \(\Delta G\) changes with changing temperature. The equivalent equation for the Helmholtz function is \[ \dfrac{\Delta A_{T_2}}{T_2} -\dfrac{\Delta A_{T_1}}{T_1} = \Delta U \left(\dfrac{1}{T_2} -\dfrac{1}{T_1} \right) \label{GH2} \] Given the following data at 298 K, calculate \(\Delta G\) at 500 K for the following reaction: \[CH_4(g) + 2 O_2(g) \rightarrow CO_2(g) + H_2O(g) \nonumber \] \(\Delta H\) and \(\Delta G_{298\, K}\) and can be calculated fairly easily. It will be assumed that \(\Delta H\) is constant over the temperature range of 298 K – 500 K. \[\Delta H = (1 \,mol)(-393.5 \,kJ/mol) + (2\, mol)(-241.8\, kJ/mol) – (1\, mol)(-74.5\, kJ/mol) = -820.6\, kJ \nonumber \] \[\Delta G_{298} = (1\, mol)(-394.4\, kJ/mol) + (2\, mol)(-228,6\, kJ/mol) – (1\, mol)(-50.5\, kJ/mol) = -801.1\, kJ \nonumber \] So using Equation \ref{GH1} with the data just calculated gives \[\dfrac{\Delta G_{500\,K}}{500 \,K} - \dfrac{-801.1\,kJ}{298\,K} = (-820.6\, kJ) \left( \dfrac{1}{500\,K} - \dfrac{1}{298\,K} \right) \nonumber \] \[ \Delta G_{500\,K} = -787.9\,kJ \nonumber \] : \(\Delta G\) became a little bit less negative at the higher temperature, which is to be expected for a reaction which is exothermic. An increase in temperature should tend to make the reaction less favorable to the formation of products, which is exactly what is seen in this case! | 3,856 | 2,012 |
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So far, we have derived a number of expressions and developed methods for evaluating how thermodynamic variables change as one variable changes while holding the rest constant. But real systems are seldom this accommodating. For example, a piece of metal (such as a railroad rail) left in the sun will undergo both an increase in temperature and an expansion due to the absorption of energy from sunlight. So both \(T\) and \(V\) are changing at the same time! If the change in a thermodynamic variable (such as \(G\)) is needed, contributions from both changes are required to be taken into account. We’ve already seen how to express this in terms of a total differential. \[ dG = \left( \dfrac{\partial G}{\partial p} \right)_T dp + \left( \dfrac{\partial G}{\partial T} \right)_p dT \label{Total2} \] Fortunately, \(G\) (like the other thermodynamic functions \(U\), \(H\), \(S\), and \(A\)) is kind enough to be a state variable. This means that we can consider the changes independently and then simply add the results. Another way to think of this is that the system may follow either of two pathways to get from the initial conditions to the final conditions: And since \(G\) has the good sense to be a state variable, the pathway connecting the initial and final states is unimportant. We are free to choose any path that is convenient to calculate the change. Calculate the entropy change for 1.00 mol of a monatomic ideal gas (C = 3/2 R) expanding from 10.0 L at 273 K to 22.0 L at 297 K. If one considers entropy to be a function of temperature and volume, one can write the total differential of entropy as \[dS = \left( \dfrac{\partial S}{\partial T} \right)_V dT + \left( \dfrac{\partial S}{\partial V} \right)_T dV \nonumber \] and thus \[\Delta S = \int_{T_1}^{T_2} \left( \dfrac{\partial S}{\partial T} \right)_V dT + \int_{V_1}^{V_2} \left( \dfrac{\partial S}{\partial V} \right)_T dV \nonumber \] The first term is the contribution due to an change: \[ \begin{align} \Delta S_{T_1 \rightarrow T_2} & = \int_{T_1}^{T_2} \left( \dfrac{\partial S}{\partial T} \right)_V dT \\ & = \int_{T_1}^{T_2} \dfrac{n C_V}{T} dT \\ & = nC_V \ln \left(\dfrac{T_2}{T_1} \right) \\ & = (1.00\, mol) \left( \dfrac{3}{2} \cdot 8.314 \dfrac{J}{mol\,K} \right) \ln \left(\dfrac{297\,K}{273\,K }\right) \\ & = 13.57 \,J/K \end{align} \nonumber \] The second term is the contribution due to an isothermal expansion: \[\Delta S_{V_1 \rightarrow V_2} = \int_{V_1}^{V_2} \left( \dfrac{\partial S}{\partial V} \right)_T dV \label{second} \] From the Maxwell relation on \(A\) \[ \left( \dfrac{\partial S}{\partial V} \right)_T = \left( \dfrac{\partial p}{\partial T} \right)_V \nonumber \] So Equation \ref{second} becomes \[ \begin{align} \Delta S_{V_1 \rightarrow V_2} & = \int_{V_1}^{V_2} \left( \dfrac{\partial p}{\partial T} \right)_V dV \\ & = \int_{V_1}^{V_2} \left( \dfrac{nR}{V} \right) dV\\ & = nR \ln \left(\dfrac{V_2}{V_1} \right) \\ &= (1.00\, mol) \left( 8.314 \dfrac{J}{mol\,K} \right) \ln \left(\dfrac{22.0\,L}{10.0\,L }\right) \\ & = 6.56\, J/K \end{align} \nonumber \] And the total entropy change is \[\begin{align} \Delta S_{tot} & = \Delta S_{V_1 \rightarrow V_2} + \Delta S_{V_1 \rightarrow V_2} \\ & = 13.57\,J/K + 6.56 \,J/K \\ & = 20.13\,J/K \end{align} \nonumber \] Thermodynamics involves many variables. But for a single component sample of matter, only two state variables are needed to describe the system and fix all of the thermodynamic properties of the system. As such, it is conceivable that two functions can be specified as functions of the same two variables. In general terms: \(z(x, y)\) and \(w(x, y)\). So an important question that can be answered is, “What happens to \(z\) if \(w\) is held constant, but \(x\) is changed?” To explore this, consider the total differential of \(z\): \[dz = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy \label{eq5} \] but \(z\) can also be considered a function of \(x\) and \(w(x, y)\). This implies that the total differential can also be written as \[dz = \left( \dfrac{\partial z}{\partial x} \right)_w dx + \left( \dfrac{\partial z}{\partial w} \right)_x dy \label{eq6} \] and these two total differentials must be equal to one another! \[ = \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy = \left( \dfrac{\partial z}{\partial x} \right)_w dx + \left( \dfrac{\partial z}{\partial w} \right)_x dw \nonumber \] If we constrain the system to a change in which \(w\) remains constant, the last term will vanish since \(dw = 0\). \[ \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x dy = \left( \dfrac{\partial z}{\partial x} \right)_w dx \label{eq10} \] but also, since \(w\) is a function \(x\) and \(y\), the total differential for \(w\) can be written \[dw = \left( \dfrac{\partial w}{\partial x} \right)_y dx + \left( \dfrac{\partial w}{\partial y} \right)_x dy \nonumber \] And it too must be zero for a process in which \(w\) is held constant. \[ 0 = \left( \dfrac{\partial w}{\partial x} \right)_y dx + \left( \dfrac{\partial w}{\partial y} \right)_x dy \nonumber \] From this expression, it can be seen that \[dy = - \left( \dfrac{\partial w}{\partial x} \right)_y \left( \dfrac{\partial y}{\partial w} \right)_x dx \nonumber \] Substituting this into the Equation \ref{eq10}, yields \[ \left( \dfrac{\partial z}{\partial x} \right)_y dx + \left( \dfrac{\partial z}{\partial y} \right)_x \left[ - \left( \dfrac{\partial w}{\partial x} \right)_y \left( \dfrac{\partial y}{\partial w} \right)_x dx \right] = \left( \dfrac{\partial z}{\partial x} \right)_w dx \label{eq20} \] which simplifies to \[ \left( \dfrac{\partial z}{\partial x} \right)_y dx - \left( \dfrac{\partial z}{\partial w} \right)_x \left( \dfrac{\partial w}{\partial x} \right)_y dx = \left( \dfrac{\partial z}{\partial x} \right)_w dx \nonumber \] So for \(dx \neq 0\), implies that \[ \left( \dfrac{\partial z}{\partial x} \right)_y - \left( \dfrac{\partial z}{\partial w} \right)_x \left( \dfrac{\partial w}{\partial x} \right)_y = \left( \dfrac{\partial z}{\partial x} \right)_w \nonumber \] or \[ \left( \dfrac{\partial z}{\partial x} \right)_y = \left( \dfrac{\partial z}{\partial x} \right)_w + \left( \dfrac{\partial z}{\partial w} \right)_x \left( \dfrac{\partial w}{\partial x} \right)_y \label{final1} \] As with partial derivative transformation types I and II, this result can be achieved in a formal, albeit less mathematically rigorous method. Consider \(z(x, w)\). This allows us to write the total differential for \(z\): \[ dz = \left( \dfrac{\partial z}{\partial x} \right)_w dx + \left( \dfrac{\partial z}{\partial w} \right)_x dw \nonumber \] Now, divide by \(dx\) and constrain to constant \(y\). \[\left.\dfrac{dz}{dx} \right\rvert_{y}= \left( \dfrac{\partial z}{\partial x} \right)_w \left.\dfrac{dx}{dx} \right\rvert_{y} + \left( \dfrac{\partial z}{\partial w} \right)_x \left.\dfrac{dw}{dx} \right\rvert_{y} \nonumber \] noting that \(dx/dx = 1\) and converting the other ratios to partial derivatives yields \[ \left( \dfrac{\partial z}{\partial x} \right)_y = \left( \dfrac{\partial z}{\partial x} \right)_w + \left( \dfrac{\partial z}{\partial w} \right)_x \left( \dfrac{\partial w}{\partial x} \right)_y \label{final2} \] which agrees with the previous result (Equation \ref{final1})! Again, the method is not mathematically rigorous, but it works so long as \(w\), \(x\), \(y\), and \(z\) are and the total differentials \(dw\), \(dx\), \(dy\), and \(dz\) are . | 7,639 | 2,013 |
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In the unlikely case that we have \(C_p\) data all the way from 0 K we could write: \[ S(T) = S(0) + \int_{0}^{T} C_p dT \nonumber \] It is tempting to set the \(S(0)\) to zero and make the entropy thus an absolute quantity. As we have seen with enthalpy, it was not really possible set \(H(0)\) to zero. All we did was define \(ΔH\) for a particular process, although if \(C_p\) data are available we could construct an enthalpy function (albeit with a floating zero point) by integration of \(C_P\) (instead of \(C_p/T\)!). Still for entropy \(S\) the situation is a bit different than for \(H\). Here we can actually put things on an scale. Both Nernst and Planck have proposed to do so. Nernst postulated that for a pure and perfect crystal \(S\) should indeed to go to zero as \(T\) goes to zero. For example, sulfur has two solid crystal structure, rhombic and monoclinic. At 368.5 K, the entropy of the phase transition from rhombic to monoclinic, \(S_{(rh)}\rightarrow S_{(mono)}\), is: \[\Delta_\text{trs}S(\text{368.5 K})=1.09\;\sf\frac{J\cdot mol}{K} \nonumber \] As the temperature is lowered to 0 K, the entropy of the phase transition approaches zero: \[\lim_{T\rightarrow 0}\Delta_rS=0 \nonumber \] This shows that the entropy of the two crystalline forms are the same. The only way is if all species have the same absolute entropy at 0 K. For energy dispersal at 0 K: For a pure, perfect crystalline solid at 0 K, there is only one energetic arrangement to represent its lowest energy state. We can use this to define a natural zero, giving entropy an absolute scale. The states that the entropy of a pure substance in a perfect crystalline form is 0 \(\sf\frac{J}{mol\cdot K}\) at 0 K: \[{\bar{S}}_{0\text{ K}}^\circ=0 \nonumber \] This is consistent with our molecular formula for entropy: \[S=k\ln{W} \nonumber \] For a perfect crystal at 0 K, the number of ways the total energy of a system can be distributed is one (\(W=1\)). The \(\ln{W}\) term goes to zero, resulting in the perfect crystal at 0 K having zero entropy. It is certainly true that for the great majority of materials we end up with a crystalline material at sufficiently low materials (although there are odd exceptions like liquid helium). However, it should be mentioned that a completely crystal can only be grown at zero Kelvin! It is not possible to grow anything at 0 K however. At any finite temperature the crystal always incorporates defects, more so if grown at higher temperatures. When cooled down very slowly the defects tend to be ejected from the lattice for the crystal to reach a new equilibrium with less defects. This tendency towards less and less disorder upon cooling is what the third law is all about. However, the ordering process often becomes impossibly slow, certainly when approaching absolute zero. This means that real crystals always have frozen in imperfections. Thus there is always residual entropy. Fortunately, the effect is often too small to be measured. This is what allows us to ignore it in many cases (but not all). We could state the Third law of thermodynamics as follows: The entropy of a perfect crystal approaches zero when T approaches zero (but perfect crystals do not exist). Another complication arises when the system undergoes a phase transition, e.g. the melting of ice. As we can write: \[Δ_{fus}H= q_P \nonumber \] If ice and water are in equilibrium with each other the process is quite reversible and so we have: \[Δ_{fus}S = \dfrac{q_{rev}}{T}= \dfrac{Δ_{fus}H}{T} \nonumber \] This means that at the melting point the curve for \(S\) makes a sudden jump by this amount because all this happens at one an the same temperature. Entropies are typically calculated from calorimetric data (\(C_P\) measurements e.g.) and tabulated in standard molar form. The standard state at any temperature is the hypothetical corresponding ideal gas at one bar for gases. In table 21.3 a number of such values are shown. There are some clear trends. E.g. when the noble gas gets heavier this induces more entropy. This is a direct consequence of the particle in the box formula: It has mass in the denominator and therefore the energy levels get more crowded together when m increases: more energy levels, more entropy. The energy levels get more crowded together when \(m\) increases: more energy levels, more entropy There are tables for \(H^\circ(T)-H^\circ(0)\), \(S^\circ(T)\) and \(G^\circ(T)-H^\circ(0)\) as a function of temperature for numerous substances. As we discussed before the plimsoll defines are standard state in terms of pressure (1 bar) and of concentration reference states, if applicable, but temperature is the one of interest. For most substances the Third Law assumption that \(\lim{S^\circ(T)}\) for \(T\rightarrow 0 = 0\) is a reasonable one but there are notable exceptions, such as carbon monoxide. In the solid form, carbon monoxide molecules should ideally be fully ordered at absolute zero, but because the sizes of the carbon atoms and the oxygen atoms are very close and the dipole of the molecule is small, it is quite possible to put in a molecule 'upside down', i.e. with the oxygen on a carbon site and vice versa. At higher temperatures, at which the crystal is formed, this lowers the Gibbs energy (\(G\)) because it increases entropy. In fact we could say that if we could put each molecule into the lattice in two different ways, the number of ways \(W_{disorder}\) we can put N molecules in into the lattice is 2 . This leads to an additional contribution to the entropy of \[S_{disorder} = k\,\ln{W_{disorder}}= N k \ln 2 = R \ln 2 = 5.7\, \frac{\text{J}}{\text{mol}\cdot\text{K}}) \nonumber \] Although at lower temperatures the entropy term in \(G = H-TS\) becomes less and less significant and the ordering of the crystal to a state of lower entropy should become a spontaneous process, in reality the kinetics are so slow that the ordering process does not happen and solid CO therefore has a non-zero entropy when approaching 0 K. In principle crystalline materials have this effect to some extent, but CO is unusual because the concentration of 'wrongly aligned' entities is of the order of 50% rather than say 1 in 10 (a typical defect concentration in say single crystal silicon). | 6,295 | 2,015 |
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In , we saw that knowing the magnitude of the equilibrium constant under a given set of conditions allows chemists to predict the extent of a reaction. Often, however, chemists must decide whether a system has reached equilibrium or if the composition of the mixture will continue to change with time. In this section, we describe how to quantitatively analyze the composition of a reaction mixture to make this determination. To determine whether a system has reached equilibrium, chemists use a quantity called the reaction quotient ( ) . The expression for the reaction quotient has precisely the same form as the equilibrium constant expression, except that may be derived from a set of values measured at time during the reaction of mixture of the reactants and the products, regardless of whether the system is at equilibrium. Therefore, for the following general reaction: \[aA+bB \rightarrow cC+dD \notag \] the reaction quotient is defined as follows:
\[Q=\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \tag{15.4.1}\] The reaction quotient Q analogous to , can be written for any reaction that involves gases by using the partial pressures of the components. To understand how information is obtained using a reaction quotient, consider the dissociation of dinitrogen tetroxide to nitrogen dioxide, \[N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)} \notag \] for which = 4.65 × 10 at 298 K. We can write for this reaction as follows:
\[Q=\dfrac{[NO_2]^2}{[N_2O_4]} \tag{15.4.2}\] The following table lists data from three experiments in which samples of the reaction mixture were obtained and analyzed at equivalent time intervals, and the corresponding values of were calculated for each. Each experiment begins with different proportions of product and reactant: As these calculations demonstrate, can have any numerical value between 0 and infinity (undefined); that is, can be greater than, less than, or equal to . Comparing the magnitudes of and enables us to determine whether a reaction mixture is already at equilibrium and, if it is not, predict how its composition will change with time to reach equilibrium (i.e., whether the reaction will proceed to the right or to the left as written). All you need to remember is that the composition of a system not at equilibrium will change in a way that makes approach . If = , for example, then the system is already at equilibrium, and no further change in the composition of the system will occur unless the conditions are changed. If < , then the ratio of the concentrations of products to the concentrations of reactants is than the ratio at equilibrium. Therefore, the reaction will proceed to the right as written, forming products at the expense of reactants. Conversely, if > , then the ratio of the concentrations of products to the concentrations of reactants is than at equilibrium, so the reaction will proceed to the left as written, forming reactants at the expense of products. These points are illustrated graphically in If < , the reaction will proceed to the right as written. If > , the reaction will proceed to the left as written. If = , then the system is at equilibrium. At elevated temperatures, methane (CH ) reacts with water to produce hydrogen and carbon monoxide in what is known as a reaction: \[CH_{4(g)}+H_2O_{(g)} \rightleftharpoons CO_{(g)}+3H_{2(g)} \notag \] = 2.4 × 10 at 900 K. Huge amounts of hydrogen are produced from natural gas in this way and are then used for the industrial synthesis of ammonia. If 1.2 × 10 mol of CH , 8.0 × 10 mol of H O, 1.6 × 10 mol of CO, and 6.0 × 10 mol of H are placed in a 2.0 L steel reactor and heated to 900 K, will the reaction be at equilibrium or will it proceed to the right to produce CO and H or to the left to form CH and H O? balanced chemical equation, , amounts of reactants and products, and volume direction of reaction Calculate the molar concentrations of the reactants and the products. Use to determine . Compare and to determine in which direction the reaction will proceed. We must first find the initial concentrations of the substances present. For example, we have 1.2 × 10 mol of CH in a 2.0 L container, so \[[CH_4]=\dfrac{1.2\times 10^{−2} mol}{2.0\; L}=6.0 \times 10^{−3} M \notag \] We can calculate the other concentrations in a similar way: We now compute and compare it with : \[Q=\dfrac{[CO,H_2]^3}{[CH_4,H_2O}=\dfrac{(8.0 \times 10^{−3})(3.0 \times 10^{−3})^3}{(6.0\times 10^{−3})(4.0 \times 10^{−3})}=9.0 \times 10^{−6} \notag \] Because = 2.4 × 10 , we see that < . Thus the ratio of the concentrations of products to the concentrations of reactants is less than the ratio for an equilibrium mixture. The reaction will therefore proceed to the right as written, forming H and CO at the expense of H O and CH . Exercise In the water–gas shift reaction introduced in Example 10, carbon monoxide produced by steam-reforming reaction of methane reacts with steam at elevated temperatures to produce more hydrogen: \[CO_{(g)}+H_2O_{(g)} \rightleftharpoons CO_{2(g)}+H_{2(g)} \notag \] = 0.64 at 900 K. If 0.010 mol of both CO and H O, 0.0080 mol of CO , and 0.012 mol of H are injected into a 4.0 L reactor and heated to 900 K, will the reaction proceed to the left or to the right as written? = 0.96 ( > ), so the reaction will proceed to the left, and CO and H O will form. By graphing a few equilibrium concentrations for a system at a given temperature and pressure, we can readily see the range of reactant and product concentrations that correspond to equilibrium conditions, for which = . Such a graph allows us to predict what will happen to a reaction when conditions change so that no longer equals , such as when a reactant concentration or a product concentration is increased or decreased. Lead carbonate decomposes to lead oxide and carbon dioxide according to the following equation:
\[PbCO_{3(s)} \rightleftharpoons PbO_{(s)}+CO_{2(g)} \tag{15.4.3}\] Because PbCO and PbO are solids, the equilibrium constant is simply = [CO ]. At a given temperature, therefore, any system that contains solid PbCO and solid PbO will have exactly the same concentration of CO at equilibrium, regardless of the ratio or the amounts of the solids present. This situation is represented in , which shows a plot of [CO ] versus the amount of PbCO added. Initially, the added PbCO decomposes completely to CO because the amount of PbCO is not sufficient to give a CO concentration equal to . Thus the left portion of the graph represents a system that is at equilibrium because it contains only CO (g) and PbO(s). In contrast, when just enough PbCO has been added to give [CO ] = , the system has reached equilibrium, and adding more PbCO has no effect on the CO concentration: the graph is a horizontal line. Thus any CO concentration that is not on the horizontal line represents a nonequilibrium state, and the system will adjust its composition to achieve equilibrium, provided enough PbCO and PbO are present. For example, the point labeled in lies above the horizontal line, so it corresponds to a [CO ] that is greater than the equilibrium concentration of CO ( > ). To reach equilibrium, the system must decrease [CO ], which it can do only by reacting CO with solid PbO to form solid PbCO . Thus the reaction in will proceed to the left as written, until [CO ] = . Conversely, the point labeled in lies below the horizontal line, so it corresponds to a [CO ] that is less than the equilibrium concentration of CO ( < ). To reach equilibrium, the system must increase [CO ], which it can do only by decomposing solid PbCO to form CO and solid PbO. The reaction in will therefore proceed to the right as written, until [CO ] = . In contrast, the reduction of cadmium oxide by hydrogen gives metallic cadmium and water vapor:
\[CdO_{(s)}+H_{2(g)} \rightleftharpoons Cd_{(s)}+H_2O_{(g)} \tag{15.4.4}\] and the equilibrium constant is [H O]/[H ]. If [H O] is doubled at equilibrium, then [H ] must also be doubled for the system to remain at equilibrium. A plot of [H O] versus [H ] at equilibrium is a straight line with a slope of ( ). Again, only those pairs of concentrations of H O and H that lie on the line correspond to equilibrium states. Any point representing a pair of concentrations that does not lie on the line corresponds to a nonequilibrium state. In such cases, the reaction in will proceed in whichever direction causes the composition of the system to move toward the equilibrium line. For example, point in lies below the line, indicating that the [H O]/[H ] ratio is less than the ratio of an equilibrium mixture ( < ). Thus the reaction in will proceed to the right as written, consuming H and producing H O, which causes the concentration ratio to move up and to the left toward the equilibrium line. Conversely, point in lies above the line, indicating that the [H O]/[H ] ratio is greater than the ratio of an equilibrium mixture ( > ). Thus the reaction in will proceed to the left as written, consuming H O and producing H , which causes the concentration ratio to move down and to the right toward the equilibrium line. In another example, solid ammonium iodide dissociates to gaseous ammonia and hydrogen iodide at elevated temperatures:
\[ NH_4I_{(s)} \rightleftharpoons NH_{3(g)}+HI_{(g)} \tag{15.4.5}\] For this system, is equal to the product of the concentrations of the two products: [NH ,HI]. If we double the concentration of NH , the concentration of HI must decrease by approximately a factor of 2 to maintain equilibrium, as shown in . As a result, for a given concentration of either HI or NH , only a equilibrium composition that contains equal concentrations of both NH and HI is possible, for which [NH ] = [HI] = . Any point that lies below and to the left of the equilibrium curve (such as point in ) corresponds to < , and the reaction in will therefore proceed to the right as written, causing the composition of the system to move toward the equilibrium line. Conversely, any point that lies above and to the right of the equilibrium curve (such as point in ) corresponds to > , and the reaction in .4 will therefore proceed to the left as written, again causing the composition of the system to move toward the equilibrium line. By graphing equilibrium concentrations for a given system at a given temperature and pressure, we can predict the direction of reaction of that mixture when the system is not at equilibrium. When a system at equilibrium is perturbed in some way, the effects of the perturbation can be predicted qualitatively using Le Chatelier’s principle (named after the French chemist Henri Louis Le Chatelier, 1850–1936). This principle can be stated as follows: Stress occurs when any change in a system affects the magnitude of or . In , for example, increasing [NH ] produces a stress on the system that requires a decrease in [HI] for the system to return to equilibrium. As a further example, consider esters, which are one of the products of an equilibrium reaction between a carboxylic acid and an alcohol. (For more information on this type of reaction, see .) Esters are responsible for the scents we associate with fruits (such as oranges and bananas), and they are also used as scents in perfumes. Applying a stress to the reaction of a carboxylic acid and an alcohol will change the composition of the system, leading to an increase or a decrease in the amount of ester produced. In and , we explore how chemists control reactions conditions to affect equilibrium concentrations. In all reactions, if a stress is applied to a system at equilibrium, the composition of the system will change to counteract the applied stress (Le Chatelier’s principle). Write an equilibrium constant expression for each reaction and use this expression to predict what will happen to the concentration of the substance in bold when the indicated change is made if the system is to maintain equilibrium. equilibrium systems and changes equilibrium constant expressions and effects of changes Write the equilibrium constant expression, remembering that pure liquids and solids do not appear in the expression. From this expression, predict the change that must occur to maintain equilibrium when the indicated changes are made. Exercise Write an equilibrium constant expression for each reaction. What must happen to the concentration of the substance in bold when the indicated change occurs if the system is to maintain equilibrium? The has the same form as the equilibrium constant expression, but it is derived from concentrations obtained at any time. When a reaction system is at equilibrium, = . Graphs derived by plotting a few equilibrium concentrations for a system at a given temperature and pressure can be used to predict the direction in which a reaction will proceed. Points that do not lie on the line or curve represent nonequilibrium states, and the system will adjust, if it can, to achieve equilibrium. states that if a stress is applied to a system at equilibrium, the composition of the system will adjust to counteract the stress. : \[Q =\dfrac{[C]^c[D]^d}{[A]^a[B]^b} \] During a set of experiments, graphs were drawn of [reactants] versus [products] at equilibrium. Using and as your guides, sketch the shape of each graph using appropriate labels. Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium? Write an equilibrium constant expression for each reaction system. Given the indicated changes, how must the concentration of the species in bold change if the system is to maintain equilibrium? The data in the following table were collected at 450°C for the reaction \(N_{2(g)}+3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\) The reaction equilibrates at a pressure of 30 atm. The pressure on the system is first increased to 100 atm and then to 600 atm. Is the system at equilibrium at each of these higher pressures? If not, in which direction will the reaction proceed to reach equilibrium? For the reaction at 200°C is 2.0. A 6.00 L flask was used to carry out the reaction at this temperature. Given the experimental data in the following table, all at 200°C, when the data for each experiment were collected, was the reaction at equilibrium? If it was not at equilibrium, in which direction will the reaction proceed? The following two reactions are carried out at 823 K: \(CoO_{(s)}+H_{2(g)} \rightleftharpoons Co_{(s)}+H_2O_{(g)}\) with \(K=67\) \(CoO_{(s)}+CO_{(g)} \rightleftharpoons Co_{(s)}+CO_{2(g)}\) with \(K=490\) Hydrogen iodide (HI) is synthesized via for which = 54.5 at 425°C. Given a 2.0 L vessel containing 1.12 × 10 mol of H and 1.8 × 10 mol of I at equilibrium, what is the concentration of HI? Excess hydrogen is added to the vessel so that the vessel now contains 3.64 × 10 mol of H . Calculate and then predict the direction in which the reaction will proceed. What are the new equilibrium concentrations? Not at equilibrium; in both cases, the sum of the equilibrium partial pressures is than the total pressure, so the reaction will proceed to the right to decrease the pressure. | 15,428 | 2,019 |
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Salts can be thought of as being derived from the neutralization of an acid and a base. A salt formed from a strong acid and a strong base will not hydrolyze (e.g., extract a proton from water). When placed in water, these salts dissociate completely, and their ions remain uncombined in solution (e.g., the NaCl salt is formed from a strong acid (HCl) and a strong base (NaOH). In discussing the titration of acids and bases, the acid or basic properties of the corresponding salts after a neutralization reaction must be addressed to calculate the final pH of a solution. To calculate the pH of a salt solution one needs to know the concentration of the salt solution, whether the salt is an acidic, basic, or neutral salt, the equation for the interaction of the ion with the water, the equilibrium expression for this interaction and the K or K value. As a quick review, the general rules for the hydrolysis of monoprotic salts are: Do not be intimidated by the salts of polyprotic acids. Yes they're bigger and "badder" then most other salts, but they can be handled the exact same way as other salts, just with a bit more math. Take for example dissociation of carbonic acid (H_2CO_3): \[H_2CO_{3(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + HCO^-_{3(aq)}\] with \(K_{a1} = 2.5 \times 10^{-4}\) \[HCO^-_{3(aq)} + H_2O_{(l)} \rightleftharpoons H_3O^+_{(aq)} + CO^{2-}_{3(aq)}\] with \(K_{a2} = 5.61 \times 10^{-11}\). This means that when calculating the values for K of CO , The K of the first hydrolysis reaction will be \(K_{b1} = \dfrac{K_w}{K_{a2}}\) since it will go in the reverse order. What is the pH of a 0.100 M \(\ce{NaHSO_4}\) solution? Sulfuric acid is a strong acid, and the \(\ce pK_{\large\textrm a_{\Large 2}}\) of \(\ce{HSO4-}\) is 1.92. First review the ions generates immediate upon dissociation of the salt. \(HSO_4^{-}\) is part of the dissociation chain of diprotic sulfuric acid (\(H_2SO_4\)) with the following steps \(K_{a2} = 1.0 \times 10^{-2}\) from Table E1. The The \(NaHSO4\) salt is completely ionized in its solution. \[\ce{NaHSO4_{(s)} \rightarrow Na+_{(sq)} + HSO4-_{(sq)}} \] The anion then further ionizes (Equation \(\ref{step2SO3}\)). We can construct an to solve for the final value of this ionization and hence \([H_3O^+]\) \[K_{a2} = \dfrac{x^2}{0.100-x} = 0.0100 \nonumber\] \[\begin{align}
\ce{[H_3O+]}&= x \nonumber \\
&= \dfrac{-0.120 + \sqrt{0.012^2 + 4\times0.00120} }{2} \nonumber\\
&= \textrm{0.0292 M} \nonumber
\end{align}\] Thus, \[\mathrm{pH = - \log \;0.0292 = 1.54} \nonumber\] Note that while the dissociation is weak, the quadratic equation cannot be avoided since 0.0292 cannot be ignore when compared to 0.1 M. Which of the following solutions are acidic, basic, or neutral? Predict whether the \(Na_2HPO_4\) salt will form an acidic or basic solution when dissolved in water. You will need values from Table E1 to address this. First review the ions generates immediate upon dissociation of the salt. From Table E1, the three equilibria values for the three acid/base reactions of the phosphate ion are: \[H_3PO_{4(sq)} \rightleftharpoons H^+_{(aq)} + H_2PO^{2-}_{4(aq)} \label{1} \] with \(K_{a1} = 6.9 \times 10^{-3}\) \[H_2PO_{4(sq)}^{-} \rightleftharpoons H^+_{(aq)} + HPO^{2-}_{4(aq)} \label{2} \] with \(K_{a2} = 6.2 \times 10^{ -8}\) \[HPO_{4(sq)}^{2-} \rightleftharpoons H^+_{(aq)} + PO^{3-}_{4(aq)} \label{3} \] with \(K_{a3} = 4.8 \times 10^{-11}\) The low \(K_a\) value for the deprotonation of \(HPO_{4(sq)}^{2-}\) means that it is a poor acid and better functions as a (Brønsted-Lowry) base: \[HPO^{2-}_{4 (aq)} + H_2O_{(l)} \rightleftharpoons H_2PO^-_{4 (aq)} + OH^-_{ (aq)} \label{3b} \] The two possible reactions that \(HPO^{2-}_{4(aq)}\) undergoes are the acid reaction in Equation \(\ref{2}\) and the basic reaction in Equation \(\ref{3b}\). Whether this ion will cause the solution to be acidic or basic depends on whether the \(K_a\) for Equation \(\ref{2}\) is bigger than the \(K_b\) for Equation \(\ref{3b}\). The \(K_b\) of this reaction is related to the \(K_{a2}\) by \[K_b = \dfrac{K_w}{ K_a} \nonumber\] \[K_b = \dfrac{1 \times 10 ^{-14}}{ 6.2 \times 10^{-8}} = 1.6 \times 10^{-7} \text{ for } HPO_4^{2-} \nonumber\] Since \(K_b > K_a\), the solution will be . To determine just how basic requires the use of an ICE table as discussed in Example \(\Page {3}\). Solving the final pH for a \(Na_2HPO_4\) salt solution is more complex than for the \(NaHSO_4\), since the latter involves a single equilibrium. Two equilibria must be simultaneously solved for \(Na_2HPO_4\). However, often time (but not always), they can be easily approximated given the relative \(pk_a\) describing the reversible reaction. Calculate the pH of the solution containing 3.875 g of \(Na_2HPO_4\) that has been dissolved in a 250 mL of water. First it is best to identify the molarity of \(Na_2HPO_4\) that was initially generated (i.e., before any hydrolysis reactions happens). \[c = \dfrac{ 3.875 \;g / \text{molecular mass}}{\text{volume}} = \dfrac{3.875\;g / 141.98\; g/mol}{250 \; mL} = 0.11\;M\nonumber \] As discussed in Example \(\Page {1}\), the low \(K_a\) value for the deprotonation of \(HPO_{4(sq)}^{2-}\) means that it is a poor acid and instead functions as a base instead. From comparing the ionization constant in Example \(\Page {1}\), the \(Na_2HPO_4\) is a far better base than acid. If we ignore the acid properties of the ion and focus on the basic we get the following reaction \[HPO^{2-}_{4 (aq)} + H_2O_{(l)} \rightleftharpoons H_2PO^-_{4 (aq)} + OH^-_{ (aq)} \] with a \(K_b\) of \(1.6 \times 10^{-7}\). Now do an to get the final \([H^+]\) concentration.
\[ K_b = \dfrac{[H_2PO^-_{4 (aq)},OH^-_{ (aq)} ]}{[HPO ^{2-}_{4 (aq)}]} = 1.6 \times 10^{-7} \] \[ \dfrac{x^2}{0.11\; M -x} = 1.6 \times 10^{-7} \] if \(x << 0.11\;M\) then \(0.11\;M -x \approx 0.11 \; M\) then \[x =1.3 \times 10^{-4} = [OH^-]\] \[pOH = -\log_{10} [OH^-] = 3.8\] \[pH = pK_w - pOH] Assuming that the reaction is at 25°, then \(pK_w = 14\) and \[pH = 14- 3.8 = 10.2\] As expected, the solution is basic. | 6,120 | 2,021 |
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The quantitative relationship of reactants and products is called . Stoichiometric problems require you to calculate the amounts of reactants required for certain amounts of products, or amounts of products produced from certain amounts of reactants. If, in a chemical reaction, one or more reactants or products are gases, gas laws must be considered for the calculation. Usually, the applications of the ideal gas law give results within 5% precision. Below, we review several important concepts that are helpful for solving Stoichiometry Problems Involving Gases. The concept is the key to both stoichiometry and gas laws. A mole is a definite amount of substance. Mole is a unit based on the number of identities (i.e. atoms, molecules, ions, or particles). A mole of anything has the same number of identities as the number of atoms in exactly 12 grams of carbon-12, the most abundant isotope of carbon. Molar volume is defined as the volume occupied by one mole of a gas. Using the ideal gas law and assuming standard pressure and temperature (STP), the volume of one mole of gas can be calculated: \[PV=nRT\] \[V = \dfrac{nRT}{P}\] \[{V = \rm \dfrac{1.00 mol\ \cdot 0.08206 \dfrac{L atm}{mol K}\cdot 273 K}{1.00 atm }}\] \[V = 22.4 \rm L\] In other words, 1 mole of a gas will occupy 22.4 L at STP, assuming ideal gas behavior. At STP, the volume of a gas is only dependent on number of moles of that gas and is independent of molar mass. With this information we can calculate the density (\( \rho \)) of a gas using only its molar mass. First, starting with the definition of density \[ \rho =\dfrac{m}{V}\label {where D = density, m=mass and V= volume}\] we rearrange for volume: \[V=\dfrac{m}{ \rho }\] We then substitute \(V\) into the ideal gas equation and rearrange for density: \[PV=nRT\] \[P\dfrac{m}{ \rho }=nRT\] \[\rho=\dfrac{mP}{nRT}\] Finally, we remember that molar mass is equal to mass divided by number of moles: \[MM =\dfrac{m}{n}\] and substitute this into our expression for density to give: \[ \rho = \dfrac{MM\cdot P}{RT}\] This equation can further be simplified if we assume STP: \[ \rho = \dfrac{MM\cdot 1 \rm atm}{(0.08206 \dfrac{\rm L atm}{\rm mol K}) \rm 273 K}\] \[ \rho = \dfrac{MM}{\rm 22.4 \dfrac{L}{mol}}\] Using this information, we can calculate the density of a gas using the gas's molar mass. Calculate the density of N gas at STP. Pressure (1 atm),temperature (273 K), the identity of the gas (N ). Density of N The molar mass of N : \[MM_{N2}=\rm2\cdot 14.0 g/mol = 28.0 g/mol\] Calculate the density of N \[ \rho = \dfrac{MM}{\rm 22.4 \dfrac{L}{mol}}\] \[ \rho = \dfrac{28.0 \rm g/mol}{\rm 22.4 \dfrac{L}{mol}}\] \[ \rho = 1.25 \rm g/L\] Calculate the density of Ne gas at 143 ºC and 4.3 atm. Pressure (4.3 atm), temperature (143 ºC ), the identity of the gas (Ne), the molar mass of Ne from the periodic table (20.2 g/mol). Density of Ne Calculate temperature in Kelvin: \[T = \rm 143 C + 273 = 416 K\] Calculate the density of Ne: \[ \rho = \dfrac{MM\cdot P}{RT}\] \[ \rho = \dfrac{\rm 20.2 g/mol \cdot 4.3 \rm atm}{(0.08206 \dfrac{\rm L atm}{\rm mol K}) \rm 416 K}\] \[ \rho = \rm 2.54 g/L\] The equations for calculating the density of a gas can rearranged to calculate the molar mass of a gas: \[MM = \dfrac{ \rho RT}{P}\label {where MM= molar mass and D = density}\] this can be further simplified if we work at STP: \[MM = \rho \cdot 22.4 L/mol\] We can use these equations to identify an unknown gas, as shown below: A unknown gas has density of 1.78 g/L at STP. What is the identify of this gas? Pressure (1.00 atm), temperature (273 K ), density of the gas (1.783 g/L) Identity of the unknown gas Since we are at STP, we can use the following equation to calculate molar mass: \[MM = \rho \cdot 22.4 L/mol\] \[MM = \rm 1.783 g/L \cdot 22.4 L/mol\] \[MM = \rm 39.9 g/mol\] The calculated molar mass is 33.9 g/mol. Examination of the periodic table reveals that Argon has a mass of 39.948 g/mol. Therefore, the unknown gas is most likely argon. Stoichiometry is the theme of the previous block of modules, and the ideal gas law is the theme of this block of modules. These subjects are related. Be prepared to solve problems requiring concepts or principles of stoichiometry and gases. For example, we can calculate the number of moles from a certain volume, temperature and pressure of a \(\ce{HCl}\) gas. When moles are dissolved in L solution, its concentration is M. Three examples are given to illustrate some calculations of stoichiometry involving gas laws and more are given in question form for you to practice. If 500 mL of \(\ce{HCl}\) gas at 300 K and 100 kPa dissolve in 100 mL of pure water, what is the concentration? Data required: R value 8.314 kPa L / (K mol). \(\begin{align}
n_{\textrm{HCl}} &= \mathrm{\dfrac{0.50\: L \times 100\: kPa}{8.314\: \dfrac{kPa\: L}{K\: mol} \times 300\: K}}\\
&= \mathrm{0.02\: mol}
\end{align}\) Concentration of \(\ce{HCl}\), \(\ce{[HCl]}\) \(\mathrm{[HCl] = \dfrac{0.02\: mol}{0.1\: L} = 0.2\: mol/L}\) Note that = 0.08205 L atm /(K mol) will not be suitable in this case. If you have difficulty, review Solutions. If 500 mL of \(\ce{HCl}\) gas at 300 K and 100 kPa dissolved in pure water requires 12.50 mL of the \(\ce{NaOH}\) solution to neutralize in a titration experiment, what is the concentration of the \(\ce{NaOH}\) solution?
Solution in Example 1 showed = 0.02 mol. From the titration experiment, we can conclude that there were 0.02 moles of \(\ce{NaOH}\) in 12.50 mL. Thus, \(\mathrm{[NaOH] = \dfrac{0.02\: mol}{0.0125\: L} = 1.60\: mol/L}\)
Think in terms of reaction, \(\begin{align}
\mathrm{HCl + NaOH \rightarrow NaCl + H_2O}
&\Leftarrow \mathrm{Reaction}\\
\mathrm{0.02\: mol \hspace{8px} 0.02\: mol \hspace{103px}}
&\Leftarrow \mathrm{Quantities\: reacted}
\end{align}\) Note that 0.02 mol of \(\ce{NaOH}\) is in 0.0125 mL solution. A 5.0-L air sample containing \(\ce{H2S}\) at STP is treated with a catalyst to promote the reaction \(\mathrm{H_2S + O_2 \rightarrow H_2O + S_{(solid)}}\). If 3.2 g of solid \(\ce{S}\) was collected, calculate the volume percentage of \(\ce{H2S}\) in the original sample. \(\mathrm{3.2\: g\: S\times\dfrac{1\: mol\: H_2S}{32\: g\: S}= 0.10\: mol\: H_2S}\) \(\begin{align}
\mathrm{V_{H_2S}} &= \mathrm{0.10\: mol \times 22.4\: L/mol}\\
&= \mathrm{2.24\: L}
\end{align}\) \(\begin{align}
\mathrm{Volume\: \%} &= \mathrm{\dfrac{2.25\: L}{5.0\: L}}\\
&= 0.45\\
&= 45 \%
\end{align}\) Data required: Atomic mass: \(\mathrm{H = 1}\); \(\mathrm{O = 16}\); \(\mathrm{S = 32}\). = 0.08205 L atm /(K mol) is now suitable values or molar volume at STP (22.4 L/mol) The volume percentage is also the mole percentage, but not the weight percentage. Hydrogen sulfide reacts with sulfur dioxide to give \(\ce{H2O}\) and \(\ce{S}\), \(\mathrm{H_2S + SO_2 \rightarrow H_2O + S_{(solid)}}\), unbalanced. If 6.0 L of \(\ce{H2S}\) gas at 750 torr produced 3.2 g of sulfur, calculate the temperature in C.
Balanced reaction: \(\mathrm{2 H_2S + SO_2 \rightarrow 2 H_2O + {3 S_{(solid)}}}\\
\mathrm{2\: mol \hspace{130px} 3\times32 = 96\: g}\) \(\mathrm{3.2\: g\: S\times\dfrac{2\: mol\: H_2S}{96\: g\: S}= 0.067\: mol\: H_2S}\) \(\mathrm{P = \dfrac{750}{760} = 0.987\: atm}\) \(\begin{align}
T =\dfrac{PV}{n R}&=\mathrm{\dfrac{0.987\: atm \times 6\: L}{0.067\: mol \times 0.08205\: \dfrac{atm\: L}{mol\: K}}}\\
&= \mathrm{1085\: K}\\
&= \mathrm{812^\circ C}
\end{align}\)
Atomic mass: \(\mathrm{H = 1.0}\); \(\mathrm{O = 16.0}\); \(\mathrm{S = 32.0}\). = 0.08205 L atm /(K mol) is OK but watch units used for pressure. When 50.0 mL of \(\ce{AgNO3}\) solution is treated with an excess amount of \(\ce{HI}\) gas to give 2.35 g of \(\ce{AgI}\), what is the concentration of the \(\ce{AgNO3}\) solution? \(\mathrm{2.35\: g\: AgI \times\dfrac{1\: mol\: Ag^+}{234.8\: g\: AgI}\times\dfrac{1\: mol\: AgNO_3}{1\: mol\: Ag^+}= 0.010\: mol\: AgNO_3}\) \(\begin{align}
\mathrm{[AgNO_3]} &= \mathrm{\dfrac{0.01\: mol\: AgNO_3}{0.050\: L}}\\
&= \mathrm{0.20\: M\: AgNO_3}
\end{align}\) A gas is involved, but there is no need to consider the gas law. At. mass: \(\mathrm{Ag = 107.9}\); \(\mathrm{N = 14.0}\); \(\mathrm{O = 16.0}\); \(\mathrm{I = 126.9}\) What volume (L) will 0.20 mol \(\ce{HI}\) occupy at 300 K and 100.0 kPa? \(\mathrm{R = 8.314\: \dfrac{kPa\: L}{K\: mol} = 0.08205\: \dfrac{atm\: L}{mol\: K}}\) \(\begin{align}
V &=\dfrac{n RT}{P}\\
&= \mathrm{\dfrac{0.20\: mol \times 8.314\, \dfrac{kPa\:L}{mol\: K} \times 300\: K}{100\: kPa}}\\
&= \mathrm{\mathrm{5\: L}
}\end{align}\) A 3.66-g sample containing \(\ce{Zn}\) (at.wt. 65.4) and \(\ce{Mg}\) (24.3) reacted with a dilute acid to produce 2.470 L \(\ce{H2}\) gas at 101.0 kPa and 300 K. Calculate the percentage of \(\ce{Zn}\) in the sample.
The number of moles of gas produced is the number of moles of metals in the sample. Once you know the number of moles, set up an equation to give the number of moles of metal in the sample. \(\begin{align}
n &= \mathrm{\dfrac{101\: kPa \times 2.470\: L}{8.3145\: \dfrac{kPa\: L}{mol\: K} \times 300\: K}}\\
&= \mathrm{0.100\: mol}
\end{align}\) Let be the mass of \(\ce{Zn}\), then the mass of \(\ce{Mg}\) is 3.66 - g. Thus, we have \(\dfrac{x}{65.4}+\dfrac{3.66 - x}{24.3}= \mathrm{0.100\: mole}\) Solving for gives = 1.96 g \(\ce{Zn}\), and the \(\mathrm{weight\: percent = 100 \times \dfrac{1.96}{3.66} = 53.6 \%}\)
Find the mole percent of \(\ce{Zn}\) in the sample. \(\mathrm{\#\: mol\: of\: Zn = \dfrac{1.96}{65.4} = 0.03\: mol}\) \(\mathrm{\#\: mol\: of\: Mg = \dfrac{1.70}{24.3} = 0.07\: mol}\) \(\mathrm{mole\: percent = 100 \times \dfrac{0.03}{0.03 + 0.07} = 30 \%}\) When a 2.00 g mixture of \(\ce{Na}\) and \(\ce{Ca}\) reacted with water, 1.164 L hydrogen was produced at 300.0 K and 100.0 kPa. What is the percentage of \(\ce{Na}\) in the sample? \(\ce{2 Na + 2 H2O \rightarrow 2 Na(OH) + H2_{\large{(g)}}}\)
\(\ce{Ca + H2O \rightarrow Ca(OH) + H2_{\large{(g)}}}\) Let x be the mass of \(\ce{Na}\), then (2.00-x) is the mass of \(\ce{Ca}\). We have the following relationship \(\mathrm{\dfrac{x\: g}{23.0\: g/mol}\times\dfrac{1\: mol\: H_2}{2\: mol\: Na}+\dfrac{(2.0 - x)\: g\: Ca}{40.1\: g\: Ca/mol}\times\dfrac{1\: mol\: H_2}{1\: mol\: Ca}=\dfrac{1.164\: L\: H_2 \times 100.0\: kPa}{8.3145\: kPa\: L\: mol^{-1}\: K^{-1}\: 300.0\: K}}\) Simplify to give \(\mathrm{\dfrac{x}{46.0}+\dfrac{2}{40.1}-\dfrac{x}{40.1}= 0.0467\:all\: in\: mol}\) Multiply all terms by (40.1 * 46.0) \(\mathrm{40.1\, x + 2 \times 46.0 - 46.0\, x = 86.1}\) Simplify \(\mathrm{-5.9\, x = 86.1 - 92.0 = -5.91}\) Thus, \(\mathrm{Mass\: of\: Na = x = 1.0\: g}\) \(\mathrm{Mass\: of\: Ca = 2.0 - x = 1.0\: g}\) \(\mathrm{Mass\: Percentage\: of\: Na = 100\times \dfrac{1}{2.0} = 50\%}\) \(\mathrm{Mole\: of\: Na = \dfrac{1}{23} = 0.0435\: mol}\) \(\mathrm{Mole\: percentage = \dfrac{\dfrac{1}{23}}{\dfrac{1}{23} + \dfrac{1}{40.1}} = 0.635 = 63.5\%}\) Compare this example with gravimetric analyses using the reaction \(\mathrm{Ag^+_{\large{(aq)}} + \sideset{ }{_{\large{(aq)}}^{-}}{Cl} \rightarrow AgCl_{\large{(s)}}}\) where \(\mathrm{\sideset{ }{_{\large{(aq)}}^{-}}{Cl}}\) comes from the disolution of two salts such as \(\ce{NaCl}\) and \(\ce{MgCl2}\). Also compare with analyses making use of the reaction \(\mathrm{Ba^{2+}_{\large{(aq)}} + SO^{2-}_{4\large{(aq)}} \rightarrow BaSO_{4\large{(s)}}}\) where the anion \(\mathrm{SO^{2-}_{4\large{(aq)}}}\) comes from the dissolution of two sulfate salts. This example is very similar to Example 7. | 11,646 | 2,023 |
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A large, one-holed rubber stopper is inserted in a porous clay cup (commonly used to isolate solutions in preparing voltaic cells), and a hose is connected to a water manometer. Freon is collected in a 1 L beaker (from a "Dust Off" aerosol, freon refill for air conditioner, etc.), and the beaker is brought up to surround the porous cup. The pressure decreases (air diffuses out of the cup faster than freon diffuses in). In a second trial, an inverted beaker of hydrogen is brought down on the inverted cup, and pressure increases in so much that it often blows water out of the manometer, as hydrogen diffuses in much faster than air diffuses out. Insert a septum in a 14/20 standard taper female end of a vacuum distillation adaptor inserted in a 50 mL round flask. Attach the vacuum adaptor to a vacuum pump. Fill syringes with freon and hydrogen, and, in turn, insert needle of each syringe through septum. To fill the syringes, insert the needle (phlebotomist style)into latex tubing between a clamp, and a stopcock on glass tubing which passes through a rubber stopper, over which a balloon of gas is attached (See Figure). Measure time to empty syringe. v1 / v2 = R1 / R2 = kt2 / kt1 = √M2 / √M1 Graham's Law of Effusion Apparatus Syringe Gas Filler | 1,271 | 2,024 |
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The analysis of a sample generates a chemical or physical signal that is proportional to the amount of analyte in the sample. This signal may be anything we can measure, such as volume or absorbance. It is convenient to divide analytical techniques into two general classes based on whether the signal is proportional to the mass or moles of analyte, or is proportional to the analyte’s concentration Consider the two graduated cylinders in Figure 3.3.1
, each of which contains a solution of 0.010 M Cu(NO ) . Cylinder 1 contains 10 mL, or \(1.0 \times 10^{-4}\) moles of Cu , and cylinder 2 contains 20 mL, or \(2.0 \times 10^{-4}\) moles of Cu . If a technique responds to the absolute amount of analyte in the sample, then the signal due to the analyte \[S_A = k_A n_A \label{3.1}\] where is the moles or grams of analyte in the sample, and is a proportionality constant. Because cylinder 2 contains twice as many moles of Cu as cylinder 1, analyzing the contents of cylinder 2 gives a signal twice as large as that for cylinder 1. A second class of analytical techniques are those that respond to the analyte’s concentration, \[S_A = k_A C_A \label{3.2}\] Since the solutions in both cylinders have the same concentration of Cu , their analysis yields identical signals. A technique that responds to the absolute amount of analyte is a technique. Mass and volume are the most common signals for a total analysis technique, and the corresponding techniques are gravimetry ( ) and titrimetry ( ). With a few exceptions, the signal for a total analysis technique is the result of one or more chemical reactions, the stoichiometry of which determines the value of in Equation \ref{3.1}. Historically, most early analytical methods used a total analysis technique. For this reason, total analysis techniques are often called “classical” techniques. Spectroscopy ( ) and electrochemistry ( ), in which an optical or an electrical signal is proportional to the relative amount of analyte in a sample, are examples of concentration techniques. The relationship between the signal and the analyte’s concentration is a theoretical function that depends on experimental conditions and the instrumentation used to measure the signal. For this reason the value of in Equation \ref{3.2} is determined experimentally. Since most concentration techniques rely on measuring an optical or electrical signal, they also are known as “instrumental” techniques. | 2,465 | 2,025 |
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The analysis of a sample generates a chemical or physical signal that is proportional to the amount of analyte in the sample. This signal may be anything we can measure, such as volume or absorbance. It is convenient to divide analytical techniques into two general classes based on whether the signal is proportional to the mass or moles of analyte, or is proportional to the analyte’s concentration Consider the two graduated cylinders in Figure 3.3.1
, each of which contains a solution of 0.010 M Cu(NO ) . Cylinder 1 contains 10 mL, or \(1.0 \times 10^{-4}\) moles of Cu , and cylinder 2 contains 20 mL, or \(2.0 \times 10^{-4}\) moles of Cu . If a technique responds to the absolute amount of analyte in the sample, then the signal due to the analyte \[S_A = k_A n_A \label{3.1}\] where is the moles or grams of analyte in the sample, and is a proportionality constant. Because cylinder 2 contains twice as many moles of Cu as cylinder 1, analyzing the contents of cylinder 2 gives a signal twice as large as that for cylinder 1. A second class of analytical techniques are those that respond to the analyte’s concentration, \[S_A = k_A C_A \label{3.2}\] Since the solutions in both cylinders have the same concentration of Cu , their analysis yields identical signals. A technique that responds to the absolute amount of analyte is a technique. Mass and volume are the most common signals for a total analysis technique, and the corresponding techniques are gravimetry ( ) and titrimetry ( ). With a few exceptions, the signal for a total analysis technique is the result of one or more chemical reactions, the stoichiometry of which determines the value of in Equation \ref{3.1}. Historically, most early analytical methods used a total analysis technique. For this reason, total analysis techniques are often called “classical” techniques. Spectroscopy ( ) and electrochemistry ( ), in which an optical or an electrical signal is proportional to the relative amount of analyte in a sample, are examples of concentration techniques. The relationship between the signal and the analyte’s concentration is a theoretical function that depends on experimental conditions and the instrumentation used to measure the signal. For this reason the value of in Equation \ref{3.2} is determined experimentally. Since most concentration techniques rely on measuring an optical or electrical signal, they also are known as “instrumental” techniques. | 2,465 | 2,026 |
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Measurements of the rates of oxidation-reduction reactions began in the late 1940s. A great deal of the early experimental work was carried out by inorganic chemists, and by the 1970s the reactivity patterns of many complexes had been uncovered. Chemists studying the mechanisms of metalloprotein electrontransfer reactions frequently seek parallels with the redox behavior of less-complicated inorganic complexes. In examining biological electron transfers, it is important to remember that metalloproteins are more than just metal ions in disguise. Virtually every property of a protein (excluding its amino-acid sequence) depends on the solution pH. Redox proteins are very large polyelectrolytes whose redox prosthetic groups are typically buried in the protein interior. One important distinction between redox reactions of proteins and redox reactions of small transition-metal complexes is the magnitude of the electron donor-to-acceptor distance. The relevant distance for small molecules, unlike redox proteins, is generally taken to be van der Waals contact. Within the last ten years, it has been convincingly demonstrated that electrons can "tunnel" at significant rates across distances of 15 Å or more in protein interiors. Experimental investigation of the factors that control the rates of biological redox reactions has not come as far as the study of the electron transfers of metal complexes, because many more variables must be dealt with (e.g., asymmetric surface charge, nonspherical shape, uncertain details of structures of proteins complexed with small molecules or other proteins). Many experimental approaches have been pursued, including the covalent attachment of redox reagents to the surfaces of metalloproteins. The simplest reactions in solution chemistry are electron self-exchange reactions (Equation 6.10), in which the reactants and products are the same (the asterisk is used to identify a specific isotope). \[\;^{\ast}A_{ox} + A_{red} \rightarrow \;^{\ast}A_{red} + A_{ox} \tag{6.10}\] The only way to establish chemically that a reaction has taken place is to introduce an isotopic label. There is no change in the free energy (\(\Delta\)G° = 0) for this type of reaction. As will become evident later on, the reason why these types of reactions are studied is because self-exchange rates and activation parameters are needed to interpret redox reactions in which a net chemical change occurs. The experimental measurement of self-exchange rates is tedious and usually only results in an order-of-magnitude estimate of the rate constant (as inferred from the experimental timescale; see Table 6.2). Most of the protein self-exchange rates reported to date have been measured by NMR line-broadening studies. Other potentially useful methods, such as Mössbauer spectroscopy and EPR, have not been widely used. An elegant example of the measurement of an electron self-exchange rate of a redox protein was reported by Dahlin The copper ion of stellacyanin was removed and then replaced with either Cu or Cu. Oxidized [ Cu] stellacyanin was allowed to react with reduced [ Cu] stellacyanin for various times (10 ms to 7 min) at 20 °C, after which the reaction was quenched by lowering the solution temperature to -120°C using a rapid-freeze apparatus: \[\;^{63}Cu^{2+} + \;^{65}Cu^{+} \rightarrow \;^{63}Cu^{+} + \;^{65}Cu^{2+} \tag{6.11}\] Subtle differences in the EPR spectra (Figure 6.20) of the two isotopic forms of stellacyanin (due to a small difference in the nuclear magnetic moments of the two isotopes) were used to monitor the progress of the reaction, yielding a rate constant of 1.2 x 10 M s . Much more common are cross reactions (Equation 6.12), where A is the oxidized reactant, B is the reduced reactant, A is the reduced product, and B is the oxidized product. \[A_{ox} + B_{red} \rightarrow A_{red} + B_{ox} \tag{6.12}\] For these reactions, \(\Delta\)G° \(\neq\) 0. The experimental measurement of cross-reaction rates is generally more straightforward than the measurement of self-exchange rates. Either the reactants are simply mixed together, or a thermodynamically unstable system is generated rapidly (via pulse radiolysis, flash photolysis, or temperature-jump relaxation) to initiate the redox reaction. Absorption spectroscopy has almost always been used to monitor the progress of protein cross reactions. The primary goal of theory, as will become evident, is to provide a relationship between \(\Delta\)G° and \(\Delta\)G for cross reactions. Both self-exchange and cross reactions can be broadly classified as inner-sphere or outer-sphere reactions. In an inner-sphere reaction, a ligand is shared between the oxidant and reductant in the transition state. An outer-sphere reaction, on the other hand, is one in which the inner coordination shells of both the oxidant and the reductant remain intact in the transition state. There is no bond breaking or bond making, and no shared ligands between redox centers. Long-range electron transfers in biology are all of the outer-sphere type. | 5,085 | 2,027 |
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One of the most important properties of metallic elements is their ability to act as Lewis acids that form complexes with a variety of Lewis bases. A metal complex consists of a central metal atom or ion that is bonded to one or more (from the Latin ligare, meaning “to bind”), which are ions or molecules that contain one or more pairs of electrons that can be shared with the metal. Metal complexes can be neutral, such as Co(NH ) Cl ; positively charged, such as [Nd(H O) ] ; or negatively charged, such as [UF ] . Electrically charged metal complexes are sometimes called . A contains one or more metal complexes. Coordination compounds are important for at least three reasons. First, most of the elements in the periodic table are metals, and almost all metals form complexes, so metal complexes are a feature of the chemistry of more than half the elements. Second, many industrial catalysts are metal complexes, and such catalysts are steadily becoming more important as a way to control reactivity. For example, a mixture of a titanium complex and an organometallic compound of aluminum is the catalyst used to produce most of the polyethylene and polypropylene “plastic” items we use every day. Finally, transition-metal complexes are essential in biochemistry. Examples include hemoglobin, an iron complex that transports oxygen in our blood; cytochromes, iron complexes that transfer electrons in our cells; and complexes of Fe, Zn, Cu, and Mo that are crucial components of certain enzymes, the catalysts for all biological reactions. Coordination compounds have been known and used since antiquity; probably the oldest is the deep blue pigment called Prussian blue: \(\ce{KFe2(CN)6}\). The chemical nature of these substances, however, was unclear for a number of reasons. For example, many compounds called “double salts” were known, such as \(\ce{AlF3·3KF}\), \(\ce{Fe(CN)2·4KCN}\), and \(\ce{ZnCl2·2CsCl}\), which were combinations of simple salts in fixed and apparently arbitrary ratios. Why should \(\ce{AlF3·3KF}\) exist but not \(\ce{AlF3·4KF}\) or \(\ce{AlF3·2KF}\)? And why should a 3:1 KF:AlF3 mixture have different chemical and physical properties than either of its components? Similarly, adducts of metal salts with neutral molecules such as ammonia were also known—for example, \(\ce{CoCl3·6NH3}\), which was first prepared sometime before 1798. Like the double salts, the compositions of these adducts exhibited fixed and apparently arbitrary ratios of the components. For example, \(\ce{CoCl3·6NH3}\), \(\ce{CoCl3·5NH3}\), \(\ce{CoCl3·4NH3}\), and \(\ce{CoCl3·3NH3}\) were all known and had very different properties, but despite all attempts, chemists could not prepare \(\ce{CoCl3·2NH3}\) or \(\ce{CoCl3·NH3}\). Although the chemical composition of such compounds was readily established by existing analytical methods, their chemical nature was puzzling and highly controversial. The major problem was that what we now call valence (i.e., the oxidation state) and coordination number were thought to be identical. As a result, highly implausible (to modern eyes at least) structures were proposed for such compounds, including the “Chattanooga choo-choo” model for CoCl ·4NH shown here. The modern theory of coordination chemistry is based largely on the work of Alfred Werner (1866–1919; Nobel Prize in Chemistry in 1913). In a series of careful experiments carried out in the late 1880s and early 1890s, he examined the properties of several series of metal halide complexes with ammonia. For example, five different “adducts” of ammonia with PtCl were known at the time: PtCl ·nNH (n = 2–6). Some of Werner’s original data on these compounds are shown in Table \(\Page {1}\). The electrical conductivity of aqueous solutions of these compounds was roughly proportional to the number of ions formed per mole, while the number of chloride ions that could be precipitated as AgCl after adding Ag (aq) was a measure of the number of “free” chloride ions present. For example, Werner’s data on PtCl ·6NH in Table \(\Page {1}\) showed that all the chloride ions were present as free chloride. In contrast, PtCl ·2NH was a neutral molecule that contained no free chloride ions. Werner, the son of a factory worker, was born in Alsace. He developed an interest in chemistry at an early age, and he did his first independent research experiments at age 18. While doing his military service in southern Germany, he attended a series of chemistry lectures, and he subsequently received his PhD at the University of Zurich in Switzerland, where he was appointed professor of chemistry at age 29. He won the Nobel Prize in Chemistry in 1913 for his work on coordination compounds, which he performed as a graduate student and first presented at age 26. Apparently, Werner was so obsessed with solving the riddle of the structure of coordination compounds that his brain continued to work on the problem even while he was asleep. In 1891, when he was only 25, he woke up in the middle of the night and, in only a few hours, had laid the foundation for modern coordination chemistry. These data led Werner to postulate that metal ions have two different kinds of valence: (1) a primary valence ( ) that corresponds to the positive charge on the metal ion and (2) a secondary valence ( ) that is the total number of ligand-metal bonds bound to the metal ion. If \(\ce{Pt}\) had a primary valence of 4 and a secondary valence of 6, Werner could explain the properties of the \(\ce{PtCl4·NH3}\) adducts by the following reactions, where the metal complex is enclosed in square brackets: \[\begin{align*} \mathrm{[Pt(NH_3)_6]Cl_4} &\rightarrow \mathrm{[Pt(NH_3)_6]^{4+}(aq)+4Cl^-(aq)} \\[4pt]
\mathrm{[Pt(NH_3)_5Cl]Cl_3} &\rightarrow \mathrm{[Pt(NH_3)_5Cl]^{3+}(aq) +3Cl^-(aq)}\\[4pt]
\mathrm{[Pt(NH_3)_4Cl_2]Cl_2} &\rightarrow \mathrm{[Pt(NH_3)_4Cl_2]^{2+}(aq) +2Cl^-(aq)}\\[4pt]
\mathrm{[Pt(NH_3)_3Cl_3]Cl} &\rightarrow \mathrm{[Pt(NH_3)_3Cl_3]^+(aq) + Cl^-(aq)}\\[4pt]
\mathrm{[Pt(NH_3)_2Cl_4]} &\rightarrow \mathrm{[Pt(NH_3)_2Cl_4]^0(aq)} \end{align*} \label{23.9}\] Further work showed that the two missing members of the series—[Pt(NH )Cl ] and [PtCl ] —could be prepared as their mono- and dipotassium salts, respectively. Similar studies established coordination numbers of 6 for Co and Cr and 4 for Pt and Pd . Werner’s studies on the analogous Co complexes also allowed him to propose a structural model for metal complexes with a coordination number of 6. Thus he found that [Co(NH ) ]Cl (yellow) and [Co(NH ) Cl]Cl (purple) were 1:3 and 1:2 electrolytes. Unexpectedly, however, two different [Co(NH ) Cl ]Cl compounds were known: one was red, and the other was green (Figure \(\Page {1a}\)). Because both compounds had the same chemical composition and the same number of groups of the same kind attached to the same metal, there had to be something different about the arrangement of the ligands around the metal ion. Werner’s key insight was that the six ligands in [Co(NH ) Cl ]Cl had to be arranged at the vertices of an octahedron because that was the only structure consistent with the existence of two, and only two, arrangements of ligands (Figure \(\Page {1b}\). His conclusion was corroborated by the existence of only two different forms of the next compound in the series: Co(NH ) Cl . In Werner’s time, many complexes of the general formula MA B were known, but no more than two different compounds with the same composition had been prepared for any metal. To confirm Werner’s reasoning, calculate the maximum number of different structures that are possible for six-coordinate MA B complexes with each of the three most symmetrical possible structures: a hexagon, a trigonal prism, and an octahedron. What does the fact that no more than two forms of any MA B complex were known tell you about the three-dimensional structures of these complexes? : three possible structures and the number of different forms known for MA B complexes : number of different arrangements of ligands for MA B complex for each structure Sketch each structure, place a B ligand at one vertex, and see how many different positions are available for the second B ligand. The three regular six-coordinate structures are shown here, with each coordination position numbered so that we can keep track of the different arrangements of ligands. For each structure, all vertices are equivalent. We begin with a symmetrical MA complex and simply replace two of the A ligands in each structure to give an MA B complex: For the hexagon, we place the first B ligand at position 1. There are now three possible places for the second B ligand: at position 2 (or 6), position 3 (or 5), or position 4. These are the only possible arrangements. The (1, 2) and (1, 6) arrangements are chemically identical because the two B ligands are adjacent to each other. The (1, 3) and (1, 5) arrangements are also identical because in both cases the two B ligands are separated by an A ligand. Turning to the trigonal prism, we place the first B ligand at position 1. Again, there are three possible choices for the second B ligand: at position 2 or 3 on the same triangular face, position 4 (on the other triangular face but adjacent to 1), or position 5 or 6 (on the other triangular face but not adjacent to 1). The (1, 2) and (1, 3) arrangements are chemically identical, as are the (1, 5) and (1, 6) arrangements. In the octahedron, however, if we place the first B ligand at position 1, then we have only two choices for the second B ligand: at position 2 (or 3 or 4 or 5) or position 6. In the latter, the two B ligands are at opposite vertices of the octahedron, with the metal lying directly between them. Although there are four possible arrangements for the former, they are chemically identical because in all cases the two B ligands are adjacent to each other. The number of possible MA B arrangements for the three geometries is thus: hexagon, 3; trigonal prism, 3; and octahedron, 2. The fact that only two different forms were known for all MA B complexes that had been prepared suggested that the correct structure was the octahedron but did not prove it. For some reason one of the three arrangements possible for the other two structures could have been less stable or harder to prepare and had simply not yet been synthesized. When combined with analogous results for other types of complexes (e.g., MA B ), however, the data were best explained by an octahedral structure for six-coordinate metal complexes. Determine the maximum number of structures that are possible for a four-coordinate MA B complex with either a square planar or a tetrahedral symmetrical structure. square planar, 2; tetrahedral, 1 The coordination numbers of metal ions in metal complexes can range from 2 to at least 9. In general, the differences in energy between different arrangements of ligands are greatest for complexes with low coordination numbers and decrease as the coordination number increases. Usually only one or two structures are possible for complexes with low coordination numbers, whereas several different energetically equivalent structures are possible for complexes with high coordination numbers (n > 6). The following presents the most commonly encountered structures for coordination numbers 2–9. Many of these structures should be familiar to you from our discussion of the valence-shell electron-pair repulsion ( ) model because they correspond to the lowest-energy arrangements of n electron pairs around a central atom. Compounds with low coordination numbers exhibit the greatest differences in energy between different arrangements of ligands. Although it is rare for most metals, this coordination number is surprisingly common for d metal ions, especially Cu , Ag , Au , and Hg . An example is the [Au(CN) ] ion, which is used to extract gold from its ores. As expected based on VSEPR considerations, these complexes have the linear L–M–L structure shown here. Although it is also rare, this coordination number is encountered with d metal ions such as Cu and Hg . Among the few known examples is the HgI ion. Three-coordinate complexes almost always have the trigonal planar structure expected from the VSEPR model. Two common structures are observed for four-coordinate metal complexes: tetrahedral and square planar. The tetrahedral structure is observed for all four-coordinate complexes of nontransition metals, such as [BeF ] , and d ions, such as [ZnCl ] . It is also found for four-coordinate complexes of the first-row transition metals, especially those with halide ligands (e.g., [FeCl ] and [FeCl ] ). In contrast, square planar structures are routinely observed for four-coordinate complexes of second- and third-row transition metals with d electron configurations, such as Rh and Pd , and they are also encountered in some complexes of Ni and Cu . This coordination number is less common than 4 and 6, but it is still found frequently in two different structures: trigonal bipyramidal and square pyramidal. Because the energies of these structures are usually rather similar for most ligands, many five-coordinate complexes have distorted structures that lie somewhere between the two extremes. This coordination number is by far the most common. The six ligands are almost always at the vertices of an octahedron or a distorted octahedron. The only other six-coordinate structure is the trigonal prism, which is very uncommon in simple metal complexes. This relatively uncommon coordination number is generally encountered for only large metals (such as the second- and third-row transition metals, lanthanides, and actinides). At least three different structures are known, two of which are derived from an octahedron or a trigonal prism by adding a ligand to one face of the polyhedron to give a “capped” octahedron or trigonal prism. By far the most common, however, is the pentagonal bipyramid. This coordination number is relatively common for larger metal ions. The simplest structure is the cube, which is rare because it does not minimize interligand repulsive interactions. Common structures are the square antiprism and the dodecahedron, both of which can be generated from the cube. This coordination number is found in larger metal ions, and the most common structure is the tricapped trigonal prism, as in [Nd(H O) ] . Transition metals form metal complexes, polyatomic species in which a metal ion is bound to one or more ligands, which are groups bound to a metal ion. Complex ions are electrically charged metal complexes, and a coordination compound contains one or more metal complexes. Metal complexes with low coordination numbers generally have only one or two possible structures, whereas those with coordination numbers greater than six can have several different structures. Coordination numbers of two and three are common for d metal ions. Tetrahedral and square planar complexes have a coordination number of four; trigonal bipyramidal and square pyramidal complexes have a coordination number of five; and octahedral complexes have a coordination number of six. At least three structures are known for a coordination number of seven, which is generally found for only large metal ions. Coordination numbers of eight and nine are also found for larger metal ions. | 15,386 | 2,028 |
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Although the is a simple and useful method for qualitatively predicting the structures of a wide range of compounds, it is infallible. It predicts, for example, that H S and PH should have structures similar to those of H O and NH , respectively. In fact, structural studies have shown that the H–S–H and H–P–H angles are more than 12° smaller than the corresponding bond angles in H O and NH . More disturbing, the VSEPR model predicts that the simple group 2 halides (MX ), which have four valence electrons, should all have linear X–M–X geometries. Instead, many of these species, including SrF and BaF , are significantly bent. A more sophisticated treatment of bonding is needed for systems such as these. In this section, we present a quantum mechanical description of bonding, in which bonding electrons are viewed as being localized between the nuclei of the bonded atoms. Valence Bond Theory has its roots in Gilbert Newton Lewis’s paper . Possibly unaware that Lewis’s model existed, Walter Heitler and Fritz London came up with the idea that resonance and wavefunctions contributed to chemical bonds, in which they used dihydrogen as an example. Their theory was equivalent to Lewis’s theory, with the difference of quantum mechanics being developed. Nonetheless, Heitler and London's theory proved to be successful, providing Linus Pauling and John C. Slater with an opportunity to assemble a general chemical theory containing all of these ideas. Valence Bond Theory was the result, which included the ideas of resonance, covalent-ionic superposition, atomic orbital overlap, and hybridization to describe chemical bonds. You learned that as two hydrogen atoms approach each other from an infinite distance, the energy of the system reaches a minimum. This region of minimum energy in the energy diagram corresponds to the formation of a covalent bond between the two atoms at an H–H distance of 74 pm. According to quantum mechanics, bonds form between atoms because their atomic orbitals overlap, with each region of overlap accommodating a maximum of two electrons with opposite spin, in accordance with the Pauli principle. In this case, a bond forms between the two hydrogen atoms when the singly occupied 1 atomic orbital of one hydrogen atom overlaps with the singly occupied 1 atomic orbital of a second hydrogen atom. Electron density between the nuclei is increased because of this orbital overlap and results in a (Figure \(\Page {1}\)). Although both Lewis and VSEPR structures also contain localized electron-pair bonds, neither description uses an atomic orbital approach to predict the stability of the bond. Doing so forms the basis for a description of chemical bonding known as valence bond theory, which is built on two assumptions: Figure \(\Page {2}\) shows an electron-pair bond formed by the overlap of two atomic orbitals, two atomic orbitals, and an and an orbital where = 2. Maximum overlap occurs between orbitals with the same spatial orientation and similar energies. An important aspect of Valence Bond theory is the concept of maximum overlap to form the strongest possible covalent bonds. Let’s examine the bonds in BeH , for example. According to the VSEPR model, BeH is a linear compound with four valence electrons and two Be–H bonds. Its bonding can also be described using an atomic orbital approach. Beryllium has a 1 2 electron configuration, and each H atom has a 1 electron configuration. Because the Be atom has a filled 2 subshell, however, it has no singly occupied orbitals available to overlap with the singly occupied 1 orbitals on the H atoms. If a singly occupied 1 orbital on hydrogen were to overlap with a filled 2 orbital on beryllium, the resulting bonding orbital would contain electrons, but the maximum allowed by quantum mechanics is . How then is beryllium able to bond to two hydrogen atoms? One way would be to add enough energy to excite one of its 2 electrons into an empty 2 orbital and reverse its spin, in a process called promotion: In this excited state, the Be atom would have two singly occupied atomic orbitals (the 2 and one of the 2 orbitals), each of which could overlap with a singly occupied 1 orbital of an H atom to form an electron-pair bond. Although this would produce BeH , the two Be–H bonds would not be equivalent: the 1 orbital of one hydrogen atom would overlap with a Be 2 orbital, and the 1 orbital of the other hydrogen atom would overlap with an orbital of a different energy, a Be 2 orbital. Experimental evidence indicates, however, that the two Be–H bonds have identical energies. To resolve this discrepancy and explain how molecules such as BeH form, scientists developed the concept of hybridization. As one can see, Valence Bond Theory can help describe how bonds are formed. However, there are some notable failures when it comes to Valence Bond Theory. One such failure is dioxygen. Valence Bond Theory fails to predict dioxygen's paramagnitism; it predicts that oxygen is diamagnetic. A species is paramagnetic if electrons are not spin paired and diamagnetic if the electrons are spin paired. Since Valence Bond theory begins with the basis that atomic orbitals overlap to create bonds and through that reasoning, one can see that electrons are spin paired when bonds overlap, dioxygen is indeed predicted to be diamagnetic if Valence Bond Theory is used. In reality, that is not the case. Also, sp d and sp both have a coordinate number of four. Thus, Valence Bond Theory cannot predict whether the molecule is a square planar or the other shape (3). One must correctly draw the Lewis structure and use VSEPR to determine the shape. | 5,707 | 2,029 |
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The elements within the same group of the periodic table tend to exhibit similar physical and chemical properties. Four major factors affect reactivity of metals: nuclear charge, atomic radius, shielding effect and sublevel arrangement (of electrons). Metal reactivity relates to ability to lose electrons (oxidize), form basic hydroxides, form ionic compounds with non-metals. In general, the bigger the atom, the greater the ability to lose electrons. The greater the shielding, the greater the ability to lose electrons. Therefore, metallic character increases going down the table, and decreases going across -- so the most active metal is towards the left and down. The word "alkali" is derived from an Arabic word meaning "ashes". Many sodium and potassium compounds were isolated from wood ashes (\(\ce{Na2CO3}\) and \(\ce{K2CO3}\) are still occasionally referred to as "soda ash" and "potash"). In the alkali group, as we go down the group we have elements Lithium (Li), Sodium (Na), Potassium (K), Rubidium (Rb), Cesium (Cs) and Francium (Fr). Several physical properties of these elements are compared in Table \(\Page {1}\). These elements have all only one electron in their outermost shells. All the elements show metallic properties and have valence +1, hence they give up electron easily. As we move down the group (from Li to Fr), the following trends are observed (Table \(\Page {1}\)): The alkali metals have the lowest \(I_1\) values of the elements This represents the relative ease with which the lone electron in the outer 's' orbital can be removed. : \[M \rightarrow M^+ + e- \nonumber \] Due to this reactivity, the alkali metals are found in nature only as compounds. The alkali metals combine directly with most nonmetals: \[2M_{(s)} + H_{2(g)} \rightarrow 2MH(s) \nonumber \] (Note: hydrogen is present in the metal hydride as the H ion) \[2M_{(s)} + S_{(s)} \rightarrow M_2S_{(s)} \nonumber \] React with chlorine to form solid \[2M_{(s)} + Cl_{2(g)} \rightarrow 2MCl_{(s)} \nonumber \] Alkali metals react with water to produce hydrogen gas and alkali metal hydroxides; this is a very exothermic reaction (Figure \(\Page {1}\)). \[2M_{(s)} + 2H_2O_{(l)} \rightarrow 2MOH_{(aq)} + H_{2(g)} \nonumber \] \[4Li_{(s)} + O_{2 (g)} \rightarrow \underbrace{2Li_2O_{(s)}}_{\text{lithium oxide}} \nonumber \] Other alkali metals form metal peroxides (contains O ion) \[2Na(s) + O_{2 (g)} \rightarrow \underbrace{Na_2O_{2(s)}}_{\text{sodium peroxide}} \nonumber \] K, Rb and Cs also form superoxides (O ion) \[K(s) + O_{2 (g)} \rightarrow \underbrace{KO_{2(s)}}_{\text{potassium superoxide}} \nonumber \] The color of a chemical is produced when a valence electron in an atom is excited from one energy level to another by visible radiation. In this case, the particular frequency of light that excites the electron is . Thus, the remaining light that you see is white light devoid of one or more wavelengths (thus appearing colored). Alkali metals, having lost their outermost electrons, have no electrons that can be excited by visible radiation. Alkali metal salts and their aqueous solution are colorless unless they contain a colored anion. When alkali metals are placed in a flame the ions are reduced (gain an electron) in the lower part of the flame. The electron is excited (jumps to a higher orbital) by the high temperature of the flame. When the excited electron falls back down to a lower orbital a photon is released. The transition of the valence electron of sodium from the 3p down to the 3s subshell results in release of a photon with a wavelength of 589 nm (yellow) Flame colors: Compared with the alkali metals, the alkaline earth metals are typically harder, more dense, melt at a higher temperature. The first ionization energies (\(I_1\)) of the alkaline earth metals are not as low as the alkali metals. The alkaline earth metals are therefore less reactive than the alkali metals (Be and Mg are the least reactive of the alkaline earth metals). Several physical properties of these elements are compared in Table \(\Page {2}\). Calcium, and elements below it, react readily with water at room temperature: \[Ca_{(s)} + 2H_2O_{(l)} \rightarrow Ca(OH)_{2(aq)} + H_{2(g)} \nonumber \] The tendency of the alkaline earths to lose their two valence electrons is demonstrated in the reactivity of Mg towards chlorine gas and oxygen: \[Mg_{(s)} + Cl_{2(g)} \rightarrow MgCl_{2(s)} \nonumber \] \[2Mg_{(s)} + O_{2(g)} \rightarrow 2MgO_{(s)} \nonumber \] The 2+ ions of the alkaline earth metals have a noble gas like electron configuration and are thus form colorless or white compounds (unless the anion is itself colored). Flame colors: ( ) | 4,708 | 2,030 |
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\( \newcommand{\Rsix}{8.31447\units{J$\,$K$\per\,$mol$\per$}} % gas constant value - 6 sig figs\) \( \newcommand{\jn}{\hspace3pt\lower.3ex{\Rule{.6pt}{2ex}{0ex}}\hspace3pt} \)
\( \newcommand{\ljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}} \hspace3pt} \)
\( \newcommand{\lljn}{\hspace3pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace1.4pt\lower.3ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise.45ex{\Rule{.6pt}{.5ex}{0ex}}\hspace-.6pt\raise1.2ex{\Rule{.6pt}{.5ex}{0ex}}\hspace3pt} \) A tall column of a gas mixture in a gravitational field, and a liquid solution in the cell of a spinning centrifuge rotor, are systems with equilibrium states that are nonuniform in pressure and composition. This section derives the ways in which pressure and composition vary spatially within these kinds of systems at equilibrium. Consider a tall column of a gas mixture in an earth-fixed lab frame. Our treatment will parallel that for a tall column of a pure gas in Sec. 8.1.4. We imagine the gas to be divided into many thin slab-shaped phases at different elevations in a rigid container, as in Fig. 8.1. We want to find the equilibrium conditions reached spontaneously when the system is isolated from its surroundings. The derivation is the same as that in Sec. 9.2.7, with the additional constraint that for each phase \(\pha\), \(\dif V\aph\) is zero in order that each phase stays at a constant elevation. The result is the relation \begin{equation} \dif S = \sum_{\pha\ne\pha'}\frac{T\aphp-T\aph}{T\aphp}\dif S\aph + \sum_i\sum_{\pha\ne\pha'}\frac{\mu_i\aphp-\mu_i\aph}{T\aphp}\dif n_i\aph \tag{9.8.1} \end{equation} In an equilibrium state, \(S\) is at a maximum and \(\dif S\) is zero for an infinitesimal change of any of the independent variables. This requires the coefficient of each term in the sums on the right side of Eq. 9.8.1 to be zero. The equation therefore tells that at equilibrium . The equation says nothing about the pressure. Just as the chemical potential of a pure substance at a given elevation is defined in this e-book as the molar Gibbs energy at that elevation (Sec. 8.1.4), the chemical potential of substance \(i\) in a mixture at elevation \(h\) is the partial molar Gibbs energy at that elevation. We define the standard potential \(\mu_i\st\gas\) of component \(i\) of the gas mixture as the chemical potential of \(i\) under standard state conditions at the reference elevation \(h{=}0\). At this elevation, the chemical potential and fugacity are related by \begin{equation} \mu_i(0) = \mu_i\st\gas + RT\ln\frac{\fug_i(0)}{p\st} \tag{9.8.2} \end{equation} If we reversibly raise a small sample of mass \(m\) of the gas mixture by an infinitesimal distance \(\dif h\), without heat and at constant \(T\) and \(V\), the fugacity \(\fug_i\) remains constant. The gravitational work during the elevation process is \(\dw'=mg\dif h\). This work contributes to the internal energy change: \(\dif U=T\dif S-p\dif V+\sum_i\mu_i\dif n_i+mg\dif h\). The total differential of the Gibbs energy of the sample is \begin{equation} \begin{split} \dif G & = \dif(U-TS+pV) \cr & = -S\dif T + V\difp + \sum_i\mu_i\dif n_i + mg\dif h \end{split} \tag{9.8.3} \end{equation} From this total differential, we write the reciprocity relation \begin{equation} \Pd{\mu_i}{h}{T,p,\allni} = \Pd{mg}{n_i}{T,p,n_{j\ne i},h} \tag{9.8.4} \end{equation} With the substitution \(m=\sum_i n_i M_i\) in the partial derivative on the right side, the partial derivative becomes \(M_i g\). At constant \(T\), \(p\), and composition, therefore, we have \(\dif\mu_i=M_i g\dif h\). Integrating over a finite elevation change from \(h=0\) to \(h=h'\), we obtain \begin{gather} \s{ \mu_i(h') - \mu_i(0) = \int_0^{h'}\!\!\!M_i g\dif h = M_i gh' } \tag{9.8.5} \cond{(\(\fug_i(h'){=}\fug_i(0)\) )} \end{gather} The general relation between \(\mu_i\), \(\fug_i\), and \(h\) that agrees with Eqs. 9.8.2 and 9.8.5 is \begin{equation} \mu_i(h) = \mu_i\st\gas + RT\ln\frac{\fug_i(h)}{p\st}+M_i gh \tag{9.8.6} \end{equation} In the equilibrium state of the tall column of gas, \(\mu_i(h)\) is equal to \(\mu_i(0)\). Equation 9.8.6 shows that this is only possible if \(\fug_i\) decreases as \(h\) increases. Equating the expressions given by this equation for \(\mu_i(h)\) and \(\mu_i(0)\), we have \begin{equation} \mu_i\st\gas + RT\ln\frac{\fug_i(h)}{p\st} + M_i gh = \mu_i\st\gas + RT\ln\frac{\fug_i(0)}{p\st} \tag{9.8.7} \end{equation} Solving for \(\fug_i(h)\) gives \begin{gather} \s{ \fug_i(h) = \fug_i(0)e^{-M_i gh/RT}} \tag{9.8.8} \cond{(gas mixture at equilibrium)} \end{gather} If the gas is an ideal gas mixture, \(\fug_i\) is the same as the partial pressure \(p_i\): \begin{gather} \s{ p_i(h) = p_i(0)e^{-M_i gh/RT}} \tag{9.8.9} \cond{(ideal gas mixture at equilibrium)} \end{gather} Equation 9.8.9 shows that each constituent of an ideal gas mixture individually obeys the barometric formula given by Eq. 8.1.13. The pressure at elevation \(h\) is found from \(p(h)=\sum_i p_i(h)\). If the constituents have different molar masses, the mole fraction composition changes with elevation. For example, in a binary ideal gas mixture the mole fraction of the constituent with the greater molar mass decreases with increasing elevation, and the mole fraction of the other constituent increases. This section derives equilibrium conditions of a dilute binary solution confined to a cell embedded in a spinning centrifuge rotor. The is the solution. The rotor’s angle of rotation with respect to a lab frame is not relevant to the state of the system, so we use a local reference frame fixed in the rotor as shown in Fig. 9.12(a). The values of heat, work, and energy changes measured in this rotating frame are different from those in a lab frame (Sec. G.9 in Appendix G). Nevertheless, the laws of thermodynamics and the relations derived from them are obeyed in the local frame when we measure the heat, work, and state functions in this frame (Sec. G.6). Note that an equilibrium state can only exist relative to the rotating local frame; an observer fixed in this frame would see no change in the state of the isolated solution over time. While the rotor rotates, however, there is no equilibrium state relative to the lab frame, because the system’s position in the frame constantly changes. We assume the centrifuge rotor rotates about the vertical \(z\) axis at a constant angular velocity \(\omega\). As shown in Fig. 9.12(a), the elevation of a point within the local frame is given by \(z\) and the radial distance from the axis of rotation is given by \(r\). In the rotating local frame, a body of mass \(m\) has exerted on it a centrifugal force \(F\sups{centr}=m\omega^2 r\) directed horizontally in the outward \(+r\) radial direction (Sec. G.9). The gravitational force in this frame, directed in the downward \(-z\) direction, is the same as the gravitational force in a lab frame. Because the height of a typical centrifuge cell is usually no greater than about one centimeter, in an equilibrium state the variation of pressure and composition between the top and bottom of the cell at any given distance from the axis of rotation is completely negligible—all the measurable variation is along the radial direction. There is also a Coriolis force that vanishes as the body’s velocity in the rotating local frame approaches zero. The centrifugal and Coriolis forces are or forces, in the sense that they are caused by the acceleration of the rotating frame rather than by interactions between particles. When we treat these forces as if they are real forces, we can use Newton’s second law of motion to relate the net force on a body and the body’s acceleration in the rotating frame (see Sec. G.6). To find conditions for equilibrium, we imagine the solution to be divided into many thin curved volume elements at different distances from the axis of rotation as depicted in Fig. 9.12(b). We treat each volume element as a uniform phase held at constant volume so that it is at a constant distance from the axis of rotation. The derivation is the same as the one used in the preceding section to obtain Eq. 9.8.1, and leads to the same conclusion: in an equilibrium state . We find the dependence of pressure on \(r\) as follows. Consider one of the thin slab-shaped volume elements of Fig. 9.12(b). The volume element is located at radial position \(r\) and its faces are perpendicular to the direction of increasing \(r\). The thickness of the volume element is \(\delta r\), the surface area of each face is \(\As\), and the mass of the solution in the volume element is \(m=\rho \As\delta r\). Expressed as components in the direction of increasing \(r\) of the forces exerted on the volume element, the force at the inner face is \(p\As\), the force at the outer face is \(-(p+\delta p)\As\), and the centrifugal force is \(m\omega^2 r = \rho \As\omega^2 r\delta r\). From Newton’s second law, the sum of these components is zero at equilibrium: \begin{equation} p \As-(p+\delta p)\As+\rho \As\omega^2 r\delta r = 0 \tag{9.8.10} \end{equation} or \(\delta p = \rho\omega^2 r\delta r\). In the limit as \(\delta r\) and \(\delta p\) are made infinitesimal, this becomes \begin{equation} \difp = \rho\omega^2 r\dif r \tag{9.8.11} \end{equation} We will assume the density \(\rho\) is uniform throughout the solution. (In the centrifugal field, this assumption is strictly true only if the solution is incompressible and its density is independent of composition.) Then integration of Eq. 9.8.11 yields \begin{equation} p''-p' = \int_{p'}^{p''}\!\difp = \rho\omega^2\!\int_{r'}^{r''}\!\!\!r\dif r = \frac{\rho \omega^2}{2}\left[\left(r''\right)^2-\left(r'\right)^2\right] \tag{9.8.12} \end{equation} where the superscripts \('\) and \(''\) denote positions at two different values of \(r\) in the cell. The pressure is seen to increase with increasing distance from the axis of rotation. Next we investigate the dependence of the solute concentration \(c\B\) on \(r\) in the equilibrium state of the binary solution. Consider a small sample of the solution of mass \(m\). Assume the extent of this sample in the radial direction is small enough for the variation of the centrifugal force field to be negligible. The reversible work in the local frame needed to move this small sample an infinitesimal distance \(\dif r\) at constant \(z\), \(T\), and \(p\), using an external force \(-F\sups{centr}\) that opposes the centrifugal force, is \begin{equation} \dw' = F\sur\dif r = (-F\sups{centr})\dif r = -m\omega^2 r\dif r \tag{9.8.13} \end{equation} This work is a contribution to the change \(\dif U\) of the internal energy. The Gibbs energy of the small sample in the local frame is a function of the independent variables \(T\), \(p\), \(n\A\), \(n\B\), and \(r\), and its total differential is \begin{equation} \begin{split} \dif G & = \dif(U-TS+pV) \cr & = -S\dif T + V\difp + \mu\A\dif n\A + \mu\B\dif n\B - m\omega^2 r\dif r \end{split} \tag{9.8.14} \end{equation} We use Eq. 9.8.14 to write the reciprocity relation \begin{equation} \Pd{\mu\B}{r}{T,p,n\A,n\B} = -\omega^2 r \Pd{m}{n\B}{T,p,n\A,r} \tag{9.8.15} \end{equation} Then, using \(m=n\A M\A+n\B M\B\), we obtain \begin{equation} \Pd{\mu\B}{r}{T,p,n\A,n\B} = -M\B \omega^2 r \tag{9.8.16} \end{equation} Thus at constant \(T\), \(p\), and composition, which are the conditions that allow the activity \(a\cbB\) to remain constant, \(\mu\B\) for the sample varies with \(r\) according to \(\dif\mu\B=-M\B \omega^2 r\dif r\). We integrate from radial position \(r'\) to position \(r''\) to obtain \begin{gather} \s{ \begin{split} \mu\B(r'') - \mu\B(r') & = - M\B \omega^2 \int_{r'}^{r''}\!\!\!r\dif r \cr & = -\onehalf M\B\omega^2\left[\left(r''\right)^2-\left(r'\right)^2\right] \end{split} } \tag{9.8.17} \cond{(\(a\cbB(r''){=}a\cbB(r')\) )} \end{gather} Let us take \(r'\) as a reference position, such as the end of the centrifuge cell farthest from the axis of rotation. We define the standard chemical potential \(\mu\st\cbB\) as the solute chemical potential under standard state conditions on a concentration basis at this position. The solute chemical potential and activity at this position are related by \begin{equation} \mu\B(r')=\mu\cbB\st+RT\ln a\cbB(r') \tag{9.8.18} \end{equation} From Eqs. 9.8.17 and 9.8.18, we obtain the following general relation between \(\mu\B\) and \(a\cbB\) at an arbitrary radial position \(r''\): \begin{equation} \mu\B(r'') = \mu\cbB\st + RT\ln a\cbB(r'') -\onehalf M\B\omega^2\left[\left(r''\right)^2-\left(r'\right)^2\right] \tag{9.8.19} \end{equation} We found earlier that when the solution is in an equilibrium state, \(\mu\B\) is independent of \(r\)—that is, \(\mu\B(r'')\) is equal to \(\mu\B(r')\) for any value of \(r''\). When we equate expressions given by Eq. 9.8.19 for \(\mu\B(r'')\) and \(\mu\B(r')\) and rearrange, we obtain the following relation between the activities at the two radial positions: \begin{gather} \s{\ln\frac{a\cbB(r'')}{a\cbB(r')} = \frac{M\B \omega^2}{2RT}\left[\left(r''\right)^2-\left(r'\right)^2\right]} \tag{9.8.20} \cond{(solution in centrifuge} \nextcond{cell at equilibrium)} \end{gather} The solute activity is related to the concentration \(c\B\) by \(a\cbB = \G\cbB \g\cbB c\B/c\st\). We assume the solution is sufficiently dilute for the activity coefficient \(\g\cbB\) to be approximated by \(1\). The pressure factor is given by \(\G\cbB \approx \exp\left[ V\B^{\infty}(p-p\st)/RT \right]\) (Table 9.6). These relations give us another expression for the logarithm of the ratio of activities: \begin{equation} \ln \frac{a\cbB(r'')}{a\cbB(r')} = \frac{V\B^{\infty}(p''-p')}{RT} + \ln\frac{c\B(r'')}{c\B(r')} \tag{9.8.21} \end{equation} We substitute for \(p''-p'\) from Eq. 9.8.12. It is also useful to make the substitution \(V\B^{\infty}=M\B v\B^{\infty}\), where \(v\B^{\infty}\) is the partial specific volume of the solute at infinite dilution. When we equate the two expressions for \(\ln[a\cbB(r'')/a\cbB(r')]\), we obtain finally \begin{gather} \s{ \ln\frac{c\B(r'')}{c\B(r')} = \frac{M\B\left(1-v\B^{\infty}\rho\right)\omega^2}{2RT} \left[\left(r''\right)^2-\left(r'\right)^2\right] } \tag{9.8.22} \cond{(solution in centrifuge} \nextcond{cell at equilibrium)} \end{gather} This equation shows that if the solution density \(\rho\) is less than the effective solute density \(1/v\B^{\infty}\), so that \(v\B^{\infty}\rho\) is less than 1, the solute concentration increases with increasing distance from the axis of rotation in the equilibrium state. If, however, \(\rho\) is greater than \(1/v\B^{\infty}\), the concentration decreases with increasing \(r\). The factor \(\left(1-v\B^{\infty}\rho\right)\) is like a buoyancy factor for the effect of the centrifugal field on the solute. Equation 9.8.22 is needed for , a method of determining the molar mass of a macromolecule. A dilute solution of the macromolecule is placed in the cell of an analytical ultracentrifuge, and the angular velocity is selected to produce a measurable solute concentration gradient at equilibrium. The solute concentration is measured optically as a function of \(r\). The equation predicts that a plot of \(\ln\left(c\B/c\st\right)\) versus \(r^2\) will be linear, with a slope equal to \(M\B\left(1-v\B^{\infty}\rho\right)\omega^2/2RT\). The partial specific volume \(v\B^{\infty}\) is found from measurements of solution density as a function of solute mass fraction (Sec. 9.2.5). By this means, the molar mass \(M\B\) of the macromolecule is evaluated. | 22,504 | 2,031 |
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Lewis formulas are structures that show the , or bonding sequence of the atoms, indicating They should also show any formal charges and unshared electrons that might be present in the molecule. Additional examples of Lewis formulas follow.\ These examples were deliberately chosen because all three molecules shown have the same molecular formula, but different connectivities, or bonding sequences. Such substances are called , or sometimes Notice that only the first structure shows the unshared electrons of chlorine. In Lewis formulas of organic compounds, it is customary to omit the lone electron pairs on the halogens unless there is a reason to show them explicitly. Lewis formulas are mostly used for covalent substances, but occasionally they also show ionic bonds that might be present in certain compounds. | 835 | 2,032 |
https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Microscopy/Scanning_Probe_Microscopy/04_Additional_SPM_Methods/01_Lateral_Force_Microscopy |
Lateral Force Microscopy (LFM) is conducted when imaging in the contact mode. During scanning in contact mode the cantilever bends not only along vertically to the surface as a result of repulsive Van der Waals interactions, but the cantilever also undergoes torsional (lateral) deformation. LFM measures the torsional bending (or twisting) of the cantilever, which is dependent on a frictional force acting on tip. As a result, this method is also known as friction force microscopy (FFM). LFM is sensitive to chemical composition or structure of the surface. This imaging mode offers nanometer-scale resolution with sensitivity to variations in surface composition, molecular organization, mechanical properties, and acid-base characteristics. For LFM imaging, the direction of scanning should be perpendicular to the long axis of the cantilever. Furthermore, the roughness of the surface makes interpretation of LFM mapping difficult, as height topography in addition to friction will cause lateral twisting of the cantilever. Therefore, LFM analysis is typically completed on smooth surfaces. Excellent source for images using phase, CFM and friction contrast. NIST Building and Fire Research Laboratory Image Gallery | 1,234 | 2,036 |
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. In addition to these individual basis; please contact What are the advantages to carrying out a reaction in solution rather than simply mixing the pure reactants? What types of compounds dissolve in polar solvents? Describe the charge distribution in liquid water. How does this distribution affect its physical properties? Must a molecule have an asymmetric charge distribution to be polar? Explain your answer. Why are many ionic substances soluble in water? Explain the phrase . What kinds of covalent compounds are soluble in water? Why do most aromatic hydrocarbons have only limited solubility in water? Would you expect their solubility to be higher, lower, or the same in ethanol compared with water? Why? Predict whether each compound will dissolve in water and explain why. Predict whether each compound will dissolve in water and explain why. Given water and toluene, predict which is the better solvent for each compound and explain your reasoning. Of water and toluene, predict which is the better solvent for each compound and explain your reasoning. Compound is divided into three equal samples. The first sample does not dissolve in water, the second sample dissolves only slightly in ethanol, and the third sample dissolves completely in toluene. What does this suggest about the polarity of ? You are given a mixture of three solid compounds— , , and —and are told that is a polar compound, is slightly polar, and is nonpolar. Suggest a method for separating these three compounds. A laboratory technician is given a sample that contains only sodium chloride, sucrose, and cyclodecanone (a ketone). You must tell the technician how to separate these three compounds from the mixture. What would you suggest? Many over-the-counter drugs are sold as ethanol/water solutions rather than as purely aqueous solutions. Give a plausible reason for this practice. What distinguishes a weak electrolyte from a strong electrolyte? Which organic groups result in aqueous solutions that conduct electricity? It is considered highly dangerous to splash barefoot in puddles during a lightning storm. Why? Which solution(s) would you expect to conduct electricity well? Explain your reasoning. Which solution(s) would you expect to conduct electricity well? Explain your reasoning. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in an aqueous solution? Explain your reasoning. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in an aqueous solution? Explain your reasoning. Which of the following is a strong electrolyte, a weak electrolyte, or a nonelectrolyte in aqueous solution? Explain your reasoning. Ionic compounds such as NaCl are held together by electrostatic interactions between oppositely charged ions in the highly ordered solid. When an ionic compound dissolves in water, the partially negatively charged oxygen atoms of the H O molecules surround the cations, and the partially positively charged hydrogen atoms in H O surround the anions. The favorable electrostatic interactions between water and the ions compensate for the loss of the electrostatic interactions between ions in the solid. An is any compound that can form ions when it dissolves in water. When a strong electrolyte dissolves in water, it dissociates completely to give the constituent ions. In contrast, when a weak electrolyte dissolves in water, it produces relatively few ions in solution. What information can be obtained from a complete ionic equation that cannot be obtained from the overall chemical equation? Predict whether mixing each pair of solutions will result in the formation of a precipitate. If so, identify the precipitate. Predict whether mixing each pair of solutions will result in the formation of a precipitate. If so, identify the precipitate. Which representation best corresponds to an aqueous solution originally containing each of the following? 1 M Ba(OH) + 1 M H SO Which representation in Problem 3 best corresponds to an aqueous solution originally containing each of the following? 3 What mass of precipitate would you expect to obtain by mixing 250 mL of a solution containing 4.88 g of Na CrO with 200 mL of a solution containing 3.84 g of AgNO ? What is the final nitrate ion concentration? Adding 10.0 mL of a dilute solution of zinc nitrate to 246 mL of 2.00 M sodium sulfide produced 0.279 g of a precipitate. How many grams of zinc(II) nitrate and sodium sulfide were consumed to produce this quantity of product? What was the concentration of each ion in the original solutions? What is the concentration of the sulfide ion in solution after the precipitation reaction, assuming no further reaction? 3.75 g Ag CrO ; 5.02 × 10 M nitrate Why was it necessary to expand on the Arrhenius definition of an acid and a base? What specific point does the Brønsted–Lowry definition address? State whether each compound is an acid, a base, or a salt. State whether each compound is an acid, a base, or a salt. Classify each compound as a strong acid, a weak acid, a strong base, or a weak base in aqueous solution. Decide whether each compound forms an aqueous solution that is strongly acidic, weakly acidic, strongly basic, or weakly basic. What is the relationship between the strength of an acid and the strength of the conjugate base derived from that acid? Would you expect the CH CO ion to be a strong base or a weak base? Why? Is the hydronium ion a strong acid or a weak acid? Explain your answer. What are the products of an acid–base reaction? Under what circumstances is one of the products a gas? Explain how an aqueous solution that is strongly basic can have a pH, which is a measure of the of a solution. Derive an equation to relate the hydrogen ion concentration to the molarity of a solution of a strong monoprotic acid. Derive an equation to relate the hydroxide ion concentration to the molarity of a solution of Given the following salts, identify the acid and the base in the neutralization reactions and then write the complete ionic equation: What is the pH of each solution? What is the hydrogen ion concentration of each substance in the indicated pH range? What is the hydrogen ion concentration of each substance in the indicated pH range? What is the pH of a solution prepared by diluting 25.00 mL of 0.879 M HCl to a volume of 555 mL? Vinegar is primarily an aqueous solution of acetic acid. Commercial vinegar typically contains 5.0 g of acetic acid in 95.0 g of water. What is the concentration of commercial vinegar? If only 3.1% of the acetic acid dissociates to CH CO and H , what is the pH of the solution? (Assume the density of the solution is 1.00 g/mL.) If a typical household cleanser is 0.50 M in strong base, what volume of 0.998 M strong monoprotic acid is needed to neutralize 50.0 mL of the cleanser? A 25.00 mL sample of a 0.9005 M solution of HCl is diluted to 500.0 mL. What is the molarity of the final solution? How many milliliters of 0.223 M NaOH are needed to neutralize 25.00 mL of this final solution? If 20.0 mL of 0.10 M NaOH are needed to neutralize 15.0 mL of gastric fluid, what is the molarity of HCl in the fluid? (Assume all the acidity is due to the presence of HCl.) What other base might be used instead of NaOH? Malonic acid (C H O ) is a diprotic acid used in the manufacture of barbiturates. How many grams of malonic acid are in a 25.00 mL sample that requires 32.68 mL of 1.124 M KOH for complete neutralization to occur? Malonic acid is a dicarboxylic acid; propose a structure for malonic acid. Describe how you would prepare 500 mL of a 1.00 M stock solution of HCl from an HCl solution that is 12.11 M. Using your stock solution, how would you prepare 500 mL of a solution that is 0.012 M in HCl? Given a stock solution that is 8.52 M in HBr, describe how you would prepare a 500 mL solution with each concentration. How many moles of solute are contained in each? A chemist needed a solution that was approximately 0.5 M in HCl but could measure only 10.00 mL samples into a 50.00 mL volumetric flask. Propose a method for preparing the solution. (Assume that concentrated HCl is 12.0 M.) Write the balanced chemical equation for each reaction. Write the balanced chemical equation for each reaction. A neutralization reaction gives calcium nitrate as one of the two products. Identify the acid and the base in this reaction. What is the second product? If the product had been cesium iodide, what would have been the acid and the base? What is the complete ionic equation for each reaction? [H O ] = [HA] M pH = 1.402 25 mL 0.13 M HCl; magnesium carbonate, MgCO , or aluminum hydroxide, Al(OH) 1.00 M solution: dilute 41.20 mL of the concentrated solution to a final volume of 500 mL. 0.012 M solution: dilute 12.0 mL of the 1.00 M stock solution to a final volume of 500 mL. The acid is nitric acid, and the base is calcium hydroxide. The other product is water. \(2HNO_3 + Ca(OH)_2 \rightarrow Ca(NO_3)_2 + 2H_2O\) The acid is hydroiodic acid, and the base is cesium hydroxide. The other product is water. \( HI + CsOH \rightarrow CsI + H_2O \) The complete ionic equations are \( 2H^+ + 2NO_3^- + Ca^{2+} + 2OH^- \rightarrow Ca^{2+} + 2NO_3^- + H_2O\)
\( H^+ + I^- + Cs^+ + OH^- \rightarrow Cs^+ + I^- + H_2O \) Which elements in the periodic table tend to be good oxidants? Which tend to be good reductants? If two compounds are mixed, one containing an element that is a poor oxidant and one with an element that is a poor reductant, do you expect a redox reaction to occur? Explain your answer. What do you predict if one is a strong oxidant and the other is a weak reductant? Why? In each redox reaction, determine which species is oxidized and which is reduced: Single-displacement reactions are a subset of redox reactions. In this subset, what is oxidized and what is reduced? Give an example of a redox reaction that is a single-displacement reaction. Balance each redox reaction under the conditions indicated. Balance each redox reaction under the conditions indicated. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction. Using the activity series, predict what happens in each situation. If a reaction occurs, write the net ionic equation; then write the complete ionic equation for the reaction. Dentists occasionally use metallic mixtures called for fillings. If an amalgam contains zinc, however, water can contaminate the amalgam as it is being manipulated, producing hydrogen gas under basic conditions. As the filling hardens, the gas can be released, causing pain and cracking the tooth. Write a balanced chemical equation for this reaction. Copper metal readily dissolves in dilute aqueous nitric acid to form blue Cu (aq) and nitric oxide gas. Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation: Classify each reaction as an acid–base reaction, a precipitation reaction, or a redox reaction, or state if there is no reaction; then complete and balance the chemical equation: Which of the representations best corresponds to a 1 M aqueous solution of each compound? Justify your answers. Na SO Which of the representations shown in Problem 1 best corresponds to a 1 M aqueous solution of each compound? Justify your answers. Would you expect a 1.0 M solution of CaCl to be a better conductor of electricity than a 1.0 M solution of NaCl? Why or why not? An alternative way to define the concentration of a solution is , abbreviated . Molality is defined as the number of moles of solute in 1 kg of . How is this different from molarity? Would you expect a 1 M solution of sucrose to be more or less concentrated than a 1 solution of sucrose? Explain your answer. What are the advantages of using solutions for quantitative calculations? If the amount of a substance required for a reaction is too small to be weighed accurately, the use of a solution of the substance, in which the solute is dispersed in a much larger mass of solvent, allows chemists to measure the quantity of the substance more accurately. Calculate the number of grams of solute in 1.000 L of each solution. Calculate the number of grams of solute in 1.000 L of each solution. If all solutions contain the same solute, which solution contains the greater mass of solute? Complete the following table for 500 mL of solution. What is the concentration of each species present in the following aqueous solutions? What is the concentration of each species present in the following aqueous solutions? What is the molar concentration of each solution? What is the molar concentration of each solution? Give the concentration of each reactant in the following equations, assuming 20.0 g of each and a solution volume of 250 mL for each reactant. An experiment required 200.0 mL of a 0.330 M solution of Na CrO . A stock solution of Na CrO containing 20.0% solute by mass with a density of 1.19 g/cm was used to prepare this solution. Describe how to prepare 200.0 mL of a 0.330 M solution of Na CrO using the stock solution. Calcium hypochlorite [Ca(OCl) ] is an effective disinfectant for clothing and bedding. If a solution has a Ca(OCl) concentration of 3.4 g per 100 mL of solution, what is the molarity of hypochlorite? Phenol (C H OH) is often used as an antiseptic in mouthwashes and throat lozenges. If a mouthwash has a phenol concentration of 1.5 g per 100 mL of solution, what is the molarity of phenol? If a tablet containing 100 mg of caffeine (C H N O ) is dissolved in water to give 10.0 oz of solution, what is the molar concentration of caffeine in the solution? A certain drug label carries instructions to add 10.0 mL of sterile water, stating that each milliliter of the resulting solution will contain 0.500 g of medication. If a patient has a prescribed dose of 900.0 mg, how many milliliters of the solution should be administered? 0.48 M ClO 1.74 × 10 M caffeine The titration procedure is an application of the use of limiting reactants. Explain why this is so. Explain how to determine the concentration of a substance using a titration. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M acetic acid with 0.10 M NaOH, and the other corresponds to the titration of 100 mL 0.10 M NaOH with 0.10 M acetic acid. Which graph corresponds to which titration? Justify your answer. Following are two graphs that illustrate how the pH of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M ammonia with 0.10 M HCl, and the other corresponds to the titration of 100 mL 0.10 M NH Cl with 0.10 M NaOH. Which graph corresponds to which titration? Justify your answer. Following are two graphs that illustrate how the electrical conductivity of a solution varies during a titration. One graph corresponds to the titration of 100 mL 0.10 M Ba(OH) with 0.10 M H SO , and the other corresponds to the titration of 100 mL of 0.10 M NaOH with 0.10 M H SO . Which graph corresponds to which titration? Justify your answer. A 10.00 mL sample of a 1.07 M solution of potassium hydrogen phthalate (KHP, formula mass = 204.22 g/mol) is diluted to 250.0 mL. What is the molarity of the final solution? How many grams of KHP are in the 10.00 mL sample? What volume of a 0.978 M solution of NaOH must be added to 25.0 mL of 0.583 M HCl to completely neutralize the acid? How many moles of NaOH are needed for the neutralization? A student was titrating 25.00 mL of a basic solution with an HCl solution that was 0.281 M. The student ran out of the HCl solution after having added 32.46 mL, so she borrowed an HCl solution that was labeled as 0.317 M. An additional 11.5 mL of the second solution was needed to complete the titration. What was the concentration of the basic solution? | 16,134 | 2,037 |
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" by Bruce A. Averill and Patricia Eldredge. . In addition to these individual basis; please contact 1. What distinguishes a nuclear reaction from a chemical reaction? Use an example of each to illustrate the differences. 2. What do chemists mean when they say a substance is ? 3. What characterizes an isotope? How is the mass of an isotope of an element related to the atomic mass of the element shown in the periodic table? 4. In a typical nucleus, why does electrostatic repulsion between protons not destabilize the nucleus? How does the neutron-to-proton ratio affect the stability of an isotope? Why are all isotopes with > 83 unstable? 5. What is the significance of a of protons or neutrons? What is the relationship between the number of stable isotopes of an element and whether the element has a magic number of protons? 6. Do you expect Bi to have a large number of stable isotopes? Ca? Explain your answers. 7. Potassium has three common isotopes, K, K, and K, but only potassium-40 is radioactive (a beta emitter). Suggest a reason for the instability of K. 8. Samarium has 11 relatively stable isotopes, but only 4 are nonradioactive. One of these 4 isotopes is Sm, which has a lower neutron-to-proton ratio than lighter, radioactive isotopes of samarium. Why is Sm more stable? 5. Isotopes with magic numbers of protons and/or neutrons tend to be especially stable. Elements with magic numbers of protons tend to have more stable isotopes than elements that do not. 7. Potassium-40 has 19 protons and 21 neutrons. Nuclei with odd numbers of both protons and neutrons tend to be unstable. In addition, the neutron-to-proton ratio is very low for an element with this mass, which decreases nuclear stability. 1. Write the nuclear symbol for each isotope using \(^A_Z \textrm X\) notation. a. chlorine-39 b. lithium-8 c. osmium-183 d. zinc-71 2.Write the nuclear symbol for each isotope using \(^A_Z \textrm X\) notation. a. lead-212 b. helium-5 c. oxygen-19 d. plutonium-242 3. Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope. a. iron-57 b. W c. potassium-39 d. Xe 4. Give the number of protons, the number of neutrons, and the neutron-to-proton ratio for each isotope. a. technetium-99 b. La c. radium-227 d. Bi 5. Which of these nuclides do you expect to be radioactive? Explain your reasoning. a. Ne b. tungsten-184 c. Ti 6. Which of these nuclides do you expect to be radioactive? Explain your reasoning. a. Ag b. V c. lutetium-176 1. a. \(^{39}_{17} \textrm{Cl}\) b. \(^{8}_{3} \textrm{Li}\) c. \(^{183}_{76} \textrm{Os}\) d. \(^{71}_{30} \textrm{Zn}\) 3. a. 26 protons; 31 neutrons; 1.19 b. 74 protons; 111 neutrons; 1.50 c. 19 protons; 20 neutrons; 1.05 d. 54 protons; 77 neutrons; 1.43 Why do scientists believe that hydrogen is the building block of all other elements? Why do scientists believe that helium-4 is the building block of the heavier elements? How does a star produce such enormous amounts of heat and light? How are elements heavier than Ni formed? Propose an explanation for the observation that elements with even atomic numbers are more abundant than elements with odd atomic numbers. During the lifetime of a star, different reactions that form different elements are used to power the fusion furnace that keeps a star “lit.” Explain the different reactions that dominate in the different stages of a star’s life cycle and their effect on the temperature of the star. A line in a popular song from the 1960s by Joni Mitchell stated, “We are stardust….” Does this statement have any merit or is it just poetic? Justify your answer. If the laws of physics were different and the primary element in the universe were boron-11 ( = 5), what would be the next four most abundant elements? Propose nuclear reactions for their formation. The raw material for all elements with > 2 is helium ( = 2), and fusion of helium nuclei will always produce nuclei with an even number of protons. Write a balanced nuclear reaction for the formation of each isotope. At the end of a star’s life cycle, it can collapse, resulting in a supernova explosion that leads to the formation of heavy elements by multiple neutron-capture events. Write a balanced nuclear reaction for the formation of each isotope during such an explosion. When a star reaches middle age, helium-4 is converted to short-lived beryllium-8 (mass = 8.00530510 amu), which reacts with another helium-4 to produce carbon-12. How much energy is released in each reaction (in megaelectronvolts)? How many atoms of helium must be “burned” in this process to produce the same amount of energy obtained from the fusion of 1 mol of hydrogen atoms to give deuterium? | 4,765 | 2,039 |
https://chem.libretexts.org/Bookshelves/Environmental_Chemistry/Environmental_Toxicology_(van_Gestel_et_al.)/05%3A_Population_Community_and_Ecosystem_Ecotoxicology/5.07%3A_Community_ecotoxicology |
: Michiel Kraak and Ivo Roessink : Kees van Gestel, Nico van den Brink, Ralf B. Schäfer You should be able to : Community ecotoxicology, species interactions, indirect effects, mesocosm, ecosystem processes. The motivation to study ecotoxicological effects at the community level is that generally the targets of environmental protection are populations, communities and ecosystems. Consequently, when scaling up research from the molecular level, via cells, organs and individual organisms towards the population, community or even ecosystem level the ecological and societal relevance of the obtained data strongly increase (Figure 1). Yet, the difficulty of obtaining data increases, due to the increasing complexity, lower reproducibility and the increasing time needed to complete the research, which typically involves higher costs. Moreover, when effects are observed in the field it may be difficult to link these to specific chemicals and to identify the drivers of the observed effects. Not surprisingly, ecotoxicological effects at the community and ecosystem level are understudied. Community ecology is defined as the study of the organization and functioning of communities, which are assemblages of interacting populations of species living within a particular area or habitat. Building on this definition, community ecotoxicology is defined as the study of the effects of toxicants on patterns of species abundance, diversity, community composition and species interactions. These species interactions are unique to the community and ecosystem level and may cause direct effects of toxicants on specific species to exert indirect effects on other species. It has been estimated that the majority of effects at these levels of biological organization are indirect rather than direct. These indirect effects are exerted via: As an example, Roessink et al. (2006) studied the impact of the fungicide triphenyltin acetate (TPT) on benthic communities in outdoor mesocosms. For several species a dose-related decrease in abundance directly after application was observed, followed by a gradual recovery coinciding with decreasing exposure concentrations, all implying direct effects of the fungicide. For some species, however, opposite results were obtained and abundance increased shortly after application, followed by a gradual decline; see the example of the Culicidae in Figure 2. In this case, these typical indirect effects were explained by a higher sensitivity of the predators and competitors of the Culicidae. Due to diminished predation and competition and higher food availability abundances of the Culicidae temporarily increased after toxicant exposure. With the decreasing exposure concentrations over time, the populations of the predators and competitors recovered, leading to a subsequent decline in numbers of the Culicidae. The indirect effects described above are thus due to species-specific sensitivities to the compound of interest, which influence the interactions between species. Yet, at higher exposure concentrations also the less sensitive species will start to be affected by the chemical. This may lead to an "arch-shaped" relationship between the number of individuals of a certain species and the concentration of a toxicant. In a mesocosm study with the insecticide lambda-cyhalothin this was observed for , which are prey for the more sensitive phantom midge (Roessink et al., 2005; Figure 3). At low exposure concentrations the indirect effects, such as release from predation by , led to an increase in abundance of the less sensitive . At intermediate exposure concentrations there was a balance between the positive indirect effects and the adverse direct effects of the toxicant. At higher exposure concentrations the adverse direct effects overruled the positive indirect effects resulting in a decline in abundance of the . These combined dose dependent direct and indirect effects are inherent to community-level experiments, but are compound and species-interaction specific. To study community ecotoxicology, experiments have to be scaled up and are therefore often performed in mesocosms, artificial ponds, ditches and streams, or even in the field, sometimes accompanied by the use of in- and exclosures. To assess the effects of toxicants on communities in such large systems requires meticulous sampling schemes, which often make use of artificial substrates and e.g. emergence traps for aquatic invertebrates with terrestrial adult life stages (see section on ). Alternatively to scaling up the experiments in community ecotoxicology, the size of the communities may be scaled down. Algae and bacteria grown on coin sized artificial substrates in the field or in experimental settings provide the unique advantage that the experimental unit is actually an entire community. Given the large scale and complexity of experiments at the community level, such experiments generally generate overwhelming amounts of data, making appropriate analysis of the results challenging. Data analysis focusing on individual responses, so-called univariate analysis, that suffice in single species experiments, obviously falls short in community ecotoxicology, where cosm or (semi-)field communities sometimes consist of over a hundred different species. Hence, multivariate analysis is often more appropriate, similar to the approaches frequently applied in field studies to identify possible drivers of patterns in species abundances. Alternative approaches are also applied, like using ecological indices such as species richness or categorizing the responses of communities into effect classes (Figure 4). To determine if species under semi-field conditions respond equally sensitive to toxicant exposure as in the laboratory, the construction and subsequent comparison of species sensitivity distributions (SSD) (see section on ) may be helpful. The analysis and interpretation of community ecotoxicity data is also challenged by the dynamic development of each individual replicate cosm, artificial pond, ditch or stream, including those from the control. From the start of the experiment, each control replicate develops independently, matures, and at the end of the experiments that generally last for several months control replicates may differ not only from the treatments, but also among each other. The challenge is then to separate the toxic signal from the natural variability in the data. In experiments that include a recovery phase, it is frequently observed that previously exposed communities do recover, but develop in another direction than the controls, which actually challenges the definition of recovery. Moreover, recovery can be decelerated or accelerated depending on the dispersal capacity of the species potentially inhabiting the cosms and the distance to nearby populations within a metapopulation (see section on ). Other crucial factors that may affect the impact of a toxicant on communities, as well as their recovery from this toxicant exposure include habitat heterogeneity and the state of the community in combination with the moment of exposure. Habitat heterogeneity may affect the distribution of toxicants over the different environmental compartments and may provide shelter to organisms. Communities generally exhibit temporal dynamics in species composition and in their contribution to ecosystem processes (see section on ), as well in the lifecycle stages of the individual species. Exponentially growing populations recover much faster than populations that reached carrying capacity and for almost all species, young individuals are up to several orders of magnitude more sensitive than adults or late instar larvae (see section on ). Hence, the timing of exposure to toxicants may seriously affect the extent of the adverse effects, as well as the recovery potential of the exposed communities. When scaling up from the community to the ecosystem level, again unique characteristics emerge: structural characteristics like biodiversity, but also ecosystem processes, quantified by functional endpoints like primary production, ecosystem respiration, nutrient cycling and decomposition. Although a good environmental quality is based on both ecosystem structure and functioning, there is definitely a bias towards ecosystem structure, both in science and in policy (see section on ). Levels of biological organisation higher than ecosystems are covered by the field of landscape ecotoxicology (see section on ) and in a more practical way by the concept of ecosystem services (see section on ). Roessink, I., Crum, S.J.H., Bransen, F., Van Leeuwen, E., Van Kerkum, F., Koelmans, A.A., Brock, T.C.M. (2006). Impact of triphenyltin acetate in microcosms simulating floodplain lakes. I. Influence of sediment quality. Ecotoxicology 15, 267-293. Roessink, I., Arts, G.H.P., Belgers, J.D.M., Bransen, F., Maund, S.J., Brock, T.C.M. (2005). Effects of lambda-cyhalothrin in two ditch mesocosm systems of different trophic status. Environmental Toxicology and Chemistry 24, 1684-1696. Clements, W.H., Newman, M.C. (2002). Community Ecotoxicology. John Wiley & Sons, Ltd. Motivate the importance of studying ecotoxicology at the community level. Define community ecotoxicology and name specific phenomena at the community and ecosystem level. Roessink et al. (2006) studied the impact of the fungicide triphenyltin acetate (TPT) on benthic communities in outdoor mesocosms. For several species they observed a dose related decrease in abundance directly after application, followed by a gradual recovery, see example in the graph below. For some species however, opposite results were obtained and the abundance increased shortly after application, followed by a gradual decline, see example in the graph below. Explain the results shown in the lower graph. Mention three ways to analyze data from experiments at the community level. Mention one advantage and three disadvantages of cosm experiments. : Martina G. Vijver : Paul J. van den Brink, Kees van Gestel To be able to : microcosms, mesocosms, realism, different biological levels It is generally anticipated that ecotoxicological tests should provide data useful for making realistic predictions of the fate and effects of chemicals in natural ecosystems (Landner et al., 1989). The ecotoxicological test, if used in an appropriate way, should be able to identify the potential environmental impact of a chemical before it has caused any damage to the ecosystem. In spite of the considerable amount of work devoted to this problem and the plethora of test methods being published, there is still reason to question whether current procedures for testing and assessing the hazard of chemicals in the environment do answer the questions we have asked. Most biologists agree that at each succeeding level of biological organization new properties appear that would not have been evident even by the most intense and careful examination of lower levels of organization (Cairns Jr., 1983). These levels of biological hierarchy might be crudely characterized as subcellular, cellular, organ, organism, population, multispecies, community, and ecosystem (Figure 1). At the lower biological level, responses are faster than those occurring at higher levels of organization. Experiments executed at the lower biological level often are performed under standard laboratory conditions (see Section on ). The laboratory setting has advantages like allowing for replication, the use of relatively easy and simplified conditions that enable outcomes that are rather robust across different laboratories, the stressor of interest being more traceable under optimal stable conditions, and easy repetition of experiments. As a consequence, at the lower biological level the responses of organisms to chemical stressors tend to be more tractable, or more causal, than those identified when studying effects at higher tiered levels. The merit to perform cosm studies, so at the higher biological level (see Figure 1), is to investigate the impact of a stressor on a variety of species, all having interactions with each other. This enables detecting both direct and indirect effects on the structure of species assemblages due to the chemicals. Indirect effects can become manifest as disruptions of species interactions, e.g. competition, predator-prey interactions and the like. A second important reason for conducting cosm studies is that abiotic interactions at the level of the ecosystem can be accounted for, allowing for measurement of effects of chemicals under more environmentally realistic exposure conditions. Conditions that likely influence the fate and behavior of chemical are sorption to sediments and plants, photolysis, changes in pH (see section on for a more detailed description), and other natural fluctuations. Microcosm or mesocosm (or cosm) studies represent a bridge between the laboratory and the natural world (examples of aquatic cosms are given in Figure 2). The difference between micro- and mesocosms is mostly restricted to size (Cooper and Barmuta, 1993). Aquatic microcosms are 10 to 10 m in size, while mesocosms are 10 to 10 m or even larger equivalent to whole or natural systems. The originality of cosms is mainly based on the combination of ecological realism, achieved by the introduction of the basic components of natural ecosystems, and facilitated access to a number of physicochemical, biological, and toxicological parameters that can be controlled to some extent. The cosm approach also makes it possible to work with treatments that can be replicated, so enabling the study of multiple environmental factors which can be manipulated. The system allows the establishment of food webs, the assessment of direct and indirect effects, and the evaluation of effects of contamination on multiple trophic and taxonomic levels in an ecologically relevant context. Cosm studies make it possible to assess effects of contaminants by looking at the parts (individuals, populations, communities) and the whole (ecosystems) simultaneously. As given in the OECD guidance document (OECD, 2004), the size to be selected for a meso- or microcosm study will depend on the objectives of the study and the type of ecosystem that is to be simulated. In general, studies in smaller systems are more suitable for short-term studies of up to three to six months and studies with smaller organisms (e.g. planktonic species). Larger systems are more appropriate for long-term studies (e.g. 6 months or longer). Numerous ecosystem-level manipulations have been conducted since the early 1970s (Hurlbert et al., 1972). The Experimental Lakes Area (ELA) situated in Ontario, Canada deserves special attention because of its significant contributions to the understanding of how natural communities respond to chemical stressors. This ELA consists of 46 natural, relatively undisturbed lakes, which were designated specially for ecosystem-level research. Many different questions have been tackled here, e.g. manipulations with nutrients (amongst others Levine and Schindler, 1999), synthetic estrogens (e.g. Kidd et al., 2014) and Wallace with pesticides in the Coweeta district (Wallace et al., 1999). It is nowadays realized that there is a need for testing more than just individual species and to take into account ecosystem elements such as fluctuations of abiotic conditions and biotic interactions when trying to understand the ecological effects of chemicals. Therefore a selection of study parameters is often considered as given by OECD (2004): A cosm approach assist in identifying and quantifying direct as well as indirect effects. Here two different types of responses are described, for more examples it is referred to the Section on . : Barmentlo et al. (2018) used an outdoor mesocosm system consisting of 65 L ponds. Using a full factorial design, they investigated the population responses of macroinvertebrate species assemblages exposed for 35 days to environmentally relevant concentrations of three commonly used agrochemicals (imidacloprid, terbuthylazine, and NPK fertilizers). A detrivorous food chain as well as an algal-driven food chain were inoculated into the cosms. At environmentally realistic concentrations of binary mixtures, the species responses could be predicted based on concentration addition (see Section on ). Overall, the effects of trinary mixtures were much more variable and counterintuitive. This was nicely illustrated by how the mayfly reacted to the various combinations of the pesticides. Compared to single substance exposures and binary mixtures, extreme low recovery of (3.6% of control recovery for both mixtures) was seen. However, after exposure to the trinary mixture, recovery of no longer deviated from the control, and therefore was was higher than expected. Unexpected effects of the mixtures were also obtained for both zooplankton species ( and sp.) As expected, the abundance of both zooplankton species was positively affected by nutrient applications, but pesticide addition did not lower their recovery. These type of unexpected results can only been identified when multiple species and multiple stressors are tested and cannot be detected in a lab-test with single species. : Van den Brink et al. (2009) studied the effects of chronic applications of a mixture of the herbicide atrazine and the insecticide lindane in indoor freshwater plankton-dominated microcosms. Both top-down and bottom-up regulation mechanisms of the species assemblage selected were affected by the pesticide mixture. Lindane exposure also caused a decrease in sensitive detritivorous macro-arthropods and herbivore arthropods. This allowed insensitive food competitors like worms, rotifers and snails to increase in abundance (although not always significantly). Atrazine inhibited algal growth and hence also affected the herbivores. A direct result of the inhibition of photosynthesis by atrazine exposure were lower dissolved oxygen and pH levels and an increase in alkalinity, nitrogen and electrical conductivity. See Figure 3 for a synthesis of all interactions observed in the study of Van den Brink et al. (2009). There is a conceptual conflict between realism and replicability when applied to mesocosms. Replicability may be achieved, in part, by a relative simplification of the system. The crucial point in designing a model system may not be to maximize the realism, but rather to make sure that ecologically relevant information can be obtained. Reliability of information on ecotoxicological effects of chemicals tested in mesocosms closely depends on the representativeness of biological processes or structures that are likely to be affected. This means that within cosms key features at both structural and functional levels should be preserved as they ensure ecological representativeness. Extrapolation from small experimental systems to the real world seems generally more problematic than the use of larger systems in which more complex interactions can be studied experimentally as well. For that reason, Caquet et al. (2000) claim that testing chemicals using mesocosms refines the classical methods of ecotoxicological risk assessment because they provide conditions for a better understanding of environmentally relevant effects of chemicals. Barmentlo S.H., Schrama M., Hunting E.R., Heutink R., Van Bodegom P.M., De Snoo G.R., Vijver M.G. (2018). Assessing combined impacts of agrochemicals: Aquatic macroinvertebrate population responses in outdoor mesocosms, Science of the Total Environment 631-632, 341-347. Caquet, T., Lagadic, L., Sheffield, S.R. (2000) Mesocosm in ecotoxicology: outdoor aquatic systems. Reviews of Environmental Contamination and Toxicology 165, 1-38. Cairns Jr. J. (1983). Are single species toxicity tests alone adequate for estimating environmental hazard? Hydrobiologica 100, 47-57. Cooper, S.D., Barmuta, L.A. (1993) Field experiments in biomonitoring. In Rosenberg, D.M., Resh, V.H. (Eds.) Freshwater Biomonitoring and Benthic Macroinvertebrates. Chapman and Hall, New York, pp. 399-441. OECD (2004). Draft Guidance Document on Simulated Freshwater Lentic Field Tests (Outdoor Microcosms and Mesocosms) (July 2004). Organization for Economic Cooperation and Development, Paris. Hurlbert, S.H., Mulla, M.S., Willson, H.R. (1972) Effects of an organophosphorus insecticide on the phytoplankton, zooplankton, and insect populations of fresh-water ponds. Ecological Monographs 42, 269-299. Kidd, K.A., Paterson, M.J., Rennie, M.D., Podemski, C.L., Findlay, D.L., Blanchfield, P.J., Liber, K. (2014). Direct and indirect responses of a freshwater food web to a potent synthetic oestrogen. Philosophical Transactions of the Royal Society B Biological Sciences 369, Article AR 20130578, DOI:10.1098/rstb.2013.0578 Landner, L., Blanck, H., Heyman, U., Lundgren, A., Notini, M., Rosemarin, A., Sundelin, B. (1989) Community Testing, Microcosm and Mesocosm Experiments: Ecotoxicological Tools with High Ecological Realism. . Springer, pp. 216-254. Levine, S.N., Schindler, D.W. (1999). Influence of nitrogen to phosphorus supply ratios and physicochemical conditions on cyanobacteria and phytoplankton species composition in the Experimental Lakes Area, Canada. Canadian Journal of Fisheries and Aquatic Sciences 56, 451-466. Newman, M.C. (2008). Ecotoxicology: The History and Present Directions. In Jørgensen, S.E., Fath, B.D. (Eds.), Ecotoxicology. Vol. 2 of Encyclopedia of Ecology, 5 vols. Oxford: Elsevier, pp.1195-1201. Van den Brink, P.J., Crum, S.J.H., Gylstra, R., Bransen, F., Cuppen, J.G.M., Brock, (2009). Effects of a herbicide - insecticide mixture in freshwater microcosms: risk assessment and ecological effect chain. Environmental Pollution 157, 237-249. Wallace, J.B., Grubaugh, J.W., Whiles, M.R. (1996). Biotic indices and stream ecosystem processes: Results from an experimental study. Ecological Applications 6, 140-151. Responses of organisms to long-term exposure can be detected at the sub-organism level by making use of biomarkers and then extrapolating these results to organism fitness and consequence at the population level. Mention two benefits of performing tests making use of biomarkers Mention at least two benefits of performing tests at the higher biological level such as community or ecosystem levels. | 22,438 | 2,040 |
https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_and_Websites_(Inorganic_Chemistry)/Descriptive_Chemistry/Main_Group_Reactions/Compounds/Aluminosilicates/Silicates |
A photograph of quartz, one of the many silicates, is shown here. Quartz alone is a very interesting mineral, used in many technologies ranging from jewry to watch, computers, to scientific instrumentation, because quartz is an important piezoelectric material. A clever application of piezoelectric effect is to use its vibration movement to deliver high viscosity ink in the drop-on-demand ink jet system. Piezoelectric materials are also used in Piezo Electric Transducers and Buzzers Techniques have also been developed to apply piezoelectric material for the conversion of static wave energy into rotation motion. Such a Piezoelectric Motor is developed for the wrist watch application. The silicates are the largest, the most interesting and the most complicated class of minerals than any other minerals. Approximately 30% of all minerals are silicates and some geologists estimate that 90% of the Earth's crust is made up of silicates, SiO based material. Thus, oxygen and silicon are the two most abundant elements in the earth's crust. Silicates is based on the basic chemical unit SiO , tetrahedron shaped anionic group. The central silicon ion has a charge of positive four while each oxygen has a charge of negative two (-2) and thus each silicon-oxygen bond is equal to one half (½ ) the total bond energy of oxygen. This condition leaves the oxygens with the option of bonding to another silicon ion and therefore linking one SiO tetrahedron to another. In the extreme case, the tetrahedra are arranged in a regular, orderly fashion forming a three-dimensional network. is such a structure (see the diagram), and its formula is SiO . If silica in the molten state is cooled very slowly it crystallizes at the freezing point. But if molten silica is cooled more rapidly, the resulting solid is a disorderly arrangement which is called , often also called . For the study of many silicate based minerals, a classification scheme is required. Otherwise, the huge amount of information becomes unmanageable. We use a simple classification scheme based on the number of \(SiO_4^{4-}\) units that are connected together by sharing the oxygen atoms. are minerals consisting of only single SiO units. The cations are some other metals. For example, the following minerals are orthosilicates: The Be and Zn ions are tetrahedrally bonded to the oxygen of the silicate in these two minerals: phenacite, Be SiO and willemite, Zn SiO . In olivine, (Fe, Mg) SiO , the cations are either Fe or Mg . This formula suggests that this mineral is a mixed salt of iron and magnesium silicates. These cations are octahedrally coordinated to the oxygen atoms of the silicate. Pure salt Fe SiO is called fayalite, and Mg SiO is called forsterite. When two SiO units are linked together, they form the group, Si O . For example, thortveitite, Sc Si O is a pyrosilicate. When two oxygen of SiO units share with other SiO units, the silicates form a ring or an infinite chain. The stoichiometry of the silicates becomes (SiO ) . Benitoite BaTi(SiO ) contain three silica rings, but these are relaxed 6-atom rings The precious stone beryl Be Al (SiO ) contain six-silica rings. Single chain silica are called . Some synthetic metasilicates Na (SiO ) have been shown to contain the simple chain silicates (SiO ) , in which the Si-O bonds of the type Si-O-Si are 168 nm, with the Si-O-Si angles of 137 . The Si=O bonds are shorter, 1.57 nm. The natural pyroxenes include enstatite, MgSiO , diopside, CaMg(SiO ) , and jadeite, NaAl(SiO ) . Double chain silicates are called , part of the double chain is shown here, same as the double chain shown in by Swaddle. These chains have a stoichiometry of (Si O ) . You can easily identify one such unit in the diagram. Tremolite, Ca Mg (Si O ) (OH) , is such a mineral. The true asbestoses such as crocidolite or blue asbestos consist of double chain silicates. Asbestoses have been identified as carcinogens, and its application has since been limited due to a ban to limit its exposure to the public. Most commercial asbestoses are chrysotile, which contain layers of silicate sheet as we shall below. Sheet slilicates are called (phyllo means leaflike). These silicates are easy to cleave (as does graphite). Talc is a typical sheet silicate, Mg (OH) (Si O . Talc is a main ingredient of the soapstone (steatite). The diagram below shows the arrangement of sheets in brucite, Mg(OH) , in which the sheets consist of corner sharing octahedrons of Mg(OH) . In chlorite, there are two types of sheets. Half of the sheets are the same as those of brucite, but half of the brucite-sheets are sandwiched between sheets of silicates. The talc consists of only the sandwiched sheets. The diagram came from a polysome series which discusses sheet silicates. Serpentine, Mg (OH) Si O , has curved sheets. The comercial asbestos chrysotile is a sheet silicate, but the sheets are rolled up like a tube. These tubes appear as fibers, and they are usually known as asbestos. As mentioned earlier, the SiO units can share every oxygen with other units to form a three dimensional network, and quartz has such a structure. A portion of such a framework is shown here. In this arrangement, the stoichiometry is reduced to SiO , which is often called . A collection of small pieces of quartz is sand. Please see an animated illustration of the ring structure by Bob Hanson. is a group of minerals. Asbestos is the name applied to six naturally occurring minerals that are mined from the earth. The different types of asbestos are: Of these six, three are used more commonly. (white) is the most common, but it is not unusual to encounter (brown / off-white), or (blue) as well. Asbestos are noncombustable fibrous material, and they have been used for terminal insulation material, brake linings, construction material, and filters. When mixed with cement, it reinforce the mechanical strength of concrete. It decomposes due to loss of water, and forms forsterite and silica at high temperature. What is the stoichiometry for an infinite chain of silicates by sharing two oxygen atoms with the neighboring units? We can draw such a chain first: In this diagram, the basic unit is bolded, and the stoichiometry is apparently SiO . Is there another type of chain that satisfies the condition given in the question? If you work with 3-dimensional models, you probably realize that this is the only way to construct a single chain. Infinite chains and a ring structures have the same stoichiometry.
Describe the basic SiO unit of silicate and its versatility in connecting to each other or one another.
Identify the units used to build a structure.
These diagrams age discussed during the lectures.
Minerals consisting of single Be , units are called orthosilicates. These form salts with divalent cations Be , Zn , Mg , Fe etc, but these divalent cations are tightly bonded to the oxygen of the Be units.
These diagrams age discussed during the lectures. | 7,048 | 2,041 |
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The chemistry of group 8 is dominated by iron, whose high abundance in Earth’s crust is due to the extremely high stability of its nucleus. Ruthenium and osmium, on the other hand, are extremely rare elements, with terrestrial abundances of only about 0.1 ppb and 5 ppb, respectively, and they were not discovered until the 19th century. Because of the high melting point of iron (1538°C), early humans could not use it for tools or weapons. The advanced techniques needed to work iron were first developed by the Hittite civilization in Asia Minor sometime before 2000 BC, and they remained a closely guarded secret that gave the Hittites military supremacy for almost a millennium. With the collapse of the Hittite civilization around 1200 BC, the technology became widely distributed, however, leading to the Iron Age.
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https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Kinetics/01%3A_Introduction_to_Reaction_Kinetics |
How can an experiment confirm that a reaction is happening in a particular way? What is the mechanism of the reaction? What intermediates are occurring, and in what order do the bond-making and bond-breaking steps take place? There are many experiments designed to illustrate how reactions happen. One of the methods used is chemical kinetics, in which the rate of a reaction is measured. By making changes in the reaction conditions and measuring the effect of the changes on the rate of reaction, we can infer what is going on at the molecular level. Kinetic studies are important in understanding reactions, and they also have practical implications. For example, in industry, reactions are conducted in reactors in which compounds are mixed together, possibly heated and stirred for a while, and then moved to the next phase of the process. It is important to know how long to hold the reaction at one stage before moving on, to make sure that reaction has finished before starting the next one. By understanding how a reaction takes place, many processes can be improved. For example, if it is known that a particular intermediate is involved in a reaction, the use of conditions (such as certain solvents) that are incompatible with that intermediate might be avoided. In addition, reagents might be added that would make certain steps in the reaction happen more easily. Not only are kinetic studies important in industry, but they are also used to understand biological processes, especially enzyme-catalyzed reactions. They also play a role in environmental and atmospheric chemistry, as part of an effort to understand a variety of issues ranging from the fate of prescription pharmaceuticals in wastewater to the cascade of reactions involved in the ozone cycle. , | 1,788 | 2,049 |
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The of a reaction (\(t_{1/2}\)), is the amount of time needed for a reactant concentration to decrease by half compared to its initial concentration. Its application is used in chemistry and medicine to predict the concentration of a substance over time. The concepts of half life plays a key role in the administration of drugs into the target, especially in the elimination phase, where half life is used to determine how quickly a drug decrease in the target after it has been absorbed in the unit of time (sec, minute, day, etc.) or elimination rate constant \(k\) (minute , hour , day ,etc.). It is important to note that the half-life is varied between different type of reactions. The following section will go over different type of reaction, as well as how its half-life reaction are derived. The last section will talk about the application of half-life in the elimination phase of pharmcokinetics. In the rate of a reaction does not depend on the substrate concentration. In other words, saturating the amount of substrate does not speed up the rate of the reaction. Below is a graph of time (\(t\)) vs. concentration (\([A]\)) in a zero order reaction, several observation can be made: the slope of this plot is a straight line with negative slope equal negative \(k\), the half-life of zero order reaction decreases as the concentration decreases. We learn that the zero-order kinetic rate law is as followed, where \([A]\) is the current concentration, \([A]_o\) is the initial concentration, and \(k\) is the reaction constant and \(t\) is time: \[ [A]= [A]_o - kt \label{1} \] We need to isolate \(t_{1/2}\) when \[[A]=\dfrac{[A]_o}{2} \nonumber \] Substituting into Equation \ref{1} \[\begin{align} \dfrac{[A]_o}{2} &= [A]_o - kt_{1/2} \nonumber \\[4pt] kt_{1/2} &= [A]_o - \dfrac{[A]_o}{2} \nonumber \\[4pt] t_{1/2} &= \dfrac{[A]_o}{2k} \label{2} \end{align} \] Equation \ref{2} show the half-life for a zero-order reaction depends on both the initial concentration and rate constant. In , the graph represents the half-life is different from zero order reaction in a way that the slope continually decreases as time progresses until it reaches zero. We can also easily see that the length of half-life will be constant, independent of concentration. For example, it takes the same amount of time for the concentration to decrease from one point to another point. In order to solve the half life of first order reactions, we recall that the rate law of a first order reaction was: \[[A]=[A]_o e^{-kt} \label{4} \] We need to isolate \(t_{1/2}\) when \[[A]=\dfrac{[A]_o}{2} \nonumber \] Substituting into Equation \ref{4} \[ \begin{align} \dfrac{[A]_0}{2} &=[A]_o e^{-kt_{1/2}} \nonumber \\[4pt] \dfrac{1}{2} &= e^{-kt_{1/2}} \nonumber \\[4pt] \ln \dfrac{1}{2} &= -kt_{1/2} \nonumber \\[4pt] t_{1/2} &= \dfrac{\ln 2}{k} \nonumber \\[4pt] &\approx \dfrac{0.693}{k} \label{5} \end{align} \] Equation \ref{5} shows that for first-order reactions, the half-life depends solely on the reaction rate constant, \(k\). We can visually see this on the graph for first order reactions when we note that the amount of time between one half life and the next are the same. Another way to see it is that the half life of a first order reaction is independent of its initial concentration. Half-life of shows concentration \([A]\) vs. time (\(t\)), which is similar to first order plots in that their slopes decrease to zero with time. However, second order reactions decrease at a much faster rate as the graph shows. We can also note that the length of half-life increase while the concentration of substrate constantly decreases, unlike zero and first order reaction. In order to solve for half life of second order reactions we need to remember that the rate law of a second order reaction is: \[\dfrac{1}{[A]} = kt + \dfrac{1}{[A]_0} \label{6} \] As in zero-order and first-order reactions, we need to isolate \(t_{1/2}\) when \[[A]=\dfrac{[A]_o}{2} \nonumber \] Substituting into Equation \ref{6} \[ \begin{align} \dfrac{2}{[A]_0} &= kt_{1/2} + \dfrac{1}{[A]_0}\nonumber \\[4pt] -kt_{1/2} &= \dfrac{1}{[A]_0} - \dfrac{2}{[A]_0} \nonumber \\[4pt] t_{1/2} &= \dfrac{1}{k[A]_0} \label{7} \end{align} \] A following example is given below to illustrate the role of half life in pharmacokinetics to determine the drugs dosage interval. Another important application of half life in pharmacokinetics is that half-life tells how tightly drugs bind to each ligands before it is undergoing decay (\(k_s\)). The smaller the value of \(k_s\), the higher the affinity binding of drug to its target ligand, which is an important aspect of drug design Examine the following graph and answer Looking at the graph, we can see the therapeutic range is the amplitude of the graph, which is 5-15 mg/L The dosage interval is the half-life of the drug, looking at the graph, the half-life is 10 hours. Even though it will get in the therapeutic range, such practice is not recommended. A patient is treating with \(\ce{^{32}P}\). How long does it takes for the radioactivity to decay by 90%? The half-life of the material is 15 days. If we want the product to decay by 90%, that means 10% is left non-decayed, so \[\dfrac{[A]_t}{[A]_o} = 0.1 \nonumber \] From ln([A] /[A] ) = -kt, plug in value of k and [A] /[A] we then have t = 50 days In first order half life, what is the best way to determine the rate constant \(k\)? Why? The best way to determine rate constant \(k\) in half-life of first order is to determine half-life by experimental data. The reason is half-life in first order order doesn't depend on initial concentration. In a first order reaction, \(\ce{A -> B}\). The half-life is 10 days. | 5,739 | 2,050 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid/Overview_of_Acids_and_Bases |
There are three major classifications of substances known as acids or bases. The Arrhenius definition states that an acid produces H in solution and a base produces OH . This theory was developed by Svante Arrhenius in 1883. Later, two more sophisticated and general theories were proposed. These are the Brønsted-Lowry and the Lewis definitions of acids and bases. The Lewis theory is discussed elsewhere. In 1884, the Swedish chemist Svante Arrhenius proposed two specific classifications of compounds; acids and bases. When dissolved in an aqueous solution, certain ions were released into the solution. An Arrhenius acid is a compound that increases the concentration of that are present when added to water. These H ions form the ion (H O ) when they combine with water molecules. This process is represented in a chemical equation by adding H O to the reactants side. \[ HCl_{(aq)} \rightarrow H^+_{(aq)} + Cl^-_{(aq)} \] In this reaction, hydrochloric acid (\(HCl\)) dissociates completely into hydrogen (H ) and chlorine (Cl ) ions when dissolved in water, thereby releasing H ions into solution. Formation of the hydronium ion equation: \[ HCl_{(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + Cl^-_{(aq)} \] The Arrhenius theory, which is the simplest and least general description of acids and bases, includes acids such as HClO and HBr and bases such as \(NaOH\) or \(Mg(OH)_2\). For example the complete dissociation of \(HBr\) gas into water results generates free \(H_3O^+\) ions. \[HBr_{(g)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + Br^-_{(aq)}\] This theory successfully describes how acids and bases react with each other to make water and salts. However, it does not explain why some substances that do not contain hydroxide ions, such as \(F^-\) and \(NO_2^-\), can make basic solutions in water. The Brønsted-Lowry definition of acids and bases addresses this problem. An Arrhenius base is a compound that increases the concentration of OH ions that are present when added to water. The dissociation is represented by the following equation: \[ NaOH \; (aq) \rightarrow Na^+ \; (aq) + OH^- \; (aq) \] In this reaction, sodium hydroxide (NaOH) disassociates into sodium (Na ) and hydroxide (OH ) ions when dissolved in water, thereby releasing OH ions into solution. Owing to the overwhelming excess of \(H_2O\) molecules in aqueous solutions, a bare hydrogen ion has no chance of surviving in water. The hydrogen ion in aqueous solution is no more than a proton, a bare nucleus. Although it carries only a single unit of positive charge, this charge is concentrated into a volume of space that is only about a hundred-millionth as large as the volume occupied by the smallest atom. (Think of a pebble sitting in the middle of a sports stadium!) The resulting extraordinarily high of the proton strongly attracts it to any part of a nearby atom or molecule in which there is an excess of negative charge. In the case of water, this will be the lone pair (unshared) electrons of the oxygen atom; the tiny proton will be buried within the lone pair and will form a shared-electron (coordinate) bond with it, creating a , \(H_3O^+\). In a sense, \(H_2O\) is acting as a base here, and the product \(H_3O^+\) is the conjugate acid of water: Although other kinds of dissolved ions have water molecules bound to them more or less tightly, the interaction between H and \(H_2O\) is so strong that writing “H ” hardly does it justice, although it is formally correct. The formula \(H_3O^+\) more adequately conveys the sense that it is both a molecule in its own right, and is also the conjugate acid of water. The equation "HA → H + A " is so much easier to write that chemists still use it to represent acid-base reactions in contexts in which the proton donor-acceptor mechanism does not need to be emphasized. Thus, it is permissible to talk about “hydrogen ions” and use the formula H in writing chemical equations as long as you remember that they are not to be taken literally in the context of aqueous solutions. The Arrhenius theory has many more limitations than the other two theories. The theory suggests that in order for a substance to release either H or OH ions, it must contain that particular ion. However, this does not explain the weak base ammonia (NH ) which, in the presence of water, releases hydroxide ions into solution, but does not contain OH itself. Hydrochloric acid is neutralized by both sodium hydroxide solution and ammonia solution. In both cases, you get a colourless solution which you can crystallize to get a white salt - either sodium chloride or ammonium chloride. These are clearly very similar reactions. The full equations are: \[ NaOH \; (aq) + HCl \; (aq) \rightarrow NaCl \; (aq) + H_2O \; (l) \] \[ NH_3 \; (aq) + HCl \; (aq) \rightarrow NH_4Cl \; (aq) \] In the sodium hydroxide case, hydrogen ions from the acid are reacting with hydroxide ions from the sodium hydroxide - in line with the Arrhenius theory. However, in the ammonia case, there are no hydroxide ions! You can get around this by saying that, when the ammonia reacts with the water, it is dissolved in to produce ammonium ions and hydroxide ions: \[ NH_3 \; (aq) + H_2O \; (l) \rightleftharpoons NH_4^+ \; (aq) + OH^- \;(aq) \] This is a reversible reaction, and in a typical dilute ammonia solution, about 99% of the ammonia remains as ammonia molecules. Nevertheless, there are hydroxide ions there, and we can squeeze this into the Arrhenius theory. However, this same reaction also happens between ammonia gas and hydrogen chloride gas. \[ NH_3 \; (g) + HCl \; (g) \rightarrow NH_4Cl \;(s) \] In this case, there are not any hydrogen ions or hydroxide ions in solution - because there isn't any solution. The Arrhenius theory wouldn't count this as an acid-base reaction, despite the fact that it is producing the same product as when the two substances were in solution. Because of this shortcoming, later theories sought to better explain the behavior of acids and bases in a new manner. In 1923, chemists Johannes Nicolaus Brønsted and Thomas Martin Lowry independently developed definitions of acids and bases based on the compounds' abilties to either donate or accept protons (H ions). In this theory, acids are defined as ; whereas bases are defined as . A compound that acts as both a Brønsted-Lowry acid and base together is called .This took the Arrhenius definition one step further, as a substance no longer needed to be composed of hydrogen (H ) or hydroxide (OH ) ions in order to be classified as an acid or base. Consider the following chemical equation: Here, hydrochloric acid (HCl) "donates" a proton (H ) to ammonia (NH ) which "accepts" it , forming a positively charged ammonium ion (NH ) and a negatively charged chloride ion (Cl ). Therefore, HCl is a Brønsted-Lowry acid (donates a proton) while the ammonia is a Brønsted-Lowry base (accepts a proton). Also, Cl is called the of the acid HCl and NH is called the of the base NH . In this theory, an (like in the Arrhenius theory) and a . A basic salt, such as Na F , generates OH ions in water by taking protons from water itself (to make HF): \[F^-_{(aq)} + H_2O_{(l)} \rightleftharpoons HF_{(aq)} + OH^-\] When a Brønsted acid dissociates, it increases the concentration of hydrogen ions in the solution, \([H^+]\); conversely, Brønsted bases dissociate by taking a proton from the solvent (water) to generate \([OH^-]\). \[HA_{(aq)} \rightleftharpoons A^-_{(aq)} + H^+_{(aq)}\] \[K_a=\dfrac{[A^-,H^+]}{[HA]}\] \[B_{(aq)} + H_2O_{(l)} \rightleftharpoons HB^+_{(aq)} + OH^-_{(aq)}\] \[K_b = \dfrac{[HB^+,OH^-]}{[B]}\] One important consequence of these equilibria is that every (\(HA\)) has a (\(A^-\)), and vice-versa. In the base, dissociation equilibrium above the conjugate acid of base \(B\) is \(HB^+\). For a given acid or base, these equilibria are linked by the water dissociation equilibrium: \[H_2O_{(l)} \rightleftharpoons H^+_{(aq)} + OH^-_{(aq)}\] with \[K_w = [H^+,OH^-]\] for which the equilibrium constant K is 1.00 x 10 at 25°C. It can be easily shown that the product of the acid and base dissociation constants K and K is K . Strong acids are molecular compounds that essentially ionize to in aqueous solution, disassociating into H ions and the additional anion; there are very few common strong acids. All other acids are "weak acids" that incompletely ionized in aqueous solution. Acids and bases that dissociate are said to be strong acids, e.g.: Here the right-handed arrow (\(\rightarrow\)) implies that the reaction goes to completion. That is, a 1.0 M solution of HClO in water actually contains 1.0 M H (aq) and 1.0 M ClO (aq), and no undissociated HClO . Conversely, weak acids such as acetic acid (CH COOH) and weak bases such as ammonia (NH ) dissociate only slightly in water - typically a few percent, depending on their concentration and exist mostly as the undissociated molecules. Strong acids such as \(HCl\) dissociate to produce such as \(Cl^-\) as conjugate bases, whereas produce weak conjugate bases. This is illustrated below for acetic acid and its conjugate base, the acetate anion. Acetic acid is a weak acid (K = 1.8 x 10 ) and acetate is a weak base (K = K /K = 5.6 x 10 ) Like acids, strong and weak bases are classified by the extent of their ionization. Strong bases disassociate almost or entirely to completion in aqueous solution. Similar to strong acids, there are very few common strong bases. Weak bases are molecular compounds where the ionization is not complete. The strength of a conjugate acid/base varies inversely with the strength or weakness of its parent acid or base. Any acid or base is technically a conjugate acid or conjugate base also; these terms are simply used to identify species in solution (i.e acetic acid is the conjugate acid of the acetate anion, a base, while acetate is the conjugate base of acetic acid, an acid). How does one define acids and bases? In chemistry, acids and bases have been defined differently by three sets of theories. One is the Arrhenius definition, which revolves around the idea that acids are substances that ionize (break off) in an aqueous solution to produce hydrogen (H ) ions while bases produce hydroxide (OH ) ions in solution. On the other hand, the Brønsted-Lowry definition defines acids as substances that donate protons (H ) whereas bases are substances that accept protons. Also, the Lewis theory of acids and bases states that acids are electron pair acceptors while bases are electron pair donors. Acids and bases can be defined by their physical and chemical observations. Since acids increase the amount of H ions present and bases increase the amount of OH ions, under the pH scale, the strength of acidity and basicity can be measured by its concentration of H ions. This scale is shown by the following formula: with [H ] being the concentration of H ions. To see how these calculations are done, refer to The pH scale is often measured on a 1 to 14 range, but this is incorrect (see for more details). Something with a pH less than 7 indicates acidic properties and greater than 7 indicates basic properties. A pH at exactly 7 is neutral. The higher the [H ], the lower the pH. The Lewis theory of acids and bases states that acids act as and bases act as . This definition doesn't mention anything about the hydrogen atom at all, unlike the other definitions. It only talks about the transfer of electron pairs. To demonstrate this theory, consider the following example. This is a reaction between ammonia (NH ) and boron trifluoride (BF ). Since there is no transfer of hydrogen atoms here, it is clear that this is a Lewis acid-base reaction. In this reaction, NH has a lone pair of electrons and BF has an incomplete octet, since boron doesn't have enough electrons around it to form an octet. Because boron only has 6 electrons around it, it can hold 2 more. BF can act as a Lewis acid and accept the pair of electrons from the nitrogen in NH which will then form a bond between the nitrogen and the boron. This is considered an acid-base reaction where NH (base) is donating the pair of electrons to BF . BF (acid) is accepting those electrons to form a new compound, H NBF . A special property of acids and bases is their ability to neutralize the other's properties. In an acid-base (or neutralization) reaction, the H ions from the acid and the OH ions from the base react to create water (H O). Another product of a neutralization reaction is an ionic compound called a salt. Therefore, the general form of an acid-base reaction is: The following are examples of neutralization reactions: 1. (NOTE: To see this reaction done experimentally, refer to the YouTube video link under the section "References".) 2. Titrations are performed with acids and bases to determine their concentrations. At the equivalence point, the number of moles of the acid will equal the number of moles of the base. This indicates that the reaction has been neutralized. Here's how the calculations are done: For instance, hydrochloric acid is titrated with sodium hydroxide: For instance, 30 mL of 1.00 M NaOH is needed to titrate 60 mL of an HCl solution. The concentration of HCl needs to be determined. At the eqivalence point: To solve for the molarity of HCl, plug in the given data into the equation above. M (60 mL HCl) = (1.00 M NaOH)(30 mL NaOH) M =0.5 M The concentration of HCl is . : There are 6 strong acids and all other acids are considered weak. HNO is one of those 6 strong acids, while NH is actuallly a weak base. The answer is A Brønsted-Lowry Base is a proton acceptor, which means it will take in an H . This eliminates HCl, H PO NH and CH NH because they are Brønsted-Lowry acids. They all give away protons. In the case of HPO , consider the following equation: Here, it is clear that HPO is the acid since it donates a proton to water to make H O and PO . Now consider the following equation: In this case, HPO is the base since it accepts a proton from water to form H PO and OH . Thus, HPO is an acid and base together, making it amphoteric. Since HPO is the only compound from the options that can act as a base, the answer is Since the number of moles of acid equals the number of moles of base at neutralization, the following equation is used to solve for the molarity of HCl: Now, plug into the equation all the information that is given: M (25 mL HCl) = (0.5 M )(50 mL NaOH) M = 1 The correct answer is . First, the number of moles of the acid needs to be calculated. This is done by using the molar mass of HBr to convert 2.79 g of HBr to moles. (2.79 g HBr)/(80.91 g/mol HBr) = 0.0345 moles HBr Since this is a neutralization reaction, the number of moles of the acid (HBr) equals the number of moles of the base (NaOH) at neutralization: The molarity of NaOH can now be determined since the amount of moles are found and the volume is given. Convert 22.72 mL to Liters first since molarity is in units of moles/L. Molarity = (0.0345 moles NaOH)/(0.02272 L NaOH) = 1.52 M The correct answer is . The Brønsted-Lowry definition says that a base accepts protons (H ions). NaOH, Ca(OH) , and KOH are all Arrhenius bases because they yield the hydroxide ion (OH ) when they ionize. However, NH does not dissociate in water like the others. Instead, it takes a proton from water and becomes NH while water becomes a hydroxide. Therefore, the correct answer is . Jim Clark ( ) | 15,529 | 2,053 |
https://chem.libretexts.org/Bookshelves/Biological_Chemistry/Concepts_in_Biophysical_Chemistry_(Tokmakoff)/06%3A_Dynamics_and_Kinetics/21%3A_Binding_and_Association/21.04%3A_Biomolecular_Kinetics |
Returning to our basic two-state scheme, we define the rate constants k and k for the association and dissociation reactions: \[ A+B \underset{k_{d}}{\stackrel{k_{a}}{\rightleftharpoons}} C \] From detailed balance, which requires that the total flux for the forward and back reactions be equal under equilibrium conditions: \[ K_a = \dfrac{k_a}{k_d} \] The units for K are M , M s for k , and s for k . For the case where we explicitly consider the AB encounter complex: \[A+B \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}}(A B) \underset{k_{-2}}{\stackrel{k_{2}}{\rightleftharpoons}} C \] Schemes of this sort are referred to as reaction–diffusion problems. Note, this corresponds to the scheme used in Michaelis–Menten kinetics for enzyme catalysis, where AB is an enzyme–substrate complex prior to the catalytic step. The kinetic equations corresponding to this scheme are often solved with the help of a steady-state approximation (∂[AB]/∂t ≈ 0), leading to \[\dfrac{d[C]}{dt} = k_a[A,B]-k_d[C] \] \[k_a = \dfrac{k_1k_2}{(k_{-1}+k_2)} \qquad \quad k_d = \dfrac{k_{-1}k_{-2}}{k_{-1}+k_2} \] Let’s look at the limiting scenarios: What if both diffusion and reaction within encounter complex matter? That is the two rates \(k_1 \approx k_2 \). \[ A+B \stackrel{k_a}{\rightleftharpoons} AB \stackrel{k_{rxn}}{\rightleftharpoons} C \] Now all the rates matter. This can be solved in the same manner that we did for diffusion to capture by a sphere, but with boundary conditions that have finite concentration of reactive species at the critical radius. The steady-state solution gives: \[ \begin{aligned} k_{eff} &= \dfrac{k_ak_{rxn}}{k_a+k_{rxn}} \\ k_{eff}^{-1} &= k_a^{-1}+k_{rxn}^{-1} \end{aligned} \] k is the effective rate of forming the product C. It depends on the association rate k (or k ) and k is an effective forward reaction rate that depends on k and k . In diffusion–reaction processes, there are two competing factors that govern the outcome of the binding process. These are another manifestation of the familiar enthalpy–entropy compensation effects we have seen before. There is a competition between enthalpically favorable contacts in the bound state and the favorable entropy for the configurational space available to the unbound partners. Overall, there must be some favorable driving force for the interaction, which can be expressed in terms of a binding potential U (R) that favors the bound state. On the other hand, for any one molecule A, the translational configuration space available to the partner B will grow as R . We can put these concepts together in a simple model. The probability of finding B at a distance R from A is \[ P(R)dR = Q^{-1}e^{-U(R)/kT}4\pi R^2dR \] where Q is a normalization constant. Then we can define a free energy along the radial coordinate \[ \begin{aligned} F(R) &= -k_BT\ln P(R)dR \\ &=U(R)-k_BT\ln R^2-\ln Q \end{aligned} \] Here F(R) applies to a single A-B pair, and therefore the free energy drops continuously as R increases. This corresponds to the infinitely dilute limit, under which circumstance the partners will never bind. However, in practice there is a finite volume and concentration for the two partners. We only need to know the distance to the nearest possible binding partner \( \langle R_{AB} \rangle \). We can then put an upper bound on the radii sampled on this free energy surface. In the simplest approximation, we can determine a cut off radius in terms of the volume available to each B, which is the inverse of the B concentration: \(\frac{4}{3}\pi r^3_c = [B]^{-1} \). Then, the probability of finding the partners in the bound state is \[ P_a = \dfrac{\int_0^{r*} e^{-F(r)/k_BT}4\pi r^2dr}{\int_0^{r_c} e^{-F(r)/k_BT}4\pi r^2dr} \] At a more molecular scale, the rates of molecular association can be related to diffusion on a potential of mean force. g(r) is the radial distribution function that describes the radial variation of B density about A, and is related to the potential of mean force W(r) through \(g(r) = exp[-W(r)/k_BT] \). Then the association rate obtained from the flux at a radius defined by the association barrier ( r = r ) is \[ k_a^{-1} = \int_{r^†}^{\infty} dr [4\pi r^2D(r)e^{-W(r)/k_BT}]^{-1} \] Here D(r) is the radial diffusion coefficient that describes the relative diffusion of A and B. The spatial dependence reflects the fact that at small r the molecules do not really diffuse independently of one another. __________________________________________ | 4,511 | 2,055 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Thermodynamic_Cycles/Carnot_Cycle |
In the early 19th century, steam engines came to play an increasingly important role in industry and transportation. However, a systematic set of theories of the conversion of thermal energy to motive power by steam engines had not yet been developed. Nicolas Léonard Sadi Carnot (1796-1832), a French military engineer, published in 1824. The book proposed a generalized theory of heat engines, as well as an idealized model of a thermodynamic system for a heat engine that is now known as the Carnot cycle. Carnot developed the foundation of the second law of thermodynamics, and is often described as the "Father of thermodynamics." The Carnot cycle consists of the following four processes: The P-V diagram of the Carnot cycle is shown in Figure \(\Page {2}\). In isothermal processes I and III, ∆U=0 because ∆T=0. In adiabatic processes II and IV, q=0. Work, heat, ∆U, and ∆H of each process in the Carnot cycle are summarized in Table \(\Page {1}\). The T-S diagram of the Carnot cycle is shown in Figure \(\Page {3}\). In isothermal processes I and III, ∆T=0. In adiabatic processes II and IV, ∆S=0 because dq=0. ∆T and ∆S of each process in the Carnot cycle are shown in Table \(\Page {2}\). The Carnot cycle is the most efficient engine possible based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures. The efficiency of the carnot engine is defined as the ratio of the energy output to the energy input. \[\begin{align*} \text{efficiency} &=\dfrac{\text{net work done by heat engine}}{\text{heat absorbed by heat engine}} =\dfrac{-w_{sys}}{q_{high}} \\[4pt] &=\dfrac{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)+nRT_{low}\ln \left(\dfrac{V_{4}}{V_{3}}\right)}{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)} \end{align*}\] Since processes II (2-3) and IV (4-1) are adiabatic, \[\left(\dfrac{T_{2}}{T_{3}}\right)^{C_{V}/R}=\dfrac{V_{3}}{V_{2}}\] and \[\left(\dfrac{T_{1}}{T_{4}}\right)^{C_{V}/R}=\dfrac{V_{4}}{V_{1}}\] And since = and = , \[\dfrac{V_{3}}{V_{4}}=\dfrac{V_{2}}{V_{1}}\] Therefore, \[\text{efficiency}=\dfrac{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)-nRT_{low}\ln\left(\dfrac{V_{2}}{V_{1}}\right)}{nRT_{high}\ln\left(\dfrac{V_{2}}{V_{1}}\right)}\] \[\boxed{\text{efficiency}=\dfrac{T_{high}-T_{low}}{T_{high}}}\] The Carnot cycle has the greatest efficiency possible of an engine (although other cycles have the same efficiency) based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures. | 2,725 | 2,056 |
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Aldehydes_and_Ketones/Reactivity_of_Aldehydes_and_Ketones/The_Triiodomethane_(Iodoform)_Reaction |
This page looks at how the triiodomethane (iodoform) reaction can be used to identify the presence of a CH CO group in aldehydes and ketones. There are two apparently quite different mixtures of reagents that can be used to do this reaction. They are, in fact, chemically equivalent. This is chemically the more obvious method. Iodine solution is added to a small amount of aldehyde or ketone, followed by just enough sodium hydroxide solution to remove the color of the iodine. If nothing happens in the cold, it may be necessary to warm the mixture very gently. A positive result is the appearance of a very pale yellow precipitate of triiodomethane (previously known as iodoform) - CHI . Apart from its color, this can be recognised by its faintly "medical" smell. It is used as an antiseptic on the sort of sticky plasters you put on minor cuts, for example. Sodium chlorate(I) is also known as sodium hypochlorite. Potassium iodide solution is added to a small amount of aldehyde or ketone, followed by sodium chlorate(I) solution. Again, if no precipitate is formed in the cold, it may be necessary to warm the mixture very gently. The positive result is the same pale yellow precipitate as before. A positive result - the pale yellow precipitate of triiodomethane (iodoform) - is given by an aldehyde or ketone containing the grouping: "R" can be a hydrogen atom or a hydrocarbon group (for example, an alkyl group). If "R" is hydrogen, then you have the aldehyde ethanal, CH CHO. We will take the reagents as being iodine and sodium hydroxide solution. The first stage involves substitution of all three hydrogens in the methyl group by iodine atoms. The presence of hydroxide ions is important for the reaction to happen - they take part in the mechanism for the reaction. In the second stage, the bond between the C I and the rest of the molecule is broken to produce triiodomethane (iodoform) and the salt of an acid. Putting all this together gives the overall equation for the reaction: Jim Clark ( ) | 2,026 | 2,057 |
https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Thermodynamics/Path_Functions/Work/Gas_Expansion |
In Gas Expansion, we assume Ideal behavior for the two types of expansions: This shows the expansion of gas at constant temperature against weight of an object's mass (m) on the piston. Temperature is held constant, therefore the change in energy is zero (U=0). So, the heat absorbed by the gas equals the work done by the ideal gas on its surroundings. Enthalpy change is also equal to zero because the change in energy zero and the pressure and volume is constant. The graphs clearly show work done (area under the curve) is greater in a reversible process. Adiabatic means when no heat exchange occurs during expansion between system and surrounding and the temperature is no longer held constant. This equation shows the relationship between PV and is useful only when it applies to ideal gas and reversible adiabatic change. The equation is very similar to Boyle's law except it has exponent (gamma) due to change in temperature. The work done by an adiabatic reversible process is given by the following equation: where T is less than T . The internal energy of the system decreases as the gas expands. The work can be calculated in two ways because the Internal energy (U) does not depend on path. The graph shows that less work is done in an adiabatic reversible process than an Isothermal reversible process. | 1,336 | 2,059 |
https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Chemistry_-_The_Central_Science_(Brown_et_al.)/01%3A_Introduction_-_Matter_and_Measurement/1.S%3A_Matter_and_Measurement_(Summary) |
An understanding of chemistry is essential for understanding much of the natural world and is central to many other disciplines. Chemistry is the study of matter and the changes material substances undergo. It is essential for understanding much of the natural world and central to many other scientific disciplines, including astronomy, geology, paleontology, biology, and medicine. Matter can be classified according to physical and chemical properties. Matter is anything that occupies space and has mass. The three states of matter are solid, liquid, and gas. A physical change involves the conversion of a substance from one state of matter to another, without changing its chemical composition. Most matter consists of mixtures of pure substances, which can be homogeneous (uniform in composition) or heterogeneous (different regions possess different compositions & properties. All matter has physical and chemical properties. Physical properties are characteristics that scientists can measure without changing the composition of the sample under study, such as mass, color, and volume (the amount of space occupied by a sample). Chemical properties describe the characteristic ability of a substance to react to form new substances; they include its flammability and susceptibility to corrosion. The natural sciences begin with observation, and this usually involves numerical measurements of quantities such as length, volume, density, and temperature. Most of these quantities have units of some kind associated with them, and these units must be retained when you use them in calculations. Measuring units can be defined in terms of a very small number of fundamental ones that, through "dimensional analysis", provide insight into their derivation and meaning. Measurements may be accurate, meaning that the measured value is the same as the true value; they may be precise, meaning that multiple measurements give nearly identical values (i.e., reproducible results); they may be both accurate and precise; or they may be neither accurate nor precise. The goal of scientists is to obtain measured values that are both accurate and precise. Dimensional analysis is used in numerical calculations, and in converting units. It can help us identify whether an equation is set up correctly (i.e. the resulting units should be as expected). Units are treated similarly to the associated numerical values, i.e., if a variable in an equation is supposed to be squared, then the associated dimensions are squared, etc. | 2,541 | 2,060 |
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