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Courses/University_of_California_Davis/Chem_4B_Lab%3A_General_Chemistry_for_Majors_II/6%3A_Determination_of_Kc_for_a_Complex_Ion_Formation_(Experiment)/6.R%3A_Determination_of_Kc_for_a_Complex_Ion_Formation_(Lab_Report) | Name: ____________________________ Lab Partner: ________________________ Date: ________________________ Lab Section: __________________ Part A: Initial concentrations of \(\ce{Fe^{3+}}\) and \(\ce{SCN^{-}}\) in Unknown Mixtures Experimental Data Tube Reagent Volumes (mL) Reagent Volumes (mL).1 Reagent Volumes (mL).2 Initial Concentrations (M) Initial Concentrations (M).1 -- 2.00 x 10-3 M \(\ce{Fe(NO3)3}\) 2.00 x 10-3 M \(\ce{KSCN}\) Water \(\ce{Fe^{3+}}\) \(\ce{SCN^{-}}\) 1 5.00 5.00 0.00 NaN NaN 2 5.00 4.00 1.00 NaN NaN 3 5.00 3.00 2.00 NaN NaN 4 5.00 2.00 3.00 NaN NaN 5 5.00 1.00 4.00 NaN NaN Show a sample dilution calculation for [\(\ce{Fe^{3+}}]\) initial in Tube #1 only Part B and C: The Standard \(\ce{FeSCN^{2+}}\) Solution (Visual Method) Given that 10.00 mL of 0.200 M \(\ce{Fe(NO3)3}\), 2.00 mL of 0.00200 M \(\ce{KSCN}\), and 8.00 mL of water Equilibrium \([\ce{FeSCN^{2+}}]\) in Standard Solution: ______________ M Note that since \([\ce{Fe^{3+}}] \gg [\ce{SCN^{-}}]\) in the Standard Solution, the reaction is forced to completion, thus causing all the \(\ce{SCN^{-}}\) to convert to \(\ce{FeSCN^{2+}}\). Show the stoichiometry and dilution calculations used to obtain this value. Equilibrium Concentrations of \(\ce{FeSCN^{2+}}\) in Mixtures: Tube Solution Depths (mm) Solution Depths (mm).1 \([\ce{FeSCN^{2+}}]_{equil}\) (M) -- Mixtures Standard -- 1 NaN NaN NaN 2 NaN NaN NaN 3 NaN NaN NaN 4 NaN NaN NaN 5 NaN NaN NaN Show a sample calculation for \([\ce{FeSCN^{2+}}]_{equil}\) in Tube #1 only. Part D: Spectrophotometric Method Calibration Curve Data Tube Absorbance \([\ce{FeSCN^{2+}}]\) (M) Blank NaN NaN 1 NaN NaN 2 NaN NaN 3 NaN NaN 4 NaN NaN Show a sample dilution calculation for \([\ce{FeSCN^{2+}}]\) in Tube #1 and 2 only. Note that since \([\ce{Fe^{3+}}] \gg [\ce{SCN^{-}}]\) in the Standard Solution, the reaction is forced to completion, thus causing all the added \(\ce{SCN^{-}}\) to be converted to \(\ce{FeSCN^{2+}}\). Plot Absorbance vs.\([\ce{FeSCN^{2+}}]\) for the standard solutions. Obtain an equation for the line. This can be used to determine the \([\ce{FeSCN^{2+}}]\) in the table below. Attach your plot to this report. Test mixtures Mixture Absorbance \([\ce{FeSCN^{2+}}]\) (M) 1 NaN NaN 2 NaN NaN 3 NaN NaN 4 NaN NaN 5 NaN NaN Show a sample calculation for \([\ce{FeSCN^{2+}}]\) in mixture 1. Linear Equation (A vs. \([\ce{FeSCN^{2+}}]\)) of Calibration Curve: ___________________ Calculations and Analysis The reaction that is assumed to occur in this experiment is: \[\ce{Fe^{3+} (aq) + SCN^{-} (aq) <=> FeSCN^{2+} (aq)} \label{eq1}\] Write the equilibrium constant expression for the reaction. Create a Reaction Table (or ICE table), as in Table 1, to demonstrate how the values below are calculated. Use the data for Mixture #1 only. Start with the known values for the initial concentrations of each species and the final value of \([\ce{FeSCN^{2+}}]\) from the data table on the previous page. Show how you find the value of the stoichiometric change in reaction concentrations that occurs, and the resulting equilibrium concentrations of the reactants. Show a sample calculation for the value of \(K_{c}\) using the data for Tube #1. Using the same method you outlined above, complete the table for all the equilibrium concentrations and value of \(K_{c}\): Tube Equilibrium Concentrations (M) Equilibrium Concentrations (M).1 Equilibrium Concentrations (M).2 \(K_{c}\) -- \(\ce{Fe^{3+}}\) \(\ce{SCN^{-}}\) \(\ce{FeSCN^{2+}}\) -- 1 NaN NaN NaN NaN 2 NaN NaN NaN NaN 3 NaN NaN NaN NaN 4 NaN NaN NaN NaN Average value of \(K_{c}\) ________________ (Use reasonable number of significant digits, based on the distribution of your \(K_{c}\) values.) Optional Analysis: Is an alternative reaction stoichiometry supported? Suppose that instead of forming \(\ce{FeSCN^{2+}}\), the reaction between \(\ce{Fe^{3+}}\) and \(\ce{SCN^{-}}\) resulted in the formation of \(\ce{Fe(SCN)_2^{+}}\). The reaction analogous to Equation \ref{eq1} would be: \[\ce{Fe^{3+} (aq) + 2SCN^{-} (aq) <=> Fe(SCN)2^{+} (aq)} \label{6}\] i.e., two moles of \(\ce{SCN^{-}}\) displace two moles of \(\ce{H2O}\) in \(\ce{Fe(H2O)6^{2+}}\) to make \(\ce{Fe(SCN)2(H2O)4^{+}}\). Write the equilibrium expression for this reaction. Create a Reaction Table for Mixture #1 only (or ICE table), as in Table 1, to clearly show how all the values below were obtained. Then show the calculation for the value of \(K_{c}\) for Tube 1. Pay special attention to the stoichiometry in this system. Begin by assuming that the equilibrium value for \([\ce{Fe(SCN)_2^{+}}]\) is be equal to \(\frac{1}{2} [\ce{FeSCN^{+2}}]\) at equilibrium obtained previously. (This is because the moles of \(\ce{SCN^{-}}\) assumed to be equal to the moles of complex ion product, \(\ce{FeSCN^{2+}}\), in our standard solutions. In the alternate reaction, moles of product, \(\ce{Fe(SCN)_2^{+}}\), equal ½ the number of moles of \(\ce{SCN^{-}}\) added in the standard solutions.) Using the method you outlined above, complete the table for all the equilibrium concentrations and values of \(K_{c}\) Tube Equilibrium Concentrations (M) Equilibrium Concentrations (M).1 Equilibrium Concentrations (M).2 \(K_{c}\) -- \(\ce{Fe^{3+}}\) \(\ce{SCN^{-}}\) \(\ce{Fe(SCN)_2{+}}\) -- 1 NaN NaN NaN NaN 2 NaN NaN NaN NaN 3 NaN NaN NaN NaN 4 NaN NaN NaN NaN 5 NaN NaN NaN NaN Average value of \(K_{c}\) ________________ (Use a reasonable number of significant digits, based on the distribution of your \(K_{c}\) values.) Based on the calculated values of \(K_{c}\) for each reaction stoichiometry, which reaction is the valid one? Briefly explain your conclusion. Compare the percent difference between the average value and the individual measurements. What does this tell you about the two possible stoichiometries? Do these reactions give consistent values of \(K_{c}\) for different initial reaction conditions? |
Courses/Smith_College/CHM_222_Chemistry_II%3A_Organic_Chemistry_(2025)/02%3A_Polar_Covalent_Bonds_Resonance_Structures_Acids_and_Bases | Chapter Objectives This chapter provides a review of the more advanced material covered in a standard introductory chemistry course through a discussion of the following topics: the use of electronegativity to determine bond polarity, and the application of this knowledge to determine whether a given molecule possesses a dipole moment. the drawing and interpretation of organic chemical structures. the concept and determination of formal charge. resonance and drawing of resonance forms the Brønsted-Lowry and Lewis definitions of acids and bases, acidity constants and acid-base reactions. intermolecular forces Chapter Videos This video provides examples of resonance delocalization. This video provides an introduction to organic acids and bases. This video provides an introduction to Ka and pKa. This video provides a worked solution to a pKa problem. 2.1: Polar Covalent Bonds - Electronegativity Because the tendency of an element to gain or lose electrons is so important in determining its chemistry, various methods have been developed to quantitatively describe this tendency. The most important method uses a measurement called electronegativity, defined as the relative ability of an atom to attract electrons to itself in a chemical compound. 2.2: Polar Covalent Bonds - Dipole Moments Mathematically, dipole moments are vectors; they possess both a magnitude and a direction. The dipole moment of a molecule is therefore the vector sum of the dipole moments of the individual bonds in the molecule. If the individual bond dipole moments cancel one another, there is no net dipole moment. 2.3: Formal Charges A formal charge is the charge assigned to an atom in a molecule, assuming that electrons in all chemical bonds are shared equally between atoms, regardless of relative electronegativity. 2.4: Resonance Resonance structures are a set of two or more Lewis Structures that collectively describe the electronic bonding a single polyatomic species including fractional bonds and fractional charges. Resonance structure are capable of describing delocalized electrons that cannot be expressed by a single Lewis formula with an integer number of covalent bonds. 2.5: Rules for Resonance Forms The above resonance structures show that the electrons are delocalized within the molecule and through this process the molecule gains extra stability. Ozone with both of its opposite formal charges creates a neutral molecule and through resonance it is a stable molecule. The extra electron that created the negative charge on one terminal oxygen can be delocalized by resonance through the other terminal oxygen. 2.6: Drawing Resonance Forms Resonance structures are used when one Lewis structure for a single molecule cannot fully describe the bonding that takes place between neighboring atoms relative to the empirical data for the actual bond lengths between those atoms. The net sum of valid resonance structures is defined as a resonance hybrid, which represents the overall delocalization of electrons within the molecule. A molecule that has several resonance structures is more stable than one with fewer. 2.7: Acids and Bases - The Brønsted-Lowry Definition In 1923, chemists Johannes Brønsted and Martin Lowry independently developed definitions of acids and bases based on compounds abilities to either donate or accept protons (H+ ions). Here, acids are defined as being able to donate protons in the form of hydrogen ions; whereas bases are defined as being able to accept protons. This took the Arrhenius definition one step further as water is no longer required to be present in the solution for acid and base reactions to occur. 2.8: Acid and Base Strength The relative acidity of different compounds or functional groups – in other words, their relative capacity to donate a proton to a common base under identical conditions – is quantified by a number called the dissociation constant, abbreviated Ka. The common base chosen for comparison is water. 2.9: Predicting Acid-Base Reactions from pKa Values pKa values can be used to predict the equilibrium of an acid-base reaction. The equilibrium will favor the side with the weaker acid. 2.10: Organic Acids and Organic Bases In the absence of pKa values, the relative strength of an organic acid can be predicted based on the stability of the conjugate base that it forms. The acid that forms the more stable conjugate base will be the stronger acid. The common factors that affect the conjugate base's stability are 1) the size and electronegativity of the the atom that has lost the proton, 2) resonance effects, 3) inductive effects, and 4) solvation effects. 2.11: Acids and Bases - The Lewis Definition A broader definition is provided by the Lewis theory of acids and bases, in which a Lewis acid is an electron-pair acceptor and a Lewis base is an electron-pair donor. This definition covers Brønsted-Lowry proton transfer reactions, but also includes reactions in which no proton transfer is involved. 2.12: Noncovalent Interactions Between Molecules In contrast to intramolecular forces, such as the covalent bonds that hold atoms together in molecules and polyatomic ions, intermolecular forces hold molecules together in a liquid or solid. Intermolecular forces are generally much weaker than covalent bonds. The most common intermolecular forces in organic chemistry are from strongest to weakest are hydrogen bonds, dipole-dipole interactions, and London Dispersion (van der Waals) forces. 2.MM: Molecular Models 2.S: Polar Covalent Bonds; Acids and Bases (Summary) |
Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Wade)_Complete_and_Semesters_I_and_II/Map%3A_Organic_Chemistry_I_(Wade)/04%3A_Structure_and_Stereochemistry_of_Alkanes/4.02%3A_Physical_Properties_of_Alkanes | Learning Objective explain & predict the physical properties of alkanes including relative bp and solubility in a mixture Overview Alkanes are not very reactive and have little biological activity; all alkanes are colorless and odorless non-polar compounds. The relative weak London dispersion forces of alkanes result in gaseous substances for short carbon chains, volatile liquids with densities around 0.7 g/mL for moderate carbon chains, and solids for long carbon chains. The differences in the physical states occurs because there is a direct relationship between the size and shape of molecules and the strength of the intermolecular forces (IMFs). Because alkanes have relatively predictable physical properties and undergo relatively few chemical reactions other than combustion, they serve as a basis of comparison for the properties of many other organic compound families. Let’s consider their physical properties first. Boiling Points Table \(\PageIndex{1}\) describes some of the properties of some of the first 10 straight-chain alkanes. Because alkane molecules are nonpolar, they are insoluble in water, which is a polar solvent, but are soluble in nonpolar and slightly polar solvents. Consequently, alkanes themselves are commonly used as solvents for organic substances of low polarity, such as fats, oils, and waxes. Nearly all alkanes have densities less than 1.0 g/mL and are therefore less dense than water (the density of H 2 O is 1.00 g/mL at 20°C). These properties explain why oil and grease do not mix with water but rather float on its surface. Molecular Name Formula Melting Point (°C) Boiling Point (°C) Density (20°C)* Physical State (at 20°C) methane CH4 –182 –164 0.668 g/L gas ethane C2H6 –183 –89 1.265 g/L gas propane C3H8 –190 –42 1.867 g/L gas butane C4H10 –138 –1 2.493 g/L gas pentane C5H12 –130 36 0.626 g/mL liquid hexane C6H14 –95 69 0.659 g/mL liquid octane C8H18 –57 125 0.703 g/mL liquid decane C10H22 –30 174 0.730 g mL liquid *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. The boiling points for the "straight chain" isomers and isoalkanes isomers are shown to demonstrate that branching decreases the surfaces area, weakens the IMFs, and lowers the boiling point. This next diagrams summarizes the physical states of the first six alkanes. The first four alkanes are gases at room temperature, and solids do not begin to appear until about \(C_{17}H_{36}\), but this is imprecise because different isomers typically have different melting and boiling points. By the time you get 17 carbons into an alkane, there are unbelievable numbers of isomers! Cycloalkanes have boiling points that are approximately 20 K higher than the corresponding straight chain alkane. There is not a significant electronegativity difference between carbon and hydrogen, thus, there is not any significant bond polarity. The molecules themselves also have very little polarity. A totally symmetrical molecule like methane is completely non-polar, meaning that the only attractions between one molecule and its neighbors will be Van der Waals dispersion forces. These forces will be very small for a molecule like methane but will increase as the molecules get bigger. Therefore, the boiling points of the alkanes increase with molecular size. Where you have isomers, the more branched the chain, the lower the boiling point tends to be. Van der Waals dispersion forces are smaller for shorter molecules and only operate over very short distances between one molecule and its neighbors. It is more difficult for short, fat molecules (with lots of branching) to lie as close together as long, thin molecules. Example: Structure dependent Boiling Points For example, the boiling points of the three isomers of \(C_5H_{12}\) are: pentane: 309.2 K 2-methylbutane: 301.0 K 2,2-dimethylpropane: 282.6 K The slightly higher boiling points for the cycloalkanes are presumably because the molecules can get closer together because the ring structure makes them tidier and less "wriggly"! Solubility Alkanes (both alkanes and cycloalkanes) are virtually insoluble in water, but dissolve in organic solvents. However, liquid alkanes are good solvents for many other non-ionic organic compounds. Solubility in Water When a molecular substance dissolves in water, the following must occur: break the intermolecular forces within the substance. In the case of the alkanes, these are the Van der Waals dispersion forces. break the intermolecular forces in the water so that the substance can fit between the water molecules. In water, the primary intermolecular attractions are hydrogen bonds. Breaking either of these attractions requires energy, although the amount of energy to break the Van der Waals dispersion forces in something like methane is relatively negligible; this is not true of the hydrogen bonds in water. As something of a simplification, a substance will dissolve if there is enough energy released when new bonds are made between the substance and the water to compensate for what is used in breaking the original attractions. The only new attractions between the alkane and the water molecules are Van der Waals forces. These forces do not release a sufficient amount of energy to compensate for the energy required to break the hydrogen bonds in water.; the alkane does not dissolve. The energy only description of solvation is an oversimplification because entropic effects are also important when things dissolve. The lack of water solubility can lead to environmental concerns when oils are spilled into natural bodies of water as shown below. Oil Spills. Crude oil coats the water’s surface in the Gulf of Mexico after the Deepwater Horizon oil rig sank following an explosion. The leak was a mile below the surface, making it difficult to estimate the size of the spill. One liter of oil can create a slick 2.5 hectares (6.3 acres) in size. This and similar spills provide a reminder that hydrocarbons and water don’t mix. Source: Photo courtesy of NASA Goddard / MODIS Rapid Response Team, http://www.nasa.gov/topics/earth/features/oilspill/oil-20100519a.html . Solubility in organic solvents In most organic solvents, the primary forces of attraction between the solvent molecules are Van der Waals - either dispersion forces or dipole-dipole attractions. Therefore, when an alkane dissolves in an organic solvent, the Van der Waals forces are broken and are replaced by new Van der Waals forces. The two processes more or less cancel each other out energetically; thus, there is no barrier to solubility. Looking Closer: Gas Densities and Fire Hazards Table \(\PageIndex{1}\) indicates that the first four members of the alkane series are gases at ordinary temperatures. Natural gas is composed chiefly of methane, which has a density of about 0.67 g/L. The density of air is about 1.29 g/L. Because natural gas is less dense than air, it rises. When a natural-gas leak is detected and shut off in a room, the gas can be removed by opening an upper window. On the other hand, bottled gas can be either propane (density 1.88 g/L) or butanes (a mixture of butane and isobutane; density about 2.5 g/L). Both are much heavier than air (density 1.2 g/L). If bottled gas escapes into a building, it collects near the floor. This presents a much more serious fire hazard than a natural-gas leak because it is more difficult to rid the room of the heavier gas. Also shown in Table \(\PageIndex{1}\) are the boiling points of the straight-chain alkanes increase with increasing molar mass. This general rule holds true for the straight-chain homologs of all organic compound families. Larger molecules have greater surface areas and consequently interact more strongly; more energy is therefore required to separate them. For a given molar mass, the boiling points of alkanes are relatively low because these nonpolar molecules have only weak dispersion forces to hold them together in the liquid state. Looking Closer: An Alkane Basis for Properties of Other Compounds An understanding of the physical properties of the alkanes is important in that petroleum and natural gas and the many products derived from them—gasoline, bottled gas, solvents, plastics, and more—are composed primarily of alkanes. This understanding is also vital because it is the basis for describing the properties of other organic and biological compound families. For example, large portions of the structures of lipids consist of nonpolar alkyl groups. Lipids include the dietary fats and fatlike compounds called phospholipids and sphingolipids that serve as structural components of living tissues. These compounds have both polar and nonpolar groups, enabling them to bridge the gap between water-soluble and water-insoluble phases. This characteristic is essential for the selective permeability of cell membranes. Tripalmitin (a), a typical fat molecule, has long hydrocarbon chains typical of most lipids. Compare these chains to hexadecane (b), an alkane with 16 carbon atoms. Exercise Without referring to a table, predict which has a higher boiling point—hexane or octane. Explain. 2. If 25 mL of hexane were added to 100 mL of water in a beaker, which of the following would you expect to happen? Explain. Hexane would dissolve in water. Hexane would not dissolve in water and would float on top. Hexane would not dissolve in water and would sink to the bottom of the container. 3. Without referring to a table or other reference, predict which member of each pair has the higher boiling point. pentane or butane heptane or nonane 4. For which member of each pair is hexane a good solvent? pentane or water sodium chloride or soybean oil Answer 1. octane because of its greater molar mass 2. b; Hexane is insoluble in water and is less dense than water so it floats on top. 3. a) pentane b) nonane 4. a) pentane b) soybean oil |
Bookshelves/Inorganic_Chemistry/Map%3A_Inorganic_Chemistry_(Housecroft)/21%3A_d-Block_Metal_Chemistry_-_The_First_Row_Metals/21.09%3A_Group_8_-_Iron/21.9B%3A_Iron(VI)_Iron(V)_and_Iron(IV) | Welcome to the Chemistry Library. This Living Library is a principal hub of the LibreTexts project , which is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning. The LibreTexts approach is highly collaborative where an Open Access textbook environment is under constant revision by students, faculty, and outside experts to supplant conventional paper-based books. Campus Bookshelves Bookshelves Learning Objects |
Courses/Louisville_Collegiate_School/General_Chemistry/LibreTexts_Louisville_Collegiate_School_Chapters_11%3A_Solutions_and_Colloids/LibreTexts%2F%2FLouisville_Collegiate_School%2F%2FChapters%2F%2F11%3A_Solutions_and_Colloids%2F%2F11.E%3A_Solutions_and_Colloids_(Exercises) | 11.2: The Dissolution Process Q11.2.1 How do solutions differ from compounds? From other mixtures? S11.2.1 A solution can vary in composition, while a compound cannot vary in composition. Solutions are homogeneous at the molecular level, while other mixtures are heterogeneous. Q11.2.2 Which of the principal characteristics of solutions can we see in the solutions of \(\ce{K2Cr2O7}\) shown in Figure: When potassium dichromate (\(\ce{K2Cr2O7}\)) is mixed with water, it forms a homogeneous orange solution. (credit: modification of work by Mark Ott) S11.2.2 The solutions are the same throughout (the color is constant throughout), and the composition of a solution of K 2 Cr 2 O 7 in water can vary. Q11.2.3 When KNO 3 is dissolved in water, the resulting solution is significantly colder than the water was originally. Is the dissolution of KNO 3 an endothermic or an exothermic process? What conclusions can you draw about the intermolecular attractions involved in the process? Is the resulting solution an ideal solution? S11.2.3 (a) The process is endothermic as the solution is consuming heat. (b) Attraction between the K + and \(\ce{NO3-}\) ions is stronger than between the ions and water molecules (the ion-ion interactions have a lower, more negative energy). Therefore, the dissolution process increases the energy of the molecular interactions, and it consumes the thermal energy of the solution to make up for the difference. (c) No, an ideal solution is formed with no appreciable heat release or consumption. Q11.2.4 Give an example of each of the following types of solutions: a gas in a liquid a gas in a gas a solid in a solid S11.2.4 (a) CO 2 in water; (b) O 2 in N 2 (air); (c) bronze (solution of tin or other metals in copper) Q11.2.5 Indicate the most important types of intermolecular attractions in each of the following solutions: The solution in Figure . NO( l ) in CO( l ) Cl 2 ( g ) in Br 2 ( l ) HCl( aq ) in benzene C 6 H 6 ( l ) Methanol CH 3 OH( l ) in H 2 O( l ) S11.2.5 (a) ion-dipole forces; (b) dipole-dipole forces; (c) dispersion forces; (d) dispersion forces; (e) hydrogen bonding Q11.2.5 Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane (C 7 H 16 , nonpolar solvent): vegetable oil (nonpolar) isopropyl alcohol (polar) potassium bromide (ionic) S11.2.5 (a) heptane; (b) water; (c) water Q11.2.6 Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes. S11.2.6 Heat is released when the total intermolecular forces (IMFs) between the solute and solvent molecules are stronger than the total IMFs in the pure solute and in the pure solvent: Breaking weaker IMFs and forming stronger IMFs releases heat. Heat is absorbed when the total IMFs in the solution are weaker than the total of those in the pure solute and in the pure solvent: Breaking stronger IMFs and forming weaker IMFs absorbs heat. Q11.2.7 Solutions of hydrogen in palladium may be formed by exposing Pd metal to H 2 gas. The concentration of hydrogen in the palladium depends on the pressure of H 2 gas applied, but in a more complex fashion than can be described by Henry’s law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal. Determine the molarity of this solution (solution density = 1.8 g/cm 3 ). Determine the molality of this solution (solution density = 1.8 g/cm 3 ). Determine the percent by mass of hydrogen atoms in this solution (solution density = 1.8 g/cm 3 ). S11.2.7 http://cnx.org/contents/ mH6aqegx @2/The-Dissolution-Process 11.3: Electrolytes Q11.3.1 Explain why the ions Na + and Cl − are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules. S11.3.2 Crystals of NaCl dissolve in water, a polar liquid with a very large dipole moment, and the individual ions become strongly solvated. Hexane is a nonpolar liquid with a dipole moment of zero and, therefore, does not significantly interact with the ions of the NaCl crystals. Q11.3.2 Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive. S11.3.2 HBr is an acid and so its molecules react with water molecules to form H 3 O + and Br − ions that provide conductivity to the solution. Though HBr is soluble in benzene, it does not react chemically but remains dissolved as neutral HBr molecules. With no ions present in the benzene solution, it is electrically nonconductive. Q11.3.3 Consider the solutions presented: (a) Which of the following sketches best represents the ions in a solution of Fe(NO 3 ) 3 ( aq )? (b) Write a balanced chemical equation showing the products of the dissolution of Fe(NO 3 ) 3 . S11.3.3 (a) Fe(NO 3 ) 3 is a strong electrolyte, thus it should completely dissociate into Fe 3+ and \(\ce{(NO3- )}\) ions. Therefore, (z) best represents the solution. (b) \(\ce{Fe(NO3)3}(s)⟶\ce{Fe^3+}(aq)+\ce{3NO3- }(aq)\) Q11.3.4 Compare the processes that occur when methanol (CH 3 OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution. S11.3.3 Methanol, \(CH_3OH\), dissolves in water in all proportions, interacting via hydrogen bonding. Methanol: \[CH_3OH_{(l)}+H_2O_{(l)}⟶CH_3OH_{(aq)}\] Hydrogen chloride, HCl, dissolves in and reacts with water to yield hydronium cations and chloride anions that are solvated by strong ion-dipole interactions. Hydrogen chloride: \[HCl{(g)}+H_2O_{(l)} \rightarrow H_3O^+_{(aq)}+Cl^−_{(aq)}\] Sodium hydroxide, NaOH, dissolves in water and dissociates to yield sodium cations and hydroxide anions that are strongly solvated by ion-dipole interactions and hydrogen bonding, respectively. Sodium hydroxide: \[NaOH_{(s)} \rightarrow Na^+_{(aq)} + OH^−_{(aq)}\] Q11.3.5 What is the expected electrical conductivity of the following solutions? NaOH( aq ) HCl( aq ) C 6 H 12 O 6 ( aq ) (glucose) NH 3 ( l ) S11.3.5 (a) high conductivity (solute is an ionic compound that will dissociate when dissolved); (b) high conductivity (solute is a strong acid and will ionize completely when dissolved); (c) nonconductive (solute is a covalent compound, neither acid nor base, unreactive towards water); (d) low conductivity (solute is a weak base and will partially ionize when dissolved) Q11.3.6 Why are most solid ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a liquid (molten) ionic compound to be electrically conductive or nonconductive? Explain. S11.3.6 A medium must contain freely mobile, charged entities to be electrically conductive. The ions present in a typical ionic solid are immobilized in a crystalline lattice and so the solid is not able to support an electrical current. When the ions are mobilized, either by melting the solid or dissolving it in water to dissociate the ions, current may flow and these forms of the ionic compound are conductive. Q11.3.7 Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions: the solutions in Figure methanol, CH 3 OH, dissolved in ethanol, C 2 H 5 OH methane, CH 4 , dissolved in benzene, C 6 H 6 the polar halocarbon CF 2 Cl 2 dissolved in the polar halocarbon CF 2 ClCFCl 2 O 2 ( l ) in N 2 ( l ) S11.3.7 (a) ion-dipole; (b) hydrogen bonds; (c) dispersion forces; (d) dipole-dipole attractions; (e) dispersion forces 11.4: Solubility Q11.4.1 Suppose you are presented with a clear solution of sodium thiosulfate, Na 2 S 2 O 3 . How could you determine whether the solution is unsaturated, saturated, or supersaturated? S11.4.1 Add a small crystal of \(Na_2S_2O_3\). It will dissolve in an unsaturated solution, remain apparently unchanged in a saturated solution, or initiate precipitation in a supersaturated solution. Q11.4.2 Supersaturated solutions of most solids in water are prepared by cooling saturated solutions. Supersaturated solutions of most gases in water are prepared by heating saturated solutions. Explain the reasons for the difference in the two procedures. S11.4.2 The solubility of solids usually decreases upon cooling a solution, while the solubility of gases usually decreases upon heating. Q11.4.3 Suggest an explanation for the observations that ethanol, C 2 H 5 OH, is completely miscible with water and that ethanethiol, C 2 H 5 SH, is soluble only to the extent of 1.5 g per 100 mL of water. S11.4.3 The hydrogen bonds between water and C 2 H 5 OH are much stronger than the intermolecular attractions between water and C 2 H 5 SH. Q11.4.4 Calculate the percent by mass of KBr in a saturated solution of KBr in water at 10 °C using the following figure for useful data, and report the computed percentage to one significant digit. This graph shows how the solubility of several solids changes with temperature. S11.4.4 At 10 °C, the solubility of KBr in water is approximately 60 g per 100 g of water. \[\%\; KBr =\dfrac{60\; g\; KBr}{(60+100)\;g\; solution} = 40\%\] Q11.4.5 Which of the following gases is expected to be most soluble in water? Explain your reasoning. CH 4 CCl 4 CHCl 3 S11.4.5 (c) CHCl 3 is expected to be most soluble in water. Of the three gases, only this one is polar and thus capable of experiencing relatively strong dipole-dipole attraction to water molecules. Q11.4.6 At 0 °C and 1.00 atm, as much as 0.70 g of O 2 can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of O 2 dissolve in 1 L of water? S11.4.6 This problem requires the application of Henry’s law. The governing equation is \(C_g = kP_g\). \[k=\dfrac{C_g}{P_g}=\dfrac{0.70\;g}{1.00\; atm} =0.70\;g\; atm^{−1}\] Under the new conditions, \(C_g=0.70\;g\;atm^{−1} \times 4.00\; atm = 2.80\; g\). Q11.4.7 Refer to following figure for the following three questions: How did the concentration of dissolved CO 2 in the beverage change when the bottle was opened? What caused this change? Is the beverage unsaturated, saturated, or supersaturated with CO 2 ? S11.4.7 (a) It decreased as some of the CO 2 gas left the solution (evidenced by effervescence). (b) Opening the bottle released the high-pressure CO 2 gas above the beverage. The reduced CO 2 gas pressure, per Henry’s law, lowers the solubility for CO 2 . (c) The dissolved CO 2 concentration will continue to slowly decrease until equilibrium is reestablished between the beverage and the very low CO 2 gas pressure in the opened bottle. Immediately after opening, the beverage, therefore, contains dissolved CO 2 at a concentration greater than its solubility, a nonequilibrium condition, and is said to be supersaturated. Q11.4.8 The Henry’s law constant for CO 2 is \(3.4 \times 10^{−2}\; M/atm\) at 25 °C. What pressure of carbon dioxide is needed to maintain a CO 2 concentration of 0.10 M in a can of lemon-lime soda? S11.4.8 \[P_g=\dfrac{C_g}{k}=\dfrac{0.10\; M}{3.4 \times 10^{−2}\;M/atm} =2.9\; atm\] Q11.4.9 The Henry’s law constant for O 2 is \(1.3\times 10^{−3}\; M/atm\) at 25 °C. What mass of oxygen would be dissolved in a 40-L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of O 2 is 0.21 atm? S11.4.9 Start with Henry's law \[C_g=kP_g\] and apply it to \(O_2\) \[C(O_2)=(1.3 \times 10^{−3}\; M/atm) (0.21\;atm)=2.7 \times 10^{−4}\;mol/L\] The total amount is \((2.7 \times 10^{−4}\; mol/L)(40\;L=1.08 \times 10^{−2} \;mol\] The mass of oxygen is \((1.08 \times 10^{−2}\; mol)(32.0\; g/mol)=0.346\;g\) or, using two significant figures, \(0.35\; g\). Q11.4.10 How many liters of HCl gas, measured at 30.0 °C and 745 torr, are required to prepare 1.25 L of a 3.20- M solution of hydrochloric acid? S11.4.10 First, calculate the moles of HCl needed. Then use the ideal gas law to find the volume required. M = mol L−1 3.20M=xmol1.25L x = 4.00 mol HCl Before using the ideal gas law, change pressure to atmospheres and convert temperature from °C to kelvin . \[1\;atmx = 760torr745torr x = 0.9803 atm V= nRTP =(4.000molHCl)(0.08206LatmK−1mol−1)(303.15K)0.9803atm=102 L HCl 102 L HCl more http://cnx.org/contents/ 2488fW6W @2/Solubility 11.5: Colligative Properties Q11.5.1 Which is/are part of the macroscopic domain of solutions and which is/are part of the microscopic domain: boiling point elevation, Henry’s law, hydrogen bond, ion-dipole attraction, molarity, nonelectrolyte, nonstoichiometric compound, osmosis, solvated ion? Q11.5.2 What is the microscopic explanation for the macroscopic behavior illustrated in [link] ? S11.5.2 The strength of the bonds between like molecules is stronger than the strength between unlike molecules. Therefore, some regions will exist in which the water molecules will exclude oil molecules and other regions will exist in which oil molecules will exclude water molecules, forming a heterogeneous region. Q11.5.3 Sketch a qualitative graph of the pressure versus time for water vapor above a sample of pure water and a sugar solution, as the liquids evaporate to half their original volume. Q11.5.4 A solution of potassium nitrate, an electrolyte, and a solution of glycerin (C 3 H 5 (OH) 3 ), a nonelectrolyte, both boil at 100.3 °C. What other physical properties of the two solutions are identical? S11.5.4 Both form homogeneous solutions; their boiling point elevations are the same, as are their lowering of vapor pressures. Osmotic pressure and the lowering of the freezing point are also the same for both solutions. Q11.5.5 What are the mole fractions of H 3 PO 4 and water in a solution of 14.5 g of H 3 PO 4 in 125 g of water? Q11.5.6 What are the mole fractions of HNO 3 and water in a concentrated solution of nitric acid (68.0% HNO 3 by mass)? S11.5.6 Find number of moles of HNO 3 and H 2 O in 100 g of the solution. Find the mole fractions for the components. The mole fraction of HNO 3 is 0.378. The mole fraction of H 2 O is 0.622. Q11.5.7 Calculate the mole fraction of each solute and solvent: 583 g of H 2 SO 4 in 1.50 kg of water—the acid solution used in an automobile battery 0.86 g of NaCl in 1.00 × 10 2 g of water—a solution of sodium chloride for intravenous injection 46.85 g of codeine, C 18 H 21 NO 3 , in 125.5 g of ethanol, C 2 H 5 OH 25 g of I 2 in 125 g of ethanol, C 2 H 5 OH S11.5.7 a. \(583\:g\:\ce{H2SO4}\times\dfrac{1\:mole\:\ce{H2SO4}}{98.08\:g\:\ce{H2SO4}}=5.94\:mole\:\ce{H2SO4}\) \(\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\) \(1.50\:kg\:\ce{H2O}\times\dfrac{1000\:g}{1\:kg}\times\dfrac{1\:mole\:\ce{H2O}}{18.02\:g\:\ce{H2O}}=83.2\:moles\:\ce{H2O}\) Q11.5.8 Calculate the mole fraction of each solute and solvent: 0.710 kg of sodium carbonate (washing soda), Na 2 CO 3 , in 10.0 kg of water—a saturated solution at 0 °C 125 g of NH 4 NO 3 in 275 g of water—a mixture used to make an instant ice pack 25 g of Cl 2 in 125 g of dichloromethane, CH 2 Cl 2 0.372 g of histamine, C 5 H 9 N, in 125 g of chloroform, CHCl 3 S11.5.8 \(X_\mathrm{Na_2CO_3}=0.0119\); \(X_\mathrm{H_2O}=0.988\); \(X_\mathrm{NH_4NO_3}=0.9927\); \(X_\mathrm{H_2O}=0.907\); \(X_\mathrm{Cl_2}=0.192\); \(X_\mathrm{CH_2CI_2}=0.808\); \(X_\mathrm{C_5H_9N}=0.00426\); \(X_\mathrm{CHCl_3}=0.997\) Q11.5.9 Calculate the mole fractions of methanol, CH 3 OH; ethanol, C 2 H 5 OH; and water in a solution that is 40% methanol, 40% ethanol, and 20% water by mass. (Assume the data are good to two significant figures.) Q11.5.10 What is the difference between a 1 M solution and a 1 m solution? S11.5.10 In a 1 M solution, the mole is contained in exactly 1 L of solution. In a 1 m solution, the mole is contained in exactly 1 kg of solvent. Q11.5.11 What is the molality of phosphoric acid, H 3 PO 4 , in a solution of 14.5 g of H 3 PO 4 in 125 g of water? Q11.5.12 What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO 3 by mass)? S11.5.12 (a) Determine the molar mass of HNO 3 . Determine the number of moles of acid in the solution. From the number of moles and the mass of solvent, determine the molality. (b) 33.7 m Q11.5.13 Calculate the molality of each of the following solutions: 583 g of H 2 SO 4 in 1.50 kg of water—the acid solution used in an automobile battery 0.86 g of NaCl in 1.00 × 10 2 g of water—a solution of sodium chloride for intravenous injection 46.85 g of codeine, C 18 H 21 NO 3 , in 125.5 g of ethanol, C 2 H 5 OH 25 g of I 2 in 125 g of ethanol, C 2 H 5 OH Q11.5.14 Calculate the molality of each of the following solutions: 0.710 kg of sodium carbonate (washing soda), Na 2 CO 3 , in 10.0 kg of water—a saturated solution at 0°C 125 g of NH 4 NO 3 in 275 g of water—a mixture used to make an instant ice pack 25 g of Cl 2 in 125 g of dichloromethane, CH 2 Cl 2 0.372 g of histamine, C 5 H 9 N, in 125 g of chloroform, CHCl 3 S11.5.14 (a) 6.70 × 10 −1 m ; (b) 5.67 m ; (c) 2.8 m ; (d) 0.0358 m Q11.5.15 The concentration of glucose, C 6 H 12 O 6 , in normal spinal fluid is \(\mathrm{\dfrac{75\:mg}{100\:g}}\). What is the molality of the solution? Q11.5.16 A 13.0% solution of K 2 CO 3 by mass has a density of 1.09 g/cm 3 . Calculate the molality of the solution. S11.5.16 1.08 m Q11.5.17 Why does 1 mol of sodium chloride depress the freezing point of 1 kg of water almost twice as much as 1 mol of glycerin? What is the boiling point of a solution of 115.0 g of sucrose, C 12 H 22 O 11 , in 350.0 g of water? S11.5.17 Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the boiling point of water and the boiling point of the solution; determine the new boiling point. 100.5 °C Q11.5.18 What is the boiling point of a solution of 9.04 g of I 2 in 75.5 g of benzene, assuming the I 2 is nonvolatile? Q11.5.19 What is the freezing temperature of a solution of 115.0 g of sucrose, C 12 H 22 O 11 , in 350.0 g of water, which freezes at 0.0 °C when pure? S11.5.19 (a) Determine the molar mass of sucrose; determine the number of moles of sucrose in the solution; convert the mass of solvent to units of kilograms; from the number of moles and the mass of solvent, determine the molality; determine the difference between the freezing temperature of water and the freezing temperature of the solution; determine the new freezing temperature. (b) −1.8 °C Q11.5.20 What is the freezing point of a solution of 9.04 g of I 2 in 75.5 g of benzene? Q11.5.21 What is the osmotic pressure of an aqueous solution of 1.64 g of Ca(NO 3 ) 2 in water at 25 °C? The volume of the solution is 275 mL. S11.5.21 (a) Determine the molar mass of Ca(NO 3 ) 2 ; determine the number of moles of Ca(NO 3 ) 2 in the solution; determine the number of moles of ions in the solution; determine the molarity of ions, then the osmotic pressure. (b) 2.67 atm Q11.5.22 What is osmotic pressure of a solution of bovine insulin (molar mass, 5700 g mol −1 ) at 18 °C if 100.0 mL of the solution contains 0.103 g of the insulin? Q11.5.23 What is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp 76.8 °C; K b = 5.02 °C/ m ) that boils at 81.5 °C at 1 atm? S11.5.24 (a) Determine the molal concentration from the change in boiling point and K b ; determine the moles of solute in the solution from the molal concentration and mass of solvent; determine the molar mass from the number of moles and the mass of solute. (b) 2.1 × 10 2 g mol −1 Q11.5.25 A sample of an organic compound (a nonelectrolyte) weighing 1.35 g lowered the freezing point of 10.0 g of benzene by 3.66 °C. Calculate the molar mass of the compound. Q11.5.26 A 1.0 m solution of HCl in benzene has a freezing point of 0.4 °C. Is HCl an electrolyte in benzene? Explain. S11.5.26 No. Pure benzene freezes at 5.5 °C, and so the observed freezing point of this solution is depressed by Δ T f = 5.5 − 0.4 = 5.1 °C. The value computed, assuming no ionization of HCl, is Δ T f = (1.0 m)(5.14 °C/ m ) = 5.1 °C. Agreement of these values supports the assumption that HCl is not ionized. Q11.5.27 A solution contains 5.00 g of urea, CO(NH 2 ) 2 , a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution? Q11.5.28 A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at −1.94 °C. Calculate the molar mass of the substance. S11.5.28 144 g mol −1 Q11.5.29 Arrange the following solutions in order by their decreasing freezing points: 0.1 m Na 3 PO 4 , 0.1 m C 2 H 5 OH, 0.01 m CO 2 , 0.15 m NaCl, and 0.2 m CaCl 2 . Q11.5.30 Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of Na 2 SO 4 , and 0.030 mol of MgCl 2 , assuming complete dissociation of these electrolytes. S11.5.30 0.870 °C Q11.5.31 How could you prepare a 3.08 m aqueous solution of glycerin, C 3 H 8 O 3 ? What is the freezing point of this solution? Q11.5.32 A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, CS 2 ( K b = 2.43 °C/ m ). If the boiling point elevation was 0.107 °C, what is the formula of a sulfur molecule in carbon disulfide? S11.5.32 S 8 Q11.5.33 In a significant experiment performed many years ago, 5.6977 g of cadmium iodide in 44.69 g of water raised the boiling point 0.181 °C. What does this suggest about the nature of a solution of CdI 2 ? Q11.5.34 Lysozyme is an enzyme that cleaves cell walls. A 0.100-L sample of a solution of lysozyme that contains 0.0750 g of the enzyme exhibits an osmotic pressure of 1.32 × 10 −3 atm at 25 °C. What is the molar mass of lysozyme? S11.5.34 1.39 × 10 4 g mol −1 Q11.5.35 The osmotic pressure of a solution containing 7.0 g of insulin per liter is 23 torr at 25 °C. What is the molar mass of insulin? Q11.5.36 The osmotic pressure of human blood is 7.6 atm at 37 °C. What mass of glucose, C 6 H 12 O 6 , is required to make 1.00 L of aqueous solution for intravenous feeding if the solution must have the same osmotic pressure as blood at body temperature, 37 °C? S11.5.36 54 g Q11.5.37 What is the freezing point of a solution of dibromobenzene, C 6 H 4 Br 2 , in 0.250 kg of benzene, if the solution boils at 83.5 °C? Q11.5.38 What is the boiling point of a solution of NaCl in water if the solution freezes at −0.93 °C? S11.5.38 100.26 °C Q11.5.39 The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C. The boiling point of ethanol is 78.35 °C, and K b for ethanol is 1.20 °C/ m . What is the molecular formula of fructose? Q11.5.40 The vapor pressure of methanol, CH 3 OH, is 94 torr at 20 °C. The vapor pressure of ethanol, C 2 H 5 OH, is 44 torr at the same temperature. Calculate the mole fraction of methanol and of ethanol in a solution of 50.0 g of methanol and 50.0 g of ethanol. Ethanol and methanol form a solution that behaves like an ideal solution. Calculate the vapor pressure of methanol and of ethanol above the solution at 20 °C. Calculate the mole fraction of methanol and of ethanol in the vapor above the solution. S11.5.40 (a) \(X_\mathrm{CH_3OH}=0.590\); \(X_\mathrm{C_2H_5OH}=0.410\); (b) Vapor pressures are: CH 3 OH: 55 torr; C 2 H 5 OH: 18 torr; (c) CH 3 OH: 0.75; C 2 H 5 OH: 0.25 Q11.5.41 The triple point of air-free water is defined as 273.15 K. Why is it important that the water be free of air? Q11.5.42 Meat can be classified as fresh (not frozen) even though it is stored at −1 °C. Why wouldn’t meat freeze at this temperature? S11.5.42 The ions and compounds present in the water in the beef lower the freezing point of the beef below −1 °C. Q11.5.43 An organic compound has a composition of 93.46% C and 6.54% H by mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at 158.4 °C. The melting point of pure camphor is 178.4 °C. K f for camphor is 37.7 °C/ m . What is the molecular formula of the solute? Show your calculations. Q11.5.44 A sample of HgCl 2 weighing 9.41 g is dissolved in 32.75 g of ethanol, C 2 H 5 OH ( K b = 1.20 °C/ m ). The boiling point elevation of the solution is 1.27 °C. Is HgCl 2 an electrolyte in ethanol? Show your calculations. S11.5.44 \(\mathrm{Δbp}=K_\ce{b}m=(1.20\:°\ce C/m)\mathrm{\left(\dfrac{9.41\:g×\dfrac{1\:mol\: HgCl_2}{271.496\:g}}{0.03275\:kg}\right)=1.27\:°\ce C}\) The observed change equals the theoretical change; therefore, no dissociation occurs. Q11.5.45 A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about −1.4 °C. What is the formula of the salt? Show your calculations. 11.6: Colloids Q11.6.1 Identify the dispersed phase and the dispersion medium in each of the following colloidal systems: starch dispersion, smoke, fog, pearl, whipped cream, floating soap, jelly, milk, and ruby. S11.6.1 Colloidal System Dispersed Phase Dispersion Medium starch dispersion starch water smoke solid particles air fog water air pearl water calcium carbonate (CaCO3) whipped cream air cream floating soap air soap jelly fruit juice pectin gel milk butterfat water ruby chromium(III) oxide (Cr2O3) aluminum oxide (Al2O3) Q11.6.2 Distinguish between dispersion methods and condensation methods for preparing colloidal systems. S11.6.2 Dispersion methods use a grinding device or some other means to bring about the subdivision of larger particles. Condensation methods bring smaller units together to form a larger unit. For example, water molecules in the vapor state come together to form very small droplets that we see as clouds. Q11.6.3 How do colloids differ from solutions with regard to dispersed particle size and homogeneity? S11.6.3 Colloidal dispersions consist of particles that are much bigger than the solutes of typical solutions. Colloidal particles are either very large molecules or aggregates of smaller species that usually are big enough to scatter light. Colloids are homogeneous on a macroscopic (visual) scale, while solutions are homogeneous on a microscopic (molecular) scale. Q11.6.4 Explain the cleansing action of soap. S11.6.4 Soap molecules have both a hydrophobic and a hydrophilic end. The charged (hydrophilic) end, which is usually associated with an alkali metal ion, ensures water solubility The hydrophobic end permits attraction to oil, grease, and other similar nonpolar substances that normally do not dissolve in water but are pulled into the solution by the soap molecules. Q11.6.5 How can it be demonstrated that colloidal particles are electrically charged? S11.6.5 If they are placed in an electrolytic cell, dispersed particles will move toward the electrode that carries a charge opposite to their own charge. At this electrode, the charged particles will be neutralized and will coagulate as a precipitate. |
Courses/Fullerton_College/Beginning_Chemistry_(Chan)/03%3A_Atoms/3.03%3A_Masses_of_Atoms_and_Molecules | Learning Objective Express the masses of atoms and molecules. Because matter is defined as anything that has mass and takes up space, it should not be surprising to learn that atoms and molecules have mass. Individual atoms and molecules, however, are very small, and the masses of individual atoms and molecules are also very small. For macroscopic objects, we use units such as grams and kilograms to state their masses, but these units are much too big to comfortably describe the masses of individual atoms and molecules. Another scale is needed. Atomic Mass Unit The atomic mass unit (u; some texts use amu, but this older style is no longer accepted) is defined as one-twelfth of the mass of a carbon-12 atom, an isotope of carbon that has six protons and six neutrons in its nucleus. By this scale, the mass of a proton is 1.00728 u, the mass of a neutron is 1.00866 u, and the mass of an electron is 0.000549 u. There will not be much error if you estimate the mass of an atom by simply counting the total number of protons and neutrons in the nucleus (i.e., identify its mass number) and ignore the electrons. Thus, the mass of carbon-12 is about 12 u, the mass of oxygen-16 is about 16 u, and the mass of uranium-238 is about 238 u. More exact masses are found in scientific references—for example, the exact mass of uranium-238 is 238.050788 u, so you can see that we are not far off by using the whole-number value as the mass of the atom. What is the mass of an element? This is somewhat more complicated because most elements exist as a mixture of isotopes, each of which has its own mass. Thus, although it is easy to speak of the mass of an atom, when talking about the mass of an element, we must take the isotopic mixture into account. Atomic Mass The atomic mass of an element is a weighted average of the masses of the isotopes that compose an element. What do we mean by a weighted average? Well, consider an element that consists of two isotopes, 50% with mass 10 u and 50% with mass 11 u. A weighted average is found by multiplying each mass by its fractional occurrence (in decimal form) and then adding all the products. The sum is the weighted average and serves as the formal atomic mass of the element. In this example, we have the following: 0 1 0.50 × 10 u = 5.0 u 0.50 × 11 u = 5.5 u Sum = 10.5 u = the atomic mass of our element Note that no atom in our hypothetical element has a mass of 10.5 u; rather, that is the average mass of the atoms, weighted by their percent occurrence. This example is similar to a real element. Boron exists as about 20% boron-10 (five protons and five neutrons in the nuclei) and about 80% boron-11 (five protons and six neutrons in the nuclei). The atomic mass of boron is calculated similarly to what we did for our hypothetical example, but the percentages are different: 0 1 0.20 × 10 u = 2.0 u 0.80 × 11 u = 8.8 u Sum = 10.8 u = the atomic mass of boron Thus, we use 10.8 u for the atomic mass of boron. Virtually all elements exist as mixtures of isotopes, so atomic masses may vary significantly from whole numbers. Table \(\PageIndex{1}\) lists the atomic masses of some elements. The atomic masses in Table \(\PageIndex{1}\) are listed to three decimal places where possible, but in most cases, only one or two decimal places are needed. Note that many of the atomic masses, especially the larger ones, are not very close to whole numbers. This is, in part, the effect of an increasing number of isotopes as the atoms increase in size. (The record number is 10 isotopes for tin.) Element Name Atomic Mass (u) Element Name.1 Atomic Mass (u).1 Aluminum 26.981 Molybdenum 95.94 Argon 39.948 Neon 20.180 Arsenic 74.922 Nickel 58.693 Barium 137.327 Nitrogen 14.007 Beryllium 9.012 Oxygen 15.999 Bismuth 208.980 Palladium 106.42 Boron 10.811 Phosphorus 30.974 Bromine 79.904 Platinum 195.084 Calcium 40.078 Potassium 39.098 Carbon 12.011 Radium NaN Chlorine 35.453 Radon NaN Cobalt 58.933 Rubidium 85.468 Copper 63.546 Scandium 44.956 Fluorine 18.998 Selenium 78.96 Gallium 69.723 Silicon 28.086 Germanium 72.64 Silver 107.868 Gold 196.967 Sodium 22.990 Helium 4.003 Strontium 87.62 Hydrogen 1.008 Sulfur 32.065 Iodine 126.904 Tantalum 180.948 Iridium 192.217 Tin 118.710 Iron 55.845 Titanium 47.867 Krypton 83.798 Tungsten 183.84 Lead 207.2 Uranium 238.029 Lithium 6.941 Xenon 131.293 Magnesium 24.305 Zinc 65.409 Manganese 54.938 Zirconium 91.224 Mercury 200.59 Molybdenum 95.94 Note: Atomic mass is given to three decimal places, if known. Note: Atomic mass is given to three decimal places, if known. Note: Atomic mass is given to three decimal places, if known. Note: Atomic mass is given to three decimal places, if known. Molecular Mass Now that we understand that atoms have mass, it is easy to extend the concept to the mass of molecules. The molecular mass is the sum of the masses of the atoms in a molecule. This may seem like a trivial extension of the concept, but it is important to count the number of each type of atom in the molecular formula. Also, although each atom in a molecule is a particular isotope, we use the weighted average, or atomic mass, for each atom in the molecule. For example, if we were to determine the molecular mass of dinitrogen trioxide, N 2 O 3 , we would need to add the atomic mass of nitrogen two times with the atomic mass of oxygen three times: 0 1 2 N masses = 2 × 14.007 u = 28.014 u 3 O masses = 3 × 15.999 u = 47.997 u Total = 76.011 u = the molecular mass of N2O3 We would not be far off if we limited our numbers to one or even two decimal places. Example \(\PageIndex{1}\) What is the molecular mass of each substance? \(\ce{NBr3}\) \(\ce{C2H6}\) Solution Add one atomic mass of nitrogen and three atomic masses of bromine: 0 1 1 N mass = 14.007 u 3 Br masses = 3 × 79.904 u = 239.712 u Total = 253.719 u = the molecular mass of NBr3 Add two atomic masses of carbon and six atomic masses of hydrogen: 0 1 2 C masses = 2 × 12.011 u = 24.022 u 6 H masses = 6 × 1.008 u = 6.048 u Total = 30.070 u = the molecular mass of C2H6 The compound \(\ce{C2H6}\) also has a common name—ethane. Exercise \(\PageIndex{1}\) What is the molecular mass of each substance? \(\ce{SO2}\) \(\ce(PF3}\) Answer a 64.063 u Answer b 87.968 u Chemistry is Everywhere: Sulfur Hexafluoride On March 20, 1995, the Japanese terrorist group Aum Shinrikyo (Sanskrit for "Supreme Truth") released some sarin gas in the Tokyo subway system; twelve people were killed, and thousands were injured (Figure \(\PageIndex{2a}\)). Sarin (molecular formula \(\ce{C4H10FPO2}\)) is a nerve toxin that was first synthesized in 1938. It is regarded as one of the most deadly toxins known, estimated to be about 500 times more potent than cyanide. Scientists and engineers who study the spread of chemical weapons such as sarin (yes, there are such scientists) would like to have a similar less dangerous chemical to study, indeed one that is nontoxic, so they are not at risk themselves. Sulfur hexafluoride is used as a model compound for sarin. \(\ce{SF6}\) (a molecular model of which is shown Figure \(\PageIndex{2b}\)) has a similar molecular mass (about 146 u) as sarin (about 140 u), so it has similar physical properties in the vapor phase. Sulfur hexafluoride is also very easy to accurately detect, even at low levels, and it is not a normal part of the atmosphere, so there is little potential for contamination from natural sources. Consequently, \(\ce{SF6}\) is also used as an aerial tracer for ventilation systems in buildings. It is nontoxic and very chemically inert, so workers do not have to take special precautions other than watching for asphyxiation. Sulfur hexafluoride also has another interesting use: a spark suppressant in high-voltage electrical equipment. High-pressure \(\ce{SF6}\) gas is used in place of older oils that may have contaminants that are not environmentally friendly (Figure \(\PageIndex{2c}\)). Key Takeaways The atomic mass unit (u) is a unit that describes the masses of individual atoms and molecules. The atomic mass is the weighted average of the masses of all isotopes of an element. The molecular mass is the sum of the masses of the atoms in a molecule. |
Courses/Oregon_Institute_of_Technology/OIT%3A_CHE_332_--_Organic_Chemistry_II_(Lund)/1%3A_Fall_term_review_sections/4%3A_Structure_Determination_I-_UV-Vis_and_Infrared_Spectroscopy_Mass_Spectrometry/4.4%3A_Ultraviolet_and_visible_spectroscopy | While interaction with infrared light causes molecules to undergo vibrational transitions, the shorter wavelength, higher energy radiation in the UV (200-400 nm) and visible (400-700 nm) range of the electromagnetic spectrum causes many organic molecules to undergo electronic transitions . What this means is that when the energy from UV or visible light is absorbed by a molecule, one of its electrons jumps from a lower energy to a higher energy molecular orbital. Electronic transitions Let’s take as our first example the simple case of molecular hydrogen, H 2 . As you may recall from section 2.1A, the molecular orbital picture for the hydrogen molecule consists of one bonding σ MO, and a higher energy antibonding σ * MO. When the molecule is in the ground state, both electrons are paired in the lower-energy bonding orbital – this is the Highest Occupied Molecular Orbital (HOMO). The antibonding σ * orbital, in turn, is the Lowest Unoccupied Molecular Orbital (LUMO). If the molecule is exposed to light of a wavelength with energy equal to Δ E, the HOMO-LUMO energy gap, this wavelength will be absorbed and the energy used to bump one of the electrons from the HOMO to the LUMO – in other words, from the σ to the σ * orbital. This is referred to as a σ - σ * transition . Δ E for this electronic transition is 258 kcal/mol, corresponding to light with a wavelength of 111 nm. When a double-bonded molecule such as ethene (common name ethylene) absorbs light, it undergoes a π - π * transition. Because π - π * energy gaps are narrower than σ - σ * gaps, ethene absorbs light at 165 nm - a longer wavelength than molecular hydrogen. The electronic transitions of both molecular hydrogen and ethene are too energetic to be accurately recorded by standard UV spectrophotometers, which generally have a range of 220 – 700 nm. Where UV-vis spectroscopy becomes useful to most organic and biological chemists is in the study of molecules with conjugated \(\pi\) systems. In these groups, the energy gap for π - π * transitions is smaller than for isolated double bonds, and thus the wavelength absorbed is longer. Molecules or parts of molecules that absorb light strongly in the UV-vis region are called chromophores . Let’s revisit the MO picture for 1,3-butadiene, the simplest conjugated system. Recall that we can draw a diagram showing the four pi MO’s that result from combining the four 2p z atomic orbitals. The lower two orbitals are bonding, while the upper two are antibonding. Comparing this MO picture to that of ethene, our isolated pi-bond example, we see that the HOMO-LUMO energy gap is indeed smaller for the conjugated system. 1,3-butadiene absorbs UV light with a wavelength of 217 nm. As conjugated pi systems become larger, the energy gap for a π - π * transition becomes increasingly narrow, and the wavelength of light absorbed correspondingly becomes longer. The absorbance due to the π - π * transition in 1,3,5-hexatriene, for example, occurs at 258 nm, corresponding to a Δ E of 111 kcal/mol. In molecules with extended pi systems, the HOMO-LUMO energy gap becomes so small that absorption occurs in the visible rather then the UV region of the electromagnetic spectrum. Beta-carotene, with its system of 11 conjugated double bonds, absorbs light with wavelengths in the blue region of the visible spectrum while allowing other visible wavelengths – mainly those in the red-yellow region - to be transmitted. This is why carrots are orange. The conjugated pi system in 4-methyl-3-penten-2-one gives rise to a strong UV absorbance at 236 nm due to a π - π * transition. However, this molecule also absorbs at 314 nm. This second absorbance is due to the transition of a non-bonding (lone pair) electron on the oxygen up to a π * antibonding MO: This is referred to as an n - π * transition . The nonbonding (n) MO’s are higher in energy than the highest bonding p orbitals, so the energy gap for an \(n \rightarrow \pi^*\) transition is smaller that that of a π - π * transition – and thus the n - π * peak is at a longer wavelength. In general, n - π * transitions are weaker (less light absorbed) than those due to π - π * transitions. Exercise 4.9 What is the energy of the photons (in kJ/mol) of light with wavelength of 470 nm, the l max of b -carotene? Exercise 4.10 Which of the following molecules would you expect absorb at a longer wavelength in the UV region of the electromagnetic spectrum? Explain your answer. Solutions Protecting yourself from sunburn Human skin can be damaged by exposure to ultraviolet light from the sun. We naturally produce a pigment, called melanin, which protects the skin by absorbing much of the ultraviolet radiation. Melanin is a complex polymer, two of the most common monomers units of which are shown below. Overexposure to the sun is still dangerous, because there is a limit to how much radiation our melanin can absorb. Most commercial sunscreens claim to offer additional protection from both UV-A and UV-B radiation: UV-A refers to wavelengths between 315-400 nm, UV-B to shorter, more harmful wavelengths between 280-315 nm. PABA ( para -aminobenzoic acid) was used in sunscreens in the past, but its relatively high polarity meant that it was not very soluble in oily lotions, and it tended to rinse away when swimming. Many sunscreens today contain, among other active ingredients, a more hydrophobic derivative of PABA called Padimate O. Looking at UV-vis spectra We have been talking in general terms about how molecules absorb UV and visible light – now let's look at some actual examples of data from a UV-vis absorbance spectrophotometer. The basic setup is the same as for IR spectroscopy: radiation with a range of wavelengths is directed through a sample of interest, and a detector records which wavelengths were absorbed and to what extent the absorption occurred. Schematic for a UV-Vis spectrophotometer (Image from Wikipedia Commons ) Below is the absorbance spectrum of an important biological molecule called nicotinamide adenine dinucleotide, abbreviated NAD + . This compound absorbs light in the UV range due to the presence of conjugated pi-bonding systems. You’ll notice that this UV spectrum is much simpler than the IR spectra we saw earlier: this one has only one peak, although many molecules have more than one. Notice also that the convention in UV-vis spectroscopy is to show the baseline at the bottom of the graph with the peaks pointing up. Wavelength values on the x-axis are generally measured in nanometers (nm) rather than in cm -1 as is the convention in IR spectroscopy. Peaks in UV spectra tend to be quite broad, often spanning well over 20 nm at half-maximal height. Typically, there are two things that we look for and record from a UV-Vis spectrum. The first is \(\lambda_{max}\), which is the wavelength at maximal light absorbance. As you can see, NAD + has \(\lambda_{max} = 260\;nm\). We also want to record how much light is absorbed at \(\lambda_{max}\). Here we use a unitless number called absorbance , abbreviated 'A'. This contains the same information as the 'percent transmittance' number used in IR spectroscopy, just expressed in slightly different terms. To calculate absorbance at a given wavelength, the computer in the spectrophotometer simply takes the intensity of light at that wavelength before it passes through the sample (I 0 ), divides this value by the intensity of the same wavelength after it passes through the sample (I), then takes the log 10 of that number: \[A = \log \dfrac{I_0}{I} \tag{4.3.1}\] You can see that the absorbance value at 260 nm (A 260 ) is about 1.0 in this spectrum. Exercise 4.11 Express A = 1.0 in terms of percent transmittance (%T, the unit usually used in IR spectroscopy (and sometimes in UV-vis as well). Solutions Here is the absorbance spectrum of the common food coloring Red #3: Here, we see that the extended system of conjugated pi bonds causes the molecule to absorb light in the visible range. Because the λ max of 524 nm falls within the green region of the spectrum, the compound appears red to our eyes. Now, take a look at the spectrum of another food coloring, Blue #1: Here, maximum absorbance is at 630 nm, in the orange range of the visible spectrum, and the compound appears blue. Applications of UV spectroscopy in organic and biological chemistry UV-vis spectroscopy has many different applications in organic and biological chemistry. One of the most basic of these applications is the use of the Beer - Lambert Law to determine the concentration of a chromophore. You most likely have performed a Beer – Lambert experiment in a previous chemistry lab. The law is simply an application of the observation that, within certain ranges, the absorbance of a chromophore at a given wavelength varies in a linear fashion with its concentration: the higher the concentration of the molecule, the greater its absorbance. If we divide the observed value of A at λ max by the concentration of the sample ( c , in mol/L), we obtain the molar absorptivity , or extinction coefficient ( ε ), which is a characteristic value for a given compound. \[\epsilon = \dfrac{A}{c} \tag{4.3.2}\] The absorbance will also depend, of course, on the path length - in other words, the distance that the beam of light travels though the sample. In most cases, sample holders are designed so that the path length is equal to 1 cm, so the units for molar absorptivity are L * mol -1 * cm -1 . If we look up the value of e for our compound at λ max , and we measure absorbance at this wavelength, we can easily calculate the concentration of our sample. As an example, for NAD + the literature value of ε at 260 nm is 18,000 L * mol -1 * cm -1 . In our NAD + spectrum we observed A 260 = 1.0, so using equation 4.4 and solving for concentration we find that our sample is 5.6 x 10 -5 M. The bases of DNA and RNA are good chromophores: Biochemists and molecular biologists often determine the concentration of a DNA sample by assuming an average value of ε = 0.020 ng -1 ×mL for double-stranded DNA at its λ max of 260 nm (notice that concentration in this application is expressed in mass/volume rather than molarity: ng/mL is often a convenient unit for DNA concentration when doing molecular biology). Exercise 4.12 50 micro liters of an aqueous sample of double stranded DNA is dissolved in 950 microliters of water. This diluted solution has a maximal absorbance of 0.326 at 260 nm. What is the concentration of the original (more concentrated) DNA sample, expressed in micrograms per microliter? Solutions Because the extinction coefficient of double stranded DNA is slightly lower than that of single stranded DNA, we can use UV spectroscopy to monitor a process known as DNA melting. If a short stretch of double stranded DNA is gradually heated up, it will begin to ‘melt’, or break apart, as the temperature increases (recall that two strands of DNA are held together by a specific pattern of hydrogen bonds formed by ‘base-pairing’). As melting proceeds, the absorbance value for the sample increases, eventually reaching a high plateau as all of the double-stranded DNA breaks apart, or ‘melts’. The mid-point of this process, called the ‘melting temperature’, provides a good indication of how tightly the two strands of DNA are able to bind to each other. Later we will see how the Beer - Lambert Law and UV spectroscopy provides us with a convenient way to follow the progress of many different enzymatic redox (oxidation-reduction) reactions. In biochemistry, oxidation of an organic molecule often occurs concurrently with reduction of nicotinamide adenine dinucleotide (NAD + , the compound whose spectrum we saw earlier in this section) to NADH: Both NAD + and NADH absorb at 260 nm. However NADH, unlike NAD + , has a second absorbance band with λ max = 340 nm and ε = 6290 L * mol -1 * cm -1 . The figure below shows the spectra of both compounds superimposed, with the NADH spectrum offset slightly on the y-axis: By monitoring the absorbance of a reaction mixture at 340 nm, we can 'watch' NADH being formed as the reaction proceeds, and calculate the rate of the reaction. UV spectroscopy is also very useful in the study of proteins. Proteins absorb light in the UV range due to the presence of the aromatic amino acids tryptophan, phenylalanine, and tyrosine, all of which are chromophores. Biochemists frequently use UV spectroscopy to study conformational changes in proteins - how they change shape in response to different conditions. When a protein undergoes a conformational shift (partial unfolding, for example), the resulting change in the environment around an aromatic amino acid chromophore can cause its UV spectrum to be altered. Khan Academy video tutorials on UV-Vis spectroscopy |
Courses/Lumen_Learning/Book%3A_US_History_I_(AY_Collection)_(Lumen)/13%3A_The_Old_South/13.4%3A_Video%3A_Slavery | In this video, John Green teaches you about America’s “peculiar institution” of slavery. I wouldn’t really call it peculiar. I’d lean more toward horrifying and depressing institution, but nobody asked me. John will talk about what life was like for a slave in the 19th century United States, and how slaves resisted oppression, to the degree that was possible. We’ll hear about cotton plantations, violent punishment of slaves, day to day slave life, and slave rebellions. Nat Turner, Harriet Tubman, and Whipped Peter all make an appearance. Slavery as an institution is arguably the darkest part of America’s history, and we’re still dealing with its aftermath 150 years after it ended. A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/ay/?p=288 All rights reserved content Slavery - Crash Course US History #13. Authored by : CrashCourse. Located at : https://youtu.be/Ajn9g5Gsv98?t=1s . License : All Rights Reserved . License Terms : Standard YouTube License |
Bookshelves/Organic_Chemistry/Organic_Chemistry_(Morsch_et_al.)/18%3A_Ethers_and_Epoxides_Thiols_and_Sulfides/18.05%3A_Cyclic_Ethers_-_Epoxides | Objectives After completing this section, you should be able to write an equation to describe the industrial preparation of ethylene oxide. list two important industrial uses of ethylene oxide. write an equation to describe the normal laboratory preparation of an epoxide. identify the epoxide produced from the reaction of a given alkene with a peroxyacid. identify the alkene, the reagent, or both, needed to prepare a given epoxide. write an equation to describe the preparation of an epoxide from an alkene via a halohydrin. Key Terms Make certain that you can define, and use in context, the key term below. epoxide (oxirane) Epoxides (also known as oxiranes ) are three-membered ring structures in which one of the vertices is an oxygen and the other two are carbons. Epoxides behave differently than other ethers due to the strain created by the three-membered ring. The most important and simplest epoxide is ethylene oxide which is prepared on an industrial scale by catalytic oxidation of ethylene by air. Ethylene oxide is used as an important chemical feedstock in the manufacturing of ethylene glycol, which is used as antifreeze, liquid coolant and solvent. In turn, ethylene glycol is used in the production of polyester and polyethylene terephthalate (PET) the raw material for plastic bottles. Nomenclature of Epoxides The name ethylene oxide is not systematic but common. Epoxides are systematically named as an oxirane ring system much like the other cyclic ethers discussed in Section 18.1 . The oxygen is numbered in the 1 position. Example \(\PageIndex{1}\) Synthesis of Epoxides As discussed in Section 8-7 , epoxides are typically prepared by the reaction of an alkene and a peroxycarboxylic acid (RCO 3 H), such as meta -chloroperbenzoic acid (MCPBA). The peroxycarboxylic acid has the unique property of having an electropositive oxygen atom on the CO 3 H group. The reaction is initiated by the electrophilic oxygen atom reacting with the nucleophilic carbon-carbon double bond. The mechanism involves a concerted reaction with a four-part, circular transition state. The result is that the originally electropositive oxygen atom ends up in the oxacyclopropane ring and the CO 3 H group becomes CO 2 H. General Reaction Stereochemical Consideration The oxygen addition to the alkene is a syn addition and is stereospecific. The substituents of a cis -alkene will appear in the cis configuration on the epoxide ring. Likewise, Substituents of a trans -alkene will appear in the trans configuration on the epoxide ring. During this reaction the sp 2 hybridized carbons of the alkene are converted to sp 3 hybridized carbons in the epoxide. Thus, there is the possibility of two new chiral carbons forming in the epoxide product. In most cases, epoxidation of an alkene will product a mixture of enantiomers in the product, due to the electrophilic oxygen being able to attack from above or below the plane of the alkene. One major exception is the epoxidation of symmetrical cis-alkene which produces a meso compound product. Example \(\PageIndex{2}\) Treatment of an alkene with X 2 & H 2 O creates a halohydrin ( Section 8-3 ). When a halohydrin is treated with a base the alcohol is deprotonated to from an alkoxide. This causes an intramolecular Williamson ether synthesis to produce an epoxide along with the elimination of HX. Note, that this reaction generally forms a mixture of enantiomeric products unless a meso compound is produced. Exercise \(\PageIndex{1}\) 1) What reagents would you use to perform the following transformation? 2) Predict the products of the following reactions: 3) Predict the product of the following reaction. Be sure to include proper stereochemistry. Answer 1) Lindlar's catalyst reduces alkynes to cis/Z alkenes. This stereochemistry is retained after epoxidation. 2) 3) |
Courses/Providence_College/CHM_331_Advanced_Analytical_Chemistry_1/06%3A_General_Properties_of_Electromagnetic_Radiation/6.01%3A_Overview_of_Spectroscopy | The focus of this chapter is on the interaction of ultraviolet, visible, and infrared radiation with matter. Because these techniques use optical materials to disperse and focus the radiation, they often are identified as optical spectroscopies. For convenience we will use the simpler term spectroscopy in place of optical spectroscopy ; however, you should understand we will consider only a limited piece of what is a much broader area of analytical techniques. Despite the difference in instrumentation, all spectroscopic techniques share several common features. Before we consider individual examples in greater detail, let’s take a moment to consider some of these similarities. As you work through the chapter, this overview will help you focus on the similarities between different spectroscopic methods of analysis. You will find it easier to understand a new analytical method when you can see its relationship to other similar methods. What is Electromagnetic Radiation? Electromagnetic radiation —light—is a form of energy whose behavior is described by the properties of both waves and particles. Some properties of electromagnetic radiation, such as its refraction when it passes from one medium to another (Figure \(\PageIndex{1}\)), are explained best when we describe light as a wave. Other properties, such as absorption and emission, are better described by treating light as a particle. The exact nature of electromagnetic radiation remains unclear, as it has since the development of quantum mechanics in the first quarter of the 20th century [Home, D.; Gribbin, J. New Scientist 1991, 2 Nov. 30–33]. Nevertheless, this dual model of wave and particle behavior provide a useful description for electromagnetic radiation. Wave Properties of Electromagnetic Radiation Electromagnetic radiation consists of oscillating electric and magnetic fields that propagate through space along a linear path and with a constant velocity. In a vacuum, electromagnetic radiation travels at the speed of light, c , which is \(2.99792 \times 10^8\) m/s. When electromagnetic radiation moves through a medium other than a vacuum, its velocity, v , is less than the speed of light in a vacuum. The difference between v and c is sufficiently small (<0.1%) that the speed of light to three significant figures, \(3.00 \times 10^8\) m/s, is accurate enough for most purposes. The oscillations in the electric field and the magnetic field are perpendicular to each other and to the direction of the wave’s propagation. Figure \(\PageIndex{2}\) shows an example of plane-polarized electromagnetic radiation, which consists of a single oscillating electric field and a single oscillating magnetic field. An electromagnetic wave is characterized by several fundamental properties, including its velocity, amplitude, frequency, phase angle, polarization, and direction of propagation [Ball, D. W. Spectroscopy 1994 , 9(5) , 24–25]. For example, the amplitude of the oscillating electric field at any point along the propagating wave is \[A_{t}=A_{e} \sin (2 \pi \nu t+\Phi) \nonumber\] where A t is the magnitude of the electric field at time t , A e is the electric field’s maximum amplitude , \(\nu\) is the wave’s frequency —the number of oscillations in the electric field per unit time—and \(\Phi\) is a phase angle that accounts for the fact that A t need not have a value of zero at t = 0. The identical equation for the magnetic field is \[A_{t}=A_{m} \sin (2 \pi \nu t+\Phi) \nonumber\] where A m is the magnetic field’s maximum amplitude. Other properties also are useful for characterizing the wave behavior of electromagnetic radiation. The wavelength , \(\lambda\), is defined as the distance between successive maxima (see Figure \(\PageIndex{2}\)). For ultraviolet and visible electromagnetic radiation the wavelength usually is expressed in nanometers (1 nm = 10 –9 m), and for infrared radiation it is expressed in microns (1 mm = 10 –6 m). The relationship between wavelength and frequency is \[\lambda = \frac {c} {\nu} \nonumber\] Another unit useful unit is the wavenumber , \(\overline{\nu}\), which is the reciprocal of wavelength \[\overline{\nu} = \frac {1} {\lambda} \nonumber\] Wavenumbers frequently are used to characterize infrared radiation, with the units given in cm –1 . When electromagnetic radiation moves between different media—for example, when it moves from air into water—its frequency, \(\nu\), remains constant. Because its velocity depends upon the medium in which it is traveling, the electromagnetic radiation’s wavelength, \(\lambda\), changes. If we replace the speed of light in a vacuum, c , with its speed in the medium, \(v\), then the wavelength is \[\lambda = \frac {v} {\nu} \nonumber\] This change in wavelength as light passes between two media explains the refraction of electromagnetic radiation shown in Figure \(\PageIndex{1}\). Example \(\PageIndex{1}\) In 1817, Josef Fraunhofer studied the spectrum of solar radiation, observing a continuous spectrum with numerous dark lines. Fraunhofer labeled the most prominent of the dark lines with letters. In 1859, Gustav Kirchhoff showed that the D line in the sun’s spectrum was due to the absorption of solar radiation by sodium atoms. The wavelength of the sodium D line is 589 nm. What are the frequency and the wavenumber for this line? Solution The frequency and wavenumber of the sodium D line are \[\nu=\frac{c}{\lambda}=\frac{3.00 \times 10^{8} \ \mathrm{m} / \mathrm{s}}{589 \times 10^{-9} \ \mathrm{m}}=5.09 \times 10^{14} \ \mathrm{s}^{-1} \nonumber\] \[\overline{\nu}=\frac{1}{\lambda}=\frac{1}{589 \times 10^{-9} \ \mathrm{m}} \times \frac{1 \ \mathrm{m}}{100 \ \mathrm{cm}}=1.70 \times 10^{4} \ \mathrm{cm}^{-1} \nonumber\] Exercise \(\PageIndex{1}\) Another historically important series of spectral lines is the Balmer series of emission lines from hydrogen. One of its lines has a wavelength of 656.3 nm. What are the frequency and the wavenumber for this line? Answer The frequency and wavenumber for the line are \[\nu=\frac{c}{\lambda}=\frac{3.00 \times 10^{8} \ \mathrm{m} / \mathrm{s}}{656.3 \times 10^{-9} \ \mathrm{m}}=4.57 \times 10^{14} \ \mathrm{s}^{-1} \nonumber\] \[\overline{\nu}=\frac{1}{\lambda}=\frac{1}{656.3 \times 10^{-9} \ \mathrm{m}} \times \frac{1 \ \mathrm{m}}{100 \ \mathrm{cm}}=1.524 \times 10^{4} \ \mathrm{cm}^{-1} \nonumber\] The Electromagnetic Spectrum The frequency and the wavelength of electromagnetic radiation vary over many orders of magnitude. For convenience, we divide electromagnetic radiation into different regions—the electromagnetic spectrum —based on the type of atomic or molecular transitions that gives rise to the absorption or emission of photons (Figure \(\PageIndex{3}\)). The boundaries between the regions of the electromagnetic spectrum are not rigid and overlap between spectral regions is possible. |
Courses/San_Francisco_State_University/General_Physical_Chemistry_I_(Gerber)/11%3A_Chemical_Kinetics_I_-_Rate_Laws/11.04%3A_Different_Rate_Laws_Predict_Different_Kinetics | Difference rate laws predict different kinetics Zero-order Kinetics If the reaction follows a zeroth order rate law, it can be expressed in terms of the time-rate of change of [A] (which will be negative since A is a reactant): \[-\dfrac{d[A]}{dt} = k \nonumber \] In this case, it is straightforward to separate the variables. Placing time variables on the right and [A] on the left \[ d[A] = - k \,dt \nonumber \] In this form, it is easy to integrate. If the concentration of A is [A] 0 at time t = 0, and the concentration of A is [A] at some arbitrary time later, the form of the integral is \[ \int _{[A]_o}^{[A]} d[A] = - k \int _{t_o}^{t}\,dt \nonumber \] which yields \[ [A] - [A]_o = -kt \nonumber \] or \[ [A] = [A]_o -kt \nonumber \] This suggests that a plot of concentration as a function of time will produce a straight line, the slope of which is –k, and the intercept of which is [A] 0 . If such a plot is linear, then the data are consistent with 0 th order kinetics. If they are not, other possibilities must be considered. Second Order Kinetics If the reaction follows a second order rate law, the some methodology can be employed. The rate can be written as \[ -\dfrac{d[A]}{dt} = k [A]^2 \label{eq1A} \] The separation of concentration and time terms (this time keeping the negative sign on the left for convenience) yields \[ -\dfrac{d[A]}{[A]^2} = k \,dt \nonumber \] The integration then becomes \[ - \int_{[A]_o}^{[A]} \dfrac{d[A]}{[A]^2} = \int_{t=0}^{t}k \,dt \label{eq1} \] And noting that \[ - \dfrac{dx}{x^2} = d \left(\dfrac{1}{x} \right) \nonumber \] the result of integration Equation \ref{eq1} is \[ \dfrac{1}{[A]} -\dfrac{1}{[A]_o} = kt \nonumber \] or \[ \dfrac{1}{[A]} = \dfrac{1}{[A]_o} + kt \nonumber \] And so a plot of \(1/[A]\) as a function of time should produce a linear plot, the slope of which is \(k\), and the intercept of which is \(1/[A]_0\). Other 2 nd order rate laws are a little bit trickier to integrate, as the integration depends on the actual stoichiometry of the reaction being investigated. For example, for a reaction of the type \[A + B \rightarrow P \nonumber \] That has rate laws given by \[ -\dfrac{d[A]}{dt} = k [A][B] \nonumber \] and \[ -\dfrac{d[B]}{dt} = k [A][B] \nonumber \] the integration will depend on the decrease of [A] and [B] (which will be related by the stoichiometry) which can be expressed in terms the concentration of the product [P]. \[[A] = [A]_o – [P] \label{eqr1} \] and \[[B] = [B]_o – [P]\label{eqr2} \] The concentration dependence on \(A\) and \(B\) can then be eliminated if the rate law is expressed in terms of the production of the product. \[ \dfrac{d[P]}{dt} = k [A][B] \label{rate2} \] Substituting the relationships for \([A]\) and \([B]\) (Equations \ref{eqr1} and \ref{eqr2}) into the rate law expression (Equation \ref{rate2}) yields \[ \dfrac{d[P]}{dt} = k ( [A]_o – [P]) ([B] = [B]_o – [P]) \label{rate3} \] Separation of concentration and time variables results in \[\dfrac{d[P]}{( [A]_o – [P]) ([B] = [B]_o – [P])} = k\,dt \nonumber \] Noting that at time \(t = 0\), \([P] = 0\), the integrated form of the rate law can be generated by solving the integral \[\int_{[A]_o}^{[A]} \dfrac{d[P]}{( [A]_o – [P]) ([B]_o – [P])} = \int_{t=0}^{t} k\,dt \nonumber \] Consulting a table of integrals reveals that for \(a \neq b\) [1] , \[ \int \dfrac{dx}{(a-x)(b-x)} = \dfrac{1}{b-a} \ln \left(\dfrac{b-x}{a-x} \right) \nonumber \] Applying the definite integral (as long as \([A]_0 \neq [B]_0\)) results in \[ \left. \dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B]_0-[P]}{[A]_0-[P]} \right) \right |_0^{[A]} = \left. k\, t \right|_0^t \nonumber \] \[ \dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B]_0-[P]}{[A]_0-[P]} \right) -\dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B]_0}{[A]_0} \right) =k\, t \label{finalint} \] Substituting Equations \ref{eqr1} and \ref{eqr2} into Equation \ref{finalint} and simplifying (combining the natural logarithm terms) yields \[\dfrac{1}{[B]_0-[A]_0} \ln \left( \dfrac{[B][A]_o}{[A][B]_o} \right) = kt \nonumber \] For this rate law, a plot of \(\ln([B]/[A])\) as a function of time will produce a straight line, the slope of which is \[ m = ([B]_0 – [A]_0)k. \nonumber \] In the limit at \([A]_0 = [B]_0\), then \([A] = [B]\) at all times, due to the stoichiometry of the reaction. As such, the rate law becomes \[ \text{rate} = k [A]^2 \nonumber \] and integrate direct like in Equation \ref{eq1A} and the integrated rate law is (as before) \[ \dfrac{1}{[A]} = \dfrac{1}{[A]_o} + kt \nonumber \] Example \(\PageIndex{2}\): Confirming Second Order Kinetics Consider the following kinetic data. Use a graph to demonstrate that the data are consistent with second order kinetics. Also, if the data are second order, determine the value of the rate constant for the reaction. 0 1 2 3 4 5 6 7 time (s) 0.000 10.000 30.000 60.000 100.000 150.000 200.000 [A] (M) 0.238 0.161 0.098 0.062 0.041 0.029 0.023 Solution The plot looks as follows: From this plot, it can be seen that the rate constant is 0.2658 M -1 s -1 . The concentration at time \(t = 0\) can also be inferred from the intercept. [1] This integral form can be generated by using the method of partial fractions. See (House, 2007) for a full derivation. Example \(\PageIndex{3}\) Dinitrogen pentoxide (N 2 O 5 ) decomposes to NO 2 and O 2 at relatively low temperatures in the following reaction: \( 2N_{2}O_{5}\left ( soln \right ) \rightarrow 4NO_{2}\left ( soln \right )+O_{2}\left ( g \right ) \) This reaction is carried out in a CCl 4 solution at 45°C. The concentrations of N 2 O 5 as a function of time are listed in the following table, together with the natural logarithms and reciprocal N 2 O 5 concentrations. Plot a graph of the concentration versus t , ln concentration versus t , and 1/concentration versus t and then determine the rate law and calculate the rate constant. Time (s) [N2O5] (M) ln[N2O5] 1/[N2O5] (M−1) 0 0.0365 −3.310 27.4 600 0.0274 −3.597 36.5 1200 0.0206 −3.882 48.5 1800 0.0157 −4.154 63.7 2400 0.0117 −4.448 85.5 3000 0.0086 −4.756 116.0 3600 0.0064 −5.051 156.0 Given: balanced chemical equation, reaction times, and concentrations Asked for: graph of data, rate law, and rate constant Strategy: A Use the data in the table to separately plot concentration, the natural logarithm of the concentration, and the reciprocal of the concentration (the vertical axis) versus time (the horizontal axis). Compare the graphs with those in Figure 13.4.2 to determine the reaction order. B Write the rate law for the reaction. Using the appropriate data from the table and the linear graph corresponding to the rate law for the reaction, calculate the slope of the plotted line to obtain the rate constant for the reaction. Solution: A Here are plots of [N 2 O 5 ] versus t , ln[N 2 O 5 ] versus t , and 1/[N 2 O 5 ] versus t : The plot of ln[N 2 O 5 ] versus t gives a straight line, whereas the plots of [N 2 O 5 ] versus t and 1/[N 2 O 5 ] versus t do not. This means that the decomposition of N 2 O 5 is first order in [N 2 O 5 ]. B The rate law for the reaction is therefore \( rate = k \left [N_{2}O_{5} \right ]\) Calculating the rate constant is straightforward because we know that the slope of the plot of ln[A] versus t for a first-order reaction is − k . We can calculate the slope using any two points that lie on the line in the plot of ln[N 2 O 5 ] versus t . Using the points for t = 0 and 3000 s, \( slope= \dfrac{ln\left [N_{2}O_{5} \right ]_{3000}-ln\left [N_{2}O_{5} \right ]_{0}}{3000\;s-0\;s} = \dfrac{\left [-4.756 \right ]-\left [-3.310 \right ]}{3000\;s} =4.820\times 10^{-4}\;s^{-1} \) Thus k = 4.820 × 10 −4 s −1 . Contributors Anonymous Modified by Joshua Halpern ( Howard University ), Scott Sinex, and Scott Johnson (PGCC) |
Courses/University_of_British_Columbia/UBC_Introductory_Chemistry/06%3A_Atomic_Structure_and_Periodic_Trends/6.11%3A_Periodic_Trends-_Atomic_Size_Ionization_Energy_and_Metallic_Character | Learning Objectives Be able to state how certain properties of atoms vary based on their relative position on the periodic table. One of the reasons the periodic table is so useful is because its structure allows us to qualitatively determine how some properties of the elements vary versus their position on the periodic table. The variations of properties versus positions on the periodic table are called periodic trends . There is no other tool in science that allows us to judge relative properties of a class of objects like this, which makes the periodic table a very useful tool. Many periodic trends are general. There may be a few points where an opposite trend is seen, but there is an overall trend when considered across a whole row or down a whole column of the periodic table. The first periodic trend we will consider is atomic radius. The atomic radius is an indication of the size of an atom. Although the concept of a definite radius of an atom is a bit fuzzy, atoms behave as if they have a certain radius. Such radii can be estimated from various experimental techniques, such as the x-ray crystallography of crystals. As you go down a column of the periodic table, the atomic radii increase. This is because the valence electron shell is getting larger and there is a larger principal quantum number, so the valence shell lies physically farther away from the nucleus. This trend can be summarized as follows: \[as\downarrow PT,atomic\; radius \uparrow \nonumber \] where PT stands for periodic table. Going across a row on the periodic table, left to right, the trend is different. Even though the valence shell maintains the same principal quantum number, the number of protons—and hence the nuclear charge—is increasing as you go across the row. The increasing positive charge casts a tighter grip on the valence electrons, so as you go across the periodic table, the atomic radii decrease. Again, we can summarize this trend as follows: \[as\rightarrow PT,atomic\; radius \downarrow \nonumber \] Figure \(\PageIndex{1}\) shows spheres representing the atoms of the s and p blocks from the periodic table to scale, showing the two trends for the atomic radius. Example \(\PageIndex{1}\): Atomic Radii Referring only to a periodic table and not to Figure \(\PageIndex{1}\), which atom is larger in each pair? Si or S S or Te Solution Si is to the left of S on the periodic table; it is larger because as you go across the row, the atoms get smaller. S is above Te on the periodic table; Te is larger because as you go down the column, the atoms get larger. Exercise \(\PageIndex{1}\): Atomic Radii Referring only to a periodic table and not to Figure \(\PageIndex{1}\), which atom is smaller, Ca or Br? Answer Br Ionization energy (IE) is the amount of energy required to remove an electron from an atom in the gas phase: \[A(g)\rightarrow A^{+}(g)+e^{-}\; \; \; \; \; \Delta H\equiv IE \nonumber \] IE is usually expressed in kJ/mol of atoms. It is always positive because the removal of an electron always requires that energy be put in (i.e., it is endothermic). IE also shows periodic trends. As you go down the periodic table, it becomes easier to remove an electron from an atom (i.e., IE decreases) because the valence electron is farther away from the nucleus. Thus, \[as\downarrow PT,\; IE\downarrow \nonumber \] However, as you go across the periodic table and the electrons get drawn closer in, it takes more energy to remove an electron; as a result, IE increases: \[as\rightarrow PT,\; IE\uparrow \nonumber \] Figure \(\PageIndex{2}\) shows values of IE versus position on the periodic table. Again, the trend is not absolute, but the general trends going across and down the periodic table should be obvious. IE also shows an interesting trend within a given atom. This is because more than one IE can be defined by removing successive electrons (if the atom has them to begin with): First Ionization Energy (IE 1 ): \[A(g) → A^+(g) + e^- \nonumber \] Second Ionization Energy (IE 2 ): \[A^{+}(g) → A^{2+}(g) + e^- \nonumber \] Third Ionization Energy (IE 3 ): \[A^{2+}(g) → A^{3+}(g) + e^- \nonumber \] and so forth. Each successive IE is larger than the previous because an electron is being removed from an atom with a progressively larger positive charge. However, IE takes a large jump when a successive ionization goes down into a new shell. For example, the following are the first three IEs for Mg, whose electron configuration is 1 s 2 2 s 2 2 p 6 3 s 2 : First Ionization Energy (IE 1 ) = 738 kJ/mol: \[\ce{Mg (g) -> Mg+ (g) + e− }\nonumber\] Second Ionization Energy (IE 2 ) = 1,450 kJ/mol: \[\ce{Mg+ (g) -> Mg^2+ (g) + e− }\nonumber\] Third Ionization Energy (IE 3 ) = 7,734 kJ/mol: \[\ce{Mg^2+ (g) -> Mg^3+ (g) + e− }\nonumber\] The second IE is twice the first, which is not a surprise: the first IE involves removing an electron from a neutral atom, while the second one involves removing an electron from a positive ion. The third IE, however, is over five times the previous one. Why is it so much larger? Because the first two electrons are removed from the 3 s subshell, but the third electron has to be removed from the n = 2 shell (specifically, the 2 p subshell, which is lower in energy than the n = 3 shell). Thus, it takes much more energy than just overcoming a larger ionic charge would suggest. It is trends like this that demonstrate that electrons within atoms are organized in groups. Example \(\PageIndex{2}\): Ionization Energies Which atom in each pair has the larger first ionization energy? Ca or Sr K or K + Solution Because Sr is below Ca on the periodic table, it is easier to remove an electron from it; thus, Ca has the higher IE. Because K + has a positive charge, it will be harder to remove another electron from it, so its IE is larger than that of K. Indeed, it will be significantly larger because the next electron in K + to be removed comes from another shell. Exercise \(\PageIndex{2}\): Ionization Energies Which atom has the lower ionization energy, C or F? Answer C The opposite of IE is described by electron affinity (EA), which is the energy change when a gas-phase atom accepts an electron: \[\ce{A (g) + e- -> A- (g) } \; \; \; \; \; \Delta H\equiv EA \nonumber \] EA is also usually expressed in kJ/mol. EA also demonstrates some periodic trends, although they are less obvious than the other periodic trends discussed previously. Generally, as you go across the periodic table, EA increases its magnitude: \[as\rightarrow PT,\; EA\uparrow \nonumber \] There is not a definitive trend as you go down the periodic table; sometimes EA increases, sometimes it decreases. Figure \(\PageIndex{3}\) shows EA values versus position on the periodic table for the s - and p -block elements. The trend is not absolute, especially considering the large positive EA values for the second column. However, the general trend going across the periodic table should be obvious. Example \(\PageIndex{3}\): Electron Affinities Predict which atom in each pair will have the highest magnitude of Electron Affinity. C or F Na or S Solution C and F are in the same row on the periodic table, but F is farther to the right. Therefore, F should have the larger magnitude of EA. Na and S are in the same row on the periodic table, but S is farther to the right. Therefore, S should have the larger magnitude of EA. Exercise \(\PageIndex{3}\): Electron Affinities Predict which atom will have the highest magnitude of Electron Affinity: As or Br. Answer Br Metallic Character The metallic character is used to define the chemical properties that metallic elements present. Generally, metals tend to lose electrons to form cations. Nonmetals tend to gain electrons to form anions. They also have a high oxidation potential—therefore they are easily oxidized and are strong reducing agents. Metals also form basic oxides; the more basic the oxide, the higher the metallic character. As you move across the table from left to right, the metallic character decreases, because the elements easily accept electrons to fill their valance shells. Therefore, these elements take on the nonmetallic character of forming anions. As you move up the table, the metallic character decreases, due to the greater pull that the nucleus has on the outer electrons. This greater pull makes it harder for the atoms to lose electrons and form cations. Uses of the Periodic Properties of Elements Predict greater or smaller atomic size and radial distribution in neutral atoms and ions. Measure and compare ionization energies. Compare electron affinities and electronegativities. Predict redox potential. Compare metallic character with other elements; ability to form cations. Predict reactions that may or may not occur due to the trends. Determine greater cell potential (sum of oxidation and reduction potential) between reactions. Complete chemical reactions according to trends. Summary Certain properties—notably atomic radius, ionization energies, and electron affinities—can be qualitatively understood by the positions of the elements on the periodic table. The major trends are summarized in the figure below. There are three factors that help in the prediction of the trends in the Periodic Table: number of protons in the nucleus, number of shells, and shielding effect. |
Courses/Williams_School/Advanced_Chemistry/15%3A_Neurochemistry | 15.1: A Drug's Life 15.2: Neurotransmitters and Antidepressants 15.3: Analgesics 15.4: No Pain, Your Gain 15.5: Narcotic Analgesics Narcotic agents are potent analgesics which are effective for the relief of severe pain. Analgesics are selective central nervous system depressants used to relieve pain. The term analgesic means "without pain". Even in therapeutic doses, narcotic analgesics can cause respiratory depression, nausea, and drowsiness. Long term administration produces tolerance, psychic, and physical dependence called addiction. 15.6: Stimulants 15.7: Antidepressant medications Antidepressant drugs are used to restore mentally depressed patients to an improved mental status. Depression results from a deficiency of norepinephrine at receptors in the brain. Mechanisms that increase their effective concentration at the receptor sites should alleviate depression. |
Courses/Southeast_Missouri_State_University/CH185%3A_General_Chemistry_(Ragain)/08%3A_Chemical_Bonding_I-_Lewis_Structures_and_Determining_Molecular_Shapes/8.07%3A_Lewis_Structures | Learning Objectives To use Lewis dot symbols to explain the stoichiometry of a compound We begin our discussion of the relationship between structure and bonding in covalent compounds by describing the interaction between two identical neutral atoms—for example, the H 2 molecule, which contains a purely covalent bond. Each hydrogen atom in H 2 contains one electron and one proton, with the electron attracted to the proton by electrostatic forces. As the two hydrogen atoms are brought together, additional interactions must be considered (Figure \(\PageIndex{1}\)): The electrons in the two atoms repel each other because they have the same charge ( The electrons in the two atoms repel each other because they have the same charge (E > 0). Similarly, the protons in adjacent atoms repel each other ( E > 0). The electron in one atom is attracted to the oppositely charged proton in the other atom and vice versa ( E < 0). Recall that it is impossible to specify precisely the position of the electron in either hydrogen atom. Hence the quantum mechanical probability distributions must be used. A plot of the potential energy of the system as a function of the internuclear distance (Figure \(\PageIndex{2}\)) shows that the system becomes more stable (the energy of the system decreases) as two hydrogen atoms move toward each other from r = ∞, until the energy reaches a minimum at r = r 0 (the observed internuclear distance in H 2 is 74 pm). Thus at intermediate distances, proton–electron attractive interactions dominate, but as the distance becomes very short, electron–electron and proton–proton repulsive interactions cause the energy of the system to increase rapidly. Notice the similarity between Figures \(\PageIndex{1}\) and \(\PageIndex{2}\), which described a system containing two oppositely charged ions . The shapes of the energy versus distance curves in the two figures are similar because they both result from attractive and repulsive forces between charged entities. At long distances, both attractive and repulsive interactions are small. As the distance between the atoms decreases, the attractive electron–proton interactions dominate, and the energy of the system decreases. At the observed bond distance, the repulsive electron–electron and proton–proton interactions just balance the attractive interactions, preventing a further decrease in the internuclear distance. At very short internuclear distances, the repulsive interactions dominate, making the system less stable than the isolated atoms. Using Lewis Dot Symbols to Describe Covalent Bonding The valence electron configurations of the constituent atoms of a covalent compound are important factors in determining its structure, stoichiometry, and properties. For example, chlorine, with seven valence electrons, is one electron short of an octet. If two chlorine atoms share their unpaired electrons by making a covalent bond and forming Cl 2 , they can each complete their valence shell: Each chlorine atom now has an octet. The electron pair being shared by the atoms is called a bonding pair; the other three pairs of electrons on each chlorine atom are called lone pairs. Lone pairs are not involved in covalent bonding. If both electrons in a covalent bond come from the same atom, the bond is called a coordinate covalent bond. Examples of this type of bonding are presented in Section 8.6 when we discuss atoms with less than an octet of electrons. We can illustrate the formation of a water molecule from two hydrogen atoms and an oxygen atom using Lewis dot symbols: The structure on the right is the Lewis electron structure , or Lewis structure , for H 2 O. With two bonding pairs and two lone pairs, the oxygen atom has now completed its octet. Moreover, by sharing a bonding pair with oxygen, each hydrogen atom now has a full valence shell of two electrons. Chemists usually indicate a bonding pair by a single line, as shown here for our two examples: The following procedure can be used to construct Lewis electron structures for more complex molecules and ions: Arrange the atoms to show specific connections. When there is a central atom, it is usually the least electronegative element in the compound. Chemists usually list this central atom first in the chemical formula (as in CCl 4 and CO 3 2 − , which both have C as the central atom), which is another clue to the compound’s structure. Hydrogen and the halogens are almost always connected to only one other atom, so they are usually terminal rather than central. Determine the total number of valence electrons in the molecule or ion. Add together the valence electrons from each atom. (Recall that the number of valence electrons is indicated by the position of the element in the periodic table.) If the species is a polyatomic ion, remember to add or subtract the number of electrons necessary to give the total charge on the ion. For CO 3 2 − , for example, we add two electrons to the total because of the −2 charge. Place a bonding pair of electrons between each pair of adjacent atoms to give a single bond. In H 2 O, for example, there is a bonding pair of electrons between oxygen and each hydrogen. Beginning with the terminal atoms, add enough electrons to each atom to give each atom an octet (two for hydrogen). These electrons will usually be lone pairs. If any electrons are left over, place them on the central atom. We will explain later that some atoms are able to accommodate more than eight electrons. If the central atom has fewer electrons than an octet, use lone pairs from terminal atoms to form multiple (double or triple) bonds to the central atom to achieve an octet. This will not change the number of electrons on the terminal atoms. Now let’s apply this procedure to some particular compounds, beginning with one we have already discussed. The central atom is usually the least electronegative element in the molecule or ion; hydrogen and the halogens are usually terminal. The \(H_2O\) Molecule Because H atoms are almost always terminal, the arrangement within the molecule must be HOH . Each H atom (group 1) has 1 valence electron, and the O atom (group 16) has 6 valence electrons, for a total of 8 valence electrons. Placing one bonding pair of electrons between the O atom and each H atom gives H:O:H, with 4 electrons left over. Each H atom has a full valence shell of 2 electrons. Adding the remaining 4 electrons to the oxygen (as two lone pairs) gives the following structure: This is the Lewis structure we drew earlier. Because it gives oxygen an octet and each hydrogen two electrons, we do not need to use step 6. The \(OCl^−\) Ion With only two atoms in the molecule, there is no central atom. Oxygen (group 16) has 6 valence electrons, and chlorine (group 17) has 7 valence electrons; we must add one more for the negative charge on the ion, giving a total of 14 valence electrons. Placing a bonding pair of electrons between O and Cl gives O:Cl, with 12 electrons left over. If we place six electrons (as three lone pairs) on each atom, we obtain the following structure: Each atom now has an octet of electrons, so steps 5 and 6 are not needed. The Lewis electron structure is drawn within brackets as is customary for an ion, with the overall charge indicated outside the brackets, and the bonding pair of electrons is indicated by a solid line. OCl − is the hypochlorite ion, the active ingredient in chlorine laundry bleach and swimming pool disinfectant. The \(CH_2O\) Molecule 1. Because carbon is less electronegative than oxygen and hydrogen is normally terminal, C must be the central atom. One possible arrangement is as follows: 2. Each hydrogen atom (group 1) has one valence electron, carbon (group 14) has 4 valence electrons, and oxygen (group 16) has 6 valence electrons, for a total of [(2)(1) + 4 + 6] = 12 valence electrons. 3. Placing a bonding pair of electrons between each pair of bonded atoms gives the following: Six electrons are used, and 6 are left over. 4. Adding all 6 remaining electrons to oxygen (as three lone pairs) gives the following: Although oxygen now has an octet and each hydrogen has 2 electrons, carbon has only 6 electrons. 5. There are no electrons left to place on the central atom. 6. To give carbon an octet of electrons, we use one of the lone pairs of electrons on oxygen to form a carbon–oxygen double bond: Both the oxygen and the carbon now have an octet of electrons, so this is an acceptable Lewis electron structure. The O has two bonding pairs and two lone pairs, and C has four bonding pairs. This is the structure of formaldehyde, which is used in embalming fluid. An alternative structure can be drawn with one H bonded to O. Formal charges , discussed later in this section, suggest that such a structure is less stable than that shown previously. Example \(\PageIndex{1}\) Write the Lewis electron structure for each species. NCl 3 S 2 2 − NOCl Given: chemical species Asked for: Lewis electron structures Strategy: Use the six-step procedure to write the Lewis electron structure for each species. Solution: Nitrogen is less electronegative than chlorine, and halogen atoms are usually terminal, so nitrogen is the central atom. The nitrogen atom (group 15) has 5 valence electrons and each chlorine atom (group 17) has 7 valence electrons, for a total of 26 valence electrons. Using 2 electrons for each N–Cl bond and adding three lone pairs to each Cl account for (3 × 2) + (3 × 2 × 3) = 24 electrons. Rule 5 leads us to place the remaining 2 electrons on the central N: Nitrogen trichloride is an unstable oily liquid once used to bleach flour; this use is now prohibited in the United States. In a diatomic molecule or ion, we do not need to worry about a central atom. Each sulfur atom (group 16) contains 6 valence electrons, and we need to add 2 electrons for the −2 charge, giving a total of 14 valence electrons. Using 2 electrons for the S–S bond, we arrange the remaining 12 electrons as three lone pairs on each sulfur, giving each S atom an octet of electrons: Because nitrogen is less electronegative than oxygen or chlorine, it is the central atom. The N atom (group 15) has 5 valence electrons, the O atom (group 16) has 6 valence electrons, and the Cl atom (group 17) has 7 valence electrons, giving a total of 18 valence electrons. Placing one bonding pair of electrons between each pair of bonded atoms uses 4 electrons and gives the following: Adding three lone pairs each to oxygen and to chlorine uses 12 more electrons, leaving 2 electrons to place as a lone pair on nitrogen: Because this Lewis structure has only 6 electrons around the central nitrogen, a lone pair of electrons on a terminal atom must be used to form a bonding pair. We could use a lone pair on either O or Cl. Because we have seen many structures in which O forms a double bond but none with a double bond to Cl, it is reasonable to select a lone pair from O to give the following: All atoms now have octet configurations. This is the Lewis electron structure of nitrosyl chloride, a highly corrosive, reddish-orange gas. Exercise \(\PageIndex{1}\) Write Lewis electron structures for CO 2 and SCl 2 , a vile-smelling, unstable red liquid that is used in the manufacture of rubber. Answer Using Lewis Electron Structures to Explain Stoichiometry Lewis dot symbols provide a simple rationalization of why elements form compounds with the observed stoichiometries. In the Lewis model, the number of bonds formed by an element in a neutral compound is the same as the number of unpaired electrons it must share with other atoms to complete its octet of electrons. For the elements of Group 17 (the halogens) , this number is one; for the elements of Group 16 (the chalcogens) , it is two; for Group 15 elements , three; and for Group 14 elements four. These requirements are illustrated by the following Lewis structures for the hydrides of the lightest members of each group: Elements may form multiple bonds to complete an octet. In ethylene, for example, each carbon contributes two electrons to the double bond, giving each carbon an octet (two electrons/bond × four bonds = eight electrons). Neutral structures with fewer or more bonds exist, but they are unusual and violate the octet rule. Allotropes of an element can have very different physical and chemical properties because of different three-dimensional arrangements of the atoms; the number of bonds formed by the component atoms, however, is always the same. As noted at the beginning of the chapter, diamond is a hard, transparent solid; graphite is a soft, black solid; and the fullerenes have open cage structures. Despite these differences, the carbon atoms in all three allotropes form four bonds, in accordance with the octet rule. Lewis structures explain why the elements of groups 14–17 form neutral compounds with four, three, two, and one bonded atom(s), respectively. Elemental phosphorus also exists in three forms: white phosphorus, a toxic, waxy substance that initially glows and then spontaneously ignites on contact with air; red phosphorus, an amorphous substance that is used commercially in safety matches, fireworks, and smoke bombs; and black phosphorus, an unreactive crystalline solid with a texture similar to graphite (Figure \(\PageIndex{3}\)). Nonetheless, the phosphorus atoms in all three forms obey the octet rule and form three bonds per phosphorus atom. Formal Charges It is sometimes possible to write more than one Lewis structure for a substance that does not violate the octet rule, as we saw for CH 2 O, but not every Lewis structure may be equally reasonable. In these situations, we can choose the most stable Lewis structure by considering the formal charge on the atoms, which is the difference between the number of valence electrons in the free atom and the number assigned to it in the Lewis electron structure. The formal charge is a way of computing the charge distribution within a Lewis structure; the sum of the formal charges on the atoms within a molecule or an ion must equal the overall charge on the molecule or ion. A formal charge does not represent a true charge on an atom in a covalent bond but is simply used to predict the most likely structure when a compound has more than one valid Lewis structure. To calculate formal charges, we assign electrons in the molecule to individual atoms according to these rules: Nonbonding electrons are assigned to the atom on which they are located. Bonding electrons are divided equally between the bonded atoms. For each atom, we then compute a formal charge: \( \begin{matrix} formal\; charge= & valence\; e^{-}- & \left ( non-bonding\; e^{-}+\frac{bonding\;e^{-}}{2} \right )\\ & ^{\left ( free\; atom \right )} & ^{\left ( atom\; in\; Lewis\; structure \right )} \end{matrix} \label{8.5.1} \) (atom in Lewis structure) To illustrate this method, let’s calculate the formal charge on the atoms in ammonia (NH 3 ) whose Lewis electron structure is as follows: A neutral nitrogen atom has five valence electrons (it is in group 15). From its Lewis electron structure, the nitrogen atom in ammonia has one lone pair and shares three bonding pairs with hydrogen atoms, so nitrogen itself is assigned a total of five electrons [2 nonbonding e − + (6 bonding e − /2)]. Substituting into Equation \(\ref{8.5.2}\), we obtain \[ formal\; charge\left ( N \right )=5\; valence\; e^{-}-\left ( 2\; non-bonding\; e^{-} +\dfrac{6\; bonding\; e^{-}}{2} \right )=0 \label{8.5.2} \] A neutral hydrogen atom has one valence electron. Each hydrogen atom in the molecule shares one pair of bonding electrons and is therefore assigned one electron [0 nonbonding e − + (2 bonding e − /2)]. Using Equation \(\ref{8.5.2}\) to calculate the formal charge on hydrogen, we obtain \[ formal\; charge\left ( H \right )=1\; valence\; e^{-}-\left ( 0\; non-bonding\; e^{-} +\dfrac{2\; bonding\; e^{-}}{2} \right )=0 \label{8.5.3} \] The hydrogen atoms in ammonia have the same number of electrons as neutral hydrogen atoms, and so their formal charge is also zero. Adding together the formal charges should give us the overall charge on the molecule or ion. In this example, the nitrogen and each hydrogen has a formal charge of zero. When summed the overall charge is zero, which is consistent with the overall charge on the NH 3 molecule. An atom, molecule, or ion has a formal charge of zero if it has the number of bonds that is typical for that species. Typically, the structure with the most charges on the atoms closest to zero is the more stable Lewis structure. In cases where there are positive or negative formal charges on various atoms, stable structures generally have negative formal charges on the more electronegative atoms and positive formal charges on the less electronegative atoms. The next example further demonstrates how to calculate formal charges. Example \(\PageIndex{2}\): The Ammonium Ion Calculate the formal charges on each atom in the NH 4 + ion. Given: chemical species Asked for: formal charges Strategy: Identify the number of valence electrons in each atom in the NH 4 + ion. Use the Lewis electron structure of NH 4 + to identify the number of bonding and nonbonding electrons associated with each atom and then use Equation \(\ref{8.5.2}\) to calculate the formal charge on each atom. Solution: The Lewis electron structure for the NH 4 + ion is as follows: The nitrogen atom shares four bonding pairs of electrons, and a neutral nitrogen atom has five valence electrons. Using Equation \(\ref{8.5.1}\), the formal charge on the nitrogen atom is therefore \[ formal\; charge\left ( N \right )=5-\left ( 0+\dfrac{8}{2} \right )=0 \nonumber \] Each hydrogen atom in has one bonding pair. The formal charge on each hydrogen atom is therefore \[ formal\; charge\left ( H \right )=1-\left ( 0+\dfrac{2}{2} \right )=0 \nonumber \] The formal charges on the atoms in the NH 4 + ion are thus Adding together the formal charges on the atoms should give us the total charge on the molecule or ion. In this case, the sum of the formal charges is 0 + 1 + 0 + 0 + 0 = +1. Exercise \(\PageIndex{2}\) Write the formal charges on all atoms in BH 4 − . Answer If an atom in a molecule or ion has the number of bonds that is typical for that atom (e.g., four bonds for carbon), its formal charge is zero. Using Formal Charges to Distinguish Viable Lewis Structures |
Bookshelves/Inorganic_Chemistry/Introduction_to_Inorganic_Chemistry_(Wikibook)/07%3A_Metals_and_Alloys_-_Mechanical_Properties/7.04%3A_Iron_and_Steel | One other very important place where the difference between the hardness of a BCC and a close-packed metal is important is in steelmaking. Between room temperature and 912 o C, iron has the BCC structure, and is a tough, hard metal ("tough as nails"). Above 912 o C, pure iron switches over to the FCC (austenite) structure, which is much more ductile. So hot iron can be bent and worked into a variety of shapes when it is very hot but still solid (it melts at 1535 o C). Rapid quenching of hot iron - e.g., when the blacksmith plunges a red hot piece directly into cold water - cools it to room temperature, but doesn't allow time for the FCC --> BCC phase transition to occur; therefore, such pieces are still relatively malleable and can be shaped. 0 The iron–iron carbide (Fe–Fe3C) phase diagram. Below 912 °C, pure iron exists as the alpha phase, ferrite, which has the BCC structure. Between 912 and 1,394 °C, pure iron exists as the gamma phase, austenite, which has the FCC structure. Carbon is more soluble in the FCC phase, which occupies area "γ" on the phase diagram, than it is in the BCC phase. The percent carbon determines the type of iron alloy that is formed upon cooling from the FCC phase, or from liquid iron: alpha iron, carbon steel (pearlite), or cast iron. Carbon is added (about 1% by weight) to iron to make "carbon steel", which is a very hard material. Carbon is rather soluble in the FCC phase of iron, but not in the BCC phase. Therefore, when the ductile FCC phase cools and turns into BCC ("tempering" the steel, which means cooling it slowly enough so the FCC to BCC transformation can occur), the iron can no longer dissolve the excess carbon. The carbon forms layers or grains of an extra phase, Fe 3 C ("cementite" - a very hard material) which are layered or dotted throughout the matrix of BCC iron grains. The effect of all these little grains of Fe 3 C is to stop the motion of dislocations, making for a harder but (with higher carbon content) increasingly brittle material. This is why knives and swords are quenched from the FCC phase, cold worked into the appropriate shapes, and then heated up again and tempered (before they are sharpened) when they are made. Cast iron objects (frying pans, radiators, etc) have a higher carbon content and are therefore very strong, but tend to fracture rather than bend because of the larger fraction of the brittle Fe 3 C phase in the alloy. 0 Upon cooling, high carbon steels phase segregate into a mixture of bcc iron (light gray) and Fe3C (dark gray) microscopic grains. |
Bookshelves/General_Chemistry/Chem1_(Lower)/08%3A_Solutions/8.05%3A__Colligative_Properties_-_Osmotic_Pressure | Learning Objectives Define a semipermeable membrane in the context of osmotic flow. Explain, in simple terms, what fundamental process "drives" osmotic flow. What is osmotic pressure, and how is it measured? Osmotic pressure can be a useful means of estimating the molecular weight of a substance, particularly if its molecular weight is quite large. Explain in your own words how this works. What is reverse osmosis, and what is its principal application? Explain the role of osmotic pressure in food preservation, and give an example. Describe the role osmosis plays in the rise of water in plants (where is the semipermeable membrane?), and why it cannot be the only cause in very tall trees. Osmosis is the process in which a liquid passes through a membrane whose pores permit the passage of solvent molecules but are too small for the larger solute molecules to pass through. Semipermeable Membranes and Osmotic flow Figure \(\PageIndex{1}\) shows a simple osmotic cell. Both compartments contain water, but the one on the right also contains a solute whose molecules (represented by green circles) are too large to pass through the membrane. Many artificial and natural substances are capable of acting as semi-permeable membranes. The walls of most plant and animal cells fall into this category. If the cell is set up so that the liquid level is initially the same in both compartments, you will soon notice that the liquid rises in the left compartment and falls in the right side, indicating that water molecules from the right compartment are migrating through the semipermeable membrane and into the left compartment. This migration of the solvent is known as osmotic flow, or simply osmosis. The escaping tendency of a substance from a phase increases with its concentration in the phase. What is the force that drives the molecules through the membrane? This is a misleading question, because there is no real “force” in the physical sense other than the thermal energies all molecules possess. Osmosis is a consequence of simple statistics: the randomly directed motions of a collection of molecules will cause more to leave a region of high concentration than return to it; the escaping tendency of a substance from a phase increases with its concentration in the phase. Diffusion and Osmotic Flow Suppose you drop a lump of sugar into a cup of tea, without stirring. Initially there will be a very high concentration of dissolved sugar at the bottom of the cup, and a very low concentration near the top. Since the molecules are in random motion, there will be more sugar molecules moving from the high concentration region to the low concentration region than in the opposite direction. The motion of a substance from a region of high concentration to one of low concentration is known as diffusion . Diffusion is a consequence of a concentration gradient (which is a measure of the difference in escaping tendency of the substance in different regions of the solution). There is really no special force on the individual molecules; diffusion is purely a consequence of statistics. Osmotic flow is simply diffusion of a solvent through a membrane impermeable to solute molecules. Now take two solutions of differing solvent concentration, and separate them by a semipermeable membrane (Figure \(\PageIndex{2}\)). Being semipermeable, the membrane is essentially invisible to the solvent molecules, so they diffuse from the high concentration region to the low concentration region just as before. This flow of solvent constitutes osmotic flow , or osmosis . Figure \(\PageIndex{2}\): Osmosis osmotic flow(a) Two sugar-water solutions of different concentrations, separated by a semipermeable membrane that passes water but not sugar. Osmosis will be to the right, since water is less concentrated there. (b) The fluid level rises until the back pressure ρgh equals the relative osmotic pressure; then, the net transfer of water is zero. (CC-BY; OpenStax). Figure \(\PageIndex{2}\) shows water molecules (blue) passing freely in both directions through the semipermeable membrane, while the larger solute molecules remain trapped in the left compartment, diluting the water and reducing its escaping tendency from this cell, compared to the water in the right side. This results in a net osmotic flow of water from the right side which continues until the increased hydrostatic pressure on the left side raises the escaping tendency of the diluted water to that of the pure water at 1 atm, at which point osmotic equilibrium is achieved. Osmotic flow is simply diffusion of a solvent through a membrane impermeable to solute molecules. In the absence of the semipermeable membrane, diffusion would continue until the concentrations of all substances are uniform throughout the liquid phase. With the semipermeable membrane in place, and if one compartment contains the pure solvent, this can never happen; no matter how much liquid flows through the membrane, the solvent in the right side will always be more concentrated than that in the left side. Osmosis will continue indefinitely until we run out of solvent, or something else stops it. Osmotic equilibrium and osmotic pressure One way to stop osmosis is to raise the hydrostatic pressure on the solution side of the membrane. This pressure squeezes the solvent molecules closer together, raising their escaping tendency from the phase. If we apply enough pressure (or let the pressure build up by osmotic flow of liquid into an enclosed region), the escaping tendency of solvent molecules from the solution will eventually rise to that of the molecules in the pure solvent, and osmotic flow will case. The pressure required to achieve osmotic equilibrium is known as the osmotic pressure . Note that the osmotic pressure is the pressure required to stop osmosis, not to sustain it. Osmotic pressure is the pressure required to stop osmotic flow It is common usage to say that a solution “has” an osmotic pressure of "x atmospheres". It is important to understand that this means nothing more than that a pressure of this value must be applied to the solution to prevent flow of pure solvent into this solution through a semipermeable membrane separating the two liquids. Osmotic Pressure and Solute Concentration The Dutch scientist Jacobus Van't Hoff (1852-1911) was one of the giants of physical chemistry. He discovered this equation after a chance encounter with a botanist friend during a walk in a park in Amsterdam; the botanist had learned that the osmotic pressure increases by about 1/273 for each degree of temperature increase. van’t Hoff immediately grasped the analogy to the ideal gas law. The osmotic pressure \(\Pi\) of a solution containing \(n\) moles of solute particles in a solution of volume \(V\) is given by the van 't Hoff equation : \[\Pi = \dfrac{nRT}{V} \label{8.4.3}\] in which \(R\) is the gas constant (0.0821 L atm mol –1 K –1 ) and \(T\) is the absolute temperature. In contrast to the need to employ solute molality to calculate the effects of a non-volatile solute on changes in the freezing and boiling points of a solution, we can use solute molarity to calculate osmotic pressures. Note that the fraction \(n/V\) corresponds to the molarity (\(M\)) of a solution of a non-dissociating solute, or to twice the molarity of a totally-dissociated solute such as \(NaCl\). In this context, molarity refers to the summed total of the concentrations of all solute species. Hence, Equation \ref{8.4.3} can be expressed as \[\Pi =MRT \label{8.4.3B}\] Recalling that \(\Pi\) is the Greek equivalent of P , the re-arranged form \(\Pi V = nRT\) of the above equation should look familiar. Much effort was expended around the end of the 19th century to explain the similarity between this relation and the ideal gas law , but in fact, the Van’t Hoff equation turns out to be only a very rough approximation of the real osmotic pressure law, which is considerably more complicated and was derived after van 't Hoff's formulation. As such, this equation gives valid results only for extremely dilute ("ideal") solutions. According to the Van't Hoff equation, an ideal solution containing 1 mole of dissolved particles per liter of solvent at 0° C will have an osmotic pressure of 22.4 atm. Example \(\PageIndex{1}\) Sea water contains dissolved salts at a total ionic concentration of about 1.13 mol L –1 . What pressure must be applied to prevent osmotic flow of pure water into sea water through a membrane permeable only to water molecules? Solution This is a simple application of Equation \ref{8.4.3B}. \[ \begin{align*} \Pi &= MRT \\[4pt] &= (1.13\; mol /L)(0.0821\; L \,atm \,mol^{–1}\; K^{–1})(298\; K) \\[4pt] &= 27.6\; atm \end{align*}\] Molecular Weight Determination by Osmotic Pressure Since all of the colligative properties of solutions depend on the concentration of the solvent, their measurement can serve as a convenient experimental tool for determining the concentration, and thus the molecular weight, of a solute. Osmotic pressure is especially useful in this regard, because a small amount of solute will produce a much larger change in this quantity than in the boiling point, freezing point, or vapor pressure. even a 10 –6 molar solution would have a measurable osmotic pressure. Molecular weight determinations are very frequently made on proteins or other high molecular weight polymers. These substances, owing to their large molecular size, tend to be only sparingly soluble in most solvents, so measurement of osmotic pressure is often the only practical way of determining their molecular weights. Example \(\PageIndex{2}\): average molecular weight The osmotic pressure of a benzene solution containing 5.0 g of polystyrene per liter was found to be 7.6 torr at 25°C. Estimate the average molecular weight of the polystyrene in this sample. Solution: osmotic pressure: \[ \begin{align*} \Pi &= \dfrac{7.6\, torr}{760\, torr\, atm^{–1}} \\[4pt] &= 0.0100 \,atm \end{align*} \] Using the form of the van 't Hoff equation (Equation \ref{8.4.3}), PV = nRT , the number of moles of polystyrene is n = (0.0100 atm)(1 L) ÷ (0.0821 L atm mol –1 K –1 )(298 K) = 4.09 x 10 –4 mol Molar mass of the polystyrene: (5.0 g) ÷ (4.09 x 10 –4 mol) = 12200 g mol –1 . The experiment to demonstrate this is quite simple: pure solvent is introduced into one side of a cell that is separated into two parts by a semipermeable membrane. The polymer solution is placed in the other side, which is enclosed and connected to a manometer or some other kind of pressure gauge. As solvent molecules diffuse into the solution cell the pressure builds up; eventually this pressure matches the osmotic pressure of the solution and the system is in osmotic equilibrium. The osmotic pressure is read from the measuring device and substituted into the van’t Hoff equation to find the number of moles of solute. |
Bookshelves/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.01%3A_Rates_of_reactions_and_rate_laws | Learning Objectives Make sure you thoroughly understand the following essential ideas: Describe the contrasting roles of thermodynamics and kinetics in understanding chemical change. Given a balanced net equation, write an expression for the rate of a reaction . Sketch a curve showing how the instantaneous rate of a reaction might change with time. Determine the order of a reaction of the form A → B + C from experimental data for the concentrations of its products at successive times. Describe the initial rate and isolation methods of determining the orders of the individual reactants in a reaction involving multiple reactants. Explain the difference between differential and integral rate laws . Sketch out a plot showing how the concentration of a component ([A] or ln [A]) that follows first-order kinetics will change with time. Indicate how the magnitude of the rate constant affects this plot Define the half-life of a reaction. Given the half-life for a first-order reaction A → products along with the initial value of [A] o , find [A] t at a subsequent time an integral number of half-lifes later. Describe the conditions under which a reaction can appear to have an order of zero. Chemical change is guided and driven by energetics, but the actual route it takes and the speed with which it occurs is the subject of "dynamics". Dynamics is itself divided into two general areas: kinetics, which deals with the rate of change and is the subject of this lesson. Mechanistics, introduced in a later lesson, is an exploration of the "road map" that links reactants to products. The energetic aspects of change are governed by the laws of thermodynamics (the "dynamics" part of this word is related to the historical origins of the field and is not a part of dynamics in the sense of these lessons.) Energetics + dynamics = chemical change Chemical change is driven by the tendency of atoms and molecules to rearrange themselves in a way that results in the maximum possible dispersion of thermal energy into the world. The observable quantity that measures this spreading and sharing of energy is the free energy of the system. As a chemical change takes place, the quantities of reactants and products change in a way that leads to a more negative free energy. When the free energy reaches its minimum possible value, there is no more net change and the system is said to be in equilibrium . The beauty of thermodynamics is that it enables us to unfailingly predict the net direction of a reaction and the composition of the equilibrium state even without conducting the experiment; the standard free energies of the reactants and products, which can be independently measured or obtained from tables, are all we need. Half the Story Thermodynamics points the way and makes it possible...but it says nothing about how long it will take to get there! It is worth noting that the concept of "time" plays no role whatsover in thermodynamics. But kinetics is all about time. The "speed" of a reaction — how long it takes to reach equilibrium — bears no relation at all to how spontaneous it is (as given by the sign and value of Δ G °) or whether it is exothermic or endothermic (given by the sign of Δ H °). Moreover, there is no way that reaction rates can be predicted in advance; each reaction must be studied individually. One reason for this is that The stoichiometric equation for the reaction says nothing about its mechanism. By mechanism, we mean, basically, "who does what to whom". Think of a reaction mechanism as something that goes on in a "black box" that joins reactants to products: The inner workings of the black box are ordinarily hidden from us, are highly unpredictable and can only be inferred by indirect means. Consider, for example, the gas-phase formation reactions of the hydrogen halides from the elements. The Balanced equation says nothing about a Reaction's Mechanism The thermodynamics of these reactions are all similar (they are all highly exothermic), but their dynamics (their kinetics and mechanisms) could not be more different. 0 1 2 \(H_{2(g)} + I_{2(g)} → 2 HI_{(g)}\) \(H_{2(g)} + Br_{2(g)} → 2 HBr_{(g)}\) \(H_{2(g)} + Cl_{2(g)} → 2 HCl_{(g)}\) Careful experiments, carried out over many years, are consistent with the simplest mechanism: a single collision between the two reactant molecules results in a rearrangement of the bonds: One might be tempted to suppose that this would proceed in a similar way, but experiments reveal that the mechanism of this reaction is far more complex. The reaction takes place in a succession of steps, some of which involve atomic H and Br. The mechanism of this reaction is different again. Although the first two reactions reach equilibrium in minutes to an hour or so at temperatures of 300 to 600 K, a mixture of hydrogen and chlorine will not react at all in the dark, but if you shine a light on the mixture, it goes off with a bang as the instantaneous reaction releases heat and expands the gas explosively. What is particularly noteworthy is that these striking differences cannot be reliably predicted from theory; they were revealed only by experimentation. The rates of chemical reactions Chemical reactions vary greatly in the speed at which they occur. Some are essentially instantaneous, while others may take years to reach equilibrium. The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed: \[\color{red} \text{rate} = \dfrac{\Delta \text{concentration}}{\Delta \text{time}} \label{2-1}\] For a reaction of the form A + B → C, the rate can be expressed in terms of the change in concentration of any of its components: \[\text{rate} = \dfrac{-\Delta[A]}{\Delta t} \label{Eq1a}\] \[\text{rate} = \dfrac{-\Delta[B]}{\Delta t} \label{Eq1b}\] \[\text{rate} = \dfrac{\Delta[C]}{\Delta t} \label{Eq1c}\] in which Δ[A] is the difference between the concentration of \(A\) over the time interv al \(\Delta t = t_2 – t_1\): \[Δ[A] = [A]_2 – [A]_1 \label{2-2}\] Notice the minus signs in Equations \(\ref{Eq1a}\) and \(\ref{Eq1a}\); the concentration of a reactant always decreases with time, so Δ[A] and Δ[B] are both negative. Since negative rates does not make much sense, rates expressed in terms of a reactant concentration are always preceded by a minus sign in order to make the rate come out positive. Consider now a reaction in which the coefficients are different: \[A + 3B → 2D \] It is clear that [B] decreases three times as rapidly as [A], so in order to avoid ambiguity when expressing the rate in terms of different components, it is customary to divide each change in concentration by the appropriate coefficient: \[\text{rate} = \dfrac{-\Delta[A]}{\Delta t}= \dfrac{-\Delta[B]}{3 \Delta t} = \dfrac{+\Delta[D]}{2\Delta t}\] Each of the above quotients is a legitimate expression of the rate of this particular reaction; they all yield the same number. Which one you employ when doing a calculation is largely a matter of convenience. Example \(\PageIndex{1}\): Oxidation of Ammonia For the oxidation of ammonia \[4 NH_3 + 3O_2 \rightarrow 2 N_2 + 6 H_2O \nonumber\] it was found that the rate of formation of N 2 was 0.27 mol L –1 s –1 . At what rate was water being formed? At what rate was ammonia being consumed? Solution: From the equation stoichiometry, Δ[H 2 O] = 6/2 Δ[N 2 ], so the rate of formation of H 2 O is 3 × (0.27 mol L –1 s –1 ) = 0.81 mol L –1 s –1 . 4 moles of NH 3 are consumed for every 2 moles of N 2 formed, so the rate of disappearance of ammonia is 2 × (0.27 mol L –1 s –1 ) = 0.54 mol L –1 s –1 . Comment : Because of the way this question is formulated, it would be acceptable to express this last value as a negative number. Instantaneous rates Most reactions slow down as the reactants are consumed. Consequently, the rates given by the expressions shown above tend to lose their meaning when measured over longer time intervals Δ t . Thus for the reaction whose progress is plotted here, the actual rate (as measured by the increasing concentration of product) varies continuously, being greatest at time zero. The instantaneous rate of a reaction is given by the slope of a tangent to the concentration-vs.-time curve. Three such rates have been identified in this plot. An instantaneous rate taken near the beginning of the reaction (t = 0) is known as an initial rate (label (1) here). As we shall soon see, initial rates play an important role in the study of reaction kinetics. If you have studied differential calculus, you will know that these tangent slopes are derivatives whose values can very at each point on the curve, so that these instantaneous rates are really limiting rates defined as \[rate = \lim_{\Delta t \rightarrow 0} \dfrac{-\Delta[A]}{\Delta t}\] However, if you does not know calculus, just bear in mind that the larger the time interval Δ t , the smaller will be the precision of the instantaneous rate. Instantaneous rates are also known as differential rates. Rate Laws and Reaction Order The relation between the rate of a reaction and the concentrations of reactants is expressed by its rate law . For example, the rate of the gas-phase decomposition of dinitrogen pentoxide \[2N_2O_5 → 4NO_2 + O_2\] has been found to be directly proportional to the concentration of \(N_2O_5\): \[\text{rate} = k [N_2O_5]\] Be very careful about confusing equilibrium constant expressions with those for rate laws. The expression for \(K_{eq}\) can always be written by inspecting the reaction equation, and it contains a term for each component (raised to the appropriate power) whose concentration changes during the reaction. For this reaction it is given by \[ K_{eq} = \dfrac{[NO_2]^2 [O_2]}{[N_2O_5]^2}\] In contrast, the expression for the rate law generally bears no necessary relation to the reaction equation, and must be determined experimentally . More generally, for a reaction of the form \[n_A A + n_B B + ... → products\] the rate law will be \[rate = [A]^a[B]^b ... \] , n B . Since the rate of a reaction has the dimensions of (concentration/time), the dimensions of the rate constant k will depend on the exponents of the concentration terms in the rate law. To make this work out properly, if we let \(p\) be the sum of the exponents of the concentration terms in the rate law \[p = a + b + ...\] then k will have the dimensions (concentration 1–p /time). Reaction order The order of a rate law is the sum of the exponents in its concentration terms. For the N 2 O 5 decomposition with the rate law k[ N 2 O 5 ], this exponent is 1 (and thus is not explicitly shown); this reaction is therefore a first order reaction. We can also say that the reaction is "first order in N 2 O 5 ". For more complicated rate laws, we can speak of the overall reaction order and also the orders with respect to each component. As an example, consider a reaction \[A + 3B + 2C → \text{products}\] whose experimental rate law is \[rate = k[A] [B]^2\] We would describe this reaction as third-order overall, first-order in A, second-order in B, and zero-order in C. Zero-order means that the rate is independent of the concentration of a particular reactant. However, enough C must be present to allow the equilibrium mixture to form. Example \(\PageIndex{2}\) The rate of oxidation of bromide ions by bromate in an acidic aqueous solution \[6H^+ + BrO_3^– + 5Br^– → 3 Br_2 + 3 H_2O \nonumber\] is found to follow the rate law \[rate = k[Br^–][BrO_3^–][H^+]^2\] What happens to the rate if, in separate experiments, [ BrO 3 – ] is doubled; the pH is increased by one unit; the solution is diluted to twice its volume, with the pH kept constant by use of a buffer? Solution Since the rate is first-order in bromate, doubling its concentration will double the reaction rate. Increasing the pH by one unit will decrease the [H + ] by a factor of 10. Since the reaction is second-order in [H + ], this will decrease the rate by a factor of 100. Dilution reduces the concentrations of both Br 2 and BrO 3 – to half their original values. Doing this to each concentration alone would reduce the rate by a factor of 2, so reducing both concentration will reduce the rate by a factor of 4, to (½)×(½) = ¼ of its initial value. How reaction orders are observed Observing rate-vs.-concentration proportionality In order to determine the value of the exponent in a rate equation term, we need to see how the rate varies with the concentration of the substance. For a single-reactant decomposition reaction of the form A → products in which the rate is – d [A]/ dt , we simply plot [A] as function of time, draw tangents at various intervals, and see how the slopes of these tangents (the instantaneous rates) depend on [A]. If doubling the concentration of A doubles the rate, then the reaction is first-order in A. If doubling the concentration results in a four-fold rate increase, the reaction is second-order in A. Example \(\PageIndex{3}\) Use the tabulated experimental data to determine the order of the reaction \[2 N_2O_5 → 4 NO_2 + O_2 \nonumber\] Time (min) p(N2O5) [N2O5] mol L-1 Rate (mol L-1 min-1) 0.0 301.6 0.0152 NaN 10.0 224.8 0.0113 3.4 × 10–4 20.0 166.7 0.0084 2.5 30.0 123.2 0.0062 1.8 40.0 92.2 0.0046 1.3 69.1 69.1 0.0035 1.0 Solution The ideal gas law can be used to convert the partial pressures of \(N_2O_5\) to molar concentrations. These are then plotted (left) to follow their decrease with time. The rates are computed from the slopes of the tangents (blue lines) and their values plotted as a function of \([N_2O_5]\) and \([N)2O)5]^2\). It is apparent that the rates are directly proportional to \([N)2O)5]^1\), indicating that this is a first-order reaction. Initial rate method When there is more than one reactant, the method described above is rarely practical, since the concentrations of the different reactants will generally fall at different rates, depending on the stoichiometry. Instead, we measure only the rate near the beginning of the reaction, before the concentrations have had time to change significantly. The experiment is then repeated with a different starting concentration of the reactant in question, but keeping the concentrations of any others the same. After the order with respect to one component is found, another series of trials is conducted in which the order of another component is found. Using the same dinitrogen pentoxide decomposition as in the previous example, we conduct a series of runs using five different initial concentrations of \(N_2O_5\). Note that we use only the tangents at the beginning of each plot — that is, at times t=0. We then plot the five initial rates of consumption of \(N_2O_5\) as a function of its molar concentration. As before, we see that these rates are directly proportional to \([N_2O_5]\). The slope of this plot gives the value of the rate constant. rate = (5.2 × 10 –3 ) [N2O5] mol L –1 s –1 Example \(\PageIndex{4}\) A study of the gas-phase reduction of nitric oxide by hydrogen \[2 NO + 2 H_2 → N_2 + 2 H_2O \nonumber\] yielded the following initial-rate data (all pressures in torr): experiment P(NO) P(H2) initial rate (torr s–1) 1 359 300 1.50 2 300 300 1.03 3 152 300 0.25 4 300 289 1.00 5 300 205 0.71 6 200 147 0.51 Find the order of the reaction with respect to each component. In looking over this data, take note of the following: The six runs recorded here fall into two groups, in which the initial pressures of H 2 and of NO, respectively, are held constant. All the data are expressed in pressures, rather than in concentrations. We can do this because the reactants are gases, whose concentrations are directly proportional to their partial pressures when T and V are held constant. And since we are only interested in comparing the ratios of pressures and rates, the units cancel out and does notmatter. It is far easier experimentally to adjust and measure pressures than concentrations. Solution Experiments 2 and 3 : Reduction of the initial partial pressure of NO by a factor of about 2 (300/152) results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide. Experiments 4 and 6 : Reducing the initial partial pressure of hydrogen by a factor of approximately 2 (289/147) causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen. The rate law is thus \[rate = k[NO]^2[H_2]\] Dealing with multiple reactants: the isolation method It is not always practical to determine orders of two or more reactants by the method illustrated in the preceding example. Fortunately, there is another way to accomplish the same task: we can use excess concentrations of all the reactants except the one we wish to investigate. For example, suppose the reaction is \[A + B + C → \text{products}\] and we need to find the order with respect to [B] in the rate law. If we set [B] o to 0.020 M and let [A] o = [C] o = 2.00M, then if the reaction goes to completion, the change in [A] and [C] will also be 0.020 M which is only 1 percent of their original values. This will often be smaller than the experimental error in determining the rates, so it can be neglected. By "flooding" the reaction mixture with one or more reactants, we are effectively isolating the one in which we are interested. |
Courses/Brevard_College/CHE_202%3A_Organic_Chemistry_II/03%3A_Amines_and_Amides/3.01%3A_Amines_-_Structures_and_Names | Learning Objectives Identify the general structure for an amine. Identify the functional group for amines. Determine the structural feature that classifies amines as primary, secondary, or tertiary. Use nomenclature systems to name amines. An amine is a derivative of ammonia in which one, two, or all three hydrogen atoms are replaced by hydrocarbon groups. Amines are classified according to the number of carbon atoms bonded directly to the nitrogen atom. A primary (1°) amine has one alkyl (or aryl) group on the nitrogen atom, a secondary (2°) amine has two, and a tertiary (3°) amine has three (Figure \(\PageIndex{1}\)). IMPORTANT: To classify alcohols, we look at the number of carbon atoms bonded to the carbon atom bearing the OH group, not the oxygen atom itself. Thus, although isopropylamine looks similar to isopropyl alcohol, the former is a primary amine, while the latter is a secondary alcohol. Summary An amine is a derivative of ammonia in which one, two, or all three hydrogen atoms are replaced by hydrocarbon groups. The amine functional group is as follows: Amines are classified as primary, secondary, or tertiary by the number of hydrocarbon groups attached to the nitrogen atom. |
Bookshelves/Analytical_Chemistry/Chemometrics_Using_R_(Harvey)/07%3A_Testing_the_Significance_of_Data/7.01%3A_Significance_Testing | Let’s consider the following problem. To determine if a medication is effective in lowering blood glucose concentrations, we collect two sets of blood samples from a patient. We collect one set of samples immediately before we administer the medication, and we collect the second set of samples several hours later. After we analyze the samples, we report their respective means and variances. How do we decide if the medication was successful in lowering the patient’s concentration of blood glucose? One way to answer this question is to construct a normal distribution curve for each sample, and to compare the two curves to each other. Three possible outcomes are shown in Figure \(\PageIndex{1}\). In Figure \(\PageIndex{1a}\), there is a complete separation of the two normal distribution curves, which suggests the two samples are significantly different from each other. In Figure \(\PageIndex{1b}\), the normal distribution curves for the two samples almost completely overlap each other, which suggests the difference between the samples is insignificant. Figure \(\PageIndex{1c}\), however, presents us with a dilemma. Although the means for the two samples seem different, the overlap of their normal distribution curves suggests that a significant number of possible outcomes could belong to either distribution. In this case the best we can do is to make a statement about the probability that the samples are significantly different from each other. The process by which we determine the probability that there is a significant difference between two samples is called significance testing or hypothesis testing. Before we discuss specific examples let's first establish a general approach to conducting and interpreting a significance test. Constructing a Significance Test The purpose of a significance test is to determine whether the difference between two or more results is sufficiently large that we are comfortable stating that the difference cannot be explained by indeterminate errors. The first step in constructing a significance test is to state the problem as a yes or no question, such as “Is this medication effective at lowering a patient’s blood glucose levels?” A null hypothesis and an alternative hypothesis define the two possible answers to our yes or no question. The null hypothesis, H 0 , is that indeterminate errors are sufficient to explain any differences between our results. The alternative hypothesis, H A , is that the differences in our results are too great to be explained by random error and that they must be determinate in nature. We test the null hypothesis, which we either retain or reject. If we reject the null hypothesis, then we must accept the alternative hypothesis and conclude that the difference is significant. Failing to reject a null hypothesis is not the same as accepting it. We retain a null hypothesis because we have insufficient evidence to prove it incorrect. It is impossible to prove that a null hypothesis is true. This is an important point and one that is easy to forget. To appreciate this point let’s use this data for the mass of 100 circulating United States pennies. Empty DataFrame Columns: [(Penny, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25), (Weight (g), 3.126, 3.140, 3.092, 3.095, 3.080, 3.065, 3.117, 3.034, 3.126, 3.057, 3.053, 3.099, 3.065, 3.059, 3.068, 3.060, 3.078, 3.125, 3.090, 3.100, 3.055, 3.105, 3.063, 3.083, 3.065), (Penny, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50), (Weight (g), 3.073, 3.084, 3.148, 3.047, 3.121, 3.116, 3.005, 3.115, 3.103, 3.086, 3.103, 3.049, 2.998, 3.063, 3.055, 3.181, 3.108, 3.114, 3.121, 3.105, 3.078, 3.147, 3.104, 3.146, 3.095), (Penny, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75), (Weight (g), 3.101, 3.049, 3.082, 3.142, 3.082, 3.066, 3.128, 3.112, 3.085, 3.086, 3.084, 3.104, 3.107, 3.093, 3.126, 3.138, 3.131, 3.120, 3.100, 3.099, 3.097, 3.091, 3.077, 3.178, 3.054), (Penny, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100), (Weight (g), 3.086, 3.123, 3.115, 3.055, 3.057, 3.097, 3.066, 3.113, 3.102, 3.033, 3.112, 3.103, 3.198, 3.103, 3.126, 3.111, 3.126, 3.052, 3.113, 3.085, 3.117, 3.142, 3.031, 3.083, 3.104)] Index: [] After looking at the data we might propose the following null and alternative hypotheses. H 0 : The mass of a circulating U.S. penny is between 2.900 g and 3.200 g H A : The mass of a circulating U.S. penny may be less than 2.900 g or more than 3.200 g To test the null hypothesis we find a penny and determine its mass. If the penny’s mass is 2.512 g then we can reject the null hypothesis and accept the alternative hypothesis. Suppose that the penny’s mass is 3.162 g. Although this result increases our confidence in the null hypothesis, it does not prove that the null hypothesis is correct because the next penny we sample might weigh less than 2.900 g or more than 3.200 g. After we state the null and the alternative hypotheses, the second step is to choose a confidence level for the analysis. The confidence level defines the probability that we will incorrectly reject the null hypothesis when it is, in fact, true. We can express this as our confidence that we are correct in rejecting the null hypothesis (e.g. 95%), or as the probability that we are incorrect in rejecting the null hypothesis. For the latter, the confidence level is given as \(\alpha\), where \[\alpha = 1 - \frac {\text{confidence interval (%)}} {100} \nonumber\] For a 95% confidence level, \(\alpha\) is 0.05. The third step is to calculate an appropriate test statistic and to compare it to a critical value. The test statistic’s critical value defines a breakpoint between values that lead us to reject or to retain the null hypothesis, which is the fourth, and final, step of a significance test. As we will see in the sections that follow, how we calculate the test statistic depends on what we are comparing. The four steps for a statistical analysis of data using a significance test: Pose a question, and state the null hypothesis, H 0 , and the alternative hypothesis, H A . Choose a confidence level for the statistical analysis. Calculate an appropriate test statistic and compare it to a critical value. Either retain the null hypothesis, or reject it and accept the alternative hypothesis. One-Tailed and Two-tailed Significance Tests Suppose we want to evaluate the accuracy of a new analytical method. We might use the method to analyze a Standard Reference Material that contains a known concentration of analyte, \(\mu\). We analyze the standard several times, obtaining a mean value, \(\overline{X}\), for the analyte’s concentration. Our null hypothesis is that there is no difference between \(\overline{X}\) and \(\mu\) \[H_0 \text{: } \overline{X} = \mu \nonumber\] If we conduct the significance test at \(\alpha = 0.05\), then we retain the null hypothesis if a 95% confidence interval around \(\overline{X}\) contains \(\mu\). If the alternative hypothesis is \[H_\text{A} \text{: } \overline{X} \neq \mu \nonumber\] then we reject the null hypothesis and accept the alternative hypothesis if \(\mu\) lies in the shaded areas at either end of the sample’s probability distribution curve (Figure \(\PageIndex{2a}\)). Each of the shaded areas accounts for 2.5% of the area under the probability distribution curve, for a total of 5%. This is a two-tailed significance test because we reject the null hypothesis for values of \(\mu\) at either extreme of the sample’s probability distribution curve. We can write the alternative hypothesis in two additional ways \[H_\text{A} \text{: } \overline{X} > \mu \nonumber\] \[H_\text{A} \text{: } \overline{X} < \mu \nonumber\] rejecting the null hypothesis if \(\mu\) falls within the shaded areas shown in Figure \(\PageIndex{2b}\) or Figure \(\PageIndex{2c}\), respectively. In each case the shaded area represents 5% of the area under the probability distribution curve. These are examples of a one-tailed significance test. For a fixed confidence level, a two-tailed significance test is the more conservative test because rejecting the null hypothesis requires a larger difference between the results we are comparing. In most situations we have no particular reason to expect that one result must be larger (or must be smaller) than the other result. This is the case, for example, when we evaluate the accuracy of a new analytical method. A two-tailed significance test, therefore, usually is the appropriate choice. We reserve a one-tailed significance test for a situation where we specifically are interested in whether one result is larger (or smaller) than the other result. For example, a one-tailed significance test is appropriate if we are evaluating a medication’s ability to lower blood glucose levels. In this case we are interested only in whether the glucose levels after we administer the medication are less than the glucose levels before we initiated treatment. If a patient’s blood glucose level is greater after we administer the medication, then we know the answer—the medication did not work—and we do not need to conduct a statistical analysis. Errors in Significance Testing Because a significance test relies on probability, its interpretation is subject to error. In a significance test, \(\alpha\) defines the probability of rejecting a null hypothesis that is true. When we conduct a significance test at \(\alpha = 0.05\), there is a 5% probability that we will incorrectly reject the null hypothesis. This is known as a type 1 error, and its risk is always equivalent to \(\alpha\). A type 1 error in a two-tailed or a one-tailed significance tests corresponds to the shaded areas under the probability distribution curves in Figure \(\PageIndex{2}\). A second type of error occurs when we retain a null hypothesis even though it is false. This is a type 2 error, and the probability of its occurrence is \(\beta\). Unfortunately, in most cases we cannot calculate or estimate the value for \(\beta\). The probability of a type 2 error, however, is inversely proportional to the probability of a type 1 error. Minimizing a type 1 error by decreasing \(\alpha\) increases the likelihood of a type 2 error. When we choose a value for \(\alpha\) we must compromise between these two types of error. Most of the examples in this text use a 95% confidence level (\(\alpha = 0.05\)) because this usually is a reasonable compromise between type 1 and type 2 errors for analytical work. It is not unusual, however, to use a more stringent (e.g. \(\alpha = 0.01\)) or a more lenient (e.g. \(\alpha = 0.10\)) confidence level when the situation calls for it. |
Courses/University_of_North_Carolina_Charlotte/CHEM_2141%3A__Survey_of_Physical_Chemistry/05%3A_Helmholtz_and_Gibbs_Energies/5.03%3A_Pressure_Dependence_of_Gibbs_Energy | The pressure and temperature dependence of \(G\) is also easy to describe. The best starting place is the definition of \(G\). \[G = U + pV -TS \label{eq1} \] Taking the total differential of \(G\) yields \[dG = dU + pdV – pdV + Vdp – TdS – SdT \nonumber \] 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. Example \(\PageIndex{1}\): Gold under Pressure The density of gold is 19.32 g/cm 3 . 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. Solution 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. \[ \begin{align} \Delta G & = \left( \dfrac{197.0\, g}{mol} \times \dfrac{1\,}{19.32\,g} \times \dfrac{1\,L}{1000\,cm^3} \right) (2.00 \,atm -1.00 \,atm) \underbrace{ \left(\dfrac{8.315 \,J}{0.08206\, atm\,L}\right)}_{\text{conversion unit}}\\ &= 1.033\,J \end{align} \nonumber \] Contributors Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay) |
Courses/Saint_Francis_University/CHEM_113%3A_Human_Chemistry_I_(Muino)/01%3A_Matter_and_Measurements/1.07%3A_Measuring_Mass_Length_and_Volume | Learning Objective Describe the units (and abbreviations) that go with various quantities. Derive new units by combining numerical prefixes with units. Mass vs. Weight One of the many interesting things about travel in outer space is the idea of weightlessness. If something is not fastened down, it will float in mid-air. Early astronauts learned that weightlessness had bad effects on bone structure. If there was no pressure on the legs, those bones would begin to lose mass. Weight provided by gravity is needed to maintain healthy bones. Specially designed equipment is now a part of every space mission so the astronauts can maintain good body fitness. Mass is a measure of the amount of matter that an object contains. The mass of an object is made in comparison to the standard mass of 1 kilogram. The kilogram was originally defined as the mass of \(1 \: \text{L}\) of liquid water at \(4^\text{o} \text{C}\) (volume of a liquid changes slightly with temperature). In the laboratory, mass is measured with a balance (see below), which must be calibrated with a standard mass so that its measurements are accurate. Other common units of mass are the gram and the milligram. A gram is 1/1000th of a kilogram, meaning that there are \(1000 \: \text{g}\) in \(1 \: \text{kg}\). A milligram is 1/1000th of a gram, so there are \(1000 \: \text{mg}\) in \(1 \: \text{g}\). The Difference Between Mass and Weight The mass of a body is a measure of its inertial property or how much matter it contains. The weight of a body is a measure of the force exerted on it by gravity or the force needed to support it. Gravity on earth gives a body a downward acceleration of about 9.8 m/s 2 . In common parlance, weight is often used as a synonym for mass in weights and measures. For instance, the verb “to weigh” means “to determine the mass of” or “to have a mass of.” The incorrect use of weight in place of mass should be phased out, and the term mass used when mass is meant. The SI unit of mass is the kilogram (kg). In science and technology, the weight of a body in a particular reference frame is defined as the force that gives the body an acceleration equal to the local acceleration of free fall in that reference frame. Thus, the SI unit of the quantity weight defined in this way (force) is the newton (N). Length Length is the measurement of the extent of something along its greatest dimension. The SI basic unit of length, or linear measure, is the meter \(\left( \text{m} \right)\). All measurements of length may be made in meters, though the prefixes listed in various tables will often be more convenient. The width of a room may be expressed as about 5 meters \(\left( \text{m} \right)\), whereas a large distance, such as the distance between New York City and Chicago, is better expressed as 1150 kilometers \(\left( \text{km} \right)\). Very small distances can be expressed in units such as the millimeter or the micrometer. The width of a typical human hair is about 10 micrometers \(\left( \mu \text{m} \right)\). Volume In addition to the fundamental units, SI also allows for derived units based on a fundamental unit or units. There are many derived units used in science. For example, the derived unit for area comes from the idea that area is defined as width times height. Because both width and height are lengths, they both have the fundamental unit of meter, so the unit of area is meter × meter, or meter 2 (m 2 ). This is sometimes spoken as "square meters." A unit with a prefix can also be used to derive a unit for area, so we can also have cm 2 , mm 2 , or km 2 as acceptable units for area. Volume is the amount of space occupied by a sample of matter. The volume of a regular object can be calculated by multiplying its length by its width and its height. Since each of those is a linear measurement, we say that units of volume are derived from units of length. One unit of volume is the cubic meter \(\left( \text{m}^3 \right)\), which is the volume occupied by a cube that measures \(1 \: \text{m}\) on each side. This very large volume is not very convenient for typical use in a chemistry laboratory. A liter \(\left( \text{L} \right)\) is the volume of a cube that measures \(10 \: \text{cm}\) \(\left( 1 \: \text{dm} \right)\) on each side. A liter is thus equal to both \(1000 \: \text{cm}^3\) \(\left( 10 \: \text{cm} \times 10 \: \text{cm} \times 10 \: \text{cm} \right)\) and to \(1 \: \text{dm}^3\). A smaller unit of volume that is commonly used is the milliliter (\(\text{mL}\) - note the capital \(\text{L}\) which is a standard practice). A milliliter is the volume of a cube that measures \(1 \: \text{cm}\) on each side. Therefore, a milliliter is equal to a cubic centimeter \(\left( \text{cm}^3 \right)\). There are \(1000 \: \text{mL}\) in \(1 \: \text{L}\), which is the same as saying that there are \(1000 \: \text{cm}^3\) in \(1 \: \text{dm}^3\). Another definition of a liter is one-tenth of a meter cubed. Because one-tenth of a meter is 10 cm, then a liter is equal to 1,000 cm 3 (Figure \(\PageIndex{3}\)). Because 1 L equals 1,000 mL, we conclude that 1 mL equals 1 cm 3 ; thus, these units are interchangeable. Units not only are multiplied together but also can be divided. For example, if you are traveling at one meter for every second of time elapsed, your velocity is 1 meter per second, or 1 m/s. The word per implies division, so velocity is determined by dividing a distance quantity by a time quantity. Other units for velocity include kilometers per hour (km/h) or even micrometers per nanosecond (μm/ns). Later, we will see other derived units that can be expressed as fractions. Key Takeaways Mass is a measure of the amount of matter that an object contains. Mass is independent of location. Weight is a measure of force that is equal to the gravitational pull on an object. Weight depends on location. Units can be multiplied and divided to generate new units for quantities like the liter for volume. |
Courses/Athabasca_University/Chemistry_360%3A_Organic_Chemistry_II/Chapter_20%3A_Carboxylic_Acids_and_Nitriles/20.01_Naming_Carboxylic_Acids_and_Nitriles | Objectives After completing this section, you should be able to write the IUPAC name of a carboxylic acid, given its Kekulé, condensed or shorthand structure. draw the condensed or shorthand structure of a carboxylic acid, given its IUPAC name. draw the structure of the following carboxylic acids, given their trivial names: formic acid and acetic acid. provide an acceptable name for a nitrile of given structure. draw the condensed or shorthand structure of a nitrile, give either its trivial or IUPAC name. Key Terms Make certain that you can define, and use in context, the key terms below. carboxylic acid nitrile Study Notes You need not memorize all the trivial names listed in the table, just remember the two names identified in Objective 3, above. Carboxylic Acids, RCO 2 H The IUPAC system of nomenclature assigns a characteristic suffix to these classes. The –e ending is removed from the name of the parent chain and is replaced -anoic acid . Since a carboxylic acid group must always lie at the end of a carbon chain, it is always is given the #1 location position in numbering and it is not necessary to include it in the name. Many carboxylic acids are called by the common names. These names were chosen by chemists to usually describe a source of where the compound is found. In common names of aldehydes, carbon atoms near the carboxyl group are often designated by Greek letters. The atom adjacent to the carbonyl function is alpha, the next removed is beta and so on. Formula Common Name Source IUPAC Name Melting Point Boiling Point HCO2H formic acid ants (L. formica) methanoic acid 8.4 ºC 101 ºC CH3CO2H acetic acid vinegar (L. acetum) ethanoic acid 16.6 ºC 118 ºC CH3CH2CO2H propionic acid milk (Gk. protus prion) propanoic acid -20.8 ºC 141 ºC CH3(CH2)2CO2H butyric acid butter (L. butyrum) butanoic acid -5.5 ºC 164 ºC CH3(CH2)3CO2H valeric acid valerian root pentanoic acid -34.5 ºC 186 ºC CH3(CH2)4CO2H caproic acid goats (L. caper) hexanoic acid -4.0 ºC 205 ºC CH3(CH2)5CO2H enanthic acid vines (Gk. oenanthe) heptanoic acid -7.5 ºC 223 ºC CH3(CH2)6CO2H caprylic acid goats (L. caper) octanoic acid 16.3 ºC 239 ºC CH3(CH2)7CO2H pelargonic acid pelargonium (an herb) nonanoic acid 12.0 ºC 253 ºC CH3(CH2)8CO2H capric acid goats (L. caper) decanoic acid 31.0 ºC 219 ºC Example (Common Names Are in Red) Naming carboxyl groups added to a ring When a carboxyl group is added to a ring the suffix -carboxylic acid is added to the name of the cyclic compound. The ring carbon attached to the carboxyl group is given the #1 location number. The acid with the carboxyl group attached directly to a benzene ring is called benzoic acid (C 6 H 5 COOH). Naming carboxylates Salts of carboxylic acids are named by writing the name of the cation followed by the name of the acid with the –ic acid ending replaced by an –ate ending. This is true for both the IUPAC and Common nomenclature systems. Naming carboxylic acids which contain other functional groups Carboxylic acids are given the highest nomenclature priority by the IUPAC system. This means that the carboxyl group is given the lowest possible location number and the appropriate nomenclature suffix is included. In the case of molecules containing carboxylic acid and alcohol functional groups the OH is named as a hydroxyl substituent. However, the l in hydroxyl is generally removed. In the case of molecules containing a carboxylic acid and aldehydes and/or ketones functional groups the carbonyl is named as a "Oxo" substituent. In the case of molecules containing a carboxylic acid an amine functional group the amine is named as an "amino" substituent. When carboxylic acids are included with an alkene the following order is followed: (Location number of the alkene)-(Prefix name for the longest carbon chain minus the -ane ending)-(an –enoic acid ending to indicate the presence of an alkene and carboxylic acid) Remember that the carboxylic acid has priority so it should get the lowest possible location number. Also, remember that cis/tran or E/Z nomenclature for the alkene needs to be included if necessary. Naming dicarboxylic acids For dicarboxylic acids the location numbers for both carboxyl groups are omitted because both functional groups are expected to occupy the ends of the parent chain. The ending –dioic acid is added to the end of the parent chain. Naming Nitriles Name the parent alkane (include the carbon atom of the nitrile as part of the parent) followed with the word -nitrile. The carbon in the nitrile is given the #1 location position. It is not necessary to include the location number in the name because it is assumed that the functional group will be on the end of the parent chain. Cycloalkanes are followed by the word -carbonitrile. The substituent name is cyano. 1-butanenitrile or 1-cyanopropane Try to name the following compounds using these conventions J J Try to draw structures for the following compounds butanedinitrile 2-methycyclohexanecarbonitrile Some common names that you should know are... acetonitrile benzonitrile Try to draw a structure for the following compound 2-methoxybenzonitrile |
Courses/Athabasca_University/Chemistry_360%3A_Organic_Chemistry_II/Chapter_24%3A_Amines_and_Heterocycles/24.01_Naming_Amines | Objectives After completing this section, you should be able to classify a given amine as being primary, secondary or tertiary. explain, briefly, the difference in meaning of the terms primary, secondary and tertiary when they are applied to the structures of amines and alcohols. determine whether a given structure represents a quaternary ammonium cation. provide an acceptable IUPAC name for an amine, given its Kekulé, condensed or shorthand structure. draw the structure of an amine, given its IUPAC name. give the name and structure of one typical heterocyclic amine (e.g., pyridine). Key Terms Make certain that you can define, and use in context, the key terms below. amine primary amine secondary amine quaternary ammonium cation tertiary amine Study Notes You should recognize that heterocyclic amines—compounds in which the nitrogen atom occurs as part of a ring—are very common in organic chemistry. Be prepared to meet such compounds throughout this, and subsequent chapters, but do not try to memorize all of the names and structures givenin the reading. You should, however, commit the structures of pyridine and pyrrole to memory. Amines are derivatives of ammonia in which one or more of the hydrogens has been replaced by an alkyl or aryl group. The nomenclature of amines is complicated by the fact that several different nomenclature systems exist, and there is no clear preference for one over the others. Furthermore, the terms primary (1º), secondary (2º) & tertiary (3º) are used to classify amines in a completely different manner than they were used for alcohols or alkyl halides. When applied to amines these terms refer to the number of alkyl (or aryl) substituents bonded to the nitrogen atom , whereas in other cases they refer to the nature of an alkyl group. The four compounds shown in the top row of the following diagram are all C 4 H 11 N isomers. The first two are classified as 1º-amines, since only one alkyl group is bonded to the nitrogen; however, the alkyl group is primary in the first example and tertiary in the second. The third and fourth compounds in the row are 2º and 3º-amines respectively. A nitrogen bonded to four alkyl groups will necessarily be positively charged, and is called a quaternary (4º)-ammonium cation. For example, (CH 3 ) 4 N (+) Br (–) is tetramethylammonium bromide. The IUPAC names are listed first and colored blue. This system names amine functions as substituents on the largest alkyl group. The simple -NH substituent found in 1º-amines is called an amino group . For 2º and 3º-amines a compound prefix (e.g. dimethylamino in the fourth example) includes the names of all but the root alkyl group. The Chemical Abstract Service has adopted a nomenclature system in which the suffix -amine is attached to the root alkyl name. For 1º-amines such as butanamine (first example) this is analogous to IUPAC alcohol nomenclature (-ol suffix). The additional nitrogen substituents in 2º and 3º-amines are designated by the prefix N- before the group name. These CA names are colored magenta in the diagram. Finally, a common system for simple amines names each alkyl substituent on nitrogen in alphabetical order, followed by the suffix -amine . These are the names given in the last row (colored black). Many aromatic and heterocyclic amines are known by unique common names, the origins of which are often unknown to the chemists that use them frequently. Since these names are not based on a rational system, it is necessary to memorize them. There is a systematic nomenclature of heterocyclic compounds, but it will not be discussed here. Exercises Exercise 24.1.1 Name the following compounds: a) b) c) d) e) f) Answer a) N-Methylpropylamine b) Dicyclopentylamine c) 1,4-Pentanediamine d) 4-Methylpyridine e) Triethylammonium Bromide f) N,N-Dimethylcyclohexylamine Exercise 24.1.2 Draw the structures corresponding to the following names: a) 3-Bromo-pentan-2-amine b) Cyclopentanamine c) Trans -3-ethylcyclohexanamine d) Sec -butyl- tert -butyl amine e) N,N-Dimethyl-3-pentanamine f) 4-Methyl-2-hexanamine g) 6-Bromo-4-amino-2-heptanol Answer a) b) c) d) e) f) g) Exercise 24.1.3 Draw the structures of the following heterocyclic compounds: a) 4-Methoxyindole b) 1,4-Dimethylpyrrole c) 3-(N,N-Dimethylamino)pyridine d) 2-Aminopyrimidine Answer a) b) c) d) |
Courses/Sacramento_City_College/SCC%3A_Chem_420_-_Organic_Chemistry_I/08%3A_Structure_and_Synthesis_of_Alkenes/8.02%3A_Physical_Properties_and_Important_Common_Names | Learning Objectives memorize the common names for vinylic and allylic groups including isoprene and styrene predict the relative physical properties of alkenes Common names The carbon atoms sharing the double bond can be referred to as the "vinyl carbons". The carbon atoms adjacent to the vinyl carbon atoms are called "allylic carbons". These carbon atoms have unique reactivity because of the potential for interaction with the pi bond. Overall, common names remove the -ane suffix and add -ylene. There are a couple of unique ones like ethenyl's common name is vinyl and 2-propenyl's common name is allyl that need to be memorized. vinyl substituent H 2 C=CH- allyl substituent H 2 C=CH-CH 2 - allene molecule H 2 C=C=CH 2 isoprene is shown below Physical Properties of Selected Alkenes Some representative alkenes—their names, structures, and physical properties—are given in the table below. IUPAC Name Molecular Formula Condensed Structural Formula Melting Point (°C) Boiling Point (°C) ethene C2H4 CH2=CH2 –169 –104 propene C3H6 CH2=CHCH3 –185 –47 1-butene C4H8 CH2=CHCH2CH3 –185 –6 1-pentene C5H10 CH2=CH(CH2)2CH3 –138 30 1-hexene C6H12 CH2=CH(CH2)3CH3 –140 63 1-heptene C7H14 CH2=CH(CH2)4CH3 –119 94 1-octene C8H16 CH2=CH(CH2)5CH3 –102 121 Polarity and Physical Properties Alkenes are non-polar hydrocarbons. The dominant intermolecular forces shared by alkenes are the London dispersion forces. These interactions are weak and temporary, so they are easily disrupted. Physical States: The physical states reflect the weak attractive forces between molecules. Ethene, propene, and butene exist as colorless gases. Alkenes with 5 to 14 carbons are liquids, and alkenes with 15 carbons or more are solids. Density: Alkenes are less dense than water with most densities in the range of 0.6 to 0.7 g/mL. Alkenes float on top of water. Solubility: Alkenes are virtually insoluble in water, but dissolve in organic solvents. The reasons for this are exactly the same as for the alkanes. Boiling Points: The boiling point of each alkene is very similar to that of the alkane with the same number of carbon atoms. Boiling points of alkenes depend on more molecular mass (chain length). The more intermolecular mass is added, the higher the boiling point. Intermolecular forces of alkenes gets stronger with increase in the size of the molecules. In each case, the alkene has a boiling point which is a small number of degrees lower than the corresponding alkane. The only attractions involved are Van der Waals dispersion forces, and these depend on the shape of the molecule and the number of electrons it contains. 0 1 Compound Boiling points (oC) Ethene -104 Propene -47 Trans-2-Butene 0.9 Cis-2-butene 3.7 Trans 1,2-dichlorobutene 155 Cis 1,2-dichlorobutene 152 1-Pentene 30 Trans-2-Pentene 36 Cis-2-Pentene 37 1-Heptene 115 3-Octene 122 3-Nonene 147 5-Decene 170 Melting Points: Melting points of alkenes depends on the packaging of the molecules so the stereochemistry of the carbon-carbon double bond has a strong influence on the relative melting points. Alkenes have similar melting points to that of alkanes, however, in cis isomers molecules are package in a U-bending shape, therefore, will display a lower melting points to that of the trans isomers. This effect is notable when comparing the melting points of fats and oils. The differences in the melting points is strongly influenced by the long hydrocarbon tails. Oils have a greater number of cis double bonds and exist as liquids at room temperature. Whereas, fats are primarily saturated and exist as solids at room temperature. 0 1 Compound Melting Points (0C) Ethene -169 Propene -185 Butene -138 1-Pentene -165 Trans-2-Pentene -135 Cis-2-Pentene -180 1-Heptene -119 3-Octene -101.9 3-Nonene -81.4 5-Decene -66.3 Exercise 1. Draw the bond-line structures for the following compounds in order of increasing boiling point: 2-methyl-2-pentene; 2-hexene; isoprene; 2-heptene. 2. Which phase will contain the most 3-octene? a) water or hexane b) water or benzene c) methanol or 1-octanol Answer 1. relative boiling points 2. a) hexane (Hydrocarbons are hydrophobic and lipophilic.) b) benzene (Hydrocarbons are hydrophobic and lipophilic.) c) 1-octanol (Hydrocarbons seek the solvent with the most carbons and fewest polar groups.) Contributors Trung Nguyen Jim Clark ( Chemguide.co.uk ) Layne A. Morsch (University of Illinois Springfield) |
Courses/Taft_College/CHEM_1510%3A_Introductory_College_Chemistry/12%3A_Chemical_Bonding/12.08%3A_Electronegativity_and_Polarity_-_Why_Oil_and_Water_Do_not_Mix | Learning Objectives Explain how polar compounds differ from nonpolar compounds. Determine if a molecule is polar or nonpolar. Given a pair of compounds, predict which would have a higher melting or boiling point. Bond Polarity The ability of an atom in a molecule to attract shared electrons is called electronegativity . When two atoms combine, the difference between their electronegativities is an indication of the type of bond that will form. If the difference between the electronegativities of the two atoms is small, neither atom can take the shared electrons completely away from the other atom, and the bond will be covalent. If the difference between the electronegativities is large, the more electronegative atom will take the bonding electrons completely away from the other atom (electron transfer will occur), and the bond will be ionic. This is why metals (low electronegativities) bonded with nonmetals (high electronegativities) typically produce ionic compounds. A bond may be so polar that an electron actually transfers from one atom to another, forming a true ionic bond. How do we judge the degree of polarity? Scientists have devised a scale called electronegativity , a scale for judging how much atoms of any element attract electrons. Electronegativity is a unitless number; the higher the number, the more an atom attracts electrons. A common scale for electronegativity is shown in Figure \(\PageIndex{1}\). The polarity of a covalent bond can be judged by determining the difference of the electronegativities of the two atoms involved in the covalent bond, as summarized in the following table: Electronegativity Difference Bond Type 0–0.4 pure covalent 0.5–2.0 polar covalent >2.0 likely ionic Nonpolar Covalent Bonds A bond in which the electronegativity difference is less than 1.9 is considered to be mostly covalent in character. However, at this point we need to distinguish between two general types of covalent bonds. A nonpolar covalent bond is a covalent bond in which the bonding electrons are shared equally between the two atoms. In a nonpolar covalent bond, the distribution of electrical charge is balanced between the two atoms. The two chlorine atoms share the pair of electrons in the single covalent bond equally, and the electron density surrounding the \(\ce{Cl_2}\) molecule is symmetrical. Also note that molecules in which the electronegativity difference is very small (<0.5) are also considered nonpolar covalent. An example would be a bond between chlorine and bromine (\(\Delta\)EN \(=3.0 - 2.8 = 0.2\)). Polar Covalent Bonds A bond in which the electronegativity difference between the atoms is between 0.5 and 2.0 is called a polar covalent bond. A polar covalent bond is a covalent bond in which the atoms have an unequal attraction for electrons and so the sharing is unequal. In a polar covalent bond, sometimes simply called a polar bond, the distribution of electrons around the molecule is no longer symmetrical. An easy way to illustrate the uneven electron distribution in a polar covalent bond is to use the Greek letter delta \(\left( \delta \right)\). The atom with the greater electronegativity acquires a partial negative charge, while the atom with the lesser electronegativity acquires a partial positive charge. The delta symbol is used to indicate that the quantity of charge is less than one. A crossed arrow can also be used to indicate the direction of greater electron density. Electronegativity differences in bonding using Pauling scale. Differences in electronegativity classify bonds as covalent, polar covalent, or ionic. Example \(\PageIndex{1}\): Bond Polarity What is the polarity of each bond? C–H O–H Solution Using Figure \(\PageIndex{1}\), we can calculate the difference of the electronegativities of the atoms involved in the bond. For the C–H bond, the difference in the electronegativities is 2.5 − 2.1 = 0.4. Thus we predict that this bond will be nonpolar covalent. For the O–H bond, the difference in electronegativities is 3.5 − 2.1 = 1.4, so we predict that this bond will be polar covalent. Exercise \(\PageIndex{1}\) What is the polarity of each bond? Rb–F P–Cl Answer a likely ionic Answer b polar covalent Molecular Polarity To determine if a molecule is polar or nonpolar, it is generally useful to look at Lewis structures. Nonpolar compounds will be symmetric, meaning all of the sides around the central atom are identical—bonded to the same element with no unshared pairs of electrons. Polar molecules are asymmetric, either containing lone pairs of electrons on a central atom or having atoms with different electronegativities bonded. This works pretty well, as long as you can visualize the molecular geometry. That's the hard part. To know how the bonds are oriented in space, you have to have a strong grasp of Lewis structures and VSEPR theory. Assuming that you do, you can look at the structure of each one and decide if it is polar or not, whether or not you know the individual atom's electronegativity . This is because you know that all bonds between dissimilar elements are polar, and in these particular examples, it doesn't matter which direction the dipole moment vectors are pointing (out or in). A polar molecule is a molecule in which one end of the molecule is slightly positive, while the other end is slightly negative. A diatomic molecule that consists of a polar covalent bond, such as \(\ce{HF}\), is a polar molecule. The two electrically charged regions on either end of the molecule are called poles, similar to a magnet having a north and a south pole. A molecule with two poles is called a dipole (see figure below). Hydrogen fluoride is a dipole. For molecules with more than two atoms, the molecular geometry must also be taken into account when determining if the molecule is polar or nonpolar. The figure below shows a comparison between carbon dioxide and water. Carbon dioxide \(\left( \ce{CO_2} \right)\) is a linear molecule. The oxygen atoms are more electronegative than the carbon atom, so there are two individual dipoles pointing outward from the \(\ce{C}\) atom to each \(\ce{O}\) atom. However, since the dipoles are of equal strength and are oriented this way, they cancel out and the overall molecular polarity of \(\ce{CO_2}\) is zero. Water is a bent molecule because of the two lone pairs on the central oxygen atom. The individual dipoles point from the \(\ce{H}\) atoms toward the \(\ce{O}\) atom. Because of the shape, the dipoles do not cancel each other out and the water molecule is polar. In the figure below, the net dipole is shown in blue and points upward. Some other molecules are shown in the figure below. Notice that a tetrahedral molecule such as \(\ce{CH_4}\) is nonpolar. However, if one of the peripheral \(\ce{H}\) atoms is replaced with another atom that has a different electronegativity, the molecule becomes polar. A trigonal planar molecule \(\left( \ce{BF_3} \right)\) may be nonpolar if all three peripheral atoms are the same, but a trigonal pyramidal molecule \(\left( \ce{NH_3} \right)\) is polar. To summarize, to be polar, a molecule must: Contain at least one polar covalent bond. Have a molecular structure such that the sum of the vectors of each bond dipole moment do not cancel. Steps to Identify Polar Molecules Draw the Lewis structure. Figure out the geometry (using VSEPR theory). Visualize or draw the geometry. Find the net dipole moment (you don't have to actually do calculations if you can visualize it). If the net dipole moment is zero, it is non-polar. Otherwise, it is polar. Properties of Polar Molecules Polar molecules tend to align when placed in an electric field with the positive end of the molecule oriented toward the negative plate and the negative end toward the positive plate (Figure \(\PageIndex{14}\)). We can use an electrically charged object to attract polar molecules, but nonpolar molecules are not attracted. Also, polar solvents are better at dissolving polar substances, and nonpolar solvents are better at dissolving nonpolar substances. While molecules can be described as "polar covalent" or "ionic", it must be noted that this is often a relative term, with one molecule simply being more polar or less polar than another. However, the following properties are typical of such molecules. Polar molecules tend to: have higher melting points than nonpolar molecules have higher boiling points than nonpolar molecules be more soluble in water (dissolve better) than nonpolar molecules have lower vapor pressures than nonpolar molecules Example \(\PageIndex{2}\): Label each of the following as polar or nonpolar. Water, H 2 O: Methanol, CH 3 OH: Hydrogen Cyanide, HCN: Oxygen, O 2 : Propane, C 3 H 8 : Solution Water is polar. Any molecule with lone pairs of electrons around the central atom is polar. Methanol is polar. This is not a symmetric molecule. The \(\ce{-OH}\) side is different from the other 3 \(\ce{-H}\) sides. Hydrogen cyanide is polar. The molecule is not symmetric. The nitrogen and hydrogen have different electronegativities, creating an uneven pull on the electrons. Oxygen is nonpolar. The molecule is symmetric. The two oxygen atoms pull on the electrons by exactly the same amount. Propane is nonpolar, because it is symmetric, with \(\ce{H}\) atoms bonded to every side around the central atoms and no unshared pairs of electrons. Exercise \(\PageIndex{2}\) Label each of the following as polar or nonpolar. \(\ce{SO3}\) \(\ce{NH3}\) Answer a nonpolar Answer b polar Contributions & Attributions StackExchange ( thomij ). |
Courses/University_of_North_Carolina_Charlotte/CHEM_2141%3A__Survey_of_Physical_Chemistry/08%3A_Optional-_Special_topics/8.02%3A_Aquatic_Chemistry | Water, a natural occurring and abundant substance that exists in solid, liquid, and gas forms on the planet Earth, has attracted the attention of artists, engineers, poets, writers, philosophers, environmentalists, scientists, and politicians. Every aspect of life involves water as food, as a medium in which to live, or as the essential ingredient of life. The food-science aspects of water range from agriculture, aquaculture, biology, biochemistry, cookery, microbiology, nutrition, photosynthesis, power generation, to zoology. Even in the narrow sense of food technology, water is intimately involved in the production, washing, preparation, manufacture, cooling, drying, and hydration of food. Water is eaten, absorbed, transported, and utilized by cells. Facts and data about water are abundant and diverse. This article can only selectively present some fundamental characteristics of water molecules and their collective properties for readers when they ponder food science at the molecular level. 8.2.1: 1. Acid-Base Chemistry of Natural Aquatic Systems 8.2.2: 2. Carbonate Equilibria in Natural Waters 8.2.3: 3. Redox Equilibria in Natural Waters 8.2.4: 4. Solids in Contact With Natural Waters 8.2.5: Fundamental Characteristics of Water The physics and chemistry of water is the backbone of engineering and sciences. The basic data for the properties of pure water, which are found in the CRC Handbook of Chemistry and Physics (1), are useful for food scientists. However, water is a universal solvent, and natural waters contain dissolved substances present in the environment. All solutes in the dilute solutions modify the water properties. 8.2.6: Natural Water Water is the most important resource. Without water life is not possible. From a chemical point of view, water, H2O, is a pure compound, but in reality, you seldom drink, see, touch or use pure water. Water from various sources contains dissolved gases, minerals, organic and inorganic substances. This photograph of Guilin shows the beauty of natural water. The rain curved an interesting landscape out of the lime stones in the area. Natural waters are often important parts of wonders of the world 8.2.7: Water Biology Since water supports life, living organisms also modify their environment, changing the nature of the water in which they live. Biology of water pollution, lists the syllabus on a course including a laboratory section. Water and biology interweave into an entangled maze waiting for explorers and curious minds. 8.2.8: Water Chemistry Water is an unusual compound with unique physical properties. As a result, its the compound of life. Yet, its the most abundant compound in the biosphere of Earth. These properties are related to its electronic structure, bonding, and chemistry. However, due to its affinity for a variety of substances, ordinary water contains other substances. Few of us has used, seen or tested pure water, based on which we discuss its chemistry. 8.2.9: Water Physics Chemical and physical properties of water are often discussed together. These properties are fundamentals of many disciplines such as hydrology, environmental studies, chemical engineering, environmental engineering, civil engineering etc. They are of interest to chemists and physicists of course. 8.2.10: Water Treatment Water treatment is a process of making water suitable for its application or returning its natural state. Thus, water treatment required before and after its application. The required treatment depends on the application. Water treatment involves science, engineering, business, and art. The treatment may include mechanical, physical, biological, and chemical methods. As with any technology, science is the foundation, and engineering makes sure that the technology works as designed. |
Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/01%3A_Structure_and_Bonding | Chapter Objectives This chapter provides a review of material covered in a standard freshman general-chemistry course through a discussion of the following topics: the differences between organic and inorganic chemistry. the shapes and significance of atomic orbitals. electron configurations. ionic and covalent bonding. molecular orbital theory. hybridization. the structure and geometry of the compounds methane, ethane, ethylene and acetylene. 1.0: Why This Chapter? The chapter introduces the fundamental concepts of structure and bonding in organic chemistry, emphasizing their importance for understanding molecular behavior and reactivity. It highlights how these concepts form the foundation for subsequent topics in the field, encouraging students to appreciate the significance of chemical structure in determining properties and functions of organic compounds. This foundation is essential for mastering organic synthesis and analysis. 1.1: Atomic Structure - The Nucleus Atoms are comprised of protons, neutrons and electrons. Protons and neutrons are found in the nucleus of the atom, while electrons are found in the electron cloud around the nucleus. The relative electrical charge of a proton is +1, a neutron has no charge, and an electron’s relative charge is -1. The number of protons in an atom’s nucleus is called the atomic number, Z. The mass number, A, is the sum of the number of protons and the number of neutrons in a nucleus. 1.2: Atomic Structure - Orbitals This section explains atomic structure and the role of orbitals in electron configuration. It describes the different types of atomic orbitals (s, p, d, f) and their shapes, energies, and how they influence chemical bonding. The concept of quantum numbers, which describe the properties of electrons in atoms, is also introduced. Understanding these fundamentals is crucial for grasping the principles of chemical bonding and reactivity in organic chemistry. 1.3: Atomic Structure - Electron Configurations The order in which electrons are placed in atomic orbitals is called the electron configuration and is governed by the aufbau principle. Electrons in the outermost shell of an atom are called valence electrons. The number of valence electrons in any atom is related to its position in the periodic table. Elements in the same periodic group have the same number of valence electrons. 1.4: Development of Chemical Bonding Theory Lewis Dot Symbols are a way of indicating the number of valence electrons in an atom. They are useful for predicting the number and types of covalent bonds within organic molecules. The molecular shape of molecules is predicted by Valence Shell Electron Pair Repulsion (VSEPR) theory. The shapes of common organic molecules are based on tetrahedral, trigonal planar or linear arrangements of electron groups. 1.5: Describing Chemical Bonds - Valence Bond Theory Covalent bonds form as valence electrons are shared between two atoms. Lewis Structures and structural formulas are common ways of showing the covalent bonding in organic molecules. Formal charge describes the changes in the number of valence electrons as an atom becomes bonded into a molecule. If the atom has a net loss of valence electrons it will have a positive formal charge. If the atom has a net gain of valence electrons it will have a negative formal charge. 1.6: sp³ Hybrid Orbitals and the Structure of Methane The four identical C-H single bonds in methane form as the result of sigma bond overlap between the sp3 hybrid orbitals of carbon and the s orbital of each hydrogen. 1.7: sp³ Hybrid Orbitals and the Structure of Ethane The C-C bond in ethane forms as the result of sigma bond overlap between a sp³ hybrid orbital on each carbon. and the s orbital of each hydrogen. The six identical C-H single bonds in form as the result of sigma bond overlap between the sp³ hybrid orbitals of carbon and the s orbital of each hydrogen. 1.8: sp² Hybrid Orbitals and the Structure of Ethylene The C=C bond in ethylene forms as the result of both a sigma bond overlap between a sp2 hybrid orbital on each carbon and a pi bond overlap of a p orbital on each carbon 1.9: sp Hybrid Orbitals and the Structure of Acetylene The carbon-carbon triple bond in acetylene forms as the result of one sigma bond overlap between a sp hybrid orbital on each carbon and two pi bond overlaps of p orbitals on each carbon. 1.10: Hybridization of Nitrogen, Oxygen, Phosphorus and Sulfur This section covers the hybridization of nitrogen, oxygen, phosphorus, and sulfur, explaining how these elements undergo hybridization to form covalent bonds. It discusses different hybridization types (sp, sp², sp³) and their geometries, linking these to molecular shapes. The concept is crucial for understanding the reactivity and bonding patterns of organic molecules. Key examples illustrate how hybridization influences molecular properties. 1.11: Describing Chemical Bonds - Molecular Orbital Theory Molecular Orbital theory (MO) is a more advanced bonding model than Valence Bond Theory, in which two atomic orbitals overlap to form two molecular orbitals – a bonding MO and an anti-bonding MO. 1.12: Drawing Chemical Structures Kekulé Formulas or structural formulas display the atoms of the molecule in the order they are bonded. Condensed structural formulas show the order of atoms like a structural formula but are written in a single line to save space. Skeleton formulas or Shorthand formulas or line-angle formulas are used to write carbon and hydrogen atoms more efficiently by replacing the letters with lines. Isomers have the same molecular formula, but different structural formulas 1.13: Chemistry Matters—Organic Foods- Risk versus Benefit Contrary to what you may hear in supermarkets or on television, all foods are organic—that is, complex mixtures of organic molecules. Even so, when applied to food, the word organic has come to mean an absence of synthetic chemicals, typically pesticides, antibiotics, and preservatives. How concerned should we be about traces of pesticides in the food we eat? Or toxins in the water we drink? Or pollutants in the air we breathe? 1.14: Key Terms 1.15: Summary The purpose of this chapter has been to get you up to speed—to review some ideas about atoms, bonds, and molecular geometry. As we’ve seen, organic chemistry is the study of carbon compounds. Although a division into organic and inorganic chemistry occurred historically, there is no scientific reason for the division. 1.16: Additional Problems |
Courses/College_of_the_Canyons/Chem_201%3A_General_Chemistry_I_OER/09%3A_Electronic_Structure_and_Periodic_Table/9.02%3A_Line_Spectra_and_Bohr_Model_of_Atoms | Learning Objectives Describe the Bohr model of the hydrogen atom Use the Rydberg equation to calculate energies of light emitted or absorbed by hydrogen atoms Line Spectra Another paradox within the classical electromagnetic theory that scientists in the late nineteenth century struggled with concerned the light emitted from atoms and molecules. When solids, liquids, or condensed gases are heated sufficiently, they radiate some of the excess energy as light. Photons produced in this manner have a range of energies, and thereby produce a continuous spectrum in which an unbroken series of wavelengths is present. Most of the light generated from stars (including our sun) is produced in this fashion. You can see all the visible wavelengths of light present in sunlight by using a prism to separate them. The sunlight also contains UV light (shorter wavelengths) and IR light (longer wavelengths) that can be detected using instruments but that are invisible to the human eye. Incandescent (glowing) solids such as tungsten filaments in incandescent lights also give off light that contains all wavelengths of visible light. These continuous spectra can often be approximated by blackbody radiation curves at some appropriate temperature. In contrast to continuous spectra, light can also occur as discrete or line spectra having very narrow line widths interspersed throughout the spectral regions. Exciting a gas at low partial pressure using an electrical current, or heating it, will produce line spectra. Fluorescent light bulbs and neon signs operate in this way (Figure \(\PageIndex{1}\)). Each element displays its own characteristic set of lines, as do molecules, although their spectra are generally much more complicated. Each emission line consists of a single wavelength of light, which implies that the light emitted by a gas consists of a set of discrete energies. For example, when an electric discharge passes through a tube containing hydrogen gas at low pressure, the H 2 molecules are broken apart into separate H atoms, and we see a blue-pink color. Passing the light through a prism produces a line spectrum, indicating that this light is composed of photons of four visible wavelengths, as shown in Figure \(\PageIndex{2}\). The origin of discrete spectra in atoms and molecules was extremely puzzling to scientists in the late nineteenth century, since according to classical electromagnetic theory, only continuous spectra should be observed. Even more puzzling, in 1885, Johann Balmer was able to derive an empirical equation that related the four visible wavelengths of light emitted by hydrogen atoms to whole integers. That equation is the following one, in which k is a constant: \[\dfrac{1}{λ}=k\left(\dfrac{1}{4}−\dfrac{1}{n^2}\right),\:n=3,\:4,\:5,\:6\] Other discrete lines for the hydrogen atom were found in the UV and IR regions. Johannes Rydberg generalized Balmer's work and developed an empirical formula that predicted all of hydrogen's emission lines, not just those restricted to the visible range, where, n 1 and n 2 are integers, n 1 < n 2 , and \(R_∞\) is the Rydberg constant (1.097 × 10 7 m −1 ). \[\dfrac{1}{λ}=R_∞\left(\dfrac{1}{n^2_1}−\dfrac{1}{n^2_2}\right)\] Even in the late nineteenth century, spectroscopy was a very precise science, and so the wavelengths of hydrogen were measured to very high accuracy, which implied that the Rydberg constant could be determined very precisely as well. That such a simple formula as the Rydberg formula could account for such precise measurements seemed astounding at the time, but it was the eventual explanation for emission spectra by Neils Bohr in 1913 that ultimately convinced scientists to abandon classical physics and spurred the development of modern quantum mechanics. Bohr Model of Atoms Following the work of Ernest Rutherford and his colleagues in the early twentieth century, the picture of atoms consisting of tiny dense nuclei surrounded by lighter and even tinier electrons continually moving about the nucleus was well established. This picture was called the planetary model, since it pictured the atom as a miniature “solar system” with the electrons orbiting the nucleus like planets orbiting the sun. The simplest atom is hydrogen, consisting of a single proton as the nucleus about which a single electron moves. The electrostatic force attracting the electron to the proton depends only on the distance between the two particles. \[ F_{gravity} = G \dfrac{ m_1 m_2}{r^2} \] with \(G\) is a gravitational constant \(m_1\) and \(m_2\) are the masses of particle 1 and 2, respectively \(r\) is the distance between the two particles The electrostatic force has the same form as the gravitational force between two mass particles except that the electrostatic force depends on the magnitudes of the charges on the particles (+1 for the proton and −1 for the electron) instead of the magnitudes of the particle masses that govern the gravitational force. \[ F_{electrostatic} = k \dfrac{ m_1 m_2}{r^2}\] with \(k\) is a constant \(m_1\) and \(m_2\) are the masses of particle 1 and 2, respectively \(r\) is the distance between the two particles Since forces can be derived from potentials, it is convenient to work with potentials instead, since they are forms of energy. The electrostatic potential is also called the Coulomb potential . Because the electrostatic potential has the same form as the gravitational potential, according to classical mechanics, the equations of motion should be similar, with the electron moving around the nucleus in circular or elliptical orbits (hence the label “planetary” model of the atom). Potentials of the form V ( r ) that depend only on the radial distance \(r\) are known as central potentials. Central potentials have spherical symmetry, and so rather than specifying the position of the electron in the usual Cartesian coordinates ( x , y , z ), it is more convenient to use polar spherical coordinates centered at the nucleus, consisting of a linear coordinate r and two angular coordinates, usually specified by the Greek letters theta (θ) and phi ( Φ ). These coordinates are similar to the ones used in GPS devices and most smart phones that track positions on our (nearly) spherical earth, with the two angular coordinates specified by the latitude and longitude, and the linear coordinate specified by sea-level elevation. Because of the spherical symmetry of central potentials, the energy and angular momentum of the classical hydrogen atom are constants, and the orbits are constrained to lie in a plane like the planets orbiting the sun. This classical mechanics description of the atom is incomplete, however, since an electron moving in an elliptical orbit would be accelerating (by changing direction) and, according to classical electromagnetism, it should continuously emit electromagnetic radiation. This loss in orbital energy should result in the electron’s orbit getting continually smaller until it spirals into the nucleus, implying that atoms are inherently unstable. In 1913, Niels Bohr attempted to resolve the atomic paradox by ignoring classical electromagnetism’s prediction that the orbiting electron in hydrogen would continuously emit light. Instead, he incorporated into the classical mechanics description of the atom Planck’s ideas of quantization and Einstein’s finding that light consists of photons whose energy is proportional to their frequency. Bohr assumed that the electron orbiting the nucleus would not normally emit any radiation (the stationary state hypothesis), but it would emit or absorb a photon if it moved to a different orbit. The energy absorbed or emitted would reflect differences in the orbital energies according to this equation: \[ |ΔE|=|E_f−E_i|=h u=\dfrac{hc}{\lambda} \label{6.3.1}\] In this equation, h is Planck’s constant and E i and E f are the initial and final orbital energies, respectively. The absolute value of the energy difference is used, since frequencies and wavelengths are always positive. Instead of allowing for continuous values for the angular momentum, energy, and orbit radius, Bohr assumed that only discrete values for these could occur (actually, quantizing any one of these would imply that the other two are also quantized). Bohr’s expression for the quantized energies is: \[E_n=−\dfrac{k}{n^2} \label{6.3.2}\] with \(n=1,2,3, ...\) In this expression, \(k\) is a constant comprising fundamental constants such as the electron mass and charge and Planck’s constant. Inserting the expression for the orbit energies into the equation for \(ΔE\) gives \[ \color{red} ΔE=k \left(\dfrac{1}{n^2_1}−\dfrac{1}{n_2^2}\right)=\dfrac{hc}{\lambda} \label{6.3.3}\] or \[ \dfrac{1}{\lambda}=\dfrac{k}{hc} \left(\dfrac{1}{n^2_1}−\dfrac{1}{n_2^2}\right) \label{6.3.4}\] The lowest few energy levels are shown in Figure \(\PageIndex{3}\). One of the fundamental laws of physics is that matter is most stable with the lowest possible energy. Thus, the electron in a hydrogen atom usually moves in the \(n = 1\) orbit, the orbit in which it has the lowest energy. When the electron is in this lowest energy orbit, the atom is said to be in its ground electronic state (or simply ground state). If the atom receives energy from an outside source, it is possible for the electron to move to an orbit with a higher \(n\) value and the atom is now in an excited electronic state (or simply an excited state) with a higher energy. When an electron transitions from an excited state (higher energy orbit) to a less excited state, or ground state, the difference in energy is emitted as a photon. Similarly, if a photon is absorbed by an atom, the energy of the photon moves an electron from a lower energy orbit up to a more excited one. We can relate the energy of electrons in atoms to what we learned previously about energy. The law of conservation of energy says that we can neither create nor destroy energy. Thus, if a certain amount of external energy is required to excite an electron from one energy level to another, that same amount of energy will be liberated when the electron returns to its initial state (Figure \(\PageIndex{4}\)). In effect, an atom can “store” energy by using it to promote an electron to a state with a higher energy and release it when the electron returns to a lower state. The energy can be released as one quantum of energy, as the electron returns to its ground state (say, from \(n = 5\) to \(n = 1\)), or it can be released as two or more smaller quanta as the electron falls to an intermediate state, then to the ground state (say, from \(n = 5\) to \(n = 4\), emitting one quantum, then to \(n = 1\), emitting a second quantum). Since Bohr’s model involved only a single electron, it could also be applied to the single electron ions He + , Li 2 + , Be 3 + , and so forth, which differ from hydrogen only in their nuclear charges, and so one-electron atoms and ions are collectively referred to as hydrogen-like or hydrogenic atoms. The energy expression for hydrogen-like atoms is a generalization of the hydrogen atom energy, in which \(Z\) is the nuclear charge (+1 for hydrogen, +2 for He, +3 for Li, and so on) and \(k\) has a value of \(2.179 \times 10^{–18}\; J\). \[ \color{red} E_n=−\dfrac{kZ^2}{n^2} \label{6.3.5}\] The sizes of the circular orbits for hydrogen-like atoms are given in terms of their radii by the following expression, in which \(a_o\) is a constant called the Bohr radius, with a value of \(5.292 \times 10^{−11}\; m\): \[ \color{red} r=\dfrac{n^2}{Z} a_0 \label{6.3.6}\] The equation also shows us that as the electron’s energy increases (as \(n\) increases), the electron is found at greater distances from the nucleus. This is implied by the inverse dependence on \(r\) in the Coulomb potential, since, as the electron moves away from the nucleus, the electrostatic attraction between it and the nucleus decreases, and it is held less tightly in the atom. Note that as \(n\) gets larger and the orbits get larger, their energies get closer to zero, and so the limits \(n⟶∞\) and \(r⟶∞\) imply that \(E = 0\) corresponds to the ionization limit where the electron is completely removed from the nucleus. Thus, for hydrogen in the ground state \(n = 1\), the ionization energy would be: \[ ΔE=E_{n⟶∞} −E_1=0+k=k \label{6.3.7}\] With three extremely puzzling paradoxes now solved (blackbody radiation, the photoelectric effect, and the hydrogen atom), and all involving Planck’s constant in a fundamental manner, it became clear to most physicists at that time that the classical theories that worked so well in the macroscopic world were fundamentally flawed and could not be extended down into the microscopic domain of atoms and molecules. Unfortunately, despite Bohr’s remarkable achievement in deriving a theoretical expression for the Rydberg constant, he was unable to extend his theory to the next simplest atom, He, which only has two electrons. Bohr’s model was severely flawed, since it was still based on the classical mechanics notion of precise orbits, a concept that was later found to be untenable in the microscopic domain, when a proper model of quantum mechanics was developed to supersede classical mechanics. Example \(\PageIndex{1}\): Calculating the Energy of an Electron in a Bohr Orbit Early researchers were very excited when they were able to predict the energy of an electron at a particular distance from the nucleus in a hydrogen atom. If a spark promotes the electron in a hydrogen atom into an orbit with \(n = 3\), what is the calculated energy, in joules, of the electron? Solution The energy of the electron is given by Equation \(\ref{6.3.5}\): \[ E=\dfrac{−kZ^2}{n^2}\] The atomic number, \(Z\), of hydrogen is 1; \(k = 2.179 \times 10^{–18}\; J\); and the electron is characterized by an n value of \(3\). Thus, \[E=\dfrac{−(2.179 \times 10^{−18}\;J)×(1)^2}{(3)^2}=−2.421 \times 10^{−19}\;J\] Exercise \(\PageIndex{1}\) The electron in Example \(\PageIndex{1}\) in the \(n=3\) state is promoted even further to an orbit with \(n = 6\). What is its new energy? Answer TBD Example \(\PageIndex{2}\): Calculating Electron Transitions in a One–electron System What is the energy (in joules) and the wavelength (in meters) of the line in the spectrum of hydrogen that represents the movement of an electron from Bohr orbit with n = 4 to the orbit with n = 6? In what part of the electromagnetic spectrum do we find this radiation? Solution In this case, the electron starts out with \(n = 4\), so \(n_1 = 4\). It comes to rest in the \(n = 6\) orbit, so \(n_2 = 6\). The difference in energy between the two states is given by this expression: \[ΔE=E_1−E_2=2.179 \times 10^{−18}\left(\dfrac{1}{n^2_1}−\dfrac{1}{n_2^2}\right)\] \[ΔE=2.179 \times 10^{−18} \left(\dfrac{1}{4^2}−\dfrac{1}{6^2}\right)\; J\] \[ΔE=2.179 \times 10^{−18} \left(\dfrac{1}{16}−\dfrac{1}{36}\right)\;J\] \[ΔE=7.566 \times 10^{−20}\;J\] This energy difference is positive , indicating a photon enters the system (is absorbed) to excite the electron from the n = 4 orbit up to the \(n = 6\) orbit. The wavelength of a photon with this energy is found by the expression \(E=hc \lambda\). Rearrangement gives: \[ \lambda=\dfrac{hc}{E}\] From the figure of electromagnetic radiation, we can see that this wavelength is found in the infrared portion of the electromagnetic spectrum. Exercise \(\PageIndex{2}\) What is the energy in joules and the wavelength in meters of the photon produced when an electron falls from the \(n = 5\) to the \(n = 3\) level in a \(He^+\) ion (\(Z = 2\) for \(He^+\))? Answer \(6.198 \times 10^{–19}\; J\) and \(3.205 \times 10^{−7}\; m\) Bohr’s model of the hydrogen atom provides insight into the behavior of matter at the microscopic level, but it is does not account for electron–electron interactions in atoms with more than one electron. It does introduce several important features of all models used to describe the distribution of electrons in an atom. These features include the following: The energies of electrons (energy levels) in an atom are quantized, described by quantum numbers: integer numbers having only specific allowed value and used to characterize the arrangement of electrons in an atom. An electron’s energy increases with increasing distance from the nucleus. The discrete energies (lines) in the spectra of the elements result from quantized electronic energies. Of these features, the most important is the postulate of quantized energy levels for an electron in an atom. As a consequence, the model laid the foundation for the quantum mechanical model of the atom. Bohr won a Nobel Prize in Physics for his contributions to our understanding of the structure of atoms and how that is related to line spectra emissions. The Bohr Atom: https://youtu.be/GuFQEOzFOgA Summary Electromagnetic radiation can be generated by exciting matter to higher energies, such as by heating it. The emitted light can be either continuous (incandescent sources like the sun) or discrete (from specific types of excited atoms). Continuous spectra often have distributions that can be approximated as blackbody radiation at some appropriate temperature. The line spectrum of hydrogen can be obtained by passing the light from an electrified tube of hydrogen gas through a prism. This line spectrum was simple enough that an empirical formula called the Rydberg formula could be derived from the spectrum. Three historically important paradoxes from the late 19th and early 20th centuries that could not be explained within the existing framework of classical mechanics and classical electromagnetism were the blackbody problem, the photoelectric effect, and the discrete spectra of atoms. The resolution of these paradoxes ultimately led to quantum theories that superseded the classical theories. Bohr incorporated Planck’s and Einstein’s quantization ideas into a model of the hydrogen atom that resolved the paradox of atom stability and discrete spectra. The Bohr model of the hydrogen atom explains the connection between the quantization of photons and the quantized emission from atoms. Bohr described the hydrogen atom in terms of an electron moving in a circular orbit about a nucleus. He postulated that the electron was restricted to certain orbits characterized by discrete energies. Transitions between these allowed orbits result in the absorption or emission of photons. When an electron moves from a higher-energy orbit to a more stable one, energy is emitted in the form of a photon. To move an electron from a stable orbit to a more excited one, a photon of energy must be absorbed. Using the Bohr model, we can calculate the energy of an electron and the radius of its orbit in any one-electron system. Glossary Bohr’s model of the hydrogen atom structural model in which an electron moves around the nucleus only in circular orbits, each with a specific allowed radius; the orbiting electron does not normally emit electromagnetic radiation, but does so when changing from one orbit to another. continuous spectrum electromagnetic radiation given off in an unbroken series of wavelengths (e.g., white light from the sun excited state state having an energy greater than the ground-state energy ground state state in which the electrons in an atom, ion, or molecule have the lowest energy possible quantum number integer number having only specific allowed values and used to characterize the arrangement of electrons in an atom line spectrum electromagnetic radiation emitted at discrete wavelengths by a specific atom (or atoms) in an excited state |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/01%3A_Modules/1.24%3A_Misc/1.14.07%3A_Boundary | The term ‘boundary’ is encountered in several contexts in thermodynamics. Indeed the definition becomes more complicated as we concentrate on the thermodynamic properties of systems. As a starting point, a boundary separates system and surroundings. A boundary is an infinitely thin surface separating system and surroundings such that the properties of system and surroundings change abruptly at the boundary. [1,2] In these terms a reaction vessel is part of the surroundings. We support this careful distinction by observing that if chemical reaction inside the system is exothermic, the liberated heat warms the reaction vessel. In this case, the boundary encloses the system. No molecules can either enter or leave the system. However heat is allowed to cross the boundary. Thus the whole universe is divided into system and surroundings [3], the only role of the boundary is to facilitate communication between system and surroundings. In these terms chemical substances, heat, and electric charge may cross a boundary between a system and surroundings. In some cases the container (e.g. reaction vessel) may be considered part of the system. In many cases [4,5] it is correct to do so and so the boundary is again a hypothetical surface separating ‘reaction solution + reaction vessel’ and the surroundings. In general terms, it is important to define the boundary for a given system. Another term for boundary is ‘envelope’ which indicates something which can be quite dynamic in terms of shape and volume rather than, for example, a glass vessel. Moreover the boundary may be selectively permeable to one or more chemical substances rather like the envelope of unit cells in living systems [6]. The term ‘boundary’ in the context of surface chemistry means a boundary phase (or, capillary phase) [7]. In such a phase there is a concentration gradient of one or more chemical substances across the boundary phase between system and surroundings. Indeed surface chemistry can be described as the chemistry of boundaries. In summary we repeat the point that in a thermodynamic analysis of experimental results, a first requirement is that the system, boundary and surroundings are carefully defined. For the most part we assume that the boundary is an infinitely thin envelope separating system and surroundings. Footnotes [1] K. S. Pitzer, Thermodynamics, McGraw-Hill, New York,1995, 3rd. edn., page 6. [2] G. N. Lewis and M. Randall, Thermodynamics, McGraw-Hill, New York, 1923, page 10. [3] D. H. Everett, Chemical Thermodynamics, Longmans, London, 1959, page 8. [4] M. L. McGlashan, Chemical Thermodynamics, Academic Press, London, 1979,page 1. [5] E. B. Smith, Basic Chemical Thermodynamics, Clarendon Press, Oxford, 3rd. edn.,1982, page 2. [6] S. E. Wood and R. Battino, Thermodynamics of Chemical Systems, Cambridge University Press, Cambridge, 1990, page 3. [7] J. N. Bronsted, Physical Chemistry, Heinemann, London, 1937, page 359. |
Courses/University_of_Arkansas_Little_Rock/Chem_1402%3A_General_Chemistry_1_(Belford)/Homework/10%3A_Gases/10.3%3A_Gas_Phase_Reactions | Gas Phase Reactions Exercise \(\PageIndex{1}\) 5.00 g of Mg is added to 50.0 mL of 0.800M HCl. A double displacement reaction occurs in a sealed container at 25.0 o C. What is the pressure of the hydrogen gas generated if the volume above the solution is 100.0 ml, and we ignore any pressure due to the evaporated water? Answer \(P_{H_{2}}=\frac{n_{H_{2}}RT}{V}=\frac{0.0200mol~H_{2}\left ( 0.08206\frac{L\cdot atm}{mol\cdot K} \right )298K}{0.100L}=4.89atm\) Exercise \(\PageIndex{2}\) Nitrogen dioxide is formed in a closed container at 90 o C and 1.00 atm when 1.50 g NO and 2.00 mole of O 2 are mixed. After the reaction is finished the pressure changes to 3.90 atm. What is the final temperature? Answer \(T_{2}=T_{1}\left ( \frac{n_{1}}{n_{2}} \right )\left ( \frac{P_{2}}{P_{1}} \right )=363.15K\left ( \frac{2.05mol}{2.025mol} \right )\left ( \frac{3.90atm}{1.00atm} \right )=1433.77K =1400K\) |
Courses/Lumen_Learning/Book%3A_US_History_II_(OS_Collection)_(Lumen)/19%3A_The_Challenges_of_the_Twenty-First_Century/19.4%3A_Video%3A_Terrorism%2C_War%2C_and_Bush | This video teaches you about the tumultuous 2000s in the United States of America, mainly the 2000s that coincide with the presidency of George W. Bush. From the controversial election in 2000, to the events of 9/11 and Bush’s prosecution of the War on Terror, the George W. Bush presidency was an eventful one. You will learn about Bush’s domestic policies like tax cutting, education reform, and the wars in Afghanistan and Iraq. The event that came to pass during Bush’s presidency are still very much affecting the United States and the world today. A YouTube element has been excluded from this version of the text. You can view it online here: http://pb.libretexts.org/2os/?p=340 All rights reserved content Terrorism, War, and Bush 43: Crash Course US History #46. Provided by : Crash Course. Located at : https://youtu.be/nlsnnhn3VWE?t=1s . License : All Rights Reserved . License Terms : Standard YouTube License |
Courses/Sacramento_City_College/SCC%3A_Chem_309_-_General_Organic_and_Biochemistry_(Bennett)/Text/18%3A_Metabolism/18.6%3A_Stage_III_of_Catabolism | Learning Objectives Describe the reactions of the citric acid cycle. Describe the function of the citric acid cycle and identify the products produced. Describe the role of the electron transport chain in energy metabolism. Describe the role of oxidative phosphorylation in energy metabolism. The acetyl group enters a cyclic sequence of reactions known collectively as the citric acid cycle (or Krebs cycle or tricarboxylic acid [TCA] cycle). The cyclical design of this complex series of reactions, which bring about the oxidation of the acetyl group of acetyl-CoA to carbon dioxide and water, was first proposed by Hans Krebs in 1937. (He was awarded the 1953 Nobel Prize in Physiology or Medicine.) Acetyl-CoA’s entrance into the citric acid cycle is the beginning of stage III of catabolism. The citric acid cycle produces adenosine triphosphate (ATP), reduced nicotinamide adenine dinucleotide (NADH), reduced flavin adenine dinucleotide (FADH 2 ), and metabolic intermediates for the synthesis of needed compounds. Steps of the Citric Acid Cycle At first glance, the citric acid cycle appears rather complex (Figure \(\PageIndex{1}\)). All the reactions, however, are familiar types in organic chemistry: hydration, oxidation, decarboxylation, and hydrolysis. Each reaction of the citric acid cycle is numbered, and in Figure \(\PageIndex{1}\), the two acetyl carbon atoms are highlighted in red. Each intermediate in the cycle is a carboxylic acid, existing as an anion at physiological pH. All the reactions occur within the mitochondria, which are small organelles within the cells of plants and animals. In the first step, acetyl-CoA enters the citric acid cycle, and the acetyl group is transferred onto oxaloacetate, yielding citrate. Note that this step releases coenzyme A. The reaction is catalyzed by citrate synthase . In the next step, aconitase catalyzes the isomerization of citrate to isocitrate. In this reaction, a tertiary alcohol, which cannot be oxidized, is converted to a secondary alcohol, which can be oxidized in the next step. Isocitrate then undergoes a reaction known as oxidative decarboxylation because the alcohol is oxidized and the molecule is shortened by one carbon atom with the release of carbon dioxide (decarboxylation). The reaction is catalyzed by isocitrate dehydrogenase , and the product of the reaction is α-ketoglutarate. An important reaction linked to this is the reduction of the coenzyme nicotinamide adenine dinucleotide (NAD + ) to NADH. The NADH is ultimately reoxidized, and the energy released is used in the synthesis of ATP, as we shall see. The fourth step is another oxidative decarboxylation. This time α-ketoglutarate is converted to succinyl-CoA, and another molecule of NAD + is reduced to NADH. The α-ketoglutarate dehydrogenase complex catalyzes this reaction. This is the only irreversible reaction in the citric acid cycle. As such, it prevents the cycle from operating in the reverse direction, in which acetyl-CoA would be synthesized from carbon dioxide. So far, in the first four steps, two carbon atoms have entered the cycle as an acetyl group, and two carbon atoms have been released as molecules of carbon dioxide. The remaining reactions of the citric acid cycle use the four carbon atoms of the succinyl group to resynthesize a molecule of oxaloacetate, which is the compound needed to combine with an incoming acetyl group and begin another round of the cycle. In the fifth reaction, the energy released by the hydrolysis of the high-energy thioester bond of succinyl-CoA is used to form guanosine triphosphate (GTP) from guanosine diphosphate (GDP) and inorganic phosphate in a reaction catalyzed by succinyl-CoA synthetase . This step is the only reaction in the citric acid cycle that directly forms a high-energy phosphate compound. GTP can readily transfer its terminal phosphate group to adenosine diphosphate (ADP) to generate ATP in the presence of nucleoside diphosphokinase . Succinate dehydrogenase then catalyzes the removal of two hydrogen atoms from succinate, forming fumarate. This oxidation-reduction reaction uses flavin adenine dinucleotide (FAD), rather than NAD + , as the oxidizing agent. Succinate dehydrogenase is the only enzyme of the citric acid cycle located within the inner mitochondrial membrane. We will see soon the importance of this. In the following step, a molecule of water is added to the double bond of fumarate to form L-malate in a reaction catalyzed by fumarase . One revolution of the cycle is completed with the oxidation of L-malate to oxaloacetate, brought about by malate dehydrogenase . This is the third oxidation-reduction reaction that uses NAD + as the oxidizing agent. Oxaloacetate can accept an acetyl group from acetyl-CoA, allowing the cycle to begin again. Video: "The Citric Acid Cycle: An Overview". In the matrix of the mitochondrion, the Citric Acid Cycle uses Acetyl CoA molecules to produce energy through eight chemical reactions. This animation provides an overview of the pathway and its products. NDSU VCell Production's animation; for more information please see Vcell, NDSU, Animations(opens in new window) [vcell.ndsu.edu] . Cellular Respiration Respiration can be defined as the process by which cells oxidize organic molecules in the presence of gaseous oxygen to produce carbon dioxide, water, and energy in the form of ATP. We have seen that two carbon atoms enter the citric acid cycle from acetyl-CoA (step 1), and two different carbon atoms exit the cycle as carbon dioxide (steps 3 and 4). Yet nowhere in our discussion of the citric acid cycle have we indicated how oxygen is used. Recall, however, that in the four oxidation-reduction steps occurring in the citric acid cycle, the coenzyme NAD + or FAD is reduced to NADH or FADH 2 , respectively. Oxygen is needed to reoxidize these coenzymes . Recall, too, that very little ATP is obtained directly from the citric acid cycle. Instead, oxygen participation and significant ATP production occur subsequent to the citric acid cycle, in two pathways that are closely linked: electron transport and oxidative phosphorylation. All the enzymes and coenzymes for the citric acid cycle, the reoxidation of NADH and FADH 2 , and the production of ATP are located in the mitochondria, which are small, oval organelles with double membranes, often referred to as the “power plants” of the cell (Figure \(\PageIndex{2}\)). A cell may contain 100–5,000 mitochondria, depending on its function, and the mitochondria can reproduce themselves if the energy requirements of the cell increase. Cellular respiration occurs in the mitochondria Figure \(\PageIndex{2}\) shows the mitochondrion’s two membranes: outer and inner . The inner membrane is extensively folded into a series of internal ridges called cristae . Thus there are two compartments in mitochondria: the intermembrane space , which lies between the membranes, and the matrix , which lies inside the inner membrane. The outer membrane is permeable, whereas the inner membrane is impermeable to most molecules and ions, although water, oxygen, and carbon dioxide can freely penetrate both membranes. The matrix contains all the enzymes of the citric acid cycle with the exception of succinate dehydrogenase, which is embedded in the inner membrane. The enzymes that are needed for the reoxidation of NADH and FADH 2 and ATP production are also located in the inner membrane. They are arranged in specific positions so that they function in a manner analogous to a bucket brigade. This highly organized sequence of oxidation-reduction enzymes is known as the electron transport chain (or respiratory chain). Electron Transport Figure \(\PageIndex{3}\) illustrates the organization of the electron transport chain. The components of the chain are organized into four complexes designated I, II, III, and IV. Each complex contains several enzymes, other proteins, and metal ions. The metal ions can be reduced and then oxidized repeatedly as electrons are passed from one component to the next. Recall that a compound is reduced when it gains electrons or hydrogen atoms and is oxidized when it loses electrons or hydrogen atoms. Electrons can enter the electron transport chain through either complex I or II. We will look first at electrons entering at complex I. These electrons come from NADH, which is formed in three reactions of the citric acid cycle. Let’s use step 8 as an example, the reaction in which L-malate is oxidized to oxaloacetate and NAD + is reduced to NADH. This reaction can be divided into two half reactions: Oxidation half-reaction : Reduction half-reaction : In the oxidation half-reaction, two hydrogen (H + ) ions and two electrons are removed from the substrate. In the reduction half-reaction, the NAD + molecule accepts both of those electrons and one of the H + ions. The other H + ion is transported from the matrix, across the inner mitochondrial membrane, and into the intermembrane space. The NADH diffuses through the matrix and is bound by complex I of the electron transport chain. In the complex, the coenzyme flavin mononucleotide (FMN) accepts both electrons from NADH. By passing the electrons along, NADH is oxidized back to NAD + and FMN is reduced to FMNH 2 (reduced form of flavin mononucleotide). Again, the reaction can be illustrated by dividing it into its respective half-reactions. Oxidation half-reaction : Reduction half-reaction : Complex I contains several proteins that have iron-sulfur (Fe·S) centers. The electrons that reduced FMN to FMNH 2 are now transferred to these proteins. The iron ions in the Fe·S centers are in the Fe(III) form at first, but by accepting an electron, each ion is reduced to the Fe(II) form. Because each Fe·S center can transfer only one electron, two centers are needed to accept the two electrons that will regenerate FMN. Oxidation half-reaction : \[\ce{FMNH_2 -> FMN + 2H+ + 2e-} \nonumber \] Reduction half-reaction : \[\ce{2Fe(III) \cdot S + 2e- -> 2Fe(II) \cdot S} \nonumber \] Electrons from FADH 2 , formed in step 6 of the citric acid cycle, enter the electron transport chain through complex II. Succinate dehydrogenase, the enzyme in the citric acid cycle that catalyzes the formation of FADH 2 from FAD is part of complex II. The electrons from FADH 2 are then transferred to an Fe·S protein. Oxidation half-reaction : \[\ce{FADH_2 -> FAD + 2H+ + 2e-} \nonumber \] Reduction half-reaction : \[\ce{2Fe(III) \cdot S + 2e- → 2Fe(II) \cdot S} \nonumber \] Electrons from complexes I and II are then transferred from the \(\ce{Fe \cdot S}\) protein to coenzyme Q (CoQ), a mobile electron carrier that acts as the electron shuttle between complexes I or II and complex III. Coenzyme Q is also called ubiquinone because it is ubiquitous in living systems. Oxidation half-reaction : \[\ce{2Fe(II) \cdot S -> 2Fe(III) \cdot S + 2e-} \nonumber \] Reduction half-reaction : Complexes III and IV include several iron-containing proteins known as cytochromes. The iron in these enzymes is located in substructures known as iron porphyrins (Figure \(\PageIndex{4}\)). Like the Fe·S centers, the characteristic feature of the cytochromes is the ability of their iron atoms to exist as either Fe(II) or Fe(III). Thus, each cytochrome in its oxidized form—Fe(III)—can accept one electron and be reduced to the Fe(II) form. This change in oxidation state is reversible, so the reduced form can donate its electron to the next cytochrome, and so on. Complex III contains cytochromes b and c, as well as Fe·S proteins, with cytochrome c acting as the electron shuttle between complex III and IV. Complex IV contains cytochromes a and a 3 in an enzyme known as cytochrome oxidase . This enzyme has the ability to transfer electrons to molecular oxygen, the last electron acceptor in the chain of electron transport reactions. In this final step, water (H 2 O) is formed. Oxidation half-reaction : \[\ce{4Cyt\, a_3–Fe(II) -> 4Cyt\, a_3–Fe(III) + 4e-} \nonumber \] Reduction half-reaction : \[\ce{O2 + 4H+ + 4e- -> 2H2O} \nonumber \] Video: Cellular Respiration (Electron Transport Chain). Cellular respiration occurs in the mitochondria and provides both animals and plants with the energy needed to power other cellular processes. This section covers the electron transport chain.NDSU Virtual Cell Animations Project animation; ror more information please see Vcell, NDSU, Animations(opens in new window) [vcell.ndsu.edu] Oxidative Phosphorylation Each intermediate compound in the electron transport chain is reduced by the addition of one or two electrons in one reaction and then subsequently restored to its original form by delivering the electron(s) to the next compound along the chain. The successive electron transfers result in energy production. But how is this energy used for the synthesis of ATP? The process that links ATP synthesis to the operation of the electron transport chain is referred to as oxidative phosphorylation. Electron transport is tightly coupled to oxidative phosphorylation. The coenzymes NADH and FADH 2 are oxidized by the respiratory chain only if ADP is simultaneously phosphorylated to ATP. The currently accepted model explaining how these two processes are linked is known as the chemiosmotic hypothesis , which was proposed by Peter Mitchell, resulting in Mitchell being awarded the 1978 Nobel Prize in Chemistry. Looking again at Figure \(\PageIndex{3}\), we see that as electrons are being transferred through the electron transport chain, hydrogen (H + ) ions are being transported across the inner mitochondrial membrane from the matrix to the intermembrane space. The concentration of H + is already higher in the intermembrane space than in the matrix, so energy is required to transport the additional H + there. This energy comes from the electron transfer reactions in the electron transport chain. But how does the extreme difference in H + concentration then lead to ATP synthesis? The buildup of H + ions in the intermembrane space results in an H + ion gradient that is a large energy source, like water behind a dam (because, given the opportunity, the protons will flow out of the intermembrane space and into the less concentrated matrix). Current research indicates that the flow of H + down this concentration gradient through a fifth enzyme complex, known as ATP synthase, leads to a change in the structure of the synthase, causing the synthesis and release of ATP. In cells that are using energy, the turnover of ATP is very high, so these cells contain high levels of ADP. They must therefore consume large quantities of oxygen continuously, so as to have the energy necessary to phosphorylate ADP to form ATP. Consider, for example, that resting skeletal muscles use about 30% of a resting adult’s oxygen consumption, but when the same muscles are working strenuously, they account for almost 90% of the total oxygen consumption of the organism. Experiment has shown that 2.5–3 ATP molecules are formed for every molecule of NADH oxidized in the electron transport chain, and 1.5–2 ATP molecules are formed for every molecule of FADH 2 oxidized. Table \(\PageIndex{1}\) summarizes the theoretical maximum yield of ATP produced by the complete oxidation of 1 mol of acetyl-CoA through the sequential action of the citric acid cycle, the electron transport chain, and oxidative phosphorylation. Reaction Comments Yield of ATP (moles) Isocitrate → α-ketoglutarate + CO2 produces 1 mol NADH NaN α-ketoglutarate → succinyl-CoA + CO2 produces 1 mol NADH NaN Succinyl-CoA → succinate produces 1 mol GTP 1.0 Succinate → fumarate produces 1 mol FADH2 NaN Malate → oxaloacetate produces 1 mol NADH NaN 1 FADH2 from the citric acid cycle yields 2 mol ATP 2.0 3 NADH from the citric acid cycle yields 3 mol ATP/NADH 9.0 Net yield of ATP: Net yield of ATP: 12.0 Key Takeaways The acetyl group of acetyl-CoA enters the citric acid cycle. For each acetyl-CoA that enters the citric acid cycle, 2 molecules of carbon dioxide, 3 molecules of NADH, 1 molecule of ATP, and 1 molecule of FADH 2 are produced. The reduced coenzymes (NADH and FADH 2 ) produced by the citric acid cycle are reoxidized by the reactions of the electron transport chain. This series of reactions also produces a pH gradient across the inner mitochondrial membrane. The pH gradient produced by the electron transport chain drives the synthesis of ATP from ADP. For each NADH reoxidized, 2.5–3 molecules of ATP are produced; for each FADH 2 reoxidized, 1.5–2 molecules of ATP are produced. |
Courses/Chabot_College/Chem_12A%3A_Organic_Chemistry_Fall_2022/06%3A_Stereoisomerism/6.06%3A_Isomerism_Summary_Diagram | Learning Objective distinguish and discern the structural and chemical relationships between isomeric compounds The various types of isomers have been introduced and explored over several chapters. It can be helpful to review, compare, and contrast all of the forms of isomerism to build our skills of discernment. A brief review of each type of isomerism follows the summary diagram. See the respective chapter for a complete explanation. Conformational Isomers The rotation of C–C single bonds both carbon chains creates conformers (the same compound shown in different rotations). Consequently, many different arrangements of the atoms are possible, each corresponding to different degrees of rotation. Differences in three-dimensional structure resulting from rotation about a σ bond are called differences in conformation, and each different arrangement is called a conformational isomer (or conformer). While complete rotation of C-C single bonds is not possible in rings. The freedom of bond movement does allow the rings to assume different conformations, such as the chair and boat for 6-membered rings. Structural (Constitutional) Isomers Unlike conformational isomers, structural isomers differ in connectivity, as illustrated below for 1-propanol and 2-propanol. Although these two alcohols have the same molecular formula C 3 H 8 O, the position of the –OH group differs creating a unique compounds with differences in their physical and chemical properties. Consider, for example, the following five structures represented by the formula C 5 H 12 . In the conversion of one structural isomer to another, at least one bond must be broken and reformed at a different position in the molecule. Structures (a) and (d) above represent the same compound, n-pentane. Structures (b) and (c) represent the same compound, 2-methylbutane. No bonds need to be broken and reformed to convert between (a) and (d) or between (b) and (c). The molecules are simply rotated 180° about a vertical axis. Structure (e) is named 2,2-dimethylpropane. There are only three structural isomers possible with the chemical formula C 5 H 12 : n-pentane, 2-methylbutane, and 2,2-dimethylpropane. Structural isomers have distinct physical and chemical properties. Stereoisomers Enantiomers are pairs of compounds that are non-superimposable images. When there are two or more chiral centers in a compounds, the diatereomers can exist. Diastereomers are stereoisomers that are NOT enantiomers. Enatiomers share all physical properties except for their interaction with plane polarized light. Diastereomers have different physical properties (melting points and boiling points and densities). Exercise 1. What kind of isomers are the following pairs? Note: It can be difficult to answer this question directly from the names. It can be helpful to draw the structures. (R)-5-chlorohexene and 6-chlorohexene (2R,3R)-dibromohexane and (2R,3S)-dibromohexane Answer 1. a. Structural Isomers b. Diastereomers |
Courses/University_of_Kansas/KU%3A_CHEM_110_GOB_Chemistry_(Sharpe_Elles)/12%3A_Organic_Chemistry_-_Alkanes/12.06%3A_Physical_Properties_of_Alkanes | Skills to Develop To identify the physical properties of alkanes and describe trends in these properties. Because alkanes have relatively predictable physical properties and undergo relatively few chemical reactions other than combustion, they serve as a basis of comparison for the properties of many other organic compound families. Let’s consider their physical properties first. Table \(\PageIndex{1}\) describes some of the properties of some of the first 10 straight-chain alkanes. Because alkane molecules are nonpolar, they are insoluble in water, which is a polar solvent, but are soluble in nonpolar and slightly polar solvents. Consequently, alkanes themselves are commonly used as solvents for organic substances of low polarity, such as fats, oils, and waxes. Nearly all alkanes have densities less than 1.0 g/mL and are therefore less dense than water (the density of H 2 O is 1.00 g/mL at 20°C). These properties explain why oil and grease do not mix with water but rather float on its surface. Molecular Name Formula Melting Point (°C) Boiling Point (°C) Density (20°C)* Physical State (at 20°C) methane CH4 –182 –164 0.668 g/L gas ethane C2H6 –183 –89 1.265 g/L gas propane C3H8 –190 –42 1.867 g/L gas butane C4H10 –138 –1 2.493 g/L gas pentane C5H12 –130 36 0.626 g/mL liquid hexane C6H14 –95 69 0.659 g/mL liquid octane C8H18 –57 125 0.703 g/mL liquid decane C10H22 –30 174 0.730 g mL liquid *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. *Note the change in units going from gases (grams per liter) to liquids (grams per milliliter). Gas densities are at 1 atm pressure. Figure \(\PageIndex{1}\) : Oil Spills. Crude oil coats the water’s surface in the Gulf of Mexico after the Deepwater Horizon oil rig sank following an explosion. The leak was a mile below the surface, making it difficult to estimate the size of the spill. One liter of oil can create a slick 2.5 hectares (6.3 acres) in size. This and similar spills provide a reminder that hydrocarbons and water don’t mix. Source: Photo courtesy of NASA Goddard / MODIS Rapid Response Team, http://www.nasa.gov/topics/earth/features/oilspill/oil-20100519a.html . Looking Closer: Gas Densities and Fire Hazards Table \(\PageIndex{1}\) indicates that the first four members of the alkane series are gases at ordinary temperatures. Natural gas is composed chiefly of methane, which has a density of about 0.67 g/L. The density of air is about 1.29 g/L. Because natural gas is less dense than air, it rises. When a natural-gas leak is detected and shut off in a room, the gas can be removed by opening an upper window. On the other hand, bottled gas can be either propane (density 1.88 g/L) or butanes (a mixture of butane and isobutane; density about 2.5 g/L). Both are much heavier than air (density 1.2 g/L). If bottled gas escapes into a building, it collects near the floor. This presents a much more serious fire hazard than a natural-gas leak because it is more difficult to rid the room of the heavier gas. Also shown in Table \(\PageIndex{1}\) are the boiling points of the straight-chain alkanes increase with increasing molar mass. This general rule holds true for the straight-chain homologs of all organic compound families. Larger molecules have greater surface areas and consequently interact more strongly; more energy is therefore required to separate them. For a given molar mass, the boiling points of alkanes are relatively low because these nonpolar molecules have only weak dispersion forces to hold them together in the liquid state. Looking Closer: An Alkane Basis for Properties of Other Compounds An understanding of the physical properties of the alkanes is important in that petroleum and natural gas and the many products derived from them—gasoline, bottled gas, solvents, plastics, and more—are composed primarily of alkanes. This understanding is also vital because it is the basis for describing the properties of other organic and biological compound families. For example, large portions of the structures of lipids consist of nonpolar alkyl groups. Lipids include the dietary fats and fatlike compounds called phospholipids and sphingolipids that serve as structural components of living tissues. These compounds have both polar and nonpolar groups, enabling them to bridge the gap between water-soluble and water-insoluble phases. This characteristic is essential for the selective permeability of cell membranes. Figure \(\PageIndex{2}\): Tripalmitin (a), a typical fat molecule, has long hydrocarbon chains typical of most lipids. Compare these chains to hexadecane (b), an alkane with 16 carbon atoms. Concept Review Exercises Without referring to a table, predict which has a higher boiling point—hexane or octane. Explain. If 25 mL of hexane were added to 100 mL of water in a beaker, which of the following would you expect to happen? Explain. Hexane would dissolve in water. Hexane would not dissolve in water and would float on top. Hexane would not dissolve in water and would sink to the bottom of the container. Answers octane because of its greater molar mass b; hexane is insoluble in water and less dense than water. Key Takeaway Alkanes are nonpolar compounds that are low boiling and insoluble in water. Exercises Without referring to a table or other reference, predict which member of each pair has the higher boiling point. pentane or butane heptane or nonane For which member of each pair is hexane a good solvent? pentane or water sodium chloride or soybean oil Answer pentane nonane |
Bookshelves/General_Chemistry/Book%3A_General_Chemistry%3A_Principles_Patterns_and_Applications_(Averill)/14%3A_Chemical_Kinetics/14.07%3A_Reaction_Rates_-_A_Microscopic_View | Learning Objectives To determine the individual steps of a simple reaction. One of the major reasons for studying chemical kinetics is to use measurements of the macroscopic properties of a system, such as the rate of change in the concentration of reactants or products with time, to discover the sequence of events that occur at the molecular level during a reaction. This molecular description is the mechanism of the reaction; it describes how individual atoms, ions, or molecules interact to form particular products. The stepwise changes are collectively called the reaction mechanism . In an internal combustion engine, for example, isooctane reacts with oxygen to give carbon dioxide and water: \( 2C_{8}H_{18}\left ( l \right ) + 25O_{2}\left ( g \right ) \rightarrow 16CO_{2}\left ( g \right ) + 18H_{2}O\left ( g \right ) \tag{14.6.1}\) For this reaction to occur in a single step, 25 dioxygen molecules and 2 isooctane molecules would have to collide simultaneously and be converted to 34 molecules of product, which is very unlikely. It is more likely that a complex series of reactions takes place in a stepwise fashion. Each individual reaction, which is called an elementary reaction Each of the complex series of reactions that take place in a stepwise fashion to convert reactants to products. , involves one, two, or (rarely) three atoms, molecules, or ions. The overall sequence of elementary reactions is the mechanism of the reaction. The sum of the individual steps, or elementary reactions, in the mechanism must give the balanced chemical equation for the overall reaction. Molecularity and the Rate-Determining Step To demonstrate how the analysis of elementary reactions helps us determine the overall reaction mechanism, we will examine the much simpler reaction of carbon monoxide with nitrogen dioxide. \( NO_{2}\left ( g \right ) + CO\left ( g \right ) \rightarrow NO\left ( g \right ) + CO_{2} \left ( g \right ) \tag{14.6.2}\) From the balanced chemical equation, one might expect the reaction to occur via a collision of one molecule of NO 2 with a molecule of CO that results in the transfer of an oxygen atom from nitrogen to carbon. The experimentally determined rate law for the reaction, however, is as follows: \( rate = k\left [ NO_{2} \right ]^{2} \tag{14.6.3} \) The fact that the reaction is second order in [NO 2 ] and independent of [CO] tells us that it does not occur by the simple collision model outlined previously. If it did, its predicted rate law would be rate = k [NO 2 ][CO]. The following two-step mechanism is consistent with the rate law if step 1 is much slower than step 2: \( \begin{matrix} step \;1 & NO_{2} + NO_{2} \overset{slow}{\rightarrow} NO_{3} + NO & elementary \;reaction\\ step \;2 & \underline{NO_{3} + CO \rightarrow NO_{2} + CO_{2} }& elementary \; reaction\\ sum & NO_{2} + CO \rightarrow NO + CO_{2} & overall reaction \end{matrix} \) According to this mechanism, the overall reaction occurs in two steps, or elementary reactions. Summing steps 1 and 2 and canceling on both sides of the equation gives the overall balanced chemical equation for the reaction. The NO 3 molecule is an intermediate A species in a reaction mechanism that does not appear in the balanced chemical equation for the overall reaction. in the reaction, a species that does not appear in the balanced chemical equation for the overall reaction. It is formed as a product of the first step but is consumed in the second step. Note the Pattern The sum of the elementary reactions in a reaction mechanism must give the overall balanced chemical equation of the reaction. Using Molecularity to Describe a Rate Law The molecularity The number of molecules that collide during any step in a reaction mechanism. of an elementary reaction is the number of molecules that collide during that step in the mechanism. If there is only a single reactant molecule in an elementary reaction, that step is designated as unimolecular ; if there are two reactant molecules, it is bimolecular ; and if there are three reactant molecules (a relatively rare situation), it is termolecular . Elementary reactions that involve the simultaneous collision of more than three molecules are highly improbable and have never been observed experimentally. (To understand why, try to make three or more marbles or pool balls collide with one another simultaneously!) Writing the rate law for an elementary reaction is straightforward because we know how many molecules must collide simultaneously for the elementary reaction to occur; hence the order of the elementary reaction is the same as its molecularity ( Table 14.6.1 ). In contrast, the rate law for the reaction cannot be determined from the balanced chemical equation for the overall reaction. The general rate law for a unimolecular elementary reaction (A → products) is rate = k [A]. For bimolecular reactions, the reaction rate depends on the number of collisions per unit time, which is proportional to the product of the concentrations of the reactants, as shown in Figure 14.6.1 For a bimolecular elementary reaction of the form A + B → products, the general rate law is rate = k [A][B]. Table 14.6.1 Common Types of Elementary Reactions and Their Rate Laws Elementary Reaction Molecularity Rate Law Reaction Order A → products unimolecular rate = k[A] first 2A → products bimolecular rate = k[A]2 second A + B → products bimolecular rate = k[A][B] second 2A + B → products termolecular rate = k[A]2[B] third A + B + C → products termolecular rate = k[A][B][C] third Identifying the Rate-Determining Step Note the important difference between writing rate laws for elementary reactions and the balanced chemical equation of the overall reaction. Because the balanced chemical equation does not necessarily reveal the individual elementary reactions by which the reaction occurs, we cannot obtain the rate law for a reaction from the overall balanced chemical equation alone. In fact, it is the rate law for the slowest overall reaction, which is the same as the rate law for the slowest step in the reaction mechanism, the rate-determining step The slowest step in a reaction mechanism. , that must give the experimentally determined rate law for the overall reaction. This statement is true if one step is substantially slower than all the others, typically by a factor of 10 or more. If two or more slow steps have comparable rates, the experimentally determined rate laws can become complex. Our discussion is limited to reactions in which one step can be identified as being substantially slower than any other. The reason for this is that any process that occurs through a sequence of steps can take place no faster than the slowest step in the sequence. In an automotive assembly line, for example, a component cannot be used faster than it is produced. Similarly, blood pressure is regulated by the flow of blood through the smallest passages, the capillaries. Because movement through capillaries constitutes the rate-determining step in blood flow, blood pressure can be regulated by medications that cause the capillaries to contract or dilate. A chemical reaction that occurs via a series of elementary reactions can take place no faster than the slowest step in the series of reactions. Rate-determining step. The phenomenon of a rate-determining step can be compared to a succession of funnels. The smallest-diameter funnel controls the rate at which the bottle is filled, whether it is the first or the last in the series. Pouring liquid into the first funnel faster than it can drain through the smallest results in an overflow. Look at the rate laws for each elementary reaction in our example as well as for the overall reaction. \( \begin{matrix} step \;1 & NO_{2} + NO_{2} \overset{k_{1}}{\rightarrow} NO_{3}+ NO & rate=k_{1}\left [ NO_{2} \right ]^{2} predicted\\ step \;2 & \underline{NO_{3} + CO \overset{k_{2}}{\rightarrow} NO_{2} + CO_{2} } & rate=k_{2}\left [ NO_{3} \right ]\left [ CO_{2} \right ] predicted\\ sum & NO_{2} + CO \overset{k}{\rightarrow} NO + CO_{2} & rate=k\left [ NO_{2} \right ]^{2} observed \end{matrix} \) The experimentally determined rate law for the reaction of NO 2 with CO is the same as the predicted rate law for step 1. This tells us that the first elementary reaction is the rate-determining step, so k for the overall reaction must equal k 1 . That is, NO 3 is formed slowly in step 1, but once it is formed, it reacts very rapidly with CO in step 2. Sometimes chemists are able to propose two or more mechanisms that are consistent with the available data. If a proposed mechanism predicts the wrong experimental rate law, however, the mechanism must be incorrect. Example 14.6.1 In an alternative mechanism for the reaction of NO 2 with CO, N 2 O 4 appears as an intermediate. \( \begin{matrix} step \;1 & NO_{2} + NO_{2} \overset{k_{1}}{\rightarrow} N_{2}O_{4} \\ step \;2 & \underline{N_{2}O_{4} + CO \overset{k_{2}}{\rightarrow} NO+ NO_{2} + CO_{2} }\\ sum & NO_{2} + CO \rightarrow NO + CO_{2} \end{matrix} \) Write the rate law for each elementary reaction. Is this mechanism consistent with the experimentally determined rate law (rate = k [NO 2 ] 2 )? Given: elementary reactions Asked for: rate law for each elementary reaction and overall rate law Strategy: A Determine the rate law for each elementary reaction in the reaction. B Determine which rate law corresponds to the experimentally determined rate law for the reaction. This rate law is the one for the rate-determining step. Solution: A The rate law for step 1 is rate = k 1 [NO 2 ] 2 ; for step 2, it is rate = k 2 [N 2 O 4 ][CO]. B If step 1 is slow (and therefore the rate-determining step), then the overall rate law for the reaction will be the same: rate = k 1 [NO 2 ] 2 . This is the same as the experimentally determined rate law. Hence this mechanism, with N 2 O 4 as an intermediate, and the one described previously, with NO 3 as an intermediate, are kinetically indistinguishable. In this case, further experiments are needed to distinguish between them. For example, the researcher could try to detect the proposed intermediates, NO 3 and N 2 O 4 , directly. Exercise A Iodine monochloride (ICl) reacts with H 2 as follows: \( 2ICl \left ( g \right ) + H_{2}\left ( g \right ) \rightarrow HCl\left ( g \right ) + I_{2} \left ( s \right ) \) The experimentally determined rate law is rate = k [ICl][H 2 ]. Write a two-step mechanism for this reaction using only bimolecular elementary reactions and show that it is consistent with the experimental rate law. (Hint: HI is an intermediate.) Answer \( \begin{matrix} step \;1 & ICl+ H_{2} \overset{k_{1}}{\rightarrow} HCl + HI & rate = k_{1}\left [ ICl \right ]\left [ H_{2} \right ] slow\\ step \;2 & \underline{HI + ICl \overset{k_{2}}{\rightarrow} HCl + I_{2} }& rate = k_{2}\left [ HI \right ]\left [ ICl \right ] fast\\ sum & 2ICl + H_{2} \rightarrow 2HCl + I_{2} \end{matrix} \) This mechanism is consistent with the experimental rate law if the first step is the rate-determining step. Exercise B The reaction between NO and H 2 occurs via a three-step process: \( \begin{matrix} step \;1 & NO + NO \overset{k_{1}}{\rightarrow} N_{2}O_{2}+ NO & fast\\ step \;2 & N_{2}O_{2} + H_{2} \overset{k_{2}}{\rightarrow} N_{2}O + H_{2}O & slow \\ step 3 & N_{2}O + H_{2} \overset{k_{3}}{\rightarrow} N_{2} + H_{2}O & fast \end{matrix} \) Write the rate law for each elementary reaction, write the balanced chemical equation for the overall reaction, and identify the rate-determining step. Is the rate law for the rate-determining step consistent with the experimentally derived rate law for the overall reaction: rate = k [NO] 2 [H 2 ] 2 ? Answer rate = k 1 [NO] 2 ; rate = k 2 [N 2 O 2 ][H 2 ]; rate = k 3 [N 2 O][H 2 ]; 2NO(g) + 2H 2 (g) → N 2 (g) + 2H 2 O(g) step 2 Yes, because the rate of formation of [N 2 O 2 ] = k 1 [NO] 2 . Substituting k 1 [NO] 2 for [N 2 O 2 ] in the rate law for step 2 gives the experimentally derived rate law for the overall chemical reaction, where k = k 1 k 2 . Chain Reactions Many reaction mechanisms, like those discussed so far, consist of only two or three elementary reactions. Many others consist of long series of elementary reactions. The most common mechanisms are chain reactions A reaction mechanism in which one or more elementary reactions that contain a highly reactive species repeat again and again during the reaction process. , in which one or more elementary reactions that contain a highly reactive species repeat again and again during the reaction process. Chain reactions occur in fuel combustion, explosions, the formation of many polymers, and the tissue changes associated with aging. They are also important in the chemistry of the atmosphere. Chain reactions are described as having three stages. The first is initiation , a step that produces one or more reactive intermediates. Often these intermediates are radicals Species that have one or more unpaired valence electrons. , species that have an unpaired valence electron. In the second stage, propagation , reactive intermediates are continuously consumed and regenerated while products are formed. Intermediates are also consumed but not regenerated in the final stage of a chain reaction, termination , usually by forming stable products. Let us look at the reaction of methane with chlorine at elevated temperatures (400°C–450°C), a chain reaction used in industry to manufacture methyl chloride (CH 3 Cl), dichloromethane (CH 2 Cl 2 ), chloroform (CHCl 3 ), and carbon tetrachloride (CCl 4 ): \( CH_{4} + Cl_{2} \rightarrow CH_{3}Cl + HCl \) \( CH_{3}Cl + Cl_{2} \rightarrow CH_{2}Cl_{2} + HCl \) \(CH_{2}Cl_{2} + Cl_{2} \rightarrow CHCl_{3} + HCl \) \(CHCl_{3} + Cl_{2} \rightarrow CCl_{4} + HCl \) Direct chlorination generally produces a mixture of all four carbon-containing products, which must then be separated by distillation. In our discussion, we will examine only the chain reactions that lead to the preparation of CH 3 Cl. In the initiation stage of this reaction, the relatively weak Cl–Cl bond cleaves at temperatures of about 400°C to produce chlorine atoms (Cl·): \( Cl_{2}\rightarrow 2Cl\cdot \) During propagation, a chlorine atom removes a hydrogen atom from a methane molecule to give HCl and CH 3 ·, the methyl radical: \( Cl\cdot + CH_{4} \rightarrow CH_{3} \cdot + HCl \) The methyl radical then reacts with a chlorine molecule to form methyl chloride and another chlorine atom, Cl·: \( CH_{3} \cdot + Cl_{2}\rightarrow CH_{3}Cl + Cl\cdot \) The sum of the propagation reactions is the same as the overall balanced chemical equation for the reaction: \( \begin{matrix} \cancel{Cl\cdot} + CH_{4} \rightarrow \cancel{CH_{3} \cdot} + HCl\\ \underline{\cancel{CH_{3} \cdot} + Cl_{2} \rightarrow CH_{3}Cl + \cancel{Cl\cdot} }\\ Cl_{2} + CH_{4} \rightarrow CH_{3}Cl + HCl \end{matrix} \) Without a chain-terminating reaction, propagation reactions would continue until either the methane or the chlorine was consumed. Because radical species react rapidly with almost anything, however, including each other, they eventually form neutral compounds, thus terminating the chain reaction in any of three ways: \( \begin{matrix} CH_{3} \cdot + Cl\cdot \rightarrow CH_{3}Cl\\ CH_{3} \cdot + CH_{3} \cdot \rightarrow H_{3}CCH_{3} \\ Cl\cdot + Cl\cdot \rightarrow Cl_{2} \end{matrix} Here is the overall chain reaction, with the desired product (CH 3 Cl) in bold: 0 1 Initiation: Cl2 → 2Cl· Propagation: Cl· + CH4 → CH3· + HCl Propagation: CH3· + Cl2 → CH3Cl + Cl· Termination: CH3· + Cl· → CH3Cl Termination: CH3· + CH3· → H3CCH3 Termination: Cl· + Cl· → Cl2 The chain reactions responsible for explosions generally have an additional feature: the existence of one or more chain branching steps , in which one radical reacts to produce two or more radicals, each of which can then go on to start a new chain reaction. Repetition of the branching step has a cascade effect such that a single initiation step generates large numbers of chain reactions. The result is a very rapid reaction or an explosion. The reaction of H 2 and O 2 , used to propel rockets, is an example of a chain branching reaction: 0 1 Initiation: H2 + O2 → HO2· + H· Propagation: HO2· + H2 → H2O + OH· Propagation: OH· + H2 → H2O + H· Termination: H· + O2 → OH· + ·O· Termination: ·O· + H2 → OH· + H· Termination: NaN Termination reactions occur when the extraordinarily reactive H· or OH· radicals react with a third species. The complexity of a chain reaction makes it unfeasible to write a rate law for the overall reaction. Summary A reaction mechanism is the microscopic path by which reactants are transformed into products. Each step is an elementary reaction . Species that are formed in one step and consumed in another are intermediates . Each elementary reaction can be described in terms of its molecularity , the number of molecules that collide in that step. The slowest step in a reaction mechanism is the rate-determining step . Chain reactions consist of three kinds of reactions: initiation, propagation, and termination. Intermediates in chain reactions are often radicals , species that have an unpaired valence electron. Key Takeaway A balanced chemical reaction does not necessarily reveal either the individual elementary reactions by which a reaction occurs or its rate law. Conceptual Problems How does the term molecularity relate to elementary reactions? How does it relate to the overall balanced chemical equation? What is the relationship between the reaction order and the molecularity of a reaction? What is the relationship between the reaction order and the balanced chemical equation? When you determine the rate law for a given reaction, why is it valid to assume that the concentration of an intermediate does not change with time during the course of the reaction? If you know the rate law for an overall reaction, how would you determine which elementary reaction is rate determining? If an intermediate is contained in the rate-determining step, how can the experimentally determined rate law for the reaction be derived from this step? Give the rate-determining step for each case. Traffic is backed up on a highway because two lanes merge into one. Gas flows from a pressurized cylinder fitted with a gas regulator and then is bubbled through a solution. A document containing text and graphics is downloaded from the Internet. Before being sent on an assignment, an aging James Bond was sent off to a health farm where part of the program’s focus was to purge his body of radicals. Why was this goal considered important to his health? Numerical Problems Cyclopropane, a mild anesthetic, rearranges to propylene via a collision that produces and destroys an energized species. The important steps in this rearrangement are as follows: where M is any molecule, including cyclopropane. Only those cyclopropane molecules with sufficient energy (denoted with an asterisk) can rearrange to propylene. Which step determines the rate constant of the overall reaction? Above approximately 500 K, the reaction between NO 2 and CO to produce CO 2 and NO follows the second-order rate law Δ[CO 2 ]/Δ t = k [NO 2 ][CO]. At lower temperatures, however, the rate law is Δ[CO 2 ]/Δ t = k′[NO 2 ] 2 , for which it is known that NO 3 is an intermediate in the mechanism. Propose a complete low-temperature mechanism for the reaction based on this rate law. Which step is the slowest? Nitramide (O 2 NNH 2 ) decomposes in aqueous solution to N 2 O and H 2 O. What is the experimental rate law (Δ[N 2 O]/Δ t ) for the decomposition of nitramide if the mechanism for the decomposition is as follows? \( \begin{matrix} O_{2}NNH_{2} \xrightarrow[k_{-1}]{k_{1}} O_{2}NNH^{-} + H^{+} & fast\\ O_{2}NNH^{-} \overset{k_{2}}{\rightarrow} N_{2}O + OH^{-} & slow\\ H^{+} + OH^{-} \overset{k_{3}}{\rightarrow} H_{2}O & fast \end{matrix} \) Assume that the rates of the forward and reverse reactions in the first equation are equal. The following reactions are given: \( \begin{matrix} A + B \xrightarrow[k_{-1}]{k_{1}} C + D\\ D + E \overset{k_{2}}{\rightarrow} F \end{matrix} \) What is the relationship between the relative magnitudes of k −1 and k 2 if these reactions have the rate law Δ[F]/Δ t = k [A][B][E]/[C]? How does the magnitude of k 1 compare to that of k 2 ? Under what conditions would you expect the rate law to be Δ[F]/Δ t = k ′[A][B]? Assume that the rates of the forward and reverse reactions in the first equation are equal. Answers The k 2 step is likely to be rate limiting; the rate cannot proceed any faster than the second step. \( rate = k_{2} \dfrac{k_{1}\left [ O_{2}NNH_{2} \right ] }{k_{-1}\left [ H^{+} \right ]} = k \dfrac{\left [ O_{2}NNH_{2} \right ] }{\left [ H^{+} \right ]} \) Contributors Anonymous Modified by Joshua B. Halpern Thumbnail from Wikimedia |
Courses/BridgeValley_Community_and_Technical_College/Fundamentals_of_Chemistry/01%3A_Measurements/1.10%3A_Density | Learning Objectives Define density. Use density as a conversion factor. Density (\(\rho\)) is a physical property found by dividing the mass of an object by its volume. Regardless of the sample size, density is always constant. For example, the density of a pure sample of tungsten is always 19.25 grams per cubic centimeter. This means that whether you have one gram or one kilogram of the sample, the density will never vary. The equation, as we already know, is as follows: \[\text{Density}=\dfrac{\text{Mass}}{\text{Volume}} \nonumber \] or just \[\rho =\dfrac{m}{V} \label{eq2} \] Based on this equation, it's clear that density can, and does, vary from element to element and substance to substance due to differences in the relationship of mass and volume. Pure water, for example, has a density of 0.998 g/cm 3 at 25° C. The average densities of some common substances are in Table \(\PageIndex{1}\). Notice that corn oil has a lower mass to volume ratio than water. This means that when added to water, corn oil will “float.” Substance Density at 25°C (g/cm3) blood 1.035 body fat 0.918 whole milk 1.030 corn oil 0.922 mayonnaise 0.910 honey 1.420 Density can be measured for all substances—solids, liquids and gases. For solids and liquids, density is often reported using the units of g/cm 3 . Densities of gases, which are significantly lower than the densities of solids and liquids, are often given using units of g/L. Example \(\PageIndex{1}\): Ethyl Alcohol Calculate the density of a 30.2 mL sample of ethyl alcohol with a mass of 23.71002 g Solution This is a direct application of Equation \ref{eq2}: \[\rho = \dfrac{23.71002\,g}{30.2\,mL} = 0.785\, g/mL \nonumber \] Exercise \(\PageIndex{1}\) Find the density (in kg/L) of a sample that has a volume of 36.5 L and a mass of 10.0 kg. If you have a 2.130 mL sample of acetic acid with mass 0.002234 kg, what is the density in kg/L? Answer a \(0.274 \,kg/L\) Answer b \(1.049 \,kg/L\) Density as a Conversion Factor Conversion factors can also be constructed for converting between different kinds of units. For example, density can be used to convert between the mass and the volume of a substance. Consider mercury, which is a liquid at room temperature and has a density of 13.6 g/mL. The density tells us that 13.6 g of mercury have a volume of 1 mL. We can write that relationship as follows: 13.6 g mercury = 1 mL mercury This relationship can be used to construct two conversion factors: \[\dfrac{13.6\:g}{1\:mL} = 1 \nonumber \] and \[\dfrac{1\:mL}{13.6\:g} = 1 \nonumber \] Which one do we use? It depends, as usual, on the units we need to cancel and introduce. For example, suppose we want to know the mass of 2.0 mL of mercury. We would use the conversion factor that has milliliters on the bottom (so that the milliliter unit cancels) and grams on top, so that our final answer has a unit of mass: \[\mathrm{2.0\:\cancel{mL}\times\dfrac{13.6\:g}{1\:\cancel{mL}}=27.2\:g=27\:g} \nonumber \] In the last step, we limit our final answer to two significant figures because the volume quantity has only two significant figures; the 1 in the volume unit is considered an exact number, so it does not affect the number of significant figures. The other conversion factor would be useful if we were given a mass and asked to find volume, as the following example illustrates. Density can be used as a conversion factor between mass and volume. Example \(\PageIndex{2}\): Mercury Thermometer Steps for Problem Solving A mercury thermometer for measuring a patient’s temperature contains 0.750 g of mercury. What is the volume of this mass of mercury? Solution Steps for Problem Solving Unit Conversion Identify the "given" information and what the problem is asking you to "find." Given: 0.750 g Find: mL List other known quantities. 13.6 g/mL (density of mercury) Prepare a concept map. NaN Calculate. \[ 0.750 \; \cancel{\rm{g}} \times \dfrac{1\; \rm{mL}}{13.6 \; \cancel{\rm{g}}} = 0.055147 ... \; \rm{mL} \approx 0.0551\; \rm{mL} \nonumber \] We have limited the final answer to three significant figures. Exercise \(\PageIndex{2}\) What is the volume of 100.0 g of air if its density is 1.3 g/L? Answer \(77 \, L\) Summary Density is defined as the mass of an object divided by its volume. Density can be used as a conversion factor between mass and volume. Contributions & Attributions |
Bookshelves/Environmental_Chemistry/Toxicology_MSDT/05%3A_Regulatory_Toxicology/5.04%3A_New_Page | Learning Objectives After completing this lesson, you will be able to: Define what is meant by “National Regulatory Toxicology”. Give an example of a national regulatory regulation. What is National Regulatory Toxicology? As the name suggests, it focuses on regulatory toxicology at the national level (e.g., United States, Canada, China). Regulations are set by authorities that apply for the entire country. The focus on the United States (US) deals with regulations established at the federal level, and all states within the US must comply with federal regulations. Example agencies that set national toxicology regulations in the U.S.: Environmental Protection Agency (EPA) Food and Drug Administration (FDA) Occupational Safety and Health Administration (OSHA) Consumer Product Safety Commission (CPSC) All agencies aim to set regulations to protect against adverse health effects from exposure to substances. Different agencies regulate on different types of substances, that are used for different purposes and affect different populations. Goal of EPA is to protect human and environmental health; deals with exposures to substances in the environment (e.g., air, water, soil). It typically pertains to exposures that the general population generally does not have control over (e.g., air pollution) as well as modernizing chemical regulation in the US, e.g. Toxic Substances Control Act (TSCA) was recently revised. TSCA was first passed in 1976. TSCA was revised with passage of the Frank R. Lautenberg Chemical Safety for the 21st Century Act in 2016. It regulates new or already existing chemicals manufactured or in use in the US. Topic 4: Key Points In this section, we explored the following main points: 1: What is National Regulatory Toxicology? 2: Examples of National Agencies in the U.S. 3: What do these Agencies regulate? 4: Highlight: EPA and TSCA Reform. Knowledge Check 1. EPA regulates chemicals that are in: Food Environmental media (e.g., water, air) Drugs and Medical Devices Shampoo Answer Environmental media (e.g., water, air) 2.US States can choose not to comply with regulations set by the FDA True False Answer False |
Courses/Westminster_College/CHE_180_-_Inorganic_Chemistry/10%3A_Chapter_10_-_The_Transition_Metals/10.7%3A_Group_9/Chemistry_of_Rhodium | World production of rhodium (from the Greek rhodon, "rose") is about 10 tons. While the metal itself has few applications, it is an important alloying agent used as a hardener for platinum and palladium. Rhodium was discovered in 1803 by William Hyde Wollaston who named it for the rose-red color of its salts. Properties Rhodium is part of the the Platinum Group Metals (PGM) whic is located in the 5th and 6th rows of the transition metal section of the periodic table and includes Ruthenium , Rhodium, Palladium , Osmium , Iridium , and Platinum . Common characteristics include resistance to wear, oxidation, and corrosion, high melting points, and oxidation states of +2 to +4. They are generally non-toxic. It shares the properties of these metals: high corrosion resistance, hardness and ductility. It is the rarest of the group, only occurring to the extent of about 1 part per 200 million in the earth's crust. |
Courses/Sacramento_City_College/SCC%3A_CHEM_330_-_Adventures_in_Chemistry_(Alviar-Agnew)/18%3A_Drugs/18.05%3A_Hormones_-_The_Regulators | Learning Objectives Identify the major classes of hormones on the basis of chemical structure. Know the function of the common hormones. Glands of the Endocrine System The endocrine system is a system of glands called endocrine glands that release chemical messenger molecules into the bloodstream. The messenger molecules of the endocrine system are called endocrine hormones. The major glands of the endocrine system are shown in Figure \(\PageIndex{1}\). The major hormones of the human body and their effects are listed in Table \(\PageIndex{1}\). Although a given hormone may travel throughout the body in the bloodstream, it will affect the activity only of its target cells; that is, cells with receptors for that particular hormone. Once the hormone binds to the receptor, a chain of events is initiated that leads to the target cell’s response. Hormones play a critical role in the regulation of physiological processes because of the target cell responses they regulate. These responses contribute to human reproduction, growth and development of body tissues, metabolism, fluid, and electrolyte balance, sleep, and many other body functions. The major hormones of the human body and their effects are identified in Table \(\PageIndex{1}\). Endocrine gland Associated hormones Chemical class Effect Pituitary Growth hormone (GH) Protein Promotes growth of body tissues NaN Prolactin (PRL) Peptide Promotes milk production NaN Thyroid-stimulating hormone (TSH) Glycoprotein Stimulates thyroid hormone release NaN Adrenocorticotropic hormone (ACTH) Peptide Stimulates hormone release by adrenal cortex NaN Follicle-stimulating hormone (FSH) Glycoprotein Stimulates gamete production NaN Luteinizing hormone (LH) Glycoprotein Stimulates androgen production by gonads NaN Antidiuretic hormone (ADH) Peptide Stimulates water reabsorption by kidneys NaN Oxytocin Peptide Stimulates uterine contractions during childbirth Thyroid Thyroxine (T4), triiodothyronine (T3) Amine Stimulate basal metabolic rate NaN Calcitonin Peptide Reduces blood Ca2+ levels Parathyroid Parathyroid hormone (PTH) Peptide Increases blood Ca2+ levels Adrenal (cortex) Aldosterone Steroid Increases blood Na+ levels NaN Cortisol, corticosterone, cortisone Steroid Increase blood glucose levels Adrenal (medulla) Epinephrine, norepinephrine Amine Stimulate fight-or-flight response Pineal Melatonin Amine Regulates sleep cycles Pancreas Insulin Protein Reduces blood glucose levels NaN Glucagon Protein Increases blood glucose levels Testes Testosterone Steroid Stimulates development of male secondary sex characteristics and sperm production Ovaries Estrogens and progesterone Steroid Stimulate development of female secondary sex characteristics and prepare the body for childbirth Types of Hormones The hormones of the human body can be divided into two major groups on the basis of their chemical structure. Hormones derived from amino acids include amines, peptides, and proteins. Those derived from lipids include steroids (Figure \(\PageIndex{1}\)). These chemical groups affect a hormone’s distribution, the type of receptors it binds to, and other aspects of its function. Amine Hormones Hormones derived from the modification of amino acids are referred to as amine hormones. Typically, the original structure of the amino acid is modified such that a –COOH, or carboxyl, group is removed, whereas the −NH3+−NH3+, or amine, group remains. Amine hormones are synthesized from the amino acids tryptophan or tyrosine. An example of a hormone derived from tryptophan is melatonin, which is secreted by the pineal gland and helps regulate circadian rhythm. Tyrosine derivatives include the metabolism-regulating thyroid hormones, as well as the catecholamines, such as epinephrine, norepinephrine, and dopamine. Epinephrine and norepinephrine are secreted by the adrenal medulla and play a role in the fight-or-flight response, whereas dopamine is secreted by the hypothalamus and inhibits the release of certain anterior pituitary hormones. Peptide and Protein Hormones Whereas the amine hormones are derived from a single amino acid, peptide and protein hormones consist of multiple amino acids that link to form an amino acid chain. Peptide hormones consist of short chains of amino acids, whereas protein hormones are longer polypeptides. Both types are synthesized like other body proteins: DNA is transcribed into mRNA, which is translated into an amino acid chain. Examples of peptide hormones include antidiuretic hormone (ADH), a pituitary hormone important in fluid balance, and atrial-natriuretic peptide, which is produced by the heart and helps to decrease blood pressure. Some examples of protein hormones include growth hormone, which is produced by the pituitary gland, and follicle-stimulating hormone (FSH), which has an attached carbohydrate group and is thus classified as a glycoprotein. FSH helps stimulate the maturation of eggs in the ovaries and sperm in the testes. Steroid Hormones Hormones are chemical messengers that are released in one tissue and transported through the circulatory system to one or more other tissues. One group of hormones is known as steroid hormones because these hormones are synthesized from cholesterol, which is also a steroid. There are two main groups of steroid hormones: adrenocortical hormones and sex hormones. Adrenocortical hormones The adrenocortical hormones, such as aldosterone and cortisol (Table \(\PageIndex{1}\)), are produced by the adrenal gland, which is located adjacent to each kidney. Aldosterone acts on most cells in the body, but it is particularly effective at enhancing the rate of reabsorption of sodium ions in the kidney tubules and increasing the secretion of potassium ions and/or hydrogen ions by the tubules. Because the concentration of sodium ions is the major factor influencing water retention in tissues, aldosterone promotes water retention and reduces urine output. Cortisol regulates several key metabolic reactions (for example, increasing glucose production and mobilizing fatty acids and amino acids). It also inhibits the inflammatory response of tissue to injury or stress. Cortisol and its analogs are therefore used pharmacologically as immunosuppressants after transplant operations and in the treatment of severe skin allergies and autoimmune diseases, such as rheumatoid arthritis. Hormone Effect NaN regulates salt metabolism; stimulates kidneys to retain sodium and excrete potassium NaN stimulates the conversion of proteins to carbohydrates NaN regulates the menstrual cycle; maintains pregnancy NaN stimulates female sex characteristics; regulates changes during the menstrual cycle NaN stimulates and maintains male sex characteristics Sex Hormones The sex hormones are a class of steroid hormones secreted by the gonads (ovaries or testes), the placenta, and the adrenal glands. Testosterone and androstenedione are the primary male sex hormones, or androgens , controlling the primary sexual characteristics of males, or the development of the male genital organs and the continuous production of sperm. Androgens are also responsible for the development of secondary male characteristics, such as facial hair, deep voice, and muscle strength. Two kinds of sex hormones are of particular importance in females: progesterone, which prepares the uterus for pregnancy and prevents the further release of eggs from the ovaries during pregnancy, and the estrogens, which are mainly responsible for the development of female secondary sexual characteristics, such as breast development and increased deposition of fat tissue in the breasts, the buttocks, and the thighs. Both males and females produce androgens and estrogens, differing in the amounts of secreted hormones rather than in the presence or absence of one or the other. Sex hormones, both natural and synthetic, are sometimes used therapeutically. For example, a woman who has had her ovaries removed may be given female hormones to compensate. Some of the earliest chemical compounds employed in cancer chemotherapy were sex hormones. For example, estrogens are one treatment option for prostate cancer because they block the release and activity of testosterone. Testosterone enhances prostate cancer growth. Sex hormones are also administered in preparation for sex-change operations, to promote the development of the proper secondary sexual characteristics. Oral contraceptives are synthetic derivatives of the female sex hormones; they work by preventing ovulation. Chemistry and Social Revolution: The Pill Hormonal contraception methods prevent pregnancy by interfering with ovulation, fertilization, and/or implantation of the fertilized egg. Oral contraceptives—combined pill (“The pill”) The pill contains the hormones estrogen and progestin. It is taken daily to keep the ovaries from releasing an egg. The pill also causes changes in the lining of the uterus and the cervical mucus to keep the sperm from joining the egg. Some women prefer the “extended cycle” pills. These have 12 weeks of pills that contain hormones (active) and 1 week of pills that don’t contain hormones (inactive). While taking extended cycle pills, women only have their period three to four times a year. Many types of oral contraceptives are available. Talk with your doctor about which is best for you. Your doctor may advise you not to take the pill if you: Are older than 35 and smoke Have a history of blood clots Have a history of breast, liver, or endometrial cancer Antibiotics may reduce how well the pill works in some women. Talk to your doctor about a backup method of birth control if you need to take antibiotics. Women should wait three weeks after giving birth to begin using birth control that contains both estrogen and progestin. These methods increase the risk of dangerous blood clots that could form after giving birth. Women who delivered by cesarean section or have other risk factors for blood clots, such as obesity, history of blood clots, smoking, or preeclampsia, should wait six weeks. The patch Also called by its brand name, Ortho Evra, this skin patch is worn on the lower abdomen, buttocks, outer arm, or upper body. It releases the hormones progestin and estrogen into the bloodstream to stop the ovaries from releasing eggs in most women. It also thickens the cervical mucus, which keeps the sperm from joining with the egg. You put on a new patch once a week for 3 weeks. You don’t use a patch the fourth week in order to have a period. Women should wait three weeks after giving birth to begin using birth control that contains both estrogen and progestin. These methods increase the risk of dangerous blood clots that could form after giving birth. Women who delivered by cesarean section or have other risk factors for blood clots, such as obesity, history of blood clots, smoking, or preeclampsia, should wait six weeks. Shot/injection The birth control shot often is called by its brand name Depo-Provera. With this method you get injections, or shots, of the hormone progestin in the buttocks or arm every 3 months. A new type is injected under the skin. The birth control shot stops the ovaries from releasing an egg in most women. It also causes changes in the cervix that keep the sperm from joining with the egg. Vaginal ring This is a thin, flexible ring that releases the hormones progestin and estrogen. It works by stopping the ovaries from releasing eggs. It also thickens the cervical mucus, which keeps the sperm from joining the egg. It is commonly called NuvaRing, its brand name. You squeeze the ring between your thumb and index finger and insert it into your vagina. You wear the ring for 3 weeks, take it out for the week that you have your period, and then put in a new ring. Women should wait three weeks after giving birth to begin using birth control that contains both estrogen and progestin. These methods increase the risk of dangerous blood clots that could form after giving birth. Women who delivered by cesarean section or have other risk factors for blood clots, such as obesity, history of blood clots, smoking, or preeclampsia, should wait six weeks. Implantable devices These devices are inserted into the body and left in place for a few years. Implantable rod This is a matchstick-size, flexible rod that is put under the skin of the upper arm. It is often called by its brand name, Implanon. The rod releases a progestin, which causes changes in the lining of the uterus and the cervical mucus to keep the sperm from joining an egg. Less often, it stops the ovaries from releasing eggs. It is effective for up to 3 years. Intrauterine devices or IUDs Hormonal IUD The hormonal IUD goes by the brand name Mirena. It is sometimes called an intrauterine system, or IUS. It releases progestin into the uterus, which keeps the ovaries from releasing an egg and causes the cervical mucus to thicken so sperm can’t reach the egg. It also affects the ability of a fertilized egg to successfully implant in the uterus. A doctor needs to put in a hormonal IUD. It can stay in your uterus for up to 5 years. Emergency Contraceptives Emergency contraceptives are used if a woman’s primary method of birth control fails. It should not be used as a regular method of birth control. The emergency contraceptive, Plan B One-Step or Next Step is also called the “morning after pill.” Emergency contraception keeps a woman from getting pregnant when she has had unprotected vaginal intercourse. “Unprotected” can mean that no method of birth control was used. It can also mean that a birth control method was used but it was used incorrectly, or did not work (like a condom breaking). Or, a woman may have forgotten to take her birth control pills. She also may have been abused or forced to have sex. These are just some of the reasons women may need emergency contraception. Emergency contraception can be taken as a single pill treatment or in two doses. A single dose treatment works as well as two doses and does not have more side effects. It works by stopping the ovaries from releasing an egg or keeping the sperm from joining with the egg. For the best chances for it to work, take the pill as soon as possible after unprotected sex. It should be taken within 72 hours after having unprotected sex. A single-pill dose or two-pill dose of emergency contraception is available over-the-counter (OTC) for women ages 17 and older. Summary A hormone is any member of a class of signaling molecules, produced by glands in multicellular organisms to regulate physiology and behavior. The hormones of the human body can be divided into two major groups on the basis of their chemical structure. Hormones derived from amino acids include amines, peptides, and proteins. Those derived from lipids include steroids. The two main groups of steroid hormones: adrenocortical hormones and sex hormones. Hormonal contraception methods prevent pregnancy by interfering with ovulation, fertilization, and/or implantation of the fertilized egg. Emergency contraceptives are used if a woman’s primary method of birth control fails. Contributors Libretext: Anatomy and Physiology (OpenStax) Libretext: Human Biology (Wakim and Grewal) Libretexts: Survey of Chemistry (Cannon) Libretexts: The Basics of GOB Chemistry (Ball et.al.) Libretexts: Contemporary Health Issues (Lumen) Wikipedia |
Courses/Athabasca_University/Chemistry_350%3A_Organic_Chemistry_I/08%3A_Alkenes-_Reactions_and_Synthesis/8.12%3A_Stereochemistry_of_Reactions-_Addition_of_HO_to_an_Achiral_Alkene | Objective After completing this section, you should be able to account for the stereochemistry of the product of the addition of water to an alkene in terms of the formation of a planar carbocation. Study Notes Organic reactions in the laboratory or in living systems can produce chiral centres. Consider reaction of 1-butene with water (acid catalyzed). Markovnikov regiochemistry occurs and the OH adds to the second carbon. However, both R and S products occur giving a racemic (50/50) mixture of 2‑butanol. How does this occur? The proton addition to 1‑butene results in a planar carbocation intermediate. A molecule of water is then equally likely to attack from the top or the bottom of this cation to produce either ( S )‑2‑butanol or ( R )‑2‑butanol, respectively. |
Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Overview_of_Intermolecular_Forces | Intermolecular forces are forces between molecules. Depending on its strength, intermolecular forces cause the forming of three physical states: solid, liquid and gas. The physical properties of melting point, boiling point, vapor pressure, evaporation, viscosity, surface tension, and solubility are related to the strength of attractive forces between molecules. These attractive forces are called Intermolecular Forces. The amplitude of interaction is important in the interpretation of the various properties listed above. Introduction There are four types of intermolecular forces. Most of the intermolecular forces are identical to bonding between atoms in a single molecule. Intermolecular forces just extend the thinking to forces between molecules and follows the patterns already set by the bonding within molecules. Coulombic Forces The forces holding ions together in ionic solids are electrostatic forces. Opposite charges attract each other. These are the strongest intermolecular forces. Ionic forces hold many ions in a crystal lattice structure. According to Coulomb’s law: \[ V=-\dfrac{q_1 q_2}{4 \pi \epsilon r} \tag{1}\] \(q\) is the charges. \(r\) is the distance of separation. Based on Coulomb’s law, we can find the potential energy between different types of molecules. Dipole Forces Polar covalent molecules are sometimes described as "dipoles", meaning that the molecule has two "poles". One end (pole) of the molecule has a partial positive charge while the other end has a partial negative charge. The molecules will orientate themselves so that the opposite charges attract principle operates effectively. In the figure below, hydrochloric acid is a polar molecule with the partial positive charge on the hydrogen and the partial negative charge on the chlorine. A network of partial + and - charges attract molecules to each other. \[ V=-\dfrac{3}{2}\dfrac{I_A I_B}{I_A+I_B}\dfrac{\alpha_A \alpha_B}{r^6} \tag{3}\] Dipole-Dipole Interactions When a polar molecule encounters another polar molecule, the positive end of one molecule is attracted to the negative end of the other polar molecule. Many molecules have dipoles, and their interaction occur by dipole-dipole interaction. For example: SO 2 ↔SO 2 . (approximate energy: 15 kJ/mol). Polar molecules have permanent dipole moments, so in this case, we consider the electrostatic interaction between the two dipoles: \[ V=-\dfrac{2}{3}\dfrac{\mu_1^2 \mu_2^2}{(4\pi\epsilon_0)^2 r^6} \tag{4}\] µ is the permanent dipole moment of the molecule 1 and 2. Ion-Dipole Interactions Ion-Dipole interaction is the interaction between an ion and polar molecules. For example, the sodium ion/water cluster interaction is approximately 50 KJ/mol. \[ Na^+ ↔ (OH_2)_n \tag{5}\] Because the interaction involves in the charge of the ion and the dipole moment of the polar molecules, we can calculate the potential energy of interaction between them using the following formula: \[ V=-\dfrac{q\mu}{(4\pi\epsilon_0) r^2} \tag{6}\] r is the distance of separation. q is the charge of the ion ( only the magnitude of the charge is shown here.) \(\mu\) is the permanent dipole moment of the polar molecule. Hydrogen Bonding The hydrogen bond is really a special case of dipole forces. A hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine. In other words - The hydrogen on one molecule attached to O or N that is attracted to an O or N of a different molecule. In the graphic below, the hydrogen is partially positive and attracted to the partially negative charge on the oxygen or nitrogen. Because oxygen has two lone pairs, two different hydrogen bonds can be made to each oxygen. This is a very specific bond as indicated. Some combinations that are not hydrogen bonds include: hydrogen to another hydrogen or hydrogen to a carbon. Induced Dipole Forces Forces between essentially non-polar molecules are the weakest of all intermolecular forces. "Temporary dipoles" are formed by the shifting of electron clouds within molecules. These temporary dipoles attract or repel the electron clouds of nearby non-polar molecules. The temporary dipoles may exist for only a fraction of a second but a force of attraction also exist for that fraction of time. The strength of induced dipole forces depends on how easily electron clouds can be distorted. Large atoms or molecules with many electrons far removed from the nucleus are more easily distorted. Review - Non-Polar Bonds London Dispersion force London Dispersion Force, also called induced dipole-induced dipole, is the weakest intermolecular force. It is the interaction between two nonpolar molecules. For example: interaction between methyl and methyl group: (-CH 3 ↔ CH 3 -). (approximate energy: 5 KJ/mol) When two non polar molecules approach each other, attractive and repulsive forces between their electrons and nuclei will lead to distortions in these electron clouds, which mean they are induced by the neighboring molecules, and this leads to the interaction. We can calculate the potential energy between the two identical nonpolar molecule as followed formula: \[ V=-\dfrac{3}{4} \dfrac{\alpha^2 I}{r^6} \tag{2}\] α is the polarizability of nonpolar molecule. r is the distance between the two molecule. I = the first ionization energy of the molecule. Negative sign indicates the attractive interaction. For two nonidentical nonpolar molecules A and B, we have the formula: Problems Two HBr molecules (u= 0.78 D) are separate by 300pm in air. Calculate the dipole-dipole interaction energy in KJ/mol if they are oriented from end to end. Calculate the potential energy of interaction of a sodium ion that is at 3 Amstrong from HCl molecule with a dipole moment of 1.08 D. Calculate the potential energy of interaction between two He atoms separated by 5 Amstrong in the air. (polarizability = 0.20x 10 -30 m 3 . I = 2373 KJ/mol) Solution 1D=3.3356x10 -30 Cm; 300pm=3 Amstrong= 3x10 -10 m. V=-2.7 KJ/mol -35 KJ/mol -0.0045 KJ/mol References Kotz, Treichel, Weaver. Chemistry and Chemical reactivity, sixth ed. Thompson, 2006. Donald Allan McQuarrie, John Douglas Simon. Physical Chemistry: a molecular approach. University Science Books, 1997. |
Courses/Los_Medanos_College/Chemistry_6_and_Chemistry_7_Combined_Laboratory_Manual_(Los_Medanos_College)/01%3A_Experiments/1.23%3A_Experiment_624_Measuring_pH_1_2 | 0 1 2 3 Student Name NaN Laboratory Date: Date Report Submitted: ___________________________ Student ID NaN Experiment Number and Title Experiment 624: Measuring pH Experiment 6 24 : Measuring pH S ection 1: Purpose and Summary Determine the pH of some solutions using ‘red cabbage’ indicator. Determine the pH of solutions using a pH meter. Understand the effect of adding an acid or a base to buffer solutions. In this experiment, the instructor will prepare a pH indicator by extracting colored substances from red cabbage. This pH indicator solution will be added to buffer solutions of pH range 1 – 13. Students will record the resultant color of each buffer solution. Students will then use the indicator to determine the pH of common solutions by comparing the color to the buffers. To measure the exact pH values of solutions, a laboratory pH meter will be used. Students will determine the hydrogen ion concentration, [H+] of the solutions from the pH value recorded. Students will determine the effect of adding HCl or NaOH to an acetic acid/acetate buffer solution. Section 2: Safety Precautions and Waste Disposal Safety Precautions: Use of eye protection is recommended for all experimental procedures. Waste Disposal: All reagents used in this experiment are generally non-toxic and may be disposed of in sinks with tap water rinses. Section 3: Procedure Part 1: Preparing the pH indicator and pH standards – INSTRUCTOR DEMO NOTE: Part 1 may be replaced by the use of universal indicating solution at the instructor’s option. 0 Place several cut-up red cabbage leaves into a 600-mL beaker and cover the leaves with laboratory water. Boil the cabbage leaves on a hot plate until the pigment has been extracted. The leaves will look whitish in color. While the indicator cools, prepare an array of buffer solutions with pH 1 – 13. Label thirteen (13) 150-mm test tubes. Fill each test tube with about a half-full of the appropriate buffer solution. Using a dropper pipet, add several drops of the cabbage indicator to each test tube. An array of colors should be observed. Students will record the color of each pH standard. 0 1 Buffer pH Color with Cabbage Indicator 1 NaN 2 NaN 3 NaN 4 NaN 5 NaN 6 NaN 7 NaN 8 NaN 9 NaN 10 NaN 11 NaN 12 NaN 13 NaN Part 2 : pH determination of household products or everyday chemicals Possible household items : white vinegar, colorless soda like Sprite TM , dish soap solution (colorless), sparkling water, alkaline water, isopropyl alcohol Everyday chemicals: 0.1 M HCl, 0.1 M acetic acid, 0.1 M ammonia, 0.1 M NaOH 0 Take a 150-mL beaker and obtain about 50 mL of cabbage indicator solution prepared in Part 1. Label ten (10) clean medium test tubes with the solutions to be tested. Transfer about 3 mL of each solution into the corresponding test tube. Using a dropper pipet, add an equal volume of cabbage indicator solution. Mix well. Compare the resulting color of the solution with the buffer array to determine the approximate pH of the solution. Record your observations on the table below. Set aside these solutions for use in the following section. 0 1 Solutions tested Approximate pH NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN Part 3 : Quantitative determination of pH This procedure describes how to use a portable Flinn pH meter (model AP8673). If the pH meter is not measuring properly (unstable, or inaccurate), refer to the manual for troubleshooting. 0 Check out a portable pH meter from the stockroom. Remove the protective cap on the electrode. Clean any salt build-up off by rinsing with laboratory water. Press the ON/OFF button once. Rinse the electrode with laboratory water and blot dry with filter paper. Take one of the solutions from Part 2 of this experiment. Immerse the electrode in the test solution. Once the display stabilizes (approx. 1 min.), record the exact pH. Remove the pH meter from the solution. Repeat steps 4 – 7 using the rest of the test solutions. When finished, rinse the electrode with laboratory water and blot dry with filter paper. Replace cap and return the pH meter to the stockroom. 0 1 Solutions tested pH NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN Part 4 : Effect of add ing an acid or a base to the pH of a buffer 0 Obtain four (4) medium test tubes and label them A – D. To test tubes A and C, add 10 mL of laboratory water. To test tubes B and D, add 5 mL of 0.1 M acetic acid (HC2H3O2) and 5 mL of 0.1 M sodium acetate (NaC2H3O2). This mixture is your acetic acid/acetate buffer solution. Stir well to mix. Using HydrionTM pH paper strips, determine the pH of the contents of test tubes A – D. Put four (small) strips of pH paper on a large watch glass. Using a stirring rod, dab a small drop of the test solution onto the pH paper. Compare the color obtained to the pH scale on the label of the pH paper casing. Record the pH values to the tenth place. To test tubes A and B, add five (5) drops of 0.1 M hydrochloric acid (HCl). Determine the pH using pH (similar to step 4 above). To test tubes C and D, add five (5) drops of 0.1 M sodium hydroxide (NaOH). Determine the pH using pH (similar to step 4 above). 0 1 2 Solutions tested pH pH after addition of 0.1 M HCl A (laboratory water) NaN NaN B (buffer solution) NaN NaN Solutions tested pH pH after addition of 0.1 M NaOH C (laboratory water) NaN NaN D (buffer solution) NaN NaN Pos t L ab Questions : Based on your results, classify each test solution as acidic, basic, or neutral. Write the mathematical equation that shows the relationship between pH value and the hydrogen ion concentration, [H + ]. What is the unit of [H + ]? Using your results in Part 3 of this experiment, calculate [H + ] for each test solution. Include units. Compare a solution with pH 10.5 and a solution with pH 4.5. Which solution contains more H + ions? Explain. Two methods were used in determining the pH of the solutions – pH paper and pH meter. Do they show similar or different pH values? What was different? Compare the pH of the buffer solutions in Part 4 of this experiment before and after the addition of 0.1 M HCl and 0.1 M NaOH. Does this confirm the expected behavior of a buffer solution against an added acid or base? Explain. |
Courses/Brevard_College/CHE_301_Biochemistry/06%3A_Nucleic_Acids/6.02%3A_Nucleotides | Learning Objectives To identify the different molecules that combine to form nucleotides. The repeating, or monomer, units that are linked together to form nucleic acids are known as nucleotides. The deoxyribonucleic acid (DNA) of a typical mammalian cell contains about 3 × 10 9 nucleotides. Nucleotides can be further broken down to phosphoric acid (H 3 PO 4 ), a pentose sugar (a sugar with five carbon atoms), and a nitrogenous base (a base containing nitrogen atoms). \[\mathrm{nucleic\: acids \underset{down\: into}{\xrightarrow{can\: be\: broken}} nucleotides \underset{down\: into}{\xrightarrow{can\: be\: broken}} H_3PO_4 + nitrogen\: base + pentose\: sugar} \nonumber \] If the pentose sugar is ribose, the nucleotide is more specifically referred to as a ribonucleotide , and the resulting nucleic acid is ribonucleic acid (RNA). If the sugar is 2-deoxyribose, the nucleotide is a deoxyribonucleotide , and the nucleic acid is DNA . The nitrogenous bases found in nucleotides are classified as pyrimidines or purines. Pyrimidines are heterocyclic amines with two nitrogen atoms in a six-member ring and include uracil, thymine, and cytosine. Purines are heterocyclic amines consisting of a pyrimidine ring fused to a five-member ring with two nitrogen atoms. Adenine and guanine are the major purines found in nucleic acids (Figure \(\PageIndex{1}\)). The formation of a bond between C1′ of the pentose sugar and N1 of the pyrimidine base or N9 of the purine base joins the pentose sugar to the nitrogenous base. In the formation of this bond, a molecule of water is removed. Table \(\PageIndex{1}\) summarizes the similarities and differences in the composition of nucleotides in DNA and RNA. The numbering convention is that primed numbers designate the atoms of the pentose ring, and unprimed numbers designate the atoms of the purine or pyrimidine ring. Composition DNA RNA purine bases adenine and guanine adenine and guanine pyrimidine bases cytosine and thymine cytosine and uracil pentose sugar 2-deoxyribose ribose inorganic acid phosphoric acid (H3PO4) H3PO4 The names and structures of the major ribonucleotides and one of the deoxyribonucleotides are given in Figure \(\PageIndex{2}\). Apart from being the monomer units of DNA and RNA, the nucleotides and some of their derivatives have other functions as well. Adenosine diphosphate (ADP) and adenosine triphosphate (ATP), shown in Figure \(\PageIndex{3}\), have a role in cell metabolism. Moreover, a number of coenzymes, including flavin adenine dinucleotide (FAD), nicotinamide adenine dinucleotide (NAD + ), and coenzyme A, contain adenine nucleotides as structural components. Summary Nucleotides are composed of phosphoric acid, a pentose sugar (ribose or deoxyribose), and a nitrogen-containing base (adenine, cytosine, guanine, thymine, or uracil). Ribonucleotides contain ribose, while deoxyribonucleotides contain deoxyribose. |
Courses/Chabot_College/Chem_12A%3A_Organic_Chemistry_Fall_2022/06%3A_Stereoisomerism/6.04%3A_Diastereomers_-_more_than_one_chiral_center | Learning Objective recognize and classify diastereomers Diastereomers are stereoisomers with two or more chiral centers that are not enantiomers. Diastereomers have different physical properties (melting points, boiling points, and densities). Depending on the reaction mechanism, diastereomers can produce different stereochemical products. Introduction So far, we have been analyzing compounds with a single chiral center. Next, we turn our attention to those which have multiple chiral centers. We'll start with some stereoisomeric four-carbon sugars with two chiral centers. To avoid confusion, we will simply refer to the different stereoisomers by capital letters. Look first at compound A below. Both chiral centers in have the R configuration (you should confirm this for yourself!). The mirror image of Compound A is compound B, which has the S configuration at both chiral centers. If we were to pick up compound A, flip it over and put it next to compound B, we would see that they are not superimposable (again, confirm this for yourself with your models!). A and B are nonsuperimposable mirror images: in other words, enantiomers. Now, look at compound C, in which the configuration is S at chiral center 1 and R at chiral center 2. Compounds A and C are stereoisomers: they have the same molecular formula and the same bond connectivity, but a different arrangement of atoms in space (recall that this is the definition of the term 'stereoisomer). However, they are not mirror images of each other (confirm this with your models!), and so they are not enantiomers. By definition, they are diastereomers of each other. Notice that compounds C and B also have a diastereomeric relationship, by the same definition. So, compounds A and B are a pair of enantiomers, and compound C is a diastereomer of both of them. Does compound C have its own enantiomer? Compound D is the mirror image of compound C, and the two are not superimposable. Therefore, C and D are a pair of enantiomers. Compound D is also a diastereomer of compounds A and B. This can also seem very confusing at first, but there some simple shortcuts to analyzing stereoisomers: Stereoisomer shortcuts If all of the chiral centers are of opposite R/S configuration between two stereoisomers, they are enantiomers. If at least one, but not all of the chiral centers are opposite between two stereoisomers, they are diastereomers. ( Note: these shortcuts to not take into account the possibility of additional stereoisomers due to alkene groups: we will come to that later) Here's another way of looking at the four stereoisomers, where one chiral center is associated with red and the other blue. Pairs of enantiomers are stacked together. We know, using the shortcut above, that the enantiomer of R R must be S S - both chiral centers are different. We also know that R S and S R are diastereomers of R R , because in each case one - but not both - chiral centers are different. Diastereomers vs. Enantiomers in Wine Chemistry Tartaric acid, C 4 H 6 O 6 , is an organic compound that can be found in grape, bananas, and in wine. The structures of tartaric acid itself is really interesting. Naturally, it is in the form of (R,R) stereocenters. Artificially, it can be in the meso form (R,S), which is achiral. R,R tartaric acid is enantiomer to is mirror image which is S,S tartaric acid and diasteromers to meso-tartaric acid (Figure 5.6.2). (R,R) and (S,S) tartaric acid have similar physical properties and reactivity. However, meso-tartaric acid have different physical properties and reactivity. For example, melting point of (R,R) & (S,S) tartaric is about 170 degree Celsius, and melting point of meso-tartaric acid is about 145 degree Celsius. Diastereomers vs. Enantiomers in Sugar Chemistry D-erythrose is a common four-carbon sugar. A note on sugar nomenclature: biochemists use a special system to refer to the stereochemistry of sugar molecules, employing names of historical origin in addition to the designators ' D ' and ' L '. You will learn about this system if you take a biochemistry class. We will use the D/L designations here to refer to different sugars, but we won't worry about learning the system. As you can see, D -erythrose is a chiral molecule: C 2 and C 3 are stereocenters, both of which have the R configuration. In addition, you should make a model to convince yourself that it is impossible to find a plane of symmetry through the molecule, regardless of the conformation. Does D-erythrose have an enantiomer? Of course it does – if it is a chiral molecule, it must. The enantiomer of erythrose is its mirror image, and is named L-erythrose (once again, you should use models to convince yourself that these mirror images of erythrose are not superimposable). Notice that both chiral centers in L-erythrose both have the S configuration. Note In a pair of enantiomers, all of the chiral centers are of the opposite configuration. What happens if we draw a stereoisomer of erythrose in which the configuration is S at C 2 and R at C 3 ? This stereoisomer, which is a sugar called D-threose, is not a mirror image of erythrose. D-threose is a diastereomer of both D-erythrose and L-erythrose. The definition of diastereomers is simple: if two molecules are stereoisomers (same molecular formula, same connectivity, different arrangement of atoms in space) but are not enantiomers, then they are diastereomers by default. In practical terms, this means that at least one - but not all - of the chiral centers are opposite in a pair of diastereomers. By definition, two molecules that are diastereomers are not mirror images of each other. L-threose, the enantiomer of D-threose, has the R configuration at C 2 and the S configuration at C 3 . L-threose is a diastereomer of both erythrose enantiomers. Erythronolide B, a precursor to the 'macrocyclic' antibiotic erythromycin, has 10 stereocenters. It’s enantiomer is that molecule in which all 10 stereocenters are inverted. In total, there are 2 10 = 1024 stereoisomers in the erythronolide B family: 1022 of these are diastereomers of the structure above, one is the enantiomer of the structure above, and the last is the structure above. We know that enantiomers have identical physical properties and equal but opposite degrees of specific rotation. Diastereomers, in theory at least, have different physical properties – we stipulate ‘in theory’ because sometimes the physical properties of two or more diastereomers are so similar that it is very difficult to separate them. In addition, the specific rotations of diastereomers are unrelated – they could be the same sign or opposite signs, and similar in magnitude or very dissimilar. Exercises 1. Draw the structures of L-galactose (the enantiomer of D-galactose) and two more diastereomers of D-glucose (one should be an epimer). 2. Determine the stereochemistry of the following molecule: Answer 1. 2. |
Courses/University_of_Ottawa/OU%3A_CHM1721B_-_Chimie_Organique_I/00%3A_Front_Matter/02%3A_InfoPage | This text is disseminated via the Open Education Resource (OER) LibreTexts Project ( https://LibreTexts.org ) and like the hundreds of other texts available within this powerful platform, it is freely available for reading, printing and "consuming." Most, but not all, pages in the library have licenses that may allow individuals to make changes, save, and print this book. Carefully consult the applicable license(s) before pursuing such effects. Instructors can adopt existing LibreTexts texts or Remix them to quickly build course-specific resources to meet the needs of their students. Unlike traditional textbooks, LibreTexts’ web based origins allow powerful integration of advanced features and new technologies to support learning. The LibreTexts mission is to unite students, faculty and scholars in a cooperative effort to develop an easy-to-use online platform for the construction, customization, and dissemination of OER content to reduce the burdens of unreasonable textbook costs to our students and society. The LibreTexts project is a multi-institutional collaborative venture to develop the next generation of open-access texts to improve postsecondary education at all levels of higher learning by developing an Open Access Resource environment. The project currently consists of 14 independently operating and interconnected libraries that are constantly being optimized by students, faculty, and outside experts to supplant conventional paper-based books. These free textbook alternatives are organized within a central environment that is both vertically (from advance to basic level) and horizontally (across different fields) integrated. The LibreTexts libraries are Powered by NICE CXOne and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. This material is based upon work supported by the National Science Foundation under Grant No. 1246120, 1525057, and 1413739. Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation nor the US Department of Education. Have questions or comments? For information about adoptions or adaptions contact [email protected] . More information on our activities can be found via Facebook ( https://facebook.com/Libretexts ), Twitter ( https://twitter.com/libretexts ), or our blog ( http://Blog.Libretexts.org ). This text was compiled on 04/21/2025 |
Bookshelves/Introductory_Chemistry/Chemistry_for_Changing_Times_(Hill_and_McCreary)/04%3A_Chemical_Bonds/4.03%3A_The_Reaction_of_Sodium_with_Chlorine | Learning Objectives Explain the bonding nature of ionic compounds. Relate microscopic bonding properties to macroscopic solid properties. Neutral atoms and their associated ions have very different physical and chemical properties. Sodium atoms form sodium metal, a soft, silvery-white metal that burns vigorously in air and reacts explosively with water. Chlorine atoms form chlorine gas, Cl 2 , a yellow-green gas that is extremely corrosive to most metals and very poisonous to animals and plants. The vigorous reaction between the elements sodium and chlorine forms the white, crystalline compound sodium chloride, common table salt, which contains sodium cations and chloride anions (Figure \(\PageIndex{1}\)). The compound composed of these ions exhibits properties entirely different from the properties of the elements sodium and chlorine. Chlorine is poisonous, but sodium chloride is essential to life; sodium atoms react vigorously with water, but sodium chloride simply dissolves in water. Ionic Bonds When two atoms approach each other, they have the potential to bond (or connect). If a metal and a nonmetal interact, then an ionic bond will result. These types of bonds involve the metal donating it(s) valence electron(s) to a nonmetal. As the electronic transfer occurs, both atoms will achieve more stable confirmations. The end result will be a less reactive compound. These type of species are composed of both cations and anions. In addition, they are crystalline and solid in nature (Figure \(\PageIndex{2}\)) . A few examples of real-world ionic compounds would be NaCl (table salt) and NaF (active ingredient in toothpaste). The formula for table salt is NaCl. It is the result of Na + ions and Cl - ions bonding together (Figure \(\PageIndex{3}\)) . If sodium metal and chlorine gas mix under the right conditions, they will form salt. The sodium loses an electron, and the chlorine gains that electron. This reaction is highly favorable because of the electrostatic attraction between the particles. In the process, a great amount of light and heat is released. The resulting salt is mostly unreactive — it is stable. It will not undergo any explosive reactions, unlike the sodium and chlorine that it is made of. Figure \(\PageIndex{2}\) NaCl crystals. (Public Domain; NASA). The reaction is represented with Lewis dot symbols below. Chloride Salts The sodium atom is donating its 1 valence electron to the chlorine atom. This creates a sodium cation and a chlorine anion. Notice that the net charge of the resulting compound is 0. Summary An ionic bond is formed when a metal donates it(s) valence electron(s) to a nonmetal. The resulting ionic compound is more stable and less reactive. References Vollhardt, K. Peter C., and Neil E. Schore. Organic Chemistry Structure and Function . New York: W. H. Freeman, 2007. Petrucci, Ralph H. General Chemistry: Principles and Modern Applications . Upper Saddle River, NJ: Pearson Education, 2007. Brown, Theodore L., Eugene H. Lemay, and Bruce E. Bursten. Chemistry: The Central Science . 6th ed. Englewood Cliffs, NJ: Prentice Hall, 1994. |
Courses/Duke_University/CHEM_210D%3A_Modern_Applications_of_Chemistry/3%3A_Textbook-_Modern_Applications_of_Chemistry/04%3A_Intermolecular_Forces/4.02%3A_Properties_of_Liquids | Learning Objectives Distinguish between adhesive and cohesive forces Define viscosity, surface tension, and capillary rise Describe the roles of intermolecular attractive forces in each of these properties/phenomena When you pour a glass of water, or fill a car with gasoline, you observe that water and gasoline flow freely. But when you pour syrup on pancakes or add oil to a car engine, you note that syrup and motor oil do not flow as readily. The viscosity of a liquid is a measure of its resistance to flow. Water, gasoline, and other liquids that flow freely have a low viscosity. Honey, syrup, motor oil, and other liquids that do not flow freely, like those shown in Figure \(\PageIndex{1}\), have higher viscosities. We can measure viscosity by measuring the rate at which a metal ball falls through a liquid (the ball falls more slowly through a more viscous liquid) or by measuring the rate at which a liquid flows through a narrow tube (more viscous liquids flow more slowly). The IMFs between the molecules of a liquid, the size and shape of the molecules, and the temperature determine how easily a liquid flows. As Table \(\PageIndex{1}\) shows, the more structurally complex are the molecules in a liquid and the stronger the IMFs between them, the more difficult it is for them to move past each other and the greater is the viscosity of the liquid. As the temperature increases, the molecules move more rapidly and their kinetic energies are better able to overcome the forces that hold them together; thus, the viscosity of the liquid decreases. Substance Formula Viscosity (mPa·s) water H2O 0.890 mercury Hg 1.526 ethanol C2H5OH 1.074 octane C8H18 0.508 ethylene glycol CH2(OH)CH2(OH) 16.1 honey variable ~2,000–10,000 motor oil variable ~50–500 The various IMFs between identical molecules of a substance are examples of cohesive forces . The molecules within a liquid are surrounded by other molecules and are attracted equally in all directions by the cohesive forces within the liquid. However, the molecules on the surface of a liquid are attracted only by about one-half as many molecules. Because of the unbalanced molecular attractions on the surface molecules, liquids contract to form a shape that minimizes the number of molecules on the surface—that is, the shape with the minimum surface area. A small drop of liquid tends to assume a spherical shape, as shown in Figure \(\PageIndex{2}\), because in a sphere, the ratio of surface area to volume is at a minimum. Larger drops are more greatly affected by gravity, air resistance, surface interactions, and so on, and as a result, are less spherical. Surface tension is defined as the energy required to increase the surface area of a liquid, or the force required to increase the length of a liquid surface by a given amount. This property results from the cohesive forces between molecules at the surface of a liquid, and it causes the surface of a liquid to behave like a stretched rubber membrane. Surface tensions of several liquids are presented in Table \(\PageIndex{2}\). Substance Formula Surface Tension (mN/m) water H2O 71.99 mercury Hg 458.48 ethanol C2H5OH 21.97 octane C8H18 21.14 ethylene glycol CH2(OH)CH2(OH) 47.99 Among common liquids, water exhibits a distinctly high surface tension due to strong hydrogen bonding between its molecules. As a result of this high surface tension, the surface of water represents a relatively “tough skin” that can withstand considerable force without breaking. A steel needle carefully placed on water will float. Some insects, like the one shown in Figure \(\PageIndex{3}\), even though they are denser than water, move on its surface because they are supported by the surface tension. The IMFs of attraction between two different molecules are called adhesive forces . Consider what happens when water comes into contact with some surface. If the adhesive forces between water molecules and the molecules of the surface are weak compared to the cohesive forces between the water molecules, the water does not “wet” the surface. For example, water does not wet waxed surfaces or many plastics such as polyethylene. Water forms drops on these surfaces because the cohesive forces within the drops are greater than the adhesive forces between the water and the plastic. Water spreads out on glass because the adhesive force between water and glass is greater than the cohesive forces within the water. When water is confined in a glass tube, its meniscus (surface) has a concave shape because the water wets the glass and creeps up the side of the tube. On the other hand, the cohesive forces between mercury atoms are much greater than the adhesive forces between mercury and glass. Mercury therefore does not wet glass, and it forms a convex meniscus when confined in a tube because the cohesive forces within the mercury tend to draw it into a drop (Figure \(\PageIndex{4}\)). If you place one end of a paper towel in spilled wine, as shown in Figure \(\PageIndex{5}\), the liquid wicks up the paper towel. A similar process occurs in a cloth towel when you use it to dry off after a shower. These are examples of capillary action —when a liquid flows within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules. The adhesive forces between the liquid and the porous material, combined with the cohesive forces within the liquid, may be strong enough to move the liquid upward against gravity. Towels soak up liquids like water because the fibers of a towel are made of molecules that are attracted to water molecules. Most cloth towels are made of cotton, and paper towels are generally made from paper pulp. Both consist of long molecules of cellulose that contain many −OH groups. Water molecules are attracted to these −OH groups and form hydrogen bonds with them, which draws the H 2 O molecules up the cellulose molecules. The water molecules are also attracted to each other, so large amounts of water are drawn up the cellulose fibers. Capillary action can also occur when one end of a small diameter tube is immersed in a liquid, as illustrated in Figure \(\PageIndex{6}\). If the liquid molecules are strongly attracted to the tube molecules, the liquid creeps up the inside of the tube until the weight of the liquid and the adhesive forces are in balance. The smaller the diameter of the tube is, the higher the liquid climbs. It is partly by capillary action occurring in plant cells called xylem that water and dissolved nutrients are brought from the soil up through the roots and into a plant. Capillary action is the basis for thin layer chromatography, a laboratory technique commonly used to separate small quantities of mixtures. You depend on a constant supply of tears to keep your eyes lubricated and on capillary action to pump tear fluid away. The height to which a liquid will rise in a capillary tube is determined by several factors as shown in the following equation: \[h=\dfrac{2T\cosθ}{rρg} \label{10.2.1} \] where h is the height of the liquid inside the capillary tube relative to the surface of the liquid outside the tube, T is the surface tension of the liquid, θ is the contact angle between the liquid and the tube, r is the radius of the tube, ρ is the density of the liquid, and g is the acceleration due to gravity, 9.8 m/s 2 . When the tube is made of a material to which the liquid molecules are strongly attracted, they will spread out completely on the surface, which corresponds to a contact angle of 0°. This is the situation for water rising in a glass tube. Example \(\PageIndex{1}\): Capillary Rise At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.25 mm? For water, T = 71.99 mN/m and ρ = 1.0 g/cm 3 . Solution The liquid will rise to a height h given by Equation \(\ref{10.2.1}\) : \[h=\dfrac{2T\cosθ}{rρg} \nonumber \] The Newton is defined as a kg m/s 2 , and so the provided surface tension is equivalent to 0.07199 kg/s 2 . The provided density must be converted into units that will cancel appropriately: ρ = 1000 kg/m 3 . The diameter of the tube in meters is 0.00025 m, so the radius is 0.000125 m. For a glass tube immersed in water, the contact angle is θ = 0°, so cos θ = 1. Finally, acceleration due to gravity on the earth is g = 9.8 m/s 2 . Substituting these values into the equation, and cancelling units, we have: \[h=\mathrm{\dfrac{2(0.07199\:kg/s^2)}{(0.000125\:m)(1000\:kg/m^3)(9.8\:m/s^2)}=0.12\:m=12\: cm} \nonumber \] Exercise \(\PageIndex{1}\) Water rises in a glass capillary tube to a height of 8.4 cm. What is the diameter of the capillary tube? Answer diameter = 0.36 mm Applications: Capillary Action is Used to Draw Blood Many medical tests require drawing a small amount of blood, for example to determine the amount of glucose in someone with diabetes or the hematocrit level in an athlete. This procedure can be easily done because of capillary action, the ability of a liquid to flow up a small tube against gravity, as shown in Figure \(\PageIndex{7}\). When your finger is pricked, a drop of blood forms and holds together due to surface tension—the unbalanced intermolecular attractions at the surface of the drop. Then, when the open end of a narrow-diameter glass tube touches the drop of blood, the adhesive forces between the molecules in the blood and those at the glass surface draw the blood up the tube. How far the blood goes up the tube depends on the diameter of the tube (and the type of fluid). A small tube has a relatively large surface area for a given volume of blood, which results in larger (relative) attractive forces, allowing the blood to be drawn farther up the tube. The liquid itself is held together by its own cohesive forces. When the weight of the liquid in the tube generates a downward force equal to the upward force associated with capillary action, the liquid stops rising. Key Concepts and Summary The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid’s viscosity (resistance to flow) and surface tension (elasticity of a liquid surface). Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for phenomena such as surface wetting and capillary rise. Key Equations \(h=\dfrac{2T\cosθ}{rρg}\) Glossary adhesive force force of attraction between molecules of different chemical identities capillary action flow of liquid within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules cohesive force force of attraction between identical molecules surface tension energy required to increase the area, or length, of a liquid surface by a given amount viscosity measure of a liquid’s resistance to flow |
Courses/can/CHEM_210_General_Chemistry_I_(Puenzo)/10%3A_Chemical_Bonding_I-_Lewis_Structures/10.09%3A_Exceptions_to_the_Octet_Rule | Learning Objectives To assign a Lewis dot symbol to elements not having an octet of electrons in their compounds. Three cases can be constructed that do not follow the octet rule, and as such, they are known as the exceptions to the octet rule. Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. However, it is hard to imagine that one rule could be followed by all molecules. There is always an exception, and in this case, three exceptions: When there are an odd number of valence electrons When there are too few valence electrons When there are too many valence electrons Exception 1: Species with Odd Numbers of Electrons The first exception to the Octet Rule is when there are an odd number of valence electrons. An example of this would be Nitrogen (II) Oxide also called nitric oxide (\(\ce{NO}\). Nitrogen has 5 valence electrons while Oxygen has 6. The total would be 11 valence electrons to be used. The Octet Rule for this molecule is fulfilled in the above example, however that is with 10 valence electrons. The last one does not know where to go. The lone electron is called an unpaired electron. But where should the unpaired electron go? The unpaired electron is usually placed in the Lewis Dot Structure so that each element in the structure will have the lowest formal charge possible. The formal charge is the perceived charge on an individual atom in a molecule when atoms do not contribute equal numbers of electrons to the bonds they participate in . No formal charge at all is the most ideal situation. An example of a stable molecule with an odd number of valence electrons would be nitric oxide. nitric oxide has 11 valence electrons. If you need more information about formal charges, see Lewis Structures. If we were to imagine nitric oxide had ten valence electrons we would come up with the Lewis Structure (Figure \(\PageIndex{1}\)): Let's look at the formal charges of Figure \(\PageIndex{2}\) based on this Lewis structure. Nitrogen normally has five valence electrons. In Figure \(\PageIndex{1}\), it has two lone pair electrons and it participates in two bonds (a double bond) with oxygen. This results in nitrogen having a formal charge of +1. Oxygen normally has six valence electrons. In Figure \(\PageIndex{1}\), oxygen has four lone pair electrons and it participates in two bonds with nitrogen. Oxygen therefore has a formal charge of 0. The overall molecule here has a formal charge of +1 (+1 for nitrogen, 0 for oxygen. +1 + 0 = +1). However, if we add the eleventh electron to nitrogen (because we want the molecule to have the lowest total formal charge), it will bring both the nitrogen and the molecule's overall charges to zero, the most ideal formal charge situation. That is exactly what is done to get the correct Lewis structure for nitric oxide (Figure \(\PageIndex{2}\)): Free Radicals There are actually very few stable molecules with odd numbers of electrons that exist, since that unpaired electron is willing to react with other unpaired electrons. Most odd electron species are highly reactive, which we call Free Radicals. Because of their instability, free radicals bond to atoms in which they can take an electron from in order to become stable, making them very chemically reactive. Radicals are found as both reactants and products, but generally react to form more stable molecules as soon as they can. In order to emphasize the existence of the unpaired electron, radicals are denoted with a dot in front of their chemical symbol as with \(\cdot OH\), the hydroxyl radical. An example of a radical you may by familiar with already is the gaseous chlorine atom, denoted \(\cdot Cl\). Interestingly, an odd Number of Valence Electrons will result in the molecule being paramagnetic. Exception 2: Incomplete Octets The second exception to the Octet Rule is when there are too few valence electrons that results in an incomplete Octet. There are even more occasions where the octet rule does not give the most correct depiction of a molecule or ion. This is also the case with incomplete octets. Species with incomplete octets are pretty rare and generally are only found in some beryllium, aluminum, and boron compounds including the boron hydrides. Let's take a look at one such hydride, BH 3 (Borane). If one were to make a Lewis structure for BH 3 following the basic strategies for drawing Lewis structures, one would probably come up with this structure (Figure \(\PageIndex{2}\)): The problem with this structure is that boron has an incomplete octet; it only has six electrons around it. Hydrogen atoms can naturally only have only 2 electrons in their outermost shell (their version of an octet), and as such there are no spare electrons to form a double bond with boron. One might surmise that the failure of this structure to form complete octets must mean that this bond should be ionic instead of covalent. However, boron has an electronegativity that is very similar to hydrogen, meaning there is likely very little ionic character in the hydrogen to boron bonds, and as such this Lewis structure, though it does not fulfill the octet rule, is likely the best structure possible for depicting BH 3 with Lewis theory. One of the things that may account for BH 3 's incomplete octet is that it is commonly a transitory species, formed temporarily in reactions that involve multiple steps. Let's take a look at another incomplete octet situation dealing with boron, BF 3 (Boron trifluorine). Like with BH 3 , the initial drawing of a Lewis structure of BF 3 will form a structure where boron has only six electrons around it (Figure \(\PageIndex{4}\)). If you look Figure \(\PageIndex{4}\), you can see that the fluorine atoms possess extra lone pairs that they can use to make additional bonds with boron, and you might think that all you have to do is make one lone pair into a bond and the structure will be correct. If we add one double bond between boron and one of the fluorines we get the following Lewis Structure (Figure \(\PageIndex{5}\)): Each fluorine has eight electrons, and the boron atom has eight as well! Each atom has a perfect octet, right? Not so fast. We must examine the formal charges of this structure. The fluorine that shares a double bond with boron has six electrons around it (four from its two lone pairs of electrons and one each from its two bonds with boron). This is one less electron than the number of valence electrons it would have naturally (Group Seven elements have seven valence electrons), so it has a formal charge of +1. The two flourines that share single bonds with boron have seven electrons around them (six from their three lone pairs and one from their single bonds with boron). This is the same amount as the number of valence electrons they would have on their own, so they both have a formal charge of zero. Finally, boron has four electrons around it (one from each of its four bonds shared with fluorine). This is one more electron than the number of valence electrons that boron would have on its own, and as such boron has a formal charge of -1. This structure is supported by the fact that the experimentally determined bond length of the boron to fluorine bonds in BF 3 is less than what would be typical for a single bond (see Bond Order and Lengths). However, this structure contradicts one of the major rules of formal charges: Negative formal charges are supposed to be found on the more electronegative atom(s) in a bond, but in the structure depicted in Figure \(\PageIndex{5}\), a positive formal charge is found on fluorine, which not only is the most electronegative element in the structure, but the most electronegative element in the entire periodic table (\(\chi=4.0\)). Boron on the other hand, with the much lower electronegativity of 2.0, has the negative formal charge in this structure. This formal charge-electronegativity disagreement makes this double-bonded structure impossible. However the large electronegativity difference here, as opposed to in BH 3 , signifies significant polar bonds between boron and fluorine, which means there is a high ionic character to this molecule. This suggests the possibility of a semi-ionic structure such as seen in Figure \(\PageIndex{6}\): None of these three structures is the "correct" structure in this instance. The most "correct" structure is most likely a resonance of all three structures: the one with the incomplete octet (Figure \(\PageIndex{4}\)), the one with the double bond (Figure \(\PageIndex{5}\)), and the one with the ionic bond (Figure \(\PageIndex{6}\)). The most contributing structure is probably the incomplete octet structure (due to Figure \(\PageIndex{5}\) being basically impossible and Figure \(\PageIndex{6}\) not matching up with the behavior and properties of BF 3 ). As you can see even when other possibilities exist, incomplete octets may best portray a molecular structure. As a side note, it is important to note that BF 3 frequently bonds with a F - ion in order to form BF 4 - rather than staying as BF 3 . This structure completes boron's octet and it is more common in nature. This exemplifies the fact that incomplete octets are rare, and other configurations are typically more favorable, including bonding with additional ions as in the case of BF 3 . Example \(\PageIndex{1}\): \(NF_3\) Draw the Lewis structure for boron trifluoride (BF 3 ). Solution 1. Add electrons (3*7) + 3 = 24 2. Draw connectivities: 3. Add octets to outer atoms: 4. Add extra electrons (24-24=0) to central atom: 5. Does central electron have octet? NO. It has 6 electrons Add a multiple bond (double bond) to see if central atom can achieve an octet: 6. The central Boron now has an octet (there would be three resonance Lewis structures) However... In this structure with a double bond the fluorine atom is sharing extra electrons with the boron. The fluorine would have a '+' partial charge, and the boron a '-' partial charge, this is inconsistent with the electronegativities of fluorine and boron. Thus, the structure of BF 3 , with single bonds, and 6 valence electrons around the central boron is the most likely structure BF 3 reacts strongly with compounds which have an unshared pair of electrons which can be used to form a bond with the boron: Exception 3: Expanded Valence Shells More common than incomplete octets are expanded octets where the central atom in a Lewis structure has more than eight electrons in its valence shell. In expanded octets, the central atom can have ten electrons, or even twelve. Molecules with expanded octets involve highly electronegative terminal atoms, and a nonmetal central atom found in the third period or below , which those terminal atoms bond to. For example, \(PCl_5\) is a legitimate compound (whereas \(NCl_5\)) is not: Expanded valence shells are observed only for elements in period 3 (i.e. n=3) and beyond The 'octet' rule is based upon available n s and n p orbitals for valence electrons (2 electrons in the s orbitals, and 6 in the p orbitals). Beginning with the n=3 principle quantum number, the d orbitals become available ( l =2). The orbital diagram for the valence shell of phosphorous is: Hence, the third period elements occasionally exceed the octet rule by using their empty d orbitals to accommodate additional electrons. Size is also an important consideration: The larger the central atom, the larger the number of electrons which can surround it Expanded valence shells occur most often when the central atom is bonded to small electronegative atoms, such as F, Cl and O. There is currently much scientific exploration and inquiry into the reason why expanded valence shells are found. The top area of interest is figuring out where the extra pair(s) of electrons are found. Many chemists think that there is not a very large energy difference between the 3p and 3d orbitals, and as such it is plausible for extra electrons to easily fill the 3d orbital when an expanded octet is more favorable than having a complete octet. This matter is still under hot debate, however and there is even debate as to what makes an expanded octet more favorable than a configuration that follows the octet rule. One of the situations where expanded octet structures are treated as more favorable than Lewis structures that follow the octet rule is when the formal charges in the expanded octet structure are smaller than in a structure that adheres to the octet rule, or when there are less formal charges in the expanded octet than in the structure a structure that adheres to the octet rule. Example \(\PageIndex{2}\): The \(SO_4^{-2}\) ion Such is the case for the sulfate ion, SO 4 -2 . A strict adherence to the octet rule forms the following Lewis structure: If we look at the formal charges on this molecule, we can see that all of the oxygen atoms have seven electrons around them (six from the three lone pairs and one from the bond with sulfur). This is one more electron than the number of valence electrons then they would have normally, and as such each of the oxygens in this structure has a formal charge of -1. Sulfur has four electrons around it in this structure (one from each of its four bonds) which is two electrons more than the number of valence electrons it would have normally, and as such it carries a formal charge of +2. If instead we made a structure for the sulfate ion with an expanded octet, it would look like this: Looking at the formal charges for this structure, the sulfur ion has six electrons around it (one from each of its bonds). This is the same amount as the number of valence electrons it would have naturally. This leaves sulfur with a formal charge of zero. The two oxygens that have double bonds to sulfur have six electrons each around them (four from the two lone pairs and one each from the two bonds with sulfur). This is the same amount of electrons as the number of valence electrons that oxygen atoms have on their own, and as such both of these oxygen atoms have a formal charge of zero. The two oxygens with the single bonds to sulfur have seven electrons around them in this structure (six from the three lone pairs and one from the bond to sulfur). That is one electron more than the number of valence electrons that oxygen would have on its own, and as such those two oxygens carry a formal charge of -1. Remember that with formal charges, the goal is to keep the formal charges (or the difference between the formal charges of each atom) as small as possible. The number of and values of the formal charges on this structure (-1 and 0 (difference of 1) in Figure \(\PageIndex{12}\), as opposed to +2 and -1 (difference of 3) in Figure \(\PageIndex{12}\)) is significantly lower than on the structure that follows the octet rule, and as such an expanded octet is plausible, and even preferred to a normal octet, in this case. Example \(\PageIndex{3}\): The \(ICl_4^-\) Ion Draw the Lewis structure for \(ICl_4^-\) ion. Solution 1. Count up the valence electrons: 7+(4*7)+1 = 36 electrons 2. Draw the connectivities: 3. Add octet of electrons to outer atoms: 4. Add extra electrons (36-32= 4 ) to central atom: 5. The ICl 4 - ion thus has 12 valence electrons around the central Iodine (in the 5 d orbitals) Expanded Lewis structures are also plausible depictions of molecules when experimentally determined bond lengths suggest partial double bond characters even when single bonds would already fully fill the octet of the central atom. Despite the cases for expanded octets, as mentioned for incomplete octets, it is important to keep in mind that, in general, the octet rule applies. Summary Following the Octet Rule for Lewis Dot Structures leads to the most accurate depictions of stable molecular and atomic structures and because of this we always want to use the octet rule when drawing Lewis Dot Structures. There are three exceptions: (1) When there are an odd number of valence electrons, (2) When there are too few valence electrons, and (3) when there are too many valence electrons References Petrucci, Ralph H.; Harwood, William S.; Herring, F. G.; Madura, Jeffrey D. General Chemistry: Principles & Modern Applications . 9th Ed. New Jersey. Pearson Education, Inc. 2007. Moore, John W.; Stanitski, Conrad L. ; Jurs, Peter C. Chemistry; The Molecular Science . 2nd Ed. 2004. |
Courses/CSU_San_Bernardino/CHEM_4300%3A_Inorganic_Chemistry_(Mink)/05%3A_Coordination_Chemistry_I_-_Structures_and_Isomers/5.01%3A_History | Coordination Chemistry Now let us direct our focus toward actual coordination chemistry, also often called complex chemistry. In the following, we will apply the previously learned concepts about atomic theory, symmetry, molecular orbital theory, and acid-base chemistry to coordination compounds. Let us first ask: How are coordination compounds, or complexes defined? They are typically Lewis-acid base adducts between a metal atom or a metal ion as the Lewis acid and one or more ligands as a Lewis base. These ligands can be inorganic ligands such as halogenide ions, water, and ammonia molecules, or organic ligands like amines or alcohols. Coordination compounds are known since the antique due to their often intense color, and are used as a pigment or dye, for example Prussian Blue (KFe[Fe(CN) 6 ] or tetrammine copper, both of which are intensely blue. You can see a few examples of coordination compounds below, namely the hexahydrate of copper (II) sulfate which is blue, iron (III) chloride, which is yellow, and nickel sulfate which is greenish-blue. Because of their intense color they have also attracted the attention of modern chemists from the early on, but the chemical bonding in these compounds remained a mystery for a relatively long time. The bonding in these compounds seemed more complex, hence the name complex compounds. History on Coordination Compounds What was the problem with the bonding in coordination compounds? Their empirical formulas could be easily determined by element analysis, but the results could not be explained with the concept of valence. In the early days of modern chemistry it was believed that the number of bonds in a compound could not exceed the valence. For example, in the compound of the formula Co(NH 3 ) 6 Cl 3 the valence of Co would be +3, therefore cobalt could not make more than three bonds. Assuming that the Co 3 + ion would make three bonds to the 3 Cl - ions, how would one involve the six NH 3 molecules in the bonding? The chemist who solved this mystery was Alfred Werner. He correctly suggested that the number of bonds would not be restricted to the valence, but that more bonds would be allowed. It would be possible that the six NH 3 molecules bind directly to Co 3 + forming a so-called complex cation. They would be in the so-called first coordination sphere around the Co. The three chloride ions would then be bound loosely to the complex cation balancing the charge of the complex cation. They would be in the second coordination sphere. Formula Writing in Coordination Chemistry How can we write a formula of a coordination compound that contains complex ions? For complex cations we write the element symbol for the metal ion first followed by the formula for the ligands. If there is more than one ligand, then we will place the ligands in parentheses, and indicate their number by a subscript behind the parentheses. The entire complex cation is placed in brackets. The formula for the anion in the second coordination sphere is placed behind the brackets and the number of anions is indicated by a subscript (Fig. 5.1.3). For complex anions, we write the formula for the counter cation first, followed by the complex anion in brackets. You can see two examples above (Fig. 5.1.3 and 5.1.4). We have complex cations of Co with six NH 3 ligands coordinated to it in the first coordination sphere. Three Cl - ions are in the second coordination sphere. Hence the formula is [Co(NH 3 ) 6 ]Cl 3 . The second example is the coordination compound with the complex anion having a 3- charge in which six cyanide anions bind to a central Fe 3 + ion. The K + ion in the second coordination sphere compensates the charge of the complex anion. Hence the formula is K 3 [Fe(CN) 6 ]. Dr. Kai Landskron ( Lehigh University ). If you like this textbook, please consider to make a donation to support the author's research at Lehigh University: Click Here to Donate . |
Courses/can/CHEM_210%3A_General_Chemistry_I_(An_Atoms_Up_Approach)/13%3A_Thermochemistry/13.03%3A__Enthalpy_H_and_Heat_of_Reaction | Learning Objectives To understand how enthalpy pertains to chemical reactions We have stated that the change in energy (Δ U ) is equal to the sum of the heat produced and the work performed. Work done by an expanding gas is called pressure-volume work , (or just PV work) . Consider, for example, a reaction that produces a gas, such as dissolving a piece of copper in concentrated nitric acid. The chemical equation for this reaction is as follows: \[Cu_{(s)} + 4HNO_{3(aq)} \rightarrow Cu(NO_3)_{2(aq)} + 2H_2O_{(l)} + 2NO_{2(g)} \label{5.4.1}\] If the reaction is carried out in a closed system that is maintained at constant pressure by a movable piston, the piston will rise as nitrogen dioxide gas is formed (Figure (\PageIndex{1}\)). The system is performing work by lifting the piston against the downward force exerted by the atmosphere (i.e., atmospheric pressure ). We find the amount of PV work done by multiplying the external pressure P by the change in volume caused by movement of the piston (Δ V ). At a constant external pressure (here, atmospheric pressure) \[w = −PΔV \label{5.4.2}\] The negative sign associated with PV work done indicates that the system loses energy when the volume increases. If the volume increases at constant pressure (Δ V > 0), the work done by the system is negative, indicating that a system has lost energy by performing work on its surroundings. Conversely, if the volume decreases (Δ V < 0), the work done by the system is positive, which means that the surroundings have performed work on the system, thereby increasing its energy. The internal energy \(U\) of a system is the sum of the kinetic energy and potential energy of all its components. It is the change in internal energy that produces heat plus work. To measure the energy changes that occur in chemical reactions, chemists usually use a related thermodynamic quantity called enthalpy ( H ) (from the Greek enthalpein , meaning “to warm”). The enthalpy of a system is defined as the sum of its internal energy \(U\) plus the product of its pressure P and volume V : \[H =U + PV \label{5.4.3}\] Because internal energy, pressure, and volume are all state functions, enthalpy is also a state function. If a chemical change occurs at constant pressure (i.e., for a given P , Δ P = 0), the change in enthalpy (Δ H ) is \[ ΔH = Δ(U + PV) = ΔU + ΔPV = ΔU + PΔV \label{5.4.4}\] Substituting q + w for ΔU (Equation \(\ref{5.4.4}\)) and − w for P Δ V (Equation \(\ref{5.4.2}\)), we obtain \[ΔH = ΔU + PΔV = q_p + w − w = q_p \label{5.4.5}\] The subscript \(p\) is used here to emphasize that this equation is true only for a process that occurs at constant pressure. From Equation \(\ref{5.4.5}\) we see that at constant pressure the change in enthalpy, Δ H of the system, defined as H final − H initial , is equal to the heat gained or lost. \[ΔH = H_{final} − H_{initial} = q_p \label{5.4.6}\] Just as with ΔU, because enthalpy is a state function, the magnitude of Δ H depends on only the initial and final states of the system, not on the path taken. Most important, the enthalpy change is the same even if the process does not occur at constant pressure. To find \(ΔH\) for a reaction , measure \(q_p\). When we study energy changes in chemical reactions, the most important quantity is usually the enthalpy of reaction (Δ H rxn ), the change in enthalpy that occurs during a reaction (such as the dissolution of a piece of copper in nitric acid). If heat flows from a system to its surroundings, the enthalpy of the system decreases, so Δ H rxn is negative. Conversely, if heat flows from the surroundings to a system, the enthalpy of the system increases, so Δ H rxn is positive. Thus Δ H rxn < 0 for an exothermic reaction , and Δ H rxn > 0 for an endothermic reaction . In chemical reactions, bond breaking requires an input of energy and is therefore an endothermic process, whereas bond making releases energy, which is an exothermic process. The sign conventions for heat flow and enthalpy changes are summarized in the following table: Reaction Type q ΔHrxn exothermic < 0 < 0 (heat flows from a system to its surroundings) endothermic > 0 > 0 (heat flows from the surroundings to a system) If Δ H rxn is negative, then the enthalpy of the products is less than the enthalpy of the reactants; that is, an exothermic reaction is energetically downhill (part (a) in Figure 5.4.2). Conversely, if Δ H rxn is positive, then the enthalpy of the products is greater than the enthalpy of the reactants; thus, an endothermic reaction is energetically uphill (part (b) in Figure \(\PageIndex{2}\)). Two important characteristics of enthalpy and changes in enthalpy are summarized in the following discussion. Bond breaking ALWAYS requires an input of energy; bond making ALWAYS releases energy.y. Reversing a reaction or a process changes the sign of Δ H . Ice absorbs heat when it melts (electrostatic interactions are broken), so liquid water must release heat when it freezes (electrostatic interactions are formed): \( \begin{matrix} heat+ H_{2}O(s) \rightarrow H_{2}O(l) & \Delta H > 0 \end{matrix} \label{5.4.7} \) \( \begin{matrix} H_{2}O(l) \rightarrow H_{2}O(s) + heat & \Delta H < 0 \end{matrix} \label{5.4.8} \) In both cases, the magnitude of the enthalpy change is the same; only the sign is different. Enthalpy is an extensive property (like mass). The magnitude of Δ H for a reaction is proportional to the amounts of the substances that react. For example, a large fire produces more heat than a single match, even though the chemical reaction—the combustion of wood—is the same in both cases. For this reason, the enthalpy change for a reaction is usually given in kilojoules per mole of a particular reactant or product. Consider Equation \(\ref{5.4.9}\), which describes the reaction of aluminum with iron(III) oxide (Fe 2 O 3 ) at constant pressure. According to the reaction stoichiometry, 2 mol of Fe, 1 mol of Al 2 O 3 , and 851.5 kJ of heat are produced for every 2 mol of Al and 1 mol of Fe 2 O 3 consumed: \[ 2Al\left (s \right )+Fe_{2}O_{3}\left (s \right ) \rightarrow 2Fe\left (s \right )+Al_{2}O_{3}\left (s \right )+ 815.5 \; kJ \label{5.4.9} \] Thus Δ H = −851.5 kJ/mol of Fe 2 O 3 . We can also describe Δ H for the reaction as −425.8 kJ/mol of Al: because 2 mol of Al are consumed in the balanced chemical equation, we divide −851.5 kJ by 2. When a value for Δ H , in kilojoules rather than kilojoules per mole, is written after the reaction, as in Equation \(\ref{5.4.10}\), it is the value of Δ H corresponding to the reaction of the molar quantities of reactants as given in the balanced chemical equation: \[ 2Al\left (s \right )+Fe_{2}O_{3}\left (s \right ) \rightarrow 2Fe\left (s \right )+Al_{2}O_{3}\left (s \right ) \;\;\;\; \Delta H_{rxn}= - 851.5 \; kJ \label{5.4.10} \] If 4 mol of Al and 2 mol of Fe 2 O 3 react, the change in enthalpy is 2 × (−851.5 kJ) = −1703 kJ. We can summarize the relationship between the amount of each substance and the enthalpy change for this reaction as follows: \[ - \dfrac{851.5 \; kJ}{2 \; mol \;Al} = - \dfrac{425.8 \; kJ}{1 \; mol \;Al} = - \dfrac{1703 \; kJ}{4 \; mol \; Al} \label{5.4.6a} \] The relationship between the magnitude of the enthalpy change and the mass of reactants is illustrated in Example \(\PageIndex{1}\). Example \(\PageIndex{1}\): Melting Icebergs Certain parts of the world, such as southern California and Saudi Arabia, are short of freshwater for drinking. One possible solution to the problem is to tow icebergs from Antarctica and then melt them as needed. If Δ H is 6.01 kJ/mol for the reaction at 0°C and constant pressure: \[H_2O_{(s)} → H_2O_{(l)}\] How much energy would be required to melt a moderately large iceberg with a mass of 1.00 million metric tons (1.00 × 10 6 metric tons)? (A metric ton is 1000 kg.) Given: energy per mole of ice and mass of iceberg Asked for: energy required to melt iceberg Strategy: Calculate the number of moles of ice contained in 1 million metric tons (1.00 × 10 6 metric tons) of ice. Calculate the energy needed to melt the ice by multiplying the number of moles of ice in the iceberg by the amount of energy required to melt 1 mol of ice. Solution: A Because enthalpy is an extensive property, the amount of energy required to melt ice depends on the amount of ice present. We are given Δ H for the process—that is, the amount of energy needed to melt 1 mol (or 18.015 g) of ice—so we need to calculate the number of moles of ice in the iceberg and multiply that number by Δ H (+6.01 kJ/mol): \[ \begin{matrix} moles \; H_{2}O & = & 1.00\times 10^{6} \; metric \; tons H_{2}O \left ( \dfrac{1000 \; \cancel{kg}}{1 \; \cancel{metric \; ton}} \right ) \left ( \dfrac{1000 \; \cancel{g}}{1 \; \cancel{kg}} \right ) \left ( \dfrac{1 \; mol \; H_{2}O}{18.015 \; \cancel{g \; H_{2}O}} \right )\\ & = & 5.55\times 10^{10} \; mol H_{2}O \end{matrix} \] B The energy needed to melt the iceberg is thus \[ \left ( \dfrac{6.01 \; kJ}{\cancel{mol \; H_{2}O}} \right )\left ( 5.55 \times 10^{10} \; \cancel{mol \; H_{2}O} \right )= 3.34 \times 10^{11} \; kJ \] Because so much energy is needed to melt the iceberg, this plan would require a relatively inexpensive source of energy to be practical. To give you some idea of the scale of such an operation, the amounts of different energy sources equivalent to the amount of energy needed to melt the iceberg are shown in the table below. Possible sources of the approximately 3.34 × 10 11 kJ needed to melt a 1.00 × 10 6 metric ton iceberg Combustion of 3.8 × 10 3 ft 3 of natural gas Combustion of 68,000 barrels of oil Combustion of 15,000 tons of coal 1.1 × 10 8 kilowatt-hours of electricity Exercise \(\PageIndex{1}\): Thermite Reaction If 17.3 g of powdered aluminum are allowed to react with excess Fe 2 O 3 , how much heat is produced? Answer: 273 kJ Enthalpies of Reaction One way to report the heat absorbed or released would be to compile a massive set of reference tables that list the enthalpy changes for all possible chemical reactions, which would require an incredible amount of effort. Fortunately, Hess’s law allows us to calculate the enthalpy change for virtually any conceivable chemical reaction using a relatively small set of tabulated data, such as the following: Enthalpy of combustion (ΔH comb ) The change in enthalpy that occurs during a combustion reaction. Enthalpy changes have been measured for the combustion of virtually any substance that will burn in oxygen; these values are usually reported as the enthalpy of combustion per mole of substance. Enthalpy of fusion (ΔH fus ) The enthalpy change that acompanies the melting (fusion) of 1 mol of a substance. The enthalpy change that accompanies the melting, or fusion, of 1 mol of a substance; these values have been measured for almost all the elements and for most simple compounds. Enthalpy of vaporization (ΔH vap ) The enthalpy change that accompanies the vaporization of 1 mol of a substance. The enthalpy change that accompanies the vaporization of 1 mol of a substance; these values have also been measured for nearly all the elements and for most volatile compounds. Enthalpy of solution (ΔH soln ) The change in enthalpy that occurs when a specified amount of solute dissolves in a given quantity of solvent. The enthalpy change when a specified amount of solute dissolves in a given quantity of solvent. Substance ΔHvap (kJ/mol) ΔHfus (kJ/mol) argon (Ar) 6.3 1.30 methane (CH4) 9.2 0.84 ethanol (CH3CH2OH) 39.3 7.60 benzene (C6H6) 31.0 10.90 water (H2O) 40.7 6.00 mercury (Hg) 59.0 2.29 iron (Fe) 340.0 14.00 The sign convention is the same for all enthalpy changes: negative if heat is released by the system and positive if heat is absorbed by the system. Summary For a chemical reaction, the enthalpy of reaction (\(ΔH_{rxn}\)) is the difference in enthalpy between products and reactants; the units of \(ΔH_{rxn}\) are kilojoules per mole. Reversing a chemical reaction reverses the sign of \(ΔH_{rxn}\). |
Courses/Maryville_College/Essential_Chemistry_for_Poisons_Potions_and_Pharmaceuticals/11%3A_Drugs/11.07%3A_Drugs_and_the_Mind | Learning Objectives Describe how neurotransmitters work. Know the different types of drugs and their functions. Art in a Cup Who knew that a cup of coffee could also be a work of art? A talented barista can make coffee look as good as it tastes. If you are a coffee drinker, you probably know that coffee can also affect your mental state. It can make you more alert and may improve your concentration. That’s because the caffeine in coffee is a psychoactive drug. In fact, caffeine is the most widely consumed psychoactive substance in the world. In North America, for example, 90 percent of adults consume caffeine daily. Psychoactive drugs are substances that change the function of the brain and result in alterations of mood, thinking, perception, and/or behavior. Psychoactive drugs may be used for many purposes, including therapeutic, ritual, or recreational purposes. Besides caffeine, other examples of psychoactive drugs include cocaine, LSD, alcohol, tobacco, codeine, and morphine. Psychoactive drugs may be legal prescription medications (e.g., codeine and morphine), legal nonprescription drugs (e.g., alcohol and tobacco), or illegal drugs (cocaine and LSD). Cannabis (or marijuana) is also a psychoactive drug, but its status is in flux, at least in the United States. Depending on the jurisdiction, cannabis may be used recreationally and/or medically, and it may be either legal or illegal. Legal prescription medications such as opioids are also used illegally by increasingly large numbers of people. Some legal drugs, such as alcohol and nicotine, are readily available almost everywhere, as illustrated by the sign pictured in Figure \(\PageIndex{2}\). Classes of Psychoactive Drugs Psychoactive drugs are divided into different classes according to their pharmacological effects. Several classes are listed below, along with examples of commonly used drugs in each class. Stimulants are drugs that stimulate the brain and increase alertness and wakefulness. Examples of stimulants include caffeine, nicotine, cocaine, and amphetamines such as Adderall. Depressants are drugs that calm the brain, reduce anxious feelings, and induce sleepiness. Examples of depressants include ethanol (in alcoholic beverages) and opioids such as codeine and heroin. Anxiolytics are drugs that have a tranquilizing effect and inhibit anxiety. Examples of anxiolytic drugs include benzodiazepines such as diazepam (Valium), barbiturates such as phenobarbital, opioids, and antidepressant drugs such as sertraline (Zoloft). Euphoriants are drugs that bring about a state of euphoria, or intense feelings of well-being and happiness. Examples of euphoriants include the so-called club drug MDMA (ecstasy), amphetamines, ethanol, and opioids such as morphine. Hallucinogens are drugs that can cause hallucinations and other perceptual anomalies. They also cause subjective changes in thoughts, emotions, and consciousness. Examples of hallucinogens include LSD, mescaline, nitrous oxide, and psilocybin. Empathogens are drugs that produce feelings of empathy, or sympathy with other people. Examples of empathogens include amphetamines and MDMA. Many psychoactive drugs have multiple effects so they may be placed in more than one class. An example is MDMA, pictured below, which may act both as a euphoriant and as an empathogen. In some people, MDMA may also have stimulant or hallucinogenic effects. As of 2016, MDMA had no accepted medical uses, but it was undergoing testing for use in the treatment of post-traumatic stress disorder and certain other types of anxiety disorders. Mechanisms of Action Psychoactive drugs generally produce their effects by affecting brain chemistry, which in turn may cause changes in a person’s mood, thinking, perception, and/or behavior. Each drug tends to have a specific action on one or more neurotransmitters or neurotransmitter receptors in the brain. Generally, they act as either agonists or antagonists. Agonists are drugs that increase the activity of particular neurotransmitters. They might act by promoting the synthesis of the neurotransmitters, reducing their reuptake from synapses, or mimicking their action by binding to receptors for the neurotransmitters. Antagonists are drugs that decrease the activity of particular neurotransmitters. They might act by interfering with the synthesis of the neurotransmitters or by blocking their receptors so the neurotransmitters cannot bind to them. Consider the example of the neurotransmitter GABA. This is one of the most common neurotransmitters in the brain, and it normally has an inhibitory effect on cells. GABA agonists, which increase its activity, include ethanol, barbiturates, and benzodiazepines, among other psychoactive drugs. All of these drugs work by promoting the activity of GABA receptors in the brain. Chemistry of the Nervous System Neurons, also called nerve cells, are electrically excitable cells that are the main functional units of the nervous system. Their function is to transmit nerve impulses. They are the only type of human cells that can carry out this function. The main parts of a neuron are labeled in Figure \(\PageIndex{4}\) and described below. The cell body is the part of a neuron that contains the cell nucleus and other cell organelles. It is usually quite compact, and may not be much wider than the nucleus. Dendrites are thin structures that are extensions of the cell body. Their function is to receive nerve impulses from other cells and carry them to the cell body. A neuron may have many dendrites, and each dendrite may branch repeatedly to form a dendrite “tree” with more than 1,000 “branches.” The end of each branch can receive nerve impulses from another cell, allowing a given neuron to communicate with tens of thousands of other cells. The axon is a long, thin extension of the cell body. It transmits nerve impulses away from the cell body and toward other cells. The axon branches at the end, forming multiple axon terminals. These are the points where nerve impulses are transmitted to other cells, often to the dendrites of other neurons. An area called a synapse occurs at each axon terminal. Synapses are complex membrane junctions that transmit signals to other cells. An axon may branch hundreds of times, but there is never more than one axon per neuron. Spread out along axons, especially the long axons of nerves, are many sections of the myelin sheath. These are lipid layers that cover sections of the axon. The myelin sheath is a very good electrical insulator, similar to the plastic or rubber that encases an electrical cord. Regularly spaced gaps between sections of myelin sheath occur along the axon. These gaps are called nodes of Ranvier , and they allow the transmission of nerve impulses along the axon. Nerve impulses skip from node to node, allowing nerve impulses to travel along the axon very rapidly. A Schwann cell (also on an axon) is a type of glial cell. Its function is to produce the myelin sheath that insulates axons in the peripheral nervous system. In the central nervous system, a different type of glial cell, called an oligodendrocyte, produces the myelin sheath. Neurotransmission is the process by which signaling molecules called neurotransmitters are released by the axon terminal of a neuron (the presynaptic neuron), and bind to and react with the receptors on the dendrites of another neuron (the postsynaptic neuron) a short distance away. Synapses are functional connections between neurons, or between neurons and other types of cells. The synaptic cleft —also called synaptic gap — is a gap between the pre- and postsynaptic cells that is about 20 nm (0.02 μ) wide. The small volume of the cleft allows neurotransmitter concentration to be raised and lowered rapidly. Neurotransmiters Neurotransmitters are endogenous chemicals that enable neurotransmission. It is a type of chemical messenger which transmits signals across a chemical synapse, such as a neuromuscular junction, from one neuron (nerve cell) to another "target" neuron, muscle cell, or gland cell. Biochemical Theories of Brain Diseases One theory that explains mental illness incorporates the various amounts of neurotransmitters. Using this theory, the lack or excess of specific neurotransmitters can be associated with depression, anxiety, bipolar disorder, or schizophrenia. Other disorders or health conditions, namely Attention Deficient Hyperactivity Disorder (ADHD) and Parkinson's Disease have also been linked with varying levels of neurotransmitters. The Noradrenaline neurotransmitter is also called norepinephrine. When evaluating antidepressants or other mental health medications, it is important to be aware of this terminology. For a brief overview of the correlation between neurotransmitters and mental health, watch the video below. Mental health conditions can result from a person's life experiences and/or genetics. Sometimes, drug usage or brain trauma can trigger mental illness. It is important to note all health conditions that are present in your family tree (biological). Also, recognizing emotional trauma and then seeking counseling/medical treatment is important for navigating a mental health condition. Depression symptoms Can be short or long-term. Experience-sadness, sleeping and eating issues, withdrawal, feelings of hopelessness, loss of interest/pleasure, lack of energy, feelings of worthlessness/guilt, slowed processing, trouble concentrating, frequent thought of suicide/death, anxiety, and unexplained health problems. Treatments for depression psychotherapy, brain stimulation, medication, exercise, light therapy. Discussions to have with dr: meds/vitamins you are on, self-medication is not answer, stopping antidepressants without assistance, and report problems with meds. Limit alcohol and refrain from illicit and scheduled drugs. FDA black-box warning for antidepressants in people under 25. Antidepressant Medications Finding the correct antidepressant Trial and Error-take for six weeks, then wean, and start another. Since the early 2000's, Mayo Clinic has been researching gene technology. Test is now offered by Assurex Health and called Genesight. https://genesight.com/gene-test-for-...surex%20Health Common Side Effects of Antidepressants This carry varies with the patient and type of medicine. Most common include weight gain, fatigue, dizziness, loss of sexual desire, nausea, dry mouth, blurred vision, agitation, insomnia, and constipation. Support and information regarding mental health conditions. Anesthetics An anesthetic or anaesthetic is a drug used to induce anesthesia - in other words, to result in a temporary loss of sensation or awareness. They may be divided into two broad classes: general anesthetics, which result in a reversible loss of consciousness, and local anesthetics, which cause a reversible loss of sensation for a limited region of the body without necessarily affecting consciousness. A wide variety of drugs are used in modern anesthetic practice. Many are rarely used outside anesthesiology, but others are used commonly in various fields of healthcare. Combinations of anesthetics are sometimes used for their synergistic and additive therapeutic effects. Adverse effects, however, may also be increased. Anesthetics are distinct from analgesics, which block only sensation of painful stimuli. General Anesthetics General anesthetics are often defined as compounds that induce a loss of consciousness in humans or loss of righting reflex in animals. Clinical definitions are also extended to include the lack of awareness to painful stimuli, sufficient to facilitate surgical applications in clinical and veterinary practice. General anesthetics do not act as analgesics and should also not be confused with sedatives. General anesthetics are a structurally diverse group of compounds whose mechanisms encompasses multiple biological targets involved in the control of neuronal pathways. The precise workings are the subject of some debate and ongoing research. General anesthetics elicit a state of general anesthesia. It remains somewhat controversial regarding how this state should be defined. General anesthetics, however, typically elicit several key reversible effects: immobility, analgesia, amnesia, unconsciousness, and reduced autonomic responsiveness to noxious stimuli. Mode of administration Drugs given to induce general anesthesia can be either as gases or vapors (inhalational anesthetics), or as injections (intravenous anesthetics or even intramuscular). All of these agents share the property of being quite hydrophobic (i.e., as liquids, they are not freely miscible—or mixable—in water, and as gases they dissolve in oils better than in water). It is possible to deliver anesthesia solely by inhalation or injection, but most commonly the two forms are combined, with an injection given to induce anesthesia and a gas used to maintain it. Inhalation General anesthetics are frequently administered as volatile liquids or gases (Figure \(\PageIndex{6}\)). Inhalational anesthetic substances are either volatile liquids or gases, and are usually delivered using an anesthesia machine. An anesthesia machine allows composing a mixture of oxygen, anesthetics and ambient air, delivering it to the patient and monitoring patient and machine parameters. Liquid anesthetics are vaporized in the machine. Many compounds have been used for inhalation anesthesia, but only a few are still in widespread use. Desflurane, isoflurane and sevoflurane are the most widely used volatile anesthetics today. They are often combined with nitrous oxide. Older, less popular, volatile anesthetics, include halothane, enflurane, and methoxyflurane. Researchers are also actively exploring the use of xenon as an anesthetic. Injection Injectable anesthetics are used for the induction and maintenance of a state of unconsciousness. Anesthetists prefer to use intravenous injections, as they are faster, generally less painful and more reliable than intramuscular or subcutaneous injections. Among the most widely used drugs are propofol, etomidate, barbiturates such as methohexital and thiopentone/thiopenta, Benzodiazepines such as midazolam, Ketamine is used in the UK as "field anaesthesia", for instance in road traffic incidents or similar situations where an operation must be conducted at the scene or when there is not enough time to move to an operating room, while preferring other anesthetics where conditions allow their use. It is more frequently used in the operative setting in the US. Benzodiazepines are sedatives and are used in combinations with other general anesthetics. Local Anesthetics The first local anesthetic to be discovered was cocaine, an alkaloid contained in large amounts in the leav es of Erythroxylon coca , a shrub grow ing in the Andes Mountains. Over 9 million kilograms of these leaves are consumed annually by the 2 million inhabitants of the highlands of Peru, who chew or suck the leaves for the sense of wellbeing it produces. Local anesthetics are drugs that block nerve conduction when applied locally to nerve conduction when applied locally to nerve tissue in appropriate concentrations. They act on any part of the nervous system and on every type of nerve fiber. For example, when they are applied to the motor cortex impulse transmission from that area stops, and when they are injected into the skin they prevent the initiation and the transmission of sensory impulses. A local anesthetic in contact with a nerve trunk can cause both sensory and motor paralysis in the area innervated. The great practical advantage of the local anesthetic is that their action is reversible: their use is followed by complete recovery in nerve function with no evidence of structural damage to nerve fibers of cells. The structures some of the typical anesthetics are shown below. These structures contain hydrophilic and hydrophobic domains that are separated by an intermediate alkyl chain. Linkage of these two domains is of either the ester or amide type. the ester link is important because this bond is readily hydrolyzed during metabolic degradation and inactivation in the body. Procaine, for example, can be divided into three main portions: the aromatic acid (para-aminobenzoic), the alcohol (ethanol), and the tertiary amino group (diethylamino). Changes in any part of the molecule alter the anesthetic potency and the toxicity of the compound. Increasing the length of the alcohol group leads to a greater anesthetic potency. It also leads to an increase in toxicity. Depressants A depressant , or central depressant , is a drug that lowers neurotransmission levels, which is to depress or reduce arousal or stimulation, in various areas of the brain. Depressants are also occasionally referred to as "downers" as they lower the level of arousal when taken. Stimulants or "uppers" increase mental and/or physical function, hence the opposite drug class of depressants is stimulants, not antidepressants. Depressants are widely used throughout the world as prescription medicines and as illicit substances. When depressants are used, effects often include ataxia, anxiolysis, pain relief, sedation or somnolence, and cognitive/memory impairment, as well as in some instances euphoria, dissociation, muscle relaxation, lowered blood pressure or heart rate, respiratory depression, and anticonvulsant effects. Depressants also act to produce anesthesia. Cannabis may sometimes be considered a depressant due to one of its components, cannabidiol. The latter is known to treat insomnia, anxiety and muscle spasms similar to other depressive drugs. However, tetrahydrocannabinol, another component, may slow brain function to a small degree while reducing reaction to stimuli, it is generally considered to be a stimulant and main psychoactive agent to sometimes cause anxiety, panic and psychosis instead. Other depressants can include drugs like Xanax (a benzodiazepine) and a number of opiates. Depressants exert their effects through a number of different pharmacological mechanisms, the most prominent of which include facilitation of GABA, and inhibition of glutamatergic or monoaminergic activity. Other examples are chemicals that modify the electrical signaling inside the body, the most prominent of these being bromides and body blockers. Alcohol Alcohol is a very prominent depressant. Alcohol can be and is more likely to be a large problem among teenagers and young adults. Symptoms of alcohol consumption at lower doses may include mild sedation and poor coordination. At higher doses, there may be slurred speech, trouble walking, and vomiting. Extreme doses may result in a respiratory depression, coma, or death. Complications may include seizures, aspiration pneumonia, injuries including suicide, and low blood sugar. Alcohol intoxication can lead to alcohol-related crime with perpetrators more likely to be intoxicated than victims. Alcohol intoxication typically begins after two or more alcoholic drinks. Risk factors include a social situation where heavy drinking is common and a person having an impulsive personality. Diagnosis is usually based on the history of events and physical examination. Verification of events by witnesses may be useful. Legally, alcohol intoxication is often defined as a blood alcohol concentration (BAC) of greater than 5.4–17.4 mmol/L (25–80 mg/dL or 0.025–0.080%). This can be measured by blood or breath testing. Alcohol is broken down in human body at a rate of about 3.3 mmol/L (15 mg/dL) per hour. Alcohol intoxication is very common, especially in the Western world. Most people who drink alcohol have at some time been intoxicated. In the United States acute intoxication directly results in about 2,200 deaths per year, and indirectly more than 30,000 deaths per year. Acute intoxication has been documented throughout history and alcohol remains one of the world's most widespread recreational drugs. Some religions consider alcohol intoxication to be a sin. Barbiturates Barbiturates are CNS depressants and are similar, in many ways, to the depressant effects of alcohol. To date, there are about 2,500 derivatives of barbituric acid of which only 15 are used medically. The first barbiturate was synthesized from barbituric acid in 1864. The original use of barbiturates was to replace drugs such as opiates, bromides, and alcohol to induce sleep. Barbiturates are effective as anxiolytics, hypnotics, and anticonvulsants, but have physical and psychological addiction potential as well as overdose potential among other possible adverse effects. They have largely been replaced by benzodiazepines (discussed below) and nonbenzodiazepines ("Z-drugs") in routine medical practice, particularly in the treatment of anxiety and insomnia, due to the significantly lower risk of addiction and overdose and the lack of an antidote for barbiturate overdose. Despite this, barbiturates are still in use for various purposes: in general anesthesia, epilepsy, treatment of acute migraines or cluster headaches, euthanasia, capital punishment, and assisted suicide. Some symptoms of an overdose typically include sluggishness, incoordination, difficulty in thinking, slowness of speech, faulty judgement, drowsiness, shallow breathing, staggering, and, in severe cases, coma or death. The lethal dosage of barbiturates varies greatly with tolerance and from one individual to another. Barbiturates in overdose with other CNS (central nervous system) depressants (e.g. alcohol, opiates, benzodiazepines) are even more dangerous due to additive CNS and respiratory depressant effects. In the case of benzodiazepines, not only do they have additive effects, barbiturates also increase the binding affinity of the benzodiazepine binding site, leading to exaggerated benzodiazepine effects. (ex. If a benzodiazepine increases the frequency of channel opening by 300%, and a barbiturate increases the duration of their opening by 300%, then the combined effects of the drugs increase the channels overall function by 900%, not 600%). Anti-anxiety Agents Anti-anxiety medications help reduce the symptoms of anxiety, such as panic attacks, or extreme fear and worry. Benzodiazepines are prescribed to quell panic attacks. Benzodiazepines are also prescribed in tandem with an antidepressant for the latent period of efficacy associated with many antidepressants for anxiety disorder. The effects of the benzodiazepines virtually all result from action of these drugs on the central nervous system, even when lethal doses are used. The most prominent of these effects are sedation, hypnosis, decreased anxiety, muscle relaxation, and anticonvulsant activity. As the dose of a benzodiazepine is increased, sedation progresses to hypnosis and hypnosis to stupor. They are used as sedatives, hypnotics, antianxiety agents (in panic disorder), anticonvulsants, muscle relaxants, in anesthesia and in alcoholism. There is risk of benzodiazepine withdrawal and rebound syndrome if BZDs are rapidly discontinued. Tolerance and dependence may occur. The risk of abuse in this class of medication is smaller than in that of barbiturates. Cognitive and behavioral adverse effects are possible. There are several useful benzodiazepines available. The skeletal structure and two examples are shown below. Benzodiazepine include: Alprazolam (Xanax), Bromazepam, Chlordiazepoxide (Librium), Clonazepam (Klonopin), Diazepam (Valium), Lorazepam (Ativan), Oxazepam, Temazepam, and Triazolam. Short half-life (or short-acting) benzodiazepines (such as Lorazepam ) and beta-blockers are used to treat the short-term symptoms of anxiety. Beta-blockers help manage physical symptoms of anxiety, such as trembling, rapid heartbeat, and sweating that people with phobias (an overwhelming and unreasonable fear of an object or situation, such as public speaking) experience in difficult situations. Taking these medications for a short period of time can help the person keep physical symptoms under control and can be used “as needed” to reduce acute anxiety. Antipsychotics, also known as neuroleptics , are a class of psychotropic medication primarily used to manage psychosis (including delusions, hallucinations, paranoia or disordered thought), principally in schizophrenia but also in a range of other psychotic disorders. They are also the mainstay together with mood stabilizers in the treatment of bipolar disorder. The phenothiazines as a class, and especially chlorpromazine, the prototype, are among the most widely used drugs in medical practice and are primarily employed in the management of patients with serious psychiatric illnesses. In addition, many members of the group have other clinically useful properties, including antiemetic, antinausea, and antihistaminic effects and the ability to potentiate analgesics, sedatives and general anesthetics. It was noted that chlorpromazine by itself did not cause a loss of consciousness but produced only a tendency to sleep and a lack of interest in what was going on. These central actions became known as neuroleptic soon after. Phenothiazine has a tricyclic structure in which two benzene rings are linked by a sulfur and a nitrogen atom (see figures below). Substitution of an electron-withdrawing group at R2 (but not at position 3 or 4) increases the efficacy of phenothiazines and other tricyclic congeners. Neuroleptic drugs reduce initiative and interest in the environment, and they reduce displays of emotion or affect. Initially there may be some slowness in response to external stimuli and drowsiness. However subject are easily aroused, capable of giving appropriate answers to direct questions, and seem to have intact intellectual functions; there is no ataxia, incoordination, or dysathria at ordinary doses. Psychotic patients become less agitated and restless, and withdrawn or autistic patients sometimes become more responsive and communicative. Aggressive and impulsive behavior diminishes. Gradually (over a period of days). psychotic symptoms of hallucinations, delusions, and disorganized or incoherent thinking tend to disappear. The most prominent observable effects of typical neuroleptic agents are strikingly similar. In low doses, operant behavior is reduced but spinal reflexes are unchanged. Exploratory behavior is diminished, and responses to a a variety of stimuli are fewer, slower, and smaller, although the ability to discriminate stimuli is retained. Conditioned avoidance behaviors are selectively inhibited, while unconditioned escape or avoidance responses are not. In 1994 an addition tot he antipsychotic drugs is risperidone (Risperdal). This drug antagonises D2 and serotonin type 2 receptors. The drug also antagonizes for other receptors such as a adrenergic and histaminergic H1 receptors. Anti Depressants Major depression is the most common of the major mental illnesses, and it must be distinguished from normal grief, sadness, and disappointment. Major depression is characterized by feelings of intense sadness and despair, mental slowing and loss of concentration, pessimistic worry, agitation, and self-depreciation. Physical changes also occur, such as weight loss, decreased libido, and disruption of hormonal circadian rhythms. Before the advent of psychotherapy in the 1950s, treatment of depression consisted of stimulants such as caffeine and amphetamines to ameliorate the depressive phases and barbiturates to allay agitation, anxiety, and insomnia. At best, such attempts at therapy may have offered transient relief to some patients. Suffering usually decreased little. Monoamine Oxidase Inhibitors Monoamine oxidase inhibitors (MAOIs) were the first effective antidepressants used. The monoamine oxidase inhibitors comprise a chemically heterogeneous group of drugs that have in common the ability to block oxidative deamination of naturally occurring monoamines. These drugs have numerous other effects, many of which are still poorly understood. For example, they lower blood pressure and were at one time used to treat hypertension. Their use in psychiatry has also become very limited as the tricyclic antidepressants have come to dominate the treatment of depression and allied conditions. Thus, MAOIs are used most often when tricyclic antidepressants give unsatisfactory results. In addition, whereas severe depression may not be the primary indication for these agents, certain neurotic illnesses with depressive features, and also with anxiety and phobias, may respond especially favorably. Serotonin Serotonin (5-hydroxytryptamine or 5-HT) is a monoamine neurotransmitter found in cardiovascular tissue, in endothelial cells, in blood cells, and in the central nervous system. The role of serotonin in neurological function is diverse, and there is little doubt that serotonin is an important CNS neurotransmitter. The monoamine serotonin is itself a precursor for melatonin production in the pineal gland. The biosynthesis of serotonin from the amino acid tryptophan is similar to that found for the catecholamines, and 5-hydroxytryptophan can cross the BBB to increase central levels of 5-HT. Although some of the serotonin is metabolized by monoamine oxidase to 5-hydroxyindole acetic acid, most of the serotonin released into the post-synaptic space is removed by the neuron through a reuptake mechanism inhibited by the tricyclic antidepressants and the newer, more selective antidepressants such as fluoxetine and sertraline. Serotonin receptors are diverse and numerous. Over the past several years, over fourteen different serotonin receptors have been cloned and sequenced through molecular biological techniques. Overall, there are seven distinct families of 5-HT receptors, with as many as five within a particular family. Only one of the 5-HT receptors is a ligand-gated ion channel (the 5-HT3 receptor), and the other six families are all G protein-coupled receptors. Tricyclic Anti-depressants Imipramine, amitriptylin, and other closely related drugs are among the drugs currently most widely used for the treatment of major depression. Because of there structure ( see below). They are often referred to as the tricyclic antidepressants. Although these compounds seem to be similar to the phenothiazines chemically, the ethylene group of imiprimine's middle ring imparts dissimilar stereochemical properties and prevents conjegation of the rings, as occurs with the phenothiazines. One might expect an effective antidepressant drug to have a stimulating or mood-elevating effect when given to a normal subject. Although this may occur with the MAOIs, it is not true of the tricyclic antidepressants. If a dose of imipramine given to a normal subject, he feels sleepy and tends to be quieter, his blood pressure falls slightly, and he feels light headed. These drug effects are usually perceived to be unpleasant, and cause a feeling of unhappiness and increased anxiety. Repeated administration for several days may lead to accentuation of these symptoms and, in addition, to difficulty in concentrating and thinking. In contrast, if the drug is given over a period of time ( two to three weeks) to depressed patients an elevated mood occurs. For this reason, the tricyclics are not prescribed on an "as-needed" basis. Selective Serotonin Reuptake Inhibitors In recent years, selective serotonin reuptake inhibitors (SSRIs) have been introduced for the treatment of depression. Prozac is the most famous drug in this class. Lilly's sales of Prozac in 1993 exceeded 1 billion US dollars. Clomiprimine, fluoxetine (Prozac), sertraline and paroxetine selectively block the reuptake of serotonin, thereby increasing the levels of serotonin in the central nervous system. Note the similarities and differences between the tricyclic antidepressants and the selective serotonin reuptake inhibitors. The SSRIs generally have fewer anticholinergic side effects, but caution is still necessary when co-administering any drugs that affect serotonergic systems (e.g., monoamine oxidase inhibitors). Some of the newer, SSRIs (e.g., clomipramine) have been useful in the treatment of obsessive-compulsive disorders. Stimulants These classes of chemicals induce alertness and stimulant the brain and nervous system. Side effects of stimulant use include wakefulness, increased speech, and motor activity, and decreased appetite. Amphetamines are a type of stimulant that incorporates a nitrogen-containing organic functional group. These compounds have been noted to block the reuptake of dopamine. Early Stimulant Use For over 5000 years, the Chinese have extracted the compound Ephedrine from locally grown Ephedra plants. Native peoples of the United States were known to isolate this compound as well. Both cultures used Ephedrine to treat asthma, hay fever, and congestion. Ephedrine was noted to have a stimulant effect on its users. Chemically similar to adrenaline, ephedrine was a more stable compound and could be taken orally. By the mid-1920s, the pharmaceutical company, Eli Lily, synthesized ephedrine and introduced it into western medicine. Figure \(\PageIndex{14}\): : https://commons.wikimedia.org/wiki/C...phedra_viridis . (Copyright, https://commons.wikimedia.org/wiki/User:Dcrjsr ) Ephedrine can decrease appetite, increase blood pressure and heart rate, and cause sleeping problems. Mental health can be affected by ephedrine use. Patients may experience hallucinogens, anxiety, and chemical dependence. Amphetamine is a potent central nervous system (CNS) stimulant of the phenethylamine class that is approved for the treatment of attention deficit hyperactivity disorder (ADHD) and narcolepsy. Amphetamine is also used off-label as a performance and cognitive enhancer, and recreationally as an aphrodisiac and euphoriant. Although it is a prescription medication in many countries, unauthorized possession and distribution of amphetamine is often tightly controlled due to the significant health risks associated with uncontrolled or heavy use. As a consequence, amphetamine is illegally manufactured in clandestine labs to be trafficked and sold to users. Based upon drug and drug precursor seizures worldwide, illicit amphetamine production and trafficking is much less prevalent than that of methamphetamine. The first pharmaceutical amphetamine was Benzedrine, a brand of inhalers used to treat a variety of conditions. At therapeutic doses, this drug causes emotional and cognitive effects such as euphoria, change in libido, increased arousal, and improved cognitive control. Likewise, it induces physical effects such as decreased reaction time, fatigue resistance, and increased muscle strength. In contrast, supratherapeutic doses of amphetamine are likely to impair cognitive function and induce rapid muscle breakdown. Very high doses can result in psychosis (e.g., delusions and paranoia), which very rarely occurs at therapeutic doses even during long-term use. As recreational doses are generally much larger than prescribed therapeutic doses, recreational use carries a far greater risk of serious side effects, such as dependence, which only rarely arises with therapeutic amphetamine use. 3,4-Methylenedioxymethamphetamine (MDMA, ecstasy, or molly) is a euphoriant, empathogen, and stimulant of the amphetamine class. Briefly used by some psychotherapists as an adjunct to therapy, the drug became popular recreationally and the DEA listed MDMA as a Schedule I controlled substance, prohibiting most medical studies and applications. MDMA is known for its entactogenic properties. The stimulant effects of MDMA include hypertension, anorexia (appetite loss), euphoria, social disinhibition, insomnia (enhanced wakefulness/inability to sleep), improved energy, increased arousal, and increased perspiration, among others. Relative to catecholaminergic transmission, MDMA enhances serotonergic transmission significantly more, when compared to classical stimulants like amphetamine. MDMA does not appear to be significantly addictive or dependence forming. Methylenedioxypyrovalerone (MDPV) is a psychoactive drug with stimulant properties that acts as a norepinephrine-dopamine reuptake inhibitor (NDRI). It was first developed in the 1960s by a team at Boehringer Ingelheim. MDPV remained an obscure stimulant until around 2004, when it was reported to be sold as a designer drug. Products labeled as bath salts containing MDPV were previously sold as recreational drugs in gas stations and convenience stores in the United States, similar to the marketing for Spice and K2 as incense. Incidents of psychological and physical harm have been attributed to MDPV use. Mephedrone is a synthetic stimulant drug of the amphetamine and cathinone classes. Slang names include drone and MCAT. It is reported to be manufactured in China and is chemically similar to the cathinone compounds found in the khat plant of eastern Africa. It comes in the form of tablets or a powder, which users can swallow, snort, or inject, producing similar effects to MDMA, amphetamines, and cocaine. Methamphetamine (contracted from N - meth yl- a lpha- m ethyl ph en et hyl amine ) is a potent psychostimulant that is used to treat attention deficit hyperactivity disorder (ADHD) and obesity. Recreationally, methamphetamine is used to increase sexual desire, lift the mood, and increase energy, allowing some users to engage in sexual activity continuously for several days straight. In low doses, methamphetamine can cause an elevated mood and increase alertness, concentration, and energy in fatigued individuals. At higher doses, it can induce psychosis, rhabdomyolysis, and cerebral hemorrhage. Methamphetamine is known to have a high potential for abuse and addiction. Recreational use of methamphetamine may result in psychosis or lead to post-withdrawal syndrome, a withdrawal syndrome that can persist for months beyond the typical withdrawal period. Unlike amphetamine and cocaine, methamphetamine is neurotoxic to humans, damaging both dopamine and serotonin neurons in the central nervous system (CNS). Entirely opposite to the long-term use of amphetamine, there is evidence that methamphetamine causes brain damage from long-term use in humans; this damage includes adverse changes in brain structure and function, such as reductions in gray matter volume in several brain regions and adverse changes in markers of metabolic integrity. Methylphenidate is a stimulant drug that is often used in the treatment of ADHD and narcolepsy and occasionally to treat obesity in combination with diet restraints and exercise. Its effects at therapeutic doses include increased focus, increased alertness, decreased appetite, decreased need for sleep and decreased impulsivity. Methylphenidate is not usually used recreationally, but when it is used, its effects are very similar to those of amphetamines. Methylphenidate is sold under a number of brand names including Ritalin. Other versions include the long lasting tablet Concerta and the long lasting transdermal patch Daytrana. Phenylpropanolamine (PPA; Accutrim; β-hydroxyamphetamine) , also known as norephedrine and norpseudoephedrine, is a psychoactive drug that is used as a stimulant, decongestant, and anorectic agent. It is commonly used in prescription and over-the-counter cough and cold preparations. In veterinary medicine, it is used to control urinary incontinence in dogs under trade names Propalin and Proin. In the United States, PPA is no longer sold without a prescription due to a proposed increased risk of stroke in younger women. In a few countries in Europe, however, it is still available either by prescription or sometimes over-the-counter. Propylhexedrine (Hexahydromethamphetamine, Obesin) is a stimulant medication, sold over-the-counter in the United States as the cold medication Benzedrex. The drug has also been used as an appetite suppressant in Europe. Propylhexedrine is not an amphetamine, though it is structurally similar; it is instead a cycloalkylamine, and thus has stimulant effects that are less potent than similarly structured amphetamines, such as methamphetamine. The abuse potential of propylhexedrine is fairly limited, due its limited routes of administration: in the United States, Benzedrex is only available as an inhalant, mixed with lavender oil and menthol. These ingredients cause unpleasant tastes, and abusers of the drug have reported unpleasant "menthol burps". Injection of the drug has been found to cause transient diplopia and brain stem dysfunction. Pseudoephedrine is used as a nasal/sinus decongestant, as a stimulant, or as a wakefulness-promoting agent. The salts pseudoephedrine hydrochloride and pseudoephedrine sulfate are found in many over-the-counter preparations, either as a single ingredient or (more commonly) in combination with antihistamines, guaifenesin, dextromethorphan, and/or paracetamol (acetaminophen) or another NSAID (such as aspirin or ibuprofen). It is also used as a precursor chemical in the illegal production of methamphetamine. Cocaine, Caffeine, and Nicotine Cocaine is a tropane alkaloid and stimulant drug obtained primarily from the leaves of two coca species, Erythroxylum coca and Erythroxylum novogranatense . It is most commonly used as a recreational drug and euphoriant. After extraction from coca leaves, cocaine may be snorted, heated until sublimated and then inhaled, or dissolved and injected into a vein. Mental effects may include an intense feeling of happiness, sexual arousal, loss of contact with reality, or agitation. Physical symptoms may include a fast heart rate, sweating, and dilated pupils. High doses can result in high blood pressure or body temperature. Effects begin within seconds to minutes of use and last between five and ninety minutes. Cocaine is addictive due to its effect on the reward pathway in the brain. A single dose of cocaine induces tolerance to the drug's effects. After a short period of use, addiction is likely. Abstention from cocaine after chronic use results in drug withdrawal, with symptoms that may include depression, decreased ability to feel pleasure and subjective fatigue. Cocaine's use increases the overall risk of death and particularly the risk of trauma, and infectious diseases, such as blood infections and AIDS. It also increases risk of stroke, heart attack, cardiac arrhythmia, lung injury (when smoked), and sudden cardiac death. Illicitly-sold cocaine is commonly adulterated with local anesthetics, levamisole, cornstarch, quinine, or sugar, which can result in additional toxicity. The Global Burden of Disease study found that cocaine use caused the death of 7.3 people per 100,000 population world-wide. Caffeine is a central nervous system (CNS) stimulant of the methylxanthine class. It is the world's most widely consumed psychoactive drug. Unlike many other psychoactive substances, it is legal and unregulated in nearly all parts of the world. There are several known mechanisms of action to explain the effects of caffeine. The most prominent is that it reversibly blocks the action of adenosine on its receptors and consequently prevents the onset of drowsiness induced by adenosine. Caffeine also stimulates certain portions of the autonomic nervous system. Methyxanthines such as caffeine, theophylline and theobromine share in common several pharmacological actions of therapeutic interest. They stimulate the central nervous system, act on the kidney to produce diuresis, stimulate cardiac muscle, and relax smooth muscle, notably bronchial muscle. Because the various xanthines differ markedly in the intensity of their action on various structures, one particular xanthine has been used more than another for a particular therapeutic effect. Since theobromine displays low potency in these pharmacological actions, it has all but disappeared from the therapeutic scene. Caffeine, theophyline, and theobromine occur naturally in plants widely distributed geographically. Caffeine is found in the coffee bean, tea leaves, guarana, and other plants. From the figure below, we can see that the methylxanthines have a structure which is very similar to adenine. Beverage/Food Milligrams Starbuck’s Grande Coffee (16 oz.) 380 Plain brewed coffee (8 oz.) 102–200 Espresso (1 oz.) 30–90 Plain, decaffeinated coffee (8 oz.) 3–12 Tea, brewed (8 oz.) 40–120 Green tea (8 oz.) 25–40 Coca-Cola Classic (12 oz.) 35 Dr. Pepper (12 oz.) 44 Jolt Cola (12 oz.) 72 Mountain Dew (12 oz.) 54 Mountain Dew, MDX (12 oz.) 71 Pepsi-Cola (12 oz.) 38 Red Bull (8.5 oz.) 80 Full Throttle (16 oz.) 144 Monster Energy (16 oz.) 160 Spike Shooter (8.4 oz.) 300 Source: MedicineNet.com. “Caffeine.” Accessed October 2, 2011. http://www.medicinenet.com/caffeine/article.htm . Health Benefits of Caffeine The most renowned effects of caffeine on the body are increased alertness and delay of fatigue and sleep. How does caffeine stimulate the brain? Watch "Video 15.6.1" to see a graphic account of a brain on caffeine. Caffeine is chemically similar to a chemical in our brains (adenosine). Caffeine interacts with adenosine’s specific protein receptor. It blocks the actions of the adenosine, and affects the levels of signaling molecules in the brain, leading to an increase in energy metabolism. At the molecular level, caffeine stimulates the brain, increasing alertness and causing a delay of fatigue and sleep. At high doses caffeine stimulates the motor cortex of the brain and interferes with the sleep-wake cycle, causing side effects such as shakiness, anxiety, and insomnia. People’s sensitivity to the adverse effects of caffeine varies and some people develop side effects at much lower doses. The many effects caffeine has on the brain do not diminish with habitual drinking of caffeinated beverages. Video 15.6.1: A Brain on Caffeine. Watch this graphic account of the brain on caffeine. Scientific studies suggest caffeine can improve endurance capacity by increasing energy available during exercise. The effect may only work in non-caffeine drinkers and it only takes 1-3 days for the body to become "caffeine-naive." Nicotine is the active chemical constituent in tobacco, which is available in many forms, including cigarettes, cigars, chewing tobacco, and smoking cessation aids such as nicotine patches, nicotine gum, and electronic cigarettes. Nicotine is used widely throughout the world for its stimulating and relaxing effects. Nicotine exerts its effects through the agonism of nicotinic acetylcholine receptor, resulting in multiple downstream effects such as increase in activity of dopaminergic neurons in the midbrain reward system, as well as the decreased expression of monoamine oxidase in the brain. Nicotine is addictive and dependence forming. Note: Hallucinogens and Dissociative Drugs from NIH-National Institute on Drug Abuse The information below is directly from the link below https://d14rmgtrwzf5a.cloudfront.net/sites/default/files/hallucinogensrrs4.pdf Hallucinogens are a class of drugs that cause hallucinations—profound distortions in a person’s perceptions of reality. Hallucinogens can be found in some plants and mushrooms (or their extracts) or can be man-made, and they are commonly divided into two broad categories: classic hallucinogens (such as LSD) and dissociative drugs (such as PCP). When under the influence of either type of drug, people often report rapid, intense emotional swings and seeing images, hearing sounds, and feeling sensations that seem real but are not. While the exact mechanisms by which hallucinogens and dissociative drugs cause their effects are not yet clearly understood, research suggests that they work at least partially by temporarily disrupting communication between neurotransmitter systems throughout the brain and spinal cord that regulate mood, sensory perception, sleep, hunger, body temperature, sexual behavior, and muscle control. Classic Hallucinogens LSD (d-lysergic acid diethylamide) —also known as acid, blotter, doses, hits, microdots, sugar cubes, trips, tabs, or window panes—is one of the most potent moodand perception-altering hallucinogenic drugs. It is a clear or white, odorless, water-soluble material synthesized from lysergic acid, a compound derived from a rye fungus. LSD is initially produced in crystalline form, which can then be used to produce tablets known as “microdots” or thin squares of gelatin called “window panes.” It can also be diluted with water or alcohol and sold in liquid form. The most common form, however, is LSD-soaked paper punched into small individual squares, known as “blotters.” Psilocybin (4-phosphoryloxyN, N-dimethyltryptamine)—also known as magic mushrooms, shrooms, boomers, or little smoke—is extracted from certain types of mushrooms found in tropical and subtropical regions of South America, Mexico, and the United States. In the past, psilocybin was ingested during religious ceremonies by indigenous cultures from Mexico and Central America. Psilocybin can either be dried or fresh and eaten raw, mixed with food, or brewed into a tea, and produces similar effects to LSD. Peyote (Mescaline) — also known as buttons, cactus, and mesc— is a small, spineless cactus with mescaline as its main ingredient. It has been used by natives in northern Mexico and the southwestern United States as a part of religious ceremonies. The top, or “crown,” of the peyote cactus has disc-shaped buttons that are cut out, dried, and usually chewed or soaked in water to produce an intoxicating liquid. Because the extract is so bitter, some users prepare a tea by boiling the plant for several hours. Mescaline can also be produced through chemical synthesis. DMT (Dimethyltryptamine) —also known as Dimitri—is a powerful hallucinogenic chemical found naturally occurring in some Amazonian plant species (see “Ayahuasca”) and also synthesized in the laboratory. Synthetic DMT usually takes the form of a white crystalline powder and is typically vaporized or smoked in a pipe. Ayahuasca—also known as hoasca, aya, and yagé—is a hallucinogenic brew made from one of several Amazonian plants containing DMT (the primary psychoactive ingredient) along with a vine containing a natural alkaloid that prevents the normal breakdown of DMT in the digestive tract. Ayahuasca tea has traditionally been used for healing and religious purposes in indigenous South American cultures, mainly in the Amazon region. Dissociative Drugs PCP (Phencyclidine) —also known as ozone, rocket fuel, love boat, hog, embalming fluid, or superweed—was originally developed in the 1950s as a general anesthetic for surgery. While it can be found in a variety of forms, including tablets or capsules, it is usually sold as a liquid or powder. PCP can be snorted, smoked, injected, or swallowed. It is sometimes smoked after being sprinkled on marijuana, tobacco, or parsley. Ketamine —also known as K, Special K, or cat Valium—is a dissociative currently used as an anesthetic for humans as well as animals. Much of the ketamine sold on the street has been diverted from veterinary offices. Although it is manufactured as an injectable liquid, ketamine is generally evaporated to form a powder that is snorted or compressed into pills for illicit use. Because ketamine is odorless and tasteless and has amnesia-inducing properties, it is sometimes added to drinks to facilitate sexual assault. 2 NIDA Research Report Series Common Hallucinogens and Dissociative Drugs *In this report, the term “hallucinogen” will refer to the classic hallucinogenic drugs LSD and Psilocybin. DXM (Dextromethorphan) — also known as robo—is a cough suppressant and expectorant ingredient in some over-the-counter (OTC) cold and cough medications that are often abused by adolescents and young adults. The most common sources of abused DXM are “extra-strength” cough syrup, which typically contains around 15 milligrams of DXM per teaspoon, and pills and gel capsules, which typically contain 15 milligrams of DXM per pill. OTC medications that contain DXM often also contain antihistamines and decongestants. Salvia divinorum —also known as diviner’s sage, Maria Pastora, Sally-D, or magic mint—is a psychoactive plant common to southern Mexico and Central and South America. Salvia is typically ingested by chewing fresh leaves or by drinking their extracted juices. The dried leaves of salvia can also be smoked or vaporized and inhaled. Short-Term General Effects of Hallucinogens Sensory and Physical Effects • Hallucinations, including seeing, hearing, touching, or smelling things in a distorted way or perceiving things that do not exist • Intensified feelings and sensory experiences (brighter colors, sharper sounds) • Mixed senses (“seeing” sounds or “hearing” colors) • Changes in sense or perception of time (time goes by slowly) Physical Effects • Increased energy and heart rate • Nausea Marijuana - NIH Medlineplus Marijuana The information below is directly from the link below https://medlineplus.gov/marijuana.html Marijuana is a green, brown, or gray mix of dried, crumbled parts from the marijuana plant. The plant contains chemicals which act on your brain and can change your mood or consciousness. Marijuana can cause both short-term and long-term effects. 0 1 Short term: Long term: While you are high, you may experience Altered senses, such as seeing brighter colors Altered sense of time, such as minutes seeming like hours Changes in mood Problems with body movement Trouble with thinking, problem-solving, and memory Increased appetite In the long term, marijuana can cause health problems, such as Problems with brain development. People who started using marijuana as teenagers may have trouble with thinking, memory, and learning. Coughing and breathing problems, if you smoke marijuana frequently Problems with child development during and after pregnancy, if a woman smokes marijuana while pregnant. Medical Marijuana The marijuana plant has chemicals that can help with some health problems. More states are making it legal to use the plant as medicine for certain medical conditions. But there isn't enough research to show that the whole plant works to treat or cure these conditions. The U.S. Food and Drug Administration (FDA) has not approved the marijuana plant as a medicine. Marijuana is still illegal at the national level. However, there have been scientific studies of cannabinoids, the chemicals in marijuana. The two main cannabinoids that are of medical interest are THC and CBD. The FDA has approved two drugs that contain THC. These drugs treat nausea caused by chemotherapy and increase appetite in patients who have severe weight loss from AIDS . There is also a liquid drug that contains CBD. It treats two forms of severe childhood epilepsy . Scientists are doing more research with marijuana and its ingredients to treat many diseases and conditions. Summary Neurotransmitters are endogenous chemicals that enable neurotransmission. It is a type of chemical messenger which transmits signals across a chemical synapse, such as a neuromuscular junction, from one neuron (nerve cell) to another "target" neuron, muscle cell, or gland cell. Strong imbalances or disruptions to neurotransmitter systems have been associated with many diseases and mental disorders. Psychoactive drugs are substances that change the function of the brain and result in alterations of mood, thinking, perception, and/or behavior. They include prescription medications such as opioid painkillers, legal substances such as nicotine and alcohol, and illegal drugs such as LSD and heroin. Psychoactive drugs are divided into different classes according to their pharmacological effects. They include stimulants, depressants, anxiolytics, euphoriants, hallucinogens, and empathogens. Many psychoactive drugs have multiple effects so they may be placed in more than one class. Psychoactive drugs generally produce their effects by affecting brain chemistry. Generally, they act either as agonists, which enhance the activity of particular neurotransmitters; or as antagonists, which decrease the activity of particular neurotransmitters. Psychoactive drugs are used for various purposes, including medical, ritual, and recreational purposes. Misuse of psychoactive drugs may lead to addiction, which is compulsive use of a drug despite negative consequences such use may entail. Sustained use of an addictive drug may produce physical or psychological dependence on the drug. Rehabilitation typically involves psychotherapy and sometimes the temporary use of other psychoactive drugs. Contributors Libretext: Human Biology (Wakim and Grewal) Edward B. Walker (Weber State University) Charles Ophardt, Professor Emeritus, Elmhurst College; Virtual Chembook NIH National Institute of Mental Health NIH National Institute on Drug Abuse NIH-Medlineplus Psychology OPENSTAX Wikipedia Marisa Alviar-Agnew ( Sacramento City College ) |
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